DEPARTMENT OF CHEMICAL ENGINEERING MANIPAL INSTITUTE OF TECHNOLOGY MANIPAL UNIVERSITY MANIPAL

V SEM LAB MANUAL CHE3111: HEAT TRANSFER LAB

1

CONTENTS Expt. No

Experiment

Page No.

General instructions

03

Scheme of Evaluation

04

1

Transient Heat Conduction

05

2

Combined Convection and Radiation Heat Transfer

09

3(A)

Heat Transfer in Bare Tube Heat Exchanger

12

3(B)

Heat Transfer in Finned Tube Heat Exchanger

15

4

Single Effect Evaporator

20

5

Drop wise and Film wise Condensation

24

6

Packed Bed Heat Exchanger

29

7

Thermal conductivity of metal rod

33

2

GENERAL INSTRUCTIONS 1. Students attending the laboratory class should come prepared to carry out their respective experiments and the calculations within the class timings. 2. Student would explain the procedure of their experiment failing which they will not be permitted to perform the experiment. 3. An observation book (Long size) should be maintained by every student for taking the observations and carrying out calculations. 4. Though the experiments are to be carried out by the respective group of students, the calculations should be done individually. 4 Record of experiment carried out; written in A 4 bond size sheets should be submitted in the subsequent class. Records submitted at any later date shall be deemed as late submission. 5 Written record should be kept in an office file and submitted. The file shall be kept with the teachers. Records of subsequent experiments, written in sheets should be brought and attached to the file. 6 The date, experiment number, title of the experiment. Objective, theory, procedure and result should be written on the front side of every page. All the observations including tables and the calculations should be on the back side of every page. 7 Line diagram of the experimental setup should be shown for each experiment. 8 Relevant theory should be written for each experiment. 9 Each record of experiment shall be valued for 10 marks out of which 1 mark is exclusively for neatness. 10 A student who was absent for an experiment, shall be permitted to carry out the experiment in a repetition class. However, the record of the experiment submitted will be credited only 80% of the marks secured. 11 After completion of course, the records of all experiments along with a contents page should be bound and brought to the class teacher for certification. The certified record book shall be submitted to the examiners at the time of examination. 12 Students should come to the laboratory, equipped with the necessary materials such as calculator, pencil, scale, eraser, appropriate graph sheets etc. 13 Wearing shoes is compulsory in the laboratory, without which student will not be permitted to perform the experiment.

3

Scheme of Evaluation for lab journal Total marks per expt. = 10

LHS Line diagram- 1 M Observations & tabular column with proper units- 1 M Specimen Calculations- 2 M Neatness- 1 M

RHS Date of experiment Experiment no.(according to manual) and name Aim Theory- correlation, concept and industrial application- 2 M Procedure in passive voice- 1 M Result with proper units- 1 M Discussion- inference, opinion- 1 M

4

EXPERIMENT: 1 TRANSIENT HEAT CONDUCTION Aim: To calculate the thermal diffusivity of material when heat is conducted under unsteady state, in the given solid bar.

Theory: Procedure: 1. Measure the length and diameter of the aluminum bar 2. Measure the distance of the 3 thermometer from the source 3. Note down the initial temperature of the solid Al bar in the 3 different thermometers. 4. Switch on the IR bulb and adjust the voltage supply to 220 V. Maintain this voltage constant throughout the experiment. 5. Start the stop watch immediately and note down the temperature of each thermometer at interval of 5 minutes till 120 minutes. 6. Plot a graph of temperature vs. time and time vs. distance 7. Find the predicted temperature using relevant equation.

Observation: Distance from the source to the 1st thermometer, x1 = . . . . . . . m Distance from the source to the 2nd thermometer, x2 = . . . . . . . m Distance from the source to the 3rd thermometer, x3 = . . . . . . . m Voltage, V = 220 V Current, I = 1.475 A Cos = 0.73 Length of the bar, L = . . . . . . . m Diameter, d = . . . . . . .m Density of the aluminum bar,  = 2700 kg/m3

5

Table 1: 1st Thermometer Temp Time 1 T1(oC) (min) 0 5 10

2nd Thermometer Temp Time 2 T2(oC) (min) 0 6 11

120

121

3rd Thermometer Temp Time 3 T3(oC) (min) 0 7 12

122

Draw graph of temperature vs. time.

Graphs: dT/dx

T1,T2,T3 oC

T oC

 (sec)

x1,x2,x3 (cms)

From the graph of T1, T2, T3 vs. , get the slope of dT1/d, dT2/d, dT3/d (3 values each). Then draw a graph of T vs. x1,x2,x3 From the graph of T vs. x1,x2,x3 get the slop dT/dx1, dT/dx2, dT/dx3 (3 values each). Then draw a graph of dT/dx vs. x1,x2,x3

6

X- N

dT/dx dT2/dx2

From the graph of dT/dx vs. x1,x2,x3 get the slope dT2/dx2 Note: Nature of graph: If straight line one slope If it is curve take slopes at 3 locations and take average. Calculations: Cross sectional area of the aluminum rod, A = d2/4 = . . . . . . . . m2 Q

=

VICos

= . . . . . . . . . .W

From the Fourier law of unsteady state heat conduction dT d2T ------ =  -----d dx2 dT/d  = -------------d2T/dx2

= ………. m2/s

Table 2: Time (t) min

Position (m)

 (m2/s)

Observed tempC

7

Result : At time of ………….. min and distance of …………… m Observed temperature = ………………C Thermal Diffusivity from the Graph = …………….. m2/sec

8

EXPERIMENT: 2 COMBINED CONVECTION AND RADIATION HEAT TRANSFER Aim: To determine the emissivity of a given solid when heat is transferred to the surroundings by natural convection and radiation. Theory:

Procedure: 1) Heat the given solid block to a temperature of about 500oC in furnace. 2) Remove the heated solid block from furnace, place it on a platform, fix a thermometer and note down the temperature. 3) Start a stop watch and note down the time for every 10oC fall of temperature till it reaches about 100oC 4) Note down the weight, length and diameter of the solid. 5) Draw a graph of temperature vs. time. Determine the slopes at three points. Observations: 1) Weight of solid m = ……. Kg 2) Length of solid L = …….. m 3) Diameter of solid d = …….. m 4) Specific heat of the solid material Cps = 0.473 J/kgo K 5) Stephen boltzman’s constant,  = 5.64 * 10-8 w/m2 K4 6) Thermal conductivity of the given solid, Ks = 43 w/m˚K 7) Ambient temperature, Ta = ………..oC Table 1: Sl.No.

Temperature (oC)

Time (sec)

9

Calculations: Tavg = (T + Ta)/2 = ……. oC Where T is temp. at which tangent is drawn  = 1/Tavg = …….oK-1 T = T – Ta = …….. oK air = PM/(RTavg) = ……. Kg/m3 NPr = (Cp)/K = ……… Cp, , K = sp. Heat, viscosity and thermal conductivity of air at Tavg 2gTL3 NGr = ---------------------2 hcL/Ks = 0.48 (NPr * NGr)0.25 0.125 (NPr * NGr)

0.33

if 104 < NPr * NGr <107 if NPr * NGr > 107

hc = ……….. w/m2˚K Area of heat transfer, A = dL + 2d2/4 = ………… m2 Total heat loss, QT = Q conv + Qradiation mCp (dT/d) = hCA(T-Ta) +  A (T4 – Ta4) Where m = mass of solid Cp = sp. heat of solid at tangent temp. dt/d = rate of change of temperature (from graph)= slope A = area of heat transfer (m2) hc = convective heat transfer coefficient (w/m2K) Ks=thermal conductivity of solid at tangent temp. (w/mK)  = Emissivity

10

Table 2: Air properties can be obtained from handbook Cp, , , K Sl.No Temp Tavg = Cp K =   (oC) (T+Ta)/2 (J/kg)) (kg/m3) (w/mK) (NS/m2) 1/Tavg (oC) (K-1)

T (oC)

Table 3: Sl.No.

NPr

NGr

NPr * NGr

hc (w/m2K)

dT/d (oC/Sec)



Graph:

Temp (oC)

Time (sec)

Result: The average emissivity of the given solid block is ……….

11

EXPERIMENT: 3A HEAT TRANSFER IN BARE TUBE HEAT EXCHANGER Aim: To determine overall heat transfer coefficient and air side film coefficient when steam is condensing inside the bare tube

Theory: Procedure : 1) Note down the height of heat exchanger and also the diameter of the tube. 2) Open the vent valve and the steam inlet valve. Regulate the vent valve such that only small quantity of steam passes through it. 3) Maintain the steam pressure to the desired value. And at this value, collect the condensate for 30 minutes in a previously filled bucket with cold water. Determine the flow rate. 4) Repeat the above step for different steam pressures and note down the flow rates of condensate. 5) Calculate the overall, and airside film coefficients using relevant equation.

Observations: 1) Height of the tube, L = . . . . . . .m 2) Diameter of the tube Outer diameter, do = .. ………. m Inner diameter, di = …………..m 3) Area, Ao = doL = ……….. m2 Table 1: Steam Mass of Time pressure condensate (Sec) (kg/m2) (kg)

Mass flow rate (kg/S)

Ta (oC)

Ts (oC)

T1  (KJ/Kg) (TsTa)oC

Q Uo (watts) (w/m2oC)

12

Calculations: 1) Mass flow rate, mo = m/t = .. kg/s 2) Ambient temperature, Ta = ………..oC 3) T1 = Ts – Ta = …………… oC 4) Overall heat transfer coefficient, Uo = (mo)/(AoT1) = ……w/m2K where  is calculated from steam tables at Pabsolute. 5) Tavg = (Tw + Ta)/2 = …………oC 6)  = 1/Tavg = ……………. K-1 PM 7)  = --------- = …………. Kg/m3 where M=mol wt. of air. RTavg Cp  8) Npr = ---------- = ……………… Get the properties of air from hand book at Tavg K gL32T2 9) NGr = ------------------------------ = ………….. 2 10) NGr * Npr = ……………………….. 11) (hoL)/K = 0.48 (NGr * Npr ) ¼ if 104 < NGr Npr < 107 0.125 (NGr * Npr) ) 1/3 if NGr * Npr >107

hi calculations: Tsat+Ta Tw = ------------------ = ……oC 2 T2 = Ts-Tw = …..oC Tf = Ts-3/4 (Ts-Tw) , use hand book for water properties at Tf Kf, f, f – the water properties are at Tf 1/4

 K f 3  f 2 g  hi  0.943    L f T2 

13

Table 2:  (K-1)

Tavg (oC)

Cp (KJ/Kg)

 (NS/m2)

 (kg/m3)

T2 (oC)

Npr

NGr

NprNGr

ho (W/m2K)

Table 3: hi calculation Tavg (oC)

T2(oC)

Tf(oC)

Kf (W/mK)

Pf (kg/m3)

 (NS/m2)

hi (w/m2K)

Compare experimentally calculated Uo and obtained from individual heat transfer coefficients 1 1 do ---- = ----- + ---Uo ho dL

xm 1 do ------ + ---- ------- where Km = thermal conductivity of the material km hi di

Km = 43 w/m˚k xm = Thickness of material Uo (experimental) = (mo λ)/(Ao ΔT1) Sl.No

Uo (predicted)

Uo (experimental)

Result: Overall heat transfer coefficient, Uo and outside natural convection heat transfer coefficients are

14

EXPERIMENT 3B HEAT TRANSFER IN FINNED TUBE HEAT EXCHANGER Aim: To determine (i) overall heat transfer coefficient (ii) air side film coefficient (iii)fin efficiency (iv) Overall fin efficiency in a finned tube condenser. Theory: Procedure : 1) Note down the height of heat exchanger ,the diameter of the tube, the length and breadth and number of fins. 2) Open the vent valve and the steam inlet valve. Regulate the vent valve such that only small quantity of steam passes through it. 3) Maintain the steam pressure to the desired value. And at this value, collect the condensate for 30 minutes in a previously filled bucket with cold water maintaining constant pressure. Determine the flow rate. 4) Repeat the above step for different steam pressures and note down the flow rates of condensate. 5) Calculate the overall, and airside film coefficients using relevant equations.

Observations: 1) Height of the tube, L = . . . . . . . m 2) Outside diameter of the tube, do = . . . . . . . m 3) Length of the fin, l = . . . . . . . m 4) Breadth of the fin, b = . . . . . . . .m 5) Thickness of the fin,  = …………. M 6) Number of fins, n = ………….. Table 1: Steam Mass of Time pressure condensate (sec) kgf/cm2 m (kg)

Mass Ta flow rate (oC) (mo(kg/s)

Tsat (oC)

 (KJ/kg)

Q watts

Uo w/m2K

15

Calculations: 1) Mass flow rate, mo = m/t = …………. Kg/s 2) Ambient temperature, Ta = …………….

C

3) T = Tsat – Ta = ………..oC mo 4) Uo = ------------- = ……….. w/m2K where Ao = dol and  is calculated from steam AoT1 tables at P absolute. Tsat + Ta 5) Tw = --------------- . . . . . . oC 2 6) T2 = Tsat – Tw = ………….oC 7) Tavg = (Tw + Ta)/ 2 = . ……….oC 8)  = 1/Tavg = ………. 1/k 9)  = (PM)/RTw = ………… kg/m3 where M=mol wt. of air.

10) Npr = (Cp)/K the properties at T avg gL32T2 11) NGr = -------------------- = ………… 2

12)

NGr * Npr = …………

13) hoL/K = 0.48 (NGr * Npr )1/4 if 104 < NGr * Npr <107 0.125 *( NGr * Npr )1/3 if NGr * Npr >107

hi calculations: Tsat+Ta Tw = ------------------ = ……oC 2 T2 = Ts-Tw = …..oC 16

Tf = Ts-3/4 (Ts-Tw) Kf, f, f – the water properties are at Tf 1/4

 K f 3  f 2 g  hi  0.943    L f T2  To determine fin efficiency:

tanh mL tan h mb f = ---------------- = ------------------------mL m.b m =  [(hoP)/(KA)] where P = perimeter of fin = 2 [b + ] (m) A = profile area = b*  (m2) K = thermal conductivity of material (w/mK)

Overall fin efficiency Ao – A’f (1 - f) overall = ----------------------------------Ao A’f = Ao + 2(l . b)n

Table 2: Tavg  (k-1) (oC)

Cp K T2   o 2 3 (KJ/kg K w/mK (NS/m ) Kg/m (oC)

Npr

NGr

Npr*NGr ho (w/m2K)

17

Table 3: ho calculation o Tavg T2(oC) Tf( C) (oC)

Kf (W/mK)

pf (kg/m3)

 (NS/m2)

hi (w/m2K)

Table 4: Sl.No.

m (m-1)

mL = mb

%f (tanhmb*100)/mb

overall (%)

To draw a graph of f vs m.b For different values of b, calculate perimeter and area. Find m and f using the formula given. Table 5:

m

b(m)

m.b

f

0.01 0.015 0.02 0.025

0.045

18

Graph:

f

m.b

Result: The overall heat transfer coefficient, airside film coefficient and the fin efficiency are reported in tables.

19

EXPERIMENT: 4

SINGLE EFFECT EVAPORATOR

Aim: To determine steam economy and the overall heat transfer coefficient of the given evaporator.

Theory: The purpose of evaporation process is the formation of a more concentrated solution or product from a dilute feed.

To obtain the concentrated product, the feed is boiled to

evaporate off water. The vapor and liquid in the boiling section are in equilibrium therefore sharing equal outlet temperatures which is the boiler temperature. The vapor then proceeds to the first effect to be condensed by cooling water and normally considered a waste product or possibly purification worthy. The concentrated product from the first effect is the final product or in large capacity operations is sent to multiple effects. The volumetric flow rates of the outlet steam, condensed vapor, and product liquid are recorded. The effect pressure and temperature, and the inlet steam pressure were known. The inlet feed temperature was assumed to be room temperature. The above data is used to calculate the systems heat loss and overall heat transfer. An overall energy balance for the system is shown in equation (1). FHf + Sls = LHL + VHv

(1)

Where, F = Feed flow rate (kg/min) Hf = Enthalpy of the feed (kj/kg) S = Steam flow rate (kg/min) L = Liquid product flow rate (kg/min) HL = Enthalpy of the liquid product (kj/kg) V = Vapor flow rate (kg/min) Hv = Enthalpy of the vapor (kj/kg) It is then desired to calculate the theoretical output steam flow rate by rearranging equation (1) to give equation (2).

20

Ms = (1 / λs) (FCp (Tb – Tf) + VHv)

(2)

Where, λs = Latent heat of the steam (kJ / kg) Cp = Heat capacity of the feed (kJ / kg K) Tb = Temperature of the boiler (K) Tf = Temperature of the feed (K) Hv = Latent heat of vapor (kJ/kg) The outlet steam flow rate found above is used to theoretically find the overall heat transferred of the system as seen in equation (3). Q = Ms * λs

(3)

The amount of heat transferred from the system is then used in equations (4) and (5) to solve for the heat transfer coefficient Q = U A (TS – Tf)

(4)

U = Q / A (TS – Tf)

(5)

Where, A = Area of the boiler (m2)

Under vacuum conditions, the vapor will boil off at a lower temperature; hence, less amount of steam is needed to obtain the desired product. Under vacuum conditions the same equations and theoretical principals at atmospheric pressure apply. However it is expected that the vapor enthalpy will change. For industrial processes the steam pressure calculations are based on the desired final product concentration. The higher the steam pressure leads to a higher product concentration. Here, a single effect evaporator was observed. The use of single effect evaporators are cost efficient only when the required capacity of operation is small.

21

Specifications:

Tube diameter

:

1/2", 16 gauge

:

12

Length of the tube

:

200 mm

Steam chest

:

200 mm dia x 200 mm length

Downcomer

:

75 mm dia x 200 mm length

Heat exchanger

:

length – 450 mm

No. of tubes

n

Diameter – 100 mm Condensate tank

:

100 mm dia x 300 mm l

Boiler

:

5 Kg/cm2

Feed tank

:

20 liters capacity, SS

Procedure: 1. Fill the Boiler up to 70% volume with clean water. Switch on the mains and then the Heaters (2KW x 3 = 6 KW). 2. Allow the pressure to reach 2 Kg/cm2 and in the meantime prepare 3% sugar solution which will be fed in to the evaporator chamber. 3% = 30gms in 1 litre water or in total 150gms sugar in 5 litre water. 3. After the feed is introduced and the steam pressure built (to 2kg/cm2), introduce the steam into the feed chamber slowly. 4. The feed solution slowly gets heated and forms vapour which goes from the coil region to condenser region. 5. Introduce condensing action by cold water into the reflex condenser. Ensure tap connection is continuous to the condenser. Note the temperature of steam inlet & outlet. 6. The Evaporated water condenses and is collected at the outlet of the condenser. Note the temperature of hot feed and vapour @ T2 (Tb)and T4 (Tv)respectively. 7. The process should be carried out for 20 minutes and the concentrated condensate or sugar syrup should be collected. 8. The steam condensate should be carefully removed and all the built pressure in Evaporator should be released. 9. The Boiler should be put off and all the steam relief valves should be opened. 22

Specimen Calculations: F

=

Initial feed kg

L

=

Final product kg

V

=

Water evaporated kg

Ms

=

Condensate collected kg

Ts

=

Steam temperature oC

Tb

=

Evaporator temperature oC

n

=

no. of the tubes

dt

=

diameter of the tube m

l

=

length of the tube m

F

=

L+V

Steam economy =

V/Ms

Q

=

UA T

A

=

n dt l

T

=

Ts – Tf oC

U

=

watts m2

Q / A T

watts / m2 K

PRECAUTIONS: 1. The level gauge valve should be opened while filling water to the boiler and then it should be closed so that steam does not enter the level gauge. 2. After switching on the heater air should be vented out from the boiler at least 3 to 4 times. 3. After feeding steam, cold water supply should be given to the condenser to prevent vapor loss. 4. After the experiment the evaporator should be drained completely.

23

EXPERIMENT: 5 DROPWISE & FILMWISE TEMPERATURES AND CONDENSATION

AIM: To determine and compare surface heat transfer co-efficient for a) Drop wise condensation b) Film wise condensation

THEORY: Condensation is the change in phase from vapor state to liquid state. It can be considered as taking place either within a bulk material or on a cooled surface and is accompanied by simultaneous heat and mass transfer.

Condensation plays a significant role in the heat rejection parts of the Rankine Power Cycle and the vapor compression refrigeration cycle, which generally involve pure substances. Dehumidification in air conditioning, production of liquefied petroleum gas, liquid Nitrogen and liquid Oxygen are examples in which condensation in a mixture takes place. Condensation on a cooled surface occurs in one of two ways- Film or Drop wise condensation.

In Film wise Condensation the liquid condensate forms a continuous film which covers the surface and takes place when the liquid wets the surface. This film flows over the surface under the action of gravity or other body surfaces, surface tension and / shear stresses due to vapor flow. Heat transfer to the solid surface takes place through the film which forms the greatest part of the thermal resistance.

In Drop wise Condensation the vapor impinges on the cool wall, reducing its energy and thereby liquidifying and forming drops which grow by direct condensation of vapor on the drops and by coalescence with neighboring drops until the drops are swept off the surface by the action of gravity or other body forces, surface tension and / shear stresses to vapor flow.

24

Drop wise condensation of steam has heat transfer co-efficient 2 to 10 times as large as film condensation. However, it has been difficult to sustain drop wise condensation commercially for long periods of time. WORKING PROCEDURE: 1) Steam Boiler should contain 70% volume of water and this can be fed from sump tank by a pump. 2) Once the water level is attained, cut the valve and switch on the Boiler switch. 3) The Boiler has a 2 KW heater and will take 15 mins to reach pressure of 1.5 Kg/ cm2 4) Once the pressure is reached, switch on the pump and allow a constant flow rate of cold water into any once chamber say copper tube. Flow can be started at 2 lpm initially. 5) Release the steam into copper tube chamber and after one minute, note the water inlet and outlet temperatures and temperature of copper tube specimen and temperature of copper tube chamber. 6) After 5 mins, cut the steam connection and collect the condensate or cut the pressure @0.5kg/ cm2. 7) Repeat the process by again building the pressure to 1.5 kg/ cm2 and feed the steam to Gold plated Rod. Cut the water connection from copper tube and feed the water to Gold plated tube. 8) After 5 minutes, note the temperatures. 9) Repeat the process for varying cold water flow rate and steam volumes. EXPERIMENTAL SET UP: A Schematic diagram of the apparatus is shown in Figure. The condensing chamber CC1 consists of the Gold Plated Copper Tube GPCT and the Plain Copper Tube PCT of identical dimensions. Boiler (1) is provided to prepare steam which will be allowed inside the chamber. Pump 2 circulates cooling water through the tubes. Inlet (suction) of the pump is connected to Sump (3). By operating Valve Vi flow control valve V2 V3 experiments to be carried out at different flow rates. Rotameter R1 & R2 measures water flow rate. T/C T1 measures the water inlet temperature to the system. The two tubes GPCT and PCT are connected in series for the water circuit. Thermocouples T2, T3, T4 and T5 enable measurement of water temperature and the entry and exit points of the tubes. Thermocouples T8, T9 measures the surface temperature. T6 and T7 measures vapour temp. Relief valve Rv is provided to prevent build up of high processes inside the system. 25

SPECIFICATIONS: 1. Outside diameter of the tube = 15mm. 2. Effective length of the tube = 150mm. EXPERIMENTAL PROCEDURE: 1.

Open valves, V1 V2 and V3

2.

Keep valves V5 and V6 closed

3.

Fill the water in the boiler by opening valve V4

4.

Fill water in the sump

5.

Switch on the electrical heater, and close V4

6.

Start the pump

7.

Allow the steam slowly to the chambers by opening Valves V5 and V6

8.

The system is allowed to attain steady state.

NOTE DOWN TEMPERATURES

T1, T2, T3, T4, T5, T6, T7, T8, T9

OBSERVATION TABLE: Sl.No. Water flow Temperatures oC Rate Q lpm T1,T2,T3,T4,T5,T6,T7, T8, T9.

Voltate

V =

Current

A =

Temperature Ta oC

Heater Voltage V

Input Current A

T2 T3 T4

Water Temp. oC

T5 26

(T8, T9)

Surface temp. oC

(T6, T7)

Vapour temperature oC

T1

Temperature of water inlet oC

FORMULAS: Heat transfer area Ao = πdoL

1.

2.

Where do is outside diameter of pipe

=15 mm

----------m

L is effective length of pipe

= 150 mm

----------m

Mass flow rate of water m = Q is m3 /sec x ρ in kg/m3 Note : 1lpm = 1/60000 m3 / s

3.

Heat picked up by water q = m x Cp x T

W

Where q is the heat picked by water in W m is mass flow rate of water in kg/s Cp is specific heat of water in J/kg oC

For Drop wise Condensation

For Film wise Condensation

TDC = (T7-T1) =

TFC = (T8-T1) =

------------------- oC

------------------- oC

4. Logarithmic Mean Temperature Difference (TLMTD) For Drop wise Condensation T1 - T2 TLMTD(DC) = ---------------T1 In

------

For Film wise Condensation T1 - T2 TLMTD(FC) = ---------------T1 In

T2 Where T1 T2

=

=

T3 - T1 oC

T3 - T7 oC

-----T2

Where T1

=

T5 - T1

T2

=

T5 - T8

27

5. Overall Heat Transfer coefficient (U)

For Drop wise Condensation UDC =

For Film wise Condensation

QDC ---------------Ao TLMTD(DC)

UFC =

QFC ---------------Ao TLMTD(FC)

6. Experimental Surface Heat Transfer coefficient

For Drop wise Condensation hDC =

QDC ---------------Ao (T6-T7)

w/m

For Film wise Condensation 2

o

C

hFC =

QFC -----------Ao (T6-T8)

w / m2

o

C

28

EXPERIMENT: 6

HEAT TRANSFER THROUGH PACKED BED Aim: To determine the inside heat transfer co-efficient (hi) of fluid flowing through packed bed. Also , to predict the outside heat transfer co-efficient (ho). Theory: Procedure: 1) Fill the sump tanks (both hot fluid and cold fluid chambers) upto 75% of the tank volume. 2) Switch on the mains and then the heaters please note there are 3 heaters each of 2 KW and all 3 would comprise to 6 KW Heating capacity. 3) Note the temperature of the both till it reaches 700C. After that switch on the cold water pump and attain a steady state and there after switch on the hot water pump. 4) Ensure that cold water is operated between 5 to 15 lpm with hot water at constant flowrate say 10 or 12 lpm.. 5) Allow the heat transfer for one minute so that the process attains steady state. 6) After one minute, please note the hot water inlet and outlet temperatures and cold water inlet and outlet temperatures.. 7) Prepare the hot water and cold water properties in a tabular column obtained from perry’s hand Book.. 8) Calculate the overall heat transfer coefficients.

Observations and Calculations: a) Packing Material

: Raschig Rings

b) Inner dia of pipe = di = 0.0273 m c) Outer dia of packed bed = d0 = 0.034 m d) Inner dia of jacket = Di = 0.0625 m e) Pipe diameter = 0.02725 m f) Area of packed bed =  di L = 0.18268 m2 g) Packed Bed Length

= 2.13 m

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Table 1: Sl No.

Cold Water Inlet Temp. Tci (0C)

Inlet Temp. Tc o (0C)

Cold Water Vol. (Vo) Time of collected collection (t) 3 (m .sec) (sec)

Hot Water Inlet Temp. Thi (0C)

Out let Temp. Th0 (0C)

1. 2. 3. 4. 5. 6.

Table 2: COLD PROPERTIES (TUBE SIDE) Sl Tavg No. (oK)

ρ (Kg/m3)

Cp J/Kg oK

 X104 (NS/m2)

K (w/m 0K)

mc (Kg/sec)

Vc (m/sec)

Q KW

NRe

NPr

T (oC)

hI

(W/m2 0K)

1. 2. 3. 4. 5. 6. SAMPLE CALCULATIONS: 1) Tavg = (Tci + Tco) / 2 =

7) NRe = (di Vρ) /  = NRe =

2) T = Tco + Tci = 3) Cp, ρ, , K are taken from PMB at Tavg

8) Npr = (Cp * ) / K Npr =

4) mc =  (Vo)x ρ ) / time = mc =

K 9) ho = 0.023 (NRe)0.8 (Npr)1/ 3 x ---------

5) Vc = (Volume) / time x area Area = (  /4) (di2 )=

di

6) Q = m Cp  T Q=

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Table 3: HOT PROPERTIES (ANNULUS SIDE) Sl Tavg No. (oC)

Cp (KJ/Kg o K)

ρ Kg/m3

 (NS/m2)

K (w/m 0K)

mH (Kg/sec)

VH (m/sec)

NRe

NPr

NN u

Q(KW)

ho 2 0

(W/m K)

1. 2. 3. 4. 5. 6.

T1 (oC)

T2 (oC)

T LMTD

U (W/m2 oK)

CALCULATIONS: 1)  T = Tho - Thi = 2) Tavg = ( Tho + Thi) / 2

3) mH = (Volume) x ρ) / time = (mc Cp  T) Cp  T ; mH = 4) V = (Volume) / time x area = m/ ρ a V= area = /4 (Deq)2 5) NRe = (Deq Vρ) /  , area = Deq= d0- Di 6) NPr = (Cp ) / K = 7) NNu = 4.758 (NRe)0.1 (NPr) 0.79

8) h0 = (NNu x K) / Deq 9) Q = mH Cp  T = 10)  T1 = Thi – Tco =  T1 = Tho– Tci =

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11)  TLMTD = ( T1  T2) / In =  TLMTD

 T2 --------- T1

 TLMTD

12) A = Area of packed bed =

 Deq L =

13. U = Q/A.  TLMTD

RESULT: 1) The outside heat transfer co-efficient was found out to be ho = 2) The inside heat transfer co-efficient was found out to be hi =

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EXPERIMENT: 7

THERMAL CONDUCTIVITY OF METAL ROD AIM: To determine the thermal conductivity of the copper rod Theory : Relevant theory about thermal conductivity. Ref: standard text book PROCEDURE : 1 Ensure water connections are proper and that the flow rate is steady, example 400 ml/min 2. Use the stop watch to monitor the steady state flow 3. Ensure variac position at zero before switching on the machine and switch ON 4. Set the voltage to the required level: for example 120 volts corresponding current will flow since the resistance of the heating coil will vary as a function of temperature 5. Wait around 45 to 60 min record the temperatures: T1 to T11 in the table given below. 6. After 2 min time interval record- the data again. All the temperatures T1 to T11 must be within 1 deg C compared with earlier data 7 Record the water flow rate one more time to ensure that it has not changed during the experimentation.

CALCULATION : Determination of thermal conductivity of copper at section AA Mass of the water (m) = (Flow rate/minute) x60/ 1000 kg/hr Specific heat of the water = 4.18 KJ/kgK Temperature Difference (T) = outlet water (T11) – Inlet water (T10) Heat carried away by water (Qw) = m CpT Diameter of the copper rod =d= 35 mm = 0.035 m Cross sectional Area of the copper rod = d2/4 Distance between two adjacent points which are 0.07 m apart where the T4 & T5 temperatures measured T c = T4 - T5 Thermal conductivity of copper rod = K cu Heat carried away from copper rod (Qw)=Kcu ATc/x where =x=0.07 m Calculate Kcu from the above equation at location AA Determination of thermal conductivity of copper at section BB Outer radius (r0)= 50 mm (position of T7) Inner radius (ri)= 25 mm (position of T6) 33

Distance between two adjacent points which are 0.07 m apart where the T6 & T7 temperatures measured    T1 = T6 – T7 = temp difference between 25 mm and 50 mm radius T2 = T3 – T4 Thermal conductivity of copper rod = K cu and Thermal conductivity of ceramic powder = K ins Heat carried away by Ceramic insulation (QI)=2Kins LT1/ln(r0/ri) Calculate Kcu from the below equation at location BB Kcu=(Qw+QI)/(AT2/x) where =x=0.07 m Determination of thermal conductivity of copper at section CC Outer radius (r0)= 50 mm (position of T8) Inner radius (ri)= 25 mm (position of T9) Distance between two adjacent points which are 0.07 m apart where the T6 & T7 temperatures measured T 3 = T8 – T9 = temp difference between 25 mm and 50 mm radius T4 = T1 – T2 Thermal conductivity of copper rod = K cu and Thermal conductivity of ceramic powder = K ins Heat carried away by Ceramic insulation (QII)=2Kins LT3/ln(r0/ri) Calculate Kcu from the below equation at location CC Kcu=(Qw+QII)/(AT4/x) where =x=0.07 m Draw a graph between these AA, BB, CC distance vs Kcu as shown in the next page

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