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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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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Figure 2.1 Schematic diagram of composite wall apparatus
Figure 2.2 Test Setup of Composite Wall Apparatus
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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(
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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
Figure 3.4 Graph of temperature versus length of the rod.
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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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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)
(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 =
(2) Discharge q = Cd A0 aHg.2
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
Do not attempt to alter the equipment as this may cause damage to the whole system.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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.
Experimental set up:
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.
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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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
(i) Film temperature, Tf =2
TaTS . ( C )
Ts = Surface temperature of the cylinder =6
765432 TTTTTT (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 =
(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
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Heat and Mass Transfer Laboratory
Dept. of Mechanical Engineering, MSEC, Bangalore.
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:
Check all the electrical connections.
Do not run the equipment if the voltage is below 180V.
Make sure that heater regulator is at the minimum position before switching on the console.
Do not attempt to alter the equipment as this may cause damage to the whole system.
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.
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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.
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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.
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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)
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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(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 =
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.
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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.
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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.
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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.
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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.
Points for Discussion :
(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.
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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.
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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.
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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.
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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:
Check all the electrical connections.
Do not run the equipment if the voltage is below 180V.
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.
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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.
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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.
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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.
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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
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(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)
CP = CPc = CPh = 4184 (KKg
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
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(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,
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) =
CP = CPc = CPh = 4184 (KKg
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
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(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 _______
Points for Discussion :
(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?
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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.
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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.
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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).
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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).
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Result: Average heat transfer co-efficient for film wise condensation is _______
Average heat transfer co-efficient for film wise condensation is _______
Points for Discussion :
(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.
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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
.
.
Experimental set up:
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.
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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
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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)
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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.
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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.
Experimental set up:
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).
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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.
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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 ___________ ,
Points for Discussion :
(1) List of various precautions observed during experimentation. (2) What is dew factor.
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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
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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.
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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
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Plot a graph of temperature v/s distance at particular time.
Tem
p.
Distance, X (m)
T theoretical
T experimental
Figure 12.3 Graph of temperature versus length of the rod.
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.
Points for Discussion :
(1) Whether the rod can modelled as a lumped system one.Comment on it.
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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