INSTRUCTOR’S MANUAL TO ACCOMPANY
ELEMENTARY PRINCIPLES OF CHEMICAL PROCESSES Third Edition
Richard M. Felder Ronald W. Rousseau with assistance from Matthew Burke, Swapnil Chhabra, Jun Gao, Gary Huvard, Concepción Jimenez-Gonzalez, Linda Holm, Norman Kaplan, Brian Keyes, Amit Khandelwal, Stephanie Manfredi, Janette Mendez-Santiago, Amy Michel, Dong Niu, Amitabh Sehgal, James Semler, Kai Wang, Esther Wilcox, Jack Winnick, Tao Wu, Jian Zhou
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INSTRUCTOR’S MANUAL to accompany ELEMENTARY PRINCIPLES OF CHEMICAL PROCESSES THIRD EDITION
RICHARD M. FELDER North Carolina State University RONALD W. ROUSSEAU Georgia Institute of Technology
JOHN WILEY & SONS New York Chichester Brisbane
Toronto
Singapore
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CONTENTS Notes to the Instructor
iv
Section/Problem Concordance
vi
Sample Assignment Schedule I
ix
Sample Assignment Schedule II
x
Sample Responses to a Creativity Exercise
xi
Transparency Masters
xvi
Compressibility charts Cox vapor pressure chart Psychrometric chart – SI units Psychrometric chart – American engineering units Enthalpy-concentration chart: H2SO4-H2O Enthalpy-concentration chart: NH3-H2O
xvii xxi xxii xxiii xxiv xxv
Problem Solutions Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 (Case Study 1) Chapter 13 (Case Study 2) Chapter 14 (Case Study 3)
2-1 3-1 4-1 5-1 6-1 7-1 8-1 9-1 10-1 11-1 12-1 13-1 14-1
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NOTES TO THE INSTRUCTOR Problem Assignments To aid in the structuring of the course, we have provided a section/problem concordance on pp. vii–ix and two sample assignment schedules on pp. x–xi. We believe there is far too much material in the textbook to attempt to cover in one semester or two quarters. The sample assignment schedules therefore cover only Chapters 1-9, and within those chapters some topics are omitted (e.g. liquid-liquid equilibrium and adsorption on solid surfaces in Chapter 6 and mechanical energy balances in Chapter 7). The missing sections can substitute for Chapters 2 and 3 in classes where the content of those chapters has been well covered in chemistry and/or physics courses), and Chapters 10 (computer flowsheeting) and 11 (transient balances) may be assigned for extra credit or covered in honors sections or subsequent courses in the curriculum. We will discuss the case studies in Chapters 12-14 separately. In the sample assignment schedules, we have designated a number of “bonus problems” which may be ignored, assigned for extra credit as add-ons to the regular assignments (those will be long assignments), or assigned for extra credit in lieu of some of the problems in the regular assignments. The bonus problems may be assigned as individual exercises or students may work on them in pairs. We have had good experience with the latter approach. Creativity Exercises The creativity exercises in the text are designed to stimulate divergent thinking and to induce the students to think about course material from different perspectives. We have used such exercises both as extra-credit assignments to individuals or pairs of students or as the foci of in-class brainstorming sessions. In all cases, we have found that they invariably lead to innovative, clever, and often amusing ideas; they give students who are by nature creative an opportunity to demonstrate their talent and they help other students develop creative problem-solving skills; and the students usually enjoy doing them. There are no “right answers” to such exercises, and so we have not included solutions in this manual. However, to provide an idea of the kind of things that students come up with, we have included on pp. xi–xv a collection of student responses to a creativity exercise given by one of the authors in a junior course on fluid dynamics. Transparency Masters Several of our colleagues have suggested that we include in the text enlarged versions of some of the figures, such as the psychometric charts, which are difficult to read in reduced format. We have chosen not to do so, since whether they are inserts or fold-outs such charts tend to be ripped off (one way or another) or otherwise lost. Instead, we have included in this manual, beginning on p. xvi, large versions of some of the most commonly used figures. These masters can be used to make transparencies for lectures; they can also be copied and distributed to the students for use in solving problem. Case Studies The case studies comprise Chapters 12 through 14 of the text. In them, we seek to (1) illustrate the development of complex chemical processes from basic principles, and to provide a broad process context for the text material; (2) raise questions that require students to think about topics strictly beyond the scope of the first course, and to seek out sources of information other than the text; (3) accustom the students to team project work. We do not organize the activities of case study
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teams, nor do we assign team leaders, although we suggest to the students that they do so. This is a risk, and sometimes it is necessary to step in and get a laggard group started. However, letting the teams shape their own working relationships and structure their own activities usually is an enlightening experience to the students. Problem Solutions The detailed solutions to 634 of the 635 chapter-end problems constitute the principal content of this manual. (The solution to the last problem of Chapter 10 is left as an exercise for the professor, or for anyone else who wants to do it.) With few exceptions, the conversion factors and physical property data needed to solve the endof-chapter problems are contained in the text. It may be presumed that conversion factors for which sources are not explicitly cited come from the front cover table; densities, latent heats, and critical constants come from Table B.1; heat capacity formulas come from Table B.2; enthalpies of combustion gases come from Tables B.8 and B.9; vapor pressures come from Table B.4 or (for water) Table B.3; and enthalpies, internal energies, and specific volumes of water at different temperatures and pressures come from Tables B.5-B.7. As the reader of the text may have discovered, we believe strongly in the systematic use of the flow charts in the solution of material and energy balance problems. When a student comes to us to ask for help with a problem, we first ask to see the labeled flow chart: no flow chart, no help. Other instructors we know demand fully labeled flow charts and solution outlines at the beginning of every problem solution, before any calculations are performed. In any case, we find that the students who can be persuaded to adopt this approach generally complete their assignments in reasonable periods of time and do well in the course; most of those who continue to resist it find themselves taking hours to do the homework problems, and do poorly in the course. Posting Problem Solutions: An Impassioned Plea It is common practice for instructors to photocopy solutions from the manual and to post them after the assignments are handed in, or, even worse, to distribute the solutions to the students. What happens then, of course, is that the solutions get into circulation and reincarnate with increasing frequency as student solutions. After one or two course offerings, the homework problems consequently lose much of their instructional value and become more exercises in stenography than engineering problem solving. In the stoichiometry course particularly, the concepts are relatively elementary: the main point is to teach the students to set up and solve problems. A great deal of classroom lecturing on concepts should therefore not be necessary, and a good deal of the class time can be spent in outlining problem solutions. The burden should be placed on the students to make sure they know how to do the problems: to ask about them in class, to make notes on solutions outlined on the board, and to fill in omitted calculations. Besides being pedagogically superior to solution-posting, this approach should cut down on the ease with which students can simply copy letter-perfect solutions instead of doing the work themselves. Errors A great deal of time and effort has been expended to make the solutions in this manual as free of errors as possible. Nevertheless, errors undoubtedly still exist. We will be grateful to any of our colleagues who send us corrections, no matter how major or minor they may be; we will provide an errata list on the text Web site (http://www2.ncsu.edu/unity/lockers/users/f/felder/public/EPCP.html) and make the corrections in subsequent printings of the text.
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SECTION/PROBLEM CONCORDANCE Key: i-Routine drill j-Application k-Longer or more challenging *-Computer solution required CHAPTER 2 Section
Problems
2.1-2.3 2.4 2.5 2.6 2.7
1-5 , 6-7 i j k j 8-10 , 11-12 , 13 , 14-15 i j j 16-17 , 18-19 , 20 * j j 21-28 , 29 *
i
j
i j k j k 30-31 , 32-37 , 38 , 39-41 , 42-44 *
CHAPTER 3 Section
Problems
3.1-3-2 3.3 3.4 3.5
1-2 , 3-8 , 9 , 10 , 11 , 12-13 i j k j 14-16 , 17-25 , 26 *, 27-31 i j k 32 , 33-46 , 47 i j k k 48 , 49-52 , 53 , 54 *
i
j
k
j
k
j
CHAPTER 4 Section
Problems
4.1-4.3a 4.3b-e 4.4 4.5 4.6a, b 4.6c 4.7a-e 4.7f 4.7g 4.8
1 , 2-5 j k k j k k 6-20 , 21 , 22 *, 22-25 , 26 , 27 * j k 28-30 , 31 j k 32-34 , 35-38 i k j 39-40 , 41 *, 42-45 k k 46 , 47-48 * j k 49-53 , 54-55 * j k 56-57 , 58-59 k k 60-62 , 63 * i j k j k j k k 64-65 , 66 , 67 *, 68-73 , 74-76 , 77-78 , 79 , 80 *
i
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CHAPTER 5 Section
Problems
5.1 5.2a,b 5.2c 5.3a-c 5.4a,b 5.4c
1 , 2-3 , 4 i j 5-6 , 7-15 j k j k j k 16-21 , 22-23 , 24-30 , 31-34 , 35-46 , 47-54 j k j k k 55-56 , 57 *, 58-60 , 61-62 , 63 * i j k j 64-65 , 66-69 , 70 , 71-73 j k j 74-77 , 78 ,79-83
i
j
k
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CHAPTER 6 Section 6.1 6.2, 6.3 6.4a 6.4b 6.4c 6.4d 6.5a,b 6.5c 6.6 6.7
Problems j k j 1-4 , 5 *, 6-8
j k k j k j k 6.9-29 , 30 , 31 *, 32-34 , 35-36 , 39-41 , 42 j 43-44 i j k 45-46 , 47-50 , 51 j k j k j k 52-57 , 58 *, 59 , 60 , 61-63 , 64 * j k j k 65-67 , 68-69 *, 70 , 71-73 i j k 74 , 75-80 , 81-83 i j k 84-85 , 86 , 87 j k j 88-91 , 92 , 93-97 j k 98-99 , 100-101
CHAPTER 7 Section
Problems
7.1, 7.2 7.3 7.4 7.5 7.6 7.7
1-2 , 3 , 4-6 , 7-8 i j 9 , 10-11 i j i j 12-13 , 14-17 , 18 , 19-23 j k 24-28 , 29 j k j k j k k 30-38 , 39 , 40-41 , 42-44 , 45-48 , 49-50 , 51 * j k 52-56 , 57-58
i
j
i
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CHAPTER 8 Section
Problems
8.1-8.3b 8.3c 8.3d 8.3e 8.4a 8.4b 8.4c 8.4d 8.4e 8.5
1-4 , 5-16 i 17-18 j k j k 19-25 , 26 , 27-31 , 32 j k 33 , 34 * i j 35 , 36-41 i j 42 , 43-44 j k j k k k 45-53 , 54 , 55-56 , 57-65 , 66 *, 67-68 i j k j 69-71 , 72-73 , 74 , 75 i j k 76-77 , 78-79 , 80 i j k j k k j 81-82 , 83-87 , 88-90 , 91 , 92 , 93-94 *, 95-99
i
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CHAPTER 9 Section
Problems
9.1 9.2 9.3, 9.4 9.5a 9.5b 9.5c 9.6a 9.6b
1-2 , 3-4 i 5-6 I j 7-9 , 10 j k j k j k 11-17 , 18 *, 19-21 , 22-23 , 24 *, 25-30 k k j k 31-34 , 35 *, 36 , 37-38 j k 39-44 , 45-47 j k j k 48-50 , 51 , 52-56 , 57-61 j k k k 62-63 , 64-67 , 68 *, 69-70
i
j
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CHAPTER 10 Section
Problems j
10.11-4 , 5 10.2, 10.3
k k j k 6 , 7 *, 8-14 *
CHAPTER 11 Section
Problems j
k
j
j
j
k
j
11.1, 11.2 1-2 , 3 , 4 , 5 *, 6-9 , 10 , 11-14 , 15-19 j k 11.320-26 , 27-30
k
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SAMPLE ASSIGNMENT SCHEDULE I
Assignment
Read (Ch., Sect.)
Problems Due
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
— 1; 2-2.6 2:3, 6, 9, 12 2.7 2:14, 17, 19, 22 3-3.3 2:30, 33; 3:3, 6 3.4-3.5 3:13, 16, 22, 29 4-4.3a 3:32, 37, 40, 49 4.3b-4.3e 3:47; 4:2, 4, 7 4.4 4:10, 11, 14 4.5 4:18, 25, 28 — 4:26, 29 TEST THROUGH SECTION 4.3 4.6a-b 4:32, 39 4.6c, 4.7a-d 4:37, 43 4.7e-g 4:49, 53 4.8 4:56, 65 5-5.2b 4:69, 73; 5:3 5.2c-5.3 5:7, 12, 15, 19 5.4 5:25, 26, 59 6-6.2 5:35, 40, 66; 6:1 6.3 6:2, 9, 12, 15 6.4a,b 6:23, 33 6.4c 6:37, 47 6.5 6:53, 66, 74 7-7.3 6:36, 70 TEST THROUGH SECTION 6.4 7.4-7.5 7.1, 6, 9, 10 7.6 7:12, 18, 25 8-8.3b 7:24, 28, 30 8.3c-8.3e 7:33, 35; 8:2 — 7:45; 8:5, 6, 8 8.4a-c 8:15, 18, 25 8.4d-e 8:29, 36, 44 — 8:46, 49 8.5 8:51, 55, 73 — 8:61 — 8:79, 81, 86 TEST THROUGH SECTION 8.4d 9-9.4 8: 95, 98; 9:1 9.5a 9:7, 12 9.5b-c 9:14 9.6a-b 9:17 9.6c-d 9:32 — 9:48 — 9:54 — 9:63
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Bonus Problems (*Computer problem)
2.13 2:20*, 36-43 2:38-43; 3.9,11 2:40*; 3.26* 3:53*, 54* 4:22* 4:27* 4:28*, 31, 35-36 4:38, 41* 4:39, 43* 4:47*, 48* 4:54*, 55* 4:58,59 4:60-62, 63*, 67* 4:79-80*; 5:4*, 22-23 5: 31-34, 47-54 5:57*, 61-62, 63*; 6:5* 6:30, 31*, 35, 42 6:51, 58*, 60, 64* 6:68-69, 71-73 6:81, 83, 86-87 6:88-101 7:3, 17 7:29, 39, 42-44 7:49-50, 51* 7:52-58 8:12 8:26, 32, 34* 8:54, 57-59 8:62-68 8:74, 80 8:88-90 8:93-94* 9:9 9:18* 9:22-34 9:35*, 37-38, 45-47 9:51, 57-61, 64-65 9:67-70
SAMPLE ASSIGMENT SCHEDULE II
Assignment
Read (Ch., Sect.)
Problems Due
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
— 1; 2-2.6 2:4, 7, 8, 11 2.7 2:15, 16, 18, 23 3-3.3 2:31, 34; 3:4, 7 3.4-3.5 3:12, 17, 23, 28 4-4.3a 3:32, 39, 43, 50 4.3b-4.3e 3:51; 4:3, 4, 6 4.4 4:9, 12, 15 4.5 4:21, 23, 28 — 4:26, 30 TEST THROUGH SECTION 4.3 4.6a-b 4:33, 40 4.6c, 4.7a-d 4:34, 45 4.7e-g 4:50, 51 4.8 4:57, 64 5-5.2b 4:70, 71; 5:2 5.2c-5.3 5:8, 11, 13, 17 5.4 5:25, 29, 58 6-6.2 5:38, 42, 67; 6:1 6.3 6:6, 8, 10, 17 6.4a,b 6:27, 35 6.4c 6:39, 46 6.5 6:52, 65, 75 7-7.3 6:41, 70 TEST THROUGH SECTION 6.4 7.4-7.5 7:2, 7, 9, 11 7.6 7:13, 18, 22 8-8.3b 7:24, 28, 30 8.3c-8.3e 7:32, 37; 8.1 — 7:47; 8:5, 6, 9 8.4a-c 8:14, 17, 24 8.4d-e 8:30, 37, 43 — 8:45, 50 8.5 8:53, 56, 69 — 8:61 — 8:78, 83, 85 TEST THROUGH SECTION 8.4d 9-9.4 8:96, 97; 9:2 9.5a 9:7, 11 9.5b-c 9:15 9.6a-b 9:16 9.6c-d 9:33 — 9:50 — 9:55 — 9:66
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Bonus Problems (*Computer problem)
2.13 2:20*, 36-43 2:38-43; 3.9,11 2:40*; 3.26* 3:53*, 54* 4.22* 4:26, 27* 4: 28*, 31, 35-36 4:38, 41* 4:39, 43* 4:47*, 48* 4:54*, 55* 4:58, 59 4:60-62, 63*, 67* 4:79-80*; 5:4*, 22-23 5: 31-34, 47-54 5:57*, 61-61, 63*; 6.5* 6:30, 31*, 35-36, 42 6:51, 58*, 60, 64* 6:68-69, 71-73 6:81, 83, 86-87 6:88-101 7:3, 17 7:29, 39, 42-44 7:49-50, 51* 7:52-58 8:12 8:26, 32, 34* 8:54, 57-59 8:62-68 8:74, 80 8:88-90 8:93-94* 9:9 9:18* 9:22-34 9:35*, 37-38, 45-47 9:51, 57-61, 64-65 9:67-70
SAMPLE RESPONSES TO A CREATIVITY EXERCISE The exercise that follows was given to a junior class in fluid dynamics. The students were given a week, and were told to do the exercise either individually or in pairs. The grading system used is explained in the statement of the exercise. Thirty-one individuals and nine pairs submitted responses, for a total of 40 responses from 49 students. Some students found their way to Perry’s Handbook and took ideas from there, which was perfectly acceptable; many were more inventive, and submitted a wide variety of clever, ingenious, and humorous responses. The average number of suggested flow measurement techniques was 26; the high was 53, and the low was 5. A summary of the collected responses with duplicates eliminated follows the exercise statement. Exercise You are faced with the task of measuring the volumetric flow rate of a liquid in a large pipeline. The liquid is in turbulent flow, and a flat velocity profile may be assumed (so that you need only measure the fluid velocity to determine the volumetric flow rate). The line is not equipped with a built-in flowmeter; however, there are taps to permit the injection or suspension of devices or substances and the withdrawal of fluid samples. The pipeline is glass and the liquid is clear. Assume that any device you want to insert in the pipe can be made leakproof if necessary, and that any technique you propose can be calibrated against known flow rates of the fluid. Come up with as many ways as you can think of to perform the measurement that might have a chance of working. (Example: insert a small salmon in the pipe, suspend a lure irresistible to salmon upstream of the insertion point, and time how long it takes the fish to traverse a measured section of the pipe.) You will get 1 point for every 5 techniques you think of (no fractional points awarded), up to a maximum of 10 points. Note, however: The techniques must be substantially different from one another to count. Giving me a pitot tube with 10 different manometer fluids, for example, will get you nowhere. Responses 1. Pitot tube. 2. Hot-wire or hot-film anemometer 3. Pass effluent through a venturi meter or orifice meter or nozzle meter or rotameter or ... (no credit for simply naming the meter if it can’t be easily inserted in the pipeline). 4. Pass effluent into a weir, measure height in notch. 5. Inject dye, measure time for it to traverse a known length. 6. Insert a solid object (such as a balloon, a bucket, a cork, a marble, a bar of Ivory soap, or the 311 book), measure time for it to traverse a known length (or travel alongside it on a bicycle or moped or pickup truck and note your speed, or attach it to a string on a spool and measure the rate of rotation of the spool). 7. Insert a series of solid objects, measure rate at which they pass a point (or frequency of collisions with the pipe wall, or rate of collection on a filter).
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8. Inject dye at fixed rate, shine light on the pipe, measure light absorbance downstream (or angle of refraction or turbidity, or put a sunbather under the pipe and measure his rate of tanning). 9. Measure the energy being consumed by the pump being used to move the fluid. 10. Put a magnet in the flow, measure the magnetic force required to hold it in place (or measure its velocity along the wall, or have two external magnetic switches triggered by its passage and time the interval between events, or measure the rate of motion of a compass needle as the magnet passes). 11. Collect effluent (or a sidestream), measure amount (volume, mass) collected in a known time interval (or the rate at which the level in the container rises, or the time required to fill a known volume or to saturate a sponge, or to water a plant or wash a pulp sample, or to saturate a plot of ground in Ethiopia where they really need it). 12. Direct effluent into a container of salt, see how much dissolves. 13. Direct effluent against a raft in a pond, measure its velocity. 14. Discharge effluent horizontally, letting it fall through a known height, and measure its horizontal displacement. 15. Discharge effluent horizontally, and measure the force it exerts on a plate. 16. Discharge effluent (or a sidestream) upward, measure height of fountain (or suspend a ping pong ball at the top, and measure its height) 17. Discharge effluent downward from a flexible hose, and measure height of nozzle above the ground. 18. Let fluid fill a balloon (or a water bed), measure time required for explosion, or volume increase in a known time interval. 19. Insert a U-tube at each of two points in the line, use as a manometer (either straight pipe between points or insert an obstacle to flow, like an orifice or a solid object). 20. Insert paddle wheel (or a water wheel at the outlet), measure rotational speed. 21. Insert propeller, measure rotational speed by counting or automatically. 22. Insert turbine-generator, measure work output (or intensity of light attached to generator). 23. Suspend solid object on a string, measure angle made by string with vertical (or horizontal displacement of object, or rotation of a lever arm, or angle at which your hand is bent back). 24. Drop in an object denser than the fluid, measure horizontal distance traversed before hitting the bottom of the channel. 25. Inject from below an object lighter than the fluid (e.g. a bubble), measure horizontal distance traversed before hitting the top of the channel.
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26. Inject from below an object heavier than the fluid, measure its horizontal displacement (or follow its trajectory using stop-motion photography). 27. Put a flexible fiber (or membrane or easily deformed plate) in the path of the flow, measure its distension in flow direction, or thickness at which flow is sufficient to break it. 28. Pluck a guitar string in the flow, time its period of vibration. 29. Attach tape to the wall, time its unraveling. 30. Feed effluent into a centrifugal pump or a lobbed-impeller flowmeter, measure rotational speed. 31. Measure height of fluid in a vertical standpipe coming from the top of the pipe. 32. Determine time required for effluent to sink a ship (or to flood out the football coach’s house, hopefully with some of his players in it). 33. Determine time required for effluent to float a duck out of a well of known depth (or to float an object of known weight and displacement). 34. Insert a solid object (e.g. a snowball, or a tootsie pop, or Alka Seltzer, or the Wicked Witch of the West), determine time required to dissolve it (or wear it away, or wash the paint off it). 35. Insert an absorbing object, measure its rate of fluid uptake. 36. Insert solid objects of different sizes, find the one such that the drag force is just sufficient to initiate motion (or measure rate at which a given object is dragged along the pipe). 37. Measure vibration intensity or amplitude of tube (or noise level, or sound of a bell clapper), either naturally occurring or after the pipe is struck. 38. Insert an iron bar, measure rate of corrosion. 39. Propel a bullet (or an arrow, or a torpedo, or Nolan Ryan’s fastball) into the pipe outlet, measure distance it travels before stopping. 40. Determine velocity of immersed submarine moving against (or with) the flow (or Mark Spitz, or a trout approaching a lure, or a seal approaching food, or a squirrel approaching an acorn, or a hungry dog approaching a dead rabbit, or a snake approaching a mouse, or a sailor approaching a mermaid, or a horny male frog approaching erotic pictures of female dancing frogs, or a 311 student after this test approaching free beer). 41. Measure vibrational speed of a trout’s tail swimming against the flow and remaining stationary (or the rate of flapping of a piece of cloth or the rate of wobble of a nutating disk or the magnitude of noise generated by “chatterbox” lure). 42. Insert balloon (or piston-fitted cylinder with piston facing upstream, or closed tube with flexible diaphragm covering opening), measure final gas pressure (or rate of pressure increase or rate of motion of piston or intensity of whistle if piston drives gas through it).
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43. Insert solid object, measure force required to hold it still (or extension or compression of a spring or elastic band, or put the object against razor blade and see how long it takes for the blade to split it). 44. Insert a solid object, determine distance required for downstream wake to disappear. 45. Put in plug, measure force required to hold it in (or distance it travels when it is ejected). 46. Measure the shear force on the pipe wall (with or without a bend in the line), or the extension of the pipe length due to shear. 47. Measure the rate of heat generation or temperature rise due to friction in the pipeline. 48. Measure rate at which air is drawn through a Buchner funnel (or pitch or intensity at which it is drawn through a whistle) by the suction created by the flowing fluid. 49. Add heat, measure temperature rise (or rate of evaporation). 50. Insert a hot object, measure its rate of cooling (or a cold object, and measure its rate of heating). 51. Cool, measure temperature drop (or rate of freezing). 52. Pass effluent through a heat exchanger, measure rate of heat transfer. 53. Add an acid or base at a known rate, measure pH downstream or determine amount needed to change litmus paper color (or add salt and measure change in electrical conductivity, or add a radioisotope and measure change in activity, or add a phosphorescent substance and measure luminescence, or add any chemical and measure its concentration by any means). 54. Add sugar at a known rate, measure rate of formation of rock candy downstream. 55. Add a second liquid of different density, measure resulting density change (or add an immiscible liquid, measure its rate of passage). 56. Add a reactant, determine amount needed to react completely with the fluid (or with another reactant on a permeable membrane in the flow channel, or inject chlorine ions and measure the rate of electroplating on a silver electrode). 57. Get a technician to drink the effluent, measure his weight gain after a fixed time (or the time required for his mouth to fill up, or the time required to drown a rat). 58. Add alcohol (or poison, or salt, or Kool-Aid mix) to the fluid at a known rate, have a technician drink it (or do it yourself), and determine the time required to feel the effects. 59. Immerse pipe outlet in water, find the depth at which the hydrostatic head is just sufficient to stop the flow. 60. Insert air tube facing upstream, determine pressure needed to initiate bubbling. 61. Place pipe in wind tunnel, find wind velocity just adequate to stop the flow.
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62. If pipe is only partially filled, put in sailboat, measure wind force needed to hold it stationary. 63. Put another pipe against outlet, find flow in second pipe that just neutralizes unknown flow. 64. Send sound wave through, time passage over known distance (or use Doppler meter, or time passage of an electrical impulse or a light wave). 65. Put bacteria in line, determine reproduction rate. 66. Put algae in pipe, measure change in COD. 67. Put a spawning fish in the line, measure how far the eggs travel in a given time interval. 68. Measure how long it takes for the effluent to put out a fire of a given size. 69. Pass the fluid spirally into a funnel, measure how long it takes for a drop of dye to disappear. 70. If the fluid is combustible, burn it in a combustion engine and measure the rate of power output. 71. Determine how long you can hold your breath, then jump in and see how far you travel before you have to breathe (or see if an animal can make it out of the pipe before drowning). 72. Add effluent to bubble bath, measure extent of generation of bubbles. 73. Insert a fish with a monitor in its heart, time how long it takes him to die. (Must kill a lot of fish to calibrate-don’t tell “Save the Whales.”) 74. Count rate of passage of molecules. 75. Insert a monkey who can insert pegs in holes at a known rate, and count the number of pegs inserted over a known distance. 76. Install a Japanese flowmeter equipped with lots of flashing lights. 77. Correlate the velocity with the rate at which the student pulls out his hair during the experiment. 78. Hire someone to do it. 79. Break into the pipeline company’s office and steal the flow rate records. 80. Look it up in the Enquirer flow rate tables. 81. Fill a balloon, throw it at your boss, and correlate his anger with the flow rate. (?) 82. Ask your local psychic.
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TRANSPARENCY MASTERS The following pages contain oversized renderings of illustrations taken from the text. The illustration numbers are listed below with their text page numbers. You are granted permission to have these illustrations reproduced as transparencies for your own use in conjunction with the textbook, Felder and Rousseau: ELEMENTARY PRINCIPLES OF CHEMICAL PROCESSES, 3rd Edition, John Wiley & Sons. Resale is expressly prohibited. Transparencies may easily be prepared using either thermal copy (ThermoFax) or electrostatic copy (Xerox) machines. See the operating manual for your particular copy machine. Transparency film can be purchased from your usual copy paper supplier. Other illustrations in the book may also serve as transparency masters. For best results, it may be necessary to enlarge the illustration to fill the sheet of copy film. Many electrostatic copiers have this capability. Figure Description Compressibility charts
Cox vapor pressure chart Psychrometric chart – SI units Psychrometric chart – Am. Engr. units Enthalpy conc. chart – H2SO4/H2O Enthalpy conc. chart – NH3/H2O
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Figure Number
Text Page
5.4-1 5.4-2 5.4-3 5.4-4 6.1-4 8.4-1 8.4-2 8.5-1 8.5-2
208 209 210 211 247 382 383 396 400
CHAPTER TWO 2.1 (a) (b) (c)
2.2 (a)
3 wk
7d
24 h
3600 s 1000 ms
1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s 3.2808
ft
554 m 4 1d 1h d kg 24 h 60 min 760 mi h
1
1 h
u 10 9 ms 18144 . 25.98 mi / h 26.0 mi / h
1 kg 10 8 cm 4 1000 g 1 m 4
m
1 h
0.0006214 mi 3600 s
340 m / s
1 m3 35.3145 ft 3
(b)
921 kg 2.20462 lb m m3 1 kg
(c)
5.37 u 10 3 kJ 1 min 1000 J min 60 s 1 kJ
3.85 u 10 4 cm 4 / min g
57.5 lb m / ft 3
1.34 u 10 -3 hp 119.93 hp 120 hp 1 J/s
2.3 Assume that a golf ball occupies the space equivalent to a 2 in u 2 in u 2 in cube. For a classroom with dimensions 40 ft u 40 ft u 40 ft : 40 u 40 u 40 ft 3 (12) 3 in 3 1 ball n balls 6.48 u 10 6 | 7 million balls 3 3 3 2 in ft The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h 1 yr 1 d
3600 s 1.86 u 105 mi 3.2808 ft 1 step 1 h 1 s 0.0006214 mi 2 ft
2.5 Distance from the earth to the moon = 238857 miles 238857 mi
1
m
1 report
0.0006214 mi
0.001 m
4 u 1011 reports
2.6 19 km 1000 m 0.0006214 mi 1000 L 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. Total Cost
Total Cost
American
European
$14,500
$21,700
Equate the two costs x
44.7 mi / gal
$1.25 1 gal x (mi) 14,500 0.04464 x gal 28 mi x (mi) $1.25 1 gal gal 44.7 mi
4.3 u 10 5 miles
2-1
21,700 0.02796 x
7 u 1016 steps
2.7 10 6 cm 3 220.83 imp. gal
5320 imp. gal 14 h 365 d plane h 1 d 1 yr ton kerosene u 10 5 1188 . plane yr
0.965 g 1 cm 3
1 kg 1 ton 1000 g 1000 kg
4.02 u 10 9 ton crude oil 1 ton kerosene plane yr 5 u 10 ton kerosene yr 7 ton crude oil 1188 . 4834 planes 5000 planes
2.8 (a) (b) (c)
2.9
2.10 2.11
32.1714 ft / s 2
25.0 lb m 25 N
1 9.8066 m / s 2
10 ton
1 lb m 5 u 10 -4 ton
50 u 15 u 2 m 3
500 lb m
(a) mdisplaced fluid
1 lb f 32.1714 lb m ft / s 2
1 kg m / s 2 1N
2.55 kg 2.6 kg 980.66 cm / s 2
1000 g 2.20462 lb m
35.3145 ft 3 1 m3
25.0 lb f
85.3 lb m 1 ft 3
1 dyne 1 g cm / s 2
1 lb f 32.174 ft 2 32.174 lb m / ft s 2 1 s
FG IJ FG 1 IJ | 25 m H K H 10K
1 m3 1 | 5 u 10 2 2 11.5 kg
1 kg 2.20462 lb m
mcylinder U f V f
9 u 10 9 dynes
U cVc U f hSr 2
4.5 u 10 6 lb f
3
U c HSr 2
Uc
Ufh
(30 cm 14.1 cm)(100 . g / cm 3 ) Uc 0.53 g / cm 3 H 30 cm U c H (30 cm)(0.53 g / cm 3 ) (b) U f 171 . g / cm 3 h (30 cm - 20.7 cm) 2.12
Vs
SR 2 H ; Vf 3
R SR 2 H Sr 2 h ; H 3 3
U fVf
U sVs U f
Uf
Us
H H
h3 H2
2
2
2
3
H3 H 3 h3
h
3
2
H
r
2
Us
s
2
Us
Uf h
R h H
FG IJ SR FG H h IJ H K 3H HK h I SR F SR H H U J G 3 H 3 H K
SR 2 H Sh Rh 3 3 H
Vf
r r h
H
Us
R
1
F hI 1 G J H HK 2-2
3
Uf
2.13
Say h m
depth of liquid y y= 1 dA y= 1– h x
2
A(m )
z
h
1m
x = 1– y
2
1 y 2
dA dy
2 1 y 2 dx
dx
1 y 2
d i
Am
2
E d i
A m2
b g
W N
z z 1
1
dA 2
1 h
1 y 2 dy
1 h
Table of integrals, or trigonometric substitution
S 2 1 h 1 1 h sin 1 1 h 2 1 h 4 m u A( m 2 ) 0.879 g 10 6 cm 2 1 kg 9.81 N 3.45 u 10 4 A cm 3 1 m3 10 3 g kg , y 1 y 2 sin 1 y
1
E Substitute for A LS W b N g 3.45 u 10 M b1 hg N2 4
b g
b g
b g
g g0
b g
1 1 h
2
b gOPQ
sin 1 1 h
32.174 lb m ft / s 2 1 slug = 32.174 lb m 1 lb f 1 poundal = 1 lb m ft / s 2 32.174 (a) (i) On the earth: 175 lb m 1 slug M 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W 5.63 u 10 3 poundals 2 s 1 lb m ft / s 2 (ii) On the moon 175 lb m 1 slug M 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W 938 poundals 2 6 s 1 lb m ft / s 2
2.14 1 lb f
1 slug ft / s 2
2-3
2.14 (cont’d) ma a
F/m
0.135 m / s
2
(b) F
2.15 (a) F
25.0 slugs
ma 1 fern = (1 bung)(32.174 ft / s 2 )
FG 1IJ H 6K
1 slug 32.174 lb m
1m 3.2808 ft
5.3623 bung ft / s 2
1 fern 5.3623 bung ft / s 2
(b) On the moon: W On the earth: W 2.16 (a) | (3)(9)
3 bung 32.174 ft 1 fern 2 6 s 5.3623 bung ft / s 2 (3)(32.174) / 5.3623 = 18 fern
(b)
27
(2.7)(8.632)
23
(c) | 2 125 127
(d)
2.365 125.2 127.5 2.17 R |
1 lb m ft / s 2 1 poundal
355 poundals
3 fern
40 u 10 4 5 u 8 u 10 4 | | 1 u 10 4 45 5u 9 4 (3.600 u 10 ) / 45 9 u 10 5 |
| 50 u 10 3 1 u 10 3 | 49 u 10 3 | 5 u 10 4 4.753 u 10 4 9 u 10 2
5 u 10 4
(7 u 10 1 )(3 u 10 5 )(6)(5 u 10 4 ) | 42 u 10 2 | 4 u 10 3 6 (3)(5 u 10 )
Rexact
3812.5 3800 38 . u 10 3
2.18 (a) A: R
731 . 72.4
0.7 o C
X
72.4 731 . 72.6 72.8 73.0 5
s
(72.4 72.8) 2 (731 . 72.8) 2 (72.6 72.8) 2 (72.8 72.8) 2 (73.0 72.8) 2 51
72.8 o C
0.3o C B: R 1031 . 97.3 58 . oC X
97.3 1014 . 98.7 1031 . 100.4 100.2 o C 5
s
(97.3 100.2) 2 (1014 . 100.2) 2 (98.7 100.2) 2 (1031 . 100.2) 2 (100.4 100.2) 2 51 2.3o C
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2-4
2.19 (a)
12
12
¦
¦ ( X 735. )
Xi
i 1
X
s
C min=
735 . 12 X 2 s 735 . 2(12 . )
711 .
C max=
X 2 s 735 . 2(12 . )
75.9
2
i 1
12 .
12 1
(b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness 2.20 (a), (b) 1 2 (a) Run 134 131 X Mean(X) 131.9 Stdev(X) 2.2 127.5 Min 136.4 Max (b) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14
X 128 131 133 130 133 129 133 135 137 133 136 138 135 139
Min 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5
3 129
4 5 6 7 8 9 10 11 12 13 14 15 133 135 131 134 130 131 136 129 130 133 130 133
Mean 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9
Max 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4
140 138 136 134 132 130 128 126 0
5
10
15
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.
2.21 (a) Q'
2.36 u 10 4 kg m 2 h
2.10462 lb 3.2808 2 ft 2 kg m2
(2 u 10 4 )(2)(9) | 12 u 10 ( 4 3) | 12 . u 10 6 lb ft 2 / s 36 u 10 3 2 . u 10 lb ft / s 0.00000148 lb ft 2 / s = 148
(b) Q' approximate | Q' exact
1 h 3600 s
2-5
CpP
2.22 N Pr
k
0.583 J / g o C 1936 lb m 0.286 W / m o C ft h
1 h 3.2808 ft 1000 g 3600 s m 2.20462 lb m
(6 u 10 1 )(2 u 10 3 )(3 u 10 3 ) 3 u 10 3 . u 10 3 . The calculator solution is 163 . u 10 3 | | 15 2 (3 u 10 1 )(4 u 10 3 )(2)
N Pr | 2.23 Re
DuU P
Re |
(5 u 10 1 )(2)(8 u 10 1 )(10 6 ) 5 u 101 ( 3) | | 2 u 10 4 the flow is turbulent 3 4 3 (3)(4 u 10)(10 )(4 u 10 )
0.48 ft 1 m 2.067 in 1 m 3 s 3.2808 ft 0.43 u 10 kg / m s 39.37 in
0.805 g cm 3
1 kg 10 6 cm 3 1000 g 1 m 3
2.24 (a) kgd p y
F P IJ FG d uU IJ 2.00 0.600G D H UD K H P K L 100 . u 10 N s / m 2.00 0.600M . kg / m )(100 . u 10 m N (100 1/ 2
1/ 3
p
5
5
3
44.426
k g (0.00500 m) (0100 . ) 100 . u 10 5 m 2 / s
OP LM (0.00500 m)(10.0 m / s)(100 . kg / m ) O PQ / s) Q N (100 . u 10 N s / m ) 1/ 3
2
2
3
5
44.426 k g
1/ 2
2
0.888 m / s
(b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) dp (m) 0.005 0.010 0.005 0.005 0.005
y 0.1 0.1 0.1 0.1 0.1
D (m2/s) 1.00E-05 1.00E-05 2.00E-05 1.00E-05 1.00E-05
P (N-s/m2) U (kg/m3) u (m/s) 1.00E-05 1 10 1.00E-05 1 10 1.00E-05 1 10 2.00E-05 1 10 1.00E-05 1 20
kg 0.889 0.620 1.427 0.796 1.240
2.25 (a) 200 crystals / min mm; 10 crystals / min mm 2
(b) r
200 crystals 0.050 in 25.4 mm 10 crystals min mm in min mm 2 238 crystals / min
b g
(c) D mm
b g
D c in
60r c
(25.4) 2 mm 2 in 2
238 crystals 1 min 4.0 crystals / s 60 s min
FG H
25.4 mm crystals 25.4 D c ; r min 1 in
b
0.050 2 in 2
g b
200 25.4 D c 10 25.4 D c
g
2
2-6
IJ K
rc
crystals
60 s
s
1 min
b g
r c 84.7 D c 108 D c
2
60r c
2.26 (a) 70.5 lb m / ft 3 ; 8.27 u 10 -7 in 2 / lb f (b) U
L8.27 u 10 / ft ) exp M N 3
(70.5 lb m
35.3145 ft 3
70.57 lb m ft
FG lb IJ H ft K F lb IJ PG H in K
(c) U
3
m
Uc
m 3
f 2
g cm 3
P'
62.43U c
N m2
1
7
in 2 lb f
9 u 10 6 N m2
m3
1000 g
6
3
10 cm
3
2.20462 lb m
1 lb m 28,317 cm 3 g 1 ft 3 453593 .
d
OP Q
113 . g / cm 3
62.43U c
12 m2 39.37 2 in 2
0.2248 lb f 1N
14.696 lb f / in 2 u 10 6 N / m 2 101325 .
145 . u 10 4 P'
id
i
d
P' 9.00 u 10 6 N / m 2 U ' 113 . exp[(120 . u 10 10 )(9.00 u 10 6 )] 113 . g / cm 3 cm d i d i 28,317 1728 in 16.39V ' expb3600t c g V '
2.27 (a) V cm 3
i
70.5 exp 8.27 u 10 7 145 . u 10 4 P ' U c 113 . exp 120 . u 10 10 P '
V ' in 3
3
3
b g 3600t cbhrg 0.06102 expb3600t cg 16.39V ' ; t s
(b) The t in the exponent has a coefficient of s-1. 2.28 (a) 3.00 mol / L, 2.00 min -1 0 C 3.00 exp[(-2.00)(0)] = 3.00 mol / L t = 1 C 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 3.00 Cint For t=0.6 min: (0.6 0) 3.00 14 . mol / L 1 0 Cexact 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L
(b) t
For C=0.10 mol/L:
1 0 (010 . 3.00) 0 112 . min 0.406 3 1 C 1 0.10 =ln = - ln = 1.70 min 2.00 3.00 2 3.00
t int t exact
(c) 3.5 C exact vs. t
3 C (mol/L)
2.5 2
(t=0.6, C =1.4)
1.5 1
(t=1.12, C=0.10)
0.5 0 0
1
2
t (min)
2-7
p*
2.29 (a) (b)
60 20 (185 166.2) 20 42 mm Hg 199.8 166.2
c
MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, * ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I=1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I=I+1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END DATA OUTPUT 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 215.5 100.0 105.0 1.8 215.0 98.7
2.30 (b) ln y b ln a (c) ln y b ln a
ln a bx y ae bx (ln y 2 ln y1 ) / ( x 2 x1 ) ln y bx
(ln 2 ln 1) / (1 2)
ln 2 0.63(1) a
4.00 y
0.693
4.00e 0.693 x
ln a b ln x y ax b (ln y 2 ln y1 ) / (ln x 2 ln x1 ) (ln 2 ln 1) / (ln 1 ln 2) ln y b ln x ln 2 ( 1) ln(1) a 2 y 2 / x
1
(d) ln( xy ) ln a b( y / x ) xy ae by / x y (a / x )e by / x [can' t get y f ( x )] b [ln( xy ) 2 ln( xy ) 1 ] / [( y / x ) 2 ( y / x ) 1 ] (ln 807.0 ln 40.2) / ( 2.0 10 . ) ln a
ln( xy ) b( y / x )
ln 807.0 3 ln( 2.0) a
2-8
2 xy
2e 3 y / x y
3
(2 / x )e 3 y / x
2.30 (cont’d) (e) ln( y 2 / x )
a ( x 2) b y [ax ( x 2) b ]1/ 2
ln a b ln( x 2) y 2 / x
b [ln( y 2 / x ) 2 ln( y 2 / x ) 1 ] / [ln( x 2) 2 ln( x 2) 1 ] (ln 807.0 ln 40.2) / (ln 2.0 ln 10 . ) 4.33 ln( y 2 / x ) b( x 2)
ln a
y2 / x
ln 807.0 4.33 ln(2.0) a
40.2( x 2) 4.33 y
6.34 x 1/ 2 ( x 2) 2.165
2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope
(c)
1 ln( y 3)
(d)
1 ( y 1) 2
1 a b b
x Plot
a ( x 3) 3 Plot
40.2
1 vs. ln( y 3)
m, Intcpt
n
x [rect. axes], slope =
1 a , intercept = b b
1 vs. ( x 3) 3 [rect. axes], slope = a , intercept = 0 ( y 1) 2
OR 2 ln( y 1) ln a 3 ln( x 3) Plot ln( y 1) vs. ln( x 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a slope = , intercept = 2 2 (e) ln y
a x b
Plot ln y vs. (f) log10 ( xy )
x [rect.] or y vs.
x [semilog ], slope = a, intercept = b
a( x 2 y 2 ) b
Plot log10 ( xy ) vs. ( x 2 y 2 ) [rect.] slope = a, intercept = b (g)
1 y OR
ax 1 y
b x x y
ax
ax 2 b Plot
b 1 x xy
a
x vs. x 2 [rect.], slope = a , intercept = b y
b 1 1 Plot vs. 2 [rect.] , slope = b, intercept = a 2 xy x x
2-9
2.32 (a) A plot of y vs. R is a line through ( R 5 , y
0.011 ) and ( R 80 , y
0169 . ).
0.18 0.16 0.14 0.12 y
0.1 0.08 0.06 0.04 0.02 0 0
20
40
60
80
100
R
y
aRb
a b
U| V| W
0.011 . 0169 2.11 u 10 3 80 5 y 0.011 2.11 u 10 3 5 4.50 u 10 4
ib g
d
2.11 u 10 3 R 4.50 u 10 4
d2.11 u 10 ib43g 4.50 u 10 0.092 kg H O kg b1200 kg hgb0.092 kg H O kgg 110 kg H O h
(b) R
3
43 y
4
2
2.33 (a) ln T b ln a (b) T T T T
2
ln a b ln I T aI b (ln T2 ln T1 ) / (ln I 2 ln I 1 ) ln T b ln I
2
(ln 120 ln 210) / (ln 40 ln 25)
ln 210 ( 119 . ) ln( 25) a
9677.6 T
119 .
9677.6I 1.19
b9677.6 / T g 85 C I b9677.6 / 85g 535 . L/s 175 C I b9677.6 / 175g 29.1 L / s 290 C I b9677.6 / 290g 19.0 L / s 1.19
9677.6I 1.19 I
1.19
o
o
1.19
o
1.19
(c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.
2-10
ln ((CA-CAe)/(CA0-CAe))
2.34 (a) Yes, because when ln[(C A C Ae ) / (C A0 C Ae )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0
50
100
150
200
0 -0.5 -1 -1.5 -2 t (m in)
Slope = -0.0093 k = 9.3 u 10-3 min 1 (b) ln[(C A C Ae ) / (C A0 C Ae )] kt C A 0.0495)e ( 9.3u10 (01823 .
CA
3
)(120)
(C A 0 C Ae )e kt C Ae
0.0495 = 9.300 u 10-2 g / L
9.300 u 10-2 g 30.5 gal 23.317 L 8.8 g L 7.4805 gal
C = m / V m = CV
2.35 (a) ft 3 and h -2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353 . u 10 2 ) ; or V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 . u 10 2 3 3 7 2 (c) V ( m ) 100 . u 10 exp(15 . u 10 t ) 2.36 PV k
C / V k ln P
C P
ln C k lnV
8.5
lnP
8 7.5 7 6.5 6 2.5
3
lnP = -1.573(lnV ) + 12.736
k
slope
3.5 lnV
( 1573 . ) 1573 . (dimensionless)
Intercept = ln C 12.736 C G GL G0 G
G G 1 0 m G GL KLC
e12 .736
3.40 u 105 mm Hg cm4.719
K L C m ln
G0 G G GL
ln K L m ln C
ln ( G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5
3 ln(G 0-G)/(G-G L )
2.37 (a)
4
2 1 0 -1 3 .5
4
4 .5
5
ln C
2-11
5 .5
2.37 (cont’d) m slope 2.483 (dimensionless) 10.045 K L
Intercept = ln K L
4.340 u 10 5 ppm-2.483
G 180 . u 103 u 10 3 . 4.340 u 105 (475) 2.483 G 1806 3.00 u 10 3 G C=475 ppm is well beyond the range of the data. 475
(b) C
2.38 (a) For runs 2, 3 and 4: Z aV b p c ln Z ln a b lnV c ln p b 0.68 ln(35 . ) ln a b ln(102 . ) c ln( 9.1) c 146 . ln(2.58) ln a b ln(102 . ) c ln(112 . ) ln(3.72) ln a b ln(175 . ) c ln(112 . ) a = 86.7 volts kPa 1.46 / (L / s) 0.678 . Slope=b, Intercept= ln a c ln p (b) When P is constant (runs 1 to 4), plot ln Z vs. lnV 2
lnZ
1.5 1 0.5 0 -1
-0.5
0
lnZ = 0.5199lnV + 1.0035
0.5
1
1.5
b
lnV
slope
0.52
Intercept = lna c ln P 10035 . When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a c lnV 2
lnZ
1.5 1 0.5 0 1.5
1.7
lnZ = -0.9972lnP + 3.4551
1.9
2.1
2.3
c
lnP
slope
0.997 10 .
Intercept = lna b lnV
3.4551
Z
Plot Z vs V b P c . Slope=a (no intercept) 7 6 5 4 3 2 1 0.05
Z = 31.096VbPc
0.1
0.15
0.2
Vb P c
a
slope
311 . volt kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2-12
2.39 (a) sxy sxx
a b y (b) a
¦x y
1 n
¦x
1 n
sx
n
1 n
[(0.4)(0.3) (2.1)(19 . ) (31 . )(3.2)] / 3 4.677
i i
i 1 n
2 i
(0.32 19 . 2 3.2 2 ) / 3 4.647
i 1 n
¦
(0.3 19 . 3.2) / 3 18 . ; sy
xi
i 1
sxy sx s y
b g
sxx sx
4.677 (18 . )(1.867) 4.647 (18 . )2
2
sxx s y sxy sx
b g
sxx sx
1 n
n
¦y
i
0.936
(4.647)(1867 . ) (4.677)(18 . ) 2 4.647 (18 . )
2
(0.4 2.1 31 . ) / 3 1867 .
i 1
0182 .
0.936 x 0182 . sxy
4.677 4.647
sxx
x y 10065 10065 . .
4
y
3
y = 0.936x + 0.182
2
y = 1.0065x
1 0 0
1
2
3
4
x
2.40 (a) 1/C vs. t. Slope= b, intercept=a slope = 0.477 L / g h;
a
Intercept = 0.082 L / g
3 2.5 2 1.5 1 0.5 0
2 1.5 C
1/C
(b) b
1 0.5 0
0
1
2
3
4
5
6
1
t
1/C = 0.4771t + 0.0823
C
2 C-fitted
3
4
5
t
(c) C 1 / (a bt ) 1 / [0.082 0.477(0)] 12.2 g / L t
(1 / C a ) / b
(1 / 0.01 0.082) / 0.477
209.5 h
(d) t=0 and C=0.01 are out of the range of the experimental data. (e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.
2-13
2.41 (a) and (c)
y
10
1 0.1
1
10
100
x
ax b ln y
(b) y
ln a b ln x; Slope = b, Intercept = ln a
ln y = 0.1684ln x + 1.1258 2
ln y
1.5 1 0.5 0 -1
0
1
2 ln x
3
4
b slope
5
0168 .
a Intercept = ln a 11258 .
3.08
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b)
400
600
ln(1-Cp/Cao)
t
k
200
Lab 1
100
600
800
-2 -4
0
k
Lab 3
0.0063 s-1
500
600
t
200
400
Lab 2
0.0111 s-1
600
800
0 -2 -4
-6 ln(1-Cp/Cao)= -0.0064t
t
400
-4
k
0
-6 ln(1-Cp/Cao) = -0.0063t
300
-2
0.0062 s-1
400
200
0
-6 ln(1-Cp/Cao) = -0.0111t
ln(1-Cp/Cao)
ln(1-Cp/Cao)
0
0
800 ln(1-Cp/Cao)
0 200 0 -1 -2 -3 -4 ln(1-Cp/Cao) = -0.0062t
t
k
Lab 4
0.0064 s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k 0.0063 s-1 (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.
2-14
2.43 yi
axi I (a )
n
¦
n
¦by
di2
i
i 1
a
n
¦ i 1
2.44
i 1
n
yi xi /
axi
g
2
dI da
n
0
¦ 2b y
i
i 1
g
axi xi
n
¦y x
i i
i 1
a
n
¦x i 1
¦x
2 i
i 1
DIMENSION X(100), Y(100) READ (5, 1) N C N = NUMBER OF DATA POINTS 1FORMAT (I10) READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2) SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J) AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X 'PROBLEM 2-39'/) WRITE (6, 4) A, B 4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X * 'RESIDUALb =', F6.3) 200SSQ = SSQ + RES ** 2 WRITE (6, 6) SSQ 6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3) STOP END $DATA 5 1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15 3.0 15.38 SOLUTION: a 6.536, b 4.206
2-15
2 i
0
2.45 (a) E(cal/mol), D0 (cm2/s) (b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0. (c) Intercept = ln D0 = -3.0151 D0 = 0.05 cm2 / s .
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
Slope = E / R = -3666 K E = (3666 K)(1.987 cal / mol K) = 7284 cal / mol
-10.0 ln D
-11.0 -12.0 -13.0 -14.0
ln D = -3666(1/T) - 3.0151
1/T
(d) Spreadsheet T 347 374.2 396.2 420.7 447.7 471.2
D 1.34E-06 2.50E-06 4.55E-06 8.52E-06 1.41E-05 2.00E-05
1/T 2.88E-03 2.67E-03 2.52E-03 2.38E-03 2.23E-03 2.12E-03 Sx Sy Syx Sxx -E/R ln D0
lnD (1/T)*(lnD) -13.5 -0.03897 -12.9 -0.03447 -12.3 -0.03105 -11.7 -0.02775 -11.2 -0.02495 -10.8 -0.02296 2.47E-03 -12.1 -3.00E-02 6.16E-06 -3666 -3.0151
D0
7284
E
0.05
2-16
(1/T)**2 8.31E-06 7.14E-06 6.37E-06 5.65E-06 4.99E-06 4.50E-06
CHAPTER THREE 3.1
16 u 6 u 2 m3 1000 kg | 2 u 10 5 2 103 | 2 u 105 kg 3 m
b
(a) m
8 oz
(b) m
2s
gb gb gd i
106 cm3 1 qt 1g 4 u 106 | | 1 u 102 g / s 3 3 32 oz 1056.68 qt cm 3 u 10 10
b
gd i
(c) Weight of a boxer | 220 lb m 12 u 220 lb m 1 stone | 220 stones Wmax t 14 lb m (d)
dictionary
SD 2 L 4
V |
314 . 4.52 ft 2 4
d
800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 3 31.5 gal
i d
i
3 u 4 u 5 u 8 u 102 u 5 u 103 u 7 4 u 3 u 10
(e) (i)V |
| 1 u 107 barrels
6 ft u 1 ft u 0.5 ft 28,317 cm3 | 3 u 3 u 104 | 1 u 105 cm3 1 ft 3
(ii)V |
150 lb m
1 ft 3 62.4 lb m
28,317 cm3 150 u 3 u 104 | | 1 u 105 cm3 3 60 1 ft
(f) SG | 105 . 3.2
(a) (i)
(ii) (b) U 3.3
(a)
995 kg 1 lb m 0.028317 m3 m3 0.45359 kg 1 ft 3 995 kg / m3
62.43 lb m / ft 3 1000 kg / m3
U H2 O u SG
62.43 lb m / ft 3 u 5.7
50 L
62.12 lb m / ft 3
62.12 lb m / ft 3 360 lb m / ft 3
0.70 u 103 kg 1 m3 35 kg m3 103 L
(b)
1150 kg m3 1000 L 1 min 27 L s 0.7 u 1000 kg 1 m3 60 s min
(c)
10 gal 1 ft 3 0.70 u 62.43 lb m # 29 lb m / min 2 min 7.481 gal 1 ft 3
3-1
3.3 (cont’d) (d) Assuming that 1 cm3 kerosene was mixed with Vg (cm3 ) gasoline
d
i d i 1dcm kerosenei 0.82dg kerosenei d0.70V 0.82idg blendi 0.78 V SG V 1dcm blend i
Vg cm3gasoline 0.70Vg g gasoline 3
0.82 0.78 0.78 0.70
g
3
g
g
Vgasoline Vkerosene
Volumetric ratio
3.4
In France: In U.S.:
3.5
50.0 kg
L
0.50 cm3 1 cm3 5 Fr
3
0.5 0 cm
0.50 cm3 gasoline / cm3 kerosene $1
$68.42 0.7 u 1.0 kg 1L 5.22 Fr 50.0 kg L 1 gal $1.20 $22.64 0.70 u 1.0 kg 3.7854 L 1 gal
VB ( ft 3 / h ), m B ( lb m / h )
V ( ft 3 / h), SG
VH ( ft 3 / h ), m H ( lb m / h )
0.850
700 lb m / h
700 lb m ft 3 1319 . ft 3 / h h 0.850 u 62.43 lb m 3 VB ft 0.879 u 62.43 lb m kg / h m B 54.88V B h ft 3 m H VH 0.659 u 62.43 4114 . VH kg / h
(a) V
d i bg d hb
VB VH
b
g
b
g
g
. ft 3 / h 1319 54.88V 4114 . V
m B m H 700 lb m B H 3 VB 114 . ft / h m B 628 lb m / h benzene VH
. ft 3 / h m H 174
. lb m / h hexane 716
(b) – No buildup of mass in unit. – U B and U H at inlet stream conditions are equal to their tabulated values (which are o
strictly valid at 20 C and 1 atm.) – Volumes of benzene and hexane are additive. – Densitometer gives correct reading.
3-2
3.6
1955 . kg H 2SO 4
(a) V
1 kg solution L u 1000 0.35kg H 2SO 4 12563 . . kg
445 L
(b) 195.5 kg H 2 SO 4
L u 1.00 kg 18255 . 195.5 kg H 2 SO 4 0.65 kg H 2 O L 0.35 kg H 2 SO 4 1.000 kg 470 445 u 100% 5.6% % error 445
Videal
3.7
b g E Weight of block bdowng Mass of oil displaced + Mass of water displaced = Mass of block U b0.542gV U b1 0.542gV U V Buoyant force up
oil
c
H 2O
From Table B.1: U c
2.26 g / cm3 , U w
100 . g / cm3 U oil
moil U oil u V 3.325 g / cm3 u 35.3 cm3 moil + flask 117.4 g 124.8 g 242 g 3.8
470 L
b g
117.4 g
b
Buoyant force up = Weight of block down Wblock ( UVg ) disp. Liq
Wdisplaced liquid
b g
b g
. Ag Expt. 1: U w 15
UB 2A g UB
U w 1.00 g/cm3
bg
UB
Let U w
for b p 1 hb 1
bg
subst. 3 for h p 1 in
b2 g, solve for h
b1
hb1
15 .
soln
density of water. Note: U A ! U w (object sinks) Ab hsi
d
Ab h p1 hb1
i
bi g
i
b
d
Ap h p1 Vd 1 Vw
WA WB h p1 pw g
W WB Vw A Ap pw g
(1)
U wVd 1 g WA WB weight of displaced water
(2)
A p h p1
Vw
15 . g / cm3 SG
d
Volume of pond water: Vw subst. 2
0.75
b g
2U B
WA WB pw gAb
h p1 hb1
B
Subst. (1) for Vd 1 , solve for h p1 hb1
Before object is jettisoned
bg
b g
0.75 g / cm3 SG
Archimedes
hU 1
hb 1
15 . 2
Volume displaced: Vd 1
hs 1
WA + WB
Uw u
U B 2 A g U soln
3.9
g
( UVg ) block
b g
Expt. 2: U soln A g
3.325 g / cm3
Ap h p1 Ab h p1 hb1
Vw WA WB Ap pw gAp
g LM 1 MN A
p
3-3
1 Ab
OP PQ
(3) (4)
i
3.9 (cont’d) hs 2
WB WA
Let V A hU2
h b2
(5)
Ab h p 2 hb 2
(6)
d
Volume displaced by boat:Vd 2
Subst. for Vd 2
E , solve for dh
WB pw gAb
p2
hb 2
i (7)
Ap h p 2 Vd 2 V A
Volume of pond water: Vw
b 5g , b 6 g & b 7 g
Vw
Ap h p 2
WB W A pw g p A g
Vw WB WA Ap pw gAp p A gAp
solve for
hp 2
h p 21
bg
subst. 8
bg
for h p 2 in 7 , solve for hb 2
i
Archimedes U WVd 2 g WB
After object is jettisoned
h p 2 hb 2
WA U Ag
volume of jettisoned object =
(8)
Vw WB WA WB Ap pw gAp p A gAp pw gAb
hb 2
(9)
(a) Change in pond level h p 2 h p1
LM 1 A gNp
b8gb 3g WA p
A
1 pW
OP Q
0 !0
VA VA 0 pW p A pW Ap
IJ F K GH
WA b pW p A g b5gF p A pW gAp
GH
I JK
the pond level falls (b)
Change in boat level h p 2 h p1
LM 1 A g MN p A p
A
the boat rises 3.10 (a) U bulk
2.93 kg CaCO 3 L CaCO 3
L OP F p F A 1I I P ! 0 1 M 1 G P G JK JK P G J p A PQ H A K M H p H A MN PQ ! 0
b9 gb4 g WA
p
1 pW Ap
O b5gF VA I M
W
0.70 L CaCO 3 L total
b
p
!0
A
p
W
b
2.05 kg / L
2.05 kg 50 L 9.807 m / s2 1N 100 . u 103 N L 1 kg m / s2 Neglected the weight of the bag itself and of the air in the filled bag.
(b) Wbag
U bulkVg
(c) The limestone would fall short of filling three bags, because – the powder would pack tighter than the original particles. – you could never recover 100% of what you fed to the mill.
3-4
3.11 (a) Wb
122.5 kg 9.807 m / s2
mb g
Ub
m f mnf
(b)
mf
xf
mb
mb
V f Vnf
b2 g,b3g
mb
Fx GH U
f
mb x f
d
Vb
I JK
U nf
f
1 / U b 1 / U nf
(c) x f
mf
Uf
mf mnf
mb x f mb (1 x f )
mb
mnf
U nf
F GH
mb Ub
I JK
1 1 xf U b U nf
mb Ub
Fx GH U
f
1 x f
f
U nf
I (V JK
lungs
Vother )
mb
b
lungs
nf
b
nf
nf
nf
other
b
lungs
b
1 / U f 1 / U nf
F1 1I GH U U JK
F 1 1 I 1 1 V V GH U U JK U U m F 1 1 I F V V I F 1 1 I F 12. 01. I J G GH U U JK GH m JK GH 103 J 11 . . K H 122.5 K x F1 1I FG 1 1 IJ GH U U JK H 0.9 11. K xf
f
1 / U b 1 / U nf
Vb
Vlungs Vother
U nf
(3)
mb 1 1 xf Ub U f U nf
(d) V f Vnf Vlungs Vother mnf
i
1 / 103 . 1 / 1.1 0.31 1 / 0.9 1 / 1.1
1 / U f 1 / U nf
(2)
mb 1 x f
1 x f
119 L
(1)
mf
(1),(2) mnf
Uf
1202 N
(1202 N - 44.0 N) 1 kg m / s2 Wb WI Uwg 0.996 kg / L u 9.807 m / s2 1N mb 122.5 kg 1.03 kg / L Vb 119 L
Vb
mf
1N 1 kg m / s2
other
b
f
f
nf
3-5
0.25
Conc. (g Ile/100 g H2O)
3.12 (a) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.987
y = 545.5x - 539.03 R2 = 0.9992
0.989
0.991
0.993
0.995
0.997
Density (g/cm3)
5455 . U 539.03
From the plot above, r (b) For U = 0.9940 g / cm3 , m Ile
150 L h
r
3.197 g Ile / 100g H 2 O
0.994 g 1000 cm3 3.197 g Ile 1 kg 3 cm L 103.197 g sol 1000 g
4.6 kg Ile / h
(c) The density of H2O increases as T decreases, therefore the density was higher than it should have been to use the calibration formula. The valve of r and hence the Ile mass flow rate calculated in part (b) would be too high. 3.13 (a)
Mass Flow Rate (kg/min)
1.20 1.00
y = 0.0743x + 0.1523 R2 = 0.9989
0.80 0.60 0.40 0.20 0.00 0.0
2.0
4.0
6.0
8.0
10.0
12.0
Rotameter Reading
b g
From the plot, R = 5.3 m 0.0743 5.3 01523 . 0.546 l g / min
3-6
3.13 (cont’d) (b) Rotamete Collection Collected r Reading Time Volume (min) (cm3) 2 1 297 2 1 301 4 1 454 4 1 448 6 0.5 300 6 0.5 298 8 0.5 371 8 0.5 377 10 0.5 440 10 0.5 453
Mass Flow Rate (kg/min) 0.297 0.301 0.454 0.448 0.600 0.596 0.742 0.754 0.880 0.906
b
Difference Duplicate (Di)
Mean Di
0.004 0.006 0.004
0.0104
0.012 0.026
g
1 0.004 0.006 0.004 0.012 0.026 0.0104 kg / min 5 . Di ) kg / min 0.610 r 0.018 kg / min 95% confidence limits: (0.610 r 174 Di
There is roughly a 95% probability that the true flow rate is between 0.532 kg / min and 0.628 kg / min . 3.14 (a) (b)
(c) (d) (e) (f) (g) (h)
15.0 kmol C 6 H 6 15.0 kmol C 6 H 6
15,000 mol C 6 H 6
78.114 kg C 6 H 6
117 . u 103 kg C 6 H 6 kmol C 6 H 6 1000 mol 15 . u 104 mol C 6 H 6 kmol lb - mole 453.6 mol
15,000 mol C 6 H 6
6 mol C 1 mol C 6 H 6
15,000 mol C 6 H 6
6 mol H 1 mol C 6 H 6
90,000 mol C 12.011 g C mol C 90,000 mol H 1.008 g H mol H
15,000 mol C 6 H 6
33.07 lb - mole C 6 H 6 90,000 mol C 90,000 mol H
1.08 u 106 g C 9.07 u 104 g H
6.022 u 1023 mol
9.03 u 1027 molecules of C6 H 6
3-7
3.15 (a) m
(b) n
175 m3 h
1000 L m3
2526 kg
1000 mol 1 min
min
0.866 kg 1h L 60 min
92.13 kg
60 s
2526 kg / min
457 mol / s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20oC and 1 atm 3.16 (a)
200.0 kg mix 0.150 kg CH 3OH kg mix
(b) m mix
3.17
M m N 2
100.0 lb - mole MA h 0.25 mol N 2 3000 kg h
kmol CH 3OH 1000 mol 32.04 kg CH 3OH
74.08 lb m MA
m / V
28.02 g N 2
0.75 mol H 2
mol N 2 kmol 0.25 kmol N 2 8.52 kg
kmol feed
936 mol CH 3OH
8715 lb m / h
2.02 g H 2
8.52 g mol mol H 2 28.02 kg N 2 2470 kg N 2 h kmol N 2
215 g 65 g 150 g
500 g / 455 mL 110 . g mL
(c) 150 g CaCO 3 / 500 g suspension 3.19
1 lb m mix
1 lb - mole MA 0.850 lb m MA
3.18 M suspension 565 g 65 g 500 g , M CaCO 3 (a) V 455 mL min , m 500 g min (b) U
1 kmol
0.300 g CaCO 3 g suspension
Assume 100 mol mix. mC2 H 5OH mC4 H 8O 2
10.0 mol C 2 H 5OH
46.07 g C 2 H 5OH
75.0 mol C 4 H 8 O 2
mol C 2 H 5OH 88.1 g C 4 H 8 O 2
461 g C 2 H 5OH
6608 g C 4 H 8 O 2 mol C 4 H 8 O 2 15.0 mol CH 3COOH 60.05 g CH 3COOH mCH 3COOH 901 g CH 3COOH mol CH 3COOH 461 g xC 2 H 5OH 0.0578 g C2 H 5OH / g mix 461 g + 6608 g + 901 g 6608 g xC 4 H 8 O 2 0.8291 g C 4 H 8 O 2 / g mix 461 g + 6608 g + 901 g 901 g xCH 3COOH 0113 . g CH 3COOH / g mix 461 g + 6608 g + 901 g 461 g + 6608 g + 901 g MW 79.7 g / mol 100 mol 25 kmol EA 100 kmol mix 79.7 kg mix m 2660 kg mix 75 kmol EA 1 kmol mix
3-8
3.20 (a) Unit Crystallizer Filter Dryer
(b) m gypsu m
Function Form solid gypsum particles from a solution Separate particles from solution Remove water from filter cake 0.35 kg C aSO 4 2 H 2 O L slurry
1 L slurry
0.35 kg CaSO4 2H2O
Vgypsum
CaSO 4 in gypsum: m
CaSO 4 in soln.: m
% recovery =
L CaSO4 2H2O
0151 . L CaSO4 2H2O 2.32 kg CaSO4 2H2O 0.35 kg gypsum 136.15 kg CaSO 4 0.277 kg CaSO 4 172.18 kg gypsum
. g L sol b1 0151
0.35 kg gypsum
(c) m
0 .35 kg C aSO 4 2 H 2 O
1.05 kg 0.209 kg CaSO 4 L 100.209 kg sol
0.05 kg sol 0.209 g CaSO 4 0.95 kg gypsum 100.209 g sol
0.277 g + 3.84 u 10 -5 g u 100% 0.277 g + 0.00186 g
FB:
45.8 L 0.90 kg min L 55.2 L 0.75 kg min L
3.84 u 10 -5 kg CaSO 4
99.3%
3.21 CSA:
0.00186 kg CaSO 4
kmol 0.5496 75 kg kmol 0.4600 90 kg
U| |V || W
kmol 0.5496 min kmol 0.4600 min
1.2
mol CSA mol FB
She was wrong. The mixer would come to a grinding halt and the motor would overheat. 3.22 (a)
150 mol EtOH 6910 g EtO H
V
SG
(b) V c
46.07 g EtOH
6910 g EtOH mol EtOH 0.600 g H 2 O 10365 g H 2 O 0.400 g EtOH
6910 g EtOH
L 789 g EtOH
(6910 +10365) g 19.1 L
L 1000 g
10365 g H 2 O
1000 g H 2 O
0.903
( 6910 10365) g mix
% error
L
L 935.18 g
(19.123 18.472 ) L u 100% 18.472 L
18.472 L 18.5 L 3.5%
3-9
19.123 L 19.1 L
3.23
0.09 mol CH 4
M
16.04 g
0.91 mol Air
29.0 g Air
27.83 g mol mol mol 700 kg kmol 0.090 kmol CH 4 2.264 kmol CH 4 h h 27.83 kg 1.00 kmol mix 2.264 kmol CH 4 0.91 kmol air 22.89 kmol air h h 0.09 kmol CH 4 5% CH 4
2.264 kmol CH 4
0.95 kmol air
h
0.05 kmol CH 4
b
43.01 kmol air h
g
Dilution air required: 43.01 - 22.89 kmol air 1000 mol 20200 mol air h h
1 kmol
20.20 kmol Air 29 kg Air Product gas: 700 kg 1286 kg h h
h
kmol Air
43.01 kmol Air 0.21 kmol O2 32.00 kg O2 h kg O2 0.225 h 1.00 kmol Air 1 kmol O 2 1286 kg total kg
3.24
mi , Ui M
xi
m m ¦ Mi V i i
¦ xi U i
A: B:
mi M , U= Vi V
xi
¦U
1 U
¦
1 M
mi2 ¦V zU i
mi Vi V 1 1 Vi Correct. ¦ M mi M M U 0.60 0.25 0.15 1.091 U 0.917 g / cm 3 0.791 1.049 1.595
¦
i
xi Ui
3.25 (a) Basis: 100 mol N 2 20 mol CH 4 N total x CO
x CH 4
(b) M
Not helpful.
100 20 64 32
R|20 u 80 25 S |T 20 u 40 25
64 mol CO 2 32 mol CO
216 mol
32 64 0.15 m ol C O / m ol , x CO 2 0.30 m ol C O 2 / m ol 216 216 20 100 0.09 mol CH 4 / mol , x N 2 0.46 mol N 2 / mol 216 216
¦y M i
i
015 . u 28 0.30 u 44 0.09 u 16 0.46 u 28
3-10
32 g / mol
3.26 (a) Samples Species
MW
k
1
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
Peak Area 3.6 2.8 2.4 1.7
Mole Mass Fraction Fraction 0.156 0.062 0.233 0.173 0.324 0.353 0.287 0.412
2
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
7.8 2.4 5.6 0.4
0.249 0.146 0.556 0.050
0.111 0.123 0.685 0.081
1.170 0.689 2.615 0.233
18.767 20.712 115.304 13.554
3
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
3.4 4.5 2.6 0.8
0.146 0.371 0.349 0.134
0.064 0.304 0.419 0.212
0.510 1.292 1.214 0.466
8.180 38.835 53.534 27.107
4
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
4.8 2.5 1.3 0.2
0.333 0.332 0.281 0.054
0.173 0.324 0.401 0.102
0.720 0.718 0.607 0.117
11.549 21.575 26.767 6.777
5
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
6.4 7.9 4.8 2.3
0.141 0.333 0.329 0.197
0.059 0.262 0.380 0.299
0.960 2.267 2.242 1.341
15.398 68.178 98.832 77.933
(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST INTEGER N, ND, ID, J READ (5, *) N CN-NUMBER OF SPECIES READ (5, *) (MW(J), K(J), J = 1, N) READ (5, *) ND DO 20 ID 1, ND READ (5, *)(A(J), J = 1, N) MOLT 0. 0 MASST 0. 0 DO 10 J = 1, N MOL(J) = MASS(J) = MOL(J) * MW(J) MOLT = MOLT + MOL(J) MASST = MASST + MASS(J) 10 CONTINUE DO 15 J = 1, N MOL(J) = MOL(J)/MOLT MASS(J) = MASS(J)/MASST 15 CONTINUE WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N) 20 CONTINUE 1 FORMAT (' SAMPLE: `, I3, /,
' SPECIES MOLE FR. MASS FR.', /,
10(3X, I3, 2(5X, F5.3), /), /)
3-11
moles
mass
0.540 0.804 1.121 0.991
8.662 24.164 49.416 57.603
END $DATA
4 16 . 04
0 . 150
30 . 07 0. 287 44 . 09 0. 467 58. 12
0 . 583
5 3. 6 2. 8 2 . 4 1. 7 7. 8 2. 4 5. 6 0. 4 3. 4 4 . 5 2. 6 0 . 8 4 . 8 2. 5 1. 3
0. 2
6. 4 7. 9 4 . 8 2 . 3 [OUTPUT] SAMPLE: 1 SPECIES
MOLE FR MASS FR
1
0.156
0.062
2
0.233
0.173
3
0.324
0.353
0.287
0.412
4 SAMPLE: 2 (ETC.)
3.27 (a)
(8.7 u 10 6 u 0.40) kg C
44 kg CO 2 12 kg C
(11 . u 10 6 u 0.26) kg C 28 kg CO 12 kg C ( 3.8 u 10 5 u 0.10) kg C
m
M
16 kg CH 4 12 kg C
1.28 u 10 7 kg CO 2 2.9 u 105 kmol CO 2
6.67 u 10 5 kg CO 2.38 u 10 4 kmol CO 5.07 u 10 4 kg CH 4 3.17 u 10 3 kmol CH 4
(1.28 u 10 7 6.67 u 10 5 5.07 u 10 4 ) kg 1 metric ton 1000 kg
¦y M i
i
0.915 u 44 0.075 u 28 0.01 u 16
13,500
metric tons yr
42.5 g / mol
3.28 (a) Basis: 1 liter of solution 1000 mL
1.03 g 5 g H 2 SO 4 mL 100 g
mol H 2 SO 4 98.08 g H 2 SO 4
3.28 (cont’d)
3-12
0.525 mol / L 0.525 molar solution
(b) t
V V
55 gal
(c) u
V A
t
L u
55 gal
3.7854 L
min
60 s
gal
87 L
min
3.7854 L 10 3 mL 1.03 g gal 1L mL
144 s
0.0500 g H 2 SO 4 g
87 L m 3 1 min min 1000 L 60 s (S u 0.06 2 / 4 ) m 2 45 m 88 s 0.513 m / s
1 lbm 453.59 g
23.6 lb m H 2 SO 4
0.513 m / s
3.29 (a) n1 (mol/min) 0.180 mol C6H14/mol 0.820 mol N2/mol
n2 (mol/min) 0.050 mol C6H14/mol 0.950 mol N2/mol 1.50 L C6H14(l)/min n 3 (mol C6H14(l)/min)
n3
150 . L 0.659 kg 1000 mol 1147 . mol / min min L 86.17 kg
UV W
|RS |T
Hexane balance: 0.180n1 0050 . n2 1147 . (mol C6 H14 / min) solve n1 838 . mol / min n2 = 72.3 mol / min Nitrogen balance: 0.820n1 0950 . n2 (mol N2 / min) (b) Hexane recovery
3.30
30 mL
n3 1147 . u 100% u 100% 76% n1 0180 . 838 .
b g
1 L 0.030 mol 172 g 0155 . g Nauseum 103 mL lL 1 mol
3-13
3.31 (a) kt is dimensionless k (min -1 ) (b) A semilog plot of C A vs. t is a straight line ln CA
ln(CA)
1 0 -1 -2 -3 -4 -5
y = -0.4137x + 0.2512 R2 = 0.9996
0.0
k
ln CAO kt
5.0 t (min)
10.0
0.414 min 1
CAO 1286 ln CAO 02512 . . lb - moles ft 3
FG 1b - molesIJ Cc mol 28.317 liter H ft K liter 1 ft t cbsg 1 min t bming t c 60 60 s
(c) C A
A
3
CA
0.06243C Ac t 3.32 (a)
(b)
(c)
3
2600 mm Hg
1000 mol
b
g
drop primes
b
g
C A mol / L
b
2.14 exp 0.00693t
g
5.30 mol / L 14.696 psi 50.3 psi 760 mm Hg
275 ft H 2 O 101.325 kPa 33.9 ft H 2 O
822.0 kPa
3.00 atm 101325 . 12 m2 u 105 N m2 1 atm 1002 cm2
30.4 N cm2
280 cm Hg 10 mm 101325 . u 106 dynes cm2 1002 cm2 (d) 1 cm 760 mm Hg 12 m2 (e) 1 atm
0.06243C Ac
C A 0 exp kt
1334 . exp 0.419t c 60
200 s C A
2.26462 lb - moles
20 cm Hg 10 mm 1 cm
1 atm 760 mm Hg
3-14
0.737 atm
3.733 u 1010
dynes m2
3.32 (cont’d) (f)
(g)
b
b25.0 14.696gpsi
760 mm Hg 14.696 psi
(h) 325 mm Hg 760 mm Hg
(i)
g
25.0 psig 760 mm Hg gauge 14.696 psig
35.0 psi 760 mm Hg 14.696 psi
3.33 (a) Pg
Ugh
h (m)
b
g
1293 mm Hg gauge
b g
2053 mm Hg abs
b
435 mm Hg gauge
g
13.546 g Hg cm3
1 cm
1540 cm CCl 4
10 mm 1595 g CCl 4 cm3 .
0.92 u 1000 kg 9.81 m / s2 m3 0111 . Pg (kPa)
h (m)
1N 1 kPa 2 3 1 kg m / s 10 N / m2
h
Pg Pg moil
68 kPa h
UV
(b) Pg Patm
0111 . u 68 7.55 m
FG 0.92 u 1000 kg IJ u FG 7.55 u S u 16 H 4 m K H 3
2
m3
IJ K
14 . u 10 6 kg
Ptop Ugh
b
g b g
68 101 115 0.92 u 1000 u 9.81 / 103 h h
5.98 m
3.34 (a) Weight of block = Sum of weights of displaced liquids U 1h1 U 2 h2 (h1 h2 ) AU b g h1 AU 1 g h2 AU 2 g U b h1 h2 (b)
Ptop Patm U1gh0 , Pbottom Patm U1g(h0 h1) U2 gh2 , Wb Ub (h1 h2 ) A
Fdown ( Patm U1gh0 ) A Ub (h1 h2 ) A , Fup [ Patm U1g(h0 h1) U2 gh2 ]A Fdown Fup Ub (h1 h2 ) A U1gh1 A U2 gh2 A Wblock Wliquid displaced
3-15
'P
3.35
bP
atm
g
Ugh Pinside
1 atm 1 atm
F
m
3.36
P
154 N 65 cm2 2
cm
. g1000 kg b105 m3
. u 104 N u 100
9.8066 m 150 m 12 m2 1N s2 1002 cm2 1 kg m / s2
. lb I FG 022481 H 1 N JK f
2250 lb f
14 . u 62.43 lb m 1 ft 3 2.3 u 106 gal 2.69 u 107 lb m UV 3 ft 7.481 gal P0 Ugh . u 62.43 lb m 32.174 ft 30 ft 1 lb f 12 ft 2 lb f 14 14.7 2 in ft 3 s2 32.174 lb m ft / s2 12 2 in 2 32.9 psi
— Structural flaw in the tank. — Tank strength inadequate for that much force. — Molasses corroded tank wall
S u 24 2 u 3 in 3 1 ft 3 8.0 u 62.43 lb m 392 lb m 3.37 (a) mhead 4 12 3 in 3 ft 3 392 lb m 32.174 ft / s 2 1 lb f W mhead g 392 lb f 32.174 lb m ft / s2 30 14.7 lb f S u 202 in 2 Fnet Finside Fout W 392 lb f 4415 lb f in 2 4 The head would blow off. 2 Fnet 4415 lb f 32.174 lb m ft / s 362 ft / s2 Initial acceleration: a mhead 392 lb m 1 lb f
b
g
(b) Vent the reactor through a valve to the outside or a hood before removing the head. 3.38 (a)
a
2m 1m
b
Pa Ugh Patm , Pb Patm If the inside pressure on the door equaled Pa , the force on the door would be F Adoor ( Pa Pb ) UghAdoor Since the pressure at every point on the door is greater than Pa , Since the pressure at every point on the door is greater than Pa , F >UghAdoor
3-16
3.38 (cont’d) (b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to fill. V 5 u 25 . u 2 ft 3 Vtub . ft 3 / min V 5 u 25 . 125 . ft 3 / min 25 | t 10 min For a full room, h
(i)
10 m
. m 1N 10 m 2 m2 1000 kg 981 F ! 2.0 u 105 N 3 2 2 m s 1 kg m / s The door will break before the room fills F!
(ii)
d i dP i
3.39 (a) Pg g
If the door holds, it will take 3 35.3145 ft 3 Vroom 5 u 15 u 10 m t fill V 12.5 ft 3 / min 1 m3 He will not have enough time.
b
g
1h 60 min
31 h
25 m H 2 O tap
b
junction
101.3 kPa 245 kPa 10.33 m H 2 O 25 5 m H 2 O 101.3 kPa 294 kPa 10.33 m H 2 O
g
(b) Air in the line. (lowers average density of the water.) (c) The line could be clogged, or there could be a leak between the junction and the tap. 3.40
Pabs Pgauge Patm
800 mm Hg 25 mm Hg 800 25 775 mm Hg
b
3.41 (a) P1 U A g h1 h2 P1 P2
bU
B
g
P2 U B gh1 U C gh2
g
b
g
U A gh1 U C U A gh2
LMb10. 0.792g g 981 cm 30.0 cm b137 . 0.792g g cm s cm N I 123.0 kPa F 1 dyne I F 101325 . kPa uG J G H 1 g cm / s K H 1.01325 u 10 dynes / cm JK
(b) P1 121 kPa +
3
2
2
3
6
2
3-17
981 cm 24.0 cm s2
OP Q
3.42 (a) U T g (500 x )
U W gR R
UT (500 x ) UW
b
g
0.866 u 500 10 424 cm 1.000 0.866 u 500 400 87 cm If h 400 cm , R = 1.000 If h 10 cm , R =
b
g
UT (500 x ) U Hg
For Hg, R c
b
g
0.866 . cm u 500 10 312 13.6 0.866 If h 400 cm , R = . cm u 500 400 637 13.6 If h 10 cm , R =
b
g
Use mercury, because the water manometer would have to be too tall. (b) If the manometer were simply filled with toluene, the level in the glass tube would be at the level in the tank. Advantages of using mercury: smaller manometer; less evaporation. (c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen, minimizing the risk of combustion.
3.43
Patm
b
P g 7.23 g P I U gJ b26 cmg dU U igb26 cmg FGH 7.23 K m
U f g 7.23 m U f
Pa Pb
f
atm
atm
w
w
F756 mmHg 1 m 1000 kg 9.81m/s GH 7.23m 100 cm m
2
3
Pa Pb
3.44 (a) 'h 900 hl
(b) 'h
Ib g JK
760 mmHg 1m 1N 26 cm 2 5 2 1 kg m/s 1.01325u10 N m 100 cm
81 . mm Hg
75 . psi 760 mm Hg 388 mm Hg hl = 900 388=512 mm 14.696 psi
388 25 u 2
338 mm Pg =
338 mm Hg
14.696 psi 760 mm Hg
3-18
6.54 psig
3.45 (a) h = L sin T (b) h
b8.7 cmg sinb15qg
3.46 (a) P
Patm Poil PHg
2.3 cm H 2 O
23 mm H 2 O
920 kg 9.81 m / s2 765 365 m3 393 mm Hg
0.10 m
1N 760 mm Hg 2 1 kg m / s 1.01325 u 105 N / m2
(b) — Nonreactive with the vapor in the apparatus. — Lighter than and immiscible with mercury. — Low rate of evaporation (low volatility).
c
3.47 (a) Let U f
h
manometer fluid density 110 . g cm 3 , U ac
c0.791 g cm h
acetone density
3
Differential manometer formula: 'P
dU
f
i
U ac gh
. 0791 . gg 981 cm h (mm) 1 cm 1 dyne g b110 cm s 10 mm 1 g cm/ s 0.02274 hb mmg V b mL sg 62 87 107 123 138 151 hb mmg 5 10 15 20 25 30 'Pb mm Hgg 0.114 0.227 0.341 0.455 0.568 0.682
b
'P mm Hg
(b) lnV
3
2
2
b g
n ln 'P ln K 6
ln(V)
5.5
y = 0.4979x + 5.2068
5 4.5 4 -2.5
-2
-1.5
-1
-0.5
0
ln( P)
b g
. ln 'P 52068 . From the plot above, ln V 04979
3-19
760 mm Hg 1.01325u106 dyne/ cm2
. . , ln K n = 04979 | 05
5.2068 K
183
ml s
bmm Hgg
0.5
3.47 (cont’d) (c) h
23 'P
b0.02274gb23g
132 mL 0.791 g s
mL
104 g s
0.523 mm Hg V 104 g s
1 mol 58.08 g
b
g
183 0.523
0.5
132 mL s
180 . mol s
. 544q R / 18 . 303 K 273 30q C 3.48 (a) T 85q F 4597 . 474q R 460 14q F (b) T 10q C 273 263 K u 18 (c) 'T
(d)
85q C 10 . qK 85q C 18 . qF 85q C 1.8q R 85q K; 153q F; 153q R 10 . qC 1q C 10 . qC
150q R 1q F 150q R 1.0q C 150q R 1.0$ K 150q F; 83.3q K; 1q R 1.8q R 1.8q R
3.49 (a) T
0.0940 u 1000$ FB 4.00 98.0$ C T = 98.0 u 1.8 + 32 = 208$ F
(b) 'T ($ C) 'T ($ F)
0.0940'T ($ FB)
15$ C 100$ L ; T2 T ($ C) aT ($ L) b
b43 15g$ C b1000 - 100g$ L
T ($ C)
0.94$ C 'T (K)
0.94 K
0.94$ C 1.8$ F 169 . $ F 'T ($ R) 169 . $R $ 1.0 C
(c) T1
a
83.3q C
43$ C 1000$ L
F $ CI ; GH $ L JK
0.0311
b
15 0.0311 u 100
0.0311T ($ L) 119 . and T ($ L)
119 . $C
32.15T ($ C) 382.6
(d) Tbp
. $ FB 3232 $ L 88.6$ C 184.6 K 332.3$ R -127.4$ F 9851
(e) 'T
50.0$ L 1.56$ C 16.6$ FB 156 . K 2.8$ F 2.8$ R
3-20
3.50
bT g (a) V b mVg
bT g
100q C
b H 2O
455q C
m AgCl
b g
aT q C b a 0.05524 mV q C 5.27 100a b b 0.2539 mV 24.88 455a b V mV 0.05524T q C 0.2539
b g
b g
T qC
b g
b g
1810 . V mV 4.596
. mV o136 . mV 1856 . q C o2508 . qC (b) 100 ln K n ln R
3.51 (a) ln T
b
g
b
g
b2508. 1856. gq C 20 s
326 . qC / s
KR n
T
ln 250.0 110.0
n
dT dt
1184 .
ln 40.0 20.0
. ln K ln 1100 . 1184 . (ln 200 . ) 1154 . K 3169 . T 3169 . R1184
FG 320 IJ H 3169 . K
(b) R
1/1.184
49.3
(c) Extrapolation error, thermocouple reading wrong. 3.52 (a) PV
0.08206nT
. g 14696 b g Pcbpsig14696 .
bg d i
, V L V c ft 3 u
P atm
b g b
n mol
g
453.59 mol n c lb - moles u , T($ K) lb moles
b Pc 14.696g u V c u 28.317
14.696
b
g
b
g
P c 14.696 u V c
0.08206 u nc u
28317 . ft 3 L T c ($ F) 32 . 27315 1.8
LM N
OP Q
453.59 (T c 32) . u 27315 1 1.8
0.08206 u 14.696 u 453.59 u n c u T c 459.7 28.317 u 18 .
b
b
g
Pc 14.696 V c 1073 . nc T c 4597 .
3-21
g
3.52 (cont’d)
b500 14.696g u 35. 10.73 u b85 459.7g
(b) ntot c
0308 . lb - mole 0.30 lb - mole CO 28 lbm CO 26 . lb m CO lb - mole lb - mole CO
mCO
(c) T c
b3000 14.696g u 35. 459.7 10.73 u 0.308
b g
3.53 (a) T q C
b
2733$ F
g
a u r ohms b
UV . a bW 33028
. a b 0 23624 100
0.308 lb - mole
a 10634 . b 25122 .
b g
T qC
FG kmol IJ n c (kmol) 1 min n c H s K min 60 s 60 Pcbmm Hgg 1 atm Pc Pbatmg 760 mm Hg 760 F m I m 1 min V c VG J V c H s K min 60 s 60
b g
. r ohms 25122 . 10634
(b) n
. b g T cbq Cg 27316
, TK
3
3
b g d b g
i
. Pc mm Hg V c m3 min 0016034 . Pc V c n c 12186 n c . T c q C 27316 60 . 760 T c 27316 60 (c) T
10.634r 25122 . r1
26159 . T1
26.95q C
r2
26157 . T2
26.93q C
r3
44.789 T3
2251 . qC
P (mm Hg)
h Patm
h (29.76 in Hg)
FG 760 mm Hg IJ H 29.92 in Hg K
h1 232 mm P1 987.9 mm Hg h2 156 mm P2
h3
9119 . mm Hg
74 mm P3 829.9 mm Hg
3-22
h 755.9
3.53 (cont’d)
b0.016034gb987.9gb947 60g 0.8331 kmol CH 26.95 27316 . b0.016034gb9119. gb195g 9.501 kmol air min
(d) n1 n2
min
. 26.93 27316 n1 n2 10.33 kmol min
n3
b
. g g b10.33gb2251. 27316 b0.016034gb829.9g
n3 T2 27316 . 0.016034 P3
(e) V3
(f)
4
0.8331 kmol CH 4 16.04 kg CH 4 min kmol 0.21 u 9.501 kmol O2 32.0 kg O2 xCH4
387 m3 min
kg CH 4 min 0.79 u 9.501 kmol N 2 29.0 kg N 2 13.36
min kmol O2 min 13.36 kg CH 4 min 0.0465 kg CH 4 kg (13.36 274) kg / min
REAL, MW, T, SLOPE, INTCPT, KO, E REAL TIME (100), CA (100), TK (100), X (100), Y(100) INTEGER IT, N, NT, J READ 5, MW, NT DO 10 IT=1, NT READ 5, TC, N TK(IT) = TC + 273.15 READ 5, (TIME (J), CA (J), J 1 , N) DO 1 J=1, N CA J CA J / MW
3.54
b g b g b g
bg bg XbJg TIMEbJg YbJg 1./CAbJg
1
CONTINUE CALL LS (X, Y, N, SLOPE, INTCPT)
b g
K IT
SLOPE
WRITE (E, 2) TK (IT), (TIME (J), CA (J), J 1 , N) WRITE (6, 3) K (IT) 10 CONTINUE DO 4 J=1, NT
3-23
kmol N 2
274
kg air min
bg YbJg XJ
bg LOGcKbJgh
1./TK J
3.54 (cont’d) 4
CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) KO EXP INTCPT
b
2
3 5
10
g
E 8.314 SLOPE WRITE (6, 5) KO, E FORMAT (' TEMPERATURE (K): ', F6.2, / * ' TIME CA', /, * ' (MIN) (MOLES)', / * 100 (IX, F5.2, 3X, F7.4, /)) FORMAT (' K (L/MOL – MIN): ', F5.3, //) FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4) END SUBROUTINE LS (X, Y, N, SLOPE, INTCPT) REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN INTEGER N, J SX=0 SY=0 SXX=0 SXY=0 DO 10 J=1,N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J)**2 SXY = SXY + X(J)*Y(J) CONTINUE AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN SLOPE = (SXY – SX*SY)/(SXX – SX**2) INTCPT = SY – SLOPE*SX RETURN END $ DATA 65.0 94.0 10.0 20.0
4 6 8.1 4.3
[OUTPUT] TEMPERATURE (K): 367.15 TIME CA (MIN) (MOLS/L) 10.00 0.1246 3-24
30.0 40.0 50.0 3.54 (cont’d)
3.0 2.2 1.8
20.00 0.0662 30.00 0.0462 40.00 0.0338
60.0
1.5
50.00 0.0277 60.00 0.0231
b
g
K L / MOL MIN : 0.707 110. 10.0 20.0 30.0 40.0 50.0 60.0
6 3.5 1.8 1.2 0.92 0.73 0.61
127.
6
ETC
bat 94qCg
TEMPERATURE (K): 383.15
b
g
K L / MOL MIN : 1.758
b g E bJ / MOLg: 0.6690E 05
K0 L / MOL MIN : 0.2329E 10
3-25
CHAPTER FOUR 4.1
a.
Continuous, Transient
b.
Input – Output = Accumulation No reactions Generation = 0, Consumption = 0
6.00 c.
4.2
t
kg kg 3.00 s s
dn dn dt dt
100 . m3 1000 kg 1 s 1 m3 3.00 kg
3.00
333 s
a.
Continuous, Steady State
b.
k
c.
Input – Output – Consumption = 0 Steady state Accumulation = 0 A is a reactant Generation = 0
0 CA
C A0
k
kg s
f CA
0
FG IJ FG IJ V FG m IJ C FG mol IJ kVC FG mol IJ C HsK H K H K H sK Hm K
m3 mol V C A0 s m3
4.3
3
A
b
a.
v kg / h m 100 kg / h 0.550 kg B / kg 0.450 kg T / kg
b
A
CA0 kV 1 V
g
0.850 kg B / kg 0.150 kg T / kg l kg / h m
A
3
g
Input – Output = 0 Steady state Accumulation = 0 No reaction Generation = 0, Consumption = 0
0.106 kg B / kg 0.894 kg T / kg
(1) Total Mass Balance: 100.0 kg / h
v m l m
(2) Benzene Balance: 0.550 u 100.0 kg B / h
v Solve (1) & (2) simultaneously m
v 0106 l 0.850m . m
59.7 kg h, m l
40.3 kg h
b.
The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output).
c.
Possible explanations a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors.
4-1
4.4
b.
c.
b
n (mol) 0.500 mol N 2 mol 0.500 mol CH 4 mol 100.0 g / s
b g bg C H g g bg C H g g
xB d.
3
8
4
10
b
b
b
b g U| R|n blb - mole DA sg S| 0.21 lb - moleO lb - mole DA V| T 0.79 lb - mole N lb - mole DAW
g
g
26.45 x E lb - mole C2 H 6 / h
n1 lb - mole H 2 O s
0.014 n kg N 2
100 x E g C2 H 6 1 lb m lb - mole C2 H 6 3600 s s 453593 h . g 30 lb m C2 H 6
n E
x E g C2 H 6 g xP
g
0.500 n mol N 2 28 g N 2 1 kg mol N 2 1000 g
nO 2
b
0.21n 2 lb - mole O 2 / s
g
g
2
2
x H 2O
2
xO 2 e.
b g
n mol
0.400 mol NO mol
b
y NO2 mol NO 2 mol
b
n N 2O 4
g
0.600 y NO2 mol N 2O 4 mol
4.5
a.
FG H
lb - mole H 2 O n1 lb - mole n1 n 2
FG H
0.21n 2 lb - mole O 2 lb - mole n1 n 2
IJ K
IJ K
b
n 0.600 y NO 2 mol N 2 O 4
g
Basis: 1000 lbm C3H8 / h fresh feed (Could also take 1 h operation as basis flow chart would be as below except that all / h would be deleted.) 1000 lb m C3H8 / h
g
b
b
n 6 lb m / h
n 7 lb m / h
g
0.02 lb m C3H 8 / lb m 0.98 lb m C3H 6 / lb m
g
0.97 lb m C3H 8 / lb m 0.03 lb m C3H 6 / lb m
Still
Compressor
Reactor
b n blb n blb n blb
g C H / hg CH / h g H / hg
b b
n1 lb m C3H 8 / h
n1 lb m C3H 8 / h 2
m
3
m
4
m
3
4
2
Note: the compressor and the off gas from the absorber are not mentioned explicitly in the process description, but their presence should be inferred.
b b
n3 lb m CH 4 / h n 4 lb m H 2 / h
g
g b
n5 lb m / h
g
Stripper
Absorber
b b n blb
n1 lb m C3H 8 / h
g g
n 2 lb m C3H 6 / h 5
4-2
g g
n 2 lb m C3H 6 / h
6
m
oil / h
g
4.5 (cont’d) b. Overall objective: To produce C3H6 from C3H8. Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3H8 to C3H6. Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other components. Stripping tower function: Recover the C3H8 and C3H6 from the solvent. Distillation column function: Separate the C3H5 from the C3H8. 4.6
a.
3 independent balances (one for each species)
b.
1 , m 3 , m 5 , x2 , y2 , y4 , z4 ) 7 unknowns ( m – 3 balances – 2 mole fraction summations 2 unknowns must be specified
c.
y2
1 x2
b gb g FGH kghA IJK F kg I F kg I 5300 G J m 1200 m G J Overall Balance: m H hK H hK F kg BIJ 1200 y 0.60m FG kg BIJ 5300 x G B Balance: 0.03m H hK H hK A Balance: 5300 x2
FG kg A IJ H h K
3 1200 0.70 m
1
3
1
z4 4.7
a.
2
4
5
1 0.70 y4
3 independent balances (one for each species)
b. Water Balance:
400 g 0.885 g H 2O min g
Acetic Acid Balance:
bg b g
R g 0.995 g H O m 2 R m min g
F g CH OOH IJ . gG b400gb0115 H min K 3
E m
356 g min
R 0.096m E 0.005m
461g min
FG g CH OOH IJ H min K 3
FG g IJ m m FG g IJ m 417 g min H min K H min K F g IJ b0.096gb461g FG g IJ 44 g min = 44 g min . gb400g b0.005gb356g G b0115 H min K H min K
C 400 Overall Balance: m c.
5
R
4-3
E
C
4.7 (cont’d) d. CH3COOH
H 2O some CH3COOH CH3COOH H 2O C4 H9OH
C4 H 9OH CH 3COOH
Extractor
Distillation Column
C4 H9OH
4.8
a.
X-large: 25 broken eggs/min 35 unbroken eggs/min
120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg
b.
120
n1 n2
b
g
25 35 n1 n2 eggs min n1 n2
b0.30gb120g c.
Large: n 1 broken eggs/min n 2 unbroken eggs/min
25 n1
50
U| n V| n W
1
2
11 39
50 large eggs min
b11 50g 0.22 22% of the large eggs (right hand) and b25 70g 36% of the extra-large eggs (left hand) n1 large eggs broken/50 large eggs
d.
are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed. 4.9
a.
b
m1 lb m strawberries 015 . lb m S / lb m 0.85 lb m W / lb m
c
m2 lb m S sugar
b.
g
b
m3 lb m W evaporated 1.00 lb m jam 0.667 lb m S / lb m 0.333 lb m W / lb m
h
3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF
4-4
g
4.10
a.
300 gal
b g
m1 lb m
0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2 O / lb m
b g
m3 lb m
0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2 O / lb m
b g m blb g
V40 gal
4 unknowns ( m1 , m2 ,V40 , m3 ) – 2 balances – 2 specific gravities 0 DF
m
2
0.400 lb m C 2 H 5 OH / lb m 0.600 lb m H 2 O / lb m
b.
300 gal
m1
1ft 3 0.877 u 62.4 lb m 7.4805 gal ft 3
Overall balance: m1 m2 m3 C2H5OH balance: 0.750m1 0.400m2 Solve (1) & (2) simultaneously m2 1646 lb m
V40 4.11
a.
b
n1 mol / s
(1) (2)
0.600m3 1646 lb m, , m3
ft 3 7.4805 gal 0.952 u 62.4lb m 1ft 3
3841 lb m
207 gal 3 unknowns ( n1 , n2 , n3 ) – 2 balances 1 DF
g
0.0403 mol C3H 8 / mol 0.9597 mol air / mol
b
2195 lb m
n 2 mol air / s
b
n3 mol / s
g
0.0205 mol C3H 8 / mol 0.9795 mol air / mol
g
0.21mol O 2 / mol 0.79 mol N 2 / mol
b.
Propane feed rate: 0.0403n1 Propane balance: 0.0403n1 Overall balance: 3722 n2
c.
b
g
150 n1 3722 mol / s 0.0205n3 n3 7317 mol / s 7317 n2 3600 mol / s
b
b
g
g
! . The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly.
4-5
4.12
a.
b
kg / h m
g
0.960 kg CH3OH / kg 0.040 kg H 2O / kg
1000 kg / h 0.500 kg CH 3OH / kg
,x ) 2 unknowns ( m – 2 balances 0 DF
0.500 kg H 2O / kg
673 kg / h
b
g 1 x b kg H O / kg g x kg CH3OH / kg 2
b.
Overall balance: 1000
673 m 327 kg / h m
b g
Methanol balance: 0.500 1000
b g b g
0.960 327 x 673 x
Molar flow rates of methanol and water: 673 kg 0.276 kg CH3OH 1000 g mol CH3OH h kg kg 32.0 g CH3OH
0.276 kg CH3OH / kg
5.80 u 103 mol CH3OH / h
673 kg 0.724 kg H 2 O 1000 g mol H 2 O 2.71 u 104 mol H 2 O / h h kg kg 18 g H 2O Mole fraction of Methanol: 5.80 u 103 0176 . mol CH3OH / mol 5.80 u 103 2.71 u 104 c. 4.13
Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state.
a.
Product Feed
Reactor
2253 kg
Reactor effluent
Purifier
2253 kg R = 388
1239 kg R = 583
W aste
b g
m w kg
R = 140
Analyzer Calibration Data 1
xp
x p = 0.000145R
0.1
1.364546
0.01 100
R
4-6
1000
4.13 (cont’d) b.
Effluent: x p
1.3645
1.3645
Product: x p
1.3645
Waste: x p Efficiency c.
b g 0.494 kg P / kg 0.000145b583g 0.861 kg P / kg . kg P / kg 0.000145b140g 0123 0.861b1239g u 100% 95.8% 0.494b2253g 0.000145 388
Mass balance on purifier: 2253 1239 mw mw P balance on purifier: Input: 0.494 kg P / kg 2253 kg 1113 kg P
1014 kg
b
gb g Output: b0.861 kg P / kggb1239 kgg b0123 . kg P / kggb1014 kgg
1192 kg P
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter. 4.14
a.
b
g
n1 lb- mole/ h . 00100 lb- mole H2O/ lb- mole . 09900 lb- mole DA/ lb- mole
b v dft / hi
b
g
h n2 lb- mole HO/ 2 2
g
n3 lb- mole/ h . lb- mole H2O / lb- mole 0100 . lb- mole DA/ lb- mole 0900
3
4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF
aR b 96.9 40.0 1.626 50 15
Assume linear relationship: v
v2 v1 R2 R1
Slope: a
va aR1
Intercept: b
v2
b g
b g
40.0 1.626 15
c
1.626 95 15.61 170 ft / h 3
h
b
3
n 2
170 ft 62 .4 lb m lb - mol h ft 3 18.0 lb m
589 lb - moles H 2 O / h
DA balance: 0.9900n1 0.900n3 Overall balance: n1 n2 n3 Solve (1) & (2) simultaneously n1 b.
15.61
g
5890 lb - moles / h, n 3
(1) (2) 6480 lb - moles / h
Bad calibration data, not at steady state, leaks, 7% value is wrong, v R relationship is not linear, extrapolation of analyzer correlation leads to error.
4-7
4.15
a.
b
kg / s m
100 kg / s 0.600 kg E / kg
g
0.900 kg E / kg 0100 . kg H 2 O / kg
0.050 kg S / kg 0.350 kg H 2 O / kg
b
kg / s m
g
, xE , xS ) 3 unknowns ( m – 3 balances 0 DF
b g x b kgS / kg g 1 x x b kg H O / kg g x E kg E / kg S
E
b.
2
g 0100 . b kgS / kgg
2m m 50.0 kg / s
Overall balance: 100
b g x b50g x
S balance: 0.050 100
b g
E balance: 0.600 100
kg E in bottom stream kg E in feed c.
b
S
S
S
0.300 kg E / kg b g b g 0.300b50g kg E in bottom stream 0.25 0.600b100g kg Ein feed
0.900 50 x E 50 x E
b g lnbag b lnbRg lnb x / x g lnb0.400 / 0100 . g . 1491 b lnb R / R g lnb38 / 15g lnbag lnb x g b lnb R g lnb0100 . g 1491 . lnb15g x aRb ln x 2
1
2
1
1
1
6.340 a 1764 u 103 .
u 103 R1.491 . x 1764 R d.
FG x IJ FG 0.900 IJ H a K H 1764 u 10 K . 1 b
3
1 1.491
. 655
Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass fractions – measure against known standard. Impurities in the stream – analyze a sample. Mixture is not all liquid – check sample. Calibration data are temperature dependent – check calibration at various temperatures. System is not at steady state – take more measurements. Scatter in data – take more measurements.
4-8
4.16
a.
4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution L of solution mol H 2SO 4 1213 . kgsolution
b.
bg
b
0.323 kg H 2SO 4 / kgsolution
5 unknowns ( v1 , v2 , v3 , m2 , m3 ) – 2 balances – 3 specific gravities 0 DF
v1 L
bg m b kg g
100 kg 0.200 kg H 2SO 4 / kg 0.800 kg H 2 O / kg . SG 1139
g
v3 L 3
0.323 kg H 2SO 4 / kg 0.677 kg H 2 O / kg SG 1.213
bg m b kg g v2 L 2
0.600 kg H 2SO 4 / kg 0.400 kg H 2 O / kg SG 1.498
Overall mass balance: 100 m2
b g
UV m m 0.677m W
m3
2
Water balance: 0.800 100 0.400m2 v1 v2 v1 v2 c.
4.17
100 kg
L 1139 . kg
44.4 kg 87.80 29.64
L 1498 . kg 2.96
3
3
44.4 kg 144 kg
87.80 L20%solution 29.64 L 60% solution
L 20%solution L 60% solution
1250 kg P 44.4 kg 60%solution L h 144 kg P 1498 . kgsolution
257 L / h
b g
m1 kg @$18 / kg 0.25 kg P / kg 0.75 kg H2O / kg
100 . kg 017 . kg P/ kg 0.83 kg H2O / kg
b g
m2 kg @$10 / kg 012 . kg P / kg 0.88 kg H 2O / kg Overall balance: m1 m2
100 . . m2 Pigment balance: 0.25m1 012
b g Solve (1) and (2) simultaneously m 0.385 kg 25% paint, m Cost of blend: 0.385b$18.00g 0.615b$10.00g $13.08 per kg . b$13.08g $14.39 per kg Selling price: 110 017 . 100 . 1
4-9
2
(1) (2) 0.615 kg12% paint
4.18
b
a.
gb
m1 kg H 2O 85% of entering water
g
100 kg 0.800 kgS / kg 0.200 kg H 2 O / kg
b g m b kg H Og m2 kgS 3
2
b
85% drying: m1
gb g 17.0 kg H O 0.800b100g 80.0 kgS
0.850 0.200 100
Sugar balance: m2
2
Overall balance: 100 17 80 m3 m3 3 kg H 2 O 0.0361 kg H 2O / kg xw 3 80 kg
b
g
17 kg H 2O 80 3 kg
m1 m2 m3 b.
3 kg H 2O
b
0.205 kg H 2O / kg wet sugar
g
1000 tons wet sugar 3 tons H 2 O 100 tons wet sugar day
30 tons H 2 O / day
1000 tons WS 0.800 tons DS 2000 lb m $0.15 365 days day ton WS ton lb m year c.
b b
g
1 xw1 x w 2 ... xw10 0.0504 kg H 2 O / kg 10 1 2 2 SD 0.00181 kg H 2 O / kg xw1 xw ... xw10 x w 9 Endpoints 0.0504 r 3 0.00181 xw
g
b
b
Lower limit
4.19
$8.8 u 10 7 per year
g
g
0.0450, Upper limit
0.0558
d.
The evaporator is probably not working according to design specifications since x w 0.0361 0.0450 .
a.
v1 m 3
c h m b kg H O g 2
1
1.00
SG
d i m b kg suspension g v3 m 3 3
d i
v2 m
3
SG
1.48
5 unknowns ( v1 , v2 , v3 , m1 , m3 ) – 1 mass balance – 1 volume balance – 3 specific gravities 0 DF
400 kg galena SG
7.44
Total mass balance: m1 400
m3
(1)
4-10
4.19 (cont’d) Assume volume additivity:
b g
m1 kg
m3 400 kg m 3 1000 kg 7440 kg
Solve (1) and (2) simultaneously m1 668 kg
v1
4.20
3
m 1000 kg
b g
m3 kg
m3 (2) 1480 kg
668 kg H 2 O, m3 1068 kg suspension
0.668 m 3 water fed to tank
b.
Specific gravity of coal < 1.48 < Specific gravity of slate
c.
The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48
a.
b
n1 mol / h
g
b
n2 mol / h
b
0.040 mol H 2 O / mol 0.960 mol DA / mol
g
x mol H 2O / mol
b
g
1 x mol DA / mol
b
n3 mol H 2 O adsorbed / h
g
g
97% of H 2O in feed
. 3.40g kg b354
Adsorption rate: n3
b
g
0.97 0.04n1 n1
97% adsorbed: 156 .
Total mole balance: n1
b g
. 1556 mol H 2 O / h
401 . mol / h
n2 n3 n 2
. Water balance: 0.040 401
4.21
5h
mol H 2O 0.0180 kg H 2 O . 1556 . 401
b
g
38.54 mol / h
b
x 38.54 x 12 1566 . . u 10 5 mol H 2O / mol
g
b.
The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%.
a.
300 lb m / h 0.55 lb m H 2SO 4 / lb m
b
0.45 lb m H 2 O / lb m
b
B lb m / h m
C lb m / h m
g
g
0.75 lb m H 2SO 4 / lb m 0.25 lb m H 2 O / lb m
0.90 lb m H 2SO 4 / lb m 010 . lb m H 2 O / lb m
B Overall balance: 300 m
m C B H2SO4 balance: 0.55 300 0.90m
b g
(1)
0.75m C
B Solve (1) and (2) simultaneously m
4-11
C 400 lb m / h, m
(2) 700 lb m / h
4.21 (cont’d) b.
500 150 A 7.78 RA 44.4 RA 25 m 70 25 800 200 B 200 B 15.0 RB 100 m RB 20 m 60 20 ln 100 ln 20 x 6.841e0.2682 Rx ln x ln 20 . Rx 4 ln x 0.2682 Rx 1923 10 4 300 44.4 400 100 44.3, mB 400 RB 333 ., mA 300 RA 7.78 15.0 1 55 ln 7.78 x 55% Rx 0.268 6.841
b b
A 150 m
g g
b
g
IJ K
FG H
c.
A m B Overall balance: m
m C
A 0.90m B H2SO4 balance: 0.01xm
d
0.75 0.01 6.841e 0.2682 Rx
15.0 RB 100 RB
d2.59 0.236e
Check: RA
4.22
a.
44.3, Rx
b
n A kmol / h
0.2682 Rx
iR
b
b
g
0.75 m A m B m B
i b7.78 R
A
b0.75 0.01xgm
g
44.4
135 . e0.2682 Rx 813 .
A
7.78 RB
e2.59 0.236e
b g
0.2682 7.78
. e j44.3 135
b g 813 .
0.2682 7.78
g 100 kg / h
b
n P kmol / h
g
0.20 kmol H 2 / kmol 0.80 kmol N 2 / kmol
g
0.50 kmol H 2 / kmol 0.50 kmol N 2 / kmol MW
b
g
b
A
015 .
. 015
. kmol H 2 / kmol 010 0.90 kmol N 2 / kmol n B kmol / h
0.75m C
g
0.20 2.016 0.80 28.012
22.813 kg / kmol
100 kg kmol 4.38 kmol / h h 22.813 kg Overall balance: n A n B 4.38 H2 balance: 010 . n A 0.50n B 0.20 4.38 n P
(1)
b g
Solve (1) and (2) simultaneously n A
4-12
(2) 3.29 kmol / h, n B
. kmol / h 110
. 333
4.22 (cont’d) b.
n P
m P 22.813
Overall balance: n A n B H2 balance: x A n A x B n B
c.
Trial 1 2 3 4 5 6 7 8 9 10 11 12
n A
b b
m P 22.813 x P m P 22.813
P xB x P m 22.813 x B x A XA 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10
g g
b b
P xP x A m 22.813 x B x A
n B
XB 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50
XP 0.10 0.20 0.30 0.40 0.50 0.60 0.10 0.20 0.30 0.40 0.50 0.60
mP 100 100 100 100 100 100 250 250 250 250 250 250
g g
nA 4.38 3.29 2.19 1.10 0.00 -1.10 10.96 8.22 5.48 2.74 0.00 -2.74
nB 0.00 1.10 2.19 3.29 4.38 5.48 0.00 2.74 5.48 8.22 10.96 13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture. d. 4.23
Results are the same as in part c. Venous blood 195.0 ml / min 175 . mg urea / ml
Arterial blood 200.0 ml / min 190 . mg urea / ml
Dialysate
b b
Dialyzing fluid 1500 ml / min a.
g
Water removal rate: 200.0 195.0 5.0 ml / min
b
g
b
g
Urea removal rate: 190 . 200.0 175 . 195.0 b.
g
v ml / min c mg urea / ml
v 1500 5.0 1505ml / min 38.8 mg urea / min 0.258 mg urea / ml c 1505 ml / min
4-13
38.8 mg urea / min
4.23 (cont’d) c. 2.7 11 . mg removed 1 min 10 3 ml 5.0 L 206 min (3.4 h) ml 38.8 mg removed 1L
b
4.24
a.
g
b
n1 kmol / min
g b
20.0 kg CO 2 / min
b
n 2 kmol / min
n3 kmol / min
g
g
0.023 kmol CO 2 / kmol
0.015 kmol CO 2 / kmol 20.0 kg CO 2 kmol 0.455 kmol CO 2 / min 44.0 kg CO 2 min Overall balance: 0.455 n 2 n3 CO2 balance: 0.455 0.015n 2 0.023n3 Solve (1) and (2) simultaneously n 2 55.6 kmol / min, n3 n1
b.
4.25
u
150 m 18 s
8.33 m / s
A
1 2 SD 4
. kmol 561 1 min s m3 D 108 . m min 0123 . kmol 60 s 8.33 m
Spectrophotometer calibration: C Dye concentration: A
0.60 cm 3
Dye injected
b
g bg
3.0 P g V L 4.26
a.
1000 L B / min
3
1
2
1
1
A 0.9 C 3
. gb018 . g b3333
g
3.333 A
3.0 P g
5.0 L
b g y b kmol SO / kmolg 1 y b kmol A / kmolg b kg / min g m x b kg SO / kgg 1 x b kg B / kg g n3 kmol / min 2
3
3
4
2
4
4
4-14
561 . kmol / min
0.600 Pg / L
1L 5.0 mg 103 P g 1L 103 cm 3 1 mg
0.600 P g / L V
b g V d m / min i n b kmol / min g y b kmol SO / kmolg 1 y b kmol A / kmolg 2 kg B / min m 1
018 . C
b
! C Pg / L
kA
(1) (2)
4.26 (cont’d)
2 , m 4 , x4 , y1 , y3 ) 8 unknowns ( n1 , n3 , v1 , m – 3 material balances – 2 analyzer readings – 1 meter reading – 1 gas density formula – 1 specific gravity 0 DF b.
Orifice meter calibration: A log plot of V vs. h is a line through the points h1 100, V1 142 and h2
d
ln V
b ln h ln a V ah ln V2 V1 ln 290 142 2
d
1
1
2.58 a
1
e 2.58
13.2 V
13.2h 0.515
Analyzer calibration: ln y bR ln a y aebR b ln a
b
ln y 2 y1
h1
R2 R1 ln y1 bR1
E
0.0600
5.00 u 10 4 e 0.0600 R
b g 207.3 m h b12.2g b150 14.7g 14.7 batmg 0.460 mol / L = 0.460 kmol / m b75 460g 18. bKg E
210 mm V1 13.2 210
U feed gas
n1
U| 90 20 || lnb0.00166g 0.0600b20g 7.60V y || |W
. 0.00166g g lnb01107
a 5.00 u 10 4
c.
207.3 m 3 0.460 kmol min m3
0.515
3
. kmol min 9534
b g expb0.0600 u 116 . g
. u 104 exp 0.0600 u 82.4 R1 82.4 y1 500
0.0702 kmol SO2 kmol
. y3 500 . u 104 R3 116
0.00100 kmol SO2 kmol
2 m
400, V2
b
d h b g 0.515 lnbh h g lnb400 100g ln a ln V b ln h lnb142g 0.515 ln 100
b
i
1000 L B 130 . kg 1300 kg / min min LB
4-15
3
i
290 .
4.26 (cont’d) A balance: 1 0.0702 95.34
b
SO2
gb g b1 0.00100gn n 88.7 kmol min x balance: b0.0702gb9534 . g( 64.0 kg / kmol) b0.00100gb88.7g(64) m 3
3
4 4
(1)
4 (1 x4 ) B balance: 1300 = m (2) 4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg Solve (1) and (2) simultaneously m 4 x4 SO2 removed = m
4.27
422 kg SO 2 / min
d.
Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a higher rate of transfer of SO2 from the gas to the liquid phase.
a.
V2 m 3 / min
b g y b kmolSO / kmol g 1 y b kmol A / kmolg
d i b kg B / min g m
n 3 kmol / min
3
d i n b kmol / min g y b kmolSO / kmolg 1 y b kmol A / kmol g V1 m 3 / min
R3
b g x b kgSO kg g 1 x b kg B / kg g 4 kg / min m
1
2
1
2
3
2
2
4
1
4
P1 , T1 , R1 , h1
b.
2 , n3 , y3 , R3 , m 4 , x4 ) 14 unknowns ( n1 ,V1 , y1 , P1 , T1 , R1 , h1 ,V2 , m – 3 material balances – 3 analyzer and orifice meter readings – 1 gas density formula (relates n1 and V1 ) 2 and V2 ) – 1 specific gravity (relates m 6 DF
b
g b1 y gn
A balance: 1 y1 n1 SO2 balance: y1n1 2 B balance: m
3
y3n3
b1 x gm 4
Calibration formulas:
(1)
3
4 x4 m 64 kgSO 2 / kmol
(2) (3)
4
y1
5.00 u 104 e0.060 R1
(4)
4 0.060 R3
(5)
y3 5.00 u 10 e V1 13.2h10.515 Gas density formula: n1
b
g
(6)
12.2 P1 14.7 / 14.7
bT 460g / 18. 1
Liquid specific gravity: SG 130 . V2
4-16
V1
(7)
b g
(8)
2 kg m m3 h 1300 kg
4.27 (cont’d) c.
T1
75 °F
y1
0.07 kmol SO2/kmol
P1
150 psig
V1
207 m3/h
h1
210 torr
n1
95.26 kmol/h
R1
82.4 x4 (kg SO2/kg)
Trial 1 2 3 4 5 6 7 8 9 10
0.10 0.10 0.10 0.10 0.10 0.20 0.20 0.20 0.20 0.20
y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h) 0.050 0.89 93.25 0.025 1.95 90.86 0.010 2.56 89.48 0.005 2.76 89.03 0.001 2.92 88.68 0.050 0.39 93.25 0.025 0.87 90.86 0.010 1.14 89.48 0.005 1.23 89.03 0.001 1.30 88.68
m4 (kg/h) m2 (kg/h) 1283.45 1155.11 2813.72 2532.35 3694.78 3325.31 3982.57 3584.31 4210.72 3789.65 641.73 513.38 1406.86 1125.49 1847.39 1477.91 1991.28 1593.03 2105.36 1684.29
3
V 2 ( m /h )
V2 vs. y3 3 .5 0 3 .0 0 2 .5 0 2 .0 0 1 .5 0 1 .0 0 0 .5 0 0 .0 0 0 .0 0 0
0 .0 2 0
0 .0 4 0
0 .0 6 0
y 3 ( k m o l S O 2 / k m o l)
x4 = 0 .1 0
x4 = 0 .2 0
For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed rate ( V2 ). For a given SO2 removal rate (y3), a higher solvent feed rate ( V2 ) tends to a more dilute SO2 solution at the outlet (lower x4). d. 4.28
Answers are the same as in part c. Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3 3 Overall mass balance m 1 Mass balance - Unit 1 m A balance - Unit 1 x1 2 Mass balance - mixing point m A balance - mixing point x2 C balance - mixing point y2
4-17
4.29
a.
100 mol / h 0.300 mol B / mol 0.250 mol T / mol 0.450 mol X / mol
b
n 2 mol / h
b g Column 1 x b mol T / molg 1 x x b mol X / mol g n b mol / h g x B 2 mol B / mol T2
B2
b
g
n 4 mol / h
0.940 mol B / mol 0.060 mol T / mol
Column 2
T2
3
0.020 mol T / mol
b
n5 mol / h
b g x b mol T / molg 1 x x b mol X / mol g T5
B5
Column 1 4 unknowns ( n2 , n3 , x B 2 , xT 2 ) –3 balances – 1 recovery of X in bot. (96%) 0 DF
b
Total mole balance: 100
b g T balance: 0.250b100g B balance: 0.300 100
Total mole balance: n2
b.
0.98n3
(1)
n 2 n3
(2)
x B 2 n 2
(3)
xT 2 n 2 0.020n3
(4)
0.940n 4
(5)
n4 n5
(6)
B balance: x B 2 n 2
0.940n 4 x B5 n5
(7)
T balance: xT 2 n 2
0.060n 4 xT 5 n5
(8)
(1) n 3 44.1 mol / h (3) x B 2 0.536 mol B / mol (5) n 4 30.95 mol / h (7) x B5 0.036 mol B / mol Overall benzene recovery:
T5
Column 2: 4 unknowns ( n3 , n4 , n5 , y x ) – 3 balances – 1 recovery of B in top (97%) 0 DF
gb g
Column 2 97% B recovery: 0.97 x B 2 n 2
g
x B5 mol B / mol
0.980 mol X / mol
Column 1 96% X recovery: 0.96 0.450 100
b
(2) n 2 55.9 mol / h (4) x T 2 0.431 mol T / mol (6) n5 24.96 mol / h (8) x T 5 0.892 mol T / mol
g
0.940 30.95 u 100% 0.300 100
b g 0.892b24.96g u 100 Overall toluene recovery: 0.250b100g
4-18
g
97%
89%
4.30
a.
g
3 kg / h m
0.035 kg S / kg
x3 kg S / kg
0.965 kg H 2 O / kg
b
w kg H 2O / h m
b.
b b
100 kg / h
1
b
g
1 x3 kg H 2O / kg
b
w kg H 2O / h 0100 . m
g
b g x b kg S / kgg 1 x b kg H O / kgg 4 kg / h m
4
g
4
4
g
b
w kg H 2O / h 0100 . m
Overall process 100 kg/h 0.035 kg S/kg 0.965 kg H2O/kg
m 10 ( kg / h) 0.050 kg S/kg 0.950 kg H2O/kg m w ( kg H 2 O / h)
b g
Salt balance: 0.035 100 Overall balance: 100 H2O yield: Yw
b b
0.050m 10
m w m 10
g
w kg H 2O recovered m 96.5 kg H 2 O in fresh feed
g
First 4 evaporators
b b
g
m 4 kg / h x 4 kg S/ kg 1 x4 kg H2 O / kg
100 kg/ h 0.035 kg S/ kg 0.965 kg H2 O / kg
b g
Yw
0.31
x4
0.0398
b
g
4 0100 . m w m 4
Salt balance: 0.035 100 c.
g
b g 4 u 0100 . m b kg H O / hg w
Overall balance: 100
2
x4 m 4
4-19
2
g
10
b
10 kg / h m
g
0.050 kg S / kg 0.950 kg H 2O / kg
b
w kg H 2O / h 0100 . m
g
4.31
b g
a.
2n1 mol
Condenser
0.97 mol B / mol 0.03 mol T / mol
b g
b g
n1 mol (89.2% of Bin feed )
n1 mol
0.97 mol B / mol 0.03 mol T / mol
0.97 mol B / mol 0.03 mol T / mol
100 mol 0.50 mol B / mol
Still
0.50 mol T / mol
b gb y b mol B / molg 1 y b mol T / mol g
n 4 mol 45% of feed to reboiler
g
B
B
b g z b mol B / molg 1 z b mol T / molg n 2 mol B
b g x b mol B / molg 1 x b mol T / molg n3 mol
Reboiler
B
B
B
Overall process:
Condenser:
3 unknowns ( n1 , n3 , x B ) Still: 5 unknowns ( n1 , n2 , n4 , y B , z B ) – 2 balances – 2 balances – 1 relationship (89.2% recovery) 3 DF 0 DF
1 unknown ( n1 ) – 0 balances 1 DF
Reboiler:
6 unknowns ( n2 , n3 , n4 , x B , y B , z B ) – 2 balances – 2 relationships (2.25 ratio & 45% vapor) 3 DF
Begin with overall process. b.
Overall process 89.2% recovery: 0.892 0.50 100
b gb g
0.97 n1
Overall balance: 100
n1 n3
B balance: 0.50 100
0.97 n1 x B n3
b g
Reboiler Composition relationship: Percent vaporized: n 4 Mole balance: n2
e j / b1 x g
yB / 1 yB xB
0.45n 2
(1)
n3 n 4
(2)
(Solve (1) and (2) simultaneously.) B balance: z B n2
2.25
B
x B n3 y B n 4
4-20
4.31 (cont’d) c. B fraction in bottoms: x B Moles of overhead: n1 Recovery of toluene: 4.32
0100 . mol B / mol Moles of bottoms: n3
46.0 mol
. gb54.02g b1 x gn u 100% b1 010 u 100% 0.50b100g 0.50b100g B
3
a.
b
m3 kg H 2O
97%
g
Bypass
Mixing point
b g
Evaporator
m1 kg
100 kg . kg S / kg 012 0.88 kg H2O / kg
54.0 mol
012 . kg S / kg 0.88 kg H2O / kg
b g
b g
m4 kg
m5 kg
0.58 kg S / kg 0.42 kg H 2O / kg
0.42 kg S / kg 0.58 kg H2O / kg
b g
m2 kg
012 . kg S / kg 0.88 kg H2O / kg
Overall process:
Evaporator:
2 unknowns ( m3 , m5 ) – 2 balances 0 DF
3 unknowns ( m1 , m3 , m4 ) – 2 balances 1 DF
b g
. 100 Overall S balance: 012 Overall mass balance: 100
Bypass:
2 unknowns ( m1 , m2 ) – 1 independent balance 1 DF
Mixing point: 3 unknowns ( m2 , m4 , m5 ) – 2 balances 1 DF
0.42m5 m3 m5
Mixing point mass balance: m4 m2
m5
. m2 Mixing point S balance: 0.58m4 012
(1)
0.42m5
(2)
Solve (1) and (2) simultaneously Bypass mass balance: 100 m1 m2 b.
m1
90.05 kg, m2
Bypass fraction: c.
m2 100
9.95 kg, m3
714 . kg, m4
18.65 kg, m5
28.6 kg product
0.095
Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a stream consisting of 90% solids could be hard to transport.
4-21
4.33
a.
b
4 kg Cr / h m
b
1 kg / h m
g
b
2 kg / h m
g
g
b g x b kg Cr / kgg 1 x b kg W / kg g 5 kg / h m
0.0515 kg Cr / kg
0.0515 kg Cr / kg
0.9485 kg W / kg
0.9485 kg W / kg
Treatment Unit
b
3 kg / h m
5
5
b g x b kg Cr / kgg 1 x b kg W / kgg 6 kg / h m 6
6
g
0.0515 kg Cr / kg 0.9485 kg W / kg
b.
1 m
b
2 6000 kg / h m
4500 kg / h maximum allowed value 3 6000 4500 1500 kg / h Bypass point mass balance: m 4 95% Cr removal: m
b
gb
g
g
0.95 0.0515 4500 220.2 kg Cr / h 5 4500 220.2 4279.8 kg / h Mass balance on treatment unit: m 0.0515 4500 220.2 Cr balance on treatment unit: x5 0.002707 kg Cr / kg 4779.8 6 1500 4279.8 5779.8 kg / h Mixing point mass balance: m
b
Mixing point Cr balance: x6
c.
b g
g
b
0.0515 1500 0.0002707 4279.8 5779.8
m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) 1000 1000 0 48.9 951 2000 2000 0 97.9 1902 3000 3000 0 147 2853 4000 4000 0 196 3804 5000 4500 500 220 4280 6000 4500 1500 220 4280 7000 4500 2500 220 4280 8000 4500 3500 220 4280 9000 4500 4500 220 4280 10000 4500 5500 220 4280
4-22
g
0.0154 kg Cr / kg
m 6 (kg/h) x5 0.00271 951 0.00271 1902 0.00271 2853 0.00271 3804 0.00271 4780 0.00271 5780 0.00271 6780 0.00271 7780 0.00271 8780 0.00271 9780
x6 0.00271 0.00271 0.00271 0.00271 0.00781 0.0154 0.0207 0.0247 0.0277 0.0301
4.33 (cont’d)
x 6 (kg Cr/kg)
m 1 vs. x 6 0.03500 0.03000 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0
2000
4000
6000
8000 10000 12000
m 1 (kg /h )
d.
4.34
Cost of additional capacity – installation and maintenance, revenue from additional recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon, regulatory limits on Cr emissions.
a.
b
175 kg H2O / s 45% of water fed to evaporator
b
1 kg / s m
g
b b
4 kg K 2SO 4 / s m
0196 . kg K 2SO 4 / kg 0.804 kg H 2O / kg
5 kg H 2O / s m
g
g
b b
6 kg K 2SO 4 / s m
Evaporator
7 kg H 2 O / s m
g
g
g
Crystallizer Filter
Filter cake
b
2 kg K 2SO 4 / s 10m
g
b g RS0.400 kg K SO / kg UV T0.600 kg H O / kg W 2 kgsoln / s m 2
Filtrate
b
3 kg / s m
4
2
g
0.400 kg K 2SO 4 / kg 0.600 kg H 2 O / kg
Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s 1, m 2) Overall process: 2 unknowns ( m - 2 balances 0 DF
Mixing point:
Evaporator:
Crystallizer:
4, m 5, m 6, m 7 ) 4 unknowns ( m – 2 balances – 1 percent evaporation 1 DF
1, m 2 Strategy: Overall balances m 5 % evaporation m 3, m 4 Balances around mixing point m 6, m 7 Balances around evaporator m
4-23
1, m 3, m 4, m 5 ) 4 unknowns ( m - 2 balances 2 DF
2, m 3, m 6, m 7 ) 4 unknowns ( m – 2 balances 2 DF
U| verify that each |V chosen subsystem involves || no more than two W unknown variables
4.34 (cont’d) Overall mass balance: m 1 175 10m 2 m 2 Overall K balance: . m 1 10m 2 0.400m 2 0196
U| V| W
Production rate of crystals 10m 2 45% evaporation: 175 kg evaporated min
0.450m 5
W balance around mixing point: 0.804m 1 0.600m 3 Mass balance around mixing point: m 1 m 3 K balance around evaporator: m 6
m 4
W balance around evaporator: m 5
175 m 7
m 4 m 5
Mole fraction of K in stream entering evaporator =
b.
1 Fresh feed rate: m
c.
b
g
3 kg recycle s m 1 kg fresh feed s m
b
m 4 m 4 m 5
221 kg / s
bg
2 Production rate of crystals 10m Recycle ratio:
m 5
. kg K s s 416
g
352.3 kg recycle . 160 220.8 kg fresh feed
Scale to 75% of capacity. Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s 46.3% K, 537% . W
d.
Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and the cooling cost for the crystallizer.
4-24
4.35
a.
Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge CO2 to the atmosphere. Absorber function: Separates CO2 from CH4. Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused.
b.
The top streams are liquids while the bottom streams are gases. The liquids are heavier than the gases so the liquids fall through the columns and the gases rise.
c.
b
n1 mol / h
g
b g n b mol CO / h g n 5 mol N 2 / h
0.010 mol CO 2 / mol 0.990 mol CH 4 / mol
100 mol / h 0.300 mol CO 2 / mol
b
Absorber
n 2 mol / h
6
g
Stripper
0.005 mol CO 2 / mol
0.700 mol CH 4 / mol
2
b
n 5 mol N 2 / h
g
0.995 mol CH 3OH / mol
b g n b mol CH OH / h g n 3 mol CO 2 / h 4
Overall:
Stripper:
3
3 unknowns ( n1 , n5 , n 6 ) – 2 balances 1 DF
Absorber:
4 unknowns ( n2 , n3 , n4 , n5 ) – 2 balances – 1 percent removal (90%) 1 DF
b0.700gb100g bmol CH / hg Overall mole balance: 100b mol / h g n n Overall CH4 balance:
4
1
Percent CO2 stripped: 0.90n3 Stripper CO2 balance: n3
n1 n6
70.71 mol / h , n2
0.990n1
6
n 6
n 6 0.005n2
Stripper CH3OH balance: n4 d.
4 unknowns ( n1 , n 2 , n3 , n 4 ) – 3 balances 1 DF
0.995n 2
. mol / h , n3 6510
32.55 mol CO 2 / h, n 4
647.7 mol CH 3OH / h ,
29.29 mol CO 2 / h
Fractional CO2 absorption: f CO 2
30.0 0.010n1 30.0
4-25
0.976 mol CO 2 absorbed / mol fed
4.35 (cont’d) Total molar flow rate of liquid feed to stripper and mole fraction of CO2: n3 n3 n 4 680 mol / h , x3 0.0478 mol CO 2 / mol n3 n 4 e.
Scale up to 1000 kg/h (=106 g/h) of product gas: MW1
b
g
b
0.01 44 g CO 2 / mol 0.99 16 g CH 4 / mol
g
16.28 g / mol
u 10 mol / h . bn g d10. u 10 g / hib16.28 g / molg 6142 u 10 mol / h) / (70.71 mol / h) . bn g b100 mol / hg (6142 6
1 new
4
4
feed new
4.36
8.69 u 104 mol / h
f.
Ta Ts The higher temperature in the stripper will help drive off the gas. Pa ! Ps The higher pressure in the absorber will help dissolve the gas in the liquid.
g.
The methanol must have a high solubility for CO2, a low solubility for CH4, and a low volatility at the stripper temperature.
a.
Basis: 100 kg beans fed
e
m kg C H 5 6 14
e
m kg C H 1 6 14
j
300 kg C 6 H14
Ex
j
Condenser
b g b g y b kg oil / kg g 1 x y b kg C H m2 kg
2
13.0 kg oil 87.0 kg S
2
2
6
b g b g 1 y b kg C H m4 kg
F
x 2 kg S / kg
14 / kg
Ev
y 4 kg oil / kg 4
g
6
14
/ kg
g
g
b g
m3 kg
0.75 kg S / kg
b
y3 kg oil / kg
b
g
0.25 y3 kg C 6 H14 / kg
Overall:
b
m6 kg oil
4 unknowns ( m1 , m3 , m6 , y3 ) – 3 balances 1 DF
Extractor:
g
3 unknowns ( m2 , x2 , y2 ) – 3 balances 0 DF
Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 ) 2 unknowns ( m1 , m5 ) – 1 balance – 2 balances 1 DF 2 DF Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 ) – 3 balances – 1 oil/hexane ratio 3 DF
Mixing Pt:
Start with extractor (0 degrees of freedom) Extractor mass balance: 300 87.0 13.0 kg
4-26
m2
4.36
(cont’d) Extractor S balance: 87.0 kg S
x2 m2
Extractor oil balance: 13.0 kg oil
y2 m2
Filter S balance: 87.0 kg S 0.75m3
b g
m3 m4 Oil / hexane ratio in filter cake:
Filter mass balance: m2 kg y3
y2 1 x2 y 2
0.25 y3
Filter oil balance: 13.0 kg oil
b
y3m3 y4 m4
g
Evaporator hexane balance: 1 y4 m4 Mixing pt. Hexane balance: m1 m5 Evaporator oil balance: y4 m4 b.
m5 300 kg C6 H14
m6
b
m6 100
g
118 . kg oil 0118 . kg oil / kg beans fed 100 kg beans fed m1 28 kg C6 H14 0.28 kg C6 H14 / kg beans fed Fresh hexanefeed 100 100 kg beans fed m5 272 kg C 6 H14 recycled Recycle ratio 9.71 kg C6 H14 recycled / kg C6 H14 fed m1 28 kg C 6 H14 fed
Yield
b b
c.
g
g
Lower heating cost for the evaporator and lower cooling cost for the condenser.
4.37
b
g
m lb m dirt 1 98 lb m dry shirts 3 lb m Whizzo 100 lb
m 2 lb m dirt 98 lb m dry shirts
b
m lb m Whizzo 2
g
b g
Tub
Filter
b g
m lb m 4 013 . lb m dirt / lb m 0.87 lb m Whizzo / lb m
m lb m 3 0.03 lb m dirt / lb m 0.97 lb m Whizzo / lb m
b g b b
m lb m 4 1 x lb m dirt / lb m 4 x lb m Whizzo / lb m 4
g
b g
m lb m 5 0.92 lb m dirt / lb m 0.08 lb m Whizzo / lb m
g
Strategy 95% dirt removal m1 ( 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) m2 , m5 (solves Part (a))
4-27
4.37
(cont’d)
b g around the filter b m , m , x g , but the tub only involves 2 b m , m g and 2 balances are Balances around the mixing point involve 3 unknowns m3 , m6 , x , as do balances 4
6
3
4
allowed for each subsystem. Balances around tub m3 , m4 Balances around mixing point m6 , x (solves Part (b)) a.
. lb dirt b0.05gb2.0g 010 Overall dirt balance: 2.0 010 . b0.92gm m 2.065 lb dirt Overall Whizzo balance: m 3 b0.08gb2.065g blb Whizzog 95% dirt removal: m1
m
5
5
2
b.
m
m
Tub dirt balance: 2 0.03m3 010 . 013 . m4 Tub Whizzo balance: 0.97m3 3 0.87m4 Solve (1) & (2) simultaneously m3 20.4 lb m , m4 19.3 lb m Mixing pt. mass balance: 317 . m6 20.4 lb m m6 17.3 lb m Mixing pt. Whizzo balance: 317 . x4 19.3 0.97 20.4 x 0.833 lb m Whizzo / lb m 833% . Whizzo, 6.7% dirt
b g b gb g
4.38
317 . lb m Whizzo
a.
2720 kg S mixer 3 Discarded C 3L kg L C 3S kg S 3300 kg S
Filter 3
C 2L kg L C 2S kg S
F 3L kg L F 3S kg S
620 kg L mixer 1 Filter 1 F 1L F 1S
mixer 2 C 1L kg L C 1S kg S kg L kg S
Filter 2 F 2L kg L F 2S kg S To holding tank
b g g U| V| W
F1L 6.2 kg L mixer filter 1: 0.01 620 F1L balance: 620 6.2 C1L C1L 6138 . kg L F2 L . F3 L F2 L 6.2 kg L mixer filter 2: 0.01 6138 balance: 6138 . F3L F2 L C3 L C2 L 613.7 kg L F3 L 61 mixer filter 3: 0.01C2 L F3L . kg L balance: 613.7 = 6.1+ C3L C3 L 607.6 kg L
b
4-28
(1) (2)
4.38 (cont’d) Solvent C1S 495 kg S m f 1: 015 . 3300 C1S F1S 2805 kg S balance: 3300 495 F1S C2 S . 495 F3S C2 S 482.6 kg S m f 2: 015 F2 S 2734.6 kg S balance: 495 F3S C2 S F2 S C3S 480.4 kg S m f 3: 015 C3S . 2720 C2 S F3S 2722.2 kg S balance: 2720 + C2S F3S C3S
b
g
b
U| |V || W
g
b
g
Holding Tank Contents 6.2 6.2 12.4 kg leaf 2805 2734.6 b.
5540 kg solvent
b g
5540 kgS
b g
QR kg
0165 . kg E / kg
Q0 kg
. kg E / kg Extraction 013 0.15kg F / kg Unit
0.835 kg W / kg
b g Q b kg Fg
Steam Stripper
0.855kg W / kg
QD kg D
b g
b g Q b kg Dg Q b kg Fg QE kg E
F
0.200 kg E / kg 0.026 kg F / kg 0.774 kg W / kg QB kg 0.013 kg E / kg 0.987 kg W / kg
D F
b
Q3 kg steam Mass of D in Product:
1 kg D
620 kg leaf
1000 kg leaf
0.62 kg D
QD
b g
Water balance around extraction unit: 0.835 5540 0.855QR QR Ethanol balance around extraction unit: 0165 . 5540 013 . 5410 QE QE 211 kg ethanol in extract
b
c.
g
b
g
b
F balance around stripper 0.015 5410 0.026Q0 Q0
b g
g
b
b g
b
4.39
a.
g
b
g
b
C 2 H 2 2 H 2 o C2 H 6 2 mol H 2 react / mol C2 H 2 react 0.5 kmol C2 H 6 formed / mol H 2 react
4-29
g
6085 kg mass of stripper bottom product
W balance around stripper 0.855 5410 QS 0.774 3121 0.987 6085 QS
b
5410 kg
3121 kg mass of stripper overhead product
E balance around stripper 013 . 5410 0.200 3121 0.013QB QB
b g
g
g
g
3796 kg steam fed to stripper
4.39 (cont’d) b. nH 2 nC2 H 2
15 . 2.0 H 2 is limiting reactant
15 . mol H 2 fed 10 . mol C2 H 2 fed 0.75 mol C2 H 2 required (theoretical) 10 . mol fed 0.75 mol required u 100% 333% % excess C2 H 2 . 0.75 mol required c.
4 u 106 tonnes C2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C2 H 6 2 kmol H 2 2.00 kg H 2 300 days 24 h 3600 s tonne 30.0 kg C2 H 6 1 kmol C2 H 6 1 kmol H 2 yr 20.6 kg H 2 / s
4.40
d.
The extra cost will be involved in separating the product from the excess reactant.
a.
4 NH 3 5 O 2 o 4 NO 6 H 2 O 5 lb - mole O 2 react 125 . lb - mole O 2 react / lb - mole NO formed 4 lb - mole NO formed
b.
dn i
100 kmol NH3 5 kmol O 2 h 4 kmol NH3
O 2 theoretical
d i
40% excess O 2 nO 2 c.
fed
b
125 kmol O 2
140 . 125 kmol O 2
g
175 kmol O 2
b50.0 kg NH gb1 kmol NH / 17 kg NH g 2.94 kmol NH . kmol O b100.0 kg O gb1 kmol O / 32 kg O g 3125 F n I 3125 F n I 5 125 . G GH n JK 2..94 106 H n JK 4 . 3
3
2
2
2
O2
NH 3
3
3
2
O2
NH 3
fed
stoich
O 2 is the limiting reactant Required NH3:
3125 . kmol O 2 4 kmol NH3 5 kmol O 2
2.50 kmol NH3
2.94 2.50 u 100% 17.6% excess NH3 2.50 nO 2 vO 2 [ 0 3125 . 5 [ [ Extent of reaction: nO 2 % excess NH3
d i
Mass of NO:
4.41
a.
b g
0
3125 . kmol O 2 4 kmol NO 30.0 kg NO 5 kmol O 2 1 kmol NO
0.625 kmol
625 mol
75.0 kg NO
By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed. Automation provides for faster and more accurate response to fluctuations in the feed stream, reducing the risk of release of H2S and SO2. It also may reduce labor costs.
4-30
4.41 (cont’d) b.
3.00 u 10 2 kmol 0.85 kmol H 2 S 1 kmol SO 2 h kmol 2 kmol H 2 S
n c
c.
127.5 kmol SO 2 / h
C a lib r a t io n C u r v e
X ( m o l H2S /m o l)
1 .2 0 1 .0 0 0 .8 0 0 .6 0 0 .4 0 0 .2 0 0 .0 0 0 .0
2 0 .0
4 0 .0
6 0 .0
8 0 .0
1 0 0 .0
R a (m V )
0.0199 Ra 0.0605
X d.
b
n c kmol SO 2 / h
b
n f kmol / h
b
g
x kmol H 2S / kmol
g
Blender
g
Flowmeter calibration:
n f aR f 100 kmol / h , R f
n f
Control valve calibration:
n c nc
UVn 15 mV W
25.0 kmol / h, R c 60.0 kmol / h , Rc
FG H
f
UV W
20 Rf 3
10.0 mV n c 25.0 mV
IJ b K
7 5 Rc 3 3
1 7 5 1 20 n f x Rc R f 0.0119 Ra 0.0605 2 3 3 2 3 5 10 R f 0.0119 Ra 0.0605 Rc 7 7
Stoichiometric feed: n c
b
n f
3.00 u 10 2 kmol / h R f
g
3 n f 20
4-31
45 mV
g
4.41 (cont’d)
b gb b g
e.
gb g
10 5 45 0.0119 76.5 0.0605 53.9 mV 7 7 7 5 n c 53.9 127.4 kmol / h 3 3 Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond range of calibration data, system had not reached steady state yet. Rc
4.42 165 mol / s
b
x mol C 2 H 4 / mol
b
b
n mol / s
g
1 x mol HBr / mol
g
0.310 mol C2 H 4 / mol . mol HBr / mol 0173 0.517 mol C 2 H 5Br / mol
g
C 2 H 4 HBr o C 2 H 5 Br C balance:
b
g
b
165 mol x mol C2 H 4 2 mol C s mol mol C2 H 4
gb g b
gb g
n 0.310 2 n 0.517 2
. gb1g b gb g nb0173
Br balance: 165 1 x 1
(2)
Solve (1) and (2) simultaneously n 108.77 mol / s, x
b g
1 x
(1)
0.545 mol C 2 H 4 / mol
0.455 mol HBr / mol
Since the C2H4/HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1), HBr is the limiting reactant .
bn g b165 mol / sgb0.455 mol HBr / molg 75.08 mol HBr . gb108.8g 75.08 b0173 u 100% Fractional conversion of HBr HBr fed
dn i 75.08 mol C H dn i b165 mol / sgb0.545 mol C H C 2 H 4 stoich
2
75.08
0.749 mol HBr react / mol fed
4
C 2 H 4 fed
2
4
/ mol
g
89.93 mol C2 H 4
89.93 75.08 19.8% 75.08 Extent of reaction: n C2 H5Br n C2 H5Br vC2 H5Br[ 108.8 0.517
% excess of C2 H 4
d
i
0
4-32
b
gb
g
bg
0 1[ [
56.2 mol / s
4.43
a.
1 O 2 o Cl 2 H 2 O 2
2HCl
Basis: 100 mol HCl fed to reactor
b g n b mol O g n b mol N g n b mol Cl g n b mol H Og n2 mol HCl
100 mol HCl
b
n1 mol air
g
0.21 mol O 2 / mol
100 mol HCl 0.5 mol O 2
2
b
g
85 mol HCl react
1 mol Cl 2
2
15 mol HCl
42.5 mol Cl 2
2 mol HCl
N 2 balance:
2
135 . u 25 n1 160.7 mol air fed
85% conversion 85 mol HCl react n2
b85gb1 2g
2
25 mol O 2
2 mol HCl
35% excess air: 0.21n1 mol O 2 fed
n6
4
6
35% excess
n5
2
5
0.79 mol N 2 / mol
bO gstoic
3
42.5 mol H 2 O
b160.7gb0.79g
n4 n4
127 mol N 2
O balance:
b160.7gb0.21g mol O
2
2 mol O 1 mol O 2
2 n3
42.5 mol H 2 O
1 mol O n3 1 mol H 2 O
12.5 mol O 2
Total moles: 5
¦nj
239.5 mol
j 2
b.
0.063
mol O 2 mol N 2 mol HCl , 0.052 , 0.530 , mol mol mol
0177 .
mol Cl 2 mol H 2 O , 0177 . mol mol
15 mol HCl 239.5 mol
As before, n1 160.7 mol air fed , n2
15 mol HCl
1 2HCl O 2 o Cl 2 H 2 O 2 ni
E
bn g
i 0
vi [
HCl: 15 100 2[ [
42.5 mol
4-33
4.43 (cont’d)
N 2 : n4
b g 0.79b160.7g
Cl 2 : n5
[
42.5 mol Cl 2
H 2 O: n6
[
42.5 mol H 2 O
c.
4.44
1 [ 12.5 mol O 2 2 127 mol N 2
0.21 160.7
O 2 : n3
These molar quantities are the same as in part (a), so the mole fractions would also be the same. Use of pure O2 would eliminate the need for an extra process to remove the N2 from the product gas, but O2 costs much more than air. The cheaper process will be the process of choice.
b g Fe O 3H SO o Fe bSO g 3H O bTiOgSO 2H O o H TiO bsg H SO H TiO bsg o TiO bsg H O
FeTiO3 2H 2SO 4 o TiO SO 4 FeSO 4 2H 2 O 2
3
2
4
2
4
2
2
2
3
2
4 3
2
3
2
4
2
Basis: 1000 kg TiO2 produced 1000 kg TiO 2
kmol TiO 2
1 kmol FeTiO 3
79.90 kg TiO 2
1 kmol TiO 2
12.52 kmol FeTiO 3 dec.
12.52 kmol FeTiO 3 decomposes
1 kmol FeTiO 3 feed
14.06 kmol FeTiO 3 fed
0.89 kmol FeTiO 3 dec. 14.06 kmol FeTiO 3
b
1 kmol Ti
47.90 kg Ti
1 kmol FeTiO 3
kmol Ti
673.5 kg Ti / M kg ore
g
0.243 M
6735 . kg Ti fed
2772 kg ore fed
b
g
Ore is made up entirely of 14.06 kmol FeTiO 3 + n kmol Fe2 O3 (Assumption!) n
2772 kg ore
638.1 kg Fe 2 O 3
14.06 kmol FeTiO 3 151.74 kg FeTiO 3 kmol FeTiO3 kmol Fe2O 3
. kg Fe2 O 3 6381
4.00 kmol Fe2O 3
159.69 kg Fe2 O 3
14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4 4012 . kmol H2SO4 1 kmol FeTiO3 1 kmol Fe2O3
b
50% excess: 15 . 4012 . kmol H 2SO 4 Mass of 80% solution: 5902.4 kg H 2 SO 4 / M
g
6018 . kmol H 2SO 4 fed
60.18 kmol H 2SO 4
a
bkg solng
98.08 kg H 2SO 4 1 kmol H 2SO 4
0.80 M
4-34
a
5902.4 kg H 2SO 4
7380 kg 80% H 2 SO 4 feed
4.45
a.
Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
dR
i
0.30 g m 3 and
10, C1
1
ln C bR ln a C b
b
g
0.0575 , ln a 48 10 C 0169 . e 0.0575 R
d
i C c(ftlb E
48, C2
2
m)
3
b g
b g
ln 2.67 0.0575 48
453.6 g 35.31 ft 3 1 m3 1 lb m
d
d2867 ft sib60 s ming
178 . a
e 1.78
0169 .
i
u 10 5 e 0.0575 R 1055 .
3
138 ft 3 lb m coal
1250 lb m min R
d
37 C c lb m SO 2 ft 3
i
u 105 eb 1055 .
8.86 u 105 lb m SO 2 138 ft 3 ft 3 1 lb m coal c.
IK
16,020C c
16,020C ' 0169 . e 0.0575 R C c lb m SO 2 ft 3 b.
2.67 g m 3
ae br
ln 2.67 0.30
C g m3
FH R
gb g
0.0575 37
0.012 0.018
8.86 u 10
5
lb m SO 2 ft 3
lb m SO 2 compliance achieved lb m coal
S O 2 o SO 2 1250 lb m coal 0.05 lb m S 64.06 lb m SO 2 min
1 lb m coal
32.06 lb m S
124.9 lb m SO 2 generated min
2867 ft 3 60 s 886 . u 105 lb m SO2 152 . lbm SO2 min in scrubbed gas s 1 min ft3 air 1250 lbm coal/min 62.5 lb m S/min
% removal d.
furnace ash
b124.9 15.2g lb
scrubbing fluid stack gas 124.9 lbm SO2 /min
scrubber scrubbed gas 15.2 lb m SO2 /min liquid effluent (124.9 – 15.2) lb m SO2 (absorbed)/min
SO 2 scrubbed min u 100% 124.9 lb m SO 2 fed to scrubber min m
88%
The regulation was avoided by diluting the stack gas with fresh air before it exited from the stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal burned is independent of the flow rate of air in the stack.
4-35
4.46
a.
A B ===== C + D nA nA [ 0
nB
nB [
yA
nC
nC [
yB
nD
nD [
yC
0 0
0
nI
nI
yD
0
¦ ni
Total nT
At equilibrium:
en en en en
bn bn
yC y D y A yB
C0 A0
gb [ gbn
B0
j [j n [j n [j n
[ nT T
C0 D0
T
g [ g
[ c n D0 [ c c
b
c
A0
B0
T
4.87 ( nT ’s cancel)
c
gh b
387 . [ 2c nC0 nD0 487 . nA0 nB0 [ c nC0nD0 487 . nA0nB0
g
0
[a[ 2c b[ c c 0]
?[ c
b.
a 387 . 1 2 b r b 4ac where b nC0 nD0 487 . nA0 nB0 2a c nC0nD0 487 . nA0nB0
e
b[
b 387 .
F b gH
1 1174 . r 2 387 .
[e e2
1174 c .
. g b1174
nA0 80, nC0 nC nA nB nC nD 4.87
2
4.87
b gb gIK [
4 387 4.87 .
e1
0.496
2.54 is also a solution but leads to a negative conversion
Fractional conversion: X A c.
g
nA0 1 nB0 1 nC0 nB0 nI 0 0
Basis: 1 mol A feed Constants: a
b
j
b
XB
g
n A0 n A n A0
[ e1 n A0
g
0.496
nJ 0 0 nC 0 0 ! [ c 70 mol 70 nC0 [ c n A0 [ c n A0 70 mol n B0 [ c 80 70 10 mol nC 0 [ c 70 mol n D0 [ c 70 mol yC y D y A yB
nD0
b gb g b gb g
70 70 nC n D n AnB n A0 70 10
4-36
4.87 n A0
170.6 mol methanol fed
4.46 (cont’d) Product gas n A nB nC
U| |V || W
yA 170.6 70 100.6 mol yB 10 mol yC 70 mol yD 70 mol
nD ntotal
4.47
0.401 mol CH3OH mol 0.040 mol CH3COOH mol 0.279 mol CH3COOCH 3 mol 0.279 mol H 2 O mol
250.6 mol
d.
Cost of reactants, selling price for product, market for product, rate of reaction, need for heating or cooling, and many other items.
a.
o CO 2 H 2 CO H 2 O m (A)
(B)
(C)
(D)
b g n b mol H Og n b mol CO g n b mol H g n b mol Ig
. mol 100 0.20 mol CO / mol . mol CO 2 / mol 010 0.40 mol H 2 O / mol 0.30 mol I / mol
n A mol CO 2
B
2
C
2
D I
6 unknowns ( n A , nB , nC , nD , n I , [ )
Degree of freedom analysis:
bg
– 4 expressions for ni [ – 1 balance on I – 1 equilibrium relationship 0 DF b.
Since two moles are prodcued for every two moles that react, ntotal out ntotal in 100 . mol
b g b g
b g
n A 0.20 [ nB 0.40 [ nC 010 . [ nD [ n I 0.30 ntot
(1) (2) (3) (4) (5)
100 . mol
At equilibrium: yD c.
nD
[
yC y D y A yB
b
. [ gb[ g b010 b0.20 [ gb0.40 [ g / molg
nC nD n A nB
0110 . mol H 2
The reaction has not reached equilibrium yet.
4-37
0.0247 exp
FG 4020 IJ [ H 1123 K
0110 . mol
4.47
(cont’d) d.
T (K) 1223 1123 1023 923 823 723 623 673 698 688
x (CO) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
x (H2O) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
1123 1123 1123 1123
0.2 0.4 0.3 0.5
0.4 0.2 0.3 0.4
x (CO2)
Keq Keq (Goal Seek) Extent of Reaction 0.6610 0.6610 0.2242 0.8858 0.8856 0.2424 1.2569 1.2569 0.2643 1.9240 1.9242 0.2905 3.2662 3.2661 0.3219 6.4187 6.4188 0.3585 15.6692 15.6692 0.3992 9.7017 9.7011 0.3785 7.8331 7.8331 0.3684 8.5171 8.5177 0.3724
0 0 0 0 0 0 0 0 0 0 0.1 0.1 0 0
0.8858 0.8858 0.8858 0.8858
0.8863 0.8857 0.8856 0.8867
0.1101 0.1100 0.1454 0.2156
y (H2) 0.224 0.242 0.264 0.291 0.322 0.358 0.399 0.378 0.368 0.372 0.110 0.110 0.145 0.216
The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of carbon monoxide and water also maximizes the extent of reaction. 4.48
a.
A 2B o C
b
ln K e1 / K e 2 1 T1 1 T2
E
4
11458
1 373 1 573
ln K e1 11458 T1
ln A0
b.
b g g lnd10.5 / 2.316 u 10 i
ln A0 E T K
ln K e
ln 10.5 11458 373 28.37 A0
4.79 u 1013
b gh
c
Ke
4.79 u 10 13 exp 11458 T K atm 2 Ke (450K) 0.0548 atm1
nA nB nC
n A0 [ nB 0 2[ nC 0 [
nT
nT 0
U| |V | 2[ |W
b b b
gb gb gb
g
y A n A0 [ nT 0 2[ y B nB 0 2[ nT 0 2[ yC nC 0 [ nT 0 2[ nT 0 n A0 nB 0 nC 0
b
g
g
g
At equilibrium,
yC 1 y A y B2 P2 c.
bn bn
C0 A0
gb [ gb n
g 2[ g
[ e nT 0 2[ e e
B0
2 2
e
bg
1 P2
bg
Ke T (substitute for K T from Part a.) e
Basis: 1 mol A (CO) n A0
1 nB 0
b
[ e 2 2[ e
1 nC 0
g
2
b1 [ gb1 2[ g e
e
2
1 4 atm 2
0 nT 0
2, P
b g
K e 423
2 atm , T
423K
0.278 atm -2 [ 2e [ e 01317 .
4-38
0
4.47 (cont’d) (For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.)
[e
, 0.844 0156 .
Reject the second solution, since it leads to a negative nB .
. g c2 2b0156 . gh y 0.500 b1 0156 y . gh c2 2b0156 . gh y 0.408 c1 2b0156 y . g c2 2b0156 . gh y 0.092 b0 0156 n n [ Fractional Conversion of CO b Ag n n yA
A
B
B
C
C
A0
A
A0
d.
0156 . mol A reacted / mol A feed
A0
Use the equations from part b. i) ii) iii) iv)
Fractional conversion decreases with increasing fraction of CO. Fractional conversion decreases with increasing fraction of CH3OH. Fractional conversion decreases with increasing temperature. Fractional conversion increases with increasing pressure.
e. *
1
2
REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI, FN, FDN, NT, CON, YA, YB, YC INTEGER NIT, INMAX TAU = 0.0001 INMAX = 10 A = 4.79E–13 E = 11458. READ (5, *) YA0, YB0, YC0, T, P KE = A * EXP(E/T) P2KE = P*P*KE C0 = YC0 – P2KE * YA0 * YB0 * YB0 C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0) C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0)) C3 = 4. * (1. + P2KE) EK = 0.0 (Assume an initial value [ e 0. 0 ) NIT = 0 FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 + EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX) GOTO 4 IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2 EK = EKPI GOTO 1 NT = 1. – 2. * EKPI YA = (YA0 – EKPI)/NT YB = (YB0 – 2. + EKPI)/NT YC = (YC0 + EKPI)/NT
4-39
4.48 (cont’d) CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP WRITE (6, 5) INMAX, EKPI FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X)) FORMAT ('DID NOT CONVERGE IN', I3, 'ITERATIONS',/, * 'CURRENT VALUE = ', F6.3) END
4 3
$DATA 0.5 RESULTS YA 0.500
0.5
YB 0.408
0.0
423.
YC 0.092
2.
CON 0.156
Note: This will only find one root — there are two others that can only be found by choosing different initial values of [ a 4.49
a.
CH 4 O 2 o HCHO H 2O
(1)
CH 4 2O 2 o CO 2 2H 2O
(2)
100 mol / s
b g n b mol O / sg n b mol HCHO / sg n b mol H O / sg n b mol CO g n1 mol CH 4 / s
0.50 mol CH 4 / mol 0.50 mol O 2 / mol
2
2
3
4
5
2
2
7 unknowns ( n1 , n 2 , n3 , n 4 , n5 , [ 1 , [ 2 ) – 5 equations for n [ , [ i
e
1
2
j
2 DF b.
n1 n2
1
n4
[ 1 [ 2[
n5
[ 2
n3
c.
50 [ 1 [ 2 50 [ 2[
1
(1) (2)
2
(3) (4)
2
(5)
Fractional conversion: Fractional yield:
n3 50
b50 n g 1
50 0.855 n3
0.900 n1
5.00 mol CH / s 4
42.75 mol HCHO / s
4-40
4.49 (cont’d)
U| || V| || W
y CH 0.0500 mol CH 4 / mol Equation 3 [ 1 42.75 4 yO 0.0275 mol O 2 / mol Equation 1 [ 2 2.25 2 Equation 2 n 2 2.75 y HCHO 0.4275 mol HCHO / mol y H O 0.4725 mol H 2 O / mol Equation 4 n 4 47.25 2 y CO 0.0225 mol CO 2 / mol Equation 5 n5 2.25 2 Selectivity:
42.75 mol HCHO / s 2.25 mol CO 2 / s
19.0 mol HCHO / mol CO 2
4-41
4.50
a.
Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely.
b.
C2H6 + Cl2 o C2H5Cl + HCl, C2H5Cl + Cl2 o C2H4Cl2 + HCl Basis: 100 mol C2H5Cl produced
c.
n1 (mol C2H6)
100 mol C2H5OH
n2 (mol C2H6)
n3 (mol C2H6) n4 (mol HCl) n5 (mol C2H5Cl2)
5 unknowns –3 atomic balances 2 D.F.
Selectivity: 100 mol C 2 H 5 Cl 14n5 (mol C 2 H 4 Cl 2 ) n5
U| n 714.3 mol C H in g n V 2n 2b100g 2n 2b7.143g|W n 114.3 mol C H out 6b714.3g 5b100g 6b114.3g n 4b7.143g n 607.1 mol HCl 2n 100 607.1 2b7.143g n 114.3 mol Cl b
. n1 13% conversion: 1 015 C balance:
H balance: Cl balance:
7.143 mol C 2 H 4 Cl 2
3
1
3
1
2
6
3
2
6
4
2
4
2
Feed Ratio: 114.3 mol Cl 2 / 714.3 mol C2 H 6
2
016 . mol Cl 2 / mol C2 H 6
Maximum possible amount of C2H5Cl: 114.3 mol Cl 2 1 mol C 2 H 5 Cl n max 114.3 mol C 2 H 5 Cl 1 mol Cl 2 Fractional yield of C2H5Cl:
4.51
nC2 H5Cl n max
100 mol 114.3 mol
0.875
d.
Some of the C2H4Cl2 is further chlorinated in an undesired side reaction: C2H5Cl2 + Cl2 o C2H4Cl3 + HCl
a.
C2H4 + H2O o C2H5OH, 2 C2H5OH o (C2H5)2O + H2O Basis: 100 mol effluent gas 100 mol 0.433 mol C 2 H 4 / mol
n (mol C H ) 1 2 4 n [mol H O (v)] 2 2 n 3 (mol I)
3 unknowns
0.025 mol C 2 H 5 OH / mol
-2 independent atomic balances
0.0014 mol (C H ) O / mol 2 5 2 0.093 mol I / mol
-1 I balance 0 D. F.
0.4476 mol H O (v) / mol 2
(1) C balance: 2n1
b
100 2 0.433 2 0.025 4 0.0014
(2) H balance: 4n1 2n2 (3) O balance: n2
b
b
g
100 4 0.433 6 0.025 10 0.0014 2 0.4476
g
g
100 0.025 0.0014 0.4476
Note; Eq. (1) 2 Eq. (3) 2 (4) I balance: n3 = 9.3
Eq. (2) 2 independent atomic balances
4-42
4.51 (cont'd) b. (1) n1 46.08 mol C 2 H 6 (3) n2 47.4 mol H 2 O Reactor feed contains 44.8% C 2 H 6 , 46.1% H 2 O, 9.1% I (4) n3 9.3 mol I
U| V| W
46.08 43.3 u 100% 6.0% 46.08 If all C2H4 were converted and the second reaction did not occur, nC2 H5OH
% conversion of C2H4:
d
d
Fractional Yield of C2H5OH: nC2 H5OH / nC2 H5OH Selectivity of C2H5OH to (C2H5)2O: 2.5 mol C 2 H 5 OH 0.14 mol (C 2 H 5 ) 2 O c. 4.52
i
max
b2.5 / 46.08g
i
max
46.08 mol
0.054
17.9 mol C 2 H 5OH / mol (C 2 H 5 ) 2 O
Keep conversion low to prevent C2H5OH from being in reactor long enough to form significant amounts of (C2H5)2O. Separate and recycle unreacted C2H4.
bg
bg
bg
bg
CaF2 s H 2 SO 4 l o CaSO 4 s 2HF g
1 metric ton acid
1000 kg acid
0.60 kg HF
1 metric ton acid
1 kg acid
600 kg HF
Basis: 100 kg Ore dissolved (not fed) 100 kg Ore d issolved 0.96 kg CaF 2 /kg 0.04 kg SiO 2/ kg nA (kg 93% H2 SO4 ) 0.93 H2 SO4 kg/ kg 0.07 H2 O kg/ kg
n1 n2 n3 n4
(kg CaSO4) (kg HF) (kg H 4SiF6 ) (kg H 2SO 4)
n5 (kg H2 O)
Atomic balance - Si:
b g
0.04 100 kg SiO 2
28.1 kg Si
n3 (kg H 4 SiF6 )
60.1 kg SiO 2
28.1 kg Si n3 146.1 kg H 4 SiF6
Atomic balance - F:
b g
0.96 100 kg CaF2
38.0 kg F
n2 (kg HF)
78.1 kg CaF2 114.0 kg F 9.72 kg H 4 SiF6 n2 146.1 kg H 4 SiF6 600 kg HF 100 kg ore diss. 41.2 kg HF
1 kg ore feed 0.95 kg ore diss.
4-43
19.0 kg F 20.0 kg HF 412 . kg HF 1533 kg ore
9.72 kg H 4 SiF6
4.53
a.
C 6 H 6 Cl 2 o C 6 H 5 Cl HCl C 6 H 5 Cl Cl 2 o C 6 H 4 Cl 2 HCl C 6 H 4 Cl 2 Cl 2 o C 6 H 3 Cl 3 HCl Convert output wt% to mol%: Basis 100 g output species C6H 6 C 6 H 5 Cl C 6 H 4 Cl 2 C 6 H 3 Cl 3
g 65.0 32.0 2.5 0.5
Mol. Wt. 78.11 112.56 147.01 181.46
mol 0.832 0.284 0.017 0.003
mol % 73.2 25.0 1.5 0.3
total 1.136 Basis: 100 mol output n 4 (mol HCl(g )) n 3 (mol I)
n1 (mol C6 H6 )
65.0 mo l C6 H6 32.0 mo l C6 H 5 Cl 2.5 mo l C 6 H 4 Cl 2 0.5 mo l C6 H 3 Cl 3
n2 (mol Cl 2) n3 (mol I) b.
4 unknowns -3 atomic balances -1 wt% Cl 2 in feed 0 D.F.
b
100 mol C H g H balance: 6b100g 6b65.0g 5b32.0g 4b2.5g 3b0.5g n n 38.5 mol HCl Cl balance: 2n 38.5 32.0 2b2.5g 3b0.5g n 38.5 mol Cl Theoretical C H 38.5 mol Cl b1 mol C H 1 mol Cl g 38.5 mol C H Excess C H : b100 38.5g 38.5 u 100% 160% excess C H Fractional Conversion: b100 65.0g 100 0.350 mol C H react / mol fed
6 65.0 32.0 2.5 0.5 n1
C balance: 6n1
6
4
2
6
6
6
4
2
6
2
6
2
6
2
6
6
6
6
6
6
6
Yield: (32.0 mol C 6 H 5 Cl) (38.5 mol C 6 H 5 Cl maximum) = 0.831
U| V| W
38.5 mol Cl 2 70.91 g Cl 2 1 g gas 2091 g gas g gas 0.357 mole Cl 2 0.98 g Cl 2 g liquid Liquid feed: 100 mol C 6 H 6 78.11 g C 6 H 6 / mol C 6 H 6 7811 g liquid Gas feed:
b
gb
g
c.
Low conversion low residence time in reactor lower chance of 2nd and 3rd reactions occurring. Large excess of C 6 H 6 Cl 2 much more likely to encounter C 6 H 6 than substituted C 6 H 6 higher selectivity.
d.
Dissolve in water to produce hydrochloric acid.
e.
Reagent grade costs much more. Use only if impurities in technical grade mixture affect the reaction rate or desired product yield.
4-44
4.54
a.
2CO 2 2CO O 2
2A 2B C
O 2 N 2 2NO
C D 2E
bn bn bn bn bn
n A0 2[ e1 yA n B 0 2[ e 2 yB nC 0 [ e1 [ e 2 y C nD0 [ e2 yD n E 0 2[ e 2 yE
nA nB nC nD nE
bn
ntotal = nT 0 [ e1
gb gb
g g
2[ e1 nT 0 [ e1 n T 0 [ e1 B 0 2[ e1 nT 0 [ e1 C 0 [ e1 [ e 2 n T 0 [ e1 D 0 1[ e 2 nT 0 [ e1 E 0 2[ e 2 A0
gb gb
gb
g g
g
n A0 n B 0 nC 0 n D 0 n E 0
T0
g
Equilibrium at 3000K and 1 atm
bn
g bn [ [ g 01071 . bn 2[ g bn [ g bn 2[ g bn [ [ gbn [ g 0.01493
y B2 y C y 2A
2[ e1
2
A0
e1
C0 2
e1
T0
e2
e1
2
y E2 yC y D
B0
A0
E f1
B0
b
e1
01071 . n A0 2[ e1 f2
b
e2
e2
g bn
D0
2
T0
e2
g b gbn
g bn [ [ g g bn 2[ g 0
[ e1 n B 0 2[ e1
0.01493 nC 0 [ e1 [ e 2
D0
[ e2
2
C0
e1
e2
2
E0
e2
U| V| b g W b g
0 Defines functions f 1 [ 1 , [ 2 and f2 [1, [ 2
b.
Given all nio’s, solve above equations for [e1 and [e2 nA, nB, nC, nD, nE yA, yB, yC, yD, yE
c.
nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 [e1 =0.0593, [e2 = 0.0208 yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d.
a11 d 1 a12 d 2 f 1 a12 f 2 a 22 f 1 d1 a11 a 22 a12 a 21
b[ g
e1 new
[ e1 d 1
a11 d 1 a12 d 2 f 1 a12 f 2 a 22 f 1 d1 a11 a 22 a12 a 21
b[ g
e1 new
[ e1 d 1
a 21 d 1 a 22 d 2 f 2 a 21 f 1 a11 f 2 d2 a11 a 22 a12 a 21
b[ g
e 2 new
[ e1 d 2
a 21 d 1 a 22 d 2 f 2 a 21 f 1 a11 f 2 d2 a11 a 22 a12 a 21
b[ g
e 2 new
(Solution given following program listing.)
4-45
[ e1 d 2
4.54 (cont’d) . 1 30 2
3
100 4 120
IMPLICIT REAL * 4(N) WRITE (6, 1) FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///) READ (5, *) NA0, NB0, NC0, ND0, NE0 IF (NA0.LT.0.0)STOP WRITE (6, 2) NA0, NB0, NC0, ND0, NE0 FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 – X1 – X1 NB = NB0 + X1 + X1 NC = NC0 + X1 – X2 ND = ND0 – X2 NE = NE0 + X2 + X2 NAS = NA ** 2 NBS = NB ** 2 NES = NE ** 2 NT = NT0 + X1 F1 = 0.1071 * NAS * NT – NBS * NC F2 = 0.01493 * NC * ND – NES A11 = –0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS A12 = NBS A21 = 0.01493 * ND A22 = –0.01493 * (NC + ND) – 4.0 * NE DEN = A11 * A22 – A12 * A21 D1 = (A12 * F2 – A22 * F1)/DEN D2 = (A21 * F1 – A11 * F2)/DEN X1C = X1 + D1 X2C = X2 + D2 WRITE (6, 3) J, X1, X2, X1C, X2C FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5) IF (ABS(D1/X1C).LT.1.0E–5.AND.ABS(D2/X2C).LT.1.0E–5) GOTO 120 X1 = X1C X2 = X2C CONTINUE WRITE (6, 4) NMAX FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/) STOP YA = NA/NT YB = NB/NT YC = NC/NT YD = ND/NT YE = NE/NT WRITE (6, 5) YA, YB, YC, YD, YE 5 FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///) GOTO 30 END $DATA 0.3333 0.00 0.3333 0.3333 0.0 0.50 0.0 0.0 0.50 0.0 0.20 0.20 0.20 0.20 0.20
4-46
4.54 (cont’d) SOLUTION TO PROBLEM 4.54 NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33 ITER = 1 X1A, X2A = 0.10000 0.10000 ITER = 2 X1A, X2A = 0.06418 0.05181 ITER = 3 X1A, X2A = 0.05969 0.02486 ITER = 4 X1A, X2A = 0.05437 0.02213 ITER = 5 X1A, X2A = 0.05931 0.02086 ITER = 6 X1A, X2A = 0.05930 0.02083
YA, YB, YC, YD, YE =
0.33 0.00 X1C, X2C = 0.06418 X1C, X2C = 0.05969 X1C, X2C = 0.05937 X1C, X2C = 0.05931 X1C, X2C = 0.05930 X1C, X2C = 0.05930
2.0270E 01 2.9501E 01
NA0, NB0, NC0, ND0, NE0 = 0.20 0.20 ITER = 1 X1A, X2A = 0.10000 p ITER = 7 X1A, X2A = –0.02244
0.20
1.1197 E 01 3.9319 E 02
0.20 0.10000
3.5100E 01
0.20 X1C, X2C = 0.00012
–0.08339 X1C, X2C = –0.02244
YA, YB, YC, YD, YE = 2.5051E 01 15868 . E 01 2.8989 E 01 3.3991E 02 4.55
0.05181 0.02986 0.02213 0.02086 0.02083 0.02083
0.00037 –0.08339
2.6693E 01
(B)
a.
mB0 (kg A / h) 1 kg B/ kg A fed to reactor ( A) m A 0 (kg A / h) x RA (kg R / kg A)
( P) m A 0 (kg A / h) x RA (kg R / kg A)
m3 (kg A / h)
R o 5
x R 3 (Kg R / kg)
99% conv.
m P (kg P / h) 0.0075 kg R / kg P
f mA0 (kg A / h) x RA (kg R / kg A)
Reactor
Splitting point 4 unknowns (mA0, mB0, f, xRA) -1 independent balance (mass) 3 D.F.
4 unknowns (xRA, mB0, m3, xRA) -1 balance (mass) -1 conversion 2 D.F
Mixing point
Total process
5 unknowns (f, m3, mP, xRA, xR3)
7 unknowns (mA0, xRA, f, mB0, m3, xR3, mP)
-2 balances (mass, R)
-5 relations
3 D.F.
2 D.F.
4-47
4.55 (cont’d) b. Mass balance on splitting point: mA0 = mB0 + f mA0
(1)
Mass balance on reactor: 2 mB0 = m3
(2)
99% conversion of R: xR3 m3 = 0.01 xRA mB0
(3)
Mass balance on mixing point: m3 + f mA0 = mP R balance on mixing point: xR3 m3 + xRA f mA0 = 0.0075 mP
(4) (5)
Given xRA and mP, solve simultaneously for mA0, mB0, f, m3, xR3 c.
mA0 = 2778 kg A/h mB0 = 2072 kg B/h fA = 0.255 kg bypass/kg fresh feed mP 4850 4850 4850 4850 4850 4850 4850 4850 4850
xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
mA0 3327 3022 2870 2778 2717 2674 2641 2616 2596
mB0 1523 1828 1980 2072 2133 2176 2209 2234 2254
f 0.54 0.40 0.31 0.25 0.21 0.19 0.16 0.15 0.13
mP 2450 2450 2450 2450 2450 2450 2450 2450 2450
xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
mA0 1663 1511 1435 1389 1359 1337 1321 1308 1298
mB0 762 914 990 1036 1066 1088 1104 1117 1127
f 0.54 0.40 0.31 0.25 0.22 0.19 0.16 0.15 0.13
f v s . x RA f (kg bypass/kg fresh feed)
d.
0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00
0.02
0.04
0.06 x R A (k g R / k g A )
4-48
0.08
0.10
0.12
4.56
a.
900 kg HCHO 1 kmol HCHO h 30.03 kg HCHO
30.0 kmol HCHO / h
n (kmol CH OH / h) 1 3
30.0 kmol HCHO / h n 2 (kmol H 2 / h) n 3 (kmol CH 3 OH / h)
% conversion:
30.0 n1
0.60 n1
50.0 kmol CH 3 OH / h
b.
n (kmol CH OH / h) 1 3
30.0 kmol HCHO / h
30.0 kmol HCHO / h
n 2 (kmol H 2 / h)
n 2 (kmol H 2 / h)
n 3 (kmol CH 3 OH / h) n (kmol CH OH / h) 3 3
Overall C balance: n1 (1) = 30.0 (1) n1 = 30.0 kmol CH3OH/h (fresh feed) Single pass conversion:
30.0 n1 n3
0.60 n3
20.0 kmol CH 3OH / h
n1 + n3 = 50.0 kmol CH3OH fed to reactor/h
4.57
c.
Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and (2) lower the quantities of unreacted methanol and so will decrease the cost of the separation. The plot would resemble a concave upward parabola with a minimum around xsp = 60%.
a.
Convert effluent composition to molar basis. Basis: 100 g effluent: 10.6 g H 2 1 mol H 2 2.02 g H 2 64.0 g CO 1 mol CO 28.01 g CO
5.25 mol H 2 2.28 mol CO
25.4 g CH 3 OH 1 mol CH 3 OH 32.04 g CH 3 OH 0.793 mol CH 3 OH
4-49
H : 0.631 mol H / mol 2 2 CO: 0.274 mol CO / mol CH OH: 0.0953 mol CH OH / mol 3 3
4.57 (cont’d) n4 (mol / min) 0.004 mol CH 3OH(v)/ mol x (mol CO/ mol) (0.896 - x) (mol H 2 / mol) Cond.
Reactor
350 mol/ min n1 (mol CO/ min) n 2 (mol H 2 / min)
0.631 mol CH 3OH(v)/ mol
n 3 (mol CH 3OH(l) / min)
0.274 mol CO/ mol CO H 2 o CH 3OH 0.0953 mol H / mol 2
Condenser
Overall process
3 unknowns (n3, n4, x)
2 unknowns (n1, n2)
-3 balances 0 degrees of freedom
-2 independent atomic balances 0 degrees of freedom
Balances around condenser
U| V| W
n 32.1 mol CH 3 OH(l) / min CO: 350 0.274 n x 3 4 n H : 350 0.631 n ( 0.996 x ) 318.7 mol recycle / min 2 4 4 CH OH: 350 0.0953 n 0.004 n x .301 molCO / mol 3 3 4 Overall balances
U|V |W
n C: n = n 1 3 1 n H: 2n = 4n 2 2 2
32.08 mol / min CO in feed 64.16 mol / min H 2 in feed
Single pass conversion of CO:
32.08 318.72 0.3009 350 0.274 u 100% 32.08 318.72 0.3009
–
32.08 0 u 100% 100% 32.08 Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.)
–
Impurities in feed. (Re-analyze feed.)
–
Leak in methanol outlet pipe before flowmeter. (Check for it.)
Overall conversion of CO: b.
25.07%
4-50
4.58
a.
Basis: 100 kmol reactor feed/hr n3 (kmol CH4 /h) 100 kmol /h
Reactor
n1 (kmol CH4 /h) 80 kmol CH4 /h n2 (kmol Cl2 /h) 20 kmol Cl2 /h
n3 (kmol CH4 /h) n4 (kmol HCl /h) 5n5 (kmol CH3Cl /h) n5 (kmol CH2Cl 2 /h)
Cond.
Solvent
Absorb n3 (kmol CH4 /h) n4 (kmol HCl/h)
n4 (kmol HCl/h)
5n5 (kmol CH3Cl /h) Still 5n5 (kmol CH3Cl /h) n5 (kmol CH2Cl 2 /h)
n5 (kmol CH2Cl 2 /h)
Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D.F. Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D.F. Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D.F. Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D.F. Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D.F. Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D.F. Atomic balances around reactor: ½ 1) C balance : 80 n 3 5n 5 n 5 ° 2) H balance : 320 4n 3 n 4 15n 5 2n 5 ¾ Solve for n 3 , n 4 , n 5 ° 3) Cl balance : 40 n 4 5n 5 2n 5 ¿ CH4 balance around mixing point: n1 = (80 – n3) Cl2 balance: n2 = 20 b.
Solve for n1
For a basis of 100 kmol/h into reactor n1 = 17.1 kmol CH4/h
n4 = 20.0 kmol HCl/h
n2 = 20.0 kmol Cl2/h
5n5 = 14.5 kmol CH3Cl/h
n3 = 62.9 kmol CH4/h c.
(1000 kg CH3Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3Cl/h Scale factor =
19.81 kmol CH 3 Cl/h 14.5 kmol CH 3Cl/h
1.366
n tot 50.6 kmol/h n (17.1)(1.366) 23.3 kmol CH 4 /h ½ Fresh feed: 1 ¾ n 2 (20.0)(1.366) 27.3 kmol Cl 2 /h ¿ 46.0 mol% CH 4 , 54.0 mole% Cl 2 Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h
4-51
4.59
a.
Basis: 100 mol fed to reactor/h 25 mol O2/h, 75 mol C2H4/h n1 (mol C 2H 4 //h) n2 (mol O 2 /h)
Seperator
reactor nC2H4 ( mol C 2H 4 /h) nO2 (mol O 2 /h)
75 mol C 2H 4 //h 25 mol O2 /h
n1 (mol C 2H 4 //h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h) n4 (mol CO 2 /h) n5 (mol H 2O /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h) n5 (mol H 2O /h)
Reactor 5 unknowns (n1 - n5) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n1- n5 2. Solve balances around mixing point to find nO2, nC2H4 (1) % Conversion n1 = .800 * 75 (2) % yield: (.200)(75) mol C 2 H 4 u
90 mol C 2 H 4 O 100 mol C 2 H 4
n 3 (production rate of C 2 H 4 O)
(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4 (4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5 (5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5 (6) O2 balance (mix pt): nO2 = 25 – n2 (7) C2H4 balance (mix pt): nC2H4 = 75 – n1 Overall conversion of C2H4: 100% b.
c.
n1 = 60.0 mol C2H4/h
n5 = 3.00 mol H2O/h
n2 = 13.75 mol O2 /h
nO2 = 11.25 mol O2/h
n3 = 13.5 mol C2H4O/h n4 = 3.00 mol CO2/h
nC2H4 = 15.0 mol C2H4/h 100% conversion of C2H4
Scale factor =
2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O h h 44.05 lbm C 2 H 4 O 13.5 mol C 2 H 4 O
nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h
4-52
3.363
lb mol / h mol / h
4.60
a.
Basis: 100 mol feed/h 100 mol/h
n1 (mol /h)
32 mol CO/h 64 mol H 2 / h 4 mol N 2 / h
.13 mol N 2 /mol
reactor
n3 (mol CH 3 OH / h)
cond.
500 mol / h x1 (mol N 2 /mol) x2 (mol CO / mol) 1-x1-x2 (mol H 2 / h)
n3 (mol / h) x1 (mol N 2 /mol) x2 (mol CO / mol) 1-x1-x2 (mol H 2 / h) Purge
Mixing point balances: total: (100) + 500 = n1 n1 = 600 mol/h N2: 4 + x1 * 500 = .13 * 600 x1 = 0.148 mol N2/mol Overall system balances: N2: 4 = .148 * n3 n3 = 27 mol/h Atomic C: 32 = n2 + x2*27 n2 = 24.3 mol CH3OH/h Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x2)*27 x2 = 0.284 mol CO/mol Overall CO conversion: 100*[32-0.284(27)]/32 = 76% Single pass CO conversion: 24.3/ (32+.284*500) = 14% b.
Recycle: To recover unconsumed CO and H2 and get a better overall conversion. Purge: to prevent buildup of N2.
4.61
a.
2N2 + 3H2 -> NH3 (1-yp) (1-fsp) n1 (mol N2) (1-yp) (1-fsp) 3n1 (mol H2) (1-yp) n2 (mol I)
1 mol (1-XI0)/4 (mol N2 / mol) 3/4 (1-XI0) (mol H2 / mol) XI0 (mol I / mol)
nr (mol) n1 (mol N2) 3n1 (mol H2) n2 (mol I)
Reactor
4-53
(1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I)
nr (mol) (1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I) 2 fsp n1 (mol NH3)
yp (1-fsp) n1 (mol N2) yp (1-fsp) 3n1 (mol H2) yp n2 (mol I)
Condenser
np (mol) 2 fsp n1 (mol NH3)
4.61 (cont’d) At mixing point: N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1 I: XI0 + (1-yp) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fspn1 Overall N2 conversion:
b.
(1 X I0 ) / 4 y p (1 f sp )n 1 (1 X I0 ) / 4
u 100%
XI0 = 0.01 fsp = 0.20 yp = 0.10 n1 = 0.884 mol N2
nr = 3.636 mol fed
n2 = 0.1 mol I
np = 0.3536 mol NH3 produced N2 conversion = 71.4%
c.
Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system.
d.
Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp results in decreasing nr, increasing np, and increasing fov. Increasing yp results in decreasing nr, decreasing np, and decreasing fov. Optimal values would result in a low value of nr and fsp, and a high value of np, this would give the highest profit. XI0 0.01 0.05 0.10 0.01 0.01 0.01 0.10 0.10 0.10
fsp 0.20 0.20 0.20 0.30 0.40 0.50 0.20 0.20 0.20
yp 0.10 0.10 0.10 0.10 0.10 0.10 0.20 0.30 0.40
nr 3.636 3.893 4.214 2.776 2.252 1.900 3.000 2.379 1.981
4-54
np 0.354 0.339 0.321 0.401 0.430 0.450 0.250 0.205 0.173
fov 71.4% 71.4% 71.4% 81.1% 87.0% 90.9% 55.6% 45.5% 38.5%
4.62
a.
i - C 4 H 10 C 4 H 8
Basis: 1-hour operation
C 8 H 18
n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10) n 1 (C 8 H18) m 4 (91% H 2 SO4 )
D
P decanter
E
Units of n : kmol Units of m: kg
n 1 (C 8 H18) n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10)
still
n 5 (n-C 4 H10) n 6 (i-C 4 H 10) n 7 (C 8 H18) m 8 (91% H 2 SO 4 )
reactor
C B
F
n 1 (C 8 H18) n 2 (n-C 4 H10 )
m 4 (kg 91% H 2 SO4 )
40000 kg A n 0 kmol 0.25 i-C4 H10 0.50 n-C4 H10 0.25 C4 H 8
n 3 (i-C 4 H 10)
Calculate moles of feed M
. g b0.25gb5610 . g b0.75gb5812
0.25 M L C4 H10 0.50 M n C4 H10 0.25 M C4 H 8 57.6 kg kmol
n0
b40000 kggb1 kmol 57.6 kgg
Overall n - C 4 H 10 balance: n2
694 kmol
b0.50gb694g
347 kmol n - C 4 H 10 in product
C 8 H 18 balance: n1
b0.25gb694g kmol C H 4
8
react 1 mol C 8 H 18 1 mol C 4 H 8
. kmol C 8 H 8 in product 1735
b
At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 n mol i - C 4 H 10
b
Note: n mol C 4 H 8
g
694g g b 5gb 0 .25 gb A
moles C 4 H 8 at A=173.5
173.5 at (A), (B) and (C) and in feed
b gb g
i - C 4 H 10 balance around first mixing point 0.25 694 n3 n3
867.5
694 kmol i - C 4 H 10 recycled from still
At C, 200 mol i - C4 H10 mol C4 H 8
b
n mol i - C4 H10
g b200gb1735. g C
4-55
34,700 kmol i - C4 H10
867.5 kmol i -C 4 H 10 at
bA g and b Bg
4.62 (cont’d) i - C 4 H 10 balance around second mixing point 867.5 n6 n6
34,700
33,800 kmol C 4 H 10 in recycle E
Recycle E: Since Streams (D) and (E) have the same composition,
b g bmoles n - C H g bmoles C H g n bmoles C H g n
n5 moles n - C 4 H 10 n2 n7 n1
4
b g bmoles i - C H g
n6 moles i - C 4 H 10
E
n3
10 D
8
18 E
6
8
18 D
3
4
n7
E
n5
16,900 kmol n - C 4 H 10
10 D
8460 kmol C 4 H 18
Hydrocarbons entering reactor:
kg I b347 16900gbkmol n - C H g FGH 58.12 kmol JK kg I F . kg IJ 1735. kmol C H FG 5610 b867.5 33800gb kmol i - C H g G 5812 . H kmol K H kmol JK kg I F 8460 kmol C H G 114.22 J 4.00 u 10 kg . H kmol K H SO solution entering reactor 4.00 u 10 kg HC 2 kg H SO baq g band leaving reactor g 1 kg HC 8.00 u 10 kg H SO baq g m b H SO in recycleg n b n - C H in recycleg 8.00 u 10 b H SO leaving reactor g n n b n - C H leaving reactor g m 7.84 u 10 kg H SO baq g in recycle E 4
10
4
8
2
10
4
6
18
6
4
6
8
2
6
4
2
m4
2
4
6
8
8
2
2
2
4
4
5
4
10
5
4
10
4
H 2 SO 4 entering reactor H 2 SO 4 in E
b g
16 . u 10 5 kg H 2 SO 4 aq recycled from decanter
b g 1480 kmol H SO in recycle d ib g d16. u 10 ib0.09gkg H O b1 kmol 18.02 kgg 799 kmol H O from decanter
16 . u 10 5 0.91 kg H 2 SO 4 1 kmol 98.08 kg 5
2
2
4
2
Summary: (Change amounts to flow rates) Product: 173.5 kmol C 8 H 18 h , 347 kmol n - C 4 H 10 h Recycle from still: 694 kmol i - C 4 H 10 h Acid recycle: 1480 kmol H 2 SO 4 h , 799 kmol H 2 O h Recycle E: 16,900 kmol n - C 4 H 10 h , 33,800 kmol L - C 4 H 10 h , 8460 kmol C 8 H 18 h, 7.84 u 10 6 kg h 91% H 2 SO 4 72,740 kmol H 2 SO 4 h , 39,150 kmol H 2 O h
4-56
4.63
a.
A balance on ith tank (input = output + consumption) Ai kC Ai C Bi mol liter min V L v L min C A , i 1 mol L vC
b
g
b
g
b
E y v, note V / v
gbg
W
C Ai kW C Ai C Bi
C A , i 1 B balance. By analogy, C B , i 1
C Bi kW C Ai C Bi
Subtract equations C Bi C Ai
C B , i 1 C A, i 1
A from balances on bi 1g tank
C B , i 2 C A, i 2 C B 0 C A 0
st
b.
C Bi C Ai
C A , i 1 2 C Ai
C B 0 C A 0 C Bi
b
C Ai C B 0 C A0 . Substitute in A balance from part (a).
g
2 0 C Ai kW C Ai C Ai C B 0 C A0 . Collect terms in C Ai , C 1Ai , C Ai .
b
g
kW C AL 1 kW C B 0 C A0 C A, i 1 2 D C AL E C AL J
0 where D
0 kW , E
b
g
1 kW C B 0 C A0 , J
C A , i 1
E E 2 4DJ
(Only + rather than r: since DJ is negative and the 2D negative solution would yield a negative concentration.)
Solution: C Ai
c. k= v= V= CA0 = CB0 = alpha = beta =
36.2 5000 2000 0.0567 0.1000 14.48 1.6270
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14
gamma -5.670E-02 -2.791E-02 -1.512E-02 -8.631E-03 -5.076E-03 -3.038E-03 -1.837E-03 -1.118E-03 -6.830E-04 -4.182E-04 -2.565E-04 -1.574E-04 -9.667E-05 -5.939E-05
CA(N) 2.791E-02 1.512E-02 8.631E-03 5.076E-03 3.038E-03 1.837E-03 1.118E-03 6.830E-04 4.182E-04 2.565E-04 1.574E-04 9.667E-05 5.939E-05 3.649E-05
xA(N) 0.5077 0.7333 0.8478 0.9105 0.9464 0.9676 0.9803 0.9880 0.9926 0.9955 0.9972 0.9983 0.9990 0.9994
(xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (xmin = 0.90, N = 4), (xmin = 0.95, N = 6), (xmin = 0.99, N = 9), (xmin = 0.999, N = 13). As xmin o 1, the required number of tanks and hence the process cost becomes infinite. d.
(i) k increases N decreases (faster reaction fewer tanks) (ii) v increases N increases (faster throughput less time spent in reactor lower conversion per reactor)
(iii) V increases N decreases (larger reactor more time spent in reactor higher conversion per reactor)
4-57
4.64
a.
Basis: 1000 g gas Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3H8
800
44.09
18.145
77.2%
87.5%
C4H10
150
58.12
2.581
11.0%
12.5%
H2O
50
18.02
2.775
11.8%
Total
1000
23.501
100%
100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ratio: 2.775 / 20.726 = 0.134 mol H 2 O / mol dry gas b.
C3H8 + 5 O2 o 3 CO2 + 4 H2O, C4H10 + 13/2 O2 o 4 CO2 + 5 H2O Theoretical O2: C3H8: C 4 H 10 :
5 kmol O 2 100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8 h 100 kg gas 44.09 kg C 3 H 8 1 kmol C 3 H 8
9.07 kmol O 2 / h
6.5 kmol O 2 100 kg gas 15 kg C 4 H 10 1 kmol C 4 H 10 h 100 kg gas 58.12 kg C 4 H 10 1 kmol C 4 H 10
1.68 kmol O 2 / h
Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h Air feed rate:
10.75 kmol O 2 1 kmol Air 1.3 kmol air fed h .21 kmol O 2 1 kmol air required
66.5 kmol air / h
The answer does not change for incomplete combustion 4.65
5 L C 6 H 14 0.659 kg C 6 H 14 1000 mol C 6 H 14 L C 6 H 14 86 kg C 6 H 14
38.3 mol C 6 H 14
4 L C 7 H 16 0.684 kg C 7 H 16 1000 mol C 7 H 16 L C 7 H 16 100 kg C 7 H 16
27.36 mol C 7 H 16
C6H14 +19/2 O2 o 6 CO2 + 7 H2O
C6H14 +13/2 O2 o 6 CO + 7 H2O
C7H16 + 11 O2 o 7 CO2 + 8 H2O
C7H16 + 15/2 O2 o 7 CO + 8 H2O
Theoretical oxygen: 38.3 mol C 6 H14
9.5 mol O 2 27.36 mol C 7 H 16 11 mol O 2 mol C 6 H 14 mol C 7 H16
O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed Percent excess air:
840 665 u 100% 665
26.3% excess air
4-58
665 mol O 2 required
4.66
CO
1 O 2 o CO 2 2
H2
1 O2 o H2O 2
130 kmol/h 0.500 kmol N2/kmol x (kmol CO/mol) (0.500–x) (kmol H2/kmol) 20% excess air
Note: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can calculate the air feed rate without determining x CO . We include its calculation for illustrative purposes.
b
A plot of x vs. R on log paper is a straight line through the points R1
bR
99.7, x 2
2
@
x
g
0.05 and
10 . .
b ln R ln a
ln x
R
g
10.0, x1
a Rb
38.3 x
0.288
. g 1303 ln a lnb10 . g 1303 . lnb99.7g 6.00 a expb 6.00g 2.49 u 10 3
b
b
g b
ln 10 . 0.05 ln 99.7 10.0
x
. 2.49 u 10 3 R 1303
moles CO mol
Theoretical O : 175 kmol 0.288 kmol CO 0.5 kmol O 2 2 h kmol kmol CO Air fed: 4.67
a.
Theoretical O 2 :
o CO 2 2H 2 O
o 3CO 2 4H 2 O
17% excess air na (kmol air/h) 0.21 O2 0.79 N2
o 4CO 2 5H 2 O
b g
b g
h
0.0060 100 kmol C 3 H 8 h 207.0 kmol O 2 h
2
100 kmol/h 0.944 CH4 0.0340 C2H6 0.0060 C3H8 0.0050 C4H10
o 2CO 2 3H 2 O
0.944 100 kmol CH 4
kmol O 0.5 kmol O 2 2 43.75 kmol H h 2 1.2 kmol air fed kmol air 250 1 kmol air required h
0.212 kmol H 2 kmol
43.75 kmol O required 1 kmol air 2 h 0.21 kmol O
CH 4 2O 2 7 C2 H 6 O2 2 C 3 H 8 5O 2 13 C 4 H 10 O 2 2
175 kmol h
b g
2 kmol O 2 0.0340 100 kmol C 2 H 6 1 kmol CH 4 h
3.5 kmol O 2 1 kmol C 2 H 6
0.0050 100 kmol C 4 H 10 5 kmol O 2 1 kmol C 3 H 3 h
6.5 kmol O 2 1 kmol C 4 H 10
b g
4-59
4.67 (cont’d) Air feed rate: n f
207.0 kmol O 2
1 kmol air
h
0.21 kmol O 2
b
gb
gb
1.17 kmol air fed kmol air req.
g
b.
na
n f 2 x1 35 . x 2 5x 3 6.5x 4 1 Pxs 100 1 0.21
c.
n f
aR f , ( n f
n a
bRa , (n a
xi
kAi
75.0 kmol / h, R f 550 kmol / h, Ra
¦x
k
i
i
¦A
60) n f 25) n a
1 k
i
i
1153 kmol air h
125 . Rf 22.0 Ra
1
¦A
i
i
Ai
xi
¦A
, i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10
i
i
4.68
Run 1 2 3
Pxs 15% 15% 15%
Rf 62 83 108
A1 248.7 305.3 294.2
A2 19.74 14.57 16.61
A3 6.35 2.56 4.78
A4 1.48 0.70 2.11
Run 1 2 3
nf 77.5 103.8 135.0
x1 0.900 0.945 0.926
x2 0.0715 0.0451 0.0523
x3 0.0230 0.0079 0.0150
x4 0.0054 0.0022 0.0066
na 934 1194 1592
d.
Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the flowmeter calibration formulas might not be linear, or the stack gas analysis could be incorrect.
a.
C4H10 + 13/2 O2 o 4 CO2 + 5 H2O Basis:
100 mol C4H10
nCO2 (mol CO2) nH2O (mol H2O) nC4H10 (mol C4H10) nO2 (mol O2) nN2 (mol N2)
Pxs (% excess air) nair (mol air) 0.21 O2 0.79 N2 D.F. analysis 6 unknowns (n, n1, n2, n3, n4, n5) -3 atomic balances (C, H, O) -1 N2 balance -1 % excess air -1 % conversion 0 D.F.
4-60
Ra 42.4 54.3 72.4
4.68 (cont’d) b.
i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2 n air
(650 mol O 2 )(1 mol air / 0.21 mol O 2 )
100% conversion n C4H10 n N2 nCO2 n H2O
0 , nO 2
3095 mol air
0
b0.79gb3095 molg 2445 mol b100 mol C H reactgb4 mol CO mol C H g b100 mol C H reactgb5 mol H O mol C H g 4
4
10
2
10
2
4
4
10
10
U| 73.1% N 400 mol CO V12.0% CO 500 mol H O |W14.9% H O 2
2
2
2
2
ii) 100% conversion nC4H10 = 0 20% excess nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2) Exit gas: 400 mol CO2
10.1% CO2
500 mol H2O
12.6% H2O
130 mol O2
3.3% O2
2934 mol N2
74.0% N 2
iii) 90% conversion nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed) 20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2) Exit gas: 10 mol C4H10
0.3% C4H10 9.1% CO2
360 mol CO2 450 mol H2O (v)
4.69
a.
11.4% H2O
195 mol O2
4.9% O2
2934 mol N2
74.3% N 2
C3H8 + 5 O2 o 3 CO2 + 4 H2O
H2 +1/2 O2 o H2O
C3H8 + 7/2 O2 o 3 CO + 4 H2O Basis: 100 mol feed gas 100 mol 0.75 mol C3H8 0.25 mol H2
n1 (mol C3H8) n2 (mol H2) n3 (mol CO2) n4 (mol CO) n5 (mol H2O) n6 (mol O2) n7 (mol N2)
n0 (mol air) 0.21 mol O2/mol 0.79 mol N2/mol Theoretical oxygen:
75 mol C 3 H 8
5 mol O 2 25 mol H 2 0.50 mol O 2 mol C 3 H 8 mol H 2
4.69 (cont’d) 4-61
387.5 mol O 2
Air feed rate: n0
387.5 mol O 2 1 kmol air 125 . kmol air fed h 0.21 kmol O 2 1 kmol air req' d.
90% propane conversion n1
2306.5 mol air
0100 . (75 mol C 3 H 8 ) = 7.5 mol C 3 H 8
(67.5 mol C 3 H 8 reacts) 85% hydrogen conversion n2
0150 . (25 mol C 3 H 8 ) = 3.75 mol H 2
0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated mol C 3 H 8 react
95% CO 2 selectivity n3
192.4 mol CO 2 0.05(67.5 mol C 3 H 8 react) 3 mol CO generated mol C 3 H 8 react
5% CO selectivity n3
FG H
H balance: (75 mol C 3 H 8 ) 8
. mol CO 101
IJ K
mol H ( 25 mol H 2 )(2) mol C 3 H 8
(7.5 mol C 3 H 8 )(8) (3.75 mol H 2 )(2) n5 ( mol H 2 O)(2) n5
2912 . mol H 2 O
mol O ) (192.4 mol CO 2 )(2) mol O 2 (101 . mol CO)(1) ( 2912 . mol H 2 O)(1) + 2n6 ( mol O 2 ) n6
1413 . mol O 2
O balance: (0.21 u 2306.5 mol O 2 )(2
N 2 balance: n7
0.79(2306.5) mol N 2
1822 mol N 2
Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol = 2468 mol CO concentration in exit gas =
b.
101 . mol CO u 10 6 2468 mol
4090 ppm
If more air is fed to the furnace, (i)
more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs)
(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced.
4-62
4.70
a.
C5H12 + 8 O2 o 5 CO2 + 6 H2O Basis: 100 moles dry product gas n1 (mol C5H12)
100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG 0.053 mol O2/mol DPG 0.091 mol CO2/mol DPG 0.853 mol N2/mol DPG n3 (mol H2O)
Excess air n2 (mol O2) 3.76n2 (mol N2)
3 unknowns (n1, n2, n3) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. Problem is overspecified b.
N2 balance: 3.76 n2 = 0.8533 (100) n2 = 22.69 mol O2 C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) n1 = 2.09 mol C5H12 H balance: 12 n1 = 12(0.0027)(100) + 2n3 n3 = 10.92 mol H2O O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 45.38 mol O = 39.72 mol O Since the 4th balance does not close, the given data cannot be correct.
c. n1 (mol C5H12) Excess air n2 (mol O2) 3.76n2 (mol N2)
100 mol dry product gas (DPG) 0.00304 mol C5H12/mol DPG 0.059 mol O2/mol DPG 0.102 mol CO2/mol DPG 0.836 mol N2/mol DPG n3 (mol H2O)
N2 balance: 3.76 n2 = 0.836 (100) n2 = 22.2 mol O2 C balance: 5 n1 = 100 (5*0.00304 + 0.102) n1 = 2.34 mol C5H12 H balance: 12 n1 = 12(0.00304)(100) + 2n3 n3 = 12.2 mol H2O O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 44.4 mol O = 44.4 mol O ¸ Fractional conversion of C5H12:
2.344 100 u 0.00304 2.344
0.870 mol react/mol fed
Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 28.75 mol O2 % excess air:
22.23 mol O 2 fed - 18.75 mol O 2 required u 100% 18.6% excess air 18.75 mol O 2 required
4-63
4.71
a.
12 L CH 3 OH 1000 ml 0.792 g mol h L ml 32.04 g
296.6 mol CH 3 OH / h
CH3OH + 3/2 O2 o CO2 +2 H2O, CH3OH + O2 o CO +2 H2O 296.6 mol CH3OH(l)/h
n 2 ( mol dry gas / h) 0.0045 mol CH3OH(v)/mol DG 0.0903 mol CO2/mol DG 0.0181 mol CO/mol DG x (mol O2/mol DG) (1–x) (mol N2/mol DG) n 3 ( mol H 2 O(v) / h)
n1 (mol O 2 / h) 3.76n1 (mol N 2 / h)
4 unknowns (n1 , n 2 , n 3 , x ) – 4 balances (C, H, O, N2) = 0 D.F. b.
Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h C balance: 296.6 = n 2 (0.0045 + 0.0903 + 0.0181) n 2 = 2627 mol/h H balance: 4 (296.6) = n 2 (4*0.0045) + 2 n 3 n 3 = 569.6 mol H2O / h
O balance : 296.6 2n1 2627[0.0045 2(0.0903) 0.0181 2(0.8871- x)] 569.6 N2 balance: 3.76 n 1 = x ( 2627) Solving simultaneously n1 Fractional conversion: % excess air:
296.6 2627(0.0045) 296.6
0.960 mol CH 3 OH react/mol fed
574.3 444.9 u 100% 29.1% 444.9 569.6 mol H 2 O (2627 569.6) mol
Mole fraction of water:
4.72
574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG
0.178 mol H 2 O/mol
c.
Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger alarm if concentrations are too high
a.
G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH 4 and xC2 H 6 are the mole fractions of methane and ethane in the fuel, then
b g b n b molg x b mol CH
gb g molgb1 mol C 1 mol CH g
ns mol xC2 H 6 mol C2 H 2 mol 2 mol C 1 mol C 2 H 6
E
s
CH 4
b bmol CH
4
4
xC2 H 6 mol C2 H 6 mol fuel xCH 4
4
mol fuel
g
g
20 85
01176 . mole C 2 H 6 mole CH 4 in fuel gas
4-64
4.72 (cont’d)
b1.134 g H Ogb1 mol 18.02 gg
mole H 2 O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not). 2
Condensation measurement:
0126 .
CH 4 2O 2 o CO 2 2H 2 O 7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2 100 mol dry gas / h
n1 (mol CH4 ) 0.1176 n1 (mol C2H6) n2 (mol CO2)
0.126 mol H2O / mol 0..874 mol dry gas / mol 0.119 mol CO2 / mol D.G. x (mol N2 / mol) (0.881-x) (mol O2 / mol D.G.)
n3 (mol O2 / h) 376 n3 (mol N2 / h)
Strategy: H balance n ; 1
C balance n 2 ;
b gb
H balance: 4n1 6 01176 . n1
UV W
N 2 balance n3 , x O balance
. gb2g n g b100gb0126
1
5.356 mol CH 4 in fuel
0.1176(5.356) = 0.630 mol C2H6 in fuel
b gb
g
C balance: 5.356 2 0.630 n2
. gn b100gb0.874gb0119
2
3.784 mol CO 2 in fuel
Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO 2 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2
b100gb0.874gx b0.881 x g . g b100gb0.874gb2g 0119 . b2gb3.784g 2n b100gb0126
N 2 balance: 3.76n3 O balance:
3
Solve simultaneously: n3 18.86 mols O 2 fed , x 0.813 5.356 mol CH 4 2 mol O 2 0.630 mol C 2 H 6 3.5 mol O 2 Theoretical O 2 : 1 mol CH 4 1 mol CH 4 12.92 mol O 2 required Desired O2 fed:
(5.356 0.630 3.784) mol fuel 7 mol air 0.21 mol O 2 = 14.36 mol O2 1 mol fuel mol air
Desired % excess air:
b.
Actual % excess air:
14.36 12.92 u 100% 11% 12.92
18.86 12.92 u 100% 46% 12.92
Actual molar feed ratio of air to fuel:
(18.86 / 0.21) mol air 9.77 mol feed
4-65
9 :1
4.73
a.
C3H8 +5 O2 o 3 CO2 + 4 H2O, C4H10 + 13/2 O2 o 4 CO2 + 5 H2O Basis 100: mol product gas n1 (mol C3H8) n2 (mol C4H10)
100 mol 0.474 mol H2O/mol x (mol CO2/mol) (0.526–x) (mol O2/mol)
n3 (mol O2) Dry product gas contains 69.4% CO2
x 0.526 x
69.4 x 30.6
0.365 mol CO 2 /mol
3 unknowns (n1, n2, n3) – 3 balances (C, H, O) = 0 D.F. O balance: 2 n3 = 152.6 n3 = 76.3 mol O2 n1 7.1 mol C 3 H 8 C balance : 3 n1 4 n 2 36.5 ½ 65.1% C 3 H 8 , 34.9% C 4 H10 ¾ H balance : 8 n1 10 n 2 94.8¿ n 2 3.8 mol C 4 H10 b.
nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H 27.8%C, 72.2% H From a: 7.10 mol C 3 H 8
3.80 mol C 4 H10 4 mol C 3 mol C mol C 4 H10 mol C 3 H 8
7.10 mol C 3 H 8 11 mol (C H) 3.80 mol C 4 H10 14 mol (C H) mol C 3 H 8 mol C 4 H10 4.74
Basis: 100 kg fuel oil Moles of C in fuel:
100 kg 0.85 kg C 1 kmol C kg 12.01 kg C
Moles of H in fuel:
100 kg 0.12 kg H 1 kmol H 12.0 kmol H kg 1 kg H
Moles of S in fuel:
100 kg 0.017 kg S 1 kmol S kg 32.064 kg S
1.3 kg non-combustible materials (NC)
4.74 (cont’d) 4-66
7.08 kmol C
0.053 kmol S
u 100% 27.8% C
100 kg fuel oil 7.08 kmol C 12.0 kmol H 0.053 kmol S 1.3 kg NC (s) 20% excess air n1 (kmol O2) 3.76 n1 (kmol N2)
n2 (kmol N2) n3 (kmol O2) n4 (kmol CO2) (8/92) n4 (kmol CO) n5 (kmol SO2) n6 (kmol H2O)
C + O2 o CO2 C + 1/2 O2 o CO 2H + 1/2 O2 o H2O S + O2 o SO2
Theoretical O2: 7.08 kmol C 1 kmol O 2 12 kmol H .5 kmol O 2 0.053 kmol S 1 kmol O 2 1 kmol C 2 kmol H 1 kmol S
10.133 kmol O 2
20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 n3 = 2.3102 kmol O2 C balance: 7.08 = n4+8n4/92 n4 = 6.514 mol CO2 8 (6.514)/92 = 0.566 mol CO S balance: n5 = 0.53 kmol SO2 H balance: 12 = 2n6 n6 = 6.00 kmol H2O N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol = 61.16 kmol 10.7% CO, 0.92% CO, 0.087% SO 2 , 9.8% H 2 O, 3.8% O 2 , 74.8% N 2
4.75
a. Basis: 5000 kg coal/h; 50 kmol air min
3000 kmol air h
5000 kg coal / h 0.75 kg C / kg 0.17 kg H / kg 0.02 kg S / kg 0.06 kg ash / kg
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2 C + 1/2 O2 --> CO
3000 kmol air / h 0.21 kmol O2 / kmol 0.79 kmol N2 / kmol
n1 (kmol O2 / h) n2 (kmol N2 / h) n3 (kmol CO2 / h) 0.1 n3 (kmol CO / h) n4 (kmol SO2 / h) n5 (kmol H2O / h)
mo kg slag / h
Theoretical O 2 : C:
0.75 5000 kg C
b g
1 kmol C
1 kmol O 2
h
12.01 kg C
1 kmol C
4.75 (cont’d)
4-67
312.2 kmol O 2 h
b g
H:
0.17 5000 kg H 1 kmol H 1 kmol H 2 O 1 kmol O 2 h 2 kmol H 2 kmol H 2 O 101 . kg H
S:
0.02 5000 kg S 1 kmol S 1 kmol O 2 = 3.1 kmol O2/h 32.06 kg S 1 kmol S h
210.4 kmol O 2 h
b g
Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O 2 h
b
g
O 2 fed = 0.21 3000 Excess air:
630 kmol O 2 h
630 525.7 u 100% 19.8% excess air 525.7
b. Balances: 0.94 0.75 5000 kg C react 1 kmol C C: h 12.01 kg C
b gb gb g n 3
266.8 kmol CO 2 h , 01 . n 3
H:
. gb5000g kg H b017
S:
(from part a)
N2 :
O:
h
n 3 0.1n 3
26.7 kmol CO h
1 kmol H 1 kmol H 2 O n 5 n5 101 . kg H 2 kmol H
b
3.1 kmol O 2 for SO 2 h
g
1 kmol SO 2 1 kmol O 2
420.8 kmol H 2 O h
n 4 n 4
31 . kmol SO 2 h
b0.79gb3000g kmol N h n n 2370 kmol N h b0.21g(3000)b2g 2n 2b266.8g 1b26.68g 2b31. g b1gb420.8g 2
2
2
2
1
n1
136.4 kmol O 2 / h
Stack gas total
3223 kmol h
Mole fractions:
c.
x CO
26.7 3224 8.3 u 10 3 mol CO mol
xSO2
31 . 3224
9.6 u 10 4 mol SO 2 mol
1 SO 2 O 2 o SO 3 2 SO 3 H 2 O o H 2SO 4 3.1 kmol SO 2 1 kmol SO 3 1 kmol H 2SO 4
98.08 kg H 2SO 4
1 kmol SO 2
kmol H 2SO 4
h
1 kmol SO 3
4-68
304 kg H 2SO 4 h
4.76 a.
Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter 100 g c.a. r. 1.147 g a.d.c. 1.207 g c.a. r.
95.03 g a.d.c
95.03 g air - dried coal; 4.97 g H 2 O lost by air drying
. ggH O b1.234 1204 2
1.234 g a.d.c. Total H 2 O 95.03 g a.d.c
4.97 g 2.31 g
2.31 g H 2 O lost in second drying step
7.28 g moisture
0.811g g b v. m. H Og . b1347 2.31 g H O 2
2
1.347 g a.d.c.
95.03 g a.d.c
0.111 g ash 1.175 g a.d.c.
8.98 g ash
b100 7.28 35.50 8.98gg
Fixed carbon
3550 . g volatile matter
48.24 g fixed carbon
7.28 g moisture 7.3% moisture 48.24 g fixed carbon 48.2% fixed carbon 35.50 g volatile matter 35.5% volatile matter 8.98 g ash 9.0% ash 100 g coal as received b. Assume volatile matter is all carbon and hydrogen. C CO 2 o CO 2 :
2H
1 mol O 2 1 mol C
1 mol C 10 3 g 1 mol air 12.01 g C 1 kg 0.21 mol O 2
0.5 mol O 2 1 O2 o H 2O : 2 2 mol H
Air required:
396.5 mol air kg C
1 mol H 10 3 g 1 mol air 1.01 g H 1 kg 0.21 mol O 2
1179 mol air kg H
1000 kg coal 0.482 kg C 396.5 mol air kg coal
1000 kg 0.355 kg v. m.
kg C 6 kg C
396.5 mol air
kg 7 kg v. m. kg C 1000 kg 0.355 kg v. m. 1 kg H 1179 mol air kg
7 kg v. m.
4-69
kg H
3.72 u 105 mol air
4.77
a.
Basis 100 mol dry fuel gas. Assume no solid or liquid products!
n1 (mol C) n2 (mol H) n3 (mol S)
100 mol dry gas C + 02 --> CO2 C + 1/2 O2 --> CO 2H + 1/2 O2 -->H2O S + O2 --> SO2
0.720 mol CO2 / mol 0.0257 mol CO / mol 0.000592 mol SO2 / mol 0.254 mol O2 / mol
n4 (mol O2) (20% excess)
n5 (mol H2O (v))
½ H balance : n 2 2 n 5 ° O balance : 2 n 4 100 [ 2(0.720) 0.0257 2 (0.000592) 2 (0.254)] n 5 ¾ ° 20 % excess O 2 : (1.20) (74.57 0.0592 0.25 n 2 ] n 4 ¿ n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O Total moles in feed: 258.4 mol (C+H+S) 28.9% C, 71.1% H, 0.023% S
4.78
Basis: 100 g oil Stack SO 2 , N 2 , O 2, CO 2, H 2O (612.5 ppm SO 2)
x n 3 mol SO 2 (N2 , O2 , CO2 , H 2 O)
0.10 (1 – x ) n 5 mol SO 2 (N2 , O2 , CO2 , H 2 O)
100 g oil 0.87 g C/g 0.10 g H/g 0.03 g S/g n 1 mol O2 3.76 n 1 mol N2 (25% excess)
furnace
n 2 mol N 2 n 3 mol O 2 n 4 mol CO2 n 5 mol SO 2 n 6 mol H 2 O
CO 2 : H 2 O:
b g
0.87 100 g C
(1 – x ) n 5 mol SO 2 (N2 , O2 , CO2 , H 2 O)
scrubber
Alkaline solution
0.90 (1 – x ) n 5 mol SO 2
1 mol C 1 mol CO 2 n4 12.01 g C 1 mol C
b g
0.10 100 g H 1 mol H 1 mol H 2 O n6 2 mol H 101 . gH
4-70
FG 7.244 mol O IJ H consumed K F 2.475 mol O IJ 4.95 mol H OG H consumed K
7.244 mol CO 2
2
2
2
4.78 (cont’d) SO 2 :
b g
0.03 100 g S
1 mol S 1 mol SO 2 n5 32.06 g S 1 mol S
b
0.0936 mol SO 2
g
FG 0.0956 mol O IJ H consumed K 2
125 . 7.244 2.475 0.0936 12.27 mol O 2
25% excess O 2 : n1
b
g
12.27 mol O 2 fed 7.244 2.475 0.0936 mol O 2 consumed
O 2 balance: n3
2.46 mol O 2 N 2 balance: n 2
b
b
3.76 12.27 mol
g
4614 . mol N 2
g
SO 2 in stack SO 2 balance around mixing point :
F H
I K
b gb
g
. 1 x 0.0936 x 0.0936 010 n5
b
0.00936 0.0842 x mol SO 2
g
Total dry gas in stack (Assume no CO2 , O2 , or N 2 is absorbed in the scrubber)
b
7.244 2.46 4614 . 0.00936 0.0842 x
bCO g bO g 2
bN g
2
2
b
g
bSO g
g
b
g
5585 . 0.0842 x mol dry gas
2
612.5 ppm SO 2 dry basis in stack gas 0.00936 0.0842 x 5585 . 0.0842 x
612.5 x 10 . u 10 6
0.295 30% bypassed
Basis: 100 mol stack gas
4.79
n 1 (mol C) n 2 (mol H) n 3 (mol S) n 4 (mol O2 ) 3.76 n 4 (mol O2 )
a.
C balance: n1 H balance: n2
C + O2 o CO2 1 2H + O2 o H 2 O 2 S + O 2 o SO 2
100 mol 0.7566 N 2 0.1024 CO2 0.0827 H 2 O 0.0575 O 2 0.000825 SO 2
. b100gb01024 g 10.24 mol C b100gb0.0827gb2g 16.54 mol H
10.24 mol C 16.54 mol H
0.62
mol C mol H
The C/H mole ratio of CH 4 is 0.25, and that of C2 H 6 is 0.333; no mixture of the two could have a C/H ratio of 0.62, so the fuel could not be the natural gas. b.
b100gb0.000825g 0.0825 mol S b10.24 mol Cgb12.0 g 1 molg 122.88 g CU| 122.88 7.35 g C g H b16.54 mol Hgb1.01 g 1 molg 16.71 g H V 216.65.71 b0.0825 mol Sgb32.07 g 1 molg 2.65 g S |W 142.24 u 100% 1.9% S
S balance: n 3
4-71
No. 4 fuel oil
4.80
a.
Basis: 1 mol CpHqOr 1 mol CpHqOr no (mol S) Xs (kg s/ kg fuel)
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2
P (% excess air) n1 (mol O2) 3.76 n1 (mol N2)
n2 (mol CO2) n3 (mol SO2) n4 (mol O2) 3.76 n1 (mol N2) n5 (mol H2O (v))
Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C) q (mol H) (1 g / mol) =
q (g H)
(12 p + q + 16 r) g fuel
r (mol O) (16 g / mol) = 16 r (g O) S in feed: no=
(12 p q 16r) g fuel
Theoretical O2:
X s (g S) 1 mol S (1 - X s ) (g fuel) 32.07 g S
X s (12 p q 16 r) (mol S) (1) 32.07(1 - X s )
p (mol C) 1 mol O 2 q (mol H) 0.5 mol O 2 ( r mol O) 1 mol O 2 1 mol C 2 mol H 2 mol O (p 1/4 q 1/2 r) mol O 2 required
% excess n1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed
(2)
C balance: n2 = p
(3)
H balance: n5 = q/2
(4)
S balance: n3 = n0
(5)
O balance: r + 2n1 = 2n2 + 2n3 + 2n4 + n5 n4 = ½ (r+2n1-2n2-2n3-n5)
(6)
Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air (1) n0 = 0.00616 mol S
(5) n3 = 0.00616 mol SO2
(2) n1 = 1.16 mol O2 fed
(6) n4 = 0.170 mol O2
(3) n2 = 0.71 mol CO2
(4) n5 = 0.55 mol H2O
(3.76*1.16) mol N2 = 4.36 mol N2 Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas Dry basis composition yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2 yO2 = (0.170 / 5.246) * 100% = 3.2% O2 yN2 = (4.36 / 5.246) * 100% = 83.1% N2 ySO2 = (0.00616 / 5.246) * 106 = 1174 ppm SO2
4-72
CHAPTER FIVE 5.1
Assume volume additivity
m
m0 m = mt
mass of tank at time t
empty tank
a.
A
0.400 0.600 U 0.703 kg L 0.730 kg L
1 U
Av. density (Eq. 5.1-1):
A mass of
A
A
UO
UD
b250 150gkg b10 3g min
14.28 kg min
0.719 kg L
bm
mass flow rate of liquid
1L / min) / min) = m(kg 14.28 kg V(L V 19.9 L min min 0.719 kg U ( kg / L) b. m0
5.2
bg
150 14.28 3 m(t) - mt
107 kg
b
g
void volume of bed: 100 cm 3 2335 . 184 cm3
porosity: 50.5 cm3 void 184 cm3 total bulk density: 600 g 184 cm3
b
50.5 cm 3
0.274 cm3 void cm3 total
3.26 g cm3
g
absolute density: 600 g 184 50.5 cm3
4.49 g cm3
5.3 C 6 H 6 (l ) B (kg / min) m VB = 20.0 L / min (kg / min) m (L / min) V
C 7 H 8 (l )
T (kg / min) m (L / min) V T 2 . m) 2 015 . m = 'V SD 'h S (55 V 0.0594 m3 / min 't 4 60 min 4 't Assume additive volumes -V V V 59.4 20.0 L / min = 39.4 L / min T B
b
m xB
g
0.879 kg 20.0 L 0.866 kg 39.4 L L min L min (0.879 kg / L)(20.0 L / min) 0.34 kg B / kg (517 . kg / min)
B UT V T UB V B m m
5-1
. kg / min 517
g
a.
b.
1 U sl
U| V| W
b
F IF I e jge jhbmgGH 11 N JK GH 11 Pa JK kg
U sl
m s2
m3
kgm s2
N m2
U sl gh
g
1 xc xc check units! Uc Ul
1 kg crystals / kg slurry kg liquid / kg slurry kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid L slurry L crystals L liquid L slurry kg slurry kg slurry kg slurry kg slurry 'P 2775 1415 kg / m3 c. i.) U sl gh 9.8066 0.200 ii.)
1 U sl
xc
xc Uc
b gb g b1- x g x FG 1 1 IJ FG 1 U HU U K HU c
l
F 1 GG 1415 kg / m H
c
3
l
sl
1 . 1000 kg / m3 12
d
I iJJK
1 Ul
F I GG 2.3d10001kg / m i 12. d10001kg / m iJJ H K msl U sl
iv.) mc
x c msl
IJ K
0.316 kg crystals / kg slurry
3
175 kg 1000 L 3 1415 kg / m m3
iii.) Vsl
1238 . L
b0.316 kg crystals / kg slurrygb175 kg slurryg
55.3 kg crystals
55.3 kg CuSO 4 5H 2 O 1 kmol 1 kmol CuSO 4 159.6 kg 249 kg 1 kmol CuSO 4 5H 2 O 1 kmol
v.) mCuSO 4 vi.) ml
c
3
b1 x gm b0.684 kg liquid / kg slurrygb175 kg slurryg c
ml Ul
vii.)
Vl
h(m) Ul(kg/m^3) Uc(kg/m^3) 'P(Pa) xc Usl(kg/m^3)
0.2 1200 2300 2353.58 0 1200.00
sl
120 kg
1000 L m3
b1.2gd1000 kg / m i 3
35.4 kg CuSO 4
120 kg liquid solution
100 L
d. 2411.24 0.05 1229.40
2471.80 0.1 1260.27
2602.52 0.2 1326.92
2747.84 0.3 1401.02
2772.61 0.316 1413.64
2910.35 0.4 1483.87
3093.28 0.5 1577.14
Effect of Slurry Density on Pressure Measurement
0.6 Solids Fraction
5.4
P0 U sl gh1 P0 U sl gh 2 'P = P1 P2 h = h1 h 2 P1 P2
0.5 0.4 0.3 0.2
'P = 2775, U = 0.316
0.1 0 2300.00
2500.00
2700.00
2900.00
Pressure Difference (Pa)
5-2
3100.00
5.4 (cont’d) e.
b
g d
b
b1- x gbkg liquid g, V dm c
l
3
liquid
d
U c kg / m
1 kg
5.7
a.
c
29.0 g
3
0.0064 m air
nRT P
b3.06L - 2.8Lg u 100%
Assume Patm
1013 . bar
b.
PV Ps Vs
n
. gbar b10 1013
nRT n n s RTs
20.0 m3
V
273K 298.2K
b
3.06 L
9.3%
2.8L
nRT n
0.0064 m3 mol
4.5 kg m 3
mol 103 g
100 . mol 0.08206 L atm 373.2 K mol K 10 atm
a.
5.8
1 kg
b. % error =
PV
3
1 atm
1 mol
V=
i
g
3 = 0.08206 m atm 313.2 K 1 kmol RT V kmol K 4.0 atm 103 mol
PV
5.6
l
Assume Patm
U
b
bV V gdm i c
5.5
1 1 xc xc Uc Ul
3
3
g i b1-Ux dgbkgkg/ mliquid i l
U sl
g
x c kg crystals
i
Basis: 1 kg slurry x c kg crystals , Vc m3 crystals
kmol K 28.02 kg N 2 20.0 m3 kmol 25 + 273.2 K 0.08314 m3 bar
g
249 kg N 2
Ts P n s T Ps Vs
. gbar b10 1013 1.013 bar
1 kmol
28.02 kg N 2 kmol 22.415 m STP 3
a.
R=
Ps Vs n sTs
1 atm 22.415 m3 1 kmol 273 K
b.
R=
Ps Vs n sTs
1 atm 760 torr 359.05 ft 3 1 lb - mole 1 atm 492 $ R
8.21 u 10 2
5-3
b g
atm m3 kmol K 555
torr ft 3 lb - mole $ R
249 kg N 2
5.9
P = 1 atm +
10 cm H 2 O
1m 1 atm 2 10 cm 10.333 m H 2 O 3
= 2.0 m T = 25$ C = 298.2 K , V 5 min m = n mol / min MW g / mol
b
a.
g
= m = m
b.
b
101 . atm
0.40 m3 min = 400 L min
g
L 28.02 PV 1.01 atm 400 min MW = Latm RT 0.08206 molK 298.2 K
400
L min
28.02 273 K 1 mol 298.2 K 22.4 L STP
b g
g mol
F mI V dm si uG J H s K Ad m i 3
5.10 Assume ideal gas behavior:
458 g min
458 g min
T2 P1 D12 nR 2 nR T1 P2 D 2
nRT P u 2 2 u1 SD 4
2
. 1013 . gbar b7.50 cmg b180 134 m sec sec 333.2 K b153 . 1013 . gbar b5.00 cmg . 100 . g atm 5 L PV b100 Assume ideal gas behavior: n 0.406 mol u2
5.11
g mol
u1
T2 P1D12 T1P2 D 22
2
60.0 m 300.2 K
2
Latm 0.08206 mol RT K 32.0 g mol Oxygen
MW 13.0 g 0.406 mol
5.12 Assume ideal gas behavior: Say m t
g t
0.009391 mol
b g
std cm3 STP min V
b g
' V liters 273K 763 mm Hg 103 cm3 1L ' t min 296.2K 760 mm Hg
I
std cm3 STP min V
5.0 9.0 12.0
139 268 370
U| |V straight line plot ||I 0.031EV 0.93 W std
b.
b g
0.010 mol N 2 22.4 liters STP 103 cm3 min 1 mole 1L
d
mol of gas in tank
gU|V n 0.009391 mol g |W m 37.0256 g b37.062 37.0256gg 3.9 g mol Helium b b
unknown: MW
std V
mass of tank, n g
m t n g 28.02 g mol m t n g 44.1 g mol
N 2 : 37.289 g CO 2 : 37.440 g
5.13 a.
300 K
i
I = 0.031 224 cm3 / min 0.93 7.9 5-4
224 cm3 / min
925.3
'V 't
/ kmol) g nbkmolgM(kg Vb Lg
b FU I si G J HU K
5.14 Assume ideal gas behavior U kg L
d
3
i V dcm
V2 cm s
1
PM
! RT
12
3
1
V1 P1 M1T2 P2 M 2 T1
12
2
LM N
cm3 758 mm Hg 28.02 g mol 323.2K s 1800 mm Hg 2.02 g mol 298.2K
a.
VH 2
b.
M
0.25M CH 4 0.75M C 3H 8
Vg
cm3 350 s
350
n P V RT
OP Q
12
881 cm3 s
b0.25gb16.05g b0.75gb44.11g
LM b758gb28.02gb323.2g OP N b1800gb37.10gb298.2g Q
37.10 g mol
12
205 cm3 s
5.15 a. Reactor
'h soap
b.
d
2 S 0.012 m 4
i
2
n CO 2
2 PV = S R 'h V 't RT
n CO 2
755 mm Hg 1 atm 11 . u 10-3 m3 / min 1 kmol 3 m atm 760 mm Hg 300 K 1000 mol 0.08206 kmol K
. m 60 s 12 . u 10 3 m3 / min 11 7.4 s min 0.044 mol / min
5.16 air m
10.0 kg / h
n air (kmol / h)
n (kmol / h) y CO (kmol CO 2 / kmol) 2
CO = 20.0 m3 / h V 2 n CO (kmol / h) 2
150 o C, 1.5 bar
Assume ideal gas behavior 10.0 kg 1 kmol n air 0.345 kmol air / h h 29.0 kg air n CO 2
PV RT
y CO 2 u 100%
15 100 kPa 20.0 m3 / h . bar 3 1 bar 423.2 K 8.314 mkmolkPa K
b
0.853 kmol CO 2 / h
0.853 kmol CO 2 / h u 100% 0.853 kmol CO 2 / h + 0.345 kmol air / h
g
5-5
712% .
5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior 3
o
1 (kg / min) m 0.70 kg H 2 O / kg 0.30 kg S / kg
311 m / min, 83 C, 1 atm n 3 (kmol / min) 0.12 kmol H 2 O / kmol 0.88 kmol dry air / kmol
n 2 (kmol air / min) (m 3 / min) V 2 167 o C, - 40 cm H 2 O gauge
4 (kg S / min) m
a.
n 3
1 atm 311 m3 365.2K min
kmol K 10.38 kmol min 0.08206 m3 atm 10.38 kmol 0.12 kmol H 2 O 18.02 kg kmol kmol min
H 2 O balance: 0.70 m1 1 m
32.2 kg min milk
b g
b
S olids balance: 0.30 32.2 kg min
4 m 4 m
b
9.6 kg S min
g
0.88 10.38 kmol min n 2
Dry air balance: n 2
2 V
g
9.13 kmol 0.08206 m3 atm min kmol K
9.13 kmol min air
440K 1033 cm H 2 O 1033 40 cm H 2 O 1 atm
b
g
3
343 m air min u air (m / min) =
air (m3 / s) V A (m2 )
343 m3 1 min min 60 s
S
4 (6
m) 2
0.20 m / s
b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor by the air instead of falling to the conveyor belt.
5.18 SG CO 2
5.19 a.
x CO 2
U CO 2 U air
PM CO2 RT PM air RT
0.75 x air
M CO 2 M air
152 .
1 0.75 0.25
Since air is 21% O 2 , x O 2 b.
44 kg / kmol 29 kg / kmol
mCO 2 = n x CO 2 M CO 2
0.0525 5.25 mole% O 2
(0.25)(0.21)
b
g
2 u 1.5 u 3 m3 0.75 kmol CO 2 44.01 kg CO 2 1 atm 12 kg m 3 atm 298 2 K kmol kmol CO . 0.08206 kmol 2 K
More needs to escape from the cylinder since the room is not sealed.
5-6
5.19 (cont’d) c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a person entered the room and closed the door, over a period of time the person could die of asphyxiation. Measures that would reduce hazards are: 1. Change the lock so the door can always be opened from the inside without a key. 2. Provide ventilation that keeps air flowing through the room. 3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount. 4. Install safety valves on the cylinder in case of leaks. 15.7 kg
5.20 n CO 2
1 kmol
0.357 kmol CO 2
44.01 kg
b
Assume ideal gas behavior, negligible temperature change T 19q C 292.2 K
a.
P1V P2 V
b
n1 RT n1 n1 0.357 RT n1 0.357
g
n1 b. Vtank
102kPa 3.27 u 103 kPa
0.0115 kmol air in tank
n1RT P1
0.0115 kmol 292.2 K 8.314 m3 kPa 103 L m3 102kPa kmol K
15700 g CO 2 + 11.5 mol air (29.0 g air / mol) 274 L
Uf c.
P1 P2
g
274 L
58.5 g / L
CO 2 sublimates large volume change due to phase change rapid pressure rise. Sublimation causes temperature drop; afterwards, T gradually rises back to room temperature, increase in T at constant V slow pressure rise.
5.21 At point of entry, P1
b10 ft H Ogb29.9 in. Hg 33.9 ft H Og 28.3 in. Hg 2
2
37.1 in. Hg .
At surface, P2
28.3 in. Hg, V2 bubble volume at entry 1 x solid x solution 0.20 0.80 Mean Slurry Density: 3 U sl U solid U solution (1.2)(1.00 g / cm ) (100 . g / cm 3 ) 0.967
cm 3 U sl g
1.03 g 2.20 lb 5 u 10 4 ton 10 6 cm 3 1 lb 264.17 gal cm 3 1000 g
a.
300 ton gal 40.0 ft 3 (STP) 534.7 o R 29.9 in Hg hr 4.3 u 10 3 ton 1000 gal 492 o R 37.1 in Hg
b.
P2 V2 P1V1
nRT V 2 nRT V1
% change =
4 3S P1 P2 4 3S
e j e j
b2.2 - 2.0g mm u 100 2.0 mm
D2 3 2
37.1 D 32 28.3
D1 3 2
10%
5-7
1.31D13
4.3 u 10 3 ton / gal
2440 ft 3 / hr
!D
D1 2 mm
2
2.2 mm
5.22 Let B = benzene n1 , n 2 , n 3 moles in the container when the sample is collected, after the helium is added, and after the gas is fed to the GC. n inj moles of gas injected n B , n air , n He moles of benzene and air in the container and moles of helium added n BGC , m BGC moles, g of benzene in the GC y B mole fraction of benzene in room air
a.
P1V1
n1 P2 V2
n2 P3 V3
n3
n1RT1 (1 { condition when sample was taken): P1 = 99 kPa, T1
mol K 99 kPa 2 L kPa 101.3 atm 306 K .08206 L atm
0.078 mol = n air n B
n 2 RT2 (2 { condition when charged with He): P2 = 500 kPa, T2
mol K 500 kPa 2 L 101.3 kPa 306 K .08206 L atm atm
mol K 400 kPa 2 L kPa 101.3 atm 296 K .08206 L atm
n BGC u
y B (ppm) =
306K
0.393 mol = n air + n B n He
n 3 RT3 (3 { final condition in lab): P3 = 400 kPa, T3
n inj = n 2 n 3 nB
306K
296K
0.325 mol = (n air n B n He ) n inj
0.068 mol n2 n inj
0.393 mol m BGC (g B) 1 mol 0.068 mol 78.0 g
nB u 106 n1
0.0741 m BGC u 106 0.078
9 am: y B
(0.950 u 106 )(0.656 u 10 6 )
1 pm: y B
(0.950 u 106 )(0.788 u 10 6 )
5 pm: y B
(0.950 u 106 )(0.910 u 10 6 )
0.0741 m BGC
0.950 u 106 m BGC
U| | 0.749 ppm VThe avg. is below the PEL | 0.864 ppm| W 0.623 ppm
b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix sufficiently and to reach thermal equilibrium. c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is less dense than the benzene and air; therefore, the sample injected in the GC may be Herich depending on where the sample was taken from the cylinder. (iv) The benzene may not be uniformly distributed in the laboratory. In some areas the benzene concentration could be well above the PEL.
5-8
b g
4 S 10 m 3 Moles of gas in balloon
5.23 Volume of balloon
b
n kmol a.
b.
4189 m3
g
3
4189 m3
492q R 3 atm 1 kmol 535q R 1 atm 22.4 m3 STP
b g
515.9 kmol
He in balloon:
m
b515.9 kmolg b4.003 kg kmolg
mg
2065 kg 9.807 m 1N 2 s 1 kg m / s2
dP dP
iV iV
gas in balloon air displaced
Fbuoyant
n gas RT n air RT
Fbuoyant
2065 kg He 20,250 N
Pair n gas Pgas
n air
172.0 kmol 29.0 kg 9.807 m 1 N 1 kmol s2 1 kg2m s
Wair displaced
Since balloon is stationary, Wtotal
Fcable
Fcable
Fbuoyant Wtotal
c. When cable is released, Fnet
b
¦F
1
48920 N
dAi = 27200 N a
1 atm 515.9 kmol 172.0 kmol 3 atm
48,920 N
0
b2065 150gkg
9.807 m 1 N s2 1 kgs2m
M tot a
27200 N 1 kg m / s2 2065 + 150 kg N
g
12.3 m s2
d. When mass of displaced air equals mass of balloon helium the balloon stops rising. Need to know how density of air varies with altitude. e. The balloon expands, displacing more air buoyant force increases balloon rises until decrease in air density at higher altitudes compensates for added volume. 5.24 Assume ideal gas behavior, Patm a. b.
1 atm
PN VN 5.7 atm 400 m3 / h 240 ft 3 h Pc 9.5 atm Mass flow rate before diversion: PN VN
240 m3 h
Pc Vc Vc
273 K 5.7 atm 1 kmol 303 K 1 atm 22.4 m3 STP
b g 5-9
44.09 kg kmol
2426
kg C 3 H 6 h
27,20
5.24 (cont’d) Monthly revenue:
b2426 kg hgb24 h daygb30 days monthgb$0.60 kgg
c.
$1,048,000 month
Mass flow rate at Noxious plant after diversion: 240 m3 hr
273 K 2.8 atm 1 kmol 306 K 1 atm 22.4 m3
Propane diverted 5.25 a. PHe PCH 4 PN 2
b2426 1180g kg h
44.09 kg 1180 kg hr kmol
1246 kg h
y He P = 0.35 (2.00 atm) = 0.70 atm y CH 4 P = 0.20 (2.00 atm) = 0.40 atm y N 2 P = 0.45 (2.00 atm) = 0.90 atm
b. Assume 1.00 mole gas 4.004 g 0.35 mol He mol
U| || 16.05 g I F 0.20 mol CH G H mol JK 3.21 g CH V|17.22 g mass fraction CH F 28.02 g IJ 12.61 g N || 0.45 mol N G H mol K W FG H
IJ K
140 . g He
4
2
c.
MW
d.
U gas
4
3.21 g 17.22 g
0186 .
2
g of gas 17.2 g / mol mol P MW m n MW V
4
d i d i b2.00 atmgb17.2 kg / kmolg V RT e0.08206 jb363.2 Kg m 3 atm kmolK
115 . kg / m3
5.26 a. It is safer to release a mixture that is too lean to ignite. If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard. b. fuel-air mixture n 1 ( mol / s) y C 3H 8 0.0403 mol C 3 H 8 / mol n C 3H 8 150 mol C3 H 8 / s
n 3 ( mol / s) 0.0205 mol C 3 H 8 / mol
diluting air n 2 ( mol / s)
n 1
150 mol C3 H 8 mol s 0.0403 mol C 3 H 8
Propane balance: 150 = 0.0205 n 3 n 3
5-10
3722 mol / s
7317 mol / s
5.26 (cont’d) Total mole balance: n 1 n 2
c.
b g
n 3 n 2
7317 3722
n 2
. n 2 13
2 V
4674 mol / s 8.314 m3 Pa 398.2 K mol K 131,000 Pa
3595 mol air / s
4674 mol / s
min
1 V
3722 mol 8.314 m3 Pa 298.2 K s mol K 110000 Pa
y2
150 mol / s n 1 n 2
b
U| |V V || V W
118 m3 / s
2
. 141
1
83.9 m3 / s
m3 diluting air m3 fuel gas
150 mol / s u 100% 18% . 3722 mol / s + 4674 mol / s
g
d. The incoming propane mixture could be higher than 4.03%. If n 2 n 2 min , fluctuations in the air flow rate would lead to temporary explosive
b g
conditions. 5.27
b
gb
Basis: 12 breaths min 500 mL air inhaled breath
g
600 mL inhaled min
o
24 C, 1 atm 6000 mL / min
lungs
n in (mol / min) 0.206 O 2 0.774 N 2 0.020 H 2 O
a.
n in
blood
37 o C, 1 atm n out (mol / min) 0.151 O 2 0.037 CO 2 0.750 N 2 0.062 H 2 O
6000 mL 1L 273K 1 mol 3 min 10 mL 297K 22.4 L STP
b g
b
gb
g
N 2 balance: 0.774 0.246
O 2 transferred to blood:
0.750n out n out
0.246 mol min 0.254 mol exhaled min
. g b mol O b0.246gb0.206g b0.254gb0151
2
g
min 32.0 g mol
0.394 g O 2 min CO 2 transferred from blood:
b0.254gb0.037g bmol CO
2
g
min 44.01 g mol
0.414 g CO 2 min H 2 O transferred from blood:
b0.254gb0.062g b0.246gb0.020g bmol H O ming 18.02 g mol 2
g H 2 O min 0195 .
5-11
5.27 (cont’d) PVin PVout
FG n IJ FG T IJ FG 0.254 mol min IJ FG 310K IJ 1078 H n K H T K H 0.246 mol min K H 297K K . mL exhaled ml inhaled . g H O lost ming b0.394 g O gained ming 0.215 g min b0.414 g CO lost ming b0195
b.
n in RTin n out RTout
Vout Vin
out
out
in
in
2
2
2
STACK
5.28
Ta (K) Ts (K) M s (g/mol) M a (g/mol) Ps (Pa) Pc (Pa)
LL(m) (M)
PM RT
Ideal gas: U
bUgLg
a.
D
b.
Ms
Pa
755 mm Hg 29.0 g mol , Ta
755 mm Hg
u
b g
UgL
Pa M a PM gL a s gL RTa RTs
stack
. gb44.1g b0.02gb32.0g b0.80gb28.0g b018
Ma
D
combust.
LM N
Pa gL M a M a R Ta Ts
310 . g mol , Ts
OP Q
655K ,
294 K , L 53 m
1 atm 53.0 m 9.807 m kmol - K 2 s 0.08206 m3 atm 760 mm Hg
LM 29.0 kg kmol 310. kg kmol OP u FG 1 N IJ 655K N 294K Q H 1 kg m / s K 2
323 N 1033 cm H 2 O m2 1.013 u 105 N m2
3.3 cm H 2 O
b.
b g
MWCCl2O 98.91 g / mol U P MW 98.91 CCl2O 3.41 U air 29.0 RT Phosgene, which is 3.41 times more dense than air, will displace air near the ground. 2 S D in L S 2 Vtube 0.635 cm - 2 0.0559 cm 15.0 cm 3.22 cm3 4 4
5.29 a. U
b g
m CCl O 2
c.
!
n CCl 2 O(l)
b
Vtube U CCl O 2
3.22 cm3
gb
g
1L 1 atm 98.91 g / mol 3 3 Latm 296.2 K 10 cm 0.08206 molK
g mol 3.22 cm3 137 . u 1000 . 3 cm 98.91 g
5-12
0.0446 mol CCl 2 O
0.0131 g
5.29 (cont’d) mol K 1 atm 2200 ft 3 28.317 L 2563 mol air 3 296.2K .08206 L atm ft
PV RT
n air n CCl 2 O
0.0446 17.4 u 10 6 2563
n air
17.4 ppm
The level of phosgene in the room exceeded the safe level by a factor of more than 100. Even if the phosgene were below the safe level, there would be an unsafe level near the floor since phosgene is denser than air, and the concentration would be much higher in the vicinity of the leak. d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or guidance from an experienced safety officer. He also should have been working under a hood and should have worn a gas mask. 5.30 CH 4 2O 2 o CO 2 2 H 2 O 7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2 C 3 H 8 5O 2 o 3CO 2 4 H 2 O 1450 m3 / h @ 15o C, 150 kPa n 1 (kmol / h) 0.86 CH 4 , 0.08 C 2 H 6 , 0.06 C 3 H 8 n 2 (kmol air / h) 8% excess, 0.21 O 2 , 0.79 N 2
n 1
1450 m3 h
273.2K 288.2K
b1013. 150gkPa 101.3 kPa
1 kmol 22.4 m3 STP
b g
152 kmol h
FG H
IJ OP K PQ
Theoretical O 2 :
LM F MN GH
IJ K
IJ K
FG H
152 kmol 2 kmol O 2 3.5 kmol O 2 5 kmol O 2 0.08 0.06 0.86 h kmol CH 4 kmol C 2 H 6 kmol C 3 H 8 air Air flow: V
b
g
108 . 349.6 kmol O 2 h
1 kmol Air 0.21 kmol O 2
5-13
b g
22.4 m3 STP kmol
349.6 kmol h O 2
b g
4.0 u 104 m3 STP h
5.31 Calibration formulas
bT dP dV dV
25.0; R T
g
0; R p
F
0; R p
A
0; R A
27g Tbq Cg 0.77R 14.2 g b 0i , d P 20.0, R 6i P bkPag 3.33R d m hi 200R 0i , d V 2.0 u 10 , R 10i V d m hi 4000R 0h , d V 1.0 u 10 , R 25i V 14 , T 35.0, R T g
T
r
gauge
3
F
F
5
A
p
F
A
A
3
F
3
A
(m 3 / h), T, P V F g CH 4 2O 2 o CO 2 2 H 2 O 7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2 C 3 H 8 5O 2 o 3CO 2 4 H 2 O 13 C 4 H 10 O 2 o 4CO 2 5H 2 O 2
n F (kmol / h) x A (mol CH 4 / mol) x B (mol C 2 H 6 / mol) x C (mol C 3 H 8 / mol) x D (mol n - C 4 H 10 / mol) x E (mol i - C 4 H 10 / mol) ( m 3 / h) (STP) V A n A (kmol / h) 0.21 mol O 2 / mol 0.79 mol N 2 / mol
n F
d
F m3 h V
i
273.2K
dP 101.3ikPa
1 kmol
g
bT 273.2gK 101.3 kPa d P 101.3i kmol 0.12031V FG IJ T + 273 b g H h K F
b g
22.4 m3 STP
g
Theoretical O 2 :
dn i
o 2 Th
b
c
Air feed: n A
dn i
o 2 Th
kmol O 2 req.
1 kmol air
h
0.21 kmol O 2
FG P IJ dn i H 100K bkmol air hg d22.4 m bSTPg kmoli 4.762 1
A V
n a
gh
n F 2x A 3.5x B 5x C 6.5 x D x E kmol O 2 req. h
x
b1 P
x
g
100 kmol feed
1 kmol req.
o 2 Th
3
b g
22.4n A m 3 STP h
RT T(C) Rp Pg(kPa) Rf xa xb xc xd xe PX(%) nF nO2, th nA Vf (m3/h) Va (m3/h) Ra 25.0 7.25 0.81 0.08 0.05 0.04 0.02 1450 22506.2 5.63 23.1 32.0 7.5 15 72.2 183.47 1004.74 64.3 5.8 0.58 0.31 0.06 0.05 0.00 1160 29697.8 7.42 7.5 20.0 19.3 23 78.9 226.4 1325.8 52.6 2.45 0.00 0.00 0.65 0.25 0.10 490 22022.3 5.51 46.5 50.0 15.8 33 28.1 155.2 983.1 10.0 1200 30439.2 7.6 21 30.4 3 6 0.02 0.4 0.35 0.1 0.13 15 53.0 248.1 1358.9 13.3 1400 29283.4 7.3 23 31.9 4 7 0.45 0.12 0.23 0.16 0.04 15 63.3 238.7 1307.3 16.7 1800 32721.2 8.2 25 33.5 5 9 0.5 0.3 0.1 0.04 0.06 15 83.4 266.7 1460.8 20.0 10 0.5 0.3 0.1 0.04 0.06 2000 37196.7 9.3 27 35.0 6 15 94.8 303.2 1660.6
5-14
5.32 NO 21 O 2 NO 2 1 mol 0.20 mol NO / mol 0.80 mol air / mol 0.21 O 2 0.79 N 2 P0 380 kPa
R| S| T
a.
n1 (mol NO) n 2 (mol O 2 ) n 3 (mol N 2 ) n 4 (mol NO 2 ) Pf (kPa)
U| V| W
Basis: 1.0 mol feed 90% NO conversion: n1
O 2 balance: n 2
0.80(0.21)
018 . mol NO 0.5 mol O 2 mol NO
N 2 balance: n 3
0.80(0.79)
0.632 mol N 2
n4 y NO yO2
018 . mol NO 1 mol NO 2 018 . mol NO 2 n f 1 mol NO mol NO 0.020 mol NO 0.022 0.91 mol mol 0.086
mol O 2 yN2 mol
Pf V n f RT Pf = P0 V n 0 RT b.
0.020 mol NO NO reacted = 0.18 mol
010 . (0.20)
n f = n0
Pf P0
P0
(1 mol)
0.695
nf n0
mol N 2 y NO 2 mol
380 kPa
360 kPa 380 kPa
FG 0.91 mol IJ H 1 mol K
0.0780 mol O 2
n1 n 2 n 3 n 4
0198 .
0.91 mol
mol NO 2 mol
346 kPa
0.95 mol
n i0 X i[
ni
E
n1 (mol NO)
0.20 [
n 2 ( mol O 2 ) (0.21)(0.80) 0.5[ n 3 (mol N 2 ) ( 0.79)(0.80) n 4 ( mol NO 2 ) [ 1 0.5[
nf
0.95 [
010 .
n1 010 mol O 2 , n 3 0.632 mol N 2 , . mol NO, n 2 0118 . n 4 010 . mol NO 2 y NO 0105 . , y O 2 0124 . , y N 2 0.665, y NO 2 NO conversion = P (atm) =
Kp
b0.20 - n g u 100% 1
3.55 atm
(y NO 2 P) ( y NO P )( y O 2 P)
50%
0.20
360 kPa 101.3 kPa atm
0105 .
(y NO 2 ) 0.5
0.5
( y NO )( y O 2 ) P
0.5
5-15
b
0.105
(0105 . ) 0.124
g b3.55g 0.5
1
0.5
151 . atm 2
5.33 Liquid composition: 49.2 kg M 1 kmol 112.6 kg
0.437 kmol M
29.6 kg D 1 kmol 147.0 kg
0.201 kmol D
0.221 kmol D / kmol
21.2 kg B 1 kmol 78.12 kg
0.271 kmol B
0.298 kmol B / kmol
100 kg liquid
0.481 kmol M / kmol
0.909 kmol a.
Basis: 1 kmol C 6 H 6 fed V1 (m 3 ) @ 40 o C, 120 kPa n1 (kmol) 0.920 HCl 0.080 Cl 2 1 kmol C 6 H 6 (78.12 kg) n 0 (kmol Cl 2 )
n 2 (kmol) 0.298 C 6 H 6 0.481 C 6 H 5Cl 0.221 C 6 H 4 Cl 2
C 6 H 6 Cl 2 o C 6 H 5Cl + HCl
1 kmol C 6 H 6
C balance:
n2
6 kmol C 1 kmol C 6 H 6
n 2 0.298 u 6 0.481 u 6 0.221 u 6
100 . kmol
1 kmol C 6 H 6
H balance:
C 6 H 5Cl Cl 2 o C 6 H 5Cl 2 + HCl
6 kmol H 1 kmol C 6 H 6
b
g
n1 0.920 (1)
n 2 0.298 u 6 0.481 u 5 0.221 u 4 n1 100 . kmol
b.
100 . kmol 1013 . kPa 0.08206 m3 atm 313.2 K kmol K 120 kPa 1 atm
n1RT P
V1
217 . m3 7812 . kg B
V1 mB
( m3 / s) V gas
d2 =
0.278 m3 / kg B
u(m / s) A(m 2 )
u
4V Sd 2 gas d2 = S u 4
B0 ( kg B) 0.278 m3 4m s 1 min 104 cm2 kg B S (10) m 60 s min m2
d(cm)
b g
B0 2.43 m
217 . m3
B0 (cm2 ) 5.90m
1 2
c. Decreased use of chlorinated products, especially solvents. 5-16
5.34 Vb ( m3 / min) @900$ C, 604 mtorr 60% DCS conversion n 1 (mol DCS / min) n 2 (mol N 2 O / min) n b (mol / min) n 3 (mol N 2 / min) n 4 (mol HCl(g) / min)
3.74 SCMM
U| V| W
n a ( mol / min) 0.220 DCS 0.780 N 2 O
SiH 2 Cl 2(g) 2 N 2 O (g) o SiO 2(s) 2 N 2(g) 2 HCl (g) a.
n a
3.74 m3 (STP) 103 mol 167 mol / min min 22.4 m3 (STP)
mol DCS I gFGH 0.220 mol JK b167 mol / ming 14.7 mol DCS / min mol DCS DCS reacted: b0.60gb0.220gb167g 22.04 mol DCS reacted / min min
b
60% conversion: n 1 = 1- 0.60
N 2 O balance: n 2
b g molminN O
0.780 167
N 2 balance: n 3 HCl balance: n 4
n B
b.
n B RT P
p DCS
x DCS P =
p N 2O
22.04 mol DCS 2 mol N 2 O mol DCS min
22.04 mol DCS 2 mol N 2 mol DCS min 22.04 mol DCS 2 mol HCl min mol DCS
n 1 n 2 n 3 n 4
B V
2
8618 . mol N 2 O / min
44.08 mol N 2 / min 44.08 mol HCl / min
189 mol / min
189 mol 62.36 L torr 0.001 m3 1173 K min mol K L 0.604 torr
2.29 u 104 m3 / min
n 1 14.7 mol DCS / min 640 mtorr = 49.8 mtorr P= n B 189 mol / min 86.2 mol N 2 O / min n 640 mtorr = 292 mtorr x N 2O P = 2 P = n B 189 mol / min
r = 3.16 u 10-8 p DCS p N 2 O = r t MW c. h(A) U SiO2
0.65
b
gb g
3.16 u 10-8 49.8 292
0.65
6.3 u 10 5
mol SiO 2 m2 s
6.3 u 10 5 mol SiO 2 60 s 120 min 60.09 g / mol 1010 A 2 6 3 m s min 2.25 u 10 g / m 1 m (Table B.1)
= 1.2 u 10 A 5
The films will be thicker closer to the entrance where the lower conversion yields higher pDCS and p N 2 O values, which in turn yields a higher deposition rate.
5-17
5.35 Basis: 100 kmol dry product gas n1 (kmol C x H y ) m1 (kg C x H y ) kmol dry gasU R|100 |V 0.105 CO S|0.053 O |W T0.842 N
(m 3 ) V 2
n 2 (kmol air) 0.21 O 2 0.79 N 2
2
2
2
o
n 3 (kmol H 2 O)
30 C, 98 kPa
a.
N 2 balance: 0.79n 2
b
O balance: 2 0.21n 2 C balance:
d
g
0.842(100) n 2 = 106.6 kmol air
b
g b
n1 kmol C x H y
i x bkmol Cg dkmol C H i x
H balance: n1y = 2n 3
g
2 0.053 n 3 n 3 100 2 0105 .
n 3 13.17
! n1 y
b
1317 . kmol H 2 O
g
b1g
n1x = 10.5 100 0105 .
y
b 2g
26.34
b g b g yx 2610.34.5 2.51 mol H / mol C fed: 0.21b106.6 kmol air g 22.4 kmol in excess = 5.3 kmol Theoretical O = b22.4 - 5.3g kmol = 17.1 kmol
Divide 2 by 1 O2
O2
2
% excess = b.
V2 m1 =
5.3 kmol O 2 u 100% 17.1 kmol O 2
31% excess air
106.6 kmol N 2 22.4 m3 (STP) 1013 . kPa 303 K kmol 98 kPa 273 K
b
g
b
g
n1x kmol C 12.0 kg n1 y kmol H 101 . kg kmol kmol
V2 2740 m3 air = m1 152.6 kg fuel
18.0
2740 m3
n1x=10.5
! m1
n1y=26.34
152.6 kg
m3 air kg fuel
5.36 3N 2 H 4 o 6xH 2 (1 2 x)N 2 (4 4 x)NH 3 a. 0 d x d 1 b.
nN2H4
50 L 0.82 kg 1 kmol L 32.06 kg
128 . kmol
LM 6x kmol H b1 2xg kmol N b4 4xg kmol NH OP 3 kmol N H N 3 kmol N H 3 kmol N H Q 128 . x + 2.13 kmol . b6x 1 2x 4 4xg 1707 3
n product
2
2
. kmol N 2 H 4 128
2
4
5-18
2
4
3
2
4
5.36 (cont’d) nproduct 2.13 2.30 2.47 2.64 2.81 2.98 3.15 3.32 3.50 3.67 3.84
Vp (L) 15447.92 16685.93 17923.94 19161.95 20399.96 21637.97 22875.98 24113.99 25352.00 26590.01 27828.02
Volume of Product Gas 30000.00
25000.00
20000.00 V (L)
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
15000.00
10000.00
5000.00
0.00 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
c.
Hydrazine is a good propellant because as it decomposes generates a large number of moles and hence a large volume of gas.
5.37 A (g A / h) m
c
h
C A (g A / m 3 )
m3 / h V air
a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste containing A poured into sink, A used as cleaning solvent.
A b. m
c. d.
FG kg A IJ H hK
yA
in
mol A mol air
yA 50 u106
d h V air
min
FG kg A IJ H hK C e j V M e j n
A m
out
gA A m3 gA A mol
F I FG GH JK H
3 m C kg A V air A h m3
air
CA
A m PV ; nair kVair RT
!
IJ K
yA
A RT m k Vair M A P
A 90 m . g/ h
A RT m kyA MA P
3
9.0 g / h
d
0.5 50 u 10
6
i
m Pa 8.314 mol 293 K K 3 101.3 u 10 Pa 104.14 g / mol
83 m3 / h
Concentration of styrene could be higher in some areas due to incomplete mixing (high concentrations of A near source); 9.0 g/h may be an underestimate; some individuals might be sensitive to concentrations PEL. e. Increase in the room temperature could increase the volatility of A and hence the rate of air : At higher T, need evaporation from the tank. T in the numerator of expression for V a greater air volume throughput for y to be PEL.
5-19
C3H 6 H 2 C3H 8
5.38 Basis: 2 mol feed gas
235$ C, P2
a. At completion, n p P2 V P1V
35.1 atm
n2
P2 T1 n1 P1 T2
g
2 np
1 mol 508K 32.0 atm 27.3 atm 2 mol 298K
n 2 T2 P1 n1 T1
35.1 atm 298K 2 mol 1.29 mol 32.0 atm 508K
1.29 2 n p n p
b
n p 2(1 n p )
2 1 1 mol
1 mol , n 2
n 2 RT2 P2 n1RT1
b. P2
U| V| W
n p (mol C 3 H 8 ) (1 - n p )(mol C 3 H 6 ) n 2 (1 - n p )(mol H 2 )
1 mol C 3 H 6 1 mol H 2 25$ C, 32 atm
0.71 mol C 3 H 8 produced
1- 0.71
0.29 mol C 3 H 6 unreacted 71% conversion of propylene
P2 (atm) 27.5 28.0 29.5 30.0 31.5 32.0 33.0 33.5 34.0 34.5 35.0 37.0 39.0 40.0
C3H8 prod. 0.99075 0.9724 0.91735 0.899 0.84395 0.8256 0.7889 0.77055 0.7522 0.73385 0.7155 0.6421 0.5687 0.532
c. %conv. 99.075 97.24 91.735 89.9 84.395 82.56 78.89 77.055 75.22 73.385 71.55 64.21 56.87 53.2
Pressure vs Fraction Conversion 120
100
80 % conversion
n2 1.009 1.028 1.083 1.101 1.156 1.174 1.211 1.229 1.248 1.266 1.285 1.358 1.431 1.468
60
%conv.
40
20
0 25.0
27.0
29.0
31.0
33.0 Pressure (atm)
5-20
35.0
37.0
39.0
41.0
5.39
Convert fuel composition to molar basis Basis: 100 g
b b
UV W
g g
97.2 mol % CH 4 5.92 mol CH 4 2.8 mol % C 2 H 6 . mol C 2 H 6 017
95 g CH 4 1 mol 16.04 g 5 g C2 H 6 1 mol 30.07 g
500 m3 / h
n 2 (kmol CO 2 / h) n 3 (kmol H 2 O / h)
n 1 (mol / h) 0.972 CH 4
n 4 (kmol O 2 / h) n 5 (kmol N 2 / h)
0.028 C 2 H 6 40$ C, 1.1 bar
(SCMH) V air 25% excess air
n 1
1 P1V RT1
. bar 500 m3 11 313K h
kmol K 0.08314 m3 bar
7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2
CH 4 2O 2 o CO 2 2 H 2 O Theoretical O 2 =
. kmol h 211
LM N
21.1 kmol 0.972 kmol CH 4 kmol h
0.028 kmol C 2 H 6 kmol
b
125 . 431 . kmol O 2 Air Feed: h
g
3.5 kmol O 2 1 kmol C 2 H 6
1 kmol Air 0.21 kmol O 2
5.40 Basis: 1 m3 gas fed @ 205q C, 1.1 bars Ac 1 m 3 @205$ C, 1.1 bar
2 kmol O 2 1 kmol CH 4
OP Q
431 .
b g
22.4 m3 STP 1 kmol
kmol O 2 h
5700 SCMH
acetone
n 3 (kmol), 10$ C, 40 bar
condenser
n1 (kmol) y1 (kmol Ac / kmol) (1 - y 1 )(kmol air / kmol) p AC 0.100 bar
y 3 (kmol Ac / kmol) (1- y 3 )(kmol air / kmol) p AC 0.379 bar
n 2 (kmol Ac(l))
a.
n1 y1
100 . m3
273K
1 kmol 110 . bars 478K 10132 bars 22.4 m3 STP .
bar 0100 . 1.1 bars
b g
0.0909 kmol Ac kmol , y 3
b
gb
g
Air balance: 0.0277 0.910 Mole balance: 0.0277 Acetone condensed
0.0277 kmol
0.379 bar 40.0 bars
(1 9.47 u 10 3 )n 3 n 3
0.0254 n 2 n 2
0.0254 kmol
0.0023 kmol Ac condensed
0.0023 kmol Ac 58.08 kg Ac 1 kmol Ac
5-21
9.47 u 10 3 kmol Ac kmol
0.133 kg acetone condensed
5.40 (cont’d)
b g
0.0254 kmol 22.4 m3 STP
Product gas volume
b.
. 283K 10132 bars 0.0149 m3 273K 40.0 bars
20.0 m3 effluent 0.0277 kmol feed 0.0909 kmol Ac 58.08 kg Ac 196 kg Ac h h 0.0149 m3 effluent kmol feed kmol Ac
5.41 Basis: 1.00 u 10 6 gal. wastewater day. Neglect evaporation of water. 1.00 u 10 6 gal / day
Effluent gas: 68$ F, 21.3 psia(assume)
n1 (lb - moles H2 O / day) 0.03n1 (lb - moles NH3 / day)
n2 (lb - moles air / day) n3 (lb - moles NH 3 / day)
300 u 10 6 ft 3 air / day
Effluent liquid
n1 (lb - moles H2 O / day) n4 (lb - moles NH3 / day)
$
20 C, 21.3 psia n2 (lb - moles air / day)
a.
Density of wastewater: Assume U
LMn N
62.4 lb m ft 3
OP Q
lb - moles H 2 O 18.02 lb m 0.03n1 lb m NH 3 17.03 lb m 1 ft 3 u day 1 lb - mole day 1 lb - mole 62.4 lb m gal 100 . u 106 day 1
n1 n2
450000 lb - moles H 2 O fed day , 0.03n1
13500 lb - moles NH 3 fed day
. psi 1 lb - mole 300 u 106 ft 3 492 $ R 213 $ day 527.7 R 14.7 psi 359 ft 3 STP
b g
93% stripping: n 3
7.4805 gal 1 ft 3
. u 106 lb - moles air day 113
0.93 u 13500 lb - moles NH 3 fed day 12555 lb - moles NH 3 day
Volumetric flow rate of effluent gas PVout PVin
n out RT Vout n in RT
Vin
n out n in
300 u 106 ft 3
d1.13 u 10
6
i
12555 lb - moles day 6
1.13 u 10 lb - moles day
day
303 u 106 ft 3 day Partial pressure of NH 3
y NH 3 P
12555 lb - moles NH 3 day
d1.129 u 10
0.234 psi
5-22
6
i
12555 lb - moles day
u 213 . psi
5.42 Basis: 2 liters fed / min Cl ads.=
2.0 L soln 1130 g 012 . g NaOH 1 mol 0.23 NaOH ads. 1 mol Cl 2 L g soln 40.0 g mol NaOH 2 mol NaOH 60 min
2 L / min @ 23$ C, 510 mm H 2 O n 1 (mol / min) y (mol Cl 2 / mol) (1 - y)(mol air / mol)
0.013
mol min
n 2 (mol air / mol)
0.013 mol Cl 2 / min Assume Patm
n 1
2L
b g b10.33 0.510g m H O
10.33 m H 2 O Pabs
273K 10.84 m H 2 O
1 mol
b g
0.013 y
0150 .
mol Cl 2 ,? specification is wrong mol
125 L / min @ 25o C, 105 kPa n 1 ( mol / min) y1 (mol H 2 O / mol) 1- y1 ( mol dry gas / mol) 0.235 mol C 2 H 6 / mol DG 0.765 mol C 2 H 4 / mol DG
R|b g S| T
10.84 m H 2 O
0.0864 mol min
min 296K 10.33 m H 2 O 22.4 L STP
Cl balance: 0.0864y
5.43
2
in
(L / min) @ 65o C, 1 atm V 3 n C 2 H 6 (mol C 2 H 6 / min) n C 2 H 4 (mol C 2 H 4 / min) n air (mol air / min) n H 2 O (mol H 2 O / min)
U| V| W
355 L / min air @ 75o C, 115 kPa n 2 (mol / min) y 2 ( mol H 2 O / mol) (1- y 2 )( mol dry air / mol)
a.
Hygrometer Calibration ln y b ln a
b. n 1 n 2 R1
b
ln y1 y 2 R 2 R1
g lnd0.2 10 i
dy
bR ln a
i
4
90 5
ln y1 bR 1
0.08942
bg
ln 10 4 0.08942 5 a
125 L 273K 105 kPa 1 mol min 298K 101 kPa 22.4 L STP
b g
355 L 273K 115 kPa min
ae bR
1 mol
b g
348K 101 kPa 22.4 L STP
86.0 o y1
0.140 , R 2
12.8 o y 2
5-23
6.395 u 105 y 6.395 u 105 e 0.08942R
5.315 mol min wet gas 14.156 mol min wet air
2.00 u 104 mol H 2 O mol
5.43 (cont’d)
C H I b5.315 mol mingFGH b1 0.140g molmolDG IJK FGH 0.235 mol J mol DG K 2
C 2 H 6 balance: n C2 H 6
6
1.07 mol C 2 H 6 min
b5.315gb0.860gb0.765g 3.50 mol C H min Dry air balance: n b14.156gd1 2.00 u 10 i 14.15 mol DA min Water balance: n b5.315gb0.140g b14.156gd1.00 u 10 i 0.746 mol H O min n b1.07 3.50 14.15g mol min 18.72 mol min , n b18.72 0.746g 19.47 mol min 19.47 mol min 22.4 L bSTPg 338K 540 liters min V C 2 H 4 balance: n C2 H 4
2
4
4
air
4
H 2O
2
dry product gas total
3
mol
Dry basis composition:
c.
273K
FG 1.07 IJ u 100% H 18.72 K
5.7% C 2 H 6 , 18.7% C 2 H 4 , 75% dry air
0.746 mol H 2 O u 1 atm 0.03832 atm 19.47 mol
p H 2O
y H 2Ol P
y H 2O
0.03832 R
FG H
1 0.03832 ln 0.08942 6.395 u 10 5
IJ K
71.5
5.44 CaCO 3 o CaO CO 2 n CO 2
1350 m3 273K 1 kmol h 1273K 22.4 m3 STP
b g
12.92 kmol CO 2
12.92 kmol CO 2 h
1 kmol CaCO 3 100.09 kg CaCO 3
h
1 kmol CO 2
1 kmol CaCO 3
1362 kg limestone
0.17 kg clay
h
0.83 kg limestone
279 kg clay h
Weight % Fe 2 O 3 kg Fe 2 O 3 kg clay
b g
279 0.07 u 100% 18% Fe 2 O 3 . 1362 279 12.92 44.1 kg limestone kg clay
b g
kg CO 2 evolved
5-24
1 kg limestone 0.95 kg CaCO 3
1362 kg limestone h
5.45 Basis:
R|864.7 g C b1 mol 12.01 gg 72.0 mol C |116.5 g H b1 mol 1.01 gg 115.3 mol H 1 kg Oil S ||13.5 g S b1 mol 32.06 gg 0.4211 mol S T5.3 g I
C O 2 o CO 2 1 C + O 2 o CO 2 S O 2 o SO 2 1 2H O 2 o H 2 O 2
5.3 g I n 1 (mol CO 2 ) n 2 (mol CO) n 3 (mol H 2 O) n 4 (mol SO 2 ) n 5 (mol O 2 ) n 6 (mol N 2 )
72.0 mol C 115.3 mol H 0.4211 mol S 5.3 g I
n a (mol), 0.21 O 2 , 0.79 N 2 15% excess air 175$ C, 180 mm Hg (gauge)
a.
Theoretical O 2 : (0.95)72.0 mol C 1 mol O 2 (0.05)72.0 mol C 0.5 mol O 2 1 mol C 1 mol C 115.3 mol H 0.25 mol O 0.4211 mol S 1 mol O 2 2 1 mol H 1 mol S Air Fed:
e
1.15 99.45 mol O
2
j
b g
544 mol Air 22.4 liter STP 1 kg oil mol
1 mol Air 0.21 mol O
544 mol Air
n
2
99.45 mol O
2
a
1 m3 448K 760 mm Hg 16.2 m3 air kg oil 3 10 liter 273K 940 mm Hg
b. S balance: n 4 0.4211 mol SO 2 H balance: 115.3 = 2n 3 n 3 57.6 mol H 2 O C balance:
b g
0.95 72.0 = n1 n1 0.05(72.0)
n2
68.4 mol CO 2
3.6 mol CO
429.8 mol N b g O balance: 0.21b544g2 = 57.6 + 3.6 + 2(68.4) + 2b0.4211g + 2n n Total moles bexcluding inertsg wet: 559.8 mols dry: 502.2 mols N 2 balance: 0.79 544 = n 6 n 6
2
5
dry basis:
3.6 mol CO 502.2 mol
wet basis:
3.6 mol CO u 10 6 559.8 mol
7.2 u 10 3
mol CO 0.4211 mol SO 2 , mol 502.2 mol
6400 ppm CO ,
5-25
5
14.8 mol O 2
8.4 u 10 4
0.4211 mol SO 2 u 10 6 559.8 mol
mol SO 2 mol
750 ppm SO 2
bg
5.46 Basis: 50.4 liters C 5 H 12 l min 50.4 L C 5 H 12 ( l ) / min
b
n 1 kmol C 5 H 12 / min
g
heater
(L / min) 15% excess air, V air n 2 kmol air 0.21 O 2 0.79 N 2 336 K, 208.6 kPa (gauge)
a f n b kmol air / ming
n 1 kmol C5 H 12 / min
Combustion chamber
2
n 3 (kmol C 5 H 12 / min) n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min) n 7 (kmol H 2 O / min) (L / min) V gas
Condenser
(L / min), 275 K, 1 atm V liq = 3.175 kg C 5O12 / min m n 3 (kmol C 5O12 / min) n 7 (kmol H 2 O(l ) / min)
C5 H 12 802 o 5CO 2 6H 2 O
a.
n 1
n 3
50.4 L 0.630 kg min
. kg 1 kmol 3175 min 72.15 kg
frac. convert = n 2 air V
L
1 kmol 72.15 kg
0.440 kmol min
0.044 kmol / min
0.440 - 0.044 kmol u 100 90% C5 H12 converted 0.440
b
0.440 kmol C 2 H 12 1.15 8 kmol O 2 min kmol C 5 H 12
b g
19.28 kmol 22.4 L STP min mol
g
1 mol air 0.21 mol O 2
19.28 kmol air min
336K
101 kPa 1000 mol 273K 309.6 kPa kmol
n 4
0.440 kmol C 2 H12 0.15(8 kmol O 2 ) min kmol C 2 H12
n 5
19.28 mol air 0.79 mol N 2 min mol air
n 6
0.90(0.440 kmol C5 H12 ) 5 kmol CO 2 min kmol C5 H 12
gas V
n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min)
173000 L min
0.528 kmol O 2
15.23 kmol N 2 / min
. kmol CO 2 / min 198
0.528 + 15.23 + 1.98 kmol 22.4 L(STP) 548 K 1000 mol min mol 273 K kmol
5-26
7.98 u 105 L / min
5.46 (cont’d) 0.9(0.440 kmol C5 H12 ) 6 kmol H 2 O min kmol C5 H 12
n 7
2.38 kmol H 2 O(l ) / min
Condensate: 0.044 kmol 72.15 kg
V C5 H 12 V H 2O
min
kmol
2.38 kmol 18.02 kg min
L
kmol
5.04 L min
0.630 kg L
42.89 L min
1 kg
Assume volume additivity (liquids are immiscible) liq 5.04 42.89 47.9 L min V b.
bg
C5 H 12 l
C 5 H 12 ( l )
bg
H 2O l
bg
H2O l
5.47 n air (kmol / min), 25$ C, 1 atm 0.21 O 2 0.79 N 2 n 0 (kmol / min)
n 1 (kmol H 2S / min) Furnace
H 2 S + 23 O 2 o SO 2 H 2 O
0.20 kmol H 2 S / mol 0.80 kmol CO 2 / mol
Reactor
n 2 (kmol H 2 S / min)
2H 2 S + SO 2 o 3S(g) 2 H 2 O
10.0 m 3 / min @ 380$ C, 205 kPa n exit (kmol / min) n 3 (kmol N 2 / min) n 4 (kmol H 2 O / min) n 5 (kmol CO 2 / min) n 6 (kmol S / min)
n exit n 1
PV RT
b0.20gn
205 kPa 10.0 m3 / min m3 kPa 653 K 8.314 kmol K 0
/ 3 = 0.0667n 0 ; n 2
0.377 kmol / min
2 n 1 = 0.133n 0
5-27
5.47 (cont’d) 0.0667 n 0 (kmol H 2 S fed) 15 . kmol O 2 1 kmol air 1 kmol H 2 S 0.21 kmol O 2 (min)
Air feed to furnace: n air
0.4764 n 0 kmol air / min 0.4764 n 0 (kmol air) 0.79 kmol N 2 (min) min
Overall N 2 balance: n 3
Overall S balance: n 6
0.200n 0 (kmol H 2S) 1 kmol S (min) 1 kmol H 2S
Overall CO 2 balance: n 5 Overall H balance:
0.3764 n 0 ( kmol N 2 / min)
0.200n 0 (kmol S / min)
0.800n 0 (kmol CO 2 / min)
0.200n 0 (kmol H 2 S) 2 kmol H (min) 1 kmol H 2 S
n 4 kmol H 2 O 2 kmol H min 1 kmol H 2 O
n 4 = 0.200n 0 (kmol H 2 O / min) n exit
b
g
n 0 0.376 0.200 + 0.200 + 0.800 = 0.377 kmol / min n 0 = 0.24 kmol / min
n air = 0.4764(0.24 kmol air / min)
0114 . kmol air / min
5.48 Basis: 100 kg ore fed 82.0 kg FeS 2 (s), 18.0 kg I.
b
gb
n FeS2 fed = 82.0 kg FeS2 1 kmol / 120.0 kg
100 kg ore 06833 . kmol FeS2 18 kg I
g
0.6833 kmol FeS2
Vout m3 (STP)
40% excess air n 1 (kmol) 0.21 O 2 0.79 N 2 V1 m 3 (STP)
n 2 (kmol SO 2 ) n 3 (kmol SO 3 ) n 4 (kmol O 2 ) n5 (kmol N 2 )
m6 (kg FeS2 ) m7 (kg Fe 2 O 3 ) 18 kg I
2 FeS 2(s) 112 O 2(g) o Fe 2 O 3(s) 4SO 2(g) 2 FeS 2(s) 152 O 2(g) o Fe 2 O 3(s) 4SO 3(g) a.
n1
0.6833 kmol FeS2 7.5 kmol O 2 1 kmol air req'd 140 . kmol air fed 2 kmol FeS2 0.21 kmol O 2 kmol air req'd
V1
b17.08 kmolgb22.4 SCM / kmolg
n2
(0.85)(0.40)0.6833 kmol FeS2 4 kmol SO 2 2 kmol FeS2
382 SCM / 100 kg ore
5-28
0.4646 kmol SO 2
17.08 kmol air
5.48 (cont’d) n3
(0.85)(0.60)0.6833 kmol FeS2 4 kmol SO 2 2 kmol FeS2
n4
b0.21 u 17.08gkmol O
n5 Vout
2
fed
0.6970 kmol SO 3
.4646 kmol SO 2 55 . kmol O 2 4 kmol SO 2
.702 kmol SO 3 7.5 kmol O 2 4 kmol SO 3
1633 . kmol O 2
b0.79 u 17.08g kmol N 13.49 kmol N = b0.4646 + 0.6970 + 1.633 + 13.49 g kmol 2
2
22.4 SCM (STP) / kmol
365 SCM / 100 kg ore fed ySO 2
0.4646 kmol SO 2 u 100% 16.285 kmol
2.9%; y SO3
4.3%; y O 2
10.0%; y N 2
82.8%
b.
e j
Product gas, T o C Converter
0.4646 kmol SO 2 SO 3 0.697 kmol 21 OO2 2 SO SO 3 . 2 kmol 1633 13.49 kmol N 2
n SO 2 (kmol) n SO 3 (kmol) n O 2 (kmol) n N 2 (kmol)
Let [ (kmol) = extent of reaction
n SO 2 0.4646 [ n SO 3 0.697 [ 21 [ n O 2 1633 . n N 2 13.49 n = 16.28 - 21 [ K p (T) =
U| |V y || y W
SO 2
1 2
P = 1 atm, T = 600o C, K p n SO 2
1 2
-1
9.53 atm 2 [
kmol fSO2 01847 .
P = 1 atm, T = 400o C, K p n SO 2
c h [h (0.4646 [ )c1633 . (0.697 [ ) 16.28 21 [
P y SO 3 P y SO 2 ( P y O 2 )
O2
0.4646 [ 0.697 [ , y SO 3 1 16.28 - 2 [ 16.28 - 21 [ 1 2[ . 1633 13.49 , y N2 1 16.28 - 2 [ 16.28 - 21 [
1 2
P
- 21
1
K p (T) P 2
0.2799 kmol
. b0.4646 01847 g kmol SO
2
0.4646 kmol SO 2 fed -1
397 atm 2 [
0.00059 kmol fSO 2
1 2
reacted
0.602
0.4587 kmol
0.987
The gases are initially heated in order to get the reaction going at a reasonable rate. Once the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium conversion of SO2.
5-29
5.48 (cont’d) c.
SO 3 leaving converter: (0.6970 + 0.4687) kmol = 1.156 kmol 1.156 kmol SO 3 1 kmol H 2 SO 4 98 kg H 2 SO 4 min kmol 1 kmol SO 3
Sulfur in ore:
0.683 kmol FeS 2 2 kmol S 32.1 kg S kmol FeS2 kmol
113.3 kg H 2SO 4 43.8 kg S
100% conv.of S:
438 . kg S
kg H 2SO 4 kg S
2.59
0.683 kmol FeS 2 2 kmol S 1 kmol H 2SO 4 98 kg kmol FeS2 1 kmol S kmol
133.9 kg H 2SO 4 43.8 kg S
113.3 kg H 2 SO 4
3.06
133.9 kg H 2SO 4
kg H 2SO 4 kg S
The sulfur is not completely converted to H2SO4 because of (i) incomplete oxidation of FeS2 in the roasting furnace, (ii) incomplete conversion of SO2 to SO3 in the converter. 5.49 N 2 O 4 2 NO 2 a.
n0
b.
n1
p NO 2
dP
gauge
i
b2.00 atmgb2.00 Lg b473Kgb0.08206 L atm mol - Kg
1.00 V
RT0 mol NO 2 , n 2
y NO 2 P
mol N 2 O 4
FG n IJ P , p Hn n K 1
1
2
N 2O4
Ideal gas equation of state PV
0.103 mol NO 2
FG n IJ P K n P n bn n g Hn n K bn n gRT n n PV / RT b1g 2
p
1
1
2 1
2
2
2
1
1
2
2
Stoichiometric equation each mole of N 2 O 4 present at equilibrium represents a loss of two moles of NO 2 from that initially present n1 2n 2 Solve (1) and (2) n1
b3g ,
2(PV / RT) 0.103
n2
0.103
b2 g
b4 g
0.103 (PV / RT)
Substitute (3) and (4) in the expression for K p , and replace P with Pgauge 1 t
gauge
t
T(K) 350 335 315 300
Pgauge(atm) 0.272 0.111 -0.097 -0.224
t
nt Kp(atm) 0.088568 5.46915 0.080821 2.131425 0.069861 0.525954 0.063037 0.164006
i
1 where n t
(1/T) ln(Kp) 0.002857 1.699123 0.002985 0.756791 0.003175 -0.64254 0.003333 -1.80785
dP
gauge
i
1 V
RT
nt
i
T
V 2L
2
y = -7367x + 22.747 R2 = 1
1 0 -1 -2 0.0028
0.003
0.0032 1/T
5-30
d
24.37 Pg 1
Variation of Kp with Temperature
ln Kp
b2n 0.103g dP n b0.103 n g 2
Kp
0.0034
5.49 (cont’d) c.
A semilog plot of K p vs.
1 is a straight line. Fitting the line to the exponential law T
yields ln K p
7367 22.747 K p T
7.567 u 109 exp
FG 7367 IJ a 7.567 u 10 H T K b = 7367K
9
atm
10.00 atm
5.50 n 1 (kmol A / h) n 2 (kmol H 2 / h)
5.00 kmol S / h n 4 (kmol A / h) n 5 (kmol H 2 / h)
3n 3 (kmol A / h) n 3 (kmol H 2 / h)
5.00 kmol S / h
n 4 (kmol A / h) n 5 (kmol H 2 / h)
Vrcy (SCMH)
Extent of reaction equations: n i
n i0 Q i[
A + H2 l S A:
n 4
H 2 : n 5 S:
3n 3 [ n [ 3
5.00 = [
! n 4 n 5 n tot
U| V| p 10.0 W
3n 3 5.00 n 3 5.00 4 n 3
A
pH2 pS Kp n 4 n 5 rcy V
pS pA p H2
b
g
5.00 4n 3 10.0 10.0 3n 3 5.00 n 3 5.00
b
gb
g
yA P =
n 4 P n tot
y H2 P = yS P =
n 3 . 0100
n 5 P n tot
3n 3 - 5.00 10.0 4 n 3 10.0 n 3 - 5.00 10.0 4 n 3 10.0
5.00 10.0 4 n 3 10.0 . kmol H 2 / h 1108
3(1108 . ) - 5.00 28.24 kmol A / h 1108 . 5.00 6.08 kmol H 2 / h
b28.24 + 6.08g kmol / h d22.4 m (STP) / kmoli 3
5-31
769 SCMH
5.51 n 4 (kmol CO / h) n 5 (kmol H 2 / h)
Reactor
Separator
100 kmol CO / h
n 1 (kmol CO / h) n 2 (kmol H 2 / h)
H xs (% H 2 excess)
a.
n 6 (kmol M / h)
n 4 (kmol CO / h) n 5 (kmol H 2 / h) n 6 (kmol M / h) T, P
n 3 (kmol H 2 / h) T (K), P (kPa)
Balances on reactor four equations in n 3 , n 4 , n 5 , and n 6 1 kmol CO CO balance n 4 100 kmol CO / h n 6 n 4 100 n 6 1 kmol M 100 kmol CO 1 kmol CO H 2 balance n 3 105 . 210 kmol H 2 / h h 1 kmol M 2 kmol CO n 5 210 kmol H 2 / h n 6 n 5 210 2 n 6 1 kmol M n T n 4 n 5 n 6 100 n 6 210 2 n 6 n 6 310 2 n 6
b g
b1g
b g
b g
b
b
K p T = 500K
g
u 10 4 1390 .
b 2g g b g b3g F 21.225 + 9143.6 7.492lnb500Kg I G exp GH +4.076 u 10500bK500Kg -1.161 u 10 b500Kg JJK -3
2
-8
9.11 u 10 7 kPa -2 Kp
yMP
d
y CO P y H 2 P
i
2
Kp P2
yM
d i
y CO y H 2
b
(1) ( 3) 2
!
n 6 310 2 n 6
g
b100 n g b210 2n g b310 2n g b310 2n g n b310 2 n g b100 n gb210 2n g 6
6
Kp P2
b
9.11 u 10 7 kPa -2 5000 kPa
g
22.775
6
6
2
6
Solving for n 6 n 6
100 n 6
n 5
210 2 n 6
n 1 n 2
rec V
bn
4
n 5
75.7 kmol M / h
n 4
24.3 kmol CO / h 58.6 kmol H 2 / h
1 kmol CO n 6 75.7 kmol CO / h 1 kmol M 2 kmol H 2 n 6 151 kmol H 2 / h 1 kmol M 3
m (STP) g 22.4 kmol
6
2
2
1860 SCMH
5-32
2
6
6
2
5.51 (cont’d) b. P(kPa) 1000 5000 10000 5000 5000 5000 5000 5000 5000
`
T(K) Hxs(%) 500 5 500 5 500 5 400 5 500 5 600 5 500 0 500 5 500 10
Kp(T)E8 9.1E+01 9.1E+01 9.1E+01 3.1E+04 9.1E+01 1.6E+00 9.1E+01 9.1E+01 9.1E+01
KpP^2 0.91 22.78 91.11 7849.77 22.78 0.41 22.78 22.78 22.78
KpP^2- n1(kmol KpcP^2 CO/h) 1.3E-05 25.55 2.3E+01 9.00 4.9E-03 86.72 3.2E-08 98.93 3.4E-03 75.68 -2.9E-04 14.58 9.8E-03 73.35 3.4E-03 75.68 -3.1E-03 77.77
ntot n6(kmol M/h) (kmol/h) KpcE8 25.55 258.90 9.1E-01 9.00 292.00 2.3E-01 86.72 136.56 9.1E+01 98.93 112.15 7.8E+03 75.68 158.64 2.3E+01 14.58 280.84 4.1E-01 73.35 153.30 2.3E+01 75.68 158.64 2.3E+01 77.77 164.45 2.3E+01
n3(kmol n4(kmol n5(kmol H2/h) H2/h) CO/h) 210 74.45 158.90 210 91.00 192.00 210 13.28 36.56 210 1.07 12.15 210 24.32 58.64 210 85.42 180.84 200 26.65 53.30 210 24.32 58.64 220 22.23 64.45 n2(kmol H2/h) 51.10 18.00 173.44 197.85 151.36 29.16 146.70 151.36 155.55
Vrec (SCMH) 5227 6339 1116 296 1858 5964 1791 1858 1942
c. Increase yield by raising pressure, lowering temperature, increasing Hxs. Increasing the pressure raises costs because more compression is needed. d. If the temperature is too low, a low reaction rate may keep the reaction from reaching equilibrium in a reasonable time period. e. Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of methanol, not steady-state measurement errors. 5.52
CO 2 CO + 21 O 2
1.0 mol CO 2 1.0 mol O 2 1.0 mol N 2 T = 3000 K, P = 5.0 atm
dp
CO p O 2
d
1 A B C 2 1 1 C D E 2 2
1/ 2
i
0.3272 atm p CO 2 1 O 2 12 N 2 NO 2 p NO 0.1222 K2 1/ 2 p N 2 p O2 K1
1/ 2
i
A CO 2 , B CO , C O 2 , D N 2 , E NO n A0
n C0
n D0
1 , n B0
n E0
5-33
0
[ 1 - extent of rxn 1 [ 2 - extent of rxn 2
5.52 (cont’d) 1 [1 [1 1 1 1 [1 [ 2 2 2 1 1 [2 2 [2 1 6 [1 3 [1 2 2
nA nB nC nD nE n tot
U| || y n n 2 b1 [ g b 6 [ g | yy b22[ [b 6 [[ gg b 6 [ g V| || yy 2b 2[ [b 6 g b[6 g [ g || W A B
A
tot
1
1
C
1
D
2
1
y B y1C2 b1 12 1g p yA
CO 2
1
b
2[ 1 2 [ 1 [ 2
12
12
12
1
K2
d
p NO pO2 p N 2
b
i
12
12
1
yE 12 12 yC y D
01222 2 [1 [ 2 .
1
p11 2 1 2
g b2 [ g 12
12
2
1
b2 [
5.53 a.
b
2[ 2 1
[2
g b2 [ g 12
12
01222 .
2
2[ 2
(2)
g b6 [ g
2 1 [1 yB yC
(1)
2
Solve (1) and (2) simultaneously with E-Z Solve [ 1
yA
0.3272
1
12
1
yiP
1
2
g b5g K p 2b1 [ gb6 [ g 0.3272b1 [ gb6 [ g 2.236[ b2 [ [ g p CO p1O22
pi
1
2
E
1
1
0.2574 mol CO 2 mol y D yE 0.0650 mol CO mol 0.3355 mol O 2 mol 1
0.20167, [ 2
012081 . ,
0.3030 mol N 2 mol 0.0390 mol NO mol
n 4 (kmol / h)
0.04 O 2 0.96 N 2
PX = C8 H 10 , TPA = C8 H 6 O, S = Solvent (m 3 / h) @105o C, 5.5 atm V 3 n 3O (kmol O 2 / h) n 3N (kmol N 2 / h) n 3W (kmol H 2 O(v) / h) (m 3 / h) at 25o C, 6.0 atm V 2 n 2 (kmol / h) 0.21 O 2 0.79 N 2
condenser
n 3W (kmol H 2 O(v) / h) (m 3 / h) V 3W
reactor
n 1 (kmol PX / h) ( n 1 n 3p ) kmol PX / h s (kg S / h) m 3 kg S / kg PX
5-34
n 3p ( kmolPX / h) 100 kmol TPA / h s (kg S / h) m
separator
100 mol TPA / s
n 3p (kmol PX / h) s (kg S / h) m
5.53 (cont’d) b. Overall C balance: n 1
c.
FG kmol PX IJ 8 kmol C H h K kmol PX
O 2 consumed =
100 kmol TPA 8 kmol C n 1 h kmol TPA
100 kmol TPA 1.5 kmol O 2 h 1 kmol TPA
150
Overall N 2 balance: 0.79n 2 Overall H 2 O balance: n 3W
2 V 3 V
n 2 RT P
bn
3W V
3W
150 kmol O 2 / h
kmol O 2 + 0.04n 4 h 0.96n 4
Overall O 2 balance: 0.21n 2
g
U| V| W
n 2 n 4
100 kmol TPA 2 kmol H 2 O h 1 kmol TPA
848 kmol 0.08206 m3 atm 298 K h kmol K 6.0 atm
n 4 RT P
100 kmol PX / h
200 kmol H 2 O / h
3450 m3 air / h
b200 + 698g kmol 0.08206 m atm 378 K 3
kmol K 5.5 atm
h
200 kmol H 2 O (l) 18.0 kg 1 m3 h kmol 1000 kg
d
b100 + 11.1g kmol PX h
5065 m3 / h
3.60 m3 H 2 O(l) / h leave condenser
i
n 1 100
d. 90% single pass conversion n 3p = 0.10 n 1 n 3p ====> n 3p recycle m
848 kmol air / h 698 kmol / h
106 kg 4 kg recycle kg PX 1 kmol PX
111 . kmol PX / h
4.71 u 104 kg recycle / h
e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the air. The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX. f. The stream can be allowed to settle and spearate into water and PX layers, which may then be separated. 5.54
n 1 (kmol CO / h), n 3 (kmol H 2 / h), 0.10n 2 (kmol H 2 / h)
Separator n 6 (kmol CO / h) n 7 (kmol H 2 / h) n 8 (kmol CO 2 / h) 2 kmol N 2 / h
0.90 n 2 2 kmol N 2 / h
n 1 , n 2 , n 3 2 kmol N 2 / h 0.300 kmol CO / kmol 0.630 kmol H 2 / kmol 0.020 kmol N 2 / kmol 0.050 kmol CO 2 / kmol
Reactor
n 1 (kmol CO / h) n 2 (kmol H 2 / h) n 3 (kmol CO 2 / h) n 4 (kmol M / h) n 5 (kmol H 2 O / h) 2 5-35 kmol N 2 / h
Separator
n 4 (kmol M / h) n 5 (kmol H 2 O / h)
5.54 (cont’d) CO + 2H 2 CH 3OH(M) CO 2 3H 2 CH 3OH + H 2 O
a.
Let [ 1 ( kmol / h)
extent of rxn 1, [ 2 ( kmol / h)
extent of rxn 2
CO: n 1 = 30 - [ 1 H 2 : n 2 = 63 - 2[ 1 3[ 2
U| M: n = [ [ || Py Py H O: n = [ V| dK i P y P Py y , dK i bP y gd P y i d i d id i N : n = 2 || n 100 - 2[ 2[ W n 2[ 2[ g n 84.65 (1) dK i P = n F n I bb[30[[ gbgb100 63 2[ 3[ g G J n H n K FG n IJ FG n IJ . (2) dK i P = FH nn IKFH nn IK [ bb[5 [[ gbgb63100 2[2[ 3[2[g g 1259 GH n JK GH n JK CO 2 : n 3 = 5 - [ 2 2 2
4
1
5
2
2
M
p 1
CO
N2
tot
p 2
H2
2
1
tot
2
1
2
2
1
2
tot
tot
2
CO 2
3
H2
4
p 1
H 2O
2
1
2
M
p 2
2
2
1
4
5
tot
tot
3
2
tot
tot
1
2
2
2
2
1
1
2 2
3
2
1
2
Solve (1) and (2) for [ 1 , [ 2 [ 1 = 25.27 kmol / h [ 2 = 0.0157 kmol / h
n 1
30.0 25.27
4.73 kmol CO / h
9.98% CO
n 2
63.0 2(25.27) 3(0.0157) 12.4 kmol H 2 / h
26.2% H 2
n 3
5.0 0.0157
10.5% CO 2
n 4
25.27 0.0157
n 5
0.0157
n total
4.98 kmol CO 2 / h
0.0157 kmol H 2 O / h
49.4 kmol / h
C balance: n 4 25.3 kmol / h O balance: n 6 2 n 8 n 4 n 5 H balance: 2n 7 b. (n 4 ) process
53.4% M 0.03% H 2 O
UV n 25.4 kmol CO / h 25.44 mol / sW n = 0.02 kmol CO / h
2(0.9 n 2 ) 4 n 4 2 n 5
6
8
123.7 n 7
237 kmol M / h
Scale Factor =
25.3 kmol M / h
237 kmol M / h 25.3 kmol / h
5-36
2
618 . mol H 2 / s
5.54 (cont’d)
237 kmol / h I F 22.4 m (STP) I gFGH 25.3 J kmol JK 18,700 SCMH kmol / h K GH F 237 kmol / h IJ 444 kmol / h Reactor effluent flow rate: b 49.4 kmol / hgG H 25.3 kmol / sK F kmol IJ FG 22.4 m (STP) IJ 9946 SCMH V G 444 H h K H kmol K 3
b
Process feed: 25.4 618 . 0.02 2.0
3
std
. kPa 9950 m 3 (STP) 473.2 K 1013 h 273.2 K 4925 kPa
Vactual c.
=V V n
354 m 3 / h 1000 L 1 kmol 444 kmol / h m 3 1000 mol
354 m 3 / h
0.8 L / mol
(5.2-36)
< 20 L / mol ====> ideal gas approximation is poor V from n using the ideal gas equation of state is likely Most obviously, the calculation of V to lead to error. In addition, the reaction equilibrium expressions are probably strictly valid only for ideal gases, so that every calculated quantity is likely to be in error.
5.55 a.
RTc PV B Bo ZB1 1 B = RT Pc V From Table B.1 for ethane: Tc 305.4 K, Pc 48.2 atm From Table 5.3 -1 Z = 0.098 0.422 0.422 0.333 Bo 0.083 1.6 0.083 1.6 Tr 308.2 K 305.4 K . . 0172 0172 4.2 0139 0.0270 B1 0139 . . 4 .2 Tr 308.2 K 305.4K RTc 0.08206 L atm 305.4 K 0.333 0.098 0.0270 B(T) = B o ZB1 Pc mol K 48.2 atm 01745 L / mol .
b
b
g
e
j
e
j
g
FG H
b
IJ K
g
2 PV mol K 2 - B = 10.0 atm V V V + 0.1745 = 0 RT 308.2K 0.08206 L atm = V
b
gb
1 r 1 - 4 0.395 mol / L 01745 . L / mol
b
g
2 0.395 mol / L
Videal
b.
g
2.343 L / mol, 0.188 L / mol
RT / P 0.08206 u 308.2 / 10.0 2.53, so the second solution is likely to be a mathematical artifact. 10.0 atm 2.343 L / mol PV 0.926 z= Latm RT 0.08206 mol 308.2K K
5-37
c.
5.56
= m
V 1000 L mol 30.0 g 1 kg MW = h 2.343 L mol 1000 g V
PV RT
1
b
RTc B B= Bo ZB1 Pc V
12.8 kg / h
g
b g 513.2 K, P 78.50 atm T bC H g 369.9 K, P 42.0 atm From Table 5.3 -1 Z bCH OH g = 0.559, Z bC H g = 0.152 From Table B.1 Tc CH 3OH c
3
c
8
c
3
Bo (CH 3OH)
0.083
Bo ( C 3 H 8 )
0.422 1.6 Tr
. 0139
B1 (CH 3OH)
0139 .
B1 (C 3 H 8 )
3
0.422 0.083 1.6 Tr
B(CH 3OH) =
0.083 0.083
. 0172 4 .2 Tr
0172 . 4 .2 Tr
RTc Bo ZB1 Pc
e
e
e
513.2K 0.422
j
0.333
j
1.6
369.9K . 0172
373.2K
513.2K 0172 .
373.2K
0.619
1.6
373.2K
373.2K
. 0139 0139 .
b
e
8
0.422
369.9K
j
j
0.516
4 .2
0.0270
4 .2
g
0.08206 L atm 513.2K 0.619 0.559 0.516 mol K 78.5 atm RTc Bo ZB1 B(C 3 H 8 ) = Pc
b
c
b
g
h
b
c
¦¦y y B i
i
B ij B mix
b
L mol
g
0.08206 L atm 369.9 K . 0.0270 0.333 0152 mol K 42.0 atm
B mix
0.4868
j
ij
B ij
j
d
0.5 B ii B jj
g
h
0.2436
L mol
i
g
0.5 0.4868 0.2436 L / mol = -0.3652 L / mol
b0.30gb0.30gb0.4868g 2b0.30gb0.70gb0.3652g b0.70gb0.70gb0.2436g
0.3166 L / mol
IJ K
FG H
2 PV mol K 2 - B = 10.0 atm V V V + 0.3166 = 0 mix RT 373.2K 0.08206 L atm = Solve for V:V V ideal
RT P
b
gb
1 r 1- 4 0.326 mol / L 0.3166 L / mol
b
g
2 0.326 mol / L
0.08206 L atm 373.2 K mol K 10.0 atm
g
2.70 L / mol, 0.359 L / mol
3.06 L / mol V virial
5-38
2.70 L / mol
5.57 a.
15.0 kmol CH 3OH / h 1000 mol 1 m3 135 m3 / h 0.30 kmol CH 3OH / kmol 1 kmol 1000 L
2.70 L / mol
= Vn V
van der Waals equation: P =
d
i
d
RT a2 2 -b V V
i
2 V - b PV 3 PV 2 b = RTV 2 aV + ab Multiply both sides by V
b
g
+ -Pb - RT V aV - ab = 0 PV 3
c3
2
P = 50.0 atm
c2
b-Pb - RTg b50.0 atmgb0.0366 L / molg c0.08206
c1
a = 133 . atm L2 / mol 2
c0
ab = - 133 . atm L2 / mol 2 0.0366 L / mol
ib
d
0.0487
hb223 Kg
201 . L atm / mol
g
3
atm L mol 3
RT P
b. V ideal
Latm molK
0.08206 L atm 223 K mol K 50.0 atm
0.366 L / mol
c. T(K)
P(atm)
223 223 223 223 223
1.0 10.0 50.0 100.0 200.0
c3
c2
1.0 10.0 50.0 100.0 200.0
c1
-18.336 -18.6654 -20.1294 -21.9594 -25.6194
c0
1.33 1.33 1.33 1.33 1.33
-0.0487 -0.0487 -0.0487 -0.0487 -0.0487
V(ideal) V f(V) % error (L/mol) (L/mol) 18.2994 18.2633 0.0000 0.2 1.8299 1.7939 0.0000 2.0 0.3660 0.3313 0.0008 10.5 0.1830 0.1532 -0.0007 19.4 0.0915 0.0835 0.0002 9.6
d. 1 eq. in 1 unknown - use Newton-Raphson.
b1g gdV i
b
Eq. (A.2-13) a Eq. (A.2-14) ad (k +1) Then V
g
b g
3 + -20.1294 V 2 133 50.0V =0 . V-.0487 wg wV
2 40.259 V + 1.33 150V solve
g d
g a
(k) d Guess V (1) V
1 2 3 4
V ideal
(k) V 0.3660 0.33714 0.33137 0.33114
5-39
0.3660 L / mol . (k +1) V 0.33714 0.33137 0.33114 0.33114
converged
b
5.58 C 3 H 8 : TC
369.9 K
5.0 m3 75 kg
Specific Volume
d
42.0 atm 4.26 u 106 Pa
PC
44.09 kg 1 kmol 1 kmol 103 mol
i
Z
0152 .
2.93 u 10 3 m3 mol
Calculate constants 0.42747
a
d8.314 m Pa mol Ki b369.9 Kg 2
3
6
4.26 u 10 Pa
0.08664
b
d8.314 m Pa mol Ki b369.9 Kg 3
4.26 u 106 Pa
b
g
b
015613 0152 0152 m 0.48508 155171 . . . .
e
D
1 0.717 1 298.2 369.9
j
2
g
2
2
0.949 m6 Pa mol 2 6.25 u 10 5 m3 mol
0.717
115 .
SRK Equation:
d8.314 m Pa mol Kib298.2 Kg d2.93 u 10 6.25 u 10 i m mol 2.93 u 10
d
3
P
3
5
3
115 . 0.949 m6 Pa mol 2 3
d
i
i
m3 mol 2.93 u 10 3 6.25 u 10 5 m3 mol
7.40 u 106 Pa 7.30 atm
P Ideal:
RT V
P
TC
3
3
8.46 u 106 Pa 8.35 atm
3
2.93 u 10 m mol (8.35 7.30) atm u 100% 14.4% 7.30 atm
Percent Error: 5.59 CO 2 :
d8.314 m Pa mol Kib298.2 Kg
304.2 K
PC
72.9 atm Z
0.225
151.2 K PC 48.0 atm Z 0.004 35.0 L / 50.0 mol 0.70 L mol P 510 . atm , V
Ar:
TC
Calculate constants (use R
L2 atm L , m 0.826 , b 0.0297 ,D 2 mol mol L2 atm L a 137 . , m 0.479 , b 0.0224 ,D 2 mol mol
CO 2 : a Ar:
bg
f T
0.08206 L atm mol K )
3.65
RT a 1 m 1 T TC V b Vb V
d
i
e
j
2
P=0
Use E-Z Solve. Initial value (ideal gas): L L atm Tideal 510 . atm 0.70 0.08206 mol mol K
b
gFGH
e 1 0.479e1
IJ FG K H
5-40
IJ K
j . j T 1512
1 0.826 1 T 304.2
435.0 K
2
2
b g
E - Z Solve Tmax
5.60 O 2 : TC
CO 2
154.4 K ; PC
455.4 K ,
bT g
431.2 K
max Ar
49.7 atm ; Z
b
g
0.021 ; T 208.2 K 65q C ; P 8.3 atm ;
0.08206 L atm mol K
250 kg h ; R m
SRK constants: a 138 . L2 atm mol 2 ; b
0.0221 L mol ; m 0.517 ; D
0.840
d i dVRT bi V dVaD bi P = 0=====> V = 2.01 L / mol E-Z Solve
f V
SRK equation:
V
250 kg kmol 103 mol 2.01 L 15,700 L h h 32.00 kg 1 kmol mol
W
5.61
e
¦F
PCO2 A - W = 0 where W = mg = 5500 kg 9.81
y
m s2
j
53900 N
PCO 2 A
a.
PCO 2
W
53900 N
A piston
S 4
. mg b015
2
b. SRK equation of state: P = For CO 2 : Tc
304.2, Pc
1 atm 1.013 u 105 N / m2
Da RT -b V V +b V
d i d
. atm 301
i
72.9 atm , Z = 0.225
a = 3.654 m atm / kmol , b = 0.02967 m3 / kmol, m 0.8263, D (25o C) 1016 . 6
2
. ge3.654 e0.08206 jb298.2 Kg b1016 j 301 . atm = eV - 0.02967 j V dV + 0.02967i m 3 atm kmolK
m6 atm kmol 2
m6 kmol 2
m3 kmol
E-Z Solve
= 0.675 m 3 / kmol =====> V
b g 0.030 m Vbafter expansiong 0.030 m b015 . mg b15 . mg 3
V before expansion
3
mCO 2
2
S 4
44.01 kg V 0.0565 m3 MW = 3 0.675 m / kmol kmol V
mCO 2 (initially) =
0.0565 m3
3.68 kg
PV 1 atm 0.030 m3 44.01 kg MW = m 3 atm 298.2 K RT kmol 0.08206 kmol K
mCO 2 (added) = 3.68 - 0.0540 kg = 3.63 kg
5-41
0.0540 kg
5.61 (cont’d) c.
W = 53,900 N
V
n o (kmol) Vo (m3 ) 1 atm, 25o C
! n (kmol) P (atm), 25o C
ho
ho
d(m)
d(m)
Given T, Vo , h, find d Vo Initial: n o Po 1 RT Sd 2 h 3.63 (kg) Final: V = Vo , n = no 4 44 (kg / kmol)
b
=V V n
h
add 3.63 kg CO 2
Vo
g
Vo 0.0825 RT
Sd 2 h 4
Vo 0.0825 RT Da 53,900 W RT 2 P= V +b A piston V - b V Sd / 4
d
Da RT V +b V-b V
i d i in b1g one equation in one unknown. Substitute expression for V
b1g Solve for d .
5.62 a. Using ideal gas assumption: Pg
nRT Patm V
35.3 lb m O 2 1 lb - mole 10.73 ft 3 psia 509.7 o R 14.7 psia = 2400 psig 32.0 lb m lb - mole o R 2.5 ft 3
b. SRK Equation of state: P =
Da RT -b V V +b V
d i d
3 = 2.5 ft 32.0 lb m / lb - mole V 35.3 lb m
For O 2 : Tc
277.9 o R, Pc
2.27
ft psi , b lb - mole 2
ft 3 lb - mole
730.4 psi, Z = 0.021
6
a = 52038 .
i
0.3537
ft 3 , m = 0.518, D 50o F lb - mole
d
i
e10.73 jd509.7 Ri b0.667ge52038. b2400 + 14.7g psi = V - 0.3537 dV + 0.3537i V d i ft 3 psi
o
ft 6 psi
lb-mole 2
o
lb-mole R
ft 3 lb-mole
= 2.139 ft 3 / lb - mole E - Z Solve V 32.0 lb m V 2.5 ft 3 mO 2 MW = 3 2.139 ft / lb - mole lb - mole V
5-42
37.4 lb m
0.667
j
ft 6 lb-mole 2
5.62 (cont’d) Ideal gas gives a conservative estimate. It calls for charging less O2 than the tank can safely hold. c. 1. 2. 3. 4.
Pressure gauge is faulty The room temperature is higher than 50qF Crack or weakness in the tank Tank was not completely evacuated before charging and O2 reacted with something in the tank 5. Faulty scale used to measure O2 6. The tank was mislabeled and did not contain pure oxygen.
5.63 a. SRK Equation of State: P =
Da RT +b V-b V V
d i d
i
d id i = PV dV - bid V + bi RTV dV + bi Dad V - bi f dV i PV RTV dDa - b P - bRTiV - Dab = 0 f dV
i
V -b V +b : multiply both sides of the equation by V
3
2
0
2
b. Problem 5.63-SRK Equation Spreadsheet Species Tc(K) Pc(atm) Z a b m
CO2 304.2 R=0.08206 m^3 atm/kmol K 72.9 0.225 3.653924 m^6 atm/kmol^2 0.029668 m^3/kmol 0.826312
f(V)=B14*E14^3-0.08206*A14*E14^2+($B$7*C14-$B$8^2*B14-$B$8*0.08206*A14)*E14-C14*$B$7*$B$8 T(K) 200 250 300 300 300
P(atm) 6.8 12.3 6.8 21.5 50.0
alpha 1.3370 1.1604 1.0115 1.0115 1.0115
V(ideal) 2.4135 1.6679 3.6203 1.1450 0.4924
V(SRK) 2.1125 1.4727 3.4972 1.0149 0.3392
f(V) 0.0003 0.0001 0.0001 0.0000 0.0001
c. E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in part b. d. REAL T, P, TC, PC, W, R, A, B, M, ALP, Y, VP, F, FP INTEGER I CHARACTER A20 GAS DATA R 10.08206/ READ (5, *) GAS WRITE (6, *) GAS 10 READ (5, *) TC, PC, W READ (5, *) T, P IF (T.LT.Q.) STOP
5-43
5.63 (cont’d) R 0.42747 *R*R/PC*TC*TC B 0.08664 *R*TC/PC W 015613 M 0.48508 W 155171 . .
b
ALP
d1. M c1 bT / TCg
0.5hi
2 .
g
VP R T / P DO 20 I 7, 15 V = VP F = R * T/(V – B) – ALP * A/V/(V + B) – P FP = ALP * A * (2. * V + B)/V/V/(V + B) ** 2 – R * T/(V – B) ** 2. VP = V – F/FP IF (ABS(VP – V)/VP.LT.0.0001) GOTO 30 20 CONTINUE WRITE (6, 2) 2 FORMAT ('DID NOT CONVERGE') STOP 30 WRITE (6, 3) T, P, VP 3 FORMAT (F6.1, 'K', 3X, F5.1, 'ATM', 3X, F5.2, 'LITER/MOL') GOTO 10 END $ DATA CARBON 304.2 200.0 250.0 300.0 –1
72.9 6.8 12.3 21.5 0.
DIOXIDE 0.225
RESULTS CARBON DIOXIDE 200.0 K 6.8 ATM 250.0 K 12.3 ATM 300.0 K 6.8 ATM 300.0 K 21.5 ATM 300.0 K 50.0 ATM 5.64 a.
b.
N 2 : TC PC
Tr 126.2 K 33.5 atm Pr
He: TC PC
Tr 5.26 K 2.26 atm Pr
2.11 LITER/MOL 1.47 LITER/MOL 3.50 LITER/MOL 1.01 LITER/MOL 0.34 LITER/MOL
U| 40 MPa 10 atm V z 12. 1178 . | 33.5 atm 1.013 MPa W . |U b200 273.2g b5.26 8g 552 V| z 16. 350 b2.26 8g 34.11 W b40 273.2g 126.2
2.48
Fig. 5.4-4
Fig. 5.4-4
n Newton’s correction
5-44
5.65 a.
d
U kg / m3
i
m (kg) V (m3 )
(MW)P RT
30 kg kmol 9.0 MPa 10 atm 3 m atm 465 K 0.08206 kmolK 1.013 MPa b.
UV W
Tr Pr
. 465 310 15 9.0 4.5 2.0
U
(MW)P zRT
5.66 Moles of CO 2 : TC PC
z
V r
VP C RTC
, Vr 1507 .
Tr1
154.4 K 49.7 atm
Pr1 Pr2
z 2 T2 P1 z1 T1 P2
127 m3 1.61 358 K
5.68 O 2 : TC PC
n1 n 2
b1600 14.7gpsi 72.9 atm
1 atm . 1507 14.7 psi
10.0 ft 3 72.9 atm lb - moleq R 1k 3 2.27 lb - moles 304.2 k 0.7302 ft atm 1.8 q R 0.80 z 0.85
1614.7 psi 10.0 ft 3 lb - moleq R 1 atm 3 0.85 2.27 lb - moles 0.7302 ft atm 14.7 psi
PV znR
Tr2
V2
2.27 lb - moles
44.01 lb m CO 2
P PC
V1
831 . kg m3
100 lb m CO 2 1 lb - mole CO 2
UV W
5.67 O : T 2 C PC
V2
0.84
69.8 kg m3 0.84
304.2 K Pr 72.9 atm
Fig. 5.4-3: Pr T
Fig. 5.4-3
69.8 kg m3
h
|UVz 1 49.7 0.02 |W 358 154.4 2.23 U| Vz 1000 49.7 20.12 |W 298 154.4 1.93
1.00 298 K 1000 atm
154.4 K 49.7 atm
FG H
V P1 P2 RT z1 z 2
IJ K
320 q F
1.00 (Fig. 5.4 - 2)
1
1 atm
779q R
0.80
b
1.61 Fig. 5.4 - 4
2
g
0.246 m3 h
Tr
b27 273.2g 154.4
Pr1
175 49.7
3.52 z1
0.95
Pr2
1.1 49.7
0.02 z 2
1.00
1.94
FG H
(Fig. 5.3-2)
mol K 175 atm 11 . atm 300.2 K 0.08206 L atm 0.95 1.00 10.0 L
5-45
IJ K
74.3 mol O 2
5.69 a.
= V 50.0 mL 44.01 g 4401 V . mL / mol n 5.00 g mol RT 82.06 mL atm 1000 K P= 186 atm mol K 440.1 mL / mol V
b. For CO 2 : Tc 304.2 K, Pc 72.9 atm T 1000 K Tr 3.2873 Tc 304.2 K VP . mL 72.9 atm mol K 4401 c Vr ideal RTc mol 304.2 K 82.06 mL atm Figure 5.4 - 4: Vr ideal P= c.
zRT V
128 . and Tr
. 128
3.29 z = 1.02
. 82.06 mL atm mol 1000 K 102 mol K 440.1 mL
190 atm
a = 3.654 u 10 6 mL2 atm / mol 2 , b = 29.67 mL / mol, m 0.8263, D (1000 K ) P=
. ge j c82.06 hb1000 Kg b01077 b440.1- 29.67g 440.1b440.1 + 29.67 g 2 3.654 u 106 mLmolatm 2
mLatm molK
mL2 mol 2
mL mol
01077 .
198 atm
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the presence of O2. b. Enough N2 needs to be added to make x O 2
10 u 10 6 . Since the O2 is so dilute at this
condition, the properties of the gas will be that of N2. . atm, Tr 2.36 Tc 126.2 K, Pc 335 n initial nO2 n O2 n2
n1
PV RT
1 atm 5000 L Latm 0.08206 molK 298.2 K
204.3 mol air
FG 0.21 mol O IJ H mol air K
10 u 10 6 n 2
2
204.3 mol
42.9 mol O 2
4.29 u 10 6 mol
5000 L 116 . u 10 -3 L / mol 6 4.29 u 10 mol mol K 335 . u 10 3 L . atm ideal VPc 116 V r RTc mol 0.08206 L atm 126.2 K not found on compressibility charts = V
Ideal gas: P =
RT V
0.08206 L atm 298.2 K mol K 116 . u 10 3 L / mol
38 . u 10 3
2.1 u 10 4 atm
The pressure required will be higher than 2.1 u 10 4 atm if z t 1, which from Fig. 5.3 - 3 is very likely. n added
ib
d
4.29 u 106 204.3 # 4.29 u 106 mol N 2 0.028 kg N 2 / mol
5-46
g
. u 105 kg N 2 120
5.70 (cont’d) c.
143 . kmol N 2
143 . kmol N 2
n initial 0.204 kmol yO 0.21 kmol O 2 / kmol 2
143 . kmol N 2 y1
N 2 at 700 kPa gauge = 7.91 atm abs. Pr n2
P2 V zRT
y1
y init n init 1.634
y2
y 1 n init 1.634
y init
0.236, Tr
7.91 atm 5000 L Latm 298.2 K 0.99 0.08206 mol K
b0.21g0.204 1634 .
y init
FG n IJ K H 1634 .
Fig 5.4-2
2.36 =======> z = 0.99
1633 . kmol
0.026
2
init
0.0033
FG y IJ FG n IJ n = H y K 4.8 Need at least 5 stages K H 1634 . F n IJ lnG H 1.634 K 5b143 . kmol N gb28.0 kg / kmolg 200 kg N ln
n
yn
143 . kmol N 2 y2
n
init
init
init
Total N 2
2
2
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is that it takes longer. 5.71
= MW a. m b.
Tc Pc
PV SPV = MW Cost ($ / h) = mS RT RT
UV W
369.9 K = 665.8o R Tr 0.85 Pr 016 42.0 atm .
= 60.4 m
PV zT
ideal m z
F 44.09 lb / lb - mol I SPV GH 0.7302 JK T m
3
ft atm lb-molo R
60.4
SPV T
Fig. 5.4-2
z = 0.91
ideal 110 . m
Delivering 10% more than they are charging for (undercharging their customer)
5-47
5.72 a.
For N 2 : Tc
After heater: Tr
n =
b. tank =
4.65 ft 3 / min
34,900 gal 133.0 K, Pc
34.5 atm
300 K 2.26 133.0 K 2514.7 psia 1 atm 34.5 atm 14.7 psia
Pr1
U| |V 5.0| |W
Fig. 5.4-3
z = 1.02
mol K 2514.7 psia 150 L 1 atm 1022 mol 1.02 300 K 14.7 psia 0.08206 L atm
After 60h: Tr1 Pr1
b.
0.418 lb - mole / min
b g
Initially: Tr1
n leak
1.02
0.418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks min h day week 0.81 62.4 lb m / ft 3
For CO: Tc
n2
U| |V z . | 12 |W
. 0.418 lb - mole 10.73 ft 3 psia 609.7 o R 102 min lb - mole o R 600 psia
4668 ft 3
n1
335 . atm
609.7 o R 2.68 227.16o R 600 psia 1 atm Pr . atm 14.7 psia 335
150 SCFM 359 SCF / lb - mole
= zRTn V P
5.73 a.
227.16o R, Pc
126.20 K
300 K 2.26 133.0 K 2258.7 psia 1 atm 34.5 atm 14.7 psia
U| |V 4.5| |W
Fig. 5.4-3
z = 1.02
mol K 2259.7 psia 150 L 1 atm 1.02 300 K 14.7 psia 0.08206 L atm n n2 173 = 1 . mol / h 60 h PV RT
918 mol
200 u 106 mol CO 1 atm 30.7 m3 1000 L Latm mol air 0.08206 mol m3 K 300 K
n2
y 2 n air
y2
t min
n2 n leak
0.25 mol 1.73 mol / h
0.25 mol
014 . h
t min would be greater because the room is not perfectly sealed c. (i) CO may not be evenly dispersed in the room air; (ii) you could walk into a high concentration area; (iii) there may be residual CO left from another tank; (iv) the tank temperature could be higher than the room temperature, and the estimate of gas escaping could be low. 5-48
5.74 CH 4 : Tc
190.7 K , Pc
45.8 atm
C 2 H 6 : Tc
305.4 K , Pc
48.2 atm
C 2 H 4 : Tc
2831 . K , Pc
50.5 atm
b0.20gb190.7g b0.30gb305.4g b0.50gb2831. g 2713. K Pseudocritical pressure: P c b0.20gb458 . g b0.30gb48.2g b0.50gb50.5g 48.9 atm U| b90 273.2gK 134 Reduced temperature: T . |V z 0.71 2713 . K 200 bars 1 atm Reduced pressure: P 4.04 | |W 48.9 atm 1.01325 bars Mean molecular weight of mixture: b0.30gM b0.50gM M b0.20gM b0.20gb16.04g b0.30gb30.07g b0.50gb28.05g Pseudocritical temperature: Tcc c
r
Figure 5.4-3
r
CH 4
C2H 6
C2 H 4
26.25 kg kmol V
znRT P
0.71 10 kg
1 kmol 0.08314 m 3 bar 26.25 kg kmol K
UV W
5.75 N 2 : Tc 126.2 K, PC 33.5 atm Tcc N 2 O: Tc 309.5 K, PC 71.7 atm Pcc M n a.
b g b g
200 bars
b
0.041 m 3 (41 L)
g
0.10 309.5 0.90 126.2 144.5 K 0.10 71.7 0.90 33.5 37.3 atm
b g b g 29.62 5.0 kgb1 kmol 29.62 kgg 0.169 kmol T b24 273.2 g 144.5 2.06
b g
0.10 44.02 0.90 28.02
169 mol
r
V r
mol K
30 L
Pr V
r
T
30 L 37.3 atm mol K 169 mol 144.5 K 0.08206 L atm
U| Vz 0.56| W
0.97 169 mol 297.2 K 0.08206 L atm
P
b.
b90 + 273g K
U|V z 0.56 b from a.g |W
273 37.3 7.32
b
g
518 K 245q C
5-49
g
133 atm 132 atm gauge
1.14 Fig. 5.4 - 3
273 atm 30 L mol K 1.14 169 mol 0.08206 L atm
b
0.97 Fig. 5.4 - 3
b g b g
UV W
5.76 CO: Tc 133.0 K, Pc 34.5 atm Tcc H 2 : Tc 33 K, Pc 12.8 atm Pcc Turbine inlet:
Turbine exit: Tr Pr in Pin V out Pout V
Tr
b150 273.2g 96.2
Pr
2000 psi 1 atm 29.0 atm 14.7 psi
373.2 96.2 3.88 1 29.0 0.03
z in nRTin Vin z out n RTout
b
g g
0.60 133.0 0.40 33 8 0.60 34.5 0.40 12.8 8
b
96.2 K 29.0 atm
U| V o z | 1.01 4.69| W 4.4
Fig. 5.4-1
z=1.0
Vout u
Pout z in Tin Pin z out Tout
15,000
ft 3 14.7 psia 1.01 423.2K min 2000 psia 1.00 373.2
126 ft 3 / min If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00 1% error
UV W
5.77 CO: Tc 133.0 K, Pc 34.5 atm Tcc CO 2 : Tc 304.2 K, Pc 72.9 atm Pcc Initial: Tr Pr Final: Pr
303.2 138.1 2.2 2014.7 524.8 3.8
b g b g
UV o z W
1889.7 524.8 3.6 z1
b
g
0.97 133.0 0.03 304.2 138.1 K 0.97 34.5 0.03 72.9 35.7 atm 524.8 psi
Fig. 5.4-3
1
b g
0.97
0.97
Total moles leaked: n1 n 2
FG P P IJ V b2000 1875gpsi 0.97 H z z K RT 1
2
1
2
30.0 L 1 atm mol K 303 K 14.7 psi 0.08206 L atm
10.6 mol leaked
b g
Moles CO leaked: 0.97 10.6 Total moles in room: Mole% CO in room =
10.3 mol CO
24.2 m3 103 L 273 K 1 mol 3 1m 303 K 22.4 L STP
b g
10.3 mol CO u 100% 10% . CO 973.4 mol
5-50
973.4 mol
CO 2H 2 o CH 3OH
5.78 Basis: 54.5 kmol CH 3OH h n 1 (kmol CO / h) 2n 1 (kmol H 2 / h) 644 K 34.5 MPa
Catalyst Bed
CO, H 2
Condenser
54.5 kmol CH 3OH (l ) / h
a.
54.5 kmol CH 3OH 1 kmol CO react
n 1
h
b g
2n 1
2 218
1 kmol CH 3OH
1 kmol CO fed 0.25 kmol CO react
b
g
436 kmol H 2 h 218 436
CO: Tc
133.0 K
Pc
34.5 atm
H 2 : Tc
33 K
Pc
12.8 atm
218 kmol h CO
654 kmol h (total feed)
Newton’s corrections Tcc Pcc Tr
b g b g 1 b3 34.5g 23 b12.8 8g 1 2 133.0 33 8 3 3
717 . K 25.4 atm
U| V| W
644 71.7 8.98 5.4-4 Fig. o z 1 34.5 MPa 10 atm 13.45 24.5 atm 1.013 MPa
Pr feed V
1.18
1.18 654 kmol 644 K 0.08206 m3 atm 1.013 MPa 120 m3 h h 34.5 MPa kmol K 10 atm Vcat
120 m3 h
1 m3 cat 25,000 m3 / h
0.0048 m3 catalyst (4.8 L)
b. CO, H 2 54.5 kmol CH 3OH (l ) / h
n 4 kmol CO / h 2n 4 kmol H 2 / h
Overall C balance n 4 Fresh feed:
feed V
54.5 mol CO h
54.5 kmol CO h 109.0 kmol H 2 h 163.5 kmol feed gas h
1.18 163.5 kmol 644 K 0.08206 m3 atm 1.013 MPa h 34.5 MPa kmol K 10 atm
5-51
29.9 m3 h
5.79
H 2 : Tc Pc
(33.3 8) K = 41.3 K
1 - butene: Tc
419.6 K
Pc
39.7 atm
(12.8 8) atm = 20.8 atm
. (413 . K) + 0.85(419.6 K) = 362.8 K Tc ' 015 . (20.8 atm) + 0.85(39.7 atm) = 36.9 atm Pc ' 015 = znRT V P
0.86 35 kmol 0.08206 m3 atm 323 K 1 h h kmol K 10 atm 60 min
CH 3 :
Tc
190.7 K Pc
458 . atm
C 2 H 4 : Tc
2831 . K Pc
50.5 atm
C 2 H 6 : Tc
305.4 K Pc
48.2 atm
UV W
Fig. 5.4-2
z = 0.86
133 . m3 / min
d
i FG 100 cmIJ S b150 m / ming H m K
F I FG IJ d i GH JK H K
4 133 . m 3 / min
3 2 m = u m A m 2 = u u Sd d = 4V V Su 4 min min
5.80
Tr ' 0.89 Pr ' 0.27
T=90 o C
. (190.7 K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K ====> Tc ' 015
10.6 cm
U| V| . W 35
. Tr ' 132
P=175 bar
. (458 . atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm =====> Pr ' Pc ' 015 5.4-3 Fig. o z = 0.67
F I FG IJ d i FG GH JK H K H
3 m = u m A m 2 = 10 m V s s s
n = 5.81
PV zRT
N2:
IJ FG 60 s IJ S b0.02 mg K H min K 4
2
. 0188
kmol K 0188 . m3 / min 175 bar 1 atm 0.67 1.013 bar .08206 m3 atm 363 K Tc
acetonitrile: Tc
126.2 K = 227.16o R Pc
335 . atm
548 K = 986.4 o R
47.7 atm
Pc
Tank 1 (acetonitrile): T1 = 550o F, P1 = 4500 psia Tr1 n 1 =
P1 V1 z 1 RT1
306 atm 0.200 ft 3 0.80 1009.7 o R
P2 V2 z 2 RT2
10.0 atm 2.00 ft 3 1.00 1009.7 o R
5-52
. kmol / min 163
. Pr1 102
Fig. 5.4-3
6.4 z1 = 0.80
lb - mole o R = 0.104 lb - mole 0.7302 ft 3 atm
Tank 2 (N 2 ): T2 = 550 o F, P2 = 10 atm Tr2 n 2 =
m3 min
4.4 , Pr2
Fig. 5.4-3
6.4 z 2 = 1.00
lb - mole o R = 0.027 lb - mole .7302 ft 3 atm
5.81 (cont’d) . I .027 I FG 0104 J 986.4 R + FGH 00131 J 227.16 R = 830 R H 0131 . K . K . I FG 0104 F 0.027 IJ 33.5 atm = 44.8 atm 47.7 atm + G J H 0131 K H 0131 . . K o
Final: Tc ' Pc '
dV i r
P=
=
ideal
o
o
T=550 Fo
o
Tr ' 122 .
' Fig. 5.4-2 VP 2.2 ft 3 44.8 atm lb - mole o R c z = 0.85 = = 1.24 RTc ' 0.131 lb - mole 830 o R 0.7302 ft 3 atm 0.85 0131 . lb - mole .7302 ft 3 atm 1009.7 o R lb - mole o R 2.2 ft 3
znRT V
37.3 atm
5.82 3.48 g Ca H b O c , 26.8o C, 499.9 kPa n c (mol C), n H (mol H), n O (mol O)
1 L @483.4 o C, 1950 kPa n p (mol) 0.387 mol CO 2 / mol 0.258 mol O 2 / mol 0.355 mol H 2 O / mol
n O 2 (mol O 2 ) 26.8o C, 499.9 kPa
a.
d
Volume of sample: 3.42 g 1 cm3 159 . g
i
2.15 cm3
O 2 in Charge:
d
1.000 L 2.15 cm 3 10 3 L km 3 L atm 0.08206 mol K
n O2
i
499.9 kPa
1 atm
300 K
101.3 kPa
0.200 mol O 2
Product
1.000 L 1950 kPa 1 atm L atm 0.08206 756.6 K 101.3 kPa mol K Balances: np
b
g
O: 2 0.200 n O
b
b
g
0.387 0.310
H: n H
2 0.355 0.310
gb
g
0.310 2 0.387 2 0.258 0.355 n O
C: n C
b
g b
0.310 mol product
0.110 mol O in sample
0.120 mol C in sample
g
0.220 mol H in sample
Assume c 1 a 0.120 0.110 1.1 b 0.220 0.110 2 Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20), etc. b.
b g
b g
MW 12.01a 1.01b 16.0c 12.01 1.1c 1.01 2c 16.0c 300 MW 350 c 10 C11 H 20 O 10
5-53
31.23c
bg
C5 H 10
5.83 Basis: 10 mL C5 H 10 l charged to reactor
15 O 2 o 5CO 2 5H 2 O 2
bg
10 mL C5 H 10 l n1 (mol C5 H 10 )
n 2 (mol air) 0.21 O 2 0.79 N 2 27 o C, 11.2 L, Po (bar)
a.
n1
b
bg
10.0 mL C5 H 10 l
Stoichiometric air: n 2 Po
n 3 (mol CO 2 ) n 4 mol H 2 O(v) n 5 (mol N 2 ) 75.3 bar (gauge), Tad
nRT V
0.745 g 1 mol mL 70.13 g 0.1062 mol C5 H 10
d Ci o
0.1062 mol C5 H 10 7.5 mol O 2
1 mol air
1 mol C 2 H 10
0.21 mol O 2
3.79 mol 0.08314 L bar 300K 11.2 L
g
mol K
3.79 mol air
8.44 bars
(We neglect the C5 H 10 that may be present in the gas phase due to evaporation) Initial gauge pressure 8.44 bar 1 bar b.
7.44 bar
U| || V| || W
0.1062 mol C 5 H 10
5 mol CO 2 0.531 mol CO 2 1 mol C 5 H 10 0.531 mol CO 2 1 mol H 2 O 4.052 mol product gas 0.531 mol H 2 O 1 mol CO 2 0.79 3.79 2.99 mol N 2
n3 n4
b g
n5
CO 2 : y 3 = 0.531 / 4.052 = 0.131 mol CO 2 / mol, Tc = 304.2 K Pc = 72.9 atm H 2 O: y 4 = 0.531 / 4.052 = 0.131 mol H 2 O / mol, N2:
y 5 = 2.99 / 4.052 = 0.738 mol N 2 / mol,
Tc = 647.4 K Pc = 218.3 atm Tc = 126.2 K Pc = 33.5 atm
. (304.2 K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K Tc ' 0131 . (72.9 atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm Pr ' 121 . Pc ' 0131 ideal V r T=
PV znR
' VP c RTc '
11.2 L 62.9 atm mol K 4.052 mol 217.8 K .08206 L atm
b75.3 1gbars 1.04
112 . L mol K 4.052 mol 0.08314 L bar
5-54
9.7 z | 1.04 (Fig. 5.4 - 3) 2439 K - 273 = 2166o C
CHAPTER SIX 6.1
a.
AB: Heat liquid - -V | constant BC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T 100 o C. CD: Heat vapor - -T increases, V increases .
b. Point B: From Perry’s Handbook, Table 3.8 U H 2 O(l), 80q C 0.9718 g ml , U H 2 O(l), 100q C 10 mL 0.9718 g / mL
bg
H 2 O l , 100q C: V B
0.9584 g / mL
0.9584 g mL
10.14 mL
Point C: H2O (v, 100qC) n
10 mL 0.9718 g
0.5393 mol 18.02 g nRTC 0.5393 mol 0.08206 L atm 373 K 16.5 L 1 atm mol K PC
mL
nRTC VC
PCVC
6.2
1 mol
a. Pfinal
243 mm Hg . Since liquid is still present, the pressure and temperature must lie on the
vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the species at the system temperature. b. Assuming ideal gas behavior for the vapor, m(vapor) m(liquid)
6.3
a.
mol K
(3.000 - 0.010) L (30 + 273.2) K 10 mL
0.08206 L atm
1.489 g mL
m(vapor) + m(liquid) = 19.5 g
x vapor =
4.59 19.48
b. ln p
B
119.39 g
760 mm Hg
mol
0.235 g vapor / g total
7.09808
1238.71 45 217
'H v 1 'H v B R T R
ln( p1* )
1 atm
14.89 g
m total
log 10 p
243 mm Hg
'H v / R T1
b
2.370 p *
d
ln p2* / p1* 1 T2
1 T1
10 2.370
i
b
6-1
g
ln 760 / 118.3
b
1 77 .0 273.2 K
K g b29.54151 273.2gK
ln 118.3
234.5 mm Hg
g
18.49
b
1 29 .5 273.2 K
g
4151K
4.59 g
6.3 (cont’d) ln p (45o C)
b
4151 18.49 p 45 273.2
g
. 234.5 2310 u 100% 234.5 c.
327.7 234.5 u 100% 234.5
b
327.7 mm Hg
39.7% error
g
1 (rect. scale) on semilog paper T 273.2 straight line: slope 7076K , intercept 2167 .
Plot p log scale vs
b
g
ln p mm Hg ' Hv R
b
7076K ' H v
7076 K
8.314 J
g
exp
1 kJ
mol K 10 3 J
LM 7076 2167 O . P N T ( C) 2732. Q o
58.8 kJ mol
ln p* = A/T(K) + B p*(mm Hg) 5 20 40 100 400 760
1/T(K) 0.002834 0.002639 0.002543 0.002410 0.002214 0.002125
ln(p*) p*(fitted) 1.609 5.03 2.996 20.01 3.689 39.26 4.605 101.05 5.991 403.81 6.633 755.13
7 6 5 4 3 2 1 0
1/T
6-2
0.003
0.0028
0.0026
0.0024
y = -7075.9x + 21.666
0.002
o
T( C) 79.7 105.8 120.0 141.8 178.5 197.3
ln(p*)
6.5
7076 2167 . p mm Hg T ( C) 273.2 o
0.0022
6.4
15% . error
FG 118.3 760IJ b45 29.5g 118.3 H 29.5 77 K
p
. mm Hg 2310
o
T( C) p*(fitted) 50 0.80 80 5.12 110 24.55 198 760.00 230 2000.00 Least confidence (Extrapolated)
6.6
a. T(°C)
1/T(K)
42.7 58.9 68.3 77.9 88.6 98.3 105.8
3.17u10-3 3.01u10-3 2.93u10-3 2.85u10-3 2.76u10-3 2.69u10-3 2.64u10-3
p*(mm Hg) =758.9 + hright -hleft 34.9 78.9 122.9 184.9 282.9 404.9 524.9
b. Plot is linear, ln p
'H v B ln p RT
5143.8 K 19.855 T
At the normal boiling point, p 760 mmHg Tb ' H v
8.314 J 5143.8 K
1 kJ
mol K
10 3 J
116q C
42.8 kJ mol
c. Yes — linearity of the ln p vs 1 / T plot over the full range implies validity.
6.7
a.
b
g
ln p a T 273.2 b y
ax b
y
b
ln p ; x 1 T 273.2
g
Perry' s Handbook, Table 3 - 8: 39.5q C , p1 400 mm Hg x1
T1
56.5q C , p2 760 mm Hg x 2
T2
50q C x 3.0941 u 10 x x1 y1 y 2 y1 x 2 x1
T
FG H
y
IJ b K
u 10 3 , y1 31980 . 3
3.0331 u 10 , y 2
5.99146 6.63332
3
g
b
6.39588 p 50q C
g
e 6.39588
599 mm Hg
b. 50q C 122q F 12 psi 760 mm Hg
Cox Chart p
c.
6.8
log p 7.02447
b
. 11610 50 224
2.7872 p 10 2.7872
g
Estimate p 35q C : Assume ln p a
b
g
ln p2 p1 1 T2
1 T1
b ln p1
a T1
625 mm Hg
14.6 psi
b
g
ln 200 50 1 45 273.2
b g
ln 50
1 25 273.2
6577.1 25 273.2
613 mm Hg
a b , interpolate given data. T K
b g
U| ln p b35q Cg 6577.1 25.97 4.630 35 273.2 |V 25.97 | p b35q Cg e 102.5 mm Hg |W
6577.1
4 .630
6-3
6.8 (cont’d) Moles in gas phase: n
150 mL
a.
m 2 S
2 F
g
4
mol
222
2.
8.0 u 10 6.9
b
273 K 102.5 mm Hg 1L 1 mol 3 35 + 273.2 K 760 mm Hg 10 mL 22.4 L STP
b g
Two intensive variable values (e.g., T & P) must be
specified to determine the state of the system. 1209.6 2.5107 p MEK 10 2.5107 324 mm Hg b. log p MEK 6.97421 55 216. Since vapor & liquid are in equilibrium p MEK p MEK 324 mm Hg y MEK
p MEK / P
324 1200 0.27 ! 0115 . The vessel does not constitute an explosion
hazard. 6.10 a. The solvent with the lower flash point is easier to ignite and more dangerous. The solvent with a flash point of 15qC should always be prevented from contacting air at room temperature. The other one should be kept from any heating sources when contacted with air. b. At the LFL, y M 0.06 p M p *M 0.06 u 760 mm Hg = 45.60 mm Hg 1473.11 T 6.85q C Antoine log 10 45.60 = 7.87863 T + 230 c. The flame may heat up the methanol-air mixture, raising its temperature above the flash point. 6.11 a. At the dew point, p ( H 2 O) = p( H 2 O) = 500 u 0.1 = 50 mm Hg T = 38.1q C from Table B.3. b. VH2O
30.0 L
273 K 500 mm Hg 1 mol 0.100 mol H2 O 18.02 g 1 cm3 (50 + 273) K 760 mm Hg 22.4 L (STP) mol mol g
c. (iv) (the gauge pressure)
6-4
134 . cm3
6.12 a.
60 mm Hg b g 755 mm Hg b577 222gmm Hg 400 mm Hg
T1
58.3q C , p1 755 mm Hg 747 52 mm Hg
T2
110q C , p2
a b T K
ln p a b
b g
b
g
b
ln p2 p1 1 T2
ln p1
a T1
g
ln 400 60
1 T1
1 110 273.2
b g
ln 60
46614 .
58.31273.2
4661.4 58.3 273.2
18.156
T=130oC=403.2 K 46614 . 18156 . T ln p 130q C 6.595 p 130q C e 6.595 7314 . mm Hg
ln p
b
b.
g
b
g
Basis: 100 mol feed gas CB denotes chlorobenzene. n1 mol @ 58.3qC, 1atm y1 (mol CB(v)/mol) (sat’d) (1-y1) (mol air/mol)
100 mol @ 130qC, 1atm y0 (mol CB(v)/mol) (sat’d) (1-y0) (mol air/mol)
n2 mol CB (l)
b
g
731 mm Hg 760 mm Hg
pCB 130q C y o
Saturation condition at inlet: y o P
0.962 mol CB mol
b
60 mm Hg 0.0789 mol CB mol g 760 mm Hg Air balance: 100b1 y g n b1 y g n b100gb1 0.962g b1 0.0789g 4.126 mol Total mole balance: 100 n n n 100 4.126 9587 . mol CBbl g pCB 58.3q C y1
Saturation condition at outlet: y1 P o
1
1
1
% condensation:
1
2
2
95.87 mol CB condensed u 100% 0.962 u 100 mol CB feed
b
g
99.7%
c. Assumptions: (1) Raoult’s law holds at initial and final conditions; (2) CB is the only condensable species (no water condenses); (3) Clausius-Clapeyron estimate is accurate at 130qC. 6.13 T
78q F = 25.56q C , Pbar
y H 2O P
b
0.87 p 25.56q C
d i
Dew Point: p Tdp
yp
29.9 in Hg = 759.5 mm Hg , hr
g
Table B.3
y H 2O
b
g
0.0293 759.5
87%
b
0.87 25544 . mm Hg 759.5 mm Hg
22.22 mm Hg
6-5
Table B.3
g
0.0293 mol H 2 O mol air Tdp
23.9q C
6.13 (cont’d) hm ha
hp
0.0293 1 0.0293
0.0302 mol H 2 O mol dry air
0.0293 mol H 2 O 18.02 g H 2 O mol dry air
b
p 2556 . qC
mol H 2 O
hm
g
mol dry air
b
29.0 g dry air
P p 2556 . qC
g
0.0182 g H 2 O g dry air
0.0293 u 100 84% 24.544 759.5 24.544
u 100%
6.14 Basis I : 1 mol humid air @ 70q F (21.1q C), 1 atm, hr hr
50% y H 2 O P Table B.3
Mass of air:
y H 2O
b
0.50 p H 2 O 21.1q C
1 mol
Volume of air:
mol H 2 O mol
0.012
0.988 mol dry air 29.0 g
1 mol
b g b273.2 21.1gK
1 mol
22.4 L STP 1 mol
28.87 g 24.13 L
Density of air
g
0.50 u 18.765 mm Hg 760.0 mm Hg
0.012 mol H 2 O 18.02 g
50%
273.2K
80% y H 2 O P Table B.3
Mass of air:
y H 2O
b
0.80 p H 2 O 21.1q C
1 mol
Density of air
80%
g
0.80 u 18.765 mm Hg 760.0 mm Hg
0.020 mol H 2 O 18.02 g
Volume of air:
0.020
mol H 2 O mol
0.980 mol dry air 29.0 g 1 mol
b g b273.2 21.1gK
1 mol
22.4 L STP 1 mol
28.78 g 24.13 L
273.2K
80% y H 2 O P Table B.3
y H 2O
24.13 L
1193 . g L
Basis III: 1 mol humid air @ 90q F (32.2q C), 1 atm, hr hr
24.13 L
1196 . g L
Basis II: 1 mol humid air @ 70q F (21.1q C), 1 atm, hr hr
28.87 g
b
0.80 p H 2 O 32.2q C
80%
g
0.80 u 36.068 mm Hg 760.0 mm Hg
6-6
0.038
mol H 2 O mol
28.78 g
6.14 (cont’d) Mass of air:
0.038 mol H 2 O 18.02 g
Volume of air: Density
1 mol
0.962 mol dry air 29.0 g 1 mol
b g b273.2 32.2gK
1 mol
28.58 g 25.04 L
22.4 L STP 1 mol
273.2K
28.58 g
25.04 L
1141 . g L
Increase in T increase in V decrease in density Increase in hr more water (MW = 18), less dry air (MW = 29) decrease in m decrease in density Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force on the ball must also be lower. Therefore, the statement is wrong. 6.15 a. hr
50% y H 2 O P Table B.3
b
0.50 p H 2 O 90q C
0.50 u 525.76 mm Hg 760.0 mm Hg
y H 2O
Dew Point: y H 2 O p
d i
p Tdp
Degrees of Superheat
b. Basis:
g
b g
0.346 760
0.346 mol H 2 O / mol
262.9 mm Hg
1 m 3 feed gas 10 3 L 273K mol 3 m 363K 22.4 L STP
n1 mol @ 25qC, 1atm y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol air/mol)
0.346 H2O mol /mol 0.654 mol air/mol
n2 mol H2O (l)
b g 23.756 P 760 Dry air balance: 0.346b33.6g n b1 0.0313g n
Saturation Condition: y1
p H* 2 O 25q C 1
g
p * ( 90q C) y H 2O
0.0313 mol H 2 O mol
1
Total mol balance: 33.6 = 12.0 + n2 n2
b
72.7q C
33.6 mol
b g
p 90q C P
Tdp
90 72.7 17.3q C of superheat
33.6 mol @ 90qC, 1atm
c. y H 2 O P
Table B.3
12.0 mol
216 . mol H 2 O condense / m 3
525.76 mmHg 0.346
6-7
1520 mm Hg = 2.00 atm
6.16 T
90q F = 32.2q C , p
29.7 in Hg = 754.4 mm Hg , hr
95%
Basis: 10 gal water condensed/min n condensed
1 ft 3 62.43 lb m 7.4805 gal ft 3
10 gal H 2 O
y2 (lb-mol H2O (v)/lb-mol) (sat’d) (1-y2) (lb-mol DA/lb-mol) 40oF (4.4oC), 754 mm Hg
y1 (lb-mol H2O (v)/lb-mol) (sat’d) (1-y1) (lb-mol DA/lb-mol) hr=95%, 90oF (32.2oC), 29.7 in Hg (754 mm Hg)
Raoult' s law: y 2 P
4.631 lb-moles H2O (l)/min
b
0.95 p 32.2q C
95% hr at inlet: y H 2 O P
y H 2O
4.631 lb - mole / min
n2 (lb - moles / min)
V1 (ft 3 / min) n1 (lb - moles / min)
Table B.3
1 lb - mol 18.02 lb m
b
g
0.95 36.068 mm Hg
g
0.045 lb - mol H 2 O lb - mol
754.4 mm Hg
b
g
Table B.3
p * 4.4q C
y2
6.292 754.4
0.00834 lb - mol H 2 O lb - mol
UV RS W T
n1 125.3 lb - moles / min Mole balance: n1 n 2 4.631 n 2 120.6 lb - moles / min Water balance: 0.045n1 0.00834n 2 4.631 125.3 lb - moles 359 ft 3 (STP) (460 + 90) o R 760 mm Hg Volume in: V = min lb - moles 492 o R 754 mm Hg 5.07 u 10 4 ft 3 / min 6.17 a. Assume no water condenses and that the vapor at 15qC can be treated as an ideal gas. 760 mm Hg
p final
(15 273) K (200 + 273) K
462.7 mm Hg ( p H 2 O ) final
0.20 u 462.7
p * (15q C) = 12.79 mm Hg < p H 2 O . Impossible condensation occurs. ( pair ) final
P
( pair ) initial
p H 2O pair
b. Basis:
Tfinal Tinitial
(0.80 u 760) mm Hg u
370.2 12.79
1 L 273 K
383 mm Hg
mol
473 K 22.4 L (STP)
0.0258 mol
6-8
288 K 473 K
370.2 mm Hg
92.6 mm Hg
6.17 (cont’d) n1 mol @ 15qC, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol dry air/mol)
0.0258 mols @ 200qC, 760 mm Hg 0.20 H2O mol /mol 0.80 mol air/mol
n2 mol H2O (l)
b
p H* 2 O 15q C
Saturation Condition: y1
12.79 mm Hg . mm Hg 3831
P
b
c.
g
g n b1 0.03339g n
Dry air balance: 0.800 0.0258
1
Mass of water condensed =
0.02135 mol
1
Total mole balance: 0.0258 = 0.02135 + n2 n2 0.00445 mol
0.03339 mol H 2 O mol
0.00445 mol
18.02 g mol
0.0802 g
6.18 Basis: 1 mol feed n2 (mol), 15.6°C, 3 atm y 2 (mol H 2 O (v)/mol)(sat'd) (1 – y 2) (mol DA/mol)
3
V1 (m ) 1 mol, 90°C, 1 atm 0.10 mol H 2O (v)/mol 0.90 mol dry air/mol
heat
100°C, 3 atm n2 (mol) 3 V2 (m )
n3 (mol) H 2 O( l ), 15.6°C, 3 atm
b
p H* 2 O 15.6q C
Saturation: y 2
g
Table B.3
P
atm 13.29 mm Hg 0.00583 3 atm 760 mm Hg
y2
b g n b1 0.00583g n H O mol balance: 0.10b1g 0.00583b0.9053g n Dry air balance: 0.90 1
2
2
Fraction H 2 O condensed: hr
y 2 P u 100% p 100q C
b
g
0.9053 mol
2
3
n3
0.0947 mol condensed 0100 . mol fed
b
g u 100%
0.00583 3 atm 1 atm
0.0947 mol
0.947 mol condense mol fed
. 175% n1 mol @ 15qC, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol dry air/mol)
0.0258 mols @ 200qC, 760 mm Hg 0.20 H2O mol /mol 0.80 mol air/mol
n2 mol H2O (l)
6-9
6.18 (cont’d)
b g
V2
0.9053 mol 22.4 L STP mol
V1
1 mol 22.4 L STP mol
V2 V1
b g
363K 1 m 3 2.98 u 10 2 m 3 feed air @ 90q C 3 273K 10 L
9.24 u 10 3 m 3 outlet air 2.98 u 10 2 m 3 feed air
6.19 Liquid H 2 O initially present:
0.0208 1 0.0208
0.310 m 3 outlet air m 3 feed air
25 L 1.00 kg
b
p H* 2 O 25q C
Saturation at outlet: y H 2 O
373K 1 atm 1 m 3 9.24 u 10 3 m 3 outlet air @ 100q C 273K 3 atm 10 3 L
g
1 kmol 18.02 kg
L
bg
1.387 kmol H 2 O l
23.76 mm Hg 15 . u 760 mm Hg
P
0.0208 mol H 2 O mol air
0.0212 mol H 2 O mol dry air
b g
15 L STP 1 mol 0.670 mol dry air min min 22.4 L STP 0.670 mol dry air 0.0212 mol H 2 O 0.0142 mol H 2 O min Evaporation Rate: min mol dry air Flow rate of dry air:
Complete Evaporation:
b g
min 1.387 kmol 10 3 mol 1h 1628 h 0.0142 mol 60 min kmol
b67.8 daysg
7.069 u 10 3 ft 3 7.481 gal S 2 5.288 u 104 gal / day 6.20 a. Daily rate of octane use = 30 (18 8) 3 4 day ft ( SG ) C8 H18
b. 'p
5.288 u 10 4 gal 1 ft 3 0.703 u 62.43 lb m day 7.481 gal ft 3 0.703 3.10 u 105 lb m C 8 H 18 / day
0.703 u 62.43 lb m ft
32.174 ft
3
c. Table B.4: pC* 8 H18 (90 o F)
s
2
1 lb f lb ft 32.174 m2 s
20.74 mm Hg
(18 - 8) ft
14.696 psi
29.921 in Hg 14.696 lb f / in 2
0.40 lb f / in 2
6.21 in Hg
poctane y octane P 760 mm Hg Octane lost to environment = octane vapor contained in the vapor space displaced by liquid during refilling.
Volume:
5.288 u 10 4 gal
1 ft 3 7.481 gal
7069 ft 3
6-10
6.20 (cont’d) (16.0 + 14.7) psi 7069 ft 3 36.77 lb - moles 10.73 ft 3 psi / (lb - mole o R) (90 + 460) o R pC H 0.40 psi Mole fraction of C 8 H 18 : y = 8 18 0.0130 lb - mole C 8 H 18 / lb - mole P (16.0 + 14.7) psi pV RT
Total moles: n
Octane lost
0.0130(36.77) lb - mole 0.479 lb - mole ( 55 lb m
25 kg)
d. A mixture of octane and air could ignite. * * (85o F) = ptol (29.44 o C) = 35.63 mmHg = ptol 6.21 a. Antoine equation ptol
ptol P
Mole fraction of toluene in gas: y yPV RT 0.0469 lb - mole tol
Toluene displaced
35.63 mmHg 760 mmHg
0.0469 lb - mole toluene / lb - mole
yntotal
1 atm
lb - mole
0.7302
1 ft 3
900 gal
3
ft atm o
lb - mole R
(85 460) o R 7.481 gal
92.13 lb m tol lb - mole
1.31 lb m toluene displaced
b. Basis: 1mol 0.0469 mol C7H8(v)/mol 0.9531 mol G/mol
nV (mol) y (mol C7H8(v)/mol) (1-y) (mol G/mol) T(oF), 5 atm
Assume G is noncondensable
nL [mol C7H8 (l)] 90% of C7H8 in feed
90% condensation n L
0.90(0.0469)(1) mol C 7 H 8
Mole balance: 1 nV 0.0422 nV Toluene balance: 0.0469(1) Raoult’s law:
ptol
0.0422 mol C 7 H 8 (l )
0.9578 mol
y (0.9578) 0.0422 y
0.004907 mol C 7 H 8 / mol
* (0.004907)(5 u 760) 18.65 mmHg = ptol (T )
yP
Antoine equation: T
B C( A log 10 p * ) A log 10 p
*
1343.943 219.377(6.95334 log 10 18.65) 6.95334 log 10 18.65
6-11
17.12 o C = 62.8 o F
6.22 a. Molar flow rate: n =
VP RT
100 m 3 h
kmol K 2 atm 3 82.06 u 10 m atm (100 + 273) K -3
6.53 kmol / h
b. Antoine Equation: 1171.530 100 + 224.366 1845 mm Hg
log 10 p *Hex (100q C) = 6.87776 p* p Hex
y Hex P
p *Hex (T )
0.150(2.00) atm 760 mm Hg
* not saturated 228 mm Hg < p Hex
atm
228 mm Hg log 10 228 = 6.87776 -
3.26601
1171.530 T + 224.366
2.35793 T
34.8q C
nV (kmol/h) y (kmol C6H14 (v)/kmol), sat’d (1-y) (kmol N2/kmol) T (oC), 2 atm
c. 6.53 kmol/h 0.15 C6H14 (v) 0.85 N2
nL (kmol C6H14 (l)/h) 80% of C6H14 in feed
n L 0.80(0.15)(6.53 kmol / h) = 0.7836 kmol C 6 H 14 (l ) / h 80% condensation: Mole balance: 6.53 nV 0.7836 nV 5.746 kmol / h Hexane balance: 015 . (6.53) y (5.746) 0.7836 y 0.03409 kmol C 6 H 14 / kmol
(0.03409)(2 u 760 mmHg) = 51.82 mmHg = p *Hex (T ) 1171530 . T 2.52 o C Antoine equation: log 10 5182 . 6.87776 T 224.366 Raoult’s law:
p Hex
yP
6.23 Let H=n-hexane a. n0 ( kmol / min) y0 (kmol H(v)/kmol (1-y0) (kmol N2/kmol) 80oC, 1 atm, 50% rel. sat’n
50% relative saturation at inlet: y o P Table B.4
yo
Condenser
n1 ( kmol / min) 0.05 kmol H(v)/kmol, sat’d 0.95 kmol N2/kmol T (oC), 1 atm 1.50 kmol H(l)/min
0.500 p H* (80 o C)
(0.500)(1068 mmHg) = 0.703 kmolH / kmol 760 mmHg
Saturation at outlet: 0.05 P
p H* (T1 ) p H* (T1 )
6-12
0.05(760 mmHg) = 38 mmHg
6.23 (cont’d) Antoine equation: log 10 38 6.87776
1171530 . T1 T1 224.366
UV RS W T
Mole balance: n 0 n1 150 . n 0 N 2 balance: (1 0.703)n 0 0.95n1 n1 N2 volume: VN 2 b.
3.24 o C
2.18 kmol / min 0.682 kmol / min
(0.95)0.682 kmol 22.4 m 3 (STP) 14.5 SCMM min kmol
Assume no condensation occurs during the compression
2.18 kmol/min 0.703 H(v) 0.297 N2 80oC, 1 atm
Compressor
V0 ( m3 / min) 2.18 kmol/min 0.703 H(v) 0.297 N2 T0 (oC), 10 atm, 50% R.S.
V1 (m 3 / min) 0.682 kmol/min 0.05 H(v), sat’d 0.95 N2 T1 (oC), 10 atm Condenser
1.5 kmol H(l)/min
50% relative saturation at condenser inlet: 0.500 p *H (T0 )
u 10 4 mmHg 0.703(7600 mmHg) p H* (T0 ) 1068 .
Saturation at outlet: 0.050(7600 mmHg) = 380 mmHg = p *H (T1 ) Volume ratio:
V1 V0
n1 RT1 / P n0 RT0 / P
n1 (T1 273.2) n0 (T0 273.2)
Antoine
Antoine
0.682 kmol / min 1321 K u 2.18 kmol / min 460 K
T0
187 o C
T1
48.2q C
m 3 out 0.22 3 m in
c. The cost of cooling to 3.24 o C (installed cost of condenser + utilities and other operating costs) vs. the cost of compressing to 10 atm and cooling at 10 atm. 6.24 a.
Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes. (SG)nonane n max
Assume T n gas
15 L C 9 H 20 (l ) 0.718 u 1.00 kg L C 9 H 20
kmol 128.25 kg
25o C, P = 1 atm 1 kmol 2 u 10 4 L 273 K 3 298 K 22.4 u 10 L(STP)
6-13
0.818 kmol
0.084 kmol C 9 H 20
6.24 (cont’d) y max
n max n gas
0.084 kmol C 9 H 20 0.818 kmol
010 . kmol C 9 H 20 / kmol (10 mole%)
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8% to 2.9%). The answer is therefore yes . The nonane will not spread uniformly—it will be high near the sump as long as liquid is present (and low far from the sump). There will always be a region where the mixture is explosive at some time during the evaporation. b. ln p *
A B T
T1
258 . o C = 299 K, p1*
5.00 mmHg
T2
66.0 o C = 339 K, p2*
40.0 mmHg
5269 5269 ln( 40.0 / 5.00) A 5269, B = ln(5.00) + 19.23 p * exp(19.23 ) 1 1 T ( K) 299 339 299 At lower explosion limit, y 0.008 kmol C 9 H 20 / kmol p * (T ) yP (0.008)(760 mm Hg) A
= 6.08 mm Hg
Formula for p
T
*
302 K = 29 o C
c. The purpose of purge is to evaporate and carry out the liquid nonane. Using steam rather than air is to make sure an explosive mixture of nonane and oxygen is never present in the tank. Before anyone goes into the tank, a sample of the contents should be drawn and analyzed for nonane. 6.25 Basis: 24 hours of breathing n0 (mol H 2 O) 23°C, 1 atm n1 (mol) @ hr = 10% 0.79 mol N 2/mol y 1 (mol H 2 O/mol) + O2 , CO2
Air inhaled: n1
Lungs
O2
CO2
37°C, 1 atm n2 (mol), saturated 0.75 mol N 2/mol y 2 (mol H 2 O/mol) + O2 , CO2
12 breaths 500 ml 1 liter min breath 10 3 ml
b
273K 1 mol 23 273 K 22.4 liter STP
g
b g
60 min 24 hr 1 hr 1 day
356 mol inhaled day
Inhaled air - -10% r. h.: y1 Inhaled air - -50% r. h.: y1
b
g
b
g
010 . p H 2O 23q C
010 . 2107 . mm Hg
P
760 mm Hg
b
g
b
0.50 p H 2O 23q C
0.50 2107 . mm Hg
P
760 mm Hg
6-14
g
2.77 u 10 3
mol H 2 O mol
139 . u 10 2
mol H 2 O mol
6.25 (cont’d) n2 y 2 n1 y1 (n0 ) 10% rh (n0 ) 50% rh
H 2 O balance: n0
(n1 y1 ) 50% (n1 y1 ) 10%
FG 356 mol IJ L(0.0139 0.00277) mol H O OFG 18.0 g IJ mol PQH 1 mol K H day K MN 2
71 g / day
Although the problem does not call for it, we could also calculate that n2 = 375 mol exhaled/day, y2 = 0.0619, and the rate of weight loss by breathing at 23oC and 50% relative humidity is n0 (18) = (n2y2 - n1y1)18 = 329 g/day. 6.26 a. To increase profits and reduce pollution. b. Assume condensation occurs. A=acetone n 1 mol @ To C, 1 at m y1 mol A(v)/ mol (sat’d) (1-y1 ) mol N2 /mo l
1 mo l @ 90o C, 1 atm 0.20 mol A(v)/ mol 0.80 mol N2 /mo l
n 2 mol A(l)
For cooling water at 20oC
d
log 10 p *A 20 o C
i
Saturation: y1 P
7.02447
d
d
11610 . 20 224
i
p *A 20 o C y1
2.26627 p *A 20 o C 184.6 760
i
184.6 mmHg
0.243 ! 0.2 , so no saturation occurs.
For refrigerant at –35oC
d
log 10 p *A 35o C
i
7.02447
11610 . 35 224
d
0.88161 p *A 35o C
i
7.61 mmHg
7.61 0.0100 d i 760 N mole balance: 1b0.8g n b1 0.01g n 0.808 mol Saturation: y1 P 2
p *A 35o C y1 1
1
Total mole balance: 1 0.808 n2 n2 Percentage acetone recovery: c. d.
0.192 mol
0192 . u 100% 1
96%
Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant The condenser temperature could never be as low as the initial cooling fluid temperature because heat is transferred between the condenser and the surrounding environment. It will lower the percentage acetone recovery.
6-15
6.27
Basis:
12500 L 1 mol 273 K 103000 Pa h 22.4 L(STP) 293 K 101325 Pa
n o (mol/h) @ 35o C, 103 KPa y0 [mol H2O(v)/mol] y1– H2 O(v)/ mol o mol y0 (mol DA/mol) ) mol DA/mo l (1-y o hr=90% h r =90%
Inlet: y o
d
hr p H* 2 O 35o C P
d
p H* 2 O 20 o C
i
i
528.5 mol / h
528.5 (mo l/h) @ 20o C, 103 KPa y0 [mol H2O(v)/mol] y1– H2 O(v)/ mol (sat’d) 1 mol y0 (mol DA/mol) (1-y1 ) mol DA/mo l
[molH2O(l)/h H2O(l)/h] n 2n2mol
0.90 u 42.175 mmHg 101325 Pa 103000 Pa 760 mmHg
0.4913 mol H 2 O / mol
17.535 mmHg 101325 Pa 0.02270 mol H 2 O / mol P 103000 Pa 760 mmHg Dry air balance: 1 0.04913 no 1 0.02270 528.5 no 543.2 mol / h Outlet: y1
b
g b
gb
g
543.2 mol 22.4 L(STP) 308 K 101325 Pa 13500 L / h h mol 273 K 103000 Pa Total balance: 543.2 528.5 n2 n2 14.7 mol / h
Inlet air:
14.7 mol 18.02 g H 2 O 1 kg h 1 mol H 2 O 1000 g
Condensation rate:
6.28
Basis:
0.265 kg / h
10000 ft 3 1 lb - mol 492 o R 29.8 in Hg min 359 ft 3 (STP) 550 o R 29.92 in Hg
n1 lb-mole/min 40oF, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] 1- y1 (lb-mole DA/mol)
24.82 lb-mole/min 90oF, 29.8 in.Hg y0 [lb-mole H2O(v)/mol 1- y0 (lb-mole DA/mol) hr = 88%
Inlet: y o Outlet: y1
d
i
b
0.88 36.07 mmHg
P
29.8 in Hg
i
p H* 2 O 40 o F P
b
g
1 in Hg 25.4 mmHg
6.274 mmHg 1 in Hg 29.8 in Hg 25.4 mmHg
Dry air balance: 24.82 1 0.0419 Total balance: 24.82
n1 lb-mole/min 65oF, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] 1- y1 (lb-mole DA/lb-mole)
n2 [lb-mole H2O(l)/min]
hr p H* 2O 90 o F
d
24.82 lb - mol / min
0.00829 lb - mol H 2 O / lb - mol
g n b1 0.00829g n 1
23.98 n2 n2
1
23.98 lb - mol / min
0.84 lb - mole / min
6-16
0.0419 lb - mol H 2 O / lb - mol
6.28 (cont’d) 0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal min lb mol 62.4 lb m 1 ft 3
Condensation rate:
Air delivered @ 65oF:
181 . gal / min
23.98 lb - mol 359 ft 3 (STP) 525o R 29.92 in Hg min 1 lb mol 492 o R 29.8 in Hg
9223 ft 3 / min
6.29 Basis: 100 mol product gas
no mol, 32oC, 1 atm yo mol H2O(v)/mol (1-yo) mol DA/mol hr=70%
100 mol, T1, 1 atm
100 mol, 25oC,1 atm
y1 mol H2O(v)/mol, (sat’d) (1-y1) mol DA/mol
y1 mol H2O(v)/mol, (1-y1) mol DA/mol hr=55%
(mol HH22O(l)) nn22lb-mol O(l)/min
Outlet: y1
d
hr p H* 2 O 25o C
i
b
g
0.55 23.756
P
b g 13.07 p bT g T 15.3 C d32 Ci 0.70b35.663g 0.0328 mol H O / mol
hr p H* 2O
b
g
Total balance: 1016 . n2
Ratio:
b
2
b
g
100 1 0.0172 no
100.0 n2
16 . mol 18.02 g 1 kg 1 mol 1000 g
g
1016 . mol
. mol (i.e. removed) 16
0.0288 kg H 2 O
100 1 0.0172 mol 29.0 g 1 kg 1 mol 1000 g
0.0288 2.85
1
760
Dry air balance: no 1 0.0328
kg dry air:
1
o
P
kg H 2 O removed :
o
* H 2O
Saturation at T1 : 0.0172 760 Inlet: y o
0.0172 mol H 2 O / mol
760
2.85 kg dry air
0.0101 kg H 2 O removed / kg dry air
6-17
6.30 a.
Room air T
22q C , P 1 atm , hr
b
Second sample T
g
9251 mm Hg 839 mm Hg
ae bH , y1
ln y
bH ln a y
b
0.01044g . g lnb01103
H 2 H1
y Basis:
b
0.01044, H1
g b gb g expb0.054827 H g
ln 0.01044 0.054827 5
7.937 u 10 3
1 m 3 delivered air
b
n o mol, 35o C, 1 at m yo mol H2 O(v)/ mol (1-yo ) mol DA/mo l H=30
01103 . mol H 2 O mol 5 , y2
01103 . , H2
48
0.054827
48 5
ln a ln y1 bH1
b.
50q C , P 839 mm Hg , saturated:
b
ln y 2 y1
0.01044 mol H 2 O mol
760 mm Hg
P H 2 O 50q C y 2
y2 P
b
b0.40g19.827 mm Hg
g
0.40 P H 2 O 22q C y1
y1 P
40% :
b
g
4.8362 a exp 4.8362
273K 1 k mol 22 273 K 22.4m 3 STP
g
b g
10 3 mol 1 kmol
7.937 u 10 3
4131 . mol air delivered
41.31 mol, T, 1 at m
41.4 mol, 22o C,1 at m
0.0104 mo l H2 O(v)/mo l, (sat’d) 0.09896 mo l DA/ mol
0.0104 mo l H2 O(v)/mo l 0.09896 mo l DA/ mol
n 1 mol H2 O(l)
Saturation condition prior to reheat stage: y H 2O P
bg
b g b0.01044gb760 mm Hgg
PH*2 O T PH 2 O T T
7.93 mm Hg Table 8.3
7.8q C
bg
Part a
30 y 0
Humidity of outside air: H
b
Overall dry air balance: n0 1 V0 Overall water balance: n0 y 0
n2
0.0411 mol H 2 O mol
. gb0.9896g b4131 b g b1 0.0411g 42.63 mol b4131 . gb0.0104g n b42.63gb0.0411g b4131 . gb0.0104g
g
4131 . 0.9896 n0
2
132 . mol H 2 O condensed Mass of condensed water
132 . mol H 2 O 18.02 g H 2 O 1 kg 1 mol H 2 O 10 3 g 0.024 kg H 2 O condensed m 3 air delivered
6-18
6.31 a.
Basis: n 0 mol feed gas . S solvent , G
solvent - free gas n1 (mol) @ Tf (qC), P4 (mm Hg) y1 [mol S(v)/mol] (sat’d) (1–y1) (mol G/mol)
n0 (mol) @ T0 (qC), P0 (mm Hg) y0 (mol S/mol) (1-y0) (mol G/mol) Td0 (qC) (dew point)
n2 (mol S (l))
p Tdo y o
n1 n2 n1
n0 y 0 f n 0 n2
(1)
d i
p T f
d i
Fractional condensation of S = f n2
b gb g
Po
p T f y1
Saturation condition at outlet: y1 Pf
Total mole balance: n 0
b g
p Tdo
b g
Inlet dew point = T0 y o Po
(2)
Pf (1) o n2
bg
Eq. 3 for n1
n1
b gP n fp bT g n0 fp T0
n 0
0
0
do
Po
n1 y1 n2
S balance: n0 y 0
(1) - (4)
b g OPFG p dT iIJ n fp bT g P PQGH P JK p L fp eTdo j OP p dT i M1 MM Po PP b1 f g p bT g LM1 fp bT g OP p dT i P N Q P MN P PQ P p T b1 f g eP do j b g LMn P MN
n 0 p Tdo
0
n 0 fp Tdo
o
f
o
0
f
do
o
f
do
f
do
o
o
f
f
o
b.
Condensation of ethylbenzene fromnitrogen Antoine constants for ethylbenzene A= 6.9572 B= 1424.3 C= 213.21 Run T0 P0 Td0 f 1 2 3 4
50 50 50 50
765 765 765 765
40 40 40 40
0.95 0.95 0.95 0.95
Tf
p* (Td0) p*(Tf)
45 40 35 20
21.492 21.492 21.492 21.492
6-19
27.62 21.49 16.56 7.08
Pf 19137 14892 11472 4904
Crefr Ccomp Ctot 2675 4700 8075 26300
107014 109689 83327 88027 64244 72319 27595 53895
(3) (4)
6.31 (cont’d) When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to c. increase the fractional condensation. When you decrease Tf, less compression is required to achieve a specified fractional condensation. d.
6.32 a.
A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr). However, since less compression is required at the lower temperature, Ccomp is lower at the lower temperature. Similarly, running at a higher Tf lowers the refrigeration cost but raises the compression cost. The sum of the two costs is a minimum at an intermediate temperature.
Basis : 120 m 3 min feed @ 1000 o C(1273K), 35 atm . Use Kay’s rule. Cmpd. H2 CO CO 2 CH 4 Tcc Pcc Tr Pr V
b g P batmg bT g
Tc K 33.2
c corr .
c
12.8
133.0 304.2 190.7
34.5 72.9 458 .
413 .
bP g
c corr
bApply Newton' s corrections for H g 2
20.8
b g b g b g b g 133.4 K 0.40b20.8g 0.35b34.5g 0.20b72.9g 0.05b458 . g 37.3 atm 1273 K 133.4 K 9.54 U Generalized compressibility charts (Fig. 5.3 - 2) V z 102 35.0 atm 37.3 atm 0.94 W . ¦yT ¦y P
i ci
i
ci
8.314 N m
1.02 35 atm
120 m 3 min
0.40 413 . 0.35 133.0 0.20 304.2 0.05 190.7
1273 K
mol K
mol 1 kmol 3.04 u 10 m 3 10 3 mol 3
1 atm 101325 N m
3
3.04 u 10 3 m 3 mol
39.5 kmol min
n1 (kmol/min), 261 K, 35atm yNaOH sat’d yH2 yCH4 (2% of feed) yCO
1.2(39.5) kmol/min MeOH(l) 39.5 kmol/min, 283K, 35 atm 0.40 mol H2/mol 0.35 mol CO/mol 0.20 mol CO2/mol 0.05 mol CH4/mol
n2 (kmol/min), liquid yMeOH yCO2 yCH4 (98% of feed)
6-20
6.32 (cont’d)
b
p MeOH 261K
Saturation at Outlet: y McOH
b.
n MeOH
n MeOH n + H 2 nCH 4
A input
10
P 4.97 u 10
y McOH
g
b
7 .878631473.11 12 2300
b
g
mm Hg
g
35 atm 760 mm Hg atm 4
mol MeOH mol n MeOH
A 0.02 of input
n MeOH 39.5 0.40 0.02 0.05 0.35
nCO
A input
E
n MeOH
0.0148 kmol min MeOH in gas
The gas may be used as a fuel. CO2 has no fuel value, so that the cost of the added energy required to pump it would be wasted.
6.33 n0 (kmol/min wet air) @ 28qC, 760 mmHg
n1 (kmol/min wet air) @ 80qC, 770
y1 (mol H2O/mol) (1-y1) (mol dry air/mol) 50% rel. sat.
y2 (mol H2O/mol) (1-y2) (mol dry air/mol) Tdew point = 40.0oC
1500 kg/min wet pulp
m 1 (kg/min wet pulp)
0.75 /(1 + 0.75) kg H2O/kg 1/1.75 kg dry pulp/kg
0.0015 kg H2O/kg 0.9985 kg dry pulp/kg
Dry pulp balance: 1500 u
1 1 0.75
50% rel. sat’n at inlet: y1 P
m 1 (1 0.0015) m 1
0.50 p H* 2 O (20 o C) y1
858 kg / min
0.50(28.349 mm Hg) / (760 mm Hg)
= 0.0187 mol H 2 O / mol o
40 C dew point at outlet: y 2 P
p H* 2 O (40 o C)
y2
(55.324 mm Hg) / ( 770 mm Hg) = 0.0718 mol H 2 O / mol
Mass balance on dry air: n 0 (1 0.0187) n1 (1 0.0718)
(1)
Mass balance on water: n 0 ( 0.0187 )(18.0 kg / kmol ) 1500( 0.75 / 1.75) Solve (1) and (2) n 0
622.8 kmol / min, n1
n 1 ( 0.0718)(18) 858( 0.0015) ( 2 )
658.4 kmol / min
Mass of water removed from pulp: [1500(0.75/1.75)–858(.0015)]kg H2O = 642 kg / min Air feed rate: V0
622.8 kmol 22.4 m 3 (STP) (273 + 28) K 1.538 u 10 4 m 3 / min kmol 273 K min
6-21
6.34
Basis: 500 lb m hr dried leather (L) n1 (lb - moles / h)@130o F, 1 atm n0 (lb - moles dry air / h)@140o F, 1 atm
y1 (lb - moles H2 O / lb - mole) (1- y1 )(lb - moles dry air / lb - mole)
0 (lb m / h) m
500 lb m / h 0.06 lb m H2 O(l) / lb m 0.94 lb m L / lb
0.61 lb m H2 O(l) / lb m 0.39 lb m L / lb m
b0.94gb500g m
Dry leather balance: 0.39m0 Humidity of outlet air: y1 P
0
b
1205 lb m wet leather hr
g
0.50 p H 2 O 130q F y1
b gb
H 2 O balance: 0.61 1205 lb m hr n1
g
a
fb
0.06 500 lb m hr
E
b1 0.0756g(517.5) lb - moles hr 478.4 lb - moles 359 ft bSTPg b140 460gq R 3
6.35 a.
hr
0.0756
mol H 2 O mol
g
0.0756n1 lb - moles H 2 O
18.02 lb m
hr
1 lb - mole
517.5 lb - moles hr
Dry air balance: n0
Vinlet
0.50(115 mm Hg) 760 mmHg
492q R
1 lb - mole
478.4 lb - moles hr
2.09 u 10 5 ft 3 hr
Basis: 1 kg dry solids n 1 (kmol)N 2, 85°C
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N 2 /mol) 70% rel. sat.
dryer 1.00 kg solids 0.78 kg Hex
n 4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80q C
b0.78 0.05gkg Antoine eq.
B
70% rel. sat.: y 2
condenser
b
0.70 p hex 80q C
n 3 (kmol) 28°C, 5.0 atm y 3 (mol Hex/mol) sat'd (1 – y 3) (mols N 2 /mol)
kmol 8.47 u 10 3 kmol Hex 86.17 kg
g b0.70g10
P
b
6.87776 1171.530 80 224 .366
760
6-22
g 0.984 mol Hex mol
6.35 (cont’d) n2
8.47 u 10 3 kmol Hex
1 kmol 0.984 kmol Hex
b1 0.984g0.0086
N 2 balance on dryer: n1
Antoine Eq.
B
b
p hex 28q C
Saturation at outlet: y 3
g
10
0.0086 kmol u 10 4 kmol 1376 .
b
6.877761171.580 28 224 .366
5 760
b
g
. u 10 4 kmol 144
n3 1 0.0452 n3
Mole balance on condenser: 0.0086 144 . u 10 4 n4 n4 Fractional hexane recovery:
0.0452 mol Hex mol
b g
P
Overall N 2 balance: 1.376 u 10 -4
g
0.0085 kmol cond. 86.17 kg 0.78 kg feed
kmol
0.0085 kmol 0.939 kg cond. kg feed
b. Basis: 1 kg dry solids
0.9n
heater
3
0.9n 3 (kmol) @ 28°C, 5.0 atm
y3 (1 – y3)
y 3 (mol Hex/mol) sat'd (1 – y 3) (mol N 2/mol)
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N2 /mol) 70% rel. sat.
n 1 (kmol)N 2 85°C dryer 1.00 kg solids 0.78 kg Hex
condenser
0.1n3
n4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80q C: .8.47 u 10 3 0.9n3 (0.0452) N2 balance on dryer: n1 0.9n3 (1 0.0452) Overall N2 balance: n1
n3 (kmol) y3 (1 – y3)
n2 (0.984)
n2 (1 0.984)
( 2)
0.3n3 (1 0.0452)
R|n Equations (1) to (3) Sn |Tn
2 3
u 10 5 kmol 1515 .
Saved fraction of nitrogen =
(3)
5
u 10 kmol 1447 . 0.0086 kmol
1
(1)
u 10 5 1.376 u 10 -4 1447 . u 100% 1.376 u 10 4
90%
Introducing the recycle leads to added costs for pumping (compression) and heating.
6-23
6.36 b. m 1 (lbm/h)
300 lbm/h wet product 0.2 . 0167 lb m T(l) / lb m 1 0.2 0.833 lb m D / lb m
0.02 / (102 . ) 0.0196 lb m T(l) 0.9804 lb m D / lb m
Dryer
n3 (lb-mole/h) @ 200
n1 (lb-mole/h)
O
F,
y3 (lb-mole T/lb-mole) y1 (lb-mole T(v)/lb-mole) (1-y3)( lb-mole N2/lb-mole) (1–y1) (lb-mole N2/lbmole) T=toluene Heater 70% r.s.,150oF, 1.2 atm D=dry solids n3 (lb-mole/h)
y3 (lb-mole T(v)/lb-mole) (1-y3) (lb-mole N2/lb-mole)
Condenser
Eq.@ 90OF, 1atm
n2 ( lb-mole T(l)/h
Strategy: Overall balance m 1 & n 2 ; Relative saturationy1;, Gas and liquid equilibriumy3 Balance over the condenser n1 & n 3
UV RS W T
m 1 u 0.0196 n 2 u 92.13 Toluene Balance: 300 u 0167 . m 1 255 lb m / h m 1 u 0.9804 Dry Solids Balance: 300 u 0833 . n 2 0.488 lb - mole / h 70% relative saturation of dryer outlet gas: pC* 7 H8 y1 P
O
O
(150 F = 6556 . C) = 10
(6.95334-
0.70 pC* 7 H8 (150 O F) y1
1343.943 ) 65.56 219 .37
0.70 pC*7 H8 P
172.47 mmHg
(0.70)(172.47) . u 760 12
lb - mole T(v) / lb - mole . 01324
Saturation at condenser outlet: pC* 7 H8 y3
O
O
(90 F = 32.22 C) = 10 pC* 7 H8 P
40.90 760
(6.95334-
1343.943 ) 32 .22 219 .37
40.90 mmHg
0.0538 mol T(v) / mol
UV RS W T
Condenser Toluene Balance: n1 u 01324 0.488 n 3 u 0.0538 . n1 Condenser N 2 Balance: n1 u (1 01324 . ) n 3 u (1 0.0538) n 3
6-24
5875 lb - mole / h . 5.387 lb - mole / h
6.36 (cont’d) Circulation rate of dry nitrogen = 5.875 u (1 - 0.1324) =
5.097 lb - mole
lb - mole
h
28.02 lb m
lb m / h 0182 .
Vinlet
6.37
b g
5.387 lb - moles 359 ft 3 STP hr 1 lb - mole
C 6 H 14
Basis: 100 mol C 6 H 14
(200 460)q R 492q R
2590 ft 3 h
19 O 2 o 6CO 2 7H 2 O 2 n1 (mol) dry gas, 1 atm 0.821 mol N 2 /mol D.G. 0.069 mol CO2 /mol D.G. 0.021 mol CO/mol D.G. 0.086 mol O 2 /mol D.G. 0.00265 mol C 6 H 14/mol D.G. n2 (mol H 2 O)
100 mol C 6 H 14 n0 (mol) air 0.21 mol O 2 /mol 0.79 mol N 2 /mol
b g n LMM0b.069g 0b.021g 6b0b.00265g gOPP n 5666 mol dry gas Q N 100 0.00265b5666g mol reacted u 100% 85.0% Conversion: C balance: 6 100
1
CO 2
1
CO
C 6 H 14
100 mol fed
N 2 balance: 0.79n0 Theoretical air:
Excess air:
b g
0.821 5666 n0
100 mol C 2 H 14
5888 4524 u 100% 4524
b g
H balance: 14 100 Dew point: y H 2 O
5888 mol air
19 mol O 2
1 mol air
2 mol C 2 H 14
0.21 mol O 2
30.2% excess air
b gb
g
2n2 5666 14 0.00265 n2 595 595 + 5666
4524 mol air
d i
p Tdp
760 mm Hg
6-25
595 mol H 2 O
d i
p Tdp
Table B.3
72.2 mm Hg Tdp
451 . qC
6.38 Basis: 1 mol outlet gas/min n0 ( mol / min) y 0 ( mol CH 4 / mol) (1 y0 ( mol C 2 H 6 / mol)
1 mol / min @ 573K, 105 kPa y1 (mol CO 2 / mol) y 2 (mol H 2 O / mol) (1 y1 y 2 ) mol N 2 / mol
n1 (mol O 2 / min) 3.76n1 (mol N 2 / min)
CH 4 2O 2 o CO 2 2H 2 O pCO 2
80 mmHg y1
80 mmHg 101325 Pa 105000 Pa 760 mmHg
b
100% O2 conversion : 2no yo 7 no 1 yo
b
C balance: no yo 2no 1 yo
. n1 N2 balance: 376
C2 H6
2
g
g
7 O 2 o 2CO 2 3H 2 O 2 0.1016 mol CO 2 / mol
(1)
n1
01016 .
(2)
1 y1 y2
b
H balance: 4no yo 6no 1 yo
g
(3) 2 y2
(4)
R|n 0.0770 mol | y 0.6924 mol CH / mol Solve equations 1 to 4 S mol O . ||n 01912 mol H O / mol . Ty 01793 Dew point: 01793 . b105000g Pa 760 mmHg 1412. mmHg T p dT i o
4
o
* H2 O
1
2
2
2
dp
dp
101325 Pa
b
6.39 Basis: 100 mol dry stack gas n P (mol C 3 H 8) n B (mol C 4H10 ) n out (mol) 0.21 O2 0.79 N2
P = 780 mm Hg Stack gas: Tdp = 46.5°C 100 mol dry gas 0.000527 mol C 3 H 8/mol 0.000527 mol C 4H 10/mol 0.0148 mol CO/mol 0.0712 mol CO 2/mol + O 2, N 2
nw (mol H2O)
C 3 H 8 5O 2 o 3CO 2 4H 2 O
C 4 H 10
6-26
g
58.8 o C Table B.3
13 O 2 o 4CO 2 5H 2 O 2
Dew point But y w
b
46.5q C y w P
nw 100 n w
g
77.6 mm Hg 780 mm Hg
p w 46.5q C y w
0.0995 n w
0.0995
mol H 2 O mol
11.0 mol H 2 O
b100g b0.000527gb3g b0.000527gb4g 0.0148 0.0712 3n 4n 8.969 b1g H balance: 8n 10n b100g b0.000527gb8g b0.000527gb10g b110 . gb2g 8n 10n 23149 . b2 g R| 56% C H . mol C H U Rn 1454 S Solve b1g & b2g simultaneously: S V . mol C H W |T 44% C H Tn 1152 bAnswers may vary r 2% due to loss of precisiong C balance: 3n p 4n B p
B
p
B
p
B
p
3
8
3
8
B
4
10
4
10
6.40 a. L1 (lb - mole C 10 H 22 / h)
L 2 (lb - mole / h) x 2 (lb - mole C 3 H 8 / lb - mole) 1 x 2 (lb - mole C 10 H 22 / lb - mole)
G 1 (lb - mole / h) y1 (lb - mole C 3 H 8 / lb - mole) 1 y1 (lb - mole N 2 / lb - mole) Basis: G 2
G 2 = 1 lb - mole / h 0.07 (lb - mole C 3 H 8 / lb - mole) 0.93 (lb - mole N 2 / lb - mole)
1 lb - mole h feed gas
b gb g
N 2 balance: 1 0.93
b
g
98.5% propane absorption G 1 y1
b1g & b2g G
1
b
g 0.93 b1 0.985gb1gb0.07g G y
G 1 1 y1 G 1 1 y1
1 1
. u 10 3 105
b1g b2g
u 10 3 mol C 3 H 8 mol 1128 .
0.93105 lb - mol h , y1
Assume G 2 L 2 streams are in equilibrium
From Cox Chart (Figure 6.1-4), p *C3 H8 (80 o F ) 160 lb / in 2
b
g
Raoult' s law: x 2 p C3H 8 80q F
b gb g
Propane balance: 0.07 1 Decane balance: L1
0.07 p x 2
b0.07gb10. atmg
G 1 y1 L 2 x 2 L 2
10.89 atm
b
gd h b d L / G h 1
2 min
0.006428
gd
mol H 2 O mol
u 10 3 0.07 0.93105 1128 .
gb
g
10.66 lb - mole h
10.7 mol liquid feed / mol gas feed
6-27
i
0.006428
10.726 lb - mole h 1 x 2 L 2 1 0.006428 10.726
b
10.89 atm
6.40 (cont’d) b. The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed rate and fractional absorption], or n C3H 8
10.726 lb - mole 0.006428 lb - mole C 3 H 3 h
lb - mole
0.06895 lb - mol C 3 H 8 h
The decane flow rate is 1.2 x 10.66 = 12.8 lb-moles C10H22/h x2
b
0.06895 lb - mole C 3 H 8 h 0.06895 + 12.8 lb - moles h
g
0.00536 lb - mole C 3 H 8 / lb - mole
c. Increasing the liquid/gas feed ratio from the minimum value decreases the size (and hence the cost) of the column, but increases the raw material (decane) and pumping costs. All three costs would have to be determined as a function of the feed ratio. 6.41 a. Basis: 100 mol/s liquid feed stream
Let B n - butane , HC = other hydrocarbons
o
100 mol/s @ 30 C, 1 atm
n4 (mol/s) @ 30qC, 1 atm
xB =12.5 mol B/s 87.5 mol other hydrocarbon/s
y4 (mol B/mol) (1-y4) (mol N2/mol)
88.125 mol/s
n3 (mol N2/s)
0.625 mol B/s (5% of B fed) 87.5 mol HC/s
p *B (30 o C) # 41 lb / in 2 Raoult' s law: y 4 P
2120 mm Hg (from Figure 6.1-4)
b
95% n-butane stripped: n 4 0.3487 Total mole balance: 100 n 3 mol gas fed mol liquid fed
b. If y 4
0.8 u 0.3487
x B p B* (30 o C) 0125 . u 2120 760 P 12.5 0.95 n 4 34.06 mol / s
x B p *B (30 o C) y 4
0.3487
g b gb g
34.08 88125 . n 3
22.20 mol / s 100 mol / s
22.20 mol / s
0.2220 mol gas fed / mol liquid fed
0.2790 , following the same steps as in Part (a),
b
g b gb g
12.5 0.95 n 4 42.56 mol / s 95% n-butane is stripped: n 4 0.2790 Total mole balance: 100 n 3 42.56 88.125 n 3 30.68 mol / s mol gas fed 30.68 mol / s 0.3068 mol gas fed / mol liquid fed mol liquid fed 100 mol / s c. Increasing the nitrogen feed rate can cause an increase of nitrogen consumption, but more butane can be recovered. Therefore, costs of nitrogen (including compression) and butane are needed to determine the most cost-effective value of the gas/liquid feed rate.
6-28
6.42 Basis: 100 mol NH 3 Preheated air 100 mol NH 3 780 kPa sat'd
N2 O2
converter
n3 n4 n5 n6
n 1 (mol) O 2 3.76 n 1 (mol) N 2 n 2 (mol) H 2 O 1 atm, 30°C h r = 0.5
a. i)
b g
P Tsat
NH 3 feed: P
Antoine: log 10 6150
b
g
820 kPa
55 wt% HNO 3 (aq ) n 8 (mol HNO 3 ) n 9 (mol H 2O)
n7 (mol H 2O)
6150 mm Hg 8.09 atm
b
g
7.55466 1002.711 Tsat 247.885 Tsat
Table B.1
VNH 3
absorber
(mol NO) (mol N 2) (mol O 2 ) (mol H 2O)
Pc Tc
18.4q C 291.6 K
UV W
. atm Pr 8.09 / 1113 . 1113 0.073 z . K Tr 2916 . / 4055 . 0.72 4055
b
0.92 100 mol
g
8.314 Pa 2916 . K mol - K 820 u 10 3 Pa
0.92
(Fig. 5.3-1)
0.272 m 3 NH 3
Air feed: NH 3 2O 2 o HNO 3 H 2 O n1
100 mol NH 3
2 mol O 2 mol NH 3
b
hr p * 30q C
Water in Air: y H 2 O
p 0.02094
200 mol O 2
g
. 0.500 u 31824 0.02094 760 n2 n2 20.36 mol H 2 O n2 4.76 200
A
( 4 .76 mol air mol O 2 )
Vair
b g
b g
4.76 200 20.36 mol 22.4 L STP 1 mol
303K
1 m3
273K 10 3 L
24.2 m 3 air
ii) Reactions: 4 NH 3 5O 2 o 4 NO 6H 2 O , 4 NH 3 3O 2 o 2 N 2 6H 2 O Balances on converter NO: n3
97 mol NH 3
4 mol NO 4 mol NH 3
6-29
97 mol NO
6.42 (cont’d)
b g
N 2 : n4
3.76 2.00 mol +
O 2 : n5
200 mol
3 mol NH 3
2 mol N 2 4 mol NH 3 5 mol O 2 4 mol NH 3
97 mol NH 3
. mol N 2 7535
3 mol O 2 76.5 mol O 2 4 mol NH 3 100 mol NH 3 6 mol H 2 O H 2 O: n6 20.36 mol + 170.4 mol H 2 O 4 mol NH 3 n total (97 7535 . 76.5 170.4) mol = 1097 mol converter effluent
3 mol NH 3
8.8% NO, 68.7% N 2 , 7.0% O 2 , 15.5% H 2 O iii) Reaction: 4 NO 3O 2 2 H 2 O o 4 HNO 3 HNO 3 bal. in absorber: n8
H 2 O in product: n9
97 mol NO react 4 mol HNO 3 4 mol NO
97 mol HNO 3
63.02 g HNO 3 mol
277.56 mol H 2 O
b
gb g
H balance on absorber: 170.4 2 2n7
b.
b
1 mol H 2 O 45 g H 2 O 55 g HNO 3 18.02 g H 2 O
gb gb
97 277.6 2 mol H
n7
155.7 mol H 2 O added
VH 2O
155.7 mol H 2 O 18.02 g H 2 O 1 cm 3 1 m3 1 mol 1 g 10 6 cm 3
V NH 3 Vair VH 2O
g bg
2.81 u 10 3 m 3 H 2 O l
97 mol HNO 3
63.02 g HNO 3 277.6 mol H 2 O 18.02 g HNO 3 mol mol 11115 g 11115 . kg
M acid in old basis
Scale factor
97 mol HNO 3
b1000 metric tonsgb1000 kg metric tong
8.997 u 10 4
11.115 kg
d8.997 u 10 id0.272 m NH i 2.45 u 10 m NH d8.997 u 10 id24.2 m airi 2.18 u 10 m air d8.997 u 10 id2.81 u 10 m H Oi 253 m H Oblg 4
3
4
3
3
4
4
6
3
3
2
6-30
3
3
3
3
2
6.43 a.
Basis: 100 mol feed gas 100 mol 0.10 mol NH3 /mol 0.90 mol G/mol
G = NH3 -free gas Absorber
n 1 (mol H2 O( l))
n 2 (mol) in equilibrium y A (mol NH 3 /mol) at 10°C(50°F) (mol H O/mol) yW 2 and 1 atm (mol G/mol) yG (mol) n3 x A (mol NH 3 /mol) (1 – x A) (mol H 2O/mol)
Composition of liquid effluent . Basis: 100 g solution Perry, Table 2.32, p. 2-99: T = 10oC (50oF), U = 0.9534 g/mL 0.120 g NH3/g solution
12.0 g NH 3 88.0 g H 2 O = 0.706 mol NH 3 , = 4.89 mol NH 3 (17.0 g / 1 mol) (18.0 g / 1 mol)
12.6 mole% NH 3 ( aq), 87.4 mole% H 2 O(l)
Composition of gas effluent T
o
50 F, x A
o . 0126
yA yW yG
p H 2O p total
. / 14.7 0.0823 mol NH 3 mol 121 . / 14.7 0.0105 mol H 2 O mol 0155 1 y A yW 0.907 mol G mol
b gb g n y n b100gb0.90g b0.907g 99.2 mol absorbed b100gb0.10g b99.2gb0.0823g . mol NH 184
G balance: 100 0.90 NH 3
g U| . 0155 psiabTable 2 - 21gV |W 14.7 psia b
. psia Table 2 - 23 121
p NH 3
Perry
% absorption
2
G
2
in
3
out
1.84 mol absorbed u 100% 18.4% 100 010 . mol fed
b gb g
b. If the slip stream or densitometer temperature were higher than the temperature in the contactor, dissolved ammonia would come out of solution and the calculated solution composition would be in error. 6.44 a. 15% oleum: Basis - 100kg 15 kg SO 3
85 kg H 2 SO 4
1 kmol H 2 SO 4 98.08 kg H 2 SO 4
84.4% SO 3
6-31
1 kmol SO 3 1 kmol H 2 SO 4
80.07 kg SO 3 1 kmol SO 3
84.4 kg
6.44 (cont’d) b.
Basis 1 kg liquid feed n o (mol), 40o C, 1.2 at m
n1 (mol), 40o C, 1.2 at m
0.90 mol SO3 /mol 0.10 mol G/ mol
y1 mol SO3 /mo l (1-y1 ) mol G/ mol Equilib riu m @ 40o C
1 kg 98% H2 SO4
m1 (kg) 15% oleu m
0.98 kg SO3 0.02 kg H2 O
0.15 kg SO3 /kg 0.85 kg H2 SO4 /kg
b
pSO3 40q C, 84.4%
g
115 . 760
151 . u 10 3 mol SO 3 mol
i)
y1
ii)
0.02 kg H 2 O 2.02 kg H 2.02 kg H 98.08 kg H 2 SO 4 18.02 kg H 2 O 0.85 m1 H 2 SO 4 2.02 kg H m1 128 . kg 98.08 kg H 2 SO 4 But since the feed solution has a mass of 1 kg, 0.28 kg SO 3 10 3 g 1 mol SO 3 absorbed 128 . 10 . kg 350 . mol kg 80.07 g 3.5 mol n0 n1 . u 10 3 n1 G balance: 0.10n0 1 151 H balance:
P 0.98 kg H 2 SO 4
b
g
d
i
E
n0 n1 V
389 . mol 0.39 mol
b g
3.89 mol 22.4 L STP 1 kg liquid feed mol
313K 1 atm 1 m 3 273K 1.2 atm 10 3 L
8.33 u 10 -2 m 3 kg liquid feed 6.45 a. Raoult’s law can be used for water and Henry’s law for nitrogen. b. Raoult’s law can be used for each component of the mixture, but Henry’s law is not valid here. c. Raoult’s law can be used for water, and Henry’s law can be used for CO2.
b g p b100q Cg
6.46 p B 100q C
T
b gh 1350.5 mm Hg c c6.95334 1343.943 b100 219.377gh 556.3 mm Hg 0.40b1350.5g 0.0711 mol Benzene mol P x p y 10b760g 0.60b556.3g 0.0439 mol Toluene mol y 10b760g
10
6.90565 1211.033 100 220.790 10
Raoult' s Law: y B
B
B
B
T
y N2
1 0.0711 0.0439
6-32
0.885 mol N 2 mol
6.47 N 2 - Henry' s law: Perry' s Chemical Engineers' Handbook, Page. 2 - 127, Table 2 - 138
b
H N 2 80q C
g
12.6 u 10 4 atm mole fraction
b0.003gd12.6 u 10 i 378 atm . mm Hg 1 atm 3551 H O - Raoult' s law: p b80q Cg 760 mm Hg p d x id p i b0.997gb0.467g 0.466 atm pN2
4
xN2 H N2
H 2O
2
H 2O
H 2O
H 2O
Total pressure: P
p N 2 p H 2O
Mole fractions: y H 2 O y N2
378 0.466 378.5 atm
p H 2O P 1 y H 2O
b
6.48 H 2 O - Raoult' s law: p H 2 O 70q C p H 2O Methane Henry' s law: p m Total pressure: P
0 / 466 / 378.5 123 . u 10 3 mol H 2 O mol gas 0.999 mol N 2 mol gas
g
233.7 mm Hg
1 atm
0.3075 atm
760 mm Hg
x H 2 O p H 2 O
b1 x gb0.3075g m
xm Hm
p m p H 2O
x m 6.66 u 10 4 (1 x m )(0.3075) 10
146 . u 10 4 mol CH 4 / mol
xm 6.49 a.
0.467 atm
Moles of water: n H 2 O
1000 cm 3
1g mol 55.49 mol 3 cm 18.02 g
Moles of nitrogen: nN2
(1 - 0.334) u 14.1 cm 3 (STP)
1 mol 1L 22.4 L (STP) 1000 cm 3
4.192 u 10 4 mol
Moles of oxygen: n O2
(0.334) 14.1 cm 3 (STP)
mol
L
22.4 L (STP) 1000 cm
3
2.102 u 10 4 mol
Mole fractions of dissolved gases: nN2 4.192 u 10 4 x N2 n H 2 O n N 2 nO 2 55.49 4.192 u 10 4 2.102 u 10 4 7.554 u 10 6 mol N 2 / mol x O2
nO 2 n H 2 O n N 2 nO 2
2.102 u 10 4 55.49 4.192 u 10 4 2.102 u 10 4
3.788 u 10 6 mol O 2 / mol
6-33
6.49 (cont’d) Henry' s law p N2
Nitrogen: H N 2
x N2 pO2
Oxygen: H O 2
x O2
0.79 1 7.554 u 10 6
u 10 5 atm / mole fraction . 1046
0.21 1 3.788 u 10 6
u 10 4 atm / mole fraction . 5544
b. Mass of oxygen dissolved in 1 liter of blood: 2.102 u 10 -4 mol 32.0 g 6.726 u 10 3 g mol 0.4 g O 2 1 L blood Mass flow rate of blood: m blood 59 L blood / min min 6.72 u 10 -3 g O 2 c. Assumptions: (1) The solubility of oxygen in blood is the same as it is in pure water (in fact, it is much greater); (2) The temperature of blood is 36.9qC. m O2
6.50 a.
bg
Basis: 1 cm 3 H 2 O l (SG)
1.0
H2O o
( SC) CO2
1 g H 2 O 1 mol 18.0 g
0.0555 mol H 2 O
b g
0.0901 cm 3 STP CO 2
0.0901
o
1 mol 22,400 cm 3 STP
b g
d4.022 u 10 i mol CO 7.246 u 10 mol CO d0.0555 + 4.022 u 10 i mol 1 atm H b20q Cg 13800 atm mole fraction 7.246 u 10 6
b.
4.022 u 10 6 mol CO 2
pCO 2
1 atm x CO2
pCO 2
x CO 2 H CO2
2
5
6
CO 2
2
mol
5
b g
For simplicity, assume n total | n H 2 O mol x CO2 nCO 2
pCO 2 H 12 oz
b35. atmg b13800 atm mole fractiong
2.536 u 10 4 mol CO 2 mol
1L 10 3 g H 2 O 1 mol H 2 O 2.536 u 10 4 mol CO 2 33.8 oz 1L 18.0 g H 2 O 1 mol H 2 O
44.0 g CO 2 1 mol CO 2
0.220 g CO 2
c. V
0.220 g CO 2
b g b273 37gK
1 mol CO 2
22.4 L STP
44.0 g CO 2
1 mol
6-34
273K
0127 . L 127 cm 3
6.51 a.
– SO2 is hazardous and should not be released directly into the atmosphere, especially if the analyzer is inside. – From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in the liquid, which increases with time. If the water were never replaced, the gas leaving the bubbler would contain 1000 ppm SO2 (nothing would be absorbed), and the mole fraction of SO2 in the liquid would have the value corresponding to 1000 ppm SO2 in the gas phase.
b
g
b
b. Calculate x mol SO 2 mol in terms of r g SO 2 100 g H 2 O
b b
Basis: 100 g H 2 O 1 mol 18.02 g r (g SO 2 ) 1 mol 64.07 g
FG H
g g
555 . mol H 2 O 0.01561r (mol SO 2 )
IJ K
mol SO 2 0.01561r mol 555 . 0.01561r From this relation and the given data, pSO 2 xSO 2
g
0 mmHg xSO2
0 mol SO 2 mol 1.4 x 10–3 2.8 x 10–3 4.2 x 10–3 5.6 x 10–3
42 85 129 176
A plot of pSO 2 vs. xSO 2 is a straight line. Fitting the line using the method of least squares (Appendix A.1) yields
dp
HSO 2 xSO2
SO 2
c.
100 ppm SO 2 ySO2 pSO2
ySO2 P
,
H SO 2
u 104 . 3136
mm Hg mole fraction
100 mol SO 2 106 mols gas
d10. u 10 ib760 mm Hgg
Henry's law xSO 2
Since xSO 2
i
4
pSO 2 H SO 2
0.0760 mm Hg
0.0760 mm Hg 3.136 u 104 mm Hg mole fraction
2.40 u 106 mol SO 2 mol is so low, we may assume for simplicity that Vfinal | Vinitial
nfinal | ninitial
bg
140 L 103 g H 2 O l 1L
1 mol 18 g
7.78 u 103 moles
7.78 u 103 mol solution 2.40 u 10 6 mol SO 2 1 mol solution 0.0187 mol SO 2 dissolved 134 . u 10 4 mol SO 2 L 140 L
nSO 2
140 L , and
0.0187 mol SO 2 dissolved
d. Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the solution to react with the absorbed SO2.
6-35
6.52 Raoult’s law + Antoine equation (S = styrene, T = toluene): 0.650(150 mm Hg)
yS P
x S pS xS
yT P
xT pT xT
10
10
b
6.92409 1420.0 T 206
0.65(150)
b
6.92409 1420.0 85.9 206
1 xS
xT
b g P b85q Cg
6.53 PB 85q C
T
b
6.95334 1343.943 T 219 .377
1 0.852
g
g
0.35(150)
b g 10 10 85.9q C (Determine using E - Z Solve or a spreadsheeet)
T
10
g
0.350(150 mm Hg)
0.65(150)
1 x S xT
xS
b
6.92409 1420.0 T 206
6.95334 1343.943 T 219 .377
0.852 mol styrene / mol
g
0148 . mol toluene / mol
b g 6.95334 1343.943 b85 219 .377 g 10 10
6.905651211.033 85 220.790
Raoult's Law: y B P
x B PB
. mm Hg 8817 . mm Hg 3451
b g b g b g b g
0.35 8817 . 0.0406 mol Benzene mol 10 760 0.65 3451 . 0.0295 mol Toluene mol 10 760 1 0.0406 0.0295 0.930 mol N 2 mol
yB yT yN 2
6.54 a. From the Cox chart, at 77q F, p*P
140 psig, p*nB
35 psig, p*iB
51 psig
Total pressure P = x p p *p + x nB p *nB + x iB p *iB 0.50(140) 0.30(35) 0.20(51) 150.8 psig P 200 psig, so the container is technically safe. b. From the Cox chart, at 140q F, pP*
* 300 psig, pnB
* 90 psig, piB
120 psig
Total pressure P = 0.50(300) 0.30(90) 0.20(120) # 200 psig The temperature in a room will never reach 140oF unless a fire breaks out, so the container is adequate.
b g P b120q Cg
6.55 a. Antoine: Pnp 120q C
ip
10 10
b
6.852211064.630 120 232 .000
b
6.78967 1020.012 120 233.097
g
g
6725 mm Hg 7960 mm Hg
When the first bubble of vapor forms, xnp
0.500 mol n - C5 H 12 (l) / mol
* Total pressure: P = xnp pnp + xip pip*
xnp
0.500 mol i - C5 H 12 (l) / mol
0.50(6725) 0.50(7960)
6-36
7342 mm Hg
6.55 (cont’d) * xnp pnp
ynp
0.500(6725) 7342
P
1 ynp
yip
0.458 mol n - C5 H 12 (v) / mol
1 0.458 0.542 mol i - C5 H 12 (v) / mol
When the last drop of liquid evaporates, 0.500 mol n - C5 H 12 (v) / mol
ynp
ynp P * pnp (120o C)
xnp xip
yip
yip P * pip (120o C)
0.5 mol i - C5 H 12 (v) / mol
0.500 P 0.500 P 6725 7960
1 P
7291 mm Hg
0.5 * 7291 mm Hg 0.542 mol n - C5 H 12 (l) / mol 6725 mm Hg 1 xnp 1 0.542 0.458 mol i - C5 H 12 (l) / mol
xnp xip
b. When the first drop of liquid forms, 0.500 mol n - C5 H 12 (v) / mol
ynp
yip
0.500 mol i - C12 H 12 (v) / mol
P = (1200 + 760) = 1960 mm Hg 0.500 P 0.500 P * * pnp (Tdp ) pip (Tdp )
xnp xip Tdp
10
b
6.852211064 .63 63.1 232 .000
10
Pip
10
xnp
0.5 * 1960 mm Hg * (631 . o C) pnp
b
g
6.78967 1020.012 63.1 233.097
1 xnp
980 10
6.78967 1020.012 /( Tdp 233.097)
631 . oC
Pnp
xip
980 6.852211064 .63/( Tdp 232 )
1756 mm Hg
g
2218 mm Hg
0.558 mol n - C5 H12 / mol
1 0.558 0.442 mol i - C5 H 12 / mol
When the last bubble of vapor condenses, xnp
0.500 mol n - C5 H 12 (l) / mol
xip
0.500 mol i - C5 H12 (l) / mol
* Total pressure: P = xnp pnp + xip pip*
1960 (0.5)10 T 62.6q C ynp
yip
* (62.6o C) xnp pnp
1 ynp
b
6.852211064 .63 T 232 .000
P 1 0.442
g (0.5)106.78967 1020.012 /( T
bp 233.097)
0.5(1731) 0.442 mol n - C5 H 12 (v) / mol 1960 0.558 mol i - C5 H 12 (v) / mol
6-37
1
6.56 B = benzene, T = toluene nv ( mol / min) at 80o C, 3 atm 10 L(STP)/min nN 2 ( mol / min)
xB [mol B(l)/mol]
yN2 (mol N2/mol) yB [mol B(v)/mol]
xT [mol T(l)/mol]
yT [mol T(v)/mol]
10.0 L(STP) / min = 0.4464 mol N 2 / min 22.4 L(STP) / mol
nN 2
b g p b80q Cg
b g 6.95334 1343.943 b80 219 .377 g 10
Antoine: p B 80q C
10
T
6.905651211.033 80 220.079
757.7 mm Hg . mm Hg 2912
a. Initially, xB = 0.500, xT = 0.500. Raoult's law:
b
g
mol B(v) mol 0166 .
b
g
0.0639 mol T(v) mol
yB
x B p B (80o C) P
0.500 757.7 3(760)
yT
xT pT (80o C) P
0.500 2912 . 3(760)
0.0639) n v N 2 balance: 0.4464 mol N 2 / min = nv (1 0166 . mol BI FG 0.5797 mol IJ FG 0166 . J H K H min mol K FG 0.5797 mol IJ FG 0.0639 mol BIJ H min K H mol K
nB0 nT0
0.5797 mol / min
mol B(v) min
0.0962 0.0370
mol T(v) min
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (= 1–xB) increases. c. Since xB decreases, yB (= xBpB*/P) also decreases. Since xT increases, yT (= xTpT*/P) also increases. 6.57 a. P
xhex phex
dT i
760 mm Hg
bp
xhep phep
0.500 10
dT i
, yi
bp
xi
yi P pi
dT i dp
¦x i
i
P
6.87776 1171.53/( Tbp 224 .366)
E-Z Solve or Goal Seek Tbp
b.
d i
xi pi Tbp
P
¦
pi
0.500 10
80.5q C yhex yi
i
dT i
1
dp
6-38
, Antoine equation for Pr 6.90240 1268.115/( Tbp 216.9 )
0.713, yhep
0.287
6.57 (cont’d) 760 mmHg
LM N10
0.30 6.87776 1171.53/( Tdp 224 .366)
E-Z Solve or Goal Seek Tdp
6.58
a.
f (T )
P
N
¦
xi pi* (T )
OP Q
0.30 10
6.90240 1268.115/( Tdp 216.9 )
711 . q C xhex
0 T , where
0.279, xhep
1
0.721
FG A B IJ T C K 10H i
pi* (T )
i
i
i 1
yi (i 1,2,, N )
xi pi* (T ) P
b. Calculation of Bubble Points A B Benzene 6.90565 1211.033 Ethylbenzene 6.95719 1424.255 Toluene 6.95334 1343.943
C 220.79 213.206 219.377
P(mmHg)= 760 xB 0.226 0.443 0.226 When x B When x EB When xT
xEB 0.443 0.226 0.226
Tbp(oC) 108.09 96.47 104.48
xT 0.331 0.331 0.548
b
g
dT i
1 pure benzene , Tbp
b
g
bp C H 6 6
1 pure ethylbenzene , Tbp
b
g
1 pure toluene , Tbp
. oC 801
dT i
dT i
bp C H 7 8
pEB pT f(T) pB 378.0 148.2 233.9 -0.086 543.1 51.6 165.2 0.11 344.0 67.3 348.6 0.07
bp C H 8 10
136.2 o C Tbp , EB ! Tbp ,T ! Tbp , B
110.6o C
Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp) than Mixture 2, and so (Tbp)1 > (Tbp)2 . Mixture 3 contains more toluene (lower bp) and less ethylbenzene (higher bp) than Mixture 1, and so (Tbp)3 < (Tbp)1. Mixture 3 contains more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp)3 > (Tbp)2
6-39
6.59
a. Basis: 150.0 L/s vapor mixture
n 1 (mol/s) @ T(oC), 1100 mm Hg 0.600 mol B(v)/mol 0.400 mol H(v)/mol
n 0 (mol/s) @ 120qC, 0.500 mol B(v)/mol 0.500 mol H(v)/mol
Gibbs phase rule: F = 2 + m - S = 2 + 2 - 2 = 2
n 2 (mol/s) x2 [mol B(l)/mol] (1- x2) [mol H(l)/mol]
Since the composition of the vapor and the pressure are given, the information is enough. Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law for butane and hexane 150.0 L 393 K
b. Molar flow rate of feed: n 0
s
mol
273 K 22.4 L (STP)
9.640 mol / s
Raoult' s law for butane: 0.600(1100) = x 2 10 6.82485 943.453/( T 239.711) Raoult's law for hexane: 0.400(1100) = (1- x 2 ) 106.877761171.530 /( T 224.366) (1) Mole balance on butane: 1(0.5) = n 1 0.6 n 2 x 2 (2) Mole balance on hexane: 1(0.5) = n1 0.4 n 2 (1 x 2 ) c. From Raoult's law, 1 =
1100(0.4) 1100(0.6) . 1171530 943.453 ) ) 10 **(6.87776 10 **(6.82485 T 239.711 T 224.366
T = 57.0 q C x2
1100(0.6) 10
6.82485 943.453/( 57 .0 239 .711)
015 . mol butane / mol
Solving (1) and (2) simultaneously n1 d.
6.60
0.778 mol C 4 H 10 / s; n 2
0.222 molC 6 H 14 / s
Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure; (2) Raoult’s law is accurate; (3) Ideal gas law is valid.
P = n-pentane, H = n-hexane 170.0 kmol/h, T1a (oC), 1 atm
85.0 kmol/h, T1b (oC), 1 0.98 mol P(l)/mol 0.02 mol H(l)/mol
n 0 (kmol/h) 0.45 kmol P(l)/kmol 0.45 kmol H(l)/kmol
n 2 (kmol/h) (l), x2 (kmol B(l)/kmol) (1- x2) (kmol H(l)/kmol)
6-40
6.60 (cont’d) a. Molar flow rate of feed: n 0 (0.45)(0.95) 85(0.98) n 0 Total mole balance : 195 85.0 n 2 n 2
195 kmol / h
110 kmol / h
Pentane balance: 195( 0.45) 85.0(0.98) 110 x 2 x 2
0.0405 mol P / mol
b. Dew point of column overhead vapor effluent: Eq. 6.4 - 7, Antoine equation 0.02(760) 0.98(760) 6.877761171.530/( T 224.366) 6.852211064.63/ ( T1a 232.000) 1a 10 10
1 T1a
37.3o C
Flow rate of column overhead vapor effluent. Assuming ideal gas behavior, Vvapor
170 kmol 0.08206 m 3 atm (273.2 + 37.3) K h kmol K 1 atm
4330 m 3 / h
Flow rate of liquid distillate product. Table B.1 U P = 0.621 g / mL, U H = 0.659 g / mL Vdistillate
0.98(85) kmol P 72.15 kg P L h kmol P 0.621 kg P +
0.02(85) kmol H 86.17 kg H L h kmol H 0.659 kg H
9.9 u 10 3 L / h
c. Reboiler temperature. 0.04 10 6.852211064.63/ ( T2 232.000) 0.96 10 6.877761171.530 /( T2 224 .366)
760 T2 = 66.6q C
Boilup composition. y2
x 2 p P* (66.6 o C) P (1 - y 2 )
0.04 10 6.852211064.63/ ( 66.6232 ) 760
0.102 mol P(v) / mol
0.898 mol H(v) / mol
d. Minimum pipe diameter
F I GH JK
m3 V s
Dmin
u max
FG mIJ u SD H sK 4
4Vvapor
S u max
2 min
(m2 )
4 4330 m 3 / h 1 h S 10 m / s 3600 s
0.39 m (39 cm)
Assumptions: Ideal gas behavior, validity of Raoult’s law and the Antoine equation, constant temperature and pressure in the pipe connecting the column and the condenser, column operates at steady state.
6-41
6.61 a.
Condenser F (mol) x 0 (mol butane/mol)
V (mol) 0.96 mol butane/mol R (mol) x 1 (mol butane/mol)
T P
Partial condenser: 40q C is the dew point of a 96% C 4 H 10 4% C 5 H 12 vapor mixture at P Pmin Total condenser: 40q C is the bubble point of a 96% C 4 H 10 - 4% C 5 H 12 liquid mixture at P Pmin yi P 1 xi Pmin Dew Point: 1
pi 40q C yi pi 40q C
¦
¦
b
(Raoult's Law)
g
4
10
i
5
12
P
2595.63 mm Hg partial condenser
¦y P ¦x p i
75 kmol / h , R V
i
Total balance: F
15 . R
75 u 15 . kmol / h 112.5 kmol / h
b b
g
Butane balance: 187.5x 0 yi xi
Raoult' s law:
d
i
*
o
p B (85 C)
1
0
yA xA yB xB
p A P
p A
p B P
p B
110.76 mm Hg
FG 6.95719 1424.255 IJ 85 213.206 K
p *EB (85o C) 10 H
187.5 kmol / h
UV P 2596 mm Hg gb gW x 0.8803 mol butane mol 112.5b0.8803g 0.96b75g x 0.9122 mol butane mol reflux
pi D AB P
FG 6.92409 1420.0 IJ 85 206 K
10 H
F
g
Equilibrium: 0.96 P x1 2830.70 Raoult' s law 0.04 P 1 x1 867.22
b
0.96 kmol butane kmol
75 112.5 187.5 kmol / h
Total balance as in b. R 112.5 kmol / h
b. p S* 85o C
g
i
Feed and product stream compositions are identical: y
6.62 a.
867.22 mmHg
b40q Cg 0.96b2830.70g 0.04b867.22g 2752.16 mm Hg b total condenser g
Bubble Point: P
c.
2830.70 mmHg
b
1 0.96 2830.70 0.04 867.22
Pmin
b. V
pi
g
FG 6.82485 943.453 IJ 40 239 .711 K 10 H FG 6.85221 1064.63 IJ 40 232 .00 K 10 H
bC H g Antoine Eq. for p bC H g Antoine Eq. for
b
¦
FG 6.90565 1211.033 IJ 85 220.790 K 10 H
15174 . mm Hg 88167 . mm Hg
6-42
D AB
6.62 (cont’d) p S*
D S,EB
110.76 15174 .
p *EB
p B*
0.730 , D B,EB
* p EB
88167 . 15174 .
5810 .
Styrene ethylbenzene is the more difficult pair to separate by distillation because D S,EB is closer to 1 than is D B,EB . c.
yi xi yj xj
D ij
d. D B , EB
D
yi xi yi (1 yi ) 1 xi
b
ij
g
x B D B , EB
yB 5810 .
1 (D B , EB 1) x B
d
D ij xi
i
1 D ij 1 xi
581 . xB , P 1 4.81x B
x B p *B (1 x B ) p *EB
bg bg
xB yB P 6.63 a.
y j 1 yi x j 1 xi
0.0 0.2 0.4 0.6 0.8 10 . mol B l mol 0.0 0.592 0.795 0.897 0.959 10 . mol B v mol 152 298 444 5900 736 882 mmHg
Since benzene is more volatile, the fraction of benzene will increase moving up the column. For ideal stages, the temperature of each stage corresponds to the bubble point temperature of the liquid. Since the fraction of benzene (the more volatile species) increases moving up the column, the temperature will decrease moving up the column.
b. Stage 1: n l y0
150 mol / h, n v
0.55 mol B mol 0.45 mol S mol ;
0.65 mol B mol 0.35 mol S mol
¦ x p bT g
Bubble point T : P P1
200 mol / h ; x1
i
(0.400 u 760) mmHg E-Z Solve
o T1 y1
i
b0.55g10
6.905651211.033/ ( T 220.79 )
b g
0.45 10 6.92409 1420 /( T 206 )
67.6 o C
bg
x1 p B T
P B balance: y 0 n v x 2 n l
b g
0.55 508
0.920 mol B mol 0.080 mol S mol 0.400 u 760 y1 n v x1n l x 2 0.910 mol B mol 0.090 mol S mol
Stage 2: Solve o T2 (0.400 u 760) mmHg 0.910 p B* (T2 ) 0.090 p S* (T2 ) E-Z
b
55.3o C
g
0.910 3310 .
0.991 mol B mol 0.009 mol S mol 760 u 0.400 B balance: y1 n v x 3 n l y 2 n v x 2 n l x 3 | 1 mol B mol | 0 mol S mol y2
c. In this process, the styrene content is less than 5% in two stages. In general, the calculation of part b would be repeated until (1–yn) is less than the specified fraction. 6-43
6.64 Basis: 100 mol/s gas feed. H=hexane. 200 mol oil/s
n l (mol/s) x +i 1 (mol H/mol)
n F (mol/s) y F (mol H/mol) 1 – y F (mol N 2/mol)
100 mol/s 0.05 mol H/mol 0.95 mol N /mol 2
n 2 (mol/s) x 2 (mol H/mol) 1 – x 2 (mol Oil/mol)
Stage i n l (mol/s) x i (mol H/mol)
b g b g b gb g
y F nF
U|V n |W y
Mole Balance: 100 200 95.025 n2 n2
b g
Hexane Balance: 0.05 100 n L
b
g
1 200 205 n L 2 Antoine
y1
b.
b
B
g
x1 p H 50q C / P
n v (mol/s) y i– 1 (mol H/mol)
99.5% of H in feed.
N 2 balance: 0.95 100 1 y F nF 99.5% absorption: 0.05 100 0.005
a.
n v (mol/s) y i (mol H/mol)
b
F
95.025 mol s
F
2.63 u 10 4 mol H(v) mol
205 mol s 0.0243 mol H(l) mol g b g 1 b100 95.025g n 97.52 mol s 2
2.63 u 10 4 95.025 x1 204.99 x1 202.48 mol s , n G
b
G
g
0.0243 403.73 / 760 0.0129 mol H(v) mol
H balance on 1st Stage: y 0 n v x 2 n l
y1 n v x1 n l x 2
0.00643 mol H(l) mol
c. The given formulas follow from Raoult’s law and a hexane balance on Stage i. d. Hexane Absorption P= y0= nGf= A=
760 0.05 95.025 6.87776
T 30
p*(T) 187.1
i 0 1 2 3
x(i) 2.43E-02 3.07E-03 5.57E-04
PR= x1= nL1= B=
y(i) 5.00E-02 5.98E-03 7.56E-04 1.37E-04
1 0.0243 204.98 1172
ye= 2.63E-04 nG= 97.52 nL= C= 224.366
T 50
p*(T) 405.3281
i 0 1 2 3 4 5
x(i) 2.43E-02 6.42E-03 1.84E-03 6.63E-04 3.60E-04
6-44
y(i) 5.00E-02 1.29E-02 3.43E-03 9.81E-04 3.53E-04 1.92E-04
202.48
T 70
p*(T) 790.5304
i 0 1 2 3 4 5 ... 21
x(i)
y(i) 5.00E-02 2.43E-02 2.52E-02 1.23E-02 1.28E-02 6.38E-03 6.63E-03 3.38E-03 3.52E-03 1.89E-03 1.96E-03 ... ... 3.80E-04 3.96E-04
6.64 (cont’d) e. If the column is long enough, the liquid flowing down eventually approaches equilibrium with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in equilibrium with the mole fraction in the entering gas is 3.80x10–4 mol H/mol, which is insufficient to bring the total hexane absorption to the desired level. To reach that level at 70oC, either the liquid feed rate must be increased or the pressure must be raised to a value for which the final mole fraction of hexane in the vapor is 2.63x10–4 or less. The solution is Pmin 951 mm Hg. 0.30 at T
6.65 a. Intersection of vapor curve with y B b. T
100q C x B
b
104q C 13% B(l), 87%T(l)
g
0.46 mol B mol liquid
b
g
0.727 mol nV 0.273 mol nL
mol vapor mol liquid
0.24 mol B mol liquid , y B n V (mol vapor) 0.46 mol B(v)/mol n L (mol liquid) 0.24 mol B(l)/mol
Basis: 1 mol 0.30 mol B(v)/mol
Balances
UV W
nL Total moles: 1 nV n L nV B: 0.30 0.46nV 0.24n L c. Intersection of liquid curve with x B
0.3 at T
6.66 a.
P
798 mm Hg, y B
0.50 mol B(v) mol
b.
P
690 mm Hg, y B
015 . mol B(l) mol
c.
P
750 mm Hg, y B
0.24 mol B(v) mol , x B
0.375
98q C 50% B(v), 50%T(v)
0.43 mol B(l) mol
nV (mol) 0.43 mol B/mol nL (mol) 0.24 mol B/mol
3 mol B 7 mol T
UV W
nV Mole bal.: 10 nV n L nL B bal.: 3 0.43nV 0.24n L
316 . mol 6.84 mol
nv nl
0.46
mol vapor mol liquid
Answers may vary due to difficulty of reading chart. d. i)
P 1000 mm Hg all liquid . Assume volume additivity of mixture components. V
3 mol B 78.11 g B 10 3 L 7 mol T 92.13 g T 10 3 L 10 . L mol B 0.879 g B mol T 0.866 g T
ii) 750 mmHg. Assume liquid volume negligible
6-45
6.66 (cont’d) 3.16 mol vapor
V
0.08206 L atm
373 K
760 mm Hg
mol K 750 mm Hg
1 atm
0.6 L
97.4 L
(Liquid volume is about 0.6 L) iii) 600 mm Hg v
10 mol vapor 0.08206 L atm
373K
760 mm Hg
mol K 600 mm Hg
1 atm
388 L
6.67 a. M = methanol n V (mol) y (mol M (v)mol) n L (mol) x (mol M (l)/mol)
n f (mol) x F (mol M (l)/mol)
UV W
n f nV n L Mole balance: x F nV x F n L MeOH balance: x F n f ynV xn L xF b. Tmin
0.4, x
0.23, y
75o C, f
0.62 f
0 , Tmax
0.4 0.23 0.62 0.23
87 o C, f
ynV xn L f
0.436
1
6.68 a.
Txy diagram (P=1 atm) 80 75 T(oC)
70
Vapor
65 liquid
60 55 50 0
0.2
0.4
0.6
Mole fraction of Acetone
b.
xA
0.47; y A
0.66
6-46
0.8
1
nV nL
xF x yx
6.68 (cont’d) c. (i) x A
0.34; y A
0.55
UV W
(ii) Mole bal.: 1 nV n L nV B bal.: 0.50 0.55nV 0.34n L (iii) U A( l ) Ul MA
0.791 g / cm 3 , U E(l)
0.789 g / cm 3 0.790 g / cm 3
46.07 g / mol
b0.34gb58.08g b1 0.34gb46.07g
Ml
0.238 mol liquid
76.2 mole% vapor
b0.34gb0.791g b1 0.34gb0.789g
58.08 g / mol, M E
0.762 mol vapor, n L
5015 . g / mol
Basis: 1 mol liquid (0.762 mol vapor / 0.238 mol liquid) = 3.2 mol vapor (1 mol)(5015 . g / mol) Liquid volume: Vl 63.48 cm 3 3 (0.790 g / cm ) Vapor volume: 3.2 mol 22400 cm 3 (STP) (65 + 273)K 88,747 cm 3 mol 273K 88,747 u 100% 99.9 volume% vapor Volume percent of vapor 88747 63.48 Vv =
d. For a basis of 1 mol fed, guess T, calculate nV as above; if nV z 0.20, pick new T. T 65 qC 64.5 qC e.
xA 0.34 0.36
Raoult' s law: yi P = xi pi* P 760 0.5 u 10
7.02447 1161/( Tbp 224 )
yA 0.55 0.56
fV 0.333 0.200
x A p *A x E p E* 0.5 u 10
8.04494 1554.3/ ( Tbp 222.65)
Tbp
66.25o C
0.5 u 10 7.02447 1161/( 66.25 224) 0.696 mol acetone / mol 760 'Tbp . 66.25 618 u 100% 7.20% error in Tbp The actual Tbp 618 . oC Tbp (real) 618 . y
xp *A P
yA
0.674
'y A y A (real)
0.696 0.674 u 100% 0.696
error in y A 316% .
Acetone and ethanol are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for acetone mole fractions that are not very close to 1.
6-47
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1oC, (Tbp)C = 61.0oC The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1oC when xC = 0 and at 61.0oC when xC = 1. (See solution to part c.) b. Txy Diagram for an Ideal Binary Solution A B C 6.90328 1163.03 227.4 Chloroform 6.90565 1211.033 220.79 Benzene 760 P(mmHg)= x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
T 80.10 78.92 77.77 76.66 75.58 74.53 73.51 72.52 71.56 70.62 69.71 68.82 67.95 67.11 66.28 65.48 64.69 63.93 63.18 62.45 61.73
y 0 0.084 0.163 0.236 0.305 0.370 0.431 0.488 0.542 0.593 0.641 0.686 0.729 0.770 0.808 0.844 0.879 0.911 0.942 0.972 1
p1 0 63.90 123.65 179.63 232.10 281.34 327.61 371.15 412.18 450.78 487.27 521.68 554.15 585.00 614.02 641.70 667.76 692.72 716.27 738.72 760
p2 760 696.13 636.28 580.34 527.86 478.59 432.30 388.79 347.85 309.20 272.79 238.38 205.83 175.10 145.94 118.36 92.17 67.35 43.75 21.33 0
p1+p2 760 760.03 759.93 759.97 759.96 759.93 759.91 759.94 760.03 759.99 760.07 760.06 759.98 760.10 759.96 760.06 759.93 760.07 760.03 760.05 760
T xy diagram (P =1 atm ) 85
75
V apor
o
T( C)
80
70 Liquid 65 60 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
M ole fraction of chloroform
6-48
0.8
0.9
1
6.69 (cont’d) d. Txy diagram (P=1 atm) 85
T(oC)
80
yc
xc
75 70 x
65
y
60 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
Mole fraction of choloroform
'T Tactual Raoult’s law: Tbp = 71o C, y = 0.58 'y y actual
71 75.3 u 100% 5.7% error in Tbp 75.3 0.58 0.60 u 100% 3.33% error in y 0.60
Benzene and chloroform are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for chloroform mole fractions that are not very close to 1. 6.70 P | 1 atm 760 mm Hg 760 0.40 u 10
d i b
g d i
x m pm* Tbp 1 x m p *P Tbp
7 .878631411/( Tbp 230 )
0.60 u 10
7 .74416 1437 .686/( Tbp 198.463)
Solve E-Z o T
79.9 o C
We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and surface tension effects on the boiling point are negligible. The liquid temperature will rise until it reaches 79.9 qC, where boiling will commence. The escaping vapor will be richer in methanol and thus the liquid composition will become richer in ethanol. The increasing fraction of the less volatile component in the residual liquid will cause the boiling temperature to rise.
6-49
6.71 Basis: 1000 kg/h product nH4 (mol H 2 /h) E = C2 H5 OH ( M = 46.05) A = CH 3 CHO ( M = 44.05)
scrubber n3 (mol/h) y A3 (mol A/mol), sat'd y E3 (mol E/mol), sat'd y H3 (mol H 2 /h) vapor, –40°C
P = 760 mm Hg Fresh feed n0 (mol E/h) nA1 (mol A/h) nE1 (mol E/h) 280°C
reactor
nA2 (mol A/h) nE2 (mol E/h) nH2 (mol H 2/h)
condenser
nC (mol/h) 0.550 A 0.450 E liquid, –40°C
Scrubbed Hydrocarbons nA4 (mol A/h) nE4 (mol E/h)
still
Product 1000 kg/h np (mol/h) 0.97 A 0.03 E
nr (mol/h) 0.05 A 0.95 E
Strategy
Calculate molar flow rate of product n p from mass flow rate and composition
x
Calculate y A3 and y E3 from Raoult’s law: y H3 the still involve fewest unknowns (n c and n r )
x x
a.
d i
x
1 y A3 y E3 . Balances about
UV W
Total mole balance about still n c , n r A balance about still A, E and H 2 balances about scrubber n A4 , n E4 , and n H4 in terms of n 3 . Overall atomic balances on C, H, and O now involve only 2 unknowns ( n 0 , n 3 )
UV W
x
Overall C balance n 0 , n 3 Overall H balance
x x x x
A balance about fresh feed-recycle mixing point n A1 E balance about fresh feed-recycle mixing point n E1 A, E, H 2 balances about condenser n A2 , n E2 , n H2 All desired quantities may now be calculated from known molar flow rates.
Molar flow rate of product M
0.97 M A 0.03 M E
b0.97gb44.05g b0.03gb46.05g
n p
1000 kg 1 kmol h 44.11 kg
22.67 kmol h
b g p b 40q Cg 0.550 p b 40 q Cg
Table B.4 (Antoine)
pA* 40q C * E
Raoult’s law y A3
* A
P
44.11 g mol
38.9 mm Hg 0.343 mm Hg 0.550(38.9) 760
6-50
0.02814 kmol A / kmol
6.71 (cont’d) y E3
b
0.450 p E* 40 q C
g
0.450( 0.343) 2.03 u 10 4 kmol E kmol 760 0.9716 kmol H 2 kmol
P 1 y A3 y E3
y H3
UV W
n r Mole balance about still: n c n p n r n c 22.67 n r A balance about still: 0.550n c 0.97(22.67) 0.05n r n c
29.5 kmol / h recycle 52.1 kmol / h
A balance about scrubber: n A4
n 3 y A3
0.02815n 3
(1)
E balance about scrubber: n E4
n 3 y E3
2.03 u 10 4 n 3
(2)
0.9716n 3
(3)
H 2 balance about scrubber: n H4
n 3 y H3
Overall C balance: n 0 (mol E) 2 mol C h
bn gb2g bn gb2g d0.97n ib2g d0.03n ib2g A4
1 mol E
n 0
E4
p
p
n A 4 n E 4 22.67
(4)
Overall H balance: 6n 0
b gb g b gb g
2n H4 4n A4 6n E4 n p 0.97 4 0.03 6
(5)
Solve (1)–(5) simultaneously (E-Z Solve): n 0
23.3 kmol E / h (fresh feed), n H 4
22.6 kmol H 2 / h (in off - gas)
n 3 = 23.3 kmol / h, n A 4 = 0.66 kmol A / h, n E 4 = 0.0047 kmol E / h
A balance about feed mixing point: n A1
0.05n r
E balance about feed mixing point: n E1
n 0 0.95n r
E balance about condenser: n E2
1.47 kmol A h
n 3 y E3 0.450n c
51.33 kmol E h 23.47 kmol E h
Ideal gas equation of state: . 5133 . g kmol 22.4 m bSTP g b273 + 280gK b147 3
Vreactor feed b. Overall conversion
h n 0 0.03n p
Single-pass conversion
1 kmol u 100%
n 0 n E1 n E2 u 100% n E1
b gb
23.33 0.03 22.67
n E4
0.0047 kmol E h
6-51
2396 m 3 h
g u 100%
23.33 513 . 235 . u 100% 51.3
Feed rate of A to scrubber: n A4 = 0.66 kmol A / h Feed rate of E to scrubber:
273K
54%
97%
6.72 a. G = dry natural gas, W = water n 3 (lb - mole G / d) n 4 (lb - mole W / d) 10 lb m W / 10 6 SCF gas 90 o F, 500 psia
Absorber n 7 (lb - mole W / d)
FG lb - mole TEG IJ H d K F lb - mole W IJ n G H d K
4.0 u 10 6 SCF / d 4 u 80 = 320 lb m W / d n1 (lb - mole G / d) n 2 [lb - mole W(v) / d]
n 5
Distillation Column
6
FG lb- mole TEGIJ H d K F lb - mole WIJ n G H d K n5 8
Overall system D. F. analysis:
5 unknowns (n1 , n 2 , n 3 , n 4 , n 7 ) 2 feed specifications (total flow rate, flow rate of water) 1 water content of dried gas 2 balances (W, G) 0 D. F.
320 lb m W 1 lb - mole
Water feed rate: n 2
d
18.0 lb m
17.78 lb - moles W / d
Dry gas feed rate: 4.0 u 106 SCF 1 lb - mole lb - moles W . u 104 lb - moles G / d 17.78 1112 n1 d d 359 SCF Overall G balance: n1
n 3 n 3
. 1112 u 10 4 lb - moles G / d
Flow rate of water in dried gas: n 4
(n 3 n 4 ) lb - moles d n3 1.112 u10 4
o n 4
359 SCF gas 10 lb m W 1 lb - mole W lb - mole
10 6 SCF
18.0 lb m
2.218 lb - mole W(l) / d
Overall W balance: n 7
(17.78 2.218) lb - moles W
18.0 lb m
d
1 lb - mole
6-52
280
lb m W u d
F 1 ft I GH 62.4 lb JK 3
m
4.5
ft 3 W d
6.72 (cont’d) b. Mole fraction of water in dried gas = n 4 n 3 n 4
yw
2.218 lb - moles W / d (2.218 + 1.112 u 10 4 ) lb - moles / d
. u 10 4 199
lb - moles W(v) lb - mole
Henry’s law: ywP = Hwxw ( x w ) max
(199 . u 10 4 )(500 psia)(1 atm / 14.7 psia) 0.398 atm / mole fraction
0.0170
lb - mole dissolved W lb - mole solution
c. Solvent/solute mole ratio 37 lb m TEG 1 lb - mole TEG 18.0 lb m W lb - mole TEG 4.434 lb m W 150.2 lb m TEG 1 lb m W lb - mole W absorbed
n5 n2 n4
n5 4.434(17.78 2.22) 69.0 lb - moles TEG / d xw = 0.80(0.0170) = 0.0136
n6 lb - mole W 5 69.0 . lb - mole W/ d = n o n6 0951 lb - mole n5 n6
Solvent stream entering absorber m
0.951 lb - moles W 18.0 lb m 69.0 lb - moles TEG d lb - mole d
150.2 lb m lb - mole
= 1.04 u 104 lb m / d W balance on absorber . 095 . 222 . ) lb - moles W/ d = 16.51 lb - moles W/ d n8 (1778 1651 . lb - moles W / d xw 019 . lb - mole W / lb - mole (16.51 + 69.9) lb - moles / d
c. The distillation column recovers the solvent for subsequent re-use in the absorber. 6.73 Basis: Given feed rates G2 G1 G3 100 mol/h 200 mol air/h n1 (mol/h) 0.999 H 2 0.96 H2 0.04 H2 S, sat'd 0.001 H 2S 1.8 atm absorber stripper 40°C L2 0°C L1 (mol/h) (mol/h) n3 n4 0.002 H 2S x 3 (mol H 2 S/mol) (1 – x 3) (mol solvent/mol) 0.998 solvent 0°C heater
6-53
G4 200 mol air/h n2 mol H 2S/mol 0.40°C, 1 at m
n3 (mol/h) x 3 (mol H 2S/mol) (1 – x 3) (mol solvent/mol) 40°C
6.73 (cont’d) Equilibrium condition: At G1, p H 2S p H 2S
x3
b0.04gb18. atmg
0.072 atm 27 atm mol fraction
H H 2S
0.072 atm
2.67 u 10 3 mole H 2 S mole
Strategy: Overall H 2 and H 2 S balances n1 , n 2 n 2 air flow rate volumetric flow rate at G4 H 2 S and solvent balances around absorber n 3 , n 4 0.998n 4 solvent flow rate
b100gb0.96g 0.999n n Overall H S balance: b100gb0.04g 0.001n n Overall H 2 balance:
1
2
961 . mol h
1
1
n1 96.1
n 2
2
3.90 mol H 2 S h
Volumetric flow rate at stripper outlet VG4
b200 + 3.90gmol h
b g b273 40gK
22.4 liters STP 1 mol
273 K
5240 L hr
H 2 S and solvent balances around absorber:
b100gb0.04g 0.002n 0.001n 0.998n n d1 2.67 u 10 i 4
4
1
3
3
Solvent flow rate
0.998n 4
n 3 x 3 n 4
1335 . n 3 1952
5820 mol solvent h
6.74 Basis: 100 g H 2 O Sat'd solution @ 60°C 100 g H 2 O 16.4 g NaHCO 3
Sat'd solution @ 30°C 100 g H 2 O 11.1 g NaHCO3 ms (g NaHCO3 ( s))
bg
NaHCO 3 balance 16.4 111 . ms ms
% crystallization
5.3 g NaHCO 3 s
5.3 g crystallized u 100% 16.4 g fed
32.3%
6.75 Basis: 1275 kg/h feed solution m1 (kg H2 O(v )/h)
875 kg/h x 0 (kg KOH/kg) (1 – x0) (kg H 2O/kg)
Sat'd solution 10°C m2 (kg H2 O(1)/h) 1.03 m2 (kg KOH/h)
m3 (kg KOH-2H 2O( s)/h) 60% of KOH in feed
6-54
|UV n W|
3
| n 4
5830 mol h
6.75 (cont’d) Analysis of feed: 2KOH H 2 SO 4 o K 2 SO 4 2H 2 O x0
bg
22.4 mL H 2 SO 4 l 1L 0.85 mol H 2 SO 4 3 5 g feed soln 10 mL L 0.427 g KOH g feed
b
gb g
60% recovery: 875 0.427 0.60 h
b
737.2 kg KOH 2H 2 O h 288.3 kg H 2 O h
56.11 kg KOH
b g
KOH balance: 0.503 1275
56.11 g KOH 1 mol KOH
224.2 g KOH h
448.9 kg KOH 92.15 kg KOH 2H 2 O
m3
2 mol KOH 1 mol H 2 SO 4
448.9 0.97m2 m2
b
g
198.4 kg h
g
Total mass balance: 875 737.2 197 . 198.4 m1 m1
147.0 kg H 2 O h evaporated
6.76 a. 45 R 0 30 g A dissolved CA 0 0.200 0.300 mL solution Plot CA vs. R CA = R / 150
CA
b. Mass of solution:
500 mol 1.10 g
550 g (160 g A, 390 g S) ml The initial solution is saturated at 10.2 qC. 160 g A 0.410 g A g S 410 . g A 100 g S @ 10.2q C Solubility @ 10.2 qC 390 g S 17.5 150 g A 1 mL soln At 0qC, R 17.5 CA 0.106 g A g soln mL soln 110 . g soln Thus 1 g of solution saturated at 0qC contains 0.106 g A & 0.894 g S. 0106 . gA Solubility @ 0qC 0118 . g A g S 118 . g A 100 g S @ 0q C 0.894 g S 390 g S 11.8 g A Mass of solid A: 160 g A 114 g A s 100 g S
bg
c.
g A remaining in soln
g A initial
0.5 u 390 g S 11.8 g A 160 114 g A 100 g S
b
g
bg
23.0 g A s
6.77 a. Table 6.5-1 shows that at 50oF (10.0oF), the salt that crystallizes is MgSO 4 7 H 2 O , which contains 48.8 wt% MgSO4. b. Basis: 1000 kg crystals/h. m 0 (g/h) sat’d solution @ 130oF
m 1 (g/h) sat’d solution @ 50oF 0.23 g MgSO4/g 0.77 g H2O/g
0.35 g MgSO4/g 0.65 g H2O/g
1000 kg MgSO4∙7H2O(s)/h
6-55
6.77 (cont’d) m 0 Mass balance: m 0 m 1 1000 kg / h MgSO 4 balance: 0.35m0 0.23m 1 0.488(1000) kg MgSO 4 / h m 1 The crystals would yield 0.488 u 1000 kg / h = 488 6.78
2150 kg feed / h 1150 kg soln / h
kg anhydrous MgSO 4 h
Basis: 1 lbm feed solution. Figure 6.5-1 a saturated KNO3 solution at 25oC contains 40 g KNO3/100 g H2O x KNO 3
40 g KNO 3 (40 + 100) g solution
0.286 g KNO 3 / g = 0.286 lb m KNO 3 / lb m x
1 lbm solution @ 80oC 0.50 lbm KNO3/lbm 0.50 lbm H2O/lbm
Mass balance: 1 lb m m1 m2 KNO3 balance: 0.50 lb m KNO3
m1(lbm) sat’d solution @ 25oC 0.286 lbm KNO3/lbm soln 0.714 lbm H2O/lbm soln m2 [lbm KNO3(s)]
0.286m1 m2
m1 = 0.700 lb m solution / lb m feed m2 0.300 lb m crystals / lb m feed
0.300 lb m crystals / lb m feed = 0.429 lb m crystals / lb m solution 0.700 lb m solution / lb m feed
Solid / liquid mass ratio =
6.79 a. Basis: 1000 kg NaCl(s)/h. Figure 6.5-1 a saturated NaCl solution at 80oC contains 39 g NaCl/100 g H2O x NaCl
39 g NaCl (39 + 100) g solution
0.281 g NaCl / g = 0.281 kg NaCl / kg m 2 [kg H 2 O(v) / h] m 1 (kg/h) sat’d solution @ 80oC 0.281 kg NaCl/kg soln 0.719 kg H2O/kg soln 1000 kg NaCl(s)/h
m 0 (kg/h) solution 0.100 kg NaCl/kg 0.900 kg H2O/kg
0 m 1 m 2 Mass balance: m 1 m 2 NaCl balance: 0.100 kg NaCl 0281 . m Solid / liquid mass ratio =
1 = 0.700 lb m solution / lb m feed m 2 0.300 lb m crystals / lbm feed m
0.300 lb m crystals / lbm feed = 0.429 lb m crystals / lb m solution 0.700 lbm solution / lbm feed
The minimum feed rate would be that for which all of the water in the feed evaporates to produce solid NaCl at the specified rate. In this case
6-56
6.79 (cont’d)
0 ) min 0100 . (m
1000 kg NaCl / h (m 0 ) min
Evaporation rate: m 2
10,000 kg / min
9000 kg H 2 O / h
Exit solution flow rate: m 1
0
b.
m 2 [kg H 2 O(v) / h] m 1 (kg/h) sat’d solution @ 80oC 0.281 kg NaCl/kg soln 0.719 kg H2O/kg soln 1000 kg NaCl(s)/h
m 0 (kg/h) solution 0.100 kg NaCl/kg 0.900 kg H2O/kg
40% solids content in slurry 1000
0 NaCl balance: 0.100m 0 Mass balance: m
kg NaCl = 0.400( m 1 ) max ( m 1 ) max h
0 0281 . (2500) m
2 m 2 2500 m
2500
kg h
7025 kg / h
4525 kg H2 O evaporate / h
6.80 Basis: 1000 kg K 2 Cr2 O 7 (s) h . Let K = K 2 Cr2 O 7 , A = dry air, S = solution, W = water. Composition of saturated solution:
0.20 kg K 0.20 kg K 01667 . kg K kg soln kg W 1+ 0.20 kg soln
b
g
n2 (mol / h) y2 (mol W(v) / mol)
m e [kg W(v) / h)
(1 y2 )(mol A/ mol) 90o C, 1 atm, Tdp 392 . oC
1 (kg / h) m f (kg / h) m
f m r (kg / h) m
CRYSTALLIZERCENTRIFUGE
0.210 kg K/ kg
0.90 kg K(s) / kg 0.10 kg soln / kg
DRYER
1000 kg K(s) / h
0.1667 kg K / kg
0.790 kg W(l) / kg
0.8333 kg W/ kg na (mol A / h)
m r (kg recycle / h) 0.1667 kg K / kg 0.8333 kg W / kg
Dryer outlet gas: y 2 P
b
g
* 39.2q C y 2 pW
f Overall K balance: 0.210m
53.01 mm Hg 760 mm Hg
1000 kg K h m f
6-57
0.0698 mol W mol
4760 kg h feed solution
6.80 (cont’d)
b
gb
1 01667 . 010 . m 1 K balance on dryer: 0.90m
g
1000 kg h m 1 1090 kg h
Mass balance around crystallizer-centrifuge
f m r m
1 m r me m e m
95% solution recycled m r
4760 1090 3670 kg h water evaporated
b0.10 u 1090g kg
h not recycled
95 kg recycled 5 kg not recycled
2070 kg h recycled Water balance on dryer
. gb1090g kg W h b0.8333gb010 18.01 u 10
3
kg mol
0.0698n 2 n 2
7.225 u 104 mol h
Dry air balance on dryer
na
b1 0.0698g7.225 u 10
4
b g
mol 22.4 L STP h 1 mol
6-58
b g
151 . u 10 6 L STP h
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm n 2w (kmol H 2 O )(sat' d)
100 kg Feed 0.07 kg Na 2 CO 3 / kg 0.93 kg H 2 O / kg
n 2c (kmol CO 2 )
Reactor Reactor
n 2a (kmol Air)
e
70 o C, 3 atm(absolute)
n1 (kmol)
Filtrate
0.70 kmol CO 2 / kmol m3 ( kg NaHCO 3 (s))
0.30 kmol Air / kmol
Filter
R|m (kg solution) U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 4
3
2
m5 (kg) 0.024 kg NaHCO 3 / kg 0.976 kg H 2 O / kg Filter cake m6 (kg) 0.86 kg NaHCO 3 (s) / kg
R|0.14 kg solution U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 3
2
Degree of freedom analysis: Reactor 6 unknowns (n1, n2, y2w, y2c, m3, m4) –4 atomic species balances (Na, C, O, H) –1 air balance –1 (Raoult's law for water) 0 DF
Filter 2 unknowns –2 balances 0 DF
Na balance on reactor 100 kg 0.07 kg Na 2 CO3 46 kg Na (m3 0.024m4 ) kg NaHCO3 23 kg Na kg 106 kg Na 2 CO3 84 kg NaHCO3 3.038 0.2738(m3 0.024m4 ) (1) Air balance: 0.300 n1
n2 a
( 2)
C balance on reactor : n1 (kmol) 0.700 kmol CO 2 kmol
12 kg C
100 kg 0.07 kg Na 2 CO 3
12 kg C
1 kmol CO 2 kg 106 kg Na 2 CO 3 12 (n2c )(12) (m3 0.024m2 )( ) 8.40n1 0.7924 12n2c 01429 . (m3 0.024m4 ) (3) 84 H balance : 1 2 2 ) (n2 w )(2) ( m3 0.024m4 )( ) 0.976m4 ( ) 18 84 18 ( 4) 10.33 2n2 w 0.01190(m3 0.024m4 ) 0.1084m4 (100)(0.93)(
6-59
6.81(cont'd) O balance (not counting O in the air): 48 16 n1 (0.700)(932) 100 (0.07)( ) 100 (0.93)( ) 106 18 48 16 (n2 w )(16) n2 c ( 32) (m3 0.024m4 )( ) 0.976m4 ( ) 84 18 22.4n1 85.84 16n2 w 32n2 c 0.5714(m3 0.024m4 ) 0.8676m4
(5)
Raoult's Law : yw P
p w* (70 o C)
n2 w
01025 . (n2 w
n2 w n2 w n 2 c n 2 a n2 c n2 a )
233.7 mm Hg (3 * 760) mm Hg (6)
Solve (1)-(6) simultaneously with E-Z solve (need a good set of starting values to converge). n1 0.8086 kmol, n2a 0.2426 kmol air, n2c 0.500 kmol CO 2 , n2w 0.0848 kmol H 2 O(v), m3 8.874 kg NaHCO 3 (s), m4 92.50 kg solution
NaHCO3 balance on filter: m3 0.024m4
0.024m5 m6 [0.86 (014 . )(0.024)]
m3 8.874
11.09
0.024m 5 0.8634m 6
(7)
m4 92 .50
Mass Balance on filter: 8.874 92.50 1014 . Solve (7) & (8) Scale factor
m5 m6
91.09 kg filtrate 10.31 kg filter cake
500 kg / h 8.867 kg
m5 m6
(8)
(0.86)(10.31) 8.867 kg NaHCO 3 (s)
56.39 h 1
(a) Gas stream leaving reactor
R| |S || T
46.7kmol / h n 2w (0.0848)(56.39) 4.78 kmol H 2 O(v) / h 0.102 kmol H 2 O(v) / kmol n 2c (0.500)(56.39) 28.2 kmol O 2 / h 0.604 kmol CO 2 / kmol n 2a (0.2426)(56.39) 13.7 kmol air / h 0.293 kmol Air / kmol
U| V| W
V2
n 2 RT P
(b) Gas feed rate: V1
(46.7 kmol / h)(0.08206 3 atm
m 3 atm )(343 K) kmol K
438 m 3 / h
56.39 u 0.8086 kmol 22.4 m 3 (STP) 1h 17.0 SCMM h kmol 60 min
6-60
6.81(cont'd) (c) Liquid feed: (100)(56.39) 5640 kg / h To calculate V , we would need to know the density of a 7 wt% aqueous Na2CO3 solution. (d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual solution would contain less than 2.4% NaHCO3. (e) Henry's law
Benefit: Higher pressure greater pCO2
higher concentration of CO 2 in solution
higher rate of reaction smaller reactor needed to get the same conversion lower cost Penalty: Higher pressure greater cost of compressing the gas (purchase cost of compressor, power consumption) 6.82 600 lb m / h
Dissolution Dissolution 0.90 MgSO4 7H 2 O Dissolution Tank Tank Tank 0.10 I 1 (lb m H 2 O / h) m
Filter I
R|m (lb soln / h) U S|0.32 kg MgSO / kg T0.68 kg H O / kg W 2
4
4
6000 lb m I / h
6 (lb m / h) m
o
2
3 (lb m so ln/ h) m 0.32 MgSO 4
110 F
0.23 lb m MgSO 4 / lb m
R|300 lb soln / h|U S|0.32 MgSO V| T0.68 H O W m
m
2
6000 lb m I / h
0.68 H 2 O
0.77 lb m H 2 O / lb m 4 (lb m MgSO 4 7 H 2 O / h m
Filter II
R|m (lb |S0.23 lb ||0.77 lb T 5
m
soln) MgSO 4 / lb m
m
H 2 O / lb m
m
U| |V || W
Crystallizer
4 (lb m MgSO 4 7 H 2 O) m
R|0.05m S|0.23 lb T0.77 lb
(lb m soln)
4 m
MgSO 4 / lb m
m
H 2 O / lb m
U| V| W
a. Heating the solution dissolves all MgSO4; filtering removes I, and cooling recrystallizes MgSO4 enabling subsequent recovery. (b) Strategy: Do D.F analysis.
6-61
6.82(cont'd)
UV W
Overall mass balance m 1 , m 4 Overall MgSO 4 balance
( MW) MgSO4
UV W
Diss. tank overall mass balance m 2 , m 6 Diss. tank MgSO 4 balance
(24.31 32.06 64.00) 120.37, ( MW) MgSO4 7H2O
(12037 . 7 * 18.01) 246.44
Overall MgSO4 balance: 60,000 lb m 0.90 lb m MgSO 4 7H 2 O 120.37 lb m MgSO 4 h lb m 246.44 lb m MgSO 4 7H 2 O (300 lb m / h)(0.32 lb m MgSO 4 / lb m ) m 4 (120.37 / 246.44) 0.05m 4 (0.23) m 4
5.257 x10 4 lb m crystals / h m 4 5.257 x104 lb m / h
Overall mass balance: 60,000 m 1
6300 105 . m 4
m 1
1494 lb m H 2 O / h
c. Diss. tank overall mass balance: Diss. tank MgSO 4 balance:
60,000 m 1 m 6 m 2 6000 54,000(120.37 / 246.44) 0.23m 6
m 2
1512 . x10 5 lb m / h
m 6
9.575x10 4 lb m / h recycle
Recycle/fresh feed ratio
9.575x10 4 lb m / h 1494 lb m / h
0.32m 2
UV W
64 lb m recycle / lb m fresh feed
6.83 a. n 1 (kmol CO 2 / h) Cryst Filter
1000 kg H 2SO 4 / h (10 wt%) 1000 kg HNO 3 / h w (kg H 2 O / h) m
2 (kg CaSO4 / h) m 3 (kg Ca(NO3 )2 / h) m
Filter cake 5 (kg / h) m 0.96 kg CaSO 4 (s) / kg
4 (kg H2O/ h) m
0.04 kg soln / kg
0 (kg CaCO 3 / h) m 0 (kg solution / h) 2m
0 (kg CaCO 3 / h) m
8 (kg soln / h) m
0 (kg solution / h) 2m
R| X (kg CaSO / kg) U| 500 X (kg H O / kg) Solution composition: S |T(1 501X )(kg Ca(NO ) / kg)V|W a
4
a
a
6-62
2
3 2
6.83 (cont’d) b. Acid is corrosive to pipes and other equipment in waste water treatment plant. c. Acid feed:
1000 kg H 2 SO 4 / h (2000 m w ) kg / h
0.10 m w
8000 kg H 2 O / h
Overall S balance: 1000 kg H 2 SO 4 h
32 kg S
5 (kg / h) (0.96 0.04 X a ) (kg CaSO 4 ) m
98 kg H 2 SO4 m8 (kg / h) X a (kg CaSO 4 ) kg
32 kg S
kg
136 kg CaSO 4
32 kg S 136 kg CaSO 4
5 (0.96 0.04 X a ) 0.2353m 8 Xa 3265 . 0.2353m
(1)
Overall N balance: 1000 kg HNO 3
14 kg N
h
63 kg HNO 3
0.04m 5 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) kg
8 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) m kg
28 kg N 164 kg Ca(NO 3 ) 2
28 kg N 164 kg Ca(NO 3 ) 2
222.2 0.00683m 5 (1 501X a ) 0171 . m 8 (1 501X a )
(2)
Overall Ca balance: 5 (kg / h) (0.96 0.04X a ) (kg CaSO 4 ) 40 kg Ca m 40 kg Ca 100 kg CaCO 3 kg 136 kg CaSO 4 5 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) 0.04m 40 kg Ca kg 164 kg Ca(NO 3 ) 2 8 (kg / h) X a (kg CaSO 4 ) m 40 kg Ca kg 136 kg CaSO 4 8 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) m 40 kg Ca kg 164 kg Ca(NO 3 ) 2
0 (kg / h) m
0.40m 0
0.294m 5 (0.96 0.04 X a ) 0.00976m 5 (1 501 X a ) 0.294m 8 X a 0.244m 8 (1 501 X a )
(3)
Overall C balance : 0 (kg / h) m 0.01m 0
12 kg C 100 kg CaCO 3 n1
n 1 (kmol CO 2 / h)
(4)
6-63
1 kmol C 1 kmol CO 2
12 kg C 1 kmol C
6.83 (cont’d) Overall H balance : 1000 (kg H 2SO4 )
2 kg H
h
98 kg H 2 SO4
1000 kg HNO3
1 kg H
h
w (kg / h) m
2 kg H
63 kg HNO3 18 kg H 2 O 5 (kg / h) 500 X a (kg H 2 O) 2 kg H 8 (kg / h) 500 X a (kg H 2 O) 2 kg H m 0.04m kg 18 kg H 2 O kg 18 kg H 2 O 5 X a 5556 8 Xa 92517 2.22m . . m
(5)
Solve eqns. (1)-(5) simultaneously, using E-Z Solve. m 0
1812.5 kg CaCO 3 (s) / h,
m 5
n1
18.1 kmol CO 2 / h(v),
Xa
Recycle stream
2 * m 0
1428.1 kg / h,
m 8
9584.9 kg soln / h,
0.00173 kg CaSO 4 / kg
3625 kg soln / h
CaSO . R| 0.00173(kg CaSO / kg) U| R||0173% S| 500 * 0.00173(kg H O / kg) V| S|86.5% H O T(1 501* 0.00173)(kg Ca(NO ) / kg)W |T13.3% Ca(NO ) 4
4 2
2
3 2
d.
3 2
From Table B.1, for CO2:
Pc 72.9 atm 304.2 K , T (40 273.2) K Tr 103 . , Tc 304.2
U| |V || W
Tc
Pr
30 atm 72.9 atm
0.411
From generalized compressibility chart (Fig. 5.4-2): z
0.86 V
0.86 0.08206 L atm 313.2 K L 0.737 mol K 30 atm mol CO 2
zRT P
Volumetric flow rate of CO2: V
e.
n1 * V
18.1 kmol CO 2
0.737 L
1000 mol
h
mol CO 2
1 kmol
1.33x10 4 L / h
Solution saturated with Ca(NO3)2:
1 501X a (kg Ca(NO 3 ) 2 / kg) 500Xa (kg H 2 O / kg)
1.526 X a
0.00079 kg CaSO 4 / kg
Let m 1 (kg HNO3/h) = feed rate of nitric acid corresponding to saturation without crystallization.
6-64
6.83 (cont’d) Overall S balance: 5 (kg / h) (0.96 (0.04)(0.00079)) (kg CaSO4 ) m
1000 kg H 2SO4
32 kg S
h
98 kg H2SO4
kg 8 (kg / h) 0.00079 (kg CaSO4 ) m kg
32 kg S 136 kg CaSO4
32 kg S 136 kg CaSO4
5 0.000186m 8 326.5 0.226m
(1')
Overall N balance: 1 (kg HNO3 ) m
14 kg N
h
63kg HNO3
5 (kg / h) (1 (501)(0.00079)) (kg Ca(NO3 ) 2 ) 0.04m kg
28 kg N 164 kg Ca(NO3 ) 2
8 (kg / h) (1 (501)(0.00079)) (kg Ca(NO3 )2 ) m kg
28 kg N 164 kg Ca(NO3 ) 2
1 000413 5 0103 8 0222 m . m . . m
(2')
Overall H balance: 1 kg HNO 3 m 1000 (kg H 2 SO 4 ) 2 kg H 1 kg H h 98 kg H 2 SO 4 h 63 kg HNO 3 5 (kg / h) 500(0.00079) (kg H 2 O) 8000 (kg / h) 2 kg H 0.04m 2 kg H 18 kg H 2 O kg 18 kg H 2 O 8 (kg / h) 500(0.00079) (kg H 2 O) m 2 kg H kg 18 kg H 2 O 1 0.00175m 5 0.0439m 8 909.30 0.0159m Solve eqns (1')-(3') simultaneously using E-Z solve: m 1
1155 . x10 4 kg / h;
5 m
x10 3 kg / h; 1424 .
Maximum ratio of nitric acid to sulfuric acid in the feed . x10 4 kg / h 1155 1000 kg / h
. kg HNO 3 / kg H 2 SO 4 115
6-65
8 m
2.484 x10 4 kg / h
(3' )
6.84 Moles of diphenyl (DP): Moles of benzene (B):
0.363 619 . 0.363
x DP
bg
bg
p B* T 'Tm
(1 x DP ) p B* T
'Tbp
'Tbp
o
m
55 . 3.6 19 . qC
o
801 82.0 q C . 185 .
RTbp2 'H mp
Tbp
2
30,765
0.0q C, 'Tfp
114.0 mm Hg
2
v
6.85 Tfp
g
g b0.0554g 3.6 K = 3.6 C T 9837 8.314b273.2 801 .g . K = 1.85 C b0.0554g 185
RTb02 x DP 'H Tb
b
0.945 120.67 mm Hg
b
U| |V 619 . mol | |W
0.0544 mol DP mol
8.314 273.2 55 .
2 RTm0 x DP 'H m
'Tbp
56.0 g 0.363 mol 154.2 g mol 550.0 ml 0.879 g 1 mol ml 78.11 g
4.6q C
xu
b
g b0.0445g
8.314 273.2 100.0 40,656
2
13 . qC
100.0 13 . 101.3 q C
30 . q C xu
'Tbp 'H mp RTbp2
b
g 8.314b373.2g 3.0 40,656
2
mol urea mol 0105 .
Initial mass of urea:
1000 g soln
1 mol soln 0.0445 mol U 60.06 g U 60.06 0.0445 18.02 0.9555 g soln mol soln mol U
b
g
b
g
134 g urea
Moles of H 2 O:
b1000 134g g
Final moles of urea:
Added Urea:
1 mol 18.02 g
m m 48.0
5.63 mol 60.06 g mol
48.0 mol H 2 O
m 5.63 mol urea 0105 . 134 g
204 g urea
6-66
b0.5150 gg b1101. g molg b0.5150 gg b1101. g molg b100.0 gg b94.10 g molg
6.86 x aI
RTm20 'TmI xs 'TmII 'H m
'Tm
x sII
b1 0.00523g mol solvent
b g
b
0.49q C 0.41q C
'H vI B , ln p s* Tbs RTb 0
b g
6380 J mol
b g
ln Ps* Tb 0 ln P0* Tbs * s
FG H
* 0
bs
6.88 m 1 (g styrene) 90 g ethylbenezene 100 g EG 90 g ethylbenzene 30 g styrene
Styrene balance: m1 m2
m 2 (g styrene) 100 g EG
30 g styrene
m2 100 m2
6.38 kJ / mol
'H v 1 'H v Tbs Tb 0 1 # R Tb 0 Tbs R Tb20
b0
Equilibrium relation:
mol solute mol solution
'H vII B RTbs
IJ K p bT g b1 x g p bT g lnb1 x g | x
b g
0.00523
2
Assume 'H vI # 'H vII ; T0 Ts # T02
b. Raoult’s Law:
0.00438
'TmI
g b0.00523g
0.49
'TmII
x sI
8.314 273.2 5.00
RTm20 xs 'Tm
ln p s* Tb 0
x sII
94.10 g solvent 0.4460 g solute 8350 . g solute mol 1 mol solvent 95.60 g solvent
0.00523 mol solute
'H m 6.87 a.
x sI
0.00438 mol solute mol
. 019
FG m IJ H 90 m K 1
1
solve simultaneously
m1
25.6 g styrene in ethylbenzene phase
m2
4.4 g styrene in ethylene glycol phase
6-67
'H v 'Tb RTb20
'Tb
RTb20 x 'H v
6.89 Basis: 100 kg/h.
A=oleic acid; C=condensed oil; P=propane
100 kg / h 0.05 kg A / kg 0.95 kg C / kg
95.0 kg C / h 2 kg A / h m 3 kg A / h m 1 kg P / h m
1 kg P / h m
a. 90% extraction: m 3
(0.09)(0.05)(100 kg / h) = 4.5 kg A / h
Balance on oleic acid: (0.05)(100) Equilibrium condition:
015 .
m 2 4.5 kg A / h m 2
0.5 / (n1 0.5) n1 4.5 / (4.5 95)
0.5 kg A / h
73.2 kg P / h
b. Operating pressure must be above the vapor pressure of propane at T=85oC=185oF * 500 psi 34 atm Figure 6.1-4 p propane c. Other less volatile hydrocarbons cost more and/or impose greater health or environmental hazards. 6.90 a. Benzene is the solvent of choice. It holds a greater amount of acetic acid for a given mass fraction of acetic acid in water. Basis: 100 kg feed.
A=Acetic acid, W=H2O, H=Hexane, B=Benzene
100 (kg)
m1 (kg)
0.30 kg A / kg
0.10 kg A / kg
0.70 kg W / kg
0.90 kg W / kg m2 (kg A)
m H (kg H)
m H (kg H) or m B (kg B)
or m B (kg B)
Balance on W:
100 * 0.70 m1 * 0.90 m1
Balance on A:
100 * 0.30 m2 77.8 * 0.10 m2
77.8 kg 22.2 kg
Equilibrium for H: KH
m2 / (m2 m H ) xA
22.2 / (22.2 m H ) 0.10
0.017 m H
22.2 / (22.2 mB ) 0.10
0.098 m B
1.30 x10 4 kg H
Equilibrium for B: KB
m2 / (m2 m B ) xA
2.20 x10 3 kg B
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and environmental considerations.
6-68
6.91
a. Basis: 100 g feed 40 g acetone, 60 g H 2 O. A = acetone, H = n - C 6 H 14 , W = water 40 g A 60 g W
e 1 (g A) 60 g W
25°C
100 g H r 1 (g A)
100 g H
25°C
75 g H
b
xA in H phase / xA in W phase
0.343 x mass fraction
Balance on A stage 1: Equilibrium condition stage 1: Balance on A stage 2: Equilibrium condition stage 2:
75 g H r 2 (g A)
g
40 e1 r1 r1 100 r1 e1
b g b60 e g
U| e 0.343V| r W U| r 0.343V| e W
1
1
1
27.8 e2 r2 r2 75 r2 e2
b g b60 e g
2
2
20.6 g A remaining u 100% 40 g A fed
% acetone not extracted
e 2 (g A) 60 g W
2
27.8 g acetone 12.2 g acetone 7.2 g acetone 20.6 g acetone
515% .
b. 40 g A 60 g W
e1 g A 60 g W
r1 g A 175 g H
175 g H
Balance on A stage 1: Equilibrium condition stage 1: % acetone not extracted c.
U| r b g 0.343V e |W b60 e g
40.0 e1 r1 r1 175 r1 e1
1
1
22.2 g A remaining u 100% 40 g A fed
40 g A 60 g W
20.6 / (m 20.6) 19.4 / (60 19.4)
17.8 g acetone 22.2g acetone
555% .
19.4 g A 60 g W 20.6 g A m (g H)
m (g H)
Equilibrium condition:
1
0.343 m
225 g hexane
d. Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane over process lifetime) – (cost of an equilibrium stage x number of stages). The most costeffective process is the one for which F is the highest.
6-69
6.92
a. P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
Broth Mixing tank
100 kg 0.015 P 0.985 Ac
m1 (kg BA) Extraction Unit I
Acid
D.F. analysis: Extraction Unit I 3 unknown (m1, m2p, m3p) –1 balance (P) –1 distribution coefficient –1 (90% transfer) 0 DF
m3P (kg P) 98.5 (kg Ac) pH=2.1 m4 (kg Alk)
Extraction II m6P (kg P) m1 (kg BA)
m5P (kg P) m4 (kg Alk) pH=5.8
Extraction Unit II (consider m1, m3p) 3 unknowns –1 balance (P) –1 distribution coefficient –1 (90% transfer) 0 DF
b. In Unit I, 90% transfer m3 P 0.90(15 . ) 135 . kg P P balance: 15 . m2 P 1.35 m2 P 015 . kg P . / (135 . m1 ) 135 m1 34.16 kg BA pH=2.1 K 25.0 . / (015 . 98.5) 015 In Unit II, 90% transfer: m5 P 0.90(m3 P ) 1.215 kg P m3 P 1215 P balance: . m6 P m6 P 0.135 kg P m6 P / (m6 P 34.16) pH=5.8 K 0.10 m4 29.65 kg Alk 1.215 / (1.215 m4 ) m1 34.16 kg BA 0.3416 kg butyl acetate / kg acidified broth 100 100 kg broth m4 29.65 kg Alk 0.2965 kg alkaline solution / kg acidified broth 100 100 kg broth Mass fraction of P in the product solution: m5 P 1215 . P xP m4 m5 P (29.65 + 1.215) kg
0.394 kg P / kg
c. (i). The first transfer (low pH) separates most of the P from the other broth constituents, which are not soluble in butyl acetate. The second transfer (high pH) moves the penicillin back into an aqueous phase without the broth impurities. (ii). Low pH favors transfer to the organic phase, and high pH favors transfer back to the aqueous phase. (iii).The penicillin always moves from the raffinate solvent to the extract solvent.
6-70
6.93
W = water, A = acetone, M = methyl isobutyl ketone xW xA xM
0.20 0.33 0.47
U| V| W
Figure 6.6-1
Phase 1: x W
0.07, x A
0.35, x M
0.58
Phase 2: x W
0.71, x A
0.25, x M
0.04
Basis: 1.2 kg of original mixture, m1=total mass in phase 1, m2=total mass in phase 2. H 2 O Balance: Acetone balance:
R| S| T
m1 0.95 kg in MIBK - rich phase 12 . * 0.20 0.07m1 0.71m2 m2 0.24 kg in water - rich phase 12 . * 0.33 0.35m1 0.25m2
6.94 Basis: Given feeds: A = acetone, W = H2O, M=MIBK Overall system composition:
U| b g V 3500 g b20 wt% A, 80 wt% M g 700 g A, 2800 g M |W 2200 g A U | 3500 g WV 25.9% A, 41.2% W, 32.9% M 2800 g M |W
5000 g 30 wt% A, 70 wt% W 1500 g A, 3500 g W
Phase 1: 31% A, 63% M, 6% W
Fig. 6.6-1
Phase 2: 21% A, 3% M, 76% W
Let m1=total mass in phase 1, m2=total mass in phase 2. H 2 O Balance: Acetone balance: 6.95
R| S| T
m1 4200 g in MIBK - rich phase 3500 0.06m1 0.76m2 m2 4270 g in water - rich phase 2200 0.31m1 0.21m2
A=acetone, W = H2O, M=MIBK 41.0 lb m / h
32 lb m / h
x A,1 , x W,1 , 0.70
x AF (lb m A / lb m ) x WF (lb m W / lb m )
2 lb m / h m x A ,2 , x W ,2 , x M ,2
1 (lb m M / h) m
Figure 6.6-1 Phase 1: x M Phase 2: x w ,2 Overall mass balance: MIBK balance:
0.700 x w ,1
0.05; x A,1
0.25 ;
0.81; x A ,2
0.81; x M ,2
0.03
UV W
m 1 32.0 lb m / h m 1 410 . lb m h m 2 m 1 410 m 2 . * 0.7 m 2 * 0.03
6-71
281 . lb m MIBK / h 19.1 lb m h
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBK System 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol W
UV W
maq,1 maq ,1 morg ,1 100 Mass balance: Acetone balance: maq ,1 * 0.275 morg ,1 * 0.375 33.33 morg ,1
417 . kg 58.3 kg
System 2: x a,org = 0.100 mol A, x m,org = 0.870 mol M, x w,org = 0.030 mol W x a,aq = 0.055 mol A, x m,aq = 0.020 mol M, x w,aq
= 0.925 mol W
UV W
m aq,2 maq ,2 morg ,2 100 Mass balance: Acetone balance: maq ,2 * 0.055 morg ,2 * 0100 . 9 morg,2 b. K a ,1
x a ,org ,1 x a ,aq ,1
0.375 136 . ; 0.275
K a ,2
x a ,org ,2 x a ,aq ,2
22.2 kg 77.8 kg
0100 . 182 . 0.055
High Ka to extract acetone from water into MIBK; low Ka to extract acetone from MIBK into water. c.
E
x a ,org / x w ,org aw,1
0.375 / 0.075 12.3; E aw,2 0.275 / 0.675
x a ,aq / x w ,aq
If water and MIBK were immiscible, x w ,org d.
0 E
aw
0100 . / 0.040 0.055 / 0.920
418 .
of
Organic phase= extract phase; aqueous phase= raffinate phase
E a ,w
( x a / x w ) org
( x a ) org / ( x a ) aq
( x a / x w ) aq
( x w ) org / ( x w ) aq
Ka Kw
When it is critically important for the raffinate to be as pure (acetone-free) as possible. 6.97 Basis: Given feed rates: A = acetone, W = water, M=MIBK
e 1
e 2
(kg / h)
x1A (kg A / kg)
x 2W (kg W / kg)
x1W (kg W / kg)
x 2M (kg M / kg)
x1M (kg M / kg) 200 kg / h 0.30 kg A / kg 0.70 kg M / kg
r1 Stage I
(kg / h)
x 2A (kg A / kg)
(kg / h)
y1A (kg A / kg) y1W (kg W / kg) y1M (kg M / kg)
300 kg W / h
6-72
Stage II Stage IIS 300 kg W / h
r2
(kg / h)
y 2A (kg A / kg) y 2W (kg W / kg) y 2M (kg M / kg)
6.97(cont'd) Overall composition of feed to Stage 1:
U| V| W
b200gb0.30g
60 kg A h 500 kg h 200 60 140 kg M h 12% A, 28% M, 60% W 300 kg W h
Figure 6.6-1
Extract: x1A Raffinate: y1A
Mass balance Acetone balance:
0.095, x1W
0.880, x1M
0.15, y1W
0.025
0.035, y1M
R| S| T
e1 500 e1 r1 r1 60 0.095e1 015 . r1
0.815
273 kg / h 227 kg / h
Overall composition of feed to Stage 2:
U| 527 kg h . g 34 kg A h b227gb015 b227gb0.815g 185 kg M h V 6.5% A, 35.1% MIBK, 58.4% W b227gb0.035g 300 308 kg W h|W Figure 6.6-1
Extract: x 2 A Raffinate: y 2 A
Mass balance: Acetone balance:
0.04, x 2 W
0.94, x 2 M
0.085, y 2 W
0.025, y 2 M
R| S| T
e2 527 e2 r2 r2 34 0.04e2 0.085r2
Acetone removed: [60 (0.085)(287)] kg A removed / h 60 kg A / h in feed
0.02 0.89
240 kg / h 287 kg / h
0.59 kg acetone removed / kg fed
Combined extract: Overall flow rate = e1 e2
273 240 513 kg / h
( x1 A e1 x 2 A e2 ) kg A
0.095 * 273 0.04 * 240 513
0.069 kg A / kg
Water:
( x1w e1 x 2 w e2 ) kg W e1 e2
0.88 * 273 0.94 * 240 513
0.908 kg W / kg
MIBK:
( x1 M e1 x 2 M e2 ) kg M (e1 e2 ) kg
Acetone:
0.025 * 273 0.02 * 240 513
6-73
0.023 kg M / kg
6.98. a. 1.50 L / min 25o C, 1atm, rh = 25% n0 (mol / min)
M (g gel) Ma (g H2O)
y0 (mol H2 O / mol) (1- y0 ) (mol dry air / mol) n 0
PV RT
(1 atm)(1.50 L / min) (0.08206 L atm / mol K)(298 K)
r.h.=25%
pH2O
0.25
pH* 2O (25o C)
Silica gel saturation condition: Water feed rate: m H
X*
12.5
p H 2O p H* 2 O
0.25 p H* 2 O (25o C )
y0
0.06134 mol / min
p
12.5 * 0.25 3125 .
0.25(23.756 mm Hg) 760 mm Hg
0.06134 mol 0.00781 mol H 2 O 18.01 g H 2 O 2O
min
mol
mol H 2 O
g H 2 O ads 100 g silica gel
0.00781
mol H 2 O mol
0.00863 g H 2 O / min
Adsorption in 2 hours (0.00863 g H 2 O / min)(120min) 1.035 g H 2 O Saturation condition:
1.035 g H 2 O M (g silica gel)
3.125 g H 2 O M 100 g silica gel
33.1 g silica gel
Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and T are constant. b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a significant fraction of the water in the entering air and relatively little oxygen and nitrogen. The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel reaches its capacity. If air were still fed to the column past this point, no further dehumidification would take place. To keep this situation from occurring, the gel is replaced at or (preferably) before the time when it becomes saturated. 6.99 a.
Let c = CCl4 Relative saturation
0.30
pc * pc (34 o C)
pc
0.30 * (169 mm Hg) 50.7 mm Hg
b. Initial moles of gas in tank: n0
P0V0 RT0
1 atm 50.0 L 1985 . mol 0.08206 L atm / mol K 307 K
Initial moles of CCl4 in tank: nc0
y c 0 n0
pc 0 n0 P0
50.7 mm Hg u 1.985 mol 760 mm Hg
6-74
0.1324 mol CCl 4
6.99 (cont’d) 50% CCl4 adsorbed nc
0.500nc 0 n0 nads
Total moles in tank: n tot
0662 mol CCl 4 (= nads) (1.985 0.0662) mol = 1.919 mol
Pressure in tank. Assume T = T0 and V = V0. n tot RT0 V0
P
nc n tot
yC
. )(0.08206)(307) FG (1919 I F 760 mm Hg IJ atmJ G H K H atm K 50.0
0.0662 mol CCl 4 1919 . mol
pc
nc nc 1853 . n tot
pc
mol CCl 4 mol
0.0345(760 mm Hg) = 26.2 mm Hg n0 nc 0
c. Moles of air in tank: na
yc
0.0345
0.001
nc nair
(1.985 0.1324) mol air = 1.853 mol air
mol CCl 4 nc mol
u 10 3 mol CCl 4 1854 .
mol 1854 .
LM n RT OP = 1854 u 10 mol . 50.0 L N V Q
yc P
735 mm Hg
0.001
3
0
tot
0
0.08206 L atm 307 K 760 mm mol K 1 atm
0.710 mm Hg X*
FG g CCl IJ H g carbon K 4
0.0762 pc X* 1 0.096 pc
Mass of CCl4 adsorbed mads
(nc 0 nc )( MW ) c
0.0762( 0.710) 1 0.096(0.710)
0.001854) mol CCl 4 (01324 .
20.3 mol CCl 4 adsorbed 20.3 g CCl 4 ads Mass of carbon required: mc g CCl 4 ads 0.0506 g carbon a.
X*
E ln X * K F p NO 2
ln(PNO2)
6.100
0.0506
0
E ln K F E ln p NO 2
1
2 ln(X*)
6-75
153.85 g 1 mol CCl 4
400 g carbon
y = 1.406x - 1.965
2 1.5 1 0.5 0 -0.5 -1 -1.5
g CCl 4 adsorbed g carbon
3
6.100 (cont’d) ln X *
X* . ln p NO2 1965 . 1406
.406 e 1.965 p 1NO 2
.406 p 1NO . 0140 2
0.140 (kg NO 2 / 100 kg gel)(mm Hg) 1.406 ; E
KF
1406 .
S * (0.05m) 2 (1m) 10 3 L 0.75 kg gel 5.89 kg gel 1m 3 L Maximum NO2 adsorbed :
b. Mass of silica gel : mg
p NO 2 in feed mads
0.010(760 mm Hg)
0.140(7.60) 1.406 kg NO 2 100 kg gel
7.60 mm Hg 5.89 kg gel
0.143 kg NO 2
Average molecular weight of feed : MW
0.01( MW ) NO2 0.99( MW ) air
(0.01)(46.01) (0.99)(29.0)
29.17
kg kmol
Mass feed rate of NO2: m
8.00 kg
1 kmol
0.01 kmol NO 2
46.01 kg NO 2
h
29.17 kg
kmol
kmol NO 2
Breakthrough time:
tb
0.143 kg NO 2 0.126 kg NO 2 / h
1.13 h
0.126
kg NO 2 h
68 min
c. The first column would start at time 0 and finish at 1.13 h, and would not be available for another run until (1.13+1.50) = 2.63 h. The second column could start at 1.13 h and finish at 2.26 h. Since the first column would still be in the regeneration stage, a third column would be needed to start at 2.26 h. It would run until 3.39 h, at which time the first column would be available for another run. The first few cycles are shown below on a Gantt chart. Run
Regenerate
Column 1 0
1.13
2.63
3.39
4.52
6.02
Column 2 1.13
2.26
3.76
4.52
5.65
Column 3 2.26
6-76
3.39
4.89
5.65
6.78
Let S=sucrose, I=trace impurities, A=activated carbon
Add mA (kg A)
mS (kg S)
mS (kg S) mI (kg I)
m I0 (kg I) R0 (color units / kg S)
R (color units / kg S)
Come to equilibrium
V (L)
V (L) mA (kg A)
mIA (kg I adsorbed)
Assume
x
no sucrose is adsorbed
solution volume (V) is not affected by addition of the carbon m a. R(color units/kg S) kCi (kg I / L) = k I (1) V x
'R = k (Ci 0 Ci )
mIA mI 0 mI
k ( mI 0 m I ) V
'R
kmIA V
'R x100% R0
km IA / V m x100 100 IA kmI 0 / V mI 0 m IA Equilibrium adsorption ratio: X i* mA Normalized percentage color removal:
% removal of color
X
% removal ( 3) 100 m IA / m I 0 = m A / mS m A / mS m X = 100X *i S X i* mI 0
Freundlich isotherm X i* X=
100mS K F E
RE
9.500 9.000
mI 0 X 100mS
E;
8.500 8.000 0.000 1.000 2.000 3.000
6-77
(3) (4)
(5) R KF ( )E k
K F' R E
y = 0.4504x + 8.0718
ln R
(2)
m IA mS m A mI 0
mI 0 X 100mS
(1),( 5)
K F Ci E mI 0 k
100
A plot of ln X vs. ln R should be linear: slope
ln v
6.101
intercept = lnK 'F
6.101 (cont’d) ln X
0.4504 ln p NO2 8.0718 X
K F'
3203, E
X
3203R 0.4504
0.4504
b. 100 kg 48% sucrose solution m S 95% reduction in color
e 8.0718 R 0.4504
480 kg
R = 0.025(20.0) = 0.50 color units / kg sucrose
K F' R E
3203(0.50) 0.4504 2344 % color reduction 97.5 2344 = mA m A / mS m A / 480
6-78
20.0 kg carbon
CHAPTER SEVEN
7.1
0.80 L 35 . u 10 4 kJ 0.30 kJ work 1h 1 kW 2.33 kW 2.3 kW h L 1 kJ heat 3600 s 1 k J s 2.33 kW 10 3 W 1.341 u 10 3 hp 312 . hp 3.1 hp 1 kW 1W
7.2
All kinetic energy dissipated by friction (a) E k
mu 2 2 5500 lbm 552 miles 2 2 h2 715 Btu
52802 ft 2 12 mile 2
12 h 2 36002 s 2
1 lbf 9.486 u 10 4 B 32.174 lbm ft / s2 0.7376 ft lb f
(b) 3 u 10 8 brakings 715 Btu 1 day 1h 1W 1 MW 4 2 day braking 24 h 3600 s 9.486 u 10 Btu / s 10 6 W
7.3
(a) Emissions: 1000 sacks Paper Plastic
2000 sacks
1000 sacks
(0.0045 0.0146) oz
(724 905) Btu sack
Plastic
2000 sacks
1 lb m
sack 16 oz sack
Energy: Paper
(0.0510 0.0516) oz
(185 464) Btu sack
2617 MW
6.41 lb m
1 lb m 16 oz
2.39 lb m
1.63 u 10 6 Btu 1.30 u 10 6 Btu
(b) For paper (double for plastic)
Materials for 400 sacks
Raw Materials Acquisition and Production
Sack Production and Use
7-1
1000 sacks
Disposal
400 sacks
7.3 (cont’d) Emissions: Paper
400 sacks
Plastic
800 sacks
1 lb m 1000 sacks 0.0510 oz sack 16 oz
1 lb m 0.0516 oz 4.5 lb m sack 16 oz 30% reduction
1 lb m 2000 sacks 0.0045 oz sack 16 oz
1 lb m 0.0146 oz 2.05 lb m sack 16 oz 14% reduction
Energy: Paper
400 sacks
Plastic
(c) .
724 Btu sack
800 sacks
3 u 10 8 persons
185 Btu sack
905 Btu
1000 sacks
sack
119 . u 10 6 Btu; 27% reduction
464 Btu
2000 sacks
sack
1 sack 1 day 1h person - day 24 h 3600 s
1.08 u 10 6 Btu; 17% reduction
649 Btu 1J 1 MW -4 1 sack 9.486 u 10 Btu 10 6 J / s
2,375 MW . (2,375 MW) = 404 MW Savings for recycling: 017 (d) Cost, toxicity, biodegradability, depletion of nonrenewable resources. 7.4
(a) Mass flow rate: m
3.00 gal 1 ft 3 (0.792)(62.43) lb m min 7.4805 gal 1 ft 3
Stream velocity: u
1 3.00 gal 1728 in 3 2 min 7.4805 gal 3 0.5 in 2
Kinetic energy: E k
mu 2 2
b g
0.330 lb m s
d7.70 u 10
3
. g ft b1225 2
2
s
2
3
f
(b) Heat losses in electrical circuits, friction in pump bearings.
7-2
1 ft 1 min . ft s 1225 12 in 60 s
ft lb f 1 1 lb f 7.70 u 103 2 s 2 32.174 lb m ft / s
. u 10 hp I iFGH 01341 .7376 ft lb / sJK
ft lb f / s
1 min 0.330 lb m s 60 s
140 . u 105 hp
7.5
(a) Mass flow rate:
b
g
42.0 m S 0.07 m
m
s 2 mu 2
E k
2
10 3 L 273 K 130 kPa 1 mol 3 573 K 101.3 kPa 22.4 L STP 1m
b g
4
42.0 2 m2 1N 1J 2 2 s 1 kg m / s N m
127.9 g 1 kg 2 s 1000 g
(b)
b g
127.9 g 1 mol 273 K 101.3 kPa 22.4 L STP s 29 g 573 K 130 kPa 1 mol
29 g mol
127.9 g s
113 J s
1 m3 4 3 10 L S (0.07)2 m2
49.32 m s
127.9 g 1 kg 49.32 2 m2 1N 1J 1558 . J/s 2 2 s 1000 g s 1 kg m / s2 N m 'E k = E k (400 $ C) - E k (300 $ C) = (155.8 - 113) J / s = 42.8 J / s 43 J / s 2 mu 2
E k
(c) Some of the heat added goes to raise T (and hence U) of the air 7.6
(a)
'E p
mg'z
1 gal
mu 2 'E p 2
(b) E k
. 1 lbf 1 ft 3 62.43 lb m 32174 ft 10 ft 834 . ft lb f 3 2 7.4805 gal 1 ft s 32.174 lbm ft / s2
b g
mg 'z u
b g
2 g 'z
12
LM2FG 32.174 ft IJ b10 ftgOP NH sK Q 2
12
25.4
ft s
(c) False 7.7 (a)
'E k positive When the pressure decreases, the volumetric flow rate increases, and hence the velocity increases. 'E negative The gas exits at a level below the entrance level. p
b g
2
5 m S 1.5 cm 2 (b) m s
1 m3 10 4 cm 2
273 K 10 bars 1 kmol 303 K 101325 . bars 22.4 m 3 STP
b g
16.0 kg CH 4 1 kmol
0.0225 kg s
d h P d AV h
Pout AVout in
in
V nRT out nRT Vin uout
7.8
'z 'E p mg
uin
Pin Pout
uout (m / s) A(m2) Pin Pout uin (m / s) A(m2) bar b g 109 bar
5ms
Pin Pout
5555 ms .
105 m3 103 L 1 kg H2 O 981 1N 1J 2.778 u 107 kW h . m 75 m h 1 m3 1L s2 1 kg m/ s2 1 N m 1J
204 . u 104 kW h h
The maximum energy to be gained equals the potential energy lost by the water, or 2.04 u 10 4 kW h 24 h 7 days h 1 day 1 week
7-3
3.43 u 10 6 kW h week (more than sufficient)
7.9
(b) Q W
'U 'E k 'E p
b g 0 b no height changeg
'E k
0 system is stationary
'E p Q W
'U , Q 0, W ! 0
(c) Q W
'U 'E k 'E p
b
g
b
g
Q 0 adiabatic , W 0 no moving parts or generated currents 'E k 0 system is stationary 'E p 0 no height change 'U (d). Q W
b b
g
g
0 'U 'E k 'E p
b
g
W 0 no moving parts or generated currents 'E k 0 system is stationary 'E p 0 no height change 'U , Q 0 Even though the system is isothermal, the occurrence of a chemical reaction assures that 'U z 0 in a non-adiabatic reactor. If the temperature went up in the adiabatic reactor, heat must be transferred from the system to keep T constant, hence Q 0 .
Q
b b
g
g
7.10 4.00 L, 30 °C, 5.00 bar V (L), T (°C), 8.00 bar (a). Closed system:
'U 'E k 'E p
RS'E |T'E
'U (b)
Constant T 'U 0 Q W
(c) Adiabatic Q
bg
S 3
0 'U
Q W
b b
0 initial / final states stationary 0 by assumption
k p
g
g
Q W
7.65 L bar
W
8.314 J 0.08314 L bar
765 J
transferred from gas to surroundings
7.65 L bar > 0, Tfinal ! 30q C
1 m2 2.83 u 10 3 m 2 4 2 10 cm (a) Downward force on piston:
7.11 A
Fd
cm 2
Patm A mpiston+weight g 1 atm 1.01325 u 105 N / m2 2.83 u 10 3 m2 atm
7-4
24.50 kg 9.81 m s
2
1N 1 kg m / s2
527 N
7.11 (cont’d) Upward force on piston: Fu
APgas
d2.83 u 10
3
i d
m 2 Pg N m 2
i
Equilibrium condition: Fu
Fd 2.83 u 10 3 m2 P0
V0
303 K 1.01325 u 105 Pa 0.08206 L atm nRT 1.40 g N 2 1 mol N 2 0.677 L P0 28.02 g 1.86 u 105 Pa 1 atm mol K
(b) For any step, 'U 'E k 'E p Step 1: Q | 0 'U Step 2: 'U that V
186 . u 10 5 N m 2
527 P0
Q W 'U 'E k 0 'E p 0
186 . u 105 Pa
Q W
W
Q W As the gas temperature changes, the pressure remains constant, so
nRT Pg must vary. This implies that the piston moves, so that W is not zero. Tfinal 'U
Overall: Tinitial
0 Q W
0
In step 1, the gas expands W ! 0 'U 0 T decreases (c) Downward force Fd
u 10 id2.83 u 10 i b4.50gb9.81gb1g . gd101325 . b100 3
5
331 N (units
as in Part (a)) Final gas pressure Pf Since T0
Tf
F A
30q C , Pf V f
Distance traversed by piston W
331 N 2.83 u 10 3 m 2 P0V0 V f 'V A
. mg b331 Ngb0142
Fd
116 . u 10 5 N m 2 V0
P0 Pf
b1.08 0.677g L
. u 10 b0.677 Lg 186 116 . u 10
5 5
Pa Pa
1 m3 10 3 L 2.83 u 10 3 m2
108 . L m 0142 .
47 N m 47 J
Since work is done by the gas on its surroundings, W
47 J Q Q W 0
47 J
(heat transferred to gas) 7.12 V H
32.00 g 4.684 cm3 mol U PV
103 L
0.1499 L mol 106 cm3 41.64 atm 0.1499 L 8.314 J / (mol K) 1706 J mol mol 0.08206 L atm / (mol K) g
7-5
2338 J mol
d
i
7.13 (a) Ref state U (b) 'U
0 liquid Bromine @ 300 K, 0.310 bar
U final U initial
0.000 28.24
d i
' H
'U ' PV
'U P'V (Pressure Constant)
0.310 bar ' H 2824 . kJ mol
79.94g L . b00516
8.314 J 1 kJ 307 . kJ mol mol 0.08341 L bar 103 J
. kJ 154 kJ b5.00 molgb30.7 kJ / molg 15358 U independent of P U b300 K, 0.205 bar g U b300 K, 0.310 bar g 28.24 kJ mol U d340 K, P i U b340 K, 1.33 bar g 29.62 kJ mol n' H
'H
(c)
28.24 kJ mol
f
'U U final U initial
E
'U
29.62 28.24 1380 kJ mol .
= P' V' V' = PV / P' V changes with pressure. At constant temperature PV (T = 300K, P = 0.205 bar) = (0.310 bar)(79.94 L / mol) 120.88 L / mol V' 0.205 bar 5.00 L 1 mol n 0.0414 mol 120.88 L 'U n'U 0.0414 mol 138 . kJ / mol 0.0571 kJ
b
gb
'U 'E k 'E p 0
Q W Q
0
g
0.0571 kJ
0
(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is being neglected; internal energy is not completely independent of pressure. 7.14 (a) By definition H U PV ; ideal gas PV
RT H
U RT
b g U bT g H bT , Pg U bT g RT H bT g independent of P
U T , P
(b) ' H 'H
'U R'T
n' H
3500
b2.5 molgb3599 cal / molg
7.15 'U 'E k 'E p
Q Ws ' Ek 'Ep Ws
'U
cal 1.987 cal 50 K mol K mol
Q W 'U
b b
3599 cal mol
8998 cal 9.0 u 10 3 cal
g
0 no change in m and u 0 no elevation change P'V since energy is transferred from the system to the surroundings
g
b
Q P'V Q
'U P'V
7-6
g
' (U PV )
'H
b b
7.16. (a) ' E k 'Ep
g
0 u1 u 2 0 0 no elevation change
g
'P 0 (the pressure is constant since restraining force is constant, and area is constrant) Ws P'V the only work done is expansion work H 34980 355 . T (J / mol), V1 = 785 cm3, T1 = 400 K, P = 125 kPa, Q = 83.8 J 125 u 103 Pa 785 cm3 1 m3 PV n= 0.0295 mol RT 8.314 m3 Pa / mol K 400 K 106 cm3 -H ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol) Q = 'H = n(H
b
.
g
2
1
2
83.8 J = 0.0295 35.5T2 - 35.5(400) T2
480 K
0.0295 mol 8.314 m3 Pa 106 cm3 480 K 941 cm3 mol K 125 u 105 Pa 1 m3 125 u 105 N (941 - 785)cm3 1 m3 P'V 19.5 J m2 106 cm3 'U P'V 'U Q 'PV 838 . J 19.5 J 64.3 J
nRT P
i) V ii ) W iii ) Q (b) 'Ep
0
7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could occur, the temperature would drop during these periods.) (b) 'U ' E p ' E R Q 't W 't 'Ep Q
0, ' E k 0.90 u 1.4 W
0, W 1 J s 1W
0, U ( t
0)
0
1.26 J s
U (J ) 126 . t Moles in tank: n U
U n
PV RT
126 . t (J) 0.0859 mol
1 atm
b
2.10 L 25 273 K
g
1 mol K 0.08206 L atm
0.0859 mol
14.67t
Thermocouple calibration: T
aE b
b g
T qC T 0 , E 0.249 T 100 , E 5.27
b g
181 . E mV 4.51
U 14.67t 0 440 880 1320 T 181 . E 4.51 25 45 65 85 (c) To keep the temperature uniform throughout the chamber. (d) Power losses in electrical lines, heat absorbed by chamber walls. (e) In a closed container, the pressure will increase with increasing temperature. However, at the low pressures of the experiment, the gas is probably close to ideal U f T only.
bg
Ideality could be tested by repeating experiment at several initial pressures same results.
7-7
7.18 (b) 'H 'E k 'E p
Q W s (The system is the liquid stream.)
c c
h
' E k 0 no change in m and u ' E p 0 no elevation change Ws 0 no moving parts or generated currents
c
' H
h
h
Q , Q ! 0
(c) 'H 'E k 'E p
Q W s (The system is the water)
c
h
' H 0 T and P ~ constant ' E k 0 no change in m and u Q 0 no ' T between system and surroundings
c
'E p
c
b
g
Q W s (The system is the oil)
c
' E k 0 no velocity change
h
Q W s Q 0 (friction loss); W s 0 (pump work).
(e) 'H 'E k 'E p
Q W s (The system is the reaction mixture)
c
h
' E k ' E p 0 given ' Ws 0 no moving parts or generated current
c
' H
h
W s , W s ! 0 for water system
(d) 'H 'E k 'E p 'H ' E p
h
h
Q , Q pos. or neg. depends on reaction
7.19 (a) molar flow:
1 mol 125 . m3 273 K 122 kPa 423 K 101.3 kPa 22.4 L STP min
b g
' H ' E k ' E p
103 L 43.4 mol min 1 m3
Q W s
c
h
' E k ' E p 0 given Ws 0 no moving parts
c
Q
' H
n' H
h
43.37 mol 1 min min
60s
3640 J
kW
mol 10 3 J / s
2.63 kW
(b) More information would be needed. The change in kinetic energy would depend on the cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet and outlet pipes would be needed to answer this question.
7-8
b g
7.20 (a) H 104 . T q C 25
H in kJ kg P=110 kPa
H out 1.04 34.0 25 9.36 kJ kg H in 1.04 30.0 25 5.20 kJ kg ' H 9.36 5.20 4.16 kJ kg ' H ' E k ' E p
n (mol/s) N2
34 oC
Q W s
c
Q =1.25 kW
h h
' E k ' E p 0 assumed Ws 0 no moving parts
c
' H n' H
Q
n
Q ' H
1.25 kW
kg 1 kJ / s 103 g 1 mol 10.7 mol s 4.16 kJ kW 1 kg 28.02 g
b g
10.7 mol 22.4 L STP V = s mol
303 K 1013 . kPa 273 K 110 kPa
2455 . L / s 246 L s
(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is assumed to depend linearly on temperature and to be independent of pressure, errors in measured temperature and in wattmeter reading.
7.21 (a) H
H 2 H 1 T2 T1 H1 aT1
aT b a b
H
0 Tref
130.2 5.2
b g
Table B.1 S . G.
b
U kJ kg
g
b
U kJ kg
b gb g
b
g
b g
5.2T q C 130.2
25q C
bg
C 6 H 14 l
H PV
U| V| W
129.8 258 . 5.2 H kJ kg 50 30 258 . 5.2 30 130.2
0.659 V
1 m3 152 . u 10 3 m 3 kg 659 kg
b5.2T 130.2gkJ / kg
1J 1 kJ 1 atm 1.0132 u 105 N / m2 1.52 u 10 3 m3 1 atm 1 kg 1 N m 103 J
g
5.2T 130.4
(b) Energy balance: Q 'Ek , 'E p , W 0
'U
20 kg [(5.2 u 20 - 130.4) - (5.2 u 80 - 130.4)] kJ 1 kg
Average rate of heat removal
6240 kJ 1 min 5 min
60 s
7-9
20.8 kW
6240 kJ
7.22
m (kg/s) 260°C, 7 bars H = 2974 kJ/kg u0 = 0
m (kg/s) 200°C, 4 bars H = 2860 kJ/kg u (m/s)
'H 'E k 'E p 'E p
Q
Q Ws
Ws
0
2 mu H out H in 'E k 'H m 2
d
d
u 2 2 H in H out
i
b
i
g
(2) 2974 2860 kJ 103 N m 1 kg m / s2 kg 1 kJ 1N
7.23 (a) 5 L/min
2.28 u 105
5 L/min 100 mm Hg (gauge)
0 mm Hg (gauge)
Qout
Qin
Since there is only one inlet stream and one outlet stream, and m in Eq. (7.4-12) may be written m 'z Q W s m 'U m ' PV ' u 2 mg 2 'U 0 given
d i
'u
Pout Pin mV
Heat input: Q in Efficiency:
V' P Q in
f
V'P
0 assume for incompressible fluid
2
'z
0
W s
0 all energy other than flow work included in heat terms
Q
(b) Flow work: V'P
m out { m ,
d i
a
m 'PV
V'P
m2 u 477 m / s s2
Q in Q out
Q in Q out 5L min
b100 0gmm Hg
5 ml O 2 min
20.2 kJ 1 ml O 2
66.7 J min 1.01 u 105 J min
1 atm 8.314 J 66.7 J min 760 mm Hg 0.08206 liter atm
10 3 J 101 . u 105 J min 1 kJ u 100 0.066%
7-10
Q W s ; 'E k , 'E p , W s
7.24 (a) 'H 'E k 'E p
b b
g
0 'H
Q
H 400q C, 1 atm 3278 kJ kg (Table B.6) H 100q C, sat' d 1 atm 2676 kJ kg (Table B.5)
g
100 kg H 2 O(v) / s
100 kg H 2 O(v) / s
o
400 o C, 1 atm
100 C, saturated Q (kW)
100 kg Q s
b3278 2676.0gkJ
10 3 J kg 1 kJ
(b) 'U 'E k 'E p
b
U 100q C, 1 atm
Q
g
Q W ; 'E k , 'E p , W 0 'U Q 2507 kJ kg U 400q C, 1 atm 2968 kJ kg (Tables B.5 & B.7)
b
g
b
m'U
'U
6.02 u 10 7 J s
g
d
100 kg 2968 2507 kJ kg 10 3 J kJ
i
4.61 u 10 7 J
The difference is the net energy needed to move the fluid through the system (flow work).
c b g h 83.9 kJ kg (Table B.5) H bsteam, 20 bars, sat' d g 2797.2 kJ kg (Table B.6)
7.25 H H 2 O l , 20q C
m [kg H 2 O(l) / h]
m [kg H 2 O(v) / h]
o
20 C
20 bar (sat' d) Q = 0.65(813 kW)
(a) 'H 'E k 'E p 'H
m (b) V
Q ' H
Q W s ; 'E k , 'E p , W s
528 kW
0 'H
Q
m 'H
528 kW
b
kg 1 kJ / s 3600 s 2797.2 83.9 kJ 1 kW 1 h
g
b701 kg hg d0.0995 m kgi A 3
701 kg h
69.7 m 3 h sat' d steam @ 20 bar
Table B.6
(c) V
nRT P
701 kg / h 18.02 g / mol
103 g / kg 485.4 K 0.08314 L bar mol K
20 bar
1 m3 103 L
78.5 m3 / h
The calculation in (b) is more accurate because the steam tables account for the effect of pressure on specific enthalpy (nonideal gas behavior). (d) Most energy released goes to raise the temperature of the combustion products, some is transferred to the boiler tubes and walls, and some is lost to the surroundings.
7-11
c bg
7.26 H H 2 O l , 24q C, 10 bar
h
100.6 kJ kg (Table B.5 for saturated liquid at 24oC; assume H
independent of P).
b
g
2776.2 kJ kg (Table B.6) ' H
H 10 bar, sat' d steam
[kg H2 O(v) / h] m
2776.2 100.6 2675.6 kJ kg
[kg H2 O(v) / h] m 15,000 m3 / h @10 bar (sat'd)
o
24 C, 10 bar Q (kW) 15000 m 3 m h
kg 01943 m3 .
7.72 u 10 4 kg h
A
b Table 8.6 g
d
Energy balance 'E p , W s
'E k
'E k
E kfinal E kinitial f mu
2
i
0 : 'H 'E k
E kinitial |0
7.72 u 10 4 kg
'E k
E kfinal
d15,000 m hi 3
2
2
015 . 2 S 4 m2
h
2
Q
1
1
h3
1J
2 3600 3 s 3
A
1 kg m 2 / s 2
A S D2 4 5.96 u 10 5 J / s 1h 7.72 u 10 4 kg 2675.6 kJ 5.96 u 10 5 J 1 kJ h kg 3600 s s 10 3 J
Q m 'H 'E k
57973 kJ s 5.80 u 10 4 kW 7.27 (a)
228 g/min 25oC
228 g/min T(oC) Q ( kW)
Energy balance: Q 'E x , 'E p , Ws 0
=0
b g
'H Q W
b g 0.263Q bW g T bq C g 25 H bJ gg 0.263QbW g 0 (b) H bbT 25g H bJ gg 3.34 T bq Cg 25
228 g 1 min ( H out H in ) J min 60 s g
H out J g
26.4 27.8 29.0 32.4 4.47 9.28 13.4
Fit to data by least squares (App. A.1)
b
7-12
¦ H i
i
24.8
bT 25g ¦ bT 25g i
i
i
2
3.34
7.27 (cont’d) (c) Q
' H
b
g
350 kg 10 3 g 1 min 3.34 40 20 J min kg 60 s g
kW s 390 kW heat input to liquid 10 3 J
(d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads. 7.28 m w [ kg H 2 O(v) / min] 3 bar, sat' d
m w [ kg H 2 O(l) / min] 27 o C
Q ( kW)
m e [ kg C 2 H 6 / min] 16 o C, 2.5 bar
(a) C H mass flow: m 2 6 e
H ei
m e [ kg C 2 H 6 / min] 93 o C, 2.5 bar
795 m 3 10 3 L 2.50 bar min m 3 289 K 2.487 u 103 kg min
1 K - mol 30.01 g 1 kg 0.08314 L - bar mol 1000 g
941 kJ kg , H ef
1073 kJ kg Energy Balance on C 2 H 6 : 'E p , W s 0, 'E k # 0 Q
LMb N
g OPQ
kJ kg Q 2.487 u 103 1073 941 min kg
b bliquid, 27q Cg
(b) H s1 3.00 bar, sat' d vapor H s2
g
'H
2.487 u 103 kJ 1 min 5.47 u 103 kW min 60 s
2724.7 kJ kg (Table B.6)
1131 . kJ kg (Table B.5)
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
d
condensing steam equals the heat transferred to the ethane 5.47 u 10 3 kW Energy balance on H 2 O: Q m Vs
Q H s2 H s1
'H
5.47 u 10 3 kJ s
b2.09 kg / sg d0.606 m kgi A 3
d
m H s2 H s1
i
kg . 2724.7 kJ 1131
b
g
i
2.09 kg s steam
1.27 m 3 s
Table B.6
Too low. Extra flow would make up for the heat losses to surroundings. (c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it would require heat flow from the ethane to the steam over some portion of the exchanger. (Observe the two outlet temperatures)
7-13
7.29
250 kg H2 O(v )/min 40 bar, 500°C H 1 (kJ/kg)
W
b b
s
Heat exchanger
250 kg/min 5 bar, T 2 (°C), H 2 (kJ/kg)
Turbine
=1500 kW
250 kg/min 5 bar, 500°C H3 (kJ/kg)
Q(kW)
g
H 2 O v , 40 bar, 500q C : H 1 3445 kJ kg (Table B.7) H 2 O v , 5 bar, 500q C : H 3 3484 kJ kg (Table B.7)
g
(a) Energy balance on turbine: 'E p 'H
d
W s m H 2 H 1
i
3445 kJ 1500 kJ s kg H
0, Q
0, 'E k # 0
W s H 2
H 1 W s m
min 60 s 3085 kJ kg 250 kg 1 min
3085 kJ kg and P 5 bars T = 310q C (Table B.7)
(b) Energy balance on heat exchanger: 'E p
d
H 3 H 2 Q 'H m
i
250 kg
d
Q W s m s H 3 H 1
Q
'H 'W s
1 min
1 kW
kg
60 s
1 kJ / s
1663 kW
0, 'E k # 0
i
Q W s
b3484 3445gkJ
250 kg min
0, 'E k # 0
b3484 3085gkJ
min
(c) Overall energy balance: 'E p 'H
0, W s
kg
1 min 1 kW 1500 kJ 1 kW s 1 kJ / s 60 s 1 kJ / s
1663 kW
b g H Obv , 5 bar, 310q Cg: V
(d) H 2 O v , 40 bar, 500q C : V1 2
2
0.0864 m 3 kg (Table B.7) 0.5318 m 3 kg (Table B.7)
u1
250 kg 1 min 0.0864 m 3 min 60 s kg
u2
250 kg min 0.5318 m 3 min 60 s kg
'E k
m 2 u2 u12 2
1 0.5 S 4 m 2 1 0.5 S 4 m 2 2
250 kg 1 1 min
min 2 0.26 kW << 1500 kW
. ms 183
2
. ms 113
. g b113. g b183
60 s
2
s
7-14
2
2
m2
1 kW s
1N 1 kg m / s
2
10 3 N m
7.30 (a) 'E p , 'E k , W s
b
0 Q
'H hA Ts To
h 8 To Ts =34.2
(b) Clothed:
g
b
300 kJ h 18 . h Ts To
g
300
kJ h
13.4q C
Nude, immersed: h 64 To Ts =34.2
316 . q C (Assuming Ts remains 34.2qC)
(c) The wind raises the effective heat transfer coefficient. (Stagnant air acts as a thermal insulator —i.e., in the absence of wind, h is low.) For a given To , the skin temperature must drop to satisfy the energy balance equation: when Ts drops, you feel cold. 7.31 Basis: 1 kg of 30°C stream 1 kg H2O(l)@30oC 3 kg H2O(l)@Tf(oC) 2 kg H2O(l)@90oC
b
g b
g
1 2 30 o C 90 $ C 70 $ C 3 3 (b) Internal Energy of feeds: U 30q C, liq. U 90q C, liq.
(a) T f
b b
U|V 376.9 kJ kg|W
g g
125.7 kJ kg
(Table B.5 - neglecting effect of P on H ) Energy Balance: Q - W = 'U + 'E p + 'E k
b
g
Q =W = 'E p = 'E k 0
b
3U f (1 kg) 125.7 kJ / kg (2 kg) 376.9 kJ / kg
7.32
U f
293.2 kJ kg T f
Diff.
70.05 70.00 u 100% 70.05
(a) Table B.6
P 5 bars
T
52.5 m3 H2O(v)/h . m(kg/h) 5 bar, T(oC)
.
Q (kW)
1518 . q C , H L
. gb140 kg hg b015
Energy balance: Q 'H
0
0.07% (Any answer of this magnitude is acceptable).
. kJ kg , H V 6401
bar, sat'd) = 0.375 m 3 / kg m V(5 (b) H 2 O evaporated
0
70.05q C (Table B.5)
.
m(kg/h) . kg H2 O( v)/kg 0.85 0.15 kg H2 O( l)/kg 5 bar, saturated, T(oC)
g
'U
21 kg h
52.5 m3 h
1
2747.5 kJ kg kg
0.375 m 3
140 kg h
21 kg h
b2747.5 640.1gkJ
1h 1 kW 12 kW kg 3600 s 1 kJ s
7-15
7.33 (a) P 5 bar
Table B.6
(b) Inlet: T=350°C, P=40 bar Outlet: T=75°C, P=5 bar uin
Vin Ain
uout
Vout Aout
1518 . o C . At 75°C the discharge is all liquid
Tsaturation
Table B.7
H in = 3095 kJ / kg , Vin = 0.0665 m 3 / kg H out = 314.3 kJ / kg , Vout = 1.03 u 10 -3 m 3 / kg
Table B.7
200 kg 1 min 0.0665 m 3 / kg S (0.075) 2 / 4 m 2 min 60 s 200 kg 1 min min 60 s
0.00103 m 3 / kg S (0.05) 2 / 4 m 2
5018 . m/s 175 . m/s
m m ( H 2 H 1 ) (u22 u12 ) 2 200 kg 1 min (314 - 3095) kJ 200 kg 1 min (1.752 - 50.18 2 ) m 2 W s Q min 60 s kg 2 min 60 s s2 13,460 kW Energy balance: Q W s | 'H 'E k
7.34 (a) Assume all heat from stream transferred to oil 1.00 u 10 4 kJ 1 min Q 167 kJ s min 60 s 100 kg oil/min 135°C m (kg H2O(v)/s) 25 bars, sat'd
Energy balance on H 2 O: Q 'E p , 'E k , W s
m
H out H in
i
962.0 kJ kg , H v , 25 bar, sat' d
167 kJ s
Time between discharges:
(b) Unit Cost of Steam:
d
m H out H in
0
H l , 25 bar, sat' d
Q
'H
100 kg oil/min 185°C m (kg H2O(l)/s) 25 bars, sat'd
b
g
0.091 kg s
1s 1 kg 1200 g 13 s discharge discharge 0.091 kg 10 3 g
$1 10 Btu 6
kg 962.0 2800.9 kJ
2800.9 kJ kg (Table B.6)
b2800.9 83.9g kJ kg
0.9486 Btu $2.6 u 10 3 / kg kJ
Yearly cost: 1000 traps 0.091 kg stream 0.10 kg last 2.6 u 10 3$ 3600 s 24 h 360 day trap s kg stream kg lost h day year $7.4 u 105 / year
7-16
7.35 Basis: Given feed rate 200 kg / h H 2 O (v ) 10 bar, sat'd H 2776.2 kJ / kg
n3 (kg / h H 2 O ( v )) 10 bar, 250 o C H 2443 kJ / kg
n2 (kg / h H 2 O (v )) 10 bar, 300 o C H 3052 kJ / kg
Q (kJ / h)
H from Table B.6 (saturated steam) or Table B.7 (superheated steam) Mass balance: 200 n 2
Energy balance: Q
'H
'E K , 'E p , W 0
(a) n 3
300 kg h
(b) Q 0
(1), (2)
(1)
n 3
(1)
b g
b
n 2
( 2)
100 kg h
406 kg h , n 3
n 2
g b
g
n 3 2943 200 2776.2 n 2 3052 Q in kJ h
7.36 (a) Tsaturation @ 1.0 bar = 99.6 °C T f H 2 O (1.0 bar, sat' d) H l
Q
(2)
3.23 u 10 4 kJ h
606 kg h 99.6 $ C
417.5 kJ / kg, H v
2675.4 kJ / kg
$
H 2 O (60 bar, 250 C) 1085.8 kJ / kg Mass balance: mv ml 100 kg Energy balance: 'H 0
(1)
'E K , Q , 'E p , W 0
mv H v ml H l m1 H 1 (1,2)
ml
70.4 kg, mv
(b) Energy balance: 'H 'E p , Q , W 0
mv H v ml H l (100 kg)(1085.8 kJ / kg) = 0 29.6 kg y v
'E k H 2 H 1
29.6 kg vapor 100 kg
'E k H 2
0.296
(2)
kg vapor kg
H 1 'E k
The system undergoes expansion, and assuming the same pipe diameter 'E k ! 0 'H 0 Since some of the water will be evaporated, the temperature will still be the saturation temperature at the given final pressure, so T is unchanged. Less water will evaporate because some of the energy that went to evaporate the water will instead be converted to kinetic energy. (c) P f
39.8 bar (pressure at which the water is still liquid, but has the same enthalpy as the
feed)
7-17
7.36 (cont’d) (d) Since enthalpy does not change, then when Pf t 39.8 bar the temperature cannot increase, because a higher temperature would increase the enthalpy. Also, when Pf t 39.8 bar , the product is only liquid no evaporation occurs. 0.4 Tf (C)
y
0.3 0.2 0.1 0 0
20
40
60
300 250 200 150 100 50 0 1
80
5
10
15
20
25
30
36 39.8 60
Pf (bar)
Pf (bar)
7.37 10 m3, n moles of steam(v), 275°C, 15 bar 10 m3, n moles of water (v+l), 1.2 bar 10.0 m3 H2O (v)
10.0 m3
min (kg) 275oC, 1.5 bar
mv [kg H2O (v)] ml [kg H2O (l)]
Q
Table B.6
(a) P=1.2 bar, saturated,
(b) Total mass of water: min = Mass Balance: mv ml
10 m 3
104.8 $ C
T2
1 kg 0.1818 m 3
55 kg
55.0
Volume additivity: Vv Vl mv
1.2 bar, saturated
7.0 kg, ml
10.0 m 3
mv (1428 . m 3 / kg) ml (0.001048 m 3 / kg)
48.0 kg condensed
(c) Table B.7 U in = 2739.2 kJ / kg; Vin = 0.1818 m 3 / kg 3 Table B.6 U l = 439.2 kJ / kg; Vl = 0.001048 m / kg U = 2512.1 kJ / kg; V = 1.428 m 3 / kg
R| S| T
v
v
Energy balance: Q = 'U = mv U v ml U l minU in 'E p , 'E k , W 0
[(7.0)(2512.1 kJ / kg) + (48.0)(439.2) - 55 kg (2739.2)] kJ = 1.12 u 10 5 kJ 7.38 (a) Assume both liquid and vapor are present in the valve effluent. 1 kg H 2 O ( v ) / s 15 bar, Tsat 150 o C
m l [ kg H 2 O ( l ) / s] m v [ kg H 2 O ( v ) / s] 1.0 bar, saturated
7-18
7.38 (cont’d) (b) Table B.6 Tsat'n (15 bar) = 198.3o C Tin 348.3o C Table B.7 H in H (348.3$ C, 15 bar) | 3149 kJ / kg Table B.6 H l (1.0 bar, sat' d) = 417.5 kJ / kg; H v (1.0 bar, sat' d) = 2675.4 kJ / kg Energy balance: 'H 0 m H m H m H 0 l
'E p , 'E k ,Q , Ws 0
m in H in
m l H l m v H v
l
v
m v m l
in
v
in
3149 kJ / kg m l (417.5) (1 m l )(2675.4)
There is no value of m l between 0 and 1 that would satisfy this equation. (For any value in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase assumption is therefore incorrect; the effluent must be pure vapor. m in m out 1 (c) Energy balance m out H out m in H in 3149 kJ / kg = H (1 bar, Tout ) Table B.7
Tout | 337 $ C
(This answer is only approximate, since 'E k is not zero in this process). 7.39 Basis: 40 lb m min circulation (a) Expansion valve R = Refrigerant 12 40 lbm R(l)/min
40 lb m / min x v lb m R ( v ) / lb m
93.3 psig, 86°F H = 27.8 Btu/lb m
(1 x v ) lb m R( l ) / lb m H v 77.8 Btu / lb m , H l
Energy balance: 'E p , W s , Q 0, neglect 'E k 'H
bg
40 X v lb m R v min
b
g
E
¦ n H ¦ n H i
i
out
bg
77.8 Btu 40 1 X v lb m R l lb m min
Xv
9.6 Btu / lb m
9.6 Btu 40 lb m min lb m
b
i
0
i
in
27.8 Btu lb m
0
g
0.267 26.7% evaporates
(b) Evaporator coil 40 lb m R( v )/min 11.8 psig, 5°F H = 77.8 Btu/lbm
40 lbm /min 0.267 R(v ) 0.733 R( l ) 11.8 psig, 5°F H v = 77.8 Btu/lbm , H l = 9.6 Btu/lbm
Energy balance: 'E p , W s
0, neglect 'E k Q
b gb
g
bg
40 lb m 77.8 Btu 40 0.267 lb m R v Q min lb m min 2000 Btu min
7-19
'H
b gb
g
bg
77.8 Btu 40 0.733 lb m R l lb m min
9.6 Btu lb m
7.39 (cont’d) (c) We may analyze the overall process in several ways, each of which leads to the same result. Let us first note that the net rate of heat input to the system is Q
Q evaporator Q condenser
2000 2500
500 Btu min
and the compressor work Wc represents the total work done on the system. The system is
b g
closed (no mass flow in or out). Consider a time interval 't min . Since the system is at steady state, the changes 'U , 'E k and 'E p over this time interval all equal zero. The total heat input is Q 't , the work input is W 't , and (Eq. 8.3-4) yields c
Q 't Wc 't
0 W c
500 Btu 1 min u 10 3 hp . 1341 . hp 118 min 60 s 9.486 u 10 4 Btu s
Q
7.40 Basis: Given feed rates n1 (mol / h)
nC3H 8 (mol C 3 H 8 / h) nC4 H10 (mol C 4 H 10 / h) 227 o C
0.2 C 3 H 8 0.8 C 4 H 10 0 o C, 1.1 atm
n2 (mol / h) 0.40 C 3 H 8 0.60 C 4 H 10 25o C, 1.1 atm
Q (kJ / h)
Molar flow rates of feed streams: 300 L 1.1 atm 1 mol n1 hr 1 atm 22.4 L STP
b g
14.7 mol h
200 L 273 K 1.1 atm 1 mol 9.00 mol h hr 298 K 1 atm 22.4 L STP 14.7 mol 0.20 mol C 3 H 8 9.00 mol 0.40 mol C 3 H 8 Propane balance n C 3H 8 h mol h mol 6.54 mol C 3 H 8 h Total mole balance: n C4 H10 (14.7 9.00 6.54) mol C 4 H 20 h 17.16 mol C 4 H 20 h
b g
n 2
Energy balance: 'E p , W s
Q 'H
¦ N H ¦ N H i
i
i
out
in
b0.40 u 9.00g mol C H
( H i
0, neglect 'E k Q
3
h
8
i
'H
6.54 mol C3 H8 20.685 kJ 17.16 mol C4 H10 h mol h
b
g
27.442 kJ mol
1.772 kJ 0.60 u 9.00 mol C4 H10 2.394 kJ 587 kJ h mol h mol
0 for components of 1st feed stream)
7-20
510 m 3 273 K 10 3 L 1 mol 3 min 291 K m 22.4 L STP
7.41 Basis: (a)
1 kmol 10 3 mol
b g
.
n 0 (kmol/min) 38°C, h r = 97% x 0 (mol H 2 O/mol) (1 – x 0) (mol dry air/mol)
. kmol min 214
21.4 kmol/min 18°C, sat'd x 1 (mol H 2 O/mol) (1 – x 1) (mol dry air)
.
n 2 (kmol H 2O(l )/mol) 18°C
b
hr PH 2 O 38q C
Inlet condition: x o
P
b
PH 2 O 18q C
Outlet condition: x1
P Dry air balance: 1 0.0634 n o
b
g
b
0.97 49.692 mm Hg 760 mm Hg
g
g
0.0634 mol H 2 O mol
15.477 mm Hg 0.0204 mol H 2 O mol 760 mm Hg 1 0.0204 21.4 n o 22.4 kmol min
g b
b
g
g
b
g
Water balance: 0.0634 22.4 n 2 0.0204 214 . n 2 0.98 kmol min 0.98 kmol 18.02 kg 18 kg / min H 2 O condenses min kmol
b
(b). Enthaphies: H air 38q C
g
b
g
0.0291 38 25
0.3783 kJ mol
b g 0.0291b18 25g 0.204 kJ mol bv, 38q Cg 2570.8 kJkg 101 kgg 18.02molg 46.33 kJ molU| || 2534.5 kJ 1 kg 18.02 g 45.67 kJ mol VTable B.5 bv, 18q Cg kg 10 g mol || 75.5 kJ 1 kg 18.02 g . kJ mol bl, 18q Cg kg 10 g mol 136 |W H air 18q C
H H 2 O H H 2 O H H 2 O
3
3
3
Energy balance: 'E , W 0, 'E # 0 p
s
b1 0.0204gd214. u 10 ib0.204g b0.0204gd214 . u 10 ib45.67g d0.98 u 10 ib136 . g b1 0.0634gd22.4 u 10 ib0.3783g b0.0634gd22.4 u 10 ib46.33g 5.67 u 10 kJ min
Q
k
'H
¦ n H ¦ n H i
out
i
i
3
i
Q
in
3
3
3
3
4
5.67 u 10 4 kJ 60 min 0.9486 Btu 1 ton cooling min h kJ 12000 Btu
7-21
270 tons of cooling
7.42 Basis: 100 mol feed n2 (mol), 63.0°C 0.98 A(v ) 0.02 B(v )
A - Acetone B - Acetic Acid
Qc (cal)
0.5 n2 (mol) 0.98 A(l ) 0.02 B(l )
100 mol, 65.0°C 0.65 A(l ) 0.35 B(l )
56.8°C
n5 (mol), 98.7°C 0.544 A(v ) 0.456 B(v )
0.5 n2 (mol) 0.98 A(l ) 0.02 B(l )
n5 (mol), 98.7°C 0.155 A(l ) 0.845 B(l )
Qr (cal)
(a) Overall balances:
UV W
Total moles: 100 0.5n2 n5 n2 A: 0.65 100 0.98 0.5n2 0155 . n5 n5
b g
b
g
120 mol 40 mol
b g b g 0.155b40g 6.2 mol A 0.845b40g 338 . mol B
Product flow rates: Overhead 0.5 120 0.98 58.8 mol A 0.5 120 0.02 12 . mol B Bottoms
'H
Overall energy balance: Q 'E , W 0 , ' E # 0 p
i
out
x
2
¦ n H ¦ n H i
i
i
in
interpolate in table
bg
B
bg
b g b g b g (b) Flow through condenser: 2b58.8g 117.6 mols A 2b12 . g 2.4 mols B
interpolate in table
B
b g
Q 58.8 0 1.2 0 6.2 1385 33.8 1312 65 271 53 257
Energy balance on condenser: Qc 'E , W 0 , ' E # 0 p
Qc
b
3
k
g
b
117.6 0 7322 2.4 0 6807
2.63 u 10 4 cal
'H
g
8.77 u 10 5 cal heat removed from condenser
Assume negligible heat transfer between system & surroundings other than Qc & Qr Qr
7.43
Q Qc
d
2.63 u 10 4 8.77 u 10 5
i
1.96 kg, P1= 10.0 bar, T1
9.03 u 105 cal heat added to reboiler
2.96 kg, P3= 7.0 bar, T3=250oC
1.00 kg, P2= 7.0 bar, T2 Q= 0
7-22
7.43 (cont’d) (a) T2 T ( P 7.0 bar, sat' d steam) = 165.0 o C H 3 ( H 2 O(v ), P = 7.0 bar, T = 250 o C) 2954 kJ kg (Table B.7) H 2 ( H 2 O(v ), P = 7.0 bar, sat' d) 2760 kJ kg (Table B.6) Energy balance 'E , Q, W , 'E # 0 p
s
k
0 2.96H 3 196 . H 1 10 . H 2 196 . H 1
'H
2.96 kg(2954 kJ / kg) - 1.0 kg(2760 kJ / kg)
H 1 (10.0 bar, T1 ) 3053 kJ / kg T1 # 300 C $
(b) The estimate is too low. If heat is being lost the entering steam temperature would have to be higher for the exiting steam to be at the given temperature. 7.44
(a)
T1 T ( P 3.0 bar, sat' d.) = 133.5$ C Vl ( P 3.0 bar, sat' d.) = 0.001074 m 3 / kg V ( P 3.0 bar, sat' d.) = 0.606 m 3 / kg
Vapor P=3 bar
v
Liquid
0.001074 m 3 1000 L 165 kg Vl 177.2 L kg m3 V space 200.0 L - 177.2 L = 22.8 L 22.8 L
mv (b) P
Pmax
1 m 1 kg 1000 L 0.606 m 3 mtotal
20.0 bar;
l l
v v
200.0 L
0.0376 kg
165.0 0.0376 165.04 kg
1m
l l
total
l
20.0 bar, sat' d.) = 0.0995 m 3 / kg
v
3
ml kg(0.001177 m 3 / kg) + (165.04 - ml ) kg(0.0995 m 3 / kg)
1000 L 164.98 kg; mv
ml
V=200.0 L Pmax=20 bar
3
T1 T ( P 20.0 bar, sat' d.) = 212.4 $ C Vl ( P 20.0 bar, sat' d.) = 0.001177 m 3 / kg; Vv ( P V m V m V m V ( m m )V total
m=165.0 kg
0.06 kg
3
0.001177 m 1000 L 164.98 kg 194.2 L; kg m3 (0.06 - 0.04) kg 1000 g mevaporated 20 g kg
Vl
(c) Energy balance Q = 'U 'E , W , ' E # 0 p
U l ( P U ( P l
s
U(P
V space
20.0 bar, sat' d) U ( P
200.0 L - 194.2 L = 5.8 L
3.0 bar, sat' d)
k
20.0 bar, sat' d.) = 906.2 kJ / kg; U v ( P 20.0 bar, sat' d.) = 2598.2 kJ / kg 3.0 bar, sat' d.) = 561.1 kJ / kg; U v ( P 3.0 bar, sat' d.) = 2543 kJ / kg
Q 0.06 kg(2598.2 kJ / kg) + 164.98 kg(906.2 kJ / kg) - 0.04 kg(2543 kJ / kg) 165.0 kg (561.1 kJ / kg) = 5.70 u 10 4 kJ Heat lost to the surroundings, energy needed to heat the walls of the tank
7-23
7.44 (cont’d) (d) (i) The specific volume of liquid increases with the temperature, hence the same mass of liquid water will occupy more space; (ii) some liquid water vaporizes, and the lower density of vapor leads to a pressure increase; (iii) the head space is smaller as a result of the changes mentioned above. (e) – Using an automatic control system that interrupts the heating at a set value of pressure – A safety valve for pressure overload. – Never leaving a tank under pressure unattended during operations that involve temperature and pressure changes. 7.45 Basis: 1 kg wet steam (a) 1 kg H2 O 20 bars 0.97 kg H2 O(v) 0.03 kg H2 O(l) H1 (kJ/kg)
1 kg H 2O,(v) 1 atm Q=0
b b
H2 (kJ/kg)
'E , 'E , Q , W =0 p
K
Q
U|VbTable B.7g 908.6 kJ kg |W H 'H 0 H 0.97b2797.2g 0.03b908.6g g g
Enthalpies: H v , 20 bars, sat' d H l , 20 bars, sat' d Energy balance on condenser:
1 kg H2 O Tamb, 1 atm
2797.2 kJ kg
2
1
3
H 2 = 2740 kJ / kg
Table B.7
T | 132 o C
(b) As the steam (which is transparent) moves away from the trap, it cools. When it reaches its saturation temperature at 1 atm, it begins to condense, so that T 100q C . The white plume is a mist formed by liquid droplets.
bg
1 quart 1 m3 1000 kg 32 oz 1057 quarts m3 (For simplicity, we assume the beverage is water)
7.46 Basis:
bg
8 oz H 2 O l
0.2365 kg H2O (l) 18°C m (kg H2O (s)) 32°F (0°C)
0.2365 kg H 2 O l
(m + 0.2365) (kg H2O (l)) 4°C
Assume P 1 atm
Internal energies (from Table B.5): U H 2 O(l), 18q C 755 . kJ / kg; U H 2 O(l), 4q C
b
g b g 168. kJ / kg; U bH O(s), 0q Cg = -348 kJ / kg Energy balance bclosed systemg: 'U ¦ n U ¦ n U 0 2
i
'E p , ' E k , Q , W
0
out
i
i
i
in
(m 0.2365) kg(16.8 kJ / kg) = 0.2365 kg(75.5 kJ / kg) + m kg (-348 kJ / kg) m 0.038 kg = 38 g ice
7-24
0 o C, H
7.47 (a) When T
0o C
0, Tref
(b) Energy Balance-Closed System: 'U 'E , ' E , Q , W k
p
0
0
25 g Fe, 175°C 25 g Fe 1000 g H2O Tf (°C)
1000 g H2O(l) 20°C
d i
b d i b g 4.13dT 175ical 4.184 J
g
U Fe T f U H 2 O T f U Fe 175q C U H 2 O 20q C, 1 atm 'U Fe
25.0 g
Table B.5 'U H 2 O
cal
1.0 L 10 3 g
eU dT i 83.9j J
1 L
g
H 2O
f
d i
432T f 1000U H 2 O T f 160 . u 10 5
Tf q C
d i
f Tf
30 2.1 u 10
40 4
2.5 u 10
7-25
0
432 T f 175 J
f
g
0 or 'U Fe 'U H 2 O
d i
f Tf 35
4
e
d i
j
1000 U H 2 O T f 83.9 J 0
34
1670 2612
Interpolate
Tf
34.6q C
7.48
I
II
H 2 O(v ) 760 mm Hg 100°C
H 2 O(v ) (760 + 50.1) mm Hg Tf 1.08 bar sat'd Tf = 101.8°C (Table 8.5) H 2 O( l ), Tf
H 2 O( l ), 100 °C T0
Tf
Energy balance - closed system: 'E p , 'E K , W , Q
v -vapor
0 mvII U vII mlII U lII mbIIU bII mvI U vI mlIU lI mbI U bI
'U
Vl Vv U l U
0
b
I 101 . bar, 100q C 1044 . 1673 419.0 2506.5
bL kgg bL kgg bL kgg bL kgg
v
Initial vapor volume: VvI
1046 . 1576 426.6 2508.6 50 kg
20.0 L 5.0 L
b
b 0.36b101.8g
5.0 L 1.044 L kg
Final energy of bar: U bII
g
g
1L
bg
14.4 L H 2 O v
8.92 kg
bg
8.61 u 10 3 kg H 2 O v
bg
4.79 kg H 2 O l
36.6 kJ kg
Assume negligible change in volume & liquid VvII
b
Final vapor mass: mvII
b -block
. bar, 101.8q Cg g II b108
Initial vapor mass: mvI = 14.4 L 1673 L kg Initial liquid mass: mlI
l -liquid
14.4 L 1576 L kg
g
14.4 L
bg
9.14 u 10 3 kg H 2 O v
Initial energy of the bar:
d
b
H 2 O evaporated
mvII mvI
g
b
g b g
b
g
b
gi
1 914 . u 10 3 2508.6 4.79 426.6 5.0 36.6 8.61 u 10 3 2506.5 4.79 419.0 5.0 kg 44.1 kJ kg 44.1 kJ / kg (a) Oven Temperature: To 122.5q C 0.36 kJ / kg o C U bI
9.14 u 10 3 kg - 8.61 u 10 3 kg = 5.30 u 10 4 kg = 0.53 g
(b) U bI 44.1 8.3 5.0 458 . kJ kg To 458 . 0.36 127.2q C (c) Meshuggeneh forgot to turn the oven on ( To 100q C ) 7-26
weight of piston atmospheric pressure area of piston
7.49 (a) Pressure in cylinder P
30.0 kg 400.0 cm2 Tsat
b100 cmg 1 bmg
9.807 N
2
2
kg
2
10 . bar 105 N m2
1 atm 1.013 bar
108 . bar
atm
1018 . qC
Heat required to bring the water and block to the boiling point Q
'U
d b
g
b
gi
d b g
b426.6 83.9gkJ 3.0 kg
7.0 kg
[0.94(1018 . 20)]kJ kg
kg 2630 kJ < 3310 kJ Sufficient heat for vaporization (b) T f
Tsat
b
mw U wl 108 . bar, sat' d U wl l, 20q C m Al U Al Tsat U Al 20q C
gi
2630 kJ
L kg , U l 426.6 kJ kg V 1046 . 1018 . q C . Table B.5 l Vv 1576 L kg , U v 2508.6 kJ kg
7.0 kg H 2 O(l ) H 426.6 kJ / kg = 1.046 L / kg V
mv (kg H 2 O(v )) 1576 L/kg, 2508.6 kJ/kg
T { 101.8°C P { 1.08 bars
1.046 L/kg, 426.6 kJ/kg ml (kg H 2 O(l ))
Q (kJ)
W (kJ)
(Since the Al block stays at the same temperature in this stage of the process, we can ignore it -i.e., U in U out ) Water balance: 7.0 ml mv (1) Work done by the piston: W
LM w P NA
atm
u
OPb A' zg Q
F' z
w piston Patm A ' z
b1.08 bar g 1576m
P' V W
8.314 J / mol K
1 kJ
0.08314 liter - bar / mol K 10 3 J
Energy balance: 'U
v
b
gb g
1.046ml 1.046 7.0 L
b170.2m
v
g
0.113ml 0.7908 kJ
Q W
'U
Q W
2508.6mv 426.6m L 426.6 7 ( 3310 2630) (170.2mv 0113 . m L 0.7908) 2679mv 426.7m L 3667 0 (2) Solving (1) and (2) simultaneously yields mv 0.302 kg , ml 6.698 kg
bg
Liquid volume Vapor volume
. L kgg b6.698 kggb1046 b0.302 kggb1576 L kgg
Piston displacement: 'z
(c) Tupper
7.01 L liquid 476 L vapor
b gb g
'V A
7.01 476 7.0 1046 . L 10 3 cm3
b
g b
1
1190 cm 1 L 400 cm2 All 3310 kJ go into the block before a measurable amount is transferred to the
water. Then 'U AL
g
Q 3.0 kg 0.94 Tu 20 kJ kg o
neglected. In fact, the bar would melt at 660 C.
7-27
3310 Tu
1194q C if melting is
7.50
100 . L H 2 O(v ), 25o C m v1 (kg)
not all the liquid UV Assume Eq. at W Tis vaporized. , P . m kg H O vaporized.
m v2 [kg H 2 O(v)] = m v1 me
f
o
f
2
e
m L2 [kg H 2 O(l)] = m L1 me
4.00 L H 2 O(l ), 25 C m L1 (kg) Q=2915 kJ
Initial conditions: Table B.5 U L1
104.8 kJ kg , VL1 1003 L kg P 0.0317 bar . 25q C, sat' d U v1 2409.9 kJ kg , V L1 43,400 L kg
T
mv1
b1.00 lg b43400 l kgg
2.304 u 10 5 kg , m LI
b4.00 lg b1.003 l kgg
3.988 kg
Energy balance: 'U
d
i d i b
g d i d
b
g
3.988 (104.8)
d
2915 kJ
g d i i dEi b 3333 d2.304 u 10 iU 3.988U
2.304 u 10 5 me U v T f 3.988 me U v T f 5
me V L Vv
F GG H
v
I J d i b A JK A 5.00 d2.304 u 10 iV 3.988V
g d i
f
v
b2g
L
Vv VL 3333 2.304 u 10 5 U v T f 3.988U L T f U U
d
i d i
d
v
L
d i
i
5.00 2.304 u 10 5 Vv 3.988VL V V v
Procedure: Assume T f Tf U v U L 2014 . 25938 . 856.7 198.3 2592.4 842.9 195.0 2590.8 828.5 196.4 25915 . 834.6 Eq 1
me
5.00 L
liters kg
5
b1g b2g f dT i
(1)
U v U L
kg
me
3333
L
Vtan k 2.304 u 10 5 me VL T f 3.988 me VL T f
bg or Eq b 2 g
ib
Q 2.304 u 10 5 me U v T f 3.988 me U L T f 2.304 u 10 5 2409.9
d i Find T
Table 8.5
U v , U L , Vv , VL f T f
Vv 123.7 1317 . 140.7 136.9
0
L
f
VL f 1159 . . u 10 2 512 193 1154 . . u 10 2 1149 . 134 . u 10 2 4.03 u 10 4 T f # 196.4q C, Pf 1151 .
2.6 u 10 3 kg 2.6 g evaporated
7-28
d i
such that f T f
0
14.4 bars
g
7.51.
Basis: 1 mol feed B = benzene T = toluene
nV (mol vapor) y B(mol B(v)/mol) (1 – y B ) (mol T(v)/mol)
1 mol @ 130°C z B (mol B(l)/mol) (1 – z B )(mol T(l)/mol)
in equilibrium at T(°C), P(mm Hg)
nL (mol liquid) x B(mol B(l)/mol) (1 – x B ) (mol T(l)/mol)
(a) 7 variables: (nV , y B , n L , x B , Q, T , P ) –2 equilibrium equations –2 material balances –1 energy balance 2 degrees of freedom. If T and P are fixed, we can calculate nV , y B , n L , x B , and Q. (b) Mass balance: nV n L 1 nV 1 n2 Benzene balance: z B nV y B n L x B
bg d C H bv g: dT C H bl g: dT C H bv g: dT C6 H 6 l : T
i d
0, H
6
6
80, H
7
8
0, H
7
8
89, H
10.85 H BL
id
4161 . , T 120, H
i d
111, H
0 , T
i d
i
i
18.58 H TL
b
T 33.25 01045 .
i
g g b g
b
g
T 37.57 01304 .
bg
b g
nV y B H BV nV 1 y B H TV n L x B H BL n L 1 x B H TL 1 z B H BL TF 1 1 z H T
b gb
Raoult' s Law:
B
TL
yB P
x B p *B
(90 C) 10
1021 mmHg
[ 6.95334 1343.943/( 90 219 .377)]
406.7 mmHg
Adding equations (8) and (9) P x B p*B + (1 x B ) pT* x B
nL
P pT*
P pT*
p*B
p*B
pT*
x B pB*
pT*
652 406.7 0.399 mol B(l) / mol 1021- 406.7
0.399(1021 mmHg) 0.625 mol B(v) / mol P 652 mmHg zB xB 0.5 0.399 0.446 mol vapor y B x B 0.625 0.399 1 nV 1 0.446 0.554 mol liquid
yB
Solving (1) and (2) nV
(7) (8)
p *B (90 o C) 10[ 6.905651211.033/( 90 220.79 )] o
(6)
F
(1 - y B ) P (1 x B ) pT* Antoine Equation. For T= 90°C and P=652 mmHg: pT*
(4) (5)
T 01674 .
52.05 H TV
0, neglect 'E k
(3)
01356 . T
45.79 H BV
111, H
49.18 , T
Energy balance: 'E p , Ws Q 'H
i
80, H
0 , T
(1) (2)
7-29
(9)
7.51 (cont’d) Substituting (3), (4), (5), and (6) in (7) Q 0.446(0.625)[01045 . (90) 33.25] 0.446(1 0.625)[01304 . (90) 37.57] 0.554(0.399)[01356 . (90)] 0.554(1 0.399)[01674 . (90)] 0.5[01356 . (130)] 0.5[01674 . (130)] Q 814 . kJ / mol (c). If P
Pmax, all the output is liquid. (d) At P=652 mmHg it is necessary to add heat to achieve the equilibrium and at P=714 mmHg, it is necessary to release heat to achieve the equilibrium. The higher the pressure, there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium vapor: enthalpy out < enthalpy in. zB 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
T 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90
P 652 714 582 590 600 610 620 630 640 650 660 670 680 690 700 710
pB 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021
pT 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7
xB 0.399 0.500 0.285 0.298 0.315 0.331 0.347 0.364 0.380 0.396 0.412 0.429 0.445 0.461 0.477 0.494
(e). Pmax = 714 mmHg, Pmin = 582 mmHg nV vs. P 1
nV
0.8 0.6 0.4 0.2 0 582
632
682
732
P (mm Hg)
nV
0.5 @ P # 640 mmHg
7-30
yB 0.625 0.715 0.500 0.516 0.535 0.554 0.572 0.589 0.606 0.622 0.638 0.653 0.668 0.682 0.696 0.710
nV 0.446 -0.001 0.998 0.925 0.840 0.758 0.680 0.605 0.532 0.460 0.389 0.318 0.247 0.176 0.103 0.029
nL 0.554 1.001 0.002 0.075 0.160 0.242 0.320 0.395 0.468 0.540 0.611 0.682 0.753 0.824 0.897 0.971
Q 8.14 -6.09 26.20 23.8 21.0 18.3 15.8 13.3 10.9 8.60 6.31 4.04 1.78 -0.50 -2.80 -5.14
'P 'u 2 g'z U 2
7.52 (a). Bernoulli equation:
d0.977 u 10
'P U
5
0
i
15 . u 10 5 Pa 1 N / m 2
m3 1.12 u 10 3 kg
Pa
bg
46.7
m2 s2
(9.8066 m / s 2 ) 6 m 58.8 m 2 / s 2
g'z
'u 2 46.7 58.8 m 2 / s 2 u22 u12 2 12.1 m 2 / s 2 2 2 5.00 m 2 / s 2 (2)(12.1) m 2 / s 2 0.800 m 2 / s 2 u2 0.894 m / s
b
Bernoulli
g
d
i
b g
d
(b). Since the fluid is incompressible, V m 3 s d1
d
u2 u1
d2
b6 cmg
0.894 m s 5.00 m s
d i b g
i
7.53 (a). V m 3 s
' P 'u 2 U 2
S d 12 u1 4 S d 22 u2 4
2.54 cm
d i b g
A1 m 2 u1 m s
(b). Bernoulli equation ('z
i
A2 m 2 u2 m s u2
u1
A1 A2
A1 4 A2
u2
4u1
0)
0 'P
P2 P1
d
U u22 u12
i
2
Multiply both sides by 1 2
2
16u1
Substitute u 2
2
Multiply top and bottom of right - hand side by A1 2 note V
P1 P2
(c) P1 P2 2
V
d
i
U Hg U H 2 O gh
b g
2 S 7.5
2 2
15
V
V 2
m4
9.8066 m
108 cm4
s2
cm4 1
2
15UV 2 2 A12
15U H 2 OV 2 2 A12
2
A1 u1
0.044 m 3 s 44 L s
7-31
F GH
I JK
2 A12 gh U Hg 1 15 U H 2O
38 cm
1m 102 cm
b13.6 1g
. 1955 u 10 3
m6 s2
7.54 (a). Point 1 - surface of fluid . P1
Point 2 - discharge pipe outlet . P2 'U U g'z
31 . . gbar b1013
1 atm , z 2
b
1.013 bar
g
10 5 N 1 m3 m 2 bar 0.792 u 10 3 kg
7 m
9.8066 m s2
7 m , u1
31 . bar , z1
Bernoulli equation
b g
b g
0ms
0 m , u2
?
2635 . m 2 s2
68.6 m 2 s 2
'u 2 2 'u
2
'P g'z U
2
b2635. 68.6g m
2
s2
332.1 m 2 s 2
2
u2 0
u22
2(332.1 m 2 s 2 )
V
S (100 . 2 ) cm 2 4
664.2 m 2 s 2 u2
258 . m/s
2580 cm 1 L 60 s 122 L / min 1 s 10 3 cm 3 1 min
(b) The friction loss term of Eq. (7.7-2), which was dropped to derive the Bernoulli equation, becomes increasingly significant as the valve is closed. 7.55 Point 1 - surface of lake . P1 Point 2 - pipe outlet . P2 u2
V A
95 gal min
FG L H
Z 2z sin 30q
IJ K
W Shaft work: s m
1 atm , z 2
1 ft 3 7.4805 gal 0
0 , u1
0
bg
z ft
1 2 S 0.5 u 1049 . in 2
b
bP P g 0.041b2 z g ft lb lb
Pressure drop: ' P U
Friction loss: F
1 atm , z1
1
g
144 in 2 1 ft 2
1 min 60 s
35.3 ft s
2
f
0.0822 z (ft lb f lb m )
m
-8 hp 0.7376 ft lb f / s 1.341 u 10
3
1 min 7.4805 gal 95 gal
hp
1 ft
3
1 ft 3
60 s
62.4 lb m
1 min
333 ft lb f lb m Kinetic energy: ' u 2 2 Potential energy: g'z
b
g
Eq. 7.7 - 2
b35.3g
2
0 2 ft 2
2 32.174 ft s2
s
2
bg
z ft
'P 'u 2 g'z F U 2
1
lb f
32.174 lb m ft / s 2 1 lb f 32.174 lb m ft / s 2
19.4 ft lb f lb m
b
z ft lb f lb m
W s 19.4 z 0.082 z 333 z m
7-32
g 290 ft
7.56 Point 1 - surface of reservoir . P1 1 atm (assume), u1 0 , z1 60 m Point 2 - discharge pipe outlet . P2 1 atm (assume), u2 ? , z 2 0 'P U
0
'u 2 2
u22 2
dV Ah
2
V 2 (m 6 / s 2 )
2
3.376V 2 9.8066 m s2
g'z
W s m
b g bN m kgg S 35
(2)
65 m
1N 1 kg m / s 2
1 2 2
cm 4
10 8 cm 4
1 N
1 m4
1 kg m / s 2
637 N m kg
s 0.80 u 10 6 W 1 N m / s V m3 W
1 m3 800 V N m kg 1000 kg
b
d i Mechanical energy balance: neglect F b Eq. 7.7 - 2g W s 3.376V 2 637 m
'P 'u 2 g'z U 2
g
127 . m 3 60 s 76.2 m 3 min s 1 min
800 T E V V
Include friction (add F ! 0 to left side of equation) V increases. 7.57 (a). Point 1: Surface at fluid in storage tank, P1
1 atm , u1
0 , z1
Point 2 (just within pipe): Entrance to washing machine. P2 u2
600 L min S 4.0 cm
b
'u 2 2
'P U
0;
g'z
9.807 m s2
u22 2
3
10 cm 4 1 L
g b7.96 m sg 2
2
c0 Hbmgh
Bernoulli Equation:
2
3
1 atm , z 2
0;
'u 2 2
317 .
1 J 1 kg m 2 / s 2
1 J 1 kg m 2 / s 2
'P 'u 2 g'z U 2
J ; g'z kg
Mechanical energy balance: W s
. J / kg 317
9.807 H (J / kg)
0 H
3.23 m
W s
600 L 0.96 kg 1 min L 60 s min
b
g
9.807 3.23 0 m
0
1 min 1m 7.96 m s 60 s 100 cm
(b). Point 1: Fluid in washing machine. P1 1 atm , u1 | 0 , z1 0 Point 2: Entrance to storage tank (within pipe). P2 1 atm , u2 'P U
b g
H m
LM 'P 'u NU 2
2
317 .
J ; F kg
g'z F
b31.7 + 31.7 + 72g J
OP Q
7.96 m s , z 2
72
3.23 m
J kg
1 kW 1.30 kW kg 10 3 J s (work applied to the system)
Rated Power 130 . kW 0.75 1.7 kW
7-33
7.58 Basis: 1000 liters of 95% solution . Assume volume additivity. xi 0.95 0.05 1 l Density of 95% solution: U 124 0.804 . kg liter U U 126 . 100 . kg b Eq. 6.1-1g i
¦
Density of 35% solution: Mass of 95% solution:
1 U
0.35 0.65 1.26 100 .
0.9278
l U 1.08 kg liter kg
1000 liters 1.24 kg 1240 kg liter
G = glycerol W = water 1240 kg (1000 L) 0.95 G 0.05 W
23 m
m1 (kg) 0.35 G 0.65 W
m2 (kg) 0.60 G 0.40 W
5 cm I.D.
UV m b gb g b gb g b0.60gbm gW m
Mass balance: 1240 m1 m2 Glycerol balance: 0.95 1240 0.35 m1
2
1740 kg
Volume of 35% solution added Final solution volume
1
b1000 1610g L
2
1740 kg 35% solution 2980 kg 60% solution
1L 1610 L 1.08 kg 2610 L
Point 1. Surface of fluid in 35% solution storage tank. P1 1 atm , u1 Point 2. Exit from discharge pipe. P2 1 atm , z 2 23 m u2
1610 L 13 min
'P U g'z
1 m 3 1 min 1 2 3 10 L 60 s S 2.5 cm 2
b g
'u 2 2
'u22 2
9.8066 m s2
23 m
0,
. g b1051
b
Mechanical energy balance Eq. 7.7 - 2 W s
m
LM 'P 'u NU 2
2
g'z F
OP Q
m2 / s2 (2)
1N 1 kg m / s 2
1740 kg 1 min 13 min 60 s
Mass flow rate: m
2
10 4 cm 2 1 m2
0 , z1
0
1051 ms .
1 N 1 kg m / s 2
225.6 N m kg , F
0.552 N m kg 50 J kg 50 N m kg
2.23 kg s
g
2.23 kg
b0.552 + 225.6 + 50gN m
s
kg
0.62 kW 0.62 kW delivered to fluid by pump.
7-34
1J
1 kW
1 N m 10 3 J s
CHAPTER EIGHT 8.1
a.
U (T ) 25.96T 0.02134T 2 J / mol U (0 o C) 0 J / mol U (100 o C) 2809 J / mol
Tref
o C) = 0) 0 o C (since U(0
b.
We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to U (100 o C) .
c.
Q W
'U 'E k 'E p 'E k
Q d.
'U
0 , 'E p
z
dU dT
V
[25.96 0.04268T ] J / (mol o C)
z
T2
'U
0
(3.0 mol)[(2809 0) J / mol] 8428 J 8400 J
F wU I GH wT JK
Cv
0, W
F GG 25.96T 0.04268 T2 OPP Q H
100
Cv (T )dT
2
(25.96 0.04268T )dT
0
T1
(3.0 mol) [25.96(100 0) 0.02134(100 2 0)] (J / mol) a.
27.0 0.0291T [J / (molq C)]
z z
100
'H
c.
z
C v dT
25
H is a state property
a.
Cv [ kJ / (mol o C)] PV RT
Q1
n'U 1
z
T2 2
OP PQ
100
2784 J mol 25
100
C p dT
d.
n
0.0891
100
25
8.3
100 25
35.3T
C p dT
25 100
'U
8428 J 8400 J
b35.3 0.0291T g[J / (molq C)] b8.314 [J / (mol K)]gb1 K 1q Cg
C p R Cv
Cv Cv
b.
0
I JJ J / mol K
(3.0 mol) 'U (J / mol)
'U
8.2
100
RdT
'H R'T
b
gb
g
2784 8.314 100 25
2160 J mol
25
0.0252 1547 . u 10 5 T 3.012 u 10 9 T 2
(2.00 atm)(3.00 L) (0.08206[atm L / (mol K)](298 K)
z
0.245 mol
1000
(0.245 mol)
z z
0.0252 dT ( kJ / mol)
6.02 kJ
25
1000
Q2
n'U 2
(0.245)
u 10 5 T ] dT [0.0252 1547 .
7.91 kJ
25 1000
Q3
n'U 3
(0.245)
u 10 5 T 3.012 u 10 9 T 2 ] dT [0.0252 1547 .
25
6.02 - 7.67 u 100% 7.67 7.91- 7.67 % error in Q2 = u 100% 7.67 % error in Q1 =
215% . 313% .
8-1
7.67 kJ
8.3 (cont’d) b.
Cv R
Cp
u 10 5 T 3.012 u 10 9 T 2 ) 0.008314 (0.0252 1547 .
C p [ kJ / (mol o C)]
u 10 5 T 3.012 u 10 9 T 2 0.0335 1547 .
z
T2
'H
Q
n C P dT T1
z
1000
(0.245 mol)
[0.0335 1547 . u 10 5 T 3.012 u 10 9 T 2 ] dT [kJ / (mol o C)]
9.65 u 10 3 J
25
Piston moves upward (gas expands). c. 8.4
a. b.
The difference is the work done on the piston by the gas in the constant pressure process.
b g b313 Kg
b g 0.1360 [kJ / (mol K)] b g b40q Cg 0.07406 32.95 u 10 b40g 25.20 u 10 b40g 77.57 u 10 b40g
dC i dC i
p C H l 6 6
0.06255 23.4 u 10 5 313 5
p C H v 6 6
2
8
12
3
0.08684 [kJ / (mol o C)] c.
dC i b g b313 Kg
d.
'H C6 H 6 bv g
e.
8.5
b g
'H Cb sg
b g
0.01118 1095 . u 10 5 313 4.891 u 10 2 313
p C s
2
0.009615 [ kJ / (mol K)]
32.95 u 10 5 2 2520 . u 10 8 3 77.57 u 10 12 4 0.07406T T T T 3 2 4 0.01118T
1.095 u 10 5 2 T 4.891 u 10 2 T 1 2
OP PQ
OP PQ
300
3171 . kJ mol 40
573
3.459 kJ / mol 313
H 2 O (v, 100 o C, 1 atm) o H 2 O (v, 350 o C, 100 bar) a. H 2926 kJ kg 2676 kJ kg 250 kJ kg
z
350
b.
H
0.03346 0.6886 u 10 5 T 0.7604 u 10 8 T 2 3.593 u 10 12 T 3 dT
100
8.845 kJ mol 491.4 kJ kg Difference results from assumption in (b) that H is independent of P. The numerical difference is 'H for H 2 O v, 350q C, 1 atm o H 2 O v, 350q C, 100 bar
b
8.6
b.
g
b
z
g
80
dC i
p n C H (l) 6 14
0.2163 kJ / (mol o C) 'H
[0.2163] dT
1190 . kJ / mol
25
The specific enthalpy of liquid n-hexane at 80oC relative to liquid n-hexane at 25oC is 11.90 kJ/mol c.
dC i
p n C H (v) [ kJ 6 14
z
/ (mol o C)]
013744 . 40.85 u 10 5 T 23.92 u 10 8 T 2 57.66 u 10 12 T 3
0
'H
[0.13744 40.85 u 10 5 T 23.92 u 10 8 T 2 57.66 u 10 12 T 3 ] dT = 110.7 kJ / mol
500
The specific enthalpy of hexane vapor at 500oC relative to hexane vapor at 0oC is 110.7 kJ/mol. The specific enthalpy of hexane vapor at 0oC relative to hexane vapor at 500oC is –110.7 kJ/mol.
8-2
8.7
b g 181. T cbq Fg 32 0.5556T cbq Fg 17.78 C bcal molq Cg 6.890 0.001436 0.5556T cbq Fg 17.78 6.864 0.0007978T cbq Fg cal 453.6 mol 1 Btu 1q C C c b Btu lb - moleq Fg C b1.00gC molq C 1 lb - mole 252 cal 1.8q F E T qC p p
p
b
g
C p Btu lb - moleq F
8.8
dC i
p CH CH OH(l) 3 2
Q
'H
bT g
p
drop primes
b g
6.864 0.0007978T q F
0.1031
. 0.1031g b01588 T
FG H
100
0.1031 0.000557T [kJ / (mol o C)]
OP Q
78.5
55.0 L 789 g 1 mol 0.000557 2 . T T 01031 s 1 L 46.07 g 2 20 kJ mol
= 941.9 u 7.636 kJ / s = 7193 kW 8.9
a.
kJ mol
Q
'H
b5,000 mol sg
z
200
0.03360 1367 . u 10 5 T 1.607 u 10 8 T 2 6.473 u 10 12 T 3 dT
100
17,650 kW b.
Q
'U
'H 'PV
b
'H nR'T
gb
gb
17,650 kJ 5.0 kmol 8.314 [kJ / (kmol K) ] 100 K
13,490 kJ c.
8.10 a. b.
The difference is the flow work done on the gas in the continuous system. Qadditional heat needed to raise temperature of vessel wall + heat that escapes from wall to surroundings. C p is a constant, i.e. C p is independent of T. Q Cp
Q m'T (16.73 - 6.14) kJ
mC p 'T C p Q m' T
(2.00 L)(3.10 K) 659 g
Table B.2 C p 8.11
H
U PV
1L
0.216 kJ / (mol o C)
86.17 g 10 3 J 1 mol
FG wH IJ FG wU IJ H wT K H wT K F wU I dU F wU I depends only on T, G H wT JK dT GH wT JK
PV RT
! H
U RT
aw wT f
= 0.223 kJ / (mol K)
0.216 kJ / (mol K)
P
!
p
But since U
1 kJ
p
8-3
R Cp p
FG wU IJ H wT K
{ Cv C p V
R p
Cv R
g
8.12 a.
dC i
75.4 kJ / (kmol o C) =75.4 kJ/(kmol.oC) V = 1230 L ,
p H O(l) 2
n
VU M
1230 L 1 kg 1 kmol 1 L 18 kg
zd
T2
n Q b.
Q t
i
Cp
T`
H 2 O(l)
68.3 kmol
dT 68.3 kmol 75.4 kJ (40 29) o C 1 h 8h 3600 s kmol o C
t
Q total
Q to the surroundings Q to water , Q to the surroundings
z
1.967 kW
. 1967 kW
40
Q total c.
n C P ( H2 O) dT
Qto water t
Q to water
t
7.212 kW E total
7.212 kW u 3 h = 21.64 kW h 7.212 kW u 1 h u $0.10 / (kW h) = $0.72
Cost keeping temperature constant for 1 hour $2.16 $0.72
d.
$2.88
If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher cost.
8.13 a.
'H N
2 (25
b.
'H H
2 (800
c.
'H CO
d.
'H O
8.14 a.
5.245 kW
21.64 kW h u $0.10 / (kW h) = $2.16
Cost heating up from 29 o C to 40 o C Cost total
68.3 kmol 75.4 kJ / (kmol o C) 11 o C 3h 3600 s / h
29
o
o
o
H H
F) o H 2 (77 o F)
2 (300
2 (970
H N
C) o N 2 (700 o C)
o
H O
m
300 kg / min n
Q
n 'H
n
z
T2
T1
2 (77
o
o
2 (0
o
F)
C)
F)
H CO
C) o CO 2 (1250 o C)
F) o O 2 (0o F)
2 (700
H N
H H
2 (1250
H O
o
2 (25
2 (800
C)
o
o
C)
F)
b20.59 0g 20.59 kJ mol b0 5021g 5021 Btu / lb - mol . g 5148 . kJ mol b63.06 1158 b539 6774g 7313 Btu / lb - mol
H CO
2 (970
o
F)
2 (300
o
C)
300 kg 1 min 1000 g 1 mol min 60 s 1 kg 28.01 g
178.5 mol / s
C p dT
z
50
(178.5 mol / s)
b
[0.02895 0.411 u 10 5 T 0.3548 u 10 8 T 2 2.22 u 10 12 T 3 ] dT [kJ / mol]
450
g
(178.5 mol / s) 12.076 [kJ / mol] = 2,156 kW b.
Q
n 'H
n H (50o C) H ( 450o C)
8.15 a.
n
250 mol / h
i)
Q
ii) =
250 mol (2676 3697) kJ 1 kg 1 h 18.02 g h 1 kg 1000 g 3600 s 1 mol
n'H
Q n'H n
z
(178.5 mol / s)(0.73 -12.815[kJ / mol]) = 2,157 kW
T2
T1
1.278 kW
C p dT
250 mol 1 h h 3600 s
z
100
600
0.6880 u 10 5 T 07604 u 108 T 2 3593 [003346 . . . u 1012 T 3 ] 1.274 kW
8-4
8.15 (cont’d)
b
g
250 mol 2.54 20.91 [kJ / mol] 1276 . kW 3600 s Method (i) is most accurate since it is not based on ideal gas assumption. The work done by the water vapor. Q
iii)
b. c.
8.16 Assume ideal gas behavior, so that pressure changes do not affect 'H . n
200 ft 3 492 o R 1.2 atm 1 lb - mol h 537 o R 1 atm 359 ft 3 (STP)
Q
n'H
(0.6125
50 kg 1.14 kJ
8.17 a.
b
b50 10gq C
dC i
p Na O 2 3
dC i dC i
p C H O(l) 6 14
105.99 g
2085 2280 u 100% 2280
b
g b
d i 2b0.026g 0.0075 3b0.017g b50 10gq C 2085 kJ
3 Cp
1 mol
molq C % error
p C
Na
O
0.1105 kJ molq C
8.6% error
g b
g
6 0.012 14 0.018 1 0.025
p CH COCH (l) 3 3
1833 Btu / h
2280 kJ
d i dC i
| 2 Cp
50,000 g 0.1105 kJ
8.18
g
lb - mole ) (2993 0) [Btu / lb - mole] h
kgq C
b.
0.6125 lb - mole / h
0.349 kJ / (mol o C) (Kopp’s Rule)
0.1230 18.6 u 10 5 T kJ (molq C)
Assume 'H mix # 0 p CH 3 COCH 3 C pm
e
p C 6 H 14 O
j
b
g
0.30 0.1230 +18.6 u 10 5 T kJ 1 mol 0.70 0.349T kJ 1 mol molq C 58.08 g molq C 102.17 g [0.003026 9.607 u 10 7 T] kJ (gq C)
z
'H
20
45
[0.003026 9.607 u 10 7 T] dT
0.07643 kJ g
8.19 Assume ideal gas behavior, 'H mix # 0 g g b g 26.68 mol 'H dC i dT 10.08 kJ / mol, 'H dC i dT 14.49 kJ / mol 2 L1 OF 1000 g IJ FG 1 mol IJ 433 kJ kg H M b14.49 kJ / mol g b10.08 kJ / molgPG 3 3 N QH 1 kg K H 26.68 g K Mw
O2
b
1 2 16.04 32.00 3 3
z
350
25
p O 2
CH 4
8-5
z
350
25
p CH 4
8.20
1000 m 3 1 min 273 K
1 kmol 0.6704 kmol s 670.4 mol / s min 60 s 303 K 22.4 m 3 STP Energy balance on air: Table B.8 for 'H 670.4 mol 0.73 kJ 1 kW Q 'H n 'H Q 489.4 kW s mol 1 kJ s n
489.4 kW heating 1 kW solar energy
Solar energy required
0.3 kW heating 1 m2
1627 kW 1000 W
Area required 8.21
b g
1 kW
900 W
1631 kW
1813 m 2
C 3 H 8 5O 2 o 3CO 2 4H 2 O n fuel
1.35 u 10 5 SCFH 1 lb - mol h 359 ft 3
n air
376 lb mol 5 lb - mol O 2 . 1 lb - mol air 115 h 1b - mol C 3 H 8 0.211b - mol O 2
376
lb - mol h 1.03 u 10 4
lb mol h
z
T2
Q = 'H = n C p dT T1
FG H
= 1.03 u 10 4
IJ K
lb mol h
z
302
[0.02894 0.4147 u 10 5 T 0.3191 u 10 8 T 2 1965 . u 10 12 T 3 ] dT
0
4
= 8.22 a.
1.03 u 10 lb - mol 8.954 kJ 453.593 mol 9.486 u 10 -1 Btu = 3.97 u 10 7 Btu / h h mol lb - mol kJ Basis: 100 mol feed (95 mol CH4 and 5 mol C2H6) 7 CH 4 2O 2 o CO 2 2H 2 O C 2 H 6 O 2 o 2CO 2 3H 2 O 2 nO2
125 .
LM 95 mol CH MN
4
2 mol O 2 5 mol C 2 H 6 3.5 mol O 2 1 mol CH 4 1 mol C 2 H 6
Product Gas: CO 2 : 95(1) + 5(2) = 105 mol CO 2
OP PQ
259.4 mol O 2
H 2 O: 95(2) + 5(3) = 205 mol H 2 O O 2 : 259.4 - 95(2) - 5(3.5) = 51.9 mol O 2 N 2 : 3.76(259.4) = 975 mol N 2 Energy balance (enthalpies from Table B.8) H 18.845 42.94 'H H CO 2 (CO , 450o C) (CO , 900 o C) 2
'H H 2O
H (H
'H O2
H (O
'H N2
H (N
Q = 'H Q
2 O,
2, 2,
2
450 o C)
H (H
450o C)
H (O
450o C)
H (N
2,
2 O,
900 o C)
1512 . 33.32
24.09 kJ / mol 18.20 kJ / mol
900o C)
13.375 28.89
15.51 kJ / mol
900 o C)
12.695 27.19
14.49 kJ / mol
2,
105(-24.09) 205(-18.20) 51.9(-15.51) 975(-14.49)
21,200 kJ / 100 mol feed
8-6
8.22 (cont’d) (40 o C) b. From Table B.5: H liq
167.5 kJ / kg; H vap (50 bars)
2794.2 kJ / kg;
= n(2794.2 -167.5) = 21200 n = 8.07 kg / 100 mol feed Q = n 'H c.
From part (b), 8.07 kg steam is produced per 100 mol feed 1250 kg steam 0.1 kmol feed 1 h 4.30 u 10 3 kmol / s n feed h 8.07 kg steam 3600 s 4.30 mol feed 1336.9 mol product gas 8.314 Pa m 3 723 K s 100 mol feed mol K 1.01325 u 10 5 Pa
Vproduct gas
3.41 m 3 / s
d. Steam produced from the waste heat boiler is used for heating, power generation, or process application. Without the waste heat boiler, the steam required will have to be produced with additional cost to the plant. Assume 'H mix # 0 'H
8.23
d i
Kopp’s rule: C p
15.0 L 879 g 1 mol L 78.11 g
'H C6 H6
8.24
10(12) 12(18) 2(25)
C10 H12 O2
e
386 J mol o C
20.0 L 1021 g 1 kJ 2.35 J (71 25) o C L 10 3 J g o C
'H C10 H12O2
'H
'H C10 H12O2 'H C6 H6
2207 1166
LM N
z
348
298
e
2.35 J g o C
j
2207 kJ
[0.06255 23.4 u 10 5 T] dT
OP Q
1166 kJ
3373 kJ
100 mol C3 H8 @ 40 o C, 250 kPa
a.
j
100 mol C3 H8 @ 240 o C, 250 kPa
VP 1 (m3 )
VP 2 (m3 ) mw kg H2 O(v) @ 300 o C, 5.0 bar
mw kg H2 O(l, sat‘d) @ 5.0 bar Vw2 (m3 )
Vw1 (m3 )
b. References: H2O (l, 0.01 oC), C3H8 (gas, 40 oC) C 3 H 8 : H in
0 kJ / mol; H out
z
240 40
H 2 O: H in c.
'H C3 H8 Q
'H
C pC H dT 19.36 kJ mol (Cp from Table B.2) 3 8
3065 kJ / kg (Table B.7); H out
19.36 kJ / mol, 'H w 100'H C 3H 8 mw 'H w
b
(2747.5 3065) kJ / kg 0 mw
From Table B.7: Vsteam 5.0 bar, 300q C
b
VC3 H8 40q C, 250 kPa 6.09 kg steam
g
2747.5 kJ / kg (Table B.6)
g
318 kJ / kg
6.09 kg
0.522 m 3 kg
0.008314 m 3 kPa (mol K) 313 K 250 kPa 3
0.522 m steam
1 mol C 3 H 8
0.0104 m 3 mol C 3 H 8
3.06 m 3 steam m 3 C 3 H 8
d.
100 mol C 3 H 8 1 kg steam 0.0104 m 3 C 3 H 8 Q mw 'H w 6.09 kg u (-318 kJ / kg) = -1940 kJ
e.
A lower outlet temperature for propane and a higher outlet temperature for steam.
8-7
8.25
a. 5500 L(ST P)/ min CH3 OH (v) 65o C
n 2 mol/min CH3 OH (v) 260o C
n 2 (mol/ min) mw kg/ min H2 O(l, sat‘d) @ 90o C
mw kg/ min H2 O(v, s at‘d) @ 300o C
Vw 2 (m3 /min )
b.
8.26 a.
Q
Vw 1 (m3 /min )
kg I F kJ I F 1 min I F 1 kW I FG113 H . min JK GH 2373.9 kg JK GH 50 sec JK GH 1 kJ / sJK
100 mol/s (30o C) 0.100 mol H2 O(v)/ mo l 0.100 mol CO/ mol 0.800 mol CO2 /mol
44.7 kW
n 2 mol/s (30o C) 0.020 mol H2 O(v)/ mo l y 2 mol (molCO/s CO/mol) (molCO CO2 /s (0.980-y 2 ) mol 2/mol)
m3 kg humid air/s (50o C)
o m4 kg humid air/s (30 (48oC) C)
H 2O(v) only
(0.002 /1.002 ) kg H2 O(v)/kg humid air (1.000 /1.002 ) kg dry air/ kg humid air
y 4 kg H2 O(v)/kg humid air (1-y 4 ) kg dry air/kg humid air
Basis: 100 mol gas mixture/s 5 unknowns: n2, m3, m4, y2, y4 – 4 independent material balances, H2O(v), CO, CO2 , dry air – 1 energy balance equation 0 degrees of freedom (all unknowns may be determined) b.
(1) CO balance:
(100)(0.100) = n 2 y 2
(2) CO 2 balance: (100)(0.800) = n 2 (1 y 2 )
|UV n W|
2
9184 . mol / s, x 2
mol CO / mol 01089 .
1000 . m4 (1 y 4 ) 1002 . (100)(0.100)(18) 0.002 (0.020)(18) m 3 m 4 y 4 (4) H 2 O balance: 9184 . 1000 1002 1000 . References: CO, CO2, H2O(v), air at 25oC ( H values from Table B.8 ) substance n in ( mol / s) n out ( mol / s) H in (kJ / mol) (3) Dry air balance: m3
H2O(v) CO CO2
10 10 80
0.169 0.146 0.193
91.84(0.020) 10 80
H out (kJ / mol) 0.169 0.146 0.193
H2O(v) dry air
m3(0.002/1.002)(1000/18) m3(1.000/1.002) (1000/29)
0.847 0.727
m4y4(1000/18) m4(1-y4) (1000/29)
0.779 0.672
(5) Energy balance:
. FG 0.002 IJ FG 1000 IJ (0.847) m FG 1000 IJ FG 1000 IJ (0.727) H 1.002 K H 18 K H 1002 K H 29 K . F 1000IJ m (1 y )(0.672)FG 1000 IJ = 91.84( 0.020)(0.169) m y (0.779)G H 18 K H 29 K
10(0.169) m3
3
4
4
4
8-8
4
8.26 (cont’d) Solve Eqs. (3)–(5) simultaneously m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2O/kg 2.55 kg humid air / s 100 mol gas / s
Mole fraction of water :
8.27 a.
29 kg DA 1 kmol H 2 O
.0963
(1-.0564) kg dry air kmol DA 18 kg H 2 O 0.0963 kmol H 2 O (1 0.0963) kmol humid air
p H 2O
Relative humidity:
kg humid air mol gas
00564 kg H 2 O .
c.
0.0255
p H* 2 O
00878 .
e48 Cj
kmol DA
kmol H 2 O kmol humid air
(0.0878)(760 mm Hg ) u 100% 83.71 mm Hg
o
kmol H 2 O
79.7%
The membrane must be permeable to water, impermeable to CO, CO2, O2, and N2, and both durable and leakproof at temperatures up to 50oC.
b
p * 57q C
g
129.82 mm Hg mol H 2 O mol . 0171 P 760 mm Hg p 3 28.5 m STP 1 mol 1270 mol h 217.2 mol H 2 O h h 0.0224 m 3 STP y H 2O
b g
1270 217.2
b g
1053
mol dry gas h
given
b3.91 kg H O hg 2
R| 89.5 mol CO h |110.5 mol CO h !S || 5.3 mol O h T847.6 mol N h 2
percentages
2
2
1270 mol/h, 620°C
425°C m (kg H2 O( l )/h), 20°C
References for enthalpy calculations:
e
j
CO, CO 2 , O 2 , N 2 at 25qC (Table B.8); H 2 O l, 0.01o C (steam tables) substance
n in
CO CO 2
89.5 110.6 5.3 847.6
O2 N2
bg H Obl g
3.91 m
H2O v
H in 18.22 27.60 19.10 18.03 3749 83.9
n out 89.5 110.6 5.3 847.6
H out 12.03 17.60 12.54 11.92
3.91 m 3330 ---
2
'H
¦ n H ¦ n H i
out
i
i
i
0 8504 3246m
0m
U| n in mol h V| H in kJ mol W UV n in kg h W H in kJ kg
2.62 kg h
in
b. When cold water contacts hot gas, heat is transferred from the hot gas to the cold water lowering the temperature of the gas (the object of the process) and raising the temperature of the water.
8-9
. gb5.294 mm Hg g b015
8.28 2qC, 15% rel. humidity p H 2 O
dy i H 2O
inhaled
n inhaled
b0.7941g b760g
0.7941 mm Hg
1.045 u 10 3 mol H 2 O mol inhaled air
5500 ml 273 K 1 liter 1 mol 3 min 275 K 10 ml 22.4 liters STP
b
p * 37q C
Saturation at 37 qC y H 2 O
g
760 mm Hg
b g
47.067 760
0.2438 mol air inhaled min
0.0619 mol H 2 O mol exhaled dry gas
0.2438 mol/min 2oC
n2 kmol/min 37oC
1.045 x 10-3 H2O 0.999 dry gas
0.0619 H2O 0.9381 dry gas n1 mol H2O(l)/min 22o C
Mass of dry gas inhaled (and exhaled)
b0.999gb0.2438g
Dry gas balance:
b0.2438gb0.999gmol dry gas
29.0 g
min
mol
0.9381 n 2 n 2
7.063 g min
0.2596 mols exhaled min
u 10 j n . b0.2438ge1045 b0.2596gb0.0619g n 0.0158 mol H O min References for enthalpy calculations: H Obl g at triple point, dry gas at 2 qC 3
H 2 O balance:
1
1
2
2
substance
m in
Dry gas H2O v
7.063 0.00459 0.285
bg H Obl g 2
Q
¦ m H ¦ m H
'H
H out 36.75 2569 —
m out 7.063 0.290 —
m in g min H in J g
n H 2 O 18.02
m H 2 O
H H 2 O from Table 8.4 H dry gas
i
i
out
8.29 a.
H in 0 2505 92.2
i
i
in
bg
75 liters C 2 H 5 OH l
966.8 J 60 min 24 hr 1.39 u 10 6 J day min 1 hr 1 day
789 g 1 mol liter 46.07 g
bg
1284 mol C 2 H 3 OH l
e
j 3054 mol H Obl g
(C p ) CH 3OH
. 01031 0.557 u 10 3 T kJ / (mol o C) (fitting the two values in Table B.2)
55 L H 2 O l
1000 g 1 mol liter 18.01 g
bg
2
(C p ) H 2O
b
0.0754 kJ molq C
g
1284 mol C2H5 OH(l) (70.0oC) 1284 mol C2H5 OH (l) (To C)
3054 mol H2O(l) (20.0oC)
3054 mol H2O(l) (To C)
ze T
b
0 1284
'U # 'H liquids
Q
0 adiabatic
g
j
. 0.557 u 10 T dT 3054 01031
gUV E Integrate, solve quadratic equation W T = 44.3 C
Q
b
3
70 o
8-10
zb T
25
g
0.0754T dT
b
105 . T 2
g
8.29 (cont’d) b. 1. 2. 3. 4. 5. 6. 7.
Heat of mixing could affect the final temperature. Heat loss to the outside (not adiabatic) Heat absorbed by the flask wall & thermometer Evaporation of the liquids will affect the final temperature. Heat capacity of ethanol may not be linear; heat capacity of water may not be constant Mistakes in measured volumes & initial temperatures of feed liquids Thermometer is wrong
8.30 a. 1515 L/s air 500o C, 835 tor, Tdp=30o C
1515 L/s air , 1 atm 110 g/s H2O(v)
110 g/s H2O, T=25o C
Let n1 (mol / s) be the molar flow rate of dry air in the air stream, and n 2 (mol / s) be the molar flow rate of H2O in the air stream. 1515 L 835 mm Hg mol K 26.2 mol / s n1 + n 2 s 773 K 62.36 L mm Hg n 2 mmHg p * (30 o C) 31824 . =y= 0.0381 mol H 2 O / mol air n1 + n 2 835 mmHg Ptotal . mol H 2 O / s n1 25.2 mol dry air / s; n 2 10 o References: H2O (l, 25 C), Air (v, 25oC) substances n in (mol / s) n out (mol / s) H in (kJ / mol) dry air
25.2
H2O(v)
14.37
zd zd
1.0
100
25 500
100
H2O(l) 'H
Cp
6.1
H2 O ( l )
H2 O ( v )
25.2 7.1
dT H vap
25
p air
z
--
z
b. c.
Q
b gb
p H O( l ) 2
vap
100
p H O( l ) 2
25
Integrate, solve : T
100
25
vap
z
z
z
g b gFGH dC i 100
25
p H O ( l ) dT 2
T
100
100
H vap
p H O(v ) 2
zd 500
100
Cp
i
H2 O ( v )
25
i i
Cp
Cp
Cp
i
air
H2 O ( l )
H2 O ( v )
dT
dT H vap dT
--
p H O ( v ) dT 2
500
139 o C
25.2 14.37 100 .
100
100
b25.2gFGH dC i dT IJK b7.1gFGH dC i dT H dC i F dC i dT H dC i dTIJ b25.2 gb14.37g b1.00gG H K T
zd zd T
0
zd T
25
dT
n out H out n in H in
0
z
i i
Cp
H out (kJ / mol)
IJ K
0
dT
IJ K
423 kW
When cold water contacts hot air, heat is transferred from the air to the cold water mist lowering the temperature of the gas and raising the temperature of the cooling water.
8-11
8.31
250 kg NH 3 10 3 g
Basis:
h
1 mol
1 kg
1h
8.48 mol NH 3 /s 25°C n 1 (mol air/s) T °C
n 2 (mol/s) 0.100 NH 3 0.900 air 600°C
Q = –7 kW
NH 3 balance: 8.48
0100 . n2 n2
b0.900gb84.8g
Air balance: n1
8.48 mol NH 3 s
17.03 g 3600 s
84.8 mol s
76.3 mol air s
bg
References for enthalphy calculations: NH 3 g , air at 25q C NH 3
H in
0.0
z
C p from
dC i dT H 25.62 kJ mol Air: C bJ mol q Cg 28.94 0.4147 u 10 T bq Cg H z C dT LMMN28.94bT 25g 0.004147FGH T2 252 IJK OPPQ molJ u 101 kJJ e2.0735 u 10 T 0.02894T 0.7248jbkJ molg H out
600
25
p NH 3
out
Table B.2
2
p
2
T
in
25
p
3
6
z
H out
100
25
C p dT
Energy balance: Q 7 kJ s
2
17.39 kJ mol 'H
E¦ out
ni H i ¦ ni H i in
b8.48 mols NH sgb25.62 kJ molg b76.3 mols air sgb17.39 kJ molg b8.48gb0.0g b76.3ge2.0735 u 10 T 0.02894T 0.7248j 3
6
E
1582 . u 10 4 T 2 2.208T 1606 8.32 a.
2
0T
2
693q C (–14,650°C)
Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane Stack gas (900oC)
100 mol/s 0.95 mol M/mol 0.05 mol E/mol
Furnace
n3 n4 n5 n6
mol mol mol mol
CO2/s H2O/s O2/s N2/s
Stack gas (ToC)
Heat Exchanger
n3 n4 n5 n6
mol mol mol mol
CO2/s H2O/s O2/s N2/s
air (245oC)
20 % excess air (20oC)
n1 mol O2/s n2 mol N2/s
n1 mol O2/s n2 mol N2/s
CH 4 2O 2 o CO 2 2H 2 O
b g
C 2 H 6 7 / 2 O 2 o 2CO 2 3H 2 O
8-12
8.32 (cont’d)
LM 95 mol M 2 mol O N s 1 mol M
2
nair
. 12
nair
1185 mol air / s
4.76 mol air 5 mol E 3.5 mol O 2 4.76 mol air mol O 2 s mol O 2 1 mol E
n1
0.21 u 1185
0.79 u 1185
n3
95 mol M 1 mol CO 2 5 mol E 2 mol CO 2 s s 1 mol M 1 mol E
105 mol CO 2 / s
n4
95 mol M 2 mol H 2 O 5 mol E 3 mol H 2 O 1 mol E 1 mol M s s
205 mol H 2 O / s
n5
249
n6
n2
249 mol O 2 / s, n2
936 mol N 2 / s
. mol O 2 95 mol M 2 mol O 2 5 mol E 35 1 mol M s s 1 mol E 936 mol N 2 / s
Energy balance on air: Table B.8: H air (20o C)
. mol H 2 O / s 415
0144 . kJ / mol, H air (245o C)
e
Q = n air H air (245o C) - H air (20o C)
j
OP Q
b
6.509 kJ / mol
. 1180 6.509 0144
g
7851 kW
Energy balance on stack gas: Q
6
FH
¦ ni
'H
7851 n3
z
i 3
T
900
dC i
z
T
900
p CO 2
dC i dT IK p i
dT n4
z
T
900
dC i
p H O ( v ) dT 2
z
n5
T
900
dC i
p O 2
dT n6
z
T
900
dC i
p N 2
dT
Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve T b.
350 m 3 (STP) mol 1000 L 1 h h 22.4 L(STP) m 3 3600 s
792 o C
4.34 mol / s
4.34 mol / s 0.0434 100 mol / s 0.0434 7851 341 kW
Scale factor =
b g
Q c 8.33 a.
Q b.
z
'H
600
0
'H
C p dT n'H
b
g b
g
100 . 4 351 . 38.4 42.0 2 36.7 40.2 43.9 335 3 150 mol 23100 J 1 kW 3465 kW s mol 1000 J / s
23100 J mol
T, y
The method of least squares (Equations A1-4 and A1-5) yields (for X
z
b g
00334 . . u 105 T q C kJ (mol q C) Q 150 1732
Cp
600
0
Cp )
0.0334 1732 . u 105 T dT
3474 kW
The estimates are exactly identical; in general, (a) would be more reliable, since a linear fit is forced in (b). 8.34 a.
ln C p b ln a
bT 1 2 ln a C p
ln C p 2 C p1 T2 T1
e
j
a exp bT 1 2 ,
T1
0.0473
ln C p1 b T1
14475 a .
e
1.4475
8-13
7.1 , C p1
U| |V C 0.235| |W
0.329 ,
p
17.3 , C p2
T2
e
0.235 exp 0.0473T 1 2
j
0.533
8.34 (cont’d) b.
z
150
e
0.235 exp 0.0473T
1800
1
20
30
40 2 200
12
jdT
b0.235gb2g RSexpe0.473T 0.0473 T
jLMNT
12
1 .0473
OPUV QW
150
1730 cal g
1800
DIMENSIONS CP(101), NPTS(2) WRITE (6, 1) FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8.37'/) NPTS(1) 51 NPTS(2) 101 DO 200K 1, 2 N NPTS (K) NM1 N – 1 NM2 N – 2 DT (150.0 – 1800.0)/FLOAT (NM1) T 1800.0 DO 20 J 1, N CP (J) 0.235*EXP(0.0473*SQRT(T)) T T DT SUMI 0.0 DO 30 J 2, NM1, 2 SUMI SUMI CP(J) SUM2 0.0 DO 40 J 3, NM2, 2 SUM2 SUM2 CP (J) DH DT*(CP(1) 4.0 SUM1 2.0 SUM2 CP(N))/3.0 WRITE (6, 2) N, DH FORMAT (1H0, 5XI3, 'bPOINT INTEGRATIONbbbDELTA(H)b ', E11.4,'bCAL/G') CONTINUE STOP END
Solution: N N
11 'H
101 'H
1731 cal g
1731 cal g
Simpson's rule with N 8.35 a.
12
U| M .W . 62.07 g / mol V Q | 'H 56.9 kJ / mol W
11 thus provides an excellent approximation
m 175 kg / min
'H
175 kg 1000 g 1 mol 56.9 kJ 1 min min kg 62.07 g mol 60 s
2670 kW
v
b. The product stream will be a mixture of vapor and liquid. c. The product stream will be a supercooled liquid. The stream goes from state A to state B as shown in the following phase diagram.
P B
A
T
8-14
8.36 a.
(T ) 68.74 o C, 'H v b
Table B.1 Tb
28.85 kJ / mol is not a function of pressure Assume: n - hexane vapor is an ideal gas, i.e. 'H
bC H g B 'H bC H g 6
6
bC H g A 'H b g o bC H g 'H
Total o
14 l, 20o C 1
T 'H v b
14 l, 68.74 o C
z z
68.74
'H 1
20 200
'H 2
0.2163 dT
68.74
6
6
b g
b.
'H
c.
U 200 o C , 2 atm
10.54 24.66 28.85 64.05 kJ / mol
64.05 kJ / mol
e
j
H PV
Assume ideal gas behavior PV U 64.05 3.93 6012 . kJ / mol
b g
'H v tb
100.00q C
e j B 'H H O el, 100 Cj H 2 O l, 50o C
e
50 o C 'H v
j
3.93 kJ / mol
e j A 'H H O e v, 100 Cj
H 2 O v, 50 o C
o
o
2
RT
40.656 kJ mol
1
z z
14 v, 68.74 o C
013744 40.85 u 10 5 T 23.92 u 10 8 T 2 57.66 u 10 9 T 3 dT .
24.66 kJ / mol 'H 1 'H 2 'H v Tb Total
Tb
2
10.54 kJ / mol
'H 2 'H
8.37
14 v, 200 o C
2
e
100 o C 'H v
j
o
o
2
100
'H 1
C pH 2 Obl g dT
3.77 kJ mol
25
25
'H 2
C pH 2 Obv g dT
. kJ mol 169
100
b
'H v 50q C
g
Steam table:
Table B.1
B
. 3.77 40.656 169
b2547.3 104.8gkJ kg
42.7 kJ mol 18.01 g 1 kg 1 mol 10 g
44.0 kJ mol
The assumption of ideal gas behavior might account for the difference between the two values. 8.38 1.75 m3 879 kg kmol min 2.0 min m 3 78.11 kg 60 s
e j B 'H C H e v, 80.1 Cj C 6 H 6 v, 580 o C
o
164.1 mol / s , Tb
1
6
6
o
e j A 'H el, 80.1 Cj
C 6 H 6 l, 25o C 2
o C 6 H 6
o
8-15
b g
801 . q C , 'H v Tb
30.765 kJ mol
8.38 (cont’d)
z z
80.1
'H 1
77.23 kJ mol
C pC6 H 6 bv g dT
580
298
'H 2 'H Q
7.699 kJ mol
C pC 6 H 6 bl gdT
353.1
54.164 kJ / mol e j b164.1 mol / sgb54.164 kJ / molg 8888 kW
'H 1 'H v 801 . o C 'H 2 n'H
'H
Antoine
8.39
UV W
35q C yCCl 4 15% relative saturation ( 'H v ) CCl 4
Table B.1
30.0
0.15
kJ Q mol
PV
B
b25q Cg
015 .
1 atm
'H
176.0 mm Hg 760 mm Hg
0.0347 mol CCl 4 mol
10 mol 0.0347 mol CCl 4 min
mol
30.0 kJ mol CCl 4
10.4 kJ min
Time to Saturation 6 kg carbon 0.40 g CCl 4 g carbon 8.40 a.
b
g
1 mol CCl 4
1 mol gas
1 min
153.84 g CCl 4
0.0347 mol CCl 4
10 mol gas
b
g
CO 2 g, 20q C o CO 2 s, 78.4q C : 'H
z
78.4
20
45.0 min
b gdT 'H sub b78.4q Cg
dC i
p CO g 2
In the absence of better heat capacity data; we use the formula given in Table B.2 (which is strictly applicable only above 0q C ). 78.4 kJ 'H | .03611 4.233 u 10 5 T 2.887 u 10 8 T 2 7.464 u 10 12 T 3 dT 20 mol
z
FG IJ H K
6030
Q
cal
4.184 u 10 3 kJ
mol
1 cal
n'H
'H
28.66 kJ mol
300 kg CO 2
10 3 g
1 mol
28.66 kJ removed
h
1 kg
44.01 g
mol CO 2
(or 6.23 u 107 cal hr or 72.4 kW ) b. According to Figure 6.1-1b, Tfusion=-56oC Q
n'H
'H
where, 'H Q
n
z
56
20
LMz dC i N
p CO (v) dT 2
56
20
e j z dC i dT 'H e56 Cj z dC i dT OQP
dC i
p CO (v) 2
v
'H v 56 o C o
78.4
56
p CO (l) dT 2
78.4
56
8-16
p CO (l) 2
. u 10 5 k J h 195
a bT
Cp
8.41 a.
U| . 0.01765T b Kg V| C bJ mol Kg 4512 . |W a 53.94 b0.01765gb500g 4512 NaCl b s, 300 Kg o NaClb s, 1073 Kg o NaClbl , 1073 Kg L b4512 O J 30.21 kJ . 0.01765T gdT P C dT 'H b1073 Kg M 'H N Q mol mol 53.94 50.41 500 300
b
z
ps
7.44 u 10 'U
Q
p
z
1073
300
b.
0.01765
n
z
m
4
10 3 J
300
1 kJ
J mol
b
1073
300
1073
Cv dT 'U m 1073 K
g
Cv |C p 'U m # 'Hm
Q | 'H
c.
8.42
n'H
200 kg 10 3 g 1 kg
2.55 u 10 8 J
t
s
1 mol
74450 J
58.44 g
mol
1 kJ
0.85 u 3000 kJ 10 3 J
'H v
136.2q C
35.98 kJ mol , Tb
Chen's rule:
LM MN
100 s
409.4 K , Pc
b0.088gb409.4 Kg
Trouton's rule: 'H v | 0.088Tb
2.55 u 10 8 J
37.0 atm , Tc
b
619.7 K (from Table B.1)
36.0 kJ mol 0.1% error
g
FG T IJ 0.0327 0.0297 log P OP HT K PQ 35.7 kJ mol (–0.7% error) 'H | FT I . G J 107 HT K F 619.7 373.2 IJ 38.2 kJ mol Watson’s correlation : 'H b100q Cg | 35.98G H 619.7 409.4 K b
Tb 0.0331
v
10
c
c
b c
0.38
v
8.43
b g b g Trouton's Rule 'H b200q Cg 0.088b200 + 273.2g = 41.6 kJ mol C H Nbl , 25q Cg o C H Nbl , 200q Cg o C H Nbv , 200q Cg
C 7 H 2 N : Kopp's Rule C p | 7 0.012 12 0.018 0.033 0.333 k J (mol q C) v
7
12
z
200
'H
25
7
b
12
7
g
12
kJ kJ 41.6 C p dT 'H r 200q C | 0.333(200 25) mol mol
8-17
100 kJ mol
8.44 a.
b g
Watson Correction: b.
1211.033 220.790 6.90565 log 100
Antoine equation: Tb q C
b g F 562.6 299.3IJ 'H b261 . q Cg 30.765G H 562.6 3531. K b
Clausius-Clapeyron: ln p
c.
0.008314
R| S| T
g
b
118 . q C ; Tb 150 mm Hg
'H v C 'H v RT
b
' H v (80.1°C)
C 6 H 6 ( l , 80.1°C)
zd zd
i
C p dT
26.1
26.1
'H 2
i
'H v 261 . qC
g
U| V| W
b
g
35.2q C
ln p 2 p1
g
1 T2 1 T1 34.3 kJ mol
' H2
C 6 H 6 (v , 80.1°C)
7.50 kJ mol
C p dT
80.1
b
l
33.6 kJ mol
C 6 H 6 (v , 26.1°C)
' H 1
80.1
g
R
ln 150 50 kJ mol K 1 308.4 K 1 285.0 K
C 6 H 6 ( l , 26.1°C)
'H 1
0.38
v
Antoine equation: Tb 50 mm Hg
'H v
261 . qC
v
4.90 kJ mol
7.50 30.765 4.90
33.4 kJ mol
8.45 a. Tout = 49.3oC. The only temperature at which a pure species can exist as both vapor and liquid at 1 atm is the normal boiling point, which from Table B.1 is 49.3oC for cyclopentane. b. Let n f , n v , and n l denote the molar flow rates of the feed, vapor product, and liquid product streams, respectively. Ideal gas equation of state n f
1550 L 273 K s
1 mol
44.66 mol C 5 H 10 (v) / s
423 K 22.4 L(STP)
55% condensation: n l
0.550(44.66 mol / s) = 24.56 mol C 5 H 10 (l) / s
Cyclopentane balance n v
(44.66 24.56) mol C 5 H 10 / s = 20.10 mol C 5 H 10 (v) / s
Reference: C5H10(l) at 49.3oC Substance
n in (mol/s)
H in (kJ/mol)
n out (mol/s)
H out (kJ/mol)
C5H10 (l)
—
—
24.56
0
C5H10 (v)
44.66
H f
20.10
H v
Hi
'H v
z
Ti
49.3o C
8-18
C p dT
8.45 (cont’d) Substituting for 'H v from Table B.1 and for C p from Table B.2 H 38.36 kJ / mol, H 27.30 kJ / mol f
v
Energy balance: Q 8.46 a.
¦n
out H out
¦n
. u 10 3 kJ / s = 116 . u 10 3 kW 116
in H in
Basis: 100 mol humid air fed n 2 (mol), 20o C, 1 atm y 2 (mol H2 O/ mol), sat’d 1-y 2 (mol d ry air/ mo l)
100 mol y 1 (mol H2 O/ mol) 1-y 1 (mol d ry air/ mo l) 50o C, 1 at m, 2o superheat
n 3 (mol H2 O(l))
There are four unknowns (n2, n3, y1 and y2) and four equations (two independent material balances, 2oC superheat and saturation at outlet). All the unknowns can be calculated. b.
b
p 48q C
2q C superheat y1
g
p
saturation at outlet y2
b
p 20q C
g
p
b gb g n b1 y g H O balance: b100gb y g bn gb y g n
dry air balance: 100 1 y1 2
1
2
2
2
b
c.
2
3
g
b
References: Air 25q C , H 2 O l, 20q C
25
Cp
100
H 2
20
=
100
20 50
H 2
20
25
H2O v
air
Cp
dT
H 2 O(l)
50
25
b
g
n in mol
n2 y 2
H 4
H in kJ mol
n3
0
b
1
g
3
0.02894 0.4147 u 10 5 T 0.3191 u 10 8 T 2 1.965 u 10 12 T 3 dT
e
j
dT 'H v 100 o C
z
50
100
dC i
p H O(v) dT 2
0.0754 dT 40.656
Cp
100
20
H 2
100 1 y1
H out H
n out
. 0.03346 0.688 u 10 5 T 0.7604 u 10 8 T 2 3593 u 10 12 T 3 dT
100
H 3
100 y1
Air
zd i z z d i z z zd i z d i 50
n2 1 y 2
n in
2
H 1
H in H
Substance
bg H Ob l g
g
air
Cp
dT
H 2 O(l)
e
j
dT 'H v 100 o C
z
20
100
dC i
p H O(v) dT 2
8-19
8.46 (cont’d) c.
'H
Q
¦ ni H i ¦ ni H i out
in
d.
100 mol 8.314 Pa m3 323 K mol K 1.01325 u 105 Pa
Vair
¦ n H ¦ n H i
Q Vair
out
i
i
i
in
100 mol 8.314 Pa m 3 323 K mol K u 10 5 Pa 101325 .
b
p 48q C
2q C superheat y1
g
p
83.71 mm Hg 760 mm Hg
b
p 20q C
saturation at outlet y2
g
p
0.110 mol H 2 O mol
17.535 mm Hg 760 mm Hg
b gb g n b1 0.023g n H O balance: b100gb0110 . g b9110 . gb0.023g n n
. dry air balance: 100 1 0110
2
. mol 9110
2
2
3
0.023 mol H 2 O mol
3
8.90 mol H 2 O 0.018 kg 1 mol
0.160 kg H 2 O condensed Q
'H
¦ ni H i ¦ ni H i out
Vair
100 mol 8.314 Pa m 3 323 K mol K 101325 u 10 5 Pa .
e. f.
480.5 kJ
in
0.160 kg H 2 O condensed 3
2.65 m air fed 480.5 kJ 3
2.65 m air fed
2.65 m 3
0.0604 kg H 2 O condensed / m 3 air fed
181 kJ / m 3 air fed
Solve equations with Maple. Q
8.47 Basis:
181 kJ 250 m 3 air fed 1 h 1 kW h 3600 s 1 kJ / s m 3 air fed 226 m 3 min
273 K
10 3 mol
b g
309 K 22.415 m 3 STP
12.6 kW
8908 mol humid air min . DA = Dry air
Q ( kJ / min)
8908 mol / min y 0 [ mol H 2 O(v) / mol] (1- y 0 )(mol DA / mol)
n1 ( mol / min) y1 [ mol H 2 O(v) / mol] (1- y1 )(mol DA / mol)
36 o C, 1 atm, 98% rel. hum.
10 o C, 1 atm, saturated
n 2 [ mol H 2 O(l) / min], 10 o C
8-20
8.47 (cont’d) a. Degree of freedom analysis 5 unknowns – (1 relative humidity + 2 material balances + 1 saturation condition at outlet + 1 energy balance) = 0 degrees of freedom.
b36Bq Cg y Table B.3
0.98 p w*
b. Inlet air: y 0 P
p (10 o C) / P
Outlet air: y1
b
0
0.98(44.563 mm Hg) 760 mm Hg
b9.209 mm Hgg b760 mm Hgg
g
Air balance: 1 0.0575 (8908 mol / min)
FG H
b1 0.0121gn
1
n1
References:
IJ K H Obl, triple point g, air b77q Fg
Substance
n in
H 2 O balance: 0.0575 8908
Air
bg H Ob l g
0.0575 mol H 2 O(v) mol 0.0121 mol H 2 O(v) mol
8499 mol / min
mol mol = 0.0121(8499 ) n 2 n 2 min min
409 mol H 2 O(l) min
2
H in
n out
H out
8396 0.3198 8396 0.4352 n in mol min
H2O v
H in kJ / mol
512
462
103
453
409
0.741
2
Air: H from Table B.8 H 2 O: H ( kJ / kg) from Table B.5 u (0.018 kg / mol) Energy balance:
Q 'H
¦
ni H i
out
¦
ni H i
. u 105 kJ 60 min 9.486 u 104 Btu 196 min
in
1h
1 ton 12000 Btu h
0.001 kJ
930 tons
8.48 Basis:
746.7 m 3 outlet gas / h 3 atm
1 kmol
100 kmol/h at 0°C, 3 atm yout (kmol C 6 H 14( v)/kmol), saturated (1 – yout) (kmol N2 /kmol) n 2 kmols/h nC6 H14 ( v), 0°C
n 1 (kmol/h) at 75°C, 3 atm yin (kmol C 6 H 14( v)/kmol), 90% sat'd (1 – yin) (kmol N 2 /kmol)
Antoine: log p v y out y in
100.0 kmol / h
b g
1 atm 22.4 m 3 STP
b g
p v 0q C
6.87776 45.24 3 760
1171530 . 224.336 T
b g
b
p v 0q C
45.24 mm Hg, p v 75q C
b g 0.0198 kmol C H kmol , 0.90 p b75q Cg b0.90gb920.44g kmol C H 0.363 3b760g kmol P P
6
14
v
6
8-21
14
g
920.44 mm Hg
8.48 (cont’d)
b
g
b
N 2 balance: n1 1 0.363
g
100 1 0.0198 n1
b
153.9 kmol h
. kmol C H bl g h gb g b100gb0.0198g n n 5389 Percent Condensation: b5389 . kmol h condenseg b0.363 u 153.9 gb kmol h in feed g u 100% C 6 H 14 balance: 153.9 0.363
2
References: N2(25oC), n-C6H14(l, 0oC) Substance n H n in
N2
bg bl g
n - C 6 H 14 r n - C 6 H 14 N 2 : H
b
98000
in
out
55800 44.75
2000
33.33
53800
0.0
z
68.7
g
C p T 25 , n C 6 H 14 (v): H
6
'H
n in mol h
i
i
z T
b g
C p" dT 'H v 68.7
C pv dT
68.7
( 2.64 u 10 6 kJ h)(1 h / 3600 s) 733 kW
¦ n H ¦ n H out
96.5%
H in kJ mol
0
Energy balance: Q
14
H out
98000 0.726
146 .
2
i
i
in
8.49 Let A denote acetone. Q ( kW)
W s
142 L / s @ 150 o C, 1.3 atm n 0 ( mol / s) y 0 [mol A(v) / mol], sat' d (1 y 0 )( mol air / mol)
25.2 kW
n1 (mol / s) @ 18 o C, 5 atm y1 [mol A(v) / mol], sat' d (1 y1 )( mol air / mol)
n 2 [ mol A(l) / s]@18 o C, 5 atm a.
Degree of freedom analysis:
b.
Ideal gas equation of state P0V0 RT0
(1) n 0
6 unknowns ( n 0 , n1 , n 2 , y 0 , y1 , Q ) –2 material balances –1 equation of state for feed gas –1 sampling result for feed gas –1 saturation condition at outlet –1 energy balance 0 degrees of freedom Raoult’s law (2) y1
p *A ( 18 o C) 5 atm
(Antoine equation for p *A )
Feed stream analysis (3)
y0
FG mol A IJ H mol K
[(4.973 4.017) g A][1 mol A / 58.05 g] [(3.00 L) P0 / RT0 ] mol feed gas 8-22
8.49 (cont’d) Air balance
Acetone balance
n 0 (1 y 0 ) (1 y1 )
(4) n1
n 0 y 0 n1 y1
(5) n 2 o
o
Reference states: A(l, –18 C), air(25 C) n in ( mol / s)
H in ( kJ / mol)
n in ( mol / s)
H in ( kJ / mol)
A(l)
n 2
0
A(v)
n 0 y 0
H A0
n1 y1
H A1
air
n 0 (1 y 0 )
H a 0
n1 (1 y1 )
H a1
Substance
(6) H A(v) (T )
z
56o C o
18 C
Table B.2
(7)
z
(C p ) A(l) dT + ( 'H v ) A +
T
56 o C
(C p ) A(v) dT
Tab le B.1
Ta ble B.2
H air (T ) from Table B.8
(8) Q
W s
¦ n
out H out
¦ n
(W s
in H in
25.2 kJ / s)
c. (1) n 0
5.32 mol feed gas / s
(3) y 0
0.147 mol A(v) / mol feed gas
(4) n1
4.57 mol outlet gas / s
(5) n 2
0.75 mol A(l) / s
(6) H A 0
56.07 kJ / mol, H A1
33.34 kJ / mol
(7) H a 0
3.666 kJ / mol, H a1
1245 kJ / mol .
(8) Q
8.50 a.
6.58 u 10 3 mol A(v) / mol outlet gas
(2) y1
90.2 kW
b g
3 m S u 35
2
cm 2
1 m2 10 4 cm 2
s
273 K
b273 40gK
850 mmHg
10 3 mol
1 kg mol
b g
760 mmHg 22.4 m 3 STP
1 kg mol
50.3 mol s H = n-hexane assume P=850 mmHg
50.3 mol/s, 850 mmHg x0 mo l H/mo l (1-x0 ) mo l a ir/ mol 40o C, Tdp =20o C
n 2 mol H(v)/ mol, sat’d @ ToC n 3 mol air/mol n 1 (mols H( l )/s) (90% of H in feed)
8-23
8.50 (cont’d) Degree-of-freedom analysis 5 unknowns (n1, n2, n3, x0 and T) – 2 independent material balances – 1 saturation condition – 1 60% recovery equation – 1 energy balance 0 degrees of freedom All unknowns can be calculated. b.
Antoine equation, Table B.4
dT i
b
p *H 25q C
25 q C x 0
dp feed
P
151 mm Hg 850 mm Hg
0178 . mol H mol
. g mols H feed b50.3gb0178
0.600
n1
60% recovery
g
bg
5.37 mols H l s
s
. g 358 . mols Hbv g s b0.400gb50.3gb0178 . g 413 . mols air s b50.3gb1 0178 n2
Air balance: n3
Mole fraction of hexane in outlet gas: n2 n2 n3
b
358 . 358 . 41.3
g
Antoine equation: p H*
bg
pH* T
bg
pH* T
850 mm Hg
67.8 mm Hg T
b
67.8 mm Hg
7.8 q C
g
Reference states: C 6 H 14 l, 7.8q C , air (25qC) Substance
bg Obl g
C 6 H14 O v C 6 H 14
8.95
H in 37.5
3.58
H out 32.7
—
—
5.37
0
41.3
0.435
41.3
–0.499
n in
Air
z
n out
68 . 74
bg
C 6 H14 O v : H
b
g
z T
C pl dT ' H v 68.74 q C
7 .8
n in mol/s H in kJ/mol
C pv dT ,
68 .74
C p from Table B.2 ' H v from Table B.1
Air: H from Table B.8 Energy balance: Q
'H
¦ n H ¦ n H i
i
i
out
c.
u A
u' A' ; A
S D2 ; D' 4
257 kJ s 1 kW cooling i
1 kJ s
in
|UV W|
1 D u' 4 u 2
8-24
12.0 m / s
257 kW
8.51 n v ( mol / min) @ 65o C, P0 (atm) y[ mol P(v) / mol], sat' d (1- y )(mol H(v) / mol) 100 mol / s @80 o C, 5.0 atm 0.500 mol P(l) / mol 0.500 mol H(l) / mol
Q ( kJ / s) n l ( mol / min) @ 65o C, P0 (atm) 0.41 mol P(l) / mol 0.59 mol H(l) / mol
a. Degree of freedom analysis 5 unknowns – 2 material balances – 2 equilibrium relations (Raoult’s law) at outlet – 1 energy balance = 0 degrees of freedom Antoine equation (Table B.4) p *P (65 o C) = 1851 mm Hg, p *H (65 o C) = 675 mm Hg Raoult' s law for pentane and hexane 0.410 p *P (65 o C) = yP0 0.590 p *H (65 o C) = (1 y ) P0
0.656 mol P(v) / mol
y
1157 mm Hg (1.52 atm)
P0
Total mole balance: 100 mol = n v n l
Pentane balance: 50 mole P = 0.656n v + 0.410n l nv RT P0
Ideal gas equation of state: Vv Fractional vaporization: f
36.6 mol
n v
36.6 mol vapor / s
n l
63.4 mol liquid / s
0.08206 L atm mol K
s
36.6 mol vapor / s 100 mol / s
0.366
mol vaporized mol fed
References: P(l), H(l) at 65 o C H in
Substance n in P(v)
P(l)
50
H(v)
H(l)
50
H out
n out
24.0 24.33
2.806 26.0
0
Liquid: H (T) =
H in kJ / mol
12.6 29.05
3.245 37.4
Vapor: H (T ) =
n in mol s
z z
Tb
65o C
T
65o C
0
C pl dT 'H v (Tb )
z
T
Tb
C pv dT
C pl dT
Tb and 'H v from Table B.1, C p from Table B.2 Energy balance:
Q
¦ n
out H out
¦ n
in H in
8-25
1040 kW
(65 + 273)K 152 . atm
667 L / s
8.52 a.
B=benzene; T=toluene n 2 mol/s 95o C 1320 mo l/s 25o C
0.735 mol B/ mo l 0.265 mol T/ mol
0.500 mol B/ mo l 0.500 mol T/ mol
n 3 mol/s 95o C 0.425 mol B/ mo l 0.575 mol T/ mol
Q
UV RS W T
Total mole balance: 1320 n 2 n 3 n 2 Benzene balance: 1320(0.500) = n 2 (0.735) n 3 (0.425) n3
319 mol / s 1001 mol / s
References: B(l, 25oC), T(l, 25oC) Substance
n in (mol / s) H in ( kJ / mol) n out (mol / s) H out ( kJ / mol)
B(l) B(v) T(l) T(v) Q
660 -660 --
¦ n H ¦ n H i
i
i
out
b.
0 -0 -i
425 234 576 85
5.361 36.39 10.43 44.19
2.06 u 10 4 kW
in
e
Antoine equation (Table B.4) p *B 95 o C Raoult' s law Benzene: Toluene:
j
e
1176 torr , p T* 95 o C
j
476.9 torr
|UV P z P' 1035 torr |W
b0.425gb1176g b0.735g P P b0.575gb476.9g b0.265g P' P'
680 torr
Analyses are inconsistent.
Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is invalid at the system conditions (not likely).
bg b5gb7.5g b12gb9.6g 17 C H Obl g — C b5gb12g b12gb18g 25 . gb113 273g 42.1 kJ mol Trouton’s rule — Eq. (8.4-3): 'H b0109 Eq. (8.4-5) 'H b0.050gb52 273g 16.25 k J mol
8.53 Kopp’s rule (Table B.10):
C 5 H 12 O s — C p 5
12
p
v
m
Basis:
235 m 3
273 K
1 kmol
b g
389 K 22.4 m 3 STP
h
10 3 mol
1h
1 kmol
3600 s
2.05 mol s
Neglect enthalpy change for the vapor transition from 116qC to 113qC.
b
g
b
g Obs, 25q Cg
b
C 5 H 12 O v , 113q C o C 5 H 12 O l , 113q C o C 5 H 12 O v , 52q C
b
g
o C 5 H 12 O s, 52q C o C 5 H 12
8-26
g
170 J mol 301 J mol
8.53 (cont’d) 'H
b
g
b
g kJ J 1 kJ 16.2 b301gb61g b170gb27g u mol mol 10 J
'H v C pl 52 113 'H m C ps 25 52 42.1
kJ mol
813 . kJ mol
3
2.05 mol 813 . kJ
n'H
'H
Required heat transfer: Q
s
1 kW
mol
1 kJ s
167 kW
8.54 Basis: 100 kg wet film a.
95 kg dry film 5 kg acetone
95 kg DF 5 kg C3 H 6 O( l ) Tf 1 = 35°C n 1 mol air Ta1 , 1.01 atm
4.5 kg acetone exit in gas phase
95 kg DF 0.5 kg C3 H 6 O( l ) Tf 2 n 1 mol air 4.5 kg C3 H 6 O( v) (40% sat'd) Ta2 = 49°C, 1.0 atm
Antoine equation (Table B.4) p C* 3H 6O 4.5 kg C 3 H 6 O
y=
0.5 kg acetone remain in film
90% A evaporation
1 kmol
10 3 mol
58.08 kg
kmol
b
bg
77.5 mol C 3 H 6 O v in exit gas
0.40 59118 . mm Hg
77.5 77.5 n1
59118 . mm Hg
b g
g n
171.6 mol 22.4 L STP
1
760 mm Hg
b
g
mol
b
g b
95 kg DF
b g
L STP
kg DF
g
References: Air 25q C , C 3 H 6 O l , 35q C , DF 35q C
b.
Substance
n in
H in
n out
DF
95
0
95
86.1
0
8.6
—
—
77.5
32.3
dC i
171.6
0.70
bg Obv g
C 6 H 14 O l C 6 H 14
171.6
Air
z
86
H A(v)
35
z
Ta1
25
dC i dT 'H p l
p air dT
zd
49 v
Cp
86
i dT , v
H out
d 0.129 dT
i 35i
1.33 T f 2 35
b
H DF
f2
'H
¦ n H ¦ n H i
i
i
out
Ta1
z
i
d
25
dC i
120q C
p air dT
z
Ta1
25
dC i
n in mol H in kJ/mol
g
z
i
126.4 T f 2 35 111 . (T f 2 35) 2623.4 1716 .
in
Ta1
n in kg H in kJ/kg
C p T 35
Energy balance
c.
405 .
d
i
127.5 T f 2 35 26234 .
p air dT
1716 .
d
i
2.78 kJ mol T f 2 35 q C
8-27
16.8q C
Ta 1
25
dC i
p air dT
0
8.54 (cont’d) T& E
34q C Ta1
d. T f 2 e.
8.55
T&E
36q C Ta1
506q C , T f 2
552q C
In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensed phase drops. In this problem, the heat transferred from the air goes to (1) vaporize 90% of the acetone in the feed; (2) raise the temperature of the remaining wet film above what it would be if the process were adiabatic. If the feed air temperature is above about 530 qC, enough heat is transferred to keep the film above its inlet temperature of 35 qC; otherwise, the film temperature drops.
b
g
200 psia | 100q F (Cox chart – Fig. 6.1-4)
Tset p a. Basis:
3.00 u 10 3 SCF 1 lb - mole h 359 SCF
8.357 lb mole h C 3 H 8
8.357 lb-mole C3H8(v)/h 200 psia, 100oF
8.357 lb-mole C3H8(l)/h 200 psia, 100oF
Q
- mole H 2 O(l) / h m(lb 70oF
- mole H 2 O(l) / h m(lb 85oF
The outlet water temperature is 85oF. It must be less than the outlet propane temperature; otherwise, heat would be transferred from the water to the propane near the outlet, causing vaporization rather than condensation of the propane. b. Energy balance on propane: Table B.1
Q
'H
n'H v
B
8.357 lb moles 18.77 kJ 0.9486 Btu 453.593 mol h mol kJ 1 lb mole
6.75 u 10 4
Energy balance on cooling water: Assume no heat loss to surroundings. Q
'H
p 'T m mC
8.07 u 10 5 Btu lb m q F h 1.0 Btu 15 q F
53,800
lb m cooling water h
8.56
m 2 [ kg H 2 O(v) / h]@30 o C, 1 atm
1000 kg/h, 30oC 0.200 kg solids/kg 0.800 kg H2O(l)/kg
m 3 ( kg / h) @ 30 o C 0.350 kg solids / kg 0.650 kg H 2 O(l) / kg m 1 [ kg H 2 O(v) / h], 1.6 bar, sat' d a.
Solids balance: 200
0.35m3
H 2 O balance: 800
m2 0.65 5714 .
b
m 1 [ kg H 2 O(l) / h], 1.6 bar, sat' d
g
m3
571.4 kg h slurry
m2
428.6 kg h H 2 O v
8-28
bg
Btu h
8.56 (cont’d) References: Solids (30qC), H 2 O (l, triple point) Substance n in n out H in H out Solids 200 0 200 0 H2O l 800 125.7 371.4 125.7 — — 428.6 2529.2 H Ov 2
bg bg
E.B. Q
'H
2791.7
m1
H 2 O , lbbars
¦ n H ¦ n H i
i
out
i
b
b
g
H H 2 O from steam tables
857.6
m1
0 1.030 u 10 6 1833m1
i
g
n kg h H kJ kg
0 m1
562 kg steam h
in
b.
b562.0 428.6g
c.
The cost of compressing and reheating the steam vs. the cost of obtaining it externally.
133 kg h additional steam
8.57 Basis: 15,000 kg feed/h.
A = acetone, B = acetic acid, C = acetic anhydride Q c (kJ/h) 2 n 1 (kg A(v )/h) 329 K
condenser
n 1 (kg A(l )/h) 303 K
n 1 (kg A(l )/h) 303 K
15000 kg/h 0.46 A 0.27 B 0.27 C 348 K, 1 atm
still
1% of A in feed n 2 (kg A(l )/h) n 3 (kg B( l )/h) Q r (kJ/h) n 4 (kg C( l )/h) 398 K reboiler
a.
b0.01gb0.46gb15,000 kg hg 69 kg A h Acetic acid balance: n b0.27gb15,000g 4050 kg B h Acetic anhydride balance: n b0.27gb15,000g 4050 kg h 6831 kg h ` Acetone balance: b0.46gb15,000g n 69 n n 2
3
4
1
1
Distillate product: 6831 kg acetone h
Bottoms product:
b.
b69 4050 4050g kg h
8169 kg h 0.8% acetone 49.6% acetic acid 49.6% acetic anhydride
Energy balance on condenser
8-29
8.57 (cont’d)
b
g
b
g
b
C 3 H 6 O v , 329 K o C 3 H 6 O l , 329 K o C 3 H 6 O l , 303 K
b
'H
z
303
g C dT b2 u 6831gkg
'H v 329 K
pl
b gb g
520.6 2.3 26
g
580.4 kJ kg
329
Q c c.
'H
n'H
h
580.4 kJ kg
7.93 u 10 6 kJ h
Overall process energy balance Reference states: A(l), B(l), C(l) at 348 K (All H m Substance n in n out H H in
out
b g — 0 6831 –103.5 115.0 69 0 — A bl, 398 Kg 4050 109.0 0 — B bl, 398 Kg 4050 113 0 — C bl, 398 Kg J C | b4 u 12g b6 u 18g b3 u 25g Acetic anhydride (l): molq C A l, 303 K
n in kg/h H in kJ/kg
1 mol 10 3 g 1 kJ 102.1 g 1 kg 10 3 J
p
bg
b
H T
Q
0)
2.3 kJ kgq C
g
C p T 348 (all substances)
'H Q c Q r
¦ n H ¦ n H i
i
i
out
in
A
i
Q r
Q c
¦ n H e7.93 u 10 i
i
6
out
. u 10 6 kJ h 813
0
(We have neglected heat losses from the still.) d.
H 2 O (saturated at | 11 bars): 'H v Q r
8.58
n H 2 O 'H v n H 2 O
1999 kJ kg (Table 8.6)
6
813 . u 10 kJ h 1999 kJ kg
4070 kg steam h
Basis: 5000 kg seawater/h a.
S = Salt n 3 (kg H 2 O(l )/h @ 4 bars) 2738 kJ/kg
5000 kg/h @ 300 K 0.035 S 0.965 H 2O( l) 113.1 kJ/kg
n 5 (kg H 2 O(l )/h @ 4 bars) 605 kJ/kg
b. S balance on 1st effect:
n 4 kg H 2 O(v )/h @ 0.2 bars 2610 kJ/kg
n 2 (kg H 2 O(v )/h @ 0.6 bars) 2654 kJ/kg n 1 (kg/h @ 0.6 bars) 0.055 S 0.945 H 2 O(l ) 360 kJ/kg
b0.035gb5000g
Mass balance on 1st effect: 5000
0.055n1 n1
3182 n 2 n 2
8-30
n 3 (kg/h @ 0.2 bars) x (kg S/kg) (1 – x) (kg H2 O(l )/hr) 252 kJ/kg n 2 (kg H 2 O(l )/h @ 0.6 bars) 360 kJ/kg
3182 kg h 1818 kg h
j
2.00 u 105 kJ h
8.58 (cont’d) Energy balance on 1st effect: 'H
b gb
n 5
n1 3182 n2 1818
c.
g b gb g b gb 2534 kg H Obv g h
g b gb g
. 0 n 2 2654 n1 360 n 5 605 2738 5000 1131
0
2
Mass balance on 2nd effect: 3182
n 3 n 4 (1)
b'H 0g bn gb2610g bn gb252g bn gb360 2654g bn gb360g E n 3182, n 1818
Energy balance on 2nd effect: 4
3
2
1
1
5.316 u 10 6
0
2
252n 3 2610n 4
(2)
Solve (1) and (2) simultaneously: n 3
1267 kg h brine solution
n 4
1915 kg h H 2 O v
bg
n 2 n 4
Production rate of fresh water
b
gb g
Overall S balance: 0.035 5000
d.
e.
b1818 1915g
1267 x x
3733 kg h fresh water
kg salt kg 0138 .
The entering steam must be at a higher temperature (and hence a higher saturation pressure) than that of the liquid to be vaporized for the required heat transfer to take place.
n 5 (kg H 2 O(v )/h) 2738 kJ/kg 5000 kg/h 0.035 S 0.965 H 2 O(l ) 113.1 kJ/kg
3733 kg/h H2 O(v ) @ 0.2 bar 2610 kJ/kg n 1 (kg brine/h @ 0.2 bar 252 kJ/kg
Q3
Mass balance: 5000
n 5 (kg H 2 O(l )/h) 605 kJ/kg
3733 n1 n1
1267 kg h
d'H 0i b3733gb2610g b1267gb252g n b605 2738g b5000gb1131. g n 4452 kg H Obv g h
Energy balance:
5
5
0
2
Which costs more: the additional 1918 kg/hr fresh steam required for the single-stage process, or the construction and maintenance of the second effect?
8-31
8.59 a.
583 kg h 0.30 5000 583 4417 kg fresh water h
Fresh water produced: n L 7 n L1 b.
Final result given in Part (d).
c.
Salt balance on i th effect: n Li x Li
bn g b x g L i 1
b0.035gb5000g
x L1 n L1 n L1
Salt balance: x L 7 n L 7
L i 1
bn g b x g L i 1
x Li
L i 1
(1)
nT Li
Energy balance on i th effect:
b g e H j
0 n vi H vi n v
'H
b g n v
b g e H j n H bn g e H j e H j e H j v
L 1
n vi H vi L 1
Li v
Li
i 1
b g effect: bn g bn g bn g
Mass balance on i 1 n Li
L 1
v i 1
L i 1
n Li H Li n L L i 1
L
L
L 1
L
L 1
b g e H j
n v
L 1
v
L 1
0 (2)
i 1
L 1
th
L i 1
b g
n Li n v
(3)
i 1
d.
Fresh steam Effect 1 Effect 2 Effect 3 Effect 4 Effect 5 Effect 6 Effect (7)
P (bar) 2.0 0.9 0.7 0.5 0.3 0.2 0.1 1.0
T (K) 393.4 369.9 363.2 354.5 342.3 333.3 319.0 300.0
nL (kg/h) --584 1518 2407 3216 3950 4562 5000
8-32
xL --0.2997 0.1153 0.0727 0.0544 0.0443 0.0384 0.0350
nV (kg/h) 981 934 889 809 734 612 438 ---
HL (kJ/kg) 504.7 405.2 376.8 340.6 289.3 251.5 191.8 113.0
HV (kJ/kg) 2706.3 2670.9 2660.1 2646.0 2625.4 2609.9 2584.8 ---
8.60 a.
b g | dC i
dC i d C i p v
20 cal (molq C) ; C v
p l
p v
v
b
R | 10 2
g molcalq C
8 cal (molq C)
b. n0 (mol N2)
n0 (mol N2)
o
3.00 L@ 93 C, 1 atm
n2 [mol A(v)] 85oC, P(atm)
n1 (mol A(l)
n3 [mol A(l)] o
85oC, P(atm)
0.70 mL, 93 C 3.00 L
n0
b
273 K 1 mol 273 93 K 22.4 L STP
70.0 mL
n1
g
b g
bg
0.90 g 1 mol mL 42 g
Energy balance 'U
0
15 . mol A l
¦ n U ¦ n U i
i
i
out
c.
0.100 mol N 2
i
0
in
b g b gb
g
References: N 2 g , A l 85q C, 1 atm
Substance n in U in n out U out N2 010 . 39.8 010 . 0 n in mol Al 15 . 160 n3 0 U in cal mol Av n 2 20050
bg bg
b g b g C b93 85g Abv , 85q Cg: U 20b90 85g 20,000 10b85 90g . gb39.8g b15 . gb160g 0 n 'U 0 n b20050g b010 A l, 93q C and N 2 g , 93q C : U
v
20050 cal mol
A( v )
v1
0.012 mol A 42 g A
d.
mol A
v1
0.012 mol A evaporate
0.51 g evaporate
Ideal gas equation of state
bn
P
0
g
n 2 RT V
0.112 mol 3.00 liters
b273 85gK
0.08206 L atm 1.097 atm mol K
Raoult’s law
b
p A 85q C
g
yAP
n2 P n 0 n2
0.012 mol 1.097 atm 0.112 mol
8-33
0.117 atm
b
89.3 mmHg
g
8.61 (a) i)
b4.4553 3.2551gkg
FG mIJ HV K
Expt 1
2.000 L
liquid
0.600
b g
kg SG L
b3.2571 3.2551gkg 0.0020 kg 2.000 L 273 K b763 500gmm Hg
ii) Expt 2 Mass of gas Moles of gas
363 K
Molecular weight
iii) Expt. 1 n
bliquid g
2.000 liters 10 3 cm 3 1 liter
0.600
2.0 g
1 mol 22.4 liters STP
760 mm Hg
b2.0 gg b0.0232 molg
liquid
b g
0.0232 mol
86 g mol
0.600 g 1 mol 14 mol cm 3 86 g
Energy balance: The data show that Cv is independent of temperature Q
'U nCv 'T
b g
Cv
b g
Cv Expt. 2 n
b vapor g
Cv
Q n'T
liquid
liquid
800 J 14 mols 2.4 K
b gb g 800 J b14 molsgb2.4 Kg
24 J mol [email protected] K 24 J mol [email protected] K
{ 24 J mol K
bg
0.0232 mol from ii
a bT Q
0.0232
z
T2
T1
(a bT )dT
LM N LM N
LM N
b 0.0232 a (T2 T1 ) (T22 T12 ) 2
b 130 . J = 0.0232 a(366.9 - 363.0) + (366.9 2 363.02 ) 2 b 130 . J = 0.0232 a(492.7 - 490.0) + (492.7 2 490.02 ) 2
b g
Cv
vapor
(J / mol K)
bg
OP U| Q |V a OP| b Q|W
4.069 0.05052
4.069 0.05052T K
iv) Liquid: C p | Cv { 24 J mol K Vapor: Assuming ideal gas behavior, C p
Cv R Cv 8.314 J mol K
b
C p J mol K v) Expt. 3 T T T T
315K , 334 K , 354 K , 379 K ,
p p p p
b763 564gmm Hg 401 mm Hg 761 mm Hg 1521 mm Hg
8-34
g
bg
4.245 0.05052T K
199 mm Hg
OP Q
8.61 (cont’d) Plot p (log scale) vs. 1 T (linear scale); straight line fit yields 3770 17.28 or p 3196 u 107 exp 3770 T ln p . T K
b
bg
vi) p
vii)
'H v R
1 A T Part v b
760 mm Hg
A 3770b Kg Part v
'H v
b g
17.28 ln 760 3770
g
2.824 u 10 3 K 1 Tb
b3770 Kgb8.314 J mol Kg 'H
3.5 L feed 273 K 1 mol s 510 K 22.4l STP Let A denote the drug
b g
(b) Basis:
v
31,300 J mol
0.0836 mol s feed gas
.
n 1 [mol A(v)/s] . n 2 [mol N 2 /s] T(K), saturated with A
0.0836 mol/s @ 510 K 0.20 A 0.80 N 2
.
Q(kW)
n 3 (mols A( l )/s), 90% of A in feed T(K)
b0.800gb0.0836 mol sg 0.0669 mol N s 90% condensation: n b0.900gb0.200 u 0.0836g 0.01505 mol Abl g s n b0100 . gb0.200 u 0.0836g 167 . u 10 mol Abv g s N 2 balance: n2
2
3
3
1
Partial pressure of A in outlet gas: pA
b
n1 P n1 n2
g
1 T
. u 10 3 mol 167 (760 mm Hg) 18.5 mm Hg 0.0686 mol
E Part (a) - (v) 17.28 lnb18.5g
3770 T 262 K
381 . u 103 K 1
bg
Reference states: N 2 , A l at 262 K nin H in nout substance 0.0669 7286 0.0669 N2 Av 0.0167 37575 167 . u 10 3 Al 0.01505
bg bg
H out 0
n in mol s 31686 H in J mol 0
8-35
354 K
bg
p A T
8.61 (cont’d)
b
g
N 2 510 K : H N 2 (510K) - H N 2 (262 K) = H N 2 (237 o C) - H N 2 ( 11o C) Table B.8
B
[6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol
b
A(v, 262K): H
g
b
g
C pl Tb 262 'H v 359 K
z
262
Tb
C pv dT
Part (a) results for Tb , C pl , C pv , 'H v
H
L T O 24b354 262g 31300 M4.245 0.05052 P 31686 J mol 2 Q N H C bT 262g 'H b354 K g z C dT 37575 J mol 2 262
354
A(v, 510K):
510
pl
b
v
Energy balance: Q
'H
Tb
¦ ni H i ¦ ni H i out
in
pv
1060 J s 1 kW cooling 1.06 kW 103 kJ s
8.62 a. Basis: 50 kg wet steaks/min D.M. = dry meat m1 (kg H 2 O(v )/min) (96% of H 2 O in feed) 60°C
50 kg/min @ –26°C 0.72 H 2O( s) 0.28 D.M. Q (kW)
m2 (kg D.M./min) m3 (kg H 2 O(l )/min) 50°C
96% vaporization: m 1 0.96 0.72 u 50 kg min
b g 34.56 kg H O bvg min m 0.04b0.72 u 50 kg ming 144 . kg H O bl g min Dry meat balance: m b0.28gb50g 14.0 kg D. M. min Reference states: Dry meat at 26q C , H Obl, 0q Cg 2
3
2
2
2
substance dry meat H 2 O s, 26q C H 2 O l , 50q C H 2 O v , 60q C
b
b b
g g
m in H in m out H out 14.0 0 14.0 105 m in kg min H in kJ kg 36.0 390 144 209 . 34.56 2599
g
kJ 76q C 105 kJ kg b g 1.38 kg Cq H Obs, 26q Cg: H Obl , 0q Cg o H Obs, 0q Cg o H Obs, 26q Cg
b
Dry meat: H 50q C 2
g
C p 50 26
2
2
8-36
2
8.62 (cont’d)
z
b g
'H
'H m 0q C C p dT
b
mol
A
0
g
b
1 mol 10 3 g 2.17 kJ 18.02 g 1 kg kgq C
6.01 kJ
26
Table B.1
g b b50 0gq C
H 2 O l, 50q C : H 2 O l , 0q C o H 2 O l , 50q C
z
0.0754 kJ mol q C
50
'H
C p dT
0
b
g
b
g
390 kJ kg
g
1 mol 1000 g 18.02 g 1 kg
A Table B.2
26q C
b
g
209 kJ kg
b
g
b
H 2 O v , 60q C : H 2 O l , 0q C o H 2 O l , 100q C o H 2 O v , 100q C o H 2 O v , 60q C 'H
0.0754 kJ
b100 0gq C
40.656
molq C
A
Ad
Table B.2
kJ mol
Table B.1 'H v
46.830 kJ 1 mol 1000 g mol 18.02 g 1 kg
zd 60
100
i
Cp
i
H 2 O(v)
A
g
dT
Table B.2
2599 kJ kg
Energy balance: Q
'H
¦
mi H i
out
¦
mi H i
in
1.06 u 10 5 kJ 1 min 1 kW 1760 kW min 60 s 1 kJ s
8.63 Basis: 20,000 kg/h ice crystallized. S = solids in juice. W = water . . m
1 (kg/h) juice 0.12 solids(S) 0.88 H 2O( l )(W) 20°C
Qf preconcentrate .m . m5 (kg/h) product Slurry(10% ice), –7°C 2 (kg/h) freezer filter x 2 (kg S/kg) 20,000 kg W( s )/h 0.45 kg S/kg . m4 kg residue/h 0.55 kg W/kg (1 – x 2) (kg W/kg) 0.45 kg S/kg 20,000 kg W( s )/h . m4 (kg/h), 0.45 S, 0.55 W 0.55 kg W( l )/kg .m (kg/h), 0°C 3 separator 0.45 kg S/kg 20,000 kg W( s )/h 0.45 kg W( l )/kg
(a) 10% ice in slurry
20000 m 4
10 m 4 90
180000 kg h concentrate leaving freezer
UV W
m 1 Overall S balance: 012 . m 1 0.45m 5 m 5 Overall mass balance: m 1 m 5 20000
27273 kg h feed 7273 kg h concentrate product
Mass balance on filter: 20000 m 4 m 5 20000 m 6
m 4 180000 m 5 7273
m 6
Mass balance on mixing point: 27273 172730 m 2 m 2
2.000 u 105 kg h preconcentrate
8-37
172730 kg h recycle
8.63 (Cont’d) S balance on mixing point: 012 . 27273 0.45 172730
b gb
g b gb
g
2.000 u 105 X 2 X 2 100%
40.5% S
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs: fresh feed and recycle streams; output; slurry leaving freezer)
bg
Refs: S, H 2 O l at 7q C m in H in m out H out m kg h 27273 108 H kJ kg 0 45% soln 172730 28 180000 20000 337 H 2O s
substance 12% soln
b b
bg
g g
b g 4.00 T b7g kJ kg 'H b T q Cg | 'H b0q Cg
Solutions: H T Ice: H
m
m
6.0095 kJ mol 337 kJ kg
D Table B.1 E.B. Q c
1452 u 107 kJ . 1h 1 kW 4030 kW h 3600 s 1 kJ s
¦ m i H i ¦ m i H i
'H
out
in
d
8.64 a. B=n-butane, I=iso-butane, hf=heating fluid. (C ) p hf 24.5 kmol/h @ 10oC, P (bar) 0.35 kmol B(l)/h
2.62 kJ / kg o C
i
24.5 kmol/h @ 180oC 0.35 kmol B(l)/h
Q ( kW) m (kg HF / h), T( o C)
m (kg HF / h), 215 o C
From the Cox chart (Figure 6.1-4)
d
p B* 10 o C
b.
i
d
22 psi, pI* 10 o C
p min
p B pI
d
i
i
32 psi
x B p B* x I pI*
28.5 psi
d
i
FG 1.01325 bar IJ H 14.696 psi K d
Hv H1 o B v, 10 o C ' o B v, 180 o C B l, 10 o C '
d
i
d
i
d
Hv H2 o I v, 10 o C ' o I v, 180 o C I l, 10 o C '
196 . bar
i
i
Assume temperature remains constant during vaporization. Assume mixture vaporizes at 10oC i.e. won’t vaporize at respective boiling points as a pure component.
8-38
8.64 (cont’d) References: B(l, 10oC), I(l, 10oC) substance nin mol / h H in kJ / mol B (l) 8575 0 B (v) --I (l) 15925 0 I (v) ---
b
z z
g
d H i d'H i dC i d H i d'H i dC i out
v
B
B
180
v
I
i
i
out
'H
i
n out mol / h
g
b
H out kJ / mol
-8575 -15925
g
-42.21 -41.01
4101 . kJ / mol
b
g
b
g
8575 42.21 15825 4101 .
i
in
u 10 6 kJ / h 1015 .
d
Q 1015 u 10 6 kJ / h = m hf 2.62 kJ / kg o C . m hf
d.
p I
10
¦ n H ¦ n H
'H
c.
I
b
g
42.21 kJ / mol
p B
10
180
out
b
i b215 45g C o
2280 kg / h
b2540 kg / hg 2.62 kJ / dkg Ci b215 45g C o
o
u 10 6 1015 u 10 6 Heat transfer rate 1131 . .
u 10 6 kJ / h 1131 . 116 . u 10 5 kJ / h
e. The heat loss leads to a pumping cost for the additional heating fluid and a greater heating cost to raise the additional fluid back to 215oC. f. Adding the insulation reduces the costs given in part (e). The insulation is probably preferable since it is a one-time cost and the other costs continue as long as the process runs. The final decision would depend on how long it would take for the savings to make up for the cost of buying and installing the insulation. 8.65 (a) Basis: 100 g of mixture, SGBenzene=0.879: SGToluene=0.866 ntotal Vtotal
dx i f
50 g 50 g mol (0.640 0.542) mol 1183 . 78.11 g / mol 92.13 g / mol 50 g 50 g 114.6 cm3 3 0.879 g / cm 0.866 g / cm3
C6H 6
0.640 mol C 6 H 6 1.183 mol
Actual feed: T
0.541 mol C 6 H 6 mol
32.5 m3 106 cm3 1183 . mol mixture 1h 9319 . mol / s 3 3 h 1 m 114.6 cm mixture 3600 s
90q C pC 6 H 6
Raoult' s law: ptot
1021 mm Hg , pC 7 H 8
407 mm Hg (from Table 6.1-1)
x C6 H 6 pC 6 H 6 x C 7 H 8 pC 7 H 8 739.2 mmHg
1 atm 760 mmHg
8-39
b0.541gb1021g b0.459gb407g 0.973 atm P0 ! 0.973 atm
8.65 (cont’d) 75q C pC 6 H 6
(b) T
Raoult's law ptank
648 mm Hg , pC 7 H 8
244 mm Hg (from Table 6.1-1)
b0.439gb648g b0.561gb244g
xC6 H 6 pC 6 H 6 xC7 H 8 pC 7 H 8
b284 137gmm Hg = 421 mmHg P 284 mm Hg 0.675 mol C H bv g mol 421 mm Hg
0.554 atm
tank
yC 6 H 6
6
6
n v (mol/s), 75°C 0.675 C 6H 6 (v ) 0.554 atm 0.325 C 7H 8 (v ) n L (mol/s), 75°C 0.439 C6 H6 (l ) 0.541 C7H8 (l )
93.19 mol/s 0.541 C 6H 6( l ) 0.459 C 7 H 8 (l ) 90°C, P0 atm
UV W
nv Mole balance: 9319 . = nv n L nL C 6 H 6 balance: 0.541 9319 . 0.675nv 0.439n L
b
gb
g
bg
40.27 mol vapor s 52.92 mol liquid s
bg
(c) Reference states: C 6 H 6 l , C 6 H 6 l at 75q C H in n out H out Substance n in C6 H 6 v 27.18 310 . C6 H 6 l 50.41 2.16 23.23 0 13.09 35.3 C7 H8 v C7 H8 l 42.78 2.64 29.69 0
bg bg bg bg
n in mol s H in kJ mol
. gb90 75g 2.16 kJ mol b g b0144 C H bl , 90q Cg: H b0176 . gb90 75g 2.64 kJ mol C H bv , 75q Cg: H b0144 . gb801 . 75g 30.77 z A C 6 H 6 l , 90q C : H 7
8
6
6
75
b
'H v 80 .1qC
g
80.1
0.074 0.330 u 103 T dT
310 . kJ mol
b
g
C 7 H 8 v , 75q C : H
. gb110.6 75g 33.47 z b0176
75
110.6
0.0942 0.380 u 10 3 T dT
35.3 kJ mol Energy balance: Q
'H
¦ n H ¦ n H i
out
i
i
in
i
1082 kJ 1 kW 1082 kW s 1 kJ s
(d) The feed composition changed; the chromatographic analysis is wrong; the heating rate changed; the system is not at steady state; Raoult’s law and/or the Antoine equation are only approximations; the vapor and liquid streams are not in equilibrium. (e) Heat is required to vaporize a liquid and heat is lost from any vessel for which T>Tambient. If insufficient heat is provided to the vessel, the temperature drops. To run the experiment isothermally, a greater heating rate is required.
8-40
8.66 a. Basis: 1 mol feed/s n V mo l vapor/s @ T, P 1 mo l/s @ TFo C
y mol A/mol (1-y) mol B/mo l
xF mol A/mol (1-xF) mol B/mo l
n L mol vapor/s @ T, P
vapor and liquid streams in equilibrium
x mol A/mol (1-x) mol B/mo l
bg b g bg bT g y x p bT g
Raoult's law x p A T 1 x p B T pA
yP
x
p A
b x gb1g F
Energy balance: 'H
bg bT g p bT g
P p B T p A
(1)
B
A
(2)
P
Mole balance: 1 n L nV nV A balance:
P x
1 n L
(4)
for nv from (4) y nV x n L Substitute o n L
¦ n H ¦ n H i
out
i
i
y xF yx
(3) (5)
0
i
in
b. ref(deg.C) = 25 Compound n-pentane n-hexane
A B C 6.85221 1064.63 232.000 6.87776 1171.53 224.366
xF Tf(deg.C) P(mm Hg) HAF(kJ/mol) HBF(kJ/mol)
0.5 110 760 16.6 18.4
0.5 110 1000 16.6 18.4
0.5 150 1000 24.4 27.0
T(deg.C) pA*(mm Hg) pB*(mm Hg) x y nL(mol/s) nV(mol/s) HAL(kJ/mol) HBL(kJ/mol) HAV(kJ/mol) HBV(kJ/mol) DH(kJ/s)
51.8 1262 432 0.395 0.656 0.598 0.402 5.2 5.8 31.4 42.4 0.00
60.0 1609 573 0.412 0.663 0.648 0.352 6.8 7.6 32.5 43.7 0.00
62.3 1715 617 0.349 0.598 0.393 0.607 7.3 8.0 32.8 44.1 0.00
al 0.195 0.216
8-41
av 0.115 0.137
bv 3.41E-04 4.09E-04
Tbp 36.07 68.74
DHv 25.77 28.85
8.66 (cont’d) c. C* C* 1 C* C* C* C* 2
20 25 3 30
PROGRAM FOR PROBLEM 8.66 IMPLICIT REAL (N) READ (5, 1) A1, B1, C1, A2, B2, C2 ANTOINE EQUATION COEFFICIENTS FOR A AND B FORMAT (8F10.4) READ (5, 1) TRA, TRB ABRITARY REFERENCE TEMPERATURES (DEG.C.) FOR A AND B READ (5, 1) CAL, TBPA, DHVA, CAV1, CAV2 READ (5, 1) CBL, TBPB, DHVB, CBV1, CBV2 CP(LIQ, KS/MBL-DEG.C.), NORMAL BOILING POINT (DEG.C), HEAT OF VAPORIZATION (KJ/MOL), COEFFICIENTS OF CP(VAP., KJ/MOL-DEG.C) CV1 CV2*T(DEG.C) READ (5, 1) XF, TF, P MOLE FRACTION OF A IN FEED, FEED TEMP.(DEG.C), EVAPORATOR PRESSURE (MMHG) WRITE (6, 2) TF, XF, P FORMAT (1H0, 'FEEDbATb', F6.1, 'bDEG.CbCONTAINSb', F6.3,' bMOLESbA/MOLEbT *OTAL'//1X'EVAPORATORbPRESSUREb ', E11.4, 'bMMbHG'/) ITER 0 DT 0.5 HAF CAL*(TF – TRA) HBF CBL*(TF – TRB) F1 XF*HAF (1.0 – XF)*HBF F2 CAL*(TBPA – TRA) DHVA – CAV1*TBPA – 0.5*CAV2*TBPA**2 F3 CBL*(TBPB – TRB) DHVB – CBV1*TBPB – 0.5*CBV2*TBPB**2 T TF INTER ITER 1 IF(ITER – 200) 30, 30, 25 WRITE (6, 3) FORMAT (1H0, 'NO CONVERGENCE') STOP PAV 10.0** (A1 – B1/(T C1)) PAV 10.0** (A2 – B2/(T C2)) XL (P – PBV)/(PAV – PBV) XV XL*PAV/P NL (XV – XF)/(XV – XL) NV 1.0 – NL IF (XL.LE.00.OR.XL.GE.1.0.OR.NL.LE.0.0.OR.NL.GE.1.0) GO TO 45 HAL CAL*(T – TRA) HBL CBL*(T – TRB) HAV F2 CAV1*T 0.5*CAV2*T**2 HBV F3 CBV1*T 0.5*CBV2*T**2
8-42
8.66(cont’d)
DELH NL *(XL*HAL (1.0 – XL)*HBL) NV*(XV*HAV (1.0 – XV)*HBV) – F1 WRITE (6, 4) T, NL, NV, DELH 4 FORMAT (1Hb, 5X' Tb ', F6.1, 3X' NLb ', F7.4, 3X' NVb ', F7.4, 3X'DELHb ',* E11.4) WRITE (6, 5) PAV, PBV, XL, HAL, HBL, XV, HAV, HBV 5 FORMAT (1Hb, 5X' PAV, PBVb ', 2F8.1, 3X' XL, HAL, HBLb ', F7.4, 2E13.4,3X' XV, HAV, HBVb ', F7.4, 2E13.4/) IF (DELH) 50, 50, 40 40 DHOLD DELH TOLD T 45 T T – DT GO TO 20 50 T (T*DHOLD – TOLD*DELH)/(DHOLD – DELH) PAV 10.0**(A1 – B1/(T C1)) PBV 10.0**(A2 – B2/(T C2)) XL (P – PBV)/(PAV – PBV) XV XL * PAV/P NL (XV – XF)/(XV – XL) NV 1.0 – NL WRITE (6, 6) T, NL, XL, NV, XV 6 FORMAT (1H0, 'PROCEDUREbCONVERGED'//3X'EVAPORATORb TEMPERATUREb ', F6. *1//3X' LIQUIDbPRODUCTb--', F6.3, 'bMOLEbCONTAININGb', F6.3, 'bMOLEbA/ *MOLEbTOTAL'//3X' VAPORbPRODUCTb--', F6.3, MOLEbCONTAININGb,' F6.3, *'bMOLEbA/MOLEb TOTAL') STOP END $DATA (Fields of 10 Columns) . 6.85221 1064.63 232.0 6.87776 117153 224.366 25.0 25.0 . . 0195 36.07 25.77 0115 0.000341 . 0.216 68.74 28.85 0137 0.000409 0.500 110.0 760.0 Solution: Tevaportor 52.2q C
d
i
d
i
nL
0.552 mol, x C5 H12
nv
0.448 mol, x C5 H12
liquid
vapor
0.383 mol C 5 H 12 mol liquid 0.644 mol C 5 H 12 mol liquid
8-43
8.67 Basis:
2500 kmol product 1 kmol condensate 10,000 kmol h fed to condenser h .25 kmol product
.
m1 , (kg/h) at T1 1090 kmol/h C 3 H 8 (v ) 7520 kmol/h i -C 4H10 (v ) 1390 kmol/h n -C 4H10 (v ) saturated vapor at Tf, P
1090 kmol/h C 3 H 8 ( l ) 7520 kmol/h i -C 4H10 ( l ) P (mm Hg) 1390 kmol/h -C H ( ) n 4 10 l Tout
.
m1 (kg/h) at 2 T2
0 o C , T1
(a) Refrigerant: Tout Antoine constants C3H 8 i C 4 H 10 n C 4 H 10
A 7.58163 6.74808 6.83029
Calculate P for Tout P
6 o C .
T2
B 1133.65 882.8 945.0
C 283.26 240.0 240.0
Tbubble pt.
¦ xi pi* b0q Cg
b
g
b
g
b
g
0109 . 3797 mm Hg 0.752 1174 mm Hg 0139 . 774 mm Hg
i
P 1404 mm Hg
d i
Tdp f Tf
Dew pt. Tf
1 P¦ i
d i
Tf
yi *
d i
Pi Tf
0 trial & error to find Tf
f Tf
5q C 0.00162 linear interpretation Tf 4q C 0.0333
bg
4.95q C
bg
Refs: C 3 H 8 l , C 4 H 10 l at 0 qC, Refrigerant @ –6qC
b g
Assume: 'H v Tb , Table B.1 substance C3H 8 i C 4 H 10 n C 4 H 10
Refrigerant
E.B.: 'H
nin
1090 19110 7520 21740 1390 22760 m 1
¦ ni H i ¦ ni H i out
H in
in
0
p
nout
H out
1090 7520 1390
0 0 0
n (kmol/h) H (kJ/kmol)
m 1
151
m (kg/h) H (kJ/kmol)
2
z
v
4 .95
p
0
0 151m 1 2.16 u 106
8-44
U| H bvaporg 'H b0q Cg V| C dTbTable B.2g W
0 m 1
UV H W
'H v
143 . u 106 kg h refrigerant
8.67 (cont’d) 40q C , T2
(b) Cooling water: Tout
¦ xi pi* b40q Cg
P
b
g
25qC
b g
b g
0109 . 11,877 0.752 3983 0139 . 2853
i
4653 mm Hg
TE
d i 1 P¦ p dyT i 0 T 45.7q C C H bl g , C H bl g @ 40qC, H Obl g @ 25qC. i
f Tf
Refs:
34qC , T1
3
i
8
'H
* i
4
f
f
10
2
0 37.7m 1 2.17 u 10
8
0 m 1
5.74 u 10 6 kg H 2 O / h
(c) Cost of refrigerant pumping and recompression, cost of cooling water pumping, cost of maintaining system at the higher pressure of part (b). 8.68 Basis: 100 mol leaving conversion reactor H 2 O(v ) 3.1 bars, sat'd
n 3 (mol O 2 ) 3.76 n 3 (mol N 2 )
H 2 O( l ) 45°C
conversion 100 mol, 600°C, 1 atm 145°C 100°C reactor 0.199 mol HCHO/mol n 4 (mol H 2 O( v)) 0.0834 mol CH 3OH/mol 0.303 mol N 2/mol n 1 (mol CH3 OH(l )) n 2 (mol CH 3 OH(l )) 0.0083 mol O 2/mol m w1 (kg H 2 O(l )) m w2 (kg H 2 O(l )) 0.050 mol H 2/mol n 8 (mol CH 3 OH(l )) 30°C 0.356 mol H 2O( v)/mol 3.1 bars, sat'd Q (kJ) CH 3 OH( l ), 1 atm, sat'd n 6a (mol HCHO) 2.5n 8 (mol CH 3 OH) distillation absorption n 6b (mol CH 3 OH( l )) (l ) n 6c (mol H 2 O( l )) sat'd, 1 atm Product solution 88°C, 1 atm n 7 (mol) m w3 (kg H 2 O(l )) Absorber off-gas 0.37 g HCHO/g (x 1 mol/min) n 5a (mol N 2 ) 0.01 g CH 3OH/g (x 2 mol/min) 30°C n 5b (mol O 2 ) 0.82 g H 3 O/g (x 3 mol/min) n 5 c (mol H 2 ) n 5d (mol H 2 O(v )), sat'd n 5 e (mol HCHO(v )), 200 ppm 27°C, 1 atm
a. Strategy C balance on conversion reactor n2 , N 2 balance on conversion reactor n3 H balance on conversion reactor n4 , (O balance on conversion reactor to check consistency) N 2 balance on absorber n5a , O 2 balance on absorber n5b H 2 balance on absorber n5e
UV W
H 2 O saturation of absorber off - gas n5d , n5b 200 ppm HCHO in absorber off - gas
8-45
8.68 (cont’d) HCHO balance on absorber n6a , CH 3 OH balance on absorber n6b Wt. fractions of product solution x1 , x 2 , x 3 HCHO balance on distillation column n7 CH 3 OH balance on distillation column n8 CH 3 OH balance on recycle mixing point n1 Energy balance on waste heat boiler mw1 , E.B. on cooler mw2 Energy balance on reboiler Q C balance on conversion reactor: n2 19.9 mol HCHO 8.34 mol CH 3 OH
28.24 mol CH 3 OH
N 2 balance on conversion reactor: 3.76n3
30.3 n3
8.06 mol O 2 , 3.76 u 8.06 30.3 mol N 2 feed
H balance on conversion reactor:
bg
bg
bg
bg bg
bg
n4 2 28.24 4 19.9 2 8.34 4 5 2 35.6 2 n4
20.7 mol H 2 O fed
O balance: 65.1 mol O in, 65.5 mol O out. Accept (precision error) N 2 balance on absorber: 30.3 n5a n5a
30.3 mol N 2
O 2 balance on absorber: 0.83 n5b n5b
0.83 mol O 2
H 2 balance on absorber: 5.00 n5c n5c
5.00 mol H 2
H 2 O saturation of off - gas: yw
b
p w* 27q C
n5d
P
g LM 26.739 mm Hg N 760 mm Hg
n5d 30.3 0.83 5.00 n5d n5e
g U| | 200 ppm HCHO in off gas: V| n 200 2| |W . n n 3613 10 b
OP Q
0.03518 3613 . n5d n5e 1
solve
n5d n5e
mol H 2 O 1318 . 7.49 u 10 3 mol HCHO
5e
5d
Moles of absorber off-gas
6
5e
n5a n5b n5c n5e
HCHO balance on absorber: 19.9
37.46 mol off - gas
n6a 7.49 u 10 3 n6a 19.89 mol HCHO
CH 3 OH balance on absorber: 8.34
n6b n6b
Product solution
8.34 mol CH 3 OH
U| |V || W
%MW
x1 . Basis - 100 g 37.0 g HCHO 1232 mol HCHO 1.0 g CH 3 OH 0.031 mol CH 3 OH x 2 x3 62.0 g H 2 O 3.441 mol H 2 O
8-46
0.262 mol HCHO mol 0.006 mol CH 3 OH mol 0.732 mol H 2 O mol
8.68 (cont’d) HCHO balance on distillation column (include the condenser + reflux stream within the system for this and the next balance): 0.262n7 n7
19.89
75.9 mol product
CH 3 OH balance on distillation column:
b g
0.006 75.9 n8 n8
8.34
7.88 mol CH 3 OH
CH 3 OH balance on recycle mixing point: n1 n8
n2 n1
28.24 7.83 20.36 mol CH 3 OH fresh feed
Summary of requested material balance results:
bg
n1
20.4 mol CH 3 OH l fresh feed
n2
75.9 mol product solution
n3
7.88 mol CH 3 OH l recycle
n4
37.5 mol absorber off - gas
bg
Waste heat boiler:
b
g
b
g
bg
Refs: HCHO v, 145q C , CH 3 OH v, 145q C ; N 2 , O 2 , H 2 , H 2 O v at 25qC for product
b
g
gas, H 2 O l, triple point for boiler water substance
nin
H in
nout
H out
HCHO CH 3 OH
19.9 8.34 30.3 0.83 5.0 35.6
22.55 32.02 17.39 18.41 16.81 20.91
19.9 8.34 30.3 0.83 5.0 35.6
0 n (mol) 0 3.51 3.60 H (kJ/mol) 3.47 4.09
mw1
566.2
mw1 2726.3 m (kg) 2 H (kJ/kg)
N2 O2 H2 H2O H2O (boiler)
E.B. 'H
¦ n H ¦ n H i
out
i
i
i
0 1814 2160mw1
in
8-47
UV H C dT W U| |V H C bT g T 25 || W UV H from steam tables W
z T
p
145
p
0 mw1
0.84 kg 3.1 bar steam
8.68 (cont’d)
b
g
Gas cooler: Same refs. as above for product gas, H 2 O l, 30q C for cooling water substance
nin
H in
nout
H out
HCHO CH 3 OH
19.9 8.34 30.3 0.83 5.0 35.6
0 0 3.51 3.60 3.47 4.09
19.9 8.34 30.3 0.83 5.0 35.6
–1.78 –2.38 2.19 2.24 2.16 2.54
mw2
0
mw2
62.76
N2 O2 H2 H2O H2O (coolant)
¦ n H ¦ n H
E.B. 'H
i
i
i
out
i
b.
Q
m (kg) H (kJ/kg)
0 1581 . 62.6mw 2
H
4.184
0 mw 2
b
g
kJ T 30 q C kgq C
2.52 kg cooling water
in
b35. gb7.88g n'H b1 atmg b27.58 molgb35.27 kJ molg
Condenser: CH 3 OH condensed E.B.:
n (mol) H (kJ/mol)
n8 2.5n8
27.58 mol CH 3 OH condensed
v
973 kJ (transferred from condenser)
3.6 u 10 4 tonne / y
10 6 g 1 yr 1d 1 metric ton 350 d 24 h
b0.37gd4.286 u 10 i b0.01gd4.286 u 10 i b0.62gd4.286 u 10 i
4.286 u 10 6 g h product soln
6
u 10 6 g HCHO h 5.281 u 10 4 mol HCHO h 1586 .
6
4.286 u 10 6 g CH 3 OH h 1338 mol CH 3 OH h
6
u 105 mol H 2 O h 2.657 u 10 6 g H 2 O h 1475 .
2.016 u 10 5 mol h Scale factor =
2.016 u 10 5 mol h 75.9 mol
U| V| |W
2657 h 1
8.69 (a) For 24°C and 50% relative humidity, from Figure 8.4-1, Absolute humidity = 0.0093 kg water / kg DA, Humid volume | 0.856 m 3 / kg DA Specific enthalpy = (48 - 0.2) kJ / kg DA = 47.8 kJ / kg DA , Dew point = 13o C, Twb
17 o C
(b) 24 o C (Tdb ) (c) 13o C (Dew point) (d) Water evaporates, causing your skin temperature to drop. Tskin | 13o C (Twb ). At 98% R.H. the rate of evaporation would be lower, Tskin would be closer to Tambient , and you would not feel as cold.
8-48
8.70
141 ft 3 . DA = dry air.
V room mDA = ha
140 ft 3
lb - mol o R 29 lb m DA 1 atm lb - mol 550 o R 0.7302 ft 3 atm
0.205 lb m H 2 O 101 . lb m DA
0.0203 lb m H 2 O / lb m DA 90 o F, ha
From the psychrometric chart, Tdb hr
Tdb
35q C
Tab
27q C
8.72 a. Tdb
hr
b. Mass of dry air: mda
c.
80.5o F
14.3 ft 3 / lb DA V m
H
44.0 011 . # 43.9 Btu / lb m
33%, ha
0.0148 kg H 2 O kg dry air
77.3o F
55% He wins
40q C, Tdew point
Mass of water:
0.0903
Twb
67%
Tdew point
8.71
101 . lb m DA
Fig. 8.4-1
20q C
hr Twb
255 . qC
1 m 3 1 kg dry air 2.2 u 10 3 kg dry air 10 3 L 0.92 m 3 n from Fig. 8.4-1
2.00 L
2.2 u 10 3 kg dry air 0.0148 kg H 2 O 10 3 g 1 kg dry air 1 kg
0.033 g H 2 O
77.4 kJ kg dry air b g b g H b20q C, saturated g | 57.5 kJ kg dry air (both values from Fig. 8.4-1) 2.2 u 10 kg dry air b57.5 77.4g kJ 10 J 'H 44 J H 40q C, 33% relative humidity | 78.0 0.65 kJ kg dry air
3
3
40o 20
kg dry air
1 kJ
d. Energy balance: closed system 2.2 u 10 3 kg dry air 10 3 g 1 mol 0.033 g H 2 O 1 mol 18 g 1 kg 29 g Q 'U n'U n 'H R'T 'H nR'T n
d
= 44 J
i
0.078 mol 8.314 J mol K
b20 40gq C
8-49
1K 1q C
0.078 mol
31 J(23 J transferred from the air)
8.73 (a)
400 kg 2.44 kg water 97.56 kg air 10 kg H 2 O min 400 kg dry air min
min
(b) ha
Fig. 8.4-1
(c) Tdb
(d)
10.0 kg water evaporates / min 0.025 kg H 2 O kg dry air , Tdb
b116 11. g
H
115 kJ kg dry air , Twb
400 kg dry air min
33q C, hr
0.0077 kg H 2 O kg dry air , H
10q C , saturated ha
b0.0250 0.0077g kg H O 2
50q C
32%, Tdew point
28.5q C
29.5 kJ kg dry air
6.92 kg H 2 O min condense
kg dry air
bg
References: Dry air at 0q C, H 2 O l at 0q C substance m in m out H in H out Air
bg
H 2O l
b
400
115
400
29.5
—
—
4.1
42
g
b
g
m air in kg dry air/min, m H 2 O in kg/min
H air in kJ/kg dry air, H H 2 O in kJ/kg
H 2 O l , 0q C o H 2 O l , 20q C : 75.4
H Q
'H
J 1 mol molq C 18 g
¦ m i H i ¦ m i H i out
(e)
in
b10 0gq C
1 kJ 103 g 42 kJ kg 103 J 1 kg 34027.8 kJ 1 min 1 kW 567 kW min 60 s 1 kJ / s
T>50°C, because the heat required to evaporate the water would be transferred from the air, causing its temperature to drop. To calculate (Tair)in, you would need to know the flow rate, heat capacity and temperature change of the solids.
8.74 a. Outside air: Tdb 87q F , hr 80% ha 0.0226 lb m H 2 O lb m D.A. , H 455 . 0.01 455 . Btu lb m D.A. Room air: Tdb 75q F , hr 40% ha 0.0075 lb m H 2 O lb m D.A. , H 26.2 0.02 26.2 Btu lb m D.A. Delivered air: Tdb 55q F , ha 0.0075 lb m H 2 O lb m D.A. H 214 . 0.02 214 . Btu lb m D.A. , V 13.07 ft 3 lb m D.A. 1,000 ft 3 1 lb m D.A. 76.5 lb m D.A. min Dry air delivered: min 13.07 ft 2 H 2 O condensed: 76.5 lb m D.A. min
b0.0226 0.0075g lb
H 2O lb m D.A.
8-50
m
12 . lb m H 2 O min condensed
8.74 (cont’d) The outside air is first cooled to a temperature at which the required amount of water is condensed, and the cold air is then reheated to 75qF. Since ha remains constant in the second step, the condition of the air following the cooling step must lie at the intersection of the ha 0.0075 line and the saturation curve T 49q F
b
g
References: Same as Fig. 8.4-2 [including H 2 O l, 32q F ] substance m in H m out H in
Air
b
g
H 2 O l, 49q F
out
76.5 45.5 76.5 21.4 m air in lb m D.A./min —
—
1.2 17.0 H in Btu/ lb D.A. air m m in lb /min, H H 2O
Q
'H
b76.5g 214. 455. + 1.2(17.0) (Btu) min
m
H 2O
in Btu/ lb m
60 min 1 ton cooling 12,000 Btu h 1h
9.1 tons cooling b. Examination of the flow chart shows that the overall energy balance must be the same as the system without recirculation, except for all extensive variables being scaled down by a factor of 1/7. (Air must be delivered to the room at the same rate, and 6/7 of it is recycled.) Consequently Net cooling: Q
9.1 tons / 7 1.3 tons
6 u 100 86% 7 Once the system reaches steady state, most of the air passing through the conditioner is cooler than the outside air, and (more importantly) much less water must be condensed (only the water in the fresh feed). Percent Saved:
c. Total recirculation could eventually lead to an unhealthy depletion of oxygen and buildup of carbon dioxide in the laboratory. 8.75 Basis: 1 kg wet chips. DA = dry air, DC = dry chips Outlet air: Tdb=38oC, Twb=29oC m2a (kg DA) m2w [kg H2O(v)]
Inlet air: 11.6 m3(STP), Tdb=100oC m1a (kg DA)
1 kg wet chips, 19oC 0.40 kg H2O(l)/kg 0.60 kg DC/kg
m3c (kg dry chips) m3w [kg H2O(l)] T (oC)
(a) Dry air: m1a =
Outlet air: Tdb 38q C, Twb
b
b g
11.6 m 3 STP DA
1 kmol 22.4 m 3 STP
b g
g
8.4-1 o H 2 29q C Fig.
8-51
29.0 kg 15.02 kg DA = m 2a 1 kmol
95.3 kJ kg D.A. , ha2
0.0223 kg H 2 O kg D.A.
8.75 (cont’d) Water in outlet air: m2 w
b
ha2 m2 a
0.0223 15.02
g
0.335 kg H 2 O
(b) H 2 O balance: 0.400 kg = 0.335 kg + m3w m3w
0.065 kg H 2 O
Moisture content of exiting chips: 0.065 kg water u 100% (1 0.335) kg chips
9.8% 15% ? meets design specification
bg
(c) References: Dry air, H 2 O l , dry chips @ 0qC. substance
min
H in
Air H 2O l dry chips
15.02 0.400 0.600
100.2 79.5 39.9
bg
H out
mout
15.02 95.3 mair in kg DA, H air in kJ/kg DA 0.065 4.184T m in kg DC, H in in kJ/kg DC 0.6 2.10T
Energy Balance: 'H Tdb 45q C hr 10% b. Twb 210 . qC hr 60%
¦m
8.76 a.
H 2 O added:
8.77
¦m
Tas
Twb
out H out
Fig. 8.4-1
Fig. 8.4-1
Tdb
in H in
0 129.3 1532 . T
210 . qC
26.8q C
15 kg air 1 kg D.A. min 1.0059 kg air
Inlet air: Tdb 50q C Tdew pt. 4q C
Fig. 8.4-1
V ha
ha
ha
0T
84.4q C
0.0059 kg H 2 O kg DA
0.0142 kg H 2 O kg DA
b0.0142 0.0059g kg H O 2
1 kg D.A.
0.92 m 3 kg D.A. , Twb
012 . kg H 2 O min
22q C
0.0050 kg H 2 O kg D.A.
11.3 m 3 1 kg D.A. 12.3 kg D.A. min min 0.92 m3 Outlet air: Twb
Tas
22q C
saturated
Evaporation:
12.3 kg D.A. min
T
22q C ha
0.0165 kg H 2 O kg D.A.
b0.0165 0.0050g kg H O 2
kg D.A.
8-52
014 . kg H 2 O min
8.78 a. Tdb 45q C Tdew point 4q C Twb
UV W
bh g
0.0050 kg H 2 O kg D.A.
a in
20.4q C, V
Twb
Fig. 8.4-1
b g
20.4q C, saturated ha
Tas
0.908 m 3 kg D.A.
0.0151 kg H 2 O kg D.A.
out
b. Basis: 1 kg entering sugar (S) solution m1 (kg D.A.) 0.0050 kg H2O/kg DA
m1 (kg D.A.) 0.0151 kg H2O(v)/kg m2 (kg) 0.20 kg S/kg 0.80 kg H2O/kg
1 kg 0.05 kg S/kg 0.95 kg H2O/kg
b gb g b0.20gm m 0.25 kg Water balance: bm gb0.0050g b1gb0.95g bm gb0.0151g b0.25gb0.80g Sugar balance: 0.05 1
2
2
1
1
m1
8.79
74 kg dry air
B
A 1 lb m D.A. ha1 (lb m H 2O) T d = 20°F h r = 70%
Inlet air (A):
C
D
Spray 1 lb m D.A. 1 lb m D.A. ha2(lb m H 2O) chamber h a3(lb m H 2O) T d = 75°F H2 O
Coil bank
Tdb 20q F hr 70%
Outlet air (D):
0.908 m 3 67 m 3 1 kg D.A.
74 kg dry air
V
UV W
Tdb 70q F hr 35%
Fig. 8.4-2
a. Inlet of spray chamber (B):
1 lb m D.A. h a3(lb m H 2O) T d = 70°F h r = 35%
ha1 | 0.0017 lb m H 2 O lb m D.A. V | 12.2 ft 3 lb m D.A.
Fig. 8.4-2
UV W
Coil bank
ha 3
0.0054 lb m H 2 O lb m D.A.
UV W
ha 0.0017 lb m H 2 O lb m D.A. Twb Tdb 75q F
49.5q F
The state of the air at (C) must lie on the same adiabatic saturation curve as does the state at (B), or Twb 49.5q F . Thus, Outlet of spray chamber (C): At point C, Tdb
b.
bh
a3
UV W
ha 0.0054 lb m H 2 O lb m D.A. hr Twb 49.5q F
52%
58.5q F
g
ha1 lb m H 2 O evaporate lb m DA
lb m DA V A ft 3 inlet air
d
8-53
i
b0.0054 0.0017g 12.2
3.0 u 10 4
lb m H 2 O ft 3 air
8.79 (cont’d) c. QBA QDC
(20 - 6.4) Btu / lb m dry air H B H A # = 1.1 Btu / ft 3 12.2 ft 3 / lb m dry air (23 - 20) Btu / lb m dry air 'H H D H C # = 0.25 Btu / ft 3 3 12.2 ft / lb m dry air
'H
d. 70% 52% 35%
C
D
A
B
58.5
20
70
75
8.80 Basis: 1 kg D.A. a.
1 kg D.A. ha1(kg H 2 O/kg D.A.) Tdb = 40°C, Tab = 18°C
1 kg D.A. ha2(kg H 2 O/kg D.A.) 20°C,
m w kg H2 O
Tdb
40q C
ha1 0.0039 kg H 2 O kg D.A. Twb 18q C Tdb 20q C ha 2 Outlet air: Twb 18q C adiabatic humidification
Inlet air:
b
g
0.0122 kg H 2 O kg D.A.
b gb g b1gbh g m b0.0122 0.0039gkg H O kg D.A.
Overall H 2 O balance: mw 1 ha1
a2
n
0.0083 kg H 2 O kg D.A. b.
ma (lb m H2 O/h) T=15o C, sat’d
1250 kg/h T=37o C, h r=50%
mc (lb m H2O/h) liquid, 12°C Qc (Btu/h)
8-54
2
8.80 (cont’d) Inlet air:
Tdb 37q C hr 50%
Moles dry air: m a Outlet air: Tdb
RSh 0.0198 kg H O kg DA TH b88.5 - 0.5g kJ kg DA 88.0 kJ kg DA
Fig. 8.4-1
a1
2
1
1250 kg
1 kg DA
h
1.0198 kg
RSh TH
Fig. 8.4-1
15q C, sat' d
1226 kg DA h
1226 kg DA h
Overall water balance m c
0.0106 kg H 2 O kg DA
a
2
42.1 kJ kg DA
b0.0198 0.0106g kg H O 2
kg DA
113 . kg H 2 O h withdrawn
bg
Reference states for enthalpy calculations: H 2 O l , dry air at 0oC. (Cp)H2O(l) = 1
b
z
g
H 2 O l , 12q C : H
12
C p dT
kJ kg o C
50.3 kJ / kg
0
Overall system energy balance: Q c
'H
¦ m H ¦ m H i
i
out
i
i
in
LM113. kg H O 50.3 kJ 1226 kg DA b42.1 88g kJ OPFG 1 h IJ FG 1 kW IJ h kg DA QH 3600 s K H 1 kJ / s K N h kg H O 2
2
155 . kW 'H
8.81 8.82 a.
b.
400 mol NH 3
78.2 kJ
31,280 kJ
mol NH 3
b
g
b
g
b
g
HCl g , 25q C , H 2 O l , 25q C o HCl 25q C, r 5 . B.11 'H 'H s 25q C, r 5 Table o 'H 64.05 kJ mol HCl
b
g HClbaq, r = fg o HClbr 5g, H Obl g 'H 'H b25q C, n 5g 'H b25q C, n fg . g kJ mol HCl 1109 . kJ / mol HCl b64.05 7514 2
s
s
8-55
8.83 Basis: 100 mol solution 20 mol NaOH, 80 mol H2O 80 mol H 2 O 20 mol NaOH
r
4.00 mol H 2 O mol NaOH
bg
Refs: NaOH(s), H 2 O l @25q C
bg bg b
b
H NaOH, r
g
g
out
H out
nout
n in mol H in kJ mol
20.0 34.43 m n in mol NaOH
34.43 kJ mol NaOH (Table B.11)
4.00
¦ ni H i ¦ ni H i
'H
H in 0.0 0.0
nin 20.0 80.0
substance NaOH s H2O l NaOH r 4.00
688.6 kJ 9.486 u 104 Btu 10 3 kJ
(20)( 34.43)
in
653.2 Btu 103 g 20.0 40.00 80.0 18.01 g 2.20462 lb m
Q
b
g
b
653.2 Btu
132.3 Btu lb m product solution
g
8.84 Basis: 1 liter solution 1 L 8 g - eq
n H 2SO4
L L
1000 mol H 2 O 18.02 kg H 2 O
2
n H 2O
r
46.49 mol H 2 O 4 mol H 2 SO 4
n H 2SO 4
d
FG 0.09808 kg IJ H 1 mol K
0.392 kg H 2 SO 4
1.230 kg solution
0.392gkg H O . b1230
n H 2O
4 mol H 2 SO 4 u
2 g - eq
1 L 1.230 kg
mtotal
1 mol
i
11.6
46.5 mol H 2 O
mol H 2 O mol H 2 SO 4
d
i
d
H 2 SO 4 aq, r = f,25o C o H 2 SO 4 aq, r = 11.6, 25o C + H 2 O l , 25o C 'H 1
'H s (r 116 . ) 'H s (r
f)
LMn N H b H SO , r 116 . , 60q Cg n RS4 mol H SO 1 4 mol H SO T
Table B.11
H 2SO 4 'H1
2
4
2
2
4
H 2SO 4
4
m
( 67.6 9619 . )
z
60
25
OP Q
28.6 kJ 1.230 kg mol H 2 SO 4
8-56
kJ mol H 2 SO 4
C p dT kJ
( mol H 2 SO 4 )
60.9 kJ mol H 2 SO 4
28.6
i
3.00 kJ kgq C
b60 25gq CUV W
d
a.
4.67 mol H 2 O r
4.67 2
2.33
mol H 2 O mol H 2SO 4
For this closed constant pressure system,
b
b. msolution
g
nH 2SO 4 'H s 25q C, r
'H
Q
2 mol H 2SO 4
2.33
mol H 2SO 4
mol
b
g b280.6 150gg 88.6 kJ +
Basis:
e j
mol HCl
z
88.6 kJ
4.67 mol H 2 O 18.0 g H 2 O mol T
25
C p dT
3.3 J
280.2 g
0
bT 25gq C
gq C
1 kJ 1000 J
0T
1120 g solution
L L
2.33 m
1 L product solution 1.12 103 g
1 L 8 mol HCl 36.47 g HCl
44.28 kJ
2 mol H 2SO 4
98.08 g H 2SO 4
0 nH 2SO 4 'H s 25q C, r
'H
8.86 a.
i
0.30 2.00 nH 2 O nH 2 O
8.85 2 mol H 2SO 4
292 g HCl
46.0 mol H2O(l, 25°C)
8.0 mo l HCl(g , 20°C, 790 mm Hg)
1120 g 292 g 828 g H 2 O
828 g H 2 O
mol
46.0 mol H 2 O
18.0 g n
1 L HCl (aq)
46.0 mol H 2 O 8.0 mol HCl
5.75 mol H 2 O mol HCl
Assume all HCl is absorbed Volume of gas:
b g
8 mol 293 K 760 mm Hg 22.4 L STP 273 K 790 mm Hg
mol
b g
185 liter STP gas feed L HCl solution
b. Ref: 25q C H in
nin
H 2O l
46.0
0.0
HCl g
8.0
0.15
8.0
59.07
b
HCl n
bg bg
g
5.75
nout
H out
substance
n in mol H in kJ mol
8-57
87q C
8.87 (cont’d)
b
g
H HCl, n
b
g
'H s 25q C, n
5.75
5.75
64.87 kJ mol
e
H HCl, 20o C
c.
Q
'H
Q
0
0.15
z
j
20
25
z
40
1 nHCl
mC p dT
25
1120 g 0.66 cal
b40 25gq C
4.184 J
kJ
cal
103 J
gq C
8 mols
0.02913 0.1341 u 10 5 T 0.9715 u 10 8 T 2 4.335 u 10 12 T 3 dT
= -0.15 kJ / mol 471 kJ L product 'H H
e j b
g
8 H 8 015 . 64.87
b
g
o 1120 g 0.66 cal T 25 C 4.184 J 1 kJ 8 mol go C cal 1000 J
192 o C
T
8.87 Basis: Given solution feed rate
.
.
n a (mol air/min) 200°C, 1.1 bars
n a (mol air/min) . n 1 (mol H 2O( v)/min) saturated @ 50°C, 1 atm
150 mol/min solution 0.001 NaOH 0.999 H 2O 25°C
b
.
n 2 (mol/min) @ 50°C 0.05 NaOH 0.95 H 2O
gb g 0.05n n 3.0 mol min H O balance: b0.999gb150g n 0.95b3.0g n 147 mol H O min n P p b50q Cg Raoult’s law: y P 92.51 mm Hg n n NaOH balance: 0.001 150
2
2
2
1
1
H 2O
1
a
1
2
Table B.4
H 2O
n1 147 P 760
b g
1061 mol 22.4 L STP min 1 mol
Vinlet air
bg
n a
1061
mol air min
. 473 K 1013 bars 37,900 L min 273 K 1.1 bars
bg 'H b25q Cg
References for enthalpy calculations: H 2 O l , NaOH s , air @ 25q C 999 mol H 2 O Table B.11 1 mol NaOH
0.1% solution @ 25qC: r 5% solution @ 50qC: r Solution mass: m
b
H 50q C
g
b
95 mol H 2 O 5 mol NaOH
19 mol H 2 O mol NaOH
b
'H s 25q C
1 mol NaOH 40.0 g 19 mol H 2 O 18.0 g 1 mol 1 mol
g
'H s 25q C m 42.81
42.47 kJ mol NaOH
s
z
g
382
42.81
kJ mol NaOH
g solution mol NaOH
50
25
C p dT
382 g 4.184 J kJ mol NaOH mol NaOH 1 gq C
8-58
b50 25gq C
1 kJ 103 J
2.85 kJ
8.87 (cont’d) Air @ 200qC: Table B.8 H
515 . kJ mol
Air (dry) @ 50qC: Table B.8 H
b
b2592 104.8g kJ
g
H 2 O v , 50q C : Table B.5 H H in
0.73 kJ mol 1 kg 18.0 g 103 g 1 mol
kg
44.81 kJ mol
H out 42.47 015 2.85 n in mol min 015 . . 147 44.81 H in kJ mol 1061 515 1061 0.73 . nin
substance NaOH aq H 2O v
b g bg
Dry air
Energy balance: Q
¦ ni H i ¦ ni H i
'H
b neglect 'E g
nout
out
n
1900 kJ min transferred to unit
in
8.88 a. Basis: 1 L 4.00 molar H2SO4 solution (S.G.
1.231)
4.00 mol H 2 SO 4 1231 392.3 838.7 g H 2 O 1 L 1231 g 1231 g 392.3 g H 2 SO 4 L 46.57 mol H 2 O B.11 . mol H 2 O / mol H 2 SO 4 Table r 1164 o 'H s
b
g
b
Ref: H 2 O l , 25q C , H 2 SO 4 25q C substance H 2O l H 2 SO 4 l H 2 SO 4 25q C, n 1164 .
bg bg
b
Q
b
'H
g
nin H in 46.57 0.0754 T 25 4.00 0
b
g
g
67.6 kJ / mol H 2 SO 4
b
nout H out n in mol H in kJ mol 4.00 67.6
g
gb
g
0 4.00 67.6 46.57 0.0754 T 25 T
52q C
(The water would not be liquid at this temperature impossible alternative!)
b
g
b
b. Ref: H 2 O l , 25q C , H 2 SO 4 25q C nin nl
substance H 2O l
bg H Ob sg 2
H 2 SO 4 (l ) H 2 SO 4 25q C, n 1164 .
b
b
'H m H 2 O, 0q C
g
g
g
H in 0.0754 0 25
b
b
g
g
n s 6.01 0.0754 0 25 4.00 0
H out
nout
4.00 67.61
n in mols H in kJ mol
6.01 kJ mol
A
Table B.1
UV n 'H 0 4.00b 67.61g n b 1885 . g b46.57 n gb 7.895gW n . g H Ob"g 547.3 g H Ob sg@0q C 2914 n" n s
46.57
l
2
l
2
8-59
l s
. mol liquid H 2 O 1618 30.39 mol ice
8.89 P2 O 5 3H 2 O o 2H 3 PO 4 mol H 3 PO 4
a.
b
g u 100% ,
n 14196 .
wt% P2 O 5
mt
B 2n
wt% H 3 PO 4
g H 3 PO 4 mol
B
b98.00g u 100%
mc
A
g total
where n
total mass .
mol P2 O 5 and mt
wt% H 3 PO 4
b
g wt% P O
2 98.00 . 14196
2
1381 . wt% P2 O 5
5
b. Basis: 1 lb m feed solution 28 wt% P2 O 5 38.67 wt% H 3 PO 4 m1 (lbm H2 O(v )), T , 3.7 psia 1 lb msolution, 125°F 0.3867 lb mH 3PO 4 0.6133 lb mH 2O
m2 (lbm solution), T 0.5800 lb mH 3PO 4/lb 0.4200 lb mH 2O/lb m
H 3 PO 4 balance: 0.3867
m
0.5800m2 m2 0.667 lb m solution
Total balance: 1 m1 m2 m1
bg
0.3333 lb m H 2 O r
bg
Evaporation ratio: 0.3333 lb m H 2 O v lb m feed solution c. Condensate:
b
P 37 . psia 0.255 bar Table B.6
Tsat
m
V
g
654 . o C =149 o F, Vliq
0.00102 m3 353145 . ft 3 / m3 kg
100 tons feed 2000 lb m 1 lb m H 2 O day
46.3 lb m min
1 ton
0.0163 ft 3 lb m
2.205 lb m / kg
1 day (24 u 60) min
3 lb m
46.3 lb m / min
7.4805 gal 5.65 gal condensate / min ft 3
Heat of condensation process:
46.3lbm H2O(v)/min
46.3lbm H2O(l)/min
(149+37)°F, 3.7 psia
149°F, 3.7 psia
.
Q (Btu/min)
8-60
0.0163
ft 3 lb m H 2 O(l)
8.89 (cont’d)
R| ||H Table B.6 S ||H |T Q m 'H
F GG H
H2 O ( l ) (149
(46.3
o
b
g
F = 65.4 o C) = (274 kJ / kg) 0.4303
LM N
lb m Btu ) (118 1141) min lb m
OP Q
I JJ kJ kg K
Btu
o o H2 O ( v ) (186 F = 85.6 C) = (2652 kJ / kg) 0.4303
lb m
1141 Btu / lb m
118 Btu / lb m
47,360 Btu / min
4.74 u 10 4 Btu min available at 149 o F
bg
bg
d. Refs: H 3 PO 4 l , H 2 O l @77q F min H in mout H out m in lb m . 100 13.95 0.667 34.13 H in Btu lb m 0.3333 1099
substance H 3 PO 4 28% H 3 PO 4 42% H 2O v
b g b g bg
b
H H 3 PO 4 , 28%
b
0.705 Btu lb m q F
5040 Btu lb - mole H 3 PO 4
b125 77gq F
0.705 Btu lb m q F
H H 3 PO 4 , 42%
g g
0.3867 lb m H 3 PO 4 1.00 lb m soln
13.95 Btu lb m soln
5040 Btu lb - mole H 3 PO 4
b186.7 77gq F
1 lb - mole H 3 PO 3 98.00 lb m H 3 PO 4
1 lb - mole H 3 PO 4 98.00 lb m H 3 PO 4
0.5800 lb m H 3 PO 4 1.00 lb m sol.
34.13 Btu lb m soln
b g H b3.7psia, 186q Fg H bl, 77q Fg b2652 104.7g kJ kg 1096 Btu lb
H H 2 O
At 27.6 psia (=1.90 bar), Table B.6 'H v 'H
¦ ni H i ¦ ni H i
out
in
2206 kJ / kg = 949 Btu / lb m
375 Btu = msteam 'H v msteam
0.395 lb m steam 100 u 2000 lb m H 3 PO 4 lb m 28% H 3 PO 4 day
375 Btu 949 Btu / lb m
0.395 lb m steam
1 day 3292 lb m steam / h 24 h
lb m steam 3292 lb m steam 118 . (46.3 u 60) lb m H 2 O evaporated / h lb m H 2 O evaporated
8-61
m
8.90 Basis: 200 kg/h feed solution.
A = NaC 2 H 3 O 2
.
n 1 (kmol H2 O(v )/h) 50°C, 16.9% of H 2O in feed 200 kg/h @ 60°C . n 0 (kmol/h) 0.20 A 0.80 H 2O
Product slurry @ 50°C . n 2 (kmol A-3H 2 O(v )/h) . n 3 (kmol solution/h) 0.154 A 0.896 H 2 O
Q (kJ/hr)
0.200 M A 0.800 M H 2 O
a. Average molecular weight of feed solution: M
b0.200gb82.0g b0.800gb18.0g Molar flow rate of feed: n0
200 kg
1 kmol
h
30.8 kg
6.49 kmol h
. gb0.80gb6.49 kmol hg b0169 b0.20gb6.49 kmol hg E n bkmol Ah 3 H Og
b. 16.9% evaporation n1 A balance:
2
2
n2 0.154n3
b gb
H 2 O balance: 0.80 6.49 kmol h
g
0.846n3 3n2 0.846n3
bg
0.877 kmol H 2 O v h 1 mole A . n3 0154 1 mole A 3 H 2 O (1)
1.30
0.877
b
n2 kmol A 3 H 2 O h
g
3 moles H 2 O 1 mole A 3 H 2 O
4.315
bg b g
bg
Solve 1 and 2 simultaneously n2 n3
1.13 kmol A 3H 2 O 136 kg A 3H 2 O
bg
154 kg NaC 2 H 3 O 2 3H 2 O s
1 kmol
Mass flow rate of product solution
h
b
gb g
bg
200 kg feed 154 kg crystals 0.877 18.0 kg H 2 O v h h h c.
b2g
113 . kmol A 3H 2 O s h 1095 kmol solution h .
Mass flow rate of crystals h
30.8 kg k
bg
30 kg solution h
bg
References for enthalpy calculations: NaC 2 H 3 O 2 s , H 2 O l @25q C
b
Feed solution: nH nH
g
n A 'H s 25q C m
b0.20g6.49 kmol A h
z
60
25
C p dT (form solution at 25q C , heat to 60q C )
171 . u 104 kJ 200 kg 3.5 kJ hr kgq C kmol A
8-62
b60 25gq C
2300 kJ h
8.90 (cont’d) Product solution: nH
b
g
n A 'H s 25q C m
b0.154g1.095 kmol A
z
50
25
. u 10 4 kJ 30 kg 3.5 kJ 171 h kgq C kmol A
h 259 kJ h Crystals: nH
n A 'H hydration m
z
b
g
H 2 O v , 50q C : n'H
25
C p dT (hydrate at 25q C , heat to 50q C )
bg
z
LM N
n 'H v
3.66 u 10 4 kJ 154 kg 1.2 kJ h kgq C kmol
OP Q
50
25
b neglect 'E g
'H
b gb
i
g
4.39 u 10 4 32.4 50 25 kJ
¦ n H ¦ n H i
out
R
b50 25gq C
C p dT (vaporize at 25q C , heat to 50q C )
0.877 kmol H 2 O h Energy balance: Q
b50 25gq C
50
1.13 kmol A 3H 2 O s h 36700 kJ h
C p dT
i
39200 kJ h
b259 36700 39200g b2300g
i
kJ h
in
60 kJ h (Transfer heat from unit) 8.91 50 mL H 2SO4
bg
84.2 mL H 2 O l
U| |V r 84.2 g H Oblg 4.678 mol H Oblg| |W
1834 . g 917 . g H2SO4 0.935 mol H 2SO4 mL 100 . g mL
2
500 . mol H 2 O mol H 2SO4
2
Ref: H 2 O , H 2SO 4 @ 25 qC H ( H 2 O(l ), 15o C) [0.0754 kJ / (mol o C)](15 25) o C = 0.754 kJ / mol
b
H H 2 SO 4 , r
g
substance
bg
H 2O l
H 2SO 4
b
H 2SO 4 r
(91.7 + 84.2) g 2.43 J kJ mol 0.935 mol H 2 SO 4 gq C ( 69.46 0.457T )( kJ / mol H 2 SO 4 ) 58.03
5.00
nin
H in
g
Energy Balance: 'H
1 kJ 10 3 J
H out
nout
4.678 –0.754 — — 0.0 0.935 0.935 — 4.00 —
bT 25gq C
n in mol H in kJ/mol
— —
b69.46 0.457T g n bmol H SO g 3
b
g
b
g
0 0.935 69.46 0.457T 4.678 0.754 T
Conditions: Adiabatic, negligible heat absorbed by the solution container. 8-63
4
144 q C
8.92 a.
mA (g A) @ TA0 (oC) nA (mol A)
nS (mol solution) @ Tmax (oC)
mB (g B) @ TB0 (oC) nB (mol B) Refs: A(l), B(l) @ 25 qC substance nin H in nout nA H A —
H out
A
—
n in mol H in J / mol
B
nB H B
—
—
S
—
nA
H S (J mol A)
—
mA (g A) , nB M A (g A / mol A)
Moles of feed materials: n A (mol A) =
mB MB
Enthalpies of feeds and product H A
m A C pA ( T A 0 25 o C), H B
r (mol B mol A) = n B n A
H S
FG J IJ H mol A K
nA
m B C pB ( TB 0 25 o C)
mB / M B mA / M A
LMn 1 M ( mol A) M MM N
A ( mol
FG J IJ H mol A K F J I u (T )( g soln) u C G H g soln C JK
A) u 'H m ( r )
(m A m B
ps
max
o
25)( o
OP PP C) P PQ
1 n A 'H m ( r ) ( m A m B ) C ps ( Tmax 25) nA
H S
Energy balance 'H
n A H S n A H A n B H B
0
bg
b
g
b
g
mA 'Hm r (m A mB )C ps (Tmax 25) m A C pA TA0 25 mB C pB TB0 25 MA
b
g
b
g
m A C pA TA0 25 mB C pB TB0 25 Tmax
25
0
bg
mA 'Hm r MA
(m A mB )C ps
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container, negligible dependence of heat capacities on temperature between 25oC and TA0 for A, 25oC and TB0 for B, and 25oC and Tmax for the solution. b.
mA
MA
mB
100.0 g 225.0 g
C ps
3.35 J (gq C)
MB
40.00 TA 0 18.01 TB 0
b
b
g UV |W
25q C C pA ? irrelevant r 40q C C pB 4.18 J (gq C)
g
'H m n 5.00
37,740 J mol A Tmax
8-64
5.00
125q C
mol H 2 O mol NaOH
8.93 Refs: Sulfuric acid and water @ 25 qC b.
substance
nin
H in
H2SO4 H2O H 2 SO 4 aq
1 r —
M A C pA T0 25 M w C pw T0 25 —
b g
b b
g g
H out
nout
— — 1
n in mol — H in J/mol — 'H m r M A rM w C ps Ts 25
bg b
g b
g
(J/mol H2SO4) 'H
bg b g b g b g 'H br g b98 18r gC bT 25g (98C 18rC )bT 1 25 (98C 18rC )bT 25g 'H br g ( 98 18r )C m
Ts
b
g
0 'H m r M A rM w C ps Ts 25 M A C pa T0 25 rM w C pw T0 25 ps
pa
s
pa
pw
pw
0
0
g
25
m
ps
c. H2O(l) H2SO4 r 0.5 1 1.5 2 3 4 5 10 25 50 100
Cp (J/mol-K) 75.4 185.6
Cp (J/g-K) 4.2 1.9
Cps 1.58 1.85 1.89 1.94 2.1 2.27 2.43 3.03 3.56 3.84 4
'H m (r ) -15,730 -28,070 -36,900 -41,920 -48,990 -54,060 -58,030 -67,030 -72,300 -73,340 -73,970
Ts 137.9 174.0 200.2 205.7 197.8 184.0 170.5 121.3 78.0 59.6 50.0
250
Ts
200 150 100 50 0 0.1
1
10
100
r
d. Some heat would be lost to the surroundings, leading to a lower final temperature. 8-65
8.94 a.
Ideal gas equation of state n A0
P0V g / RT0
b
gd
Vl ( L) u SG B u 1 kg / L 10 3 g / kg
Total moles of B: n B 0 ( mol B)
i
(2)
M B (g / mol B)
n Av n Al
Total moles of A: n Ao Henry’s Law: r
(1)
FG mol A(l) IJ H mol B K
(3)
ks pA
bc
n Al n B0
0
c1T
g n V RT Av
(4)
g
Solve (3) and (4) for nAl and nAv.
b
nB 0 RT c0 c1T Vg
n Al
g (5)
LM1 n RT bc c T gOP MN V PQ n LM1 n RT bc c T gOP MN V PQ B0
0
1
g
Ao
n Av
B0
0
(6)
1
g
Ideal gas equation of state n Av RT Vg
P
n A0 RT Vg nB 0 RT c0 c1T
( 6)
b
(7)
g
b g bg
Refs: A g , B l @ 298 K substance
bg Bb l g
Ag
Solution U 1
U eq
M A CvA T0 298
n Av
M A CvA T 298
M B CvB
—
—
n Al
U 1 (kJ/mol A)
U in
n Ao nB0
—
b bT
0
g 298g
—
b
g b
b
g
n in mol U in kJ/mol
g
1 'U s n Al M A n B 0 M B Cvs T 298 n Al
E.B.: 'U
0
¦ n U ¦ n U i
out
0
neq
nin
cn
Av CvA
T
i
i
i
in
b
g hb g b d'U i bn C n C gbT 298g n C bn M n M gC
gb
g
n Al M A nB M B Cvs T 298 n Al 'U s n Ao CvA n B CvB T0 298 298
n Al
s
Av
Ao
vA
Al
vA
B
A
8-66
0
vB
B
B
vs
8.94 (cont’d) b.
c.
Vt 20.0
MA 47.0
CvA 0.831
MB 26.0
CvB 3.85
SGB 1.76
Vl 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0
T0 300 300 300 300 330 330 330 330
P0 1.0 5.0 10.0 20.0 1.0 5.0 10.0 20.0
Vg 17.0 17.0 17.0 17.0 17.0 17.0 17.0 17.0
nB0 203.1 203.1 203.1 203.1 203.1 203.1 203.1 203.1
nA0 0.691 3.453 6.906 13.811 0.628 3.139 6.278 12.555
C*
c0 c1 0.00154 -1.60E-06 T 301.4 307.0 313.9 327.6 331.3 336.4 342.8 355.3
nA(v) 0.526 2.624 5.234 10.414 0.473 2.359 4.709 9.381
Dus -174000
Cvs 3.80
nA(l) 0.164 0.828 1.671 3.397 0.155 0.779 1.569 3.174
P 0.8 3.9 7.9 16.5 0.8 3.8 7.8 16.1
REAL R, NB, T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS REAL NA0, T, DEN, P, NAL, NAV, NUM, TN INTEGER K R = 0.08206 1 READ (5, *) NB IF (NB.LT.0) STOP READ (1, *) T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS WRITE (6, 900) NA0 = P0 * VG/R/T0 T = 1.1 * T0 K=1 10 DEN = VG/R/T/NB + C + D * T P = NA0/NB/DEN NAL = (C + D * T) * NA0/DEN NAV = VG/R/T/NB * NA0/DEN NUM = NAL * (–DUS) + (NA0 * CVA + NB * CVB) * (TO – 298) DEN = NAV * CVA + (NAL * MA + NB * MB) * CVS TN = 298 + NUM/DEN WRITE (6, 901) T, P, NAV, NAL, TN IF (ABS(T – TN).LT.0.01) GOTO 20 K=K+1 T = TN IF (K.LT.15) GOTO 10 WRITE (6, 902) STOP 20 WRITE (6, 903) GOTO 1 900 FORMAT ('T(assumed) P Nav Nal T(calc.)'/ * ' (K) (atm) (mols) (mols) (K)') 901 FORMAT (F9.2, 2X, F6.3, 2X, F7.3, 2X, F7.3, 2X, F7.3, 2) 902 FORMAT (' *** DID NOT CONVERGE ***') 903 FORMAT ('CONVERGENCE'/) END $ DATA 300 291 10.0 15.0 1.54E–3 –2.6E–6 –74 35.0 18.0 0.0291 0.0754 4.2E–03
8-67
Tcalc 301.4 307.0 313.9 327.6 331.3 336.4 342.8 355.3
8.94 (cont’d) 300 291 35.0 –1
50.0 18.0
Program Output T (assumed) (K) 321.10 296.54 296.57
15.0 0.0291
1.54E–3 0.0754
–2.6E–6 4.2E–03
–74
P (atm) 8.019 7.415 7.416
Nav (mols) 4.579 4.571 4.571
Nal (mols) 1.703 1.711 1.711
T(calc.) (K) 296.542 296.568 296.568
P (atm) 40.093 39.676 39.680
Nav (mols) 22.895 22.885 22.885
Nal (mols) 8.573 8.523 8.523
T(calc.) (K) 316.912 316.942 316.942
Convergence T (assumed) (K) 320.10 316.91 316.94
8.95 Q=0
350 mL 85% H2SO4 ma(g), 60 oF, U=1.78
30% H2SO4 ms(g), T(oF)
H2O, Vw(mL), mw(g), 60 oF
a. Vw
350 mL feed
. 178 g 1 mL feed
. g H 2 O added 1 mL water 0.85(70 / 30) 015 g feed 1 g water
1140 mL H 2 O b.
Fig. 8.5 -1 H a | 103 Btu / lb m ;
Water: H water | 65 Btu / lb m
Mass Balance: mp=mf+mw=(350 mL)(1.78 g/mL)+(1142 mL)(1 g/mL)=623+1142=1765 g Energy Balance:
'H
H product c.
T ( H
mf H f mw H w mp (623)( 103) (1140)( 65) 5.7 Btu / lb m 1765
0 mp H product ma H a mw H w H s
5.7 Btu / lb m ,30%) | 160o F
d. When acid is added slowly to water, the rate of temperature change is slow: few isotherms are crossed on Fig. 8.5-1 when xacid increases by, say, 0.10. On the other hand, a change from xacid=1 to xacid=0.9 can lead to a temperature increase of 200°F or more.
8-68
8.96 a. 2.30 lb m 15.0 wt% H 2 SO 4 10 Btu / lb @ 77 o F H 1
m
U| || V| o m ( lb || W adiabatic mixing
3
m2 (lb m ) 80.0 wt% H 2 SO 4 120 Btu / lb @ 60o F H 2
60.0 wt% H 2 SO4 @ To F, H 3
|UV |RSm mass balance: 2.30b0.150g m b0.800g m (0.600) W| |m T 2.30 + m2
Total mass balance: H 2 SO 4
m
m)
m3
2
2
b. Adiabatic mixing Q = 'H
3
3
517 . lb m (80%) 7.47 lb m (60%)
0
. gb 120g b7.47gH b2.30gb10g b517
0 H 3
3
. Btu / lb m 861
E Figure 8.5 - 1
T = 140 o F c.
d
i
H 60 wt%, 77 o F
130 Btu / lb m
d
i
Q m3 H 60 wt%, 77 o F H 3
b7.475gb130 861. g
328 Btu
d. Add the concentrated solution to the dilute solution . The rate of temperature rise is much lower (isotherms are crossed at a lower rate) when moving from left to right on Figure 8.5-1. 8.97 a.
b.
x NH 3
0.30
Fig. 8.5-2
y NH 3
0.96 lb m NH 3 lb m vapor , T
Basis: 1 lb m system mass
0.90 lb m liquid
. lb m vapor 010 Mass fractions: zNH 3
b0.27 0.096glb
m
NH 3
1 lb m 1 0.37
Enthalpy: H
0.90 lb m liquid
25 Btu
1 lb m
1 lb m liquid
8-69
80q F
x NH 3 0.30
0.27 lb m NH 3 0.63 lb m H 2 O
x NH 3 0.96
0.096 lb m NH 3 0.004 lb m H 2 O
0.37 lb m NH 3 lb m 0.63 lb m H 2 O lb m 0.10 lb m vapor
670 Btu
1 lb m
1 lb m vapor
44 Btu lb m
8.98
T
140q F
Fig. 8.5-2
Vapor: 80% NH 3 , 20% H 2 O Liquid: 14% NH 3 , 86% H 2 O
C A B
Basis: 250 g system mass mv ( g vapor), mL ( g liquid) Mass Balance: mv mL
Liquid: mNH 3
.60
.80 x NH3
250
NH3 Balance: 0.80m g 014 . mL Vapor: mNH 3
.14
(0.60)( 250) mv
175 g, mL
75g
b0.80gb175 gg 140 g NH , 35 g H O . gb75 gg 10.5 g NH , 64.5 g H O Liquid b014 3
2
3
2
8.99 Basis: 200 lb m feed h m v (lb m h) xv(lbm NH3(g)/lbm) H v ( Btu lb m )
200 lbm/h 0.70 lbm NH3(aq)/lbm 0.30 lbm H2O(l)/lbm
m l (lb m h)
H f
xl[lbm NH3(aq)/lbm]
50 Btu lb m
in equilibrium at 80oF
H l ( Btu lb m )
Q ( Btu h) Figure 8.5-2 Mass fraction of NH 3 in vapor: xv
0.96 lb m NH 3 lb m
Mass fraction of NH 3 in liquid: xl
0.30 lb m NH 3 lb m
Specific enthalpies: H v
650 Btu lb m , H l
UV W
m v Mass balance: 200 m v m l m l Ammonia balance: 0.70 200 0.96m v 0.30m l
b gb g
30 Btu lb m 120 lb m h vapor
80 lb m h liquid
Energy balance: Neglect 'E k . Q
'H
¦ m H i
out
i
m f H f
120 lb m h 86,000
650 Btu 80 lb m lb m h
Btu h
8-70
30 Btu 200 lb m lb m h
50 Btu lb m
CHAPTER NINE 4 NH 3 (g) 5O 2 (g) o 4NO(g) + 6H 2 O(g) 'H o 904.7 kJ / mol
9.1
r
a.
When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25qC and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25qC and 1 atm, the change in enthalpy is -904.7 kJ.
b.
Exothermic at 25qC. The reactor must be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the molecular bonds of the reactants is less than the energy released when the product bonds are formed.
c.
5 O 2 (g) o 2NO(g) + 3H 2 O(g) 2 Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half. 904.7 'H ro 452.4 kJ / mol 2 5 3 NO(g) + H 2 O(g) o NH 3 (g) O 2 (g) 4 2 Reversing the reaction reverses the sign of the heat of reaction. Also reducing the stoichiometric coefficients to one-fourth reduces the heat of reaction to one-fourth. ( 904.7) 'H ro 226.2 kJ / mol 4
d.
e.
2 NH 3 (g)
NH 3 m n NH 3 Q
340 g / s 340 g
= 'H
1 mol
20.0 mol / s 17.03 g o n NH 3 'H 20.0 mol NH 3 904.7 kJ r
s
Q NH 3
s
1 mol NH 3
18 . u 10 4 kJ / s
The reactor pressure is low enough to have a negligible effect on enthalpy. f.
9.2
Yes. Pure water can only exist as vapor at 1 atm above 100qC, but in a mixture of gases, it can exist as vapor at lower temperatures. C 9 H 20 (l) 14O 2 (g) o 9CO 2 (g) +10H 2 O(l) 'H o 6124 kJ / mol r
a.
When 1 g-mole of C9H20(l) and 14 g-moles of O2(g) at 25qC and 1 atm react to form 9 g-moles of CO2(g) and 10 g-moles of water vapor at 25qC and 1 atm, the change in enthalpy is -6124 kJ.
b.
Exothermic at 25qC. The reactor must be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the molecular bonds of the reactants is less than the energy released when the product bonds are formed.
c.
Q
= 'H
0 n C9 H 20 'H r
25.0 mol C 9 H 20
Q C9 H 20
s
6124 kJ
1 kW
1 mol C 9 H 20 1 kJ / s
9-1
153 . u 105 kW
9.2 (cont'd) Heat Output = 1.53u105 kW. The reactor pressure is low enough to have a negligible effect on enthalpy. d.
C 9 H 20 (g) 14O 2 (g) o 9CO 2 (g) +10H 2 O(l) 'H o 6171 kJ / mol
(1)
C 9 H 20 (l) 14O 2 (g) o 9CO 2 (g) +10H 2 O(l) 'H o 6124 kJ / mol
(2)
r
r
(2) (1) C 9 H 20 (l) o C 9 H 20 (g) 6124 kJ / mol ( 6171 kJ / mol) = 47 kJ / mol 'H o (C H ,25$ C) v
e. 9.4 (cont'd)
9.3
a.
9
20
Yes. Pure n-nonane can only exist as vapor at 1 atm above 150.6qC, but in a mixture of gases, it can exist as a vapor at lower temperatures.
Exothermic. The reactor will have to be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the reactant bonds is less than the energy released when the product bonds are formed.
b.
b g 192 O bgg o 6CO bgg 7H Obgg b1g 'H ? 19 'H 1.791 u 10 Btu lb - mole C H blg O bg g o 6CO bgg 7 H Oblg b2g 'H 2 e 'H j 13,550 Btu lb - mole C H bgg o C H blg b3g 'H H Oblg o H Obg g b4g 'H e'H j 18,934 Btu lb - mole b1g b2g b3g 7 u b4g 'H 'H 'H 7'H 1.672 u 10 Btu lb - mole C 6 H 14 g 6
14
6
14
2
2
2
6
2
2
14
o r
2
2
3
2
v
4
v
o r
2
C 2 H 14
H 2O
Hess's law
1
c.
M O2 =32.0
m 120 lb m / s Q
'H
bg
nO 2 'H ro vO 2
n
2
3
3.75 lb - mole / s.
s
a.
b
6.27 u 106 Btu / s from reactor
1 lb - mole O 2
bg
bg
bg
CaC 2 s 5H 2 O l o CaO s 2CO 2 g 5H 2 g , 'H ro 9.4
6
4
. u 10 6 Btu 3.75 lb - mole 1672
bg
6
g
69.36 kJ kmol
Endothermic. The reactor will have to be heated to keep the temperature constant. The temperature would decrease under adiabatic conditions. The energy required to break the reactant bonds is more than the energy released when the product bonds are formed.
b.
'U ro
'H ro RT
LM OP MM ¦ Q ¦ Q PP MN PQ i
gaseous products
i
69.36
gaseous reactants
52.0 kJ mol
9-2
kJ 8.314 J 1 kJ 298 K mol mol K 10 3 J
b7 0g
9.4 (cont’d)
'U ro is the change in internal energy when 1 g - mole of CaC2 (s) and 5 g - moles of H2 O(l) at 25$ C and 1 atm react to form 1 g - mole of CaO(s), 2 g - moles of CO2 (g) and 5 g - moles of H2 (g) at 25$ C and 1 atm. c.
'U
Q
nCaC 2 'U ro
150 g CaC 2
1 mol
52.0 kJ
121.7 kJ
64.10 g 1 mol CaC 2
vCaC 2
Heat must be transferred to the reactor. 9.5
a.
Given reaction
(1) – (2)
Hess's law
'H ro
'H ro1 'H ro2
b1226 18,935g Btu lb - mole 17,709 Btu lb - mole
b.
Given reaction
(1) – (2)
Hess's law
'H ro
'H ro1 'H ro2
b121,740 104,040g Btu lb - mole 17,700 Btu lb - mole
9.6
9.7
bg
Hess's law
b326.2 kJ molg 2b2858. kJ molg
'H ro
a.
Reaction (3)
b.
Reactions (1) and (2) are easy to carry out experimentally, but it would be very hard to decompose methanol with only reaction (3) occurring.
bg
(1) 2 u 2
bg
Table B.1
bg
a.
N 2 g O 2 g o 2NO g ,
b.
n C 5 H 12 g
bg
6
o f
14
2
e j 'H fo
NO(g)
180.74 kJ mol
11 O 2 g o 5CO g 6H 2 O l 2 6 'H fo 'H fo
CO(g)
o f
CO 2
n C 5 H 12 g
H 2O l
2
o r
d.
'H ro
bg bg bg 'H 5e 'H j e j bg e j b g . g b 146.4g kJ mol 2121.2 kJ mol b5gb110.52g b6gb28584 19 C H blg O bgg o 6CO bgg 7H Obgg 2 'H 6e 'H j 7e 'H j b g e 'H j b g . g b 173.0g kJ mol 3855 kJ mol b6gb393.5g 7b24183 o r
c.
F B I 2G 90.37 kJ molJ GH JK
245.4 kJ mol
2
o f
2
H 2O g
o f
C 6 H 14 l
Na 2SO 4 (l) 4CO(g) o Na 2S(l) 4CO 2 (g) 'H ro
e'H j
4e 'H j e j e'H j ( 373.2 6.7) b4gb 3935 . g b 1384.5 24.3g 4( 110.52 o f
Na 2S( l )
4 'H fo
CO 2 ( g )
o f
9-3
Na 2SO 4 ( l )
o f
CO(g )
kJ mol
138.2 kJ mol
9.8
a.
e'H j e'H j
'H ro1
o f
'H ro2 b.
c.
e 'H j 385.76 52.28 333.48 kJ mol e j 276.2 92.31 333.48 35.03 kJ mol b g e 'H j b g e 'H j b1g b2g 385.76 35.03 420.79 kJ mol
C 2 H 2 Cl 4 ( l )
o f
'H
o f
C 2 HCl 3 l
Given reaction
Q
'H fo
o f
C2 H 4 (g)
o f
HCl g
300 mol C 2 HCl 3
420.79 kJ
h
mol
C 2 H 2 Cl 4 ( l )
C 2 H 2 Cl 4 (l )
b
126 . u 105 kJ h
35 kW
g
Heat is evolved. 9.9
b.
c.
5 'H co 1299.6 kJ mol O 2 (g) o 2CO 2 (g) + H 2 O(l) 2 The enthalpy change when 1 g-mole of C2H2(g) and 2.5 g-moles of O2(g) at 25qC and 1 atm react to form 2 g-moles of CO2(g) and 1 g-mole of H2O(l) at 25qC and 1 atm is -1299.6 kJ. C 2 H 2 ( g)
e j
'H co
2 'H fo
Table B.1
B
b
e j
'H fo
g b
b g e 'H f j C H b g g o
H 2O l
2
g b
kJ g mol
2 393.5 285.84 226.75
d'H i
(i) 'H ro
o f
Table B.1
B
C2 H 6 ( g )
o c
Table B.1
B
C 2 H 2 ( g)
d
'H fo
i
C2H 2 (g)
2
1299.6
kJ mol
bg
C2 H2 g
kJ b84.67g b226.75g mol
d'H i
(ii) 'H ro
d.
CO 2 ( g )
d
2 'H co
i
3114 .
H2 (g)
d
'H co
kJ mol
i
bg
C2H 6 g
kJ . ) b 1559.9g b1299.6g 2(28584 mol
5 O 2 (g) o 2CO 2 (g) + H 2 O(l) 2
1 O 2 (g) o H 2 O(l) 2 7 C 2 H 6 (g) O 2 (g) o 2CO 2 (g) + 3H 2 O(l) 2 H 2 ( g)
3114 .
kJ mol
(1) 'H co1
1299.6 kJ mol
(2) 'H co2
. kJ mol 28584
(3) 'H co3
1559.9 kJ mol
The acetylene dehydrogenation reaction is (1) + 2 u (2) (3) Hess's law
'H ro
'H co1 2 u 'H co2 'H co3
b1299.6 2(285.84) (1559.9)g kJ mol
9-4
311.4 kJ / mol
9.10
a.
bg
C 8 H 18 l
bg
bg
25 O 2 (g) o 8CO 2 g 9H 2 O g 2
'H ro
4850 kJ / mol
When 1 g-mole of C8H18(l) and 12.5 g-moles of O2(g) at 25qC and 1 atm react to form 8 g-moles of CO2(g) and 9 g-moles of H2O(g), the change in enthalpy equals -4850 kJ. b.
Energy balance on reaction system (not including heated water):
'E k , 'E p , W
0Q
Q
'U co
'H co
114.2 g
LM OP MM ¦ Q ¦ Q PP MN PQ i
5079 kJ mol
1 mol C 8 H 18
gaseous reactants
8.314 J
1 kJ
b8 9 12.5g
298 K
3
mol K 10 J
5090 kJ mol
% difference = 'H co
'U co (kJ)
i
gaseous products
c.
89.4 kJ
5079 kJ mol
'U co RT
'H co
g
1 mol
2.01 g C 8 H 18 consumed 1 mol C8 H 18
'U 89.4 kJ
b
75.4 u 10 3 kJ 21.34q C 18.0 u 10 3 kg mol.$ C
1.00 kg
mH 2O (Cp ) H 2 O(l) 'T
g
75.4 u 10 3 kJ / mol.$ C
(Cp ) H 2 O(l) from Table B.2 Q
b
n mol C 8 H 18 consumed 'U co kJ mol
'U
e j e 'H j 8 'H fo
( 5090) ( 4850) u 100 = 4.7 % 5090
b g 9e'H ff j H Obg g e 'H f jC H blg o
CO 2 g
o f
bg
C8 H 18 l
b
o
2
8
g b
g
18
. 5090 kJ / mol 8 393.5 9 24183
234.5 kJ / mol
There is no practical way to react carbon and hydrogen such that 2,3,3-trimethylpentane is the only product.
9-5
9.11
a.
bg
bg
n C 4 H10 g o i C 4 H10 g Basis: 1 mol feed gas
0.930 mol n-C4H10
(nn- C4H10)out
0.050 mol i-C4H10
( ni-C4H10)out
0.020 mol HCl
0.020 mol HCl
149qC
Q(kJ/mol)
(n n-CH 4 H10 ) out
0.930(1 0.400)
(n i-CH 4 H10 ) out
0.050 0.930 u 0.400
0.560 mol
(n n-C4H10 ) out (n n-C4H10 ) in
[
Q n-C
e'H j
'H ro
c.
References: n C 4 H10
o f
i C 4 H 10
substance
Table B.1
n in
n C 4 H 10 4
b kJ molg
n C 4 H 10
1
i C 4 H 10
z
149
25
Table B.2
B
Cp
O kJ dT P PQ mol
Q = 'H ['H r o
i
For 325 mol/h fed, Q =
b
g
g
kJ mol
9.8 kJ mol
b kJ molg
(mol) 0.600
H 1
0.400
H 2
H 2
14.29 kJ mol
i
134.5 124.7
H out
n out
¦ n H ¦ n H out
'H r 149q C
b
o r
10
H in
H 1
LM MN
0.370 mol
e j 'H bgg, i C H bgg at 25q C
'H fo
(mol)
H 1
0.420 mol
0.560 0.930 1
4 H10
b.
d.
149qC
i
i
LM MN
z
b gb
149
25
Table B.2
B
C p dT
OP kJ PQ mol
g b gb
14.14 kJ mol
g
9.8 1 14.142 1 14.287 kJ
9.95 kJ
in
9.95 kJ 325 mol feed 1h 1 kW 1 mol feed h 3600 s 1 kJ / s
9.95 kJ (0.930 0.560) mol n - C 4 H 10 react
9-6
26.9 kJ / mol
0.90 kW
9.12
a.
1 m3 at 298K, 3.00 torr
Products at 1375K, 3.00 torr
n0 (mol) 0.111 mol SiH4/mol 0.8889 mol O2/mol
n1 (mol O2) n2 (mol SiO2) n3 (mol H2)
SiH 4 (g) + O 2 (g) o SiO 2 (s) + 2H 2 (g)
1 m3
Ideal Gas Equation of state : no
273 K 3.00 torr 1 mol 298 K 760 torr 22.4 u 10 -3 m 3
. m 01017
nio Q i [
ni
SiH 4 : 0 = 0.1111(0.1017 mol) [ [ O 2 : n1 SiO 2 : n2
. 0.8889(01017 mol) [
[
0.0113 mol
0.0790 mol O 2
0.0113 mol SiO 2
H 2 : n3 = 2[ = 0.0226 mol H 2 b.
'H ro
( 'H fo )SiO 2 (s) ( 'H fo )SiH 4 ( g) = [851 ( 61.9)] kJ mol
789.1 kJ / mol
References : SiH 4 (g), O 2 (g), SiO 2 (g), H 2 (g) at 298 K Substance nin nout H H in
SiH 4 O2
out
(mol h) (kJ mol) (mol h) (kJ mol) 0 0.013 0.904 0 0.0790 H 1
SiO 2
H2
H 2 H
0.0113 0.0226
3
Table B.8
O 2 (g,1375K): H 1
B
H O 2 (1375K) = 46.14 kJ / mol
z
1375
SiO 2 (s,1375K): H 2
79.18 kJ / mol
(C p )SiO 2 (s) dT
298 Table B.8
H 2 (g,1375K): H 3 c.
Q
'H
B
H H 2 (1375K) = 41.08 kJ / mol
[ 'H ro ¦ ni H i ¦ ni H i out
= 3.45 kJ 27.5 m Q m3 h
3.45 kJ / m 3 feed
in
3
1h
1 kW
3600 s 1 kJ / s
9-7
0.0264 kW (transferred from reactor)
9.13
a.
bg bg
bg
bg
Fe 2 O 3 s 3C s o 2 Fe s + 3CO g , 'H r (77 $ F) Basis:
2.111 u 105 Btu lb - mole
2000 lb m Fe 1 lb - mole
3581 . lb - moles Fe produced 55.85 lb m 53.72 lb - moles CO produced
17.9 lb - moles Fe 2 O 3 fed 53.72 lb - moles C fed
17.9 lb-moles Fe2O3 (s) 77q F
35.81 lb-moles Fe (l) 2800q F
53.72 lb-moles C 77q F
53.72 lb-moles CO(g) 570q F Q (Btu/ton Fe)
b.
bg bg bg
Substance
b
b g Febl,2800q Fg CObg,570q Fg
g
Fe(l,2800$ F): H 1 CO(g,570$ F): H 2
'H
H out
nout
(lb - moles) (Btu lb - mole) (lb - moles) (Btu lb - mole) 17.91 0
C s,77q F
Q
H in
nin
Fe 2 O 3 s,77q F
c.
bg
References: Fe 2 O 3 s , C s , Fe s , CO g at 77q F
53.72
0
. 3581
H
53.72
H 2
z
2794
77
dC i
b gdT 'Hm b2794q Fg
p Fe s
H CO (570$ F)
nFe 'H ro Q Fe
i
dC i
2800
2794
b gdT
p Fe l
28400 Btu lb - mole
3486 Btu lb - mole
A
FH interpolating I from Table B.9 K
¦ n H ¦ n H out
z
1
i
i
i
in
. ge2.111 u 10 j b3581 . gb28400g b53.72gb3486g 0 b3581 5
2
d.
4.98 u 106 Btu / ton Fe produced
Effect of any pressure changes on enthalpy are neglected. Specific heat of Fe(s) is assumed to vary linearly with temperature from 77qF to 570qF. Specific heat of Fe(l) is assumed to remain constant with temperature. Reaction is complete. No vaporization occurs.
9-8
9.14
a.
bg
C 7 H 16 g o C 6 H 5CH 3 (g) 4 H 2 (g) Basis: 1 mol C7H16 1 mol C7H16
1 mol C6H5CH3
400qC
4 mol H2 400q C
Q (kJ/mol)
bg b g
References: C s , H 2 g at 25q C
b.
H in
substance nin
H out
nout
bmolg bkJ molg bmolg bkJ molg
C 7 H 16
b
C 7 H 16
1
H 1
C7 H8
1
H2
4
g
g,400q C : H 1
( 'H f$ ) C7 H 16 ( g)
LM MN
z
H2 H
3
Table B.2
400
25
B
C p dT
OP PQ
( 187.8 + 104.1) kJ / mol = 83.7 kJ / mol
b
g
C 6 H 5 CH 3 g,400q C : H 2
( 'H f$ ) C 6 H 5CH 3 ( g)
L M MN
z
Table B.2
400
25
B
C p dT
OP PQ
(+50 + 60.2) kJ / mol = 110.2 kJ / mol
b
Table B.8
g
H 2 g,400q C : H 3 c.
B
H H 2 (400$ C)
10.89 kJ mol
1 mol C 7 H 16 reacted [ = 1 mol Q
'H
['H ro
¦ n H ¦ n H i
out
i
i
i
in
ob1gb237.8g b1gb60.2g b4gb10.86g b1gb104.1g tkJ d.
'H r ( 400 $ C) =
237.3 kJ 1 mol C 7 H16 react
237.3 kJ / mol
9-9
237.3 kJ (transferred to reactor)
9.15
a.
bCH g Obgg o CH bgg H bgg CObgg 3 2
4
2
Moles charged: (Assume ideal gas) 2.00 liters 273 K 350 mm Hg
1 mol
b g
b g
0.01286 mol CH 3 2 O
873 K 760 mm Hg 22.4 liters STP Let x
b g
fraction CH 3 2 O decomposed (Clearly x<1 since Pf 3 P0 ) 0.01286(1 – x ) mol (CH3 )2O 0.01286 x mol CH4 0.01286 x mol H2 0.01286 x mol CO
0.01286 mol (CH 3)2 O 600°C, 350 mm Hg
Total moles in tank at t
b.
Pf V
n f RT
P0V
n0 RT
nf
Pf
n0
P0
b g
b
0.01286 1 x 3x
2h
b
0.01286 1 2 x
g
875 x 350
0.01286
600°C 875 mm Hg
g
0.01286 1 2 x mol
0.75 75% decomposed
af b g b g
References:C s , H 2 g , O 2 g at 25$ F H in
nin
substance
H out
nout
( mol) (mol) ( kJ / mol) (kJ / mol) 0.01286 H 1 H 1 0.25 u 0.01286 H 0.75 u 0.01286
b g bg CH bg g H bg g CObgg CH 3 2 O g 4
2
2
0.75 u 0.01286
0.75 u 0.01286
bCH g O(g,600 C): H $
3 2
( 'H fo ) bCH 3 g
1
O 2
118 kJ mol
CH 4 (g,600 C): H 2 $
( 'H fo ) CH 4
L M MN
z
LM MN
z
given
873
298
B
C p dT
Table B.2
600
25
B
C p dT
H 3 H 4
OP J 1 kJ PQ mol u 10 J 3
OP 1 PQ 10
3
( 180.16 62.40) kJ / mol
74.85 29.46 45.39 kJ mol
Table B.8
H 2 (g,600$ C): H 3
B
H H 2 (600$ C)
16.81 kJ mol Table B.1 Table B.8
Table B.8
CO(g,600$ C): H 4
( 'H fo ) CO
B H CO (600 $ C)
110.52 17.57 kJ mol
B
92.95 kJ mol
For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are : c.
'H
[ 'H
¦n
out H out
¦n
( n CH 4 ) out ( n CH 4 ) in
Q CH 4
b
g
in H in
1.5507 ( 1.5143) kJ
0.75 u 0.01286 mol 1
b
[ 'H r 600q C 'H r 600q C
g
0.009645 mol
0.0364 kJ 0.009645
9-10
0.0364 kJ
3.77 kJ / mol
9.15 (cont’d)
b
'U r 600q C
g
b
g
¦Q
'H r 600q C RT [
i
gaseous products
3.77 kJ mol d. 9.16
a.
Q
b
[ 'U r 600q C
SO 2 (g)
Basis :
g
8.314 J
¦Q
i
]
gaseous reactants
1 kJ
873 K
3
b1 1 1 1g
18.3 kJ mol
mol K 10 J
(0.009645 mol)( 18.3 kJ / mol)
0.176 kJ (transferred from reactor)
1 O 2 (g) o SO 3 (g) 2
l00 kg SO 3
10 3 mol SO 3
min
80.07 kg SO 3
1249 mol SO 3 min
n 0 mol SO2 /min 450°C 100% excess ni1 mol O2 /min 3.76 n 1 mol O2 /min 450°C
1249 mol SO 3/min s n 0 mol SO2 /min n 3 mol O 2/min 3.76 n 1 mol O2 550°C / i
mw (kg H 2O( l) /h) 25°C
mw (kg H 2O(l) /h) 40°C
Assume low enough pressure for H to be independent of P. SO 3 balance :
bGeneration
output
n0 (mol SO 2 fed) 0.65 mol SO 2 react 1 mol SO 3 produced
g n0
1922 mol SO 2
0.5 mol O 2 reqd
min
1 mol SO 2
b g
N 2 balance : 3.76 1922
b
7227 mol / min in & out
b1 +1g mol O
g
b
.
.
Extent of reaction : [
2
fed
1 mol O 2 reqd
673 mol SO g b2gb1922g b2gb1922g b3gb1249g b2gb673g 2n 1922 1 0.65 mol s
65% conversion : n2
b.
2
min out
3
n3
mol SO 3 min
.
(n SO 2 ) out (n SO 2 ) in
( 'H fo )SO 3 ( g) ( 'H fo )SO 2 ( g)
1922 mol O 2 min fed
1298 mol / min out
Q SO 2
673 1922 1
1249 mol / min
Table B.1
'H ro
1249
1 mol SO 2 react
1922 mol SO 2 / min fed
100% excess air: n1
O balance:
1 mol SO 2 fed
min
B
395.18 ( 296.9)
9-11
99.28 kJ / mol
9.16 (cont'd)
bg bg bg
bg
References : SO 2 g , O 2 g , N 2 g , SO 3 g at 25$ C H in
nin
Substance
H out
( mol / min) ( kJ / mol) ( mol / min) ( kJ / mol) H 1 H 4 1922 673 H 2 H 5 1922 1298 H H 7227 7227
SO 2 O2 N2
3
SO 3
z
SO 2 (g,450 C) : H 1 $
6
1249
H 7
Table B.2
B
450
C p dT
25
19.62 kJ / mol Table B.8
B
$
H 2
H O 2 (450$ C)
$
H 3
H N 2 (450 C)
O 2 (g,450 C)
nout
13.36 kJ / mol
Table B.8
N 2 (g,450 C) Out :
z
SO 2 (g,450 C) : H 4 $
12.69 kJ / mol
Table B.2
550
25
B
24.79 kJ / mol
C p dT
Table B.8
$
H 5
H O 2 (550 $ C)
$
H 6
H N 2 (550$ C)
O 2 (g,550 C)
B
$
B
16.71 kJ / mol
Table B.8
N 2 (g,550 C)
z
SO 3 (g,550$ C) : H 7 Q
' H
[ 'H ro
B
1581 . kJ / mol
Table B.2
B
550
35.34 kJ / mol
C p dT
25
¦ n H ¦ n H i
i
i
out
i
in
b1249gb98.28g b673gb24.796g b179.8gb16.711g b7227gb15.808g b1249gb35.336g b1922gb19.623g 1922b13.362 g b7227gb12.691g . u 104 kJ / min 8111 c.
Assume system is adiabatic, so that Q lost from reactor Q
'H
LM MM A e N Table B.5
8.111 u 10 4 d.
j
e
m w H w l, 40$ C H w l, 25$ C kJ min
m w
A
Table B.5
OP jP PQ
Q gained by cooling water
FG kg IJ 167.5 104.8 kJ m H min K kg
w
1290 kg min cooling water
If elemental species were taken as references, the heats of formation of each molecular species would have to be taken into account in the enthalpy calculations and the heat of reaction term would not have been included in the calculation of 'H .
9-12
bg
9.17
CO(g) H 2 O v o H 2 (g) CO 2 (g) , Table B.1
'H ro a.
e j 'H fo
CO 2 ( g) 3
e j 'H fo
CO(g)
e j 'H fo
B
bg
H 2O v
4115 .
b g
3
kJ mol
b g
Basis : 2.5 m STP product gas h 1000 mol 22.4 m STP n 0 (mol CO/h) 25°C n 2 (mol H 2 O(v)/h) 150°C
Q r (kW)
H balance on reactor : 2n2
2
44.64 mol CO 1 mol H 2 O h
1 mol CO
b66.96 44.64g mol h u 100%
% excess steam
44.64 mol h
CO 2 balance on condenser : n3 H 2 balance on condenser: n4
b
pw 15q C
g
bg
66.96 mol H 2 O v h
44.64 mol H 2 O
50% excess steam
b0.40gb1116. mol hg 44.64 mol CO b0.40gb111.6 mol hg 44.64 mol H h
2
h
2
Saturation of condenser outlet gas: y H 2O
Q c (kW)
b0.40gb1116. mol hg 44.64 mol CO h 111.6 b2gb0.40g b2 gb0.20g mol h n
Steam theoretically required
n 3 (mol CO2 /h) n 4 (mol H 2 /h) n 5 (mol H 2 O(v)/h),sat'd 15°C, 1 atm n 6 (mol H 2 O( l )/h) 15°C, 1 atm
111.6 mol/h condenser 0.40 mol H 2/mol 0.40 mol CO 2/mol 0.20 mol H 2O( v)/h 500°C
reactor
C balance on reactor : n1
1116 . mol h
b
g
n5 mol H 2 O h 44.64 + 44.64 + n5 mol h
12.788 mm Hg n5 760 mm Hg
b gb g H O balance on condenser: b111.6gb0.20gmol H O h 153 . n p
2
2
n6 b.
6
20.8 mol H 2 O h condensed = 0.374 kg / h
Energy balance on condenser $
References : H 2 ( g), CO 2 (g) at 25 C, H 2 O at reference point of steam tables
Substance CO 2 (g) H 2 (g)
bg bg
H 2O v H 2O l
n in n out H in H out mol / h kJ / mol mol / h kJ / mol 44.64 H 1 H 4 44.64 44.64 44.64 H 2 H 5 22.32 153 H 3 H 6 . 20.80 H 7
Enthalpies for CO2 and H2 from Table B.8
CO 2 (g,500 $ C) : H 1 H 2 ( g,500 $ C) : H 2
H CO 2 (500 $ C)
2134 . kJ / mol
H H 2 (500 $ C) 1383 . kJ / mol
9-13
bg
. mol H 2 O v h 153
9.17 (cont’d)
H 2 O(v,500 $ C) : H 3 CO 2 (g,15$ C) : H 4
H CO2 (15$ C)
H 2 ( g,15$ C) : H 5
H H 2 (15$ C)
H 2 O(v,15$ C) : H 6
2529
H 2 O(l,15$ C) : H 7
62.9
Q
FG H
i
i
out
i
FG H
IJ K
IJ K
45.52 kJ mol
. kJ mol 113
b49.22 29718. g kJ
Energy balance on reactor : References : H 2 (g), C(s), O 2 (g) at 25q C Substance CO(g) H 2 O ( v)
bg bg
H2 g
H in
nin
H out
nout
( mol / h) ( kJ / mol) ( mol / h) ( kJ / mol) H 1 44.64 66.96 H2 22.32 H 3 H 44.64 4
CO 2 g
CO(g,25$ C) : H 1
( 'H f$ ) CO
H 5
44.64 Table B.1
110.52 kJ / mol
H 2 O(v,150 $ C) : H 2 = ( 'H f$ ) H 2 O(v) H H 2 O (150 $ C) H 2 O(v,500 $ C) : H 3 = ( 'H f$ ) H 2 O(v) H H 2 O (500 $ C) H 2 (g,500 $ C) : H 4
H H 2 (500 $ C)
CO 2 (g,500 $ C) : H 5 Q
'H
i
Table B.8
i
i
in
i
bheat transferred from reactor g
Tables B.1, B.8
Tables B.1, B.8
237.56 kJ mol 224.82 kJ mol
1383 . kJ / mol
( 'H $f ) CO 2 H CO2 (500 $ C)
¦ n H ¦ n H out
d.
1h 1 kW 0.812 kW 3600 s 1 kJ s
h
in
bheat transferred from condenser g c.
62.86 kJ mol
0.432 kJ / mol
kJ 18.0 kg u kg 10 3 mol i
IJ K
0.552 kJ / mol
kJ 18.0 kg u kg 10 3 mol
¦ n H ¦ n H
'H
FG H
18 kg kJ u kg 10 3 mol
3488
Tables B.1, B.8
372.16 kJ / mol
2101383 1h 1 kW . ( 20839.96) kJ h 3600 s 1 kJ s
0.0483 kW
Benefits Preheating CO more heat transferred from reactor (possibly generate additional steam for plant) Cooling CO lower cooling cost in condenser.
9-14
9.1 8
b. References : FeO(s), CO(g), Fe(s), CO 2 ( g) at 25o C nin nout H in H out ( mol) ( kJ / mol) ( mol) ( kJ / mol) 0 n1 H 1 1.00 n0 H0 n2 H 2 n H
Substance FeO CO Fe
CO 2
Q [ 'H ro
¦n
out H out
¦n
3
3
n4
H 4
in H in
Q [ 'H ro n1 H 1 n2 H 2 n3 H 3 n4 H 4 n0 H 0 (100 . n1 ) n1 1 X Fractional Conversion : X . 100 CO consumed : n2
(1 n1 ) mol FeO consumed
1 mol CO 1 mol FeO consumed
n0 (1 n1 )
(1 n1 ) mol CO
n0 X
Fe produced : n3 =
(1 n1 ) mol FeO consumed
1 mol Fe
1 mol FeO consumed 1 mol CO 2 (1 n1 ) mol FeO consumed CO 2 produced : n4 = 1 mol FeO consumed Extent of reaction : [
z
(nCO ) out (nCO ) in
n2 n0
Q CO
1
(1 n1 ) mol Fe = X (1 n1 ) mol CO 2
X
T
H i
C pi dT
for i
0,1,2,3,4
25
H 0
0.02761 (T0 298) 2.51 u 10 6 (T0 2 298 2 ) H ( 8.451 0.02761 T 2.51 u 10 6 T 2 ) kJ / mol 0
u 10 6 T 2 3188 u 10 2 / T ) kJ / mol ( 17.0814 0.0528 T 31215 . .
H 1
(0.02761 ( T 298) 2.51 u 10 6 (T 2 298 2 )
H 2 H 3
8.451 0.02761 T 2.51 u 10 6 T 2 ) kJ / mol
u 10 5 (T 2 298 2 ) 0.01728 (T 298) 1335 .
H 3 H 4
0
0.0528 (T 298) 31215 . . u 10 6 (T 2 298 2 ) 3188 u 10 2 (1 / T 1 / 298)
H 1 H 2
0
u 10 5 T 2 ) kJ / mol ( 6.335 0.01728 T 1335 .
0.04326(T 298) 0.573 u 10 5 (T 2 298 2 ) 818 . u 10 2 (1 / T 1 / 298) H 0.04326 T 0.573 u 10 5 T 2 818 ( 16145 . . u 10 2 / T ) kJ / mol 4
9-15
X
9.18 (cont'd) c. n0
2.0 mol CO, T0 n1
1 0.7
Summary: H 0 H 3
550 K, and X
2 0.7 13 . , n3
0.3, n2
1520 . kJ / mol, H 1
7.207 kJ / mol, H 4
'H ro Q
350 K, T
0.700 mol FeO reacted / mol FeO fed
0.7, n4
1348 . kJ / mol, H 2
0.7, [
0.7
7.494 kJ / mol,
1087 . kJ / mol
1.648 kJ / mol
(0.7)( 1.648) (0.3)(1.348) (1.3)(7.494) (0.7)(7.207) (0.7)(1087 . ) (2)(1520 . )
Q
1186 . kJ / mol
d. no
X T 1 298 1 400 1 500 1 600 1 700 1 800 1 900 1 1000
Xi
no To 1 298 1 400 1 500 1 600 1 700 1 800 1 900 1 1000
X 1 1 1 1 1 1 1 1
T 700 700 700 700 700 700 700 700
Xi
no
1 1 1 1 1 1 1 1
To 400 400 400 400 400 400 400 400
n1 0 0 0 0 0 0 0 0
n2 n3 n4 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1
H0 2.995 2.995 2.995 2.995 2.995 2.995 2.995 2.995
n2 n3 n4 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1
H0
0 0 0 0 0 0 0 0
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
n2 n3 n4 1 0 0 0.9 0.1 0 0.8 0.2 0 0.7 0.3 0 0.6 0.4 0 0.5 0.5 1 0.4 0.6 1 0.3 0.7 1 0.2 0.8 1 0.1 0.9 1 0 1 1
1 1 1 1 1 1 1 1 n1 1 1 1 1 1 1 1 1
To 400 400 400 400 400 400 400 400 400 400 400
X 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
T 500 500 500 500 500 500 500 500 500 500 500
Xi
1 1 1 1 1 1 1 1 1 1 1 no 0.5 0.5 0.5 1 1
To 400 400 400 400 400
X 0.5 0.5 0.5 0.5 0.5
T 400 400 400 400 400
Xi 0.5 0.5 0.5 0.5 0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
n1
n1 n2 0.5 0.5 0.5 0.5 0.5
n3 n4 0.5 1 0.5 0.5 0.5 0.5
9-16
H1
H2
0 5.335 10.737 16.254 21.864 27.555 33.321 39.159
0 2.995 5.982 9.019 12.11 15.24 18.43 21.67
H3 0 2.713 5.643 8.839 12.303 16.033 20.031 24.295
H4 0 4.121 8.553 13.237 18.113 23.152 28.339 33.663
Q -19.48 -12.64 -5.279 2.601 10.941 19.71 28.895 38.483
0 2.995 5.982 9.019 12.11 15.24 18.43 21.67
H1 21.864 21.864 21.864 21.864 21.864 21.864 21.864 21.864
H2 12.11 12.11 12.11 12.11 12.11 12.11 12.11 12.11
H3 12.303 12.303 12.303 12.303 12.303 12.303 12.303 12.303
H4 18.113 18.113 18.113 18.113 18.113 18.113 18.113 18.113
Q 13.936 10.941 7.954 4.917 1.83 -1.308 -4.495 -7.733
H0 2.995 2.995 2.995 2.995 2.995 2.995 2.995 2.995 2.995 2.995 2.995
H1 10.737 10.737 10.737 10.737 10.737 10.737 10.737 10.737 10.737 10.737 10.737
H2 5.55 5.55 5.55 5.55 5.55 5.55 5.55 5.55 5.55 5.55 5.55
H3 5.643 5.643 5.643 5.643 5.643 5.643 5.643 5.643 5.643 5.643 5.643
H4 8.533 8.533 8.533 8.533 8.533 8.533 8.533 8.533 8.533 8.533 8.533
Q -3.188 -3.399 -3.61 -3.821 -4.032 -4.244 -4.455 -4.666 -4.877 -5.088 -5.299
H0 2.995 2.995 2.995 2.995 2.995
H1 5.335 5.335 5.335 5.335 5.335
H2 2.995 2.995 2.995 2.995 2.995
H3 2.713 2.713 2.713 2.713 2.713
H4 4.121 4.121 4.121 4.121 4.121
Q -3.653 -3.653 -3.653 -3.653 -3.653
9.18 (cont'd) 400 400 400 400 400 400
0.5 0.5 0.5 0.5 0.5 0.5
400 400 400 400 400 400
0.5 0.5 0.5 0.5 0.5 0.5
0.5 0.5 0.5 0.5 0.5 0.5
0.5 0.5 0.5 0.5 0.5 0.5
2.995 2.995 2.995 2.995 2.995 2.995
50 40 30 20 10 0 -10 -20 -30
5.335 5.335 5.335 5.335 5.335 5.335
2.995 2.995 2.995 2.995 2.995 2.995
2.713 2.713 2.713 2.713 2.713 2.713
10 5 0 -5 -10 500
1000
0
1500
500
0 -1
-3
Q
Q
-2
-4 -5 -6 0
0.2
1000
1500
T o (K )
T (K )
0.4
0.6
0.8
1
0 -0.5 -1 -1.5 -2 -2.5 -3 -3.5 -4 0
0.5
1
X
a.
-3.653 -3.653 -3.653 -3.653 -3.653 -3.653
15
0
9.19
4.121 4.121 4.121 4.121 4.121 4.121
20
Q
Q
1 1 1 2 2 2
1.5
2
2.5
no
Fermentor capacity : 550,000 gal Solution volume : (0.9 u 550,000)
495,000 gal
R|0.071 lb C H OH / lb solution S| T0.86 lb H O / lb solution m
2
5
m
Final reaction mixture : 0.069 lb m (yeast, other species) / lb m solution 2
Mass of tank contents :
m
1 ft 3
65.52 lb m
7.4805 gal
1 ft 3
495,000 gal
Mass of ethanol produced : 5
4335593 lb m
4.336 u 106 lb m solution 0.071 lb m C 2 H 5OH
3.078 u 10 lb m C 2 H 5OH
307827 lb m C 2 H 5OH
lb m solution 1 lb - mole C 2 H 5OH 46.1 lb m C 2 H 5OH
6677 lb - mole C 2 H 5OH
1 ft 3 C 2 H 5OH
7.4805 gal
49.67 lb m C 2 H 5OH
1 ft 3
Makeup water required : 495,000 gal
3.078 u 105 lb m C 2 H 5
46,360 gal C 2 H 5OH
46,360 gal C 2 H 5OH 25 gal mash 2.6 gal C 2 H 5OH
9-17
4.9 u 104 gal
9.19 (cont'd)
46,360 gal C2 H5OH 1 bu 1 acre 1 batch 24 h 330 days acres 175 . u 105 Acres reqd. : 1 batch 2.6 gal C2 H5OH 101 bu 8 h 1 day 1 year year b.
C12 H 22 O11 (s) 12O 2 (g) o 12CO 2 (g) 11H 2 O(l) 'H co 'H co 12 'H fo (CO2 ) 11'H fo ( H 2 O) 'H fo (C12 H 22 O11 )
. kJ / mol 56491
'H fo (C12 H 22 O11 ) 221714 . kJ / mol C12 H 22 O11 (s) H 2 O(l) o 4C2 H 5OH(l) 4CO2 (g) 'H ro
c.
4 'H fo (C 2 H 5 OH) 4 'H fo (CO 2 ) 'H fo (C12 H 22 O11 ) 'H fo ( H 2 O) = 184.5kJ / mol 1815 . kJ 453.6 mol 0.9486 Btu 'H ro 7.811 u 104 Btu / lb - mole 1 mol 1 lb - mole 1 kJ Moles of maltose :
4.336 u 106 lb m solution 0.071 lb C 2 H 5OH 1 lb - mole C2 H 5 OH 1 lb - mole C12 H 22 O11 46.1 lb C2 H 5OH 4 lb - mole C2 H 5 OH 1 lb m solution 1669 lb - moles C12 H 22 O11 [ nC10H22O11
1669 lb - moles
Q = ['H r mC p (95$ F - 85$ F) Btu Btu ) (4.336 u 106 lb m )(0.95 $ )(10$ F) lb - mole lb - F 7 = 8.9 u 10 Btu ( heat transferred from reactor)
= (1669 lb - moles)(7.811 u 104
9.20
d.
Brazil has a shortage of natural reserves of petroleum, unlike Venezuela.
a.
4NH 3 5O 2 o 4NO 6H 2 O, 3 O 2 o N 2 3H 2 O 2 References: N 2 g , H 2 g , O 2 (g), at 25q C 2NH 3
bg bg
NH 3 Air NO
H in
nin
Substance
nout
H out
(mol min) (kJ mol) (mol min) (kJ mol) H 1 100 900 H2 90 H 3
H 2O
150
N2
716
H 4 H
O2
69
H 6
H i
'H foi
b
z
T
25
C pi dT
g
NH 3 g, 25q C : H 1
b
g
5
Air g, 150q C : H 2
Table B.1
( 'H fo ) NH 3
B
46.19 kJ mol
Table B.8
B
3.67 kJ mol
9-18
9.20 (cont'd)
b
g
NO g, 700q C : H 3
b
90.37 Table B.1, Table B.8
g
B
H 2 O g, 700q C : H 4
b
g
Q
g
'H
B
C p dT
111.97 kJ mol
216.91 kJ mol
B
20.59 kJ mol
Table B.8
O 2 g, 700q C : H 6 b.
25
Table B.1,Table B.2
Table B.8
N 2 g, 700q C : H 5
b
z
700
B
21.86 kJ mol
¦ n H ¦ n H i
i
out
i
i
4890 kJ min u (1 min / 60s) 815 . kW
in
(heat transferred from the reactor)
9.21
c.
If molecular species had been chosen as references for enthalpy calculations, the extents of each reaction would have to be calculated and Equation 9.5-1b used to determine 'H . The value of Q would remain unchanged.
a.
Basis: 1 mol feed I=Inert 1 mol at 310qC 0.537 C2H4 (v) 0.367 H20 (v) 0.096 N2(g)
Products at 310qC n1 (mol C2H4 (v)) n2 (mol H2O(v)) 0.096 mol N2 (g) n3 (mol C2H5OH (v)) n4 (mol (C2H5)2O) (v))
C 2 H 4 ( v) H 2 O(v) C 2 H 5OH(v)
b
g
2C 2 H 5OH(v) C 2 H 5 2 O(v) H 2 O(v) 5% ethylene conversion:
b0.537gb0.05g n1
0.02685 mol C 2 H 4 consumed
b0.95gb0.537g
0.510 mol C 2 H 4
90% ethanol yield: n3
0.02685 mol C 2 H 4 consumed 0.9 mol C 2 H 5OH
C balance :
0.02417 mol C 2 H 5OH
1 mol C 2 H 4
b2gb0.537g b2gb0.510g b2gb0.02417g 4n
n4
4
O balance : 0.367
n2 0.02417 1415 . u 10 3 n2
af b g b g
bg
0.3414 mol H 2 O
References: C s , H 2 g , O 2 g at 25$ C, I g at 310 $ C
9-19
b
g
1.415 u 10 3 mol C 2 H 5 2 O
9.21 (cont'd) substance
H in
nin
0.3414
0
0.096
C 2 H 5 OH
0.02417 1.415 u 10 3
H 4
b52.28 16.41g
2
4
o f C2 H 4
1
g
b
0.510
2
5 2
H 2 O g, 310q C : H 2
z
300
25
C p dT
g
( 'H fo ) C 2 H 5OH(g)
bC H g Obg, 310q Cg: H e'H j 5 2
o f
4
(C 2 H 5 )O(l)
z
310
25
'H
¦ n H ¦ n H i
i
out
C p dT
b
i
i
0 H
3
68.69 kJ mol
. 9.93g b24183
b235.31 24.16g
Table B.1 Table B.8
Table B.1 Table B.2
g
'H v 25q C
204.2 kJ mol Energy balance: Q
2
Table B.1 for 'H fo Table B.2 for Cp
( 'H fo ) H 2 O(v) H H 2 O(v) (310$ C)
C 2 H 5OH g, 310q C : H 3
2
(kJ / mol) H 1 H
0.096
bC H g O C H bg, 310q Cg : H ( 'H )
b.
(mol)
I
H2O
b
H out
(mol) (kJ / mol) 0.537 H 1 0.367 H
C2H4
2
nout
z
310
25
C p dT
231.90 kJ mol
21115 . kJ mol
b272.8 26.05 42.52g
1.3 kJ 1.3 kJ transferred from reactor mol feed
in
To suppress the undesired side reaction. Separation of unconsumed reactants from products and recycle of ethylene. 9.22
C 6 H 5CH 3 O 2 o C 6 H 5CHO H 2 O C 6 H 5CH 3 9O 2 o 7CO 2 4H 2 O Basis: 100 lb-mole of C 6 H 5CH 3 fed to reactor.
100 lb-moles C 6 H5 CH 3 n0 (lb-moles O 2 ) 3.76n 0 (lb-moles N 2 ) 350°F, 1 atm V0 (ft3 )
reactor Q(Btu) jacket
mw(lbm H2 O( l )), 80°F
Vp (ft3 ) at 379°F, 1 atm n1 (lb-moles C6 H5 CH3 ) n2 (lb-moles O 2 ) 3.76n0 (lb-moles N 2 ) n3 (lb-moles C 6 H5 CHO) n4 (lb-moles CO2 ) n5 (lb-moles H2 O) mw(lbm H2 O( l )), 105°F
Strategy: All material and energy balances will be performed for the assumed basis of 100 lb-mole C 6 H 5CH 3 . The calculated quantities will then be scaled to the known flow rate of water in
b
g
the product gas 29.3 lb m 4 h .
9-20
9.22 (cont'd)
Plan of attack: % excess air n0 13% C6 H5CHO formation n3 0.5% CO2 formation n4 C balance n1 H balance n5 O balance n2
Ideal gas equation of state V0 Ideal gas equation of state V p E.B. on reactor Q E.B. on jacket mw Scale V0 , Vp , Q, mw by n5 actual / n5 basis
b g
100% excess air: n0
100 lb - moles C 6 H 5CH 3
b
N 2 feed & output
g
b1 1gmole O
1 mol O 2 reqd 1 mole C 6 H 5CH 3
b g
3.76 200 lb - moles N 2
fed 1 mol O 2 reqd 2
b g
200 lb - moles O 2
752 lb - moles N 2
100 lb - moles C6 H5CH 3 0.13 mol C6 H5CH 3 react 1 mole C6 H5CH 3 1 mole C6 H5CH 3 fed 1 mole C6 H5CH 3
13% o C6 H5CHO n3
= 13 lb - moles C7 H 6 O
b100gb0.005glb - moles C H CH 6
0.5% o CO2 n4
5
3
react
7 moles CO2 35 . lb - moles CO2 1 mole C6 H5CH 3
mol C mole C7H8
C balance: H balance: O balance:
a100faB7f lb - moles C 7n a13fa7f a3.5fa1f n 86.5 lb - moles C H CH b100gb8glb - moles H b86.5gb8g b13gb6g 2n n 15.0 lb - moles H Obvg b200gb2glb - moles O 2n b13gb1g b35. gb2g b15gb1g n 182.5 lb - moles O 1
1
5
b100 + 200 + 752glb - moles
b g b350 460g R
FG 86.5 182.5 H
Vp
O2
C 7 H 8O
CO 2
H 2O
$
359 ft 3 STP 1 lb - moles
$
492 R
IJ K
N2
Energy balance on reactor (excluding cooling jacket)
bg b g b g b g
e
6.218 u 105 ft 3
b379 460g R
1 lb - mole
492 $ R
$
j
References : C s , H 2 g , O 2 g , N 2 g at 25$ C 77 $ F nin
H in
nout
H out
C 6 H 5CH 3
100
86.5
O2
200
182.5
N2
752
H 1 H 2 H
H 4 H 5 H
C 6 H 5CHO
13
CO 2
3.5
H 7 H
H 2O
15
H 9
substance
blb - molesg bBtu lb - moleg blb - molesg bBtu lb - moleg 3
9-21
752
3
2
359 ft 3
13 3.5 15 752 lb - moles
5
2
2
Ideal gas law – outlet: C7 H 8
5
2
Ideal gas law inlet:
V0
6
6
8
6.443 u 105 ft 3
9.22 (cont'd) Enthalpies:
LM B OP 430.28 Btu lb - mole Btu 31 C H CH (g,T): H bT g 'H b kJ molg u bT 77g FP MM 1 kJ mol 1b - moleq F PQ N Table B.1
6
5
C 6 H 5CH 3 (g,350$ F): H 1 C H CH (g,379 $ F): H 6
5
$
o f
3
2.998 u 104 Btu lb - mole 3.088 u 104 Btu lb - mole
4
3
bg
b
C 6 H 5 CHO(g,T): H T H 7
g
$
17200 31 T 77 F Btu lb - mole
7.83 u 103 Btu lb - mole Table B.9
e
$
e
$
e
$
e
$
j
H O 2 (350 F)
j
H N 2 (350 F)
j
H O 2 (379 $ F)
j
H N 2 (379 $ F)
O 2 g,350 F : H 2
B
$
1.972 u 10 3 Btu / lb mole
Table B.9
N 2 g,350 F : H 3
B
$
u 10 3 Btu / lb mole 1911 .
Table B.9
O 2 g,379 F : H 5
B
2.186 u 103 Btu / lb mole
Table B.9
N 2 g,379 F : H 6
B
2.116 u 10 3 Btu / lb mole Table B.1 and B.9
e j H Obg,379q Fg: H
CO 2 g,379 F : H 8 $
2
( 'H f$ ) CO2 ( g)
H CO2 (379 F)
( 'H f$ ) H 2 O ( g)
H H 2 O (379 F)
$
B
1664 u 105 Btu / lb mole .
Table B.1 and B.9
9
$
B
1016 u 105 Btu / lb mole .
Energy Balance : Q
'H
¦ n H ¦ n H i
i
i
out
i
2.376 u 10 6 Btu
in
Energy balance on cooling jacket:
Q
'H
mw Q
z
105
80
b g
mw lb m u 1.0
bn g bn g
5 actual
V0 Vp
b.
Q
w m
b
29.3 lb m H 2 O
g
bg
9.504 u 10 4 lb m H 2 O l
1b - mole H 2 O
1
18.016 lb m H 2 O 15.0 lb - moles H 2 O
d6.218 u 10 ft id0.02711 h i d6.443 u 10 ft id0.02711 h i d2.376 u 10
1.0 Btu (lb m $ F)
Btu $ u 105 80 F mw lb m q F
4h
5 basis
a.
b g dT
p H O l 2
2.376 u 10 4 Btu , C p
2.376 u 10 6 Btu
Scale factor:
dC i
0.02711 h 1
b g h b product g
5
3
1
169 . u 10 4 ft 3 h feed
5
3
1
175 . u 10 4 ft 3
id
i
6.44 u 10 4 Btu / h
id
i
2577 lb m 1 ft 3 7.4805 gal 1 h . gal H 2 O min 515 h 62.4 lb m 1 ft 3 60 min
6
Btu 0.02711 h 1
. u 10 d9504
Btu 0.02711 h 1
4
9-22
9.23
a.
CaCO 3 ( s) o CaO(s) +CO 2 (g) CaO(s) 900qC CaCO3(s) 25qC
CO2(g) 900qC Q (kJ)
1000 kg
Basis : 1000 kg CaCO 3
10.0 kmol CaO(s) produced
1 mol
10.0 kmol CaCO 3 10.0 kmol CO 2 (g) produced
0.100 kg
10.0 kmol CaCO 3 (s) fed
References: Ca(s), C(s), O2(g) at 25qC
nin nout H in H out Substance (mol) (kJ / mol) (mol) (kJ / mol) CaCO 3 10.0 H 1 CaO H2 10.0 CO 2 10.0 H 3 Table B.1
CaCO 3 (s, 25 C) : H 1 o
B
( 'H fo ) CaCO 3 ( s)
z
1206.9 kJ / mol Table B.1, Table B.2
1173
CaO(s, 900 o C) : H 2
( 'H fo ) CaO ( s)
B
C p dT
( 635.6 48.54) kJ / mol
587.06 kJ / mol
298 Table B.1, Table B.8
CO 2 (g, 900 o C) : H 3 Energy balance: Q
( 'H fo ) CO2 ( g) H CO2 (900 o C)
'H
F n H n H I GH ¦ ¦ JK i
out
b.
i
i
i
B
( 393.5 42.94) kJ / mol
2.7 u 10 6 kJ
in
Basis : 1000 kg CaCO3 fed 10.0 kmol CaCO3 CaCO 3 ( s) o CaO(s) + CO 2 (g) 2CO + O 2 o 2CO 2 10 kmol CaCO3 25 oC 200 kmol at 900oC 0.75 N2 0.020 O2 0.090 CO 0.14 CO2
10 kmol CaCO 3 react n1
Product gas at 900oC n2 (kmol CO2 ) n3 (kmol N2 ) n4 (kmol CO) n1 [kmol CaO(s)]
10.0 kmol CaO
9-23
350.56 kJ / mol
9.23 (cont'd) n2
(0.14)(200)
n3
(0.75)(200)
10.0 kmol CaCO 3 react 1 kmol CO 2 1 kmol O 2
4 kmol O 2 react 2 kmol CO 2 1 kmol O 2
46 kmol CO 2
150 kmol N 2
C balance: (10.0)(1) + (200)(0.09)(1) + (200)(0.14)(1) = 46(1) + n4 (1) n4
10.0 kmol CO
References : Ca(s), C(s), O 2 (g), N 2 (g) at 25$ C H in
nin
H out
nout
Substance (mol) (kJ / mol) (mol) (kJ / mol) CaCO 3 10.0 H 1 CaO
10
587.06
CO 2
28
46
CO
18
350.56 H
350.56 H
O2
4.0
N2
150
10
1
H 2 H
1
H
150
3
3
Table B.1, Table B.8
CO(g, 900 o C) : H 1
( 'H fo ) CO( g) H CO (900 o C)
O 2 (g, 900 C) : H 2
H O 2 (900 o C)
B
( 110.52 27.49) kJ / mol
83.03 kJ / mol
Table B.8 o
B
28.89 kJ / mol
Table B.8
N 2 (g, 900o C) : H 3 Q
'H
H N 2 (900 o C)
F n H n H I GH ¦ ¦ JK i
i
i
out
i
9.24
a.
27.19 kJ / mol
0.44 u 106 kJ
in
% reduction in heat requirement c.
B
2.7 u 106 0.44 u 10 6 2.7 u 10 6
u 100
838% .
The hot combustion gases raise the temperature of the limestone, so that less heat from the outside is needed to do so. Additional thermal energy is provided by the combustion of CO.
A+Bo C
(1)
2C o D + B
(2)
Basis: 1 mol of feed gas 1.0 mol x AO (mol A / mol)
n A (mol A)
x BO (mol B / mol)
n B (mol B)
x IO (mol I / mol)
nC (mol C) n D (mol D) n I (mol I) T ( $ C)
Fractional conversion:
fA
mol A consumed mol A feed
9-24
x AO n A nA x AO
x AO (1 f A )
9.24 (cont'd)
C generated: n0 nC
x A0 (mol A fed)
f A (mol A consumed) YC (mol C generated) mol A fed
mol A consumed
x AO f A YC
D generated: nD = 0.5 u mol C consumed = (1 2) u (mol A consumed mol C out) nD
(1 2)( x AO f A nC )
Balance on B: mol B out = mol B in mol B consumed in (1) + mol B generated in (2) = mol B in mol A consumed in (1) + mol D generated in (2) n B x BO x AO f A n D Balance on I: mol I out = mol I in n I
x IO
b.
Species Formula A C2H4(v) B H2O(v) C2H5OH(v) C D C4H10)O(v I N2(g)
c. 9.25
a.
Tf 310
Tp 310
Species A B C D I
n(in) (mol) 0.537 0.367 0 0 0.096
Q(kJ) =
-1.31
DHf 52.28 -241.83 -235.31 -246.75 0
a 0.04075 0.03346 0.06134 0.08945 0.02900
b 1.15E-04 6.88E-06 1.57E-04 4.03E-04 2.20E-05
xB0 0.367
xI0 0.096
xA0 0.537 H(in) (kJ/mol) 68.7 -231.9 -211.2 -204.2 9.4
n(out) (mol) 0.510 0.341 0.024 0.001 0.096
c -6.89E-08 7.60E-09 -8.75E-08 -2.24E-07 5.72E-09 fA 0.05
d 1.77E-11 -3.59E-12 1.98E-11 0 -2.87E-12 YC 0.90
H(out) (kJ/mol) 68.7 -231.9 -211.2 -204.2 9.4
For T f = 125o C, Q = 7.90 kJ . Raising Tp, lowering fA, and raising YC all increase Q. CH 4 ( g) O 2 ( g) o HCHO(g) + H 2 O(g) n3 (mol HCHO) n4 (mol H2O)
10 L, 200 kPa n0 (mol feed gas) at 25qC 0.851 mol CH4/mol 0.15 mol O2 /mol
n5 (mol CH4) T (qC), P(kPa), 10L Q (kJ)
9-25
9.25 (cont'd) Basis : n0
200 kPa 1000 Pa 10 L 10 3 m 3 1 kPa
1 mol K 8.314 m 3 Pa
1L
298 K
0.8072 mol feed gas mixture 0.8072 mol feed gas mixture (0.85)(0.8072) = 0.6861 mol CH 4 , (0.15)(0.8072) = 0.1211 mol O 2
1 mol CH 4 0.1211 mol O 2 fed 1 mol O 2 fed
CH 4 consumed : n5
(0.6861 01211 . ) mol CH 4
0.5650 mol CH 4
1 mol HCHO
HCHO produced : n3
mol CH 4 . 01211
01211 . mol CH 4 consumed
1 mol CH 4 consumed 1 mol H 2 O
H 2 O produced : n4
0.1211 mol CH 4 consumed
1 mol CH 4 consumed
Extent of reaction : [
( nO 2 ) out ( nO 2 ) in
0 0.1211
Q O2
1
0.1211 mol HCHO 01211 . mol H 2 O
0.1211 mol
References : CH 4 (g), O 2 (g), HCHO(g), H 2 O(g), at 25o C Substance CH 4
z T
U i
U in
nin
nout
mol kJ mol mol kJ mol 0.6861 0 0.5650 U 1
O2
01211 .
0
HCHO
01211 .
H 2O
01211 .
z
U out
U 2 U
3
T
(Cv ) i dT
25
(C p R ) i dT
1,2,3
i
25
Using (C p ) i from Table B.2 and R = 8.314 u 10 3 kJ / mol K: U 1 U
(0.02599 T 2.7345 u 10 5 T 2 0.1220 u 10 8 T 3 2.75 u 10 12 T 4 0.6670) kJ / mol
U 3
(0.02515 T 0.3440 u 10 5 T 2 0.2535 u 10 8 T 3 0.8983 u 10 12 T 4 0.6309) kJ / mol
(0.02597 T 2.1340 u 10 5 T 2 2.1735 u 10 12 T 4 0.6623) kJ / mol
2
Q
100 J 85 s s
1 kJ
8.5 kJ
1000 J
Table B.1
'H ro
('H fo ) HCHO ('H fo ) H2O ('H fo ) CH4
B
. ) (24183 . ) (74.85)g kJ / mol b(11590
28288 . kJ / mol 'U ro
'H ro RT (
¦
Qi
gaseous products
282.88 kJ / mol
¦Q
i
)
gaseous reactants
8.314 J
298 K (1 + 1 1 1)
1 kJ 10 3 J
mol K 9-26
282.88 kJ / mol
9.25 (cont'd)
Energy Balance : Q ['U ro
¦ (n
i ) out (U i ) out
¦ (n
i ) in (U i ) in
. kJ / mol) +0.5650 U 1 01211 . . (0.1211)(28288 U 2 01211 U 3 Substitute for U 1 through U 3 and Q
u 10 5 T 2 0.09963 u 10 8 T 3 1926 u 10 12 T 4 4329 0 0.02088 T 1845 . . . kJ / mol Solve for T using E - Z Solve T 1091o C 1364 K 1L 0.8072 mol 8.314 m3 Pa 1364 K mol K 10 L 10 3 m3
P nRT / V
915 u 103 Pa 915 kPa
Add heat to raise the reactants to a temperature at which the reaction rate is significant. b. c.
9.26
Side reaction : CH 4 2O2 o CO2 2H 2 O. T would have been higher (more negative heat of reaction for combustion of methane), volume and total moles would be the same, therefore P nRT / V would be greater.
bg
a.
bg
1 O 2 (g) o C 2 H 4 O g 2 C 2 H 4 3O 2 o 2CO 2 2H 2 O C2 H 4 g
Basis: 2 mol C 2 H 4 fed to reactor
n 6 (mol CO2 ) n 7 (mol H 2 O(l )) 25°C
Qr (kJ) heat n1 (mol C 2H 4) n2 (mol O 2 ) 25°C
reactor
2 mol C2 H4 1 mol O2 450°C
n 3 (mol C 2H 4) n 4 (mol O 2 )
separation process n 3 (mol C 2H 4) n 4 (mol O 2 ) n 5 (mol C 2H 4O) n 6 (mol CO2 ) n 7 (mol H 2 O) 450°C
25% conversion 0.500 mol C 2 H 4 consumed n 3 70% yield n5
150 . mol C 2 H 4
0.500 mol C 2 H 4 consumed 0.700 mol C 2 H 4 O
C balance on reactor: Water formed: n 7 O balance on reactor:
1 mol C 2 H 4
. g b2gb0.350g n b2gb2g b2gb150 0.300 mol CO 2
b2gb1g
1 mol H 2 O 1 mol CO 2
n 5 (mol C 2H 4O(g)) 25°C
6
n6
0.350 mol C 2 H 4 O
0.300 mol CO 2
0.300 mol H 2 O
0.375 mol O b gb g 0.300 b2gb0.350g n 0.500 mol C H n b2gb0.300g b0.300g b0.350g n 0.625 mol O
2n 4 0.350 2 0.300 0.300 n4
Overall C balance: 2n1
n 6 2n5
Overall O balance: 2n 2
2n 6 n 7
1
5
2
2
2
9-27
4
2
9.26 (cont'd)
Feed stream: 44.4% C 2 H 4 , 55.6% O 2
Reactor inlet: 66.7% C 2 H 4 , 33.3% O 2
Recycle stream: 80.0% C 2 H 4 , 20.0% O 2 Reactor outlet: 53.1% C 2 H 4 , 13.3% O 2 , 12.4% C 2 H 4 O, 10.6% CO 2 , 10.6% H 2 O 0.350 mol C 2 H 4 O 44.05 g
Mass of ethylene oxide b.
1 kg
bg
0.0154 kg
10 3 g
1 mol
bg bg
References for enthalpy calculations : C s , H 2 g , O 2 g at 25q C
bg
H i T
'H ofi
z z
T
25
'H 0f
C p dT for C 2 H 4
T 273
298
C p dT for C 2 H 4 O
'H ofi H i ( table B.8)
bg
for O 2 , CO 2 , H 2 O g
bg
'H of for H 2 O l
Overall Process H in
nin
Substance
Reactor H out
nout
(mol) (kJ / mol) (mol) (kJ / mol) 52.28 0.500
C2 H4
H out
(mol) (kJ / mol) (mol) (kJ / mol) 79.26 150 79.26 . 2
C2 H 4
O2
1
0350 .
5100 .
C2 H 4 O
0.350
19.99
0300 .
3935 .
CO 2
0.300
374.66
0300 .
28584 .
H2O g
0.300
226.72
0625 .
C2 H4 O CO2 H2 O l
0
bg
¦ n H ¦ n H
'H
Energy balance on process: Q
i
i
i
out
1337 .
0.375
1337 .
248 kJ
i
in
¦ n H ¦ n H
'H
Energy balance on reactor: Q
i
out
c.
nout
O2
bg
H in
nin
substance
i
i
236 kJ
i
in
Scale to 1500 kg C 2 H 4 O day : C 2 H 4 O production for initial basis Scale factor
(0.350 mol)(
1500 kg day 0.01542 kg
44.05 kg 10 3 mol
)
0.01542 kg C 2 H 4 O
9.73 u 10 4 day 1
U| V| M b0.500gb28.05 g C H molg b0.625gb32.0 g O molg W = 34.025 u 10 kg kgje9.73 u 10 day j 3310 kg day (44.4% C H , 55.6% O ) day j 1 day 1 hr 1 kW 279 kW
In initial basis, fresh feed contains 0.500 mol C 2 H 4 0.625 mol O 2
2
Qprocess
e34.025 u 10 b248 kJ ge9.73 u 10
Qreactor
b236 kJge9.73 u 10
Fresh feed rate
4
2
3
3
4
4
1
2
1
24 hr 3600 s 1 kJ s
4
day 1
j
1 day
1 hr
1 kW
24 hr 3600 s 1 kJ s
9-28
265 kW
4
2
9.27
a.
Basis:
1200 lb m C 9 H 12
1 lb - mole
h 120 lb m Overall process : n 1 (lb-moles/h) 0.75 C 3H 6 0.25 C 4H 10
10.0 lb - moles cumene produced h n 3 (lb-moles C3 H 6 /h) n 4 (lb-moles C4 H10 /h)
n 2 (lb-moles C6 H 6 /h)
bg
10.0 lb-moles C9 H12 /h
bg
bg
b
Benzene balance: n 2
binput
consumption
g
output consumption
h
1 mole C 9 H12 produced
10.0 lb - moles C 6 H 6
78.1 lb m C 6 H 6
h
1 lb - mole
1 mole C 3 H 6
h
1 mole C 9 H12
g
b
Mass flow rate of C 3 H 6 / C 4 H 10 feed
UV n 16.67 lb - moles h 2.50 lb - moles C H h n gW b0.75gb16.67glb - moles C H 42.08 lb C H 1
3
b0.25gb16.67glb - moles C H 4
10
m
58.12 lb m C 4 H 10 1 lb - mole
h
1 mole fresh feed 10.0 lb-moles C9H12/h 2.50 lb-moles C3H6/h 4.17 lb-moles C4H10/h 30.0 lb-moles C6H6/h 400oF
40 lb - moles C 6 H 6 h
46.7 lb-moles/h 21.4% C9H12 5.4% C3H6 8.9% C4H10 64.3% C6H6
.67 lb - moles h UV 637.5% C H hW 62.5% C H
2.50 lb - moles C 3 H 6 h 4.17 lb - moles C 4 H 10
3
6
4
10
Heat exchanger : Reactor effluent at 400°F 10.0 lb-moles C9H12 /h 2.50 lb-moles C3H6 /h 4.17 lb-moles C4H10 /h 30.0 lb-moles C6H6 /h
200°F
40.0 lb-moles C6H6 /h 77°F
T (°F)
Energy balance: 'H (Assume adiabatic)
0
9-29 ni H i , out H i , in
¦ e
6
768 lb m h
a3 1fmoles fed to reactor
Overhead from T1
3
1 lb - mole
10.0 lb - moles fresh feed
40.0 lb-moles C6H6/h
6
6
h
h
16.67 lb-moles/h @ 77oF 0.75 C3H6 0.25 C4H10
b.
3
3
Reactor : Benzene feed rate
781 lb m C 6 H 6 h
10.0 lb - moles C 9 H12
0.75n1 n3 10 20% C 3 H 6 unreacted n3 0.20 0.75n1
39520 Btu lb - mole
10.0 lb - moles C 9 H12 produced 1 mole C 6 H 6 consumed
Propylene balance: 0.75n1 n3
binput
g
'H r 77q F
C 3 H 6 l C 6 H 6 l o C 9 H 12 l ,
j ¦ n C bT i
pi
out
Tin
g
i
0
9.27 (cont'd) 0
Energy balance: 'H
¦ n e H i
i , out
H i , in
j ¦ n C bT i
pi
out
Tin
g
0
i
(Assume adiabatic)
LM10 lb - moles C H h N 9
C 4 H10
B
b gb
120 lb m
12
1 lb - mole
$
1b m
gb ge
C 3H 6
OPe200 F 400 Fj b2.B50gb42.08gb0.57ge200 F 400 FQ
0.40 Btu
j b A gb
$
$
$
gb ge
j
4.17 5812 0.55 200$ F 400$ F 30.0 7811 . . 0.45 200 $ F 400$ F
b A gb
C6H6 in effluent
gb ge
j
40.0 7811 . 0.45 T 77 $ F
0 T
323q F
C6H6 fed to reactor
(Refer to flow chart of Part b: T 323q F ) References : C 3 H 6 l , C 4 H 10 l , C 6 H 6 l , C 9 H 12 l at 77q F H i Btu lb - mole C pi Btu lb m q F M i lb m lb - mole T 77 q F
b
bg
g
bg
b
bg
H in
n in
Substance
bg g b
gb
gb g H out
n out
(lb - mole / h) (Btu / lb - mole) (lb - mole / h) (Btu / lb - mole) 0 2.50 7750 12.0
C3H 6 C 4 H10
4.17
0
4.17
10330
C6H 6
40.0
C 9 H12
8650
30.0
11350
10.0
15530
Energy balance on reactor : n C 9 H12 'H ro Q 'H ni H i vC 9 H12 out
¦
¦ n H i
i
in
b10.0gb39520g b2.50gb7750g b4.17gb10330g b30.0gb11350g b10.0gb15530g b1g b40.0gb8650g 183000 Btu h b heat removalg
9.28
Basis :
a.
100 kg C 8 H 8 10 3 g h
1 kg
1 mol 104.15 g
960 mol h styrene produced
C 8 H 10 (g) o C 8 H 8 (g) H 2 (g) Overall system n 2 (mol H2 /h)
Fresh feed n 1 (mol C8H10/h)
960 mol C8H8 /h
Fresh feed rate: n1
bC H 8
10
balance
g
H 2 balance : n 2
960 mol C 8 H 8 1 mol C 8 H10
h 960 mol C 8 H10 h
1 mol C 8 H 8 1 mol H 2 1 mol C 8 H10
9-30
960 mol C 8 H10 h fresh feed 960 mol H 2 h
9.28 (cont'd)
Reactor : .
n 3 (mol C8H10 /h) n 4 (mol H2O( v )/h) 600°C
n5 (mol C8H10 /h) v n 4 (mol H2O(v)/s) 960 (mol C8H8 /s) 960 (mol H2 /s) 560°C
Qc (kJ/h)
35% 1-pass conversion n 3
b
0.35n3 mol C 8 H 10 react
2740 mol C 8 H10
h h fed to reactor
a2740 960f
Recycle rate
g
1 mol C 8 H 8 1 mol C 8 H 10
960 mol C 8 H 8 h
1780 mol C 8 H10 h recycled
Reactor feed mixing point 2740 mol C8H10(v)/h 500oC 2740 mol C8H10(v)/h n4 [mol H2O(v)/h] 600oC
n4 [mol H2O(v)/h] 700oC
2740 'H C8H10 n 4 'H H 2 O
Energy balance: 'H
b Neglect Q, 'E g k
LM OP J 1 kJ u .30T gdT P MM b 118 0 mol C 10 J P N Q
z
'H C8 H10 'H H 2 O
b g
0 kJ h
600
Table B.8
28.3 kJ mol
3
$
500
Cp
3.9 kJ mol
P 1 bar
a2740fa28.3f n a3.9f
1.99 u 10 4 mol H 2 O / h
0 n 4
4
bg
Ethylbenzene preheater A : b.
bg
960 mol fresh feed 1780 mol recycled 2740 mol EB l at 25q C h h h 2740 mol EB v at 500q C h 136 500 'H C pi dT 'H v 136q C C pv dT 20.2 36.0 77.7 kJ mol 133.9 kJ mol
bg
z
a
25
Q A
'H
f
z
a
136
2740 mol C 8 H10
133.9 kJ
h
mol C 8 H10
f
b
3.67 u 10 5 kJ h preheater
bg 19400 mol h H Obl, 25q Cg o 19400 mol h H Ob v, 700q C, 1 atmg Table B.5 H bl, 25q Cg 104.8 kJ kg ; Table B.7 H b v, 700q C, 1 atm | 1 bar g 3928 kJ kg Steam generator F : 2
2
9-31
g
9.28 (cont'd) Q F
19400 mol H 2 O 18.0 g
'H
h
1.34 u 10
6
1 kg 3
a3928 104.8fkJ
1 mol 10 g
b
kJ h steam generator
kg
g
bg
Reactor C :
bg
bg bg
bg
References: C8 H 8 v , C8 H 10 v , H 2 g , H 2 O v at 600q C
e
H i 560 $ C
j
zd 560
600
i
C pv i dT for C 8 H 10 , C 8 H 8
| H (T) for H 2 , H 2 O (interpolating from Table B.8) n in n out Substance H H out
in
C 8 H10
(mol h ) (kJ mol) (mol h ) (kJ mol) 11.68 0 1780 2740 0
19900
C8H8
19900
960
1.56 10.86
H2
960
119 .
H2O
Energy balance : Q c
'H
960 mol C 8 H 8 produced
124.5 kJ
h
1 mol C 8 H 8
a
5.61 u 10 4 kJ h reactor c.
¦ n H ¦ n H i
out
i
i
in
H2
1 O2 o H 2O 2 CH 3 OH O2 , N 2 H2 product gas 145°C separation units
a. n f (mol/h) at 145°C, 1 atm 0.42 mol CH 3 OH/mol 0.58 mol air/mol 0.21 mol O2 /mol air 0.79 mol N2 /mol air n s mol H2 O(v )/h saturated at 145°C
b.
i
This is a poorly designed process as shown. The reactor effluents are cooled to 25$ C , and then all but the hydrogen are reheated after separation. Probably less cooling is needed, and in any case provisions for heat exchange should be included in the design. CH 3OH o HCHO H 2 ,
9.29
f
reactor
reactor product gas, 600°C n 1 (mol CH3 OH/h) waste n 2 (mol O 2 /h) heat n 3 (mol N 2 /h) boiler n 4 (mol HCHO/h) 0.37 kg HCHO/h n 5 (mol H 2 /h) 0.63 kg H 2O/h n 6 (mol H 2 O/h) mb (kg H2 O(v)/h) mb (kg H2 O(v)/h) 30°C sat'd at 3.1 bars
In the absence of data to the contrary, we assume that the separation of methanol from formaldehyde is complete. Methanol vaporizer:
bg
e j
The product stream, which contains 42 mole % CH 3OH v , is saturated at Tm $ C and 1 atm.
9-32
9.29 (cont'd)
b g b gb
pm Tm 0.42 760 mmHg
ym P
o p m Antoine equation
c.
g
b g
319.2 mmHg = pm Tm
319.2 mmHg Tm
44.1$ C
Moles HCHO formed : 36 u 106 kg solution 0.37 kg HCHO 350 days
1 kg solution
1 kmol
1 day
30.03 kg HCHO
24 h
52.80
kmol HCHO h
52.80 kmol HCHO h
but if all the HCHO is recovered, then this equals n4 , or n4 70% conversion : 52.80 kmol HCHO
1 kmol CH 3OH react
h n f
1 kmol CH 3OH fed
1 kmol feed gas
1 kmol HCHO formed 0.70 kmol CH 3OH react 0.42 kmol CH 3OH
n f
179.59 kmol h
Methanol unreacted: n1
b0.42gb179.59gkmol CH OH fed b1 0.70g kmol CH OH fed 3
3
h
1 kmol CH 3OH fed
b179.6 kmol hgb0.58gb0.79g
N 2 balance: n3
22.63
kmol CH 3OH h
82.29 kmol N 2 h
Four reactor stream variables remain unknown — n s , n2 , n5 , and n6 — and four relations are available — H and O balances, the given H 2 content of the product gas (5%), and the energy balance. The solution is tedious but straightforward.
b
gb gb g
H balance: 179.6 0.42 4 2ns n s
b
b22.63gb4g b52.8gb2g 2n 2n 5
n 5 n 6 52.80
gb gb g b
(1)
g
b gb g
O balance: 179.6 0.42 1 179.6 (0.58) 0.21 2 n s n s
(2)
bg b g b g b g
References : C s , H 2 g , O 2 g , N 2 g at 25q C H
'H fo
z
Table B.2 T
25
(22.63)(1) 2n 2 (52.80)(1) n 6
2n 2 n 6 43.75
n 5 22.63 n 2 82.29 52.89 n 5 n 6
H 2 content:
6
B
C p dT
or Table B.8 for O 2 , N 2 and H 2
9-33
0.05 19n 5 n 2 n 6
157.72
(3)
9.29 (cont'd) substance
n in kmol / h
H in kJ / kmol
n out kmol / h
H out kJ / kmol
75.43 . 2188 82.29
195220
22.63
163200
3620 3510
n2 82.29
ns
237740
n6 52.80
18410 17390 220920 88800
n5
16810
CH 3 OH O2 N2 H 2O HCHO H2
Energy Balance : 'H
¦n H ¦n H i
i
out
i
0 18410n2 16810n5 220920n6 237704ns
i
7.406 u 106
(4)
in
We now have four equations in four unknowns. Solve using E-Z Solve.
bg
n s
58.8 kmol H 2 O v h
18.02 kg 1 kmol
n2
2.26 kmol O 2 h , n5
1060 kg steam fed h
. kmol H 2 h , n6 1358
98.00 kmol H 2 O h
Summarizing, the product gas component flow rates are 22.63 kmol CH3OH/h, 2.26 kmol O2/h, 82.29 kmol N2/h, 52.80 kmol HCHO/h, 13.58 kmol H2/h, and 98.02 kmol H2O/h d.
272 kmol h product gas 8% CH 3 OH, 0.8% O 2 , 30% N 2 , 19% HCHO, 5% H 2 , 37% H 2 O
Energy balance on waste heat boiler. Since we have already calculated specific enthalpies of all components of the product gas at the boiler inlet (at 600qC), and for all but two of them at the boiler outlet (at 145qC), we will use the same reference states for the boiler calculation
bg b g b g b g H Oblg at triple point for boiler water
Reference States: C s , H 2 g , O 2 g , N 2 g at 25q C for reactor gas 2
Substance CH 3 OH O2 N2 H 2O HCHO
n in kmol / h 22.63 2.26 82.29 98.02 52.80 13.58
H2 H 2O
mb ( kg / h)
H in kJ / kmol 163200 18410
n out mol 22.63 2.26
H out kJ / mol 195220 3620
17390 220920 88800
82.29 98.02
3510 237730 111350
16800 125.7 (kJ / kg)
Energy Balance : 'H
¦ n H ¦ n H i
out
b
i
i
in
i
0
g
mb 2726.1 125.7 4.92 u 10 6 mb
0
1892 kg steam h
9-34
52.80 13.58 mb (kg / h)
3550 2726.1 ( kJ / kg )
9.30
a.
C 2 H 4 HCl o C 2 H 5Cl
bg
Basis:
1600 kg C 2 H 5Cl l
103 g
1 mol
h
1 kg
64.52 g
C
n 3 (mol HCl(g)/h) n 4 (mol C 2H 4( g)/h) condenser n 5 (mol C 2H 6( g)/h) n 6 (mol C 2H 5Cl( g)/h) 50°C
A n 1 (mol HCl(g)/h) 0°C
n 3 (mol HCl(g)/h) n 4 (mol C 2H 4( g)/h) n 5 (mol C 2H 6( g)/h) 0°C
n 6 (mol C 2H 5Cl( g)/h) reactor
B
24800 mol h C 2 H 5Cl
n 2 (mol/h) at 0°C 0.93 C 2H 4 0.07 C 2H 6
D
( n 6 – 24,800) (mol C 2H 5 Cl( l)/h) 0°C
24,800 mol C2 H5 Cl(l )/h
Product composition data: n3
0.015n1
n4
0.015 0.93n2
n5
0.07n2
b
g
b1g b2 g b3g
0.01395n2
Overall Cl balance :
b
n1 mol HCl h
g
1 mol Cl 1 mol HCl
bn gb1g b24800gb1g
b4g
3
Solve (4) simultaneously with (1) n1
25180 mol h
2518 . kmol HCl fed / h
bg
378 mol HCl g h
n3 Overall C balance :
b gb g b gb g
n2 0.93 2 n2 0.07 2
b gb
g
2n4 2n5 2 24800
LM N
From Eqs. (2) and (3) 2n2 0.93 0.07 0.0139 0.07
b.
c.
n2
27070 mol fed h
n3
378 mol HCl h
n4
0.01395 27070 = 378 mol C 2 H 4 h
n5
0.07 27070 = 1895 mol C 2 H 6 h
b
b
g
OP b2gb24800g Q
27.07 kmol h of Feed B
g
bg
bg
U| 2.65 kmol / h of Product C V| 14.3% HCl, 14.3% C H , 71.4% C H W 2
bg
bg
References : C 2 H 4 g , C 2 H 6 g , C 2 H 5 Cl g , HCl g at 0 $ C
e
j
e
j
C 2 H 4 g, 50$ C : H C 2 H 4 g, 50$ C : H
z z
50
C p dT
0
50
C p dT
0
Table B.2
Table B.2
2.181 kJ mol 2.512 kJ mol
9-35
4
2
6
9-30 (cont’d)
e
z
j
HCl g, 50$ C : H
Table B.2
50
1.456 kJ mol
C p dT
0
e j ' H e0 Cj 24.7 kJ mol C H Cleg, 50 Cj: H C dT 2.709 kJ mol C 2 H 5Cl l, 0$ C : H 2
z
$
5
$
v
50
pv
0
HCl
nin mol 25180
H in kJ / mol 0
nout mol 378
H out kJ / mol 1456 .
C2 H 4
25175
0
378
2.181
C2 H 6
1895
0
1895
2.512
C 2 H 5Cl
n6 24800
24.7
n6
2.709
substance
Energy balance: 'H
0
e j
nA 'H r 0$ C
QA
¦ n H ¦ n H i
i
i
out
b25180 378gmol HCl react
64.5 kJ
h
1 mol HCl
b
gb
2.627n 6 n 6 248000 24.7
g
b gb
g b gb
g b gb
. 378 1456 378 2.181 1895 2.512 0 n6
g
282475 mol C 2 H 5 Cl h in reactor effluent
282475 mol condensed 24800 mol product h h kmol recycled 257.6 h
C 2 H 5Cl recycled
d.
0
i
in
257600
mol h
C p is a linear function of temperature. 'H v is independent of temperature. 100% condensation of ethylbenzene in the heat exchanger is assumed. Heat of mixing and influence of pressure on enthalpy is neglected. Reactor is adiabatic. No C2H4 or C2H6 is absorbed in the ethyl chloride product.
9.31
a.
4NH3(g) + 5O2(g) Æ 4NO(g) +6H2O(g)
'H ro
904.7 kJ / mol
Basis : 10 mol/s Feed gas 4 mol / s NH 3 6 mol / s O 2
n3 (mol O 2 ) n4 (mol NO)
Tin = 200o C
n5 (mol H 2 O) Tout
O 2 consumed :
5 mol O 2
4 mol NH 3 fed
4 mol NH 3
s
NO produced : n 4
5 mol / s n 3
4 mol NO produced 4 mol NH 3 fed 4 mol NH 3
s
9-36
(6 1) mol O 2 / s
= 4 mol NO / s
1 mol O 2 /
9.31 (cont'd) 6 mol H 2 O produced 4 mol NH 3 fed
H 2 O produced : n5
4 mol NH 3
Extent of reaction : [ =
s
(n NH 3 ) out (n NH 3 ) in
04
Q NH 3
4
= 6 mol H 2 O / s
1 mol / s
b.
c.
Well-insulated reactor, so no heat loss No absorption of heat by container wall Neglect kinetic and potential energy changes; No shaft work No side reactions. References : NH 3 ( g), O 2 (g), NO(g), H 2 O(g) at 25o C, 1atm Substance NH 3 (g) O 2 ( g) NO(g)
nin nout H in H out ( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol) 4.00 H 1 6.00 100 . H2 H 3 4.00 H 4
H 2 O(g)
z
H 5
6.00
Table B.2
200
H 1
(C p ) NH 3 dT
B
Table B.8
6.74 kJ / mol,
H 2
H O 2 (200 o C)
B
5.31 kJ / mol
25
Using (C p ) i from Table B.2 : H 3
(0.0291 Tout 0.5790 u 10 5 Tout 2 0.2025 u 10 8 Tout 3 0.3278 u 10 12 Tout 4 0.7311) kJ / mol
H 4
(0.0295 Tout 0.4094 u 10 5 Tout 2 0.0975 u 10 8 Tout 3 0.0913 u 10 12 Tout 4 0.7400) kJ / mol
H 5
(0.03346 Tout 0.3440 u 10 5 Tout 2 0.2535 u 10 8 Tout 3 0.8983 u 10 12 Tout 4 0.8387) kJ / mol
Energy Balance: 'H
'H
[ 'H ro
5
¦
0
(ni ) out ( H i ) out
i 3
'H
E
'H
2
¦ (n
i ) in ( Hi ) in
i 1
[ 'H ro (1.00) H 3 (4.00) H 4 (6.00) H 5 (4.00) H 1 (6.00) H 2
o Substitute for [ , 'H r , and H 1 through H 6
(0.3479 Tout 4.28 u 10 5 Tout 2 2.114 u 10 8 Tout 3 4.697 u 10 12 Tout 4 ) 972.24 kJ / mol
E - Z Solve Tout
2261 o C
d.
z T
If only the first term from Table B.2 is used, H i
(C pi )dT
C pi (T 25)
25
H 1 H
4
0.03515(200 25) 0.0295(Tout
615 . kJ / mol, H 2
25), H 5
5.31 kJ / mol, H 3
0.03346(Tout 25)
9-37
0.0291(Tout 25),
9-31 (cont’d) E. B. 'H
[ 'H ro (1.00) H 3 (4.00) H 4 (6.00) H 5 (4.00) H 1 (6.00) H 2
E
o Substitute for [ ( = 1 mol / s), 'H r (
0 = 0.3479 Tout 969.86 Tout e.
9.32
0
904 .7 kJ / mol) and H 1 through H 6
2788 o C % error =
2788 o C 2261o C 2261 o C
u 100
23%
If the higher temperature were used as the basis, the reactor design would be safer (but more expensive). Basis : 100 lb m coke fed 84 lb m C 7.00 lb - moles C fed 7.00 lb - moles CO 2 fed 7.00 lb-moles CO2 400°F 7.00 lb-moles(84 lbm)C/hr 16 lb mash/hr 77°F 585,900 Btu
a.
bg
bg bg e77 Fj e'H j b g 2e'H j b g 393.50 b2gb 282.99g kJ
n 1 (lb-moles CO) n 2 (lb-moles CO2 ) 1830°F n 3 lb-moles C( s )/hr 16 lb mash/hr 1830°F
C s CO 2 g o 2CO g , 'H ro
o c
$
25q C
o c
CO 2 g
CO g
0.9486 Btu 453.6 mols
mol Let x
1 lb - mole
74,210 Btu lb - mole
fractional conversion of C and CO 2 :
E
b
7.00 x lb - moles C reacted
n1
1 kJ
b g 7.00b1 x g lb - moles Cbsg
g
2 lb - moles CO formed 1 lb - mole C reacted
14.0 x lb - moles CO
7.00 1 x lb - moles CO 2
n2 n3
bg
bg bg
$ References for enthalpy calculations: C s , CO 2 g , CO g , ash at 77 F
b g CO bg,1830q Fg: H CObg,1830q Fg: H CO 2 g,400q F : H 2
b
g
Solid 1830q F : H
H CO 2 (400$ F)
Table B.9
H CO 2 (1830$ F) H CO (1830$ F) 0.24 Btu lb m $ F
3130 Btu lb - mole
Table B.9
20,880 Btu lb - mole
Table B.9
13,280 Btu lb - mole
b1830 77gq F
420 Btu lb m
Mass of solids (emerging)
b g
7.00 1 x lb - moles C
12.0 lb m 16 lb m 1 lb - mole
9-38
b100 84 xg lb
m
9.32 (cont'd) substance
nin nout H in H out (lb moles) (Btu lb - mole) (lb moles) (Btu lb - mole) 7.00 3130 7.00 1 x 20,890
b g
CO 2 CO
(lb m )
(Btu lb m )
14.0 x (lb m )
13,280 (Btu lb m )
solid
100
0
100 84 x
420
Extent of reaction: n CO
( n CO ) o Q CO[ 14.0 x
2[ [ ( lb - moles) = 7.0 x
Energy balance: Q
[ 'H ro
'H
¦ n H ¦ n H i
i
i
out
585,900 Btu
a fa f a14.0 x fa13,280 f a100 84 x fa420 f a7.00 fa3130 f E 7.0 x (lb - moles)
74,210 Btu lb - mole
7.00 1 x 20,880
0.801 80.1% conversion
x b.
i
in
Advantages of CO. Gases are easier to store and transport than solids, and the product of the combustion is CO2, which is a much lower environmental hazard than are the products of coke combustion. Disadvantages of CO. It is highly toxic and dangerous if it leaks or is not completely burned, and it has a lower heating value than coke. Also, it costs something to produce it from coke.
9.33
Basis :
17.1 m 3 10 3 L 273 K 5.00 atm
1 mol
a f
h 1 m 3 298 K 1.00 atm 22.4 L STP CO g 2 H 2 g o CH 3OH g ,
bg
'H ro
bg e'H j o f
bg b g e 'H j
CH 3OH g
o f
CO(g)
3497 mol h feed
90.68 kJ mol
3497 mol/h n 1 (mol CH 3 OH /h) 0.333 mol CO/mol n 2 (mol CO/h) 0.667 mol H 2/mol n 3 (mol H 2 /h) 25°C, 5 atm 127°C, 5 atm Q = –17.05 kW Let f fractional conversion of CO (which also equals the fractional conversion of H 2 , since CO and H 2 are fed in stoichiometric proportion). f mol react 1166 f mol CO react mol feed 1166 f mol CO react 1 mol CH 3 OH CH 3OH produced : n1 1166 f mol CH 3 OH h 1 mol CO CO remaining : n 2 1166 1 f mol CO h CO reacted :
3497 0. 333 mol CO feed
a f
9-39
9.33 (cont'd)
b3497gb0.667g mol H fed 1166 f mol CO react 2332b1 f g mol H h Reference states : CO(g), H bgg , CH OHbgg at 25qC
H 2 remaining : n3
2
2 mol H 2 react 1 mol CO react
2
2
H in
n in
Substance
3
H out
n out
bmol hg bkJ molg bmol hg bkJ molg 1166a1 f f 1166 0 H 2332 0 2332a1 f f H
CO
1
H2
2
CH 3 OH
H 3
1166 f
Table B.8
e
$
e
$
j
H CO (127 $ C)
j
H H 2 (127 $ C)
CO g,127 C : H 1
B
2.99 kJ mol
Table B.8
H 2 g,127 C : H 2
z
'H
B
C p dT
25
Energy balance : Q
2.943 kJ mol
Table B.2
122
CH 3OH(g,127 C): H 3 $
B
[ 'H ro
5.009 kJ / mol
¦ n H ¦ n H i
i
out
17.05 kJ 3600 s s
1h
b
1102 u 10 5 f .
n2 n3
n tot
9.34
a.
(1166 f )( 90.68)
7.173 u 10 4 f
i
b
kJ 1166 1 f h
g b2.993g 1166 f b5.009g bkJ hg
2332 1 f
n1
i
in
b
g b2.99g g
0.651 mol CO or H 2 converted mol fed
b g 759.1 mol h 1166b1 0.651g 406.9 mol h 2332b1 0.651g 813.9 mol h E
1166 0.651
1980
mol Vout h
bg bg
b g
1980 mol 22.4 L STP h
bg
1 mol
bg
CH 4 g 4S g o CS2 g 2 H 2 S g , 'H r 700q C
400 K 1.00 atm
1 m3
273 K 5.00 atm 103 L
13.0 m 3 h
274 kJ mol
Basis : 1 mol of feed 1 mol at 700°C 0.20 mol CH 4/mol 0.80 mol S/mol
Reactor
Q = –41 kJ
Product gas at 800°C n1 (mol CS2) n2 (mol H 2S) n3 (mol CH 4) n4 (mol S (v))
Let f fractional conversion of CH 4 (which also equals fractional conversion of S, since the species are fed in stoichiometric proportion)
9-40
9.34 (cont'd) 0.20 f , Extent of reaction = [ (mol) = 0.20 f
Moles CH 4 reacted
b
g
n3
0.20 1 f mol CH 4
n4
0.80 mol S fed
n1 n2
b
0.20 f mol CH 4 react
g
4 mol S react 1 mol CH 4 react
0.20 f mol CH 4 react
1 mol CS 2 1 mol CH 4
0.20 f mol CH 4 react
2 mol H 2 S 1 mol CH 4
b
g
0.80 1 f mol S
0.20 f mol CS 2 0.40 f mol H 2 S
bg
References: CH 4 (g), S g , CS2 (g), H 2 S(g) at 700qC (temperature at which 'H r is known) substance nin
H in
bmolg bkJ molg bmolg bkJ molg 0.20b1 f g 0.20 0 H 0.80 0 0.80b1 f g H
CH 4
1
S
2
CS 2
0.20 f
H 2S
0.40 f
H out
H out
nout
C pi 800 700
b
H 3 H 4
g
CH 4 g, 800q C : H 1 7.14 kJ / mol S g, 800q C : H 3.64 kJ / mol
b
g
b b
g g
2
CS 2 g, 800q C : H 3 H S g, 800q C : H 2
4
3.18 kJ / mol 4.48 kJ / mol
Energy balance on reactor: Q
'H
[ 'H r
¦ n H ¦ n H i
out
i
i
i
in
41
kJ s
b0.20 f gb274.0g 0.20b1 f gb7.140g 0.80b1 f gb3.640g 0.20 f b3180 . g 0.40 f b4.480g b1g f
0.800
b. 0.04 mol CH4 0.16 mol S(l ) 0.16 mol CS2 0.32 mol H2 S 200qC 0.20 mol CH4 0.80 mol S(l ) 150°C
0.20 mol CH4 0.80 mol S( g) T (°C)
Q (kJ)
preheater
0.04 mol CH4 0.16 mol S(g ) 0.16 mol CS2 0.32 mol H2S 800°C
9-41
0.20 mol CH4 0.80 mol S(l ) 700°C
9.34 (cont'd) System: Heat exchanger-preheater combination. Assume the heat exchanger is adiabatic, so that the only heat transferred to the system from its surroundings is Q for the preheater.
bg
References : CH 4 (g), S l , CS2 (g), H 2S(g) at 200qC Substance
n in
H in
n out
H out
0.20
H 1 H
0.20
H 7
0.04
0
0.16 0.16
0 H8 0
0.32
0
bmolg bkJ molg bmolg bkJ molg
bCH g bCH g Sblg Sbgg
4 150q ,700q
4 800q,200q
0.04
2
H 3 H
0.80 0.16
CS 2
0.16
H 2S
0.32
0.80
4
H 5 H 6
a f dC i a f aT 200f for Salf dC i a f FGH 444.6 200IJK 'H bT g dC i b g aT 444.6f for Sbgg
H i
C pi T 200 for all substances but S p Sl
p Sl
b b
v 83.7 k J mol
Tb
g g
CH 4 g, 150q C : H 1 CH g, 800q C : H
b b
4
g g
4
2
b
103.83 kJ / mol
b g ¦ n H ¦ n H i
g g
b
g
6
19.08 kJ / mol 26.88 kJ / mol
CH 4 g, 700q C : H 7 35.7 kJ / mol S g, 700q C : H 8 100.19 kJ / mol
1.47 kJ / mol
out
b b
CS 2 g, 800q C : H 5 H S g, 800q C : H
42.84 kJ / mol
Energy balance: Q kJ c.
p Sg
3.57 kJ / mol
2
S l, 150q C : H 3 S g, 800q C : H
b
i
i
i
Q
g
59.2 kJ 59.2 kJ mol feed
in
The energy economy might be improved by insulating the reactor better. The reactor effluent will emerge at a higher temperature and transfer more heat to the fresh feed in the first preheater, lowering (and possibly eliminating) the heat requirement in the second preheater.
9-42
9.35
Basis : 1 mol C 2 H 6 fed to reactor n (mols) @ T (K), P atm n C 2H6 (mol C 2H 6) n C 2H4 (mol C 2H 4) n H 2 (mol H 2)
1 mol C H 2 6 1273 K, P atm
a.
x C2H 4 x H 2
C2 H 6 C2 H 4 H 2 , K p
b
7.28 u 10 6 exp[ 17,000 / T ( K )]
P
x C2 H 6
Fractional conversion = f mols C 2 H 6 react mol fed
U| x b1 f gbmol C H g|| f b mol C H g V| x f b mol H g || x 1 f b molsg W
[ (mol)
f
nH2 n
2
2
C2H 4
2
f
x C2H 4 x H 2
Kp
e1 f jK
2
f2P 1 f 1 f
2
b
F K I GH P K JK
12
f P f
p
bg
bg
gb
g
f
2
1 f
2
P
b2 g
p
2
p
H2
b1 f g P b1 f g b1 f g
P Kp
x C2 H 4 2
b.
6
4
g
1 f mol C 2 H 6 1 f mol f mol C 2 H 4 1 f mol f mol H 2 1 f mol
C2 H 6
n C2 H 6 n C2 H 4
(1)
bg
References : C 2 H 6 g , C 2 H 4 g , H 2 g at 1273 K Energy balance:
e H j e H j i
i
b
g ¦ n H ¦ n H
0 [ 'H r 1273 K
'H
i
b
0 inlet temperature
in
out
z
i
i
out
T
1273
reference temperature
g
C pi dT energy balance
b
i
in
g b
f 'H r 1273 K kJ 1 f
z
g dC i T
1273
p
C2H 6
dT f
z
T
1273
dC i
p C H dT 2 4
rearrange, reverse limits and change signs of integrals
1 f f
b
g
z
z
dC i dT dC i dT b3g C dT d i 'H r 1273K
1273
z
p C H 2 4
T
1273
T
1273
T
p C H 2 6
bg
I T
1 f f
bg
I T 1 f
bg
fI T f
b g b4g
1 1 I T
9-43
p H 2
f
z
T
1273
dC i
p H dT 2
0
9.35 (cont'd) 145600
bg
IT
zb
g
1273
T
z
1273
T
bg
I T
c.
z
1273
e26.90 4.167 u 10 T jdT . 0.1392T gdT b1135
9.419 0.1147T dT
3
T
3052 36.2T 0.05943T 2 127240 113 . T 0.0696T 2
F K I F K I GH 1 K JK 1 I1bT g GH 1 K JK 1 I1bT g \ bT g 0 I bT g given by expression of Part b. K bT g given by Eq. (1) 12
12
p
p
p
p
p
d.
P (atm) 0.01 0.05 0.1 0.5 1 5 10
T (K) 794 847.4 872.3 932.8 960.3 1026 1055
f 0.518 0.47 0.446 0.388 0.36 0.292 0.261
Kp (atm) 0.0037 0.0141 0.025 0.0886 0.1492 0.4646 0.7283
Phi
Psi
0.93152 -0.0001115 1.12964 -0.0002618 1.24028 0.00097743 1.57826 3.41E-05 1.77566 4.69E-05 2.42913 -2.57E-05 2.83692 -7.54E-05
Plot of T vs ln P
Plot of f vs. ln P
1100
0.6 0.5
1000
f
T(K)
0.4 900
0.3 0.2
800
0.1 0
700 -3
-2
-1
0
1
-3
2
-2
ln P(atm)
e.
-1
0
ln P(atm)
C **PROGRAM FOR PROBLEM 9-35 WRITE (5, 1) 1 FORMAT ('1', 20X, 'SOLUTION TO PROBLEM 9-35'//) T 1200.0 TLAST 0.0 PSIL 0.0
9-44
1
2
9.35 (cont'd) C **DECREMENT BY 50 DEG. AND LOOK FOR A SIGN IN PSI DO 10I 1, 20 CALL PSICAL (T, PHI, PSI) IF ((PSIL*PSI).LT.0.0) GO TO 40 TLAST T PSIL PSI T T – 50. 10 CONTINUE 40 IF (T.GE.0.0) GO TO 45 WRITE (3, 2) 2 FORMAT (1X, 'T LESS THAN ZERO -- ERROR') STOP C **APPLY REGULA-FALSI 45 DO 50 I 1, 20 IF (I.NE.1) T2L T2 T2 (T*PSIL-TLAST*PSI)/(PSIL-PSI) IF (ABS(T2-T2L).LT.0.01) GO TO 99 CALL PSICAL (T2, PHIT, PSIT) IF (PSIT.EQ.0) GO TO 99 IF ((PBIT*PBIL).GT.0.0) PSIL PSIT IF ((PSIT*PSIL).GT.0.0) TLAST T2 IF ((PSIT*PSI).GT.0.0) PSI PSIT IF ((PSIT*PSI).GT.0.0) T T2 50 CONTINUE IF (I.EQ.20) WRITE (3, 3) 3 FORMAT ('0', 'REGULA-FALSI DID NOT CONVERGE IN 20 ITERATIONS') 93 STOP END SUBROUTINE PSICAL (T, PHI, PSI) REAL KF PHI (3052 36.2*T 36.2*T 0.05943*T**2)/(127240. – 11.35*T * – 0.0636*T**2) KP 7.28E6*EXP(-17000./T) FBI SQRT((KP/(1. KP)) – 1./12. PHI) WRITE (3, 1) T, PSI 1 FORMAT (6X, 'T ', F6.2, 4X, 'PSI ', E11,4) RETURN END OUTPUT: SOLUTION TO PROBLEM 9-35 T 1200.00 PSI 0.8226E 00 T 1150.00 PSI 0.7048E 00 T 1100.00 PSI T 1050.00 PSI
0.5551E 00 0.3696E 00
T
950.00
PSI
. 01619 E 00 0.3950E 01
T
959.80
PSI
0.1824 E 02
T
960.25
PSI
0.7671E 04
T
960.27
PSI
0.3278E 05
T 1000.00 PSI
Solution: T
960.3 K, f
0.360 mol C 2 H 6 reacted mol fed
9-45
2CH 4 o C 2 H 2 3H 2
9.36
C 2 H 2 o 2C(s) + H 2
bg
Basis: 10 mol CH 4 g fed s 10.0 mol CH 4 (g) / s
n1 (mol CH 4 / s)
o
1500 C
n 2 (mol C 2 H 2 / s) n 3 (mol C(s) / s) 1500 o C 975 kW
a.
b
60% conversion n1
g
10 1 0.600
4.00 mol CH 4 s
b g 4b1g 2n n 2n n 6 10b4 g 4b4g 2n 2n 2n 2n
C balance: 10 1 H balance:
2
4
2
2
3
(1)
4
2
24
3
(2)
bg
References for enthalpy calculations : C(s), H 2 g at 25qC
e
Hi
'H fo
Substance
bg bg H bg g Cbsg
CH 4 g
C2 H 2 g 2
Energy Balance: Q
j
i
b
g
C pi 1500 25 , i
CH 4 , C 2 H 2 , C, H 2
H in
n in
( mol s) ( kJ mol) ( mol s) (kJ mol) 10 41.68 4 4168 . 303.45 n 2 45.72 n 3
'H 975 kJ / s n 2
Solve (1) - (3) simultaneously n 3 n 4
b.
32.45
n 4
¦ n H ¦ n H i
out
Yield of acetylene
H out
n out
i
i
i
(3)
in
2.50 mol C 2 H 2 / s 9.50 mol H 2 / s 100 . mol C / s
2.50 mol C 2 H 2 s 6.00 mol CH 4 consumed s
0.417 mol C 2 H 2 mol CH 4 consumed
If no side reaction, n1
10.0(1 0.600)
n 3
0 n 2
4.00 mol CH 4 / s
3.00 mol C 2 H 2 / s, n 4
9.00 mol H 2 / s
Yield of acetylene
3.00 mol C 2 H 2 s 6.00 mol CH 4 consumed s
Reactor Efficiency
0.417 0.500
0.834
9-46
0.500 mol C 2 H 2 mol CH 4 consumed
9.37
bg bg bg bg CObgg H Ob vg o CO bgg H bgg
C 3 H 8 g 3H 2 O v o 3CO g 7H 2 g 2
2
2
Basis : 1 mol C 3 H 8 fed Heating gas 4.94 m3 at 1400°C, 1 atm n g (mol)
n g (mol), 900°C
4.94 m 3
ng
10 3 L 1m
273 K
3
Product gas, 800°C n 1 (mol C 3H 8) = 0 n 2 (mol H 2O) n 3 (mol CO) n 4 (mol CO2 ) n 5 (mol H 2)
a
1 mol C 3H 8(g ) 6 mol H 2 O( g ) 125°C
1 mol
1673 K 22.4 L
35.99 mol heating gas
Let [ 1 and [ 2 be the extents of the two reactions. n1 0
n1
1 [1 [1
n2
6 3[ 1 [ 2 n 2
1 mol [1 1
[1 1
3[ 1 [ 2 n 3
n3
3 [ 2
n4
[2
n5
7[ 1 [ 2 n5
[1 1
7[2
3[2
bg bg
References : C(s), H 2 g , O 2 g at 25qC, heating gas at 900qC
z T
H i
'H fio C pi dT
for C 3 H 8
25
Table B.8 for CO 2 , H 2 , H 2 O, CO
z T
C p dT
b
g
C p T 900 for heating gas
900
n in
H in
n out
H out
C3H8
mol 1
kJ / mol 95.39
mol 0
kJ / mol
H 2O
6
238.43 3 [ 2
212.78
CO
3[2
86.39
CO 2
356.15
H2
[2 7[2 35.99
Substance
heating gas 35.99
200.00
22.85 0
Energy Balance :
¦ n H ¦ n H i
i
i
i
out
in
n4
1 mol CO 2 , n5
0 [2
2.00 mol n 2
1 mol H 2 O, n 3
1 mol CO,
9 mol H 2 7.7 mol % H 2 O, 7.7% CO, 15.4% CO 2 , 69.2% H 2
9-47
9.38
a.
Any C consumed in reaction (2) is lost to reaction (1). Without the energy released by reaction (2) to compensate for the energy consumed by reaction (1), the temperature in the adiabatic reactor and hence the reaction rate would drop.
b.
Basis : 1.00 kg coal fed (+0.500 kg H20) 0.500 kg H20
1.0 kg coal 0.105 kg H2O/kg coal 0.226 kg ash/kg coal 0.669 kg combustible / kg coal
R| |S0.812 kg C / kg combustible ||0.134 kg O / kg combustible T0.054 kg H / kg combustible
nf1 (mol C) nf2 (mol O) nf3 (mol H) nf4 (mol H2O) 0.226 kg ash
U| |V || W
n f 1 = [ (1.00)(0.669)(0.812) kg C][1 mol C / 12.01 u 10 3 kg] = 45.23 mol C n f 2 = (1.00)(0.669)(0.134) / 16.0 u 10 3
5.6 mol O
n f 3 = (1.00)(0.669)(0.054) / 1.01 u 10 3
35.77 mol H
n f 4 = [ (0.500 + 0.105) kg][1 mol H 2 O / 18.016 u 10 3 kg] = 33.58 mol H 2 O n0 (mol O2) 25qC
Product gas at 2500qC n1 (mol CO2) n2 (mol CO) n3 (mol H2) n4 (mol H2O)
1 kg coal + H2O, 25qC 45.23 mol C 5.60 mol O 35.77 mol H 33.58 mol H2O 0.226 mol kg ash
0.226 kg slag 2500qC
Reactive oxygen (O) available
(2n 0 5.60) mol O
Oxygen consumed by H ( 2H + O o H 2 O) :
35.77 mol H 1 mol O 2 mol H
Reactive O remaining = (2n 0 5.60) 17.88 CO 2 formed ( C + 2O o CO 2 ) : n1 C balance : 45.23 = n1 n 2
O balance : 2n 0 5.60 3358 .
(2n 0 12.28) mol O
(2n 0 12.28) mol O 1 mol CO 2 2 mol O
n1 n0 6.24
17.88 mol O
(n 0 6.24) mol CO 2
(51.5 n 0 ) mol CO
n2
2n1 n 2 n 4
n1 n0 6.24
n4
(n 0 0.16) mol H 2 O
n2 51.5 n0
H balance : 35.77 + 2(33.58) = 2n 3 2n 4
n4 n0 0.16
9-48
n3
(51.3 n 0 ) mol H 2
9.38 (cont'd) c.
1 kg coal contains 45.23 mol C and 35.77 mol H 1 kg coal + nO 2 o 45.23 CO 2 (35.77 / 2) mol H 2 O (l) 'H r
21,400 kJ = 45.23( 'H fo ) CO 2 (35.77 / 2)( 'H fo ) H 2 O(l) ( 'H fo ) coal
( 'H fo ) coal
1510 kJ / kg Re ferences : Coal, ash, CO, CO 2 , H 2 , H 2 O (l) at 25 o C H in
nin
Substance
(mol) ( kJ / mol) n 0 6.24
(mol)
CO 2 H2
H 2O
. 3358
0
Coal Ash(slag)
1 kg (in coal)
0 0
CO
H out
n out
. n0 515
( kJ / mol) H 1 H
51.3 n 0 . n 0 016
H 3 H
0.266 kg
H 5 ( kJ / kg)
2
4
H i
C pi (2500 25), i
H 1
0.0508(2475)
125.7 kJ / mol CO 2
H 2
0.0332(2475)
82.17 kJ / mol CO
H 3
0.0300(2475)
74.25 kJ / mol H 2
H 4
(C p ) H 2 O(l) (100 25) 'H v (100 o C) + (C p ) H 2 O(v) (2500 100)
1,3
(0.0754)(75) 40.656 kJ / mol + (0.0395)(2400) = 141.1 kJ / mol H 2 O H 5
( 'H m ) ash 1.4(2475)
710 1.4(2475)
4175 kJ / kg ash
Reaction 1 : 1 kg coal + n1 O 2 o 45.23 mol CO 2 +17.88 mol H 2 O(l) o 'H r1
21,400 kJ / 45.23 mol CO 2
Reaction 2 : 1 kg coal + n 2 O 2 o 45.23 mol CO + 17.88 mol H 2 O(l) o 'H r2
45.23( 'H fo ) CO 17.88( 'H fo ) H 2 O(l) ( 'H fo ) Coal
8600 kJ / 45.23 mol CO
Energy Balance 'H
(n 0 6.24) mol CO 2 u (
¦n
out
H out
¦n
8600 kJ 21,400 kJ ) (51.5 n 0 ) mol CO u ( ) 45.23 mol CO 2 45.23 mol CO
in H in
0
4731 . (n 0 6.24) 190.1(51.5 n 0 ) (n 0 6.24)125.7 (51.5 n 0 )82.17 (51.3 n0 )74.25 + (n 0 016 . )1411 . (0.266)(4175) n0
0
9.0 mol O 2
9-49
9.39
Mass of H 2 SO 4
3 m3 103 L 1 mol H 2 SO 4 1 m3 L
Mass of solution
3 m3 103 L 103 mL 1 m3 L
Moles of H 2 O
FG mol H O IJ H mol H SO K e'H j b 2
f
4
2.941 u 105 g H 2 SO 4
H 2SO 4 aq., r 53.6
e'H j o f
g
bg
H 2SO 4 l
A
53.6 mol H 2 O mol H 2 SO 4
e j
'H s
b
H 2SO 4 aq ., r 53.6
A
Table B.1
H
FG 98.02 gIJ H 1 mol K
1.064 g . u 10 6 g solution 3192 1 mL 1 mol u 10 6 2.941 u 10 5 )g H 2 O( (3192 . ) 1.61 u 10 5 mol H 2 O 18.02 g
161 . u 10 5 mol H 2 O 3000 mol H 2 SO 4
2
n
3000 mol H 2 SO 4
kJ b811.32 73.39g mol
g
884.7 kJ mol
Table B.11
(3000 mol H 2 SO 4 )(-884.7 kJ / mol H 2 SO 4 ) = -2.65 u 10 6 kJ
9.40
e'H j
HCl (aq): 'H fo
o f
bg
HCl g
e
'H so
j
Tables B.1, B11
. 92.31 7514
f
167.45 kJ mol
Tables B1, B.11
e
NaOH (aq): 'H fo
j
'H fo
bg
NaOH s
e
'H so
j
B
426.6 42.89
f Table B.1
NaCl (aq):
e
'H fo
b g
'H fo
b g
j
bg
NaCl s
e
'H so
b g
j
B
bg
HCl aq NaOH aq o NaCl aq H 2 O l 'H ro 406.1 28584 . 167.45 469.49
b
g b HClbgg NaOHbsg o NaClbsg H Oblg
Given
B
. 4.87 4110
f
g
469.49 kJ mol
406.1 kJ mol
55.0 kJ mol
2
¦
'H ro
v i 'H fo
products
¦
v i 'H fo
reactants
b
g b
g
411.0 285.84 92.31 426.6
kJ mol
177.9 kJ mol
The difference between the two calculated values equals 'H 'H 'H .
{e j s
9.41
a.
e j s
NaCl
b g
HCl
e j s
NaOH
b g
}
b g
bg
H 2 SO 4 aq 2NaOH aq o Na 2 SO 4 aq 2H 2 O l Basis: 1 mol H 2 SO 4 soln
0.10 mol H 2 SO 4 u 98.08 g mol
g
0.90 mol H 2 O u 18.02 g mol
16.22 g H 2 O
26.03 g soln 1 cm 3
0.10 mol H 2 SO 4
127 . g
b
b
g
9.808 g H 2 SO 4
U| V| W
20.49 cm 3
2 mol NaOH 1 mol H 2 SO 4
1 liter caustic soln 10 3 cm 3 3 mol NaOH
9-50
1L
b g
66.67 cm 3 NaOH aq
9.41 (cont'd) 66.67 cm 3 NaOH(aq)
Volume ratio b.
3.25 cm 3 caustic solution / cm 3 acid solution
3
20.49 cm H 2 SO 4 (aq)
b g
9 mol H 2 O / 1 mol H 2 SO 4
H 2 SO 4 aq : r
kJ 877 kJ mol H SO b811.32 65.23g mol NaOH(aq) : The solution fed contains e66.67 cm je113 . g cm j 75.34 g , and (0.2 mol NaOH)b40.00 g molg 8.00 g NaOH b75.34 8.00g g H O b67.39 g H Ogb1 mol 18.02 gg 3.74 mol H O
e'H j o f
soln
e'H j o f
b g e 'H f j H SO baq., r 9 g o
H 2 SO 4 l
2
2
4
3
2
r
soln
e'H j Na SO baq g: e'H j e'H j 2
4
o f
o f
o f
o f
soln
3
2
2
3.74 mol H 2 O 0.20 mol NaOH
e'H j
18.7 mol H 2 O / mol NaOH
b g e 'H s j NaOH bsgbaq., r o
NaOH s
bg
Na 2SO 4 s
e
'H fo
j
b g
Na 2SO 4 aq
18.7
g
kJ b426.6 42.8g mol
kJ . g b1384.5 117 mol
(n H 2SO4 ) fed Q H 2SO4 [ 0
Extent of reaction: (n H 2SO ) final 4
469.4 kJ mol NaOH
1385.7 kJ mol Na 2 SO 4
0.10mol (1)[ [
010 . mol
Energy Balance: 'H
Q
['H ro
[ ( 'H fo ) Na 2SO4 ( aq) 2( 'H fo ) H 2 O ( l) ( 'H fo ) H 2SO 4 ( aq) 2( 'H fo ) NaOH ( aq)
. ) ( 876.55) (2)( 469.4) = (0.10 mol) 1385.7 + 2( 28584 9.42
a.
kJ mol
14.2 kJ
Table B.1, given
NaCl(aq): 'H fo
e'H j o f
bg
NaCl s
e
'H so
j
B
f
b411.0 4.87gkJ / mol
406.1 kJ mol
NaOH(aq): Table B.1
B b426.6 42.89gkJ / mol e'H j b g e'H j 1 1 NaClbaq g H Oblg o H bgg Cl bgg NaOHbaq g 2 2 'H 469.5 b 406.1g b 28584 . g kJ mol 222.44 kJ mol 'H fo 2
o f
o s
NaOH s 2
f
469.5 kJ mol
2
o r
b.
8500 ktonne Cl 2 yr
10 3 tonne 10 3 kg 10 3 g 1 ktonne
1 tonne
10 3 J 2.778 u 10 7 kW h 1 kJ
1J
4
1 kg
1 mol Cl 2
222.44 kJ
70.91 g Cl 2
0.5 mol Cl 2
1 MW h 10 3 kW h
9-51
148 . u 10 7 MW h / yr
9.43
a.
bg
bg
b
CaCl 2 s 10H 2 O l o CaCl 2 aq , r
bg
bg
b
g
10
g 6H Obsg
CaCl 2 6H 2 O s 4 H 2 O l o CaCl 2 aq , r 2
2
o r1
o r1
o f
b
Basis: 1 mol NH 4
2
32.41 kJ mol
2
o f
CaCl 2 aq , r 10
bg
g SO produced 4
2
1 mol (NH4)2SO4 (aq) 25qC
b g b
2NH 3 g H 2 SO 4 aq o NH 4
g SO baqg 4
2
References : Elements at 25qC
b
z
g
b
'H fo C p dT 25
g
e'H j b g SO baq, 25q Cg: H e'H jb
H 2 SO 4 aq , 25q C : H
bNH
I FG 4619 . J kJ / mol K H . 183
75
NH 3 g, 75q C : H
4 2
CaCl 2 s
CaCl 2 aq , r 10
2 mol NH3 (g) 75qC 1mol H2SO4 (aq) 25qC
a.
64.85 kJ mol
'H ro2
o r2
o f
9.44
10
'H ro1
(3) b1g b2g CaCl bsg 6H Oblg o CaCl 'H 'H 'H b Hess' s law g 97.26 kJ mol From (1), 'H e'H j b g e'H j b g e 'H j b g b64.85 794.96g kJ mol 859.81 kJ mol o r3
b.
b1g b2 g
o f
H 2SO 4 aq o f
4
907.51 kJ mol H 2 SO 4 (Ta.ble B.1)
g
NH 4
44.36 kJ mol (Table B.1, B.2)
b
11731 . kJ mol NH 4
g SO b aq g 2
4
g SO 2
4
(Table B.1)
Energy balance: Q
'H
¦ n H ¦ n H b1gb11731. g b2gb44.36g b1gb907.51g kJ i
i
out
177 kJ 177
b.
i
i
in
kJ withdrawn mol NH 4 2 SO 4 produced
b
g
1 mole % (NH 4 ) 2 SO 4 solution
1 mol (NH 4 ) 2 SO 4
132 g mol
99 mol H 2 O 18 g
132 g (NH 4 ) 2 SO 4
1782 g H 2 O mol 1914 g solution The heat transferred from the reactor in part (a) now goes to heat the product solution from 25$ C to Tfinal 177 kJ = c.
1.914g
1 kg
=
4.184 kJ (T 25) $ C
3
10 g
$
kg C
Tfinal
47.1$ C
In a real reactor, the final solution temperature will be less than the value calculated in part b, due to heat loss to the surroundings. The final temperature will therefore be less than 47.1oC.
9-52
9.45
a.
b g
b g
b g
b g
H 2 SO 4 aq 2 NaOH aq o Na 2 SO 4 aq 2 H 2 O aq 1 mol H 2SO 4 49 mol H 2O 25°C
Basis : 1 mol H 2 SO 4 fed
1 mol Na 2SO 4 89 mol H 2O 40°C
2 mol NaOH 38 mol H 2O 25°C
bg b g bg b g H SO eaq , r 49, 25 Cj: L O 'H b r 49gPb kJ molg 49e 'H j nH b1 mol H SO gMe 'H j b g Q N b1g 811.3 73.3 884.6 kJ 49e'H j b g NaOHeaq , r 19, 25 Cj: L O 'H b r 19gPb kJ molg 38e 'H j nH b2 mol NaOH gMe 'H j bg N Q b2g 426.6 42.8 938.8 kJ 38e'H j b g Na SO eaq , r 89, 40 Cj:
Reference states : Na s , H 2 g , S s , O 2 g at 25qC 2
$
4
2
o f
4
o s
H 2SO 4 l
o f
o f
bg
H 2O l
H 2O l
$
o f
o s
NaOH s
o f
o f
2
H 2O l
$
4
1 kmol Na 2 SO 4
142.0 kg
b1 mol Na SO 2
4
gLMNe'H j o f
b
89 kmol H 2 O 18.02 kg
0.142 kg,
1 kmol nH
bg
H 2O l
Na 2SO 4
kg 1746 kg 1604 . .
1 kmol
e
'H so
j
Na 2SO 4
g
OP 89e'H j Q o f
b g mC p b40 25g
H 2O l
'H fo 1384.5 kJ mol Table B.1 'H so 1.2 kJ mol 4 .814 kJ (kg$ C) m=1.746 kg, C p | C p H 2O l
FH IK
nH
bg
e
1276 kJ 89 'H fo 'H
Energy balance: Q
j
bg
H 2O l
¦
ni H i
out
Mass of acid fed 1 mol H 2 SO 4
98.08 g H 2 SO 4 1 mol
b.
Q M acid
¦
285.84 kJ mol
e
ni H i
547.4 2 'H fo
in
j
49 mol H 2 O 18.02 g H 2 O 1 mol
bg
H 2O l
24.3 kJ
981 g = 0.981 kg
24.3 kJ 24.8 kJ / kg acid transferred from reactor contents 0.981 kg acid
If the reactor is adiabatic, the heat transferred from the reactor of Part(a) instead goes to heat the product solution from 40qC to T f 3
24.3 u 10 J
1746 . kg 4.184 kJ $
kg C
9-53
dT
f
i
$
40 C
Tf
43 $ C
9.46
a.
b g
b g
b g
bg
H 2 SO 4 aq 2NaOH aq o Na 2 SO 4 aq 2H 2 O l H 2SO4 solution: : 75 ml of 4M H 2SO 4 solution
4 mol H 2 SO 4
1L
1 L acid soln
3
75 mL
0.30 mol H 2SO 4
10 mL
b75 mLgb1.23 g mLg 92.25 g, (0.3 mol H SO )b98.08 g molg 29.42 g H SO b92.25 29.42g g H O b62.83 g H Ogb1 mol 18.02 gg 3.49 mol H O 2
2
r o f
soln
e'H j o f
2
2
e j
'H fo
bg
H 2SO 4 l
4
2
3.49 mol H 2 O 0.30 mol H 2 SO 4
e'H j
4
11.63 mol H 2 O / mol H 2 SO 4 Table B.1, Table B.11
B
b
kJ b811.32 67.42g mol
g
H 2SO 4 aq ., r 11.63
878.74 kJ mol H 2 SO 4 NaOH solution required: 0.30 m ol H 2 SO 4 2 m ol NaOH
1 L NaOH (aq)
10 3 m L
1 m ol H 2 SO 4
12 m ol NaOH
1L
. g mLg b50.00 mLgb137
68.5 g
12 mol NaOH
50 mL
1L
1 L NaOH(aq) 10 3 mL
b
g
40 g/ mol NaOH
0.60 mol NaOH
b
gb
68.5 24.00 g H 2 O 44.5 g H 2 O 1 mol 18.02 g r o f
soln
e'H j o f
b g e'Hs j NaOHbsgbaq., r o
NaOH s
g
24.00 g NaOH
2.47 mol H 2 O
4.12 mol H 2 O mol NaOH
2.47 mol H 2 O 0.6 mol NaOH
e'H j
b g
50.00 m L NaOH aq
4.12
kJ . g b426.6 3510 mol
g
46170 . kJ mol NaOH
b g e'H j e'H j Na 2SO 4 aq : o f
mtotal
o f
soln
b g e'H f j Na SO baq g o
Na 2SO 4 s
2
4
kJ . g b1384.5 117 mol
1385.7 kJ mol Na 2SO 4
total mass of reactants or products = (92.25g H 2SO4 soln + 68.5g NaOH) = 160.75g = 0.161 kg
Extent of reaction: (nH 2SO 4 ) final
(nH 2SO 4 ) fed Q H 2SO 4 [ 0
0.30 mol (1)[ [
0.30 mol
Standard heat of reaction
'H ro
e'H j o f
b g 2e'H f j H Oblg e'H f j H SO baq g 2e 'H f j NaOHbaq g
Na 2SO 4 aq
o
o
2
o
2
4
Energy Balance : Q 'H ['H ro mtotal C p (T 25) $ C
F GH
(0.30 mol)(1552 . kJ / mol) (0161 . kg) 4.184 b.
Volumes are additive. Heat transferred to and through the container wall is negligible.
9-54
I (T 25) C = 0 T J kg C K kJ
$
$
94 $ C
9.47
Basis : 50,000 mol flue gas/h
50,000 mol/h 0.00300 SO2 0.997 N 2 50°C
n4 (mol SO2 /h) n5 (mol N 2 /h) 35°C
n1 (mol solution/h) 0.100 (NH 4 ) 2 SO 3 0.900 H 2O( l ) 25°C
n2 (mol NH4 HSO3 /h) 1.5n2 (mol (NH 4)2 SO3 /h) n3 (mol H 2 O(l )/h) 35°C
a fb g 15.0 mol SO h n a0.997fb50,000 mol hg 49,850 mol N h N balance: NH balance: b2gb0100 . gbn g n b15 . gb2gn n 20n U| n 5400 mol h V 270 mol NH HSO S balance: 0100 . n b0.00300gb50,000g 150 . n 15 . n |W n 270 mol NH HSO produced 1 mol H O consumed H O balance: n b0900 . gb5400g h 2 mol NH HSO produced 4725 mol H Oblg h
90% SO2 removal: n 4
0.100 0.00300 50,000 mol h
2
2
5
2
+ 4
1
2
2
1
1
2
2
4
3
1
2
4
2
2
3
3
h
2
4
3
2
Heat of reaction: e 'H j e j b g e 'H j b g b g e 'H j 2b 760g b890g b 296.90g b 285.84g kJ mol 47.3 kJ mol References : N bgg, SO bgg, b NH g SO baq g, NH HSO baq g, H Oblg at 25qC . kJ mol ( C from Table B.2) SO eg, 50 Cj: H z dC i dT 101 'H ro
2 'H fo
o f
NH 4 HSO 4 aq
2
2
NH 4
4 2
o f
SO 3 aq 2
3
4
o f
SO 2 ( g )
3
H 2 O(l)
2
50
$
2
e
zd
j
SO 2 g, 35$ C : H
e j N eg, 35 Cj: $
2
Cp
25
N 2 g, 50$ C : H H
p
p SO 2
25 35
i
SO 2
0.40 kJ mol
dT
073 . kJ mol (Table B.8) 0.292 kJ mol
Entering solution: H
0
Effluent solution at 35qC
b g
m g h
nH
270 mol NH 4 HSO 3
99 g
h
mol
b
g
1.5 u 270 mol NH 4 2 SO 3 116 g 4725 mol H 2 O 18 g g 159,000 h 1 mol h h mol
mC p 'T
159,000 g 4 J h gq C
Extent of reaction: (n NH 4 HSO 3 ) out
a35 25fq C
1 kJ 10 3 J
6360 kJ / h
(n NH 4 HSO3 )in Q NH 4 HSO 3 [ 270 mol / h = 0 + 2[ [ = 135 mol / h
9-55
9.47 (cont'd) Energy balance: Q Q 9.48
a.
135 mol
47.3 kJ
h
mol
effluent solution
6360
bg
[ 'H ro
'H
¦ n H ¦ n H i
i
out
SO 2 out
i
in N 2 out
a fa f a
fa
15 0.40 49,850 0.292
a
fa f a
fa
i
f
fa f
50,000 0.003 1.01 49,850 0.73
bg
22,000 kJ
1h
h
1 kW
3600 s 1 kJ s
6.11 kW
bg
CH 4 g + 2O 2 g o CO 2 (g) 2H 2 O v
at 25qC
Table B.1 HHV
HHV
B
'H co
890.36 kJ / mol, LHV
B
e j
2 'H v
b
g
890.36 2 44.01 kJ mol
H 2O
802.34 kJ mol CH 4
bg
7 C 2 H 4 g + O 2 (g) o 2CO 2 (g) 3H 2 O(v) 2 HHV
a
f
a
f
1559.9 3 44.01 kJ mol 1427.87 kJ mol C 2 H 6
1559.9 kJ / mol, LHV
bg
C 3 H 8 g + 5O 2 (g) o 3CO 2 (g) 4H 2 O(v) HHV
b HHV g
2220.0 4 44.01 kJ mol
2220.0 kJ / mol, LHV
natural gas
2043.96 kJ mol C 3 H 8
b0.875gb890.36 kJ molg b0.070gb1559.9 kJ molg b0.020gb2200.00 kJ molg 933 kJ mol
b LHV g
natural gas
b0.875gb802.34 kJ molg b0.070gb1427.87 kJ molg b0.020gb2043.96 kJ molg = 843 kJ mol
b.
g I g I gFGH mol JK b0.070 mol C H gFGH 30.07 mol JK g I g I 1 kg F F b0.020 mol C H gG 44.09 H mol JK b0.035 mol N gGH 28.02 mol JK ] u 10 g 0.01800 kg
b
1 mol natural gas [ 0.875 mol CH 4 16.04 3
c.
9.49
2
8
6
2
843 kJ
1 mol
mol
0.01800 kg
3
46800 kJ kg
The enthalpy change when 1 kg of the natural gas at 25oC is burned completely with oxygen at 25oC and the products CO2(g) and H2O(v) are brought back to 25oC.
Table B.1
bg
C s + O 2 (g) o CO 2 (g), 'H co
B
e j 'H fo
CO 2 ( g)
393.5 kJ 1 mol 10 3 g mol 12.01 g 1 kg
Table B.1
bg
S s + O 2 (g) o SO 2 (g),
'H co
B
e j 'H fo
MSO2 32 .064
296.90 kJ mol
SO 2
B
9261 kJ / kg S
Table B.1
bg
bg
1 H 2 g + O 2 (g) o H 2 O l , 'H co 2
32,764 kJ kg C
B
e j 'H fo
M H 2 1. 008
bg
H 2O l
9-56
285.84 kJ mol H 2
B
141,790 kJ kg H
9.49 (cont'd) a.
x0 (kg O) 2 kg H total H – H in H 2 O ; latter is kg coal 16 kg O
H available for combustion Eq. (9.6-3) HHV
FG H
32,764C 141,790 H
A
in water
IJ K
O 9261S 8
This formula does not take into account the heats of formation of the chemical constituents of coal. b.
C
0. 758 , H
1 kg coal
0. 082 , S
b
0. 016 HHV
Dulong
31,646 kJ kg coal
0.016 kg S 64.07 kg SO 2 formed
Diluting the stack gas lowers the mole fraction of SO2, but does not reduce SO2 emission rates. The dilution does not affect the kg SO2/kJ ratio, so there is nothing to be gained by it.
bg
9.50
g
0. 0320 kg SO 2 kg coal 32.06 kg S burned 0.0320 kg SO 2 kg coal 1.01 u 10 6 kg SO 2 kJ 31,646 kJ kg coal
I c.
0. 051, O
'H co
CH 4 + 2O 2 o CO 2 2H 2 O l , HHV
b
g
890.36 kJ mol Table B.1
bg
7 C 2 H 6 + O 2 o 2CO 2 3H 2 O l , HHV 1559. 9 kJ mol 2 1 CO + O 2 o CO 2 , HHV 282. 99 kJ mol 2 2.000 L 273.2K 2323 mm Hg 1 mol Initial moles charged: 25 + 273.2 K 760 mm Hg 22.4 L STP (Assume ideal gas)
a
f
a f
0.25 mol
Average mol. wt.: (4.929 g) (0.25 mol) = 19.72 g / mol Let x1 MW
g b gh 19.72 x b16.04 g mol CH g x b30.07g b1 x x gb28.01g 19.72 b1g 963.7 kJ mol x b890.36g x b1559.9g b1 x x gb282.99g 963.7 b2g mol CH 4 mol gas , x2 1
HHV
c b
mol C 2 H 6 mol gas 1 x1 x2 mol CO mol gas 4
2
1
2
1
2
1
2
Solving (1) & (2) simultaneously yields x1 9.51
a.
0.725 mol CH 4 mol, x 2
0.188 mol C 2 H 6 mol, 1 x1 x 2
Basis : 1mol/s fuel gas CH 4 (g) 2O 2 (g) o CO 2 (g) 2H 2 O(v), 'H co C 2 H 6 (g)
7 O 2 (g) o 2CO 2 (g) 3H 2 O(v), 'H co 2
0.087 mol CO mol
890.36 kJ / mol 1559.9 kJ / mol
Excess O2, 25qC n2 , mol CO 2 n3 , mol H 2 O n4 , mol O 2 25qC
1 mol/s fuel gas, 25qC 85% CH4 15% C2H6
9-57
9.51 (cont’d)
1 mol / s fuel gas 0.85 mol CH 4 / s , 0.15 mol C 2 H 6 / s Theoretical oxygen
2 mol O 2
0.85 mol CH 4
1 mol CH 4
s
3.5 mol O 2
0.15 mol C 2 H 6
1 mol C 2 H 6
s
2.225 mol O 2 / s
Assume 10% excess O 2 O 2 fed = 1.1 u 2.225 = 2.448 mol O 2 / s . gb2g n b0.85gb1g b015 H balance : 2n . gb6g n b0.85gb4g b015 10% excess O n b01. gb2.225g mol O C balance : n 2
115 . mol CO 2 / s
2
3
2.15 mol H 2 O / s
3
2
4
Extents of reaction: [ 1
2
/s
0.223 mol O 2 / s
0.85 mol / s, [ 2
n CH 4
bg
. mol / s 015
n C 2 H 6
bg bg bg
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C (We will use the values of
'H co
bg
given in Table B.1, which are based on H 2 O l as a
combustion product, and so must choose the liquid as a reference state for water)
e
'H v 25o C
nout
H out
CH 4
mol 0.85
kJ mol 0
mol
kJ mol
C2 H 6
015 .
0
O2
2.225
0
0.223
0
CO 2
115 .
0
H 2O v
2.15
H1
bg
H 1
H in
nin
Substance
j
Energy Balance : Q nCH 4 'H co
e j
44.01 kJ / mol
CH 4
e j
nC 2 H 6 'H co
C2 H 6
¦ n H ¦ n H i
out
i
i
i
in
. mol / s C H gb 1559.9 kJ molg b0.85 mol / s CH gb890.36 kJ molg b015 b2.15 mol / s H Ogb44.01 kJ / molg 896 kW 4
2
6
2
Q b.
896 kW (transferred from reactor)
Constant Volume Process. The flowchart and stoichiometry and material balance calculations are the same as in part (a), except that amounts replace flow rates (mol instead of mol/s, etc.) 1 mol fuel gas 0.85 mol CH 4 , 0.15 mol C 2 H 6 Theoretical oxygen 2.225 mol O 2 Assume 10% excess O 2 O 2 fed = 1.1 u 2.225 = 2.448 mol O 2
. gb2g n b0.85gb1g b015 H balance : 2n b0.85gb4g b015 . gb6g n 10% excess O n b0.1gb2.225g mol O C balance : n2
3
2
115 . mol CO 2
2
4
3
2.15 mol H 2 O
2
0.223 mol O 2
9-58
9.51 (cont'd)
bg
bg bg bg
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C For a constant volume process the heat released or absorbed is determined by the internal energy of reaction. nin nout U in U out Substance mol kJ mol mol kJ mol 0.85 0 CH 4 C2 H 6
015 .
0
O2
2.225
0
0.223
0
CO 2
115 .
H 2O v
2.15
0 U 1
bg
e
U 1
'U v 25 o C
j
e
j
'H v 25 o C RT
Eq. (9.1-5) 'U co
'H co RT (
¦Q
i
e
j
o c
i
1 kJ
mol K 1000 J
. 4153
kJ mol
)
298 K (1 + 2 1 2)
1 kJ
890.36
3
10 J
8.314 J 298 K (3 + 2 35 . 1)
1 kJ
b1559.9 kJ molg mol K
C2H 6
298 K
gaseous reactants
.314 J b890.36 kJ molg 8mol K
CH 4
e'U j
¦Q
gaseous products
'U co
8.314 J
44.01 kJ / mol
3
10 J
kJ mol
156114 .
kJ mol
Energy balance: Q
e
j n e'U j ¦ n U ¦ n U . mol / s C H gb 156114 . kJ mol g b0.85 mol / s CH gb890.36 kJ molg b015 . kJ / mol g 902 kJ b2.15 mol / s H Ogb4153 n CH 4 'U co
'U
CH 4
C2 H 6
o c
C2 H 6
i
i
i
out
4
i
in
2
6
2
Q c.
9.52
a.
902 kJ (transferred from reactor)
Since the O2 (and N2 if air were used) are at 25qC at both the inlet and outlet of this process, their specific enthalpies or internal energies are zero and their amounts therefore have no effect on the calculated values of 'H and 'U . n fuel ( 'H co )
W s Q l (Rate of heat release due to combustion = shaft work + rate of heat loss)
V (gal)
28.317 L
h
7.4805 gal 100 hp
0.700 kg 10 3 g 49 kJ L
1 kg
g
1J/s
1 kJ
3600 s
1.341 u 10 3 hp
10 3 J
h
15 u 10 6 kJ V 298 h
2.5 gal / h
b.
The work delivered would be less since more of the energy released by combustion would go into heating the exhaust gas.
c.
Heat loss increases as Ta decreases. Lubricating oil becomes thicker, so more energy goes to overcoming friction.
9-59
9.53
a.
Energy balance: 'U
a
f
b
n lb m fuel burned
0
b
g
'U co (Btu)
gb
lb m
ga
b
0.00215 'U co 4.62 lb m 0.900 Btu lb m q F 87.06q F 77.00q F 'U co b.
g
0
764.0
kJ mol
mCv Tout 77q F
f
0
19500 Btu lb m
The reaction for which we determined 'U co is 1 lb m oil + aO 2 ( g) o bCO 2 ( g) + cH 2 O(v)
(1)
The higher heating value is 'H r for the reaction 1 lb m oil + aO 2 ( g) o bCO 2 ( g) + cH 2 O(l) o Eq. (9.1-5) on p. 441 'H c1
(2)
o RT (b c a) 'U c1
o Eq. (9.6-1) on p. 462 'H c2
o 'H c1 c'H v (H 2 O, 77$ F)
( HHV )
( LHV )
To calculate the higher heating value, we therefore need
9.54
a.
a
lb - moles of O 2 that react with 1 lb m fuel oil
b
lb - moles of CO 2 formed when 1 lb m fuel oil is burned
c
lb - moles of H 2 O formed when 1 lb m fuel oil is burned
bg
bg
3
2
e'H j
'H ro
CH 3 OH v + O 2 (g) o CO 2 (g) 2H 2 O l
o c
bg
CH 3OH v
Basis : 1 mol CH 3OH fed and burned 1 mol CH 3OH( l ) 25°C, 1.1 atm
vaporizer
Q2(kJ)
1 mol CH 3OH( v) 100°C 1 atm
Q 1 (kJ)
reactor
n 0 (mol O2 ) 3.76n 0 (mol N2 ) 100°C
Overall C balance:
1 mol CH 3OH
1 mol C
% excess air
b
b20.83gb0.809g n 4.482 mol O b1 mol CH OHgb15. mol O mol CH OHg 0
3
2
(4.482 15 . ) mol O 2 u 100% 15 . mol O 2
3
nw uP nw n p
a
15 . mol O 2
200% excess air
gb
(An atomic O balance 9.96 mol O
20.83 mol dry gas
2
H balance: 1 mol CH 3OH 4 mol H 1 mol CH 3OH
pw
gb g
n p 0.048 1 n p
1 mol CH 3OH
N 2 balance: 3.76n0 Theoretical O 2 :
b
Effluent at 300°C, 1 atm n p (mol dry gas) 0.048 mol CO 2/mol D.G. 0.143 mol O 2/mol D.G. 0.809 mol N 2/mol D.G. n w (mol H 2O)
g
bg
nw 2 nw
2 mol H 2 O
9.96 mol O , so that the results are consistent.)
2 mol H 2 O u 760 mm Hg 2 20.83 mol
f
9-60
66.58 mm Hg
d i
pw Tdp
Table B.3
Tdp
44.1q C
9.54 (cont'd) b.
Energy balance on vaporizer:
OP kJ C C dT PP mol 40.33 kJ A A A Q References : CH OHa vf, N (g), O (g), CO (g), H Oal f at 25q C n'H
'H
Q1
LM 1 mol M MN
Substance CH 3 OH
Table B.2
Table B.1
3
2
n in
H in
(mol) 1.00
(kJ / mol) 3.603
z
100
pv
64.7
Table B.2
2
2
2
H out
n out (mol)
(kJ / mol)
2.187
16.85
8.118
4.482
2.235
2.98
8.470
CO 2
1.00
11578 .
H 2O
2.00
5358 .
O2
af
dT 'H v
pl
25
16.85
N2
H T
z
64.7
H i for N 2 , O 2 , CO 2 (Table B.8) 'H v 25$ C H i for H 2 O v (Eq. 9.6 - 2a on p. 462, Table B.8)
z
d
T
25
af
i
af
C p dT for CH 3 OH v (Table B.2)
bg
(Note: H 2 O l was chosen as the reference state since the given value of 'H co presumes liquid water as the product.) Extent of reaction: (nCH 3OH ) out
(nCH 3OH ) in Q CH 3OH [ 0 1 mol [ [
Energy balance on reactor: Q2
['H co
¦ n H ¦ n H i
out
i
b1gb764.0g b16.85gb8.118gb4.482gb2.235g kJ
i
i
in
bTable B.1g
534 kJ 534 kJ transferred from reactor
9.55
a.
3
CH 4 2O 2 o CO 2 2H 2 O
CH 4 O 2 o CO 2H 2 O
Basis : 1000 mol CH 4 h fed
2
Q(kJ/h) 1000 mol CH4 /h 25°C
Stack gas, 400°C n1 (mol CH4 /h) n2 (mol O2 /h) 3.76n0 (mol N2 /h) n3 (mol CO/h) 10 n3 (mol CO/h) n4 (mol H2 O/h)
n0 (mol O2 /h) 3.76n0 (mol N2 /h) 100°C
a f
90% combustion n1 0.10 1000 100 mol CH 4 h Theoretical O2 required = 2000 mol/h 10% excess O2 O 2 fed = 1.1(2000 mol / h) = 2200 mol / h
9-61
1 mol
9.55 (cont’d) C balance: 1000 mol CH 4 h 1 mol C mol CH 4
b
gb
10n 3
g b100gb1g n b1g 10n b1g n 3
3
3
. mol CO h 818
818 mol CO 2 h
a fa f a100fa4f 2n n 1800 mol H O h O balance: a2200 fa2 f 2 n a81.8fa1f a818fa2 f a1800fa1f n 441 mol O References :Casf, H bgg, O bgg, N bgg at 25 C
H balance: 1000 4
4
4
2
2
2
2
H in
n out
H out
1000
74.85
100
57.62
O2
2200
2.24
441
1172 .
N2
8272
2.19
8272
1115 .
CO
818 .
99.27
CO 2
818
377.2
H 2O
1800
228.63
CH 4
Table B.1
Bo
'H f
z
$
2
n in
Substance
H
2
h
2
bmol hg bkJ molg bmol hg bkJ molg
Table B.2
B
T
C p dT for CH 4
25
Table B.8
B
'H fo H i (T) for others Energy balance: Q
'H
¦ n H ¦ n H i
i
out
b.
i
7.344 u 10 5 kJ h
a
204 kW
f
in
(iii)
A (increases) Q A %XS A Q B (more energy required to heat additional O and N to 400 C, therefore less energy transferred.) S A Q A (reaction to form CO2 has a greater heat of combustion and so releases
(iv)
more thermal energy) Tstack Q (more energy required to heat combustion products)
(i) (ii)
9.56
i
Tair
2
2
CO 2 CO
A
B
CH 4 2O 2 o CO 2 2H 2 O, C 2 H 6
7 O 2 o 2 CO 2 3H 2 O 2
Basis : 100 mol stack gas. Assume ideal gas behavior. n1 (mol CH4 ) n2 (mol C2 H6 ) Vf (m3 at 25°C, 1 atm)
100 mol at 800°C, 1 atm 0.0532 mol CO 2 /mol 0.0160 mol CO/mol 0.0732 mol O2 /mol 0.1224 mol H 2 O/mol 0.7352 mol N2 /mol
n3 (mol O2 ) 3.76n3 (mol N2 ) 200°C, 1 atm
9-62
o
9.56 (cont’d) a. N 2 balance: 3.76n3
b100gb0.7352gmol N n 19.55 mol O fed 3.72 mol CH C balance: n b1g n b2 g b100gb0.0532gb1g b100gb0.0160gb1gU| n V H balance: n b4g n b6g b100gb0.1224gb2g W| n 1.60 mol C H . gmol fuel gas 22.4 LbSTPg 298.2 K 1 m b3.72 160 0130 . m V 1
2
3
2
1
2
2
2
3
f
3.72 mol CH 4
2 mol O 2 1 mol CH 4
1.60 mol C 2 H 6
3
3.5 mol O 2 1 mol C 2 H 6
UV 69.9 mole% CH 1.60 mol C H W 30.1 mole% C H b19.55 13.04gmol O in excess u 100% 50% excess air 3.72 mol CH 4
Fuel composition:
2
4
2
6
6
2
% Excess air:
13.04 mol O 2 required
bg b g b g b g
References : C s , H 2 g , O 2 g , N 2 g at 25q C
b.
n in
H in
n out
H out
CH 4
mol 3.72
kJ / mol 74.85
mol
kJ / mol
C2 H 6
1.60
84.67
O2
19.55
5.31
7.32
25.35
N2
. 7352
. 513
. 7352
. 2386
CO
1.60
86.39
CO 2
5.32
356.1
H 2O
Substance
Table B.1
H
Bo
'H f
z
12.24 212.78
Table B.2, for CH 4 , C2 H6
B
T
C p dT
25
Table B.8
B = 'H fo H i (T) for O 2 , N 2 , CO, CO 2 , H 2 O v
bg
Energy balance: Q
'H
¦ n H ¦ n H i
out
i
i
in
2764 kJ i
0.130 m 3 fuel
9-63
6
273.2 K 103 L
1 mol
Theoretical O 2
4
1
2
2.13 u 104 kJ m 3 fuel
13.04 mol O 2
9.57
b0.730gb50000glb
Basis : 50000 lb m coal fed h
mC
1b - mole C
h
3039 1b - mole C h
12.01 lb m
. 2327 lb - moles H h (does not include H in water) b0.047gb50000g 101 b0.037gb50000g 32.07 57.7 lb - moles S h b0.068gb50000g 18.02 189 lb - moles H O h . gb50000g 5900 lb ash h b0118 2
m
a.
50,000 lb m coal/h 3039 lb-moles C/h 2327 lb-moles H/h 57.7 lb-moles S/h 189 lb-moles H 2O/h 5900 lb m ash/h 77°F, 1 atm (assume)
Stack gas at 600°F, 1 atm (assume) n 2 (lb-moles CO2 /h) n 3 (lb-moles H2 O/h) n 4 (lb-moles SO 2 /h) n 5 (lb-moles O2 /h) n 6 (lb-moles N2 /h) m 7 (lb m fly ash/h)
n 1 (lb-moles air/h) 0.210 O 2 0.790 N 2 77°F, 1 atm (assume)
m 8 (lbm slag/h) at 600°F 0.287 lb mC/lb m 0.016 lb m S/lb m 0.697 lb mash/lb m
Feed rate of air :
b
O 2 required to oxidize carbon C + O 2 o CO 2
3039 lb - moles C 1 lb - mole O 2
g
h
1 lb - mole C
3039 lb - moles O 2 h Air fed: n1
1.5 u 3039 lb - moles O 2 fed
1 mole air
h
0.210 mole O 2
b
8 30% ash in coal emerges in slag 0.697m m 7
b g
0.700 5900
b
g
b
gb
g
M CO 2 44.01
2978 lb - moles CO 2 h
b
g b gb g
1352.5 lb - moles H 2 O h
1.31 u 10 5 lb m CO 2 h
2n 3
M H 2 O 18.02
2.44 u 10 4 lb m H 2 O h
b0.790g21710 lb - moles h 17150 lb - moles N 57.7blb - moles S hg b1gn 0.016 b2540g 32.06
N 2 balance: n 6 S balance: n 4
2
h
M N 2 28.02
4.81 u 105 lb m N 2 h
4
56.4 lb - moles SO 2 h
b gb g b gb
M SO 2 64.2
bcoal g
b air g
3620 lb m SO 2 h
gb g b2978gb2g b1352.5gb1g b56.4gb2g 2b n g
O balance: 189 1 0.21 21710 2 n 5
2540 lb m slag / h
n 2 0.287 2540 12.01
H balance: 2327 lb - moles H h 189 2 n 3
g
0.30 5900 lb m h m 8
4130 lb m fly ash h
C balance: 3039 lb - moles C h n 2
21710 lb - moles air h
bCO g 2
943 lb - moles O 2 h 30200 lb m O 2 h
9-64
eH O j 2
bSO g 2
5 O2
9.57 (cont'd)
Summary of component mass flow rates Stack gas at 600q F, 1 atm 2978 lb - moles CO 2 h
131000 lb m CO 2 h
1352.5 lb - moles H 2 O h 24400 lb m H 2 O h 56.4 lb - moles SO 2 h
3620 lb m SO 2 h
943 lb - moles O 2 h
30200 lb m O 2 h
17150 lb - moles N 2 h
48100 lb m N 2 h
674,350 lbm stack gas/h
4130 lb m fly ash h
b
gb g
Check: 50000 21710 29
b679600g
in
b
b.
g
676900
674350 2540
out
(0.4% roundoff error)
out
22480 lb - moles h at 600q F , 1 atm (excluding fly ash)
Total molar flow rate V
in
a f
22480 lb - moles 359 ft 3 STP h
1060q R
1.74 u 10 7 ft 3 h
492q R
1 lb - mole
References: Coal components, air at 77qF ¦ ni H i
0
in
674350 lb m h
Stack gas: nH Slag: nH
2540 lb m
0.22 Btu
h
lb m q F
Energy balance: Q
'H
b600 77gq F
7.063 Btu 1 lb - mole lb - moleq F 28.02 lb m
b600 77gq F b
8.90 u 10 7 Btu h
2.92 u 105 Btu h
g ¦ n H ¦ n H
n coal burned 'H co 77q F
i
i
i
out
5 u 104 lb m
18 . u 10 4 Btu
h
lb m
i
in
e
j
8.90 u 107 2.92 u 105 Btu h
8.11 u 108 Btu h
b0.35ge8.11 u 10 jBtu 8
Power generated c.
Q
e8.11 u 10
Q HHV
8
j b5000 lb
1W
3600 s 9.486 u 10
h Btu h
1 hr
m
1. 62 u 10 4 Btu lb m 1.80 u 104 Btu lb m
coal h
g
4
1 MW Btu s 10 6 W
831 . MW
162 . u 104 Btu lb m coal
0. 901
Some of the heat of combustion goes to vaporize water and heat the stack gas. d.
Q HHV would be closer to 1. Use heat exchange between the entering air and the stack gas.
9-65
9.58
b.
Basis : 1 mol fuel gas/s n 0 (mol O 2 s) 3.76n 0 (mol N 2 s)
o
Stack gas, Ts ( C)
o
Ta ( C)
n O2 (mol O 2 / s) 3.76n 0 (mol N 2 / s) o
1 mol / s @ 25 C x m (mol CH 4 / mol)
n CO (mol CO / s) rn CO (mol CO 2 / s)
x a (mol Ar / mol)
n H 2 O (mol H 2 O / s) n Ar (mol Ar / s)
(1 x m x a ) (mol C 2 H 6 / mol)
CH 4 2O 2 o CO 2 2 H 2 O C2 H 6
7 2
O 2 o 2CO 2 3H 2 O
Pxs ) 2 x m 3.5(1 x m x a ) 100 x m 2(1 x m x a ) C balance: x m 2(1 x m x a ) (1 r )n CO n CO (1 r ) (1
Percent excess air: n 0
H balance: 4 x m 6(1 x m x a ) O balance: 2n 0
2n H 2 O n H 2 O
2 x m 3(1 x m x a )
2n O 2 n CO 2 r n CO n H 2 O n O 2
n 0 n CO (1 2r ) / 2 n H 2 O / 2
References : C(s), H2(g), O2(g), N2(g) at 25qC Substance nin H in nout
CH 4 C2 H 6
xm (1 xm x A )
0 0
A O2
xA no
xA nO2
N2 CO
3.76no
0 H 1 H
CO 2 H 2O
H
3
H 4 H
3.76no nCO
2
z
H out
5
H 6 H
r nCO nH 2 O
7
H 8
Ta or Ts Table B.2
H i
c.
( 'H f ) i
B
C p ,i dT
25
Given : x m no
0.85, x a
0.05, Px s
2.153, n CO
5%, r
0.0955, n H 2 O
150 o C, Ts
10.0, Ta 2.00, nO 2
700 o C
. 01500
H 1 (kJ / mol) = 8.091, H 2 = 29.588, H 3 = 0.702, H 4 = 3.279, H = 166.72, H = 8.567, H = 345.35, H = 433.82 5
6
Energy balance: Q
¦ n
7
out H out
¦ n
8
in H in
9-66
655 kW
9.58 (cont'd)
0.0 5 0.0 5 1 0.0 5 1 0.0 5 1 0.0 5 1 0.0 5 1 0.1 5 0.1 10 0.1 20 0.1 50 0.1 100
r
Ta
Ts
10 10 10 10 10 10
150 150 150 150 150 150
700 700 700 700 700 700
10 10 10 10 10
150 150 150 150 150
700 700 700 700 700
Q 0
-
-200 0
996 90 813 - 22 631 90 869 - 99
0.1 0.2 0.3 0.4 0.5
5 5 5 5 5
10 10 10 10 10
150 150 150 150 150
700 700 700 700 700
893 8 - 1 860
0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
5 5 5 5 5 5 5 5 5
1 2 3 4 5 10 20 50 100
150 150 150 150 150 150 150 150 150
700 700 700 700 700 700 700 700 700
-722 -796 -834 -856 -871 -905 -924 -936 -941
0.1 0.1 0.1 0.1 0.1
5 5 5 5 5
10 10 10 10 10
25 100 150 200 250
700 700 700 700 700
-852 -883 -905 -926 -948
0.1 0.1 0.1 0.1 0.1 0.1
5 5 5 5 5 5
10 10 10 10 10 10
150 500 150 600 150 700 150 800 150 900 150 1000
-1014 -960 -905 -848 -790 -731
9-67
0.2
0.4
0.6
0.8
1
1.2
80
100
120
0.4
0.5
0.6
80
100
120
200
250
300
800
1000
1200
-400 Q
-600 -800 -1000 -1200 Xm
0 - 0 200 Q 400 600 -
20
40
-800 1000
60
Pxs
-850 -860 0
0.1
0.2
0.3
-870 Q -880 -890 -900 -910 x
0 0
20
40
60
-200 -400 Q
Pxs
-600 -800 -1000 r
-800 -850 Q
Xa
0
50
100
150
-900 -950 -1000 Ta
0 0
200
400
600
-500 Q
d.
-1000 -1500 Ts
9.59
a.
Basis: H
207.4 liters 273.2 K 1.1 atm s
1 mol
b g
10.0 mols s fuel gas to furnace
278.2 K 1.0 atm 22.4 liters STP
C 6 H 14 ; M
CH 4 mw (kg H 2O( l)/s) 25°C
Qc (kW) n0 (mol/s) y 0 (mol C 6H 14/mol) (1 – y 0) (mol CH 4/mol) 60°C, 1.2 atm Tdp = 55°C
10.0 mol/s at 5°C, 1.1 atm y2 (mol C 6H 14/mol) (1 – y2) (mol CH 4/mol) sat'd with C 6H 14
condenser
nb (mol C 6H 14( l)/s)
mw (kg H 2O( v)/s) 10 bars, sat'd
Stack gas at 400°C, 1 atm n3 (mol O 2/s) n4 (mol N 2/s) n5 (mol CO 2/s) n6 (mol H 2O( v)/s)
reactor
na (mol air/s) @ 200°C 0.21 mol O /mol 2 0.79 mol N /mol 2 100% excess
Antoine Eq.
Tdp
p DH
55q C y 0 P y0
B
b55q Cg
483.3 mm Hg 1.2 u 760 mm Hg
483.3 mm Hg
0.530 mol C 6 H 14 mol 0.470 mol CH 4 mol
Saturation at condenser outlet: p H 5q C 58.89 mm Hg y2 P 1.1 u 760 mm Hg
b g
0.070 mol C 6 H 14 mol
b
Methane balance on condenser: n0 1 y0 Hexane balance on condenser: n0 y0
Volume of condensate
g
b
g
10.0 1 y2
n b 10.0 y2
bg
9.78 mol C 6 H 14 l
y2 0.070
19.78 mol s
n0
9.78 mol C 6 H 14 s condensed
n b
n0 19.78 y0 0.530 y2 0.070
cm 3
86.17 g
s
y0 0.530
0.93% mol CH 4 mol
1L 3
0.659 g 10 cm
mol
A
A
Table B.1
3600 s 3
1h
Table B.1
4600 L C 6 H 14 (l ) h
b.
e
j
e
References : CH 4 g, 5$ C , C 6 H 14 l, 5$ C Substance CH 4
bg bl g
C 6 H 14 v
bg
H out
nout
(mol / s) (kJ / mol) (mol / s) (kJ / mol) 9.30 1985 9.30 0 . 10.48
41212 .
0.70
32.940
9.78
0
C 6 H 14
CH 4 g : H
H in
nin
j
z
Table B.2
T
5
B
bg
C 6 H 14 v : H
C p dT
Condenser energy balance: Q c
'H
¦ n H ¦ n H i
out
9-68
i
i
in
i
z
Table B.1
Tb
5
B
Table B.1
B
C pR dT 'H v
427 kW
z
T
Tb
C pv dT
9.59 (cont'd) CH 4 2O 2 o CO 2 2H 2 O , C 6 H 14 Theoretical O 2 :
2
O 2 o 6CO 2 7H 2 O
9.30 mol CH 4
2 mol O 2
s
1 mol CH 4
b g 2 u bO g balance: 0.79b240.95g n n
100% excess O 2 N2
19
2 theor.
fed
4
0.70 mol C 6 H 14
9.5 mol O 2
s
1 mol C 6 H 14
0.21na
2 u 25.3 na
25.3 mol O 2 s
240.95 mol air s
190.35 mol N 2 s
4
C balance: 9.30 mol CH 4
1 mol C
s
1 mol CH 4
n5
. mol CO 2 s 135
0.70 mol C 6 H 14
b
6 mol C
n5 mol CO 2
g
1 mol C 1 mol CO 2
1 mol C 6 H 14
H balance:
b9.30 mol CH sgb4 mol H mol CH g b0.70gb14g n b2g n 235. mol H O 1 Since combustion is complete, bO g bO g bO g n 25.3 mol O s 2 References : Cbsg, H bgg, O bgg, N bgg at 25q C for reactor side, H Oblg at triple point for 4
4
6
2 remaining
2
2
6
2 excess
2
2 fed
2
3
2
2
steam side (reference state for steam tables) H in
n in
Substance
H out
n out
mol / s 9.30
kJ / mol mol / s 75.553
kJ / mol
C 6 H 14 v
0.70
170.07
O2
50.6
5.31
25.3
. 1172
N2
190.35
. 513
190.35
. 1115
CO 2
. 135
377.15
H 2O v
235 .
228.60
CH 4
bg
bg H Ob boiler water g 2
bg
m w (kg / s) 104.8
Table B.1 and B.2
B =
H T
z
m w ( kg / s) 2776.2
T
'H fo C p dT
for CH 4 , C 6 H 14
25 Table B.1 and B.8
B =
bg
bg
'H fo H i T for O 2 , N 2 , CO 2 , H 2 O v
Energy balance on reactor (assume adiabatic): 'H
¦ n H ¦ n H i
out
i
i
i
b
g
0 8468 m w 2776.2 104.8
in
9-69
0 m w
3.2 kg steam s
9.60
a.
CH 4 2O2 o CO2 2H 2O
Basis: 450 kmol CH 4 fed h n a ( kmol air / h)@25 o C
o
0.21 kmol O 2 / kmol 0.79 kmol N 2 / kmol
Stack gas@300 C n 1 (kmol CO 2 / h) n 2 (kmol H 2 O / h)
o
Q ( kJ / h)
450 kmol CH 4 / h @ 25 C
n 3 (kmol O 2 / h) n 4 (kmol N 2 / h)
m w [kg H 2 O(l) / h]
m w [kg H 2 O(v) / h]
o
17 bar, 250 o C
25 C
450 kmol CH 4 h
Air fed: n a
2 kmol O 2 req' d 1.2 kmol O 2 fed 1 kmol air 1 kmol CH 4 1 kmol O 2 req' d 0.21 kmol O 2
5143 kmol air h 450 kmol / h CH 4 react n1
450 kmol CO 2 h , n 2
. u 10 b0.79ge5143
N 2 balance: n 4
6
mol h
j
900 kmol H 2 O h
4060 kmol N 2 h
Molecular O 2 balance:
b0.21gb5143g mol Oh y 450 kmol CO h U | 900 kmol H O h | |V y y 4060 kmol N h | | y 180 kmol O h | W n 3
2
fed
450 kmol CH 4 react 2 mol O 2 h 1 mol CH 4
CO 2
0.0805
2
H 2O
. 0161
2
N2
0.726
2
O2
0.0322
2
180 kmol O 2 h
5590 kmol / h Mean heat capacity of stack gas Cp
¦y C i
pi
. gb0.0343g b0.726gb0.0297g b0.0322gb0.0312g b0.0805gb0.0423g b0161
0.0315 kJ mol $ C Energy balance on furnace (combustion side only)
bg
bg bg
bg
bg
References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25$ C n in H in n out H out Substance (kmol / h) (kJ / kmol) (kJ / h) CH 4
450
0
Air
5143
0
Stack gas
H p
Extent of reaction:
[
n CH 4
450 kmol / h
9-70
9.60 (cont’d)
H p
n 2 ( 'H v ) H
o 2 O(25 C)
n stack gas (C p ) stack gas (Tstack gas 25o C)
3 5590 kmol 10 3 mol = 180 kmol H 2 O 10 mol 44.01 kJ h 1 kmol mol h 1 kmol 7 = 5.63 u 10 kJ / h
Q
[ ( 'H co ) CH 4
'H
¦ n H ¦ n H i
i
out
=
i
0.0315 kJ (300- 25) o C mol o C
i
in
FG 450 kmol IJ FG1000 mol IJ FG 890.36 kJ IJ 5.63 u 10 H h K H kmol K H mol K
7
kJ h
3.44 u 10 8
kJ h
Energy balance on steam boiler Table B.7 Table B.6
Q
LM FG IJ OP LMb2914 105g kJ OP kg Q N H KQ N
kg kJ m w 'H w + 3.44 u 10 8 m w h h m w 123 . u 10 5 kg steam / h
b.
n a (mol air/h) at
45 kmol CH 25°C
4
/h
mw (kg H 2O/h) Liquid, 25°C
T a (°C)
Stack gas n 1 (mol CO 2/h) n 1 (mol CO 2/h) air n 2 (mol H O/h) n 2 (mol H 2O/h) 2 furnace n 3 (mol O 2/h) n 3 (mol O 2/h) preheater n 4 (mol N 2/h) n 4 (mol N 2/h) 300°C 150°C mw (kg H 2O/h) vapor, 17 bars n a (mol air/h) at 25°C 250°C 0.21 O 2 0.79 N 2
E.B. on overall process: The material balances and the energy balance are identical to those of part (a), except that the stack gas exits at 150oC instead of 300oC.
b g b g b g b g bg b g H Oalf at triple point (steam table reference) (steam tube side)
References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25 $ C furnace side 2
Substance
n in (kmol / h)
H in (kJ / kmol)
n out H out (kJ / h)
CH 4 Air Stack gas
450 5143
0 0
H p
H 2O
H p
n 2 ( 'H v ) H
o 2 O(25 C)
m w ( kg / h) 105 kJ / kg
m w ( kg / h) 2914 kJ / kg
n stack gas (C p ) stack gas (Tstack gas 25o C)
3 5590 kmol 10 3 mol = 180 kmol H 2 O 10 mol 44.01 kJ h 1 kmol mol h 1 kmol 7 = 2.99 u 10 kJ / h
9-71
0.0315 kJ (150- 25) o C mol o C
9.60 (cont’d)
[ ( 'H co ) CH 4
'H
¦ n H ¦ n H i
i
i
out
0
i
in
FG 450 kmol IJ FG1000 mol IJ FG 890.36 kJ IJ 2.99 u 10 kJ H h K H kmol K H mol K h L F kg I OL kJ O Mm G J P Mb2914 105g P 0 m = 1.32 u 10 H K h kg N QN Q Energy balance on preheater: 'H d 'H i d 'H i 0 5590 kmol 10 mol 0.0315 kJ b150 300g C nC 'T b 'H g 7
w
w
stack gas
stack gas
c.
b 'H g
stack gas
b 'H g
H air
5133 . kJ / mol
h
$
$
mol C
1 kmol
Table B.8
2.64 u 10 7
kJ h
2.64 u 10 7 kJ / h 1 kmol = 5.133 kJ 5143 kmol / h 10 3 mol mol
n a H air (Ta ) H air (Ta )
air
kg steam / h
air
3
p
5
199 o C
Ta
The energy balance on the furnace includes the term
¦n
in H in .
If the air is preheated and the
stack gas temperature remains the same, this term and hence Q become more negative, meaning that more heat is transferred to the boiler water and more steam is produced. The stack gas is a logical heating medium since it is available at a high temperature and costs nothing.
9.61
Basis: 40000 kg coal h
b0.76 u 40000gkg C
10 3 g 1 mol C
h
Assume coal enters at 25q C
1 kg
. j b0.05 u 4000gkg H h e10 101 b0.08 u 4000gkg O h e10 16.0j . u 40000g 4400 kg ash h b011
12.01 g
3
198 . u 10 6 mol H h
3
2.00 u 10 5 mol O h
4400 kg ash/h, 450°C Q to steam
40,000 kg coal/h 2.531 u10 6 mol C/h 1.98 u10 6 mol H/h 2.00 u10 5 mol O/h 4400 kg ash/h 25°C
Flue gas at 260°C n3 (mol dry gas/h) 0.078 mol CO 2 /mol D.G. 0.012 mol CO/mol D.G. 0.114 mol O 2/mol D.G. 0.796 mol N 2/mol D.G. n4 (mol H2 O/h)
furnace
Preheated air at Ta (°C)
a.
2.531 u 10 6 mol C h
air at 30°C, 1 atm, hr = 30% n1 (mol O2 /h) 3.76n1 (mol N2 /h) n2 (mol H2 O/h) preheater Cooled flue gas at 150°C n3 (mol dry gas/h) 0.078 CO 2 0.012 CO 0.114 O 2 0.796 N 2 n4 (mol H2 O/h)
Overall system balances C balance: 2.531 u 10 6 N 2 balance: 3.76n1
0.078n 3 0.012n 3 n 3
b0.796ge2.812 u 10 j n 7
1
7
224 u 10 mol N 2 h
9-72
2.812 u 10 7 mol h dry flue gas
b ge
5.95 u 10 6 mol O 2 h 3.76 5.95 u 10 6
j
9.61 (cont'd) 30% relative humidity (inlet air): Table B.3
0.30 p H 2 O
y H 2O P
n 2
b30q Cg 5.95 u 10
b760 mm Hgg
n 2 6
7
2.24 u 10 n 2
b
B
0.300 31824 . mm Hg
3.61 u 10 5 mol H 2 O h
Volumetric flow rate of inlet air: V
e5.95 u 10
6
b g
j
224 u 10 7 3.61 u 10 5 mol 22.4 liters STP h
Air/fuel ratio:
10 liters
161 . SCM air kg coal
e
j
H balance: 198 . u 10 6 mol H h 2 3.61 u 10 5 mol H h H in coal
2n 4 n 4
1.351 u 10 6 mol H 2 O h
H in water vapor
1357 u 10 6 mol H 2 O h .
H 2 O content of stack gas b.
6.43 u 10 5 SCMH
3
1 mol
6.43 u 10 5 m 3 air h 40000 kg coal h
1 m3
u 10 . e1357
6
j
2.812 u 10 7 mol h
u 100%
4.6% H 2 O
Energy balance on stack gas in preheater
bg
References : CO 2 , CO, O 2 , N 2 , H 2 O v at 25 $ C n in mol h 2.193 u 10 6 0.337 u 10 6 3.706 u 10 6 22.38 u 10 6 1.357 u 10 6
Substance CO 2 CO O2 N2 H 2O
H in kJ mol 4.942 3669 3758 3655 4266
H i (T ) from Table B.8 for inlet Q
¦ n H ¦ n H i
i
out
i
i
n out mol h 2.193 u 10 6 0.337 u 10 6 3.206 u 10 6 72.38 u 10 6 1.351 u 10 6
H i (T ) =
b
z
H out kJ mol 9.738 6.961 7.193 6.918 8135
Table B.2
B Cp
dT for outlet
g
101 . u 10 8 kJ h Heat transferred from stack gas
in
Air preheating 1.01 u10 8 kJ/hr
2.83 u10 7 mol dry air/h 3.61 u10 5 mol H 2 O/h 30°C
2.83 u10 7 mol dry air/h 3.61 u10 5 mol H 2 O/h T a (°C)
(We assume preheater is adiabatic, so that Qstack gas Energy balance on air: Q
'H 1.01 u 10
8
kJ hr
¦
z
Ta
z
Qair )
Ta
ni (C p ) i dT
30
30
9-73
Table B.2
n dry air
z
T
Table B.2
a B B (C p ) dry air dT n H 2 O (C p ) H 2 O dT
30
g
9.61 (cont'd) 101 . u 10 8 8.31 u 10 5 (Ta 30) 59.92(Ta2 30 2 ) 0.031(Ta3 30 3 ) 1.42 u 10 5 (Ta4 30 4 ) Ta 150 $ C c.
4400 kg ash/h at 450°C
40,000 kg coal/h 25°C
Flue gas at 260°C 2193u 106 mol CO 2 /h 0.337 u10 6 mo l CO/h 3.206 u 106 mo l O2 /h 22.38 u 106 mo l N2 /h 1.351 u 106 mo l H2 O( v)/h
5.95 u10 6 mo l O2 /h 2.24 u10 7 mo l N2 /h 3.61 u10 5 mo l H2 O( v)/h 150°C(= 149.8°C)
m (kg H 2 O(l )/h) 50°C
m (kg H 2 O(v )/h) 30 bars, sat'd
bg
References for energy balance on furnace: CO 2 , CO, O 2 , N 2 , H 2 O l , coal at 25q C
bg
(Must choose H 2 O l since we are given the higher heating value of the coal.) H in 0 3.758 3.655 48.28
substance n in 40000 Coal Ash 5.95 u 10 6 O2 N2 2.24 u 10 6 CO 2 CO 3.61 u 10 5 H 2O
n out 4400 3.206 u 10 6 2.24 u 10 7 2.193 u 10 6 0.337 u 10 6 1.351 u 10 6
H out n kg h 412.25 H kJ kg 7.193 6.918 n mol h 9.738 H kJ mol 6.961 52.14
b b b b
g
g g g
(Furnace only — exclude boiler water) Heat transferred from furnace Q
n coal 'H io
¦ n H ¦ n H i
i
out
i
i
in
FG 4 u 10 kg IJ FG 2.5 u 10 kJ IJ FG 2.74 u 10 H h KH kg K G H 4
8.76u 10
4
8
3
1.22 u 10 8
A
H of preheated air
I kJ JJ kg K
kJ h 8
8
Heat transferred to boiler water: 0.60(8.76x10 kJ/h) = 5.25x10 kJ/h
b g m FGH kgh IJK H cH Oblg, 30b, sat' dh H eH Oblg, 50 Cj L O kJ kJ h m M2802.3 209.3P MN A A PQ kg m 2.02 u 10 kg steam h
Energy balance on boiler: Q kJ h 5.25 u 10 8
2
2
5
Table B.6
Table B.5
9-74
$
9.62
1
CO O 2 o CO 2 , 'H co
Basis : 1 mol CO burned. 1 mol CO n 0 mol O2 3.76n 0 mol N2 25°C
a.
Oxygen in product gas: n1
1 mol CO2 (n 0 – 0.5) mol O2 3.76n 0 mol N2 1400°C
b
g
n0 mol O 2 fed
1 mol CO react 0.5 mol O 2 1 mol CO
References: CO, CO 2 , O 2 , N 2 at 25$ C n in n out H in Substance mol mol kJ mol
b g b
CO O2 N2 CO 2
b g N bg,1400q Cg: H CO bg,1400q Cg: H
H O 2 (1400$ C)
n 0 0.5 3.76n 0 1
H 1 H 2 H 3
0 0 0
g
B
47.07 kJ mol
Table B.8
H N 2 (1400$ C)
2
2
1 n0 3.76n 0
g b g b
H out kJ mol
n0 0.5
Table B.8
O 2 g,1400q C : H 1 2
282.99 kJ mol
2
B
44.51 kJ mol
Table B.8
B
H CO 2 (1400$ C)
3
7189 . kJ mol
E.B.: nCO 'H co
'H
¦ n H ¦ n H i
i
i
out
n0
b
g
b1 mol COgb0.5 mol O
2
mol CO
g
a.
0
0.500 mol O 2
1.094 mol fed 0.500 mol reqd. u 100% 119% excess oxygen 0.500 mol Increase %XS air Tad would decrease, since the heat liberated by combustion would go into heating a larger quantity of gas (i.e., the additional N 2 and unconsumed O 2 ).
Excess oxygen:
9.63
g
1.094 mol O 2
Theoretical O 2
b.
b
282.99 47.07 n0 0.5 44.51 3.76n0 71.89
i
in
Basis : 100 mol natural gas 82 mol CH 4 , 18 mol C 2 H 6 CH (g) 2O (g) o CO (g) 2H O(v), 'H o 890.36 kJ / mol 4
2
2
2
c
7 C 2 H 6 (g) O 2 (g) o 2CO 2 (g) 3H 2 O(v), 'H co 2 82 mol CH4 18 mol C2H6 298 K
1559.9 kJ / mol
Stack gas at T(qC) n2 (mol CO2) n3 (mol H2O (v)) n4 (mol O2) n5 (mol N2)
n0 (mol air) at 423 K 0.21 O2 (20% XS) 0.79 N2
9-75
9.63 (cont’d) 2 mol O 2
Theoretical oxygen
82 mol CH 4
1 mol CH 4
1.2 u 227 mol O 2
Air fed : n1
1 mol air 0.21 mol O 2
b82.00gb1g b18.00gb2g n b82.00gb4g b18.00gb6g n
C balance : n2
b0.79gb1297.14g
Extents of reaction: [ 1
nCH 4
bg
227 mol O 2
1 mol C 2 H 6 1297.14 mol air
218.00 mol H 2 O
3
b0.2gb227g mol O
20% excess air, complete combustion n4 N 2 balance : n5
18 mol C 2 H 6
118.00 mol CO 2
2
H balance : 2n3
3.5 mol O 2
45.40 mol O 2
2
1024.63 mol N 2
82 mol, [ 1
nC2 H 6
18 mol
bg bg bg
bg
Reference states: CH 4 g , C2 H 6 g , N 2 g , O 2 g , H 2 O l at 298 K (We will use the values of 'H co given in Table B.1, which are based on H 2 O l as a combustion
bg
product, and so must choose the liquid as a reference state for water.)
bg
b
H i T
g
C pi T 298 K for all species but water
b
g
b
g
'H v,H 2 O 298 K C p ,H 2 O T 298 K for water Substance
n in mol
H in kJ mol
CH 4 C2 H 6 O2 N2 CO 2 H 2O v
82.00 18.00 272.40 1024.63
0 4.14 3.91
bg
Energy balance : 'H
e
[ 1 'H co
b
j
CH 4
H out kJ mol
n out mol
45.40 0.0331 T 298 1024.63 0.0313 T 298 118.00 0.0500 T 298 218.00 44.013 0.0385 T 298
b b b
b
g g g
g
0
e
[ 2 'H co
j
C2 H 6
gb
¦ n H ¦ n H i
i
out
i
i
0
in
g b gb g b
gb g b
g gb
82.00 mol CH 4 890.36 kJ mol 18.00 mol C 2 H 6 1559.90 kJ mol 45.40 0.0331 1024.63 0.0313 118.00 0.0500 218.00 0.0385 T 298
b.
b gb g b gb b218.00gb44.01g (272.40)( 4.14) (1024.63)( 3.91)
Solving for T using E - Z Solve T
gb
g
0
2317 K
Increase % excess air Tout decreases. (Heat of combustion has more gas to heat) % methane increases Tout might decrease. (lower heat of combustion, but heat released goes into heating fewer moles of gas.)
9-76
bg
bg
af
bg
C 3 H 6 O g 4O 2 g o 3CO 2 g 3H 2 O l , 'H io
9.64
Basis :
b g
103 mol
min
22.4 m STP
1410 m 3 STP feed gas
3
1821.4 kJ mol
1 min
b g
1049 mol s feed gas
60 s
Stochiometric proportion:
b
1 mol C 3 H 6 O 4 mol O 2 4 u 3.76 15.04 mol N 2 1 4 15.04 yC 3 H 6O
1 mol C 3 H 6 O 20.04 mol
0. 0499
mol C 3 H 6 O , yO 2 mol
Preheat Feed gas 1049 mol/s 0.0499 C 3 H 6 O 0.1996 O22 0.1496 0.7505 N 2 T f (°C), 150 mm Hg Rel. satn = 12.2%
p b.
20.04 mol
0.1996 mol O 2 mol
Reaction
Cooling
Feed gas 562°C
12.2% y C 3H 6O P
b0.0499gb1500 mm Hgg
Q2 (kW)
d i
0122 . p C 3H 6O T f
Table B.4
613.52 mm Hg
0.122
Product gas 350°C
Product gas n 1 (mol CO 2 /s) n 2 (mol H 2O/s) n 3 (mol N 2/s) T a (°C)
Q1 (kW)
a. Relative saturation
4 20.04
g
50.0 o C
Tf
b gb g 52.34 mol C H O s b1049gb0.1996g 209.4 mol O s b1049gb0.7505g 787.3 mol N s n b52.34gb3g 157.0 mol CO s U 14.25 mole% CO Product contains n b52.34gb3g 157.0 mol H O s|V 14.25% H O |W 71.5% N 787.3 mol N s n References : C H Obgg, O , N , H Oblg, CO at 25 C Feed contains 1049 mol s 0.0499 C 3 H 6 O mol
3
6
2
2
1
2
2
2
3
2
2
3
6
2
2
2
2
C3H 6O O2 N2 CO 2 H 2O
52.34 209.4 787.3
67.66 17.72 16.65
$
2
n in H in n out ( mols) (kJ / mol) ( mols) (562 $ C)
Substance
2
H out ( kJ / mol) Ta
787.3 157.0 157.0
0.031 Ta 25 0.050 Ta 25 44.013 0.038 Ta 25
H H 2 O
'H v 25$ C H H 2 O ( T)
b b
d
b
g g
g
i
Energy balance on reactor:
'H
b
nC 3H 6O 'H co
g FGH
¦ n H ¦ n H i
out
5234 mol s 18211 .
i
i
IJ K
b g
0 kJ s
i
in
b
g
b
g
kJ 38.22 Ta 25 157.0 44.013 2.036 u 10 4 mol
9-77
0 Ta
2871q C
9.64 (cont'd) c. Preheating step: References: C 3 H 6 g , O 2 , N 2 at 25q C
bg
n in H in n out H out ( mol / s) (kJ / mol) (mol / s) ( kJ / mol) (50 $ C) (562 $ C) 52.34 315 52.34 67.66 . 209.4 0.826 209.4 17.72 787.3 0.775 787.3 16.65
Substance C3H 6O O2 N2 E.B. Q 1
¦ n H ¦ n H i
i
i
out
1.94 u 10 4 kW
i
in
af
Cooling step. References: CO 2 (g), H 2 O v , N 2 (g) at 25$ C n H in n out H out Substance in ( mol) (kJ / mol) (mol) (kJ / mol)
e2871 Cj
e350 Cj
$
CO 2 H 2O N2 E.B. Q2
157.0 157.0 787.3
142.3 108.15 88.23
¦ n H ¦ n H i
i
out
i
i
$
157.0 157.0 787.3
16.25 12.35 10.08
9.64 u 10 4 kW
in
Exchange heat between the reactor feed and product gases. 9.65
a.
Basis : 1 mol C5H12 (l)
C5 H 12 (l) 8O 2 (g) o 5CO 2 (g) 6H 2 O(v),
'H co
1 mol C5H12 (l)
n2(mol CO2) n3 (mol H2O (v)) n4 (mol O2) Tad(oC)
n0 (mol O2) , 75qC 30% excess
Theoretical oxygen 30% excess n0 C balance: n2 H balance: 2n3
3509.5 kJ / mol
1 mol C 5 H 12
8 mol O 2 1 mol C 5 H 12
8 mol O 2
1.3 u 8 10.4 mol O 2
b1gb5g n 5 mol CO b1gb12g n 6 mol H O 2
2
3
2
30% excess O2, complete combustion n4
bg b g
bg
b0.3gb8g mol O
2
Reference states: C5 H 12 l , O 2 g , H 2 O l , CO 2 (g) at 25o C
2.4 mol O 2
bg
(We will use the values of 'H c0 given in Table B.1, which are based on H 2 O l as a combustion product, and so must choose the liquid as a reference state for water)
9-78
9.65 (cont'd) substance C 5 H 12 O2 CO 2
H in
nin
nout
H out
mol 1.00
kJ mol mol kJ mol 0 10.40 2.40 H 1 H 2 5.00 H 3
H 2O
z
6.00
H 4
T
H i
(C p ) i dT
i
25
e
j
2,3
z T
'H v 25 o C (C p ) H 2 O(v) dT for H 2 O(v) 25 Table B.8
H 1
B
H O 2 (75 o C)
=
148 . kJ / mol
Substituting ( C p ) i from Table B.2 : kJ mol kJ 0.9158) mol
H 2
(0.0291 Tad 0.579 u 10 5 Tad 2 0.2025 u 10 8 Tad 3 0.3278 u 10 12 Tad 4 0.7311)
H 3
(0.03611 Tad 2.1165 u 10 5 Tad 2 0.9623 u 10 8 Tad 3 1866 . u 10 12 Tad 4
H 4
. . ) 44.01 (0.03346 Tad 0.3440 u 10 5 Tad 2 0.2535 u 10 8 Tad 3 08983 u 10 12 Tad 4 0838
kJ mol
. (0.03346 Tad 0.3440 u 10 5 Tad 2 0.2535 u 10 8 Tad 3 0.8983 u 10 12 Tad 4 ) 4317
kJ mol
H 4
Energy balance : 'H
e j
nC5H12 'H co
0
C5 H 12 ( l)
¦ n H ¦ n H i
i
i
out
i
0
in
(1 mol C5 H 12 )( 3509.5 kJ / mol) (2.40) H 2 (5.00) H 3 (6.00) H 4 (10.40)( H 1 )
0
Substitute for H 1 through H 4
'H
(0.4512 Tad 14.036 u 10 5 Tad 2 3.777 u 10 8 Tad 3 4.727 u 10 12 Tad 4 ) 3272.20 kJ / mol = 0
f (Tad ) Check :
3272.20 0.4512 Tad 14.036 u 10 5 Tad 2 3.777 u 10 8 Tad 3 4.727 u 10 12 Tad 4 3272.20
4.727 u 10 12
6.922 u 1014
Solving for Tad using E - Z Solve Tad
b.
Terms 1 2 3
4414 o C
Tad % Error 7252 64.3% 3481 –21.1% 3938 –10.8%
9-79
0
9.65 (cont’d) c.
d.
T 7252 5634 4680 4426 4414
f(T) 6.05E+03 1.73E+03 3.10E+02 1.41E+01 3.11E-02
f'(T) Tnew 3.74 5634 1.82 4680 1.22 4426 1.11 4414 1.11 4414
The polynomial formulas are only applicable for T d 1500qC
9.66
5.5 L/s at 25qC, 1.1 atm n 1(mol CH4/s)
Adiabatic 25% excess air Reactor n 2 (mol O2/s) 3.76 n 2 (mol N2/s) n 4 (mol CO2/s) 150qC, 1.1 atm
n 3 (mol O2/s) 3.76 n 2 (mol N2/s) n 5 (mol H2O/s) T(qC), 1.05 atm
2CH 4 2O 2 o CO 2 2 H 2 O 550 . L 273 K 1.1 atm
Fuel feed rate :
s
2 u 0.247
Theoretical O 2
mol
298 K 1.0 atm 22.4 L(STP)
0.247 mol CH 4 / s
0.494 mol O 2 / s
25% excess air n 2 125 . (0.494 ) 0.6175 mol O 2 / s , 3.76 u 0.6175 2.32 mol N 2 / s Complete combustion [
n1 = 0.247 mol / s, n 4
n 3
0.247 mol CO 2 / s, n 5
0.6175 mol O 2 fed / s 0.494 mol consumed / s 0.124 mol O 2 / s
Re ferences: CH 4 , O 2 , N 2 , CO 2 , H 2 O at 25 o C n in n out H in H out Substance ( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol) CH 4 O2 N2 CO 2 H 2O H 1
H (O 2, 150 o C)
H 2
H ( N 2, 150 o C)
( 'H co ) CH 4
z
0.247 0.6175 2.32 Table B.8
Table B.8
0 H 1 H 2
3.78 kJ / mol 3.66 kJ / mol
890.36 kJ / mol
T
H i
C pi dT , i
35
25
9-80
0124 . 2.32 0.247 0.497
H 3 H 4 H 5 H 6
0.494 mol H 2 O / s
9.66 (cont'd)
z T
H b a.
( 'H v ) H
$ 2 O(25 C)
(C p ) H 2 O(v) dT 25
Energy Balance 'H [ ( 'H o ) c
CH 4
¦ n
out H out
Table B.2 for C pi ( T ), ( 'H v ) H O 2
¦ n
0
in H in
44.01 kJ / mol
0.247( 890.36) 0.494(44.01) 0.0963(T 25) 174 . u 10 5 (T 2 25 2 ) 0.305 u 10 8 (T 3 25 3 ) 1.61 u 10 12 (T 4 25 4 ) 0.6175(3.78) 2.32(3.66) 0 211.4 0.0963Tad 1.74 u 10 5 Tad 2 0.305 u 10 8 Tad 3 161 . u 10 12 Tad 4 T b.
1672 C
In product gas, T
1672 o C, P
. u 760 105
0.155 mol H 2 O / mol Table B.3
pH* 2 O (Tdp ) pH* 2 O
Raoult's law: y H 2 O P
a.
798 mmHg
0.494 mol / s (0124 . 2.32 0.247 0.494 ) mol / s
y H 2O
9.67
0
o
(0.155)(798)
124 mmHg
Tdp
56$ C
CH 4 (l) 2O 2 (g) o CO 2 (g) + 2H 2 O(v) Basis : 1 mol CH 4 1 mol CH 4
Theoretical oxygen
30 % excess air 130 . (2.00)
2 mol O 2 2.00 mol O 2 1 mol CH 4 2.60 mol O 2 , 3.76 u 2.60
9.78 mol N 2
n2 (mol CO2) n3 (mol H2O) n4 (mol O2 )
1 mol CH4 2.60 mol O2 9.78 mol N2 25q C, 4.00 atm Complete combustion n 2
1.00 mol CO 2 , n 3
2.00 mol O 2 consumed n4
2.00 mol H 2 O
(2.60 2.00) mol O 2
Internal energy of reaction: Eq. (9.1-5)
'U co
'H co
0.60 mol O 2
I F J G RT G ¦ Q ¦ Q J JK GH i
gaseous products
e
'U co
z
j
CH 4
T
U
25
FG 890.36 kJ IJ 8.314 J H mol K mol K
Ideal Gas
(Cv )dT
z
298 K (1 + 2 1 2)
T
(C p Rg )dT
25
If C p is independent of T U
(C p Rg )(T 25o C)
9-81
i
gaseous reactants
1 kJ 10 3 J
890.36
kJ mol
9.67 (cont’d) b.
bg bg bg
bg
Reference states: CH 4 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
bg
(We will use the values of 'H c0 given in Table B.1, which are based on H 2 O l as a combustion product, and so must choose the liquid as a reference state for water.) U in
nin
Substance
nout
U out
mol kJ mol mol 100 0 .
O2
2.60
0
N2
9.78
0
kJ mol 0.60 U 1 9.78 U
CO 2
1.00
H 2O v
2.00
CH 4
bg
2
U 3 U 4
Part a
U i
B
(C p Rg )(T 25) for all species except H 2 O(v) 'U v 25 o C (C p Rg )(T 25) 'H v 25 o C Rg Tref (C p Rg )(T 25) for H 2 O(v)
e
j
e
j
8.314 u 10 3 kJ / mol yields
Substituting given values of (C p ) i and Rg U 1
(0.033 8.314 u 10 3 )(T 25) kJ / mol
(0.02469T 0.6172) kJ / mol
U 2
(0.032 8.314 u 10 3 )( T 25) kJ / mol
(0.02369T 0.5922) kJ / mol
U 3
(0.052 8.314 u 10 3 )(T 25) kJ / mol
(0.04369T 10922 . ) kJ / mol
LM44.01 kJ FG 8.314 u 10 N mol H
U 4 U 4
OP Q
IJ K
kJ kJ (298 K) + (0.040 8.314 u 10 3 )( T 25) mol K mol kJ kJ kJ . (0.03167T 40.74) 4153 + (0.052 8.314 u 10 3 )(T 25) mol mol mol 3
Energy Balance Q
j ¦ n U ¦ n U (100 . )b 890.36 kJ / molg (0.60)U e
n CH 4 'U co Q
CH 4
i
i
out
i
0
i
in
1
(9.87)U 2 (100 . )U 3 (2.00)U 4
0
Substituting U 1 through U 4
0.3557 T 816.19
0 T
Ideal Gas Equation of State c.
Pf Pi
2295 o C Tf Pf Ti
FG (2295 273) K IJ u 4.00 atm H (25 273) K K
– Heat loss to and through reactor wall – Tank would expand at high temperatures and pressures
9-82
34.5 atm
9.68
b.
1 mol natural gas yCH 4 (mol CH 4 / mol)
nCO 2 (mol CO 2 )
yC2 H 6 (mol C 2 H 6 / mol)
nH 2 O (mol H 2 O)
yC3H 8 (mol C 3 H 8 / mol)
nN 2 (mol N 2 ) nO 2 mol O 2 )
Humid air na (mol air) ywo (mol H20(v)/mol) (1-ywo) (mol dry air/mol) 0.21 mol O2/mol DA 0.79 mol N2/mol DA Basis : 1 g-mole natural gas CH 4 (g) 2O 2 (g) o CO 2 (g) H 2 O(v) 7 O 2 (g) o 2CO 2 (g) 3H 2 O(v) 2 C 3 H 8 (g) 5O 2 (g) o 3CO 2 (g) 4 H 2 O(v) C 2 H 6 (g)
Theoretical oxygen : 2 mol O2 1 mol CH 4
yCH4 (mol CH 4 )
3.5 mol O2 1 mol C2 H 6
yC2H6 (mol C2 H 6 )
5 mol O2 1 mol C3 H8
yC3H8 (mol C 3 H 8 )
( 2 yCH 4 35 . yC 2 H 6 + 5 y C 3 H 8 ) Excess oxygen: 0.21n a (1 y wo )
FG H
na = 1
IJ K
FG1 P IJ ( 2 y H 100 K xs
CH 4
3.5 y C 2 H 6 + 5 y C 3H 8 ) mol O 2
Pxs 1 ( 2 y CH 4 3.5 y C 2 H 6 + 5 y C 3H 8 ) mol air 100 0.21(1 y w0 )
Feed components (n O 2 ) in
0.21n a (1 y wo ), (n N 2 ) in
0.79n a (1 y wo ), (n H 2O ) in
n a y wo
N 2 in product gas: n N 2 = (n N 2 ) in mol N 2 CO2 in product gas : nCO2
1 mol CO2 nCH4 (mol CH4 ) 2 mol CO2 nC2H6 (mol C2 H6 ) 3 mol CO2 nC3H8 (mol C3 H8 ) 1 mol CH4 1 mol C2 H 6 1 mol C3 H8 (nCH4 2nC2H6 3nC3H8 ) mol CO2
H 2 O in product gas : nH2O
1 mol H 2 O nCH4 (mol CH 4 ) 3 mol H 2 O nC2H6 (mol C2 H 6 ) 4 mol O2 nC3H8 (mol C3 H8 ) 1 mol CH 4 1 mol C2 H 6 1 mol C3 H8 [2nCH4 3nC2H6 4nC3H8 + na (1- ywo )] mol H 2 O
O 2 in product gas : n O 2
Pxs ( 2n CH 4 35 . n C 2 H 6 + 5 nC3 H 8 ) mol O 2 100
9-83
9.68 (cont’d) c. References : C(s), H 2 (g) at 25o C
z T
H CH 4 ( T)
( ' H of ) CH 4 (C p ) CH 4 dT 25
Using ( ' H fo )CH 4 from Table B.1 and (C p )CH 4 from Table B.2 . kJ / mol + H CH (T) 7485 4
F GG H
z
I JJ K
T
u 108 T 2 1100 . . u 1012 T 3 ) dT kJ / mol (0.03431+5.469 u 105 T 03661
25
H CH (T ) [7572 . + 3.431 u 10 2 T + 2.734 u 10 5 T 2 0122 . u 10 8 T 3 2.75 u 10 12 T 4 ] kJ / mol 4
7
¦
6
(ni ) out ( Hi ) out
i 4
H i
¦(n )
i in ( Hi ) in
i 1 3
ai bi T ci T 2 d i T ei T 4
6
¦
H out
n out
mol kJ / mol mol kJ / mol n1 H 1 n2 H 2 n3 H3 n4 H 4 n7 H 7 n5 H 5 n8 H 8 n6 n9 H 9 n10 H 10
CH 4 C2 H 6 C3H 8 O2 N2 CO 2 H 2O 'H
H in
n in
Substance
3
¦
(ni ) in ( Hi ) in
i 1
i 1
7
'H
(ni ) in H i (Tf )
¦
¦(n )
i in Hi (Ta )
i 4
(ni ) out (ai bi T ci T 2 d i T 3 ei T 4 ) out
i 4 7
'H
6
7
¦ (n )
i out ai
i 1
¦ (n )
i out bi T
i 4
3
7
¦ (n )
¦ (n )
i out ci T
i 1
6
i in Hi (Tf
2
)
i 1
¦ (n )
3
¦
(ni ) in H i (Tf )
i 1 7
¦(n )
i out d i T
i 1
i in Hi (Ta )
i 4
2
D 0 D 1T D 2 T D 3T D 4 T 4 7
where D 0
¦
3
(ni ) out ai
i 1 7
D1
3
¦
(ni ) in H i (Tf )
i 1
¦ (n )
¦ (n )
D2
¦ (n )
D4
i out bi
i 4
¦(n )
i out ci
i 1 7
i out d i
i 1
¦ (n )
i out ei
i 1
.
9-84
i in Hi (Ta )
7
i 1 7
D3
6
3
6
¦ (n )
i in Hi (Ta )
i 4 7
¦ (n )
i out ei T
i 1
4
9.68 (cont’d) d.
Run 1 yCH4 0.75 yC2H6 0.21 yC3H8 0.04 Tf 40 Ta 150 Pxs 25 ywo 0.0306 nO2i 3.04 nN2 11.44 nH2Oi 0.46 HCH4 -74.3 HC2H6 -83.9 HC3H8 -102.7 HO2i 3.6 HN2i 3.8 HH2Oi -237.6 nCO2 1.29 nH2O 2.75 nO2 0.61 nN2 11.44 Tad 1743.1 alph0 -1052 alph1 0.4892 alph2 0.0001 alph3 -3.00E-08 alph4 3.00E-12 Delta H 3.00E-07
Species CH4 C2H6 C3H8 O2 N2 H20 CO2
a -75.72 -85.95 -105.6 -0.731 -0.728 -242.7 -394.4
Run 2 0.86 0.1 0.04 40 150 25 0.0306 2.84 10.67 0.43 -74.3 -83.9 -102.7 3.6 3.8 -237.6 1.18 2.61 0.57 10.67 1737.7 -978.9 0.4567 1.00E-04 -3.00E-08 3.00E-12 9.00E-06
Run 3 0.75 0.21 0.04 150 150 25 0.0306 3.04 11.44 0.46 -70 -77 -93 3.6 3.8 -237.6 1.29 2.75 0.61 11.44 1750.7 -1057 0.4892 0.0001 -3.00E-08 3.00E-12 -4.00E-07
Run 4 0.75 0.21 0.04 40 250 25 0.0306 3.04 11.44 0.46 -74.3 -83.9 -102.7 6.6 6.9 -234.1 1.29 2.75 0.61 11.44 1812.1 -1099 0.4892 0.0001 -3.00E-08 3.00E-12 -1.00E-04
Run 5 0.75 0.21 0.04 40 150 100 0.0306 4.87 18.31 0.73 -74.3 -83.9 -102.7 3.6 3.8 -237.6 1.29 3.02 2.44 18.31 1237.5 -1093 0.7512 0.0001 -4.00E-08 4.00E-12 -1.00E-05
b x 10^2 3.431 4.937 6.803 2.9 2.91 3.346 3.611
c x 10^5 2.734 6.96 11.3 0.11 0.579 0.344 2.117
d x 10^8 0.122 -1.939 -4.37 0.191 -0.203 0.254 -0.962
e x 10^12 -2.75 1.82 7.928 -0.718 0.328 -0.898 1.866
9-85
Run 6 0.75 0.21 0.04 40 150 25 0.1 3.04 11.44 1.61 -74.3 -83.9 -102.7 3.6 3.8 -237.6 1.29 3.9 0.61 11.44 1633.6 -1058 0.5278 0.0001 -2.00E-08 2.00E-12 6.00E-04
9.69
n 14 (mol CH4 /h) 25°C Preheaters
Absorber off-gas n 8 (mol H 2 /h) n 12 (mol N 2 /h) 0.988 n 9 (mol CO/h) 0.950 n 6 (mol CH4 /h) 0.006 n 7 (mol C 2H 2/h)
n 13 (mol C(s )/h) Converter converter product quench Tad (°C)
Feed gas, 650°C n 14 (mol CH4 /h) 0.96n 15 (mol O2 /h) 0.04n 15 (mol N2 /h)
Converter product 38°C
0.917 n 1 (mol DMF/h) Lean solvent filter absorber
n 6 (mol CH4 /h) n 7 (mol C 2H 2/h) n 8 (mol H 2 /h) n 9 (mol CO/h) n 10 (mol CO2 /h) n 11 (mol H 2 O/h) n 12 (mol N 2 /h) n 13 (mol C(s )/h)
n 15 (mol/h) 0.96 mol O2 /mol 0.04 mol N2 /mol 25°C
Basis: 5000 kg/h Product gas n 1 (mol/h) 0.991 mol C2 H2 ( g)/mol 0.00059 mol H2 O/mol 0.00841 mol CO2 /mol
n 6 (mol CH4 /h) n 7 (mol C 2H 2/h) n 8 (mol H 2 /h) n 9 (mol CO/h) n 10 (mol CO2 /h) n 11 (mol H 2 O/h) n 12 (mol N 2 /h)
Rich solvent
n 1 (mol/h) 0.0155 mol C2 H2 /mol 0.0063 mol CO2 /mol 0.00055 mol CO/mol 0.00055 mol CH4 /mol 0.0596 mol H2 O/mol 0.917 mol DMF/mol
Average M.W. of product gas:
b
g
b
g
b
g
0.991 26.04 0.00059 18.016 0.00841 44.01
M
Molar flow rate of product gas: n0
5000 kg 103 g day
1 kg
1 mol
1 day
26.19 g
24 h
7955 mol h
Stripper balances: C 2 H 2 n1 , CO n2 , CH 4 n3 , H 2 O n4 , CO 2 n5 Absorber balances: CH 4 n6 , C 2 H 2 n7 , CO n9 , CO 2 n10 , H 2 O n11 13 , n14 ,
converter H balance n8
Converter O balance n15 , converter N 2 balance n12 Stripper balances: C 2 H 2 : 0.0155n1
b
g
0.991 7955 mol h n1
b
5.086 u 105 mol h
ge j n n 79.7 mol CO h CH : b0.00055ge5.086 u 10 j n n 79.7 mol CH h H O: b0.0596ge5.086 u 10 j n b0.00059gb7955g n = 30308 mol H O h CO : b0.0068ge5.086 u 10 j n b0.00841gb7955g n = 3392 mol CO h CO: 0.00055 5.086 u 105
5
4
5
2
5
2
Absorber balances CH 4 : n6
2
b
2
3
3
4
4
4
5
5
ge
0.950n6 0.00055 5.086 u 105
9-86
j
Stripper off-gas n 2 (mol CO/h) n 3 (mol CH4 /h) n 4 (mol H 2O(v )/h) n 5 (mol CO2 /h)
2619 . g mol
Material balances -- plan of attack (refer to flow chart):
RS5.67% soot formationUV n Tconverter C balance W
stripper
n6 5595 mol CH 4 h
2
2
9.69 (cont'd)
b0.0155ge5.086 u 10 j 0.006n n 7931 mol C H h CO: n 0.988n b0.00055ge5.086 u 10 j n = 23311 mol CO h CO : n b0.0068ge5.086 u 10 j 3458 mol CO h H O: n b0.0596ge5.086 u 10 j 30313 mol H O h n b0.0567gn (mol CH ) 1 mol C n 0.0567n b1g Soot formation: h 1 mol CH 5
C 2 H 2 : n7 9
2
2
7
7
5
9
2
5
11
13
2
9
5
10
2
2
14
4
13
14
4
Converter C balance:
b5595 mol CH hgb1 mol C mol CH g b7931gb2g b23311gb1g b3458gb1g n n n 48226 b2g 2899 mol C bsg h , n 51120 mol CH h Solve (1) & (2) simultaneously n 51120 mol CH 4 mol H Converter H balance: b5595gb4g b7931gb2g 2n b30313gb2g h 1 mol Ch n14
4
14
4
13
13
13
14
CH 4
4
4
C2 H2
8
4
n8
b
gb g
Converter O balance: 0.96n15 2
H 2O
H2
52816 mol H 2 h 23311 mol CO
1 mol O
h
1 mol CO
b
CO 2
gb g b
H 2O
gb g
3458 2 30313 1
n15 31531 mol h
b gb
g
Converter N 2 balance: 0.04 31531 n12 n12 a.
Feed stream flow rates VCH 4 VO 2
b.
51120 mol CH 4 h
b
31531 mol O 2 N 2 h
Gas feed to absorber 5595 mol 7931 mol 23311 mol 3458 mol 30313 mol 52816 mol 1261 mol
b g
0.0244 m 3 STP 1 mol
g
1261 mol N 2 h
1145 SCMH CH 4
b g
0.0244 m 3 STP 1 mol
b
706 SCMH O 2 N 2
g
U| || |V || || W
CH 4 h C2 H 2 h CO h CO 2 h 4.5 mole% CH 4 , 6.4 % C 2 H 2 , 18.7% CO , H 2 O h 125 kmol h , 2.8% CO 2 , 24.3% H 2 O , 42.4% H 2 , 1.0% N 2 H2 h N2 h
1.2469 u 10 5 mol h
Absorber off-gas 52816 mol H 2 h 1261 mol N 2 h 23031 mol CO h 64.1 mole% H 2 , 1.5% N 2 , 27.9% CO, 5315 mol CH 4 h 82.5 kmol h, 6.4% CH , 0.06% C H 4 2 2 41.6 mol C 2 H 2 h 8.2471 u 10 4
U| || V| | mol h|W
9-87
9.69 (cont'd) Stripper off-gas 279.7 mol CO h 279.7 mol CH 4 h 30308 mol H 2 O h 34.3 kmol h, 0.82% CO, 0.82% CH 4 , 88.5% H 2 O, 9.9% CO 2 3392 mol CO h 3.4259 u 10 4
U| |V || mol h W
c.
DMF recirculation rate
d.
Overall product yield
mol I F 1 kmol I FG JG J 466 kmol DMF h H h K H 10 mol K b0.991gb7955g mol C H in product gas 0154 mol C H . 0.917 5.086 u 10 5
3
2
2
2
51120 mol CH 4 in feed h
2
mol CH 4
The theoretical maximum yield would be obtained if only the reaction 2CH 4 o C 2 H 2 3H 2 occurred, the reaction went to completion, and all the C 2 H 2 formed were recovered in the product gas. This yield is (1 mol C2H2/2 mol CH4) = 0.500 mol C2H2/2 mol CH4. The ratio of the actual yield to the theoretical yield is 0.154/0.500 = 0.308. e.
Methane preheater Q CH 4
'H
n14
z
Table B.2 650
25
B
d i Cp
CH 4
dT
51120 mol 32824 J 1h 1 kJ h mol 3601 s 10 3 J
466 kW
Oxygen preheater Table B.8
Q O 2
'H
Table B.8
B
B
0.96n15 H (O 2 ,650 $ C) 0.04n15 H ( N 2 ,650 $ C)
kJ OF 1 h I FG 31531 mol IJ LMb0.96 u 20135 . 0.04 u 18.99g PG J H K h N mol C QH 3600 s K References : Cbsg, H bgg, O bgg, N bgg at 25q C Substance n H b650q Cg n H bT g $
f.
2
in
2
in
2
out
out
z
51120
42.026
5595
O2
30270
20.125
N2
1261
18.988
1261
C2 H 2
7931
H2
52816
CO
23311 110.52 C p dT
CO 2
3458
H 2O
. C p dT 30313 24183
Cs
2899
9-88
74.85
out Ta
CH 4
bg
176 kW
25
z z z z z z z Ta
35
C p dT
C p dT
226.75
Ta
25
C p dT
C p dT
393.5 C p dT
C p dT
b g H b kJ molg n mol h
9.69 (cont’d)
z T
H i
'H i0 C pi dT
kJ mol
¦ n H
i
¦ n H
i
i
in
i
out
25 kJ mol qC
. 1575 u 10 6 kJ h
z
LM5595dC i 1261dC i 7931dC i N OP 1 kJ dT 52816dC i 23311dC i 3458dC i 3013dC i b g Q 10 J 1 kJ dC i b g u 10 J dT 9.888 u 10 6 kJ h
z
Tout
p CH 4
25
p H 2
p CO
p N 2
p CO 2
p C H 3 2
p H O v 2
3
Tad 273
p C s
298
3
We will apply the heat capacity formulas of Table B.2, recognizing that we will probably push at least some of them above their upper temperature limits
¦ n H i
out
¦ n H i
i
z
Tad 273
298
z
Tad
e3902 1.2185 5.9885 u 10 F 32.411 0.031744T 14179 u 10 I . GH JKdT T
9.888 u 10 6 kJ h
i
25
4
6
2
1.000 u 10 7 3943Ta 0.6251Ta2 1.996 u 10 4 Ta3 2.5405 u 10 8 Ta4
out
Energy balance: 'H
¦ n H ¦ n H i
out
b g
f Tc
E-Z Solve
j
T 2 1.0162 u 10 7 T 3 dT
i
i
i
0
in
. u 10 4 Tc3 2.5405 u 10 8 Tc4 8.485 u 106 3943Tc 0.6251Tc2 1996 Tc
. u 10 6 1418 Ta 273
2032 o C.
9-89
1418 . u 106 Tc 273
0
9.70 a. n1[ kg W(v) / d]
W = H2O
100o C
o
24,000 kg sludge / d, 22 C 0.35 solids, 0.65 W(l)
DRYER
Q 2
F n2 (kg conc. sludge / d), 100o C 0.75 solids, 0.25 W(l)
INCINERATOR
Waste gas
n3 [kg W(v) / d] 4B, sat'd C n3 [kg W(l) / d]
BOILER
n3 [kg W(l) / d] 4B, sat'd
Q 3 ( kJ / d)
20o C
D
0.10
kmol C2 H 6
110oC
kmol
Q 4 (kJ / d)
Q1
n4 ( kg oil / d) 0.87 C 0.10 H 0.0084 S 0.0216 ash
n6 (kg gas / d) kmol CH 4 0.90 kmol
Stack gas CO2 , H 2O(v) 125o C
SO2 O2 , N 2 ash
n7 (kg air / d) 25o C
Q 0 ( kJ / d)
E n5 (kg air / d) 25o C
9-90
9.70 (cont'd) Solids balance on dryer: 0.35 u 24,000 kg / d 0.75n2 n2
11200 kg / d n1 11200 n1
Mass Balance on dryer: 24,000
F 11.2 tonnes / d (conc. sludge) 12,800 kg / d
Energy balance on sludge side of dryer: References : H 2 O(l,22 $ C), Solids(22 $ C) nin nout H in H out (mol d ) (kJ mol) (mol d) (kJ mol) 8400 0 8400 H 1 15600 0 2800 H 2 12800 H 3
Substance Solids H 2 O(l) H 2O(v)
H 1 H
2.5(100 22) (419.1 92.2)
326.9 kJ / kg
H 1 ( H
(2676 92.2)
2584 kJ / kg
2
water
195.0 kJ / kg
from Table B.5)
¦ m H ¦ m H Q
Q 2
i
i
out
i
in 7
3.56 u 10 0.55
Q steam
3.56 u 10 7 kJ day
2
i
6.47 u 10 7 kJ / d Q 3
2.91 u 10 7 kJ / d
Energy balance on steam side of dryer: 6.47 u 10 7
FG IJ H K
'H v for H 2 O(sat'd, )
B kJ kg = n3 u 2133 d d
FG kJ IJ F 1 tonne I n H kg K GH 10 kg JK 3
3
30.3 tonnes / d (boiler feedwater)
Energy balance on steam side of boiler: (30300
Q1
kg kJ )(2737.6 83.9) d kg
8.04 u 10 7 kJ / d
62% efficiency Fuel heating value needed = n4
130 . u 108 kJ / d 3.75 u 104 kJ / kg
8.04 u 10 7 0.62
13 . u 108 kJ / d
3458 kg / d D = 3.5 tonnes / day (fuel oil)
Air feed to boiler furnace: C + O 2 o CO 2 , 4H + O 2 o 2H 2 O, (nO 2 ) theo
3458
S + O 2 o SO 2
LM N
kJ kgC 1 kmol C 1 kmol O 2 1 1 1 1 (0.87 )( )( ) + (0.10)( )( ) (0.0084)( )( ) kg kg 1 4 12 kg 1 kmol C 32 1
338 kmol O 2 / d
Air fed (25% excess) = 1.25(4.76
kmol air kmol O2 kmol air )(338 ) 2011 kmol O2 d d
2011 kmol 29 kg 1 tonne E = 58.3 tonnes/ d (air to boiler) d kmol 103 kg
9-91
OP Q
9.70 (cont’d) 2011 kmol 29.25 kJ (125 25) $ C
Energy balance on boiler air preheater: Q o
d
kmol $ C
Q o
588 . u 10 6 kJ / d
Supplementary fuel for incinerator: n6
11.2 tonne sludge 195 SCM d
MWgas M gas
tonne
1 kmol 22.4 SCM
0.90 MWCH 4 0.10 MWC2 H 6
97.5 kmol d
(0.90)(16) (0.10)(30)
17.4 kg kmol
= 1.7 tonne / d (natural gas) (97.5)(17.4) G
CH 4 2O 2 o CO 2 + 2H 2 O, C 2 H 6
7 O 2 o 2CO 2 + 3H 2 O 2
Air feed to incinerator: 11200 75 kg sol 19000 kJ 3.2 kg air 1 kmol (O 2 ) th for sludge: d kg kg sol 104 kJ 29 kg (O 2 ) th for gas: 97.5 n7
2(1761 210)
1.88 u 10 4
kmol O 2 kmol air u 4.76 d kmol O 2
OP Q
210 kmol O 2 d
. u 104 kmol air / d 188
kmol kg = 545 tonnes / d ( incinerator air) u 29 H d kmol
Air preheater : Q 4 b.
LM N
kmol CH 4 2 kmol O 2 kmol 0.90 u (010 . )(3.5) d kmol kmol CH 4
1761 kmol O 2 d
1.88 u 104 kmol
29.2 kJ $
(110 25) $ C
Q 4
4.7 u 10 7 kJ / d
d kmol C Cost of fuel oil, natural gas, fuel oil and air preheating, pumping and compression, piping, utilities, operating personnel, instrumentation and control, environmental monitoring. Lowering environmental hazard might justify lack of profit.
c.
Put hot product gases from boiler and/or incinerator through heat exchangers to preheat both air streams. Make use of steam from dryer.
d.
Sulfur dioxide, possibly NO2, fly ash in boiler stack gas, volatile toxic and odorous compounds in gas effluents from dryer and incinerator.
9-92
CHAPTER TEN 10.1 b. Assume no combustion n 1 (mol gas),T1 (°C) x 1 (mol CH4 /mol) x 2 (mol C2 H6 /mol) 1 – x 1 – x 2 (mol C3 H8 /mol)
n 3 (mol), 200°C y 1 (mol CH4 /mol) y 2 (mol C2 H6 /mol) y 3 (mol C3 H8 /mol) 1 – y 1 – y 2 – y 3 (mol air/mol)
n 2 (mol air), T2 (°C)
Q (kJ)
b b
g
11 variables n1 , n2 , n3 , x1 , x 2 , y1 , y 2 , y 3 , T1 , T2 , Q 5 relations 4 material balances and 1 energy balance 6 degrees of freedom
g
ln , n , x , x , T , T q
A feasible set of design variables:
1
2
1
2
1
2
Calculate n3 from total mole balance, y1 , y 2 , and y 3 from component balances, Q from energy balance. An infeasible set:
ln , n , n , x , x , T q 1
2
3
1
2
1
Specifying n1 and n2 determines n3 (from a total mole balance) c.
n 2 (mol gas), T 2 , P y 2 (mol C 6 H 14/mol) 1 – y 2 (mol N 2 /mol)
n 1 (mol gas), T 1 , P y 1 (mol C 6 H 14/mol) 1 – y 1 (mol N 2 /mol)
n 3 (mol C 6 H 14( l )/mol), T 2 , P
Q (kJ)
b d
g
n1 , n2 , n3 , y1 , y 2 , T1 , T2 , Q, P 9 variables 4 relations 2 material, 1 energy, and 1 equilibrium: y 2 P 5 degrees of freedom A feasible set:
b gi
PC*6 H14 T2
ln, y , T , P, n q 1
1
3
Calculate n2 from total balance, y 2 from C 6 H 14 balance, T2 from Raoult’s law: [ y2 P
b g
PC 6 H 4 T2 ], Q from energy balance
An infeasible set:
ln , y , n , P, T q 2
2
3
2
Once y 2 and P are specified, T2 is determined from Raoult’s law
10- 1
b
g
10.2 10 variables n1 , n2 , n3 , n4 , x1 , x 2 , x 3 , x 4 , T , P 2 material balances 2 equilibrium relations: [ x 3 P x 4 PB* T , 1 x 3 P 6 degrees of freedom
bgb
g b1 x g P bT g] 4
* C
ln , n , n , x , x , Tq
a. A straightforward set:
1
3
4
1
4
Calculate n2 from total material balance, P from sum of Raoult's laws: P
bg b
g bg
x 4 p B T 1 x 4 Pc T
x 3 from Raoult's law, x 2 from B balance
b. An iterative set:
ln , n , n , x , x , x q 1
2
3
1
2
3
Calculate n4 from total mole balance, x 4 from B balance. Guess P, calculate T from Raoult's law for B, P from Raoult’s law for C, iterate until pressure checks. c. An impossible set:
ln , n , n , n , T , Pq 1
2
3
4
Once n1 , n2 , and n3 are specified, a total mole balance determines n4 .
bg
bg
bg
bg
10.3 2BaSO 4 s 4C s o 2BaS s 4CO 2 g a.
100 kg ore, T 0 (K) xb (kg BaSO4 /kg)
n 1 (kg C) n 2 (kg BaS) n 3 (kg CO2 ) n 4 (kg other solids) T f (K)
n 0 (kg coal), T 0 (K) xc (kg C/kg) Pex (% excess coal)
Q (kJ)
d
i
11 variables n0 , n1 , n2 , n3 , n4 , x b , x c , T0 , T f , Q, Pex 5 material balances C, BaS, CO 2 , BaSO 4 , other solids 1 energy balance 1 reaction 1 relation defining Pex in terms of n0 , x b , and x c
b
g
5 degrees of freedom
n
b. Design set: x b , x c , T0 , T f , Pex
s
Calculate n0 from x b , x c , and Pex ; n1 through n4 from material balances, Q from energy balance
10-2
10.3 (cont’d)
l
q
c. Design set: x B , x c , T0 , n2 , Q
Specifying x B determines n2 impossible design set.
l
q
d. Design set: x B , x c , T0 , Pex , Q
Calculate n2 from x B , n3 from x B n0 from x B , x c and Pex n1 from C material balance, n4 from total material balance T f from energy balance (trial-and-error probably required) 10.4 2C 2 H 5 OH O 2 o 2CH 3 CHO 2H 2 O 2CH 3 COH O 2 o 2CH 3 CHOOH n f (mol solution), T 0 x ef (mol EtOH/mol) 1 – x ef (mol H 2 O/mol)
n e (mol EtOH), T n ah (mol CH 3 CHO) n ea (mol CH 3 COOH) n w (mol H 2O) n ax (mol O 2) n n (mol N 2)
n w (mol air), Pxs , T 0 0.79 n air (mol N 2 ) 0.21 n air (mol O 2 ) (Pxs = % excess air)
a.
Q (kJ)
d
13 variables n f , naw , ne , neh , nea , n w , nex , n0 , x ef , T0 , T , Q, Pxs 6 material balances 1 energy balance 1 relation between Pxs , n f , x ef , and nair 2 reactions 7 degrees of freedom
n
b. Design set: n f , x ef , Pxs , ne , nah , T0 , T
i
s
Calculate nair from n f , x ef and Pxs ; nn from N 2 balance; naa and nw from n f , x ef , ne , nah and material balances; nex from O atomic balance; Q from energy balance
n
c. Design set: n f , x ef , T0 , nair , Q, ne , n w
s
Calculate Pxs from n f , x ef and nair ; n’s from material balances; T from energy balance (generally nonlinear in T)
l
q
d. Design set: nair , nn , . Once nair is specified, an N 2 balance fixes nn
10- 3
10.5 a.
n 1 (mol CO) n 2 (mol H2 ) n 3 (mol C3 H6 )
reactor
n 4 (mol C3 H6 ) n 5 (mol CO) n 6 (mol H2 ) n 7 (mol C7 H8 O) n 8 (mol C4 H7 OH) n 9 (kg catalyst)
Flash tank
n10 (kg catalyst)
n 16(mol C3 H6 ) n 11(mol C3 H6 ) n 12(mol CO) n 17(mol CO) Separation n 18(mol H2 ) n 13(mol H2 ) n 14(mol C7 H8 O) n 15(mol C4 H7 OH) n 19(mol C7 H8 O) n 20(mol C4 H7 OH) n 21(mol H2 )
Hydrogenator
n 22(mol H2 )
n 20(mol C4 H7 OH)
b
g
b
g
b
g
b
g
Reactor:
10 variables n1 n16 6 material balances 2 reactions 6 degrees of freedom
Flash Tank:
12 variables n4 n15 6 material balances 6 degrees of freedom
Separation:
10 variables n11 n20 5 material balances 5 degrees of freedom
Hydrogenator:
5 variables n19 n23 3 material balances 1 reaction 3 degrees of freedom
Process:
20 Local degrees of freedom 14 ties 6 overall degrees of freedom
The last answer is what one gets by observing that 14 variables were counted two times each in summing the local degrees of freedom. However, one relation also was counted twice: the catalyst material balances on the reactor and flash tank are each n9 n10 . We must therefore add one degree of freedom to compensate for having subtracted the same relation twice, to finally obtain 7 overall degrees of freedom (A student who gets this one has done very well indeed!) b. The catalyst circulation rate is not included in any equations other than the catalyst balance (n9 = n10). It may therefore not be determined unless either n9 or n10 is specified.
10-4
b
10.6 n C 4 H 10 o i C 4 H 10 n B i B n 1 (mol n-B)
mixer
n 2 (mol n-B) n 3 (mol i-B)
g
reactor
n 4 (mol n-B) n 5 (mol i-B)
n 6 (mol) x 6 (mol n-B/mol) (1 – x 6)(mol i-B/mol)
still
n r (mol) x r (mol n-B/mol) (1 – x r)(mol i-B/mol)
b
a. Mixer:
5 variables n1 , n2 , n3 , nr , x r 2 material balances 3 degrees of freedom
b
g
g
Reactor:
4 variables n2 , n3 , n3 , n5 2 material balances 1 reaction 3 degrees of freedom
Still:
6 variables n4 , n5 , n6 , x 6 , nr , x r 2 material balances 4 degrees of freedom
Process:
10 Local degrees of freedom 6 ties 4 overall degrees of freedom
b. n1
b
100 mol n C 4 H 10 , x 6
b gb g
0.115 mol n C 4 H 10 mol , x r
100 mol n - B fed
35% S.P. conversion: n4
n4
b
4 0.885 4 mol C n6 . n6 0115
Mixer n-B balance: 100 0.85nT
Still n – B balance:
n2
b1g
0.65n2 n4
b1g
65 0.5525nr
b2 g
b2g
. gb100g 0.85n n 179.83 mol b0115 mol recycle . b179.83 mol recycleg b100 mol fresh feedg 179 mol fresh feed
n6 x 6 nr x r 65 0.5525nr
Recycle ratio
0.85 mol n C 4 H 10 mol
100 mol overhead gb g b gb g . gmol n - B unreacted 100 mol n - B fed b100gb0115 u 100% 88.5%
Overall C balance: 100 4 Overall conversion
g
r
10.6 (cont’d) 10- 5
r
c. nr n2
k 1 k 2 k 3 100.0 132.3 1515 . 185.0 212.5 228.8
100 0.85n r n r 1 0.85
b
n3
g
15.0 19.85 22.73 120.25 138.1 148.7
0.65n 2
n4 n5
n2 n 3 n 4
UV W
n 4 n5 n 6 n r n6 n 4 0.115n 6 0.85n r nr
Error:
0.595 0.595 1
q nrb
94.21 102.8 80.76 88.55
132.3
1515 . 163.0
179.83 163.0 u 100 9.3% error 179.83
1515 . 132.3 132.3 100.0
d. w
79.75 67.69
0.595
1470 .
b
g c b
ghb g
g
1470 132.3 1 1470 1515 . . .
Error:
179.8 179.8 u 100 01% . error 179.8
3
179.8
e. Successive substitution, Iteration 32: nr = 179.8319 Æ nr = 179.8319 Wegstein, Iteration 3: nr = 179.8319 Æ nr = 179.8319 10.7
S1 SF
Split
S2
a. 1 2 3 4 5 6 7 8
A X1 =
nA nB nC nD T(deg.C)
B
C 0.6 Molar flow rates (mol/h) SF S1 85.5 51.3 52.5 31.5 12.0 7.2 0.0 0.0 315 315
Formula in C4: = $B$1*B4 Formula in D4: = B4-C4
10.7 (cont’d) 10-6
D
S2 34.2 21.0 4.8 0.0 315
b.
C **CHAPTER 10 -- PROBLEM 7 DIMENSION SF(8), S1(8), S2(8) FLOW 150. N 3 SF(1) 0.35*FLOW SF(2) 0.57*FLOW SF(3) 0.08*FLOW SF(8) 315. X1 0.60 CALL SPLIT (SF, S1, S2, X1, N) WRITE (6, 900)' STREAM 1', S1(1), S1(2), S1(3), S1(B) WRITE (6, 900)' STREAM 2', S2(1), S2(2), S2(3), S2(B) 900 FORMAT (A10, F8.2,' mols/h n-octane', /, * 10X, F8.2,' mols/h iso-octane', /, * 10X, F8.2,' mols/h inerts', /, * 10X, F8.2,' K') END C C C
100
SUBROUTINE SPLIT SUBROUTINE SPLIT (SF, S1, S2, X1, N) DIMENSION SF(8), S1(8), S2(8) D0 100 J 1, N S1(J) X1*SF(J) S2(J) SF(J) – S1(J) S1(8) SF (8) S2(8) SF (8) RETURN END
Program Output: Stream 1 3150 . mols h n-octane 51.30 mols h iso-octane 7.20 mols h inerts 315.00 K Stream 2 21.00 mols h n-octane 34.20 mols h iso-octane 4.80 mols h inerts 315.00 K
10- 7
10.8 a. Let Bz = benzene, Tl = toluene Antoine equations: p *Bz 10 6.905651211.033/( T 220.790) (= 1350.491)
pTl* Raoult' s law: x Bz
10 6.953341343.943/( T 219.377) (= 556.3212)
( P pTl* ) / ( p *Bz - pTl* ) (= 0.307) , y Bz
Total mole balance: 100 = nv nl Benzene balance: 40 = y Bz nv x Bz nl
UV W
* / P ( 0.518) x Bz p Bz
Fractional benzene vaporization: f B
40 100x Bz (= 44.13), nl y Bz x Bz nv y Bz / 40 (= 0.571)
Fractional toluene vaporization: f T
nv (1 y Bz ) / 60 (= 0.354)
nv
100 nv (= 55.87)
The specific enthalpies are calculated by integrating heat capacities and (for vapors) adding the heat of vaporization. Q ¦ nout H out ¦ nin H in (= 1097.9) b. Once the spreadsheet has been prepared, the goalseek tool can be used to determine the bubble-point temperature (find the temperature for which nv=0) and the dew-point temperature (find the temperature for which nl =0). The solutions are Tbp 96.9 o C, Tdp 103.2 o C
C ** CHAPTER 10 PROBLEM B DIMENSION SF(3), SL(3), SV(3) DATA A1, B1, C1/6.90565, 1211.033, 220.790/ DATA A2, B2, C2/6.95334, 1343.943, 219.377/ DATA CP1, CP2, HV1, HV2/ 0.160, 0.190, 30.765, 33.47/ COMMON A1, B1, C1, A2, B2, C2, CP1, CP2, NV1, NV2 FLOW 1.0 SF(1) 0.30*FLOW SF(2) 0.70*FLOW T 363.0 P 512.0 CALL FLASH2 (SF, SL, SV, T, P, Q) WRITE (6, 900) 'Liquid Stream', SL(1), SL(2), SL(3) WRITE (6, 900) 'Vapor Stream', SV(1), SV(2), SV(3) 900 FORMAT (A15, F7.4,' mol/s Benzene',/, * 15X, F7.4, mol/s Toluene',/, * 15X, F7.2, 'K') WRITE (6, 901) Q 10.8 (cont’d) c.
10-8
901 C
C
C
C
C
C
FORMAT ('Heat Required', F7.2,' kW') END SUBROUTINE FLASN2 (SF, SL, SV, T, P, Q) REAL NF, NL, NV DIMESION SF(3), SL(3), SV(3) COMMON A1, B1, C1, C2, CP1, CP2, NV1, NV2 Vapor Pressure PV1 10.**(A1 – B1/(T – 273.15 C1)) PV2 10.**(A2 – B2/(T – 273.15 C2)) Product fractions XL1 (P – PV2)/(PV1 – PVS) XV1 XL1*PM/P Feed Variables NF SF(1) SF(2) XF1 SF(1)/NF Product flows NL NF*(XF1 – XV1)/(XL1 – XV1) NV NF – NL SL(1) XL1*NL SL(2) NL – SL(1) SY(1) XY1*NY SY(2) NV – SY(1) SL(3) T SV(3) T Energy Balance Q CP1*SF(1)*SF(1) CP2*SF(2) Q Q*(T – SF(3)) (NV1*XV1 HV2*(1 – XV1))*NV RETURN END
b1g XF b I g NF XLb I g NL XV b I g NV I 1,2 n 1 b2g Energy Balance: Q bT TF g ¦ CPb I g c XLb I g NL XV b I g NV h NL NV
10.9 a. Mass Balance: NF
N
I 1
N
bg
b g b3g
NV ¦ HV I XV 1 I 1
b g
where: XL N
Raoult’s law: P
1
¦ XLb I g
N 1
b g
XV N
I 1
¦ XV b I g
N 1 I 1
¦ XLb I g PV b I g N
I 1
bg
XV I P
b4 g XLb I g PV b I g
1
10.9 (cont’d) 10- 9
I
1,2, N 1
b5g
b g 10
d Ab I g Bb I g cCb I g T hi I 1,2, N 1 3 3b N 1g N 4 variables b NF , NL, NV , XF ( I ), XL( I ), XV ( I ), PV ( I ), TF , T , P, Qg
where: PV I
N mass balance 1 energy balances N equilibrium relations N Antoine equations N 3 degrees of freedom
b gr
m
Design Set TF , T , P , NF , XF I
Eliminate NL form (2) using (1) Eliminate XV(I) form (2) using (5) Solve (2) for XL(I) XL I XF I NF NF NV PV I P 1
bg
bg
c bg
d
hi b6g
Sum (6) ove I to Eliminate XL(I)
b g
f NV
N
b g d NF NV c PV b I g P 1hi
1 NF ¦ XF I I 1
0
Use Newton's Method to solve (7) for NV Calulate NL from (1) XL(I) from (2) XV(I) from (5) Q from (3) C ** CHAPTER 10 - - PROBLEM 9 DIMENSION SF(8), SL(8), SV(8) DIMENSION A(7), B(7), C(7), CP(7), HV(7) COMMON A, B, C, CP, NV DATA A/6.85221, 6.87776, 6.402040, 0., 0., 0., 0./ DATA B/1064.63, 1171.530, 1268.115, 0., 0., 0., 0./ DATA C/232.00, 224.366, 216.900, 0., 0., 0., 0./ DATA CP/0.188, 0.216, 0.213, 0., 0., 0., 0./ DATA NV/25.77, 28.85, 31.69, 0., 0., 0., 0./ FLOW 1.0 N*3 SF(1) 0.348*FLOW SF(2) 0.300*FLOW SF(3) 0.352*FLOW SF(4) 363 SL(4) 338 SV(4) 338 P*611 CALL FLASHN (SF, SL, SV, N, P, Q) WRITE (6, 900)' Liquid Stream', (SL(I), I 1, N 1) WRITE (6, 900)' Vapor Stream', (SV(I), I 1, N 1) 10.9 (cont’d) b.
10-10
b 7g
900
901 C
C
100 200
C 300 C
500
400 900 C
FORMAT (A15, F7.4,' mols/s n-pentane', /, * 15X, F7.4,' mols/s n-hexane', /, * 15X, F7.4,' mols/s n-hephane', /, * 15X, F7.2,' K') WRITE (6, 901) Q FORMAT ('Heat Required', F7.2, 'kW') END SUBROUTINE FLASHIN (SF, SL, SV, N, P, Q) REAL NF, NL, NV, NVP DIMENSION SF(8), SL(8), SV(8) DIMENSION XF(7), XL(7), XV(7), PV(7) DIMENSION A(7), B(7), C(7), CP(7), HV(7) COMMON A, B, C, CP, HV TOL 1,5 – 6 Feed Variables NF 0. DO 100 I 1, N NF NF SF(I) DO 200 I 1, N XF(I) SF(I)/NF TF SF (N 1) T SL (N 1) TC T – 273.15 Vapor Pressures DO 300 I 1, N PV(I) 10.**(A(I) – B(I)/(TC C(I))) Find NV -- Initial Guess NF/2 NVP NF/2 DO 400 ITER 1, 10 NV NVP F –1. FP 0. DO 500 I 1, N PPM1 PV(I)/P – 1. F F NF*XF(I)/(NF NV*PPM1) FP = FP – PPM1*XF(I)/(NF NV*PPM1)**2. NVP NV – F/FP IF (ABS((NVP – NV)/NVP).LT.TOL) GOTO 600 CONTINUE WRITE (6, 900) FORMAT ('FLASHN did not converge on NV') STOP Other Variables
10.9 (cont’d)
10- 11
600
700
800
NL NF – NVP DO 700 I 1, N XL(I) XF(I)*NF/(NF NV**(PV(I)/P – 1)) SL(I) XL(I)*NL XV(I) XL(I)*PV(I)/P SV(I) SF(I) – SL(I) Q1 0. Q2 0. DO 800 I 1, N Q1 Q1 CP(I)*SF(I) Q2 Q2 HV(I)*XV(I) Q Q1*(T – TF) Q2*NVP RETURN END
Program Output: Liquid Stream 0.0563 mols 0.1000 mols 0.2011 mols 338.00 K Vapor Stream 0.2944 mols 0.2000 mols 0.1509 mols 338.00 K Heat Required 13.01 kW
s n-pentane s n-hexane s n-heptane s n-pentane s n-hexane s n-heptane
10.10 a. Q (kW)
n v (mol / s) x v ( mol A(v) / mol) 1 x v ( mol B(g) / mol) T (K), P(mm Hg)
n F (mol / s) x F (mol A(v) / mol) 1 x F (mol B(g) / mol) TF (K), P(mm Hg) n l (mol A(l) / s)
10.10 (cont’d) 10 variables (n F , x F , TF , P , n v , x v , T , n l , p *,A , Q ) –2 material balances 10-12
–1 –1 –1 5
Antoine equation Raoult’s law energy balance degrees of freedom
b.
References: A(l), B(g) at 25oC Substance nin H in
n out
A(l)
—
—
n l
A(v)
n F x F
n v x v
B(g)
n F (1 x F )
H 1 H 2
n v (1 x v )
H out H 3
H 4 H
5
Given n F and x F (or n AF and n BF ), TF , P , y c (fractional condensation), Fractional condensation n l y c n F x F Mole balance n v A balance x v
n F n l (n F x F n l ) / n v
Raoult' s law p *A
xv P
B C A log 10 p *A 'H v C pv (TF 25), H 2 C pg (TF 25), H 3
Antoine' s equation T Enthalpies: H 1
H 4 Energy balance: Q
'H v C pv (T 25), H 5
C pl (T 25),
C pg (T 25)
¦ n out H out ¦ n in H in
c. nAF 0.704 nV 0.3664 Cpv 0.050
nBF 0.296 xV 0.1921 Cpg 0.030
nF 1.00 A 7.87863 H1 37.02
xF 0.704 B 1473.11 H2 1.05
TF 333 C 230 H3 0.2183
P 760 pA* 146.0 H4 35.41
yc 0.90 T 300.8 H5 0.0839
nL 0.6336 Cpl 0.078 Q –23.7
Greater fractional methanol condensation (yc) lower temperature (T). (yc = 0.10 T = 328oC.)
10.10 (cont’d)
10- 13
e.
C **CHAPTER 10 -- PROBLEM 10 DIMENSION SF(3), SV(3), SL(2) COMMON A, B, C, CPL, HV, CPV, CPG DATA A, B, C / 7.87863, 1473.11, 230.0/ DATA CPL, HV, CPV, CPG,/ 0.078, 35.27, 0.050, 0.029/ FLOW = 1.0 SF(1) = 0.704*FLOW SF(2) = FLOW – SF(1) YC = 0.90 P = 1. SF(3) = 333. CALL CNDNS (SF, SV, SL, P, YC, Q) WRITE (6, 900) SV(3) WRITE (6, 401) 'Vapor Stream', SV(1), SV(2) WRITE (6, 401) 'Liquid Stream', SL(1) WRITE (6, 902)Q 900 FORMAT ('Condenser Temperature', F7.2,' K') 901 FORMAT (A15, F7.3,' 'mols/s Methyl Alcohol', /, * 15X, F7.3, 'mols/s air') 902 FORMAT ('Heat Removal Rate', F7.2,' kW') END C SUBROUTINE CNDNS (SF, SV, SL, P, YC, Q) REAL NF, NL, NV DIMENSION SF(3), SV(3), SL(2) COMMON A, B, C, CPL, HV, CPV, CPG C Inlet Stream Variables NF = SF(1) + SF(2) TF = SF(3) XF = SF(1)/NF C Solve Equations NL = YC * XF * NF NV = NF - NL XV = (XF*NF - NL)/NV PV = P * XV * 760. T = B/(A - LOG(N)/LOG (10.)) - C T = T + 273.15 Q = ((CPV * XV + CPG * (1 - XY)) * NV + CPL * NL) * (T - TF) - NL * HV C Output Variables SL(1) = NL S2(2) = T SV(1) = XV*NV SV(2) = NV - SV(1) SV(3) = T RETURN END
10-14
10.11
K 1 A1 K 2 A2 K 3 A3 K m Am
0
a. Extent of reaction equations:
b g b g SPb I g SF b I g NU b I g [ I
[
[ SF IX X ] NU IX
1,2, N
Energy Balance: Reference states are molecular species at 298K.
b
g TP SPb N 1g ¦ HF b I g NU b I g
SF N 1
TF 'H r
N
I 1
b
g
N
bg bg
N
bg bg
Q [ 'H r TP 298 ¦ SP I CP I (TF 298) ¦ SF I CP I I 1
I 1
b. C 3 H 8 5O 2 o 3CO 2 4H 2 O Subscripts: 1 = C3H8, 2 = O2, 3 = N2, 4 = CO2, 5 = H2O 270 m 3 h
1 atm mol K 1000 liter h 3.348 mol C 3 H 8 s [=SF(1)] 3 m 3600 s 273K 0.08206 liter atm
b
3.348 mol C3 H 8 1.2 5 mol O2 sec mol C3 H 8
g
20.09 mol O2 s [= SF(2)] 7554 . mol N 2 s [= SF(3)]
X C3 H8
0.90 n C3 H8
010 . (3.348)
[
b g
b g
[ SF IX X ] NU IX
Nu nin (SF) X Xi nout (SP) Cp Tin Hin Tout Hout HF DHr Q
0.3348 mol C 3 H 8 / s in product gas [= SP(1)]
= –(3.348 mol/s)(0.90)/(–1) = 3.013 mol/s
1-C3H8 -1 3.348
2-O2 -5 20.09
3-N2 0 75.54
4-CO2 5-H2O(v) 3 4
0.3348 0.1431
5.024 0.033
75.54 0.0308
9.0396 0.0495
12.0528 0.0375
17.9
4.1
3.9
6.2
4.7
107.6 -103.8
24.8 0
23.2 0
37.2 -393.5
28.2 -241.83
0.90 3.01
423 1050
-2044 -4006
For the given conditions, Q
4006 kJ / s . As Tstack increases, more heat goes into the
stack gas so less is transferred out of the reactor: that is, Q becomes less negative.
10- 15
10.11 (cont’d) C **CHAPTER 10 PROBLEM 11 DIMENSION SF(8), SP(8), CP(7), HF(7) REAL NU(7) DATA NU/–1., –5, 0., 3., 4., 0., 0./ DATA CP/0.1431, 0.0330, 0.0308, 0.0495, 0.0375, 0., 0./ DATA HF/–103.8, 0., 0., –393.5, –241.83, 0., 0./ COMMON CP, HF SF(1) = 3.348 SF(2) = 20.09 SF(3) = 75.54 SF(4) = 0. SF(5) = 0. SF(6) = 423. SP(6) = 1050. IX = 1 X = 0.90 N=5 CALL REACTS (SF, SP, NU, N, X, IX, Q) WRITE (6, 900) (SP(I), I = 1, N + 1), Q 900 FORMAT ('Product Stream', F7.3, ' mols/s propane', /, * 15X, F7.3,' mols/s oxygen', /, * 15X, F7.3,' mols/s nitrogen', /, * * 15X, F7.3,' mols/s carbon dioxide', /, * 15X, F7.3,' mols/s water', /, * 15X, F7.2,'K', /, Heat required', F8.2, 'kW') END C SUBROUTINE REACTS (SF, SP, NU, N, X, IX, Q) DIMENSION SF(8), SP(8), CP(7), HF(7) REAL NU(7) COMMON CP, HF C Extent of Reaction EXT = –SF(IX)*X/NU(IX) C Solve Material Balances DO 100 I = 1, N 100 SP(I) = SF(I) + EXT = NU(I) C Heat of Reaction HR = 0 DO 200 I = 1, N 200 HR = HR + NF(I)*NU(I) C Product Enthalpy (ref * inlet) HP = 0. DO 300 I = 1, N
10-16
10.11 (cont’d) 300
HP = HP + SP(I)*CP(I) HP = HP + (SP(N + 1) – SF (N + 1)) Q = EXT * HR + HP RETURN END
10.12 a. Extent of reaction equations: [ SF IX X NU IX
b g b g SPb I g SF b I g NU b I g [
1, N
I
Energy Balance: Reference states are molecular species at feed stream temperature. Q
'H
N
bg bg
N
b g z CPb I gdT
0 0 [ ¦ NU I HF I ¦ SP I
['H r ¦ nout H out
i 1
I 1
T
Tfeed
CP(I) = ACP(I) + BCP(I)*T + CCP(I)*T2 + DCP(I)*T3
bg
f T
bg
N
bg
[ ¦ NU I * HF I AP (T Tfeed ) I 1
where: AP
BP 2
(T 2 Tfeed ) 2
CP DP 3 4
(T 3 Tfeed
(T 4 Tfeed ) ) 3 4
¦ SPb I g ACPb I g , N
0
and similarly for BP, CP, & DP
I 1
Use goalseek to solve f (T )
0 for T [= SP(N+1)]
b. 2CO O 2 o 2CO 2 Temporary basis: 2 mol CO fed
b
2 mol CO 1.25 1 mol O2 2 mol CO
g
125 . mol O2 4.70 mol N 2
Total moles fed = (2.00 + 1.25 + 4.70) mol = 7.95 mol Scale to given basis: (23.0
kmol 1h 10 3 mol )( ) )( h 3600 s 1 kmol 7.95 mol
mol CO fed s . SF (1) 1607 0.8036 SF (2) 1.004 mol O 2 fed s SF (3) 3.777 mol N 2 fed s
10- 17
10.12 (cont’d) Solution to Problem 10.12
Nu nin (SF) X Xi nout (SP) ACP BCP CCP DCP AP BP CP DP Tfeed DHF DHr T f(T)
1-CO -2 1.607
2-O2 -1 1.004
3-N2 0 3.777
4-CO2 2 0
0.45 0.36 0.88385 0.02895 4.11E-06 3.55E-09 -2.22E-12
0.642425 3.777 0.72315 0.0291 0.029 0.03611 1.16E-05 2.20E-06 4.23E-05 -6.08E-09 5.72E-09 -2.89E-08 1.31E-12 -2.87E-12 7.46E-12
0.1799 5.00E-05 -2.90E-11 -6.57E-12 650 -110.52
0
0
-393.5
-566 1560 -4.7E-08
The adiabatic reaction temperature is 1560 o C . As X increases, T increases. (The reaction is exothermic, so more reaction means more heat released.) d.
C ** CHAPTER 10 -- PROBLEM 12 DIMENSION SF(8), SP(B), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7) COMMON ACP, BCP, CCP, DCP, NF DATA NU / –2., –1., 0., 2., 0., 0., 0./ DATA ACP/ 28.95E-3, 29.10E-3, 29.00E-3, 36.11E-3, 0., 0., 0./ DATA BCP/ 0.4110E-5, 1.158E-5, 0.2199E-5, 4.233E-6, 0., 0., 0./ DATA CCP/ 0.3548E-B, –0.6076E-8, 0.5723E-8, –2.887E-8, 0., 0., 0./ DATA DCP/ –2.220 E-12, 1.311E-12, –2.871E-12, 7.464E-12, 0., 0., 0./ DATA HF / –110.52, 0., 0., –393.5, 0., 0., 0./ SF(1) = 1.607 SF(2) = 1.004 SF(3) = 3.777 SF(4) = 0. SF(5) = 650. IX = 1 X = 0.45 N=4 CALL REACTAD (SF, SP, NU, N, X, IX) WRITE (6, 900) (SP(I), I = 1, N + 1) 10-18
10.12 (cont’d) 900
C
C C 100 C 200 C
300 C
400 900
FORMAT ('Product Stream', F7.3, ' mols/s carbon monoxide', /, * 15X, F7.3, 'mols/s oxygen', /. * 15X, F7.3, 'mols/s nitrogen', /. * * 15X, F7.3, 'mols/s carbon dioxide', /, 15X, F7.2, 'C') END SUBROUTINE REACTAD (SF, SP, NU, N, X, IX) DIMENSION SF(8), SP(8), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7) COMMON ACP, BCP, CCP, DCP, NF TOL = 1.E-6 Extent of Reaction EXT = –SF(IX)*X/NU(IX) Solve Material Balances DO 100 I = 1, N SP(I) = SF(I) + EXT*NU(I) Heat of Reaction HR = 0 DO 200 I = 1, N HR = HR + HF(I) * NU(I) HR = HR * EXT Product Heat Capacity AP = 0. BP = 0. CP = 0. DP = 0. DO 300 I = 1, N AP = AP + SP(I)*ACP(I) BP = BP + BP(I)*BCP(I) CP = CP + SP(I)*CCP(I) DP = DP + SP(I)*DCP(I) Find T TIN = SF (N + 1) TP = TIN D0 400 ITER = 1, 10 T = TP F = HR FP = 0. F = F +T*(AP + T*(BP/2. + T*(CP/3. + T*DP/4.))) * –TIN*(AP + TIN*(BP/2. + TIN*(CP/3. + TIN*DP/4.))) FP = FP + AP + T *(BP + T*(CP + T*DP)) TP = T – F/FP IF(ABS((TP – T)/T).LT.TOL) GOTO 500 CONTINUE WRITE (6, 900) FORMAT ('REACTED did not converge') STOP 10- 19
10.12 (cont’d) 500
SP(N + 1) = T RETURN END
Program Output: 0.884 mol/s carbon monoxide 0.642 mol/s oxygen 3.777 mol/s nitrogen 0.723 mol/s carbon dioxide T = 1560.43 C
10-20
10.13 37.5 mol C2H4O a.
Separator 50 mol C2H4 50 mol O2
208.3333 mol C2H4 50 mol O2
Reactor
166.6667 18.75 37.5 8.333333 8.333333
mol C2H4 mol O2 mol C2H4O mol CO2 mol H2O
8.333333 18.75 8.333333 8.333333
mol C2H4 mol O2 mol CO2 mol H2O
Xsp = 0.2 Ysp = 0.9 158.3333 mol C2H4 (Ra)
158.3333 mol C2H4 (Rc)
Rc-Ra =
0
Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator. Rc is recalculated recycle rate. Use goalseek to find the value of Ra that drives (Rc-Ra) to zero. b.
Xsp 0.2 0.2 0.3 0.3
Ysp 0.72 1 0.75333 1
Yo 0.6 0.833 0.674 0.896
no 158.33 158.33 99.25 99.25
The second reaction consumes six times more oxygen per mole of ethylene consumed. The lower the single pass ethylene oxide yield, the more oxygen is consumed in the second reaction. At a certain yield for a specified ethylene conversion, all the oxygen in the feed is consumed. A yield lower than this value would be physically impossible.
10-21 21
10.14
C ** CHAPTER 10 -- PROBLEM 14 DIMENSION XA(3), XC(3) N=2 EPS = 0.001 KMAX = 20 IPR = 1 XA(1) = 2.0 XA(2) = 2.0 CALL CONVG (XA, XC, N, KMAX, EPS, IPR) END C SUBROUTINE FUNCGEN(N, XA, XC) DIMENSION XA(3), XC(3) XC(1) = 0.5*(3. – XA(2) + (XA(1) + XA(2))**0.5 XC(2) = 4. – 5./(XA(1) + XA(2)) RETURN END C SUBROUTINE CONVG (XA, XC, N, KMAX, EPS, IPR) DIMENSION XA(3), XC(3), XAH(3), XCM(3) K=1 CALL FUNCGEN (N, XA, XC) IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N) DO 100 I = 1, N XAM(I) = XA(I) XA(I) = XC(I) 100 XCM(I) = XC(I) 110 K=K+1 CALL FUNCGEN (N, XA, XC) IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N) D0 200 I = 1, N IF (ABS ((XA(I) - XC(I))/XC(I)).GE.EPS) GOTO 300 200 CONTINUE C Convergence RETURN 300 IF(K.EQ.KMAX) GOTO 500 DO 400 I = 1, N W = (XC(I) – XCM(I))/(XA(I) – XAM(I)) Q = W/(W – 1.) IF (Q.GT.0.5) Q = 0.5 IF (Q.LT.–5) Q = –5. XCM(I) = XC(I) XAM(I) = XA(I) 400 XA(I) = Q = XAM(I) + (1. – Q)*XCM(I) GOTO 110 500 WRITE (6, 900) 900 FORMAT (' CONVG did not converge') STOP END
10.14 (cont’d) 10-22
Elementary Principles of Chemical Processes
23
C
SUBROUTINE IPRNT (K, XA, XC, N) DIMENSION XA(3), XC(3) IF (K.EQ.1) WRITE (6, 400) IF (K.NE.1) WRITE (6, *) DO 100 I = 1, N 100 WRITE (6, 901) K, I, XA(I), XC(I) RETURN 900 FORMAT (' K Var Assumed Calculated') 901 FORMAT (I4, I4, 2E15.6) END Program Output: K Var Assumed Calculated 1 1 0.200000E + 01 0.150000E + 01 1 2 0.200000E + 01 0.275000E + 01 2 2
1 2
0.150000E + 01 0.115578E + 01 0.275000E + 01 0.282353E + 01
3
1
3 8 8
2
0.395135E + 00 0.482384E + 00 0.283152E + 01 0.245041E + 01
1 2
0.113575E + 01 0.113289E + 01 0.269023E + 01 0.269315E + 01
4 9
1 2
0.113199E + 01 0.113180E + 01 0.269186E + 01 0.269241E + 01
10- 23
CHAPTER ELEVEN 11.1 a.
The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is: Mp
xp
M
1 M p / M Therefore, the leakage rate of hydrogen peroxide is m b.
Balance on mass: Accumulation = input – output
E
dM dt t
0 m 1 m
0, M
M 0 (mass in tank when leakage begins)
Balance on H 2 O 2 : Accumulation = input – output – consumption
E dM p dt t 11.2 a.
0 x p0 m 1 m
0, M p
FG M IJ kM H MK p
p
M p0
Balance on H3PO4: Accumulation = input . g / ml. Density of H3PO4: U 1834 Molecular weight of H3PO4: M 98.00 g / mol . dn p Accumulation = (kmol / min) dt 20.0 L 1000 ml 1.834 g mol 1 kmol Input = min L ml 98.00 g 1000 mol
0.3743 kmol / min
E
dn p
0.3743
dt t
z
0, n p0
z
np
b.
t
dn p
7.5
xp
c.
150 u 0.05 7.5 kmol
. 015
0.3743 dt n p
7.5 0.3743t (kmol H 3PO 4 in tank )
0
np
np
n
n0 n p n p 0
7.5 0.3743t t 150 0.3743t
7.5 0.3743t 150 0.3743t
kmol H 3PO 4 kmol
. min 471
11-1
11.3 a.
w m
b
a bt t
g b
w 0, m
750 & t
g
b
w kg h 1000 m
w 5, m
g
bg
750 50t h
Balance on methanol: Accumulation = Input – Output M kg CH 3OH in tank dM dt
b
f m w m
E b
dM 450 50t kg h dt t 0, M 750 kg
z zb
M
t
dM
b.
750
g
g
450 50t dt
0
E
450t 25t 2
M 750
E
M
g
1200 kg h 750 50t kg h
750 450t 25t 2
Check the solution in two ways: (1) t 0, M 750 kg satisfies the initial condition; (2)
c.
dM dt M t
d.
dM dt
450 50t
0t
0
reproduces the mass balance.
450 50 9 h M
450 r
b450g 4b25gb750g t = –1.54 h, 19.54 h 2b25g 2
1 m3
t
2775 kg (maximum)
750 450t 25t 2
3.40 m 3 103 liter 0.792 kg
M
750 450(9 ) 25(9 )2
2693 450 r
1 liter
2693 kg (capacity of tank)
750 450t 25t 2
b450g 4b25gb750 2693g t 2b 25g 2
719 . h,10.81 h
Expressions for M(t) are:
R|750 + 450t - 25t b0 d t d 719 . and 10.81 d t d 19.54g (tank is filling or draining) M(t) = S2693 (tank is overflowing) ( 719 . d t d 10.81) ||T0 (tank is empty, draining (19.54 d t d 20.54) 2
as fast as methanol is fed to it)
11-2
11.3 (cont’d) 3000 2500 M(kg)
2000 1500 1000 500 0 0
5
10
15
20
t(h)
11.4 a.
10.0 ft 3
Air initially in tank: N 0
492q R
1 lb - mole
b g
532q R 359 ft 3 STP
0.0258 lb - mole
Air in tank after 15 s: Pf V
N f RT
P0V
N 0 RT
Nf
0.0258 lb - mole 114.7 psia
P0
14.7 psia 15 s
b
g
0.0117 lb - moles s ; t
z z
N
Integrate balance:
t
dN
0.0258
0.0117 lb - mole air s
n dt N
0, N
0.0258 lb - mole
b
0.0258 0.0117t lb - mole air
0
Check the solution in two ways: (1) t = 0, N = 0.0258 lb - mole satisfies the initial condition ( 2)
d.
0.2013 lb - mole
Balance on air in tank: Accumulation = input dN dt
c.
Pf
b0.2013 0.0258g lb - mole air
Rate of addition: n
b.
N0
dN dt
0.0117 lb - mole air / s reproduces the mass balance
t 120 s N
O 2 in tank
b
gb g
0.0258 0.0117 120
b g
0.21 143 .
143 . lb - moles air
0.30 lb - mole O 2
11-3
g
11.5 a.
Since the temperature and pressure of the gas are constant, a volume balance on the gas is equivalent to a mole balance (conversion factors cancel). Accumulation = Input – Output dV 540 m 3 1h Q w m 3 min dt h 60 min
e
t
0, V
b
3.00 u 103 m 3 t
z zb
V
t
b.
0 corresponds to 8:00 AM
g
e j
9.00 Q w dt V m 3
dV
3.00 u103
0
Let Q w i
tabulated value of Q w at t
z
240
LM MN
10 Q w dt # Q w1 Q w 25 4 3 i 0 2488 m 3 V
j
z t
3.00 u 103 9.00t Q w dt t in minutes 0
b g
10 i 1
24
24
i
O PQ
¦ Q w i 2 ¦ Q w i P
2, 4 ,
b g
g
3.00 u 103 9.00 240 2488
i 3, 5,
1, 2, , 25
b
g b
2672 m 3
c.
Measure the height of the float roof (proportional to volume). The feed rate decreased, or the withdrawal rate increased between data points, or the storage tank has a leak, or Simpson’s rule introduced an error.
d.
REAL VW(25), T, V, V0, H INTEGER I DATA V0, H/3.0E3, 10./ READ (5, *) (VW(I), I = 1, 25) V= V0 T=0. WRITE (6, 1) WRITE (6, 2) T, V DO 10 I = 2, 25 T = H * (I – 1) V = V + 9.00 * H – 0.5 * H * (VW(I – 1) + VW(I)) WRITE (6, 2) T, V 10 CONTINUE 1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)') 2 FORMAT (F8.2, 7X, F6.0) END $DATA 11.4 11.9 12.1 11.8 11.5 11.3 Results: TIME (MIN) VOLUME (CUBIC METERS) 0.00 3000. 10.00 2974. 20.00 2944. 230.00 2683. 240.00 2674. Vtrapezoid
2674 m 3 ; VSimpson
2672 m 3 ;
2674 2672 u 100% 2672
Simpson’s rule is more accurate.
11-4
g
10 114 . 9.8 4 124.6 2 113.4 3
0.07%
11.6 a.
b.
b
Q out L min
g
bg
kV L
0.200V Q out
Q out
V 300 Q out 60
20.0 L min Vs
100 L
Balance on water: Accumulation = input – output (L/min). (Balance volume directly since density is constant) dV 20.0 0.200V dt t 0, V 300
c.
dV dt
0
200 0.200Vs Vs
100 L
V
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is 20.0 0.200(300) 40.0. As t increases, V decreases. dV / dt 20.0 0.200V becomes less negative, approaches zero as t o f . The curve is therefore concave up.
t
z
V
d.
dV 20 . 0 0.200V 300
t
dt
0
FG H
1 20.0 0.200V ln 40.0 0.200
0.5 0.005V V
z
b g
. 100 101
b
IJ K
t
g
exp 0.200t V
b
b
100.0 200.0 exp 0.200t
g
101 L 1% from steady state
b
g
101 100 200 exp 0.200t t
b
ln 1 200
g
0.200
11-5
26.5 min
g
11.7 a.
A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points ( t D 2385 kg week ) and ( t 6 weeks, D 755 kg week ). ln D2 D1 t 2 t1
b
b
g
ln 755 2385
ln D1 bt1
ln a
E
b g b
e8.007
8.007 a
3000
Inventory balance: Accumulation = –output
b
z z I
t
dI
18 ,000
11.8 a.
gb g
ln 2385 0.230 1
dI 3000e 0.230t kg week dt t 0, I 18,000 kg
c.
0.230
6 1
3000e 0.230t
D
b.
ae bt
bt ln a D
ln D
t
g
3000e 0.230t dt I 18,000
0
f I
3000 0.230t e 0.230
t 0
I
4957 13,043e 0.230t
4957 kg
Total moles in room: N
Molar throughput rate: n
SO 2 balance ( t
1100 m 3
273 K
103 mol
b g
295 K 22.4 m 3 STP 700 m 3
273 K
103 mol
b g
295 K 22.4 m 3 STP
min
45,440 mol
28,920 mol min
0 is the instant after the SO 2 is released into the room):
b gb
N mol x mol SO 2 mol
g
mol SO 2 in room
Accumulation = –output.
b g
d Nx dt
nx
dx 45, 440 dt
N n 28,920
t
b.
0.6364 x
0, x
15 . mol SO 2 45,440 mol
330 . u 10 5 mol SO 2 mol
The plot of x vs. t begins at (t=0, x=3.30u10-5). When t=0, the slope (dx/dt) is . u 10 5 210 . u 10 5 . As t increases, x decreases. 0.6364 u 330 dx dt 0.6364 x becomes less negative, approaches zero as t o f . The curve is therefore concave up.
11-6
1 week,
x
11.8 (cont’d)
0 t
c.
Separate variables and integrate the balance equation:
z x
3.30 u10
5
dx x
z t
0.6364 dt ln
0
x 330 . u 10 5
0.6364t x
330 . u 10 5 e 0.6364t
Check the solution in two ways: (1) t = 0, x = 3.30 u 10-5 mol SO 2 / mol satisfies the initial condition; (2)
d.
e.
dx dt
0.6364 u 330 . u 10 5 e 0.6364 t 45,440 moles x mol SO 2
CSO2
1100 m
3
mol
i)
t
2 min CSO 2
ii)
x
10 6 t
382 . u 10 7
e
0.6364 x reproduces the mass balance.
1 m3 3
10 L
mol SO 2 liter
ln 10 6 3.30 u 10 5 0.6364
4131 . u 10 2 x 13632 . u 10 6 e 0.6364t mol SO 2 / L
j
55 . min
The room air composition may not be uniform, so the actual concentration of the SO2 in parts of the room may still be higher than the safe level. Also, “safe” is on the average; someone would be particularly sensitive to SO2.
11-7
11.9 a.
Balance on CO: Accumulation=-output N ( mol ) x ( mol CO / mol) = total moles of CO in the laboratory PQ p kmol Molar flow rate of entering and leaving gas: n ( ) h RT PQ p kmol kmol CO Rate at which CO leaves: n ( )x = x h kmol RT CO balance: Accumulation = -output PQ p dx P d ( Nx ) Q p x x dt NRT dt RT
FG H
FG H
E PV
dx dt
0, x
t
z
c.
dx x 0.01
Q p V
kmol CO kmol
0.01
Q p V
z
tr
dt t r
0
e
tr
d.
x
b g
V ln 100 x Q p
350 m 3 350 ln 100 u 35 u 10 6 700
V
IJ K
NRT
x
b.
IJ K
j
2.83 hrs
The room air composition may not be uniform, so the actual concentration of CO in parts of the room may still be higher than the safe level. Also, “safe” is on the average; someone could be particularly sensitive to CO. Precautionary steps: Purge the laboratory longer than the calculated purge time. Use a CO detector to measure the real concentration of CO in the laboratory and make sure it is lower than the safe level everywhere in the laboratory.
11.10 a.
Total mass balance: Accumulation = input – output
b
dM dt b.
m kg min m
g
0 ? M is a constant
200 kg
Sodium nitrate balance: Accumulation = - output x = mass fraction of NaNO 3
b g E
d xM dt dx dt
m x M t 0, x
b
kg min xm
g
m x 200 90 200 0.45
11-8
11.10 (cont’d) c.
0.45
m
50 kg / min m 100 kg / min 200 kg / min
x
m
0 t(min)
dx m x 0 , x decreases when t increases dt 200 dx becomes less negative until x reaches 0; dt Each curve is concave up and approaches x = 0 as t o f; dx becomes more negative x decreases faster. dt
increases m
z x
d.
dx x 0.45
z t
0
m x dt ln 0.45 M
FG H
m tx 200
0.45 exp
mt 200
IJ K
Check the solution: (1) t = 0, x = 0.45 satisfies the initial condition; m mt m dx 0.45 u x satisfies the mass balance. exp( ) (2) dt 200 200 200 0.45 0.4
m
0.35
50 kg / m in m
0.3
100 kg / m in m
x
0.25
200 kg / m in
0.2 0.15 0.1 0.05 0 0
5
10
15
20
t(min)
e.
100 kg min t m 90% x f
0.045 t
99% x f
0.0045 t
99.9% x f
d
i
2 ln x f 0.45
4.6 min 9.2 min
0.00045 t
138 . min
11-9
25
11.11 a.
e je
Mass of tracer in tank: V m 3 C kg m 3
j
Tracer balance: Accumulation = –output. If perfectly mixed, C out
b g
d VC
b
Q C kg min
dt
dC dt
V is constant
g
t
C
Q C V m0 0, C V
b.
z
c.
Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption
dC C
C
m0 V
z
C tank
FG H
Q C dt ln 0V m0 V t
e
of perfect mixing) through t
b
2 1
E e30 m
V
11.12 a.
3
min
FG IJ H K
Qt C V
m0 Q t exp V V
j e
0.223 u 10 3 & t
1, C
j
0.050 u 10 3 .
2, C
1495 min 1 .
j e1.495 min j 1
. m3 201
In tent at any time, P=14.7 psia, V=40.0 ft3, T=68qF=528qR N
b.
g
ln 0.050 0.223
Q V
IJ K
PV RT
m(liquid)
10.73
40.0 ft 3
14.7 psia ft 3 psia o
lb - mole R
528 o R
Molar throughout rate: 60 ft 3 492q R 16.0 psia 1 lb - mole n in n out n min 528q R 14.7 psia 359 ft 3 STP Moles of O2 in tank= N (lb - mole) u
FG lb - mole O IJ H lb - mole K
b g
0.1038 lb - mole
0.1695 lb - mole min
2
Balance on O2: Accumulation = input – output
b g
d Nx dt
c.
z
0.35n xn 01038 .
dx 0.21 0.35 x x
0.35 x 0.14
x
0.27 t
z
t
dx dt
163 . dt ln
0
e 1.63t x
LM FG N H
b
b
dx 163 . 0.35 x 01695 . 0.35 x dt t 0, x 0.21
b
b0.35 xg
g
0.35 0.21
g
1.63t
0.35 014 . e 1.63t
1 0.35 0.27 ln 163 . 0.35 0.21
IJ OP KQ
0.343 min (or 20.6 s)
11-10
g
11.13 a.
b gb
Mass of isotope at any time V liters C mg isotope liter Balance on isotope: Accumulation = –consumption
b g kCFGH Lmg sIJKV bLg
dC kC dt t 0, C C 0
Cancel V
d VC dt
g
Separate variables and integrate
z z C
C0
dC C
0
0
C
0.01C0
12
k
ln 2 2.6 hr
2.6 hr k
t1 2
0
kdt ln
0.5C0 t 1 2
C
b.
FG C IJ kt t lnbC C g k HC K lnb0.5g ln 2 t
t
t=-ln(C/C0)/k
k
0.267 hr 1 t
b g
ln 0.01 0.267
17.2 hr
11.14 A o products a.
Mole balance on A: Accumulation = –consumption
b g
d C AV t
0, C A
b.
z
CA
CA0
bV constant; cancelsg
kC AV
dt
C A0
dC A CA
z
t
kdt ln
0
FG C IJ HC K A
b g
kt C A
A0
C A0 exp kt
Plot C A (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies
b
assumption of first-order) through t kt ln C A 0
ln C A
b
ln 0.0185 0.0262
k
. 120.0 213
g
213 . , CA
g b
0.0262 & t
3.53 u 10 3 min 1 k
120.0, C A
. u 10 3 min 1 35
11.15 2 A o 2 B C a.
Mole balance on A: Accumulation = –consumption
b g
d C AV
kC 2AV
dt t
0, C A
z
bV constant; cancelsg
C A0
CA
dC A
CA0
C 2A
z
1 1 kdt 0 C A C A0 t
kt C A
11-11
LM 1 NC
A0
O kt P Q
1
g
0.0185 .
11.15 (cont’d) b.
CA
0.5C A 0
nA
0.5n A 0
b0.5n b0.5n
nB nC
A0 A0
total moles c.
1 1 0.5C A 0 C A 0
gb g mol A react.gb1 mol C 2 mol A react.g mol A react. 2 mol B 2 mol A react.
125 . n A 0 P1 2
125 .
n A 0 RT V
n A0 V
P0 t1 2 RT
0.5n A0 0.25n A0
125 . P0
Plot t 1 2 vs. 1 P0 on rectangular paper. Data fall on straight line (verifying 2nd order
d
decomposition) through t 1 2 RT k
Slope:
1 0.135 & t 1 2
b1015 Kgb0.08206 L atm mol Kg
1 0.683
0.582 L mol s
143.2 s atm
F GH
FG IJ H K
i
209, 1 P0
143.2 s atm
I JK
t 1 2 P0 RT E exp ln k 0 P0 RT RT
t1 2
i d
1060, 1 P0
1060 209 1 0135 . 1 0.683 k
d.
1 ; but C A0 kC A 0
kt 1 2 t 1 2
ln
1 E 1 k0 R T
Plot t 1 2 P0 RT (log scale) vs. 1 T (rect. scale) on semilog paper.
bg
t 1 2 s , P0
1 atm, R
bg
0.08206 L atm / (mol K), T K
d
Data fall on straight line through t 1 2 P0 RT
dt
1 2 P0
e.
0.6383, 1 T
b
g
ln 0.6383 74.0
E R ln
RT
1 k0
T C A0
b
g
ln 0.6383
980 K k
i
1 900 &
i
1 1050
29,940 K
1 1050 1 900
74.0, 1 T
29,940 1050
FG H
k 0 exp
b
R=8.314 J/ (mol ∙K)
28.96 k 0 E RT
IJ K
E
2.49 u 10 5 J mol
3.79 u 1012 L (mol s)
0.204 L (mol s)
g
0.70 120 . atm
b0.08206 L atm mol Kgb980 Kg
1045 . u 10 2 mol L
90% conversion CA
0.10C A0 t
LM N
1 1 1 k C A C A0 4222 s
OP Q
LM N
1 1 1 3 0.204 1.045 u 10 u 10 2 1045 .
70.4 min
11-12
OP Q
RT kP0
11.16 A o B a.
Mole balance on A: Accumulation = –consumption(V constant) dC A dt t
z
k 1C A 1 k2CA
0, C A
C A0
1 k2C A dC A CA0 k1C A
b.
CA
b
z
b
C k 1 ln A 2 C A C A 0 k 1 C A0 k 1
t
dt
0
g
b
Plot t C A C A0 vs. ln C A / C A 0 y
t C A0 C A
b
g bC
A0
g
b
g
k2 C 1 C A0 C A ln A k1 k 1 C A0
t t
g
C A on rectangular paper:
x
k 1 ln C A C A 0 2 k 1 C A0 C A k1
g ;
b
g
slope
1
FG H
intercept y1
IJ FG K H
x1
y2
IJ K
x2
Data fall on straight line through 116.28, 0.2111 & 130.01, 0.2496
130.01 116.28 0.2496 0.2111
1 k1
b
2.80 u 10 3 L (mol s)
g
0115 . L mol
g
b
k2 k1
356.62 k 1
130.01 356.62 0.2496
4100 . k2
11.17 CO Cl 2 COCl 2 a.
3.00 L
273 K
. mol gas b g 012035 0.60b0.12035 molg 3.00 L 0.02407 mol L CO U |Vinitial concentrations 0.40b012035 . molg 3.00 L 0.01605 mol L Cl | W 0.02407 C bt g U | Since 1 mol COCl formed requires 1 mol of each reactant 0.01605 C bt g V| W
303.8 K 22.4 L STP
bC g dC i C bt g C bt g
CO i Cl 2
CO Cl 2
b.
i
2
p
2
p
Mole balance on Phosgene: Accumulation = generation
d i
d VC p
V=3.00 L
8.75CCO CCl 2
d1 58.6C
dt
c.
1 mol
Cl 2
34.3C p
i
t
z
0
24.3C i . d1941 dC d0.02407 C id0.01605 C i
g
p
p
d1.941 24.3C i
0.75 0.01605 2
0.01204
dt
p
t
b
p
11-13
p
id
2.92 0.02407 C p 0.01605 C p
2
Cl 2 limiting; 75% conversion C p 1 2.92
d
dC p
0, C p
0
0.01204 mol L
2
i
11.17 (cont’d) d.
11.18 a.
REAL F(51), SUM1, SUM2, SIMP INTEGER I, J, NPD(3), N, NM1, NM2 DATA NPD/5, 21, 51/ FN(C) = (1.441 – 24.3 * C) ** 2/(0.02407 – C)/(0.01605 – C) DO 10 I = 1, 3 N = NPD(I) NM1 = N – 1 NM2 = N – 2 DO 20 J = 1, N C = 0.01204 * FLOAT(J – 1)/FLOAT(NM1) F(J) = FN(C) 20 CONTINUE SUM1 = 0. DO 30 J = 2, NM1, 2 SUM = SUM1 + F(S) 30 CONTINUE SUM2 = 0. DO 40 J = 3, NM2, 2 SUM2 = SUM2 + F(J) 40 CONTINUE SIMP = 0.01204/FLOAT(NM1)/3.0 * (F(1) + F(N) + 4.0 * SUM1 + 2.0 * SUM2) T = SIMP/2.92 WRITE (6, 1) N, T 10 CONTINUE 1 FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES') END RESULTS 5 POINTS — 91.0 MINUTES 21 POINTS — 90.4 MINUTES 51 POINTS — 90.4 MINUTES t 90.4 minutes
e j e
Moles of CO 2 in liquid phase at any time V cm 3 C A mols cm 3
j
Balance on CO 2 in liquid phase: Accumulation = input
b g
d VC A dt
e
kS C *A C A
FjG molsIJ dCdt kSV eC H s K t 0, C 0 A
0
C *A
ln
j
A
dC A
CA
CA
yV
y A P is constant, C *A
Separate variables and integrate. Since p A
z
* A
CA
C *A C A C *A
1 C A C *A
z
t
0
kS dt ln C *A C A V
e
C kS expb g t 1 *A V CA
j
CA CA 0
kS t V
e kSt V C A
11-14
p A H is also a constant.
e
C *A 1 e kSt V
j
11.18 (cont’d) b.
t
LM MN
C V ln 1 *A kS CA
V
OP PQ
5000 cm 3 , k
5L
78.5 cm 2 , C A
0.020 cm s , S
0.62 u 10 3 mol / cm 3
a0.30fa20 atmf d9230 atm cm moli 0.65 u 10 mol cm e5000 cm j lnF1 0.62 u 10 I 9800 s 2.7 hr b0.02 cm sge78.5 cm j GH 0.65 u 10 JK
C *A
3
t
3
3
yAP H
3
3
3
2
(We assume, in the absence of more information, that the gas-liquid interfacial surface area equals the cross sectional area of the tank. If the liquid is well agitated, S may in fact be much greater than this value, leading to a significantly lower t than that to be calculated) 11.19 A o B a.
Total Mass Balance: Accumulation input dM d ( UV ) Uv dt dt dV dt t
E
v 0, V
0
A Balance: Accumulation input – consumption dN A C A 0 v ( kC A )V C =N /V A A dt
dN A dt t
b.
c.
dN A dt
Steady State:
z z z V
t
0
0
dV
NA
0
V vdt
dN A C A 0 v kN A
C A0 v k
vt
z
t
dt
0
FG H
IJ t C v kN C v K C v 1 expb kt g t o f N k
C v kN A 1 ln A 0 k C A 0 v A0
NA
CA
0 NA
NA V
A0
e kt
A
A0
A
C A 0 [1 exp( kt )] kt
11-15
C A0 v k
C Ao v kN A
0, N A
0
11.19 (cont’d) When the feed rate of A equals the rate at which A reacts, NA reaches a steady value. NA would never reach the steady value in a real reactor. The reasons are: t o f , V o f. (1) In our calculation, V = vt But in a real reactor, the volume is limited by the reactor volume; (2) The steady value can only be reached at t o f. In a real reactor, the reaction time is finite. d.
C A0 [1 exp( kt )] t of kt
lim C A
C A0 t of kt
lim
t of
lim
0
From part c, t o f, N A o a finite number, V o f C A 11.20 a.
dT dt M
MC v
Q W (3.00 L)(1.00 kg / L) = 3.00 kg
Cv
Cp
W dT dt
NA o0 V
(0.0754 kJ / mol o C)(1 mol / 0.018 kg) = 4.184 kJ / kg o C
0
0.0797Q (kJ / s) t = 0, T = 18 o C
b. c.
11.21 a.
z z 100o C
240 s
18 C
0
dT o
0.0797Q dt Q
100 18 240 u 0.0797
4.287
kJ s
4.29 kW
Stove output is much greater. Only a small fraction of energy goes to heat the water. Some energy heats the kettle. Some energy is lost to the surroundings (air). Energy balance: MC v
dT dt
Q W
M
20.0 kg o
o
C v | C p ( 0.0754 kJ / mol C)(1 mol / 0.0180 kg) = 4.184 kJ / (kg C) Q 0.97 (2.50) 2.425 kJ s W 0
a f
dT dt
b g
0.0290 q C s , t
0, T
25q C
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
z z T
b.
t
dT
o
25 C
c.
T
0.0290dt T
bg
25q C 0.0290t s
0
100q C t
b100 25g 0.0290
2585 s 43.1 min
No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal boiling point).
11-16
11.22 a. Energy balance on the bar MCv
dTb dt
b
Q W
UA Tb Tw
B
g
Table B.1
e60 cm je7.7 g cm j 3
M Cv
0.46 kJ (kg q C), Tw 0.050 J (min cm q C)
A
2 2 3 2 10 3 10 cm
25q C
2
a fa f a fa f a fa f
b
gb
2
0.02635 Tb 25 q C min 0, Tb
t dTb dt
462 g
U
dTb dt
b.
3
g
95q C
d
i
0.02635 Tbf 25 Tbf
0
2
112 cm
25q C
95 85 75
Tb( oC)
65 55 45 35 25 15 5 0 t
z
z
Tb
c.
t
dTb T 25 95 b
0.02635dt
0
FG T 25IJ 0.02635t H 95 25K T bt g 25 70 expb 0.02635t g ln
b
b
Check the solution in three ways: (1) t = 0, Tb
25 70 95o C satisfies the initial condition;
dTb 70 u 0.02635e 0.02635t 0.02635(Tb 25) reproduces the mass balance; dt (3) t o f, Tb 25o C confirms the steady state condition. (2)
Tb
30q C t
100 min
11-17
11.23 12.0 kg/min 25oC
12.0 kg/min T (oC) Q (kJ/min) = UA (Tsteam-T)
a.
dT dt
Energy Balance: MCv
b
g
b
p 25 T UA Tsteam T mC
g
M 760 kg m 12.0 kg min
dT / dt Cv | C p UA
0, T
25o C
2.30 kJ (min q C)
. kJ (min q C) 115
a
f
Tsteam sat' d; 7.5bars
b.
150 . 0.0224T ( o C min), t
Steady State:
dT dt
167.8q C
0 150 . 0.0224Ts Ts
67q C
T( oC)
67
25 0 t
z
Tf
c.
dT 150 . 0.0224T 25
U changed. Let x dT dt
z
FG H
IJ K
1 150 . 0.0224T T ln 0.0224 0.94
150 . 0.94 exp( 0.0224t ) 0.0224
49.8q C (UA) new . The differential equation becomes:
0.3947 0.096 x (0.01579 5.721x )T
55
25
dt t
0
40 min. T
t d.
z t
0.3947 0.096 x (0.01579 5.721 u 104 x )T x
'U U initial
1 0.01579 5.721 u 104 14.27 kJ / (mino C)
'(UA) (UA)initial
z
40
dT
dt
0
L 0.3947 0.096x e0.01579 5.721 u 10 xj u 55 OP ln M x M 0.3947 0.096 x e0.01579 5.721 u 10 x j u 25 P PQ NM
14.27 115 . u 100% 115 .
4
4
241% .
11-18
40
11.24 a.
Energy balance: MCv
dT Q W dt W 0, Cv M
. J g q C 177 350 g, Q 40.2W 40.2 J s
b gU|V T |W T
dT 0.0649 q C s dt t 0, T 20q C b.
bg
20 0.0649t s 40q C t
. min 308 s 51
The benzene temperature will continue to rise until it reaches Tb 801 . q C ; thereafter the heat input will serve to vaporize benzene isothermally. 801 . 20 Time to reach Tb neglect evaporation : t 926 s 0.0649 Time remaining: 40 minutes 60 s min 926 s 1474 s 30.765 kJ mol 1 mol 78.11 g 1000 J kJ 393 J g Evaporation: 'H v
b
b
g
b
gb
g
gb
g
g s . b40.2 J sg / b393 J gg 0102 . g sgb1474 sg 200 g Benzene remaining 350 g b0102 Evaporation rate
c.
11.25 a.
1. Used a dirty flask. Chemicals remaining in the flask could react with benzene. Use a clean flask. 2. Put an open flask on the burner. Benzene vaporizes toxicity, fire hazard. Use a covered container or work under a hood. 3. Left the burner unattended. 4. Looked down into the flask with the boiling chemicals. Damage eyes. Wear goggles. 5. Rubbed his eyes with his hand. Wash with water. 6. Picked up flask with bare hands. Use lab gloves. 7. Put hot flask on partner’s homework. Fire hazard. Moles of air in room: n
60 m 3 273 K
1 kg - mole
2.58 kg - moles 283 K 22.4 m 3 STP dT Q W Energy balance on room air: nCv dt Q m s 'H v H 2 O, 3bars, sat' d 30.0 T T0 W 0 dT s 'H v 30.0 T T0 m nCv dt N 2.58 kg - moles
b g
b
g
b
b
dT dt t
g
g
Cv 20.8 kJ (kg - moleq C) 'H v 2163 kJ kg from Table B.6 T0
b
g
0q C
b
40.3m s 0.559 T q C hr 0, T
g
10q C
(Note: a real process of this type would involve air escaping from the room and a constant pressure being maintained. We simplify the analysis by assuming n is constant.)
11-19
11.25 (cont’d) 0 40.3m s 0.559T
b. At steady-state, dT dt 24q C m s
T c.
0.333 kg hr
Separate variables and integrate the balance equation:
z
Tf
10
z
dT 40.3m s 0.559T
t
0
dt
z
0.333
m s
E
t
Integral energy balance 'U
Q
bt
b.
LM MN
b60 20gq C
4.00 u 104 kJ
20 min
1 kW
60 s
dT dt
Q
. kW 333
1 kJ s
dT dt
250 kg
M
Cv 4 .00 kJ kgq C
z z
T
Integrate:
t
dT
20o C
t
z t
0.001 Q dT T
20 o C Qdt
0
0
b
g
b
g
2 34 37 41 47 54 62 70 80 90 100
b g
jb
e
T 600 s 20o C + 0.001 oC / kJ 34830 kJ
c.
Past 600 s, Q
100
b
g
10 kW t 600 s 60 s
LM M T 20 0.001 Qdt 20 0.001M Qdt MM N 0.001 F t 600 I T 54.8 t bs g G 6 H6 2 JK
z 0
2
85q C t
2
54.8q C
OP t P dt 6 P PP Q
600
t
0
600
34830
34830 kJ
t 6
z z
t
T
g
b
g
12000 T 24.8
850 s 14 min, 10 s explosion at 10:14 10 s
11-20
bg
0.001Q t 0, T
Evaluate the integral by Simpson's Rule (Appendix A.3) 600 s 30 33 4 33 35 39 44 50 58 66 75 85 95 Qdt 3 0
z
4.8 hr
20 min
4.00 u 104 kJ 1 min
Differential energy balance: MCv
b g OP b g PQ
13.4 0.559 23 1 ln 0.559 13.4 0.559 10
kgq C
Required power input: Q
t
g
0 to t
250 kg 4.00 kJ
MCv 'T
dT
23
10 13.4 0.559T
T f 23q C
11.26 a.
0.559T 40.3
0 m s
20q C
11.27 a. Total Mass Balance: Accumulation=Input– Output
E dM tot dt
d( UV) dt
i m o m
U=constant
8.00U 4.00U
dV 4.00 L / s dt t 0, V0 400 L
KCl Balance: Accumulation=Input-Output
dM KCl dt
i , KCl m o, KCl m
dV dC V C dt dt b.
8 4C
d( CV ) dt
dV dt
4
100 . u 8.00 4.00C dC 8 8C dt V t 0, C 0 0 g / L
(i)The plot of V vs. t begins at (t=0, V=400). The slope (=dV/dt) is 4 (a positive constant). V increases linearly with increasing t until V reaches 2000. Then the tank begins to overflow and V stays constant at 2000.
V
2000
400 0 t
(ii) The plot of C vs. t begins at (t=0, C=0). When t=0, the slope (=dC/dt) is (8-0)/400=0.02. As t increases, C increases and V increases (or stays constant) dC/dt=(8-8C)/V becomes less positive, approaches zero as to f. The curve is therefore concave down.
C
1
0 t
c.
dV dt
4
z
V
dV
400
z
t
4 dt V 0
400 4t
11-21
11.27 (cont’d) 8 8C 1 C dC V 400 4 t V dt 50 0.5t C dC t dt C t ln(1 C ) 0 2 ln(50 0.5t ) 0 0 1 C 0 50 + 0.5t 50 0.5t ln(1- C)-1 2 ln ln(1 0.01t )2 50 1 1 (1 0.01t )2 C 1 1- C (1 0.01t )2
dC dt
z
z
When the tank overflows, V
400 4t
C = 1-
2000 t 1
b1 + 0.01 u 400g
2
400 s 0.96 g / L
11.28 a. Salt Balance on the 1st tank: Accumulation=-Output
E
d(CS1V1 ) dt
CS1v
dCS1 dt
CS1
v V1
0.08CS1
CS1 ( 0 ) 1500 500
3g/L
Salt Balance on the 2nd tank: Accumulation=Input-Output
E
d(CS2V2 ) dt
CS 1v CS 2 v
dCS2 dt CS 2 ( 0 )
( CS 1 CS 2 )
v V2
0.08( CS1 CS 2 )
v V3
0.04( CS 2 CS 3 )
0 g/L
Salt Balance on the 3rd tank: Accumulation=Input-Output
E
d(CS3V3 ) dt
CS 2 v CS 3v
dCS3 dt CS 3 ( 0 )
( CS 2 CS 3 ) 0 g/L
b.
CS1, CS2, CS3
3
CS1
CS2 CS3
0 t
11-22
11.28 (cont’d) The plot of CS1 vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is 0.08 u 3 0.24 . As t increases, CS1 decreases dCS1/dt=-0.08CS1 becomes less negative, approaches zero as to f. The curve is therefore concave up. The plot of CS2 vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 0.08( 3 0) 0.24 . As t increases, CS2 increases, CS1 decreases (CS2 < CS1) dCS2/dt =0.08(CS1-CS2) becomes less positive until dCS2/dt changes to negative (CS2 > CS1). Then CS2 decreases with increasing t as well as CS1. Finally dCS2/dt approaches zero as tof. Therefore, CS2 increases until it reaches a maximum value, then it decreases. The plot of CS3 vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 0.04( 0 0) 0 . As t increases, CS2 increases (CS3 < CS2) dCS3/dt =0.04(CS2-CS3) becomes positive CS2 increases with increasing t until dCS3/dt changes to negative (CS3 > CS1). Finally dCS3/dt approaches zero as tof. Therefore, CS3 increases until it reaches a maximum value then it decreases. c. 3
CS1, CS2, CS3 (g/L)
2.5 2 CS1 1.5 CS2
1
CS3
0.5 0 0
20
40
60
80
100
120
140
160
11.29 a. (i) Rate of generation of B in the 1st reaction: rB1
2r1
0.2C A
t (s)
(ii) Rate of consumption of B in the 2nd reaction: rB 2 b. Mole Balance on A: Accumulation=-Consumption
E
d ( C AV ) dt
01 . C AV
dC A 01 . CA dt t 0, C A0 100 . mol / L
Mole Balance on B: Accumulation= Generation-Consumption
E
d ( CBV ) dt
0.2C AV 0.2CB2V
dCB 0.2C A 0.2CB2 dt t 0, CB 0 0 mol / L
11-23
r2
0.2C B2
11.29 (cont’d) c. 2
CA, CB, CC
CC
1 CB CA 0 t
. u 1 01 . . The plot of CA vs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is 01 As t increases, CA decreases dCA/dt=-0.1CA becomes less negative, approaches zero as tof. CAo0 as tof. The curve is therefore concave up. The plot of CB vs. t begins at (t=0, CB=0). When t=0, the slope (=dCB/dt) is 0.2(1 0) As t increases, CB increases, CA decreases
( CB2
< CA) dCB/dt
=0.2(CA- CB2
0.2 .
) becomes less positive
( CB2
until dCB/dt changes to negative > CA). Then CB decreases with increasing t as well as CA. Finally dCB/dt approaches zero as tof. Therefore, CB increases first until it reaches a maximum value, then it decreases. CBo0 as tof. The plot of CC vs. t begins at (t=0, CC=0). When t=0, the slope (=dCC/dt) is 0.2( 0) increases, CB increases dCc/dt
=0.2 CB2
0 . As t
becomes positive also increases with increasing t
CC increases faster until CB decreases with increasing t dCc/dt =0.2 CB2 becomes less positive, approaches zero as tof so CC increases more slowly. Finally CCo2 as tof. The curve is therefore S-shaped.
CA, CB, CC (mol/L)
d. 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0
CC
CB CA
0
10
20
30 t (s)
11-24
40
50
11.30 a. When x =1, y =1 . y
ax x 1, y 1 a 1 a 1 b 1 b xb
b.Raoult’s Law:
pC5 H12
p *C5 H12 ( 46 o C ) 10
Antoine Equation: o
xp *C5 H12 ( 46 C )
y
xp *C5 H12 ( 46 o C ) y
yP
0.7 u 1053 760
P
( 6.85221
1064.63 ) 46 232.00
xp *C5 H12 ( 46o C ) P 1053 mm Hg
0.970
R| y ax x=0.70, y=0.970 0.970 0.70a (1) R|a S 0.70 b S| x b TFrom part (a), a = 1+ b(2) T|b
. 1078 0.078
c. Mole Balance on Residual Liquid: Accumulation=-Output
E
dN L nV dt t 0, N L 100 mol Balance on Pentane: Accumulation=-Output
E
d( N L x) dt
nV y x
ax dx dN L NL nV xb dt dt dN L / dt nV
E
dx dt t
FG H
nV ax x NL x b
IJ K
0, x = 0.70
d.Energy Balance: Consumption=Input
E
nV 'H vap
Q t
From part (c),
'H vap 27.0 kJ /mol
0, N L
dN L nV dt nV Q 27.0 NL Qt 100 27.0
100 mol
NL
nV
b
Q 27.0 kJ / mol
100 nV t 100
Substitute this expression into the equation for dx/dt from part (c):
11-25
g
Qt 27.0
11.30 (cont’d)
dx dt
FG H
nV ax x NL x b
IJ K
FG H
Q 27.0 ax x xb Qt 100 27.0
IJ K
x(0) = 0.70 e. 1 0.9 0.8
y (Q=1.5 kJ/s)
0.7 x, y
0.6
x (Q=1.5 kJ/s)
0.5 0.4
x (Q=3 kJ/s)
0.3
y (Q=3 kJ/s)
0.2 0.1 0 0
200
400
600
800
1000 1200 1400 1600 1800
t(s)
f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The higher the heating rate, the faster x and y decrease.
11-26
CHAPTER THIRTEEN Problem
13.1
Methanol
Production
Rate
430,000 metric tons/year 51,114 kg/h 1,597 kmol/h Process stoichiometry: CH4 + H20 ---> CH30H + H2 So that the required feed rates (with given assumptions) are CH4 Feed Rate = Steam Feed Rate =
Problem
1,597 kmol/h 1,597 kmol/h
cubic
meters/min
13.2
1 P = r.h. = (pH20/p*H20 p*H20 = 0.0424 pH20 = 0.02968 yH20 0.029292 Basis: 1 mole of component moles N2 0.79 0.21 02 0.030176 Water Total 1.03 component N2 02 Total
Basis:
atm @ 30C) x 100% = bar bar
70%
dry air (79 mole% N2, 21 mole% 02) Mw mass mole frac 28 22.12 32 6.72 0.0293 18 0.54 29.38 1.0000
moles Mw 0.79 0.21 1.00
mass 22.12 6.72 . 28.84 Difference in avg molecular weight is due to presence of water; the difference is slight. 1 km01 of CH4 burned
flow rate of air/km01 Problem
596 standard 28,75lkg/h
28 32
nat gas burned
13.3
Composition of effluent gas from burners mole frac mass frac Component km01 kg 3.2 0.0103 02 0.100 0.0088 N2 7.900 0.6990 221.2 0.7139 44.0 0.1420 co2 1.000 0.0885 0.1337 H20 2.302 0.2037 41.4 1.0000 309.8 1.0000 total 11.302 p*H20 @ 150C = 4.74 bar < pH20 Therefore, there is no condensation in cooling the exhaust gases to 15OC, which means the effluent gas and stack gas have the same composition. Volumetric
flow rate effluent gas stack gas density of air = density of stack gas = specific gravity =
1,143 m3/kmol CH4 burned 392 m3/kmol CH4 burned 1.1471 kg/m3 0.7899 kg/m3 1 0.6886 relative to ambient air
13-1
I
Problem
13.4
Reformer Temperature Reformer Temperature Reformer Pressure Equilibrium Temperature
855 1128 15.8 1128
C K atm K
1.6 Mpa
CH4 + Hz0 ---> CO + 3Hz Production
rates CH4 H20 CO CO2 H2
feed rate CH4 x (1 - fractional conversion) feed rate H20 - fractional conversion x feed rate CH4 fractional conversion x feed rate CH4 feed rate CO2 feed rate + 3 x feed rate CH4 x fractional conversion
(a) methane:steam of 3:l Stoichiometric
Feed I (kmol/h)( 1600 4800 0 0 0 6400
Table: CH4 H20 co co2 H2 Total
(kmol/h) 170 3,370 1,430 0 4,291 9,261
574.3668 KP~ Ratio1 574.366846 fractional conversion of CH4* =
=
Product MolFrac (kg/h) 0.0183 2,716 0.3639 60,655 0.1544 40,047 0.0000 0 0.4633 8,582 1.0000 112,000
MassFrac 0.0242 0.5416 0.3576 0.0000 0.0766 1.0000
lO"(-(11,769/T(K))+l3.1927)
( (Yco x YH2A3) / (YCH4 x YH20) 6.8939 -3.57E-07 Converge* = ((Kpl/Ratiol)-1)lOO * the Goal Seek tool is used to adjust the fractional conversion until Converge is close to zero Methane Conversion =
) p2
1
Product Flow Rate =-I (b) methane:steam of 1:l Stoichiometric Table: CH4 H20 co co2 H2 Total
fractional
Feed I (kmol/h)l 1600 1600 0 0 0 3200
574.3668 Kpl Ratio1 1653.61852 conversion of CH4 =
(kmol/h) 471 471 1,129 0 3,387 5,458 =
lO"(-(11,769/T(K))+l3.1927)
=
( kc0 x YH2A3)
/ (YCH4
x YH20)
Jp
2
0.7055 -65.266061Converge = ((Kpl/Ratiol)-1)lOO
Methane Conversion --I-[ Product Flow Rate ='
Product MolFrac (kg/h) MassFrac 0.0863 7,538 0.1386 0.0863 8,480 0.1559 0.5810 0.2068 31,608 0.0000 0 0.0000 0.6205 6,773 0.1245 54.400 1.0000 1.0000
5,458 kmol/h 54,400 kg/h 13-2
Problem 13.4 (cont'd) methane:steam
of 2:l
Stoichiometric
Table: CH4 H20 co co2 H2 Total
Feed I (kmol/h)l, 1600 3200 0 0 0 4800
Product (kmol/h) MolFrac MassFrac (kg/h) 0.0476 248 0.0330 3,960 1,848 0.2462 33,256 0.3997 0.1802 37,869 0.4552 1,352 0 0.0000 0 .o.oooo 4,057 0.5406 8,115 0.0975 7,505 1.0000 83,200 1.0000
574.3668 = lO"(-(11,769/T(K))+l3.1927) Kpl Ratio1 874.51201 = ( (ycO x YH2^3) / (YCH4 x YH20) )p2 fractional conversion of CH4 = 0.84529344 -34.321446 Converge = ((Kpl/Ratiol)-I)100 Methane
Conversion
=1
Product Flow Rate =
methane:steam
of
Stoichiometric
7,505 kmol/h 83,200 kg/h
4:l
Table: CH4 H20 co co2 H2 Total
Feed I (kmol/h)l 1600 6400 0 0 0 8000
Product MassFrac (kmol/h) MolFrac (kg/h) 0.0147 130 0.0118 2,074 4,930 0.4506 88,733 0.6302 1,470 0.1344 41,171 0.2924 0.0000 0 0.0000 0 0.4032 8,822 0.0627 4,411 10,941 1.0000 140,800 1.0000
= 574.3668 lO"(-(11,769/T(K))+l3.1927) Kpl = Ratio1 411.494126 ( (Yco x YH2^3) / (YCH4 x YH~o) jp2 fractional conversion of CH4 = 0.91899524 39.5808125 Converge = ((Kpl/Ratiol)-I)100 Methane
Conversion = 1
Product Flow Rate =I[
Summary
Moles Steam:Mole CH4 1 2 3 4
Methane H2 Conversion Produced 6,773 70.6% 84.5% 8,115 89.4% 8,582 91.9% 8,822
13-3
H2:CO 3.0000 3.0000 3.0000 3.0000
Problem
13.5
Reformer Temperature Reformer Temperature Reformer Pressure Equilibrium Temperature
Production
1
855 1128 15.79 1128
C K atm K
1.6 Mpa
CH, + Hz0 ---> CO + 3H, CO f Hz0 ---> COz f H, rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 - production rate of CO2 co fractional conversion x feed rate CH4 - production rate of CO2 co2 feed rate CO2 + production rate of CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion + production rate of CO2
methane:steam of 3:l Stoichiometric Feed I (kmol/h)[ CH4 1600 H20 4800 co 0 co2 0 H2 0 Total 6400
(kmol/h) 176 3,156 1,205 219 4,492 9,249
Product MolFrac (kg/h) 0.0190 2,811 0.3413 56,814 0.1303 33,740 0.0237 9,650 0.4857 8,985 1.0000 112,000
MassFrac 0.0251 0.5073 0.3013 0.0862 0.0802 1.0000
574.4 = lO"(-(11,769/T(K))+l3.1927) Kpl Ratio1 574.4 = ( (Yco x YHZh3) / (YCHQ x YH20) )P2 0.2590 = 10"(1,197.8/T(K)-1.6485) KP~ 0.2590 = Ratio2 (Y co2 x YtI2) / (Yco x YH20) fractional conversion of CH4* 0.89021022 Converge 1" -4.3538E-05 = ((Kpl/Ratiol)-1)*100 moles of CO2 formed** 219.322711 = ((Kp2/Ratio2)-l)*lOO Converge 2"" -0.00028801 * Use Goal Seek tool to adjust CH4 conversion so that Converge 1 goes to zero. ** Use Goal Seek tool to adjust CO2 formation so that Converge 2 goes to zero. CH4 Conv = CH4 to CO =
-1 Product Flow Rate =mI
CO with water-gas shift reaction:CO
without water-gas shift reaction = 184.381
13-4
Problem 13.5 (cont'd)
methane:steam of 1:l Stoichiometric Feed I (Iunol/h)l CH4 1600 H20 1600 co 0 co2 0 H2 0 Total 3200
*(kmol/h) 482 445 1,082 37 3,392 5,437
Product MolFrac (kg/h) 0.0886 7,706 0.0818 8,007 0.1990 30,286 0.0068 1,617 0.6239 6,784 1.0000 54,400
MassFrac 0.1416 0.1472 0.5567 0.0297 0.1247 1.0000
= KP~ 574.4 lO"(-(11,769/T(K))+l3.1927) = Ratio1 1662.1 ( (Yco x Y,2^3 11 (YCH4 x YH20) ) p2 = KP~ 0.2590 10A(1,197.8/T(K)-1.6485) = Ratio2 0.2590 (Y co2 x Y,,) / (Yco x YIi20) fractional conversion of CH4* 0.69900166 Converge 1" -65.4436238 = ((Kpl/Ratiol)-1)*100 moles of CO2 formed** 36.7477842 Converge 2** -5.79043-05 = ((Kp2/Ratio2)-I)*100 CH4 Conv = CH4 to CO = Product Flow Rate =I1
methane:steam of 2:l Stoichiometric Feed I (kmol/h)l CH4 1600 3200 H20 co 0 0 co2 H2 0 Total 4800
(kmol/h) 257 1,726 1,213 130 4,160 7,487
= 574.4 Kpl = 876.6 Ratio1 = 0.2590 KP~ = Ratio2 0.2590 fractional conversion of CH4* Converge l* -34.4757385 moles of CO2 formed** Converge 2** -1.52263-05
5:;4: g;Fh
Product MolFrac (kg/h) 0.0343 4,108 0.2306 31,074 0.1620 33,961 0.0174 5,737 0.5557 8,320 1.0000 83,200
lO^(-(11,769/T(K))+13.1927) ( (yco x ~~2~3) 1 (YcH4 x YHZO) ) P2 10"(1,197.8/T(K)-1.6485) (Yco2 x YH2) / (Yco x YH20) 0.83954105 = ((Kpl/Ratiol)-I)*100 130.381595 = ((Kp2/Ratio2)-l)*lOO
CH4 Conv = CH4 to CO = Product Flow Rate -Im[
13-5
MassFrac 0.0494 0.3735 0.4082 0.0690 0.1000 1.0000
Problem 13.5 (cont'd)
methane:steam of 4:l Stoichiometric Feed (kmol/h)
CH4 H20 co co2 H2 Total
1600 6400 0 0 0 8000
I
1
,(kmol/h) 133 4,634 1,169 299 4,700 10,934
Product MolFrac (kg/h) 0.0122 2,126 0.4238 83,418 0.1069 32,722 0.0273 13,134 0.4298 9,400 140,800 1.0000
MassFrac 0.0151 0.5925 0.2324 0.0933 0.0668 1.0000
574.4 = lO"(-(11,769/T(K))+l3.1927) Kpl Ratio1 410.9 = ( (Yco x ~~2~x1 ! (YCH4 x YH20) ) p2 0.2590 = 10"(1,197.8/T(K)-1.6485) Kp2 Ratio2 0.2590 = (Y co2 x YH2) / (Yco x YH20) fractional conversion of CH4* 0.91695885 Converge 1* 39.77214561 = ((Kpl/Ratiol)-l)*lOO moles of CO2 formed** 298.507525 Converge 2** -l.O4E-05 = ((Kp2/Ratio2)-l)*lOO CH4 Conv = CH4 to CO = Product Flow Rate =mI
Moles Methane H2 H2:CO Steam:Mole CH4 Conversion Production Production CO:H2 1 69.9% 3,392 3.1359 0.3188883 84.0% 4,160 3.4300 0.2915462 2 91.7% 4,492 3.7280 0.2682379 3 4,700 4.0217 0.2486487 4 89.0%
13-6
Problem
13.6
Reformer Reformer
Temperature Temperature Pressure
855 C 1128 K ,15.79 atm
1.6 MPa
Production
rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 - production rate of CO2 CO fractional conversion x feed rate CH4 - production rate of CO2 CO2 feed rate CO2 + production rate of CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion + production rate of CO2 Stoichiometric Table: COMPONENT FEED PRODUCT MolFrac CH4 1600 176 0.0190 4800 H20 3,156 0.3413 co 0 1,205 0.1303 co2 0 219 0.0237 H2 0 4,492 0.4857 Total 6400 9,249 1.0000 574.4 = Kpl 574.4 = Ratio1 0.2590 = Kp2 Ratio2 0.2590 = fractional conversion of CH4* Converge 1* -4.35353-05 moles of CO2 formed** Converge 2** -0.00028798 Reformer Reformer Pressure Stoichiometric CH4 H20 co co2 H2 Total
T T
lO"(-(11,769/T(K))+l3.1927) ( (Yco x Y,,^3) / (YCHI x YHZO) )P2 10"(1,197.8/T(K)-1.6485) (Y CO2 x YH2) / (YCO x YH20) 0.8902102 ((Kpl/Ratiol)-l)*lOO 219.=32271 = ((Kp2/Ratio2)-l)*lOO
750 c 1023 K 15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 576 3,510 759 266 3,338 8,448
MOLFRAC 0.0681 0.4155 0.0898 0.0314 0.3951
Kpl 48.79 Ratio1 48.79 Kp2 0.3329 Ratio2 0.3329 fractional conversion of CH4* 0.64015 Converge 1" - 6 . 4 5 1 9 3 - 0 5 = moles of CO2 formed** 265.59039 Converge 2** 0.000608333 =
13-7
nCO/nCH4 nH2/nCH4 nCO/nC02
0.474156 2.086444 2.856465
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-1)*100
Problem 13.6 (cont'd)
Reformer Reformer
Stoichiometric
Temperature Temperature Pressure
800 C 1073 K .15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
CH4 H20 co co2 H2 Total
PRODUCT 357 3,312 999 244 3,975 8,887
MOLFRAC 0.0401 0.3727 0.1124 0.0275 0.4472
Kpl 167.6 Ratio1 167.6 Kp2 0.2936 Ratio2 0.2936 fractional conversion of CH4* 0.7771058 Converge l* 0.000238464 ((Kpl/Ratiol)-l)*lOO moles of CO2 formed** 244: 4398 Converge 2** -2.48073-05 = ((Kp2/Ratio2)-l)*lOO
Reformer Reformer
Stoichiometric CH4 H20 co co2 H2 Total
Temperature Temperature Pressure
900 c 1173 K 15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 88 3,086 1,311 201 4,738 9,424
MOLFRAC 0.0093 0.3275 0.1391 0.0214 0.5027 1.0000
Kpl 1443.6 Ratio1 1443.6 KP~ 0.2359 Ratio2 0.2359 fractional conversion of CH4* 0.9451022 Converge l* -4.1853-05 = moles of CO2 formed** 201.39252 Converge 2"" -0.00020159 =
13-8
((Kpl/Ratiol)-I)*100 ((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer Reformer Stoichiometric
Temperature Temperature Pressure
950 c ' 1223 K 15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
CH4 H20 co co2 H2 Total
PRODUCT 39 3,054 1,377 185 4,869 9,523
MOLFRAC 0.0040 0.3207 0.1445 0.0194 0.5113 1.0000
Kpl 3712.3 Ratio1 3712.3 KP~ 0.2142 Ratio2 0.2142 fractional conversion of CH4* 0.9759079 Converge l* 0.000662212 = moles of CO2 formed** 184.93694 Converge 2** -2.0226E-05 = 1.60 MPa T (Cl nCO/nCH4
nH2/nCH4 nCO/nCO2
750 0.474 2.086 2.856
800 0.624 2.484 4.087
13-9
855 0.753 2.808 5.494
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO 900 0.819 2.961 6.509
950 0.860 3.043 7.443
Problem 13.6 (cont'd)
Reformer Reformer
Temperature Temperature Pressure
855 C 1128 K *11.84 atm
1.2 MPa
Production
rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 - production rate of Co2 CO fractional conversion x feed rate CH4 - production rate of CO2 CO2 feed rate CO2 + production rate of CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion + production rate of CO2 Stoichiometric Table: COMPONENT FEED PRODUCT MolFrac CH4 1600 116 0.0124 H20 4800 3,098 0.3307 co 0 1,266 0.1352 co2 0 218 0.0232 H2 0 4,670 0.4985 6400 9,368 Total 1.0000 574.4 = Kpl Ratio1 574.4 = 0.2590 = Kp2 Ratio2 0.2590 = fractional conversion of CH4* Converge 1* 5.08455E-05 moles of CO2 formed** Converge 2** -4.7046E-05
Reformer T Pressure Stoichiometric CH4 H20 co co2 H2 Total
lO"(-(11,769/T(K))+13.1927) ( (Yco x YH2&3) / (YCH4 x YH20) )P2 10"(1,197.8/T(K)-1.6485) (Y CO2 x yi-12) / (YCO x YH20) 0.9275941 ((Kpl/Ratiol)-1)*100 217.=65358 = ((Kp2/Ratio2)-1)*100
1023 K 11.84 atm
1.2 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 472 3,405 861 267 3,651 8,656
MOLFRAC 0.0545 0.3934 0.0994 0.0309 0.4218
KP~ 48.79 Ratio1 48.79 0.3329 KP~ 0.3329 Ratio2 fractional conversion of CH4* 0.7049568 Converge 1* -0.00022618 moles of CO2 formed** 267 .=23789 Converge 2** 0.000779089 =
13-10
nCO/nCH4 0.537933 nH2/nCH4 2.281894 3.2207 nCO/nCO2
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer Temperature Reformer Temperature Pressure Stoichiometric
800 c 1073 K *11.84 atm
Table: FEED 1600 4800 0 0 0 6400
CH4 H20 co co2 H2 Total
PRODUCT 265 3,221 1,092 243 4,249 9,071
MOLFRW 0.0292 0.3551 0.1204 0.0268 0.4685
KP~ 167.6 Ratio1 167.6 KP~ 0.2936 Ratio2 0.2936 fractional conversion of CH4* 0.8346471 Converge 1* -0.00043805 moles of CO2 formed** 243.14362 Converge 2** -2.3364E-05 =
Reformer Temperature Reformer Temperature Pressure Stoichiometric CH4 H20 co co2 H2 Total
1.2 MPa
900 c 1173 K 11.84 atm
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO
1.2 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 54 3,054 1,346 200 4,839 9,492
MOLFRAC 0.0057 0.3217 0.1418 0.0211 0.5098 1.0000
KP~ 1443.6 Ratio1 1443.6 Kp2 0.2359 Ratio2 0.2359 fractional conversion of CH4* 0.966348 Converge l* -0.00083839 = moles of CO2 formed** 200.31077 Converge 2** -2.46013-05 =
13-11
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO
