Department of Aeronautical engineering
School of Mechanical engineering
Vel Tech Dr RR & SR Technical University
Course Material
U6AEA23- Heat Transfer
1
U6AEA23
HEAT TRANSFER
UNIT I Heat Conduction
9
Basic Modes of Heat Transfer – One dimensional steady state heat conduction: Composite Medium – Critical
thickness – Effect of variation of thermal Conductivity – Extended Surfaces – Unsteady state. Heat Conduction:
Lumped System Analysis – Heat Transfer in Semi infinite and infinite solids – Use of Transient – Temperature
charts – Application of numerical techniques.
UNIT II
Convective Heat Transfer
9
Introduction – Free convection in atmosphere free convection on a vertical flat plate – Empirical relation in free
convection – Forced convection – Laminar and turbulent convective heat transfer analysis in flows between
parallel plates, over a flat plate and in a circular pipe. Empirical relations, application of numerical techniques in
problem solving.
UNIT III
Radiative Heat Transfer
9
Introduction to Physical mechanism – Radiation properties – Radiation shape factors – Heat exchange between
non – black bodies – Radiation shields.
UNIT IV
Heat Exchangers
9
Classification – Temperature Distribution – Overall heat transfer coefficient, Heat Exchange Analysis – LMTD
Method and E-NTU Method, problems using LMTD and E-NTUmethds.
UNIT V
Heat Transfer Problems In Aerospace Engineering
9
High-Speed flow Heat Transfer, Heat Transfer problems in gas turbine combustion chambers – Rocket thrust
chambers – Aerodynamic heating – Ablative heat transfer, Heat transfer problems in nozzles.
TEXT BOOKS
1.
Yunus A. Cengel., “Heat Transfer – A practical approach”, Second Edition, Tata McGraw-Hill, 2002.
2.
Incropera. F.P.and Dewitt.D.P. “Introduction to Heat Transfer”, John Wiley and Sons – 2002.
REFERENCE BOOKS
1.
Lienhard, J.H., “A Heat Transfer Text Book”, Prentice Hall Inc., 1981.
2.
Holman, J.P. “Heat Transfer”, McGraw-Hill Book Co., Inc., New York, 6th Edn., 1991.
3. Sachdeva, S.C., “Fundamentals of Engineering Heat & Mass Transfer”, Wiley Eastern Ltd., New Delhi,
1981.
4.Mathur, M. and Sharma, R.P. “Gas Turbine and Jet and Rocket Propulsion”, Standard Publishers, New
Delhi 1988.
UNIT – I
Basic Modes of Heat Transfer
2
One dimensional steady state heat conduction:
Composite Medium
Critical thickness
Effect of variation of thermal Conductivity
Extended Surfaces
Unsteady state. Heat Conduction: Lumped System Analysis
Heat Transfer in Semi infinite and infinite solids
Use of Transient Temperature charts
Application of numerical techniques.
CONDUCTION
PART – A
3
1. Define Heat Transfer.
Heat transfer can be defined as the transmission of energy from one region to another
region due to temperature difference.
2. What are the modes of Heat Transfer?
 Conduction
 Convection
 Radiation
3. Define Conduction.
Heat conduction is a mechanism of heat transfer from a region of high temperature to
a region of low temperature within a medium (solid, liquid or gases) or between different
medium in direct physical contact.
In condition energy exchange takes place by the kinematic motion or direct impact of
molecules. Pure conduction is found only in solids.
4. Define Convection.
Convection is a process of heat transfer that will occur between a solid surface and a
fluid medium when they are at different temperatures.
Convection is possible only in the presence of fluid medium.
5. Define Radiation.
The heat transfer from one body to another without any transmitting medium is
known as radiation. It is an electromagnetic wave phenomenon.
6. State Fourier’s Law of conduction.
The rate of heat conduction is proportional to the area measured – normal to the
direction of heat flow and to the temperature gradient in that direction.
dT
dx
dT
Q  - KA
dx
Q - A
where A – are in m2
dT
dx
- Temperature gradient in K/m
K – Thermal conductivity W/mK.
7. Define Thermal Conductivity.
4
Thermal conductivity is defined as the ability of a substance to conduct heat.
8. Write down the equation for conduction of heat through a slab or plane wall.
Heat transfer Q 
Toverall
R
Where
 T = T1 – T2
L
- Thermal resistance of slab
R
KA
L = Thickness of slab
K = Thermal conductivity of slab
A = Area
9. Write down the equation for conduction of heat through a hollow cylinder.
Heat transfer Q 
Toverall
R
Where
 T = T1 – T2
R
r 
1
in  2  thermal resistance of slab
2 LK
 r1 
L – Length of cylinder
K – Thermal conductivity
r2 – Outer radius
r1 – inner radius
10. Write down equation for conduction of heat through hollow sphere.
Heat transfer Q 
Toverall
R
Where
 T = T1 – T2
R
r2  r1
- Thermal resistance of hollow sphere.
4 K (r1r2 )
11. State Newton’s law of cooling or convection law.
Heat transfer by convection is given by Newton’s law of cooling
Q = hA (Ts - T)
Where
A – Area exposed to heat transfer in m2
h - heat transfer coefficient in W/m2K
Ts – Temperature of the surface in K
T - Temperature of the fluid in K.
5
12. Write down the equation for heat transfer through a composite plane wall.
Heat transfer Q 
Toverall
R
Where
 T = Ta– Tb
R
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K3 A hb A
L – Thickness of slab
ha – heat transfer coefficient at inner diameter
hb – heat transfer coefficient at outer side.
13. Write down the equation for heat transfer through composite pipes or cylinder.
Heat transfer Q 
Toverall
R
Where
 T = Ta– Tb
r 
r 
In  2  In  1  L2
r
r
1
1
1
R
  1  2 
.
2 L ha r1
K1
K2
hb r3
14. Write down one dimensional, steady state conduction equation without internal heat
generation.
 2T
0
x 2
15. Write down steady state, two dimensional conduction equation without heat
generation.
 2T  2T

0
x 2 y 2
16. Write down the general equation for one dimensional steady state heat transfer in
slab or plane wall without heat generation.
 2T  2T  2T 1 T



x 2 y 2 z 2  t
17. Define overall heat transfer co-efficient.
The overall heat transfer by combined modes is usually expressed in terms of an
overall conductance or overall heat transfer co-efficient ‘U’.
Heat transfer Q = UA T.
6
18. Write down the general equation for one dimensional steady state heat transfer in
slab with heat generation.
 2T  2T  2T q 1 T


 
x 2 y 2 z 2 K  t
19. What is critical radius of insulation (or) critical thickness.
Critical radius = rc
Critical thickness = rc – r1
Addition of insulating material on a surface does not reduce the amount of heat
transfer rate always. In fact under certain circumstances it actually increases the heat loss up
to certain thickness of insulation. The radius of insulation for which the heat transfer is
maximum is called critical radius of insulation, and the corresponding thickness is called
critical thickness.
20. Define fins (or) Extended surfaces.
It is possible to increase the heat transfer rate by increasing the surface of heat
transfer. The surfaces used for increasing heat transfer are called extended surfaces or
sometimes known as fins.
21. State the applications of fins.
The main application of fins are
1.
2.
3.
4.
Cooling of electronic components
Cooling of motor cycle engines.
Cooling of transformers
Cooling of small capacity compressors
22. Define Fin efficiency.
The efficiency of a fin is defined as the ratio of actual heat transfer by the fin to the
maximum possible heat transferred by the fin.
 fin 
Q fin
Qmax
23. Define Fin effectiveness.
Fin effectiveness is the ratio of heat transfer with fin to that without fin
Fin effectiveness =
Q with fin
Qwithout fin
24. What is meant by steady state heat conduction?
7
If the temperature of a body does not vary with time, it is said to be in a steady state
and that type of conduction is known as steady state heat conduction.
25. What is meant by Transient heat conduction or unsteady state conduction?
If the temperature of a body varies with time, it is said to be in a transient state and
that type of conduction is known as transient heat conduction or unsteady state conduction.
26. What is Periodic heat flow?
In periodic heat flow, the temperature varies on a regular basis.
Example:
1. Cylinder of an IC engine.
2. Surface of earth during a period of 24 hours.
27. What is non periodic heat flow?
In non periodic heat flow, the temperature at any point within the system varies non
linearly with time.
Examples :
1. Heating of an ingot in a furnace.
2. Cooling of bars.
28. What is meant by Newtonian heating or cooling process?
The process in which the internal resistance is assumed as negligible in comparison
with its surface resistance is known as Newtonian heating or cooling process.
29. What is meant by Lumped heat analysis?
In a Newtonian heating or cooling process the temperature throughout the solid is
considered to be uniform at a given time. Such an analysis is called Lumped heat capacity
analysis.
30. What is meant by Semi-infinite solids?
In a semi infinite solid, at any instant of time, there is always a point where the effect
of heating or cooling at one of its boundaries is not felt at all. At this point the temperature
remains unchanged. In semi infinite solids, the biot number value is .
31. What is meant by infinite solid?
A solid which extends itself infinitely in all directions of space is known as infinite solid.
8
In semi infinite solids, the biot number value is in between 0.1 and 100.
0.1 < Bi < 100.
32. Define Biot number.
It is defined as the ratio of internal conductive resistance to the surface convective
resistance.
Bi =
Bi =
Internal conductive resistance
Surface convective resistance
hLL
.
K
33. What is the significance of Biot number?
Biot number is used to find Lumped heat analysis, semi infinite solids and infinite solids
If Bi < 0.1 L  Lumped heat analysis
Bi =   Semi infinite solids
0.1 < Bi < 100  Infinite solids.
34. Explain the significance of Fourier number.
It is defined as the ratio of characteristic body dimension to temperature wave
penetration depth in time.
Fourier Number =
Characteristic body dimension
Temperature wave penetration
depth in time
It signifies the degree of penetration of heating or cooling effect of a solid.
35. What are the factors affecting the thermal conductivity?
1.
2.
3.
4.
5.
Moisture
Density of material
Pressure
Temperature
Structure of material
36. Explain the significance of thermal diffusivity.
The physical significance of thermal diffusivity is that it tells us how fast heat is
propagated or it diffuses through a material during changes of temperature with time.
37. What are Heisler charts?
In Heisler chart, the solutions for temperature distributions and heat flows in plane
walls, long cylinders and spheres with finite internal and surface resistance are presented.
9
Heisler charts are nothing but a analytical solutions in the form of graphs.
PART – B
1. A wall of 0.6m thickness having thermal conductivity of 1.2 w/Mk. The wall is to be
insulated with a material having an average thermal conductivity of 0.3 W/mK. Inner
and outer surface temperatures are 1000 C and 10C. Heat transfer rate is 1400 W/m2
calculate the thickness of insulation.
Given Data
Thickness of wall L1 = 0.6 m
Thermal conductivity of wall K1 = 1.2 W/mK.
Thermal conductivity of insulation K2 = 0.3 W/mK.
Inner surface Temperature
T1 = 1000C + 273 = 1273 K
Outer surface Temperature
T3 = 10C + 273 = 283 K
Heat transfer per unit area Q/A = 1400 W/m2.
Solution:
Let the thickness of insulation be L2
We know
Q
Toverall
R
[From equation (13)] (or) [HMT Data book page No. 34]
Where
 T = Ta– Tb (or) T1 – T3
L
L
L
1
1
R
 1  2  3 
ha A K1 A K 2 A K3 A hb A
Q
[T1  T3 ]
L
L1
L
1
1

 2  3 
ha A K1 A K 2 A K 3 A hb A
Heat transfer coefficient ha, hb and thickness L3 are not given. So neglect that terms.
 Q=

T1  T3 
L1
L
 2
K1 A K 2 A
Q  T1  T3 


A L1 L2

K1 K 2
1273  283
0.6 L2

1.2 0.3
L2  0.0621 m
1400 
10
2. The wall of a cold room is composed of three layer. The outer layer is brick 30cm
thick. The middle layer is cork 20 cm thick, the inside layer is cement 15 cm thick. The
temperatures of the outside air is 25C and on the inside air is -20C. The film coefficient for outside air and brick is 55.4 W/m2K. Film co-efficient for inside air and
cement is 17 W/m2K. Find heat flow rate.
Take
K for brick
= 2.5 W/mK
K for cork
= 0.05 W/mK
K for cement = 0.28 W/mK
Given Data
Thickness of brick L3 = 30 cm = 0.3 m
Thickness of cork L2 = 20 cm = 0.2 m
Thickness of cement L1 = 15 cm = 0.15 m
Inside air temperature Ta = -20C + 273 = 253 K
Outside air temperature Tb = 25C + 273 = 298 K
Film co-efficient for inner side ha = 17 W/m2K
Film co-efficient for outside hb = 55.4 W/m2K
Kbrick = K3 = 2.5 W/mK
Kcork = K2 = 0.05 W/mK.
Kcement = K1 = 0.08 W/mK.
Solution:
Heat flow through composite wall is given by
Q
Toverall
R
[From equation (13)] (or) [HMT Data book page No. 34]
Where
 T = Ta– Tb
L
L
L
1
1
R
 1  2  3 
ha A K1 A K 2 A K3 A hb A
11
Q
[Ta  Tb ]
L
L1
L
1
1

 2  3 
ha A K1 A K 2 A K 3 A hb A
 Q/ A 
 Q/ A 
Ta  Tb 
L
L
L
1
1
 1  2  3 
ha K1 K 2 K 3 hb
253  298
1 0.15 0.2 0.3
1




17 0.28 0.05 2.5 55.4
Q / A  9.5 W / m 2
The negative sign indicates that the heat flows from the outside into the cold room.
3. A wall is constructed of several layers. The first layer consists of masonry brick 20
cm. thick of thermal conductivity 0.66 W/mK, the second layer consists of 3 cm thick
mortar of thermal conductivity 0.6 W/mK, the third layer consists of 8 cm thick lime
stone of thermal conductivity 0.58 W/mK and the outer layer consists of 1.2 cm thick
plaster of thermal conductivity 0.6 W/mK. The heat transfer coefficient on the interior
and exterior of the wall are 5.6 W/m2K and 11 W/m2K respectively. Interior room
temperature is 22C and outside air temperature is -5C.
Calculate
a)
b)
c)
d)
Overall heat transfer coefficient
Overall thermal resistance
The rate of heat transfer
The temperature at the junction between the mortar and the limestone.
Given Data
Thickness of masonry L1 = 20cm = 0.20 m
Thermal conductivity K1 = 0.66 W/mK
Thickness of mortar L2 = 3cm = 0.03 m
Thermal conductivity of mortar K2 = 0.6 W/mK
Thickness of limestone L3 = 8 cm = 0.08 m
Thermal conductivity K3 = 0.58 W/mK
Thickness of Plaster L4 = 1.2 cm = 0.012 m
Thermal conductivity K4 = 0.6 W/mK
Interior heat transfer coefficient ha = 5.6 W/m2K
Exterior heat transfer co-efficient hb = 11 W/m2K
Interior room temperature Ta = 22C + 273 = 295 K
Outside air temperature Tb = -5C + 273 = 268 K.
Solution:
Heat flow through composite wall is given by
12
Q
Toverall
R
[From equation (13)] (or) [HMT Data book page No. 34]
Where
 T = Ta– Tb
R
L
L
L
L
1
1
 1  2  3  4 
ha A K1 A K 2 A K 3 A K 4 A hb A
Q
Ta  Tb
L
L1
L2
L
1
1


 3  4 
ha A K1 A K 2 A K 3 A K 4 A hb A
Q/ A 
295  268
1 0.20 0.03 0.08 0.012 1





5.6 0.66 0.6 0.58
0.6 11
Heat transfer per unit area Q/A = 34.56 W/m2
We know
Heat transfer Q = UA (Ta – Tb) [From equation (14)]
Where U – overall heat transfer co-efficient
U 
Q
A  (Ta  Tb )
U 
34.56
295  268
Overall heat transfer co - efficient U = 1.28 W/m2 K
We know
Overall Thermal resistance (R)
R
L
L
L
L
1
1
 1  2  3  4 
ha A K1 A K 2 A K3 A K 4 A hb A
For unit Area
R
=
1 L1 L2 L3 L4 1





ha K1 K 2 K 3 K 4 hb
1 0.20 0.03 0.08 0.012 1





56 0.66 0.6 0.58
0.6 11
R 0.78 K / W
Interface temperature between mortar and the limestone T3
Interface temperatures relation
13
Q
Ta  T1 T1  T2 T2  T3 T3  T4 T4  T5 T5  Tb





Ra
R1
R2
R3
R4
Rb
Q
Ta  T1
Ra
Q=
295-T1
1/ ha A
Q/ A

1 
 Ra 

h
a A

295  T1
1/ ha
295  T1
1/ 5.6
 T1  288.8 K
 34.56 
Q
Q
T1  T2
R1
288.8  T2
L1
K1 A
Q/ A 
 34.56 

L1 
 R1 

k1 A 

288.8  T2
L1
K1
288.8  T2
0.20
0.66
 T2  278.3 K
Q =
Q
T2  T3
R2
278.3  T3
L2
K2 A
Q/ A 
 34.56 

L2 
 R2 

K2 A 

278.3  T3
L2
K2
278.3  T3
0.03
0.6
 T3  276.5 K
Temperature between Mortar and limestone (T3 is 276.5 K)
14
4. A steam to liquid heat exchanger area of 25.2 m2 is constructed with 0.5cm nickel and
0.1 cm plating of copper on the steam sides. The resistivity of a water-scale deposit on
the steam side is 0.0015 K/W. The steam and liquid surface conductance are 5400
W/m2K ad 560 W/m2K respectively. The heated steam is at 110C and heated liquid is
at 70C.
Calculate
1. Overall steam to liquid heat transfer co-efficient
2. Temperature drop across the scale deposit
Take
K(Copper) = 350 W/mK and K (Nickel) = 55 W/mK.
Given
Area A = 25.2 m2
Thickness of Nickel L1 = 0.5 cm = 0.5  10-2 m
Thickness of Copper L2 = 0.1 cm = 0.1  10-2 m
Resistivity of scale R3 = 0.0015 K/W
Liquid surface conductance ha = 560 W/m2K
Steam surface conductance hb = 5400 W/m2K
Steam temperature Tb = 110C + 273 = 383 K
Liquid temperature Ta = 70C + 273 = 343 K
K2 (Copper) = 350 W/mK
K1 (Nickel) = 55 W/mK
Solution:
Heat transfer through composite wall is given by
Q
Toverall
R
[From equation (13)] (or) [HMT Data book page No. 34]
Where
 T = Ta– Tb = 343 – 383 = -40 K
R
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K 3 A hb A
= R a  R1  R2  R3  Rb
R3 value is given, R3 = 0.0015 K/W
R
=
L
L
1
1
 1  2  0.0015 
ha A K1 A K 2 A
hb A
1
0.5  10-2 0.1102
1
+

 0.0015 
560  25.2
55  25.2 350  25.2
5400  25.2
 R  1.58  103 K / W
Q
Toverall
R
15
Q
40
1.58 103
Heat transfer Q = - 25.2  103 W
[-ve sign indicates that the heat flows from, outside to inside]
we know
Heat transfer Q = UA (Ta – Tb) [From equation No. (14)]
 U=

Q
A(Ta  Tb )
25.2  103
25.2  (40)
Overall heat transfer co - efficient U = 25 W/m2 K
Temperature drop (T3 – T4) across the scale is given by
Q
T
Rscale
 T= T3  T4 
T
0.0015
 T  37.8C
25.2 103 
5. A surface wall is made up of 3 layers one of fire brick, one of insulating brick and one
of red brick. The inner and outer surface temperatures are 900C and 30C
respectively. The respective co-efficient of thermal conductivity of the layers are 1.2,
0.14 and 0.9 W/mK and the thickness of 20cm, 8 cm and 11 cm. Assuming close bonding
of the layers at the interfaces. Find the heat loss per square meter and interface
temperatures.
Given
Inner temperature T1 = 900C + 273 = 1173 K
Outer temperature T4 = 30C + 273 = 303 K
Thermal conductivity of fire brick K1 = 1.2 W/mK
Thermal conductivity of insulating brick K2 = 0.14 W/mK
Thermal conductivity of red brick K3 = 0.9 W/mK
Thickness of fire brick L1 = 20 cm = 0.2 m
Thickness of insulating brick L2 = 8 cm = 0.08 m
Thickness of red brick L3 = 11 cm = 0.11 m
Solution:
(i)
Heat loss per square metre (Q/A)
Heat transfer Q 
Toverall
R
[From equation (13)] (or)
Where
 T = Ta– Tb = T1 – T4
16
[HMT Data book page No. 34]
R
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K 3 A hb A
 Q=
T1  T4
L
L1
L
1
1

 2  3 
ha A K1 A K 2 A K 3 A hb A
[Convective heat transfer co-efficient ha, hb are not given.
So neglect that terms]
T1  T4
L
L1
L
 2  3
K1 A K 2 A K 3 A
 Q=
T1  T4
L1 L2 L3


K1 K 2 K 3
Q/ A 

1173  303
0.2 0.08 0.11


1.2 0.14 0.9
Q / A  1011.2546 W / m 2
(ii) Interface temperatures (T2 and T3)
We know that, interface temperatures relation
Q
T1  T4 T1  T2 T2  T3 T3  T4



......( A)
R
R1
R2
R3
( A)  Q 
T1  T2
R1
where
R1 
L1
K1 A
Q
Q/A =
T1  T2
L1
K1 A
T1  T2
L1
K1
1011.2546 
1173  T2
0.2
1.2
T2  1004.457 K
Similarly,
Q
T2  T3
R2
where
17
R2 
L2
K2 A
Q
T2  T3
L2
K2 A
 Q/A =
T2  T3
L2
K2
1004.457  T3
0.08
0.14
T3  426.597 K
1011.2546 
6. A furnace wall made up of 7.5 cm of fire plate and 0.65 cm of mild steel plate. Inside
surface exposed to hot gas at 650C and outside air temperature 27C. The convective
heat transfer co-efficient for inner side is 60 W/m2K. The convective heat transfer coefficient for outer side is 8W/m2K. Calculate the heat lost per square meter area of the
furnace wall and also find outside surface temperature.
Given Data
Thickness of fire plate L1 = 7.5 cm = 0.075 m
Thickness of mild steel L2 = 0.65 cm = 0.0065 m
Inside hot gas temperature Ta = 650C + 273 = 923 K
Outside air temperature Tb = 27C + 273 = 300K
Convective heat transfer co-efficient for
Inner side ha = 60W/m2K
Convective heat transfer co-efficient for
Outer side hb = 8 W/m2K.
Solution:
(i)
Heat lost per square meter area (Q/A)
Thermal conductivity for fire plate
K1 = 1035  10-3 W/mK
[From HMT data book page No.11]
Thermal conductivity for mild steel plate
K2 = 53.6W/mK
[From HMT data book page No.1]
Heat flow Q 
Toverall
R
Where
 T = Ta– Tb
18
R
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K 3 A hb A
Ta  Tb
L
L1
L
1
1

 2  3 
ha A K1 A K 2 A K 3 A hb A
 Q=
[The term L3 is not given so neglect that term]
 Q=
Ta  Tb
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K 3 A hb A
The term L 3 is not given so neglect that term]
Q=
Q/ A 
Ta  Tb
L1
L
1
1

 2 
ha A K1 A K 2 A hb A
923  300
1 0.075 0.0065 1



60 1.035
53.6
8
Q / A  2907.79 W / m 2
(ii)
Outside surface temperature T3
We know that, Interface temperatures relation
Q
Ta  Tb Ta  T1 T1  T2 T2  T3 T3  Tb




......( A)
R
Ra
R1
R2
Rb
( A)  Q 
T3  Tb
Rb
where
Rb 
1
hb A
Q
 Q/A =
T3  Tb
1
hb A
T3  Tb
1
hb
T3  300
1
8
T3  663.473 K
 2907.79 
19
7. A mild steel tank of wall thickness 10mm contains water at 90C. Calculate the rate of
heat loss per m2 of tank surface area when the atmospheric temperature is 15C. The
thermal conductivity of mild steel is 50 W/mK and the heat transfer co-efficient for
inside and outside the tank is 2800 and 11 W/m2K respectively. Calculate also the
temperature of the outside surface of the tank.
Given Data
Thickness of wall L1 = 10mm = 0.01 m
Inside temperature of water Ta = 90 + 273 = 363 K
Atmospheric temperature Tb = 15C + 273 = 288 K
Heat transfer co-efficient for inside ha = 2800 W/m2K
Heat transfer co-efficient for outside hb = 11 W/m2K
Thermal conductivity of mild steel K = 50 W/mK
To find
Rate of heat loss per m2 of tank surface area (Q/A)
Tank outside surface temperature (T2)
(i)
(ii)
Solution:
Heat loss
Q
Toverall
R
Where
 T = Ta– Tb
R
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K3 A hb A
[L3, L2 not given so neglect L2 and L3 terms]
 R=
Q=
Q/A =
Q/ A
L
1
1
 1 
ha A K1 A hb A
Ta  Tb
L
1
1
 1 
ha A K1 A hb A
Ta  Tb
1 L1 1


ha K1 hb
363  288
1
0.01 1


2800 50 11
Q / A  819.9 W / m 2
We know
20
Q
Ta  Tb Ta  T1 T1  T2 T2  Tb



......( A)
R
Ra
R1
Rb
( A)  Q 
Q
 Q/A =
Ta  T1
Ra
where R a 
1
ha A
363  T1
1
ha A
363 - T1
1
ha
363  T1
1
2800
T1  362.7 K
 819.9 
( A)  Q 
Q
 Q/A =
T1  T2
R1
where R1 
L1
K1 A
T1  T2
L1
K1 A
T1  T2
L1
K1
362.7  T2
0.01
50
T2  362.5 K
 819.9 
8. A composite slab is made of three layers 15 cm, 10 cm and 12 cm thickness
respectively. The first layer is made of material with K = 1.45 W/mK, for 60% of the
area and the rest of material with K = 2.5 W/mK. The second layer is made of material
with K = 12.5 W/mK for 50% of area and rest of material with K = 18.5 W/mK. The
third layer is made of single material of K = 0.76 W/mK. The composite slab is exposed
on one side to warn at 26C and cold air at -20C. The inside heat transfer co-efficient is
15 W/m2K. The outside heat transfer co-efficient is 20 W/m2K determine heat flow rate
and interface temperatures.
Given Data
L1 = 15 cm = 0.15 m
L2 = 10 cm = 0.1 m
L3 = 12 cm = 0.12 m
K1a = 1.45 W/mK,
K1b = 2.5 W/mK
K2a = 12.5 W/mK
K2b = 18.5 W/mK
A1a = .60
A1b = .40
A2a = .50
A2b = .50
21
K3 = 0.76 W/mK
Ta = 26C + 273 = 299 K
Tb = -20C + 273 = 253 K
ha = 15 W/m2K
hb = 20 W/m2K
Solution :
Heat flow Q 
Toverall
R
Where
 T = Ta– Tb
L
L
11 L
1
R   1  2  3  
A  ha K1 K 2 K3 hb 
L
L
L
1
1
=
 1  2  3 
Aa ha A1 K1 A2 K 2 A3 K 3 Ab hb
R = R a  R1  R2  R3  Rb
Q =
Ta  Tb
....( A)
R a  R1  R2  R3  Rb
Where
1
1
Ra 

Aa ha 115
Ra  0.066 K / W
R1 
R1a  R1b
R1a  R1b
R1a 
.....(1)
L1
0.15

K1a  A1a 1.45  0.6
R1a  0.1724 K/W
R1b 
L1
0.15

K1b  A1b 2.5  0.4
R1b  0.15 K/W
Substitute R1a and R1b value in (1)
0.1724  0.15
0.1724  0.15
R1  0.08 K / W
(1)  R1 
Similarly,
22
R2 
R2 a  R2b
R2 a  R2b
R 2a 
.....(2)
L2
0.1

K 2 a  A2 a 12.5  0.5
R2 a  0.016 K/W
R 2b 
L2
0.1

K 2b  A2b 18.5  0.5
R2b  0.0108 K/W
0.016  0.0108
0.016  0.0108
R2  0.0064 K / W
(2)  R2 
R3 
L3
0.12

A3 K 3 1 0.76
 A3  1m 2 
R3  0.15789 K/W
Rb 
1
1

Aa hb 1 20
 A b  1m 2 
Rb  0.05 K/W
299  253
0.066  0.08  0.0064  0.15789  0.05
Q  127.67 W
( A)  Q 
(ii) Interface temperatures (T1, T2, T3 and T4)
We know
Q
Ta  Tb Ta  T1 T1  T2 T2  T3



R
Ra
R1
R2

(B)  Q 
T3  T4
R3
=
T4  Tb
.....(B)
Rb
Ta  T1
Ra
299  T1
0.066
299  T1
127.67 
0.066
=
T1  290.57 K
23
T1  T2
R1
(B)  Q 
290.57  T2
0.08
127.67 
T2  280.35 K
(B)  Q 
T2  T3
R2
280.35  T3
0.0064
T3  279.532 K
127.67 
(B)  Q 
T3  T4
R3
279.532  T4
0.15789
T4  259.374 K
127.67 
9. An external wall of a house is made up of 10 cm common brick (K = 0.7 W/mK)
followed by a 4 cm layer of zibsum plaster (K = 0.48 W/mK). What thickness of loosely
packed insulation (K = 0.065 W/mK) should be added to reduce the heat loss through
the wall by 80%.
Given Data
Thickness of brick L1 = 10 cm = 0.1 m
Thermal conductivity of brick K1 = 0.7 W/mK
Thickness of zibsum L2 = 4 cm = 0.04 m
Thermal conductivity of zibsum K2 = 0.48 W/mK
Thermal conductivity of insulation K3 = 0.065 W/mK
To find
Thickness of insulation to reduce the heat loss through the wall by 80% (L3)
Solution:
Heat flow rate Q 
Toverall
R
[From HMT data book Page No.34]
Where
R
1  1 L1 L2 L3 1 


 
 
A  ha K1 K 2 K 3 hb 
[The terms ha , hb is not given so neglect that terms].
R =
1  L1 L2 L3 

 

A  K1 K 2 K 3 
24
Considering two slabs
T
Q=
[Assume heat transfer (Q) = 100 W]
L1 L2

K1 K 2
T
0.1 0.04

0.7 0.48
T  22.619 K
100 =

A = 1m2 
Heat loss is reduced by 80% due to insulation, so heat transfer is 20 W.
T

1 L1 L2 L3 

 

A  K1 K 2 K 3 
22.619
20 =
L 
1  0.1 0.04

 3
1  0.7 0.48 0.065 
Q=
[ A = 1m2 ]
L3  0.0588 m
10. A furnace wall consists of steel plate of 20 mm thick, thermal conductivity 16.2
W/mK lined on inside with silica bricks 150 mm thick with conductivity 2.2 W/mK and
on the outside with magnesia brick 200 mm thick, of conductivity 5.1 W/mK. The inside
and outside surfaces of the wall are maintained at 650C and 150C respectively.
Calculate the heat loss from the wall per unit area. If the heat loss is reduced to
2850W/m2 by providing an air gap between steel and silica bricks, find the necessary
width of air gap if the thermal conductivity of air may be taken as 0.030 W/mK.
Given Data
Steel plate thickness L1 = 20 mm = 0.02 m
Thermal conductivity of steel K1 = 16.2 W/mK
Thickness of the silica L2 = 150 mm = 0.150 m
Thermal conductivity of silica K2 = 2.2 W/mK
Thickness of the magnesia L3 = 200 mm = 0.2 m
Thermal conductivity of magnesia K3 = 5.1 W/mK
Inner surface temperature T1 = 650C + 273 = 923
Outer surface temperature T4 = 150C + 273 = 423 K
Heat loss reduced due to air gap is 2850 W/m2
Thermal conductivity of the air gap Kair = 0.030 W/mK
Solution :
Heat transfer through composite wall is given by [without considering air gap]
25
Q
T
R
Where
 T = T1– T4
R
L
L
L
1
1
 1  2  3 
ha A K1 A K 2 A K 3 A hb A
 Q=
T1  T4
L
L1
L
1
1

 2  3 
ha A K1 A K 2 A K 3 A hb A
Neglecting unknown terms (ha and hb)
Q=
T1  T4
L
L1
L
 2  3
K1 A K 2 A K 3 A
923  423
0.02
0.150
0.2


16.2  1 2.2  1 5.11
500
Q=
0.1086
 A = 1m2 
Q
Q  4602.6 W / m 2
Heat loss is reduced to 2850 W/m2 due to air gap. So the new thermal resistance is
T1
Rnew
T T
 2850  1 4
Rnew
923  423
Rnew 
2850
Rnew  0.1754 K/W
Q
Thermal resistance of air gap
Rair = Rnew – R
= 0.1754 – 0.1086


L
L1
L
 2  3  0.1086
 R
K1 A K 2 A K3 A


Rair  0.066 K/W
We know
Rair 
Lair
K air  A
0.066 
Lair
0.030 1
 A = 1m 2 
 Lair  1.98 103 m
Thickness of the air gap = 1.98  10-3 m
26
11. A thick walled tube of stainless steel [K = 77.85 kJ/hr mC] 25 mm ID and 50 mm
OD is covered with a 25 mm layer of asbestos [K = 0.88 kJ/hr mC]. If the inside wall
temperature of the pipe is maintained at 550C and the outside of the insulator at 45C.
Calculate the heat loss per meter length of the pipe.
Given Data
Inner diameter of steel d1 = 25 mm
Inner radius r1 = 12.5 mm  0.0125 m
Outer diameter D2 = 50 mm
Outer radius r2 = 25 mm  0.025 m
Radius r3 = r2 + 25 mm = 50 mm  0.05 m
Thermal conductivity of stainless steel
K1 = 77.85 kJ/hr mC =
77.85
3600
kJ/sec mC
= 0.0216 kJ/sec mC  0.0216 kW/mC
K1  21.625 W / mC
Similarly,
Thermal conductivity of asbestos K2 = 0.88 kJ/hr mC
K 2  0.244 W / mC
Ta  550C
Tb  45C
Solution :
Heat flow through composite cylinder is given by
Q
Toverall
R
[From equation No.(19) or HMT
data book Page No.35]
Where
 T = Ta– Tb


r 
r 
In  2  In  3 


r
r
1  1
1 
R
  1  2
2 L  h a r1
K1
K2
hb r3 




Convective heat transfer co-efficient are not given so neglect ha and hb terms.
27
Q =
Ta  Tb
  r2 
 r3  
 In   In   
r
1   r1 
  2 
2 L  K1
K2 




Ta  Tb
 Q/L =
  r2 
 r3  
 In   In   
r
1   r1 
  2 
2  K1
K2 




550 - 45
Q/L =
  0.025 
 0.05  
In
In 
1   0.0125 
0.025  


 
2  21.625
0.244 


Q / L  1103.9 W/m
12. A steel tube (K = 43.26 W/mK) of 5.08 cm inner diameter and 7.62 cm outer
diameter is covered with 2.5 cm layer of insulation (K = 0.208 W/mK) the inside surface
of the tube receivers heat from a hot gas at the temperature of 316C with heat transfer
co-efficient of 28 W/m2K. While the outer surface exposed to the ambient air at 30C
with heat transfer co-efficient of 17 W/m2K. Calculate heat loss for 3 m length of the
tube.
Given
Steel tube thermal conductivity K1 = 43.26 W/mK
Inner diameter of steel d1 = 5.08 cm = 0.0508 m
Inner radius r1 = 0.0254 m
Outer diameter of steel d2 = 7.62 cm = 0.0762 m
Outer radius r2 = 0.0381 m
Radius r3 = r2 + thickness of insulation
Radius r3 = 0.0381 + 0.025 m
r3 = 0.0631 m
Thermal conductivity of insulation K2 = 0.208 W/mK
Hot gas temperature Ta = 316C + 273 = 589 K
Ambient air temperature Tb = 30C + 273 = 303 K
Heat transfer co-efficient at inner side ha = 28 W/m2K
Heat transfer co-efficient at outer side hb = 17 W/m2K
Length L = 3 m
Solution :
Heat flow Q 
Toverall
R
[From equation No.(19) or HMT
data book Page No.35]
Where
28
 T = Ta– Tb
 1
r 
r  1
r  1
1
1 

In  2  
In  3  
In  4  


2 L  h a r1 K1  r1  K 2  r2  K 3  r3  hb r4 
Ta  Tb
Q =
r 
r  1
r  1
1  1
1
1 

In  2  
In  3  
In  4  


2 L  h a r1 K1  r1  K 2  r2  K 3  r3  hb r4 
R
1
[The terms K3 and r4 are not given, so neglect that terms]
Q =
Ta  Tb
 1
r  1
r 
1
1 

In  2  
In  3  


2 L  h a r1 K1  r1  K 2  r2  hb r3 
1
Q =
589 - 303
1

1
1
1
1

 0.0381 
 0.0631 

In 
+
In 




2  3  28  0.0254 43.26  0.0254  0.208  0.0381  17  0.0631 
Q  1129.42 W
Heat loss Q = 1129.42 W.
13. A hollow sphere (K = 65 W/mK) of 120 mm inner diameter and 350 mm outer
diameter is covered 10 mm layer of insulation (K = 10 W/mK). The inside and outside
temperatures are 500C and 50C respectively. Calculate the rate of heat flow through
this sphere.
Given
Thermal conductivity of sphere K1 = 65 W/mK
Inner diameter of sphere d1 = 120 mm
Radius r1 = 60 mm = 0.060 m
Outer diameter of sphere d2 = 350 mm
Radius r2 = 175 mm = 0.175 m
Radius r3 = r2 + thickness of insulation
R3 = 0.175 + 0.010
r3  0.185m
Thermal conductivity of insulation K2 = 10 W/mK
Inside temperature Ta = 500C + 273 = 773 K
Outside temperature Tb = 50C + 273 = 323 K
Solution:
29
Heat loss through hollow sphere is given by
Q
Toverall
R
[From equation No.(19) or HMT
data book Page No.34 & 35]
Where
 T = Ta– Tb
1  1
1 1 1  1  1 1 
1 



  
  
2
4  h a r1
K1  r1 r2  K 2  r2 r3  hb r32 
Ta  Tb
Q =

1
1
1 1 1  1  1 1 
1 



  
  
2
4  h a r1
K1  r1 r2  K 2  r2 r3  hb r32 
R
ha, hb not given so neglect that terms.
Ta  Tb
Q
 1  1 1  1  1 1 
       
 K1  r1 r2  k2  r2 r3  
773 -323

1 1  1
1  1  1
1 



4  65  0.060 0.175  10  0.175 0.185  
1
4
 Q = 28361 W
Heat transfer = Q = 28361 W
Radius r3 = r2 + thickness of insulation
= 0.0455 + 90  10-3 m
r3 = 0.1355 m
Radius r4 = r3 + thickness of insulation
= 0.1355 + 40  10-3 m
r4 = 0.1755 m
Thermal conductivity of pipe K1 = 47 W/mK
Thermal conductivity of insulation (I) K2 = 0.5 W/mK
Thermal conductivity of insulation (II) K3 = 0.25 W/mK
Outside temperature T4 = 20C + 273 = 293 K
Solution :
Heat flow through composite cylinder is given by
Q
Toverall
R
[From equation No.(19) or HMT
data book Page No.35]
Where
 T = Ta– Tb (or) T1 –T4
30


r 
r 
r 
In  2  In  3  In  4 


r
r
r
1  1
1 
R
  1  2  3
2 L  h a r1
K1
K2
K3 hb r4 




T1  T4
Q =


r 
r 
r 
In  2  In  3  In  4 


r
r
r
1  1
1 
  1  2  3
2 L  h a r1
K1
K2
K3
hb r4 




Heat transfer coefficients ha,h are not given.
b
So neglect that terms.
 Q=
T1  T4
  r2 
 r4  
 r3 
 In   In   In   
r
r
1   r1 
  2    3 
2 L  K1
K2
K3 




523 - 293
Q =
  0.0455 
 0.1355 
 0.1755  
In
In 
In 


1   0.040 
0.0455
   0.1355  

 
2 L 
47
0.5
0.25



 Q / L  448.8 W/m
Heat transfer Q/L = 448.8 W/m.
14. A hollow sphere has inside surface temperature of 300C and then outside surface
temperature of 30C. If K = 18 W/mK. Calculate (i) heat lost by conduction for inside
diameter of 5 cm and outside diameter of 15 cm (ii) heat lost by conduction, if equation
for a plain wall area is equal to sphere area.
Given Data :
T1 = 300C + 273 = 573 K
T2 = 30C + 273 = 303 K
K1 = 18 W/mK
d1 = 5 cm = 0.05 m
r1 = 0.025 m
d2 = 15 cm = 0.15 m
r2 = 0.075 m
31
Solution:
(i) Heat lost (Q)
Heat flow Q 
Toverall
R
[From HMT data book Page
No.34 & 35]
Where
 T = Ta– Tb (or) T1 – T2
 1
1 1 1 
1 



  
2
K1  r1 r2  hb r2 2 
 h a r1
T1  T2
Q =
1  1
1 1 1 
1 



  
2
4  h a r1
K1  r1 r2  hb r2 2 
R
1
4
[The terms ha, hb not given so neglect that terms].
T1  T2
Q
 1  1 1 
   
 K1  r1 r2  
573 -303
 Q=
1 1  1
1 

4 18  0.025 0.075  
1
4
 Q = 2290.22 W
(ii) Heat loss (If the area is equal to the plain wall area) Q1
L = r2 – r1
= 0.075 – 0.025
L  0.05 m
A1  A2
2
4 r12  4 r2 2
=
2
A

A = 4 r 2 
A  2 (r 21  r 2 2 )
We know
T
Q1 
R
32
T1  T2
L
KA
T1  T2
Q1 
L
K1  2  r12  r 2 2  
573 - 303
Q1 
0.05
18  2 (0.0252  0.0752 )
Q1 



R=
L

for plain wall
KA

Q1  3817.03W
Derive an expression of Critical Radius of Insulation For A Cylinder.
Consider a cylinder having thermal conductivity K. Let r1 and r0 inner and outer radii
of insulation.
Heat transfer Q 
Ti  T
r 
In  0 
 r1 
2 KL
[From equation No.(3)]
Considering h be the outside heat transfer co-efficient.
Q =
Ti  T
r 
In  0 
 r1   1
2 KL A 0h
Here A 0  2 r0L
Q
Ti  T
r 
In  0 
 r1   1
2 KL 2 r0Lh
To find the critical radius of insulation, differentiate Q with respect to r0 and equate it
to zero.
33
 1
1 
0  (Ti  T ) 


2 KLr0 2 hLr0 2 
dQ



dr0
r 
1
1
In  0  
2 KL  r1  2 hLr0
since (Ti  T )  0

1
1

0
2 KLr0 2 hLr0 2
 r0 
K
 rc
h
15. A wire of 6 mm diameter with 2 mm thick insulation (K = 0.11 W/mK). If the
convective heat transfer co-efficient between the insulating surface and air is 25 W/m2L,
find the critical thickness of insulation. And also find the percentage of change in the
heat transfer rate if the critical radius is used.
Given Data
d1= 6 mm
r1 = 3 mm = 0.003 m
r2 = r1 + 2 = 3 + 2 = 5 mm = 0.005 m
K = 0.11 W/mK
hb = 25 W/m2K
Solution:
1. Critical radius rc 
rc 
K
h
[From equation No.(21)]
0.11
 4.4  103 m
25
rc  4.4  103 m
Critical thickness = rc – r1
 4.4  103  0.003
 1.4  103 m
Critical thickness t c = 1.4  10-3 (or) 1.4 mm
2. Heat transfer through an insulated wire is given by
Ta  Tb
Q1 
  r2 

 In  

1   r1 
1 

2 L  K1
hbr2 




34
From HMT data book Page No.35
=
Q1 =
2 L (Ta  Tb )
  0.005 

 In  0.003 

1

 

25  0.005 
 0.11


2 L (Ta  Tb )
12.64
Heat flow through an insulated wire when critical radius is used is given by
Q2 
Ta  Tb
  rc 

 In  

1   r1 
1 

2 L  K1
hbrc 




r2  rc 
2 L (Ta  Tb )
 4.4  10 3 
In 

1
 0.003  
0.11
25  4.4  10 3
2 L (Ta  Tb )
Q2 =
12.572
=
 Percentage of increase in heat flow by using
Q2  Q1
 100
Q1
1
1

 100
 12.57 12.64
1
12.64
 0.55%
Critical radius =
Internal Heat Generation – Formulae used
For plane wall :
qL
1. Surface temperature Tw  T 
2h
qL2
2. Maximum temperature Tmax  Tmax 
8K
35
where
T - Fluid temperature, K
q - Heat generation, W/m3
L – Thickness, m
h - Heat transfer co-efficient, W/m2K
K – Thermal conductivity, W/mK.
For Cylinder
Q
1. Heat generation q 
V
2. Maximum temperature Tmax  Tw 
3. Surface temperature Tw  T 
qr 2
4K
rq
2h
Where
V – Volume - r2 L
r – radius – m
For sphere
qr 2
1. Temperature at the centre Tc  Tw 
6K
16. A current of 200 A is passed through a stainless steel wire (K = 19 W/mK) 3 mm in
diameter. The resistivity of the steel may be taken as 70  cm and the length of the
wire is submerged in a liquid at 110C with heat transfer co-efficient h = 4 kW/m2C.
Calculate the centre temperature of the wire.
Given
Current A = 200 A
Thermal conductivity K = 19 W/mK
Diameter d = 3 mm = 3  10-3 m
Resistivity = 70  - cm
Liquid temperature Tw = 110C + 273 = 383 K
Heat transfer co-efficient h = 4 kW/m2C
= 4  10-3 W/m2C
Solution:
The maximum temperature in the wire occurs at the centre.
qr 2
Tmax  Tc  Tw 
........(A) [From Equation No.12]
4K
36
Resistivity  Length
Area
-6
2
70  10  10  1
Re sis tance of wire R =
=

3  10 
4
3
2
R  0.099 
We know that
Q = I2R
= (200)2  (0.099)
Q = 3960 W
Q
3960
Heat generated q  

V
d2  L
4
3960
q
  3  103 2  1
4
q  560  106 W / m3
Substituting q value in Equation (A)
Tmax
560  106  (1.5  103 )2
 Tc  383 
4  19
Tc  399.5 K
17. A sphere of 100 mm diameter, having thermal conductivity of 0.18 W/mK. The outer
surface temperature is 8C and 250 W/m2 of energy is released due to heat source.
Calculate
1. Heat generated
2. Temperature at the centre of the sphere.
Given
Diameter of sphere d = 100 mm
r = 50 mm = 0.050 m
Thermal conductivity K = 0.18 W/mK
Surface temperature Tw = 8C + 273 = 281 K
Energy released Q = 250 W/m2
Solution:
Heat generated q 
Q
V
37
 q/ A 
Q/ A
V
 q/ A 
Q/ A
V
 q/ A 
250
4 / 3 r 3

Here Q/A = 250 W/m 2 
q
250

2
4 r
4 / 3 r 3
q

Here Q/A = 250 W/m 2 
250  4    (0.050)2
4 / 3  (0.50)3
q = 15,000 W/m3
Temperature at the centre of the sphere
qr 2
Tc  Tw 
[From Equation No.16]
6K
15000  (0.050)2
= 281 +
6  0.18
Tc  315.7 K
18. One end of the long solid rod of 50 mm diameter is inserted into a furnace with the
other end is projecting the atmosphere at 25C. Once the steady state is reached, the
temperature of the rod is measured at two points 20 cm apart are found to be 150C
and 100C. The convective heat transfer co-efficient between the rod and the
surrounding air is 30 W/m2K. Calculate the thermal conductivity of the rod material.
Given Data:
Atmospheric Temperature T = 25C + 273 = 298 K
Distance x = 20 cm = 0.20 m
Base temperature Tb = 150C + 273 = 423 K
Intermediate temperature T = 100C + 273 = 373 K
Heat transfer co-efficient h = 30 W/m2K.
Solution:
Since the rod is long, it is treated as long fin. So, temperature distribution
T  T
[From HMT data book (CPK)
 e mx
Tb  T
Page No.41]
38
373 - 298
 e m(0.20)
423 - 298
 0.6 = e-m (0.20)
 In (0.6)= -m  (0.20)

 - 0.51 = -m  (0.20)
m = 2.55 m-1
We know that,
hP
m
KA
[From HMT data book
(CPK) Page No.41]
hP
.............(A)
KA
h – heat transfer co-efficient = 30 W/m2K
P – Perimeter = d =   0.050
2.55 =
P  0.157 m
A  Area 
=

4
d2

4
(0.050)2
A  1.96  103 m2
30  0.157
K  1.96  10 3
30  0.157
 6.50 =
K  1.96  10 -3
(A)  2.55 
K = 369.7 W/mK
19. An aluminium alloy fin of 7 mm thick and 50 mm long protrudes from a wall, which
is maintained at 120C. The ambient air temperature is 22C. The heat transfer
coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mK
respectively. Determine
1. Temperature at the end of the fin.
2. Temperature at the middle of the fin.
3. Total heat dissipated by the fin.
Given
Thickness t = 7mm = 0.007 m
Length L= 50 mm = 0.050 m
Base temperature Tb = 120C + 273 = 393 K
39
Ambient temperature T = 22 + 273 = 295 K
Heat transfer co-efficient h = 140 W/m2K
Thermal conductivity K = 55 W/mK.
Solution:
Length of the fin is 50 mm. So, this is short fin type problem. Assume end is
insulated.
We know
Temperature distribution [Short fin, end insulated]
T  T cos h m [L -x]

.......(A)
Tb  T
cos h (mL)
[From HMT data book Page No.41]
(i) Temperature at the end of the fin, Put x = L
T - T cos h m [L-L]

Tb  T
cos h (mL)
(A) 
T - T
1

Tb  T cos h (mL)

...(1)
where
hP
KA
P = Perimeter = 2  L (Approx)
= 2  0.050
m=
P = 0.1 m
A – Area = Length  thickness = 0.050  0.007
A  3.5  104 m2
 m=

hP
KA
140  0.1
55  3.5  104
m  26.96
40
(1)

T - T
1

Tb  T cos h (26.9  0.050)

T - T
1

Tb  T 2.05
T - 295
1

393 - 295 2.05
 T - 295 = 47.8

 T = 342.8 K
Temperature at the end of the fin Tx L  342.8 K
(ii) Temperature of the middle of the fin,
Put x = L/2 in Equation (A)
T - T cos hm [L-L/2]
(A) 

Tb  T
cos h (mL)
0.050 

cos h 26.9 0.050 T - T
2 



Tb  T
cos h 26.9  (0.050) 
T- 295
1.234

393 - 295 2.049
T - 295

 0.6025
393 -295

T  354.04 K
Temperature at the middle of the fin
Tx L / 2  354.04 K
(iii) Total heat dissipated
[From HMT data book Page No.41]
 Q = (hPKA)1/2 (Tb  T )tan h (mL)
 [140  0.1  55  3.5  10-4 ]1/ 2  (393  295)
 tan h (26.9  0.050)
Q = 44.4 W
20. Ten thin brass fins (K = 100 W/mK), 0.75 mm thick are placed axially on a 1 m long
and 60 mm diameter engine cylinder which is surrounded by 35C. The fins are
extended 1.5 cm from the cylinder surface and the heat transfer co-efficient between
cylinder and atmospheric air is 15 W/m2K. Calculate the rate of heat transfer and the
temperature at the end of fins when the cylinder surface is at 160C.
41
Given
Number of fins = 10
Thermal conductivity K = 100 W/mK
Thickness of the fin t = 0.75 mm = 0.75  10-3 m
Length of engine cylinder = 1m
Diameter of the cylinder d = 60 mm = 0.060 m
Atmosphere temperature T = 35C + 273 = 300 K
Length of the fin L = 1.5 cm = 1.5  10-2 m
Heat transfer co-efficient h = 15 W/m2K
Cylinder surface temperature
i.e. Base temperature Tb = 160C + 273 = 433 K
Solution
Assuming that the fin end is insulated and length of the fin is 1.5 cm. So this is short
fin end insulated type problem.
We know
Heat transferred Q = (hPKA)1/2 (Tb - T) tan h (mL)….(A)
[From HMT data book Page No.41]
Where
P – Perimeter = 2  Length of the cylinder
=21
P  2m
A = Area = length of the cylinder  thickness
= 1  0.75  10-3 m
A  0.75  10 3 m2
m
=
hP
KA
15  2
100  0.75  10-3
m  20
 Q = (hPKA)1/2 (Tb  T )tan h (mL)
 [15  2  100  0.75  103 ]1/ 2  (433  300)
 tan h (20  1.5  102 )
Q = 1.5  133  0.29
Q = 58.1 W
Heat transferred per fin = 58.1 W
The heat transfer for 10 fins = 58.1  10
Q1  581 W
....(B)
42
Heat transfer from unfinned surface due to convection is
Q2  h A T
= h  ( dL - 10  t  L) (Tb  T )
[ Area of unfinned surface = Area of cylinder - Area of
fin]
= 15  [  0.060  1]  [10  0.75  10 3  1.5  10 2 ]
[433 - 300]
Q2  375.8 W ..........(C)
So, Total heat transfer Q = Q1 + Q2
Q = 581 + 375.8
Total heat transfer Q  956.8 W
We know that,
Temperature distribution [short fin, end insulated]
T  T
cos h m [L-x]

Tb  T
cos h (mL)
[From HMT data book Page No.41]
Temperature at the end of fin, so put x = L
21. Aluminium fins 1.5 cm wide and 10 mm thick are placed on a 2.5 cm diameter tube
to dissipate the heat. The tube surface temperature is 170C ambient temperature is
20C. Calculate the heat loss per fin. Take h = 130 W/m2 C and K = 200 W/m2 C for
aluminium.
Given
Wide of the fin b = 1.5 cm = 1.5  10-2 m
Thickness t = 10 mm = 10  10-3 m
Diameter of the tube d = 2.5 cm = 2.5  10-2 m
Surface temperature Tb = 170C + 273 = 443 K
Ambient temperature T = 20C + 273 = 293 K
Heat transfer co-efficient h = 130 W/m2 C
Thermal conductivity K = 200 W/mC
Solution
Assume fin end is insulated, so this is short fin end insulated type problem.
Heat transfer [short fin, end insulated]
Q = (hPKA)1/2 (Tb - T) tan h (mL) ……..(1)
[From HMT data book Page No.41]
Where
A – Area = Breadth  thickness
43
 1.5  10 2  10  10 3
A  1.5  10 4 m2
P  Perimeter  2(b  t)
= 2[(1.5  10 2 )  (10  10 3 )]
P = 0.05 m
m=
=
hP
KA
130  0.05
200  1.5  10-4
m = 14.7 m-1
(1)  Q = [130  0.05  200  1.5  104 ]1/ 2
 (443-293)  tan h (14.7  1.5  10-2 )
Q  14.3 W
22. A straight rectangular fin has a length of 35 mm, thickness of 1.4 mm. The thermal
conductivity is 55W/mC. The fin is exposed to a convection environment at 20C and h
= 500 W/m2C. Calculate the heat loss for a base temperature of 150C.
Given
Length L = 35 mm = 0.035 m
Thickness t = 1.4 mm = 0.0014 m
Thermal conductivity K = 55 W/mC
Fluid temperature T = 20C + 273 = 293 K
Base temperature Tb = 150C + 273 = 423 K
Heat transfer co-efficient h = 500 W/m2K.
Solution
Length of the fin is 35 mm, so this is short fin type problem. Assume end is insulated.
Heat transferred [Short fin, end insulated]
Q = (hPKA)1/2 (Tb - T) tan h (mL) …….(1)
[From HMT data book Page No.41]
Where
P – Perimeter = 2  Length (Approximately)
= 2  0.035
P = 0.07 m
A – Area = Length  thickness
= 0.035  0.0014
A  4.9  105 m2
44
m
=
hP
KA
500  0.07
55  4.9  10 -5
m = 113.9 m-1
Substituting h, p, K, A, Tb, T, m, L values in equation (1)
(1)  Q = [500  0.07  55  4.0  10-5 ]1/ 2
 (423 - 293)  tan h (113.9  0.035)
Q = 39.8 W
23. A heating unit made in the form of a cylinder is 6 cm diameter and 1.2 m long. It is
provided with 20 longitudinal fins 3 mm thick which protrude 50 mm from the surface
of the cylinder. The temperature at the base of the fin is 80C. The ambient
temperatures is 25C. The film heat transfer co-efficient from the cylinder and fins to
the surrounding air is 10 W/m2K. Calculate the rate of heat transfer from the finned
wall to the surrounding. Take K = 90 W/mK.
Given
Diameter of the cylinder d = 6 cm = 0.06 m
Length of the cylinder = 1.2 m
Number of fins = 20
Thickness of fin (t) = 3 mm = 0.003 m
Length of fin L = 50 mm = 0.050 m
Base temperature Tb = 80C + 273 = 353 K
Ambient temperature T = 25C + 273 = 298 K
Film heat transfer co-efficient h = 10 W/m2K
Thermal conductivity K = 90 W/mK.
Solution
Length of the fins is 50 mm. Assume end is insulated. So this is short fin end
insulated type problem.
We know
Heat transferred [short fin, end insulated]
Q = (hPKA)1/2 (Tb - T) tan h (mL) ……..(1)
[From HMT data book Page No. 41]
Where
P – Perimeter = 2  Length of the cylinder
45
= 2  1.2
P  2.4 m
A – Area = Length of the cylinder  thickness of fin
= 1.2  0.003
A  3.6  103 m2
m
=
hP
KA
10  2.4
90  3.6  10 -3
m = 8.6 m-1
(1)  Q = [10  2.4  90  3.6  103 ]1/ 2
 (353 - 298)  tan h (8.6  0.050)
Q  62.16 W
Heat transferred per fin = 62.16 W
Number of fins = 20
So, Total heat transferred Q1 = 62.16  20
Q1 = 1243.28 W
Heat transfer from unfinned surface due to convection is Q2 = h A T
 h   dL  20  t  L  (Tb  T )
[
Area of unfinned surface = Area of cylinder - Area
of fin]
= 10 [  0.06  1.2 - 20  0.003  0.050]
[353 - 298]
Q2  122.75 W
So,
Total heat transfer Q = Q1 + Q2
Q = 1243.28 + 122.75
Total heat transfer Q  1366 W
24. An aluminium cube 6 cm on a side is originally at a temperature of 500C. It is
suddenly immersed in a liquid at 10C for which h is 120 W/m2K. Estimate the time
required for the cube to reach a temperature of 250C. For aluminium  = 2700 kg/m3,
46
C = 900 J/kg K, K = 204 W/mK.
Given
Thickness of cube L = 6 cm = 0.06 m
Initial temperature T0 = 500C + 273 = 773 K
Final temperature T = 10C + 273 = 283 K
Intermediate temperature T = 250C + 273 = 523 K
Heat transfer co-efficient h = 120 w/m2K
Density  = 2700 kg/m3
Specific heat C = 900 J/Kg k
Thermal conductivity K = 204 W/mK
Solution
For Cube,
Characteristic length L c 

L
6
0.06
6
Lc  0.01 m
We know
hLc
K
120  0.01

204
Bi = 5.88  10 3  0.1
Biot number Bi 
Biot number value is less than 0.1. So this is lumped heat analysis type problem
For lumped parameter system,

 hA

t 

T  T
C  V   
 e  
....(1)
T0  T
[From HMT data book Page No.48]
We know,
V
Characteristics length L c 
A
47


h
t 

T-T
C L  
(1) 
 e   c 
T0  T

120

t 

523 - 283

 e  9000.012700 
773 - 283
-120
 In (0.489) =
t
900  0.01 2700

t = 144.86 s
Time required for the cube to reach 250C is 144.86 s.
25. A copper plate 2 mm thick is heated up to 400C and quenched into water at 30C.
Find the time required for the plate to reach the temperature of 50C. Heat transfer coefficient is 100 W/m2K. Density of copper is 8800 kg/m3. Specific heat of copper = 0.36
kJ/kg K.
Plate dimensions = 30  30 cm.
Given
Thickness of plate L = 2 mm = 0.002 m
Initial temperature T0 = 400C + 273 = 673 K
Final temperature T = 30C + 273 = 303 K
Intermediate temperature T = 50C + 273 = 323 K
Heat transfer co-efficient h = 100 W/m2K
Density  = 8800 kg/m3
Specific heat C= 360 J/kg k
Plate dimensions = 30  30 cm
To find
Time required for the plate to reach 50C.
[From HMT data book Page No.2]
Solution:
Thermal conductivity of the copper K = 386 W/mK
For slab,
L
Characteristic length L c 
2
0.002
=
2
Lc  0.001 m
We know,
Biot number Bi 
hLc
K
48
100  0.001
386
Bi = 2.59  10 4  0.1

Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
……….(1)
 e  
T0  T
[From HMT data book Page No.48]
We know,
V
Characteristics length Lc =
A

h

t 

T-T
 C  Lc   
(1) 
e
T0  T



100

t 

323 - 303
 e  3600.0018800 
673 - 303
t = 92.43 s
Time required for the plate to reach 50C is 92.43 s.
26. A 12 cm diameter long bar initially at a uniform temperature of 40 C is placed in a
medium at 650C with a convective co-efficient of 22 W/m2K. Determine the time
required for the center to reach 255C. For the material of the bar, K = 20 W/mK,
Density = 580 kg/m3, specific heat = 1050 J/kg K.
Given:
Diameter of bar, D = 12 cm = 0.12 m
Radius of bar, R = 6 cm = 0.06 m
Initial temperature T0 = 40C + 273 = 313 K
Final temperature T = 650C + 273 = 923 K
Intermediate temperature T = 255C + 273 = 528 K
Heat transfer co-efficient h = 22 W/m2K
Thermal conductivity K = 20 W/mK
Density  = 580 kg/m3
Specific heat C = 1050 J/kg k
Solution
For cylinder,
Characteristic Length L c 
R
2
49
=
0.06
2
Lc  0.03 m
We know,
hLc
K
22  0.03

20
Biot number Bi 
Bi = 0.033 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
……….(1)
 e  
T0  T
[From HMT data book Page No.48]
We know,
V
Characteristics length Lc =
A

h

t 

T-T
C L  
(1) 
 e   c 
T0  T

22t



528 - 923

 e 10500.03580 
313 - 923

t = 360.8 s
Time required for the cube to reach 255C is 360.8 s.
27. A steel ball (specific heat = 0.46 kJ/kgK. and thermal conductivity = 35 W/mK)
having 5 cm diameter and initially at a uniform temperature of 450C is suddenly
placed in a control environment in which the temperature is maintained at 100C.
Calculate the time required for the balls to attained a temperature of 150C. Take h =
10W/m2K.
Given
Specific heat C = 0.46 kJ/kg K = 460 J/kg K
Thermal conductivity K = 35 W/mK
Diameter of the sphere D = 5 cm = 0.05 m
Radius of the sphere R = 0.025 m
Initial temperature T0 = 450C + 273 = 723 K
Final temperature T = 100C + 273 = 373 K
Intermediate temperature T = 150C + 273 = 423 K
50
Heat transfer co-efficient h = 10 W/m2K
To find
Time required for the ball to reach 150C
[From HMT data book Page No.1]
Solution
Density of steel is 7833 kg/m3
  7833 kg / m3
For sphere,
Characteristic Length L c 
R
3
0.025
3
=
Lc  8.33  103 m
We know,
hLc
K
10  8.3  10 3

35
Biot number Bi 
Bi = 2.38  10-3 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,


 hA
t 

T  T
C  V  
……….(1)
 e  
T0  T
[From HMT data book Page No.48]
We know,
Characteristics length Lc =

h
V
A

t 

T-T
C L  
(1) 
 e   c 
T0  T

10

t 

423 - 373
3
 e  4608.3310 7833 
723 - 373
423 - 373
10
 In

t
723 - 373 460  8.33  10 3  7833


t = 5840.54 s
Time required for the ball to reach 150C is 5840.54 s.
28. An aluminium sphere mass 5.5 kg and initially at a temperature of 290o is suddenly
immersed in a fluid at 15C with heat transfer co-efficient 58 W/m3K. Estimate the time
required to cool the aluminium to 95C. For aluminium take  = 2700 kg/m3, C = 900
51
J/kg K, K = 205 W/mK.
Given
Mass, m = 5.5 kg
Initial temperature T0 = 290C + 273 = 563 K
Final temperature T = 15C + 273 = 288 K
Intermediate temperature T = 95C + 273 =368 K
Heat transfer co-efficient h = 58 W/m2K
Thermal conductivity K = 205 W/mK
Density  = 2700 kg/m3
Specific heat C = 900 J/kg K.
Solution
We know,
mass
m

volume V
m
Density  =

V=
=

5.5
2700
V  2.037  10 3 m3
We know,
4
 R3
3
3V 3  2.03  103
R3 

4
4
Volume of sphere V 

R  0.0786 m
For sphere,
R
Characteristic Length L c 
3
0.0786
=
3
Lc  0.0262 m
We know,
hLc
K
58  0.0262

205
Biot number Bi 
Bi = 7.41  10-3 < 0.1
52
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
……….(1)
 e  
T0  T
[From HMT data book Page No.48]
We know,
Characteristics length Lc =

V
A

h
t 

T-T
C L  
(1) 
 e   c 
T0  T

58

t 

368 - 288

 e  9000.02622700 
563 - 288
58
 368 - 288 
 In 

t

 563 - 288  900  0.0262  2700

t = 1355.36 s
Time required to cool the aluminium to 95C is 1355.6 s.
29. Alloy steel ball of 2 mm diameter heated to 800C is quenched in a bath at 100C.
The material properties of the ball are K = 205 kJ/m hr K,  = 7860 kg/m3, C = 0.45
kJ/kg K, h = 150 KJ/ hr m2 K. Determine (i) Temperature of ball after 10 second and
(ii) Time for ball to cool to 400C.
Given
Diameter of the ball D = 12 mm = 0.012 m
Radius of the ball R = 0.006m
Initial temperature T0 = 800C + 273 = 1073 K
Final temperature T = 100C + 273 = 373 K
Thermal conductivity K = 205 kJ/m hr K
205  1000J

3600 s mK
 56.94 W / mK
[ J/s = W]
3
Density  = 7860 kg/m
Specific heat C = 0.45 kJ/kg K
= 450 J/kg K
Heat transfer co-efficient h = 150 kJ/hr m2 K
53
150  1000J
3600 s m2K
 41.66 W / m2K

Solution
Case (i) Temperature of ball after 10 sec.
For sphere,
R
3
0.006
=
3
Characteristic Length L c 
Lc  0.002 m
We know,
hLc
K
41.667  0.002

56.94
Biot number Bi 
Bi = 1.46  10-3 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,


 hA
t 

T  T
C  V  
……….(1)
 e  
T0  T
[From HMT data book Page No.48]
We know,
V
Characteristics length Lc =
A

h

t 

T-T
 C  Lc   
(1) 
e
T0  T



41.667
..........(2)

10 

T - 373
 e  4500.0027860 
1073 - 373
T = 1032.95 K
Case (ii) Time for ball to cool to 400C
T = 400C + 273 = 673 K
54

h

t 

T-T
C L  
(2) 
 e   c 
T0  T

41.667
.......(2)

t 

673 - 373

 e  4500.0027860 
1073 - 373
41.667
 673 - 373 
 In 

t

1073 - 373  450  0.002  7860

t = 143.849 s
30. A large wall 2 cm thick has uniform temperature 30C initially and the wall
temperature is suddenly raised and maintained at 400C. Find
1. The temperature at a depth of 0.8 cm from the surface of the wall after 10 s.
2. Instantaneous heat flow rate through that surface per m2 per hour.
Take  = 0.008 m2/hr, K = 6 W/mC.
Given
Thickness L = 2 cm = 0.02 m
Initial temperature Ti = 30C + 273 = 303 K
Surface temperature T0 = 400C + 273 = 673 K
Thermal diffusivity  = 0.008 m2/h
= 2.22  10-6 m2/s
Thermal conductivity K = 6 W/mC.
Case (i)
Depth  0.8 cm = 0.8  10-2 m
= 0.008 m
Time t = 10 s
Case (ii)
Time t = 1 h = 3600 s
Solution

In this problem heat transfer co-efficient h is not given. So take it as . i.e. h  .
We know that,
hLc
Biot number Bi =
K
h=
Bi  
Bi value is . So this is semi infinite solid type problem.
Case (i)
55
For semi infinite solid.
Tx  T0
 x 
 erf 

Ti  T0
 2 at 
[From HMT data book Page No. 50]
Tx  T0

 erf (X) .......(1)
Ti  T0
Where,
x
X
2 at
Put x = 0.008 m, t = 10 s,  = 2.22  10-6 m2/s.

X=
0.008
2 2.22  10-6  10
X = 0.848
X = 0.848, corresponding erf (X) is 0.7706

erf (X) = 0.7706
[Refer HMT data book Page No.52]
Tx -T0
(1) 
 0.7706
Ti  T0
Tx - 673
 0.7706
303 - 673
T - 673
 x
 0.7706
- 370


Tx = 387.85 K
Case (ii)
Instantaneous heat flow
qx 
K  T0  Ti 
a t
e
  x2 


 4 t 
[From HMT data book Page No.50]
t = 3600 s (Given)
56
 qx 
6 (673  303)
  2.22  10-6  3600
e


 (0.008)2


6
 42.2210 3600 
qx  13982.37 W / m2
Intermediate temperature Tx = 387.85 K
Heat flux qx = 13982.37 W/m2.
31. A large cast iron at 750C is taken out from a furnace and its one of its surface is
suddenly lowered and maintained at 45C. Calculate the following:
1. The time required to reach the temperature 350C at a depth of 45 mm from the
surface.
2. Instantaneous heat flow rate at a depth of 45 mm and on surface after 30
minutes.
3. Total heat energy after 2 hr for ingot,
Take  = 0.06 m2/hr, K = 48.5 W/mK.
Given
Initial temperature Ti = 750C + 273 = 1023 K
Surface temperature T0 = 45C + 273 = 318 K
Intermediate temperature Tx = 350C + 273 = 623 K
Depth x = 45 mm = 0.045 m
Thermal diffusivity  = 0.06 m2/hr = 1.66  10-5 m2/s
Thermal conductivity K = 48.5 W/mK.
Solution
In this problem heat transfer co-efficient h is not given. So take it as , i.e. h  .
We know that,
Biot number Bi =
h=

hLc
K
Bi  
Bi value is . So this is semi infinite solid type problem.
1. For semi infinite solid.
Tx  T0
 x 
 erf 

Ti  T0
 2 at 
[From HMT data book Page No. 50]
57

Tx  T0
 erf (X) where,
Ti  T0
x
X
2 at

623  318
 erf (X)
1023  318

0.432 = erf (X)

erf (X) = 0.432
erf (X) = 0.432, corresponding X is 0.41

X  0.41
We know
X


x
2 at
0.045
0.41 =
2 1.66  10-5  t
(0.045)2
2
(0.41) 
(2)2  1.66  105  t

t = 181.42 s
Time required to reach 350C is 181.42 s.
2. Instantaneous heat flow
qx 
K  T0  Ti 
a t
e
  x2 


 4 t 
[From HMT data book Page No.50]
t = 30 minutes (Given)
t = 1800 s
 qx 
48.5 (318  1023)
  1.66  10-5  1800
e


 (0.045)2


5
 41.6610 1800 
qx  109725.4 W / m2
[Negative sign shows that heat lost from the ingot].
58
3. Total heat energy
q  2K[T0  Ti ]
t

7200
  1.66  10 5
[Time is given, 2 hr = 7200 s]
 2  48.5(318  1023) 
q  803.5  106 J/ m2
[Negative sign shows that heat lost from the ingot]
32. A large steel plate 5 cm thick is initially at a uniform temperature of 400C. It is
suddenly exposed on both sides to a surrounding at 60C with convective heat transfer
co-efficient of 285 W/m2K. Calculate the centre line temperature and the temperature
inside the plate 1.25 cm from themed plane after 3 minutes.
Take K for steel = 42.5 W/mK,  for steel = 0.043 m2/hr.
Given
Thickness L = 5 cm = 0.05 m
Initial temperature Ti = 400C + 273 = 673 K
Final temperature T = 60C + 273 = 333 K
Distance x = 1.25 mm = 0.0125 m
Time t = 3 minutes = 180 s
Heat transfer co-efficient h = 285 W/m2K
Thermal diffusivity  = 0.043 m2/hr
= 1.19  10-5 m2/s.
Thermal conductivity K = 42.5 W/mK.
Solution
For Plate :
L
2
0.05
=
2
Characteristic Length L c 
Lc  0.025 m
We know,
hLc
K
285  0.025

42.5
 Bi  0.1675
Biot number Bi 
59
0.1 < Bi < 100, So this is infinite solid type problem.
Infinite Solids
Case (i)
[To calculate centre line temperature (or) Mid plane temperature for infinite plate,
refer HMT data book Page No.59 Heisler chart].
t
X axis  Fourier number = 2
Lc
=
1.19  10-5  180
(0.025)2
X axis  Fourier number = 3.42
Curve 

hLc
K
285  0.025
 0.167
42.5
Curve 
hLc
 0.167
K
X axis value is 3.42, curve value is 0.167, corresponding Y axis value is 0.64
Y axis =
T0  T
 0.64
Ti  T
T0  T
 0.64
Ti  T


T0  T
 0.64
Ti  T
T0  333
 0.64
673  333
 T0  550.6 K
Center line temperature T0  550.6 K
Case (ii)
Temperature (Tx) at a distance of 0.0125 m from mid plane
[Refer HMT data book Page No.60, Heisler chart]
60
X axis  Biot number Bi 
Curve 
hLc
 0.167
K
x 0.0125

 0.5
Lc
0.025
X axis value is 0.167, curve value is 0.5, corresponding Y axis value is 0.97.
Tx  T
 0.97
T0  T
Y axis =
Tx  T
 0.97
T0  T

Tx  T
 0.97
T0  T

Tx  333
 0.97
550.6  333

Tx  544 K
Temperature inside the plate 1.25 cm from the mid plane is 544 K.
33. A 10 cm diameter apple approximately spherical in shape is taken from a 20C
environment and placed in a refrigerator where temperature is 5C and average heat
transfer coefficient is 6 W/m2K. Calculate the temperature at the centre of the apple
after a period of 1 hour. The physical properties of apple are density = 998 kg/m3.
Specific heat = 4180 J/kg K, Thermal conductivity = 0.6 W/mK.
Given:
Diameter of sphere D = 10 cm = 0.10 m
Radius of sphere R = 5 cm = 0.05 m
Initial temperature Ti = 20C + 273 = 293 K
Final temperature T = 5C + 273 = 278 K
Time t = 1 hour = 3600 s
Density  = 998 kg/m3
Heat transfer co-efficient h = 6 W/m2K
Specific heat C = 4180 J/kg K
Thermal conductivity K = 0.6 W/mK
K
0.6
Thermal diffusivity  =

C 998  4180
  1.43  107 m2 / s.
Solution
For Sphere,
61
Characteristic Length L c 
=
R
3
0.05
3
Lc  0.016 m
We know,
hLc
K
6  0.016

0.6
Biot number Bi 
 Bi  0.16
0.1 < Bi < 100, So this is infinite solid type problem.
Infinite Solids
[To calculate centre line temperature for sphere, refer HMT data book Page No.63].
X axis =
t
R2
1.43  10-7  3600
=
(0.05)2
X axis = 0.20
Curve 

hR
K
6  0.05
 0.5
0.6
Curve  0.5
X axis value is 0.20, curve value is 0.5, corresponding Y axis value is 0.86.
T0  T
 0.86
Ti  T

Y axis =

T0  T
 0.86
Ti  T

T0  278
 0.86
293  278

T0  290.9 K
Center line temperature T0 = 290.9 K.
62
34. A long steel cylinder 12 cm diameter and initially at 20C is placed into furnace at
820C with h = 140 W/m2K. Calculate the time required for the axis temperature to
reach 800C. Also calculate the corresponding temperature at a radius of 5.4 cm at that
time. Physical properties of steel are K = 21 W/mK,  = 6.11  10-6 m2/s.
Given:
Diameter of cylinder D = 12 cm = 0.12 m
Radius of sphere R = 6 cm = 0.06 m
Initial temperature Ti = 20C + 273 = 293 K
Final temperature T = 820C + 273 = 1093 K
Heat transfer co-efficient h = 140 W/m2K
Axis temperature


(or)
 T0  800C  273  1073 K
Centre line temperature 

Intermediate radius r = 5.4 cm = 0.054 m
Thermal diffusivity  = 6.11  10-6 m2/s.
Thermal conductivity K = 21W/mK
To find
1. Time (t) required for the axis temperature to reach 800C.
2. Corresponding temperature (Tt) at a radius of 5.4 cm.
Solution
For Cylinder,
Characteristic Length Lc 
R 0.06

2
2
Lc  0.03 m
We know,
hLc
K
140  0.03

21
Biot number Bi =

Bi  0.2
0.1 < Bi <100, So this is infinite solid type problem.
Infinite Solids
Case (i)
Axis temperature


(or)
 T0  800C
Centre line temperature 
To = 800C + 273 = 1073 K
Time (t) ?
63
[Refer HMT data book Page No.61. Heisler chart]
hR
K
140  0.06
=
 0.4
21
Curve 
Y axis =
=
T0  T
Ti  T
1073 - 1093
293 - 1093
Y axis = 0.025
Curve value is 0.4, Y axis 0.025, corresponding X axis value is 5.
T0  T
 0.025
Ti  T


t
5
R2
5  (0.06)2
t=
(6.11 10 -6 )
X axis =
t  2945.9 s
Case (ii)
Intermediate radius r – 5.4 cm = 0.054 m
[Refer HMT data book Page No.62]
r 0.054

 0.9
R 0.06
hR
X axis =
K
140  0.06
=
 0.4
21
Curve 
Curve value is 0.9, X axis value is 0.4, corresponding Y axis value is 0.84.
64
Tr  T
 0.84
T0  T

Y axis =

Tr  T
 0.84
T0  T

Tr  1093
 0.84
1073  1093

Tr  1076.2 K
1. Time required for the axis temperature to reach 800C is 2945.9 s.
2. Temperature (Tr) at a radius of 5.4 cm is 1076.2 K
65
UNIT – II
Introduction
Free convection in atmosphere free convection on a vertical flat plate
Empirical relation in free convection
Forced convection
Laminar and turbulent convective heat transfer analysis in flows
between parallel plates, over a flat plate and in a circular pipe.
Empirical relations,
Application of numerical techniques in problem solving.
66
CONVECTION
PART – A
1. What is dimensional analysis?
Dimensional analysis is a mathematical method which makes use of the study of the
dimensions for solving several engineering problems. This method can be applied to all types
of fluid resistances, heat flow problems in fluid mechanics and thermodynamics.
2. State Buckingham  theorem.
Buckingham  theorem states as Follows: “If there are n variables in a dimensionally
homogeneous equation and if these contain m fundamental dimensions, then the variables are
arranged into (n – m) dimensionless terms. These dimensionless terms are called  terms.
3. What are all the advantages of dimensional analysis?
1. It expresses the functional relationship between the variables in dimensional terms.
2. It enables getting up a theoretical solution in a simplified dimensionless form.
3. The results of one series of tests can be applied to a large number of other similar
problems with the help of dimensional analysis.
4. What are all the limitations of dimensional analysis?
1.
The complete information is not provided by dimensional analysis. It only indicates
that there is some relationship between the parameters.
2. No information is given about the internal mechanism of physical phenomenon.
3. Dimensional analysis does not give any clue regarding the selection of variables.
5. Define Reynolds number (Re).
It is defined as the ratio of inertia force to viscous force.
Inertia force
Re 
Viscous force
6. Define prandtl number (Pr).
It is the ratio of the momentum diffusivity of the thermal diffusivity.
Momentum diffusivity
Pr 
Thermal diffusivity
7. Define Nusselt number (Nu).
It is defined as the ratio of the heat flow by convection process under an unit
temperature gradient to the heat flow rate by conduction under an unit temperature gradient
through a stationary thickness (L) of metre.
67
Nusselt number (Nu) =
Qconv
.
Qcond
8. Define Grash of number (Gr).
It is defined as the ratio of product of inertia force and buoyancy force to the square of
viscous force.
Gr 
Inertia force  Buyoyancy force
(Viscous force)2
9. Define Stanton number (St).
It is the ratio of nusselt number to the product of Reynolds number and prandtl
number.
Nu
St 
Re Pr
10. What is meant by Newtonion and non – Newtonion fluids?
The fluids which obey the Newton’s Law of viscosity are called Newtonion fluids and
those which do not obey are called non – newtonion fluids.
11. What is meant by laminar flow and turbulent flow?
Laminar flow: Laminar flow is sometimes called stream line flow. In this type of flow, the
fluid moves in layers and each fluid particle follows a smooth continuous path. The fluid
particles in each layer remain in an orderly sequence without mixing with each other.
Turbulent flow: In addition to the laminar type of flow, a distinct irregular flow is
frequency observed in nature. This type of flow is called turbulent flow. The path of any
individual particle is zig – zag and irregular. Fig. shows the instantaneous velocity in laminar
and turbulent flow.
12. What is hydrodynamic boundary layer?
In hydrodynamic boundary layer, velocity of the fluid is less than 99% of free stream
velocity.
13. What is thermal boundary layer?
In thermal boundary layer, temperature of the fluid is less than 99% of free stream
velocity.
14. Define convection.
Convection is a process of heat transfer that will occur between a solid surface and a
fluid medium when they are at different temperatures.
68
15. State Newton’s law of convection.
Heat transfer from the moving fluid to solid surface is given by the equation
Q = h A (Tw – T)
This equation is referred to as Newton’s law of cooling.
Where
h – Local heat transfer coefficient in W/m2K.
A – Surface area in m2
Tw – Surface (or) Wall temperature in K
T - Temperature of fluid in K.
16. What is meant by free or natural convection?
If the fluid motion is produced due to change in density resulting from temperature
gradients, the mode of heat transfer is said to be free or natural convection.
17. What is forced convection?
If the fluid motion is artificially created by means of an external force like a blower or
fan, that type of heat transfer is known as forced convection.
18. According to Newton’s law of cooling the amount of heat transfer from a solid surface of
area A at temperature Tw to a fluid at a temperature T is given by ________.
Ans : Q = h A (Tw – T)
19. What is the form of equation used to calculate heat transfer for flow through
cylindrical pipes?
Nu = 0.023 (Re)0.8 (Pr)n
n = 0.4 for heating of fluids
n = 0.3 for cooling of fluids
20. What are the dimensionless parameters used in forced convection?
1. Reynolds number (Re)
2. Nusdselt number (Nu)
3. Prandtl number (Pr)
21. Define boundary layer thickness.
The thickness of the boundary layer has been defined as the distance from the surface
at which the local velocity or temperature reaches 99% of the external velocity or
temperature.
69
PART – B
1. Air at 20C, at a pressure of 1 bar is flowing over a flat plate at a velocity of 3 m/s. if
the plate maintained at 60C, calculate the heat transfer per unit width of the plate.
Assuming the length of the plate along the flow of air is 2m.
Given : Fluid temperature T = 20C,
Pressure p
= 1 bar,
Velocity U
= 3 m/s,
Plate surface temperature Tw = 60C,
Width W
= 1 m,
Length L
= 2m.
Solution : We know,
Film temperature Tf 
60  20
2
Tf  40C
Tw  T
2

Properties of air at 40C:
Density  = 1.129 Kg/m3
Thermal conductivity K = 26.56  103 W / mK,
Kinematic viscosity v = 16.96  106 m2 / s.
Prandtl number
Pr = 0.699
We know,
UL
Reynolds number Re =
v
32

16.96  10 6
 35.377  10 4
Re  35.377  104  5  105
Reynolds number value is less than 5  105, so this is laminar flow.
For flat plate, Laminar flow,
Local Nusselt Number Nux = 0.332 (Re)0.5 (Pr)0.333
70
Nux  0.332 (35.377  10 4 )0.5  (0.699)0.333
Nux  175.27
We know that,
Local Nusselt Number Nu x 
 175.27 
hs  L
K
hs  2
26.56  103
Local heat transfer coefficient hx = 2.327 W/m2K
We know,
Average heat transfer coefficient h = 2  hx
h  2  2.327
h = 4.65 W/m2K
Heat transfer Q = h A (Tw - T)
 4.65  2 (60  20)
[ Area  width  length  1 2  2]
Q  372 Watts.
2. Air at 20C at atmospheric pressure flows over a flat plate at a velocity of 3 m/s. if the
plate is 1 m wide and 80C, calculate the following at x = 300 mm.
1. Hydrodynamic boundary layer thickness,
2. Thermal boundary layer thickness,
3. Local friction coefficient,
4. Average friction coefficient,
5. Local heat transfer coefficient
6. Average heat transfer coefficient,
7. Heat transfer.
Given: Fluid temperature T = 20C
Velocity
U = 3 m/s
Wide
W=1m
Surface temperature
Tw = 80C
Distance x = 300 mm = 0.3 m
Solution: We know
Film temperature Tf 
Tw  T
2
71
80  20
2
Tf  50C

Properties of air at 50C
Density  = 1.093 kg/m3
Kinematic viscosity v = 17.95  10 -6m 2 / s
Pr andt l number Pr =0.698
Thermal conductivity K = 28.26  10 -3 W / mK
We know,
Reynolds number Re =
UL
v
3  0.3
17.95  106
Re  5.01 104  5  105

Since Re < 5  105, flow is laminar
For Flat plate, laminar flow,
1. Hydrodynamic boundary layer thickness:
 hx  5  x  (Re)0.5
= 5  0.3  (5.01 10 4 )0.5
 hx  6.7  10 3 m
2. Thermal boundary layer thickness:
 TX   hx (Pr)0.333


  TX  6.7  10 3 (0.698)0.333
 TX  7.5  10 3 m
3. Local Friction coefficient:
Cfx  0.664(Re)0.5
= 0.664 (5.01 10 4 ) 0.5
Cfx = 2.96  10 -3
4. Average friction coefficient:
72
CfL  1.328 (Re)-0.5
= 1.328 (5.01 10 4 )0.5
= 5.9  10-3
CfL  5.9  103
5. Local heat transfer coefficient (hx):
Local Nusselt Number
Nux = 0.332 (Re)0.5 (Pr)0.333
 0.332 (5.01 104 ) (0.698)0.333
Nux  65.9
We know
Local Nusselt Number
h L
Nux  x
K
hx  0.3
65.9 

23.26  10 3
 hx  6.20 W/m2K
x = L = 0.3m
Local heat transfer coefficient h x  6.20 W / m2K
6. Average heat transfer coefficient (h):
h  2  hx
 2  6.20
h  12.41 W / m2K
7. Heat transfer:
We know that,
Q  h A(Tw  T )
= 12.41 (1 0.3) (80-20)
Q = 23.38 Watts
3. Air at 30C flows over a flat plate at a velocity of 2 m/s. The plate is 2 m long and 1.5
m wide. Calculate the following:
1. Boundary layer thickness at the trailing edge of the plate,
2. Total drag force,
3. Total mass flow rate through the boundary layer between x = 40 cm and x = 85
cm.
Given: Fluid temperature T = 30C
Velocity
U = 2 m/s
Length
L =2m
73
Wide W
W = 1.5 m
To find:
1. Boundary layer thickness
2. Total drag force.
3. Total mass flow rate through the boundary layer between x = 40 cm and x = 85
cm.
Solution: Properties of air at 30C
  1.165 kg/m3
v  16  10 6 m2 / s
Pr  0.701
K  26.75  10  3 W / mK
We know,
UL
Reynolds number Re 
v
2 2

16  10 6
Re  2.5  105  5  105
Since Re<5  105 ,flow is laminar
For flat plate, laminar flow, [from HMT data book, Page No.99]
Hydrodynamic boundary layer thickness
 hx  5  x  (Re)0.5
= 5  2  (2.5  105 )0.5
 hx  0.02 m
Thermal boundary layer thickness,
 tx hx  (Pr)0.333
=0.02  (0.701)-0.333
 TX  0.0225 m
We know,
Average friction coefficient,
CfL  1.328 (Re)0.5
= 1.328  (2.5  105 )0.5
CfL  2.65  10-3
We know
74
CfL 
t
U2
2
t
1.165  (2)2
2
 Average shear stress t = 6.1 10 -3N / m 2
 2.65  10-3 
Drag force = Area  Average shear stress
= 2  1.5  6.1 10-3
Drag force = 0.018 N
Drag force on two sides of the plate
= 0.018  2
= 0.036 N
Total mass flow rate between x = 40 cm and x= 85 cm.
5
m   U  hx  85   hx  40 
8
Hydrodynamic boundary layer thickness
 hx 0.5  5  x  (Re)0.5
U  x 
= 5  0.85  

 v 
 2  0.85 
 5  0.85  
6
16  10 
 HX0.85  0.0130 m
 hx=0.40
0.5
0.5
= 5  x  (Re)-0.5
 U x 
 5  0.40  

 v 
0.5
 2  0.40 
 5  0.40  
6 
 16  10 
 HX0.40  8.9  10 3 m
0.5
5
(1)  m=  1.165  2 0.0130  8.9  10 3 
8
m = 5.97  10 -3Kg / s,
4. Air at 30C, Flows over a flat plate at a velocity of 4 m/s. The plate measures 50  30
cm and is maintained at a uniform temperature of 90C. Compare the heat loss from
75
the plate when the air flows
(a) Parallel to 50 cm,
(b) Parallel to 30 cm
Also calculate the percentage of heat loss.
Given: Fluid temperature T = 30C
Velocity U
= 4 m/s
Plate dimensions = 50 cm  30 cm
 0.50  0.30 m2
Surface temperature Tw = 90C
Solution: Film temperature Tf 
90  30
2
Tf  60C
Tw  T
2

Properties of air at 60C,
  1.060 Kg/m3
  18.97  106 m2 / s
Pr  0.696
K = 28.96  10-3 W/mK
Case (i) : When the flow is parallel to 50 cm.
UL
Reynolds number Re 
v
4  0.50

18.97  10 6
Re  1.05  105  5  105
Since Re <5  105 ,flow is laminar
Local nusselt number NUx
NUx =0.332 1.05  105 
0.5
= 0.332(Re)0.5(Pr)0.333
 (0.696)0.333
Local nusselt number NUx =95.35
We know
hL
NUx  x
K
hx  0.50
95.35 
28.96  10 3
Local heat transfer coefficient h x  5.52 W/m2K
76
We know
Average heat transfer coefficient h  2  hx
 h  2  5.52
h  11.04 W/m2K
Heat transfer Q1  h A(Tw  T )
 11.04  (0.5  0.3)  (90  30)
Q1  99.36 W
Case (ii) : When the flow is parallel to 30 cm side.
UL
Reynolds number Re =
v
4  0.3

18.97  10 6
Re = 6.3  10 4  5  105
Since Re<5  105 , flow is laminar
For flat plate, laminar flow,
Local Nusselt Number
NUx  0.332 (Re)0.5 (0.696)0.333
 0.332 (6.32  10 4 )0.5 (0.696)0.333
NUx  74.008
We know that, NUx
=
hx  0.30
28.96  10 3
 hx  7.141 W/m2K
hxL
K
74.008 
Local heat transfer coefficient h x  7.141 W/m2K
Average heat transfer coefficient h = 2hx
h  2  7.14
h  14.28 W/m2K
We know
Heat transfer Q2  h  A  (Tw  T )
 h  L  W (Tw  T )
 14.28  0.3  0.5  (363  303)
Q2  128.5W
77
Case (iii):
% heat loss =
Q2  Q1
 100
Q1
128.5-99.36
 100
99.36
% heat loss = 29.3%
=
5. Air at 40C is flows over a flat plate of 0.9 m at a velocity of 3 m/s. Calculate the
following:
1. Overall drag coefficient
2. Average shear stress,
3. Compare the average shear stress with local shear stress (shear stress at the
trailing edge)
Given : Fluid temperature T = 40C
Length L = 0.9 m
Velocity U = 3 m/s.
Solution:
Properties of air at 40C:
  1.128 Kg/m3
 = 16.96  10-6 m2 / s
Pr  0.699
K  26.56  10-3 W/mK
We know,
UL
Reynolds number Re 
v
3  0.9

16.96  10 6
Re  1.59  105  5  105
Since Re< 5  105 , flow is laminar
For plate, laminar flow,
Drag coefficient (or) Average skin friction coefficient
CfL  1.328  (Re)0.5
 1.328  (1.59  105 )0.5
CfL  3.3  103
We know
78
Average friction coefficient CfL 

U2
2
  CfL 
U
2
2
3.3  10-3  1.128  (3)2
=
2
Average shear stress  = 0.016 N/m2
We know,
Local skin friction coefficient
Cfx  0.664  (Re)0.5
 0.664  (1.59  105 )0.5
Cfx  1.66  103
we know
Local skin friction coefficient Cfx 
 1.66  103 

U2

2
1.128  (3)2
2
3
2
 x  8.4  10 N/ m
Local shear stress  x  8.4  10 3 N/ m2
Local shear stress  x
8.4  10 3 N/ m2

Average shear stress 
0.016 N / m2
 0.52
6. Air at 290C flows over a flat plate at a velocity of 6 m/s. The plate is 1m long and 0.5
m wide. The pressure of the air is 6 kN/2. If the plate is maintained at a temperature of
70C, estimate the rate of heat removed form the plate.
Given : Fluid temperature T = 290C
Velocity U
= 6 m/s.
Length L
=1m
Wide W
= 0.5 m
Pressure of air P = 6 kN/m2
 6  103 N/ m2
Plate surface temperature Tw = 70C
To find: Heat removed from the plate
79
Solution:
We know
Film temperature Tf 
70  290
2
Tf  180C
Tw  T
2

Properties of air at 180C (At atmospheric pressure)
  0.799 Kg/m3
 = 32.49  10-6 m2 / s
Pr  0.681
K  37.80  10-3 W/mK
Note: Pressure other than atmospheric pressure is given, so kinematic viscosity will vary with
pressure. Pr, K, Cp are same for all pressures.
P
Kinematic viscosity    atm  atm
Pgiven
1 bar
6  103N/ m2
Atmospheric pressure = 1 bar 
   32.49  106

 32.49  106 
105 N/ m2
6  103 N/ m3
 1 bar = 1 105N/ m2 
Kinematic viscosity v = 5.145  10-4m2 / s.
We know,
Reynolds number Re 
UL
v
6 1
5.145  10 4
Re  1.10  10 4  5  105

Since Re< 5  105 , flow is laminar
For plate, laminar flow,
Local nusselt number
80
NUx  0.332 (Re)0.5 (Pr)0.333
 0.332 (1.10  10 4 )0.5 (0.681)0.333
NUx  30.63
We know
hL
NUx = x
K
hx  1
30.63 
37.80  103
[
L = 1 m]
Local heat transfer coefficient h x  1.15 W/m2K
We know
Average heat transfer coefficient h = 2hx
h  2  1.15
h  2.31 W/m2K
We know
Heat transferred Q  h A (T  Tw )
 2.31 (1 0.5)  (563  343)
Q  254.1 W
Heat transfer from both side of the plate = 2  254.1
= 508.2 W.
7. Air at 40C flows over a flat plate, 0.8 m long at a velocity of 50 m/s. The plate surface
is maintained at 300C. Determine the heat transferred from the entire plate length to
air taking into consideration both laminar and turbulent portion of the boundary layer.
Also calculate the percentage error if the boundary layer is assumed to be turbulent
nature from the very leading edge of the plate.
Given : Fluid temperature T = 40C
Length
L = 0.8 m
Velocity
U = 50 m/s
Plate surface temperature Tw = 300C
To find :
1. Heat transferred for:
i. Entire plate is considered as combination of both laminar and turbulent flow.
ii. Entire plate is considered as turbulent flow.
2. Percentage error.
Solution: We know
81
Film temperature Tf 
300  40
 443 K
2
Tf  170C
Tw  T
T
2

Pr operties of air at 170C:
 = 0.790 Kg/m3
  31.10  10 6 m2 / s
Pr  0.6815
K  37  10 3 W/mK
We know
Reynolds number Re=
UL
v
50  0.8
 1.26  106
6
31.10  10
Re = 1.26  106  5  105

Re  5  105 ,so this is turbulent flow
Case (i): Laminar – turbulent combined. [It means, flow is laminar upto Reynolds number
value is 5  105, after that flow is turbulent]
Average nusselt number = Nu = (Pr)0.333 (Re)0.8 – 871
Nu = (0.6815)0.333 [0.037 (1.26  106)0.8 – 871
Average nusselt number Nu = 1705.3
hL
We know Nu =
K
h  0.8
1705.3 
37  10 3
h  78.8 W / m2K
Average heat transfer coefficient
h=78.8 W/m2K
Head transfer Q1  h  A  (Tw  T )
 h  L  W  (Tw  T )
= 78.8  0.8  1 (300 - 40)
Q1  16390.4 W
Case (ii) : Entire plate is turbulent flow:
82
Local nusselt number} Nux = 0.0296  (Re)0.8  (Pr)0.333
NUx = 0.0296  (1.26 106)0.8  (0.6815)0.333
NUx = 1977.57
We know
h L
NUx  x
K
h  0.8
1977.57  x
37  103
hx  91.46 W/m2K
Local heat transfer coefficient hx = 91.46 W/m2K
Average heat transfer coefficient (for turbulent flow)
h = 1.24  hx
= 1.24 91.46
Average heat transfer coefficient} h = 113.41 W/m2K
We know Heat transfer Q2 = h  A  (Tw + T)
= h  L  W  (Tw + T)
= 113.41  0.8  1 (300 – 40)
Q2 = 23589.2 W
Q  Q1
2. Percentage error = 2
Q1
23589.2 - 16390.4
 100
16390.4
= 43.9%
=
8. Air at 20C flows over a flat plate at 60C with a free stream velocity of 6 m/s.
Determine the value of the average convective heat transfer coefficient upto a length of
1 m in the flow direction.
Given : Fluid temperature T = 20C
Plate temperature Tw = 60C
Velocity
U = 6 m/s
Length
L =1m
To find : Average heat transfer coefficient
Solution : We know
83
Film temperature Tf 
Tw  T
2
60+20
2
Tf  40C

Properties of air at 40C:
Density  = 1.128 Kg/m3
Thermal conductivity K = 26.56  10 -3 W/mK
Kinematic viscosity v = 16.96  10 -6m 2 / s
Pr andtl number
Pr = 0.699
We know
Reynolds number Re =
UL
v
6 1
16.96  10 6
Re  3.53  105  5  105

Since Re < 5  105 , flow is laminar
For flat plate, laminar flow
Local nusselt number} Nux = 0.332  (Re)0.5  (Pr)0.333
= 0.332  (3.53  105)0.5  (0.699)0.333
NUx = 175.27
We know,
Local nusselt number} NUx 
hx  L
K
hx  1
26.56  10 3
Local nusselt number} NUx  4.65 W/m 2K
175.27 
Average heat transfer coefficient} h = 2  hx
 2  4.65
h  9.31 W/m2K
9. Air at 25C at the atmospheric pressure is flowing over a flat plate at 3 m/s. If the
plate is 1 m wide and the temperature Tw = 75C. Calculate the following at a location of
1m from leading edge.
i. Hydrodynamic boundary layer thickness,
ii. Local friction coefficient,
iii. Thermal boundary layer thickness,
iv.
Local heat transfer coefficient
84
Given : Fluid temperature T = 25C
Velocity
U = 3 m/s
Wide
W=1m
Plate surface temperature Tw = 75C
Distance
=1m
To find:
1.
2.
3.
4.
Hydrodynamic boundary layer thickness.
Local friction coefficient
Thermal boundary layer thickness
Local heat transfer coefficient
Solution: We know
Film temperature Tf 
Tw  T
2
75  25
 323 K = 50C
2
Tf  50C

Pr operties of air at 50C:
Density  = 1.093
Kinematic viscosity     10 6 m2 / s
Prandtl number Pr = 0.698
Thermal conductivity K = 28.26  13 -3 W/m K
We know,
UL
[ x = L 1m]
Reynolds number Re=
v
3 1

 1.67  105
6
17.95  10
Re  1.67  105  5  105
Since Re < 5  105 ,flow is laminar
For flat plate, laminar flow,
1. Hydrodynamic boundary layer thickness,
 hx  5  x  (Re)0.5
= 5  1 (1.67  105 )0.5
 hx  0.0122 m
85
2. Local friction coefficient
Cfx  0.644 (Re)-0.5
= 0.644 (1.67  105 )0.5
Cfx  1.62  10 3
3. Thermal boundary layer thickness,
 TX   hx  (Pr)0.333
 0.0122  (0.698)0.333
 TX  0.01375
4. Local heat transfer coefficient (hx):
We know
Local nusselt number} NUx = 0.332 (Re)0.5 (Pr)0.333
= 0.332 (1.67  105)0.5 (0.698)0.333
NUx = 120.415
h L
We know, NUx  x
K
hx  1
 120.415 =
 x = L = 1m
28.26  10-3
Local heat transfer coefficient} hx  3.4 W / m2K
10. Atmospheric air at 300 K with a velocity of 2.5 m/s flows over a flat plate of length L
= 2m and width W = 1m maintained at uniform temperature of 400 K. Calculate the
local heat transfer coefficient at 1 m length and the average heat transfer coefficient
from L = 0 to L = 2m. Also find the heat transfer,
Given: Fluid temperature T = 300 K
Velocity
U = 2.5 m/s
Total Length L = 2 m
Width
W=1m
Surface temperature Tw = 400 K
To find:
1. Local heat transfer coefficient at L = 1 m
2. Average heat transfer coefficient at L = 2 m
3. Heat transfer Q
Solution:
86
Case (i): Local heat transfer coefficient at L = 1m
Film temperature Tf 
Tw  T
2
400  300
 350 K
2
Tf  77C

Properties of air at 77C  80C:
 = 1 Kg/m3
 = 21.09  10-6 m2 / s
Pr = 0.692
K = 30.47  10-3 W/mK
We know
Reynolds number Re =
2.5  1
21.09  10 6
Re  118539.45  5  105
UL
v

Since Re < 5  105 ,flow is laminar.
For flat plate, laminar flow,
Local Nusselt number} NUx = 0.332 (Re)0.5 (Pr)0.333
= 0.332 (118539.5)0.5 (0.692)0.333
NUx = 101.18
We know,
hL
Local nusselt number} NUx  x
K
hx  1
101.18 =
30.47  103
hx = 3.0832 W/m2K
 Local heat transfer coefficient} hx = 3.08 W/m2K
Case (ii): Average heat transfer coefficient at L = 2m
UL
Reynolds number Re =
v
87
2.5  2
21.09  10 6
Re  237079.18 < 5  10 5
Re 
Since Re < 5  105 ,flow is laminar.
For flat plate, laminar flow,
NUx = 0.332 (Re)0.5 (Pr)0.333
= 0.332 (237079.18)0.5 (0.692)0.333
NUx = 143
h L
We know that, NUx  x
K
hx  2
 143 =
30.47  10 3
Local heat transfer coefficient} hx = 2.17 W/m2K
We know that,
Average heat transfer coefficient} h = 2  hx
h = 2  2.17
h = 4.35 W/m2K
Average heat transfer coefficient} h = 4.35 W/m2K
Case (iii) : Heat transfer Q = h A (Tw - T)
= 4.35  2  1 (400 – 300)
 L = 2m; W= 1m
Q = 870 W.
11. For a particular engine, the underside of the crank case can be idealized as a flat
plat measuring 80 cm  20 cm. The engine runs at 80 km/hr and the crank case is cooled
by air flowing past it at the same speed. Calculate the loss of heat from the crank case
surface of temperature 75C to the ambient air temperature 25C. Assume the
boundary layer becomes turbulent from the loading edge itself.
Given : Area A = 80 cm  20 cm
= 1600 cm2 = 0.16m2
Velocity U = 80 Km/hr
80  103 m

3600s
 22.22 m/s
Surface temperature Tw  75 C
Ambient air temperature T  25 C
88
Flow is turbulent from the leading edge, i.e,. flow is fully turbulent.
To find:
1. Heat loss
Film temperature Tf 
Tw  T 75  25

2
2
Tf  50C
Properties of air at 50C:
 = 1.093 Kg/m3
 = 17.95  10 -6 m2 / s
Pr  0.698
K  28.26  10 3 W/mK
We know
UL
v
Re ynolds number Re =
22.22  0.8
17.95  10 6
Re = 9  105

[
L = 0.8m]
Re  9  105  5  105
Since Re>5  105 , Flow is turbulent
For flat plate, turbulent flow,
[Fully turbulent from leading edge – given]
Local Nusselt number} NUx = 0.0296 (Re)0.8 (Pt)0.333
= 0.0296 [9  105]0.8 (0.698)0.33
NUx = 1524.6
We know that, NUx 
h xL
K
hx  0.8
28.26  10 3
hx  53.85 W/m2K
1524.6 
[
L = 0.8m]
Local heat transfer coefficient} hx = 53.85 W/m2K
For turbulent flow, flat plate
Average heat transfer coefficient} h = 1.24 hx
h = 1.24  53.85
h = 66.78 W/m2K
89
We know,
Heat loss Q = h A (Tw - T)
= 66.78  0.16 (75 – 25)
Q = 534.2 W
Formula used for Flow over cylinders and spheres
TW  T
2
Where T - Fluid temperature C
Tw – Plate surface temperature C
1. Film temperature Tf 
2. Reynolds number NUx 
UD
v
Where U – Velocity, m/s
D - Diameter, m
 - Kinematic viscosity, m2/s
3. Nusselt number NU = C (Re)m (Pr)0.333
4. Nusselt number NU =
hD
K
5. Heat transfer Q = h  A  (Tw - T)
Where A =  DL
For sphere:
Nusselt number NU = 0.37 (Re)0.6
Heat transfer Q = h A (Tw - T)
Where A 4r2
12. Air at 15C, 30 km/h flows over a cylinder of 400 mm diameter and 1500 mm height
with surface temperature of 45C. Calculate the heat loss.
Given : Fluid temperature T = 15C
Velocity
U = 30 Km/h
3
30  10 m

3600 s
U  8.33 m/s
Diameter D = 400 mm = 0.4 m
Length L = 1500 mm = 1.5 m
Plate surface temperature Tw = 45C
90
To find: Heat loss.
Solution: We know
Film temperature Tf 
45  15
2
Tf  30C
Tw  T
2

Properties of air at 30C : [From HMT data book, Page No.22]
Density  = 1.165 Kg/m3
Kinematic viscosity v = 16  10-6 m2/s
Prandtl Number Pr = 0.701
Thermal conductivity K = 26.75  10-3 W/mK
We know
Reynolds Number Re =

8.33  0.4
16  10 6
UD
v
ReD  2.08  105
We know
Nusselt Number Nu = C (Re)m (Pr)0.333
[From HMT data book, Page No.105]
ReD value is 2.08  105, so C value is 0.0266 and m value is 0.805.
[From HMT data book, Page No.105]
 NU = 0.0266  (2.08  105)0.805  (0.701)0.333
NU  451.3
We know that,
Nusselt Number NU =
hD
K
h  0.4
26.75  10-3
h = 30.18 W/m2K
 451.3 =

91
Heat transfer coefficient h = 30.18 W/m2K
Heat transfer Q = hA (Tw  T )
= h    D  L  (Tw  T )
[
A =  DL]
= 30.18    0.4  1.5  (45 -15)
Q = 1706.6 W
13. Air at 30C, 0.2 m/s flows across a 120W electric bulb at 130C. Find heat transfer
and power lost due to convection if bulb diameter is 70 mm.
Given : Fluid temperature T = 30C
Velocity
U = 0.2 m/s
Heat energy Q1 = 120 W
Surface temperature Tw = 130C
Diameter
D = 70 mm = 0.070 m
To find:
1. Heat Transfer
2. Power lost due to convection
Solution:
1. Film temperature Tf 
Tw  T
2
130  30
2
Tf  80C

Properties of air at 80C:
 = 1 Kg/m3
 = 21.09  10-6 m2 / s
Pr = 0.692
K = 30.47  10-3 W/mK
We know
Reynolds number Re =
UD

0.2  0.070

 663.82
21.09  103
Re  663.82
We know
92
Nusselt Number Nu = 0.37 (Re)0.6
= 0.37 (663.82)0.6
Nu = 18.25
We know
hD
Nusselt number Nu 
K
h  0.070
 18.25 =
30.47  10-3
 h = 7.94 W/m2K
Heat transfer coefficient h = 7.94 W/m2K
We know
Heat transfer Q2 = h A (Tw - T)
 h  4 r 2 [Tw  T ] [ A = 4 r 2 ]
2
 0.070 
 7.94  4    
  (130  30)
 2 
Heat transfer Q 2  12.22 W
2. % of heat lost =
Q2
 100
Q1
12.22
 100
120
 10.18%

14. Air at 40C flows over a tube with a velocity of 30 m/s. The tube surface
temperature is 120C. Calculate the heat transfer for the following cases.
1. Tube could be square with a side of 6 cm.
2. Tube is circular cylinder of diameter 6 cm
Given : Fluid temperature T = 40C
Velocity
U = 30 m/s
Tube surface temperature Tw = 120C
To find: Heat transfer coefficient (h)
Solution: We know
Film temperature Tf 

Tw  T
2
120  40
2
93
Tf  80C
Properties of air at 80C:
 = 1 Kg/m3
 = 21.09  10-6 m2 / s
Pr  0.692
K = 30.47  10-3 W/mK
Case (i): Tube is considered as square of side 6 cm
i.e., L = 6cm = 0.06m
UL
Reynolds number Re =
v
30  0.06
21.09  10 6
Re  0.853  105

Nusselt Number Nu = C  (Re)n (Pr)0.333
For square, n = 0.675
C = 0.092
 Nu = 0.092 (0.853  10 5 )0.675  (0.692)0.333
 Nu = 173.3
We know that, NU =
hL
K
h  0.06
30.47  103
Heat transfer coefficient h = 88 W/m2K
173.3 
Case (ii)
Tube diameter D = 6cm = 0.06 m
Reynolds number Re =
UD

30  0.06
21.09  106
Re  0.853  105

Nusselt number Nu = C (ReD )m (Pr)0.333
Re value is 0.853  105 , so corresponding C and m values are 0.0266
respectively.
94
and 0.805
 Nu = 0.0266  (0.853  105 )0.805  (0.692)0.333
Nu = 219.3
We know Nu =
hD
K
h  0.06
30.47  10-3
 h = 111.3 W/m2K
 219.3
 Heat transfer coefficient h = 111.3 W/m2K
Formulae Used for Flow Over Bank of Tubes
Sn
Sn  D
Where Sn – Transverse pitch, m.
1. Maximum velocity Umax = U 
2. Reynolds Number Re =
Umax  D

3. Nusselt Number, NU = 1.13  (Pr)0.33 [C Ren]
[From HMT data book, Page No.114]
15. In a surface condenser, water flows through staggered tubes while the air is passed
in cross flow over the tubes. The temperature and velocity of air are 30C and 8 m/s
respectively. The longitudinal and transverse pitches are 22 mm and 20 mm
respectively. The tube outside diameter is 18 mm and tube surface temperature is 90C.
Calculate the heat transfer coefficient.
Given: Fluid temperature
T = 30C
Velocity
U = 8 m/s
Longitudinal pitch, Sp = 22mm = 0.022 m
Transverse pitch, Sn = 20mm = 0.020 m
Diameter D
= 18mm = 0.018 m
Tube surface temperature Tw = 90C
Solution:
95
Film temperature Tf 
Tw  T
2
90  30
2
Tf  60C
Properties of air at 60C:

 = 1.060 Kg/m3
 = 18.97  10-6 m2 / s
Pr  0.696
K = 28.96  10-3 W / mK
We know
Maximum velocity Umax = U 
Sn
Sn  D
0.020
0.020  0.018
= 80 m/s
 Umax  8 
Umax
We know
Reynolds Number Re =
Umax  D

80  0.018
18.97  10 6
Re  7.5  10 4

Sn 0.020

 1.11
D 0.018
Sn
 1.11
D
Sp 0.022

 1.22
D 0.018
Sp
 1.22
D
S
Sn
 1.11. p  1.22, corresponding C, n values are 0.518 and 0.556 respectively.
D
D
[From HMT data book, Page No.114]
C = 0.518
n = 0.556
We know,
96
Nusselt Number Nu = 1.13 (Pr) 0.333[C (Re)n]
[From HMT data book, Page No.114]
 Nu = 1.13  (0.696)0.333  [0.518  (7.5  10 4 )0.556 ]
Nu = 266.3
We know
hD
Nusselt Number Nu 
K
h  0.018

266.3 =
28.96  10-3
Heat transfer coefficient h = 428.6 W/m2K.
Formulae used for flow through Cylinders (Internal flow)
1. Bulk mean temperature
T  Tmo
Tm  mi
2
Tmi = Inlet temperature C,
Where
Tmo = Outlet temperature C.
UD
2. Reynolds Number Re 

If Reynolds number value is less than 2300, flow is laminar. If Reynolds number values is
greater than 2300, flow is turbulent.
3. Laminar Flow:
Nusselt Number NU – 3.66
[From HMT data book, Page No.116]
4. Turbulent Flow (General Equation)
Nusselt Number Nu = 0.023 (Re)0.8 (Pr)n
n = 0.4 – Heating process
n = 0.3 – Cooling process
[From HMT data book, Page No.119]
This equation is valid for
0.6 < Pr < 160,
Re < 10000
L
 60
D
For turbulent flow,
Nu  0.036 (Re) (Pr)
0.8
0.33
D
L
 
0.055
This equation is valid for
97
10 
L
 400
D
5. Equivalent diameter for rectangular section,
Dh (or) Dc 
4A 4(L  W)

P
2(L  W)
Where A – Area, m2,
P – Perimeter, m
L – Length, m,
W – Width, m.
6. Equivalent diameter for hollow cylinder
4A
Dh (or) Dc =
P

4  D02  Di2 
4
=
 D0  Di 
Where D0  Outer diameter
Di - Inner diameter
7. Heat transfer
Q = h A (Tw – Tm) where A =   D  L
(or)
Q = m Cp (Tmo – Tmi)
Where Tw – Tube wall temperature C,
Tm – Mean temperature C.
Tmi – Inlet temperature C
Tmo – Outlet temperature C.
8. Mass flow rate
m -   A  U Kg/s
Where  - Density, Kg/m3
 2 2
D,m
4
U – Velocity, m/s
A – Area,
16. When 0.6 Kg of water per minute is passed through a tube of 2 cm diameter, it is
found to be heated from 20C to 60C. The heating is achieved by condensing steam on
the surface of the tube and subsequently the surface temperature of the tube is
maintained at 90C. Determine the length of the tube required for fully developed flow.
98
0.6
kg / s
60
= 0.01 Kg/s
Diameter D = 2 cm = 0.02 m
Given : Mass m = 0.6 Kg/min =
Inlet temperature Tmi  20C
Outlet temperature Tmo  60C
Tube surface temperature Tw  90C
To find: length of the tube (L)
Solution:
Bulk mean temperature Tm 
20  60
2
Tm  40C
Tmi  Tmo
2

Properties of water at 40C:
 = 995 Kg/m3
 = 0.657  10 -6m2 / s
Pr = 4.340
K = 628  10 -3 W/mK
CP  4.178 KJ/KgK = 4178 J/KgK
Mass flow rate m =  A U
m
 U=
A
0.01
=

995  (0.02)2
4
Velocity U = 0.031 m/s
Let us first determine the type of flow
UD
Re 

0.031 0.02
 Re 
0.657  10 6
Re  943.6
Since Re < 2300, flow is laminar
For laminar flow,
Nusselt number NU = 3.66
99
We know
NU 
hD
K
h  0.02
628  10-3
 h = 114.9 W/m2K
Heat transfer Q = mCP t
 3.66 =
= mCP (Tmo  Tmi )
= 0.01 4178  (60-20)
Q = 1671.2 W
We know that Q = h A 
= h    D  L  (Tw  Tm )
= 1671.2 =114.9    0.02  L  (90-40)
L = 4.62m
17. Water at 50C enters 50 mm diameter and 4 m long tube with a velocity of 0.8 m/s.
The tube wall is maintained at a constant temperature of 90C. Determine the heat
transfer coefficient and the total amount of heat transferred if exist water temperature
is 70C.
Given:
Inner temperature of water Tmi = 50C
Diameter
D = 50mm = 0.05 m
Length
L =4m
Velocity
U = 0.8 m/s
Total wall temperature
Tw = 90C
Exit temperature of water Tmo = 70C
To find:
1. Heat transfer coefficient (h)
2. Heat transfer (Q)
Solution:
100
Bulk mean temperature Tm 
50  70
2
Tm  60C
Tmi  Tmo
2

Properties of water at 60C:
 = 985 Kg/m3
 = 0.478  10-6m2 / s
Pr  3.020
K = 651.3  10-3 W/mK
Let us first determine the type of flow:
UD
Re 

0.8  0.05
=
0.478  10 -6
Re  8.36  104
Since Re > 2300, flow is turbulent
L
4

 80
D 0.05
L
 80 > 60
D
Re = 8.36  104  10,000
Pr  3.020  0.6 < Pr < 160
L
ratio is greater than 60. Re value is greater than 10,000 and Pr value is in between 0.6 and
D
160 so,
Nusselt number NU = 0.023 (Re)0.8 (Pr)n
[Inlet temperature 50C, Exit temperature 70C
 Heating Process, So n = 0.4]
 Nu = 0.023  (8.36  10 4 )0.8  (3.020)0.4
Nu = 310
We know that Nu=
310 
hD
K
h  0.05
651.3  10 3
Heat transfer coefficient h = 4039.3 W/m2K
Heat transfer Q = h A (Tw – Tm)
101
= h    D  L  (Tw  Tm )
= 4093.3    0.05  4  (90 - 60)
Q = 76139 W
18. What flows through 0.8 cm diameter, 3m long tube at an average temperature of
40C. The flow velocity is 0.65 m/s and tube wall temperature is 140C. Calculate the
average heat transfer coefficient.
Given : Diameter of tube D = 0.8 cm = 0.008 m
Length
L
=3m
Average temperature Tm = 40C
Velocity U
= 0.65 m/s
Tube wall temperature Tw = 140C
To find: Heat transfer coefficient (h)
 = 995 Kg/m3
 = 0.657  10-6 m2 / s
Pr  4.340
K  628  10 3 W/mK
We know Re =
UD

0.65  0.008
0.657  10 6
Re  7914.76
Since Re > 2300, flow is turbulent.
L
3

 375
D 0.008
L
10 <  400
D
L
ratio is in between 10 and 400, Re < 10000, so Nusselt Number Nu = 0.036 (Re)0.8 (Pr)0.33
D

D
L
 
0.055
 0.008 
 Nu = 0.036 (7914.76)0.8 (4.340)0.33  

 3 
 Nu = 55.44
We know
102
0.055
Nusselt number NU=
hD
K
h  0.008
628  103
Heat transfer coefficient h = 4352.3 W/m2K
55.44 
19. Air at 15C, 35 m/s, flows through a hollow cylinder of 4 cm inner diameter and 6
cm outer diameter and leaves at 45C. Tube wall is maintained at 60C. Calculate the
heat transfer coefficient between the air and the inner tube.
Given: Inner temperature of air Tmi = 15C
Velocity U
= 35 m/s
Inner diameter Di
= 4 cm = 0.04m
Outer diameter Do
= 6 cm = 0.06m
Exit temperature of air Tmo = 45C
Tube wall temperature Tw = 60C
To find: Heat transfer coefficient (h)
Solution: We know
Mean temperature Tm 
Tmi  Tmo
2
15  45
2
Tm  30C
Properties of air at 30C

 = 1.165 Kg/m3
 = 16  10-6 m2 / s
Pr = 0.701
K = 26.75  10-3 W/mK
Hydraulic of Equivalent diameter
103
4A
De 

P
4

D2  Di2 
4
 Do  Di 
D2  Di2 
=
Do  Di
=
(Do  Di ) (Do  Di )
(Do  Di )
= D o  Di
= 0.06 - 0.04
De = 0.02 m
Reynolds Number Re=
UDc

35  0.02
=
16  10-6
Re  43750
Since Re > 2300, flow is turbulent
For turbulent flow, general equation is (Re > 10000)
Nu = 0.023 (Re)0.8 (Pr)n
This is heating process so, n = 0.4
 Nu = 0.023  (43750)0.8  (0.701)0.4
Nu = 102.9
hDe
K
h  0.02
102.9 
26.75  10 3
 h = 137.7 W/m2K.
We know Nu =
20. Air at 30C, 6 m/s flows over a rectangular section of size 300  800 mm. Calculate
the heat leakage per meter length per unit temperature difference.
Given : Air temperature Tm = 30C
Velocity U = 6 m/s
Area A
= 300  800 mm2
2
A = 0.24 m
104
To find:
1. Heat leakage per metre length per unit temperature difference.
Solution:
Properties of air at 30C
 = 1.165 Kg/m3
 = 16  10-4 m2 / s
Pr = 0.701
K = 26.75  10 -3 W / mK
Equivalent diameter for 300  800 mm2 cross section is given by
4A 4  (0.3  0.8)

P
2 (0.3  0.8)
Where P - Perimeter = 2 (L+W)
 De  0.436 m
De 
We know
Reynolds Number Re =
UDe

6  0.436
16  10 6
Re = 16.3  10 4
Since Re > 2300, flow is turbulent.

For turbulent flow general equation is (Re > 10000)
Nu = 0.023 (Re)0.8 (Pr)n
Assuming the pipe wall temperature to be higher than a temperature. So heating
process  n = 0.4
 Nu = 0.023 (16.3  10 4 )0.8 (0.701)0.4
Nu  294.96
We know
hDe
Nusselt Number Nu =
K
h  0.436
 294.96 =
26.75  10 -3
Heat transfer coefficient  h = 18.09 W/m2K
Heat leakage per unit per length per unit temperature difference
Q=hP
105
= 18.09  2  (0.3 + 0.8
Q = 39.79 W
21. Air at 333K, 1.5 bar pressure, flow through 12 cm diameter tube. The surface
temperature of the tube is maintained at 400K and mass flow rate is 75 kg/hr.
Calculate the heat transfer rate for 1.5 m length of the tube.
Given : Air temperature Tm = 333 K = 60C
Diameter D
= 12 cm = 0.12 m
Surface temperature Tw
= 400 K = 127C
75 Kg
Mass flow rate m
= 75 kg/hr =
3600 s
m = 0.020 Kg/s
Length L = 1.5 m
To find:
1. Heat transfer rate (Q)
Solution:
Since the pressure is not much above atmospheric, physical properties of air may be taken at
atmospheric condition
Properties of air at 60C
 = 1.060 Kg/m3
 = 18.97  10-6 m2 / s
Pr = 0.696
K = 28.96  10-3 W/mK
Reynolds number Re =
UD

We know
Mass flow rate m p  U
106
0.020 = 1.060 

4
 D2  U
 0.020 = 1.060 

4
 (0.12)2  U
 U = 1.668 m/s
UD
(1)  Re =

1.668  0.12
18.97  10-6
Re = 10551.3

Since Re > 2300, so flow is turbulent
For turbulent flow, general equation is (Re>10000)
Nu  0.023  (Re)0.8  (0.696)0.4
Nu = 32.9
hD
We know Nu =
K
h  0.12
 32.9 =
28.96  10 -3
 h = 7.94 W/m2K
Heat transfer rate Q = h A (Tw  Tm )
 h  (  D  L)  (Tw  Tm )
 7.94  (  0.12  1.5)  (127  60)
Q  300.82 W
22. 250 Kg/hr of air are cooled from 100C to 30C by flowing through a 3.5 cm inner
diameter pipe coil bent in to a helix of 0.6 m diameter. Calculate the value of air side
heat transfer coefficient if the properties of air at 65C are
K = 0.0298 W/mK
 = 0.003 Kg/hr – m
Pr = 0.7
 = 1.044 Kg/m3
Given : Mass flow rate in = 205 kg/hr
205

Kg / s in = 0.056 Kg/s
3600
Inlet temperature of air Tmi = 100C
Outlet temperature of air Tmo = 30C
Diameter D = 3.5 cm = 0.035 m
107
Mean temperature Tm 
Tmi  Tmo
 65C
2
To find: Heat transfer coefficient (h)
Solution:
Reynolds Number Re =
UD


Kinematic viscosity  

0.003
Kg / s  m
3600
1.044 Kg/m3
v  7.98  10 7 m2 / s
Mass flow rate in =  A U
0.056  1.044 
0.056  1.044 

4

4
 D2  U
 (0.035)2  U
 U = 55.7 m/s
UD
(1)  Re =

55.7  0.035
7.98  10-7
Re = 2.44  106
=
Since Re > 2300, flow is turbulent
For turbulent flow, general equation is (Re > 10000)
Nu  0.023  (Re)0.8  (Pr)0.3
This is cooling process, so n = 0.3
 Nu = 0.023  (2.44  10 6 )0.8  (0.7)0.3
Nu  2661.7
We know that, Nu 
2661.7 
h  0.035
0.0298
hD
K
Heat transfer coefficient h = 2266.2 W/m2K
108
23. In a long annulus (3.125 cm ID and 5 cm OD) the air is heated by maintaining the
temperature of the outer surface of inner tube at 50C. The air enters at 16C and
leaves at 32C. Its flow rate is 30 m/s. Estimate the heat transfer coefficient between air
and the inner tube.
Given : Inner diameter Di = 3.125 cm = 0.03125 m
Outer diameter Do = 5 cm = 0.05 m
Tube wall temperature Tw = 50C
Inner temperature of air Tmi = 16C
Outer temperature of air tmo = 32C
Flow rate U = 30 m/s
To find: Heat transfer coefficient (h)
Solution:
Mean temperature Tm =
16  32
2
Tm  24C
Tmi  Tmo
2

Properties of air at 24C:
 = 1.614 Kg/m3
 = 15.9  10 -6 m2 / s
Pr = 0.707
K = 26.3  10 -3 W / mK
We know,
Hydraulic or equivalent diameter

4  D2  Di2 
4A
4
Dh 

P
 Do  Di 

Do  Di  Do  Di 
(Do  Di )
 Do  Di
= 0.05 – 0.03125
Dh = 0.01875 m
Reynolds number Re =
UDh

30  0.01875
15.9  10 6
Re = 35.3  10-6

109
Since Re > 2300, flow is turbulent
For turbulent flow, general equation is (Re > 10000)
Nu = 0.023 (Re)0.8 (Pr)n
This is heating process. So n = 0.4
 Nu = 0.023  (35.3  103 )0.8 (0.707)0.4
Nu  87.19
hDh
We know Nu =
K
h  0.01875
 87.19=
26.3  10-3
 h = 122.3 W/m2K
24. Engine oil flows through a 50 mm diameter tube at an average temperature of
147C. The flow velocity is 80 cm/s. Calculate the average heat transfer coefficient if the
tube wall is maintained at a temperature of 200C and it is 2 m long.
Given : Diameter D = 50 mm
= 0.050 m
Average temperature Tm
= 147C
Velocity
U
= 80 cm/s = 0.80 m/s
Tube wall temperature Tw = 200C
Length
L
= 2m
To find: Average heat transfer coefficient (h)
Solution : Properties of engine oil at 147C
 = 816 Kg/m3
 = 7  10-6 m2 / s
Pr = 116
K = 133.8  10-3 W/mK
We know
Reynolds number Re =
UD

0.8  0.05
7  106
Re = 5714.2

Since Re < 2300 flow is turbulent
110
L
2

 40
D 0.050
L
10   400
D
For turbulent flow, (Re < 10000)
0.8
Nusselt number Nu = 0.036 (Re) (Pr)
0.33
D
L
 
 0.050 
Nu  0.036 (5714.2)0.8  (116)0.33  

 2 
Nu  142.8
0.055
0.055
hD
K
h  0.050
 142.8 =
133.8  10 -3
 h = 382.3 W/m2K
We know Nu =
25. A system for heating water from an inlet temperature of 20C to an outlet
temperature of 40C involves passing the water through a 2.5cm diameter steel pipe.
The pipe surface temperature is maintained at 110C by condensing steam on its
surface. For a water mass flow rate of 0.5 kg/min, find the length of the tube desired.
Given : Inlet temperature Tmi = 20C
Outlet temperature Tmo = 40C
Diameter D = 2.5 cm = 0.025 m
Piper surface temperature Tw = 110C
Mass flow rate m = 0.5 Kg/min = 8.33  10-3 Kg/s
To find: Length of the tube (L)
Solution: We know
111
Bulk mean temperature Tm 
Tmi  Tmo
2
20  40
2
Tm  30C

Properties of water at 30C
 = 997 Kg/m3
 = 0.857  10 -6 m2 / s
Pr = 5.5
K = 610  10 -3 W/mK
CP  4.178 KJ/Kg K = 4178 J/Kg K
We know
UD
Reynolds number Nu =

We know that,
Mass flow rate in =  AU

8.33  10-3    D2  U
4

8.33  10 3  997   (0.025)2  U
4
 U = 0.017 m/s
UD
(1)  Re =

0.017  0.025

0.857  10 6
Re  495
Since Re < 2300, flow is laminar
For laminar flow,
Nusselt number Nu = 3.66
hD
We know that, Nu =
K
h  0.025
3.66 
610  10 3
Heat transfer coefficient h = 89.3 W/m2K
112
Heat transfer Q = m CP T
 m CP (Tmo  Tmi )
 8.33  10 3  4178 (40  20)
Q  696.05 W
Heat transfer Q = h A (Tw  Tm )
 h    D  L (Tw  Tm )
696.05  89.3    0.025  L  (110  30)
 L = 1.24 m
Formulae used for free convection
Tw  T
2
where Tw – Surface temperature in C
T - Fluid temperature in C
1. Film temperature Tf 
2. Coefficient of thermal expansion
1

Tf in K
3. Nusselt Number Nu =
hL
K
Where h – Heat transfer coefficient W/m2K
L – Length, m
K – Thermal conductivity, W/mK
4. Grashof number for vertical plate
g    L3  T
Gr 
v2
Where L - Length of the plate,
T - Tw  T
 - Kinematic viscosity, m2 / s,
 - Coefficient of thermal expansion.
5. If GrPr value is less than 109, flow is laminar. If GrPr value is greater than 109, flow is
turbulent.
i.e., GrPr > 109,  Laminar flow
GrPr > 109,  Turbulent flow
6. For laminar flow (Vertical plate):
113
Nusselt number Nu = 0.59 (GrPr)0.25
This expression is valid for,
104 < Gr Pr < 109
7. For turbulent flow (Vertical plate):
Nusselt Number Nu = 0.10 [Gr Pr]0.333
8. Heat transfer (vertical plate):
Q = h A (Tw - T)
9. Grashof number for horizontal plate:
g    Lc 3  T
Gr 
v2
W
Where Lc – Characteristic length =
2
W – Width of the plate.
10. For horizontal plate, upper surface heated,
Nusselt number Nu = 0.54 [Gr Pr]0.25
This expression is valid for
2  104  Gr Pr < 8  106
Nusselt number Nu = 0.15 [Gr Pr]0.333
This expression is valid for 8  106  Gr Pr <1011
11. For horizontal plate, lower surface heated
Nusselt Number Nu = 0.27 [Gr Pr]0.25
This expression is valid for 105 < Gr Pr < 1011
12. Heat transfer (Horizontal plate)
Q = (hu + hj)  A  (Tw - T)
Where hu – Upper surface heated, heat transfer coefficient W/m2 K
Hi – Lower surface heated, heat transfer coefficient, W/m2K
13. For horizontal cylinder
Nusselt number Nu = C [Gr Pr]m
14. For horizontal cylinder,
Heat transfer Q = h A (Tw - T)
114
Where A - DL
15.For sphere,
Nusselt number Nu = 2 + 0.43 [Gr Pr]0.25
Heat transfer Q = h  A  (Tw - T)
Where A - 4r2
16. Boundary layer thickness
 x  [3.93  (Pr)0.5 (0.952+Pr)0.25  (Gr)0.25 ]  x
26. A vertical plate of 0.75 m height is at 170 C and is exposed to air at a temperature
of 105C and one atmosphere calculate:
1. Mean heat transfer coefficient,
2. Rate of heat transfer per unit width of the plate
Given :
Length L
= 0.75 m
Wall temperature Tw
= 170C
Fluid temperature T = 105C
To find:
1. Heat transfer coefficient (h)
2. Heat transfer (Q) per unit width
Solution: Velocity (U) is not given. So this is natural convection type problem.
T  T
Film temperature Tf  w
2
170  105

2
Tf  137.5C
Pr operties of air at Tf = 137.5C = 140C
Density  = 0.854 Kg/m3
Kinematic viscosity  = 27.80  10 -6 m2 / s
Prandtl number Pr = 0.684
Thermal conductivity K = 34.89  103 W / mK
We know that
115
Coefficient of thermal expansion}  =
 =
1
Tf in K
1
137.5+273
1
410.5
  2.4  10 3 K 1

We know
g    L3  T
v2
9.81 2.4  10 -3  (0.75)3  (170  105)
 Gr =
(27.80  10 6 )2
Grahsof number Gr =
 Gr = 8.35  108
 Gr Pr = 8.35  108  0.684
Gr Pr = 5.71 108
Since Gr Pr < 109, flow is laminar
Gr Pr value is in between 104 and 109 i.e., 104 < Gr Pr < 109
So, Nusselt Number
Nu = 0.59 (Gr Pr)0.25
 = 0.59 (5.71 108 )0.25
Nu  91.21
We know
hL
K
h  0.75
 91.21 =
34.89  10-3
 h = 4.24 W/m2K
Nusselt number Nu =
Heat transfer coefficient h = 4.24 W/m2K
We know
Heat transfer Q = h A (Tw  T )
= 4.24  1 0.75  (170-105)
[
Q = 206.8 W
W = 1m]
116
27. A large vertical plate 4 m height is maintained at 606C and exposed to atmospheric
air at 106C. Calculate the heat transfer is the plate is 10 m wide.
Given :
Vertical plate length (or) Height L = 4 m
Wall temperature Tw = 606C
Air temperature T = 106C
Wide W
= 10 m
To find: Heat transfer (Q)
Solution:
Film temperature Tf 
Tw  T
2
606  106
2
Tf  356C

Properties of air at 356C = 350C
 = 0.566 Kg/m3
  55.46  10-6 m2 / s
Pr = 0.676
K = 49.08  10 -3 W/mK
Coefficient of thermal expansion}  =
1
Tf in K
1
1

356  273 629
 = 1.58  10-3K 1

g    L3  T
v2
9.81 2.4  10-3  (4)3  (606  106)
 Gr =
(55.46  10 6 )2
Grashof number Gr =
Gr = 1.61  1011
Gr Pr = 1.61  1011  0.676
Gr Pr = 1.08  1011
Since Gr Pr > 109, flow is turbulent
For turbulent flow,
Nusselt number Nu = 0.10 [Gr Pr]0.333
117
 Nu = 0.10 [1.08  1011]0.333
Nu = 471.20
We know that,
hL
K
h 4
 472.20 =
49.08  10-3
Heat transfer coefficient h = 5.78 W/m2K
Nusselt number Nu 
Heat transfer Q = h A T
 h  W  L  (Tw  T )
 5.78  10  4  (606  106)
Q  115600 W
Q = 115.6  103 W
28. A thin 100 cm long and 10 cm wide horizontal plate is maintained at a uniform
temperature of 150C in a large tank full of water at 75C. Estimate the rate of heat to
be supplied to the plate to maintain constant plate temperature as heat is dissipated
from either side of plate.
Given :
Length of horizontal plate L = 100 cm = 1m
Wide W
= 10 cm = 0.10 m
Plate temperature Tw = 150C
Fluid temperature T = 75C
To find: Heat loss (Q) from either side of plate
Solution:
Film temperature Tf 
Tw  T
2
150  75
2
Tf  112.5C

Properties of water at 112.5C
 = 951 Kg/m3
 = 0.264  10 -6 m2 / s
Pr = 1.55
K = 683  10 3 W/mK
118
Coefficient of thermal expansion}  =
1
T in K
f

1
112.5  273
  2.59  103 K 1
Grashof Number Gr =
g    L3  T
v2
For horizontal plate,
Characteristic length Lc 
W 0.10

2
2
Lc = 0.05 m
9.81 2.59  10-3  (0.05)3  (150  75)
(1)  Gr =
(0.264  106 )2
Gr = 3.41 109
Gr Pr = 3.41 109  1.55
Gr Pr = 5.29  109
Gr Pr value is in between 8  106 and 1011
i.e., 8  106 < Gr Pr < 1011
For horizontal plate, upper surface heated:
Nusselt number Nu = 0.15 (Gr Pr)0.333
 Nu = 0.15 [5.29  109 ]0.333 
 Nu = 259.41
We know that,
Nusselt number Nu =
huLc
K
hu  0.05
683  10 3
hu = 3543.6 W/m2K
259.41 
Upper surface heated, heat transfer coefficient hu = 3543.6 W/m2K
For horizontal plate, lower surface heated:
Nusselt number Nu = 0.27 [Gr Pr]0.25
119
 Nu = 0.27 [5.29  10 9 ]0.25
Nu = 72.8
We know that,
Nusselt number Nu =
h1Lc
K
h1Lc
K
h1  0.05
72.8 
683  10 3
h1  994.6 W/m 2K
72.8 
Lower surface heated, heat transfer coefficient h1 = 994.6 W/m2K
Total heat transfer Q = (hu + h1)  A  T
= (hu + h1)  W  L  (Tw - T)
= (3543.6 + 994.6)  0.10  (150 – 75)
Q = 34036.5 W
29. A hot plate 20 cm in height and 60 cm wide is exposed to the ambient air at 30C.
Assuming the temperature of the plate is maintained at 110C. Find the beat loss from
both surface of the plate. Assume horizontal plate.
Given:
Height (or) Length of the Plate L
= 20 cm = 0.20 m
Wide W
= 60 cm = 0.60 m
Fluid temperature T
= 30C
Plate surface temperature Tw = 110C
To find:
Heat loss from both the surface of the plate (Q)
Solution:
Film temperature Tf 
Tw  T
2
110  30
2
Tf  70C

Properties of air at 70C:
 = 1.029 Kg/m3
120
 = 20.02  10-6 m2 / s
Pr = 0.694
K = 29.66  10-3 W/mK
We know
Coefficient of thermal expansion}  =
1
1

Tf in K 70  273
1
343
 2.91 103 K 1

 = 2.91 103K 1
We know
Grashof number Gr =
g    L3  T
v2
Where Lc  Characteristic length =
W
2
0.60
 0.30 m
2
Lc  0.30 m
Lc 
(1)  Gr =
9.81 2.91 10-3  (0.30)3  (110  30)
(20.02  10 6 )2
Gr = 1.5384  108
Gr Pr = 1.5384  108  0.694
Gr Pr = 1.0676  108
Gr Pr value is in between 8  106 and 1011
i.e., 8  106 < Gr Pr < 1011
For horizontal plate, Upper surface heated,
Nusselt number Nu = 0.15 (Gr Pr)0.333
121
0.333
 Nu = 0.15 1.0676  108 
Nu = 70.72
We know that,
hL
Nusselt number Nu = u c
K
hu  0.30
70.72 
29.66  10 3
hu  6.99 W/m2K
Upper surface heated, heat transfer coefficient hu = 6.99 W/m2K
For horizontal plate, lower surface heated:
Nusselt number Nu = 0.27 (Gr Pr)0.25
= 0.277 [1.06  108]0.25
Nu = 28.15
We know that,
hL
Nusselt number Nu = 1 c
K
h1  0.30
28.15 
29.66  10 3
 h1 = 2.78 W/m2K
Lower surface heated, heat transfer coefficient h = 2.78 W/m2K
Total heat transfer Q = (hu  h1)  A  T
 (hu  h1 )  W  L  (Tw  T )
 Q = (6.99+2.78)  0.60  0.20  (110-30)
Q = 93.82 W
30. A vertical pipe 80 mm diameter and 2 m height is maintained at a consent
temperature of 120C. The pipe is surrounded by still atmospheric air at 30C. Find
heat loss by natural convection.
Given :
Vertical pipe diameter D
Height (or) Length L
Surface temperature Tw
Air temperature T
= 80 mm = 0.080 m
=2m
= 120 C
= 30C
To find: Heat loss (Q)
Solution: We know
122
Film temperature Tf 
Tw  T
2
120 + 30
2
Tf  75C

Properties of air at 75C
 = 1.0145 Kg/m3
 = 20.55  10 -6 m2 / s
Pr = 0.693
K = 30.06  10 -3 W/mK
We know  =
1
Tf in K
1
 2.87  10 3 K 1
75 + 273
 = 2.87  10-3K 1

We know
g    L3  T
Gr =
v2
Gr =
9.81 2.87  10 -3  (2)3  (120  30)
(20.55  10 6 )2
Gr = 4.80  1010
Gr Pr = 4.80  1010  0.693
Gr Pr = 3.32  1010
Since Gr Pr > 109 , flow is turbulent
For turbulent flow,
Nu = 0.10 [Gr Pr]0.333
= 0.10 [3.32  1010]0.333
Nu = 318.8
We know that,
123
Nusselt Number Nu =
hL
K
h2
30.06  10 3
Heat transfer coefficient, h = 4.79 W/m 2K
Heat loss Q = h  A  T
= h    D  L  (Tw  T )
318.8 
 4.79    0.080  2  (120  30)
Q  216.7 W
124
UNIT – III
Introduction to Physical mechanism
Radiation properties
Radiation shape factors
Heat exchange between non- black bodies
Radiation shields.
125
RADIATION
PART – A
1. Define Radiation.
The heat transfer from one body to another without any transmitting medium is
known as radiation. It is an electromagnetic wave phenomenon.
2. Define emissive power [E]
The emissive power is defined as the total amount of radiation emitted by a body per
unit time and unit area. It is expressed in W/m2.
3. Define monochromatic emissive power. [Eb]
The energy emitted by the surface at a given length per unit time per unit area in all
directions is known as monochromatic emissive power.
4. What is meant by absorptivity?
Absorptivity is defined as the ratio between radiation absorbed and incident radiation.
Radiation absorbed
Absorptivity  
Incident radiation
5. What is meant by reflectivity?
Reflectivity is defined as the ratio of radiation reflected to the incident radiation.
Reflectivity  
Radiation reflected
Incident radiation
Absorptivity  
Radiation absorbed
Incident radiation
6. What is meant by transmissivity?
Transmissivity is defined as the ratio of radiation transmitted to the incident radiation.
Transmissivity  
Radiation transmitted
Incident radiation
7. What is black body?
Black body is an ideal surface having the following properties.
1. A black body absorbs all incident radiation, regardless of wave length and direction.
126
2. For a prescribed temperature and wave length, no surface can emit more energy than
black body.
8. State Planck’s distribution law.
The relationship between the monochromatic emissive power of a black body and
wave length of a radiation at a particular temperature is given by the following expression, by
Planck.
Eb 
C1 5
 C2 


e   T  1
Where Eb = Monochromatic emissive power W/m2
 = Wave length – m
c1 = 0.374  10-15 W m2
c2 = 14.4  10-3 mK
9. State Wien’s displacement law.
The Wien’s law gives the relationship between temperature and wave length
corresponding to the maximum spectral emissive power of the black body at that temperature.

T = c3
Where c3 = 2.9  10-3


mas
mas
[Radiation constant]
T = 2.9  10-3 mK
10. State Stefan – Boltzmann law.
The emissive power of a black body is proportional to the fourth power of absolute
temperature.
Eb  T 4
Where
Eb
=
 T4
Eb
=
Emissive power, w/m2

=
Stefan. Boltzmann constant
T
=
=
5.67  10-8 W/m2 K 4
Temperature, K
11. Define Emissivity.
127
It is defined as the ability of the surface of a body to radiate heat. It is also defined as
the ratio of emissive power of any body to the emissive power of a black body of equal
temperature.
Emissivity  
E
Eb
12. What is meant by gray body?
If a body absorbs a definite percentage of incident radiation irrespective of their wave
length, the body is known as gray body. The emissive power of a gray body is always less
than that of the black body.
13. State Kirchoff’s law of radiation.
This law states that the ratio of total emissive power to the absorbtivity is constant for
all surfaces which are in thermal equilibrium with the surroundings. This can be written as
E1
1

E2
2

E3
3
It also states that the emissivity of the body is always equal to its absorptivity when
the body remains in thermal equilibrium with its surroundings.
1 = E1; 2 = E2 and so on.
14. Define intensity of radiation (Ib).
It is defined as the rate of energy leaving a space in a given direction per unit solid
angle per unit area of the emitting surface normal to the mean direction in space.
E
In  b

15. State Lambert’s cosine law.
It states that the total emissive power Eb from a radiating plane surface in any
direction proportional to the cosine of the angle of emission
Eb

cos 
16. What is the purpose of radiation shield?
Radiation shields constructed from low emissivity (high reflective) materials. It is
128
used to reduce the net radiation transfer between two surfaces.
17. Define irradiation (G)
It is defined as the total radiation incident upon a surface per unit time per unit area. It
is expressed in W/m2.
18. What is radiosity (J)
It is used to indicate the total radiation leaving a surface per unit time per unit area. It
is expressed in W/m2.
19. What are the assumptions made to calculate radiation exchange between the
surfaces?
1. All surfaces are considered to be either black or gray
2. Radiation and reflection process are assumed to be diffuse.
3. The absorptivity of a surface is taken equal to its emissivity and independent of
temperature of the source of the incident radiation.
20. What is meant by shape factor?
The shape factor is defined as the fraction of the radiative energy that is diffused from
on surface element and strikes the other surface directly with no intervening reflections. It is
represented by Fij. Other names for radiation shape factor are view factor, angle factor and
configuration factor.
PART – B
1. A black body at 3000 K emits radiation. Calculate the following:
Monochromatic emissive power at 7 m wave length.
Wave length at which emission is maximum.
Maximum emissive power.
Total emissive power,
Calculate the total emissive of the furnace if it is assumed as a real surface
having emissivity equal to 0.85.
Given: Surface temperature T = 3000K
i)
ii)
iii)
iv)
v)
Solution:
129
1. Monochromatic Emissive Power :
From Planck’s distribution law, we know
C1 5
Eb 
 C2 


e   T  1
[From HMT data book, Page No.71]
Where
c1 = 0.374  10-15 W m2
c2 = 14.4  10-3 mK
 = 1  10-6 m
[Given]

Eb
0.374  1015 [1 10 6 ]5

 144  103 
1 106  3000 

 1
Eb  3.10  1012 W/m2
2. Maximum wave length (max)
From Wien’s law, we know
max T  2.9  10 3 mK

2.9  10 3
3000
= 0.966  10 -6m
max =
max
3. Maximum emissive power (Eb) max:
Maximum emissive power
(Eb)max = 1.307  10-5 T5
= 1.307  10-5  (3000)5
(Eb)max = 3.17  1012 W/m2
4. Total emissive power (Eb):
From Stefan – Boltzmann law, we know that
Eb
=  T4
[From HMT data book Page No.71]
Where 
 Eb
Eb
= Stefan – Boltzmann constant
= 5.67  10-8 W/m2K4
= (5.67  10-8) (3000)4
= 4.59  106 W/m2
130
5. Total emissive power of a real surface:
(Eb)real =   T4
Where  =
Emissivity = 0.85
(Eb)real = 0.85  5.67  108  (3000)4
(Eb )real  3.90  106 W / m2
2. A black body of 1200 cm2 emits radiation at 1000 K. Calculate the following:
1.
2.
3.
4.
Total rate of energy emission
Intensity of normal radiation
Wave length of maximum monochromatic emissive power.
Intensity of radiation along a direction at 60 to the normal.
Solution:
From Stefan – Boltzmann law.
1. Energy emission Eb =  T4
[From HMT data book, Page No.71]
Eb = 5.67  108  (1000)4

Eb = 5.67  103 W/m2
Here

Area
= 1200  10-4 m2,
Eb = 5.67103  1200  10-4
Eb = 6804 W
2. Intensity of normal radiation
In 
Eb
=

56.7  103 W / m2

In = 18,048 W/m2
3. From Wien’s law, we know that
max T = 2.9  10-3 mK

2.9  10-3
3000
= 2.9  10-6 m
max =
max
max = 2.9 
[
1 = 10 -6 m]
131
3. Assuming sun to be black body emitting radiation at 6000 K at a mean distance of 12
 1010 m from the earth. The diameter of the sun is 1.5  109 m and that of the earth is
13.2  106 m. Calculation the following.
1. Total energy emitted by the sun.
2. The emission received per m2 just outside the earth’s atmosphere.
3. The total energy received by the earth if no radiation is blocked by the earth’s
atmosphere.
4. The energy received by a 2  2 m solar collector whose normal is inclined at 45 to
the sun. The energy loss through the atmosphere is 50% and the diffuse radiation is
20% of direct radiation.
Given: Surface temperature T = 6000 K
Distance between earth and sun R = 12  1010 m
Diameter on the sun D1 = 1.5  109 m
Diameter of the earth D2 = 13.2  106 m
Solution:
1. Energy emitted by sun Eb

=  T4
Eb = 5.67  10-8  (6000)4
 = Stefan - Boltzmann constant
[
Eb
= 5.67  10-8 W / m2 K 4 ]
= 73.4  10 6 W/m 2
Area of sun A 1  4 R12
 1.5  109 
= 4  

2


2
A 1  7  1018 m 2
 Energy emitted by the sun
Eb
= 73.4  106  7  1018
Eb  5.14  1026 W
2. The emission received per m2 just outside the earth’s atmosphere:
The distance between earth and sun
R = 12  1010 m
132
Area, A = 4 R 2
= 4    (12  1010 )2
A = 1.80  1023 m2
 The radiation received outside the earth atmosphere per
m2
Eb
A
5.14  1026
=
1.80  1023
= 2855.5 W/m2
=
3. Energy received by the earth:
Earth area =
=

4

4
(D2 )2
 [13.2  106 ]2
Earth area = 1.36  10 4m2
Energy received by the earth
 2855.5  1.36  104
 3.88  1017 W
4. The energy received by a 2  2 m solar collector;
Energy loss through the atmosphere is 50%. So energy reaching the earth.
 100 - 50 = 50%
= 0.50
Energy received by the earth
 0.50  2855.5
 1427.7 W/m2
......(1)
Diffuse radiation is 20%
 0.20  1427.7 = 285.5 W/m2
Diffuse radiation = 285.5 W/m2
.........(2)
Total radiation reaching the collection
 142.7  285.5
 1713.2 W/m2
133
Plate area
= A  cos 
= 2  2  cos 45
= 2.82 m2
Energy received by the collector
 2.82  1713.2
 4831.2 W
4. A large enclosure is maintained at a uniform temperature of 3000 K. Calculate the
following:
1. Emissive power
2. The wave length 1 below which 20 percent of the emission is concentrated and the
wave length 2 above which 20 percent of the emission is concentrated.
3. The maximum wave length.
4. Spectral emissive power.
5. The irradiation incident.
Given : Surface temperature T = 3000 K
1. Emissive power Eb =  T4
 5.67  108  (3000)4
Eb  4.59  106 W/m2
2. The wave length 1 corresponds to the upper limit, containing 20% of emitted radiation.
Eb(0-1T)

 0.20 , corresponding
 T4
1T = 2666 K
[From HMT data book, Page No.72]
 1T  2666 K
 1 =

2666
3000
1  0.88
The wave length 2 corresponds to the lower limit, containing 20% of emitted
radiation.
134
Eb(0-1T)

 T4
Eb(0-1T)

 (1  0.20)
 0.80, corresponding
 T4
2 T  6888 K
2 =

6888
3000
2  2.2 
3. Maximum wave length (max):
max T = 2.9  10-3 mK
2.9  10
3000
= 9.6  10-7m
-3
max =
max  0.96  106 m
4. Spectral Emissive Power:
From Planck’s distribution law, we know
Eb 
C1max 5
 C2 


e   T  1
[From HMT data book, Page No.71]
C1  0.374  10 15 W m2
where
C2  14.4  10 3 mK
 0.374  10   0.96  10 

15

Eb

Eb = 3.1  1012 W/m2
6
5


14.4  10 3


6
0.96  10  3000  1
e
5. Irradiation:
The irradiation incident on a small object placed within the enclosure may be treated
as equal to emission from a black body at the enclosure surface temperature.
So, G = Eb = 4.59  106 W/m2 .
5. The sun emits maximum radiation at  = 0.52. Assuming the sun to be a black body,
calculate the surface temperature of the sun. Also calculate the monochromatic emissive
135
power of the sun’s surface.
Given : max  0.52  0.52  10 6 m
To find : 1. Surface temperature T.
2. Monochromatic emissive power Eb.
Solution:
1. From Wien’s law, we know
max T = 2.9  103 mK
[From HMT data book, Page No.71]

T=
2.9  10-3
0.52  106
T = 5576 K
2. Monochromatic emissive power (Eb):
From Planck’s law, we know
c1 5
Eb 
 C2 


e   T  1
[From HMT data book, Page No.71]
where
C1  0.374  10 15 W m 2
C2  14.4  10 3 mK
 = 0.52  10-6 m

E b
0.374  10 15  (0.52  10 6 )5



14.4  10 3


6
0.52  10  5576  1
e
Eb  6.9  1013 W / m2
6. A furnace wall emits radiation at 2000 K. Treating it as black body radiation,
calculate
1. Monochromatic radiant flux density at 1m wave length.
2. Wave length at which emission is maximum and the corresponding emissive
power.
3. Total emissive power
Given: Temperature T = 2000 K;  = 1 m = 1  10-6
Solution:
136
1. Monochromatic emissive power (Eb):
Eb 
c1 5
 C2 


e   T  1
[From HMT data book, Page No.71]
C1  0.374  10 15 W m2
where
C2  14.4  103 mK
 = 1  m = 1 10-6 m

Eb 
[Given]
0.374  10 15  (1 10 6 )5
 14.4  10 3 


1 10 6  2000  1
e
Eb  2.79  1011 W / m2
2. Maximum Wave Length (max):
From Wien’s Law, we know that
max T = 2.9  10 3 mK
[From HMT data book, Page No.71]
2.9  10 3
max 
T
2.9  10 3
=
 1.45  10 6 m
2000
max  1.45 
Corresponding emissive power
c1max 5
E b 
 C2 


 T
e  max  1

0.374  10 15  1.45  10 6 
5


14.4  10 3


6
1.45  10  2000 1
e
 4.09  1011 W / m2
3. Total emissive power (Eb):
137
From Stefan – Boltzmann law, we know
Eb =  T4
Where  - Stefan – Boltzmann constant
 5.67  10 8 W / m2 K 4
Eb = 5.67  10 -8  (2000)4
Eb  907.2  103 W/m2
7. The temperature of a black surface 0.25 m2 of area is 650C. Calculate,
1. The total rate of energy emission
2. The intensity of normal radiation.
The wavelength of maximum monochromatic emissive power.
Given :
A = 0.25 m2
T = 650 + 273 = 923 K
To find : 1. Eb ; 2. In ; 3. max
Solution:
1. We know
Emissive power Eb =  T4
= 5.67  108  (923)4
Eb = 41151.8 W/m2
Area = 0.25 m2
Here

Eb = 41151.8 W/m2  0.25 m2
Eb  10.28  103 Watts
2. We know
Intensity  In 
Eb
=

10.28  103

In = 3274.7 W
3. From Wien’s law,
max T = 2.9  10 3 m
max
2.9  10-3
=
923
max  3.13  10 6 m
138
8. Calculate the heat exchange by radiation between the surfaces of two long cylinders
having radii 120mm and 60mm respectively. The axis of the cylinder are parallel to
each other. The inner cylinder is maintained at a temperature of 130C and emissivity
of 0.6. Outer cylinder is maintained at a temperature of 30C and emissivity of 0.5.
Given : r1 = 60 mm
= 0.060 m
r2 = 120 mm
= 0.12
T1 = 130C + 273
= 403
1 = 0.6
T2 = 30C + 273 = 303 K
2 = 0.5
To find : Heat exchange (Q)
Solution: Heat exchange between two large concentric cylinder is given by
Q    A T14  T2 4 
.....(1)
[From equation No.27]
1
where  

1 A1  1

  1
1 A 2   2 
1
[ A   DL]
1  D1L 2  1


1
0.6  D2L 2  0.5 
 =
=
1
[
1 0.12  1


1
0.6 0.24  0.5 
L1  L 2  1]
  0.46
(1) 
Q12  0.46  5.67  10 8    D1  L  (403)4  (303) 4 
= 0.46  5.67  10 8    0.12  1 (403) 4  (303) 4 
Q12  176.47 W
9. Two concentric spheres 30 cm and 40 cm in diameter with the space between them
evacuated are used to store liquid air at - 130C in a room at 25C. The surfaces of the
spheres are flushed with aluminium of emissivity  = 0.05. Calculate the rate of
evaporation of liquid air if the latent heat of vaporization of liquid air is 220 kJ/kg.
Given: Inner diameter D1 = 30 cm
139
= 0.30 m
Inner radius
r1 = 0.15 m
Outer diameter
D2 = 40 cm
= 0.40 m
Outer radius
r2 = 0.20 m
T1 = - 130C + 273
= 143 K
T2 = 25C + 273
= 298 K
 = 0.05
Latent heat of vapourisation = 220kJ /kg
= 220 103 J / kg
To find: Rate of evaporation
Solution: This is heat exchange between large concentric sphere problem.
Heat transfer Q12    A1 [T14  T24 ] ......(1)
1
Where  

1 A1  1

  1
1 A 2   2 

1
4 r12  1
1


 1
2 
0.05 4 r2  0.05 
[
=
1   2  0.05;A  4 r 2 ]
1
1
(0.15)2  1


 1
2 
0.05 (0.20)  0.05 
  0.032
10. A pipe of outside diameter 30 cm having emissivity 0.6 and at a temperature of 600
K runs centrally in a brick of 40 cm side square section having emissivity 0.8 and at a
temperature of 300K. Calculate the following:
1. Heat exchange per metre length.
2. Convective heat transfer coefficient when surrounding of duct is 280 K.
Given:
Pipe diameter
Surface area
D1 = 30 cm
D1 = 0.30 m
A1 =  D1L
=   0.30  1
A1 = 0.942 m2
140
1
T1
= 0.6
= 600 K
[ L  1 m]
Brick duct side = 40 cm = 0.40 m
Surface area A2 = (0.4  1)  4
[length L = 1m; No. of sides = 4]
A 2  1.6 m2
 2  0.8
T2  300 K
To find: 1. Heat exchange (Q)
2. Convective heat transfer coefficient (h) when
T = 280 K
Solution:
Case 1: We know that
Heat exchange Q12      A1 T14  T2 4  ....(1)
1

1 A1  1

  1
1 A 2   2 
1
=
1 0.942  1


 1

0.6
1.6  0.8 
where  
  0.55
(1)  Q12  0.55    A1  [T14  T2 4 ]
= 0.55  5.67  10 8  0.942  (600)4  (300)4 
Heat exchange Q12  3569.2 W / m
......(2)
Case (ii) :
Heat transfer by convection Q
= hA (T - T)
Q12 = h  A  (T2 - T)
Q12 = h  1  (300 – 280)
Q12  20h
.....(3)
Equating (2) and (3),
3569.2 = 20h
Heat transfer coefficient = 178.46 W/m2 K
141
11. Emissivities of two large parallel plates maintained at 800C and 300C are 0.5
respectively. Find net radiant hat exchange per square metre for these plates. Find the
percentage reduction in heat transfer when a polished aluminium radiation shield of
emissivity 0.06 is placed between them. Also find the temperature of the shield.
Given : T1 = 800C + 273
= 1073 K
T2 = 300C + 273
= 573 K
1 = 0.3
2 = 0.5
Shield emissivity 3 = 0.06
To find:
1. Net radiant heat exchange per square metre. (Q/A)
2. Percentage reduction in heat loss due to radiation shield.
3. Temperature of the shield (T3).
Solution: Heat exchange between two large parallel plates without radiation shield is given
by
Q12    A T14  T2 4 

1
1

1
1
2
=
1
1
1
1

1
0.3 0.5
  0.230
(1)  Q12  0.230    A  [T14  T2 4 ]
= 0.230  5.67  10 8  A  (1073)4  (573)4 
Q12
 15,879.9 W/m2
A
Heat transfer square metre without radiation shield
Q12
 15.87 k W/m2
A
......(1)
Heat exchange between plate 1 and radiation shield 3 is given by
142
Q13   A [T14  T2 4 ]
(1) 

1
1

Q13 
1
1

1
3
  A [T14  T3 4 ]
1
1

1
.....(A)
1
3
Heat exchange between radiation shield 3 and plate 2 is given by
Q32   A [T3 4  T2 4 ]
=
Where
1
3

Q32 

1
1
2
  A T3 4  T2 4 
1
3
We know

=
1

2
.....(B)
1
Q13 = Q32
 A [T14  T3 4 ]
1
1

1

1
3

 A [T3 4  T2 4 ]
1
1
3

1
2
1
[T14  T3 4 ]
[T3 4  T2 4 ]

1
1
1
1

1

1
0.3 0.06
0.06 0.5
(1073)4  (T3 4 ) (T3 4 )  (573)4

19
17.6
4
4
17.6 (1073)  (T3 ) 
=
+ (573)4
19
= 0.926 (1073)4  (T3 )4   (573)4
=


T3 4

T3 4

T3 4 = 0.926 (1073)4  0.926  (T3 )4   (573)4
(T3 )4  0.926 (T3 4 )  1.33  1012
(1.926) (T3 )4 = 1.33  1012
(T3 )4 = 6.90  1011
T3  911.5 K
143
Radiation shield temperature T3  911.5 K
Substituting T3 value in equation (A) (or) equation (B), Heat transfer with radiation shield
5.67  10 8  A  (1073)4  (911.5)4 

Q13 
1
1

1
0.3 0.06
Q13
 1895.76 W/m2
A
Heat transfer with radiation shield

Q13
 1.89kW / m2 ......(2)
A
Reduction in heat loss due to radiation shield

Q without shield  Q with shield Q12  Q13

Q without shield
Q12
15.87 - 1.89
15.87
= 0.88 = 88%
=
12. A pipe of diameter 30 cm, carrying steam runs in a large room and is exposed to air
at a temperature of 25C. The surface temperature of the pipe is 300C. Calculate the
loss of heat of surrounding per meter length of pipe due to thermal radiation. The
emissivity of the pipe surface is 0.8.
What would be the loss of heat due to radiation of the pipe is enclosed in a 55 cm
diameter brick of emissivity 0.91?
Given :
Case 1: Diameter of pipe D1 = 30 cm = 0.30 m
Surface temperature T1 = 300C + 273
= 573 K
Air temperature
T2 = 25C + 273
= 298 K
Emissivity of the pipe
1 = 0.8
Case 2: Outer diameter
D2 = 55 cm = 0.55m
Emissivity
2 = 0.91
To find: 1. Loss of heat per metre length (Q/L).
2. Reduction in heat loss.
144
Solution:
Case 1:
Heat transfer Q  1  A T14  T2 4 
 1     DL  T14  T2 4 
[
A =  DL
Q = 0.8  5.67  10-8    0.30  L (573)4  (298)4 

Q/L = 4271.3 W/m
Heat loss per metre length = 4271.3 W/m
Case 2: When the 30 cm dia pipe is enclosed in a 55 cm diameter pipe, heat exchange
between two large concentric cylinder is given by
Q    A1 T14  T2 4 
1

1 A1  1

  1
1 A 2   2 
1
=
1  D1L 2  1


 1

0.8  D2L 2  0.91 
where  
=
=
1
D  1
1

 1
 1
0.8 D2  0.91 
1
1 0.30  1


 1

0.8 0.55  0.91 
  0.76
Substituting emissivity  value in equation (A),
(A)  Q  0.76  5.67  10 8    D1  L1 (573)4  (298)4 
Q
= 0.76  5.67  10 8    0.30  (573)4  (298)4 
L
Q
 4057.8 W / m
L
Reduction in heat loss
145
= 4271.3 – 4057.8
= 21.3.4
13. Emissivities of two large parallel plates maintained at T1 K and T2 K are 0.6 and 0.6
respectively. Heat transfer is reduced 75 times when a polished aluminium radiation
shields of emissivity 0.04 are placed in between them. Calculate the number of shields
required.
Given: 1 = 0.6
2 = 0.6
Heat transfer reduced = 75 times
Emissivity of radiation shield, s = 3 = 0.04
To find: Number of screens require.
Solution: Heat transfer with n shield is given by
A [T14  T2 4 ]
Qin =
.......(1)
1 1  2n 
    (n  1)
1  2   s 
Heat transfer without shield, i.e., n=0
(1) 
A [T14  T2 4 ]
1 1
 1
Q12 =
1
.......(2)
2
Heat transfer is reduced 75 times

Q without shield
 75
Q with shield

Q12
 75
Q13
A [T14  T2 4 ]
1 1
 1
1  2
A [T14  T2 4 ]
(2)

(1)
1
1
1

1


1 2n
 (n  1)
2 s
1 2n
2 s
1
1

 75
 (n  1)
1
2
 75
1
146



1
1
2n

 (n  1)
0.6 0.6 0.04
= 75
1
1

1
0.6 0.6
3.33  50n  (n  1)
 75
2.33
50n n-1 = 171.67
49n - 1 = 171.67
49n
= 172.67
n
= 3.52  4
n=4
14. Find the relative heat transfer between two large plane at temperature 1000 K and
500 K when they are
1. Black bodies
2. Gray bodies with emissivities of each surface is 0.7.
Given: T1 = 1000 K
T2 = 500 K
1 = 0.7
2 = 0.7
Solution :
Case 1: Heat exchange between two large parallel plate is given by
Q   A  T14  T2 4 
For black bodies,   1
Q = A  T14  T2 4 
Q
 5.67  108 (1000)4  (500)4 
A
Q
 53.15  103 W / m2
A
Case 2: Q   A  T14  T2 4 
147

1
1
 =

1
1
2
1
1
1
1

1
0.7 0.7
  0.538
 Q  0.538  A  5.67  10 8 (1000)4  (500) 4 
Q
 28.6  103 W / m2
A
15. The inner sphere of liquid oxygen container is 40 cm diameter and outer sphere is 50
cm diameter. Both have emissivities 0.05. Determine the rate at which the liquid oxygen
would evaporate at -183C when the outer sphere at 20C. Latent heat of oxygen is 210
kJ/kg.
Given : Inner diameter
D1 = 40 cm = 0.40 m
Inner radius
r1 = 0.20 m
Outer diameter
D2 = 50 cm = 0.50 m
Outer radius r2 = 0.25 m
Emissivity
1 = 0.05
2 = 0.05
Inner temperature T1 = -183C + 273 = 90K
Outer temperature T2 = 20C + 273
= 293 K
Latent heat of oxygen = 210 kJ / kg
= 210  103 J/kg
To find : Rate of evaporation
Solution :
This is heat exchange between two large concentric spheres problem.
......(1)
Heat transfer Q    A1 T14  T2 4 
[From equation No.27]
1
where  

1 A1  1

  1
1 A 2   2 
148
1
=
4 r  1
1


 1

0.05 4 r  0.05 
1
=
2
r  1
1

 12 
 1
0.05 r2  0.05 
1

1
(0.20)2  1


 1
2 
0.05 (0.25)  0.05 
2
1
2
2
[
A = 4 r 2 ]
  0.031
(1)  Q12  0.031 5.67  10 8  4   (90)4  (293) 4 
Q12  6.45 W
[Negative sign indicates heat is transferred from outer surface to inner surface.]
Rate of evaporation =
Heat transfer
Latent heat
6.45 W

210  103 J/kg

6.45 J/s
210  103 J / kg
Rate of evaporation = 3.07  10-5kg / s
16. Emissivities of two large parallel plates maintained at 800C are 0.3 and 0.5
respectively. Find the net radiant heat exchange per square metre of the plates. If a
polished aluminium shield ( = 0.05) is placed between them. Find the percentage of
reduction in heat transfer.
Given : T1 = 800C + 273 = 1073 K
T2 = 300C + 273 = 573 K
1 = 0.3
2 = 0.5
Radiation shield emissivity 3 = 0.05
To find:
149
Q 
1. Net radiant heat exchange per square metre  12 
 A 
2. Percentage of reduction in heat loss due to radiation shield.
Solution:
Case 1 : Heat transfer without radiation shield:
Heat exchange between two large parallel plats without radiation shield is given by
Q12    A T14  T2 4 

1
1

1
1
2
=
1
1
1
1

1
0.3 0.5
  0.230
 Q12  0.230  5.67  10 8  A  (1073)4  (573)4 
Heat transfer
Q12
 15.8  103 W / m2
A
Case 2: Heat transfer with radiation shield:
Heat exchange between plate 1 and radiation shield 3 is given by
Q13   A [T14  T2 4 ]

where
1
1

Q13 

1
1
3
1
 A [T14  T3 4 ]
1
1

1
3
.....(A)
1
Heat exchange between radiation shield 3 and plate 2 is given by
Q32   A [T3 4  T24 ]
150
 =
Where
1
3

Q32 

1
1
 A T3 4  T2 4 
1
3
We know

=



1
2
.....(B)
1
Q13 = Q32
 A [T14  T3 4 ]
1
1

1
2

1
3

 A [T3 4  T2 4 ]
1
1
3

1
2
1
(1073)4 - (T3 4 ) (T3 4 )  (573)4 ]
=

1
1
1
1

1

1
0.3 0.05
0.3 0.05
(1073)4  (T3 4 ) (T3 4 )  (573)4

22.3
21
13
4
2.78  10  21T3  22.3T3 4  2.4  1013

3.02  1013  43.3T3 4
Shield temperature  T3  913.8 K
Substitute T3 value in equation (A) or (B).
Substituting T3 value in equation (A) (or) equation (B),
8
4
4
5.67  10  A  (1073)  (913.8) 



Q 
1
1
radiation shield  13

1
Heat transfer with
0.3
Q13
 1594.6 W / m2
A
0.05
......(2)
Re duction in heat loss  Qwithout shield  Qwith shield

due to radiation shield 
Q without shield
Q  Q13
 12
Q12
15.8  103 - 1594.6
15.8  103
= 0.899 = 89.9%
=
151
17. The amount of radiant energy falling on a 50 cm  50 cm horizontal thin metal plate
insulated to the bottom is 3600kJ /m2 hr. If the emissivity of the plate surface is 0.8 and
the ambient air temperature is 30C, find the equilibrium temperature of the plate.
Given :
Area A = 50 cm  50 cm
= 0.5  0.5 m
A  0.25 m2
Radiant energy Q = 3600 kJ / m2 hr
3600  103 J
3600 m2s
= 10 3J / s  m 2
=
= 1000
Here Area




W
m2
J

 W
s

= 0.25 m2
Q = 1000 
W
 0.25m2
m2
Q = 250 W
Emissivity  = 0.8
Ambient air temperature T2 = 30C + 273
= 303 K
To find : Plate temperature T1
Solution : We know
Heat transfer Q    A T14  T2 4 
250  0.8  5.67  10 8  0.25  T14  (303)4 
250  1.13  10 8 T14  (303)4 
T14  (303)4 = 2.2  1010
T1  417.8 K
18. Calculate the shape factors for the configuration shown in fig.
1. A black body inside a black enclosure.
152
2. A tube with cross section of an equilateral triangle.
3. Hemispherical surface and a plane surface
Solution:
Case 1:
[All radiation emitted from the black surface 2 is absorbed by the enclosing surface
1.]
We know
F1-1 + F1 – 2 = 1
….(1)
By reciprocity theorem
A1F12 = A2F21
AF

F12  2 21
A1
(1)  F1-1  1  F12
A2
F21
A1
A
F11  1  2
A1
F11  1 
F11  1 
[
F21  1]
A2
A1
F21  1
Re sult : F1-1  1 
A2
A
, F1-2  2 , F21  1
A1
A1
Case 2 : We know
F1-1  F12  F13  1
F1-1  0
[For Flat surface shape factor (F1-1 )  0]....(2)

F1 - 2  F1  3  1
F12  F13 [Since symmetry triangle]
(2) 
F1 - 2  0.5
F1 - 3  0.5
Now considering radiation from surface 2,
153
F21  F22  F23  1
F2 - 2  0
F21  F23  1
F23  1  F21
.....(3)
By reciprocity theorem, we know
A1F12 = A2F21
A
F21  1 F12
A2
F21  F12
(3) 
[
A1  A 2 ]
F23  1  F21
 1  F12 [
F21  F12 ]
 1  0.5 [
F12  0.5]
F23  0.5
Result: F1 – 1 = 0,
F1 – 2 = 0.5,
F1 – 3 = 0.5
F21 = F12 = 0.5
F22 = 0
F2 – 3 = 0.5
Case 3: We know
F1 – 1 + F1 – 2 = 1
By reciprocity theorem,
A1 F1 – 2 = A2 F2 – 1
A
F12  2 F21
...... (4)
A1
F2 - 1  1
[Since all radiation emitting from the black surface 2 are absorbed by the enclosing surface 1]
A
F1 - 2  2
[ F2 - 1  1]
A1
(4) 
F1 - 2 
 r2 1
  0.5
2 r 2 2
154
F1 - 2  0.5
We know F1 - 1  F12  1

F1 - 1  0.5  1
F1 - 1  0.5
19. Two black square plates of size 2 by 2 m are placed parallel to each other at a
distance of 0.5 m. One plate is maintained at a temperature of 1000C and the other at
500C. Find the heat exchange between the plates.
Given: Area A = 2  2 = 4 m2
T1 = 1000C + 273
= 1273 K
T2 = 500C + 273
= 773 K
Distance = 0.5 m
To find : Heat transfer (Q)
Solution : We know
Heat transfer general equation is
where Q12 
 T14  T2 4 
1  1
1 2
1


A11 A1F12 A1 2
[From equation No.(6)]
For black body
1   2  1
 Q12   [T14  T2 4 ]  A1F12
= 5.67  10 8 (1273)4  (773)4   4  F12
Q12  5.14  105 F12
......(1)
Where F12 – Shape factor for square plates
In order to find shape factor F12, refer HMT data book, Page No.76.
155
Smaller side
Distance between planes
2
=
0.5
X axis =
X axis = 4
Curve  2
[Since given is square plates]
X axis value is 4, curve is 2. So corresponding Y axis value is 0.62.
i.e., F12  0.62
(1)  Q12  5.14  105  0.62
Q12  3.18  105 W
20. Two circular discs of diameter 0.3 m each are placed parallel to each other at a
distance of 0.2 m. one is disc is maintained at a temperature of 750C and the other at
350C and their corresponding emissivities are 0.3 and 0.6. Calculate heat exchange
between the discs.
Given : D1 = 0.3 m
D2 = 0.3 m
A1  A 2
=
=

4

4
D2
(0.3)2
A1  A 2  0.070 m2
T1 = 750C + 273 = 1023 K
T2 = 350C + 273 = 623 K
1 = 0.3
2 = 0.6
Distance between discs = 0.2 m.
To find : Heat exchange between discs (Q),
Solution:
Heat transfer by radiation general equation is
156
 T14  T2 4 
Q12 
1  1
1 2
1


A11 A1F12 A1 2

[From equation (6)]
5.67  108 (1023)4  (623)4 
1  0.3
1
1  0.6


0.070  0.3 0.070F12 0.070  0.6
5.35  104
........(1)
1
42.85 
0.070 F12
Where F12 – Shape factor for disc
Q12 
In order to find shape factor F12, refer HMT data book, Page No.76.
Diameter
Distance between discs
0.3
=
0.2
X axis =
X axis = 1.5
Curve  1
[Since given is disc]
X axis value is 1.5, curve is 1. So, corresponding Y axis value is 0.28.

(1) 
F12  0.28
5.35  10 4
Q12 
1
42.85 
0.070  0.28
Q12  569.9 W
21. Two parallel rectangular surfaces 1 m  2m are opposite to each other at a distance
of 4m. The surfaces are black and at 300C and 200C. Calculate the heat exchange by
radiation between two surfaces.
Given: Area A = 2  2 = 2 m2
Distance = 4 m
T1 = 300C + 273
= 573 K
T2 = 200C + 273
= 473 K
To find: Heat exchange (Q12)
157
Solution : We know, Heat transfer general equation is
 T14  T2 4 
Q12 
1  1
1 2
1


A11 A1F12 A1 2
For Black surface,
1   2  1

Q12   T14  T2 4   A1F12
......(1)
Where F12 – Shape factor for parallel rectangles
In order to find shape factor refer HMT data book, Page No.77 and 78.
b Longer side

c
Dis tance
2
=  0.5
4
a 1
Y 
  0.25
c 4
X
From graph, we know,
F12  0.04
(1)  Q12  5.67  108 (573)4  (473)4   2  0.04
Q12  261.9 W
22. Two parallel plates of size 3 m  2 m are placed parallel to each other at a distance
of 1 m. One plate is maintained at a temperature of 550C and the other at 250C and
the emissivities are 0.35 and 0.55 respectively. The plates are located in a large room
whose walls are at 35C. If the plates located exchange heat with each other and with
the room, calculate.
1. Heat lost by the plates.
2. Heat received by the room.
Given:
Size of the plates = 3 m  2 m
Distance between plates
=1m
158
First plate temperature
T1
= 550C + 273 = 823 K
Second plate temperature
T2
= 250C + 273 = 523 K
Emissivity of first plate
1
= 0.35
Emissivity of second plate 2
= 0.55
Room temperature
T3 = 35C + 273 = 308 K
To find: 1. Heat lost by the plates
2. Heat received by the room.
Solution: In this problem, heat exchange take place between two plates and the room. So this
is three surface problem and the corresponding radiation network is given below.
Area A1 = 3  2 = 6 m2
A1  A 2  6m2
Since the room is large A 3  
From electrical network diagram.
1  1 1  0.35

 0.309
1A1 0.35  6
1   2 1  0.55

 0.136
 2 A 2 0.55  6
1 3
0
3 A3
Apply
[
A 3  ]
1 3
1-1
1 2
 0,
 0.309,
 0.136
3 A3
1A1
2A2
diagram.
To find shape factor F12 refer HMT data book, Page No.78.
b 3
 3
c 1
a 2
Y 
 2
c 1
X
X value is 3, Y value is 2, corresponding shape factor
[From table]
F12 = 0.47
159
values in electrical network
F12  0.47
We know that,
F11 + F12 + F13 = 1
But,
F11 = 0
 F13  1  F12

F13  1  0.47
F13  0.53
Similarly, F21 + F22 + F23 = 1
We know
F22 = 0
 F23  1  F21

F23  1  F12
F13 = 1 - 0.47
F23  0.53
From electrical network diagram,
1
1

 0.314
A1F13 6  0.53
....(1)
1
1

 0.314
A 2F23 6  0.53
....(2)
1
1

 0.354
A1F12 6  0.47
....(3)
From Stefan – Boltzmann law, we know
Eb   T 4
Eb1   T14
= 5.67  10 -8 823 
4
Eb1  26.01 103 W / m2
.....(4)
Eb2   T2 4
= 5.67  10-8 823 
4
Eb2  4.24  103 W / m2
.....(5)
Eb3   T3 4
= 5.67  10-8 308 
4
Eb3  J3  510.25 W / m2
.....(6)
160
[From diagram]
The radiosities, J1 and J2 can be calculated by using Kirchoff’s law.
 The sum of current entering the node J1 is zero.
At Node J1:
Eb1  J1 J2  J1 Eb3  J1


0
1
1
0.309
A1F12
A1F13
[From diagram]
26.01 103  J1 J2  J1 510.25  J1


0
0.309
0.354
0.314
J1
J2
J1
J1
 84.17  103 


 1625 
0
0.309 0.354 0.354
0.354

-9.24J1  2.82J2  85.79  103
.....(7)

At node j2
J1  J2 Eb3  J2 Eb2  J2


 0 -+*
1
1
0.136
A1F12
A 2F23
J1  J2 510.25  J2 4.24  103  J2


0
0.354
0.314
0.136
J1
J2
J2
J2
510.25
4.24  103





0
0.354 0.354 0.314 0.314
0.136
0.136

2.82J1  13.3J2  32.8  103
....(8)
Solving equation (7) and (8),

-9.24J1  2.82J2  85.79  103 .....(7)

2.82J1  13.3J2  32.8  103
.....(8)
J2  4.73  103 W / m2
J1  10.73  103 W / m2
Heat lost by plate (1) is given by
161
Q1 
Eb1  J1
 1  1 


 1A1 
26.01 103  10.73  103
Q1 
1  0.35
0.35  6
Q1  49.36  103 W
Heat lost by plate 2 is given by
E J
Q2  b2 2
 1 2 


 2A2 
Q2 
4.24  103  4.73  103
1  0.55
6  0.55
Q2  3.59  103 W
Total heat lost by the plates
Q = Q1 + Q2
= 49.36  103 – 3.59  103
Q  45.76  103 W
......(9)
Heat received by the room
Q

J1  J3 J2  J3

1
1
A1F13
A1F12
10.73  103  510.25 4.24  103  510.25

0.314
0.314
[ Eb1  J1  512.9]
Q = 45.9  103 W
.....(10)
From equation (9), (10), we came to know heat lost by the plates is equal to heat
received by the room.
162
23. Two black square plates of size 1 by 1 m are placed parallel to each other at a
distance of 0.4 m. One plate is maintained at a temperature of 900C and the other at
400. Find the net heat exchange of energy due to radiation between the two plates.
Given: Area A = 1  1 = 1 m2
Distance = 0.4 m
T1 = 900C + 273
= 1173 K
T2 = 400C + 273
= 673 K
To find: Heat exchange (Q)
Solution: Heat transfer by radiation general equation is
 T14  T2 4 
Q12 
1  1
1 2
1


A11 A1F12 A1 2
[From equation No.(6)]
For black body,
1   2  1

Q12   T14  T2 4  A1F12
= 5.67  10-8 (1173)4  (673)4  F12
Q12 = 95.7  103F12
.......(1)
Where F12 – shape factor for square plates.
In order to find shape factor F12, refer HMT data book, Page No.76.
Smaller side
X axis =
Distance between planes
1
=
0.4
X axis = 2.5
Curve 2 [since given is square plate]
X axis value is 2.5, curve is 2, so corresponding Y axis value is 0.42.
i.e., F12 = 0.42
(1)  Q12  95.7  103  0.42
Q12  40  103 W
163
24. Two circular discs of diameter 20 cm each are placed 2 m apart. Calculate the
radiant heat exchange for these discs if there are maintained at 800C and 300C
respectively and the corresponding emissivities are 0.3 and 0.5.
Given :
D1 = 20 cm = 0.2 m
D2 = 0.2 m
T1 = 800C + 273
= 1073 K
T2 = 300C + 273
= 573 K
1 = 0.3
2 = 0.5
To find: Heat exchange (Q)
A1 

D 
4
2
1

(0.2)2  0.031 m2
4
A1 = 0.031 m2
A
2
2 = 0.031 m
 D1  D2 
Heat transfer by radiation generation equation is
Solution: Area =
 T14  T2 4 
Q12 
1  1
1 2
1


A11 A1F12 A 2 2

Q12 
5.67  108 (1073)4  (573)4 
1  0.3
1
1  0.5


0.031 0.3 0.31 F12 0.031 0.5
69  103
1
107.45 
0.031 F12
.......(1)
Where F12 = Shape factor for disc.
In order to find shape factor, F12 refer HMT data book, Page No.76.
Diameter
X axis =
Distance between disc
0.2
=
2
X axis = 0.1
164
Curve 1 [since given is disc]
X axis value is 0.1, curve is 1, so corresponding Y axis value is 0.01.
 F12 = 0.01
F12 = 0.01
(1)  Q12 
69  103
1
0.031 0.01
Q12 = 20.7 Watts.
107.45 
25. A long cylindrical heater 30 in diameter is maintained at 700C. It has surface
emissivity of 0.8. The heater is located in a large room whose wall are 35C. Find the
radiant heat transfer. Find the percentage of reduction in heat transfer if the heater is
completely covered by radiation shield ( = 0.05) and diameter 40 mm.
Given : Diameter of cylinder D1=30mm=0.030 mm
Temperature
T1=700C + 273 = 973 K
Emissivity
1 = 0.8
Room temperature T2 = 35C + 273 = 308 K
Radiation Shield :
Emissivity 3 = 0.05
Diameter D3 = 40 mm = 0.040 m
Solution:
Case 1 : Heat transfer without shield:
Heat transfer by radiation general equation is
 T14  T2 4 
Q12 
1  1
1 2
1


A11 A1F12 A 2 2
Where A1   DL    0.030  1  0.094 m
A1  0.094 m2
Since room is large A 2  
F12 = Shape factor
Small body enclosed by large body  F12 = 1
165
(1)  Q12 
[Refer HMT data book, Page No.73]
5.67  10 8 (973)4  (308)4 
1  0.8
1  0.5

0
0.094  0.8 0.094  1


1 2
Since A12  , A   0 

2 2

Heat transfer without shield
Q12  3783.2 W
........(2)
Case 2: Heat transfer with shield:
Heat transfer between heater (1) and radiation shield (3) is given by
 T14  T3 4 
Q13 
1 3
1  1
1


A11 A1F13 A 3 3
 T14  T2 4 
Q12 
1  1
1 2
1


A11 A1F12 A 2 2
Where A 3   D3L    0.040  1
A 3  0.094 m2
Shape factor for concentric long cylinder F13 = 1
[Refer HMT data book, Page No. 73]
(1)  Q13 
5.67  108 (973)4  T3 4 
1  0.8
1
1  0.5

+
0.094  0.8 0.094  1 0.125  0.05
Q13  3.43  1010 (973)4  T3 4 
........(3)
Heat exchange between radiation shield (3) and Room (2) is given by
Q32 
 T3 4  T2 4 
1 3
1 2
1


A 3 3 A 3F32 A 2 2
Since room is large, A2 = 
166
1  2
0
A 2 2
Shape factor for small body enclosed by large body
F32 = 1
[Refer HMT data book, Page No.73]

 Q32 
5.67  10 8 T3 4  (308)4 
1  0.05
1

+0
0.125  0.5 0.125  1
Q32  3.54  10 10 T3 4  (308)4 
........(4)
 3.43  10-10 (973)4  T3 4   3.54  10 10 T3 4  (308)4 
 307.4 - 3.43  10 10 T3 4  3.54  10 10 T3 4  3.18
310.58= 6.97  10-10 T3 4

T3  817 K
Substitute T3 value in (3) or (4).
Heat transfer with radiation shield
Q13  3.43  10 10 (973)4  (817)4 
Q13  154.6 W
Re duction heat 
 Q without shield  Q with shield
loss due to

Q without shield
radiation shield 

Q12  Q13
Q12

3783.2  154.6
 95.9%
3783.2
26. A gas is enclosed in a body at a temperature of 727C. The mean beam length of the
gas body is 3 m. The partial pressure of water vapour is 0.2 atm and the total pressure is
2 atm. Calculate the emissivity of water vapour.
Given : Temperature T = 727C + 273 = 1000K
Mean beam length Lm = 3m
Partial pressure of water vapour PH2  0.2 atm.
0
Total pressure P = 2 atm.
To find : Emissivity of water vapour (H2 )
o
167
Solution: PH2o  Lm  0.2  3
PH2oLm  0.6 m atm
From HMT data book, Page No.92, we can find emissivity of H2o.
From graph,
Emissivity of H2o = 0.3
H2o  0.3
To find correction factor for H20:
PH2 0  P 0.2  2

 1.1
2
2
PH2 0  P
 1.1, PH2 0 Lm  0.6
2
From HMT data book, Page No.94, we can find correction factor for H2o
PH2O  P
2

From graph,
Correction factor for H2o = 1.36
C H2o  1.36 ...........(2)
So, Emissivity of H2o, H2 0  0.3  1.36
H 0  0.408
2
27. A gas mixture contains 20% CO2 and 10% H2o by volume. The total pressure is 2
atm. The temperature of the gas is 927C. The mean beam length is 0.3 m. Calculate the
emissivity of the mixture.
Given : Partial pressure of CO2, PCO2 = 20% = 0.20 atm
Partial pressure of H2o, PH2 0 = 10% = 0.10 atm.
Total pressure P
Temperature T
Mean beam length Lm
= 2 atm
= 927C + 273
= 1200 K
= 0.3 m
To find: Emissivity of mixture (mix).
168
Solution : To find emissivity of CO2
PCO2  Lm  0.2  0.3
PCO2  Lm  0.06 m - atm
From HMT data book, Page No.90, we can find emissivity of CO2.
From graph, Emissivity of CO2 = 0.09
 CO  0.09
2
To find correction factor for CO2
Total pressure, P = 2 atm
PCO2 Lm = 0.06 m - atm.
From HMT data book, Page No.91, we can find correction factor for CO2
From graph, correction factor for CO2 is 1.25
CCO2  1.25
 CO  CCO  0.09  1.25
2
2
 CO  CCO  0.1125
2
2
To find emissivity of H2o :
PH2o  Lm  0.1 0.3
PH2oLm  0.03 m - atm
From HMT data book, Page No.92, we can find emissivity of H2o.
From graph Emissivity of H2o = 0.048
H o  0.048
2
To find correction factor for H2o :
169
PH2o  P
2
PH2o  P

0.1  2
 1.05
2
 1.05,
2
PH2o Lm  0.03 m - atm
From HMT data book, Page No.92 we can find emission of H20
From graph,
Correction factor for H2o = 1.39
CH2O  1.39
H O  CH O  0.048  1.39
2
2
H O  CH O  0.066
2
2
Correction factor for mixture of CO2 and H2O:
PH2o
PH2o  PCO2

PH2o
PH2o  PCO2
0.1
 1.05
0.1  0.2
 0.333
PCO2  Lm  PH2O  Lm  0.06  0.03
PCO2  Lm  PH2O  Lm  0.09
From HMT data book, Page No.95, we can find correction factor for mixture of CO2 and
H2o.
From graph,
  0.002
Total emissivity of gascous mixture
 max   co2 CCO2  H2O CH2O  
 max  0.1125  0.066  0.002
[From equation (1), (2) and (3)]
 max  0.1765
28. A furnace of 25 m2 area and 12 m2 volume is maintained at a temperature of 925C
170
over is entire volume. The total pressure of the combustion gases is 3 atm, the partial
pressure of water vapour is 0.1 atm and that of CO2 is 0.25 atm.
Calculate the emissivity of the gaseous mixture.
Given :
Area A
Volume V
Temperature T
Total pressure P
= 25 m2
= 12 m3
= 925 + 273
= 1198 K
= 3 atm
Partial pressure of water vapour, PH2O = 0.1 atm.
Partial pressure of CO2 PCO2  0.25 atm.
To find: Emissivity of mixture ( mix ).
Solution : We know Mean beam length for gaseous mixture.
V
Lm  3.6 
A
12
= 3.6 
25
Lm  1.72 m
To find emissivity of CO2
PCO2  Lm  0.25  1.72
PCO2  Lm  0.43 m-atm.
From HMT data book, Page No.90, we can find emissivity of CO2.
 CO  CCO  0.18
2
2
From graph,
Emissivity of CO2 = 0.15
 CO2  0.15
To find correction factor for CO2:
Total pressure P = 3 atm.
PCO2 Lm  0.43 m-atm
From HMT data book, Page No.91, we can find correction factor for CO2.
From graph, we find CCO2  1.2
CCO2  1.2
171
 CO2  CCO2  0.15  1.72
.....(1)
To find emissivity of H2O:
PH2O  Lm  0.1 1.72
PH2O  Lm  0.172
From HMT data book, Page No.92, we can find emissivity of H2O.
From graph,
Emissivity of H2O = 0.15
H2O  0.15
To find correction factor for H2O:
PH2O  P
2
PH2O  P
2
PCO2
0.1  3
 1.55
2

 1.55,PH2O Lm  0.172.
From HMT data book, Page No.94, we can find correction factor for H2O.
Lm  PH2O Lm  0.602
From graph, we find
CH2O  1.58
CH2O  1.58
 H2O  CH2O  0.15  1.58
 H O  CH O  0.237 ........(2)
2
2
Correction Factor for mixture of CO2 and H2O:
PH2O
PH2O  PCO2

0.1
 0.285
0.1  0.25
172
PH2O
PH2O  PCO2
 0.285
PCO2  Lm  PH2O  Lm  0.25  1.72  0.1 1.72
= 0.602.
From HMT data book, Page No.95 we can find correction factor for mixture of CO 2
and H O
2
From graph, we find  = 0.045.
  0.045
..........(3)
Total emissivity of the gaseous mixture is
 mix   CO2 CCO2  H2O CH2O  
 mix  0.18  0.237  0.045
[From equation (1), (2) and (3)]
 mix  0.372
173
UNIT – IV
Classification
Temperature Distribution
Overall heat transfer coefficient
Heat Exchange Analysis – LMTD Method and E-NTU Method,
problems using LMTD and E-NTUmethds
174
Heat Exchangers
PART – A
1. What is heat exchanger?
A heat exchanger is defined as an equipment which transfers the heat from a hot fluid
to a cold fluid.
2. What are the types of heat exchangers?
The types of heat exchangers are as follows
1.
2.
3.
4.
5.
6.
7.
8.
Direct contact heat exchangers
Indirect contact heat exchangers
Surface heat exchangers
Parallel flow heat exchangers
Counter flow heat exchangers
Cross flow heat exchangers
Shell and tube heat exchangers
Compact heat exchangers.
3. What is meant by Direct heat exchanger (or) open heat exchanger?
In direct contact heat exchanger, the heat exchange takes place by direct mixing of hot
and cold fluids.
4. What is meant by Indirect contact heat exchanger?
In this type of heat exchangers, the transfer of heat between two fluids could be
carried out by transmission through a wall which separates the two fluids.
5. What is meant by Regenerators?
In this type of heat exchangers, hot and cold fluids flow alternately through the same
space.
Examples: IC engines, gas turbines.
6. What is meant by Recupcradors (or) surface heat exchangers?
This is the most common type of heat exchangers in which the hot and cold fluid do
not come into direct contact with each other but are separated by a tube wall or a surface.
7. What is meant by parallel flow heat exchanger?
In this type of heat exchanger, hot and cold fluids move in the same direction.
8. What is meant by counter flow heat exchanger?
In this type of heat exchanger hot and cold fluids move in parallel but opposite
directions.
175
9. What is meant by cross flow heat exchanger?
In this type of heat exchanger, hot and cold fluids move at right angles to each other.
10. What is meant by shell and tube heat exchanger?
In this type of heat exchanger, one of the fluids move through a bundle of tubes
enclosed by a shell. The other fluid is forced through the shell and it moves over the outside
surface of the tubes.
11. What is meant by compact heat exchangers? [Nov 1996 MU]
There are many special purpose heat exchangers called compact heat exchangers.
They are generally employed when convective heat transfer coefficient associated with one of
the fluids is much smaller than that associated with the other fluid.
12. What is meant by LMTD?
We know that the temperature difference between the hot and cold fluids in the heat
exchanger varies from point in addition various modes of heat transfer are involved.
Therefore based on concept of appropriate mean temperature difference, also called
logarithmic mean temperature difference, also called logarithmic mean temperature
difference, the total heat transfer rate in the heat exchanger is expressed as
Q = U A (T)m
Where
U – Overall heat transfer coefficient W/m2K
A – Area m2
(T)m – Logarithmic mean temperature difference.
13. What is meant by Fouling factor?
We know the surfaces of a heat exchangers do not remain clean after it has been in
use for some time. The surfaces become fouled with scaling or deposits. The effect of these
deposits the value of overall heat transfer coefficient. This effect is taken care of by
introducing an additional thermal resistance called the fouling resistance.
14. What is meant by effectiveness?
The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the
maximum possible heat transfer.
Effectiveness  
Actual heat transfer
Maximum possible heat transfer
Q
=
Qmax
176
PART – B
1. A condenser is to designed to condense 600 kg/h of dry saturated steam at a pressure
of 0.12 bar. A square array of 400 tubes, each of 8 mm diameter is to be used. The tube
surface is maintained at 30C. Calculate the heat transfer coefficient and the length of
each tube.
Given :
m  600 kg/h =
600
kg / s  0.166 kg/s
3600
m = 0.166 kg/s
Pressure P – 0.12 bar
No. of tubes = 400
Diameter D = 8mm = 8  10-3m
Surface temperature Tw  30C
Solution
Properties of steam at 0.12 bar
From R.S. Khurmi steam table Page No.7
Tsat  49.45C
hfg  2384.3 kj/kg
hfg = 2384.9  103 j / kg
We know
Film temperature Tf 
Tw  Tsat
2
30  49.45
2
Tf  39.72C  40C

Properties of saturated water at 40C
From HMT data book Page No.13
 - 995 kg/m3
 = .657  10-6 m2 / s
k  628.7  10 3 W/mk
 =   = 995  0.657  10 -6
 = 653.7  10 -6 Ns/m2
177
with 400 tubes a 20  20 tube of square array could be formed
N  400  20
i.e.
N  20
For horizontal bank of tubes heat transfer coefficient.
 K 3  2g hfg 
h = 0.728 

  D (Tsat  Tw ) 
0.25
From HMT data book Page No.150
 (628  10-3 )3  (995)2  9.81 2384.3  103 
h = 0.728 

6
3
 653.7  10  20  8  10  (49.45  30) 
0.25
h = 5304.75 W/m2K
We know
Heat transfer
Q  hA(Tsat  Tw )
No. of tubes = 400
Q = 400  h    D  L  (Tsat  Tw )
Q  400  5304.75    8  10 3  1 (49.45-30)
Q = 1.05  10 6  L........1
We know
Q  m  hfg
= 0.166  2384.3  103
Q = 0.3957  10 6 W
= 0.3957  10 6  1.05  10 6 L
L  0.37 m
Problems on Parallel flow and Counter flow heat exchangers
From HMT data book Page No.135
Formulae used
1. Heat transfer Q = UA (T)m
Where
U – Overall heat transfer coefficient, W/m2K
A – Area, m2
178
(T)m – Logarithmic Mean Temperature Difference. LMTD
For parallel flow
( T)m 
(T1  t1 )  (T2  t 2 )
T t 
In  1 1 
 T2  t 2 
In Counter flow
(T  t )  (T2  t 2 )
( T)m  1 1
T t 
In  1 1 
 T2  t 2 
Where
T1 – Entry temperature of hot fluid C
T2 – Exit temperature of hot fluid C
T1 – Entry temperature of cold fluid C
T2 – Exit temperature of cold fluid C
2. Heat lost by hot fluid = Heat gained by cold fluid
Qh = Qc
mhCph (T1  T2 )  mc Cpc (t 2  t1 )
Mh – Mass flow rate of hot fluid, kg/s
Mc – Mass flow rate of cold fluid kg/s
Cph – Specific heat of hot fluid J/kg K
Cpc – Specific heat of cold fluid J/kg L
3. Surface area of tube
A = D1 L
Where D1 Inner din
4. Q = m  hfg
Where hfg – Enthalpy of evaporation j/kg K
5. Mass flow rate
m =  AC
2. In a counter flow double pipe heat exchanger, oil is cooled from 85C to 55C by
water entering at 25C. The mass flow rate of oil is 9,800 kg/h and specific heat of oil is
2000 j/kg K. the mass flow rate of water is 8,000 kg/h and specific heat of water is 4180
j/kg K. Determine the heat exchanger area and heat transfer rate for an overall heat
transfer coefficient of 280 W/m2 K.
Given:
179
Hot fluid – oil
T1,T2
Cold fluid – water
t1, t2
Entry temperature of oil T1 = 85C
Exit temperature of oil T2 = 55C
Entry temperature of water t1 = 25C
Mass flow rate of oil (Hot fluid) mh = 9,800 kg/h
9,800
kg / s
3600
mh  2.72 kg/s
Specific heat of oil Cph = 2000 j/kg K
Mass flow of water (Cold fluid mc = 8,000 kg/h)
8,000
kg / s
3600
me  2.22 kg / s
Specific heat of water Cpc – 4180 j/kg K
Overall heat transfer coefficient U = 280 W/m2K
To find
1. Heat exchanger area (A)
2. Heat transfer rate (Q)
Solution
We know that
Heat lost by oil Hot fluid = Heat gained by water cold fluid
Qh = Qc
mhCph (T1  T2 )  mc Cpc (t 2  t1 )
2.72  2000 (85-55)=2.22  4180  (t 2  25)
163.2  103  9279.6 t 2  231.9  10 3
t 2  42.5C
Exit temperature of water t 2  42.5 C
Heat transfer Q =mc Cpc (t 2  t1 ) (or) mhCph (T1  T2 )
Q  2.22  4180  (42.5  25)
Q  162  103 W
Q  UA (T)m........1
From HMT data book Page No.154
180
For counter flow
( T)m 
(T1  t1 ) - (T2  t 2 )
T t 
In  1 1 
 T2  t 2 
From HMT data book Page No.154
(85  42.5) - (55  25)
( T)m 
 85  42.5 
In 
 55  25 
( T)m  35.8C
Substitute (T)m U and Q values in Equation (1)
(1)  Q = UA (T)m
 162  103  280  A  35.8
 A = 16.16 m2
3. Water flows at the rate of 65 kg/min through a double pipe counter flow heat
exchanger. Water is heated from 50C to 75C by an oil flowing through the tube. The
specific heat of the oil is 1.780 kj/kg K. The oil enters at 115C and leaves at 70C. The
overall heat transfer coefficient is 340 W/m2 K. Calculate the following
1. Heat exchanger area
2. Rate of heat transfer
Given :
Hot fluid – oil
T1, T2
Cold fluid – water
t1, t2
Mass flow rate of water cold fluid mc = 65 kg/min
65
kg / s
60
mc  1.08 kg/s
Entry temperature of water t1 = 50C
Exit temperature of water t2 = 75C
Specific heat of oil (Hot fluid) Cph = 1.780 kj/kg K
= 1.780  103 j/kg K
Entry temperature of oil T1 = 115C
Exit temperature of oil T2
= 70C
Overall heat transfer coefficient U = 340 W/m2/K
181
To find
1. Heat exchanger area (A)
2. Heat transfer rate (Q)
Solution
We know
Heat transfer Q = mc Cpc (t 2  t1 ) (or)
mhCph (T1  T2 )
Q  mc Cpc (t 2  t1 )
Q  1.08  4186  (75  50)
Specific heat of water Cph  4186 j/kg K
Q = 113  103 W
We know
Heat transfer Q = U  A (T)m
From HMT data book Page No., 154
Where
(T)m – Logarithmic Mean Temperature Difference. LMTD
For Counter flow
(T  t )  (T2  t 2 )
( T)m  1 1
T t 
In  1 1 
 T2  t 2 
From HMT data book Page No.154
(115  75)  (70  50)
(T)m 
115  75 
In 
 70  50 
( T)m  28.8C
Substitute (T)m Q and U values in Equation (1)
(1)  Q =UA (T)m
 113  103  340  A  28.8
 A = 11.54 m2
182
4. In a double pipe heat exchanger hot fluid with a specific heat of 2300 j/kg K enters at
380C and leaves at 300C. cold fluid enters at 25C and leaves at 210C. Calculate the
heat exchanger area required for
1. Parallel flow
2. Counter flow
Take overall heat transfer coefficient is 750 w/m2 K and mass flow rate of hot fluid is 1
kg/s.
Given :
Specific heat of hot fluid Cph = 2300 j/kg K
Entry temperature of hot fluid T1 = 380C
Exit temperature of hot fluid T2 = 300C
Entry temperature of cold fluid t1 = 25C
Exit temperature of cold fluid t2 = 210C
Overall heat transfer coefficient U = 750 W/m2K
Mass flow rate of fluid mh = 1 kg/s
Solution
Case (i)
For parallel flow
( T)m 
(T1  t1 )  (T2  t 2 )
T t 
In  1 1 
 T2  t 2 
From HMT data book Page No.154
(380  25)  (300  210)
(T)m 
 380  25 
In 
 300  210 
( T)m  193.1C
Heat transfer Q = mc Cpc (t 2  t1 ) (or)
mhCph (T1  T2 )
Q  mc Cpc (t 2  t1 )
= 1 2300  380  300
Q = 184  103 W
From HMT data book Page No.154
We know that
183
Q  U  A (T)m
Heat transfer 184  103  750  A  193.1
Area for parallel flow A = 1.27 m2
Case (ii)
For counter flow
( T)m 
(T1  t 2 )  (T2  t1 )
T  t 
In  1 2 
 T2  t1 
From HMT data book Page No.154
(380  210)  (300  25)
(T)m 
 380  210 
In 
 300  25 
( T)m  218.3C
We know that,
Heat transfer Q = UA (T)m
 184  103  750  A  218.3
Area for counter flow A = 1.12 m2
5. In a counter flow single pass heat exchanger is used to cool the engine oil from 150C
with water available at 23c as the cooling medium. The specific heat of oil is 2125 J/kg
K. The flow rate of cooling water through the inner tube of 0.4 m diameter is 2.2 Kg/s.
the flow rate of oil through the outer tube of 0.75m diameter is 2.4 kg/s. If the value of
the overall heat transfer coefficient is 240 W/m2 how long must the heat exchanger be to
meet its cooling requirement?
Given :
Hot fluid oil
T1, T2
Cold fluid water
(t1, t2)
Entry temperature of oil T1 = 150C
Exit temperature of oil T2
= 55
Entry temperature of water t1 = 23C
Specific heat of oil hot fluid Cph
= 2125 J/Kg K
Inner Diameter D1
= 0.4 m
Flow rate of water cooling fluid mc = 2.2 kg/s
Outer diameter D2
= 0.75 m
Flow rate of oil Hot fluid mh
= 2.4 kg/s
Over all heat transfer coefficient U= 240 W/m2K
Solution
184
We know
Heat lost by oil Hot fluid = Heat gained by water (cold fluid)

Qh  Qc



mhCph (T1  T2 )  mc Cpc (t 2  t1 )
2.4  2125 (150 - 55) = 2.2  4186  (t 2  23)
[Specific heat of water Cpc = 4186 J/Kg K]
434.5  103 = 9209.2 t 2 - 211  103

t 2  75.6C
Exit temperature of water t 2  75.6C
We know
Heat transfer Q = mc Cpc (t2 – t1) (or)
MhCph (T1 – T2)

Q = 2.2  4186  (75.6 - 53)
Q = 484.4  103 W
We know
Heat transfer Q = UA (T)m …………(1)
[From HMT data book Page No.154]
where
(T)m – Logarithmic Mean Temperature Difference (LMTD).
For Counter flow,
(T  t )  (T2  t1)
( T)m  1 2
T  t 
In  1 2 
 T2  t1 
From HMT data book Page No.154
185
( T)m 
(150  75.6)  (55  23)
150  75.6 
In 
 55  23 
( T)m  50C
Substitute (T)m U and Q values in equation 1
Q = U  A (T)m
484.4  103  240  A  50.2
A  40.20 m2
We know
Area A =   D1  L
40.20    0.4  L
L  31.9 m
6. Saturated steam at 126C is condensing on the outer tube surface of a single pass heat
exchanger. The heat exchanger heats 1050 kg/h of water from 20C to 95C. The overall
heat transfer coefficient is 1800 W/m2K. Calculate the following.
1. Area of heat exchanger
2. Rate of condensation of steam
Take hfg = 2185 kj/kg
Given :
Hot fluid – steam
T1, T2
Cold fluid – water
t1, t2
Saturated steam temperature T1 = T2 = 126C
Mass flow rate of water mc = 1050 kg/h

1050 kg
3600 s
mc  0.29 kg/s
Entry temperature of water t1 = 20C
Exit temperature of water t2 = 95C
Over all transfer coefficient U = 1800 W/m2K
Enthalpy of evaporation hfg = 2185 kg/jg
= 2185  103 j/kg
Solution
We know
Heat transfer
186
Q  mc Cpc (t 2  t1 )
Q  0.29  4186  (95  20)
[ Specific heat of water Cpc  4186 J/kg K]
Q = 91 103 W
We know
Heat transfer
Q  mh  hfg
91 103  mh  2185  103
Rate of condensation of steam mh = 0.0416 kg/s
We know that
Heat transfer Q = UA (T)m ………….1
From HMT data book Page No.154
Where
( T)m  Logarithmic Mean Temperature Difference LMTD
For parallel flow
(T)m 
(T1  t1 )  (T2  t 2 )
T t 
In  1 1 
 T2  t 2 
(126 - 20) - (126-95)
=
126-20 
In 
126-95 
(T)m  61C
Substitute (T)m Q, U values in equation (1)
(1)  q = UA (T)m
 91 103  1800  A  61
Area A = 0.828 m2
7. An oil cooler of the form of tubular heat exchanger cools oil from a temperature of
90C to 35C by a large pool of stagnant water assumed constant temperature of 28C.
The tube length is 32 m and diameter is 28 mm. The specific heat and specific gravity of
the oil are 2.45 kj/kg K and 0.8 respectively. The velocity of the oil is 62 cm/s. Calculate
the over all heat transfer coefficient.
Given :
Hot fluid – Oil
T1, T2
Cold fluid – Water
t1, t2
187
Entry temperature of oil T1 = 90C
Exit temperature of oil T2
= 35C
Entry and Exit temperature of water t1=t2=28C
Tube length L = 32 m
Diameter D = 28 mm = 0.028 m
Specific heat of oil Cph = 2.45 kj/kg/K
Cph = 2.45  103 j/kg K
Specific gravity of oil = 0.8
Velocity of oil C = 62 cm/s = 0.62 m/s.
To find
We know
Specific gravity of oil =
=
0.8 =
Density of oil
Density of water
Q
w
0
1000
Density of oil  0 = 800 kg/m3
Mass flow rate of oil
mh   0  A  C
 800 
 800 


4
(D2 )  0.62
(0.028)2  0.62
4
mh  0.305 kg/s
We know
Heat transfer
Q  mc Cpc (t 2  t1 )
0.305  2.45  103  90  35 
Q = 41 103 W
We know
Heat transfer Q = U A (T)m ………..1
(From HMT data book Page No.154)
Where
(T)m – Logarithmic Mean Temperature Difference LMTD.
For parallel flow
188
( T)m 
(T1  t1 )  (T2  t 2 )
T t 
In  1 1 
 T2  t 2 
From HMT data book Page No.154
(90-28)-(35 -28)
=
 90-28 
In 
 35-28 
(T)m  25.2C
Substitute (T)m Q value in equation 1
(1)  q = U A (T)m
41 103  U   DL  (T)m
41 103  U    0.028  32  25.2
U = 577.9
Overall heat transfer coefficient U = 577.9 W/m2K
Result
U = 577.9 W/m2K
Problems on cross flow heat exchangers (or) shell and tube heat exchangers.
From HMT data book Page No.154
Formulae used
1. Q = FU A (T)m
Where
(Counter Flow)
F – Correction factor – From data book
U – Overall heat transfer coefficient W/m2K
(T)m – Logarithmic mean temperature difference
For counter flow
(T  t )  (T2  t1)
( T)m  1 2
T  t 
In  1 2 
 T2  t1 
where
T1 – Entry temperature of hot fluid C
T2 – Exit temperature of hot fluid C
T1 – Entry temperature of cold fluid C
189
T2 – Exit temperature of cold fluid C
2. Heat lost by hot fluid = Heat gained by cold fluid
Qh = Qc
 mhCph  T1  T2   mcCph (t 2  t1)
8. In a cross heat exchangers both fluids unmixed hot fluid with a specific heat of 2300
J/kg K enters at 380C and leaves at 300C cold fluids enters at 25C and leaves at
210C. Calculate the required surface of heat exchanger. Take overall heat transfer
coefficient is 750 W/m2 K. Mass flow rate of hot fluid is 1 kg/s
Given :
Specific heat of hot fluid Cph = 23000 J/kg K
Entry temperature of hot fluid T1
= 380C
Exit temperature of heat fluid T2
= 300C
Entry temperature of cold fluid t1
= 25C
Exit temperature of cold fluid t2
= 210C
Overall heat transfer coefficient U = 750 W/m2K
Mass flow rate of hot fluid mh
= 1 kg/s
To find
Heat exchanger area (A)
Solution:
This is cross flow both fluids unmixed type heat exchanger. For cross flow heat exchanger.
Q = FU A (T)m (Counter flow)……….1
From HMT data book Page No.154
Where
F – correction factor
(T)m – Logarithmic Mean Temperature Difference for Counter Flow
For Counter flow
( T)m 
(T1  t 2 )  (T2  t1)
T  t 
In  1 2 
 T2  t1 
190
=
(380-210)-(300 -25)
 380-210 
In 
 300-25 
(T)m  218.3C
We know
Heat transfer
 Q = mhCph (T1  T2 )
Q  1 1200 (380-300)
Q = 184  103 W
To find correction factor E refer HMT data book Page No.164.
Single pass cross flow heat exchanger – Both fluids unmixed.
From graph
Xaxis Value P =
t 2  t1 210  25

 0.52
T1  t1 380  25
Curve value = R=
T1  T2 380  300

 0.432
t 2  t1
210  25
Xaxis Value is 0.52 curve
Value is 0.432 corresponding Yaxis value is 0.97
i.e. F = 0.07
Substitute Q, F (T)m and U value in Equation (1)
(1)  Q = FU A (T)m
184  103  0.97  750  A  218.3
A  1.15 m2
9. In a refrigerating plant water is cooled from 20C to 7C by brine solution entering at
-2C and leaving at 3C. The design heat load is 5500 W and the overall heat transfer
coefficient is 800 W/m2 K. What area required when using a shell and tube heat
exchanger with the water making one shell pass and the brine making two tube passes.
Given:
Hot fluid – Water
Cold fluid – brine solution
191
(T1, T2)
(t1, t2)
Entry temperature of water T1 = 20C
Exit temperature of water T2 = 7C
Entry temperature of brine solution t1 = -2C
Exit temperature of brine solution t2 = 3C
Heat load Q = 5500 W
Overall heat transfer coefficient U = 8000 W/m2 K
To find
Area required A
Solution
Shell and tube heat exchanger – one shell pass and two tube passes
For shell and tube heat exchanger or cross heat exchanger.
Q = F U A (T)m (Counter flow)
(From HMT data book Page No.154)
Where
F – Correction factor
(T)m – Logarithmic mean temperature difference for counter flow
For counter flow
(T  t )  (T2  t1)
( T)m  1 2
T  t 
In  1 2 
 T2  t1 
(20-3)-(7 +2)
=
 20-3 
In 
 7+2 
(T)m  12.57C
To find correction factor F refer HMT data book Page No.161
One shell pass and two tube passes
From graph
Xaxis Value P =
t 2  t1
32
5


T1  t1 20  2 22
P  0.22
Curve value = R=
T1  T2 20  7 13


t 2  t1
32
5
R = 2.6
Xaxis value is 0.22 curve value is 2.6 corresponding Yaxis value is 0.94
Substitute (T)m Q, U and F value is Equation (1)
192
1  Q  F U A (T)m
5500  0.94  800  A  12.57
A  0.58 m2
10. Saturated steam at 120C is condensing in shell and tube heat exchanger. The
cooling water enters the tuber at 25C and leaves at 80C. Calculate the logarithmic
meant temperature difference if the arrangement is
(a) Counter Flow (b) Parallel Flow (c) Cross Flow
Given :
Hot fluid steam
(T1, T2)
Cold fluid water
(t1, t2)
Saturated steam temperature T1
Entry temperature of water t1
Exit temperature of water t2
= T2 = 120C
= 25C
= 80C
To find
(T)m for parallel flow counter flow and cross flow
Solution:
Case (i)
For parallel flow
( T)m 
=
[From HMT data book Page No.154]
(T1  t1 )  (T2  t 2 )
T t 
In  1 1 
 T2  t 2 
(120-25)-(120-80)
120-25 
In 
120-80 
(T)m for parallel flow = 63.5C
Case (ii)
For Counter Flow
193
( T)m 
=
(T1  t 2 )  (T2  t1 )
T  t 
In  1 2 
 T2  t1 
(120-80)-(120-25)
120-80 
In 
120-25 
(T)m for Counter flow = 63.5C
Case (iii)
For cross flow
(T)m = F (T)m for Counter flow
( T)m  F  63.5 ...........2
Where
F = Correction factor
Refer HMT data book Page No.163
Correction factor for single pass cross flow heat exchanger one fluid mixed other unmixed.
Xaxis Value P =
t 2  t1 80  25

T1  t1 120  25
P  0.578
Curve value = R=
T1  T2 120  120

t 2  t1
80  25
R=0
Xaxis value is 0.578 curve value is 0
So corresponding Yaxis value is 1
Correction Faction F = 1
(3)  ( T)m  F  63.5C  1 63.5
(T)m for cross flow = 63.5 .........3
From (1) (2) and (4) we came to know when one of the fluids in a heat exchanger changes
phase, the logarithmic mean temperature difference and rate of heat transfer will remain same
for parallel flow counter flow and cross flow.
Solved problems – NTU method
Note NTU method is used to determine the inlet or exit temperatures of heat exchanger.
11. A parallel flow heat exchanger is used to cool. 4.2 kg/min of hot liquid of specific
194
heat 3.5 kj/kg K at 130C. A cooling water of specific heat 4.18 kj/kg K is used for
cooling purpose at a temperature of 15C. The mass flow rate of cooling water is 17
kg/min calculate the following.
1. Outlet temperature of liquid
2. Outlet temperature of water
3. Effectiveness of heat exchanger
Take care,
Overall heat transfer coefficient is 1100 W/m2 K.
Heat exchanger area is 0.30 m2
Given :
Mass flow rate of hot liquid mh
mh = 0.07 kg/s
specific heat of hot liquid Cph
= 4.2 kg/min
= 3.5 kj/kg K
Cph  3.5  103 j / kg K
Inlet temperature of hot liquid T1
= 130C
Specific heat of water Cph
= 4.18 kj/kg K
Cph
= 4.18  103 j/kg K
Inlet temperature of cooling water t1 = 15C
Mass flow rate of cooling water mc = 17 kg/min
mc  0.28 kg/s
Overall heat transfer coefficient U = 1100 w/m2K
Area A = 030 m2
To find
1. Outlet temperature of liquid (T2)
2. Outlet temperature of water (t2)
3. Effectiveness of heat exchanger ()
Solution:
Capacity rate of hot liquid C  mh  Cph
0.07  3.5  103
C  245 W/K .......1
195
Capacity rate of liquid C = mh  Cph
= 0.28  4.18  103
C = 1170.4 W/K ..........2
From (1) and (2)
Cmin = 245 w/k
Cmax  1170.4 w/k
Cmin
245

 0.209
Cmax 1170.4
Cmin
 0.209..........3
Cmax
UA
Cmin
From HMT data book Page No.155
Number of transfer units NTU =
1100  0.30
245
NTU  1.34 .........4
NTU 
To find effectiveness  refer HMT data book Page No.165
Parallel Flow heat exchanger
From graph
Xaxis  NTU  1.34
Curve 
Cmin
 0.209
Cmax
Corresponding Yaxis value is 64%
i.e.  = 0.64
We know
Maximum possible heat transfer
Qmax  Cmin (T1  t1 )
= 245 (130-15)
Qmax  28,175 W
Actual heat transfer rate
Q    Qmax
= 0.64  28.175
Q = 18,032 W
196
We know that
Heat transfer
Q  mc Cpc (t 2  t1 )
 18,032 = 0.28  4.18  103  t 2  15 
 18,032 = 1170.4 t 2  17556
 t 2  30.40C
Outlet temperature of water t 2  30.40C
We know that
Heat transfer
Q  mc Cpc (T1  T2 )
 18,032 = 0.07  3.5  103 130  T2 
 18,032 = 31850  245 T2
 T2  56.4C
Outlet temperature of liquid T2  56.4C
12. In a counter flow heat exchanger water at 20C flowing at the rate of 1200 kg/h it is
heated by oil of specific heat 2100 J/kg K flowing at the rate of 520 kg/h at inlet
temperature of 95C. Determine the following.
1. Total heat transfer
2. Outlet temperature of water
3. Outlet temperature of oil
Take
Overall heat transfer coefficient is 1000 W/m2 K. Heat exchanger area is 1m2
Given:
Cold fluid – Water
Hot fluid – Oil
Inlet temperature of water t1 = 20C
Mass flow rate of water mc = 1200 kg/h
mc = 0.33 kg/s
Specific heat of oil
Cph = 2100 J/kg K
Mass flow rate of oil = mh = 520 kg/h
520

kg / s
3600
mh  0.144 kg / s
197
Inter temperature of oil T1 = 95C
Overall heat transfer coefficient U = 1000 W/m2K
Heat exchanger area A = 1m2
To find
1. Total heat transfer (Q)
2. Outlet temperature of water (T2)
3. Outlet temperature of oil (t2)
Solution
Capacity rate of oil C  mh  Cph
 0.144  2100
C = 302.4 W/K .......1
Capacity rate of water
C  mh  Cph
= 0.33  4186
C = 1381.3 W/K ........2
Specific heat of water Cpc = 4186 J/kg K
From Equation (1) and (2)
Cmin  302.4 W/K
Cmax = 1381.3 W/K
Cmin
302.4
=
 0.218
Cmax
1381.3
Cmin
 0.218 ......3
Cmax
UA
Cmin
From HMT data book Page No.155
100  1
302.4
Number of transfer units NTU =
NTU -3.3 ........4
To find effectiveness  refer HMT data book Page No.166
(Counter Flow heat exchanger)
198
From graph
Xaxis  NTU = 3.3
Curve 
Cmin
 0.218
Cmax
Corresponding Yaxis value is 0.95
i.e.  = 0.095
We know
Maximum possible heat transfer
Qmax  Cmin  T1  t1 
= 302.4 (95-20)
Qmax  22,680 W
We know
Actual heat transfer rate
Q    Qmax
= 0.925  22,680
Q = 21,546 W
We know that
Heat transfer
Q  mc Cpc  t 2  t1 
21,546  0.33  4186 (t 2  20)
Cpc = 4186 J/kg K
21,546 - 1381.38 t 2  27.627.6
t 2  35.5C
Outlet temperature of water t 2  35.5 C
We know that
Heat transfer
199
Q  mc Cpc  T1  T2 
21,546  0.144  2100 (95  T2 )
21,546 = 28,728 - 3024 t 2
T2  23.75C
Outlet temperature of oil T2  23.75C
13. In a cross flow both fluids unmixed heat exchanger, water at 6C flowing at the rate
of 1.25 kg/s. It is used to cool 1.2 kg/s of air that is initially at a temperature of 50C.
Calculate the following.
1. Exit temperature of air
2. Exit temperature of water
Assume overall heat transfer coefficient is 130 W/m2K and area is 23 m2.
Given :
Cold fluid – water
Hot fluid – air
Inlet temperature of water t1 = 6C
Mss flow rate of water mc
= 1.25 kg/s
Mass flow rate of air mh
= 1.2 kg/s
Initial temperature of air T1 = 50C
Overall heat transfer coefficient U = 130 W/m2 K
Surface area A = 23 m2
To find
1. Exit temperature of air (T2)
2. Exit temperature of water (t2)
Solution
We know
Specific heat of water Cpc = 4186 J/kg K
Specific heat of air Cph = 1010 J/kg K (constant)
We know
Capacity rate of water
C  mc  Cpc
= 1.25  4186
C = 5232.5 W.K ............1
Capacity rate of air
C  mh  Cpc
= 1.2  1010
C = 1212 W.K ............2
From Equation (1) and (2) we know
200
Cmin  1212 W / K
Cmax = 5232.5 W/K
Cmin
1212

 0.23
Cmax 5232.5
Cmin
 0.23 .......3
Cmax
UA
Cmin
Number of transfer units NTU =
=
130  23
1212
NTU = 2.46 ...........4
(To find effectiveness  refer HMT data book Page No.169)
(cross flow both fluids unmixed)
From graph
Xaxis  NTU  2.46
Curve 
Cmin
 0.23
Cmax
Corresponding Yaxis valueis 0.85
i.e.
 = 0.85
Maximum heat transfer
Qmax  Cmin (T1  t1 )
= 1212 (50-6)
Qmax  53,328 W
Actual heat transfer rate
Q    Qmax
= 0.85  53,328
Q = 45,328 W
201
We know
Heat transfer
Q  mc Cpc (t 2  t1 )
45,328  1.25  4186 (t 2  6)
45,328  5232.5 t 2  31,395
t 2  14.6C
Outlet temperature of water t 2  14.6C
we know
Heat transfer
Q = mhCph (T1  T2 )
45,328  1.2  1010 (50  T2 )
45,328  60,600  1212 T2
T2 = 12.6C
Outlet transfer of air T2  12.6C
202
UNIT – V
High-Speed flow Heat Transfer,
Heat Transfer problems in gas turbine combustion chambers
Rocket thrust chambers
Aerodynamic heating
Ablative heat transfer
Heat transfer problems in nozzles
203
Heat Transfer Problems In Aerospace Engineering
PART – A
1. Define Jet propulsion.
When oxygen is obtained from the surrounding atmosphere for combustion process,
the system is called as Jet propulsion.
2. List out the components of aircraft gas turbine.
The components of aircraft gas turbines are:
i. Compressor
ii. Combustion chamber
iii. Turbine
iv. Tail pipe (or) Nozzle
v. After burner
3. List the different types of jet engines.
i. Turbo jet engine
ii. Turbo prop engine
iii. Ram jet engine
iv. Pulse jet engine
v. Turbo fan engine.
4. Define Thrust.
The force which propels the aircraft forward at a given speed is called thrust (or) propulsive
force.
5. Define specific Impulse.
Specific impulse is defined as the thrust developed per unit weight flow rate through
propulsive device.
F
_____
Isp =
W
6. Define thrust power (or) propulsive power.
It is define as the power developed by the thrust of the engine is called thrust power
which is the thrust force times the distance moved by air craft per unit time.
204
Pth = F . U
Pth = m (Cj – U ) x u
Where
M – flow rate of air fuel mix
Cj – Jet velocity
U – flight velocity
7. Define specific fuel consumption (SFC).
Specific fuel consumption is defined as the fuel consumption rate per unit thrust.
mf
TSFC = ____
F
8. Define propulsive efficiency.
It is the ratio of propulsive or thrust power to power output of the engine.
Propulsive power (or) thrust power
p = ____________________________________________
Power output of the engine
9. Define thermal efficiency.
It is the ratio of power output of the engine to power input to the engine through the
fuel.
Power output of the engine
th =
___________________________________
Power input to the engine
½ m (Cj2 – u2)
th = ___________________
m f . Qf
where, m – mass flow rate of air fuel on mixture
Cj – Jet velocity
u – flight velocity
mf – mass rate of fuel
Qf – calorific value of fuel
10. Define overall efficiency.
It is defined as the ratio of propulsive power to power input to the engine through
205
fuel.
Propulsive Power
0 = ________________________________ __________________
Power input to the engine through fuel
m (Cj – u) . u
0 = _________________
mf Qf
0 = p x nth
11. Give an expression that relatives overall efficiency and TSFC.
u
0 =
________________
TSFC x Qf
12. Define propulsive efficiency as applied to jet propulsion system.
Thrust power
_______________________________
Propulsive efficiency =
Thrust Power + K.E losses
13. Define air standard efficiency of ideal cycle.
Air standard efficiency is defined as the ratio of work to the heat supplied
Qr
j = 1 - _____
Qs
1
1
____
_________
j = 1 =1t
(r-1)/r
14. Define Ram effect.
The pressure rise takes place due to the conversion of kinetic energy of incoming air
into pressure energy by the diffuser. This type of compression is called as Ram effect.
15. Give the expression for ideal efficiency of ram jet engine.
ideal = 1 -
1
1
_____
___________
t
where r – pressure ratio
=1-
r (r-1/r)
206
PART – B
1. A turboprop engine operates at an altitude of 3000 meters above mean sea level and
an aircraft speed of 525 kmph. The data for the engine is given below:
Inlet diffuser efficiency
= 0.875
Compressor efficiency
= 0.790
Velocity of air at compressor entry
= 90 m/s
Temperature rise through the compressor= 2300C
Properties of air:  = 1.4, cp = 1.005 kJ/kg K
From the above data calculate (a) pressure rise through the inlet diffuser (b) pressure
ratio developed by the compressor (c) power required by the compressor per unit flow rate of
air and (d) the air standard efficiency of the engine.
Solution. At Z = 3000 m
Ti = 268.65K, pi = 0.701, pi = 0.909 kg/m3
ai = 328.7 m/s
u = 525/3.6 = 145.833 m/s
Mi = u/ai = 145.833/328.7 = 0.4436
Toi
-1
--- =1+ ---- Mi2 = 1+ 0.2 x 0.44362= 1.03936
Ti
2
Toi = To1 = 1.03936 x 268.65 = 279.244 K
T1 = T01 – c12/2cp = 279.244-8100/2x1005 = 275.194 K
For the inlet diffuser,
T1s – Ti = D(T1 – Ti)
T1s
T1
p1s (-1)/ p1 ( -1)/
----- = 1 +D --- -1 = ------ = ---Ti
Ti
pi
pi
( 1) / 
 p1 
 275.194 
= ! + 0.875 
 1  1.0213
 
 268.65

 pi 
3.5
p1/pi = (1.0213) = 1.0766
p1 = 1.0766  0.701 = 0.7547 bar.
(a)
(b)
Pressure rise through the inlet diffuser
P1-Pi= 0.7547-0.7010=0.0537bar Ans
Compressor efficiency is given by
207


c  T01 roc  1 /   1 T02  T01 
r0 c 
 1 / 

roc 

c
T01
T02  T01   1
0.79  230
 1  1.6507
279.224
P02
3.5
 1.6507   5.779 Ans
P01
( c) power required by the compressor is
ma c p T02  T01   11.005  230
 231.15kw /  kg / s  . Ans
(d) Air standard efficiency of the engine for the pressure ratio of 5.779 is
 j  1
1
1
 1
0.286
 r0c   1 / 
 5.770 
 j   0.3942 Ans.
2. The diameter of the propeller of an aircraft is 2.5m; It flies at a speed of 500 kmph at
an altitude of 8000m. For a flight to jet speed ratio of 0.75 determine (a) the flow rate of
air through the propeller, (b) thrust produced, ( c) Specific thrust, (d) specific impulse
and (e) the thrust power.
Solution: Area of cross-section of the propeller disc
A

4
d2 

4
 2.52  4.908m 2
Air density at Z= 8000m is
 =0.525kg/m3
Flight speed u =500 138.89m/s
σ=u/cj=0.75
c j =138.89/0.75=185.18m/s
(a) velocity of air flow at the propeller disc is
208
1
u  c j 
2
c  0.5 138.89  185.18   162.35m / s
c
Theoretical value of the flow rate is given by
ma   Ac  0.525  4.908 162.035
ma  417.516kg / s Ans
(b) F=ma  c j  u 
F  417.516 185.18  138.89  103
F  19.3268kN Ans
F 19326.8
(c) Fs 

 46.29 N  kg / s  Ans
ma 417.516
(d )
Is 
F
F
46.29


 4.718s Ans.
wa ma g
9.81
(e) Thrust power is P=F  u
P=19.3268 138.89=2684.3kW Ans
3. An aircraft flies at 960 kmph. One of its turbojet engines takes in 40 kg/s of air and
expands the gases to the ambient pressure. The air –fuel ratio is 50 and the lower
calorific value of the fuel is 43 MJ/kg. For maximum thrust power determine (a) jet
velocity (b) thrust (c) specific thrust (d) thrust power (e) propulsive, thermal and
overall efficiencies and (f) TSFC).
Solution: u=960 X 1000/3600=266.7m/s
(a)
For maximum thrust power   u / c j  0.5
c j  266.7 / 0.5  533.4m / s
 mf 
(b)in  ma  m f  ma 1 
 ma 
1

m  40 1    40.8kg / s
 50 
F  mc j  ma u
F   40.8  533.4  40  266.7   103
F  11.094 KN Ans.
( c) specific thrust based on air intake
209
FS  F / ma
Fs  11094 / 40  277.35 N  kg / s  Ans
(d )Thrust power p= F  u
p=11.094  266.7=2958.77kw Ans
(e) p 
 th 
1
2

 0.666or 66.6% Ans
1 c j /u 1 3
1
m  c2 j  u 2 
2
mfQf
mf 
 th 
1
40
ma 
 0.8kg / s
50
50
0.5  40.8  533.42  266.7 2 
0.8  43 106
th  0.1265 or 12.65% Ans
0   p  th  0.666  0.1265
0  0.0842 or 8.42% Ans
TSFO=m j  3600 / F
 0.8  3600 /11094
 0.2596kg / kN Ans
4. A turbojet engine propels an aircraft at a Mach number of 0.8 in level flight at an
altitude of 10 km. The data for the engine is given below.
Stagnation temperature at the turbine inlet= 1200K
Stagnation temperature rise through the compressor =175K
Calorific value of the fuel=43MJ/kg,
Compressor efficiency = 0.75
Combustion chamber efficiency = 0.975
Turbine efficiency = 0.81
Mechanical efficiency of the power transmission between turbine and compressor =0.98
Exhaust nozzle efficiency = 0.97.
Specific impulse =25 seconds.
Assuming the same properties for air and combustion gases calculate:
(a) fuel-air ratio,
(b) Compressor pressure ratio
(c) Turbine pressure ratio
(d) Exhaust nozzle pressure ratio, and
210
(e) Mach number of exhaust jet.
Solution:
Temperature and velocity of sound at the engine entry at Z=10km are.
Ti  223.15 K , ai  299.6m / s
  1 2 
T01  Toi  1 
M i  Ti
2


 1.4  1

T01  1 
 0.82  223.15  251.71k
2


T02  T01  175, T02  175  251.71  426.71K
( a) For the combustion chamber,
m
a
 m f  c pT03  ma c pT02   Bm f Q f
1 
1

1
T

T02   Bm f Q f / cp
03


f
f 
1
0.975  43 106
1200

426.71

 1200


f
1005
f
(b)
mf
ma
 0.01908 Ans
For the compressor,
T P 
T02  T01  01  02 
c  P01 
 1 / 
1
Stagnation pressure ratio across the compressor
=  P02 / P01  roc therefore
   1

T T 
roc  1  c 02 01 
T01 

 0.75 175 
roc  1 

251.71 

3.5
 1.5214 
3.5
roc  4.344 Ans
211
( c) Compressor power required = power supplied by the turbine; there fore
m  ma  m f  cp T03  T04   T02  T01
T03  T04  175 / 0.98  1.01908  175.228

1 
T03 1 

 r0T 
 1 / y
 r  175.228
 /   1
175.228 

ToT  1 

 1200  0.81 
Is 
 2.005 Ans
u1 
  1
g 

u / ai  M i
u  M i ai  0.8  299.6  239.68m / s
(d) 25 
239.68  1 
1
9.81   
  u / ce  0.4942
ce  239.68 / 0.4942  484.93m / s
For the exhaust or propelling nozzle.
Toe  T04  Te  c 2 e / 2c p
T04  1200  175.228  1024.772k
Te  1024.772  484.932 / 2 1005
Te  907.778 K
 1
T04  Te  n T04  Tes nT04 1     1 
 rn 
 1
1024.772  907.778  0.97  1024.772 1     1 
 rn 
rn 
p04
 1.55 Ans
Pe
( e) velocity of sound at the nozzle exit
212
ae   RTe 
1/ 2
 1.4  287  907.778
1/ 2
 603.94m / s
The exit Mach number is given by
M e  ce / ae  484.93 / 603.94
M e  0.803 Ans
The Mach number corresponding to the isentropic flow will be higher than this value.
Using isentropic flow tables for =1.4
M = 0.816 at pe/P04=1.55=0.645
5. A ramjet engine operates at M=1.5 at a altitude of 6500m. The diameter of the inlet
diffuser at entry is 50 cm and the stagnation temperature at the nozzle entry is 1600K.
The calorific value of the fuel used is 40 MJ/kg. the properties of the combustion gases
are same as those of air ( =1.4, R=287 J/Kg K.. The velocity of air at the diffuser exit is
negligible.
Calculate (a) the efficiency of the ideal cycle (b) flight speed (c) air flow rate (d)
nozzle jet Mach number (h) propulsive efficiency (i) and thrust. Assume the following
values:  D  0.90, B  0.98, j  0.96 stagnation pressure loss in the combustion
chamber =002p02
Solution:
Refer to Figure
At
z=6500m the properties of air are
T1  245.90 K , P1  0.440, a1  314.50m / s
1  0.624kg / m
3
(a)ideal cycle efficiency

i 1 

2 1 
  1 M 21 
1
2
1 

i  1 
 2   0.310. Ans
 1.4  1 1.5 
u
(b) M 1 
a1
u  1.5  314.50  471.75m / s
u  1698.3kmph. Ans
( c) Area of cross section of the diffuser inlet
213
A1 

 0.52  0.1963m 2
4
ma  1uA1
ma  0.624  471.48  0.1963
ma  57.752kg / sAns
(d) For negligible velocity at the diffuser exit p02=p2
 1
c T  T   rD 
 D  p 2s 1 
1 2
 1 2
c1
M 1
2
2
 rD 
 1
 1 D
rD  1.405 
3.5
 1
2
M 21  1  0.9 
1.4  1
 1.52
2
 3.2875. Ans
p2  p02  rD.P1  3.2875  0.440  1.446
T01
 1 2
 1
M 1  1  0.2 1.52  1.45
T1
2
T02  T01  1.45  245.90  356.55 K
M a c p T03  T02    Bm f Q f
(e) f 
mf
ma
 c p T03  T02  B.Qf
f  1.005 1600  356.55  / 0.98  40000
f  0.03188 Ans
(f)
P03  P02  0.02 p02  0.98 p02
P03  0.98 1.446  1.417bar
Nozzle pressure ratio
p
1.417
rj  03 
 3.22 Ans
p4 0.440
(g)The mach number at the nozzle exit for a pressure ratio of 3.22 in an isentropic expansion
would be M4s =1.41; however on account of irreversible expansion ( j =0.96) the exit
velocity and mach number will be slightly lower.
214
T04
 1 2
 1
M 4 S  1  0.2 1.412  1.3976
T4 s
2
T4 s  1600 /1.3976  1144.82 K
T04  T4   j T04  T4 s   0.96 1600  1144.82 
T4  1600  4363973  1163.027 K
T04  T4  c 2 4 / 2c p
C 2 4  2c p T04  T4   2 1005  436.973
c4  937.185m / s
a4  1.4  287 1163.027 
1/ 2
M4 
 683.59m / s
c4 937.185

 1.371Ans
a4 683.596
h)   u / c 4  471.48 / 937.185  0.503
p 
2
2  0.503

 0.6693 Ans
1   1  0.503
(i)m f  57.752  0.03188  1.841kg / s
m  ma  m f  57.752  1.841  59.593Kg / s
F  mc4  ma u
f   59.593  937.185  57.752  471.75 10 3
F  28.614kN Ans.
6. A turbo jet engine takes in 50 kg/sec of air and propels an aircraft with uniform flight
speed 880 km/hr. Isentrophic enthalphy change for nozzle is 188 kj/kg and its velocity
co-efficient is 0.96. the fuel air ratio is 1.2%. combustion efficiency is 95%, calorific
value of fuel is 44000 kj/kg. find out,
i.
ii.
iii.
iv.
Thermal efficiency of the engine
Fuel flow in kg/hr
Propulsive efficiency
Overall efficiency
Given data :
215
ma  50 kgs; u=
880
 244.444m / s
3.6
 n  188.10 j / kh; Cu  0.96;
mf
 0.012
ma
nCB  0.95; C.V=44000 kj/kg
mf
 0.012  50 = 0.6 kg/s
ma
Ce  Cv 2h
 0.96 2  188  1000 = 588.66 m/s
Fuel flow in kg/hr = 0.6 X 3600
= 2160 kg/hr
Since the expansion in the nozzle I isentropic,
Ci Cc  588.66 m/s
np=
2u
 58.6827%
cj  u
m  ma  mf  50.6 kg/s
Therefore the velocity of jet.
Thrust power = [mcj-mau]u
= 50.6 X 588.66 – 50 X 244.444) 244.444
= 4293.4134 kw
Overall efficiency
Thrust power
Heat supplied
4293.4134  103
= no  17.1188%
0.6  44000  103  0.95
Thermal efficiency of the engine nth 
no
 29.1719%
np
Result :
i.
ii.
iii.
iv.
Thermal efficiency of the engine nth = 29.1719%
Fuel flow in kg/hr =
m1 = 2160 kg/hr
Propulsive efficiency
=
np = 58.6827%
Overall efficiency =
n0 = 17.1188%
7. A turbo jet has a speed of 750 km/hr while flying at an altitude of 10000m. the
216
propulsive efficiency of the jet is 50% and the overall efficiency of the turbine plant is
16%. The density of the air at 10000m altitude is 0.173 kg/m3. the drag on the plane is
6250 N. calorific value of the fuel is 48000 kj/kg. Calculate,
i.
Absolute velocity of the jet
ii.
Diameter of the jet and
iii.
Power output of the unit in kw.
Given data :
u
750
 208.333 m/s
3.6
Z=10000m; np  0.5; n0  0.16
P  0.173 kg/m3 ; F=D=6250N
C.V =48000 kj/kg
2u
2  208.3333
np 
 cj  u 
= 833.3333 m/s
cj  u
0.5
 C j  625m / s
n0 
Thrust power
Heatsupplied
n0 
F u
mf  c.v
 mf 
6250  208.3333
= 0.1695421 kg/s
0.16  48000  1000
abolute velocity of the jet = C j  u  416.66667 m/s
F  (ma  mf )C j  mau = maC j  mau  mf C j
 ma

F  (mf  C j )
(C j  u )
6250  (0.165421 625)
= 14.745686 kg/s
416.66667
 m=ma  mf = 14.915228 kg/s
217
Volume of gas / sec =
m 14.915228

86.21519m3 / s
P
0.173
But volume Q=Area of the jet  C1
 Aj 
Q
 0.1379443m3
Cj
 Diameter of the jet di  0.419m
Propulsive efficiency =
 Power output of the engine =
Thrust power
Power output of the engine
6250  208.33333
0.5
= 2604.16666kw
Result :
i.
ii.
iii.
Absolute velocity of the jet
c = 416.66667 m/s
Diameter of the jet
dj = 0.419 m
Power output of the engine Pout = 2604.1666 kw
8. For a turbo jet with a flight velocity of 800 km/h at an ambient of 60 kPa, the
properties of gas entering the nozzle are 300 kPa and 200C. The mass flow rate of air is
20kg/sec. assuming air (Y=1.4 and R=287 J/kg) as working fluid, find a) thrust
developed, b) thrust power and c) propulsive efficiency.
Given Data :
u
800
 22.2222 m/s; Pa  60 kPa
3.6
P4  300 kPa; T4  200  273  473k
ma  20kg / s; y=1.4
R=287J/kgk
Assuming that, the expansion of gases in the nozzle is isentropic.
218

Te  P 


T4  P4 
j
e
y 1
y
0.4
 60  1.4
Te  
  473  298.645 k asume Pe  Pa
 300 
Ce  2Cp (T4  Te )
 2  1004.5(473  298.645)
= 591.84389 m/s
Since m is not given =  m  ma
ma (C j  u )
Thrust developed = 20(591.84389 – 222.2222)
F = 7392.4334 N
=FXu
Thrust power = 1642.763 kw
Propulsive efficiency = np 
2u
 54.595%
Cj  u
9. The diameter of the propeller of an aircraft is 2.5 m. it flies at a speed of 500 kmph at
an altitude of 8000 m for flight to jet speed ratio = 0.75. determine i) The flow of air
through the propeller, ii) Thrust produced, iii) Specific thrust, iv) Specific impulse, v)
Thrust power.
Given data :
d  2.5m ; u=
500
 138.8889m / s
3.6
  0.75; Z=8000 m
=
u
u
; C j   185.1851m / s
cj

velocity of air flow at the propeller disc,
c j+u
 162.037 m/s
2
From gas Table Z = 8000 m, p = 0.525 m3.
From continuity equation, ma = pAc kg
c=
219
 0.525 

(2.5)2  162.037 = 417.5835 kg/s
4
Thrust produced F=ma (C j  u ) = 19332.5292 N
Thrust power = F  U = 2685.0737 kW
Specific thrust Fsp 
Isp 
F
 46.2962 N(kg/s)
ma
F
F
 sp
Wa
g
Specific impulse = 4.71928 sec
Result :
i.
Flow of air through the propeller ma = 417.5835 kg/s
ii.
Thrust produced F = 19.3325 KN
iii.
Specific thrust Fsp = 46.296n N/(kg/s)
iv.
Specific impulse Isp = 4.71928 sec.
v.
Thrust power P = 2685.0737 kw.
10. A turbo jet engine is traveling at a speed of 236.11 m/sec under conditions 288K
and 1.013 bar. The ram efficiency is 85%. Calculate the total pressure and total
temperature of air after inlet diffuser. Take Cpa = 1.005, Ya=1.4.
For the above engine, the total temperature and total pressure at the inlet to the
nozzle is 806k and 2.124 bar. The flow through nozzle is adiabatic with jet efficiency
(total-to-static) 95%. Calculate the net specific thrust and thrust specific fuel
consumption (TSFC) if fuel air ratio is 0.0122. take Cpg=1.16 and yg 1.33.
Given data :
U = 236.11 m/s
Ti = 288 K; Pi = 1.013 bar.
nR = 0.85; Cpa = 1.005; ya = 1.4
t04=806k; P04=2.124 bar
(nt  s )f 
mf
 0.0122; Cpg  1.16 y Yg  1.33
ma
Cpa (T01  Ta ) 
u2
2
From the figure, T01 
(236.11)2
 288 = 315.7352 k
2  1005
a-1 isentroic Diffusion
220
T01  P01 
 
Ta  Pa 
y 1
y
T 
 P01   01 
 Ta 
y
y 1Pa
1.4
 315.7352
 1.4
= 
K
 288

P01  1.397546 bar
P01  Pa
P01  Pa
Ram efficiency P01 = 1.33986 bar
4-e 
nR 
Isentropic expansion
P 
Te
 e 
04
T
 P04 
y 1
y
0.33
 1.013  1.33
Te  
  806
 2.124 
Te=670.7384 K
T  Te
n j  04
T04  Te
806  Te
 Te  677.5015 k
806  670.7384
C2
Cpg (T04  Te )  j
2
From the figure
0.95 
 C j  Cpg (T04  Te ) = 546 m/s
The specific gross thrust is given by,
Fg = Fnet + u
Fnet = 552.6616 – 236.11
=
316.5516 N(kg/s) of air flow
Total specific fuel consumption
TSFC 
f  3600
= 0.138745 kg of fuel / N-hr
Fnet
Result :
221
i.
ii.
iii.
iv.
Pressure after the inlet diffuser P01=1.33986 bar.
Temperature after the inlet diffuser T01=315.7352 K
Net specific thrust Fnet=316.5516 N(kg/s) of air.
TSFC = 0.138745 kg fuel N-hr.
11. A rocket flies at 10,080 kmph with an effective exhaust jet velocity of 1400 m/s and
propellant flow rate of 5.0 kg/s. If the heat of reaction of the propellants is 6500 kJ/kg of
the propellant mixture determine:
a) Propulsion efficiency and propulsion power,
b) engine output and thermal efficiency, and
c) overall efficiency.
Solution: (a) cj = 1400 m/s
u = 10,080  1000/3600 = 2800 m/s
 = u/ cj = 2800/1400 = 2.0
2
22
p = ---------- = -------- = 0.80
1 + 2
1+4
Ans.
F = mp  cj = 5  1400 = 7000 N
Propulsive power = F  u
Pp = 7000  2800  10-6
Pp = 19.6 MW
Ans.
PEn = Pp / p
(b)
PEn = 19.6 / 0.8 = 24.5 MW
Ans.
1
Alternatively, PEn = --- mp (c2j + u2)
2
= 0.5  5 (1.42 + 2.82)  106 watts
= 2.5  9.8 = 24.5 MW
Ans.
PEn
 th = -------mp QR
mp QR = 5  6500  1000 watts
mp QR = 32.5 MW
 th = 24.5 / 32.5 = 0.7538
o
Ans.
= p   th = 0.8  0.7538 = 0.603 Ans.
222
12. Determine the maximum velocity of a rocket and the altitude attained from the
following data:
Mass ratio
= 0.15
Burn out time
= 75 s
Effective jet velocity = 2500 m/s
What are the values of the velocity and altitude losses due to gravity? Ignore drag and
assume vertical trajectory.
Solution: The maximum velocity is given by
1
up = cj ln --------  g tp
MR
up
1
= 2500 ln -------  9.81  75
0.15
up = 4742.8  735.75 = 4007.05 m/s
Ans.
The altitude gain is given by
Z = Z p + Zc
1
1
1
Zp = cj tp 1 + 1  ---- ln
----   ---- g tp2

MR
2
 = 1  MR = 1  0.15 = 0.85
1
9.81  752  0.5
2
---- g tp = --------------------- = 27.59 km
2
1000
2500  75
1
Zp = --------------- 1 + 1  -----1000
0.85
1
ln -----0.15
 27.59
Zp = 124.72  27.59 = 97.13 km
1 up2
0.5
4007.032
Zc = ---- ----- = -------  -------------- = 818.36 km
2 g
9.81
1000
Z = 97.13 + 818.36 = 915.49 km
Ans.
Velocity loss due to gravity is
223
g tp = 735.75 m/s
Ans.
Altitude loss due to gravity is
1
---- g tp2 = 27.59 km
2
Ans.
13. A missile has a maximum flight speed to jet speed ratio of 0.2105 and specific
impulse equal to 203.88 seconds. Determine for a burn out time of 8 seconds.
(a) effective jet velocity,
(b) mass ratio and propellant mass fraction,
(c) maximum flight speed, and
(d) altitude gain during powered and coasting
flights.
Solution: (a) cj = g Is
cj = 9.81  203.88 = 2000.06 m/s
Ans.
1
up = cj ln ____  g tp
MR
(b)
up
1
____
cj
= ln ____  g tp / cj
MR
1
0.2105 = ln
____
8
 9.81 
MR
_________
2000.06
1
ln
____
= 0.2105 + 0.0392 = 0.2497
MR
1
_____
= 1.281, MR = 0.780
Ans.
MR
 = 1  MR = 1  0.780 = 0.220 Ans.
(c)
up / cj = 0.2105
up = 0.2105  2000.06 = 2121 m/s
up = 1515.65 kmph
(d)
Zp = cj tp
1
1 + 1  ___
1
Ans.
1

ln
1
_____

MR
____
2
1
1
224
g tp2
Zp = 2000.06  8
1 + 1  ____ _
0.22
ln ______  ___  9.81  82
0.780
2
Zp = 1593.94m = 1.594 km
1
Ans.
up2
Zc = ____ _____
2
g
Zc = 0.5  4212 / 9.81  1000
Zc = 9.0336 km
Ans.
14. A rocket files at 10.080 kmph with an effective jet velocity of 1400 m/s and the
propellant flow rate of 5 kg/s. if the heat of reaction of the propellants is 6500 kj/kg of
the propellent mixture, determine,
a) Propulsion efficiency and propulsion power
b) Engine output and thermal efficiency
c) Overall efficiency
Given Data:
u
10800
 2800m / s; C j  1400m / s
3.6
mp  5kg; C.V=6500 kj/kg
=
U
2
Cj
Propulsion efficiency np 
2 4
 0.8  80%
 2 15
Propulsion power = mp  C j  u
 5  1400  2800
Pp  19.6MW
nth 
C 2j  u 2
2  c.v
=
(1400)2  (2800)2
2  6500  103
From equation, 0.753846
n0=np X nth
225
 0.8  0.753846  0.060307
1
mp (c 2j  u 2 )
2
1
Engine ouput power =  5  [(1400)2  (2800)2 ] = 24.5 MW
2
Result:
a) Propulsion efficiency = 80%
Propulsion power Pp = 19.6 MW
b) Engine Output power = 24.5 MW
Thermal efficiency nth = 75.3846 %
c) Overall efficiency n0 = 60.307%
15. The following conditions refer to a rocker propellant flow rate = 193 kg/s.
Thrust chamber pressure 27 bar and temperature 3000k.
Nozzle exit diameter = 600 mm.
Nozzle exit pressure = 1.1 bar and ambient pressure = 1.013 bar.
Thrust produced = 380 kN.
a)
b)
c)
Find the effective jet velocity, actual jet velocity, specific impulse and the specific
propellant consumption.
Recalculate the values of thrust and specific impulse for an altitude of 20000 m.
The rocket speed is 2500 Kmph and the heat of reaction of the propellant is 6500
KJ/kg, find for case a) i.e. ambient pressure of 1.013 bar find np, nth, and n0.
Given data :
Propellant flow rate
=
Thrust chamber pressure =
Nozzle exit diameter
=
Ambient pressure
Thrust produced
Rocket Speed
=
Heat of reaction of propellant
193 kg/s
27 bar
1.1 bar
=
1.013 bar
=
380 N
2500 kmph
=
6500 kj/kg
mp = 193kg/s; ds = 0.6m; Pe1.1 bar.
Pa = 1.013 bar; F= 380KN
i) Thrust F = mp x cj
 cj
F
1968.9119m / s.
mp
Also
F = mpCe + (Pa - Pa) Ae
226

380 x 103 =193 x ce + (1.1-013) x 105x  0.6 2
4
ii) ce – 1956.1664m/s
iii) Specific impulse I sp 
F
380  10 3 1


mp  g 193  9.81 ls

iv) Specific propellant consumption
1
Is
 4.9824  10 3
1
S
Result :
i) Effective jet velocity cj
= 1968.9119 m/s
ii) Actual jet velocity ce
= 1956.1664 m/s
iii) Specific impulse Isp
= 200.7045 sec s
iv) Specific propellant consumption = 4.9824x10-3 1/secs
b) Z= 20,000m
From gas table Z= 20000 m.
Pa = 0.0548 bar
Thrust F = mp x Ce + (Pe – Pa) Ae

= 193 x1956.1664+(1.1-0.0548) x105   0.6 2
4
= 407.0924 KN

F
407.0924  10 3

mp  g
193  9.81
Specific impulse = 215.01397 sec s
Result :
i)
ii)
Thrust F = 407.0294 KN
Specific impulse
Isp = 215.013 sec s
C) Flight speed u 
2500
 694.444m / s
3.6
Thrust power Pp = Fx u= 380 x103 x 694.444

1
Power lost in exhaust gases  mp c j  u
2
1
2
  193  1968.9119  694.444 
2

= 263888.872KW
2
 156741.9033KN
Engine Out power = Pp + Power lost in exhaust gases
PE = 420630.7753 KN
227
Propulsive efficiency

propulsive power
Engine output power
263888.872
 0.62736
420630.7753
Engine output power
nth 
Energy sup plied by the fuel

=
420630.7753
193  6500
nth  0.3352975
Overall efficiency n0 = np x n th
= 0.21035
Result :
i)
ii)
iii)
Propulsive efficiency np = 62.736%
Thermal efficiency nth = 33.529%
Overall efficiency no
= 21.035%
16. A rocket has the following data:
Propellant flow rate = 5 kg/s
Nozzle exit diameter =10 cm
Nozzle exit pressure = 1.02 bar
Ambient pressure = 1.013 bar
Thrust chamber pressure = 20 bar
Thrust = 7 KN
Determine the effective jet velocity, actual jet velocity specific impulse and specific
propellant consumption.
Given:
mp = 5kg/s; de = 0 1m; Pe = 1.02bar; chamber pressure
Po = 20 bar and F = 7000N
Solution:
We know that F = mp x Cj
 c j 
F
 1400m / s
mp
We know that the thrust
228
F  mp  Ce   Pe  Pa  Ae
Ce 
F   Pe  Pa 
mp

2
7000   1.02  1.013   10 5   0.1 
4


F
5

7000  5.49778
5
Ce  1398.9m / s.
Actual jet velocity
We know that
I sp 
Specific impulse

F
mp  g
7000
5  9.81
I sp  142.7115 sec S
SPC=
I
1

I s 142.71
Specific propellant consumption  7.0071  10 3
1
sec
Result:
i)
ii)
Effective jet velocity Cj = 1400m/s
Actual jet velocity Ca =1398.9 m/s
iii)
Specific propellant consumption SPC = 7.0071x10-3
1
sec
17. A rocket nozzle has a throat of 18cm2 and combustor pressure 25 bar. If the specific
impulse is 127.42 and the rate of flow of propellant is 44.145 N/S, determine the thrust
co-efficient, propellant weight flow co-efficient specific propellant consumption and
characteristic velocity.
Given :
A=18 cm2; combustor pressure Po = 25 bar = Pe
Isp = 127.42 sec; wp = 44.145N/S
Specific impulse
229
I sp 
F
WP
 F  ISp  WP
=127.42  44.145
F=5624.9559N
From equation 7.11, the thrust co-efficient
CF 
F 5624.9559  10 4
 1.25
Po A 25  10 5  18
CW 
Wp

Po A
44.145
25  10 5  18  10 4
=9.81  10-3
1
S
1
1

I s 127.42
1
 7.848  10 3
sec
Specific propellant consumption
We know that
F  mp  C j 
Cj 
W
g
Cj
Fg
W
Thrust F 
5624.9559  9.81
44.145
=1250m/s
From equation 7.14 the characteristic velocity
Vchar 
Cj
CF

1250
1.25
Vchar  1000 m / s
18. A rocket engine has the following data:
Effective jet velocity = 1400 m/s
Flight to jet speed ratio = 0.82
Oxidizer flow rate
= 3.5 kg/s
Fuel flow rate
= 1.2 kg/s
Heat of reaction per kg of the exhaust gas = 2600 KJ/kg. Calculate the thrust, specific
impulse, propulsive efficiency, thermal efficiency and overall efficiency.
Given
230
Cj = 1400m/s;  = 0.82;mo
mf = 1.2kg/s; QR = 2600x103 J/kg
Solution:
mp = m o + mf
mp = 4.7kg/s

u
Cj
Propellant flow rate
u =  x cj = 0.82 x 1400
u = 1148 m/s
Thrust F = mp x Cj
F = 6580 N
Specific impulse
I sp 
F
mp  g
=142.7115 sec.
Specific impulse = 142.7115 sec
np 
2
1 2
Propulsive efficiency

2  0.82
2
1   0.82 
 98.062%
nth 
=
C j2  u2
2  QR
 1400 2   1148 2
2  2600  10 3
Thermal efficiency nth = 63.036%
Overall efficiency n0 = np x nth
Result:
i) Thrust F
ii) Specific impulsive Isp
=6580 N
=142.7115 secs
iii) Propulsive efficiency np =98.062%
iv) Thermal efficiency nth
=63.036%
v) Overall efficiency n0
=61.814%
******************
231