1. Reg. No_________________
Roll. No____________
VINAYAKA MISSIONS UNIVERSITY
SALEM
HEAT TRANSFER LAB
(Final Year B.E./B.Tech. Students)
LAB MANUAL
VINAYAKA MISSIONS KIRUPANANDA VARIYAR
ENGINEERING COLLEGE,
VINAYAKA MISSIONS UNIVERSITY,
SALEM-636 308.

2. (For Private Circulation only)
2

3. VINAYAKA MISSIONS UNIVERSITY
V.M.K.V. ENGINEERING COLLEGE
SALEM-636 308
DEPARTMENT OF MECHANICAL ENGINEERING
NAME :
Roll No. :
REG. No. :
YEAR & SEMESTER :
SECTION :
LAB MANUAL
HEAT TRANSFER LAB
(For Private Circulation Only)
3

4. 4

5. CONTENTS
EX.
No.
DATE EXERCISE
PAGE
No.
STAFF
SIGN.
5

6. 6

7. HEAT TRANSFER LAB
LIST OF EXPERIMENTS
1. Thermal conductivity by guarded hot plate method
2. Heat exchanger test – parallel and counter flow
3. Heat exchanger test – shell and tube heat exchanger
4. Emissivity measurement
5. COP of a refrigerator
6. Heat transfer from fins-natural and forced convection
7. Thermal conductivity of insulating material
8. Heat transfer through composite walls
9. Heat transfer by free and forced convection
10. Stefan – Boltzman apparatus
11. Boiler trial
7

8. Tabulation
Sl.
No.
Inner Heater Outer Heater
Cooling
Plate
Thermal
Conductivity
W/mk
V
Volts
I
Amp
T1
°C
T2
°C
V2
Volts
I2
Amp
T3
°C
T4
°C
T5
°C
T6
°C
Formula
Thermal Conductivity ( )
( )
1
/
h C
W x L
K W mk
A T T
=
−
Where
W1 = Input to the inner heater in watts = V × I
L = Specimen thickness in metre
A = Area =
2
4
D
π
in m2
D = Diameter of the heater plate in metre
X = Width of gap between the heater plates in metre
1 2 3 4( )
4
h
T T T T
T
+ + +
= °C =
5 6( )
2
C
T T
T
+
=
8

9. Ex. No: 1 Date:
THERMAL CONDUCTIVITY BY TWO SLAB GUARDED HOT
PLATE METHOD
Aim
To determine the thermal conductivity of the given specimen by using guarded hot
plate method.
Apparatus Required
Thermal conductivity apparatus
Central heater
Guarded heated ring
Description
The heater plate is surrounded by a heating ring for stabilizing the temperature of
the primary heater and prevents heat loss completely around its edges. The primary and
guard heater are made up of mica sheets in which is wound closely with equal space
nichrome wire and packed with upper and lower mica sheets. These heaters together form
a flat which together with upper and lower copper plates and rings form the heater plate
assembly. Two thermo couples are used to measure the hot face temperature at the upper
and lower central heater assembly copper plates two thermocouples are used to check the
balance in both the heater inputs.
Specimens are held between the heater and cooling unit on each side of the
apparatus. Measure the temperature of the upper cooling plate and lower cooling plate
respectively. The heater plate assembly together with the with cooling plates and specimen
held in position by 3 vertical studs and nuts on a base plate are shown in the assembly
drawing. The cooling chamber is a composite assembly of grooced aluminum casting and
aluminum cover with entry and exit adaptors for water inlet and outlet.
9

10. 10

11. Procedure
The specimen is placed on either side of the heating plate assembly, uniformly
touching the cooling plates. Then the outer container is filled with lose fill insulation such
as glass wool (supplied in small cloth packets). The cooling circuit is opened then
calculated input is given to central and guard heaters through separate single phase supply
lines with a dimmer, stat in each line and it is adjusted to maintain the desired temperature.
The guard heater input is adjusted in such away that there is no radial heat flow which is
checked form thermocouple reading and is adjusted accordingly. The input to the central
heater and the thermocouple readings are reordered in every 10 minutes till a reasonably
steady state condition is reached.
Result:
Thus the thermal conductivity of the specimen is determined.
11

12. Tabulation
Type
of
flow
Hot Water Cold Water
LMTD
°C
Heat
transfer
co-
efficient
kJ/hr m2
k
Effectiv
eness (e)
Time
taken
for 1
liter
water
flow in
sec.
Inlet
temp
T3°C
Outl
et
temp
T4°C
Time
taken
for 1
liter
water
flow
in sec.
Inlet
temp
T1°C
Outlet
temp
T2°C
12

13. Ex. No: 2 Date:
PARALLEL AND COUNTER FLOW HEAT EXCHANGER
Aim
To study and compare the temperature distribution, heat transfer rate and overall
heat transfer coefficient in parallel and counter flow heat exchanger. To calculate the
effectiveness of parallel flow and counter flow heat exchanger.
Apparatus Required
Stop watch
Measuring tape
Thermometers (0-100) °C
Description
The apparatus consist of concentric tube that exchanger. The hot fluid (i.e. hot
water) is obtained form an electric geyser and it flows through the inner tube. The cold
fluid is cold water and can be admitted at any one of the ends enabling the heat exchanger
to run as a parallel flow apparatus or a counter flow apparatus. This can be done by
operating the different values provided. Temperatures of the fluids can be measured using
thermometers. Flow rate can be measured using stop watch and measuring flask. The out
tube is provided with adequate asbestos rope insulation to minimize the heat loss to the
surroundings.
Procedure
Parallel flow
1. Parallel flow means the direction of cold and hot water flow is the same
2. Adjust the valves of pipes and maintain the same desired direction of flow
3. On the heater and slightly open the inlet valves of cold and hot water and
measure the mass flow of cold and hot water at outlet per litre.
4. Take inlet and outlet temperature of cold and hot water
5. Increase the valve opening and measure the above readings in steps of 200
ml/min. to 1 lit/min.
6. The measured values are tabulated and the required results are obtained.
13

14. Formula:
Heat transfer from hot water, 3 4( ) /h h hQ m C T T kJ hr= × × −
Where,
mh = Mass flow rate of hot water kg/hr
Ch = Specific heat of hot water = 4.187 kJ/kgK
T3 = Inlet temperature of hot water, °C
T4 = Outlet temperature of hot water, °C
Heat transfer from cold water 2 1( ) /C C CQ m C T T kJ hr= × × −
Where, mc = Mass flow rate of cold water kg/hr =
6
1
1
3600 10 C
t
ρ− 
× × × 
 
ρc = Density of cold water
Cc = Specific heat of cold water = 4.187 kJ/kgK
T1 = Inlet temperature of hot water, °C
T2 = Outlet temperature of hot water, °C
Logarithmic Mean Temperature Difference (LMTD),
log ( )
i o
m
c i o
T T
T
T T
∆ − ∆
∆ =
∆ − ∆
iT∆ = Temperature difference at inlet, °C
0T∆ = Temperature difference at outlet, °C
Overall Heat Transfer Coefficient, CQ
U LMTD
A
= ×
Where,
A = Area of the tube in m2
D = Outer diameter of the tube = 12.5mm
L = Length of the tube = 1200mm
Effectiveness,
2 1 2 1
3 1 3 1
( ) ( )
( ) ( )
C C
C C
m Cp T T T T
E
m Cp T T T T
× − −
= =
× − −
14

15. Counter flow
1. Counter flow means the direction of cold and hot water is in opposite direction
2. Adjust the valves of pipe and maintain the flow of cold and hot water in the
opposite direction to each other
3. Take the readings of the inlet and outlet temperatures of cold and hot water at
various levels.
4. The measured values are tabulated and the required results are obtained.
15

16. 16

17. Result
Thus the temperature distribution, heat transfer rate, overall heat transfer co
efficient and effectiveness of the parallel flow and counter flow heat exchangers are
calculated.
17

18. Tabulation
Cold
Water
temp °C
Steam
Inlet
Temp
T3°C
Cond
ensat
e
Quali
ty
T4°C
Condensate
Quality
Heat
Transfer
to water
KJ/hr
Cold Water
flow rate
LMTD
°C
Heat
Transfer
Co-
efficient
KJ/hr
m2
°C
Effecti
veness
(E)
Inlet
T1
Out
let
T2
ml tC
Sec
ml/sec Kg/h
r
18

19. Ex. No: 3 Date:
SHELL AND TUBE HEAT EXCHANGER
Aim
To determine the overall heat transfer coefficient to determine the heat exchanger
effectiveness.
Apparatus Required
Stop watch
Thermometers (0°C-100°C)
50 ml beaker
Description
In the present setup the unit operates as a condenser with steam condensing over
the tubes and cooling water flowing through the tubes. The design of condensers for such
varied applications as steam power plants, chemical processing plants and nuclear power
plants and for space vehicles involved a variety of heat transfer and fluid flow problems
associated with the condensation of vapours.
The banks of smooth horizontal round tubes configurations and with the high
vapour velocities which are normally associated with steam condenser, the overall heat
transfer co-efficient is primarily a function of cooling water velocity for clean, bright, new
horizontal tubes with no contamination wither on the steam side or on the water side. If the
tubes are of a material other than admiralty metal or have a wall thickness other than
18BWG correction factors indicated in table one should be employed.
The condenser is a horizontal shell and tube heat exchanger with steam condensing
over the tubes. Cooling water flows through tow tube passes. Steam pressure and
temperature at inlet to the exchanger are monitored. The condensate leaves at the bottom
through a valve. The condensate temperature is also monitored. Water inlet and outlet
temperatures are measure by dial thermometers. A manometer is provided for evaluating
the pressure drop in the cooling water circuit. Water flow rate is measured by either
measuring the quantity of water collected in bucket in a known time or by means of rot
meter quantity of steam condensed is measured by collecting the condensate in a
measuring jar.
19

20. Formula
mw = Mass flow rate of water in kg/hr = w
w
w
T
q
ρ××× −6
103600}{
Heat transfer to water, )( 12 TTCmQ Pccw −×= in kJ/hr
T1 = Inlet temperature of cold water in °C
T2 = Outlet temperature of cold water in °C
CPc = Specific heat of cold water = 4.187 kJ/kgK
ρw = Density of water = 1000 kg/m3
Heat given out by steam = c
c
c
s L
T
Q
Q ρ××××= −
360010 6
Where,
L = Latent heat of steam
Qc = Quantity of condensate collected in ml.
tc = Time for collecting condensate in seconds
ρc = Density of steam at temperature T3 °C
Logarithmic Mean Temperature Difference (LMTD) =
43
13
12
)(log
TT
TT
TT
T
e
m
−
−
∆−∆
=∆
Heat transfer = NLd t ××× 0π
Where, N = Number of tubes
d0 = Outside diameter of tube in metre
L = Effective length in metre
Overall heat transfer coefficient = LMTD
A
Qw
× kJ/hrm2
K
Where, A = Area of tube in m2
D = Outer diameter of tube 12.5mm
L = Length of tube 1200mm
Effectiveness, )( 13
min
TT
C
qc
−×=ε
Where, T3 = Temperature of steam at inlet, °C
T1 = Temperature of water at inlet, °C
20

21. Procedure
1. Switch on the boiler heaters
2. In the pump and adjust the mass flow of cold water to 25ml/sec and wait until
the boiler pressure reduces to 0.5kgfcm2.
3. When the boiler pressure is 0.5kgf/cm2
4. Then note the steam inlet temperature and coldwater inlet and outlet temp, and
time taken for 50ml condensate and steam outlet temp.
5. Then varying the mass flow of the cold water to 45, 65,…105 ml/sec.
corresponding readings are tabulated.
21

22. 22

23. Result
Thus the overall heat transfer co-efficient and effectiveness are determined for the
given shell and tube heat exchanger.
23

24. Tabulation
Sl.
No.
Black Plate Test Plate Ambient
Temperature
°C
Emissivity
of test
plate
V1
volts
I1
Amp
T1
°C
V2
volts
I2
Amp
T2
°C
24

25. Ex. No: 4 Date:
EMISSIVITY MEASUREMENT
Aim
To determine the emissivity of the given specimen.
Apparatus Required
Emissivity apparatus
Description
The experimental set up consists of two circular aluminum plates identical in size
and is provided with heating coils sandwiched. The plates are mounted on brackets and are
kept in an enclosure so as to provide undisturbed natural convection surroundings. The
heat input to the heater is varied by separate dimmer stats and is measured by using an
ammeter and a voltmeter with the help of double through switches. The temperature of the
plates is measured by thermocouples. Separate wires are connected to diametrically
opposite points to get the average surface temperature of plates. Another thermocouple is
kept in the enclosure to read the ambient temperature of enclosure. Plate 1 is blackened by
a thick layer of lamp black to form the idealized black surface where as the plate whose
emissivity is to be determined.
The heater inputs to the two plates are dissipated from the plates by conduction,
convection and radiation. The experimental set up is designed in such a way that under
steady state conditions the heat dissipation by conduction and convection is same for both
plates, when the surface temperatures are same and the difference in the heater input
readings is because of the difference in radiation characteristics due to their different
emissivity’s.
25

26. Formula
)()(
4
3
4
21 TTEEAWW sb −×−××=− σ
Where,
W1 = Heater input to black plate = V1×I1 Watts
W2 = Heater input to test plate = V2×I2 Watts
A = Area of two plates =
2
4
2 d××
π
m2
Diameter of the black and test plate (d) = 160mm
Ts = Surface temperature of discs = T1 + 273K (or) T2 + 273K
T3 = Temperature of enclosure = T3 + 273K
Eb = Emissivity of black plate = 1
σ = Stefan Boltzmann constant = 5.67 × 10-8
W/m2
K4
By using Stefan Boltzmann Law:
86.0
)()(
4
3
4
21
TTEEA
WW sb −×−××
=−
σ
26

27. Procedure
1. Give power supply to T.P. (230 v signal phase) and adjust the reading in it
equal to room temperature by roating the compensation knob (normally this is
pre-adjusted).
2. Select the proper range of voltage on voltmeter
3. Gradually increase the input to the heater to black plate and adjust it to some
value viz., 0,50,75 watts. And adjust the heater input to test plate slightly less
than the black plate 27, 35, 55 watts etc.,
4. Check the temperatures of the two plates with small time intervals and adjust
the input of test plate only, by the dimmer stat so that the two plates will be
maintained at the same temperature.
5. This will require some trial and error and one has to wait sufficiently (more
than one hour or so) to obtain the study state condition.
6. After attaining the steady state condition record the temperatures, and
voltmeter and ammeter readings for both the plates.
7. The same procedure is repeated for various surface temperatures in increasing
order.
27

28. 28

29. Graph:
Graph is drawn between Surface temperature (X axis) Vs Emissivity (Y axis)
Result
Thus the emissivity of the given specimen is calculated and corresponding graph
has been drawn.
29
Emissivity
Surface Temperature in Kelvin

30. Tabulation
Sl.
No.
Compressor
work time
taken for 30
revolution
in second t1
Fan work
time taken for
10 revolution
in seconds t2
Time
taken for
one lit
water flow
in seconds
t3
Water Temperature
°C
Co-efficient
of
performance
(COP)
Inlet
T1
Outlet
T2
T1-T2
Formula
Heat extracted from water Q = mw × CPw × (T1 - T2) kJ/hr
Where,
w
w
w
w
T
q
m ρ×××= −6
103600}{ kg/hr
T1 = Inlet temperature of water in °C
T2 = Outlet temperature of water in °C
CPw = Specific heat of water = 4.186kJ/kgK
ρ = Density of water = 1000kg/m3
Coefficient of Performance,
W
Q
COP =
Where, W = W1+W2
W1 = Compressor work,
W2 = Fan work
30

31. Ex. No: 5 Date:
PERFORMANCE TEST ON A REFRIGERATOR
Aim
To determine the co-efficient of performance of the refrigerator
Apparatus Required
Refrigerator test rig
Thermometer
Measuring jar
Stop watch
Description
The refrigeration is the process of cooling a space less than the surrounding
temperature. The working substance used in the refrigerator is known as refrigerant. The
refrigerant passes through the compressor to increase the pressure. Then it flows to
condenser where it is condensed at constant pressure. The pressurized liquid refrigerant
expands in the expansion valve. Thus the low pressure and low temperature liquid
evaporates in the evaporator by absorbing the latnt heat of evaporation form the space to
be cooled at constant pressure. Now the vapour again passes through the compressor and
the cycle is repeated again.
Procedure
Before starting the unit, ensure that valves are closed. The water is allowed to flow
at constant rate into the container by opening inlet supply tap. The refrigeration can work
with either solenoid valve or thermostat. First the fan is switched on. Then the following
procedure is followed.
The solenoid valve with thermostat expansion is started, the valve s2, s4 and s5 are
closed and solenoid valve switch is put on. The valves s2, s4 and s5 are closed if the
capillary is put off. then the thermostat is rotated in clockwise direction. The thermostat is
inserted in their respective places. The refrigerator is allowed to run for stabilization. The
no of revolutions of the disc in the energy meter of the compressor and fan are noted down
for a known time.
31

32. 32

33. Result:
Thus the coefficient of performance of the given refrigerator is calculated
33

34. Tabulation
Experiment Power
supply Fin. Temperature °C
Monometer
Reading
Ambient
Temperatu
re °C
Rate of
Heat
Transfer
Q (J/sec)
Effecti
veness
Amp Volt T1 T2 T3 T4 T5 h1 h2
Natural
Convection
Forced
Convection
34

35. Ex. No: 6 Date:
HEAT TRANSFER THROUGH PINFIN
Aim
To study the temperature distribution along the length of a pinfin in natural and
forced convection and to determine the heat transfer rate from the fin and the fin
effectiveness.
Description
A brass fin of circular cross section is fitted across a long rectangular duct. The
other end of the duct is connected to the suction side of a bowler and the air flows past the
fin perpendicular to its axis. One end of the fin projects outside the duct and is heated by a
heater. Temperature at five points of the fin measured by chromel alumel thermocouples
embedded on the fine. The air flow rat is measured by an orifice meter fitted on the
delivery side of the bowler.
Procedure
Natural convection
1. Start heating the fin by switching on the meter element and adjust the voltage
on dimmer stat, to 50volts by increasing slowly from 0 onwards
2. Note down the thermocouple reading I to 5
3. When steady state condition is reached record the final reading I to 5 and also
record the ambient temperature reading T6
4. Repeat the same experiment with 100 volts and 120 volts.
Forced Convention
1. Start the heating of fin by switching on the heater and adjust dimmer stat
voltage equal to 100 volts.
2. Start the blowe rand adjust the difference of level in the monmieter H=….cm
with the help of value.
3. Note down the thermocouple reading I to 5 at the time interval of 5 minutes
4. When the steady state steady is reached, note the ambined temp reading 6.
5. Repeat the same experiment with different value of 4 volts.
35

36. Formula
Natural Convection
1. T average =
5
54321 TTTTT ++++
°C
2. Tmf = mean film temperature =
2
6TTavg +
3. Grash Number Gr =
where g = acceleration due to gravity = 9.81 m/sec2
β = , ΔT = Taverage- T6
T6 = Temperature of direct fluid
γ = kinematic viscosity m2
/sec (From Data book in table for Tmf
o
C)
d = diameter of the fin 1.27 x 10-2
m
4. Heat transfer coefficient h =
where Nu = Nusselt Number
= 1.10 x (Gr x Pr)1/6
for 10-1
< Gr x Pr < 104
36

37. = 0.53 x (Gr x Pr)1/4
for 104
< Gr x Pr < 109
= 0.13 x (Gr x Pr)1/3
for 109
< Gr x Pr < 1012
Ka = Thermal conductivity of air W/mK ( From Data book in table for Tmf
o
C)
37

38. 38

39. 5. m = m-1
Where h = Heat transfer coefficient W/m2
K
P = Circumference of the fin = π × d in metre
K = Thermal conductivity of the material (given specimen), W/mK
kbrass = 110.5 W/mK,
ksteel = 46.5 W/mK
kAl = 232.6 W/mK
A = Cross sectional area of the fin, A= m2
6. The rate of heat transfer from the fin (Q) =
Where, K = thermal conductivity of air (at mean film temperature)
7. Efficiency of the fin =
Where, L = Length of fin = 1.5 cm
8. Predicted temperature at each point
At starting point T1, T1 (Pr)=
T2 (Pr) = + T6
T3 (Pr) = + T6
T4 (Pr) = + T6
T5 (Pr) = + T6
39

40. Where, X2 = 2.5 cm, X3 = 5 cm, X4 = 7.5 cm, X5 = 10 cm, L = 15cm
40
Length of the fin
cm
Temperature in
o
C
Predicted
Observed
Length of the fin cm
Temperatureino
C Predicted
Observed

41. 41

42. Forced Convection
1. T average =
o
C
2. Tmf = mean film temperature = o
C
3. Discharge of air, )(2
a
w
md hgACQ
ρ
ρ
×××××= m3
/sec
Where, Cd = Co-efficient of discharge of air = 0.64
A = Cross sectional area of the orifice =
2
4
d
π
in m2
g = Acceleration due to gravity = 9.81m/s2
hm = Difference in manometer reading in metre
ρw = Density of water = 1000kg/m3
ρa = Diameter of the orifice = 1.8×10-2
metre
4. Velocity of air in duct (VD) at T6
o
C
D
D
A
Q
V = m/sec
Q = Discharge of air in m3
/sec
A = Duct area = 0.15 × 0.10 m2
5. Velocity of air (Vmf) at mean film temperature Tmf
o
C
273
273
6 ×
×
×=
T
T
VV
mf
Dmf m/sec
6. Heat transfer coefficient, h =
Where, d = Diameter of the fin, 1.27 10-2
m
Nu = 0.615 × (Re)0.466
when 40<Re<4000
7.
AK
Ph
m
×
×
= m-1
Where, h = heat transfer coefficient, W/m2
K
P = Circumference of the fin = π × d, m
42

43. K = Thermal conductivity of the material (Given specimen)
A = Cross sectional area of the fin =
2
4
d
π
in m2
43

44. 44

45. 8. The rate of heat transfer from the fin, )tanh()( 21 mlTTKPhQ ×−×××=
Nu = 0.174(Rc)0.618
when 40 < Rc >40000
Re = Reynold’s number =
d = Diameter of the orifice= 1.8 x 10-2
m
γ = Kinematic viscosity (m2
/sec) at Tmf
9. Efficiency of the fin =
Where, L=Length of = 1.5 cm
10. Predicted temperature at each point
At starting point T1, T1=15
T2 (Pr) = + T6
T3 (Pr) = + T6
T4 (Pr) = + T6
T5 (Pr) = + T6
Where, X2 = 2.5 cm, X3 = 5 cm, X4 = 7.5 cm, X5 = 10 cm, L = 15cm
45
Length of the fin
Temperatureino
C
Predicted
Observed

46. 46

47. Temperature Distribution
Result
Thus the temperature distribution along the length of a pinfin in natural and forced
convention is determined and effectiveness is also determined.
47