CHAPTER TWO FIRST LAW OF THERMODYNAMICS Table of Contents INTRODUCTION ..................................................................................................................... 1 Background ...................................................................................................................................1 Learning Objectives........................................................................................................................2
THE FIRST LAW ...................................................................................................................... 2 Mechanical Energy .........................................................................................................................2 Conservation of Energy: The First Law ............................................................................................3 Definitions .....................................................................................................................................5 Types of Thermodynamic Systems .......................................................................................................................5 Types of Thermodynamics Processes ..................................................................................................................6 Types of Biochemical Reactions (exclude from handout) ....................................................................................7
More on Notation and Changes in Internal Energy (exclude from student handout) .........................7 General Representation of Internal Energy .........................................................................................................7 Fixed Number of Moles ........................................................................................................................................8 Changing in Number of Moles .............................................................................................................................8
MORE ON HEAT TRANSFER .................................................................................................... 9 General..........................................................................................................................................9 Convection ..................................................................................................................................10 Conduction ..................................................................................................................................10 Radiation .....................................................................................................................................11 Example Problem 2.1: Heat Transfer from a Human Body.............................................................12
MORE ON WORK ................................................................................................................. 14 General........................................................................................................................................14 Thermodynamic Work .......................................................................................................................................14 Power .................................................................................................................................................................15 Notation .............................................................................................................................................................16
Expansion/Compression Work .....................................................................................................16 Quasi-equilibrium Processes ........................................................................................................17 Use of Ideal Gas Equation.............................................................................................................18 General Formulation ..........................................................................................................................................18 Isothermal Expansion .........................................................................................................................................19
Example Problem 2.2: Application of 1st Law for Isothermal Conditions ........................................19
INTERNAL ENERGY AND TEMPERATURE ............................................................................... 21 Background .................................................................................................................................21
Heat Capacities ............................................................................................................................21 General Approach ..............................................................................................................................................21 Heat Capacity at Constant Volume ....................................................................................................................22 Heat Capacity at Constant Pressure ...................................................................................................................24 Relationship between Cvm and Cpm .....................................................................................................................26 Incompressible Substance..................................................................................................................................26 Variation with Temperature...............................................................................................................................27
Enthalpy ......................................................................................................................................27 Mathematical Features ......................................................................................................................................27 More on Cp (exclude from student handout) .....................................................................................................28
Application of Heat Capacities to Engineering Problems ................................................................29 Example Problem 2.3: Application of 1st Law for Isobaric Conditions..............................................30 Example Problem 2.4: Application of 1st Law for Varying P, T and V. ..............................................33 Problem Definition and Solution Results ...........................................................................................................33 Exact and Inexact Differentials...........................................................................................................................38 More Analyses (exclude from student handout) ...............................................................................................39
Additional Discussion ...................................................................................................................41
ENTHALPY OF PHASE TRANSITIONS...................................................................................... 42 Introduction ................................................................................................................................42 Standard Enthalpies of Transitions ...............................................................................................44 Introduction .......................................................................................................................................................44 Vaporization and Notation .................................................................................................................................45 Fusion and Sublimation ......................................................................................................................................46 Tabular Values ....................................................................................................................................................47
Changes with Temperature ..........................................................................................................47 Example Problem 2.5: Cooling by Perspiration .............................................................................50
ENTHALPY OF CHEMICAL REACTIONS ................................................................................... 51 Introduction ................................................................................................................................51 General Concepts ...............................................................................................................................................51 Molar Extent of Reaction ...................................................................................................................................53
Enthalpy of Reaction ....................................................................................................................55 Standard Enthalpy of Formation ...................................................................................................56 Reference State ..................................................................................................................................................56 Experimentally Derived Values ..........................................................................................................................56 Hess’ Law............................................................................................................................................................57
Changes with Temperature ..........................................................................................................57 Example Problem 2.6: Standard Enthalpy of Reaction of Urea .......................................................58 Enthalpy of Combustion ...............................................................................................................61 Metabolic Rate and Thermoregulation .........................................................................................62 Introduction .......................................................................................................................................................62 Animal Calorimetry ............................................................................................................................................63 Human Metabolic Rate ......................................................................................................................................67
Example Problem 2.7: Change in Room Air Temperature .................................................................................67
HEAT TRANSFER BY RADIATON ............................................................................................ 70 Basic Concepts .............................................................................................................................70 Blackbody ...........................................................................................................................................................70 Early Theoretical Framework .............................................................................................................................71 Spectral Flux Density ..........................................................................................................................................76 Solar and Terrestrial Radiation...........................................................................................................................78 Representation Using Single Curve: Excluded from Student Handout .............................................................78 Example Problem 2.8: Molar Photon Energy .....................................................................................................79
Radiant Flux Density ....................................................................................................................80 Blackbody radiation ...........................................................................................................................................80 Gray bodies ........................................................................................................................................................81
Radiation Properties ....................................................................................................................81 Measured Solar and Terrestrial Radiation .........................................................................................................81 Interactions with Incident Radiation ..................................................................................................................82 More Discussion .................................................................................................................................................84 Kirchhoff’s law ....................................................................................................................................................84 Absorptivity values .............................................................................................................................................84 Reflectivity values ..............................................................................................................................................85
RADIATATIVE BALANCE FOR THERMAL EQUILIBRIUM .......................................................... 85 Earth’s Orbit around the Sun ........................................................................................................85 Radiation and Thermal Equilibrium Relationships .........................................................................86 Extraterrestrial Radiation ...................................................................................................................................86 Thermal Equilibrium for Non-absorbing Atmospheric Gases ............................................................................87
Example Problem 2.9: Thermal Equilibrium and Earth’s Temperature............................................88 Discussion....................................................................................................................................89
THERMODYNAMIC PROCESSES OF PLANT CANOPIES ............................................................ 90 Photosynthesis and Transpiration ................................................................................................90 Transpiration and the First Law ....................................................................................................93 System Definition ...............................................................................................................................................93 Use of Bowen Ratio ............................................................................................................................................93
Example Problem 2.10: Transpiration Depth from Plant Canopies ................................................94
PROBLEM ASSIGMENTS ....................................................................................................... 99
Working Notes
CHAPTER TWO1
FIRST LAW OF THERMODYNAMICS Indeed the phenomena of nature, whether mechanical, chemical or vital, consist almost entirely in a continual conversion of attraction through space (potential energy), living force (kinetic energy) and heat into one another. Thus it is that order is maintained in the universe – nothing is deranged, nothing ever lost, but the entire machinery, complicated as it is, works smoothly and harmoniously. And though, as in the awful vision of Ezekiel, “wheel may be in the middle of wheel”, and everything may appear complicated and involved in the apparent confusion and intricacy of an almost endless variety of causes, effects, conversion, and arrangements, yet is the most perfect regularity preserved – the whole being governed by the sovereign will of God. James P. Joule, Co-discover of the 1st Law, 1818-1889
INTRODUCTION Background Early developments in thermodynamics were tied to steam engines. As discussed in Chapter 1, steam engines had a dramatic impact on technological developments in the nineteenth century. Prior to these developments, coal was almost exclusively used to provide heat during winter. The expansion of its use to stream engines required a greater understanding on the nature of heat transfer and its relationship to mechanical work. A major advancement in heat transfer was made by Fourier in 1811. He proposed that heat flow is proportional to the gradient of temperature. Another important development was the analysis of heat engines by Carnot in 1824. Contributions of both Fourier and Carnot are particularly noteworthy because the conservation of energy wasn’t recognized for another two decades. The First Law of thermodynamics was largely formulated in the 1840s. A surgeon named Robert Mayer was arguably the first scientist to observe that energy is generally conserved in 1842. However, his contributions were widely ignored. He apparently lived a troubled life that included attempts of suicide and hospitalization for mental illness. Another noteworthy contributions in the 1840’s were by James Joule, a son of a prosperous brewer. He had the freedom to conduct careful experiments and was able to establish that heat generated in an electric circuit is proportional to the square of the current multiplied by the resistance of the conductor. Joule also estimated a mechanical equivalent of heat of 772.55 ft-lb of mechanical energy per 1 BTU (British Thermal Unit) of heat. This number is engraved on his tombstone. Because of his work in linking heat and work within the same framework, Joule is widely
1
© 2018. Bruce N. Wilson and the Regents of the University of Minnesota. All Rights Reserved. These notes have not been professionally reviewed.
Last Modified: March, 2018
2-1
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
recognized as a co-discover of the First Law. Rudolf Clausius (1822-1888) and Lord Kelvin (William Thomson, 1824-1907) were influential in formulating the First (and Second) Law that we are familiar with today. In addition to the First Law, this chapter will also introduce radiation as one of the forms of heat transfer. With the exception of nuclear and volcanic energy, all energy on Earth comes from the sun through radiation. It is of fundamental importance to environmental processes related to global weather patterns, hurricanes and thunderstorms; photosynthesis and plant growth; terrestrial evaporation and transpiration; and cold climate conditions related to snowmelt and glacial movement. Insights into these processes require information on the distribution of radiation energy among electro-magnetic waves of different frequencies. Sir Isaac Newton (1642-1727) used prisms to conclude that “white” light is composed of a mixture of colors ranging from frequencies corresponding to the colors of violet to red. These colors were a nuisance in the use of microscopes, field glasses, and telescopes, and the development of lenses to fix these problems resulted in the discovery of additional frequencies. Johann Ritter (1776-1810) discovered the ultraviolet range of light by the response of silver chloride (important to photography) to light beyond the blue and violet range. Gustaf Kirchhoff (1824-1887) discovered that each element sends out light of frequencies that are characteristics of the element and that it loses this frequency when passed through the element vapors. An early application of this discovery was that sunlight passes through sodium at solar surface because it lacks the frequency corresponding to this element. Learning Objectives After reading the lecture materials and completing the homework assignments, you are expected to understand and to be able to apply: • • • • • •
The First Law of thermodynamics, Enthalpy corresponding to temperature changes and experimentally-derived heat capacities coefficients Enthalpy of phase transitions corresponding to vaporization, fusion and sublimation Enthalpy of chemical reaction using the standard enthalpy of formulation, Radiation and the heating of Earth, and Evapotranspiration from plant canopies.
THE FIRST LAW Mechanical Energy Let’s revisit the definition of mechanical work using Newton’s second law where force is equal to the rate of change in momentum (𝐹𝐹⃗ = 𝑚𝑚 𝑑𝑑𝑣𝑣⃗⁄𝑑𝑑𝑑𝑑 for constant mass). Since our focus is to illustrate energy-work concepts, we will simplify by considering movement in only the vertical z direction. Work is then defined by multiplying both sides by dz and integrating over a distance of zf-zi, that is,
Last Modified: March, 2018
2-2
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
zf zf 𝑑𝑑𝑑𝑑 w = � 𝐹𝐹𝑧𝑧 𝑑𝑑𝑑𝑑 = 𝑚𝑚 � � � 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 zi zi
(2.1.1)
Let’s now consider work only done by gravity in accelerating our system, where Fz = -mg. By also using the chain rule of calculus and v=dz/dt, we obtain zf
zf
vf 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 −mg � 𝑑𝑑𝑑𝑑 = 𝑚𝑚 � � � � � 𝑑𝑑𝑑𝑑 = m � 𝑣𝑣 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 zi zi vi
(2.1.2a)
By evaluating the integrals, we conclude that
𝑚𝑚𝑣𝑣𝑓𝑓2 𝑚𝑚𝑣𝑣𝑖𝑖2 −( mg zf − mg zi ) = −(PEf − PEi ) = − = 𝐾𝐾𝐸𝐸𝑓𝑓 − 𝐾𝐾𝐸𝐸𝑖𝑖 2 2
(2.1.2b)
where KE is the kinetic energy defined as KE = mv2/2. It is the energy associated with the state of motion of the system. PE is the gravitational potential energy defined as PE = mgz. It is the energy of the system because of its position within a gravitational field. If the final velocity is greater than the initial velocity, the work is positive, and there has been a reduction in the potential energy (zf < zi). Eq. 2.1.2a is a special form of the work-energy theorem evaluated for work done by gravity. Mechanical energy is defined as the sum of potential and kinetic energies. The conservation of mechanical energy can be obtained from Eq. 2.1.2b as ΔKE + ΔPE = 0
(2.1.3)
where ΔPE = PEf – PEi and ΔKE = KEf- KEi. Mechanical energy is defined for conservative forces, and it doesn’t include frictional and drag forces. Conservation of Energy: The First Law Some engineering problems with negligible friction forces and heat transfer can be solved by using the conservation of mechanical energy. However, these problems could have alternatively been solved by directly applying Newton’s second law of motion. The engineer has simply replaced the solution approach using Newton’s second law of motion with an alternative form of this law written in terms of mechanical energies. The First Law is a more general and powerful statement than the conservation of mechanical energy. It ties together the seemingly different processes of heat transfer and work as forms of transfer of a fundamental property of the Universe called energy. In fact, combining work, heat transfer and other important processes within a single energy-based framework is one of the greatest contributions in all of science. Let’s revisit the energies for the universe, system and surroundings shown below that were first introduced in Chapter 1.
Last Modified: March, 2018
2-3
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.1. Important Components of the First Law. The First Law of thermodynamics is succinctly stated by Rudolf Clausius as "Die Energie der Welt ist constant" (the energy of the Universe is a constant). To conserve total energy, a change in the energy of the system must be balanced by an equal and opposite energy change in the surroundings. This change is consequence of the transfer of energy between the system and the surroundings. Processes of energy transfer are shown in the schematic in Fig. 2.1. In comparison to the First Law of Chapter 1, we have included the transfer of energy by the flow of matter into and out-of the system. Energy transfer determined by temperatures of the system and the surroundings is called heat (q) transfer. All other non-mass energy transfer is done by work (w). The conservation of energy for the system can then be written as ΔE = − ΔEsur = q + w + ΔEtransport = 𝑞𝑞 + 𝑤𝑤 + �𝑚𝑚̇𝑖𝑖𝑖𝑖 𝐸𝐸𝑠𝑠,𝑠𝑠𝑠𝑠𝑠𝑠 Δ𝑡𝑡 − 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 𝐸𝐸𝑠𝑠 Δ𝑡𝑡�
(2.1.5)
where Δ𝐸𝐸𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑚𝑚̇𝑖𝑖𝑖𝑖 𝐸𝐸𝑠𝑠,𝑠𝑠𝑠𝑠𝑠𝑠 Δ𝑡𝑡 − 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 𝐸𝐸𝑠𝑠 Δ𝑡𝑡 is the transport of energy by the flow of matter, 𝑚𝑚̇𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 is the mass flow rate (mass per time) into and out of the system, respectively, Es and Es,sur are the energies per unit mass of matter leaving the system and entering from the surroundings, respectively, and Δt is the time interval over which the change in energy is of interest. The mass flow rate is defined by the product of the density of matter, its velocity and the surface area of flow. For most of the systems discussed in this course, the system boundaries don’t allow the flow of matter through them, or they have conditions of 𝑚𝑚̇𝑖𝑖𝑖𝑖 𝐸𝐸𝑠𝑠,𝑠𝑠𝑠𝑠𝑠𝑠 Δ𝑡𝑡 ≈ 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 𝐸𝐸𝑠𝑠 Δ𝑡𝑡. ΔEtransport is then zero or negligible, and the transfer of energy is limited to heat transfer and work. To review our notation from Chapter 1, heat transfer is positive when the transfer is from the surroundings into the system, causing an increase in the system energy, and as negative when the transfer is from the system into the surroundings, causing a decrease in its energy. Similarly, work is positive when the surroundings does work on the system and negative when the system does work on the surroundings. In Fig. 2.1, the system energy is evaluated with kinetic energy (KE), gravitational potential energy (PE), internal energy (U) and other possible energies (O). Changes in kinetic energy
Last Modified: March, 2018
2-4
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
and potential energy are defined by changes in the position and motion of the system. These forms of energies are external to the changes within the system itself. As discussed in Chapter 1, internal energy is the energy associated with the motion, interactions and bonding of atoms, ions and molecules in the system. Internal energy is an extensive property of the system, and therefore an infinitesimal change is represented by an exact differential. By focusing on energy changes within the system, the First Law can then be simplified for a system of fixed mass as ΔU = ΔUtpr + ΔUphase + ΔUr = 𝑞𝑞 + 𝑤𝑤
(2.1.6)
Infinitesimal changes in heat transfer and work are represented by inexact differentials. We have divided the change in internal energy into possible changes with temperature (ΔUtpr), phase transitions (ΔUphase), and chemical reactions (ΔUr). Changes between two equilibrium states are shown graphically in Fig. 2.3. Each type of change is introduced by holding constant the internal energy associated with the other two. Sometimes, especially in older textbooks, ΔUtpr, ΔUphase, and ΔUr are considered by using terminology of heat energy (or even sensible heat), latent heat and heat of reaction, respectively. Our terminology provides an easier theoretical framework for discussing thermodynamics and limits the definition of “heat” to the transfer of energy by a temperature difference.
Figure 2.3. Heat Transfer and Work Done Between Two Equilibrium States. Definitions Types of Thermodynamic Systems Let’s consider the following systems:
Last Modified: March, 2018
2-5
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
No energy transfer q= w = 0
Energy transfer
Energy transfer q
System
w min
System
No mass exchange
Isolated System
q
No mass exchange
Closed System
w mout System
Mass exchange
Open System
Figure 2.4. Types of System Bound in Energy and Mass Exchange at Their Boundaries. Open system: Exchanges mass and energy with its surroundings. Hence the mass flow rates of Eq. 2.1.2 are not zero. Closed system: Has no mass flow into and out of the system and therefore it only exchanges energy with its surroundings. The mass flow rates are zero. Isolated system: Does not exchange mass or energy with the surroundings. The Universe is an isolated system. Types of Thermodynamics Processes Let’s consider the following system and surroundings: Tsur
Possible q
T
Tsur Possible w
T
System
System
q=0
T =constant
Adiabatic Process
Isothermal Process
Figure 2.5. Adiabatic and Isothermal Processes. Adiabatic process: Boundaries do not allow for heat transfer, even if there is a temperature difference between system and the surroundings. Adiabatic conditions are often as an approximation to rapidly changing system properties where the heat transfer is negligible in comparison to other energy terms.
Last Modified: March, 2018
2-6
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Isothermal process: Process where the temperature of the system is constant. Isothermal conditions are frequently an adequate approximation when a gas is given sufficient time (slow change) to expand (or contract) to maintain a constant temperature. Many of the processes in animals are approximately isothermal. For example, humans maintain a nearly constant body temperature of approximately 37 C. Isobaric process: Process where the pressure of the system is constant. Many environmental processes occur under nearly constant ambient pressure of the atmosphere, which is often taken as P = 1 atm = 101.325 kPa. Isochoric process: Process where the volume of the system is constant. Types of Biochemical Reactions (exclude from handout) Endothermic reaction: Endothermic reactions have heat transfer from the surroundings into the system. Reactions of this type are less common than exothermic reactions. Exothermic reaction: Chemical processes in a system that release heat into the surroundings are exothermic reactions. Examples include all combustion in which organic compounds are oxidized by O2 gas with a byproduct of CO2 gas. The conversion of glucose is an important example and is shown below (s – solid, g – gas, l – liquid): C6 H12 O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O
For most organisms, the root source of energy is from the sun. Organic molecules are created by photosynthesis. As shown in the previous reaction, these organic molecules can be oxidized to provide the energy needs of living organisms. Nicotinamide adenine dinucleotide (NADH) is an important initial energy carrier in cells. Cellular oxidation-reduction reactions transfer energy from NADH into mobile adenosine triphosphate (ATP) and into ion gradients across membranes. Because of these processes, we have the following additional definitions for biochemical reactions. Metabolism: Collection of chemical reactions that trap, store, and utilize energy in biological cells. It is the sum of catabolic and anabolic processes. Catabolism: Collection of metabolic pathways that breakdown molecules into smaller units by cellular respiration. Usually these processes releases energy. Anabolism: Collection of metabolic pathways that builds components of cell by the biosynthesis of small and large molecules. These reactions require energy. Energy from catabolic reactions is used to drive anabolic reactions. More on Changes in Internal Energy (exclude from student handout) General Representation of Internal Energy In Chapter 1, internal energy was represented as function of temperature and volume for a fixed mass (or a fixed number of moles). This representation is appropriate for considering
Last Modified: March, 2018
2-7
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
the change in internal energy with a change in temperature for a closed system. However for phase transitions and chemical reactions, changes in the number of moles are of fundamental importance. For vaporization, the number of liquid moles decreases and the number of gas moles increases; whereas for chemical reactions, the number of reactant moles decreases and number of product moles increases. A more general representation of internal energy to consider these important processes is U = U(T, V, n1 , 𝑛𝑛2 , … , 𝑛𝑛𝑘𝑘 )
(2.1.8)
where n1, n2, … , nk are the number of moles for each of the k components in the system. A differential change is then defined as ∂U ∂U ∂U dU = � � dT + � � dV + � � dn1 + ⋯ ∂T V,nj ∂V T,nj ∂n1 T,V,n j≠1 ∂U + � � dnk ∂nk T,V,n
(2.1.9)
j≠k
where subscript nj refers to constant number of moles for all components and nj≠i refers to constant number of moles for all components other than the ith component. After entropy is introduced in Chapter 4, we will reconsider change in internal energy using entropy instead of temperature. Fixed Number of Moles Let’s consider the change in the internal energy for a fixed number of moles corresponding to dn1 = dn2 = … = dnk = 0 in Eq. 2.1.6. We will use the notation of Utpr =U(T,V) for the change internal energy with temperature and volume. This change is defined as ∂U ∂U ∂U dU = dUtpr = � � dT + � � dV = Cv dT + � � dV ∂T V,nj ∂V T,nj ∂V T,nj
(2.1.10a)
Although changes in temperature internal energy are theoretically a function of changes in temperature and volume, but for ideal gases, ∂U/∂V = 0 and for solids and liquids dV ≈ 0 so that dU is effectively only a function of temperature for many engineering systems, that is, dUtpr = �
∂U � dT = Cv dT ∂T V,nj
(2.1.10b)
Changing in Number of Moles
Parameters to characterize phase transitions and chemical reactions are typically evaluated under conditions of no net changes in temperature and pressure. For phase transition, we are interested in the change in internal energy of nβ moles of phase β (e.g., liquid) to nα = -nβ of phase α (e.g., gas). We will use the notation of Uphase = U(nβ,nα) for the change in internal energy with phase transitions. This change can be written as
Last Modified: March, 2018
2-8
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
∂U 𝑑𝑑𝑑𝑑 = dUphase = � � � ∂nβ
T,V,nα
∂U − � � � dnβ ∂nα T,V,n
(2.1.11)
β
For phase transitions, -dnα = dnβ. The change in internal energy for phase transition is directly related to the difference in the change in the internal energy with the number of moles of phases α and β. The change in internal energy of chemical reactions is directly related to changes in internal energy with the number of reactants and products for conditions of constant temperature and pressure and no phase transitions. We will use the notation of Ur=U(n1,n2, … ,nk) for these conditions. Since the number of moles of products increases and the number of moles of reactant decrease, we can write the change in internal energy for chemical reactions as p
∂U 𝑑𝑑𝑑𝑑 = dUr = �� � � ∂nj
T,V,ni≠j
j=1
r
dnj �
prod
− �� � j=1
∂U � ∂nj
T,V,ni≠j
dnj �
where p is the number of products and r is the number of reactants.
(2.1.12)
react
By using our three types of internal energy, the general change in internal energy of Eq. 2.1.9 can alternatively be written as dU = dUtpr + dUphase + dUr
(2.1.13)
Let’s finish this section by considering our solution for vaporization (i.e. Eq. 2.1.11) with that obtained from the First Law under conditions of constant temperature, pressure and no chemical reactions. We then have ∂U dUvap = δq + δw = � � � ∂ngas
T,V,nliq
∂U − � � ∂nliq
T,V,ngas
� dngas
(2.1.14a)
After we achieve a new equilibrium state, we have δq=δw=dUvap = 0. This condition corresponds to dynamic equilibrium between vaporization and condensation of molecules. As discussed in greater detail in Chapter 5, this balance occurs when ∂U � � ∂ngas
T,V,nliq
=�
∂U � ∂nliq
(2.1.14b)
T,V,ngas
The above partial derivatives are important and provide the foundation for the definition of chemical potential given in Chapter 5. At equilibrium, the chemical potential of the gas phase equals that of the liquid.
MORE ON HEAT TRANSFER General
Last Modified: March, 2018
2-9
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Relationships for heat transfer are often written in terms of heat flux. Heat flux is the rate of heat transfer per unit area. Heat transfer is obtained by integrating over the surface area and over the time interval of interest. This integration will usually be done by assuming that the heat flux is constant with space and constant for the time between equilibrium states. The heat transfer between initial and final equilibrium states is obtained from its rate as q=
tf
∫
q dt ≈ q ∆t
(2.2.1)
ti
where 𝑞𝑞̇ is the rate of heat transfer with units of J s-1 or W, tf and ti are the initial and final times of the transfer, and Δt=tf-ti. Convection Convection is a process of heat transfer that is driven transport processes of gas or liquid velocity (or bulk movement) moving over the surface area of the system. The concept is shown below for velocity over a flat plate.
Figure 2.6. Convection Processes. Convection is often predicted using Newton’s law of cooling defined as 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑘𝑘𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (𝑇𝑇𝑝𝑝 − 𝑇𝑇𝑓𝑓 )
2.2.2
where 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is the rate of heat transfer by convection, kconv is the heat transfer coefficient of convection, Aconv is the surface area, Tp is the temperature (K) of the plate and Tf is the temperature (K) of the fluid. Note that the rate of heat transfer is positive for a temperature of the plate is greater than that of the fluid. For a system corresponding to the plate temperature, heat transfer is from the system to surroundings, and you will need to use a negative heat transfer in the First Law. Conduction Conduction is a process of heat transfer by the interactions of more energetic molecules at higher temperatures with those at lower temperatures. These interactions occur in solids,
Last Modified: March, 2018
2-10
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
liquids and gases. To distinguish conduction from convection for fluids, conduction corresponds to the condition of no net transport by a fluid velocity. This concept is shown below for the heat transfer through a wall.
x Inside House
T1
Outside House
A T2
q
q Δx
Figure 2.7. Conduction Process. Conduction is often predicted by the well-known Fourier’s Law, where the rate of heat transfer is directly proportional to the temperature gradient. Fourier’s Law can be written as 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = − 𝑘𝑘𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑑𝑑𝑑𝑑 𝑇𝑇2 − 𝑇𝑇1 ≈ −𝑘𝑘𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴𝑐𝑐𝑐𝑐𝑛𝑛𝑑𝑑 � � 𝑑𝑑𝑑𝑑 𝑥𝑥2 − 𝑥𝑥1
2.2.3a
where 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is the rate of heat transfer for conduction, kcond is thermal conductivity, Acond is the surface area, dT/dx is the temperature gradient, and T and x are temperature (K) and distance as defined in the above figure. Heat transfer in Fig. 2.7 is positive from higher temperature to a lower temperature. If the system is outside the house then the sign of heat transfer is correct; however, if the system is inside the house, then you need to use a negative heat transfer in the First Law. The R-value is often used as a measure of the thermal conductivity of insulation properties of clothing and materials used in the construction. R-values are determining by directly measuring the heat transfer rate though materials of thickness Δx for a known temperature difference (ΔT). The R-value (rq) is defined as 𝑟𝑟𝑞𝑞 =
ΔT −Δ𝑥𝑥 = 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ∕ 𝐴𝐴 𝑘𝑘𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2.2.3b
The rate of heat conduction is then defined using R-value as 1 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = � � 𝐴𝐴 Δ𝑇𝑇 𝑟𝑟𝑞𝑞
2.2.3c
Radiation
Last Modified: March, 2018
2-11
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Radiation is a process where energy is transported by photons (or as electromagnetic waves) as a consequence of changes in the configurations of atoms, including the detachment of electrons and changes in the transition states of rotational and vibrational energies. Radiation is of fundamental importance in understanding energy transfer in many biological processes. It will be discussed in detail later in this chapter and also in Chapter 3. To introduce the concept of heat transfer by radiation, consider two large plates of the same area and material as shown below.
A = Area
T1 q rad ,1 = f (T14 )
q rad , 2 = f (T24 ) T2 Figure 2.8 Radiation Process. Under certain conditions, the rate of energy transfer from the lower plate is defined by using the Stefan-Boltzmann law for each plate to obtain 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝜎𝜎 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟 ( 𝜀𝜀2 𝑇𝑇24 − 𝜀𝜀1 𝑇𝑇14 )
2.2.4
where 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 is the rate of radiation heat transfer, ε2 and ε1 are the emissivities of the radiating matter for the lower and upper plates, respectively,, σ is the Stefan-Boltzmann constant (5.67 x 10-8 W m-2 K-4) and T is the temperature (K) of the two bodies. For the same emissivities, Eq. 2.2.4 is positive if T2>T1. If the lower plate corresponds to the system temperature, then you will need to use a negative heat transfer in the First Law. Processes of heat transfer by radiation are considerably different than convection and conduction. For example, the net heat transfer can be from a cooler to a warmer body if the emissivity of the warmer plate is smaller than that of the cooler plate. We will use the subscript “rad” for these processes to avoid possible confusion in the solution of problems. Example Problem 2.1: Heat Transfer from a Human Body Problem Statement: For inactive nude human body resting on a bench, determine the heat transfer rate from the body by: (1) Radiation, (2) Convection and (3) Conduction An illustrative sketch of the problem and key parameters are shown below. The human body is our system.
Last Modified: March, 2018
2-12
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
The core temperature of humans is approximately 37 C, but the surface skin temperature is closer to Tskin = 34 C for the conditions of this problem. Temperatures of the air and the walls are Tair = Twall = 23 C. The surface area of humans can be estimated using mass and height 2. We will use a surface area of Askin = 1.8 m2. For this problem, the effective percentage of this surface area for radiation is estimated as 65%, 85% for convection, and 5% for conduction (direct contact area with bench). Rigorous analysis of the effective surface area of the walls, floor and ceiling corresponding to incoming radiation to our human is beyond the scope of this problem. To simplify, we will use the same surface area for incoming and outgoing radiation. Emissivity of the skin and walls is taken as ε = 0.95, the heat transfer coefficient for convection is taken as 3 J s-1 m-2 K-1, and the R-value for conduction is taken as rq = 0.166 m2 K s J-1. Preliminary calculations are given below. 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 34 + 273.15 = 304.15 𝐾𝐾 𝑎𝑎𝑎𝑎𝑎𝑎 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑇𝑇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 23 + 273.15 = 296.15 𝐾𝐾 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑓𝑓𝐸𝐸𝐸𝐸,𝑟𝑟𝑟𝑟𝑟𝑟 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.65(1.8) = 1.17 𝑚𝑚2 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑓𝑓𝐸𝐸𝐸𝐸,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.85 (1.8) = 1.53 𝑚𝑚2 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑓𝑓𝐸𝐸𝐸𝐸,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.05 (1.8) = 0.09 𝑚𝑚2
Solution: For radiation, we have:
4 4 ) 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝜀𝜀 𝜎𝜎 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟 (𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝐽𝐽 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 = (0.95) �5.67𝑥𝑥10−8 � (1.17 𝑚𝑚2 )( (304.15 𝐾𝐾)4 − (296.15 𝐾𝐾)4 ) = 54.5 𝐽𝐽𝑠𝑠 −1 𝑠𝑠 𝑚𝑚2 𝐾𝐾 4
For convection:
𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑘𝑘𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 �𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 � The following relationship for skin surface was used: 𝐴𝐴𝑠𝑠 = 0.2 𝑚𝑚0.425 ℎ0.725 , where As has units of m2, m has units of kg, and h has units of m.
2
Last Modified: March, 2018
2-13
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = �3
𝐽𝐽 � (1.53 𝑚𝑚2 )( 304.15 𝐾𝐾 − 296.15 𝐾𝐾) = 36.7 𝐽𝐽𝑠𝑠 −1 𝑠𝑠 𝑚𝑚2 𝐾𝐾
For conduction:
𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 �𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 � ∕ 𝑟𝑟𝑞𝑞 (1.53 𝑚𝑚2 )( 304.15 𝐾𝐾 − 296.15 𝐾𝐾) 𝑞𝑞̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = 4.32 𝐽𝐽𝑠𝑠 −1 0.16 𝑚𝑚2 𝐾𝐾 𝑠𝑠 𝐽𝐽−1
Since our thermodynamic system is the human body, all of these heat transfer rates are negative. The largest mode of heat transfer is by radiation.
MORE ON WORK General Thermodynamic Work In Chapter 1, work was defined as the energy transfer by coherent or non-random motion of particles and heat transfer as that done by a motion that includes an incoherent or random component. We have also previously defined mechanical work as 2
𝑤𝑤 = � 𝐹𝐹⃗ ⋅ 𝑑𝑑𝑥𝑥⃗
2.3.1
1
Thermodynamic work requires the type of motion of mechanical work be placed within our framework of a system and a surroundings. Within this framework, we will define thermodynamic work as interactions between the system and surroundings such that the sole external effect could have been done by the coherent motion associated with the raising of a weight. To illustrate the definition of thermodynamic work, we will use the conditions shown in Fig. 2.9. Here we are first interested in the transfer of the energy from a system with a battery and motor to a fan operating in the surroundings. The transfer of energy from the change in internal energy of the battery is not done by heat transfer. The transfer of energy by thermodynamic work can be obtained by replacing the fan with a pulley supporting a weight. This weight is raised by the rotation of the motor’s shaft from which we can compute the thermodynamic work. Let’s further explore the definition of thermodynamic work by moving the motor out of the system as shown in the right-sided schematic of Fig. 2.9. We are now interested in thermodynamic work obtained by the flow of electrical energy across the system boundaries. This transfer of energy results in the sole effect could be the raising of a weight for a perfect motor. The issue of a perfect motor requires an analysis of the surroundings, and it is not really a concern for us. We can rationalize that someday, perhaps in the very distant future, a nearly perfect motor could be designed.
Last Modified: March, 2018
2-14
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.9. Insight into Thermodynamic Work. Power Relationships for work are sometimes given as rate of work done. Rate of work done is called power. Similar to heat transfer rate, the work is obtained from power between initial and final equilibrium states as 𝑡𝑡𝑓𝑓
𝑤𝑤 = � 𝑤𝑤̇ 𝑑𝑑𝑑𝑑 ≈ 𝑤𝑤̇ Δ𝑡𝑡
2.3.2
𝑡𝑡𝑖𝑖
where 𝑤𝑤̇ is the power with units of J s-1 or W and tf and ti are the initial and final times of the transfer. The last expression on the right-hand side is only valid if the power is approximately constant. Let’s discuss the work and power for the rotating shifts for the systems of Fig. 2.9. The analysis of the key components is shown below.
Shaft
+
Ft
Shaft Electrical Source
ω
Motor
r
Figure 2-10. Schematic of Work Done by a Shaft. Angular velocity (ω) and torque (τω) are features of interest in the design of machine elements with shafts. For a radius of r, we have the following definitions 𝜏𝜏𝜔𝜔 = 𝐹𝐹𝑡𝑡 𝑟𝑟 𝑣𝑣 = 𝜔𝜔 𝑟𝑟 Last Modified: March, 2018
2.3.3a 2.3.3b
2-15
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where Ft is the tangential force and v is the velocity at the boundary of the shaft. By using the definition of power, we obtain 𝑤𝑤̇ =
Notation
𝑤𝑤 𝐹𝐹𝑡𝑡 Δ𝑥𝑥 𝜏𝜏𝜔𝜔 = = 𝐹𝐹𝑡𝑡 𝑣𝑣 = � � ( 𝜔𝜔 𝑟𝑟 ) = 𝜏𝜏𝜔𝜔 𝜔𝜔 Δ𝑡𝑡 Δ𝑡𝑡 𝑟𝑟
2.3.3c
As previously discussed, work is defined as positive for work done by the surroundings on the system. This notation is consistent with that used for heat transfer. However, prior to 1990, many engineering textbooks defined the work done by the system on the surroundings as positive. Engineers are often interested in the amount of useful work available from the system, which is negative using our notation but is positive using the old notation. The First Law can be analyzed using either notation as long as it is consistently applied (e.g. ∆𝑈𝑈 = 𝑞𝑞 − 𝑤𝑤 �). For some of the materials in the lecture notes, the interpretation of results is more intuitive using the notation of 𝑤𝑤 � = −𝑤𝑤, where the hat symbol is used to indicate work done by the system on the surroundings. Expansion/Compression Work The most important type of work in this course is the work done by pressure forces resulting in the expansion or compression of the system. The system of interest is shown in the following figure.
Δz
A
zf
ΔV
zi Fp = P A System Figure 2.15. Schematic Illustrating Expansion Work.
By using the definition of mechanical work, and for approximately constant pressure, we obtain that the work done between the initial and final position as w = −Fp Δz = −(P A)Δz = −P ΔV
2.4.1
where P is the pressure, A is the surface area and Δz = zf – zi. The negative sign is used in the above equation because it is work done by the system on the surroundings. Expansion work doesn’t include any friction forces between the piston and the cylinder. For a non-constant pressure, we then obtain the following definition of work
Last Modified: March, 2018
2-16
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Vf
w = − � 𝑃𝑃 𝑑𝑑𝑑𝑑
2.4.2
Vi
where volume is defined by exact differentials. Quasi-equilibrium Processes Internal energy is defined by exact differentials, and therefore the change in internal energy is independent of the path. In contrast, expansion work is defined by inexact differentials and is a function of the path between initial and final states. To compute this work (and heat transfer), the path is often assumed to be obtained by a quasi-equilibrium (or quasi-static) process. Quasi-equilibrium processes are defined such that the path can be treated at thermodynamic equilibrium at every step of the change between two endpoint states. Similar to ideal gases, quasi-equilibrium processes are an idealized representation of actual thermodynamic processes. They provide insight into thermodynamic processes and are widely used in engineering designs. Insight into quasi-equilibrium processes for expansion work can be obtained by considering the gas under compression as series of equilibrium steps shown in Figs. 2.16a and 2.16b. For the non quasi-equilibrium process, the piston is moved rapidly in the downward direction for step. For this type of movement, the internal pressure in the cylinder does not have sufficient time to adjust its pressure over the entire chamber. The pressure at the top of the piston is then different than at its bottom, and there is not a uniform pressure to use in the ideal gas equation. A noticeable time interval is required for the system to establish equilibrium conditions. After the new equilibrium conditions is established the piston is then again moved rapidly in the downward direction. Let’s consider a piston that is pressed downward slowly as shown in Fig. 2.16b. The system now has time to adjust the pressure throughout the chamber at each step of the process. The internal pressure is effectively the same throughout the entire system during the compression. The time interval required to establish new equilibrium conditions is negligible. Because of problems in defining “pressure” (and other characteristics) for rapid moving piston, our definition of expansion work is only valid for quasi-equilibrium processes. The solution for quasi-equilibrium processes can be obtained by integrated Eq. 2.4.2 (which assumes equilibrium at each infinitesimal volume).
Last Modified: March, 2018
2-17
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.16a. Non Quasi-equilibrium Process.
Figure 2.16b. Quasi-equilibrium Process. Gases generally respond quickly to changes in the boundaries of their system. As shown in Chapter 1, the root-mean-square speed of common gas molecules is of the magnitude of 400 m s-1 (~ 1000 mph) at room temperatures. The impact of changes in boundary location can then be quickly converted into changes in thermodynamic properties such as pressure, temperature and internal energy. Another feature of equilibrium condition is that there is no work acting on the system at equilibrium. Therefore for quasi-equilibrium condition, the pressure force of the gas inside the piston is essentially equal to the external pressure force of the surroundings on the system. Use of Ideal Gas Equation General Formulation
Last Modified: March, 2018
2-18
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Assumption of quasi-equilibrium process allows us to use the ideal gas equation to represent pressure within the system. For an ideal gas, P = nRT/V, we then have the following relationship for expansion work (for constant number of moles): Vf
w=− � � Vi
Isothermal Expansion
Vf 𝑛𝑛𝑛𝑛𝑛𝑛 𝑇𝑇 � 𝑑𝑑𝑑𝑑 = − n R � � � 𝑑𝑑𝑑𝑑 𝑉𝑉 𝑉𝑉 Vi
2.4.3
Isothermal expansion refers to expansion work that occurs while maintaining a constant gas temperature in the system. This can be done by controlling the heat transfer to closely balance the work done. In practice, it corresponds to gases that are expanding slowly. By using expansion work with the ideal gas equation, isothermal expansion is defined for an ideal gas as Vf
w=− � � Vi
Vf 𝑉𝑉𝑓𝑓 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 � 𝑑𝑑𝑑𝑑 = −nRT � = −𝑛𝑛𝑛𝑛𝑛𝑛 ln � � 𝑉𝑉 𝑉𝑉𝑖𝑖 Vi 𝑉𝑉
2.4.4
An isotherm is often used to indicate a line on a graph of constant temperature. Isothermal expansion work corresponds to the area under this curve. Example Problem 2.2: Application of 1st Law for Isothermal Conditions Problem Statement: Consider 450 g of air that is compressed slowly in a piston-cylinder assembly from 100 kPa at 26 C to final pressure of 620 kPa. Molar mass of air is approximately 29 g mol-1 (as shown in Chapter 3). During the compression, the heat transfer to surroundings is at rate so that the process is isothermal. We will further assume a quasiequilibrium process for an ideal gas. There are no phase transitions or chemical reactions within our system. The process is shown in the following schematic.
P Pi = 100 kPa Ti= 26 C
Pi, Ti, Vi
Pf= 620 kPa Tf = 26 C
Air - 450 g mm = 29 g mol-1
Pf, Tf, Vf
620 kPa
Pf
100 kPa
Pi
q
V Vf
Vi
Determine:
Last Modified: March, 2018
2-19
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
(1) Change in the internal energy of the gas, (2) Compression work done, and (3) Quantity of heat transferred. Solution: Change in internal energy. Let’s consider the isothermal process for an ideal gas. In Chapter 1, we have previously consider the internal energy as Um = a R T
where a = 3/2, a=5/2, and a=3 for monatomic, diatomic, and polyatomic gases, respectively. As discussed in Chapter 1, the internal energy of ideal gas is only a function of temperature. Since the initial and final temperatures of the gas are the same, and there is no phase transitions or chemical reactions, we will conclude that internal energy is constant, that is, ΔU = Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + Δ𝑈𝑈𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 + Δ𝑈𝑈𝑟𝑟 = Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 = 0
Solution: Compression work. For isothermal processes, we know that Vf w = −n RT ln � � Vi
For this problem, we have R = 8.314 J K-1 mol-1 T = 26 + 273.15 = 299.15 K m 450 g n= = = 15.51 mol mm 29 g mol−1 Vf Pi Pf Vf = nRT = Pi Vi → = Vi Pf
and therefore we can compute the work done as Pi 100 𝑘𝑘𝑘𝑘𝑘𝑘 w = −nRT ln � � = −(15.51 mol)(8.314 JK −1 mol−1 )(299.15 K) ln � � Pf 620 𝑘𝑘𝑘𝑘𝑘𝑘
w = 70416 J = 70.416 kJ
The work is positive because it is the surroundings that does work on the system to compress the gas. Solution: Heat transfer. Heat transfer can be determined from the First Law of thermodynamics. For no change in internal energy, we have ΔU = w + q = 0
and therefore
q = - w = -70.426 kJ
Last Modified: March, 2018
2-20
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
that is, 70.426 kJ is need to be transferred from the system to the surroundings to have isothermal compression.
INTERNAL ENERGY AND TEMPERATURE Background In Chapter 1, we used the kinetic theory of gas to derive relationships between the internal energy to the gas’s temperature. These relationships were developed to provide insight into the ideal gas equation and internal energy. They are too imprecise for design purposes. From this analysis, we conclude that internal energy is dependent on: • Characteristics of the substance, and • Temperature of the substance To better understand the observed internal energies of solids, Einstein proposed that energies of the oscillators were limited to discrete energy levels as required by quantum mechanics. Theoretical internal energies are then in good agreement with those observed when “Einstein” temperature is properly selected. Details of Einstein’s approach are given in Chapter 7. Instead of relying solely on theoretical models, engineers have long used experiments to study and to define parameters for different substances to quantify changes in the internal energies related to temperature change, phase change and chemical reactions. These parameters are usually summarized in tables that engineers use in their analysis and design. The important parameters for this course are given in the downloadable Thermodynamic Data Handout. Heat transfer is a relatively easy to measure experimentally. Work can also be defined by experimentally limiting it to expansion work under conditions of constant pressure (w=PΔV) or of constant volume (no expansion work). The measurable heat transfer is then related changes in internal energy from the First Law as q = ΔU − w = ΔUtpr + ΔUphase + ΔUr + PΔV
2.5.1)
The focus in this section is the development and use of relationships to relate changes in internal energy for a system without phase transitions and chemical reactions (i.e. ΔUphase = ΔUr = 0). The measured heat transfer is then used to define parameters tied to changes with temperature differences, and the subscript “tpr” is used. In subsequent sections, parameters for phase change are obtained from Eq. 2.5.1 where experimentally ΔUtpr = ΔUr = 0 and for chemical reaction where ΔUtpr=ΔUphase = 0. Some of the relationships developed in this section are general, that is, they are applicable to all three types of changes in internal energy. Heat Capacities General Approach
Last Modified: March, 2018
2-21
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Heat capacity is defined as the amount of heat transfer necessary to raise the temperature of a system by one degree for a system with no phase transitions or chemical reactions. We can therefore use the following general definition of heat capacity C=
q ΔT
C =
Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑤𝑤 Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑃𝑃� Δ𝑉𝑉 = = ΔT ΔT ΔT
2.5.2a
As previously discussed, the heat capacity is related to the internal energy using the First Law with only expansion work, that is, w = -∫PdV. For no phase transitions or chemical reactions, the heat capacity can then be evaluated as 2.5.2b
where 𝑃𝑃� is the average pressure for a change in volume. In this section, we are interested in measuring C under conditions of either constant volume or constant pressure. We are interested in (1) how the heat capacity can be determined experimentally and (2) how they can be used to solve engineering problems. Specific heat capacity (Cs = C/m) is defined as the heat capacity per unit mass. Molar heat capacity (Cm = C/n) is defined as the heat capacity per mole of substance. Heat Capacity at Constant Volume To illustrate the computations of heat capacity at constant volume, we will consider the hypothetical experiment shown below for a system of fixed (constant) volume (no phase transitions or chemical reactions). Here the heat transfer of 400 J is measured to change the temperature by 4 K for 4 moles of a gas. The heat capacity at constant volume is then determined by dividing the heat transfer by the change in temperature, that is, Cv = 100 J K-1. We will use the notation of Cvs for specific heat at constant volume (Cvs = Cv/m) and Cvm for molar heat capacity at constant volume (Cvm = Cv/n). Therefore, the molar heat capacity at constant volume for our hypothetical experiment is obtained by simply dividing the experimental determined Cv by the number of moles, or Cvm = 25 J K-1mol-1. This example illustrate how heat capacity can be determined experimentally using a constant volume system.
Last Modified: March, 2018
2-22
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Fasteners to Prevent Expansion Vf
w = − ∫ PdV = 0 Vi
Cv =
n = 4 moles ∆T= 4 K
q = 400 J
q 400 J = = 100 J K −1 ∆T 4K
Cvm =
Cv 100 J K −1 = = 25 J K −1mol −1 4 mol n
Constant Volume: ∆V=0
Figure 2.17. Illustration of Heat Capacity at Constant Volume To use the experimentally-derived Cv to solving engineering problems, let’s evaluate the heat transfer using the First Law under conditions of constant volume (i.e., V=0). The heat capacity can then be related to the change in the internal energy as Cv =
Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑃𝑃� Δ𝑉𝑉 Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 = ΔT ΔT
2.5.3
where Cv is now defined using a fundamental property of the system (internal energy) that is useful in solving thermodynamic problems. If we consider the limit as ΔT → 0, we obtain Cv = �
∂U � ∂T 𝑉𝑉
2.5.4
where internal energy for our system is tied only to temperature for ideal gases and incompressible liquids and solids. For these conditions, the change in internal energy for our system can be determined directly as Tf
𝑇𝑇𝑓𝑓
Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 = � 𝐶𝐶𝑣𝑣 𝑑𝑑𝑑𝑑 = � 𝑛𝑛 𝐶𝐶𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 ≈ 𝑛𝑛 𝐶𝐶𝑝𝑝𝑝𝑝 Δ𝑇𝑇 Ti
2.5.6
𝑇𝑇𝑖𝑖
for a system of fixed volume with no phase transitions or chemical reactions. For a constant Cv, internal energy is the defined as Utpr = Cv T, where the integration constant is taken as zero at T=0. The qualifier “at constant volume” describes the experimental method used to measure heat capacity. By using the kinetic theory of gases, crude theoretical relationships between internal energy and temperature were obtained using “3nR/2”, “5nR/2” or “3nR” terms. These terms were tied to fundamental characteristics of gases, and their values were not limited to a particular type of measurement system. We are able to improve the determination of ΔUtpr by replacing our crude kinetic-theory terms with the experimentally derived Cv. However, it is still useful to view Cv as a fundamental characteristic of substances that is not limited to the type of systems used to measure it. Experimentally-derived values of Cv are generally used to compute the change in internal energy for systems of both constant and
Last Modified: March, 2018
2-23
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
varying volumes. As discussed later, this use is valid for ideal gases and incompressible liquids and solids. Heat Capacity at Constant Pressure Thermodynamic systems often have one or more expanding (or contracting) boundaries surrounded by ambient air (open vessels or parcels of air). Ambient air pressure is nearly constant and equal to the atmospheric air pressure of approximately 100 kPa. For these systems, expansion work is done under conditions of constant pressure. Illustrations of three important processes are shown in Fig. 2.18. For each of these processes, it is more convenient to evaluate their thermodynamic properties using conditions of constant pressure. We will start our discussion by considering heat capacity at constant pressure. Possible q Increased Vapor
Gas Product
T Initial
System
Gases
Initial
Initial
Evaporation
Chemical Reaction
Figure 2.18. Systems under Ambient Air Conditions. To illustrate the experimental approach to determine the heat capacity at constant pressure, we will re-do our hypothetical experiment of the previous section, but now we will allow the system to expand with the change in temperature. An illustration of the revised experimental approach is shown in Fig. 2.19 where a heat transfer of 533 J is needed to change the temperature of the same gas by 4 K. An additional heat transfer of 133 J is needed for our expanding system because of the work done by the system on the surroundings. Heat capacity at constant pressure is determined by dividing the heat transfer by change in temperature, that is, Cp = 133.25 J K-1. The specific heat at constant pressure is Cps = Cp/m and the molar heat capacity at constant pressure is Cpm = Cp/n, which for our example is Cpm=33.31 J K-1 mol-1. The experimentally derived Cpm can now be placed in a table or other formats and used to solve engineering problems.
Last Modified: March, 2018
2-24
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.19. Illustration of Heat Capacity at Constant Pressure. To apply the experimentally-derived Cp to thermodynamic, the heat transfer will be evaluated using the First Law under conditions of constant pressure. We then obtain Cp =
Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑃𝑃 Δ𝑉𝑉 Δ𝐻𝐻𝑡𝑡𝑡𝑡𝑡𝑡 = ΔT Δ𝑇𝑇
2.5.7
where Cp is the heat capacity at constant pressure. Heat capacity here is now defined using Htpr, which is another property (not as fundamental as internal energy) of the system. This property is called enthalpy. The subscript of “tpr” (for temperature) has been used to indicate this definition doesn’t include possible changes in enthalpy with the number of moles from phase transitions or chemical reactions. Enthalpy is defined in general as 𝐻𝐻 = 𝑈𝑈 + 𝑃𝑃𝑃𝑃
2.5.8
More on enthalpy is given in the next subsection. For the limit as ΔT → 0, the heat capacity at constant pressure is defined as ∂H Cp = � � ∂T 𝑃𝑃
2.5.9
where the partial derivative is used because enthalpy is, in general, also a function of pressure. As discussed in greater detail later, enthalpy for our system is tied only to temperature for ideal gases and incompressible liquids and solids. By using dHtpr = Cp dT for ideal gases, liquids and solids, we obtain the change in enthalpy as 𝑇𝑇𝑓𝑓
ΔHtpr = � 𝐶𝐶𝑝𝑝 𝑑𝑑𝑑𝑑
2.5.10a
Ti
You can obtain Cpm for different substances from the Thermodynamic Data Handout available at the BBE 3043 website. These values are for a temperature of 25 C and a standard pressure of 100 kPa. We will often assume that Cpm is constant over the range of temperatures in our problems. Changes in temperature enthalpy are then obtained.as
Last Modified: March, 2018
2-25
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑇𝑇𝑓𝑓
ΔHtpr = � 𝑛𝑛 𝐶𝐶𝑝𝑝 𝑑𝑑𝑑𝑑 ≈ 𝑛𝑛 𝐶𝐶𝑝𝑝𝑝𝑝 Δ𝑇𝑇
2.5.10b
Ti
where Cpm values is obtained from the Thermodynamic Data Handout. Additional discussion on the use of Eqs. 2.6.10 is discussed later in this section. Relationship between Cvm and Cpm Let’s consider the relationship between Cvm and Cpm for ideal gases. We will use the following useful form of Cp evaluated for conditions of constant pressure with a fixed mass (n=constant). We then obtain: dHtpr = Cp 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑃𝑃𝑃𝑃𝑃𝑃 = 𝐶𝐶𝑣𝑣 𝑑𝑑𝑑𝑑 + 𝑃𝑃𝑃𝑃𝑃𝑃
2.5.12
Cp 𝑑𝑑𝑑𝑑 = 𝐶𝐶𝑣𝑣 𝑑𝑑𝑑𝑑 + 𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑
2.5.13
where we have also used Cv = dUtpr/dT (valid for ideal gases for constant and varying volumes as discussed later). For an ideal gas (PV = nRT) undergoing changes for constant pressure with a fixed mass (n=constant), we have d(PV) = PdV = nRdT. We can then simplify as
By cancelling dT and dividing by n, we obtain the following relationship between the molar heat capacities at constant pressure and constant volume as Cpm = 𝐶𝐶𝑣𝑣𝑣𝑣 + 𝑅𝑅
2.5.14a
Cvm = 𝐶𝐶𝑝𝑝𝑝𝑝 − 𝑅𝑅
2.5.14b
or, alternatively, Cvm can be computed from Cpm values in the Thermodynamic Data Handout as
which is valid for ideal gases. Heat capacity is generally tabulated using Cpm instead of Cvm. Not only is enthalpy often measured directly using ambient air conditions, it is also a convenient property for computing heat transfer (or change in temperature for a given q) for the common condition of constant pressure. Heat transfer for this condition is simply equal to the change in enthalpy. An alternative approach to obtain heat transfer from the First Law by computing separately (1) the change in internal energy using Cvm obtained from Eq. 2.5.14b and (2) the expansion work under constant pressure. The heat transfer by this approach is the same as that obtained using enthalpy, but the one-step enthalpy-based approach is simpler. Incompressible Substance Many liquids and solids can be evaluated as incompressible substances. For incompressible substances, the internal energy is only a function of temperature. We conclude that the heat capacity at constant volume is defined as
Last Modified: March, 2018
2-26
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Cv =
dUtpr dT
2.5.15a
Furthermore, the heat capacities at constant volume and pressure are equal for an incompressible substance, that is, Cv = Cp
2.5.15b
Variation with Temperature Heat capacities for gases vary with temperature (for example, see the internal energy relationship of hydrogen gas given in Chapter 1 by Fig. 1.25). To represent this variation, the heat capacities are often defined using power functions with temperature. For example, the molar heat capacity at constant pressure is sometimes represented by the following function: Cpm = a + b T + c T 2 + d T 3
2.5.16
where the coefficients a, b, c, and d derived from data analysis. Coefficients for a few gases are shown in Table 2.1 (from Sandler, 2006; NIST, 2017). Table 2.1 Coefficients for Temperature Dependent Molar Heat Capacities.
Type
Gas
Monatomic
He H2 N2 O2 H2O CH4 CO2
Diatomic
Polyatomic
a
b
c
d
J K-1 mol-1 20.786 29.088 28.883 25.460 32.218 19.875 22.243
J K-2 mol-1 4.851 x 10-13 -0.192 x 10-2 -0.157 x 10-2 1.519 x 10-2 0.192 x 10-2 5.021 x 10-2 5.977 x 10-2
J K-3 mol-1 -1.583 x 10-16 0.400 x 10-5 0.808 x 10-5 -0.715 x 10-5 1.055 x 10-5 1.268 x 10-5 -3.499 x 10-5
J K-4 mol-1 1.525 x 10-20 -0.870 x 10-9 -2.871 x 10-9 1.311 x 10-9 -3.593 x 10-9 -11.004 x 10-9 7.464 x 10-9
The error obtained by using a constant Cpm, instead of Eq. 2.5.16, is explored in Homework Problems 2.11 and 2.12. This error is acceptable for many engineering designs. The vast majority of the problems in this course is solved by using a constant Cpm. Eq. 2.5.9 is used only for a large difference in final and initial temperatures or if a very precise solution is needed. Enthalpy Mathematical Features We now wish to explore more fully the characteristics and definition of enthalpy. Let’s first investigate its differential characteristics. Enthalpy is defined by exact differentials. Since enthalpy is defined as H = U + PV, we obtain dH = dU + d(PV) = dU + PdV + VdP
Last Modified: March, 2018
2-27
2.5.17
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
The change in enthalpy between initial and final states can be written as ΔH = ∫ dU + ∫ PdV + ∫ VdP
2.5.18
For a system under constant pressure, dP = 0. We then obtain
and
dH = dU + PdV
2.5.19a
ΔH = ΔU + ∫ PdV
2.5.19b
Since dH is an exact differential, the change in enthalpy is only dependent on the difference in H between the final and initial states. This has important implications on possible solution approaches and is discussed in greater detail later. Enthalpy is also an extensive property. The enthalpy of the system can then be determined as the sum of the enthalpy of its components. Molar enthalpy is defined as enthalpy per mole that is Hm = H/n. Specific enthalpy is defined as enthalpy per mass that is Hs = H/m. More on Cp (exclude from student handout) Similar to Cv, heat capacity at constant pressure is an empirically derived characteristic of the substance, where the qualifier “at constant pressure” describes the experimental method to obtain this characteristics. For an ideal gas, it can be used to compute the temperature enthalpy for systems of both constant and varying pressure. The validity for varying pressure may seem odd because the temperature enthalpy using Cp is defined for conditions of constant pressure (see Eq. 2.5.7); whereas the general definition dH includes a VdP term that is not zero for a system with changing pressure. Let’s consider the enthalpy of an ideal gas defined as Htpr = Utpr + P V = Cv T + n R T
2.5.20
where Utpr = n CvmT and PV = nRT for an ideal gas. We conclude that enthalpy is not a function of pressure for ideal gases. Additional insight can be obtained from definition of differential change for enthalpy. By following the logic given in Chapter 1 for differential change in internal energy, we can select two of the following three properties of temperature, volume, and pressure to represent the properties of the system (using an equation of state). For enthalpy, temperature and pressure are usually selected. We then have the following general function for enthalpy: H = H(T, P)
2.5.21
where T is temperature and P is pressure. By using the definition of total differential, we then obtain
Last Modified: March, 2018
2-28
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
dH = �
∂H ∂H ∂H � dT + � � dP = Cp dT + � � dP ∂T P ∂P T ∂P T
2.5.22
where the definition of heat capacity at constant pressure has been used. For an ideal gas, we have shown that Htpr is only a function of temperature (Eq. 2.5.20), that is, (∂Htpr/∂P)T = 0. A more general representation of (∂Htpr/∂P)T is given in Chapter 5. The differential change in temperature enthalpy for constant pressure (dP=0) systems or for ideal gases (in constant or varying pressure system) is obtained as (dH = dHtpr) dHtpr = Cp dT
2.5.23
Application of Heat Capacities to Engineering Problems A summary of the typical formulation of engineering design problems based on heat capacities is given in Figure 2.20. Here we have an ideal gas with no phase transitions or phase changes so the number of moles is constant between the initial and final equilibrium states. We will only consider expansion work. In general, the initial pressure, temperature, and volume are known (one of them is often computed using the ideal gas equation). Possible unknowns are ΔUtpr, q and w between the two equilibrium states. In addition, the engineer may be interested in the final pressure, temperature and volume of the system. If these are unknown, then the engineers has the potential of 6 unknown characteristics.
Figure 2.20. Typical Formulation of Engineering Design Problems. Let’s now consider the number of relationships that we have available to us. We need six relationships to solve for six unknowns. One of our relationships is the First Law. For an ideal gas, we have the ideal-gas equation and the change in internal energy with temperature using Cvm. A fourth relationship is obtained using definition of expansion work. However, determining expansion work requires a relationship between pressure and volume or between temperature and volume that effectively increases our unknowns to seven! We can obtain an additional equation if the process is adiabatic (q= 0), isothermal (P=nRTi/V), isobaric
Last Modified: March, 2018
2-29
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
(P=constant), or isochoric (V=constant). If one of these processes is applicable, we have five equations for the seven unknowns. The engineer needs to reduce the unknowns to five (1) by using relationships obtained from other courses (such as a heat transfer course), (2) by measuring one or more of the final state variables (Tf, Pf, Vf) and/or q, w or ΔUtpr or (3) by setting the final state variables based on design specifications. Let’s now simplify the discussion for solving problems in this course. We are primarily interested in expansion work. Thermodynamic processes for many biological and environmental systems occur under constant pressure (isobaric process). We are then able to solve for q for a ΔT or for ΔT for a given q as q = ΔHtpr ≈ n Cpm Δ𝑇𝑇 q ΔT = n Cpm
2.5.30a 2.5.30b
Once again, these relationships are only valid for constant pressure systems. For example, q from Eq. 2.5.30a is zero for isothermal systems (ΔT=0), but we have previously shown in Example Problem 2.2 that it is actually q = -w. If we are interested in the change in internal energy for an ideal gas (where PdV = nRdT), we can use the definition of enthalpy to obtain ΔUtpr = ΔHtpr − PΔV = n Cpm Δ𝑇𝑇 − 𝑛𝑛𝑛𝑛Δ𝑇𝑇 = 𝑛𝑛�𝐶𝐶𝑝𝑝𝑝𝑝 − 𝑅𝑅�Δ𝑇𝑇
2.5.30c
q = ΔUtpr ≈ n Cvm Δ𝑇𝑇 q ΔT = n Cvm
2.5.31a
We will occasionally be interested in thermodynamic processes in rigid containers. Rigid containers correspond to systems of constant volume (isochoric process). The heat transfer and change in temperature for rigid containers (for ΔV= 0) are obtained as
2.5.31b
Clearly, we need an estimate of Cpm to determine the heat transfer for some (many) systems and Cvm for others (e.g. rigid containers). Values for Cpm for different substances are given in the Thermodynamic Data Handout. But what about values for Cvm? These values are frequently not tabulated. Solids and liquids can often be represented as incompressible substances for which the heat capacities are equal. As previously shown for ideal gases, we can estimate Cvm directly from values in the Thermodynamic Data Handout as 𝐶𝐶𝑣𝑣𝑣𝑣 = 𝐶𝐶𝑝𝑝𝑝𝑝 − 𝑅𝑅.
Example Problem 2.3: Application of 1st Law for Isobaric Conditions
Problem Statement. A vertical piston-cylinder assembly with an initial volume of 32 L is filled with 50 g of nitrogen gas. A piston is weighted so that it maintains a constant pressure on the system of 140 kPa. We will compress the gas in the cylinder until the final volume is 90% of the initial value. Estimate the (1) final temperature of nitrogen, (2) amount of energy transferred by heating, (3) the amount of energy transferred by work and (4) the change in internal energy. Assume a constant molar heat capacity and a quasi-equilibrium process of an
Last Modified: March, 2018
2-30
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
ideal gas. At least two different methods are available to compute the some of the unknowns, and they are given to show that you obtain the same value with the different methods.
P = 140 kPa
P
Ti
Tf? Vi = 32 L w?
System Boundaries
P = 140 kPa
∆U? m = 50 g
Vf = 0.9Vi
q?
Vf
Vi
V
Tabular values. From the Data Handout, we obtain R = 8.314 J K-1 mol-1 = 8.314 Pa m3 K-1 mol-1 mm = 28 g mol-1 Cpm = 29.125 J K-1 mol-1 and therefore the number of moles is n=
m 50 g = = 1.785 mol mm 28 g mol−1
Ti =
P Vi nR
Solution: Temperature at Final State. For a quasi-equilibrium process, the system is near equilibrium conditions at each step of the path. Since at equilibrium there is no work (or heat transfer) on the system, we conclude that the piston pressure equals the external pressure, that is, 140 kPa. We can then estimate the initial temperature of the gas from the ideal gas equation, that is,
Ti =
Pa m3 � (32 L) �1000 L� kPa = 301.76 K (28.6 C) 1.785 mol(8.314 Pa m3 mol−1 K −1 )
140 kPa �1000
The final temperature can be obtained from the ideal gas equation, that is, Pa m3 (0.9)(32 � L) � P Vf 1000 L� kPa Tf = = = 271.6 K nR 1.785 mol(8.314 Pa m3 mol−1 K −1 ) Δ𝑇𝑇 = Tf − Ti = 271.6 K − 301.76 K = −30.16 K 140 kPa �1000
An alternative (and more fun) method for obtaining Tf and ΔT is to use relationships that can be derived from the condition of constant pressure. For a constant pressure system we have
Last Modified: March, 2018
2-31
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
P =
n R Ti n R Tf Vf = → Tf = Ti � � Vi Vf Vi
and the change in temperature and final temperature are defined as 𝑉𝑉𝑓𝑓 Ti P − 1� = � � (Vf − Vi ) = ( ) (Vf − Vi ) 𝑉𝑉𝑖𝑖 Vi nR 0.9Vi ΔT = 301.76 K � − 1� = 301.76 K (−0.1) = −30.16 K Vi Tf = 𝑇𝑇𝑖𝑖 + Δ𝑇𝑇 = 301.76 − 30.17 = 271.6 𝐾𝐾 Δ𝑇𝑇 = 𝑇𝑇𝑓𝑓 − 𝑇𝑇𝑖𝑖 = 𝑇𝑇𝑖𝑖 �
Note that the change in temperature is independent of the characteristics of N2. The same change in temperature would be obtained for any ideal gas (ΔT = PΔV/nR for constant pressure). We didn’t need to compute the final and initial temperatures to evaluate the internal energy change, work and heat transfer, but they were included in our analysis to provide insight on the compression process. Solution: Heat transfer. From the First Law of thermodynamics with expansion work under constant pressure, we have 𝑞𝑞 = Δ𝑈𝑈 − 𝑤𝑤 = Δ𝑈𝑈 + 𝑃𝑃 Δ𝑉𝑉 = Δ𝐻𝐻
Using the definition of heat capacity at constant pressure we obtain 𝐻𝐻𝑓𝑓
𝑇𝑇𝑓𝑓
𝑞𝑞 = � 𝑑𝑑𝐻𝐻𝑡𝑡𝑡𝑡𝑡𝑡 = � 𝑛𝑛 𝐶𝐶𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 ≈ 𝑛𝑛 𝐶𝐶𝑝𝑝𝑝𝑝 �𝑇𝑇𝑓𝑓 − 𝑇𝑇𝑖𝑖 � = 𝑛𝑛 𝐶𝐶𝑝𝑝𝑝𝑝 Δ𝑇𝑇 𝐻𝐻𝑖𝑖
𝑇𝑇𝑖𝑖
𝐽𝐽 𝑞𝑞 ≈ (1.785 𝑚𝑚𝑚𝑚𝑚𝑚) 29.125 � � (−30.17 𝐾𝐾) = −1569 𝐽𝐽 = −1.569 𝑘𝑘𝑘𝑘 𝐾𝐾 𝑚𝑚𝑚𝑚𝑚𝑚
Therefore 1.57 kJ of energy is transferred to the surroundings by heating.
Solution: Work done. Since we have a constant pressure system, we can simply use the change in volume to compute work (1 kJ = 1 kPa m3), that is, kJ m−3 m3 w = −PΔV = −( 140 kPa � � ( 0.9 (32 L) − 32 L) � � = 0.448 kJ kPa 1000 L
As an alternative method for computing work, let’s use derivative of PV for the ideal gas equation for a constant pressure, that is, 𝑑𝑑(𝑃𝑃𝑃𝑃) = 𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛
w = −PΔV = −nRΔT = −1.785 mol �8.314 w = 448 J = 0.448 kJ
J � (271.58 − 301.75) K K mol
which is identical to our previous value. Expansion work is only a function of the number of moles, R and ΔT. If we replace N2 in our cylinder with the same number of moles of O2, He,
Last Modified: March, 2018
2-32
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
H2, CO2 or any other gas that behaves as an ideal gas, the expansion work is identical (heat transfer and internal energy would change). Solution: Changes in Internal Energy. From the First Law, we have ΔU = q + w = −1.560 kJ + 0.448 kJ = −1.121 kJ
As an alternative method, let’s compute the change in internal energy using the heat capacity at constant volume. For an ideal gas, we have J J Cvm = Cpm − R = 29.125 � � − 8.314 � � = 20.811 J K −1 mol−1 K mol K mol
Using the definition of heat capacity at constant volume we obtain 𝑈𝑈𝑓𝑓
𝑇𝑇𝑓𝑓
Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 = � 𝑑𝑑𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 = � 𝑛𝑛 𝐶𝐶𝑣𝑣𝑣𝑣 𝑑𝑑𝑑𝑑 ≈ 𝑛𝑛 𝐶𝐶𝑣𝑣𝑣𝑣 �𝑇𝑇𝑓𝑓 − 𝑇𝑇𝑖𝑖 � = 𝑛𝑛 𝐶𝐶𝑣𝑣𝑣𝑣 Δ𝑇𝑇 𝑈𝑈𝑖𝑖
𝑇𝑇𝑖𝑖
𝐽𝐽 𝑞𝑞 ≈ (1.785 𝑚𝑚𝑚𝑚𝑚𝑚) 20.811 � � (−30.17 𝐾𝐾) = −1121 𝐽𝐽 = −1.121 𝑘𝑘𝑘𝑘 𝐾𝐾 𝑚𝑚𝑚𝑚𝑚𝑚
Even though the volume of the system is not constant, the change in the internal energy using the molar heat capacity at constant volume is equal to that obtained from the First Law. More on this result is given later in the chapter. Summary. The components of the First Law of thermodynamics are shown in the following schematic. The consequence of compression work is to increase the internal energy by 0.45 kJ. However to maintain a constant pressure of 140 kPa, 1.57 kJ of heat transfer is necessary from the system to the surroundings. Therefore the net change in the internal energy is a decrease of 1.12 kJ.
w 0.45 kJ Ui q -1.57 kJ
∆U =q+w -1.12 kJ
Example Problem 2.4: Application of 1st Law for Varying P, T and V. Problem Definition and Solution Results Problem Statement. Let’s consider a vertical piston-cylinder assembly designed so that the 500 g of oxygen can be compressed using a non-constant piston pressure. For this assembly, we are interested in thermodynamic characteristics for the two different paths shown below
Last Modified: March, 2018
2-33
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
with the same initial states of Pi = 100 kPa and Vi = 0.3 m3 and final states of Pf = 400 kPa and Vf = 0.1 m3. Isotherms corresponding to temperatures of 231 K, 308 K and 400 K are also shown in the figure. We will assume a quasi-equilibrium process and ideal gas characteristics. For steps corresponding to an incremental change in volume of 0.02 m3, we will compute for each path: (1) temperature, (2) work done, (3) change in internal energy, (3) heat transfer, (4) temperature enthalpy (not a function of pressure for ideal gases), and (5) quantity of heat transfer divided by temperature. We will compare the results between Paths #1 and #2 to gain insight on which characteristics are represented by exact differentials. 700
Path #1:
400 K
xo = 0.1 m3, a = 400 kPa b= -7500 kPa m-3
600
Pressure (kPa)
308 K 500
Vf Pf
400
P = a + b ( V − xo ) 2
231 K 300 200
Path #2:
100 0
Pi
xo = 0.3 m3, a = 100 kPa b= 7500 kPa m-3 0.08
0.1
0.12
0.14
Vi 0.16
0.18 0.2 0.22 Volume (m3)
0.24
0.26
0.28
0.3
0.32
Thermodynamic parameters. From the Data Handout, we obtain R = 8.314 J K-1 mol-1 = 8.314 Pa m3 K-1 mol-1 mm = 32 g mol-1 Cpm = 29.355 J K-1 mol-1 The number of moles are defined as n=
m 500 g = = 15.625 mol mm 32 g mol−1
For a more rigorous analysis, we will use a temperature-dependent molar heat capacities defined as 𝐶𝐶𝑝𝑝𝑝𝑝 = 𝑎𝑎 + 𝑏𝑏 𝑇𝑇 + 𝑐𝑐 𝑇𝑇 2 + 𝑑𝑑 𝑇𝑇 3 𝐶𝐶𝑣𝑣𝑣𝑣 = 𝐶𝐶𝑝𝑝𝑝𝑝 − 𝑅𝑅 = (𝑎𝑎 − 𝑅𝑅) + 𝑏𝑏 𝑇𝑇 + 𝑐𝑐 𝑇𝑇 2 + 𝑑𝑑 𝑇𝑇 3
where, for oxygen, we have previously given in Table 2.1 values of a, b, c and d of 25.46, 0.01519, -7.153 x 10-6, and 1.31 x 10-9, respectively. Units for these coefficients are defined
Last Modified: March, 2018
2-34
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
corresponding to those necessary to obtain molar heat capacity in units of J K-1 mol-1. To convert to units of kJ K-1, we will multiply the above relationships by the number of moles and use 1000 J = 1 kJ (conversion factor of 0.015625 mol kJ J-1) to obtain 𝐶𝐶𝑝𝑝 = 0.3978 + (2.373𝑥𝑥10−4 ) 𝑇𝑇 − (1.1176𝑥𝑥10−7 )𝑇𝑇 2 + (2.048𝑥𝑥10−11 ) 𝑇𝑇 3 𝐶𝐶𝑣𝑣 = 0.2679 + (2.373𝑥𝑥10−4 ) 𝑇𝑇 − (1.1176𝑥𝑥10−7 )𝑇𝑇 2 + (2.048𝑥𝑥10−11 ) 𝑇𝑇 3
Integration Relationships. The change is our variables of interest is obtained for compression steps in our cylinder of ΔV=-0.02 m3. These changes are determined by direct integration of appropriate relationships. Subscripts “j” and “j+1” refer to values at the beginning and end of the step, respectively. The subscript “tpr” for ΔU and ΔH has not been included to simplify the notation. Changes in internal energy and enthalpy are defined as: ΔUj,j+1 = �
Tj+1
Tj
ΔHj,j+1 = �
Tj+1
Tj
𝑇𝑇
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 𝑗𝑗+1 𝐶𝐶𝑣𝑣 𝑑𝑑𝑑𝑑 = � 0.2679 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � 2 3 4 𝑇𝑇 𝐶𝐶𝑝𝑝 𝑑𝑑𝑑𝑑 = � 0.3978 𝑇𝑇 +
2.373𝑥𝑥10 2
−4
𝑇𝑇 2 −
1.1176𝑥𝑥10 3
−7
𝑇𝑇 3 +
2.048𝑥𝑥10 4
−11
T4�
Work done is computed by the changing pressure over the volumetric change. Heat transfer is obtained using the First Law. Expansion work and heat transfer are then defined as wj,j+1 = − �
Vj+1
Vj
𝑃𝑃𝑃𝑃𝑃𝑃 = − �
Vj+1
Vj
= − � aΔV +
𝑞𝑞𝑗𝑗,𝑗𝑗+1 = ΔUj,j+1 − wj,j+1
(𝑎𝑎 + 𝑏𝑏(𝑉𝑉 − 𝑥𝑥𝑜𝑜 )2 )𝑑𝑑𝑑𝑑
b 3 3 ��𝑉𝑉𝑗𝑗+1 − 𝑥𝑥𝑜𝑜 � − �𝑉𝑉𝑗𝑗 − 𝑥𝑥𝑜𝑜 � �� 3
We are also interested in the heat transfer normalized by the temperature. In Chapter 4, under certain conditions, we use this ratio as a definition of entropy. The method given in this section is a rough approximation. More rigorous equations are given in the next section and in Chapter 4. The change in the ratio of heat transfer and temperature is ΔS = �
𝛿𝛿𝛿𝛿 q = � 𝑇𝑇 T
where 𝑇𝑇� is the average temperature corresponding to the heat transfer.
Path #1 Results. Let’s evaluate the characteristics of thermodynamic system for a change of volume from 0.3 m3 to 0.28 m3. For Path #1, we have a= 400 kPa, b = -7500 kPa m-3, and xo = 0.1 m3. 2
Pj+1 = a + b �Vj+1 − xo � = 400 − 7500(0.28 − 0.1)2 = 157 kPa
Last Modified: March, 2018
2-35
Biological and Environmental Thermodynamics
𝑗𝑗
𝑇𝑇𝑗𝑗+1 𝑇𝑇𝑗𝑗
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Pa 100 kPa �1000 � (0.3 𝑚𝑚3 ) P Vj kPa Tj = = = 230.9 K nR 15.625 mol(8.314 Pa m3 mol−1 K −1 ) Pa 157 kPa �1000 � (0.28 𝑚𝑚3 ) P Vj+1 kPa Tj+1 = = = 338.4 K nR 15.625 mol(8.314 Pa m3 mol−1 K −1 ) b 3 3 wj,j+1 = − � aΔV + ��𝑉𝑉𝑗𝑗+1 − 𝑥𝑥𝑜𝑜 � − �𝑉𝑉𝑗𝑗 − 𝑥𝑥𝑜𝑜 � �� 3 wj,j+1 = − � 400 (0.02) −
ΔUj,j+1
7500 ((0.28 − 0.1)3 − (0.3 − 0.1)3 )� 3
338.4
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 = � 0.2679 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � = 35.1 𝑘𝑘𝑘𝑘 2 3 4 230.9 338.4
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 ΔHj,j+1 = � 0.3978 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � = 49.1 𝑘𝑘𝑘𝑘 2 3 4 230.9 q j,j+1 = ΔUj,j+1 q j,j+1 32.5 𝑘𝑘𝑘𝑘 ΔSj,j+1 = = �𝑇𝑇𝑗𝑗+1 + 𝑇𝑇𝑗𝑗 �⁄2 (338.4 𝐾𝐾 + 230.9 𝐾𝐾)⁄2 A summary of the computation steps for Path #1 is shown below. Volume
P
T
w
ΔHtpr
ΔUtpr
q
ΔS
m3
kPa
K
kJ
kJ
kJ
kJ
kJ K-1
0.30
100
230.9
0.28
157
338.4
2.58
49.1
35.1
32.5
0.114
0.26
208
416.3
3.66
36.8
26.7
23.0
0.061
0.24
253
467.4
4.62
24.7
18.0
13.4
0.030
0.22
292
494.5
5.46
13.2
9.7
4.3
0.009
0.20
325
500.4
6.18
2.9
2.1
-4.1
-0.008
0.18
352
487.7
6.78
-6.2
-4.5
-11.3
-0.023
0.16
373
459.4
7.26
-13.8
-10.1
-17.4
-0.037
0.14
388
418.1
7.62
-19.9
-14.5
-22.2
-0.050
0.12
397
366.7
7.86
-24.4
-17.7
-25.6
-0.065
0.10
400
307.9
7.98 60
-27.4 34.9
-19.8 24.9
-27.7 -35.1
-0.082
Total
-0.051
For the initial steps, the relative pressure increases are greater than the decreases in volume (T increases because PV increases), but later in the compression, the relative volume decreases are greater than the pressure increases (T decreases because PV decreases). Summation totals are of greatest interest to us. Of course the total change in internal energy is equal to the sum of work and heat transfer. Most of the problems in BBE 3043 are for expansion work (only) under constant pressure. The heat transfer for these problems is equal to the temperature
Last Modified: March, 2018
2-36
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
enthalpy. Heat transfer is not equal to the temperature enthalpy for systems under varying pressure conditions. Path #2 Results. For Path #2, we have a= 100 kPa, b = 7500 kPa m-3, and xo = 0.3 m3. We can now repeat our calculations for this path. Examples of these calculations are shown below. 2
Pj+1 = a + b �Vj+1 − xo � = 100 + 7500(0.28 − 0.3)2 = 103 kPa Pa 103 kPa �1000 � (0.28 𝑚𝑚3 ) P Vj+1 kPa Tj+1 = = = 220.0 K nR 15.625 mol(8.314 Pa m3 mol−1 K −1 ) 7500 ((0.28 − 0.3)3 − (0.3 − 0.3)3 )� wj,j+1 = − � 100 (0.02) + 3
ΔUj,j+1
222
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 = � 0.2679 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � = −2.83 𝑘𝑘𝑘𝑘 2 3 4 230.9 222
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 ΔHj,j+1 = � 0.3978 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � = −4.0 𝑘𝑘𝑘𝑘 2 3 4 230.9 q j,j+1 = −2.83 q j,j+1 −4.85 𝑘𝑘𝑘𝑘 ΔSj,j+1 = = �𝑇𝑇𝑗𝑗+1 + 𝑇𝑇𝑗𝑗 �⁄2 (222 𝐾𝐾 + 230.9 𝐾𝐾)⁄2 The remaining steps are shown in the table below. Volume
P
T
w
ΔHtpr
ΔUtpr
q
ΔS
m3
kPa
K
kJ
kJ
kJ
kJ
kJ K-1
0.30
100
230.9
0.28
103
222.0
2.0
-4.0
-2.8
-4.8
-0.021
0.26
112
224.2
2.1
1.0
0.7
-1.5
-0.007
0.24
127
234.6
2.4
4.7
3.3
0.9
0.004
0.22
148
250.6
2.7
7.2
5.1
2.4
0.010
0.20
175
269.4
3.2
8.5
6.1
2.8
0.011
0.18
208
288.2
3.8
8.6
6.1
2.3
0.008
0.16
247
304.2
4.5
7.3
5.3
0.7
0.002
0.14
292
314.7
5.4
4.8
3.5
-1.9
-0.006
0.12
343
316.8
6.3
1.0
0.7
-5.6
-0.018
0.10
400
307.9
7.4
-4.1 34.9
-3.0 24.9
-10.4 -15.1
-0.033
Total
40.0
-0.050
The initial and final temperatures are identical to those obtained for Path #1 (which is of course necessary for the same pressures and volumes). For Path #2, the changes in PV are initially negative (T decreases) and are positive (T increases) later in the compression process. Although the work and heat transfer for Path #2 are smaller than that of Path #1, the total change in internal energy is again equal to their sum and is the same as Path #1.
Last Modified: March, 2018
2-37
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Exact and Inexact Differentials An important component of this problem is to examine which of the thermodynamic characteristics are represented by exact differentials and which by inexact differentials. Let’s first compute values using the endpoint states only. The initial state corresponds to Pi = 100 kPa, Vi = 0.3 m3 and Ti = 230.9 K and final states of Pf = 400 kPa, Vf = 0.1 m3, and Tf = 307.9 K. We can then compute ΔU and ΔH between initial and final states as ΔUtpr
ΔHtpr
307.9
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 = � 0.2679 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � = 24.9 𝑘𝑘𝑘𝑘 2 3 4 230.9 307.9
2.373𝑥𝑥10−4 2 1.1176𝑥𝑥10−7 3 2.048𝑥𝑥10−11 4 = � 0.3978 𝑇𝑇 + 𝑇𝑇 − 𝑇𝑇 + T � = 34.9 𝑘𝑘𝑘𝑘 2 3 4 230.9
We can also compute the heat transfer between initial and final states using q
= ΔUtpr − w
where work is defined as w = − � a(Vf − Vi ) +
b 3 ��𝑉𝑉𝑓𝑓 − 𝑥𝑥𝑜𝑜 � − (𝑉𝑉𝑖𝑖 − 𝑥𝑥𝑜𝑜 )3 �� 3
Work, however, is a function of initial and final volumes and on the coefficients of a, b, and xo. Since these coefficients are path dependent, work is also path dependent. Since ΔU is the same for both paths, heat transfer is different for the two paths as well. A summary of our calculations is shown below. Characteristic ΔU (kJ) ΔHtpr (kJ) ΔS (kJ K-1) w (kJ) q (kJ)
Path #1 24.9 34.9 -0.051 60.0 -35.1
Path #2 24.9 34.9 -0.050 40.0 -15.1
Endpoint Only 24.9 34.9 See next section Path dependent Path dependent
Based on these results, we conclude the work and heat transfer are defined by inexact differentials. Under certain conditions discussed in Chapter 4, ΔS is a change in entropy of the system. A more rigorous analysis of this change is given in the next subsection. The results of the example problem support the assertion that the internal energy, enthalpy and entropy are defined by exact differentials. As discussed briefly in Chapter 1, the Second Law requires that the change in entropy is greater than or equal to zero. You may then be concerned that the change in entropy for this problem is negative. Clearly we will need to be careful about entropy concepts discussed in detail in Chapter 4. Suffice it to say for now that the Second Law applies to the change in entropy of the Universe, and for this problem, we have only computed the change in entropy of our system.
Last Modified: March, 2018
2-38
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
More Analyses (exclude from student handout) If we had incorrectly used q = ΔHtpr for our varying pressure problem, heat transfer for Path 2 would have been 34.9 kJ; whereas the correct heat transfer from the First law is -15.1 kJ. The percent error obtained from the constant pressure solution is approximately 300%! Clearly heat transfer cannot be obtained by simply using the change in enthalpy for a system with varying pressure. Let’s revisit the general relationship for the change in enthalpy (i.e., dH = dU+PdV+VdP). By using the First Law for expansion work of q = ΔU + ∫PdV, we have ΔH = ΔU + ∫ PdV + ∫ VdP = q + ∫ VdP
The heat transfer for a varying pressure system of an ideal gas is therefore obtained as q = ΔHtpr − ∫ V dP
We will use this equation to compute heat transfer and compare these results to those previously given using the First law. The relationship between volume and pressure can be written as P−a P−a V = xo ± � = xo − � b b
where the negative sign is needed to have a reduction of volume for an increase in pressure. We then obtain �
Pj+1
Pj
𝑉𝑉 𝑑𝑑𝑑𝑑 = xo Δ𝑃𝑃 +
1
√𝑏𝑏
�
Pj+1
Pj
(𝑃𝑃 − 𝑎𝑎)1⁄2 𝑑𝑑𝑑𝑑 = xo Δ𝑃𝑃 −
2
3 √𝑏𝑏
� �Pj+1 − a�
3⁄2
− �Pj − a�
3⁄2
Let’s also consider more carefully our use of dHtpr = n Cpm dT to compute the change in enthalpy for varying pressure systems. This approach looks troublesome because of the nonzero VdP term for these systems (see Eq. 2.5.12). We have previously claimed that it is valid for ideal gases (see Eq. 2.5.23). We can evaluate our claim by comparing values computed for this problem (using dHtpr = n Cpm dT ) to those obtained directly from our definition of enthalpy, that is, ΔHj,j+1 = ΔUj,j+1 + �
Vj+1
Vj
𝑃𝑃 𝑑𝑑𝑑𝑑 + �
Pj+1
Pj
𝑉𝑉 𝑑𝑑𝑑𝑑 = ΔUj,j+1 − wj,j+1 + �
Pj+1
Pj
The results of this analysis for Path #2 are shown below:
𝑉𝑉 𝑑𝑑𝑑𝑑 ΔS
ΔHtpr
q
� 𝒒𝒒/𝑻𝑻
∫VdP
1st Law
dH-VdP
Cpm Values
H=U+PV
m 0.3
kJ
kJ
kJ
kJ
kJ
kJ K-
kJ K-1
0.28
0.86
-4.84
-4.84
-3.98
-3.98
-0.021
0.1167
0.26
2.42
-1.46
-1.46
0.96
0.96
-0.007
0.0585
Volume 3
Last Modified: March, 2018
2-39
∫δq/T
Biological and Environmental Thermodynamics
�
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
0.24
3.74
0.94
0.94
4.68
4.68
0.004
0.0277
0.22
4.82
2.37
2.37
7.19
7.19
0.010
0.0072
0.2
5.66
2.84
2.84
8.50
8.50
0.011
-0.0085
0.18
6.26
2.30
2.30
8.56
8.56
0.008
-0.0221
0.16
6.62
0.73
0.73
7.35
7.35
0.002
-0.0350
0.14
6.74
-1.91
-1.91
4.83
4.83
-0.006
-0.0483
0.12
6.62
-5.62
-5.62
1.00
1.00
-0.018
-0.0632
0.1
6.26
-10.38
-10.38
-4.12
-4.12
-0.033
-0.0812
50
-15.05
-15.05
34.95
34.95
-0.050
-0.0481
Total
As expected, the heat transfer obtained by properly subtracting the ∫VdP from ΔH is equal to that obtained by the First law. The use of enthalpy as a simple one-step solution for computing heat transfer is no longer valid. It is now as easy to compute heat transfer directly from the First Law. Computed values of ΔHtpr using dHtpr = n Cpm dT are equal to those obtained using dH = dU+PdV+VdP. The change in enthalpy for varying pressure systems can be obtained directly from the heat capacity at constant pressure for ideal gases. However since heat transfer cannot be obtained by the simple temperature enthalpy for these system, the importance of this conclusion is much less important than similar conclusions obtained for Cv discussed in the next section. We estimated ΔS using the average of the temperatures at the beginning and end of the volume interval. This is a reasonable approximation for relatively mild changes in temperatures. However, the change in temperature for some of the intervals can be quite large. A more precise estimate of ΔS can be obtained using 𝛿𝛿𝛿𝛿 𝑑𝑑𝑑𝑑 − 𝛿𝛿𝛿𝛿 =� 𝑇𝑇 𝑇𝑇 Tj+1 Vj+1 (𝑛𝑛𝑛𝑛𝑛𝑛 ⁄ ) Tj+1 Vj+1 𝑑𝑑𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑃𝑃 𝑑𝑑𝑑𝑑 𝐶𝐶𝑣𝑣 𝑑𝑑𝑑𝑑 𝑉𝑉 𝑑𝑑𝑑𝑑 =� =� +� = Cv ln � � + nR ln � � 𝑇𝑇 𝑇𝑇 𝑇𝑇 Tj Vj Tj Vj
ΔS = � ΔSj,j+1
where a constant Cv is used over the temperature interval using the following relationships obtained from the Thermodynamic Data Handout.
Cv = n�Cpm − R� = 15.625 mol (28.355 − 8.314)J K −1 mol−1 = 328.8 J K −1 = 0.329 kJ K −1
An example calculation of ΔS is given below for the first interval of Path #2 (using nR = 0.1299 kJ K-1). Values for the other intervals are given in the above table. There are substantial different values between using the more precise approach given here and that obtained from the average-temperature methods of the previous section. Nonetheless, the total change in entropy is approximately equal for the two methods (ΔS = 0.050 and ΔS = 0.0481 kJ K-1). More information on methods to compute changes in entropy are given in Chapter 4.
Last Modified: March, 2018
2-40
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
ΔSj,j+1 = Cv ln �
Tj+1 Vj+1 222 0.28 � + nR ln � � = 0.329 ln � � + 0.13 ln � � = 0.117 kJ K −1 Tj Vj 230.9 0.30
ΔSj,j+1 = Cv ln �
Tj+1 Vj+1 307.9 0.1 � + nR ln � � = 0.329 ln � � + 0.13 ln � � = −0.0481 kJ K −1 Tj Vj 230.9 0.3
For exact differential, the change in the function can be evaluated by only using endpoint conditions. For the initial conditions of Ti = 230.9 K and Vi = 0.3 m3 and final conditions of Tf = 307.9 K and Vf = 0.1 m3, we obtain
The solution obtained solely from the endpoint conditions is equal to that obtained for Path #2. If the integrated solution is applied to Path #1. The ΔS is also equal to -0.0481 kJ K-1. Additional Discussion
For the design of a constant volume system, we can straightforwardly compute the change in internal energy using dU = Cv dT. But what if our design is for an expanding or contracting (varying volume) system? For Example Problem 2.2, the change in internal energy using Cv was shown to equal that obtained from the First law, even though this system was clearly contracting between initial and final states. Insight into why we can use dU = Cv dT even for systems with varying volume has already been explored in Chapter 1. In Chapter 1, we defined the change in internal energy (with no phase transitions or chemical reactions) as ∂U ∂U ∂U dU = � � 𝑑𝑑𝑑𝑑 + � � 𝑑𝑑𝑑𝑑 = 𝐶𝐶𝑣𝑣 𝑑𝑑𝑑𝑑 + � � 𝑑𝑑𝑑𝑑 ∂T 𝑉𝑉 ∂V 𝑇𝑇 ∂V 𝑇𝑇
2.5.36
where partial derivatives are the changes in internal energy with respect to temperature (for a given volume) and volume (for a given temperature). We have also used our definition of heat capacity at constant volume of Cv= (∂U/∂T)V. Internal energy is, in its more general representation, a function of temperature and volume. For a system of constant volume, dV=0. The second term on the right-hand side of Eq. 2.5.36 is then zero. We can then compute the change in internal energy as previously done in the example problems as 𝑈𝑈𝑓𝑓
𝑇𝑇𝑓𝑓
ΔU = ΔUtpr = � 𝑑𝑑𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 = � 𝑛𝑛 𝐶𝐶𝑣𝑣𝑣𝑣 𝑑𝑑𝑑𝑑 𝑈𝑈𝑖𝑖
2.5.37
𝑇𝑇𝑖𝑖
Let’s consider the system where dV is not zero. For the second term on the right-hand side of Eq. 2.5.36 to be zero, we now need to remember that internal energy is only a function of temperature for ideal gases. For example in Chapter 1, we used Ut,m = a+bT in describing the ideal gas equation. When internal energy is only a function of temperature, we know that ∂U � � =0 ∂T 𝑉𝑉
Last Modified: March, 2018
2.5.38
2-41
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For ideal gases, the term (∂U/∂V)T dV is then always zero regardless of whether or not dV is zero. The change in internal energy can be computed directly from Eq. 2.5.37 as done in Example Problem 2.2. We conclude that Eq. 2.5.37 can be used to compute the change in internal energy for non-ideal gases for systems of constant volume and for ideal gases for systems of constant and varying volumes. Students sometime wonder why we didn’t develop relationships to adjust Cp for pressure. For Example Problem 2.2, the Cpm was obtained from the Thermodynamic Handout that corresponds to Po= 100 kPa; whereas the system was compressed using P = 140 kPa. Wouldn’t pressure work, and consequently Cpm, be greater for a larger pressure? Let’s consider the response of ideal gases. For a fixed mass system undergoing a constantpressure change, we have previously shown that PΔV = nRΔT (Eq. 2.5.27). Let’s now consider the pressure work for two different pressures (say P1 = 100 kPa and P2 = 140 kPa) for the same temperature difference (same change in internal energies). Since n, R, and ΔT are constant, we conclude that P1 ΔV1 = P2 Δ𝑉𝑉2
2.5.39
that is, the expansion work is equal for the two different pressures. An increase in pressure corresponds to an equivalent decrease in expansion volume. Consequently, Cp values are not a function of pressure for ideal gases.
ENTHALPY OF PHASE TRANSITIONS Introduction We now wish to expand our application of the First Law to include phase transitions. Phase transition is the conversion of a substance to a different phase. There can be more than one phase for solids. Only the most basic features of phase transitions will be given here. A more rigorous analysis is given in Chapter 5. Let’s consider the heating of water and the corresponding change in temperature of water from an initial state with ice cubes to a final state of boiling water (see Fig. 2.23). Although heat transfer into the system is positive, the initial water-ice mixture maintains a temperature of approximately 0 C until the ice is melted. Since there is no change in internal energy with temperature, the heat transfer must be tied to a change in the internal energy for the phase transition from solid to liquid, that is, the internal energy of liquid water is greater than that of ice. Since the impact of heat transfer cannot be “sensed” by a changing temperature, this type of transfer of energy has historically been considered to be “hidden”. The change in internal energy was referred to as latent heat. After the ice is melted, the impact of additional heat transfer results in an increase in the temperature of the liquid. Since this impact can be easily sensed by us, this type of heat transfer has historically been called sensible heat transfer. Once the liquid starts to boil, the temperature of the liquid again is constant and the internal energy change with temperature is again zero. The heat transfer is used to drive liquid
Last Modified: March, 2018
2-42
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
molecules apart to form water vapor. Once again, this process is sometimes referred to as latent heat. Although engineering problems can be properly solved using the terminology of latent and sensible heat, they are not needed to understand thermodynamic processes and provide a less straightforward framework for analyzing systems. Latent heat is usually defined by using the heat transfer corresponding to phase transitions. Similar to heat capacity, this heat transfer changes the characteristics of matter. In the distant past, the processes of this change were “hidden” from us, but the connection is now well established. “Latent heat” is also ambiguous as to whether it refers to energy transfer for the change in the internal energy or the change in internal energy and expansion work. We can avoid this unnecessary confusion by consistently applying general enthalpy concepts. Enthalpy is a well-establish property of thermodynamics. It is defined by exact differentials and is a thermodynamic potential (as discussed in Chapter 5). Both of these attributes are valuable in understanding the response of biological and environmental systems.
Figure 2.23. Temperature Changes of Water with Phase Transitions. Possible phase transitions are summarized in Fig. 2.24. Vaporization (or evaporation) is the phase transition from a liquid to a gas and condensation is the reverse of vaporization (transition from a gas to liquids). Fusion (or melting) is the phase transition from solid to liquid and freezing is the reverse of fusion (transition from a liquid to solid). Sublimation is the phase transition from solid to gas (dry ice, snow to vapor) and vapor deposition is the reverse of sublimation (transition from a gas to solid). Vapors are gases and therefore behave as gases.
Last Modified: March, 2018
2-43
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Vapor
Solid
Liquid Freezing
Condensation
Fusion (melting)
Vaporization
Sublimation Vapor Deposition
Figure 2.24. Terminology Used for Phase Transitions. Standard Enthalpies of Transitions Introduction Similar to experiments done for heat capacities, let’s consider a possible experiment that measures the heat transfer into the system to determine the change in internal energy and expansion work with phase transitions as shown in Fig. 2.25. Conceptually, data collected with this type of experiment could then be used to estimate the change in enthalpy (actual measurement is based on the theory given in Chapter 5). However for phase transition, we are not interested in the change in the temperature of the system but the change in the number of moles of each phase, that is, q/Δn instead of q/ΔT as used for heat capacities (which is obviously of no value if ΔT=0). Enthalpies of transitions are therefore reported in units of energy per mole (or per mass) of phase change. For our illustration in Fig. 2.25, heat transfer into the system is used to vaporize 10 moles of liquid water and to expand the system until the water vapor pressure equals the external (atmospheric) pressure. A change of internal of energy and expansion work of 377 kJ and – 30 kJ, respectively, requires a heat transfer of 407 kJ. The measured molar enthalpy of vaporization would then be 40.7 kJ mol-1. Of course the increase in the number of moles of vapor in Fig. 2.25 is equal to the decrease in the moles of the liquid phase. However, changes in liquid volume are negligible in comparison to changes in water vapor (see Problem 1.12).
Figure 2.25. Conceptual Representation of Molar Enthalpy of Vaporization.
Last Modified: March, 2018
2-44
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For vaporization, the initial and final states correspond to the liquid and vapor phases, respectively. Since enthalpy is defined by an exact differential, the change in enthalpy for vaporization can be evaluated from basic calculus as ΔHphase = �
𝐻𝐻𝑔𝑔𝑔𝑔𝑔𝑔
Hliq
𝑑𝑑𝑑𝑑 = 𝐻𝐻𝑔𝑔𝑔𝑔𝑔𝑔 − 𝐻𝐻𝑙𝑙𝑙𝑙𝑙𝑙 = − �
𝐻𝐻𝑙𝑙𝑙𝑙𝑙𝑙
𝐻𝐻𝑔𝑔𝑔𝑔𝑔𝑔
𝑑𝑑𝑑𝑑 = −(𝐻𝐻𝑙𝑙𝑙𝑙𝑙𝑙 − 𝐻𝐻𝑔𝑔𝑔𝑔𝑔𝑔 )
2.6.1
where for condensation the initial state corresponds to the vapor phase and the final state to the liquid phase. We therefore conclude that the enthalpy of transition of particular direction is equal to the negative value of the transition in the reverse direction, that is, Condensation Enthalpy =- Vaporization Enthalpy Freezing Enthalpy = - Fusion Enthalpy Vapor deposition Enthalpy = - Sublimation Enthalpy A positive ΔH indicates that heat transfer from the surroundings to the system is necessary to drive the system. For example, vaporization requires heat transfer from the surroundings to provide the energy necessary for molecules to break away from their liquid phase. These concepts are shown in the schematic below for melting and freezing processes of water.
Figure 2.26. Sign Notation for Enthalpy of Phase Transitions. Standard Enthalpy Vaporization and Notation The standard enthalpy of vaporization is defined as Heat transfer required to evaporate (vaporize) one mole of molecules of a pure substance at a constant pressure of 1 bar (100 kPa). Although the definition of standard enthalpy does not require that data be collected at a standard temperature, enthalpies of transitions are dependent on temperature. A corresponding temperature is therefore needed for the reported standard enthalpy of substances, usually given at their transition temperature. Molar enthalpy of vaporization is used to refer to values at non-standard pressures.
Last Modified: March, 2018
2-45
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑜𝑜 We will use the notation of Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 for standard enthalpy of vaporization. For expansion work by ideal gases under constant pressure and temperature, we have d(PV) = P dV = d(nRT) = RT dn. Standard enthalpy of vaporization is defined as o ΔTvap Hm =
Δ𝑈𝑈𝑣𝑣𝑣𝑣𝑣𝑣 𝑃𝑃Δ𝑉𝑉 q 𝑜𝑜 𝑜𝑜 𝑜𝑜 = + = Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝑈𝑈𝑚𝑚 + 𝑅𝑅𝑅𝑅 = 𝐻𝐻𝑚𝑚,𝑔𝑔𝑔𝑔𝑔𝑔 − 𝐻𝐻𝑚𝑚,𝑙𝑙𝑙𝑙𝑙𝑙 Δn Δ𝑛𝑛 Δ𝑛𝑛
2.6.3
which has a component related to the change in the internal energy and an expansion work component. Eq. 2.6.3 uses the First Law for conditions of constant pressure and ΔUtpr=ΔUphase = 0. The “°” is used to indicate that it corresponds to the standard pressure. The symbol T is a holder for the temperature at which the data has been recorded or the temperature corresponding to the system of interest. For example, the standard enthalpy of 𝐶𝐶 𝑜𝑜 𝑂𝑂 𝐶𝐶 𝑜𝑜 -1 vaporization for water at a temperature of 100 C is Δ10𝑂𝑂 𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 = 40.7 kJ mol and is Δ𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 = 45.07 kJ mol-1 at 0 C. We will use the specific enthalpy of vaporization as the enthalpy of vaporization per mass (instead of moles) and adopt the notation of Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑠𝑠𝑜𝑜 . Since 1 mole of water = 18 g, the specific enthalpy of vaporization for water at temperature of 0 C is defined 𝐶𝐶 𝑜𝑜 𝐻𝐻𝑠𝑠 =45.07/18 = 2.5 kJ g-1 or 2500 J g-1. The change in internal energy for as Δ𝑂𝑂𝑣𝑣𝑣𝑣𝑣𝑣 vaporization can be determined from a measured enthalpy value by rearranging Eq. 2.6.3 as 𝑜𝑜 𝑜𝑜 Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝑈𝑈𝑚𝑚 = Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 − 𝑅𝑅𝑅𝑅.
Similar to Cpm of the previous section, we need to be careful to distinguish between the tabular values obtained from experiments and the application of these values to 𝑜𝑜 thermodynamic problems. The notation of Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 is used for measured change in enthalpy that are summarized in table or equivalent relationships. It is a property of the substance. To solve thermodynamic problems that have vaporization, the notation for this enthalpy change is represented by Δ𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣 = 𝑛𝑛 Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 = 𝑚𝑚 Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑠𝑠 , where n and m are the number of moles or mass, respectively, that are vaporized. Fusion and Sublimation Similar to vaporization, the standard enthalpy of fusion is defined as Heat transfer required to melt one mole of molecules of a pure substance at a constant pressure of 1 bar (100 kPa). 𝑇𝑇 𝑜𝑜 Similar to vaporization, we will use the notation of Δ𝑓𝑓𝑓𝑓𝑓𝑓 𝐻𝐻𝑚𝑚 for standard enthalpy of fusion. 0 𝐶𝐶 𝑜𝑜 For a temperature of 0 C, it is defined for water as Δ𝑓𝑓𝑓𝑓𝑓𝑓 𝐻𝐻𝑚𝑚 =6.01 kJ mol-1. Once again, a positive value indicates that heat transfer is into the system for melting.
The standard enthalpy of sublimation is defined as: Heat transfer required to convert one mole of molecules of a pure substance at a constant pressure of 1 bar (100 kPa) from a solid to a gas.
Last Modified: March, 2018
2-46
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑜𝑜 We will use the notation of Δ𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝐻𝐻𝑚𝑚 for standard enthalpy of sublimation. Standard enthalpy of sublimation equals the sum of that for fusion and vaporization. For a temperature of 0 C of water, we then obtain C o C o o Δ0sub Hm = Δ0vap Hm + Δ0fusC Hm = 45.07 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 + 6.01 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 = 51.08 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1
Tabular Values
Values for standard enthalpy of transitions are shown in the following tables. Most of the values are reported at their transition temperatures (freezing point and boiling point). Table 2.2. Standard Enthalpies of Transitions for Different Substances.
Substance
Chemical Symbol
T: Freezing (K)
Ammonia Benzene Ethanol Helium Methane Water
NH3 C6H6 C2H5OH He CH4 H2O
195.3 278.7 158.7 3.5 90.7 273.15
∆Tfus H om kJ mol-1 5.65 9.87 4.60 0.02 0.94 6.01
T: Boiling (K) 239.7 353.3 351.5 4.22 111.7 373.2
∆Tvap H om
kJ mol-1 23.4 30.8 38.6 0.08 8.2 40.7
Changes with Temperature We are often interested in converting the above tabular enthalpy of vaporization to different conditions. A relatively rigorous analysis of the dependency of Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 with temperature and pressure is given in Chapter 5. However, greater insight into changes with temperature (only) and in the usefulness of exact differential is obtained by vaporization of water for two different paths. The change in enthalpy is defined by an exact differential, and it is independent of the path. We can therefore use the equality of the solution for each path to determine the relationship between temperature and the standard enthalpy of vaporization. Let’s consider the change in enthalpy corresponding to n moles of water at the liquid state at 25 C to n moles of vapor at a temperature of 35 C (Δn vapor = n moles). We will consider two possible paths between the initial and final states as shown conceptually in Fig. 2.29. For Path #1, all of the water is evaporated at 25 C and the vapor temperature increases to 35 C. For Path #2, the temperature of liquid water is first increased to 35 C and then all of the water is evaporated. We have to consider both the change in enthalpy for phase transition and the enthalpy changes with temperature. These two paths were selected to obtain simple solutions. It is likely that the actual path won’t correspond to either of them.
Last Modified: March, 2018
2-47
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Path 1
Final
Hf
Enthalpy
H2 O (g) n moles, 35 C
∆H1 =∆H2
Hi Path 2
H2 O (l) n moles, 25 C
Initial
Initial
Final
Progression of Change
Figure 2.29. Framework Temperature Dependent Enthalpy Relationship. Let’s now consider the change in enthalpy for the following path where the n moles of water are first vaporized at 25 C and then the temperature of the water vapor is increased to 35 C. Final State 35 C
Initial State 25 C Vaporization at 25 C H2 O (l) n moles
n moles
Increase gas temperature H2 O (g) n moles
∆25vapC H°
Figure 2.30a. ΔH Change for Liquid → Gas and then 25 C → 35 C. The change in enthalpy for this pathway can be computed as ΔH1 = n
C o Δ25 vap Hm
Tf
C o + � 𝑛𝑛 �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 𝑑𝑑𝑑𝑑 ≈ n Δ25 vap Hm + 𝑛𝑛 �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 Δ𝑇𝑇
2.6.4
Ti
where ΔH1 is the change in enthalpy for the first path. We have assumed a negligible change in Cpm (water vapor) with temperature. Let’s now consider the alternative path between initial and final states as shown below. For this path, the temperature of the liquid water is first increased to 35 C. The liquid water is then vaporized at 35 C to obtain the conditions corresponding to the final state.
Last Modified: March, 2018
2-48
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Initial State 25 C
Final State 35 C Increase liquid temperature H2 O (l) n moles, 35 C
H2 O (l) n moles
n moles
Vaporization at 35 C
∆35vapC H°
Figure 2.30b. ΔH Change for 25 C → 35 C and then Liquid → Gas. The change in enthalpy here is Tf
C o 35 C o ΔH2 = � 𝑛𝑛 �𝐶𝐶𝑝𝑝𝑚𝑚 �𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑 + n Δ35 vap Hm ≈ 𝑛𝑛 �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑙𝑙𝑙𝑙𝑙𝑙 Δ𝑇𝑇 + n Δvap Hm
2.6.5
Ti
where ΔH2 is the change in enthalpy for the second path and where again we have assumed a constant Cpm (liquid water) with temperature. Since enthalpy corresponds to an exact differential, the change in enthalpy is independent of the path. If we set the change by the two paths equal to each other (ΔH1=ΔH2), we obtain n�
Tf
Ti
�𝐶𝐶𝑝𝑝𝑝𝑝 �𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑 + n
C o Δ35 vap Hm
= n
C o Δ25 vap Hm
+ n�
Tf
Ti
�𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 𝑑𝑑𝑑𝑑
2.6.6
If we divide both sides by n and rearrange terms, we can solve for the standard enthalpy of vaporization at 35 C as C o Δ35 vap Hm
=
C o Δ25 vap Hm
≈
+�
Tf
Ti 25 C o Δvap Hm
[ �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 −�𝐶𝐶𝑝𝑝𝑝𝑝 �𝑙𝑙𝑙𝑙𝑙𝑙 ] 𝑑𝑑𝑑𝑑
2.6.7
+ [ �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 −�𝐶𝐶𝑝𝑝𝑝𝑝 �𝑙𝑙𝑙𝑙𝑙𝑙 ] Δ𝑇𝑇
where for water (Cpm)gas = 33.58 J K-1 mol-1 and (Cpm)liq = 75.29 J K-1 mol-1. Let’s generalize by representing the temperature Ti = To for the tabular standard enthalpy of vaporization and the temperature of interest as Tf = T. We then obtain T
o o o ΔTvap Hm ≈ Δvap Hm + [ �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 −�𝐶𝐶𝑝𝑝𝑝𝑝 �𝑙𝑙𝑙𝑙𝑙𝑙 ] (𝑇𝑇 − 𝑇𝑇𝑜𝑜 )
2.6.8
Let’s consider the usefulness of enthalpies defined by exact differentials by placing a cup of water at a temperature of 25 C into a room at a temperature of 35 C. The water in an open cup would eventually evaporate and the resulting water vapors would eventually reach a temperature of 35 C. The actual path between these initial and final states can be quite complex. For example, there would be some evaporation at 25 C, at 26 C, at 27 C and at other temperature to our final state temperature of 35 C (assuming that there is still liquid water in the cup). For each of these temperatures, we need to compute the heat transfer to
Last Modified: March, 2018
2-49
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
increase the liquid temperature from 25 C, the heat transfer to evaporate the mass, and the heat transfer to increase the gas temperature to 35 C. Additional equations for heat and mass transport are needed to determine the actual evaporation rates at each temperature. Fortunately, this level of complexity is not needed for our solution. Since enthalpy is defined by exact differentials, the change in enthalpy for the actual path is the same as our idealized Path 1 (all of the water evaporates at 25 C) or as our idealized Path 2 (all of the water evaporates at 35 C). For a system under constant pressure, we can therefore determine the heat transfer between our endpoints by selecting a path that corresponds to one of these simpler solutions. We don’t need to understand the complex processes corresponding to the actual path if the endpoint conditions are known. Example Problem 2.5: Cooling by Perspiration Problem Statement. Perspiration is an important mechanism for cooling the body. During vigorous exercise, as much as 1 kg to 2 kg of water (1 to 2 L) of water can be produced per hour. We will assume that 0.5 kg of perspired water is evaporated at a constant temperature of 20 C and that all of the heat transfer from the skin is used to the evaporate water for a closed system (no mass transfer). We wish to calculate the (1) heat removal from the body as a result of the evaporation and (2) the expansion work and change in the internal energy of the system. A definition schematic of the system is shown below.
Initial
Vaporization
Expansion work
Final 0.5 kg vapor T = 20 C
0.5 kg liquid T = 20 C
q
Skin
Tabular Values. From the Data Handout, we obtain mm = 18 g mol-1 (Cpm )gas = 33.58 J K-1 mol-1 (Cpm )liq = 75.29 J K-1 mol-1 As previously given C o −1 Δ100 vap Hm = 40.7 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙
Preliminary Calculations. Temperature and number of moles: T = 20 + 273.15 = 293.15 K m 0.5 kg (1000 g kg −1 ) n= = = 27.8 mol mm 18 g mol−1 Last Modified: March, 2018
2-50
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Standard enthalpy of vaporization at 20 C C o 100 C o Δ20 vap Hm ≈ Δvap Hm + [ �𝐶𝐶𝑝𝑝𝑝𝑝 �𝑔𝑔𝑔𝑔𝑔𝑔 −�𝐶𝐶𝑝𝑝𝑝𝑝 �𝑙𝑙𝑙𝑙𝑙𝑙 ] Δ𝑇𝑇
𝐽𝐽 𝑘𝑘𝑘𝑘 C o −1 Δ20 + (33.58 − 75.29) � �� � ( −80 𝐾𝐾) vap Hm ≈ 40.7 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 𝐾𝐾 𝑚𝑚𝑚𝑚𝑚𝑚 1000 𝐽𝐽 C o −1 Δ20 vap Hm = 44.04 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙
Solution: Heat Transfer. From the First Law for constant-pressure expansion work, we have for this problem q = ΔUvap + 𝑃𝑃Δ𝑉𝑉 = Δ𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣
By using the standard enthalpy of vaporization to determine ΔHvap, we obtain C o −1 𝑞𝑞 = Δ𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣 Δ20 vap Hm = 27.8 𝑚𝑚𝑚𝑚𝑚𝑚 (44.04 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 ) = 1223 𝑘𝑘𝑘𝑘
We conclude that 1223 kJ of heat transfer from the skin into the system is needed to evaporate 0.5 kg of perspiration. This removal of heat from the body is an important cooling mechanism for humans. Solution: Expansion work and Internal Energy. The change in enthalpy includes a change in internal energy as well as expansion work. We will assume that the water vapor behaves as an ideal gas where PV = nRT. For a constant temperature and pressure, we then obtain d(PV) = PdV = RT dn. The change in enthalpy can then be written as ΔHvap = Δ𝑈𝑈vap + ∫ 𝑃𝑃 𝑑𝑑𝑑𝑑 = Δ𝑈𝑈vap + 𝑅𝑅𝑅𝑅 �
𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣,𝑓𝑓
𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣,𝑖𝑖
𝑑𝑑𝑑𝑑 = Δ𝑈𝑈vap + 𝑅𝑅𝑅𝑅(𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣,𝑓𝑓 − 𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣,𝑖𝑖 )
where the initial number of vapor moles is zero. We can then evaluate the expansion work as PΔV = RT Δnvap = �8.314
J � (293.15 𝐾𝐾)(27.8 − 0)𝑚𝑚𝑚𝑚𝑚𝑚 = 67701 𝐽𝐽 = 67.7 𝑘𝑘𝑘𝑘 K mol
Expansion work is defined as w = - PΔV = - 67.7 kJ, that is, work is done on the surroundings to create a volume for 27.8 moles of water vapor. Of the heat transfer of 1223 kJ, 67.7 kJ of the energy transfer was needed for this work or roughly 5.5% of the heat transfer (67.7/1223). The remaining heat transfer is used to change the internal energy of the system. The change in the internal energy is simply ΔUvap = q + w = 1223 k𝐽𝐽 − 67.7 𝑘𝑘𝑘𝑘 = 1155.3 𝑘𝑘𝑘𝑘
ENTHALPY OF CHEMICAL REACTIONS
Introduction General Concepts
Last Modified: March, 2018
2-51
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Biological systems are often driven by the heat transfer as a result of chemical reactions. Chemical reactions require a change in the composition of constituent atoms among various molecules. Reconfiguration possibilities are restricted by the mass balance requirements corresponding to the chemical reaction equation. The redistribution of atoms among molecules also requires a redistribution of internal energy. Changes in the binding energy can result in an increase or decrease of the overall internal energy of systems. These changes are represented by ΔUr = Uprod – Ureact, where Uprod and Ureact are the internal energies of the products and reactants, respectively. Similar to heat capacities, chemical reactions can be analyzed for a given temperature and volume or a given temperature and pressure, where the former is used to analyze the change in internal energy and the latter the change in enthalpy. Let’s start by considering the chemical reaction of urea (a widely used fertilizer) with oxygen. The chemical reaction is defined as (NH2)2CO(s) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l) + N2 (g)
2.7.1
where the stoichiometric coefficients (necessary for mass conservation) are defined as υurea = 1, υo2 = 3⁄2 , 𝜐𝜐𝑐𝑐𝑐𝑐2 = 1, 𝜐𝜐ℎ2𝑜𝑜 = 2, 𝑎𝑎𝑎𝑎𝑎𝑎 𝜐𝜐𝑛𝑛2 = 1
This chemical reaction results in the heat transfer from the system to the surroundings as shown in Figure 2.31. Specialized equipment can be used to measure this transfer. An example of this type of equipment is an adiabatic bomb calorimeter. More on its use is given in Homework Problem 2.32. As shown later in Example Problem 2.5, the urea reaction increases the volume of gases resulting in expansion work for a system under constant pressure. Enthalpy is therefore a useful measure of the heat transfer because it includes both changes in internal energies and expansion work.
Reactants (Initial)
Vf
Products (Final)
w = − ∫ PdV = −P∆V Vi
Expansion work CO2(g) N2 (g)
3/2 O2(g) (NH2)2CO (s) q
2 H2O (l)
Figure 2.31. Conceptual Representation of Reaction Enthalpy. Enthalpy of chemical reactions is determined from reactions carried out under a constant pressure and where the products are returned to the same temperature as the reactants. For the
Last Modified: March, 2018
2-52
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
closed system under quasi-equilibrium and constant pressure, the heat transfer is equal to the change in this enthalpy, that is, 𝑞𝑞 = Δ𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡 + Δ𝑈𝑈𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 + Δ𝑈𝑈𝑟𝑟 − 𝑤𝑤 = Δ𝑈𝑈𝑟𝑟 + 𝑃𝑃Δ𝑉𝑉 = Δ𝐻𝐻𝑟𝑟
2.7.2
where ΔUr and ΔHr are the changes in the internal energy and in the enthalpy, respectively, for the chemical reaction. Since the initial and final temperatures are the same and there are no phase transitions, ΔUtpr = ΔUphase = 0. The change in enthalpy is then determined by the change internal energy resulting from a new configuration of atoms of the products and the expansion work resulting from the net increase or decrease of gases. Heat transfer for each reaction of interest could conceptually be obtained from an experiment similar to that shown above for urea. Reaction enthalpy could then be computed and summarized in tables. However, there are a very large number of interesting chemical reactions and the reporting of the reaction enthalpy for each of them would be cumbersome (and unnecessary). An alternative approach is to determine the enthalpy for each of the compounds and appropriately combined them to determine the overall reaction enthalpy. For example, since the internal energy of O2 is the same as a reactant with urea and as a reactant with methane (for the same initial pressure and temperature), we can avoid duplication of effort if we determine this internal energy once and then use it for all reactions of O2. Our approach utilizes enthalpy’s extensive properties (individual components can be added to determine the total) and that the change in enthalpy is independent of the path. After the enthalpies of compounds are defined, they can be appropriately combined to determine the reaction enthalpy of chemical reactions using them. Molar Extent of Reaction Similar to the molar heat capacity and standard enthalpy of transitions, we are interested in analyzing the reaction enthalpy in units of energy per mole. The number of moles used in molar heat capacity and standard enthalpy of transitions was easily defined from the mass of the system and the mass change with phase transitions, respectively. Selection of the “number of moles” to obtain a molar energy unit needs to be done carefully for chemical reactions. This point will be illustrated using the chemical reaction equation for urea (Eq. 2.7.1). Let’s consider our urea reaction with 2 moles of consumed urea. To conserve mass, we need to maintain the same number of hydrogen, carbon, and nitrogen atoms before and after the reaction. We then conclude that 2 moles of CO2 and N2 and 4 moles of H2O must be produced by the reaction. To maintain the balance of oxygen, 3 moles of O2 must have also been used in the reaction. A summary of the changes in moles (Δn) is shown below. Δnurea = υurea (2 moles) = (−1)(2 moles) = −2 moles Δno2 = υo2 (2 moles) = (−3/2)(2 moles) = −3 moles Δnco2 = υco2 (2 moles) = (1)(2 moles) = 2 moles Δnh2o = υh2o (2 moles) = (2)(2 moles) = 4 moles Last Modified: March, 2018
2-53
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Δnn2 = υn2 (2 moles) = (1)(2 moles) = 2 moles
where υurea through υN2 are the stoichiometric coefficients. There is a decrease in the number of moles for the reactants and an increase in the number for the products. At an initial glance, it might seem impossible to select a “number of moles” for defining a molar energy unit because the number of moles varies among chemical compounds, ranging from 2 moles for CO2 to 4 moles for H2O. However, notice that a common quantity of 2 moles is used to compute the change in the number of moles for each compound. This common quantity is called the molar extent of reaction (Δξ). The change in molar enthalpy is defined as the enthalpy change per molar extent of reaction. Let’s extend the concepts established for the reaction of urea to the more general representation of chemical reactions (as given in Chapter 1) of 𝑣𝑣𝐴𝐴 𝐴𝐴 + 𝜈𝜈𝐵𝐵 𝐵𝐵 + ⋯ → 𝜈𝜈𝐿𝐿 𝐿𝐿 + 𝜈𝜈𝑀𝑀 𝑀𝑀 + ⋯.
2.7.3a
where υA through υM are the stoichiometric coefficients. The molar extent of reaction is used to compute the change in the number of moles in reaction from the stoichiometric coefficients. For a given molar extent of reaction of Δξ, the change in the number of moles for each substance is defined as ΔnA = υA Δξ , ΔnB = υB Δξ , … ΔnL = υL Δξ, ΔnM = υM Δξ , …
2.7.3b
where Δn is the change in the number of moles as a consequence of a given molar extent of reaction. By notation convention, we can define the change for the reactants as negative and the change in the products as positive (alternatively, the reactant stoichiometric coefficients are sometimes defined as negative). The molar extent of reaction is constant for all substances. Clearly from the above relationships, the stoichiometric coefficient for substance j is defined as υj =
Δnj Δ𝜉𝜉
Δξ =
ΔnO2 −12 mol = = 8 mol −υO2 −3/2
2.7.4
A rigorous discussion on the definition of the molar extent of reaction will have to wait until we have a better understanding of the Second Law. However, it is possible for us to provide some insight on Δξ. Let’s assume that you want a gas mixture of 10 moles of CO2 and 10 moles of N2. Based on the chemical reaction for urea, you intend to obtain this mixture by the reaction of 10 moles of urea with 15 moles of O2. This plan assumes that all of the moles of urea is used. For our hypothetical example, let’s assume that after your reaction you measured a reduction in the number of O2 of only 12 moles. From your observed data and the stoichiometric coefficients, we can compute the molar extent of reaction, and the corresponding moles of the other compounds (because of mass balance), as:
Last Modified: March, 2018
2.7.5a
2-54
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Δnurea = −υurea Δξ = −(1)(8 𝑚𝑚𝑚𝑚𝑚𝑚) = −8 𝑚𝑚𝑚𝑚𝑚𝑚, ΔnCO2 = (1)(8 𝑚𝑚𝑚𝑚𝑚𝑚) = 8 𝑚𝑚𝑚𝑚𝑚𝑚, ΔnN2 = 8 mol and ΔnH2O = 16 mol
2.7.5b
Even though adequate O2 was available, not all of the urea was used in the reaction. The actual molar extent of the reaction is only 8 moles. Instead of obtaining your target of 10 moles of CO2 and N2, you were only able to obtain 8 moles of each of them. Incomplete consumption of reactant occurs because chemical equilibrium between the reactants and the products. For metabolic processes given later in the chapter, the molar extent of reaction is obtained (effectively) by measuring changes in the number of moles of reactants and products as computed by Eq. 2.7.5a. However, a more insightful approach is to use theoretical relationships obtained from the Second Law of thermodynamics. This approach is discussed in detail in Chapter 6. Let’s now consider reactions for powerful combustion processes. Here the organic reactant (i.e., “fuel”) is completely consumed (essentially) by the reaction. The molar extent of reaction is then equal to the number of moles of the organic reactant. This concept can be illustrated for the combustion of methane (natural gas is mostly methane), represented by the following reaction CH 4 (g ) + 2 O 2 (g ) → CO 2 (g ) + 2 H 2O (l)
2.7.6
If we have a system with 10 moles of methane then the molar extent of reaction would be Δξ = 10 moles. Twenty moles of oxygen gas would then also be consumed by the combustion process, and we would have products of 10 moles of carbon dioxide and 20 moles of water. Enthalpy of Reaction Since change for exact differentials is equal to the difference between the endpoints, we can readily evaluate the change in enthalpy for chemical reactions at constant pressure and temperature, that is, Δ𝐻𝐻𝑟𝑟 = 𝐻𝐻𝑓𝑓 − 𝐻𝐻𝑖𝑖 = 𝐻𝐻𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − 𝐻𝐻𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
2.7.7
where Hprod and Hreact is the enthalpies of the products and reactants, respectively. Enthalpies of the products and the reactants can be obtained by the sum of their components. The change in enthalpy for a chemical reaction can then be written as Δ𝐻𝐻𝑟𝑟 = ��Δ𝑛𝑛𝑗𝑗 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑑𝑑
− ��Δ𝑛𝑛𝑗𝑗 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
= ��𝜈𝜈𝑗𝑗 Δξ 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
− ��𝜈𝜈𝑗𝑗 Δξ 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
2.7.8 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
where Δnj = υj Δξ is the change in the number of moles for each product/reactant, υj are the stoichiometric coefficients, Δξ is the molar extent of reaction and Hmj is the molar enthalpies of the components in the reaction, that is, Hmj = Hj/Δnj. If we divide both sides by the molar extent of reaction, we obtain the reaction enthalpy defined as
Last Modified: March, 2018
2-55
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 =
Δ𝐻𝐻𝑟𝑟 = ��𝜈𝜈𝑗𝑗 𝐻𝐻𝑚𝑚𝑚𝑚 � − ��𝜈𝜈𝑗𝑗 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 Δ𝜉𝜉 𝑗𝑗
2.7.9
𝑗𝑗
where ΔrTHm is the molar enthalpy of reaction at temperature T and pressure of the reaction. It is defined per molar extent of reaction. The final step is to determine the molar reaction enthalpies of its components. However, the determination of these enthalpies needs to be done thoughtfully and carefully. The concept of standard enthalpy of formation (ΔfTHom) is used to systematically evaluate the molar reaction enthalpies. Details on this concept are given in the next section. Standard enthalpy of reaction is defined from the standard enthalpy of formulation as 𝑜𝑜 𝑜𝑜 Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 = ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑜𝑜 − ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
2.7.10
Stoichiometric coefficients are sometimes defined as a positive for the products (increasing) and negative for the reactants (decreasing). Negative changes in the number of moles are then defined directly from Eq. 2.7.2, and the standard enthalpy of reaction can then be computed using a single summation term. We will, however, use the notation of positive stoichiometric coefficients for both products and reactants and subtract the reactant enthalpy from the product enthalpy to more clearly show a change in enthalpy between the products and reactants. Standard Enthalpy of Formation Reference State Standard enthalpy of formation is defined as the change in enthalpy upon forming one mole of material from the element’s most stable pure state. The most stable pure state is called the reference state. There is only one reference state for each element. In contrast, more than one standard state can exist for elements. For example for carbon, there is a standard state for graphite (also the reference state) and a standard state for diamond (not a reference state). Standard enthalpy of formation is defined for a standard pressure of 1 bar (100 kPa), and usually at a temperature of 25 C. Standard enthalpies of formation of elements at their reference state are, by definition, equal to zero. For example, the most stable state for carbon is graphite (solid, but not diamond), for hydrogen is H2, and oxygen is O2. If you look in the thermodynamic data handout, you will find that the standard enthalpy of formation for these substances is given as: 𝑜𝑜 Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚 =0
2.7.11
Experimentally Derived Values Standard enthalpy of formation for some substances can be determined directly from measurements. For example, the enthalpy of reaction for the combustion of carbon (graphite)
Last Modified: March, 2018
2-56
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
with oxygen can be measured in laboratory experiments. The chemical reaction is shown below. C (graphite) + O2 (g) → CO2 (g)
2.7.12
At a temperature of 25 C and pressure of 100 kPa, the standard reaction enthalpy for this -1 𝐶𝐶 𝑜𝑜 reaction has been measured as Δ25 𝑟𝑟 𝐻𝐻𝑚𝑚 = -393.51 kJ mol . This is an exothermal reaction that results in considerable heat transfer from the system to the surroundings. Let’s consider the definition of standard reaction enthalpy for the above reaction. We have 𝑜𝑜 𝐶𝐶 𝑜𝑜 Δ25 𝐻𝐻𝑚𝑚 = −393.51 𝑘𝑘𝑘𝑘 𝑚𝑚𝑜𝑜𝑜𝑜 −1 = ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑟𝑟
=
𝑜𝑜 (1)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝐶𝐶𝑂𝑂 2
−
𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑜𝑜 (1)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝐶𝐶
−
𝑜𝑜 − ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 �
𝑗𝑗 𝑜𝑜 (1)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝑂𝑂 2
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
2.7.13
Since the carbon and oxygen of the reactants are at their reference states, their enthalpies of formation are zero. We then obtain the standard enthalpy of formation for CO2 as 𝐶𝐶 𝑜𝑜 Δ25 𝐻𝐻𝑚𝑚 = −393.51 𝑘𝑘𝑘𝑘 𝑚𝑚𝑜𝑜−1 𝑟𝑟
2.7.14
which corresponds to the value given in the thermodynamic handout. After standard enthalpies of formation for compounds have been determined, they can be used to determine the standard enthalpy of formation for other compounds from measured data. For example in Homework Problem 2.33, the standard enthalpy of formation for carbon monoxide is 𝑜𝑜 computed using the Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚 for CO2 computed in our example.
Hess’ Law
Our definition of standard enthalpy of reaction follows directly from Hess’ law. Hess’ law states that the standard enthalpy of reaction can be computed using the sum of the enthalpies of individual reactions into which the overall reaction may be divided. This law follows directly from the characteristics of an exact differential where the change is independent of the path between endpoints. Application of Hess’ law is used to solve Homework Problems 2.36 and 2.37. Changes with Temperature Derivation The standard enthalpy of formation in the thermodynamic data handout can be used to compute the standard enthalpies of reaction at 25 C. You will be interested in the enthalpies at other temperatures. A schematic for this conversion is shown in Fig. 2.32. Conceptually, the change in the enthalpy with temperature is computed for each of the compounds using their molar heat capacities. After the standard enthalpy of formation has been modified to the design temperature, we can then easily compute the standard reaction enthalpy at this temperature.
Last Modified: March, 2018
2-57
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Design Needs (T)
υA A + υB B + … → υL L+ υM M+… ∆H om, M =
T
∫ C p, m, M dT
To
Available Data (To)
υA A + υB B + … → υL L+ υM M+… Figure 2.32. Changing Reaction Enthalpy with Temperature.
We will start with our general definition of reaction enthalpy at standard pressure, that is, 𝑜𝑜 𝑜𝑜 Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 = ��𝜈𝜈𝑗𝑗 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑜𝑜 − ��𝜈𝜈𝑗𝑗 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑜𝑜 = ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑜𝑜 − ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
For changes with temperature, it useful to take the derivative with respect to temperature, which can be written as 𝑜𝑜 𝑜𝑜 𝑜𝑜 ) 𝑑𝑑�𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑑𝑑�𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑑𝑑(Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 = � �𝜈𝜈𝑗𝑗 � − � �𝜈𝜈𝑗𝑗 � 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑗𝑗
2.7.19
𝑗𝑗
Imagine a system at a temperature corresponding to tabular value (To = 25 C for our thermodynamic handout). Let’s consider a general infinitesimal change in standard enthalpy with temperature (for a constant pressure) using the definition of molar heat capacity at constant pressure 𝑜𝑜 𝑑𝑑�𝐻𝐻𝑚𝑚𝑚𝑚 � = 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 𝑑𝑑𝑑𝑑
2.7.20
and therefore we obtain 𝑜𝑜 ) 𝑑𝑑(Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 = ��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − ��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 𝑗𝑗
2.7.21
𝑗𝑗
where ΔrCpm is called the molar heat capacity of reaction. It is defined for a system of products and reactants. If we integrate between a temperature of known values (To) and the temperature of interest T, or 𝑜𝑜 Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚
=
𝑇𝑇 𝑜𝑜 Δ𝑟𝑟𝑜𝑜 𝐻𝐻𝑚𝑚
𝑇𝑇
𝑇𝑇
𝑜𝑜 + � Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 ≈ Δ𝑟𝑟𝑜𝑜 𝐻𝐻𝑚𝑚 + Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 Δ𝑇𝑇 𝑇𝑇𝑜𝑜
2.7.22
Example Problem 2.6: Standard Enthalpy of Reaction of Urea
Last Modified: March, 2018
2-58
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Problem Statement. For the following reaction of urea with oxygen at a constant pressure of 100 kPa: (NH2)2CO(s) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l) + N2 (g) Determine: (1) Standard enthalpy of reaction at 25 C and molar heat transfer (2) Change in molar internal energy and the molar expansion work (3) Standard enthalpy of reaction at 80 C. Thermodynamic Data. From the thermodynamic data handout, we obtained for T=25 C: C o -1 Urea: Cpm = 93.14 J K-1 mol-1; ∆25 f H m = -333.1 kJ mol C o -1 O2 (g): Cpm = 29.36 J K-1 mol-1; ∆25 f H m = 0 kJ mol
C o -1 CO2 (g): Cpm = 37.11 J K-1 mol-1; ∆25 f H m = -393.51 kJ mol 25 C N2 (g): Cpm = 29.13 J K-1 mol-1; ∆ f H om = 0 kJ mol-1
H2O (l): Cpm = 75.29 J K-1 mol-1;
.15 ∆298 H° = -285.83 kJ mol-1 f
Solution: Standard Enthalpy of Reaction at 25 C. Standard enthalpy of reaction is defined as 𝑜𝑜 𝐶𝐶 𝑜𝑜 Δ25 𝐻𝐻𝑚𝑚 = ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑟𝑟 𝑗𝑗
For the products, we have 𝑜𝑜 ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑜𝑜 ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑗𝑗
𝑜𝑜 ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑜𝑜 − ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚𝑚𝑚 � 𝑗𝑗
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑜𝑜 𝑜𝑜 𝑜𝑜 = (1)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝐶𝐶𝑂𝑂 + (2)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝐻𝐻 + (1)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝑁𝑁 2 2 2 𝑂𝑂
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
= −(1)(393.51) − (2)(285.83) + (1)(0) = −965.17 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑜𝑜 𝑜𝑜 = (1)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + (3 ∕ 2)Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝑂𝑂 2
For the reactants
𝑜𝑜 ��𝜈𝜈𝑗𝑗 Δ𝑓𝑓𝑇𝑇 𝐻𝐻𝑚𝑚𝑚𝑚 �
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
= −(1)(331.1) + (3⁄2)(0) = −331.1 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1
We therefore obtain the following standard enthalpy of reaction: 𝐶𝐶 𝑜𝑜 Δ25 𝐻𝐻𝑚𝑚 = −965.17 − (−331.1) = −632.1 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝑟𝑟
For a constant pressure system, we have previously shown that 𝑞𝑞 = Δ𝐻𝐻 = Δ𝑈𝑈 + 𝑃𝑃Δ𝑉𝑉 Last Modified: March, 2018
2-59
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
If we divide both sides by the molar extent of reaction (Δξ), the molar heat transfer is simply 𝑞𝑞𝑚𝑚 =
𝑞𝑞 𝐶𝐶 𝑜𝑜 = Δ25 𝐻𝐻𝑚𝑚 = −632.1 kJ mol−1 𝑟𝑟 Δξ
Solution: Molar Expansion Work and Molar Internal Energy. Expansion work is related to the gas components of the chemical reaction. We need to divide the number of moles of gases into their components, that is, 𝑛𝑛 = 𝑛𝑛𝑂𝑂2 + 𝑛𝑛𝐶𝐶𝑂𝑂2 + 𝑛𝑛𝑁𝑁2
where for complete combustion no2 is zero after the reaction. Before the reaction, nco2 and nn2 are zero. We will use ideal gases where PV = nRT and use the solution obtained for a constant temperature expansion, that is, RTi=RTf = constant.. For constant pressure, PdV = d(PV) = RT dn. The expansion work for our reaction can then be evaluated as ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑅𝑅𝑅𝑅 ∫ 𝑑𝑑�𝑛𝑛𝑂𝑂2 + 𝑛𝑛𝐶𝐶𝑂𝑂2 + 𝑛𝑛𝑁𝑁2 � = 𝑅𝑅𝑅𝑅 (𝑛𝑛𝑂𝑂2 + 𝑛𝑛𝐶𝐶𝑂𝑂2 + 𝑛𝑛𝑁𝑁2 )
where Δn = nf – ni refers to the change in the number of moles for each of the gas components. The change in number of moles for O2 is negative. Since reaction enthalpy is defined by molar extent of reaction (Δξ), we obtain an appropriate measure of the expansion work as 𝑛𝑛𝑂𝑂 𝑛𝑛𝐶𝐶𝑂𝑂2 𝑛𝑛𝑁𝑁2 𝑃𝑃 Δ𝑉𝑉 = 𝑅𝑅𝑅𝑅 � 2 + + � = 𝑅𝑅𝑅𝑅 (− 𝜈𝜈𝑂𝑂2 + 𝜈𝜈𝐶𝐶𝑂𝑂2 + 𝜈𝜈𝑁𝑁2 ) Δξ Δ𝜉𝜉 Δ𝜉𝜉 Δ𝜉𝜉
and therefore for this problem
𝑃𝑃 Δ𝑉𝑉 3 = 8.314 𝐽𝐽 𝐾𝐾 −1 𝑚𝑚𝑚𝑚𝑙𝑙 −1 (298.15 𝐾𝐾) �− + 1 + 1 � = 1239 J mol−1 = 1.239 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 Δξ 2
which is less than one percent of the reaction enthalpy. The expansion work per molar extent of reaction is defined as wm = - (PΔV)/ Δξ = - 1.24 kJ mol-1. The net volume of gases increased as a consequence of the reaction resulting in expansion work of the system on the surroundings. The change in internal energy per molar extent of reaction (ΔrUm) can then be defined as C o Δr Um = q m + wm = Δ25 Hm − r
P ΔV Δξ
Δr Um = (−632.1 − 1.24) kJ mol−1 = −633.3 kJ mol−1
The change in internal energy by the chemical reaction was greater than the heat transferred from the system because of expansion work. Solution: Standard Enthalpy of Reaction at 80 C. From Part 1, we have determined the standard reaction enthalpy at 25 C as
Last Modified: March, 2018
2-60
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝐶𝐶 𝑜𝑜 Δ25 𝐻𝐻𝑚𝑚 = −632.1 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝑟𝑟
Let’s now compute the molar heat capacity of reaction defined as: Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 = ��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − ��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑗𝑗
𝑗𝑗
For the products, we have
��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = (1)𝐶𝐶𝑝𝑝𝑝𝑝,𝐶𝐶𝑂𝑂2 + (2)𝐶𝐶𝑝𝑝𝑝𝑝,𝐻𝐻2 𝑂𝑂 + (1)𝐶𝐶𝑝𝑝𝑝𝑝,𝑁𝑁2 𝑗𝑗
��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = (1)(37.11) + (2)(75.29) + (1)(29.13) = 216.82 𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝐾𝐾 −1 𝑗𝑗
For the reactants
��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = (1)𝐶𝐶𝑝𝑝𝑝𝑝,𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + (3 ∕ 2)𝐶𝐶𝑝𝑝𝑝𝑝,𝑂𝑂2 𝑗𝑗
��𝜈𝜈𝑗𝑗 𝐶𝐶𝑝𝑝𝑝𝑝,𝑗𝑗 �𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = (1)(93.14) + (3 ∕ 2)(29.36) = 137.18 𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝐾𝐾 −1 𝑗𝑗
We then obtain:
Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 = 216.82 𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝐾𝐾 −1 − 137.18 𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝐾𝐾 −1 = 79.64 𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝐾𝐾 −1
The standard enthalpy of reaction at To = 80 C = 353.15 K is approximately 𝐶𝐶 𝑜𝑜 Δ80 𝐻𝐻𝑚𝑚 𝑟𝑟 𝐶𝐶 𝑜𝑜 Δ80 𝐻𝐻𝑚𝑚 𝑟𝑟
=
𝐶𝐶 𝑜𝑜 Δ25 𝑟𝑟 𝐻𝐻𝑚𝑚
+�
353.15
298.15
≈ −632.1 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙
−1
𝐶𝐶 𝑜𝑜 Δ80 𝐻𝐻𝑚𝑚 ≈ −627.7 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝑟𝑟
𝐶𝐶 𝑜𝑜 Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 ≈ Δ25 𝑟𝑟 𝐻𝐻𝑚𝑚 + Δ𝑟𝑟 𝐶𝐶𝑝𝑝𝑝𝑝 Δ𝑇𝑇
𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1 𝑘𝑘𝑘𝑘 (353.15 − 298.15) 𝐾𝐾 ( + 79.64 ) 𝐾𝐾 1000 𝐽𝐽
The change in standard reaction enthalpy was less than 1%. The change in reaction enthalpy with temperature is generally much less than other properties (such as the change in reaction entropy given in Chapter 6). From its definition, ΔrCpm = 0 if the molar heat capacities and the stoichiometric coefficients are equal. Enthalpy of Combustion Combustion of organic compounds is often the chemical reaction of greatest interest in the design and understanding of mechanical and biological systems. Combustions are chemical reactions where organic compounds are completely oxidized by O2 (g). For common organic compounds containing C, H, and O, the products are CO2 (g) and H2O (l). For compounds that also contain N, N2 (g) is also a product. Combustion processes are important sources of energy for cells as well as mechanical engines.
Last Modified: March, 2018
2-61
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Ignition energy is often needed to start combustion of engine fuels. This energy allows the elemental bonds to be broken between atoms for a few organic molecules. When these atoms recombine with the oxygen atoms, additional energy is generated that releases other atoms from their molecules. The reaction will then proceed until all (effectively) of the fuel is consumed (assuming adequate oxygen). If the chemical reaction is written such that the stoichiometric coefficient for the fuel is υfuel=1, then the molar extent of reaction is equal to the number of moles of the fuel, that is, Δ𝜉𝜉 =
Δ𝑛𝑛𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = Δ𝑛𝑛𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑛𝑛𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝜈𝜈𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
2.7.24
where nfuel is the number of fuel moles in the system and, for these powerful reactions, it is equal to the actual number of fuel moles consumed (Δnfuel). Standard reaction enthalpies for complete combustion reactions (organic reactant + O2) can be computed and stored in tables and databases for your use (you don’t need to compute them from their products and reactant components). Some examples are given in the Thermodynamic Data Handout. They are called standard enthalpy of combustion (sometimes called heat of combustion) and we will use the notation of ΔcTHm°. It is defined for a standard pressure as o o ) ΔTc Hm = ( ΔTr Hm combustion =
( ΔTr H o )combustion 𝑛𝑛𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
2.7.25
The standard enthalpy of combustion is equal to the standard reaction enthalpy for Δξ = nfuel, that is, the standard enthalpies of reactions for one mole of organic reactant. For expansion work under constant pressure, we can easily compute the heat transfer as simply q = ΔU+PΔV= nfuel ΔcTHmo. We have already analyzed the following examples of combustion in this section and the results for the combustion of carbon monoxide of Homework Problem 2.33. Once again, the 𝑜𝑜 numerical values of standard enthalpies of reaction and combustion are equal but Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 is 𝑜𝑜 defined per molar extent of reaction and Δ𝑇𝑇𝑐𝑐 𝐻𝐻𝑚𝑚 is defined by mole of organic reactant. Standard enthalpies of combustion for sugars, fats and proteins are given in the next section. 𝐶𝐶 (𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔ℎ𝑖𝑖𝑖𝑖𝑖𝑖) + O2 (g) → CO2 (g)
o −1 Δ25C c Hm,c = − 110.5 kJ mol (NH2 )2 CO (s) + 3⁄2 O2 (g) → CO2 (g) + 2 H2 O (l) + N2 (g) o −1 Δ25C c Hm,urea = − 632 kJ mol 𝐶𝐶𝐶𝐶 + 1⁄2 O2 (g) → CO2 (g) o −1 Δ25C c Hm,co = − 110.5 kJ mol
Metabolic Rate and Thermoregulation Introduction
Biochemical reactions are the source of the metabolic rate of humans and other animals. The energy from these reactions are defined by the standard enthalpy of reaction for the
Last Modified: March, 2018
2-62
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
combustion of carbohydrates (including sugar), fats, proteins and other organic compounds. As shown by Homework Problem 2.26, biochemical reactions require thermoregulation processes to avoid high body temperatures that can be lethal to humans (and other animals). Thermoregulation is achieved by the heat transfer of convection, conduction, and radiation and by the evaporation of perspiration. As the body temperature increase, the blood flow to the skin increases causing greater heat transfer from the body and larger evaporation rate. If the temperature of the surroundings is greater than the body temperature (and emissivities are equal), evaporation is the only process available to regulate body temperature. We are often interested in understanding and in designing systems that include humans and other animals. For conditions of constant pressure, the First Law for these system is ΔU + PΔV = ΔHtpr + ΔHvap + ΔHr = q
2.7.26a
where the system has all of three possible changes in its internal energy given in this chapter. Temperature changes in rooms or building are important in ventilation designs. This change can be obtained from Eq. 2.7.26a as ΔT =
ΔHtpr q − ΔHvap − ΔHr = n Cpm n Cpm
2.7.26b
where the reaction enthalpy is negative for metabolic processes and vaporization enthalpy is a component of thermoregulation. Animal Calorimetry Animal calorimetry is the measurement of the heat transfer between animals and their environment and is dependent on the bio-chemical reactions of metabolism. A simplified representation of the measurement approaches is given in Fig. 2.33. The direct measurement of heat transfer (with consideration of vaporization) from animals is called direct calorimetry. The simplest of direct calorimeters measures the heat transfer using principles similar to those of an adiabatic bomb calorimeter (see Homework Problem 2.32). Here heat transfer from animals is obtained by measuring changes in temperature of a substance with well-defined thermodynamic characteristics. Thermoregulation by evaporation is typically assessed with a separate set of measurements. For a rigid chamber, expansion work needs to be computed separately to define enthalpy values. Experiments using direct calorimetry need to be done carefully to account for evaporation from fountains and fecal pans.
Last Modified: March, 2018
2-63
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.33. Simplified Representation of Calorimetric Methods. Indirect calorimetry uses the reaction enthalpy for the combustion of carbohydrates (including sugars), fats, protein and other organic compounds using metabolic chemical reactions to determine the equivalent heat transfer. This approach is also shown in Fig. 2.33. In contrast to direct calorimetry, the heat transfer corresponds to metabolic processes within living tissues of animals. The heat transfer leaving animals only equals this rate for negligible change in the body temperature and no tissue gain or production of products like milk or eggs. Indirect calorimetry is more widely used than direct calorimetry and is closely tied to our enthalpy of reaction relationships. Since enthalpy of reaction varies with the source, molar extents of reaction for sugar, fats and other food are needed to determine the net enthalpy of reaction (where measured changes for the reactants are negative). The measured the respiratory quotient (Rq) shown in Fig. 2.33 is useful in partitioning metabolism among the different sources of food. Table 2.3. Summary of Illustrative Metabolic Reactions.
Source
o Δ25C r Hm
o Δ25C r HmO2
o Δ25C r HmCO2
o Δ25C r HvO2
o Δ25C r HvCO2
kJ L-1CO2
Rq
-2803
-467
-467
-18.9
-18.9
1.00
-17.7
-24.8
0.71
kJ mol-1
kJ mol-1O2
kJ mol-1CO2
C6 H12 O6 (s, glucose) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (l) C6H12O6
C57 H104 O6 (l, triolein) + 80 O2 (g) → 57 CO2 (g) + 52 H2 O (l) C57H104O6
-35099
-439
-616
kJ L-1O2
C11 H14 N2 O3 (s, glyclphenylalanine ) + 13 O2 (g) → 11CO2 (g) + 7H2 O (l) + N2 (g) C11H14N2O3
-5645
Last Modified: March, 2018
-434
-513
2-64
-17.5
-20.7
0.84
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
CH3 CH(OH)COOH (s, lactic acid ) + 3 O2 (g) → 3CO2 (g) + 3H2 O (l) C3H6O3
-1344
-448
-448
-18
-18
1.00
Generic Metabolic Reactions (Walsberg and Hoffmann, 2005) Carbohydrates
-473
-473
-19.1
-19.1
1.00
Lipids (fats)
-444
-624
-17.9
-25.1
0.71
Proteins
-430
-531
-17.3
-21.4
0.81
Let’s further explore the metabolic rate obtained from indirect calorimetric methods by considering the standard enthalpies of reaction for different food sources. Table 2.3 shows 𝑜𝑜 ′𝑠𝑠 for glucose (sugar), triolein (fat), chemical reactions and corresponding Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 glyclphenylalanine (peptide used to represent proteins) and lactic acid. The stoichiometric coefficients are defined relative to a value of one for the source (organic reactant). We will begin our analysis by considering the combustion of a single food source (such as a sugar or fat) where the standard enthalpy of reaction and stoichiometric coefficients are as defined in Table 2.3. For indirect calorimetry, changes in the moles of oxygen and/or carbon dioxide are directly measured. From our definition of stoichiometric coefficients (Eq. 2.7.4), the molar extent of reaction can be defined by the measured change in the number of O2 or CO2 moles (same result for either measurement). The reaction enthalpy of metabolism is then defined as Δξ = −
Δno2 Δnco2 = νo2 νco2
o ΔHr = Δξ ΔTr Hm = −Δno2 �
2.7.27a 𝑜𝑜 𝑜𝑜 ΔTr 𝐻𝐻𝑚𝑚 ΔTr 𝐻𝐻𝑚𝑚 � = Δnco2 � � 𝜈𝜈𝑜𝑜2 𝜈𝜈𝑐𝑐𝑐𝑐2
2.7.27b
where Δn of reactants are negative values. Instead of using the molar extent of reaction, it is easier to determine the reaction enthalpy by using the direct measurement of the changes in number of moles or in the gas volume of oxygen or carbon dioxide. Alternative forms of the standard enthalpy of reaction are given in the equations below, where molar volume of Vmo2 = Vmco2 = 24.8 L mol-1are defined from the ideal gas equation (i.e., RT/P) at standard ambient temperature and pressure (SATP). 𝑜𝑜 ΔHr ΔTr 𝐻𝐻𝑚𝑚 = = − Δ𝑛𝑛𝑜𝑜2 𝜈𝜈𝑜𝑜2 T 𝑜𝑜 ΔH Δ r r 𝐻𝐻𝑚𝑚 o ΔTr Hmco2 = = Δ𝑛𝑛𝑐𝑐𝑐𝑐2 𝜈𝜈𝑐𝑐𝑐𝑐2 𝑜𝑜 T 𝑜𝑜 ΔH −Δn ΔTr 𝐻𝐻𝑚𝑚𝑚𝑚2 r O2 Δr 𝐻𝐻𝑚𝑚𝑚𝑚2 T o Δr Hvo2 = = = −Δ𝑉𝑉𝑂𝑂2 −Δ𝑛𝑛𝑜𝑜2 (𝑅𝑅𝑅𝑅⁄𝑃𝑃) 𝑉𝑉𝑚𝑚𝑚𝑚2 T 𝑜𝑜 T 𝑜𝑜 ΔH Δn Δ 𝐻𝐻 Δ r co2 r 𝑚𝑚𝑚𝑚𝑚𝑚2 r 𝐻𝐻𝑚𝑚𝑚𝑚𝑚𝑚2 o ΔTr Hvco2 = = = Δ𝑉𝑉𝑐𝑐𝑐𝑐2 Δ𝑛𝑛𝑐𝑐𝑐𝑐2 (𝑅𝑅𝑅𝑅⁄𝑃𝑃) 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚2 o ΔTr Hmo2
2.7.27c 2.7.27d 2.7.27e 2.7.27f
Values for different food sources obtained from Eqs. 2.7.27c through 2.7.27f are summarized in Table 2.3. You will compute these values as part of the assignments of Homework
Last Modified: March, 2018
2-65
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Problems 2.35 and 2.37. To compute ΔHr for a given food source, you only need to multiply the appropriate table value by the measured changes in number of moles or in gas volumes. Let’s consider the case where the only source of metabolism is the fat triolein. If the measured concentrations is a decrease in O2 of 20 L O2 h-1 (and therefore an increase of CO2 of 14.2 L CO2 h-1), the metabolic rate is simply (-17.7 kJ L-1 O2) (20 L O2 h-1) = -354 kJ h-1. Insight into types of consumed food sources can be obtained by examining the measured respiratory quotient (Rq) defined as Rq =
Δnco2 Δξ νco2 νco2 = = −Δno2 Δξ νo2 𝜈𝜈𝑜𝑜2
2.7.27g
where, once again, the measured Δno2 is negative. The measured Rqobs = Δnco2/(-Δno2) can be compared to the Rq = υco2/υo2 values in the Table 2.3 for different food sources of carbohydrates, fats and proteins. A measured Rqobs = 1 corresponds to bio-chemical reactions dominated by carbohydrates; whereas a measured Rqobs = 0.85 corresponds to bio-chemical reactions of more than one food source. The use of the measured respiratory quotient to determine the metabolic rate for two different food sources is illustrated in Homework Problem 2.38. This concept is also illustrated in Fig. 2.33. Here the measured Rqobs is used to determine the separate molar extents of reaction for sugar and fat. The total reaction enthalpy of metabolism is then computed by adding the reaction enthalpy of each food source. A general relationship for relating metabolic rate to measured oxygen consumption and carbon dioxide production is Brouwer equation (see Nienaber et al., 2009). The most general form of this equation also includes methane production and nitrogen excretion rate. For our purposes, we will only use measured oxygen and carbon dioxide rates. The reaction enthalpy from metabolic processes is then defined from Brouwer equation as Δr H = −(14.6 Vo2 + 4.53 Vco2 )
2.7.28
where ΔrH is the effective reaction enthalpy for metabolic processes involving multiple food sources in units of kJ, Vo2 is the consumption volume of O2 in liters at SATP and Vco2 is the production volume of CO2 in liters at SATP. The use of this equation is illustrated in Homework Problem 2.39. Let’s consider the expansion work for the metabolic reactions. From Example Problem 2.5, we can use the following relationship to compute molar expansion work for the reactions in Table 2.3: −𝑤𝑤𝑚𝑚 =
P ΔV = R T (−νo2 + 𝜈𝜈𝑐𝑐𝑐𝑐2 + 𝜈𝜈𝑛𝑛2 ) Δ𝜉𝜉
2.7.29
We are only interested in the stoichiometric coefficient for N2 for metabolic reactions of proteins and related compounds. For glucose and lactic acid reactions where the respiratory quotient is one, the expansion work term is zero and the change in internal energy equals change in enthalpy. As shown in Homework Problem 2.35 for triolein and
Last Modified: March, 2018
2-66
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
glyclphenylalanine, the expansion work term is less than 1% of the standard enthalpy of combustion. We will therefore not make adjustments for expansion work to estimate the change in internal energy for metabolic reactions, that is, o o o ΔTr Um = ΔTr Hm − (P ΔV⁄Δξ) ≈ ΔTr Hm
2.7.30
Human Metabolic Rate
Values for reaction enthalpies of combustion processes obtained from calorimetric methods are often summarized in relatively simple relationships for design problems. Relationships for the design of pig and turkey facilities are given in Homework Problems 2.42 and 2.43. In this section, we will focus on relationships for human by first considering the basal human metabolic rate. The basal human metabolic rate is defined as the rate of energy expenditures for an undisturbed state in a thermally neutral environment without actively digesting food. For our purposes, an effective resting metabolic rate (corresponding to a relaxed sitting position) of Mifflin-St Jeor can be estimated by using the following basal rate equation (based on data from 498 individuals between 19 and 78) rest Δr Ḣ body = −0.174 (10 m + 625 ht − 5 age + Csex )
2.7.31
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 where the Δr 𝐻𝐻̇𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 is the effective metabolic rate corresponding to the heat transfer in units -1 -1 of kJ h person , m is the mass of the human in kg, ht is height in m, age is person age in years, and Csex is a factor related to gender. It is equal to 5 for males and -161 for females. The metabolic rate increases with physical activities. Approximate estimates for these activities can be obtained by multiplying the metabolic rate of Eq. 2.7.31 by factors given in Table 2.4.
The evaporation rate of sweat increases with body temperature. For exercising healthy young males (with a body surface area of approximately 1.85 m2), the sweat rate increases roughly by 240 g m-2 h-1 for ΔTbody = 1 C. Estimates of the equivalent heat transfer for vaporization are shown in Table 2.3 as a percentage of the negative reaction enthalpy. Vaporization enthalpies vary substantially based on a number of factors. It increases with warmer surroundings. Table 2.4. Increase in Metabolic Rate with Activities Relative to Resting. Activity Studying Moderate Walking Moderate Labor
Metabolic Factor 1.3 1.5 1.6
% Evap 55% 65% 65%
Activity Dancing Fast Walking Heavy Labor
Metabolic Factor 2.5 3.0 4.0
% Evap 75% 73% 70%
Example Problem 2.7: Change in Room Air Temperature Problem Statement. Consider a study room with dimensions of 15 m by 20 m with a ceiling height of 3 m. The room is completely insulated (q= 0) and has 10 males and 10 females who
Last Modified: March, 2018
2-67
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
are happily studying thermodynamics. The men have an average height of 1.8 m, average mass of 75 kg, and an average age of 20 years. The women have an average height of 1.65 m, average mass of 60 kg, and an average age of 20 years. After studying for one-hour, what is the change in the room’s temperature? What is the total mass of water evaporated from the students? You can estimate the thermodynamic parameters using a room temperature of 25 C.
Thermodynamic Data: For this problem, we will use the following characteristics for air and water (R = 8.314 J K-1 mol-1): Air: mm = 29 g mol-1, Cpm = 29 J K-1 mol-1; ρair = 1.3 kg m-3 25 C o H2O: mm = 18 g mol-1, ∆ vap H m = 44 kJ mol-1 Cvm = Cpm − R = 29 − 8.314 = 20.686 J K −1 mol−1
From Table 2.4: Metabolic Studying Factor: 1.3 Evaporation fraction is the equivalent heat transfer for vaporization (qvap) relative to reaction enthalpy of metabolic processes. q vap 𝑓𝑓𝑣𝑣𝑣𝑣𝑣𝑣 = = −0.55 Δ𝐻𝐻𝑟𝑟
Non-expansion Work Relationships: For this problem, the room is of constant volume and therefore the expansion work is zero. For q=0 and PΔV = 0, the First Law can be written as ΔU = ΔUtpr + ΔUvap + ΔUr = q + w = 0
From the ideal gas equation, we know that PdV = RT dn for vaporization (see Example Problem 2.4). We can then estimate the molar change in internal energy for Δn moles of vaporization as:
Last Modified: March, 2018
2-68
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
R T Δnvap P ΔV C o C o = Δ25 = Δ25 vap Um + vap Um + R T Δ𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣 Δ𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣 1 kJ 8.314 �J x 1000 J� C o 25 C o (25 + 273.5) 𝐾𝐾) = 41.5 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 Δ25 vap Um = Δvap Hm − RT = (44 − K mol 𝐶𝐶 o Δ 25 vapHm
C o = Δ25 vap Um +
As previously discussed, we will neglect the expansion work term for the metabolic reactions 𝑜𝑜 𝑜𝑜 Δ𝑇𝑇𝑟𝑟 𝑈𝑈𝑚𝑚 ≈ Δ𝑇𝑇𝑟𝑟 𝐻𝐻𝑚𝑚 and use ΔUr = ΔHr − P ΔV ≈ ΔHr
The change in internal energy of vaporization is defined from fvap. For our constant volume room, w=0 and we obtain (using the First law) ΔUvap = q vap + w = q vap = fvap ΔHr = −0.55 ΔHr = − 0.55 ΔUr
where fvap = -0.55 has been used.
Solution: The number of moles of air in the room is obtained as 𝑘𝑘𝑘𝑘 (15 𝑚𝑚)(3 𝑚𝑚)(20 𝑚𝑚) = 1170 𝑘𝑘𝑘𝑘 𝑚𝑚3 g 1170 kg x1000 m kg n= = = 40,344.8 mol mm 29 g mol−1 m = ρair 𝑉𝑉 = 1.3
The reaction enthalpy corresponding to metabolism is defined per student as 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 rest Δr H𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 1.3 �Δr Ḣ body �Δt = −1.3 � 0.174 (10 m + 625 ht − 5 age + Csex )� (1 h)
Female students: 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Δr H𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = − 0.226 (10 (60 kg) + 625(1.65 𝑚𝑚) − 5(20 𝑦𝑦) − 161) = −310.0
Male students: 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Δr H𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = − 0.226(10 (75 kg) + 625(1.8 𝑚𝑚) − 5(20 𝑦𝑦) + 5) = − 402.6
𝑘𝑘𝑘𝑘 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
These rates correspond to burning an equivalent number of daily food calories of approximately 1770 and 2300 for females and males, respectively. Change in reaction enthalpy and internal energy: study study ΔUr = ΔHr = Δr H𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (# female) + Δr H𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 (# male) ΔUr = −310(10 females) − 402.6 (10 males) = −7126 kJ
Change in internal energy from liquid water to vapor:
ΔUvap = −0.55 ΔUr = −(−7126 𝑘𝑘𝑘𝑘)(0.55) = 3919 𝑘𝑘𝑘𝑘 Last Modified: March, 2018
2-69
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Internal energy for related temperature: As previously shown, the 1st Law can be rearranged as ΔUtpr = −ΔUvap − ΔUr = −(−7126 𝑘𝑘𝑘𝑘) − 3919 𝑘𝑘𝑘𝑘 = 3207 𝑘𝑘𝑘𝑘
Change in Room Temperature: ΔUtpr = n Cvm ΔT
1000 𝐽𝐽 3207 𝑘𝑘𝑘𝑘 ΔUtpr 𝑘𝑘𝑘𝑘 ΔT = = = 3.8 𝐾𝐾 = 3.8 𝐶𝐶 𝑛𝑛 𝐶𝐶𝑣𝑣𝑣𝑣 (40,345 𝑚𝑚𝑚𝑚𝑚𝑚) 20.686 𝐽𝐽 𝐾𝐾 −1 𝑚𝑚𝑚𝑚𝑙𝑙 −1
Mass of water:
o ΔUvap = Δ𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣 Δvap Um ΔUvap 3919 𝑘𝑘𝑘𝑘 Δ𝑛𝑛𝑣𝑣𝑣𝑣𝑣𝑣 = = = 94.4 𝑚𝑚𝑚𝑚𝑚𝑚 o Δvap Um 41.5 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 g Δmvap = Δnvap mm = (94.4 mol) 18 = 1699 g = 1.7 kg mol
The change in temperature is likely too large to maintain a comfortable environment for the students. Either a window needs to be opened or ventilation system is needed for this room. The design of a ventilation system for this problem is given in Example Problem 3.5 in Chapter 3.
HEAT TRANSFER BY RADIATON Basic Concepts Blackbody To gain insight into the nature of radiation, Kirchhoff conceived of a blackbody that sends out radiation at all frequencies and absorbs all radiations (hence black because it reflects none). Black bodies were used experimentally to study radiation by using soot-covered interior walls to form a blackened cavity. The hole of the cavity served as a surrogate of a black body. Black bodies are similar to ideal gases. These idealized substances provide a useful framework for deriving theoretical relationships and are often an adequate approximation to real material.
Last Modified: March, 2018
2-70
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Blackened cavity ~ Black Body Figure 2.34. Black Body as Idealized Source of Radiation. Radiation is typically represented using radiant flux density or radiant power per unit area, defined as 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 𝑅𝑅̇ =
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 𝐽𝐽 𝑚𝑚−2 𝑠𝑠 −1 = 𝑊𝑊 𝑚𝑚−2 (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴) (𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇)
2.8.1
which is a function of the temperature of the body To determine the radiant power we need to integrate over the appropriate area: 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 = ∫ 𝑅𝑅̇ 𝑑𝑑𝑑𝑑
2.8.2
where 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 is the radiant power for any arbitrary body. To determine the radiant heat transfer (not net) for body i: 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 = � 𝑡𝑡
𝑡𝑡+Δ𝑡𝑡
𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑
2.8.3
The net heat transfer requires a balance between the incoming and outgoing radiation from the body. Irradiance is as the radiant flux density incident (upon) on a surface arriving from all directions (J m-2 s-1 = W m-2). Radiant emittance is the radiant flux density leaving a surface Early Theoretical Framework By the late 19th century, there was general consensus that radiation was caused by the vibrations of molecules and atoms of solids but the proper theoretical framework was not yet developed. A simple harmonic oscillator was proposed to represent the vibration of electrons. We will introduce this topic by considering continuously vibrated springs. As discussed later in the chapter, this approach needed to be modified to account for quantum mechanics. Let’s first review the notation of a simple spring shown below.
Last Modified: March, 2018
2-71
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
x negative
x positive
Relax State
x =0 Fx = 0
Fx positive
Fx negative
x
x
x
Figure 2.35a. Position Dependent Forces of Springs. Clearly the force is negative after the spring is extended from the relaxed and is positive if the spring is compressed. We will limit our discussion to the magnitude for a vector acting in the x-direction only. Force for a spring varies with its position with x as defined by Hooke’s law, or, Fsp = −k sp x
2.8.4a3
where Fsp is the spring force acting on the block and ksp is the spring constant (units of mass per time2). We now expand our analysis for a continuously vibrating spring. For illustration purposes, we will rotate our spring so that movement is in the vertical direction. Our idealized system has no gravity or friction forces. As shown below, our continuous vibrating spring has different locations with time. We start at its relax state of z = 0. The maximum and minimum values of z correspond to z = zmax and z = -zmin. As shown in the right-sided figure, the spring will clearly return to these locations as it continuously vibrates. The time required for it to repeat a cycle is related to the frequency of the spring.
Figure 2.35b. Continuously Vibrating Spring. Since force is the product of mass and acceleration, our spring force can then be written as
Last Modified: March, 2018
2-72
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Fsp
d2 z = m 2 = −k sp z dt
2.8.4b
where m is the oscillating mass, z is the displacement distance shown in the above figure and ksp is the spring constant. The above equation can be rearranged into the following secondorder linear ordinary differential equation with constant coefficients k sp d2 z + � �z = 0 dt 2 m
2.8.4c
A good course on ordinary differential equation will cover the general solution to this type of equation. For our purposes, we will simply suggest a solution of the form of z = a sin (2πbt), where “a” and “b” are unknown coefficients and t is time. This is certainly a reasonable suggestion given that our continuously vibrating spring periodically returns to the same vertical positions. By trying this relationship, the spring equation can be evaluated as k sp −a (2πb)2 sin( 2π b t) + � � a sin( 2πb t) = 0 m
2.8.4d
We conclude that our suggested function is a valid solution if we rearrange term and define our frequency response (υw) as b = νw =
k 1 � sp 2π m
2.8.4e
When t = 1/(4υw), we obtain a sine value of one and minus one for t =3/(4υw) We conclude that “a” is the amplitude defined as 𝒜𝒜𝑠𝑠𝑠𝑠 = zmax = -zmin. Our equation for simple harmonic motion is then defined as z = 𝒜𝒜sp sin( 2π υw t)
2.8.4f
Graphical solution is shown below. The period is defined as λ = 1/υw and corresponds to the time interval for the mass returns to the relaxed state (and other locations).
Last Modified: March, 2018
2-73
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.35c. Position Locations for Vibrating Springs. From the Maxwell’s elegant electromagnetic theory, the wavelength and frequency (defined as the number of times per second the value at a fixed point repeats) of electromagnetic waves of radiation are related as c = λ υw
2.8.4g
where c is the speed of light (299,792,458 m s-1 ≈ 300,000,000 m s-1), λ is the wave length [L], and υw is the frequency [cycles per time - Hz]. As easily shown with prisms, light is composed of electro-magnetic waves of different wavelengths. The range of wavelengths of radiation for a blackbody is shown below. The wavelength varies from a length much smaller than the diameter of a hydrogen atom to a length much larger than a football field. Wavelengths for the visible spectrum are much longer than gamma waves or X rays, but they are still quite small, approximately 1000 times smaller than the typical width of human hair.
Last Modified: March, 2018
2-74
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.35d. Range in Wavelengths for Blackbodies. Many biological and environmental processes are dependent on the energy associated with a particular range of wavelengths. For these processes, it is useful to define Energy per wavelength 2πc = ( nλ )� ξ̅p � = � 4 � � ξ̅p � (Area)(Time) λ
2.8.4g
where the first term in the brackets is loosely the number of waves per unit volume of a blackbody cavity per wavelength between frequencies of λ and λ+dλ (with adjustments to obtain flux units), and the second bracket term is the average energy per wave. In Chapter 1, we discussed the equipartition of energy corresponding to RT/2 per mole for each degree of freedom, which is equivalent to kBT/2 per molecule for each degree of freedoms. For a simple one-dimensional harmonic oscillator with two degrees of freedom corresponding to a “spring” for each direction, the average energy per wave is: ξ̅p = k B T
2.8.4h
This approach closely matched experimental data for large wavelengths but poorly represented radiation for small wave lengths. This so-called ultraviolet catastrophe was successfully resolved using the concepts from quantum mechanics. These results are given later in the chapter.
Last Modified: March, 2018
2-75
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Spectral Flux Density Let’s now expand upon on radiation concepts introduced with vibrating springs. We have previously discussed that black-body radiation can be represented by electro-magnetic waves with wavelengths varying from very small (smaller than diameter of hydrogen atoms) to very large (longer than football fields). The radiation energy corresponding to individual wavelengths is important in understanding photosynthesis, climate change and other biological and environmental processes. Spectral flux density is used to represent the radiant flux density per wavelength. Insight into the definition of spectral flux density can be obtained by cumulating the energy for increasing wavelengths. We can start by using zero energy if there is no wavelength (i.e., λ equals zero). We can add the contribution to radiant energy by computing the energy for each Δλ as we move to larger wavelengths. We will use the symbol of 𝑅𝑅� (𝜆𝜆) for this cumulative radiant energy. This concept is shown in Figure 2.36 where the range in wavelengths varies between gamma rays to long radio waves. Let’s examine the slope of a point on the curve. This slope is the spectral flux density and is defined mathematically as 𝑅𝑅̇𝜆𝜆 = lim � Δ𝜆𝜆→0
𝑅𝑅� (𝜆𝜆 + Δ𝜆𝜆) − 𝑅𝑅� (𝜆𝜆) 𝑑𝑑𝑅𝑅� 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � = → Δ𝜆𝜆 𝑑𝑑𝑑𝑑 (𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴)(𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇)(𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊ℎ)
2.8.4
Spectral flux density, 𝑅𝑅̇𝜆𝜆 , is defined such that of 𝑅𝑅̇ (𝜆𝜆)𝑑𝑑 𝜆𝜆 is the radiant power per unit area between the wavelength of and λ+dλ. Radiant flux density is then obtained by integrating over all wavelengths.
Figure 2.36. Illustration of Spectral Density Function.
Last Modified: March, 2018
2-76
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
We are interested in developing a theory that can be used to determine the spectral flux density for a black body. A good starting point was the definition of the energy per wave previously given where the average energy per wave was initially estimated as kBT. However, Planck drastically improved the representation by assuming that energy is not continuous, but has a discrete set of values defined as ξp = 𝜄𝜄 ℎ𝑝𝑝 𝜐𝜐𝑤𝑤
𝜄𝜄 = 0, 1, 2, …
2.8.6
where ξp is the energy, ι is a positive integer called the quantum number, and hP is Planck’s constant (= 6.63x10-34 J s). Einstein realized that radiation is emitted or absorbed when the electron moves between different energy levels of e2 and e1, that is, hPυw = e2-e1. Photons are released by the movement of electrons between these different levels. A graphical illustration of Einstein’s concept for one form of photon generation is shown below. Also shown below is the extension of this concept within the framework of Bohr’s model of electron orbits. Before Excited State
During
After n=3
e2
n=2 n=1 photon=hυ
Incident photon=hυ
∆e=hυ
photon=hυ
e1 Ground State
Einstein’s Approach
Bohr’s Simple Atom
Figure 2.37. Illustration of Discrete Energy Levels. For the quantum representation of energy and by using the relationship between speed of light ( c ), frequency (υw) and wavelength (λ) of c = υwλ, the average energy for each frequency is defined as 𝜉𝜉𝑝𝑝̅ =
ℎ𝑝𝑝 𝜐𝜐𝑤𝑤 ℎ𝑝𝑝 𝑐𝑐 ∕ 𝜆𝜆 = ℎ 𝜐𝜐 ℎ 𝑐𝑐 exp � 𝑃𝑃 𝑤𝑤 � − 1 exp � 𝑃𝑃 � − 1 𝑘𝑘𝐵𝐵 𝑇𝑇 𝜆𝜆 𝑘𝑘𝐵𝐵 𝑇𝑇
2.8.7
By using this relationship for the frequency energy, the spectral flux density can be defined as ℎ𝑝𝑝 𝑐𝑐 ⁄𝜆𝜆 2 𝜋𝜋 ℎ𝑝𝑝 𝑐𝑐 2 2 𝜋𝜋 𝑐𝑐 1 ̇ 𝑅𝑅𝜆𝜆𝜆𝜆 = � 4 � � �=� �� � 5 ℎ 𝑐𝑐 ℎ 𝑐𝑐 𝜆𝜆 𝜆𝜆 exp � 𝑃𝑃 � − 1 exp � 𝑃𝑃 � − 1 𝜆𝜆 𝑘𝑘𝐵𝐵 𝑇𝑇 𝜆𝜆 𝑘𝑘𝐵𝐵 𝑇𝑇
2.8.8
For hP = 6.62607x10-34 J s, c = 299,792,458 m s-1 and kB = 1.38066 x 10-23 J K-1, we can simplify as
Last Modified: March, 2018
2-77
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
C1 = 2π h P c 2 = 3.742 x 10 −16 J m 2s −1 = 3.742 x 108 W µm 4 m −2 C2 =
2.8.9
hP c = 1.4388 x 10 − 2 m K = 1.4388 x 10 4 µm K kB
2.8.10
which simplifies the equation as function of the wavelength and temperature as 1 𝐶𝐶1 𝑅𝑅̇𝜆𝜆𝜆𝜆 = � 5 � � � 𝐶𝐶2 𝜆𝜆 exp � � − 1 𝜆𝜆 𝑇𝑇
2.8.11
Solar and Terrestrial Radiation
120
60
Solar (6000 K) Photosynthetically Active Radiation (0.4 to 0.7 µm)
80
40
Earth (288 K)
40
20
0 0.1
Ultraviolet (< 0.4 µm)
10 Wavelength (µm)
1
Violet (0.40-0.43µm)
Green (0.49-0.56µm)
Blue (0.43-0.49µm)
Orange (0.59-0.63µm)
0
Earth Spectral Flux Density (W/m2/μm)
Solar Spectral Flux Density (MW/m2/μm)
The spectral flux density for the temperature of the sun (T = 6000 K) and for the temperature of the earth (T = 288 K) are shown below.
100
Infrared (> 0.76 µm)
Yellow Red (560 - 590µm) (0.63-0.76µm)
Figure 2.38. Spectral Flux Densities for Temperatures of the Sun and Earth Most of the solar radiation corresponds to relatively short wavelengths (less than 4 m) and most of the terrestrial radiation corresponds to relatively long wavelengths (greater than 4 m). Note that the scale for the earth is considerably smaller. We are interested in determining the total energy as a function of temperature. We are also interested in the amount of solar radiation reaching the earth. In the above figure, the range of wavelengths used for photosynthesis is shown. This component is called photosynthetically active radiation (PAR). Many animals have evolved such that their eyes see those frequencies where sunlight has its maximum intensity. Representation Using Single Curve: Excluded from Student Handout
Last Modified: March, 2018
2-78
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
As shown for the Sun and Earth, the spectral density function is dependent on temperature, that is, a different curve is obtained for each temperature. However, a single curve as a function of λT can be constructed for a radiation variable defined as
R λb
R λT, b =
T
5
=
1 (λT )
5
(
C1 ) C2 exp( ) − 1 λT
2.8.12
A plot of this relationship is shown below in the left-sided figure. For a given temperature, the spectral density function can then be determined from this graph. Of greater usefulness to us, however, is the cumulative faction of spectral density function shown in the right-sided figure. For a given temperature (T= constant), this fraction can be defined from the left-sided figure because λT
∫ R λT, b d( λT)
FλT = 0 ∞
∫ R λT, b d (λT)
5 ∫ (R λb / T ) d( λT)
= 0 ∞
T
∫ R λb d λ
=0
∫ (R λb / T ) d (λT) 5
2.8.13
R b
0
0
1.0
0.015
0.9
Fraction of Radiant Flux Density
Spectral Flux Density per T5 (nW m-2 μm-1 K-5)
λT
0.010
0.005
0.000
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
0
2500 5000 7500 10000 12500 15000
λ T (μm K)
0
2500
5000
7500 10000 12500 15000
λ T (μm K)
Figure 2.39. Single Curve Representation of Spectral Flux Density. Example Problem 2.8: Molar Photon Energy Let’s compute the energy ξp of one quantum of energy (m=1) of green light (λ=5.5x10-7 m). The energy of single quantum is (using υw = c/λ) defined as
Last Modified: March, 2018
2-79
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝜉𝜉𝑝𝑝 = ℎ𝑃𝑃 𝜐𝜐𝑤𝑤 =
ℎ𝑝𝑝 𝑐𝑐 𝜆𝜆
By using hP = 6.63x10-34 J s and c = 3 x108 m s-1, we obtain 𝜉𝜉𝑝𝑝 =
(6.63𝑥𝑥10−34 𝐽𝐽 𝑠𝑠)(3𝑥𝑥108 𝑚𝑚 𝑠𝑠 −1 ) = 3.6 𝑥𝑥10−19 𝐽𝐽 𝑝𝑝ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛−1 5.5 𝑥𝑥 107 𝑚𝑚
Energy content for a mole of photons corresponds to Avagadro’s number of them. We then obtain 𝜉𝜉𝑝𝑝𝑝𝑝 = (6.023 𝑥𝑥 1023 𝑝𝑝ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑙𝑙 −1 )(3.6 𝑥𝑥10−19 𝐽𝐽 𝑝𝑝ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛−1 ) = 2.17 𝑥𝑥 105 𝐽𝐽 𝑚𝑚𝑚𝑚𝑙𝑙 −1
This is often called the molar photon energy. Radiant Flux Density Blackbody radiation
We are now ready to derive the radiant flux density. From the spectral flux density, the radiant flux density can be defined by integrating over all frequencies, that is ∞
𝑅𝑅̇ = � 𝑅𝑅̇𝜆𝜆 𝑑𝑑𝑑𝑑
2.8.14
0
If we use the spectral flux density relationship developed for a black body in the previous section, we obtain ∞ ∞ 1 1 𝑅𝑅̇𝑏𝑏 = � 𝑅𝑅̇𝜆𝜆𝜆𝜆 𝑑𝑑𝑑𝑑 = 2 𝜋𝜋 ℎ𝑝𝑝 𝑐𝑐 2 � � 5 � � � 𝑑𝑑𝑑𝑑 ℎ𝑃𝑃 𝑐𝑐 𝜆𝜆 0 0 exp � �−1 𝜆𝜆 𝑘𝑘𝐵𝐵 𝑇𝑇
2.8.15
where the subscript “b” has been used to indicate radiant flux density for a black body. The integration is easier if we define an integration variable of y = hPc/(λkBT) and dλ = [hPc/(kBTy2)] dy. We then obtain ∞
𝑅𝑅̇𝑏𝑏 = � 𝑅𝑅̇𝜆𝜆𝜆𝜆 𝑑𝑑𝑑𝑑 = 0
2 𝜋𝜋 𝑘𝑘𝐵𝐵4 𝑇𝑇 4 ∞ 𝑦𝑦 3 2 𝜋𝜋 5 𝑘𝑘𝐵𝐵4 4 � � 𝑦𝑦 � 𝑑𝑑𝑑𝑑 = � 3 2� T 𝑒𝑒 − 1 ℎ𝑃𝑃3 𝑐𝑐 2 15 ℎ 𝑐𝑐 0 𝑃𝑃
2.8.16
where the integral can be evaluated as (π4/15). Since all of the terms between parentheses are constants, we will lump them together into the Stefan-Boltzmann constant of σ = 2π5kB4/(15hP3c2) = 5.6704 x 10-8 J m-2 K-4 s-1 to define our radiant flux density as 𝑅𝑅̇𝑏𝑏 = 𝜎𝜎 T 4
2.8.17
where 𝑅𝑅̇𝑏𝑏 is the radiant flux density of a black body and T is the absolute-scale temperature.
The emittance from the Sun corresponds to a temperature of approximately 6000 K. The radiant density flux for a black body at the temperature of the sun is defined as
Last Modified: March, 2018
2-80
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑅𝑅̇𝑏𝑏 (𝑇𝑇 = 6000 𝐾𝐾) = 𝜎𝜎 T 4 = �5.6704x10−8 = 7.35x107 W m−2
J
� (6000 K)4 = 7.35x107 J m−2 s −1
m2 K 4 s
and for the Earth at a temperature of approximately 288 K 𝑅𝑅̇𝑏𝑏 (𝑇𝑇 = 288 𝐾𝐾) = 𝜎𝜎 T 4 = �5.6704x10−8
Gray bodies
J
� (288 K)4 = 390 J m−2 s−1 = 390 W m−2
m2 K 4 s
The radiant flux density for a non-black body is done using the emissivity defined as 𝜀𝜀 =
𝑅𝑅̇ 𝑅𝑅̇𝑏𝑏
2.8.18
and therefore the radiant flux density is defined as 𝑅𝑅̇ = 𝜀𝜀 𝑅𝑅̇𝑏𝑏 = 𝜀𝜀 𝜎𝜎 𝑇𝑇 4
2.8.19
In general, emissivity is a function of temperature, emission angle and wavelength. Dependence on emission angle and wavelengths is discussed in greater detail in Chapter 3. Gray bodies are defined for substances where the emissivity is constant for all wavelengths. This is the assumption used in this chapter. Reported values for emissivity are given in Table 2.5. Table 2.5. Total Emissivity for Selected Substances. Surface Clay Concrete Wood Polished Stainless Steel Polished Silver Plaster Aluminum Paints
ε 0.91 0.94 0.85 0.13 0.05 0.91 0.52
Surface Ice Liquid water Snow Clouds Dry soil Typical plant values Typical clear skies
ε 0.97 0.96 0.82 0.98 0.90– 0.95 0.90 - 0.99 0.60 - 0.75
Radiation Properties Measured Solar and Terrestrial Radiation Radiation data have been measured by satellites at the edge of the atmosphere and by instruments located at the ground surface. An example of these measurements is shown below. Let’s first discuss the spectral flux density measured at the top of the atmosphere for the irradiance from the Sun. These values are well approximated by the spectral flux densities corresponding to a T= 5900 K, adjusted for the distance between the Sun and Earth. Also shown in the figure is the solar radiation measured at the ground surface. These values are smaller because of (1) the reflection of solar radiation back into space and (2) the absorption
Last Modified: March, 2018
2-81
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
of radiation by atmospheric gases as it propagates to the Earth’s surface. Additional details of the absorption of gases are shown in the lower figure. Different gases absorb radiation at different wavelengths. For example, O2 and O3 (ozone) absorb radiation at wavelengths corresponding to solar radiation. These gases reduce the energy from potentially dangerous X-rays and other small-wavelength radiation. Carbon dioxide and water vapor absorb radiation at wavelengths more common for terrestrial radiation. This impact can be seen by the observed radiation leaving the top of the atmosphere. As shown by the right-sided graph, the spectral flux density for a temperature corresponding to Earth is well approximated by the observed values for wavelengths with little absorptivity by CO2 or H2O. Observed Incoming Solar
40
Atm Window
Predicted, T = 5900 K
2000
30 Predicted T = 294 K
1500
Observed Ground 20
1000 10
500
0
0 PAR
Absorptivity (%)
Spectral Flux Density (W m-2 μm-1)
Spectral Flux Density (W m-2 μm-1)
2500
1
Wavelength (μm)
10
Observed Outgoing100 Terrestrial
100 50
0
O2
O3
O2
H2 O
CO2
H2 O
CO2
H2 O
Figure 2.42. Observed and Predicted Solar and Terrestrial Radiation Based on these observations, we will discuss the following topics related to radiation: (1) Solar radiation reaching Earth’s atmosphere (2) Absorptivity, Transmissivity, and Reflectivity of Radiation. Interactions with Incident Radiation Let’s consider the response between radiation and matter as shown below. We have already discussed the emittance of radiation using Stefan-Boltzmann radiation laws. The radiation incident on the surface can either by reflected, absorbed or transmitted. These properties are shown in Fig. 2.43. In this chapter, we will consider the total reflectivity, absorptivity and
Last Modified: March, 2018
2-82
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
transmissivity for all wavelengths. Our more general formulation given in Chapter 3 includes wavelength and directional dependence on reflectivity, absorptivity, and transmissivity.
Figure 2.43. Interactions with Incident Radiation. Absorptivity [fa]: Fraction of the incident flux density absorbed by the material. It results in heat transfer to the material. The absorptivity over the entire range of wavelengths can be written as 𝑓𝑓𝑎𝑎 =
𝑅𝑅̇𝑎𝑎 𝑅𝑅̇𝑖𝑖
𝑓𝑓𝑡𝑡 =
𝑅𝑅̇𝑡𝑡 𝑅𝑅̇𝑖𝑖
𝑓𝑓𝑟𝑟 =
𝑅𝑅̇𝑟𝑟 𝑅𝑅̇𝑖𝑖
2.8.20
Transmissivity [ft]: Fraction of the incident flux density transmitted by the material. The transmissivity over the entire range of wavelengths can be written as 2.8.21
Reflectivity [fr]: Fraction of the incident flux density reflected by the material. The fraction of solar radiation reflected from a material (especially at the Earth’s surface) is also called albedo. The reflectivity over the entire range of wavelengths can be written as 2.8.22
Scattering (exclude from student handout): For gases and other types of particle-laden media, transmissivity and reflectivity are components of scattering processes. Scattering includes reflection and refraction and other non-absorbing processes. In contrast to reflectivity that deflects radiation in one direction, scattering is done by particles and molecules that defect radiation in all directions. Scattering is a two-step process. First radiant energy is
Last Modified: March, 2018
2-83
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
converted into internal energy and then the internal energy is converted back into radiant energy. In this conversion process, the frequency and direction of propagation can be changed. Scattering is a function of the size of the particles and radiation wavelength. Rayleigh scattering refers to selective deflection of radiation for a particular wavelength. For example, nitrogen and oxygen are more effective at scattering wavelength corresponding to blue and violet colors (why the sky appears blue). Mie scattering refers to the scattering of all visible wavelengths. Cloud droplets are an important example of Mie scattering and it is why clouds appear white. Very deep clouds can reduce the passage of radiation creating a darker appearance. More Discussion Since the incident radiation has only three possible pathways, we conclude that 𝑓𝑓𝑎𝑎 + 𝑓𝑓𝑟𝑟 + 𝑓𝑓𝑡𝑡 = 1
2.8.23
𝑓𝑓𝑡𝑡 = 1 − 𝑓𝑓𝑎𝑎 − 𝑓𝑓𝑟𝑟
2.8.24
𝑓𝑓𝑎𝑎 = 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑟𝑟 = 𝑓𝑓𝑡𝑡 = 0
2.8.25
Values for absorptivity and reflectivity are commonly available in tables and charts. The transmissivity value can then be determined as
For a black body, all of the incident radiation is absorbed, and therefore
Kirchhoff’s law
Kirchhoff’s law states that for a system in thermodynamic equilibrium, the emissivity equals the absorptivity, that is, 𝜀𝜀 = 𝑓𝑓𝑎𝑎
2.8.26
which, of course, for a black body both the emissivity and absorptivity equal one. Insight into this law can be obtained by considering thermodynamic equilibrium in a medium of uniform temperature and where the incident radiation is from a black body. At equilibrium, the emitted radiation equals that absorbed. We then have 𝜀𝜀 𝜎𝜎𝑇𝑇𝑒𝑒4 = 𝑓𝑓𝑎𝑎 �𝑅𝑅̇𝑏𝑏𝑏𝑏 � = 𝑓𝑓𝑎𝑎 ( 𝜎𝜎 𝑇𝑇𝑖𝑖4 )
2.8.27
For uniform temperatures, the emittance temperature (Te) equals the temperature of the incoming radiation (Ti). For our conditions, we then conclude that ε = fa. Although Kirchoff’s law is derived for equilibrium conditions, it widely used as an approximation for other conditions. More rigorous discussion of Kirchhoff’s law is given in most textbooks on radiation including Chapter 5 of Thomas and Stammes (1999). Absorptivity values
Last Modified: March, 2018
2-84
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Absorptivity is tied to the energy levels of molecules as determined by their chemical structure. These energy levels for solids or liquids allow absorption (and emissions) of radiation as a nearly continuous process with wavelengths. Absorptivity of gases is tied more closely to wavelengths. Gas absorptivities (and emissions) can be relatively large for some wavelengths and non-existent for others. In general, polyatomic gases have vibrational and rotational energy states and therefore have more interesting absorption characteristics. Absorptivity of gases has been shown in our previous figure. It is discussed in greater detail in Chapter 3. Reflectivity values Some measured values of reflectivity for terrestrial surface are shown below. Table 2.6. Reflectivity for Different Surfaces. Cover Clean snow Black soil, dry Black soil, moist Grey soil, dry Grey soil, moist Sand Tops of oak
fr 0.90 0.14 0.08 0.26 0.11 0.35 0.19
Cover Alfalfa Green grass Cotton Water bodies (0-30 lat) – year Water bodies (30-60 lat) Dec Water bodies (30-60 lat) Jun Water bodies (30-60 lat) year
fr 0.23 0.26 0.21 0.05 0.17 0.05 0.08
The reflectivity of water varies with wave activities, turbidity, and other water body characteristics. One of the most important factors is the angle of the rays striking the water. For an angle of 5 degrees, the reflectivity is as high as 90%; whereas the reflectivity is as low as 2% when the angle is 60 degrees. The values in the table are from Cogley (1979). For latitudes closer to the equator, the monthly variability in fr is small. Reflectivity values and variability with month are greater for the larger latitudes.
RADIATATIVE BALANCE FOR THERMAL EQUILIBRIUM Earth’s Orbit around the Sun The most important source of energy for us is the radiation from the Sun. The average solar radiant flux density at the edge of the Earth’s atmosphere is commonly called the solar constant. Determining the solar constant requires knowledge of the Earth’s orbit. A schematic of this orbit is shown below. The Earth’s elliptical orbit is nearly circular with an eccentricity of 0.017. Along the main axis, the distance the Earth is closest to the sun is a distance of approximately 147 million km and the farthest distance is 152 million km. The dates corresponding to these distances vary slightly with year.
Last Modified: March, 2018
2-85
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Eccentricity = 1 −
(Minor ) 2 = 0.017 (Major ) 2
147 Million km
152 Million km
Aphelion ~ July 3
Sun
Perihelion ~ January 3
Earth
Figure 2.44. Orbit of Earth around the Sun. The Earth is tilted at an angle of 23.45 degrees. This tilt plays an important role in the distribution of solar radiation on the surface of the Earth. We will first explore the implications of the radii of the Earth and Sun and between them on the amount of solar energy. The distribution of solar radiation on Earth is given in Appendix 2-A. Radiation and Thermal Equilibrium Relationships Extraterrestrial Radiation A schematic illustrating key concepts used to develop relationships for establishing thermal equilibrium using radiative heat transfer for Earth are shown below.
Figure 2.45. Key Concepts for Thermal Equilibrium. Let’s start by computing the total solar radiant power (energy per time) using the radius of the Sun. Since the surface area of a sphere at that radius is As,sun = 4π rs2, we obtain 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑠𝑠 = 𝑅𝑅̇𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴𝑠𝑠,𝑠𝑠𝑠𝑠𝑠𝑠 = (𝜀𝜀𝑠𝑠 𝜎𝜎𝑇𝑇𝑠𝑠4 )(4 𝜋𝜋 𝑟𝑟𝑠𝑠2 ) Last Modified: March, 2018
2-86
2.9.1
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
The solar constant is the solar radiant flux density to reach earth. It can be computed using the total radiant power over a surface area corresponding to a radius equal to the mean distance between the Earth and the Sun (As,s-e = 4πr2s-e). The solar constant for the mean orbit distance is then obtained as 𝑆𝑆𝑐𝑐̅ =
𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑠𝑠 (𝜀𝜀𝑠𝑠 𝜎𝜎𝑇𝑇𝑠𝑠4 )(4 𝜋𝜋 𝑟𝑟𝑠𝑠2 ) 𝑟𝑟𝑠𝑠2 4 = = 𝜀𝜀 𝜎𝜎𝑇𝑇 � � 𝑠𝑠 𝑠𝑠 2 2 𝐴𝐴𝑠𝑠,𝑠𝑠−𝑒𝑒 4 𝜋𝜋 𝑟𝑟𝑠𝑠−𝑒𝑒 𝑟𝑟𝑠𝑠−𝑒𝑒
2.9.2
Let’s now consider the net radiant flux density absorbed by the Earth (and atmosphere). Net radiation is equal to the incoming radiation (solar constant) minus the radiation reflected back into the atmosphere. The net radiant power absorbed by the Earth corresponds to the product of the net radiant flux density and the projected area of the Earth. The projected area is the area of a circle for the Earth’s radius. The net radiant power (𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑖𝑖 ) is then obtained as 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑖𝑖 = 𝑆𝑆𝑐𝑐̅ (1 − 𝑓𝑓𝑟𝑟 )(𝜋𝜋 𝑟𝑟𝑒𝑒2 )
2.9.3
The global radiant flux density is defined as the average solar radiant energy per area of the Earth’s surface. This form is useful in determining the solar energy available for use by plants, evaporation and melting, and solar heating systems. To determine global flux density, we need to divide the net radiant power by the surface area of Earth (As,earth = 4πre2). We then obtain 𝑅𝑅̇𝑖𝑖𝑖𝑖 =
𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑖𝑖 𝑆𝑆𝑐𝑐̅ (1 − 𝑓𝑓𝑟𝑟 )(𝜋𝜋 𝑟𝑟𝑒𝑒2 ) 𝑆𝑆𝑐𝑐̅ (1 − 𝑓𝑓𝑟𝑟 ) = = 4 𝜋𝜋 𝑟𝑟𝑒𝑒2 4 π re2 4
2.9.4
where 𝑅𝑅̇𝑖𝑖𝑖𝑖 is the global radiant flux density.
Thermal Equilibrium for Non-absorbing Atmospheric Gases After a sufficiently long time, the Earth should reach a thermal equilibrium with the radiation from the sun. At thermal equilibrium, the heat transfer entering the Earth by solar radiation is balanced by the radiation heat transfer leaving by corresponding to the Earth’s (including its atmosphere) emittance. From the First Law’s perspective, we have Δ𝑈𝑈̇ = 0 = 𝑞𝑞̇ + 𝑤𝑤̇ = 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑖𝑖 − 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑒𝑒 + 𝑞𝑞̇ 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒
2.9.5
𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑒𝑒 = 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑖𝑖
2.9.6
Where the net rate of change in internal energy is zero, and since the pressure at the top of the atmosphere is zero, the rate of expansion work is also zero. The heat transfer is divided into the radiant power entering the Earth from the Sun (𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑖𝑖 as defined by Eq. 2.9.3), radiant power leaving from Earth’s emittance (𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑒𝑒 = 𝑅𝑅̇𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ 𝐴𝐴𝑠𝑠,𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ ) and other sources of heat transfer (𝑞𝑞̇ 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 ) that are equal to zero for thermal equilibrium. As shown in Homework Problem 2.52, 𝑞𝑞̇ 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 ≈ 0 is a valid condition of Venus, Mars, and Earth, but not for that of Jupiter. Thermal equilibrium conditions then correspond to
Last Modified: March, 2018
2-87
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For non-absorbing atmospheric gases, the radiant power from Earth’s emittance is defined by 𝑞𝑞̇ 𝑟𝑟𝑟𝑟𝑟𝑟,𝑒𝑒 = 𝑅𝑅̇𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ 𝐴𝐴𝑠𝑠,𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ℎ . By also using Eq. 2.9.3, we obtain (𝜀𝜀𝑒𝑒 𝜎𝜎 𝑇𝑇𝑒𝑒4 )(4 𝜋𝜋 𝑟𝑟𝑒𝑒2 ) = 𝑆𝑆𝑐𝑐̅ (1 − 𝑓𝑓𝑟𝑟 )(𝜋𝜋 𝑟𝑟𝑒𝑒2 ) = 𝜀𝜀𝑠𝑠 𝜎𝜎𝑇𝑇𝑠𝑠4 �
𝑟𝑟𝑠𝑠2 � (1 − 𝑓𝑓𝑟𝑟 )(𝜋𝜋 𝑟𝑟𝑒𝑒2 ) 𝑟𝑟𝑠𝑠2−𝑒𝑒
2.9.7
where Te is the temperature of the Earth’s surface. We can then solve for this temperature at thermal equilibrium as
Te = Ts �
rs ε (1 − fr ) � s 2 rs−e εe
2.9.8
Example Problem 2.9: Thermal Equilibrium and Earth’s Temperature Problem Statement. For approximately spherical bodies of Earth and the sun, estimate the (1) solar constant, (2) the temperature of the earth at thermal equilibrium, and (3) the global radiant flux density for heating the earth and atmosphere. We will use the following characteristics: Sun radius = rs = 6.96 x 108 m Distance between Sun and Earth = rs-e = 1.496 x 1011 m Sun temperature = Ts = 5778 K Emissivity of sun = εs = 1.0 (excellent approximation) Emissivity of earth = εe = 1.0 (good approximation for wavelengths in infrared range) Reflectivity of the atmosphere is approximately 21% and Earth’s surface is 9% Total reflectivity of atmosphere and Earth’s surface: fr = 0.3
Earth
Sun
εe =1 fr = 0.3 Pi T = 5778 K εs = 1
T? Mean Distance = 1.496 x1011 m
re = 6.37x106 m
rs = 6.96x108 m
Solution for Extraterrestrial Radiation. The solar constant for the mean orbit distance is
Last Modified: March, 2018
2-88
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
q̇ rad,s rs2 4 = εs σ Ts � 2 � 2 4 π rs−e rs−e (6.96 𝑥𝑥 108 𝑚𝑚)2 𝑊𝑊 4 �Sc = (1) � 5.6704 𝑥𝑥 10−8 � (5778 𝐾𝐾) ( ) (1.496 𝑥𝑥 1011 𝑚𝑚)2 𝑚𝑚2 𝐾𝐾 4 S�c = 1368 W m−2 = 4.92 MJ m−2 h−1 S�c =
We then obtain the following estimate of the global radiant flux density for the atmosphere and surface of fr = 0.3, as 𝑅𝑅̇𝑖𝑖𝑖𝑖 =
𝑆𝑆𝑐𝑐̅ (1 − 𝑓𝑓𝑟𝑟 ) 1368 (1 − 0.3) = = 240 W m−2 4 4
The irradiance global radiant flux density is simply (1368/4) = 342 W m-2.
Solution: Average temperature of Earth. The average temperature of Earth for thermal equilibrium using radiative heat transfer is
Te = Ts �
rs ε (1 − fr ) � s 2 rs−e εe
Te = (5778 K) �
(1) (1 − fr ) 6.96 𝑥𝑥 108 𝑚𝑚 � 2 (1.496 x 1011 𝑚𝑚) 1
𝑇𝑇𝑒𝑒 = 254.9 𝐾𝐾 = −18.2 𝐶𝐶
Discussion
The measured temperature of the earth is approximately 288 K (15 C). A temperature of -18 C computed in the example problem is too cold to support our most important biotic communities. The absorption of terrestrial radiation by the so-called greenhouse gas are why our temperature is warmer than the equilibrium temperature. The temperature of the Earth surface needs to increase to 288 K to achieve an outgoing radiation equal to that of incoming global radiant flux density of 240 W m-2 because of the role of greenhouse gases. Thermal equilibrium conditions are shown in Fig. 2.46 using the simple thermal equilibrium of this problem (without considering absorptivity of atmospheric gases) and with absorptivity of these gases. Equilibrium conditions of the left-sided figure correspond to those values of the example problem. Thirty percent of the solar radiation is reflected. The transmitted radiation is balanced by the emitted terrestrial radiation at a temperature of 254 K. The actual thermal equilibrium conditions with absorbing atmospheric gases are shown in the right-sided figure. Conditions are more complicated with these gases, but the incoming and outgoing radiation are still equal to each other (342 W m-2). Only 50% of the solar radiation is transmitted to the Earth’s surface, that is, 20% of the incoming solar radiation is absorbed by the atmospheric gases. For an Earth’s temperature of 288 K, the emitted radiation is 390 W
Last Modified: March, 2018
2-89
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
m-2. Only 5% (approximately) of this radiation is transmitted to outer space, the rest in absorbed by the atmospheric gases. An additional 103 W m-2 of energy is absorbed by these gas from convection and phase change processes. The total emitted radiation from the atmospheric gases (and clouds) is 542 W m-2. A significant fraction of this radiation is absorbed by Earth. Understanding the impact of increasing the greenhouse gases requires greater knowledge of the structure of the atmosphere and the selective absorptivity of wavelengths for these gases. These concepts are discussed in Chapter 3.
Figure 2.46. Thermal Equilibrium Without and With Absorptivity of Atmospheric Gases.
THERMODYNAMIC PROCESSES OF PLANT CANOPIES Photosynthesis and Transpiration Photosynthesis and transpiration within plant canopies are vital processes for life. Both are largely driven by solar radiation. Photosynthesis is the process that converts solar radiation and inorganic raw materials into organic compounds. It provide a mechanism for the conversion of solar energy into forms available for metabolic activities of animals. Transpiration is the process by which water is evaporated from the air spaces in plant leaves. It is an important component of the water cycle. Approximately 70% of U.S. precipitation is returned to the hydrologic cycle by evaporation from lakes, rivers and soil and by transpiration from plants. It is critically important in the design of irrigation systems, in the analysis of water supply for municipalities and in maintaining flow rates and aquatic communities in streams and rivers.
Last Modified: March, 2018
2-90
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
The simplest form of photosynthesis can be written as 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆ℎ𝑡𝑡
6CO2 + 6 H2 O �⎯⎯⎯⎯⎯⎯� C6 𝐻𝐻12 𝑂𝑂6 + 6𝑂𝑂2
where C6H12O6 is simple sugar (glucose or fructose). The standard enthalpy of reaction for the photosynthesis of glucose can be written as 𝑜𝑜 𝑜𝑜 C o Δ25 Hm = �νc6h12o6 Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝑐𝑐6ℎ12𝑜𝑜6 + νo2 Δ𝑓𝑓25 𝐶𝐶 𝐻𝐻𝑚𝑚,𝑜𝑜6 � r 25 𝐶𝐶 𝑜𝑜 25 𝐶𝐶 𝑜𝑜 − �νco2 Δ𝑓𝑓 𝐻𝐻𝑚𝑚,𝑐𝑐𝑐𝑐2 + νh2o Δ𝑓𝑓 𝐻𝐻𝑚𝑚,ℎ20 � = 2803 kJ mol−1
2.10.1
which is equal to that previously given for the oxidation of glucose (see Table 2.3) for a reversal of the products and reactants compounds. The standard enthalpy for this photosynthetic reactions is therefore equal to a positive 2803 kJ mol-1 (heat transfer into a constant pressure system). The critical role of solar radiation in the amount of biomass in plants is illustrated in Fig. 2.47. Cumulative biomass for eleven different agronomic plants is shown to vary linearly with cumulative photosynthetic radiation. These results are for crops that has sufficient water and nutrients, are not limited by competition from weeds or by other pests, and have temperatures suitable for plant growth. For these conditions, the daily biomass before plant maturity can be estimated as ΔBm,A = 2.83 fpar 𝑞𝑞𝑠𝑠𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖,𝐴𝐴
2.10.2
where ΔBm,A is the daily production of dry mass per area for the canopy in units of g m-2, qsun,in,A is the solar radiation per unit area reaching the leaves within the canopy in units of MJ m-2, and fpar is the fraction of the solar radiation (absorbed by plants) that is available for photosynthesis. For canopy where the surface area of leaves (one sided) is three times greater that the soil area, fpar is approximately 0.45.
Figure 2.47. Relationships between Biomass and Photosynthetic Radiation.
Last Modified: March, 2018
2-91
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Additional insight into photosynthesis and transpiration can be obtained by considering the processes within the leaves in greater detail. A simplified schematic of plant leaves is shown in Fig. 2.47. The epidermis is the outer, protective cell which secretes a waxy cutin. Most of the space between the upper and lower epidermis consists of thin-walled cells called mesophyll that are full of chloroplasts and are the primary site of photosynthesis. Water and nutrients are delivered from the soil to the plant leaves by conducting tissue called xylem. Scattered over the epidermis are many small openings called stomata. There approximately 5000 stomata per cm2 for a corn leaf. These openings are essential for movement of water vapor (transpiration), oxygen, and carbon dioxide. The size of the openings is controlled by guard cells. The guard cells expand or contract based on turgor pressure resulting in smaller openings for periods of no photosynthesis (at night), and when the plant experiences water stress because of inadequate supply of water delivered from the roots. The evaporation of water within plant leaves and the movement of water vapor through the stomata is called transpiration. This process is fundamentally related to the First Law. O2, H2O
CO2
Upper Epidermis Xylem (H2O)
Stomata
CO2
Mesophyll (Chloroplasts)
Lower Epidermis
O2, H2O CO2
CO2
Stomata O2, H2O
Transpiration (Vaporization)
Figure 2.48. Components of a Plant Leaf.
Last Modified: March, 2018
2-92
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.49. Energy Components for Transpiration Processes. Transpiration and the First Law System Definition The First Law will be applied to the plant canopy shown in Fig. 2.49. The atmospheric pressure is effectively constant during transpiration. Total heat transfer (qt) is divided into a component from convection and conduction from the canopy to the atmosphere and soil (q) and radiation (qrad) from solar and terrestrial sources. The First Law of thermodynamics can then be written as ΔU = q t + w = q + q rad + wo − P ΔV
2.10.3
ΔU + PΔV = ΔH = ΔHvap + ΔHtpr + ΔHr = q + q rad
2.10.4
ΔHvap − q = q rad − ΔHtpr − ΔHr
2.10.5
where PΔV is expansion work (constant pressure) and wo = 0 is the other non-expansion work. We can rearrange this equation using the definition of enthalpy for constant pressure as
where ΔH is the total change in enthalpy, and it generally includes changes as the result of a change in temperature (ΔHtpr), change in phase transitions (ΔHvap) and the change from chemical reactions (ΔHr). We can rewrite the First Law as
Use of Bowen Ratio
We will use the simple Bowen ratio to assist us in our calculations. The Bowen ratio is defined as the ratio of heat transfer by conduction and convection (q) out of the system (to the surroundings of air) and the energy released by vaporization (ΔHvap): 𝛽𝛽 =
−𝑞𝑞 Δ𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣
Last Modified: March, 2018
2.10.6
2-93
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Bowen ratio is dependent on many factors. In the Example Problem 2.9 (and the homework problem), its value will be given to you. By using a known Bowen ratio, the convective/conductive heat transfer can be written as 𝑞𝑞 = −𝛽𝛽 Δ𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣
2.10.7
and the First Law can then be evaluated as ΔHvap + β ΔHvap = (1 + β)ΔHvap = q rad − ΔHtpr − ΔHr
2.10.8
We can then solve for the enthalpy used to evaporate (vaporize) in the plant canopy as ΔHvap =
q rad − ΔHtpr − Δ𝐻𝐻𝑟𝑟 1+β
2.10.9
which has units of energy (kJ).
Example Problem 2.10: Transpiration Depth from Plant Canopies Problem Statement We are interested in the transpiration from a well-watered alfalfa field where the plant canopy represents the system, and the air and the soil represents the surroundings. The field is located in Waseca, MN. Let’s consider our system for a clear day on July 15 for a particular year where the average air temperature during transpiration is 25 C and the corresponding average canopy temperature is 28 C. Heat transfer to the soil and radiant heat transfer during non-daylight periods are assumed to be negligible. We wish to compute (1) the net daily radiation energy entering the system, and (2) the daily transpiration depth. Selection of parameter values Key parameters are summarized in the figure below. They are: ∆28vapC H om = 43.7 kJ mol-1 at 28 C, mm = 18 g mol-1 Stefan-Boltzmann Constant = 5.67 x10-8 J m-2 K-4 s-1, Density (liquid water) = 996 kg m-3 Emissivity of alfalfa (plants): εp = 0.95, Emissivity of air: εa = 0.75 Reflectivity of sky: fra = 0.10, Absorptivity of sky: fa= 0.15 Reflectivity of alfalfa (plants): frp = 0.20, Bowen ratio for well-watered plants: β = 0.30 Representative area in the middle of canopy: Ap = 100 m2 Canopy height = 1.25 m, Atmospheric pressure = 101 kPa Molar heat capacity at constant pressure for air: Cpm,a = 29.15 J K-1 mol-1 Change in temperature of air in canopy = 4 C Fraction of solar radiation in plant canopy available for photosynthesis: fpar = 0.45 Photosynthesis represented by glucose reaction shown below: 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑔𝑔ℎ𝑡𝑡
C o 6CO2 + 6 H2 O �⎯⎯⎯⎯⎯⎯� C6 𝐻𝐻12 𝑂𝑂6 + 6𝑂𝑂2 ; Δ25 Hm = 2803 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 −1 ; 𝑚𝑚𝑚𝑚 = 180.16 𝑔𝑔 𝑚𝑚𝑚𝑚𝑙𝑙 −1 r
Last Modified: March, 2018
2-94
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Results from Appendix 2-A In Appendix 2-A, we have computed the day length for July 15 for Waseca, Minnesota as Δt = 15.01 h
and the solar irradiance at the top of the atmosphere for the day of July 15 at the latitude of Waseca as q sun,A = 4.194x 107 J m−2 = 41941 kJ m−2
Net solar radiation
Transmissivity can be defined using the reflectivity of the atmosphere (fra) and the absorptivity of the atmosphere (faa) as fta = 1 − fra − faa = 1 − 0.10 − 0.15 = 0.75
and therefore the solar radiation on the system is
q sun,canopy,A = fta q sun,A = (0.75)41941 = 31456 kJ m−2
The solar energy entering the system of a plant canopy can be computed using the reflectivity of the plant (frp) as q sun,in,A = �1 − frp �q sun,canopy,A = (1 − 0.2)31456 = 25165 kJ m−2 = 25.16 MJ m−2
Net terrestrial radiation
Last Modified: March, 2018
2-95
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Net radiant flux density from the plant canopy and the surrounding air (terrestrial) can be computed from the emissivities and temperatures. This corresponds to the heat transfer by radiation, or Ṙ t,net = Ṙ ts − Ṙ tp = εa σTa4 − εp σTp4 = σ � εa Ta4 − εp Tp4 �
For the values of this problem Ta = 25 + 273.15 = 298.15 K and Tp = 28 + 273.15 = 301.15 K, Ṙ t,net = �5.67 x 10−8
J
� [ 0.75(298.15)4 − 0.95(301.15)4 ] = −107 J m−2 s −1
m2 K 4 s
To estimate the net terrestrial energy during the daylight hours, we use the number of daylight hours computed in the previous section. We then obtain 3600 s kJ q terr,net,A = Ṙ t,net Δt sun = −107 (15.01 h) � �� � = −5782 kJ m−2 h 1000 J
where the radiant heat transfer during the non-daylight periods is neglected. Net Solar and Terrestrial Radiation
The net energy flow into the system is equal to the sum of the solar and terrestrial energies. This is the heat transfer by radiation. We obtain q rad = �q sun,in,A + q terr,net,A �Ap = (25165 − 5782)
Change in Enthalpy with Temperature
𝑘𝑘𝑘𝑘 (100 m2 ) = 1,938,200 kJ 2 𝑚𝑚
Within the plant canopy, temperature of the air and the plants (mostly water by mass) are typically different. In the morning, the air is typically cooler than the plants but in the late afternoon the air temperature is often warmer than that of the plants. For our purposes, the volume of the liquid water in plants and the change in its temperature during the transpiration are negligible. We will then focus on the change in enthalpy of the canopy air. The number of moles of this air can be estimated from the ideal gas equation. For a temperature of T = 28 C = 301.15 K, a pressure of P = 101 kPa, and a volume of V = 100 m2 (1.25 m) = 125 m3, we obtain n=
PV = RT
(101000 Pa) 125 m3 = 5042.4 mol Pa m3 (8.314 )(301.15 K) K mol
By using ΔT = 4 C = 4 K, the change in enthalpy of the canopy by temperature change is ΔHtpr ≈ n Cpm ΔT = (5042 mol) �29.15
Change in Enthalpy with Photosynthetic Reactions
Last Modified: March, 2018
2-96
J 1 kJ � (4 K) � � = 587.9 kJ mol K 1000 J
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
The biomass obtained by photosynthesis can be estimated using solar radiation reaching the plants within the canopy using ΔBm,A = 2.83 fpar 𝑞𝑞𝑠𝑠𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖,𝐴𝐴 = 2.83 (0.45)(25.16 MJ m−2 ) = 32.0 g m−2
2.10.2
which corresponds to an equivalent number of moles of glucose of Δnglucose
(3207 g m−2 ) 100 m2 = = 17.8 mol 180.16 𝑔𝑔 𝑚𝑚𝑚𝑚𝑙𝑙 −1
Since the stoichiometric coefficient for glucose, we conclude that Δ𝜉𝜉 = Δ𝑛𝑛𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ⁄𝜈𝜈𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 17.8 𝑚𝑚𝑚𝑚𝑚𝑚. By standard enthalpy for this photosynthetic reaction, we obtain C o ΔHr = Δξ Δ25 Hm = (17.8 mol)(2803 kJ mol−1 ) = 49,850 kJ r
Change in Enthalpy of Vaporization of Plant Canopy
As previously derived, we can determine the enthalpy for transpiration as ΔHvap =
q rad − ΔHtpr − Δ𝐻𝐻𝑟𝑟 1938300 − 588 − 49,850 = = 1,452,000 kJ 1+β 1.3
We conclude from the First Law of thermodynamics that the 1,439,000 kJ of energy is used to evaporate water (and provide for expansion work) by transpiration from the plant canopy of 100 m2. The change in enthalpy by temperature difference is clearly a small component of the energy balance. If there is insufficient water for this evaporation, then the Bowen ratio would be larger. Mass of Evaporated Water and Transpiration Depth To compute the mass of vaporization, we need to determine the specific enthalpy of vaporization at 28 C. Specific enthalpy can be obtained by dividing the molar enthalpy by the molar mass. Therefore we obtain C Δ28 vap Hs
C o Δ28 43.7 kJ mol−1 vap Hm = = = 2.428 kJ g −1 −1 mm 18 g mol
where the standard enthalpy of vaporization is used for our pressure of 101 kPa. The mass of water vaporized per unit area is defined as Δmvap =
ΔHvap
C Δ28 vap Hs
=
1452000 kJ = 598,000 g = 598 kg 2.428 kJ g −1
The transpiration volume is then obtained as Vvap =
Δmvap 598 kg = = 0.600 m3 −3 ρh2o 996 kg m
The transpiration depth is equal to the volume per unit area, or
Last Modified: March, 2018
2-97
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
hvap
Vvap 0.600 m3 = = = 0.00600 m = 6 mm = 0.24 in Ap 100 m2
Last Modified: March, 2018
2-98
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
PROBLEM ASSIGMENTS Format All homework assignments must be done in a neat and organized manner. Clearly identify the solution for those problems with several mathematical steps by underlining or boxing in your answer. Points will be deducted from problems that are sloppy and difficult to follow. Problems 2.1.
Consider an elevator of mass of 2300 kg that is at rest at a position 5 m above the base of the elevator shaft. It is lifted to a height 80 m above the base of the shaft. The cable holding it breaks, and it falls freely to the base of the shaft (where it is brought to rest by a strong spring). Assume the entire process to be frictionless. Calculate (1) the gravitational potential energy of the elevator at its initial and highest positions, relative to the base of the shaft, (2) the work done in lifting the elevator to its highest position, and (3) the kinetic energy and velocity of the elevator just before it strikes the spring.
2.2.
A billiard ball of mass = 0.15 kg is dropped from a four story building, corresponding to a height of 12 m. If mechanical energy is conserved, what are the (1) kinetic energy and (2) the velocity of the ball just before it strikes the ground surface? Assume that the ball remains intact after impact and that the collision is elastic (kinetic energy is conserved). By using the definitions given for the kinetic theory of gases in Chapter 1, what is the impulse? If the collision duration is 0.01 s, what is the average force of the collision?
2.3.
The heat transfer rate and the temperature difference have been measured for different types of mineral-fiber insulation materials by Tye et al. (1980). For fiberglass insulation, they measured a heat transfer rate of 2.5 J s-1 over an area of 0.19 m2 and a thickness of insulation of 0.105 m (Δx = - 0.105 m). The temperature difference between the two sides of the material was determined to be 27.7 K. Assuming heat transfer by conduction, what is the thermal conductivity in units of J s-1m-1K-1 and the Rvalue in units of K m2 s J-1? R-values for insulation in the United States are reported using units of F ft2 h BTU-1. By using the conversion that 1 K m2s J-1 = 5.68 F ft2 h BTU-1, what R-value would you use to label this fiberglass material?
2.4.
Consider a spring with a spring constant of 50 N m-1 and a mass of 0.05 kg. What is the natural frequency for the spring in a simple harmonic motion? For an amplitude of 0.1 m, compute and plot the location of the spring with time between 0 and 0.5 seconds using 0.01 seconds increments. Identify on the graph the wavelength for the spring motion. In addition for each time increment, compute the work done, the elastic potential energy, the spring velocity, the kinetic energy, and mechanical energy.
2.5.
Consider a human lung as your system. A typical volume of a human lung is 5 x 10-4 m3 (0.5 L) and a typical air pressure is 101 kPa. How much pressure work is done by the muscles (of the surroundings) to exhale air from a typical human lung? Convert this work into an equivalent number of textbooks raised to a height of Δz = 1 m. Use an average mass of textbooks of 1 kg.
Last Modified: March, 2018
2-99
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
2.6
Measured pressures and volumes are given below for a cylinder-piston system where gas expanded slowly under quasi-equilibrium conditions. The first value in the table corresponds to the initial state and the last value to the final state. Plot the observed pressure as a function of volume. What is the work done by the gas during the expansion? P bar 12 10 8 6 4 2
2.7
V m3 0.045 0.052 0.063 0.079 0.109 0.189
Consider the following relationship between pressure and volume for a cylinder-piston system under quasi-equilibrium conditions between initial state of Pi = 50 kPa and Vi = 0.5 m3 and a final state of Pf = 200 kPa and Vf = 0.1 m3. 𝑃𝑃 = 𝑎𝑎 + 𝑏𝑏 𝑒𝑒 𝑐𝑐 (𝑉𝑉−𝑥𝑥𝑜𝑜 ) = 𝑎𝑎 + 𝑏𝑏 exp� 𝑐𝑐(𝑉𝑉 − 𝑥𝑥𝑜𝑜 )�
You wish to analysis two possible paths between the endpoint states. For Path 1, the coefficients are defined as a = -72.4 kPa, b = 122.4 kPa, c = -2 m-3 and xo = 0.5 m3. For Path 2, the coefficients are defined as a = 322.4 kPa, b = -122.4 kPa, c = 2 m-3 and xo = 0.1 m3. Part I: You are required to compute the work for each path using only the conditions of the initial and final states. Is work defined by an exact or inexact differential? Part II: You are also required to plot the pressure for Path #1 and Path #2 for volumes between 0.5 m3 and 0.1 m3. You can compute the pressure for both paths using ΔV = 0.02 m3, that is, find the pressure for V=0.5 m3, V=0.48 m3, V=0.46 m3, ... , V=0.1 m3. Part III. For a system with n = 10 moles of nitrogen gas, compute gas temperature for each volume obtained using ΔV = 0.02 m3 for the two different paths. Also compute the work done between each of the consecutive volume for both Paths #1 and #2. Compare the total work done for the two paths obtained from Part III with the solution obtained for Part I. 2.8
You have determined the following linear relationship between pressure and temperature for a specially designed cylinder-piston system: 𝑃𝑃 = 200 + 10 𝑇𝑇
where the coefficient “200” has units of Pa and the coefficient of “10” has units of
Last Modified: March, 2018
2-100
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Pa K-1. For an initial temperature of 200 K and a final temperature of 350 K, what is the work done per mole of ideal gas? 2.9
A colleague in your consulting firm has studied the relationship between pressure and temperature for a particular cylinder-piston system with 50 moles of O2. She obtained the following equation: 𝑃𝑃 = 10 √𝑇𝑇
where the coefficient “10” has units of kPa K-1/2. For an initial temperature of 200 K and a final temperature of 400 K, how much work is done? 2.10.
What is the kinetic energy of a system of 1 kg of water just prior to landing in a pool after falling 150 m over a waterfall? What would be the increase in temperature of the system if all of the kinetic energy is converted into internal energy? You can assume that water is an incompressible substance. You can treat the 1 kg of water as an isolated system once it enters the pool at the bottom of the waterfall.
2.11.
As discussed in the chapter, the molar heat capacity at constant pressure is sometimes related to temperature using the equation given below. Coefficients for helium (monatomic gas), nitrogen (diatomic gas) and carbon dioxide (polyatomic gas) are also shown below. Compute and plot the molar heat capacity at constant pressure for temperature between 10 and 150 C in 10 C increments. What is the largest percent error in Cpm between the equation values and Thermodynamic Data Handout value (given for 25 C) for each of these gases? For this problem, use the Cpm obtained from the equation below as the standard for computing percent error (% error = 100% x (Cpm-Cpm,25 C)/Cpm).
C pm = a + b T + c T 2 + d T 3
Gas He N2 CO2 2.12.
a -1
J K mol 20.786 28.883 22.243
b -1
-2
c -1
J K mol 4.851 x 10-13 -0.157 x 10-2 5.977 x 10-2
-3
d -1
J K mol -1.583 x 10-16 0.808 x 10-5 -3.499 x 10-5
-4
J K mol-1 1.525 x 10-20 -2.871 x 10-9 7.464 x 10-9
We are interested in the error obtained in change in the molar enthalpy by assuming a constant Cpm. Consider the helium, nitrogen and carbon dioxide gases of Problem 2.10 for temperature between (1) 15 C to 35 C, (2) 15 C and 60 C, and (3) 50 C and 100 C. For each of these temperature ranges and for each of the gases, compute the molar enthalpy using a (a) constant Cpm from the Thermodynamic Data Handout (corresponding to 25 C) and (b) using Cpm defined by the relationship of
C pm = a + b T + c T 2 + d T 3
Last Modified: March, 2018
2-101
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
What are the percent errors obtained by using the simpler constant Cpm to compute change in molar enthalpy for each of the temperature intervals and each of the gases? For this problem, use the more accurate solution obtained from the relationship (Method b) as the standard for computing the percent error. 2.13.
In Problem 1.20 of Chapter 1, you computed the theoretical change in internal energies per mole between 10 C and 40 C for a helium (He, monatomic gas) using Um=3RT/2, nitrogen gas (N2, diatomic gas) using Um=5RT/2 and carbon dioxide (CO2, polyatomic gas) using Um=3RT. Re-compute the changes in molar internal energy using the Thermodynamic Data Handout that is based on measured values. You can assume a constant heat capacity between the two temperatures. What are the percent errors in the internal energies obtained using the theoretical values of Chapter 1 and that obtained using the thermodynamic data obtained from experimental measurements?
2.14.
Consider 1 kg of air compressed slowly in a piston-cylinder assembly from 150 kPa to a final pressure of 900 kPa at constant temperature of 38 C. Determine (a) the change in the internal energy of the gas, (b) the work interaction in kJ, and (c) the quantity of heat transfer in J kg-1. You can use a molar mass of air of 29 g mol-1.
2.15.
A piston-cylinder assembly is filled with 2 kg of oxygen gas that has an initial temperature of 30 C. Under conditions of constant pressure, heat transfer into the cylinder between initial and final states has been determined to be 20 kJ. What is the temperature of the oxygen gas at the final equilibrium state? How much expansion work was done in the process?
2.16.
A vertical piston-cylinder assembly with a volume of 28 L is filled with 45 g of nitrogen gas. The piston is weighed so that the pressure on the nitrogen is always maintained at 135 kPa. Heat transfer is allowed to take place until the volume is 80% of its initial value. Determine (a) the energy transfer by heating, (b) final temperature of the nitrogen at equilibrium, (c) the change in internal energy and (d) the expansion work done in the process.
2.17.
Consider a rigid tank with 7 kg of methane gas. The gas is heated from 7 to 174 C. By assuming an ideal gas, determine (1) the work done by the gas and (2) the heat transferred. Also compute the change in enthalpy for the temperature difference (i.e. ΔHtpr) from a constant Cpm over the range of temperatures. Why is the heat transfer not equal to this change in enthalpy?
2.18.
A rigid insulated tank is divided into equal volumes by a partition, 450 g of carbon dioxide is introduced into one side of the partitioned tank, and the other side remains evacuated. For this initial equilibrium state, the pressure and temperature of the gas are measured and recorded as 105 kPa and 345 C. The partition is then pulled out, and the gas is allowed to expand into the entire tank. Expansion in this type of system is called Joule expansion. Determine the final pressure and temperature at a new equilibrium state for an ideal gas.
Last Modified: March, 2018
2-102
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
2.19.
Many students find the outcome of Joule expansion (Problem 2.18) to be counterintuitive. This outcome (examined experimentally by Joule) played an important role in defining ideal gases. Ideal gases are defined such that “no change in temperature occurs when air is allowed to expand in such a manner as not to develop mechanical power” (i.e., w=0). Let’s consider the change in temperature for the system of Problem 2.18 for a non-ideal gas using van der Waals equation shown below (and given in Chapter 1 for the complete set of lecture notes). The translational molar internal energy is then defined as 3 na Um = � � RT − 2 V where for carbon dioxide, the parameter “a” has been determined to be a = 0.000365 kPa m6 mol-2. You can use an initial volume of 0.5 m3. What is the change in temperature for the conditions of Problem 2.18 using the more rigorous non-ideal gas equation?
2.20
A pressure-release valve has been installed on a 0.5 m3 rigid tank. The valve releases gas so that the tank pressure doesn’t exceed 300 kPa. Initially, the tank is filled with nitrogen gas to a pressure of 200 kPa and a temperature of 50 C. Consider a heat transfer into the tank of 200 kJ. How many moles of nitrogen is released from the tank? What is the temperature of nitrogen in the tank? You can neglect possible work with the release of nitrogen.
2.21.
Let’s revisit the 10 moles of nitrogen gas placed in a cylinder of Problem 2.7. For each step of changing volumes, compute (1) change in enthalpy, (2) change in internal energy, (3) heat transfer and (4) change in entropy estimated as 𝑞𝑞/𝑇𝑇�. You can use the temperatures and work done computed in Problem 2.7. Perform these calculations for both Paths #1 and #2. Assume constant molar heat capacity. For a given path, is the net heat transfer equal to the total change in enthalpy? Which of the thermodynamic characteristics appear to be defined by exact differentials and which ones by inexact differentials?
2.22.
You are interested in ΔUtpr, ΔHtpr and ∫δq/T for the system of Problem 2.21; however for this problem, the pressure-volume path between the endpoint states is unknown. Since the actual path is unknown, a friend suggested that you take the simplest possible path defined by assuming a linear relationship between pressure and volume. By using a linear relationship between endpoint pressure and volume conditions, repeat the calculations done in Problem 2.21 (and Problem 2.7). Since the path is the simplest, and not the actual path, you are not interested in the work done and heat transfer, but these characteristics are needed to determine ∫δq/T. Compare the computed ΔUtpr, ΔHtpr and ∫δq/T obtained using the linear relationship with the results of Problem 2.21. Could have you determine these changes obtained for the actual Paths #1 or #2 using your simple linear relationship (yes or no)?
2.23.
A particular automobile tire has an approximately constant volume of 0.05 m3. Assume that the heat transfer into the tire is 0.025 kJ mile-1. For an initial temperature of 20 C and an initial pressure of 210 kPa, how many miles need to be driven to increase the tire
Last Modified: March, 2018
2-103
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
pressure to 235 kPa? You can use a molar heat capacity at constant pressure of air of Cpm = 28.25 J K-1 mol-1. 2.25.
What is the heat transfer required (1) to melt 200 g of ice at 0 C, (2) to increase its temperature from 0 C to 100 C and (3) to evaporate it at 100 C? You can assume constant heat capacities with temperatures and that water is only evaporated at T=100 C. How much more heat transfer is needed to evaporate the 200 g of water than to raise its temperature from 0 to 100 C?
2.26.
Recommended caloric intake to support metabolic processes is approximately 2400 calories per day for a typical college student (definition of calories for food is actually kilocalories in thermodynamics). By using 1 food calorie = 4.1868 kJ, convert these calories into a change of the internal energy from metabolic biochemical reactions (-ΔUr) in units of MJ per day. If the human body of mass 65 kg is an isolated system that has the heat capacity of liquid water, what would be the change in temperature in the human body as a consequence of its metabolic processes?
2.27.
An important mechanism for controlling the temperature of the human body is perspiration and subsequent evaporation. Convert the standard enthalpy of vaporization given in the lecture notes of 40.7 kJ mol-1 at 100 C to 37 C (body temperature). Assuming that the body generates –ΔUr =10 MJ d-1 from metabolic processes (see Problem 2.26), how much water in kg needs to be evaporated per day to maintain the body at constant temperature of 37 C? What is the percent difference in mass if you had computed the evaporated mass using the standard enthalpy of vaporization at 100 C? 𝐶𝐶 𝑜𝑜 Convert the evaporated mass of liquid water to liters for Δ37 𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 . How many one-halfliter glasses of water needs to be drunk per day to meet this evaporative demand? The recommend intake of water by the Institute of Medicine is approximately 3 liters for men and 2.2 liters for women. As shown by Example Problem 2.1, the human body also has heat transfer by radiation, convection and conduction.
2.28.
Methane has a similar molar mass (mm = 16 g mol-1) to water, but has standard enthalpy 𝑜𝑜 of vaporization of only 8.2 kJ mol-1. If Δ𝑇𝑇𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 of water was equal to that of methane (still use water properties for all other characteristics), how many additional one-halfliter glasses of water per day is needed to meet the evaporative demand of Problem 2.27? You don’t need to adjust 8.2 kJ mol-1 to 37 C.
2.29.
The change in internal energy and heat transfer resulting from the condensation of water vapor is an important process in the development of thunderstorms and hurricanes. Consider a depth of rainfall of 25.4 mm (1 inch) for Minneapolis, MN. Minneapolis has an area of 151 km2 within its city limits. If the rain falls uniformly over the land area of Minneapolis, what is the total mass of precipitation in kg? Using the standard enthalpy of condensation at 25 C, how much energy is available for heat transfer with the condensation of this mass of water from vapor to rain droplets (liquid phase)? The Monticello, MN nuclear plant generates approximately 600 megawatts of electrical
Last Modified: March, 2018
2-104
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
power. How many days must the Monticello plant generate power to obtain the energy corresponding to the phase transition of our 1-inch storm? 2.30.
Consider the phase transition of vaporization and fusion as consequence of the following reactions: H2 O (l) → H2 O (g) and H2 O (s) → H2 O (l)
For vaporization, compute the standard enthalpy of reaction at 25 C using the Thermodynamic Data in the Handout. How does this value compare to the standard enthalpy of vaporization given in the lecture notes? For fusion, show that the molar heat capacity of reaction (i.e. ΔrCpm) can be used to compute the standard enthalpy of fusion at different temperature using the equation of 𝑇𝑇
2.31.
𝑇𝑇 𝑜𝑜 o o Δ𝑓𝑓𝑓𝑓𝑓𝑓 Hm ≈ Δ𝑓𝑓𝑓𝑓𝑓𝑓 Hm + � �Cpm �
liq
− �Cpm � � (T − To ) s
Let’s consider an alternative solution to the determination of the standard enthalpy of reaction for urea (see Lecture Example Problem 2.5) at T= 80 C. The reaction of interest is: (NH2)2CO(s) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l) + N2 (g) Instead of using the molar heat capacity of reaction as done in lecture, you need to adjust the standard enthalpy of formation for change in temperature for each of the compounds separately, using their individual molar heat capacities. For example, the standard enthalpy of formation for urea would be evaluated as
∆80f C H mo ,urea ≈ ∆25f C H mo ,urea + C pm,urea ∆T After adjusting each of the reactants’ and products’ standard enthalpies of formation to T= 80 C, compute the standard enthalpy of reaction at T= 80 C? How well does it compare to the value computed in Lecture Example Problem 2.5? 2.32.
The chemical reaction for the combustion of ethanol is shown below. You wish to determine the standard enthalpy of reaction from measured data and from our definition using standard enthalpy of formation. C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l) Part I. The measured data for the reaction can be obtained using an adiabatic bomb calorimeter. A schematic is shown below. Adiabatic bomb calorimeters are constructed from rigid and insulated containers that have an inner chamber where combustion occurs. Heat transfer occurs between the inner chamber and a liquid surrounding the chamber with a heat capacity of 3570 J K-1. The inner chamber is also rigid so that the heat transfer from the reaction to the surrounding liquid is the change in the internal energy (no expansion work) from the ethanol reaction.
Last Modified: March, 2018
2-105
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For our experiment, the combustion of 0.7 g of ethanol (C2H5OH) caused the temperature of the liquid to increase from 25 to 30.814 C. What is the molar internal energy for the combustion of ethanol? To determine the enthalpy of reaction, you need to compute the expansion work that would have occurred under constant pressure for the ethanol reaction at a temperature of 25 C. The molar expansion work is determined using (as derived in Example Problem 2.5) P ΔV wm = − = −R T �−νO2 + νCO2 � Δξ Based on these experimental results, what value would you report for the standard enthalpy of reaction for the combustion of ethanol? 𝐶𝐶 𝑜𝑜 Part II: Also compute the standard enthalpy of the ethanol reaction (Δ25 𝐻𝐻𝑚𝑚 ) at 25 C 𝑟𝑟 25 𝐶𝐶 𝑜𝑜 using the Δ𝑓𝑓 𝐻𝐻𝑚𝑚 values given in the thermodynamic handout. Is this value in good agreement with that obtained experimentally in Part I? Compare your calculated value 𝐶𝐶 𝑜𝑜 25 𝐶𝐶 𝑜𝑜 of Δ25 𝐻𝐻𝑚𝑚 ) 𝑟𝑟 𝐻𝐻𝑚𝑚 with the reported value of the standard enthalpy of combustion (Δ𝑐𝑐 for liquid ethanol given in the thermodynamic handout. Are the values in reasonably close agreement?
2.33.
The reaction for the combustion of carbon monoxide is shown below: CO (g) + (1/2) O2 (g) → CO2 (g) For an experiment at a temperature of 25 C and standard pressure, the standard reaction 𝐶𝐶 𝑜𝑜 𝐻𝐻𝑚𝑚 = - 282.98 kJ mol-1. Determine enthalpy for this reaction has been measured as ∆25 𝑟𝑟 the standard enthalpy of formation for CO. Is your computed value in good agreement with that given in the thermodynamic data handout?
2.34.
From the table of standard enthalpies of formation given below, calculate the standard reaction enthalpy for the following chemical reaction: Ca3P2 (s) + 6 HCl (g) → 3 CaCl2 (s) + 2 PH3 (g) Compound Ca3P2 (s) PH3 (g)
Last Modified: March, 2018
ΔfTHo (kJ mol-1) - 504 +9
2-106
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
HCl (g) CaCl2 (s) 2.35.
- 92 - 794
Metabolism is largely dependent on the combustion of sugar, fats, and proteins. For this problems, we are interested in the standard enthalpies of combustion of fats (using triolein) and proteins (using the polypeptide of glyclphenylalanine as a surrogate). The chemical reaction for the combustion of these compounds are shown below. NIST 𝑜𝑜 reports a standard enthalpy of formation for triolein of Δ𝑓𝑓25𝐶𝐶 𝐻𝐻𝑚𝑚 = -2193.7 kJ mol-1 and 𝑜𝑜 𝑜𝑜 an equivalent Δ𝑓𝑓25𝐶𝐶 𝐻𝐻𝑚𝑚 for glyclphenylalanine of Δ𝑓𝑓25𝐶𝐶 𝐻𝐻𝑚𝑚 = - 684.3 kJ mol-1. C57 H104 O6 (l, triolein) + 80 O2 (g) → 57 CO2 (g) + 52 H2 O (l) C11 H14 N2 O3 (s, glyclphenylalanine ) + 13 O2 (g) → 11CO2 (g) + 7H2 O (l) + N2 (g)
To illustrate the reported range in thermodynamic properties for complex organic 𝑜𝑜 compounds, NIST also reports Δ𝑓𝑓25𝐶𝐶 𝐻𝐻𝑚𝑚 = -1592.2 kJ mol-1 for triolein (you don’t need to do anything with this alternative value).
2.36.
Part I. For both reactions, what are the respiratory quotients and the standard enthalpy 𝑜𝑜 of reactions (i. e. , Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚 )? You can use a temperature of 25 C. Also for both 𝑜𝑜 reactions, compute the change in molar internal energies (i. e. , Δ25𝐶𝐶 𝑟𝑟 𝑈𝑈𝑚𝑚 ). For this 𝐶𝐶 𝑜𝑜 calculation, you will first need to compute the molar work. Is the ratio of Δ25 𝑟𝑟 𝑈𝑈𝑚𝑚 to 25𝐶𝐶 𝑜𝑜 25 𝐶𝐶 𝑜𝑜 Δ𝑟𝑟 𝐻𝐻𝑚𝑚 nearly one (yes or no)? Is the error likely small if we use Δ𝑟𝑟 𝐻𝐻𝑚𝑚 instead of 𝑜𝑜 Δ25𝐶𝐶 𝑟𝑟 𝑈𝑈𝑚𝑚 in the thermodynamic analysis of rooms or buildings of constant volume (yes or no)? Part II: Metabolic rate of animals is often estimated by measuring the amount of oxygen consumed and the amount of carbon dioxide produced. To make these 𝑜𝑜 calculations easier, convert the Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚 values of Part I to kJ per mole of O2 and kJ per 𝑜𝑜 25𝐶𝐶 𝑜𝑜 mole of CO2 (i. e. , Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚𝑚𝑚2 𝑎𝑎𝑛𝑛𝑑𝑑 Δ𝑟𝑟 𝐻𝐻𝑚𝑚𝑚𝑚𝑚𝑚2 in lecture notes). As shown by Homework Problem 1.11, the volume of 1 mole of O2 or CO2 at SATP is 24.8 L. What are the standard enthalpies of combustion in units of kJ L-1 O2 and kJ L-1 CO2 (i.e., 𝑜𝑜 25𝐶𝐶 𝑜𝑜 Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑣𝑣𝑣𝑣2 𝑎𝑎𝑎𝑎𝑎𝑎 Δ𝑟𝑟 𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣2 in lecture notes)?
We wish to investigate whether the change in enthalpy of chemical reactions is independent of path (as done for changes in enthalpy with temperature and phase transitions) by considering the production of methane gas from coal (approximated as graphite) at a temperature of T= 220 C. The sequence of three reactions given below for Path #1 corresponds more closely to the actual path. Path #2 is the chemical reaction more commonly used to represent this process. If you consider the moles of reactants as negative and those of the products as positive, the sum of the moles for reaction sequence of Path #1 results in the chemical reaction given for Path #2.
Path #1: Sequence of three separate reactions whose sum equals the reaction of Path #2. 2 C (s) + 2 H2O (g) → 2 CO (g) + 2 H2 (g) CO (g) + H2O (g) → CO2 (g) + H2 (g) CO (g) + 3 H2 (g) → CH4 (g) + H2O (g)
Last Modified: March, 2018
2-107
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Path #2: Single chemical reaction from coal to methane: 2C (s) + 2H2O (g) → CH4 (g) + CO2 (g) For Path #1, you are required to compute the standard reaction enthalpy at 220 C for all three reactions using the Thermodynamic Data Handout. Sum these three standard enthalpy of reactions to determine the total change in enthalpy from carbon to methane gas by Path #1. You are also required to compute the standard reaction enthalpy at 220 C for Path #2. For this example, is the total change in reaction enthalpy independent of path? These results support Hess’ law. 2.37.
The energy derived from sugar is an important chemical reaction for biological cells. For cells with an abundance of oxygen, glucose is oxidized as C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) However during exercise, oxygen levels in muscle cells may be insufficient to complete the above reaction. In this case, glucose is converted in lactic acid (C3H6O3) by a process called glycolysis. This reaction is shown below. C6H12O6 (s) → 2 CH3CH(OH)COOH (s) 𝑜𝑜 Part I. Compute the standard enthalpy of reaction (Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚 ) for the first glucose reaction using the standard enthalpies of formation given in the Thermodynamic Data 𝑜𝑜 Handout. Compare your computed Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚 to the standard enthalpy of combustion 25𝐶𝐶 𝑜𝑜 (Δ𝑐𝑐 𝐻𝐻𝑚𝑚 ) for glucose (also given in the Thermodynamic Data Handout) by using the 𝑜𝑜 25𝐶𝐶 𝑜𝑜 ratio of Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚 to Δ𝑐𝑐 𝐻𝐻𝑚𝑚 . Is this ratio nearly one (yes or no)?
Part II: The standard enthalpy of formation for lactic acid is not available in the Thermodynamic Data Handout. However, Atkins and de Paula (2006) have reported the molar enthalpy of reaction for following reaction of lactic acid as ΔrTHm° = -1344 kJ mol-1. CH3CH(OH)COOH (s) + 3 O2 (g) → 3 CO2 (g) + 3 H2O (l) From the results of Problem 2.36, you have shown that the standard enthalpy of reaction can be computed by solving the reaction for three separate reaction. Apply this principle to determine the standard enthalpy of reaction for glycolysis using the following sequence of reactions C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) 6 CO2 (g) + 6 H2O (l) → 2 CH3CH(OH)COOH (s) + 6 O2 (g) What is the ratio of the standard enthalpy of reaction of the oxidation of glucose to that obtained by glycolysis? Which of the two reactions provide more energy for other biological activities?
Last Modified: March, 2018
2-108
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
2.38.
𝑜𝑜 Part III. Convert the Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚 obtained in Part I to enthalpy of reaction per mole of O2 𝑜𝑜 (i. e. , Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑚𝑚𝑚𝑚2 in lecture notes). Compute the enthalpy of reaction per L of consumed 𝑜𝑜 O2 at SATP (i.e., Δ25𝐶𝐶 𝑟𝑟 𝐻𝐻𝑣𝑣𝑣𝑣2 in lecture notes).
Consider a hypothetical animal whose metabolism can be well approximated by the combustion of glucose and triolein. These chemical reactions have been analyzed in Homework Problems 2.35 and 2.37, and thermodynamic results are summarized in Table 2.3. For our animal, the consumption rate of oxygen has been measured to be 20 L O2 h-1 , the production rate of CO2 has been measured to be 16 L CO2 h-1 and the production rate of water vapor by vaporization has been measured to be 90 L H2O h-1. All of these measurements correspond to standard ambient temperature and pressure of T=25 C and Po = 100 kPa for ideal gases.
Part I. Compute the metabolic rate in units of MJ d-1 and the molar extents of reaction rate (mol h-1) for both glucose and triolein? Part II. Compute the fraction of the metabolic rate obtained in Part I that is used to evaporate water. You can use the standard enthalpy of vaporization corresponding to 39 𝐶𝐶 𝑜𝑜 −1 C (animal temperature) of Δ39 . 𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 = 43.2 𝑘𝑘𝑘𝑘 𝑚𝑚𝑜𝑜𝑙𝑙 Part III. Compare the metabolic rate of Part I to that obtained by Brouwer’s relationship (i.e., Eq. 2.7.28). What is the percent error that you obtained using Brouwer’s relationship? 2.39.
For four piglets, Professor Larry Jacobson at the University of Minnesota measured a consumption rate of 30.83 L O2 h-1 and a production rate of CO2 of 20.08 L CO2 h-1 defined using the STP of a temperature of T = 0 C and a pressure of P = 101.325 kPa (1 atm). What are the consumption and production rates in units of moles of O2 and moles of CO2 per hour? in units of L O2 and L CO2 per hour adjusted for standard ambient temperature and pressures SATP of T= 25 C and Po=100 kPa? What is the observed respiratory quotient? By using Brouwer’s relationship (see Eq. 2.7.28), what is the metabolic rate in units of MJ d-1 pig-1?
2.40.
Consider an experiment for an animal that is consuming carbohydrates, fats, and proteins. You can use the metabolic reactions given in Table 2.3 for glucose, triolein and glyclphenylalanine to determine those for carbohydrates, fats, and proteins, respectively. The experiment measured a change in the number of moles of oxygen, carbon dioxide and nitrogen of Δno2 = - 0.6 mol O2 h-1, Δnco2 =0.5 mol CO2 h-1, and Δnn2 = 0.01 mol N2 h-1. What is the metabolic rate for the animal in units of MJ d-1? What are the molar extents of reaction rate (per mol h-1) for carbohydrates, fats and proteins?
2.41.
You wish to determine the metabolic rate of animal that consumes carbohydrates, fats, and proteins. Previous studies have shown that the calorie consumption from fats for this animal is approximately twice that of proteins (that is, ΔHr,fat = 2 ΔHr,prot). You can use the general metabolic reactions and respiratory quotient given in Table 2.3 for our food sources. You have experimentally measured a change in the number of moles of oxygen and carbon dioxide of Δno2 = - 0.6 mol O2 h-1 and Δnco2 =0.55 mol CO2 h-1 when the animal is resting and Δno2 = - 1.0 mol O2 h-1 and Δnco2 =0.8 mol CO2 h-1 when the
Last Modified: March, 2018
2-109
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
animal is exercising. What are the resting and exercising metabolic rate for the animal in units of MJ d-1? What fractions of these total metabolic rates are from carbohydrates, from fats and from proteins? Compute these fractions for both resting and exercising conditions. 2.42
You have a pig barn that has no heat transfer with the surroundings. The barn contains 50 pigs and has a width of 10 m, a length of 18 m and a height of 6 m. The average mass of pigs is 15 kg. The reaction enthalpy rate of metabolism per pig, for masses between 10 and 100 kg, can be approximated as Δr Ḣ pig = −14.95 m0.6
where m is the mass of the pig in kg and Δ𝑟𝑟 Ḣ pig is the reaction enthalpy rate with units of J s-1 per pig (the coefficient “14.95” has units of J s-1 pig-1 kg-0.6). The percentage of (negative) reaction enthalpy used for vaporization is approximately 50%. What is the change in the air temperature in the barn after 90 minutes? What is the total mass of water evaporated from the pigs? For air within the barn, you can use mm = 29 g mol-1, Cpm = 29 J K-1 mol-1; and ρair = 1.3 kg m-3. You use standard enthalpy of vaporization 𝐶𝐶 𝑜𝑜 −1 corresponding to 39 C (pig temperature) of Δ39 . You can neglect 𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 = 43.2 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 expansion work component of the reaction enthalpy. 2.43.
You have a turkey barn that has a constant heat transfer rate into the building of q = 50,000 kJ h-1. The barn contains 10,000 birds with an average mass of 5 kg. The reaction enthalpy of metabolism per bird can be estimated using Δr Ḣ bird = −35.5 m0.77
where m is the mass of the bird in kg and Δ𝑟𝑟 Ḣ bird is the reaction enthalpy rate with units of kJ h-1 per bird (the coefficient “35.5” has units of kJ h-1 bird-1 kg-0.77). Forty percent of the (negative) reaction enthalpy is used to evaporate water from turkeys. What air flow rate is needed (units of kmol of air per hour) from ventilation fans to prevent the temperature of the building to increase by 5 C? What is the rate of water evaporated from the birds (kmol h-1)? You can use mm = 29 g mol-1 and Cpm = 29 J K-1 mol-1 for air and a standard enthalpy of vaporization at the bird temperature of 𝐶𝐶 𝑜𝑜 −1 Δ41 . Consider the barn to be a rigid system. You can neglect the 𝑣𝑣𝑣𝑣𝑣𝑣 𝐻𝐻𝑚𝑚 = 43.2 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑙𝑙 expansion work component of the reaction enthalpy. 2.44.
Propane gas tanks are widely used to supply fuel for outdoor grills. The combustion of propane for grilling is done under conditions of nearly constant atmospheric pressure. Grill gas tanks are typically filled with 9 kg (20 lbs) of propane. What is the possible heat transfer in MJ for the complete combustion of propane in one of these tanks? This heat transfer can be used for cooking, and it will also increase the air temperature of the surroundings.
2.45.
You are planning on taking food in a picnic basket on a hot summer day and wish to avoid food spoilage. You have determined that there is no spoilage if the temperature in the basket at the end of the day is 20 C. A friend (who has completed a heat transfer
Last Modified: March, 2018
2-110
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
course) has estimated that the heat transfer of 22.5 kJ will occur during the day given the temperature difference between the outside air and inside the picnic basket. You are required to determine the mass of ice to place into the basket. Your picnic basket is not rigid system and can expand with changes in temperature. It has a representative volume of 0.25 m3. You can assume a constant pressure process and perform your computations without any food in the basket. You can further assume that the initial temperature of the air (and the ice) is 0 C. You can neglect evaporation of liquid water, use a constant air pressure of 101 kPa, and a molar heat capacity at constant pressure for air of 29.15 J K-1 mol-1. 2.47.
For an electromagnetic wave of wavelength of 625 nm, what is the frequency of this wave? What region of the electromagnetic spectrum is it found in? What is its molar photon energy?
2.48.
Our eyes can perceive green light (λ = 5.5 x 10-7 m) wave length when the rate of incidence of radiation is 2 x 10-16 J s-1. Calculate the number of photons of green light that must fall on the retina per second to produce vision.
2.49.
Compute the energy of one quantum of red light (λ = 700 nm). How many joules of energy are in a mole of these photons? How many red light quanta are in one joule of energy?
2.50.
Photosynthetically active radiation (PAR) wavelengths lie between 400 and 700 nm. Compute the radiant flux density for PAR by numerically integrating the area under the spectral flux density over these wavelengths for the temperature of the Sun of 5778 K (neglecting the impact of absorptivity of atmospheric gases). First divide the range in PAR wavelengths into a minimum 10 intervals. For each interval, compute the spectral flux density using the midpoint wavelength. You can approximate the area of each interval as rectangles. Also compute the total radiant flux density corresponding to all wavelengths for T = 5778 K. What fraction of the total radiant flux density lies within the wave lengths of PAR?
2.51.
The energy content of the proven reserves of coal, oil and gas has been estimated to be 36 x 1021 J. The real reserve of fossil fuels (all fossil fuels under the earth’s surface) is often estimated as five times greater than the proven reserves. Part I: Let’s obtain insight into the magnitude of solar energy by using a solar constant of 1368 W m-2 and the radius of the Earth’s land surface of 6.37 x 106 m. Assume that the net transmissivity of the atmosphere is 0.5, that 1% of the projected Earth’s surface is available for capturing solar radiation, and that this surface receives sunlight corresponding to the solar constant for an equivalent 6 hours per day. How many years would it take for solar radiation landing on this surface to obtain the energy equivalence of the proven reserves of fossil fuels? of real reserves of fossil fuels? Part II. Hall and Rao (1999) has estimated that the global production of biomass by photosynthesis as 2 x 1011 tonnes of organic matter per year, which is equivalent to 4 x
Last Modified: March, 2018
2-111
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
1021 J y-1. Annual food intake by humans has been estimated as an equivalent of 16 x 1018 J y-1. For proven reserves of coal, oil and gas, how many years of net photosynthesis – adjusted for the biomass necessary to feed humans - would it take to generate the energy equivalent of these reserves? How many years of net photosynthesis – again adjusted for the biomass necessary to feed humans - would it take to generate the energy equivalent of all fossil fuels under the Earth’s surface? 2.52.
Characteristics of Earth, Venus, Mars and Jupiter are given below. The subscript “o” refers to characteristics on the surface of the planet, and rs-p and rp are the distance from the sun to the planet and the radius of the planet, respectively. The symbol Tp,meas is the approximate temperature of the planet (surface and atmosphere) and To,meas is the approximate temperature of the planet surface. You can assume that the emissivities of the sun and planets are one. For Venus, Mars, and Jupiter, compute the solar constant and the global radiant flux density for a sun temperature of T=5778 K. Also compute the equilibrium temperatures for these planets. For which planets (include Earth based on Example Problem 2.8) are the equilibrium temperatures in good agreement with Tp,meas? with To,meas? Does the atmosphere play an important role on the surface temperatures? rs-p km
rp km
fr
Planet
g m s-2
Po kPa
ρo kg m3
Tp,meas K
To,meas K
Venus
8.9
1.0820 x 108
6052
0.77
9200
65
230
750
Mars
3.7
2.0660 x 108
3396
0.15
0.636
0.02
220
218
Jupiter
23.1
7.7860 x 108
71492
0.58
NA
NA
130
NA
Earth
9.8
1.4960 x 108
6378
0.30
101
1.217
250
280
2.53.
Compute the solar irradiance at the top of the atmosphere for Crookston, MN (latitude = 47.75 degree N, longitude = 96.5 degree W) for July 1.
2.54.
For the solar irradiance determined in Problem 2.53, compute the transpiration depth for a well-watered alfalfa field located near Crookston, MN for a clear-sky day of July 1, where the average air temperature during transpiration is 25 C and the average difference between the system (plants) and the surrounding (air) is 5 C. You can assume emissivity, reflectivities, absorptivities, and Bowen ratio are equal to those values used in the example problem solved in lecture and perform your calculations on a representative area of the canopy of 100 m2.
2.55.
For a location in southern Minnesota with a day length of 15 hours, the solar irradiance at the top of the atmosphere has been determined to be 50000 kJ m-2. You are interested in the transpiration depth for a well-watered alfalfa field for a clear-sky day of July 1, where the average air temperature during transpiration is 26 C for the plant canopy and 20 C for the surrounding air. Use a transmissivity of atmosphere of fta = 0.7; emissivities of εp = 0.9 for alfalfa and εa = 0.7 for air; reflectivity of frp = 0.22 for alfalfa (plant), and a Bowen ratio of β = 0.25. The temperature change of the air within the
Last Modified: March, 2018
2-112
Biological and Environmental Thermodynamics
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
canopy is 3 C. Perform your calculations on a representative area of the canopy of 100 m2, a canopy height of 1.1 m, air pressure of 100 kPa, and molar heat capacity at constant pressure of air of 29.15 J K-1 mol-1. You can assume that the fpar =0.4 and represent the reaction enthalpy using the photosynthesis of glucose. What is the transpiration depth? What is the depth if you neglect ΔHtpr? What is the depth if you neglect ΔHtpr and ΔHr? 2.56.
Transpiration from plant canopies is often estimated using measured evaporation depths from a standard Class A pan. This type of pan is a cylinder with a surface area of 1.14 m2 and is typically filled with water to a depth of 0.2 m. For one week (7 days) in southern Minnesota, the evaporation depth of 56.4 mm was measured from one of these pans. The incoming solar radiation (at the pan) for that week was measured to be 181,000 kJ m-2. The reflectivity of the pan and water is approximately 0.4. Use an average air temperature and pan water temperature during the week of 30 C and 18 C, respectively. Assume the emissivity of air of εa = 0.8 and the emissivity of the evaporation pan (with water) of εw = 0.9. Use a change in the temperature of the water in the evaporation pan of 10 C during the evaporation process. Determine the heat transfer by into (or out-of) the evaporation pan by conduction and convection processes. What is the Bowen ratio for the evaporation pan?
Last Modified: March, 2018
2-113
Biological and Environmental Thermodynamics
