HMT 2015 Manual

Heat and Mass Transfer Laboratory

Dept. of Mechanical Engineering, MSEC, Bangalore.

1

Department of Mechanical Engineering

M S Engineering College

Bengaluru-520110

HEAT AND MASS TRANSFER

LAB MANUAL

SUBJECT CODE:10MEL67

Student name :

USN :

Signature : Faculty Signature



2

GENERAL GUIDELINES

1. Students should wear Identity card all the time. Students without the same will not be allowed to

enter either the class room or the lab.

2. Students should be in time for the Laboratory class and subsequent classes thereafter.

3. Students should keep the Classroom and Laboratories clean and tidy.

4. Students should bring the observation book as well as the laboratory record book completed in

all respect to the laboratory.

5. Students are supposed to fill in the columns in the LOG BOOK at the time of entering the Labs.

6. Students who fail to get minimum 50% marks in Internal assessment of Practical will fall in

NSSR category and / or not eligible to take up Examination.

7. Students are to maintain absolute discipline and decorum, so as to promote the fair name of their

department and college in all its activities

8. Students are informed to clarify their doubts in the respective subjects with the faculty by taking

prior appointments..

9. Students having less the 85% attendance in any subject (both Theory and Practical) will not be

allowed to take up the university examination.

10. Parents are to follow the progress of their wards by being in touch with the college authorities at

regular intervals.

11. Writing on desks and walls is strictly prohibited, failing which the students will be punished.

12. Attendance of the students will be displayed on the department notice board at the end of the

each month.

13. Students are not supposed to alter the configuration of the system/ or any software on the

system.

14. Final examination of 3 hrs duration.

15. Those students who have less than 85% attendance should sign the undertaking given by their

class teachers.

Note: Please do not submit incomplete report.

Student signature



3

LIST OF EXPERIMENTS

*DOC- Date Of Conducted

*DOS-Date Of Submitted

Sl

No Name Of The Experiment Page

No DOC* DOS* Marks

Faculty

Sign

Part A

01 Determination of Thermal

Conductivity of a metal Rod.

02

Determination of Overall heat

Transfer Coefficient of a Composite

wall

03 Determination of Effectiveness on

metallic Fin

04

Determination of heat transfer

Coefficient in a free Convection on a

vertical tube

05

Determination of heat Transfer

Coefficient in a Forced Convection

Flow through a pipe

06 Determination of Emissivity of a

Surface.

Part B

07 Determination of Steffan- Boltzman

Constant.

08

Determination of LMTD and

Effectiveness in a Parallel Flow and

counter Flown heat Exchanger

09 Experiments on Boiling of Liquid and

Condensation of vapour

10 Performance Test on Vapour

Compression Refrigeration.

11 Performance Test on Vapour

Compression Air Conditioner

12 Experiment on Transient Conduction

heat Transfer.



4

EXPERIMENT NO. 1

Thermal Conductivity Of Metal Rod Date:

Introduction:

Thermal conductivity is the physical property of material denoting the ease with a particular

substance can accomplish the transmission of thermal energy by molecular motion. Thermal

conductivity of a material is found, to depend on the chemical composition of the substances of

which it is a composed, the phase (i.e. gas, liquid or solid) in which its crystalline structure if a

solid, the temperature & pressure to which it is subjected and whether or not it is homogeneous

material.

Thermal energy in solids may be conducted in two modes.

They are:

- Lattice vibration:

- transport by free electrons.

In good electrical conductors a rather large number of free electrons move about in a lattice

structure of the material. Just as these electrons may transport may transport electric charge, they

may also carry thermal energy from a high temperature region to low temperature region. In fact,

these electrons are frequently referred as the electron gas. Energy may also be transmitted as

vibrational energy in the lattice structure of the material. In general, however, this latter mode of

energy transfer is not as large as the electron transport and it is for this reason that good

electrical conductors are almost always good heat conductors, for eg: ALUMINIUM, COPPER

& SILVER.

With the increase in temperature, however the increased lattice vibrations come in the way of

electron transport by free electrons and for most of the pure metals the thermal conductivity

decreases with the increase in the temperature.

Experimental set up: The experimental set up consists of a metal bar, one end of which is

heated by an electrical heater while the other end projects inside a cooling water jacket. The

heater is provided with a dimmerstat for controlling the heat input .The middle portion is

surrounded by a cylindrical shell filled with insulating powder and thermocouples are placed on

the bar for temperature measurement. Water under the constant head is circulated through the

jacket and its flow rate and temperature rise are measured using rotameters and thermocouples.

Apparatus: Metal rod, Heater, Rotameter, Thermocouples, Power supply panel, etc.

Experimentation:

Aim: To find out the THERMAL CONDUCTIVITY of given metal rod.

Procedure:

Give necessary electrical and water connections to the instrument. Switch on the MCB and console ON to activate the control panel. Give input to the heater by slowly rotating the heater regulator. Start the cooling water supply through the water jacket (make sure not to exceed 3lpm). Note the temperature at different points, when steady state is reached. Repeat the experiment for different heater input.



5

After the experiment is over, switch off the electrical connections, allow the water to flow for some time in the water jacket and then stop it.

Specifications:

Room temperature (T) =---------o C

Diameter of metal rod (d)=----------m

Length of the metal bar (L)= _____m

Distance between each thermocouples, X = ______m

Figure 1.1 Graphical representation of Thermal Conductivity of Metal Rod.

Figure 1.2 Setup of Thermal Conductivity of Metal Rod.



6

Tabular Column

Calculations:

Cross sectional area of metal rod: A= 4

. 2d = ------------------ m

2

d = Diameter of the rod in m.

Mass flowrate of water, m = 60

R---------------kg/ sec.

Where,

R = Rotameter reading in lpm.

Heat input , QI = V I =-------------------- W

Where V = Voltage. (Volts)

I = Current (amps)

Heat absorbed by water, QW = m Cp (T8-T9) =------------------W

Where m = mass flow rate of water. (Kg/s)

Cp = specific heat of water = 4.178 (KJ/ kg K) (T8 - T9) = difference between of water at inlet and outlet.(K)

Sl.

no

Mass flow

rate

Volt

V

Amp

I

Room

Temp

T1

Surface temp. of metal rod

Water temp.

Q I

V I ( W)

Q W

(W)

Thermal

conductivity

k

(W/m-k)

lpm

Kg/s

T2

T3

T4

T5

T6

T7

inlet

T8

outlet

T9



7

4. Plot the graph of temperature T v/s distance X (Fig 1.2), from the best fitting straight line for

temperatures T2, T3, T4, T5, T6 and T7, Average temperature gradient (slope). dX

dTis calculated.

dT

dX

T C

X in m

T2

T3

T4T5

T6

T7

Figure. 1.3 Graph of Temperature versus Distance

5. Heat transfer through metal rod is given by QW = -kA dX

dT =--------------W

Hence, k = -

dX

dTA

WQ=---------------------W/mK

Where k = Thermal conductivity in W/m-K

A = Cross section area of metal rod

dX

dT= Temperature gradient Slope from graph T vs X

Result: The thermal conductivity of given material was found to be k = W/ m-K.

Precautions:

Do not give heater input without the supply of water. Input should be given very slowly. Run the water in the jacket for about 5 min after the experiment. Do not run the equipment if the voltage is below 180V. Check all the electrical connections before running.



8

Before starting and after finishing the experiment the heater controller should be in off position.

Do not attempt to alter the equipment as this may cause damage to the whole system.

Point for Discussion

(1) Compare the experimental value of k with the standard value at the average temperature of the set of observations.

(2) Is there any specific reason why the water jacket is circulated on one end of the rod. Comment on it.



9



10

EXPERIMENT NO. 2

Composite Wall Apparatus Date:

Introduction:

In engineering applications, we deal with many problems. Heat Transfer through composite

walls is one of them. It is the transport of energy between two or more bodies of different

thermal conductivity arranged in series or parallel. For example, a fastener joining two mediums

also acts as one of the layers between these mediums. Hence, the thermal conductivity of the

fastener is also very much necessary in determining the overall heat transfer through the

medium. An attempt has been made to show the concept of heat transfers through composite

walls.

Apparatus: Given Composite wall, Heater, Set of Thermocouples, Power Supply panel,

Channel Temperature Indicator and necessary connections.

Experimental set up: The experimental set up consists of three slabs of different materials of

same thickness clamped in the centre using screw rod. At the centre of the composite wall, a

heater is fitted. End losses from the composite wall are minimized by providing thick insulation

all rounds to ensure unidirectional heat flow.

Thermocouples are fitted at the interface of the plates at different points as to obtain average

temperature for each surface. Heat conducted through the composite wall is taken away by

atmospheric air.

Experimentation:

Aim: To Determine the overall heat Transfer coefficient of a composite wall. Procedure:

Symmetrically arrange the plates and ensure perfect contact between the plates. Switch ON mains and the CONSOLE. Set the heater regulator to the known value. Wait for sufficient time to allow temperature to reach steady values. Note down the Temperatures 1 to 7 using the channel selector and digital temperature

indicator.

Note down the ammeter and voltmeter readings. Calculate the overall conductance using the procedure given below. Repeat the experiment for different heat input.

Specification:

1. Mild Steel _______ mm thick of _______mm dia.

2. Wood ________ mm thick of ________ mm dia.

3. Copper _______ mm thick of ________ mm dia.

4. Heater _______ watts.

5. Thermocouples ______ Nos.



11

Figure 2.1 Schematic diagram of composite wall apparatus

Figure 2.2 Test Setup of Composite Wall Apparatus



12

Tabular Column :

Sl No Voltage

V Volts

Current

I Amps

Temperatures ( 0C ) Water inlet

temperature

T8(0C )

Water outlet

temperature

T9(0C ) T1 T2 T3 T4 T5 T6 T7

Calculations:

(a) Heat flow through composite wall Q = V x I=----------------- W

Q = 1

3211 )(

L

TTAk =

2

5422 )(

L

TTAk =

3

7633 )(

L

TTAk ----------------(1)

k1 = Thermal conductivity of copper (W/m K). L1 = thickness of copper.

k2 = Thermal conductivity of Wood (W/m K). L2 = thickness of wood.

k3= Thermal conductivity of Mild steel (W/m K). L3 = thickness of Mild Steel.

From equation ..(1)

k1 = )( 321

1

TTA

QL

=-------------------------------

Where A= 4

. 2dd = Diameter of the slab

k2 = )( 542

2

TTA

QL

=--------------------------------

A1 = A2 = A3 = A.

k3 = )( 763

3

TTA

QL

=--------------------------------



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(b) Overall heat transfer coefficient (Uo)

Uo = 1

1

k

L+

2

2

k

L+

3

3

k

L=----------------------------

Unit (Km

W

2)

Tabulation :-

Sl.

No

k1

(Km

W

)

k2

(Km

W

)

k3

(Km

W

)

Overall heat transfer

co-efficient (U0)

(Km

W

2)

Result: The Overall heat transfer co-efficient of given composite slab was found to be

U0 = _____W/ m2-K.

Precautions:

1. Check all the electrical connections.

2. Do not run the equipment if the voltage is below 180V.

3. Do not attempt to alter the equipment as this may cause damage to the whole system.

Points for discussion:-

1) Comment on the experimental set up,its validity and accuracy. 2) Overall heat transfer co-efficient can also be calculated without the requirement of

intermediate temperatures.comment on it.



14



15

EXPERIMENT NO. 3

Heat Transfer through Pin Fin Apparatus Date:

Introduction:

A spine or pin-fin is an extended surface of cylindrical or conical shape used for increasing the

heat transfer rates from the surfaces, whenever it is not possible to increase the rate of heat

transfer either by increasing heat transfer co-efficient or by increasing the temperature difference

between the surface and surrounding fluids.

The fins are commonly used on engine heads of scooter, motorcycles, as well as small capacity

compressors. The pin type fins are also used on the condenser of a domestic refrigerator.

Experimental set up:

The experimental set up consists of a blower unit having a rectangular duct inside which the pin

fin is fitted. One end of the pin fin is connected to heater. Five thermocouples are embedded on

the fin surface.Blower along with the orifice is used to measure flow of air through the duct.

Input to heater is given through dimmerstat and measured by voltmeter and ammeter. Air flow is

controlled by the gate valve and is measured with the help of orifice meter and the manometer

fitted on the board.

Experimentation:

Aim: To determine the efficiency and effectiveness of the fin by natural convection and forced

convection using pin fin apparatus.

Procedure :

(a) Natural Convection:

Switch on the MCB and then console on switch to activate the control panel. Switch on the heater and regulate the power input using the heater regulator. Wait for reasonable time to allow temperatures to reach steady state. Measure the voltage, current and temperatures from T1 toT6 at known time interval. Calculate the effectiveness & efficiency of the fin using the procedure given. Repeat the experiment for different values of power input to the heater.

(b) Forced Convection:

Switch on the MCB and then console on switch to activate the control panel. Switch on the heater and regulate the power input using the heater regulator. Switch on the blower unit and adjust the flow of air using gate valve of blower to a

desired difference in manometer (for forced flow only otherwise skip to step 4).

Wait for reasonable time to allow temperatures to reach steady state. Measure the voltage, current and temperatures at known time interval. Calculate the effectiveness & efficiency of the fin using the procedure given. Repeat the experiment for different values of power input to the heater and blower air

flow rates.



16

Pin fin

Gate valve

orificeGI Pipe

To u-TubeManometer

T T T T T1 2 3 4 5

Temperature Indicator

AmmeterVoltmeter

Dimmerstat

Blower

Fig 3.1(a) Pin fin apparatus

Fig 3.2(b) Pin fin enlarged

.

Figure 3.2 Setup of Heat Transfer through Pin Fin Apparatus



17

Tabular Column (Natural convection):-

Sl.

No.

Voltage

( V )

Volts

Current

( I )

Amps

Power

Q=VI

Watts

Temperature at different point

in the outer surface of cylinder

hnatural W/m

2-K

T1 T2 T3 T4 T5 T6

Calculations:

(1) Experimental Efficiency of pin fin(exp) A graph of T(x) versus x is plotted and the curve is plotted and the curve is extrapolated to

intersect the temperature axis.Then the value of T(x) @ x =0 and dX

dT@ x = 0 are read from the

graph.

Tem

p.

Distance, X (m)

dX

dT

X = 0

Figure 3.3 Graph of temperature versus length of the rod.

From the graph,T(x)@ x = 0 = ________ , dX

dT@ x = 0 = _______ .



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The efficiency of the pin fin experimentally is given by

exp = maxfin

fin

Q

Q

Where, Qfin = - kA dX

dT@ x = 0

A = Area of cross section of metal rod, A= 4

. 2d = ________ m

2

d = Diameter of the pin fin . ( m.)

(Qfin)max = h.AS.( T0 T) + S.( T04 T

4)

now AS = .d.l. (m2)

(2) Theoritical Efficiency of pin fin(theo)

theo = ml

mltanh

Where, m = Ak

Ph

f .

.

h = heat transfer coefficient of fin

P = Perimeter of fin (D)

kf = thermal conductivity of pin fin = 110 (Km

W

)

A= Area of fin= 4

. 2d (d = Diameter of the pin fin, m), m2

l = Length of pin fin, m

To find heat transfer coefficient of fin(h)

(i) Film temperature, Tf =2

TaTs ( C )

Where, Ts = Surface temperature of the cylinder =5

65432 TTTTT (K)

Ta =T1 = ambient temperature (K)

Properties of air at the film temperature (Tf) is obtained from data book, hence determine

Density, = Specific heat, Cp =

Kinematic viscosity, = Thermal conductivity, k =

Prandtl number, Pr =



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(ii) Grassoff number Gr =2

3...

dg

Where,

= Co-efficient of thermal expansion = )273(

1

fT. (K

-l )

T = Tf -T1. (K)

d = Diameter of the pin fin. (m)

= Kinematic viscosity. (m2 /s)

Ra. No = Gr Pr =

From heat transfer Data hand book corresponding to this value of Rayleigh number Ra, the

nusselt number(Nu) is given by

For 10-1

< Gr Pr < 104, Nu = 0.68 +444.0

5625.0

25.0

Pr

492.01

Pr)(67.0

Gr

For 104 < Gr Pr < 109, Nu = 0.59 (Gr Pr)0.25

For 109 < Gr Pr < 1012, Nu = 0.13 (Gr Pr)0.33

Hence,

Nu =k

lh.. Or

h = l

kNu..

Where, h = natural heat transfer co-efficient, h. (Km

W

2)

(3) Effectiveness of Fin,

Pk

Ah

ml

f .

.

)tanh(



20

Tabular Column (Forced convection) :-

Sl.

No.

Voltage

( V )

Volts

Current

( I )

Amps

Power

W=VI

Watts

Temperature at different

point in the outer surface

of cylinder

Manometer

Readings

Hw=h1- h2

hnat

(Km

W

2)

T1 T2 T3 T4 T5 T6

Calculations:

(1) Experimental Efficiency of pin fin

A graph of T(x) versus x is plotted and the curve is plotted and the curve is extrapolated to

intersect the temperature axis.Then the value of T(x) @ x =0 and dX

dT@ x = 0 are read from the

graph.

Tem

p.

Distance, X (m)

dX

dT

X = 0


From the graph,T(x)@ x = 0 = ________ , dX

dT@ x = 0 = _______ .

The efficiency of the pin fin experimentally is given by



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exp = maxfin

fin

Q

Q

Where, Qfin = - kA dX

dT@ x = 0

A = Area of cross section of metal rod, A= 4

. 2d = ________ m

2

d = Diameter of the pin fin . ( m.)

(Qfin)max = h.AS.( T0 T) + S.( T04 T

4)

now AS = .d.l. (m2)

(2) Theoritical Efficiency of pin fin(theo)

theo = ml

mltanh

Where, m = Ak

Ph

f .

.

h = heat transfer coefficient of fin

P = Perimeter of fin (D)

kf = thermal conductivity of pin fin = 110 (Km

W

)

A= Area of fin= 4

. 2d (d = Diameter of the pin fin, m), m2

l = Length of pin fin, m

To find heat transfer coefficient of fin(h)


TaTs . ( C )

where, Ts = Surface temperature of the cylinder =5

65432 TTTTT . (K)

Ta = T1= ambient temperature. (K)

Properties of air at the film temperature (Tf) is obtained from data book,hence determine




(2) Discharge q = Cd A0 aHg.2



22

Where,

Cd = 0.65

A0 = 4

2

od

do = Diameter of orifice

Ha = Head of air = w Hw/ a

w = 1000 kg/m3

a = 1.2 kg/m3

(3) Velocity v =A

q. (m

2)

(4) Reynolds No. Re = vd /

Where,

d = Diameter of the pin fin. (m)

v = Velocity of the air. (m/s)

From heat transfer Data hand book corresponding to this value of Reynolds number, the nusselt

number(Nu) is given by

Nu = 0.618 (Re )0.466

, if 40 < Re < 4000

Nu = 0.174 (Re )0.618

, if 4000 < Re < 40,000.

Hence,

Nu =k

lh.. Or

h = l

kNu..

Where, hnat = natural heat transfer co-efficient, h. (Km

W

2)

(3) Effectiveness of Fin,

Pk

Ah

ml

f .

.

)tanh(

Result:- Natural convection theo = ______ , exp = _______ ,

forced convection theo = ______ , exp = _______,

Precautions:

Check all the electrical connections.

Do not run the equipment if the voltage is below 180V.

Do not obstruct flow of air while experiment is going on.

Do not turn the heater regulator to the maximum as soon as the equipment is started.




23

Points for discussion:-

(1) If the readings are taken when the steady state is yet to be reached,how will the accuracy of

the results will be affected.

(2) From the observations taken in this experiment, explain how would you determine the

thermal conductivity of the material of the fin.



24

EXPERIMENT NO. 4

Free or Natural Convection Date:

Introduction:

There are certain situations in which the fluid motion is produced due to change in density

resulting from temperature gradients. The mechanism of heat transfer in these situations is called

free or natural convection. Free convection is the principal mode of heat transfer from pipes,

transmission lines, refrigerating coils, hot radiators etc. The movement of fluid in free

convection is due to the fact that the fluid particles in the immediate vicinity of the hot object

become warmer than the surrounding fluid resulting in a local change of density. The colder

fluid creating convection currents would replace the warmer fluid. These currents originate when

a body force (gravitational, centrifugal, electrostatic etc) acts on a fluid in which there are

density gradients. The force, which induces these convection currents, is called a buoyancy force

that is due to the presence of a density gradient within the fluid and a body force. Grashoffs

number a dimensionless quantity plays a very important role in natural convection.


The apparatus consists of a brass tube fitted in a rectangular duct in a vertical position. The duct

is open at the top and bottom and forms an enclosure and serves the purpose of undisturbed

surrounding. An electric heating element is kept in the vertical tube which in turn heats the tube

surface. The heat is lost from the tube to the surrounding air by natural convection. The

temperature of the vertical tube is measured by six thermocouples. The heat input to the heater is

measured by an Ammeter and a Voltmeter and is varied by a dimmerstat. The tube surface is

polished to minimize the radiation losses.

Experimentation:

Aim: To determine the heat transfer co-efficient for natural convection using the given

apparatus.

Procedure:

Keep the tube in the vertical position. Switch on MCB and then CONSOLE ON switch. Switch on the heater and set the voltage (say 40V) using heater regulator and the digital

voltmeter.

Wait for sufficient time to allow temperature to reach steady values. Note down the Temperatures using the channel selector and digital temperature indicator. Note down the ammeter and voltmeter readings. Calculate the convection heat transfer co-efficient using the procedure given below.

Repeat the experiment for different heat inputs.



25

Specifications:

Diameter of chromium plated brass tube =

Length of brass tube =

Brass tube wall thickness =

Distance between thermocouples =

Thermocouples (T2- T7)

230 V50 Hz

HeatingElementBrass

Tube

Wattmeter

Dimmerstat

Enclosure

Figure 4.1 Schematic Diagram of test set up for free convection



26

Figure 4.2 Photograph of test set up for natural convection.

Tabular column:

Sl.

No.

Voltage

( V )

Volts

Current

(I)

Amps

Power

W=VI

Watts

Temperature at different point in the

outer surface of cylinder

Heat transfer co-

efficient W/m2-K

T1 T2 T3 T4 T5 T6 T7 hexp htheo

Calculations:

I. Experimental Method



27

Experimental heat transfer co-efficient, hexp =)( aSS TTA

Q

.

Where

Q = Rate of heating, = V x I, watts.

A =dl, Where,

d = Diameter of cylinder rod. (m)

l = length of cylinder. (m)

II. Analytical (Theoretical) Method


TaTS . ( C )

Ts = Surface temperature of the cylinder =6

765432 TTTTTT (K)

Ta = T1 =ambient temperature. (K)





(ii) Grassoff number Gr = 2

3...

clg

Where,

= Co-efficient of thermal expansion = )273(

1

fT. ( K

-l )

T = Tf -T1. ( K)

lc = characteristic length = length of the vertical cylinder . (m)

= Kinematic viscosity. (m2 /s)

Ra.No = Gr Pr =

From heat transfer Data hand book corresponding to this value of Rayleigh number Ra, the

nusselt number(Nu) is given by

For 10-1

< Gr x Pr < 104, search a suitable correlation from data hand book.

For 104 < Gr x Pr < 10

9, Nu = 0.59 (Gr Pr)0.25

For 109 < Gr x Pr < 10

12, Nu = 0.13 (Gr Pr) 0.33



28

Hence,

Nu =k

lhtheo. . or

htheo = l

kNu..

Where, hnat = natural heat transfer co-efficient, h. (Km

W

2)

The heat transfer co-efficient for the vertical tube was found to be _______ Km

W

2

Result : The co-efficient of heat transfer by experiment is

hexp =

The co-efficient of heat transfer by theoretical is

htheo =

Precautions:



Make sure that heater regulator is at the minimum position before switching on the console.


Points for Discussion :

1. List of various precautions observed during experimentation. 2. Comment upon the values of heat transfer coefficient obtained experimentally and the

one by using correlations.



29



30

EXPERIMENT NO. 5

Forced convection Date:

Introduction: Convection is a process of Heat transfer involving motion of fluid

molecules.When the motion of fluid molecules is induced by some external agency such as

pump or a blower, the process is called forced convection. The intensity of the mixing motion is

generally high in forced convection and consequently the heat transfer coefficients are higher

than free convection. The heat transfer coefficient h can be obtained from Newton law of

convection.

Q = hAs (Ts - T)

Where,

As - Surface area. (m2).

(Ts - T) - Temperature difference between the surface and fluid. (K).

h - Surface heat transfer co-efficient (ranges from 10 - 500 Km

W

2for forced convection).

Experimental set up: The experimental set up consists of a blower unit fitted with the test pipe.

The test section is surrounded by band heater. Seven thermocouples are embedded on the test

section and two thermo couples are placed in the air stream at the entrance and exit of the test

section to measure the air inlet and outlet temperatures. Test pipe is connected to the delivery

side of the blower along with the orifice to measure flow of air through the pipe. Input to heater

is given through dimmerstat and measured by voltmeter and ammeter. Air flow is controlled by

the gate valve and is measured with the help of orifice meter and the manometer fitted on the

board.

Experimentation:

Aim: To determine the Forced Convection heat transfer coefficient for flow through the given

Horizontal tube cylinder.

Procedure:

Keep the tube in the vertical position. Switch on MCB and then CONSOLE ON switch. Switch on the heater and set the voltage (say 40V) using heater regulator and the digital

voltmeter.

Switch on the blower unit and adjust the flow of air using gate valve of blower to a desired difference in manometer.

Wait for sufficient time to allow temperature to reach steady values. Note down the Temperatures using the channel selector and digital temperature indicator. Note down the ammeter and voltmeter readings. Calculate the convection heat transfer co-efficient using the procedure given below.

Repeat the experiment for different heat inputs.



31

Specifications:

1 Coefficient of discharge of orifice meter Cd = _______ 2 Length of the hallow pipe L= ______ 3 Orifice diameter do = _______ 4 Inside pipe diameter d = ______

Ammeter

Dimmerstat

Temperature Indicator Voltmeter

Blower

Test SectionTo U-Tube Manometer

Air Exit

Figure 5.1 Schematic Diagram of forced convection test set up.



32

Figure 5.2 Photograph of test set up for Forced convection.

I. Experimental Method

Experimental heat transfer co-efficient, hexp = Q /As (Ts - Ta)

Where

Q = Rate of heating, = V x I, watts.

As = dl, Where,

d = Diameter of cylinder rod

l = length of cylinder

II. Analytical (Theoretical) Method

(1) Bulk mean temperature, Tm = T1+T7 / 2. ( C )

Where, T1 = Air inlet temperature. (K)

T7 = Air outlet temperature. (K)


Density, = Specific heat, Cp = Kinematic viscosity, = Thermal conductivity, k = Prandtl number, Pr



33

(2) Discharge q = Cd A0 aHg.2

Where,

Cd = Co-efficient of discharge = 0.65

A0 = 4

.2

od

do = Diameter of orifice

Ha = Head of air = w Hw/ a

w = 1000 kg/m3

a = 1.2 kg/m3

(3) Velocity v =A

q. (m

2)

(4) Reynolds No. Re = vd /

Where,

d = Diameter of the cylinder, (m)

v = Velocity of the air. (m/s)

From heat transfer Data hand book corresponding to this value of Reynolds number, the nusselt

number(Nu) is given by

Nu = 0.618 (Re )0.466

, if 40 < Re < 4000

Nu = 0.174 (Re )0.618

, if 4000 < Re < 40,000.

Hence,

Nu =k

dhtheo. . Or

htheo= d

kNu..

Where, htheo = Theoritical heat transfer co-efficient, (Km

W

2)

Result : The co-efficient of heat transfer by experiment is

hexp =

The co-efficient of heat transfer by theoretical is

htheo =


(1) List of various precautions observed during experimentation.

(2) Comment upon the values of heat transfer coefficient obtained experimentally and the

one by using correlations.



34



35

EXPERIMENT NO. 6

Emissivity Measurement Date:

Introduction:

Radiation is one of the modes of heat transfer, which does not require any material medium for

its propagation. All bodies can emit radiation & have also the capacity to absorb all or a part of

the radiation coming from the surrounding towards it. The mechanism is assumed to be

electromagnetic in nature and is a result of temperature difference. Thermodynamic

considerations show that an ideal radiator or black body will emit energy at a rate proportional to

the fourth power of the absolute temperature of the body. Other types of surfaces such as glossy

painted surface or a polished metal plate do not radiate as much energy as the black body,

however the total radiation emitted by these bodies still generally follow the fourth power

proportionality. To take account of the gray nature of such surfaces, the factor called emissivity

(), which relates the radiation of the gray surface to that of an ideal black surface, is used. The emissivity of the surface is the ratio of the emissive power of the surface to the emissive power

of the black surface at the same temperature. Emissivity is the property of the surface and

depends upon the nature of the surface and temperature.

Experimental set up:-The experimental set up consists of an Enclosure which houses a pair of

metal discs which are geometrically similar made of same material. One of the discs is painted

black (black body) whose emissivity is known and the other disc (grey body) whose emissivity

has to be determined. The geometric and material similarity of the two discs ensures that the

conduction and convection losses between the disc and the ambient will remain same if both the

discs are at the same temperatures. Three thermocouples are placed on the surface of each disc.

Apparatus: - blackbody, test body, heater, digital voltmeter Ammeter, control body.

Experimentation:

Aim: - To determine the emissivity of grey body at temperature.

Procedure:

Give necessary electrical connections and switch on the MCB and switch on the console on to activate the control panel.

Switch On the heater of the Gray body and set the voltage using the heater regulator and digital voltmeter.

Switch On the heater of the Black body and set the voltage or current (say higher than gray body) using the heater regulator and digital voltmeter.

Wait to attain the steady state. Note down the temperatures at different points and also the voltmeter and ammeter

readings.

Tabulate the readings and calculate the surface emissivity of the non black surface.



36

Ammeter

DimmerstatBlack Body

Temperature Indicator Voltmeter

DimmerstatGrey Body

TemperatureSelector Switch

Black Body Grey Body

T5 T6 T7 T3 T4T2

Figure 6.1 Line diagram of Emissivity test set up apparatus.

Figure 6.2 Photograph of Emissivity apparatus.



37

Specifications:

Diameter of black body = _______

Diameter of Grey body = ________

Tabular column:-

Sl.

No

Grey body readings Black body readings Room

Temp

T1

Encloure

Temp

T8

Emiss

i-

vity

Volts

V2

current

I2

Surface temp

Volts

V1

Current

I1

Surface temp

T2 T3 T4 T5 T6 T7

Formulae & Calculations

Heat supplied to grey body Qg = Vg Ig

Also, Qg = gA( Tg4-T8

4).................................... (4)

Where, Ag = area of the grey body

b = emissivity of grey body = Stefan boltzman constant = 5.66910-8.

Tg = Avg. grey body temperature =

3

432 TTT

T8 = Enclosure Temperature.

Heat supplied to black body- Qb= Vb Ib

Also, Qg = bA ( Tb4-T8

4) .................................... (5)

Where, Ab= area of the black body

b = emissivity of black body = Stefan boltzman constant = 5.66910-8.

Tb = Avg. black body temperature =

3

765 TTT

T8 = Enclosure Temperature.



38

Procedure:

Give necessary electrical connections and switch on the MCB and switch on the console on to activate the control panel.

Switch On the heater of the Gray body and set the voltage using the heater regulator and digital voltmeter.

Switch On the heater of the Black body and set the voltage or current (say higher than gray body) using the heater regulator and digital voltmeter.

Wait to attain the steady state. Note down the temperatures at different points and also the voltmeter and ammeter

readings.

Tabulate the readings and calculate the surface emissivity of the gray body surface.

b

g

Q

Q=

)(

)(

8

8

TTA

TTA

bbb

ggg

Result: - The experiment is successfully completed, the results tabulated in tabular column.

The emissivity of given apparatus is

Percussions:

Allways keep dimmerstate at zero before starting.

Increase power input slowly.

Do not touch the black plate.


(1) The value of emissivity obtained is total is total emissivity or monochromatic emissivity.What is the difference between the two.

(2) How and under what conditions emissivity of a surface equals its absorptivity.



39



40

EXPERIMENT NO. 7

Stefan Boltzman Apparatus Dtae:

Introduction:

The most commonly used relationship in radiation heat transfer is the Stefan Boltzmans law which relates the heat transfer rate to the temperatures of hat and cold surfaces.

Q = A (Th4 Tc

4)

Where,

Q= rate of heat transfer, watts

= Stefan Boltzmans constant = 5.669 x 10-8 watts/m2 K4 A = Surface area, m

2

Th = Temperature of the hot body, K

Tc = Temperature of the cold body, K

The above equation is applicable only to black bodies (for example a piece of metal covered

with carbon black approximates this behavior) and is valid only for thermal radiation. Other

types of bodies (like a glossy painted surface or a polished metal plate) do not radiate as much

energy as the black body but still the total radiation emitted generally follow temperature

proportionality.

Experimental setup: The experimental set up consists of a hemispherical shell which is fixed to a non-conducting

plate. The outer surface of the hemisphere is enclosed in a metal water jacket used to heat the

copper shell to any desired temperature. The shell shape is chosen as hemispherical to see that

the water in the jacket is drained off easily. Thermocouples are fixed at various points on the

surface of the shell to measure the shell surface temperature. The disc is mounted in an insulated

bakelite sleeve fits in a hole drilled in the centre of the base plate. Thermocouple is attached to

the disc to measure its temperature at any time.

Apparatus: Copper hemispherical enclosure, thermocouple jacket to hold hot water.

Experimentation:

Aim: To determine the value of Stefan Boltzman constant.

Procedure: Fill water slowly into the overhead water heater. Switch on the supply mains and console. Switch on the heater and regulate the power input using the heater regulator. (say 60 85

C)

After water attains the maximum temperature, open the valve of the heater and dump to the enclosure jacket.

Wait for about few seconds to allow hemispherical enclosure to attain uniform

temperature the chamber will soon reach the equilibrium. Note the enclosure temperature.



41

Insert the Test specimen with the sleeve into its position and record the temperature at different instants of time using the stop watch.

Plot the variation of specimen temperature with time and get the slope of temperature versus time variation at the time t = 0 sec

Calculate the Stefan Boltzmans constant using the equations provided. Repeat the experiment 3 to 4 times and calculate the average value to obtain the better

results.

Specifications:

1. Mass of disc = 2. Diameter of rod = 3. Specific heat = 4. Emitter temperature Te = _______

0C

TemperatureIndicator

Water Jacket

Heater

Water Tank

TemperatureSelectionSwitch

CopperHemisphere

Disc(Copper)

T2

T3

T7

T4

T5

T6

Figure 7.1 Stefan boltzman apparatus.



42

Figure 7.2 Photograph of Stefan boltzman apparatus.

Tabular column:

SL.NO TEMPERATURES Time in sec for which T7 is

noted as

in

42 Km

W

T1 T2 T3 T4 T5 T6 0 30 60 90 120

1.

2.

3.

4.



43

Formulae and Calculations:-

1) Stefan boltzman constant ():-

= 474

..

TTA

dt

dTCm

Scu

Pcucu

Where, mcu = mass of copper. (kg)

CPcu = Specific heat of copper. (KKg

J

)

TS = 5

65432 TTTTT

TS = Average temperature.

dt

dT= slope from graph Temperature (T) Vs time (t) .

T in deg

time t in sec

dT

dt

Figure 7.3 Graph of Temperature of disc versus time.

Result: The experiment was conducted successfully .The Stefan boltzman constant was found to

be ________

Precautions:



Do not switch on the heater without water in the overhead tank.

Do not turn the heater regulator to the maximum as soon as the equipment is started.



44

Do not attempt to alter the equipment as this may cause damage to the whole system. Points for Discussion:

(1) What errors will creep in if the temperature recordings of the disc are not taken quickly.. (2) Comment why the shell is chosen as Hemisphere.



45



46

EXPERIMENT NO. 8

Parallel And Counter Flow Heat Exchanger Date:

Introduction:

Heat exchangers are devices in which heat is transferred from one fluid to another. The fluids

may be in direct contact with each other or separated by a solid wall. Heat Exchangers can be

classified based on its principle of operation and the direction of flow. The temperature of the

fluids change in the direction of flow and consequently there occurs a change in the thermal

head causing the flow of heat. The temperatures profiles at the two fluids in parallel and counter

flow are curved and has logarithmic variations. LMTD is less then the arithmetic mean

temperature difference. So, it is always safer for the designer to use LMTD so as to provide

larger heating surface for a certain amount of heat transfer.

Experimental set-up: A typical set up of concentric tube heat exchanger is shown in figure

below. An electric geyser is provided to supply hot water which flows through the inner tube.

Cold water is admitted at either end enabling the same heat exchanger to run in a parallel flow or

a counter flow mode. This is achieved by operating different valves provided in the water line.

The outer tube is adequately insulated to minimize the heat loss to the surroundings.

Thermocouples are used to measure the inlet and outer temperatures of the cold and hot fluids.

The flow rates of the two fluids are measured by measuring flask.

Apparatus: Heat exchanger hearing inner copper tube through which water flows, heaters,

rotameter, blower, channel selector and digital temperature, display.

Experimentation:

Aim: To determine rate of heat transfer and overall heat transfer co-efficient in parallel and

counter flow heat exchanger.

Procedure:

1) Switch ON mains and the CONSOLE. 2) Start the flow on the hot water side. 3) Start the flow through annulus also. 4) Set the exchanger for parallel or counter flow using the change over mechanism. 5) Switch ON the heater of the geyser. 6) Set the flow rate of the hot water (say 1 to 4 Lpm) using the rotameter of the hot water. 7) Set the flow rate of the cold water (say 3 to 8 Lpm) using the rotameter of the cold water. 8) Wait for sufficient time to allow temperature to reach steady values. 9) Note down the Temperatures 1 to 4 using the channel selector and digital temperature

indicator.

10) Note down the flow rates of the water and tabulate. 11) Now, change the direction of flow for the same flow rates and repeat the steps 9 to 11.



47

12) Repeat the experiment for different flow rates of water.

Figure 8.1(a) Line diagram for parallel flow heater exchanger.

Figure 8.1(b) Line diagram for counter flow heater exchanger.



48

Figure 8.2 Photograph of Heat exchanger.

Specifications:

a) Inner dia of Copper tube di = _____mm

b) Outer dia of Copper tube do = _____mm

c) Length of the tube , L = _____mm

a) Inner dia of G.I tube Di = _____mm

b) Outer dia of G.I. Tube Do = _____mm

c) Length of the tube, L = _____mm

CONSTANT VALUES AND RELATIONS

(a) Parallel Flow

Tabular column

Sl.

No

Hot Water Side Cold Water side

Flow rate

(LPM)

T2 in 0C

inlet

T3 in 0C

outlet

Flow rate

(LPM)

T4 in 0C

inlet

T5 in 0C

outlet



49

(1) Heat transfer rate from hot water, Qh = Cp (mh) .Twh

Where,

Mass flow rate (mh) =

CP = CPc = CPh = 4184 (KKg

J

)

Twh =T3 T2

(2) Heat transfer rate from hot water, Qc = Cp (mc).Twc

Where,

Mass flow rate (mh)


J

)

Twc = T5 T4

(3) Q = 2

ch QQ

(4) Log mean temperature difference ( LMTD )

LMTD =

2

1

21

ln

Where,

1 = Thi -Tci = T2 T4

2 = Tho - Tco = T3 T5

(5) Overall Heat Transfer co-efficient (U)

U = LMTDA

Q

S , (

Km

W

2)

Where, AS = do.L

(6) Effectiveness () = 42

32

TT

TT

(7) Theoretical effectiveness

(8) C = Cmin / Cmax

NTU = No. of Transfer units is given by



50

(9)

(a) Counter Flow

Counter flow

Sl.

No

Hot Water Side Cold Water side

Flow rate

(LPM)

T2 in 0C

inlet

T3 in 0C

outlet

Flow rate

(LPM)

T5 in 0C

inlet

T4 in 0C

outlet

(1) Heat transfer rate from hot water, Qh = Cp (mh) .Twh Where,



J

)

Twh =T3 T2 (2) Heat transfer rate from hot water, Qc = Cp (mc).Twc

Where,



J

)

Twc = T4 T5 ,

Q = 2

ch QQ

(4) Log mean temperature difference (LMTD)

LMTD =

2

1

21

ln

Where,

1 = Thi -Tci = T2 T5

2 = Tho - Tco = T3 T4

(5) Overall Heat Transfer co-efficient (U)

U = LMTDA

Q

S



51

(6) Effectiveness () = 52

32

TT

TT

(7)Theoretical effectiveness

(8) C = Cmin / Cmax

NTU = No. of Transfer units is given by

(9)

Result:

For Parallel Flow

i) The Heat transfer obtained is ______ watts

ii) Overall Heat Transfer is _______

iii) Effectiveness is ______

iv) % Error is _______

For Counter Flow

i) The Heat transfer obtained is ______ watts

ii) Overall Heat Transfer is ______

ii) Effectiveness is ______

iv) % Error is _______


(1) Which mode of exchanger is better and why?. (2) If LMTD comes out to be zero,then what is the rule adopted to find the heat transfer

co-efficient.

(3) If max

min

C

C= 0, What type of heat exchanger it will represent?



52



53

EXPERIMENT NO. 9

Drop wise and Film wise Condensation On Vertical Tubes. Date:

Introduction:

Boilers and condenser which are used as heat exchanger posses unique characteristics of heat

transfer mechanism on the condensing and boiling side.

When a vapour strikes a surface temp below the corresponding saturation temperature the

vapour will immediately condense into the liquid phase. The porous of condensation may take

place into two different type namely:

a. Film wise condensation

b. Drop wise condensation

When the condensate tends to wet the surface it is called film wise condensation. When

condensate does not wet the surface and when it forms droplets on the surface, it is known as

drop wise condensation.

Experimental setup:- The experimental set up consists of an electric generator to generate

steam instantaneously and the steam is made to condense on two vertical hollow copper

cylinders with one coated with chromium(drop wise).On one cylinder the condensation will be

film-wise while on the other which is coated, it will be drop wise. The flow of water is regulated

by means of a pump. The schematic for the experiment is shown in the figure 9.1.Provisions are

made to measure the cooling water flow rate, condensate flow rate, the outer surface temperature

of the cylinder, the pressure of steam condensing and the inlet and exit temperature of the

cooling water.

Apparatus: Gas tube, steam generator copper coated internal tube, stop watch.

Experimentation:

Aim: To determine the average heat transfer co-efficient for film wise and dropwise

condensation and comparison of both.

Procedure:

(1) First fill the water in stainless steel tank and then switch on the main and console.

(2) Switch on the supply pump starter. (3) Change the ball valve position of conducting film wise condensation experiment. (4) Switch on the heater and maintain a steady water flow, say 4LPM, through the rata meter. (5) Wait for sometime till steady steam is generated. Then the stem is passed through a

separator to supply only dry stream to the cylinder.

(6) The stem starts condensing on outer surface of the condenser tube and gets collected as a condensate at the bottom of the cylinder.

(7) Note down the different temperatures of inlet and outlet water supply, stem pressure, flow rate of water.

(8) Now change the ball valve position to drop wise condensation and conduct the experiment for same flow rate.

(9) Repeat the above procedure for different flow rates and different steam pressures.



54

Steam inlet

Plain Condensor

Plated Condensor

Glass Tube

Condensed WaterOutlet

Water inlet Water outlet

Water inlet Water outlet

T

T T

T

T

6

2 3

4

5T2

Figure 9.1 Schematic of film and drop wise condensation unit.

Figure 9.2 Photograph of test set up of stefan boltzman unit.



55

Specimen calculations for film-wise condensation:

Outer diameter of condensing surface = d = _______,

Length of the condensing surface = L = _______,

Surface temperature = TS = ________

Pressure of steam condensing = _________

Saturation temperature at condensing pressure = Tsat = _______ (From tables)

Inlet temperature of cooling water = Twi = ______

Exit temperature of cooling water = Two = ______

Volume flow rate of cooling water = Vw =______

Volume flow rate of condensate = Vs = Volume of steam admitted per time.

= ________

Mass flow rate of water = mw = w Vw = ______ Specific volume of condensate = vs = vf at condensing pressure.

Condensate flow rate = msteam = s

s

v

V

Heat carried away by cooling water = Qw = mw.Cpw[ Two- Twi]

= __________

Heat given out by steam due to condensation = Qsteam = msteam.hfg

Mean film temperature = Tm = 2

ssat TT

Properties of condensate at Tm are: k =______ , = _______ , hfg = _____ ,

Hence, Qsteam = _________.

Condensation heat transfer co-efficient = hav = )( ssat

steam

TTdl

Q

= __________

To find hav using standard correlation:

Reynolds number for the condensate = Re = d

msteam4 = _________

If Re < 1800, Condensate flow is laminar.

Hence hav =

33.0

33.0

2

23

Re

47.1

gk

= _______________ (Km

W

2).



56

Specimen calculations for drop-wise condensation:

Surface temperature = TS = ________

Pressure of steam condensing = _________

Saturation temperature at condensing pressure = Tsat = _______ (From tables)

Inlet temperature of cooling water = Twi = ______

Exit temperature of cooling water = Two = ______

Volume flow rate of cooling water = Vw =______

Volume flow rate of condensate = Vs = Volume of steam admitted per time.

= ________

Mass flow rate of water = mw = w Vw = ______ Specific volume of condensate = vs = vf at condensing pressure.

Condensate mass flow rate = msteam = s

s

v

V

Heat carried away by cooling water = Qw = mw.Cpw[ Two- Twi]

= __________

Heat given out by steam due to condensation = Qsteam = msteam.hfg

Mean film temperature = Tm = 2

ssat TT

Properties of condensate at Tm are: k =______ , = _______ , hfg = _____ ,

Hence, Qsteam = _________.

Condensation heat transfer co-efficient = hav = )( ssat

steam

TTdl

Q

= __________

To find hav using standard correlation for drop wise condensation:

Reynolds number for the condensate = Re = d

msteam4 = _________

If Re < 1800, Condensate flow is laminar.

Hence hav =

33.0

33.0

2

23

Re

47.1

gk

= _______________ (Km

W

2).



57

Result: Average heat transfer co-efficient for film wise condensation is _______

Average heat transfer co-efficient for film wise condensation is _______


(1) List of various precautions especially that of steam generator observed during experimentation.

(2) Which is more effective among film wise and drop wise condensation and why.



58

EXPERIMENT NO. 10

Vapour Compression Refrigeration Test Rig Date:

Introduction: Refrigeration is a process by which the temperature of a given space is reduced

below that of the atmosphere or surroundings. Refrigeration can be realized by several methods,

for example, Ice refrigeration, Dry ice refrigeration, Evaporative refrigeration, Air refrigeration,

Vapour compression refrigeration etc. The modem refrigeration uses the vapour compression

method. In this method, a closed system (the refrigerant) experiences a thermodynamic cycle ;

by virtue of doing network on the system in such a cycle, it is possible to extract heat from a low

temperature source (the refrigerated space) and to reject heat to a higher temperature sink (the

atmosphere or cooling water).

The performance of refrigerators and heat pumps is expressed in terms of the coefficient of

performance (COP), defined as

COP = inputrequired

outputDesired

.

.


Fig. (1) Shows the schematic of the vapour compression refrigeration set up and its (T-S)

Diagram.The cycle consists of the following processes:

(l-2) : Isentropic compression (in ideal cycle) of vapour refrigerant from lower pressure to higher

pressure, temperature increases. (But in actual practice, the process is not isentropic because of

losses and inefficiency).

(2-3): Condensation of high pressure and temperature of saturated/super heated vapour to liquid

at high pressure.

Apparatus: Refrigerant, Compressor, Condenser, Cooling fan, Throttle valve, Capillary tube,

Evaporator.

Experimentation:

Aim: To conduct a test on Vapour Compression Refrigerator and find its coefficient of

performance.

Procedure:

(a). Switch-on the Mains and the Console.

(b) Keep either the Throttle Valve or the capillary Tube open-when the capillary tube is open,

the throttle valve should be closed and vice versa. Both devices have the same expansion (or

throttling) effect.

(c) Switch-on the motor which drives the compressor and the fan (which cools the condenser).

(d) The refrigerant passes through the vapour compression cycle as mentioned earlier resulting

in cooling in the evaporator chamber or freezer.

(e) Wait for about 5 minutes and note the Temperatures T1 to T5 and Pressures P1 and P2.



59

Specifications:

1 Energy meter Constant for compressor C = rev/ kwh 2 Refrigerant used in this test rig = _______

Fig 10.1 (a) Fig 10.1 (b)

Figure 10.1 (a) Schematic of vapour compression Refrigeration cycle

(b) T-S Diagram for the cycle

(3-4) : Constant enthalpy process in throttling valve or capillary tube-expansion of high pressure

liquid refrigerant to lower pressure liquid vapour mixture temperature decreases.

(4-l) : Vapourization of refrigerant in evaporator at constant lower pressure. Cooling of

atmosphere or surroundings takes plate due to absorption of latent heat in vapourization.

In the ideal cycle the refrigerant enters the throttling valve/capillary as a liquid at (3) and leaves

(4) at constant enthalpy as liquid-vapour mixture. The refrigerant enters the evaporator and

extracts heat from the refrigerated space at constant temperature (T1) and lower pressure in the

vapourization process (4-1). The refrigerant may leave the evaporator as a two phase mixture or



60

as a saturated vapour or as a slightly superheated vapour as shown in this T-S Diagram. Ideally,

the refrigerant is compressed to the condenser pressure in the isentropic process (1-2). Heat is

rejected from the condenser to the atmosphere or cooling water in the process (2-3).

Figure 10.2 Photograph of Refrigeration test rig.

T1 = Temperature at Compressor Inlet, ( OC )

T2 = Temperature at Compressor Outlet, ( OC )

T3 = Temperature at Condenser Outlet, ( OC )

T4 = Temperature at Evaporator Inlet , ( OC )

T5 = Temperature inside Freezer, ( OC )

P1 = Pressure Upstream of the Compressor, ( 2cm

Kg)



61

P2 = Pressure Downstream of the Compressor, ( 2cm

Kg)

Using the measured temperatures, pressures and power input to the compressor, the co-efficient

of performance and the capacity of the refrigerator can be determined using the procedure given

below:

Fig.10.1 (a) Shows the schematic of the heat transfer process in the vapour compression

refrigeration cycle. In a steady state, the heat balance is given by,

QL+ Win = QH . Eq (1)

QL = Heat removed by the evaporator from the refrigerated system,

= Heat gained by the refrigerant in the evaporator.

= h1 - h4 per Kg of refrigerant.

QH = Heat transferred from the refrigerant in the condenser.

= h2 - h3 per Kg of refrigerant.

Win = Work done by the compressor on the refrigerant.

= h2 h1 per Kg of refrigerant.

h1= Enthalpy of the refrigerant at exit of the evaporator.

h2 = Enthalpy of the refrigerant at exit of the compressor.

h3 = Enthalpy of the refrigerant at exit of condenser.

h4 = Enthalpy of the refrigerant at exit of the throttle valve.

The values of enthalpies of the refrigerant at different states can be obtained from tables/ charts

using the measured values of pressures and temperatures. The work done in the compressor can

be directly obtained from the energy meter in the panel. The co-efficient of performance (COP)

of the refrigerant system is given by,

COP = in

H

W

Q=

42

41

hh

hh

.......... Eq. (2)

Where, QH = Amount of heat extracted in the refrigerator

Win= Net work done.

Result : The COP of the refrigerator came out to be _______ .

Point for discussion

(1) Discuss the suitability of different Compressor speeds for typical duties and others. (2) Can refrigerator equipment can be used as heater. If yes how it is done.



62



63

EXPERIMENT NO. 11

Air Conditioning Test Rig Date:

Introduction: The science of Air Conditioning deals with maintaining a desirable internal air

conditions irrespective of external atmospheric conditions. The factors involved in any air

conditioning installation are:

(a) Temperature

(b) Humidity

(c) Air movement and circulation

(d) Air filtering, cleaning and purification.

The simultaneous control of these factors within the required limits is essential for

human comfort or for any industrial application of the air conditioning system.

In any air conditioning system, temperature and humidity are

controlled by thermodynamic processes. Depending on the season, the air conditioning

processes involve cooling, heating, humidification and dehumidification of air. Other aspects

such as air movements, circulation, purification, etc. are obtained by installing suitable fans,

blowers, ducting and filters.

This equipment is designed to demonstrate different air conditioning processes such as

cooling, heating, humidification, etc. required for different seasons of the year.


The schematic for studying and analyzing the air conditioning process is shown below

Nomenclature for the set up:

2. Tdb1 and Tdb2 : Dry bulb thermometers. 3. Twb1 and Twb2 : Wet bulb thermometers. 4. Steam inlet: Used to increase the humidity of air. 5. H1 : Heater 1 to increase the dry bulb temperature of air. 6. H2 : Heater 1 to increase the dry bulb temperature of air. 7. CR : Compressor of the vapor compression refrigeration plant. 8. CDR : Condenser of the vapor compression refrigeration plant. 9. CC : Evaporator of the refrigeration plant (also the cooling coil of the air

conditioner).



64

ExpansionValve

CDR

CR

AirDuct

Air in

Air out

12

3

4

CC

Steam Inlet

H 1

H 2

T db1

T wb1

T db2

T wb2

Figure 11.1 Shematic of Air conditioning test set up.

Figure 11.2 Photograph of Air conditioning test set up.



65

Brief description of the set up:

The experimental set up consists of a duct of rectangular cross section.The air is made

to flow through this duct by means of a blower. Provisions are made in the duct

(i) To heat the air (Heater 1) (ii) To add moisture to the air in the form of steam which is generated in a steam boiler. (iii) To cool the air to a temperature below that of surroundings by employing a vapor

compression refrigeration plant and to dehumidify the air

(iv) To heat the air to the desired dry bulb temperature after dehumidifying by using heater (Heater 2)

Dry bulb and wet bulb temperatures (Tdb1, Tdb1, Twb1 and Twb2 ) are

provided to measure dry bulb and wet bulb temperatures of air before and after the air

conditioning process.The velocity of air at entry and exit of the duct are measured using an

anemometer.

Apparatus: Cooling Coil, Air Heaters-2 Sets, Steam Generator, Duct

Experimentation:

Aim: To study the effect of Heating and humidification of air conditioning process

Procedure:

The purpose of the experiment is to increase the humidity and dry bulb temperature of the

incoming air.Hence for this experiment the refrigeration system should be switched off and the

steam boilers and heaters H1 and H2 should be on.After steady state is reached the following

measurements are made and noted.

Tdb1 = _____ Twb1 = _____

Tdb2 = _____ Twb2 = _____

The velocity of air at the exit section of the duct = v2 =

Cross sectional area of the duct = A2 = ____

The inlet and exit conditions of the air are represented by the psychrometric charts as shown in

the figure 11.1 (a)

From Psychrometric chart, we have

hh

hv

Volume flow rate of air in (s

m3) = Vair = A2.v2

Heater energy meter constant = ________

Time taken for one revolution of energy meter = ______ s

Temperature of steam supply = Tsteam = _____ 0C

Mass flow rate of dry air in (s

Kg )= ma =

2v

Vair =______

Amount of moisture added to air = mw = = __________.



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Heat supplied by the heating coil = ..1..tan..

3600.....

revtakenfortimetconsmeterEnergy

meterenergyofsrevolutionofNumber

Hence Qcoil =

Energy balance equation for the process gives

Qcoil + ma(hsteam = QAir + QLoss .(1) From steam tables hsteam = hg at Tsteam = ________

QLoss = _________

Sensible heat added to air = QS = ma[hA-h1]

= ______

Sensible heat factor = SHF = air

S

Q

Q= __________

T db1

T wb1

T db2

T wb2

T db

1

2

A

h1

h2

hA

Figure 11.3 Shematic of Psychrometric chart.

Result: Sensible heat factor for heating and humidification process is ___________ ,


(1) List of various precautions observed during experimentation. (2) What is dew factor.



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68

EXPERIMENT NO. 12

Transient Heat Conduction Date:

Introduction: A Heat transfer process which is time dependent is designated as an unsteady

state or transient heat transfer. There are large number of situations where changes in condition

result in transient temperature distribution.Unsteady state heat transfer generally occurs before

steady state operating conditions.

Transient temperature distribution results in manufacture of bricks, Cooking and freezing of

food and in heat and cold treatment of metals.

Experimental set-up: The experimental setup consists of a metal rod, one end of which is

heated by an electrical heater while the other end projects inside a cooling water jacket. The

whole rod is insulated with asbestos rope. Seven Thermocouples are placed on the rod for

temperature measurement.

The heater is provided with a Dimmerstat for controlling the heat input where as water under the

constant head is circulated through the jacket and its flow rate and temperature rise are measured

using Rotameter and thermocouples.

Apparatus: Transient heat conduction set-up, thermocouples, stop watch/clock.

Experimentation:

Aim :To obtain temperature profile for conduction through a cylindrical rod heated by a constant

source and to compare theoretically predicted temperature distribution under transient

conditions.

Procedure:

(1) Heat the metal rod (Stainless steel) by electrical heating at constant rate by adjusting the current and voltage.

(2) Adjust the flow of water to one liters/min using rotameter. (3) Note the readings of the Ammeter and the Voltmeter. (4) Note down the initial temperature at seven different positions, say at distances 65mm,

130mm, 195mm etc. from the heat source.

(5) Start the stopwatch and note the temperature indicated by the thermocouples at a various points recorded at 5 minutes interval.

(6) Continue to note the temperatures till constant temperature is attained. (7) The calculations are performed to obtain temperature profile and compared with

theoretical values obtained.

(8) Temperature profile: Theoritical temperature and experimentally observed temmperature versus distance from the hot source is drawn on a graph sheet for 15 & 30 minutes time

intervals.



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T1 T2 T3 T4 T6 T7T5

T8

T9

Heater

Water in

waterout

0.65m 0.65m

Figure 12.1 Transient test set up.

Figure 12.2 Photograph of Transient conduction test set up

Specifications:

1 Diameter of rod d = mm 2 Distance between each thermocouple X = mm 3 Test length of the rod L = mm



70

Tabular column:

Time

in

sec.

At

distance

near

Heater.

X1=0 m

At

X2=0.065m

At

X3=0.13m

At

X4=0.195m

At

X5= 0.26m

At

X6=0.325m

At

X=0.39

m

Water

out T8

Water

inT9

Calculations:

Voltmeter reading V: (volts)

Ammeter reading I : (amps)

Total heat supplied Q = V x I = (watts)

Thermal diffusivity of the material is calculated by

PC

k

, (

s

m 2)

= Density of material (Copper) = 8954 3m

Kg

CP = Specific heat of material (Copper) = 381 KKg

J

k = Thermal conductivity (Copper = 386Km

W

)

A= Area of cross section of metal rod =4

. 2d = ________ m

2

Where, d = Diameter of the rod in. m.



71

Calculate the Value of Z using the formula:

Z = t

X

2

From this, Calculate the error function value of Z from the Data Hand book i.e., erf (z). Now,the

Temperature distribution for a body subjected to constant heat flux condition is given by

T(X) Ti = k

t

Xtq

4exp2

2

- k

ZerfXq )](1[ .

taking particular time (say, t = 900 seconds) and substitute in the above Theoretical equation.

Ti = Initial temperature before starting the experiment.

t = At any particular times (say, t = 900 seconds)

X = Different distances at which thermocouples are placed.

Temperature (OC) T experimental = Values are obtained directly from the experiment at

different times.

Serial

no.

Time(t)

seconds

Distance X (m) Experimental

Temperature

Z

Erf(Z)

Theoretical

Temperature

1

000

X1 = 0

X1 = 0.065

X1 = 0.13

X1 = 0.195

X1 = 0.26

X1 = 0.325

X1 = 0.390

2

300

X1 = 0

X1 = 0.065

X1 = 0.13

X1 = 0.195

X1 = 0.26

X1 = 0.325

X1 = 0.390

3

600

X1 = 0

X1 = 0.065

X1 = 0.13

X1 = 0.195

X1 = 0.26

X1 = 0.325

X1 = 0.390



72

Plot a graph of temperature v/s distance at particular time.

Tem

p.

Distance, X (m)

T theoretical

T experimental


Result: The heat conduction experiment is conducted under transient conditions and the

temperature conduction through the rod heated by a constant heat flux is obtained and compared

with theoretical values.


(1) Whether the rod can modelled as a lumped system one.Comment on it.



73



74

Viva Questions

1. Distinguish between steady-state and unsteady state Heat transfer.

2. Define thermal conductivity of a material.

3.Give examples of natural convection and forced convection.

4.write the laplace equation and fourier equation and diffusion equation.

5.Define Thermal diffusivity.

6.Define Conduction shape factor.

7.What is meant by Thermal resistance ?

8.What is convection Heat Transfer Coefficient(h)? On what factors does it depend.

9.What is the range of h for free and forced convections?

10. What is meant by critical radius of insulation? 11. What is an extended surface ? 12. What is mode of heat transfer through a fin ? 13. Name few commonly used fins. 14. Define fin efficiency. 15. What is meant by Transient heat transfer? 16. What is the physical significance of the Biot modulus ? 17. What is meant by semi-infinite model? 18. What is meant by Stability criteria in heat transfer? 19. How does Thermal radiation differ from other types of Electromagnetic radiation ? 20. Define the terms - Reflectivity, Transmitivity and Absorptivity. 21. What is meant by opaque body? 22. Define emissivity of a body. 23. What is a Black body? 24. What is Stefan - Boltzmann Law? 25. What is Kirchhoff's Law? 26. What is Plank's Law? 27. What is Wiens displacement Law? 28. Define intensity of radiation. 29. What is a grey body? 30. Define the terms - Irradiation and Radiosity. 31. What is the purpose of radiation shield ? 32. Define convection heat transfer coefficient . 33. What is the physical significance of Nusselt's number?

34. What is Buckingham - Theorem? 35. Define local friction Coefficient and average friction Coefficient. 36. Define the friction factor f. 37. Distinguish between natural convection and forced convection. 38. What are the forces responsible for fluid motion in natural convection ? 39. Define Grasshoff's number and what is its physical significance? 40. What is film temperature? 41. What is meant by filmwise and dropwise condensation ? 42. What is the Reynolds number of a condensed film? 43. What is pool boiling? 44. Distinguish between nucleate pool boiling and film boiling. 45. Discuss the various regimes of boiling. 46. Distinguish between parallel flow and counter flow heat exchanger.which is effective.



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47. Classify heat exchangers. 48. Distinguish between shell type and cross flow heat exchangers. 49. Define LMTD.What i

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