DT Q A DX DT Q KA DX: Overall

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HEAT AND MASS TRANSFER
UNIT-1
CONDUCTION
PART-A
1. State Fourier’s law of condition
The rate of heat conduction is propagation to the area measured normal to the direction of heat flow and to the
temperature gradient in that direction.
dT
QA
dx
dT
Q  KA
dx
Where A – area in m2
dT
 Temperature gradient in K/m
dx
K – Temperature conductivity W/mK
2. Defined Thermal Conductivity
Thermal conductivity is defined as the ability of a substance to conduct heat.
3. Write down the equation for conduction of heat through a slab or plan wall.
T overall
Heat transfer Q 
R
where T  T1  T2
1
R  Thermal resistance of slab
KA
L  Thickness of slab
K  Thermal conductivity of slab
A  Area
4. Write down the equation for conduction of heat through a hollow cylinder
Toverall
Heat transfer Q 
R
where T  T1  T2
1 r 
R in  2  thermal resistance of slab
2LK  r1 
L  Length of cylinder
K  Thermal conductivity
r2  Outer radius
r1  inner radius
5. State Newton’s law of cooling or conservation law.
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Heat transfer by conservation given by Newton‘s law of cooling
Q  hA  Ts  T 
Where A – Area exposed to heat transfer in m2
h – heat transfer co efficient in W/m2K
Ts – Temperature of the surface in K
T∞ - Temperature of the fluid in K
6. Write down the general equation for one dimension steady state heat transfer in slab or plane wall with
and without heat generation.
 2 T  2 T  2 T 1 T
  
x 2 y 2 z 2  t
 2 T  2 T  2 T 1 T
  
x 2 y 2 z 2  t
7. Define overall beat 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
8. Write down the equation for heat transfer through composite pipes or cylinder.
Heat transfer
T overall
Q
R
Where
T  Ta  Tb
r  r 
In  2  In  2  L 2
  1  1 
1 1 r r 1
R
2L h n r1 K1 K2 h b r3
9. What is critical radius of insulation or critical thickness?
Critical radius  rc
Critical Thickness  rc  r1
Additional of insulation material on a surface does not reduce the amount of heat transfer rate always. In fact
under certain circumferences it actually increases the heat loose up certain thickness of insulation. The radius of
insulation for which the heat transfer is maximum is called critical radius insulation and the coprresponding
thickness is called critical thickness.
10. Define external surfaces
It is possible to increase the heat transfer rate by increasing the surface of heat transfer. The surface used for
increasing heat transfer are called extended surface used or sometimes known as fin.

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11. State applications of fins.
The main application of fins are,
1. Cooling of electronic components
2. Cooling of motor cycle engines
3. Cooling of transformers
4, Cooling of small capacity compressors.
12. 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
Qlim
min 
Qmax
13. Define Fin effectiveness.
Fin effectiveness is the ratio of heat transfer with fin to that without fin.
Q with fin
Fin effectiveness 
Q without fin
14. List the mode of heat transfer.
1) Conduction
2)Convection
3) Radiation,
15. List down the types of boundary condition.
1) Prescribed temperature
2) Prescribed heat flux
3) Convection boundary condition
16. What is meant by lumped heat analysis?
In Newton heating or cooling process the temperature through the solid is considered to be uniform at a given
time. Such an analysis is called Lumped heat capacity analysis.
17. What is meant by semi infinite solids?
In semi infinite solids, at ant instant of time, there is always a point where the effect of heating or cooling at one
of its boundaries is not fell at all. At this point the temperature remains unchanged. In semi-infinite solids, the
biot number value is infinite.
18. Explain the significance of Fourier number.

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It is defined as the ratio of characteristic body dimension to temperature wave penetration depth in time. lt
signifies the degree of penetration of heating or cooling effect of a solid
19. What are the factors affecting in thermal conductivity?
1. Moisture
2. Density of materials,
3. Pressure
4. Temperature
5. Structure of material
20.Explain the significance of thermal diffusivity?
The physics significance of thermal diffusivity is that tells us how fast heat is propagated or it diffuses through
a material during changes of temperature with time.
PART - B
1. Consider a 1.2 m high and 2 m wide double-pane window consisting of two 3 mm thick layers of glass
(k = 0.78W/mK) separate by a 12 mm wide stagnant air space (k = 0.026W/mK). Determine the steady
rate of heat transfer through this double-pane window and the temperature of its inner surface when the
room is maintained at 24°C while the temperature of the outdoors is -5°C. Take the convection heat
transfer coefficients on the inner and outer surfaces of the window to be 10 W/m2K and 25 W/m2K
respectively.
Given data
Thermal conductivity of glass
 K 1  K3  0.78 W / mK
Thermal conductivity of air K2 = 0.026 W/mK
Ta
T1 T2 T3 T4
Tb
Air Glass
K2 K3
3mm 12mm 3mm
Height of double pane window = 1.2 m
Width of double pane window = 2 m

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Area of double pane window = 1.2 × 2 = 2.4 m2
Interior room temperature, T a = 24° C
Outside air temperature, Tb = -5° c
Thickness of glass, L1 = L3 = 3mm = 0.003m
Thickness of air, L2 = 12mm = 0.012m
Interior convective heat transfer coefficient,
ha = 100 W/m2K.
Outer convective heat transfer coefficient,
hb = 25W /m2K
To find:
(a) Rate of heat transfer, Q
(b) Inner surface temperature, T i
Solution:
Heat flow through composite wall is given by.
T
Q where T  Ta  Tb
R
1 L L L 1
R  1  2  2 
h a A K1 A K 2 A K 3 A h b A
Ta  Tb
Q or
R convet  R12  R 23  R conv 2
Ta  Tb
Q
1 L L L 1
 1  2  2 
h a A K1 A K 2 A K 3 A h b A

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1 1
R convert    0.04167C / W
h 0 A 10  2.4
L1 0.003
R12    0.0016C / W
K1A 0.78  2.4
L2 0.012
R 23    0.1923C / W
K 2 A 0.026  2.4
L3 0.003
R 34    0.0016C / W
K 3 A 0.78  2.4
1 1
R conv 2   0.01667C / W
h 0 A 25  2.4
24   5 
Q
0.4167  0.0016  0.1923  0.0016  0.01667
29

0.25384
Q  114.24 W
24  T1
Q
R conv1
2. Derive the general 3 dimension heat conduction equation in Cartesian coordinates.
Consider a small rectangular element of sides dx, dy and dz shown in figure
The energy balance of this rectangular element is obtained from first law of thermodynamics.
 Net heat 
conducted into  Heat  Heat 
  generated  Stored 
     
 element from   
all the coordinate   with in  in the 
  element  element 
directions
 

Net heat conducted into element from all the coordinates direction
Let qx be the heat flux in a direction of face ABCD and q x+dx be the heat flux in a direction of face
EFGH.
The rate of heat flow into the element in x direction through the face ABCD is
T
Qx  q x dy dz   k x dy dz
x
Where k – Thermal conductivity, W/mK

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T
 Temperature gradient
x
The rate of heat flow out of the element in x direction through the face EFGH is

Q x  dx  Q x   Q x  dx
x
T   T 
 k dydz  k x dydz  dx
x x  x 
T   T 
Q x  dx  k x dydz  kc dx dy dz
x x  x 
Substiting
T  T   T  
Q x  Q x  dx  k x dy dz   k x dydz   k x  dx dy dz 
x  x x  x  
T   T 
 k x dy dz9  k x dy dz  kx dx dy dz
x x  x 
  T 
 Q x  Q x  dx  kx dx dy dz
x  x 
similarly
  T 
Q y  Q y  dy  k y  dx dy dz
y  y 
  T 
Q z  Q z  dz   k z dx dy dz
z  z 
Adding the equation
  T 
Net heat conductance  kx dx dy dz 
x  x 
  T    T 
 ky  dx dy dz   k z dx dy dz
y  y  z  z 
   T    T    T  
  k x   k y   k z   dx dy dz
 x  x  y  y  z  z  
Net heat conducted into element from all the coordinates direction
   T    T    T  
  k x   k y   k z   dx dy dz
 x  x  y  y  z  z  
Heat stored in the element
We know that
Heat stored  Mass of  Specific  Risein 

       
in the    the   heat of the    temperature 
element  element  element  of element 
       

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T
 m  Cp 
t
T
  dx dy dz  C p  Mass  Density  Volume
t
Heat stored in  T
    Cp dx dy dz
 the element  t
Heat generated within the element
Heat generated within the element is given by
Q  q dx dy dz
Substituting equations in
   T    T    T   T
  k x   k y   k z   dx dy dz + q dx dy dz = C dx dy dz
 x  x  y  y  z  z   t
  T    T    T  T
  kx    ky   k z   q   C
x  x  y  y  z  z  t
Considering the material is isotropic. So,
Kx= ky= kz= k = Constant.
 2T 2T 2T  T

  2  2  2  k  q   C
 x y z  t
Divided by k,
 2 T  2 T  2 T q  Cp T
   
x 2 y 2 z 2 k k t
 2 T  2 T  2 T q 1 T
    ........ 1.10 
x 2 y 2 z 2 k  t
It is a general three dimensional heat conduction e1uation in Cartesian co-ordinates
k
Where,   Thermal diffusivity   m2 / s
 Cp
Thermal diffusivity is nothing but how fast heat is diffused through a material during changes of
temperature with time.
Case (i) : No heat sources
In the absence of internal heat generation, equation (1.10) reduces to
 2 T  2 T  2 T 1 T
   ............. 1.11
x 2 y2 z 2  t

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This equation is known as diffusion equation (or) Fourier‘s equation.
Case (ii): Steady state conditions
T
In steady state condition, the temperature does not change with time. So,  0 . The heat condition
t
equation (1.10) reduces to
2T 2T 2T q

    0 ......... 1.12 
x 2 y 2 z 2 k
 or 
q
2T  0
k
The equation is known as Poisson‘s equation.
In the absence of internal heat generations, equation (1.12)
Becomes:
2T 2T 2T

  0 ........... 1.13
x 2 y 2 z 2
 or 
2 P  0
This equation is known as Laplace equation.
Case (iii): One dimensional steady state heat conduction
If the temperature varies only in the x direction, the equation (1.10) reduces to
2T q
 0 .......... 1.14 
x 2 k
In the heat absence of internal heat generation, equation (1.14)
Becomes:
2T
0 .... 1.15 
x 2
Case: (iv): Two dimensional steady state heat conduction
If the temperature varies only in the x and y directions the equation (1.10) becomes:
2T 2T q
  0 ........ 1.16 
x 2 y2 k
In the absence of internal heat generation, equation, (1.16) reduces to
2T 2T
 0 .......... 1.17 
x 2 y 2
Case (v): Unsteady state, one dimensional, without internal heat generation

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In steady state, the temperature changes with time.
T
i.e.,  0. So, the general conduction equation (1.10) reduces to
t
 2 T 1 T
 ................. 1.18 
x 2  t
3. A cylinder 1m long and 5cm in diameter is placed in an atmosphere at 45℃. It is provided with 10
longitudinal straight fins of material having k = 120 W/mk. The height of 0.76 mm thick fins is 1.27 cm
from the cylinder surface. The heat transfer co-efficient between cylinder and atmosphere air is 17
W/m2K. Calculate the rate of heat transfer and the temperature at the end of fins if surface temperature
of cylinder is 150℃. (16)
Length of engine cylinder, Lcy= 1m
Diameter of the cylinder, d =-5cm = 0.05m
Atmospheric temperature, T∞= 45℃ + 273 = 318k
Number of fins = 10
Thermal conductivity of fin, k = 120 W/mK
Thickness of the fin, t = 0.76 mm = 0.76 × 10 -3m
Length (height) of the fin, Lf= 1.27 cm = 1.27 × 10-2m
Heat transfer co-efficient, h = 17 W/m2K
Cylinder surface temperature
Or
Base temperature, Tb= 150℃ + 273 = 423K
To find:
1. Rate of heat transfer, Q
2. Temperature at the end of the fin.
Solution:
Length of the fin is 1.27 cm. So, this is short fin. Assume that the fin end is insulated.

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We know that,
Heat transferred, Q1   hPkA   Tb  T  tan h  mLf  ........ 1

1/ 2
Where,
P- Parameter = 2 × Length of the cylinder
=2×1
P  2m
A- area = Length of the cylinder × thickness
= 1 × 0.76 × 10-3
A  0.76 103 m2
hP
m
kA
17  2

120  0.76  103
m  19.30m 1
(1)⟹ Q1 = (h PkA)1/2 (Tb - T∞) tan h (mLf)
  423  318  tan h 19.30  1.27  102 

1/ 2
 17 120  0.76  103 
 1.76 105  0.240
Q1  44.3W
Heat transferred per fin = 44.3 W
Heat transferred for 10 fins = 44.3 × 10
= 443 W
Q1  443W .........  2 

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Heat transfer from unfinned surface due to convection is
Q2 = hA ∆T
= h (πd Lcy- 10 × t × Lf )× (Tb-T∞)
[∵ Area of unfinned surface = Area of cylinder – Area of finn
 17   d0.05 1  10  0.76 103 1.27 102     423  318 
Q2  280.21W
So total heat transfer, Q = Q1+ Q2
= 443 + 280.21
Q  723.21W
We know that,
Temperature distribution [Short fin, end insulated]
T  T cos h  m  Lf  x 

Tb  T cos h  mLf 
We need temperature at the end of fin. So, put ∞.
T  T cos h  m  L  L f  
 
Tb  T cos h  mL f 
T  T 1
 
Tb  T cos h 19.30 1.27 10 2 
T  318 1

423  318 1.030
T  318
 0.970
105
 T  419.94 K
Result:
1. Heat transfer, Q = 723.21 W
2. Temperature at the end of the fin, T = 419.94 K
4. Explain the mechanism of heat conduction in solids and gases. (4)
Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low
temperature within medium (solid, liquid or gases) or different medium in direction physical contact.
(CONTINUATION OF ABOVE) in conduction, energy exchange take place by the kinematic motion of direct
impact of molecules pure conduction found only in solids
5. At a certain instant of time, the temperature distribution in a long cylindrical tube is, T = 800 + 1000-
5000 r2 where, T is in ℃ and r in m. The inner and outer radio of tube are respectively 30 cm and 50 cm.
The tube material has a thermal conductivity of 58 W/mK and thermal diffusivity of 0.004 m 2/hr.

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Determine the rate heat flow at inside and outside surfaces per unit length rate of heat storage per unit
length and rate of change temperature at inner and outer surfaces.
Given Data: in cylindrical tube,
T = 800 + 1000 r – 5000 r2
Inner radius, r1 = 30 cm = 30 × 10-2m
Outer radius, r2 = 50 cm = 50 × 10-3m
Thermal conductivity, K = 58 W/mK
Thermal diffusivity, ∝ = 0.004 m2/hr
0.004
 1.11106 m2 / hr
3600
To find:
(i) Rate of heat flow at inside and outside surfaces per unit length.
(ii) Rate of heat storage per unit length.
(iii) Rate of change of temperature at inner and outer surfaces.
@Solution:
(i) Rate of heat flow at inside surface per unit length,
 dT 
Qin  KA i    0.3
 dr ri
 d  800  1000r  5000r 2  
Qin  58  2  0.3  1    0.3
 dr 
ri
Qin  109.33 2000  21.86 104 W
Rate of heat flow at outside surface per unit length, Qout
 dT 
 K A 0    0.5
 dr  r0
 d  800  1000  5000r  
Qout  58  3.14     0.5
 dr  r0
 58  3.14  4000
Qout  72.84  104 W
Rate of heat storage per unit length,
Qstored  Qin  Qout

  21.86  72.84  104
Qstored  50.98 104 W
T = 800 + 1000 r – 5000 r2

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dT
 1000  1000r
dr
d2T
 10, 000
dr 2
Rate of change of temperature at inner surfaces, at r i=0.3m\
d 2 T 1 dT 1 dT
  .
dr 2 r dr  dt
1 1  dT 
1000  1000  10000  0.3  6  
0.3 1.11 10  dt ri  0.3
 dT 
   0.01851  C / s
 dt ri  0.3
Rate of change of temperature at outer surfaces
d 2 T 1 dT 1  dT  1 1  dT 
  .   10000  100  5000  2  0.5  6  
dr 2
r dr   d t r  0.5 0.5 1.11 10  dt ro  0.5
o
 dT  18, 000
    0.02 C / s
 dt ro  0.5 9 105
6. With neat shetches, explain the different fin profiles. (4)
It is possible to increase the heat transfer rate by increment the surface of heat transfer. The
surface used for increasing the transfer are called extended surfaces or fins.
Types of fins
Some common types of fin configuration are shown in fig:
(i) Uniform Straight fin

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(ii) Tappered straight fin
(iii) Splines
(iv) Annular fin

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Commonly there are three types of fin
1. Identify long fin
2. Short fin (end is insulated)
3. Short fin (end is not insulated)
7.A circulferential rectangular fins of 140 mm wide, and 5mm thick are fitted on a 200mm diameter tube.
The fin base temperature is 170℃ and the ambient temperature is 25℃. Estimate fin efficiency and heat
loss per fin.
Take Thermal cionductivity, k = 200 W/mK.
Heat transfer co-efficient, h = 140 W/m2K.
Given:
Wide, L = 140 mm = 0.140 m
Thickness, t = 5 mm = 0.005 m
Diameter, d= 200 mm ⟹ r = 100 mm = 0.1 m
Fin base temperature, Tb = 170℃ + 273 = 443 K
Ambient temperature, T∞= 25℃ + 273 – 298 K
Thermal conductivity, k = 220 W/mK.
Heat transfer co-efficient, h = 140 W/m2K.
To find:
1. fin efficiency, η

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2. Heat loss Q
Solution:
A rectangular fin is long and wide. So, heat loss is calculated by using fin efficiency curves.
[From HMT data book page no.50 (Sixth edition)]
Correlated length, Lc = L + t/2
0.005
 0.140 
2
Lc  0.1425 m
R2c = r1 + Lc
 0.100  0.1425
r2c  0.2425 m
As= 2π [r2c2 – r12]
As = 2 π [(0.2425)2 – (0.100)2]
As  0.30650 m2
Am = t[r2c – r1]
= 0.005 [(0.2425) –(0.100)]
Am  7.125 104 m2
From the graph, we know that,
[HMT data book page no.50]

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0.5
 h 
X axis  L C
1.5
 
 kA m 
0.5
1.5  140 
  0.1425   4 
 230  7.125 10 
X axis  1.60
r2C 0.2425
Curve ⟶   2.425
r1 0.1
Xaxis value is 1.60
Curve value is 2.425
By using these values, we can find fin efficiency, η from graph
fin efficiency,   28%
Heat transfer, Q = η As h [Tb - T∞)
[From HMT data book page no]
 Q  0.28  0.30650 140   443  298

Q  1742.99 W
Result:
1. Fin efficiency, η = 28%
2. Heat loss, Q = 1742.99
8.A furnace wall is mode up of three layers of thickness 25 cm. 10 cm and 15 cm with thermal
conductivities of 1.65 K and 9.2 w/mK respectively. The inside is expressed to gases at 1250℃m with a
convection coefficient of 25 W/m2℃ and the inside surface is at 1100℃, the outside surface is expressed to
air at 25℃ with convection coefficient of 12 W/m2K. Determine (i) the unknown thermal conductivity, (ii)
The overall heat transfer coefficient, (iiii) All the surface temperatures. (16)
Given:
Thickness, L1 = 25 cm = 0.25 m
L2 = 10 cm = 0.10 m

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L3 = 15 cm = 0.15 m
Thermal conductivity, k1 = 1.65 W/mK
K2 = k
K3 = 9.2 W/mK
Inside gas temperature, T a = 1250℃ + 273
= 1523 K
Outside air temperature, Tb = 25℃ + 273
= 298 K
Inside heat transfer co-efficient, ha = 25 W/m2K
Outside heat transfer co-efficient, hb = 12 W/m2K
Inner surface temperature, T 1= 1100℃ + 273
= 1373K
To find: (i) Unknown thermal conductivity, k2
(ii) Overall heat transfer co-efficient, U
(iii) Surface temperatures T2, T3, T4
@Solution;
We know that,
Heat transfer Q = ha A (Ta – Tt)
Q
  h a  Ta  T1 
A
 25 1523  1373
Q
 3750 W / m 2
A
We know that,
Toverall
Heat flow, Q 
R
[From HMT data book, page no. 43 & 44]
Where, ∆T = Ta - Tb

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1 L L L 1
R  1  2  3 
h a A k1 A k 2 A k 3 A h b A
Ta  Tb
 Q
1 L1 L L 1
  2  3 
Q Ta  Tb
 
A 1 L1 L 2 L3 1
   
ha k 2 k 2 k3 h b
1523  298
3750 
1 0.25 0.10 0.15 1
   
25 1.65 k 2 9.2 12
1225
 3750 
0.10
0.2911 
k2
 0.1 
 3750  0.2911    1225
 k2 
0.1
 0.2911   0.3266
k2
 Thermal conductivity, k 2  2.816 W / mK
We know that,
1
Overall heat transfer co-efficient, U 
R total
1 L1 L 2 L3 1
 R total      Take A  1m 2 
h a k1 k 2 k 3 h b
1 0.25 0.1 0.15 1
    
25 1.65 2.816 9.2 12
R load  0.3267 W / m 2
1
U
R total
1

0.3267
U  3.06W / m 2 K
We know that,
Ta  Tb Ta  T1 T1  T2 T2  T3 T3  T4 T4  Tb
Q     
R Ra R1 R2 R3 Rb

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T1  T2 L
1  Q where R 1  1
R1 k1 A
T1  T2
 Q
L1
k1 A
Q T1  T2
 
A L1
k1
13730  T2
3750 
0.25
1.65
 T2  804.8K
T2  T3 L
 2  Q where R 2  2
R2 k2A
T2  T3

L2
k2A
Q T2  T3

A L2
k2
804.8  T3
3750 
0.10
2.816
 T3  671.45K
T3  T4 L3
 3  Q where R 3 
R3 k3A
T3  T4

L3
k3A
Q T3  T4

A L3
k3
671.45  T4
3750 
0.15
9.2
 T4  610.30 K
9. Pin fans are provided to increase the heat transfer rate from a hot surface. Which of the following
arrangements will give higher heat transfer rate?(1) 6-fins of 10 cm length, (2) 12-fins of 5 cm length. Take k of
fin material = 200 W/mK and h = 20 W/m ℃ cross sectional area of the fin = 2 cm2; Perimeter of fin = 4cm ; Fin base
2
temperature = 230℃;Surrounding air temperature = 30℃
Given Data:
Thermal conductivity of fin material, k = 200 W/mK
Heat transfer coefficient, h = 20 W/m2℃ = 20 W/m2K

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Cross sectional area, A = 2 cm2 = 2 × 10-4m
Perimeter of fin, P = 0.04 m
Fin base temperature, Tb = 230℃ + 273 = 503K
Air temperature, T∞ = 30℃ + 273 = 303 K
To find: (i) Q1, Heat transfer rate (6 fins of 10 cm length)
(ii) Q2, Heat transfer rate (12 fins of 5 cm length)
@Solution: We know that,
hP 20  0.04
m 
kA 200  2 104
m  4.4721
We know that,
Heat transfer rate, Q = n[(h P k A)1/2 (Tk - T∞) tanh (mL)]
(i) n = 6; L = 0.1m
m L = 4.4721 × 01 [From HMT data book, page no. 49]
m L  0.44721
Q1  n  h P k A   Tb  T  tanh  0.447  
1/ 2
 
 6  20  0.04  200  2  104    503  303  tanh  0.447  
1/ 2
 
Q1  90W
(ii) Number of fins, n = 12, Length of fins L = 5 cm = 0.05 m
m L = 4.4721 × 0.05
m L  0.2235
Q2 = n [(h P k A)1/2 (Tb - T∞) tanh (0.2235)]
 12  20  0.04  200  2  104    503  300   tanh  0.2235

1/ 2
 
Q2  94.34 W
Since Q2 > Q1. The second case is better.
10. A steel ball 50 mm in diameter and at 900℃ is placed in still air 30℃. Calculate the initial rate of cooling
of ball in ℃/min. Take 𝛒 = 7800kg/m3; C = 2kJ/kg ℃; h = 30 W/m2 ℃. Neglect the internal resistance of the
ball. (8)
Given Data:
Steel ball diameter, D = 50 mm = 0.05 m

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Initial temperature, T0 = 900℃ + 273 = 1173 K
Air temperature, T∞ = 30℃ + 273 = 303 K
Density, ρ = 7800 kg/m3
Specific heat, Cp = 2 kJ/kg℃ = 2 × 103 J/kg℃
Heat transfer co-efficient, h = 30 W/m2℃
Time, t = 60 seconds
To find: Initial rate of cooling of ball in ℃/min.
@Solution:
T  T  h As t 
 exp  
T0  T   V C p 
 From HMT data book, Page no.5
h As t 30  4R 2  t
 
 V Cp  4 R 3  C
p
3
30  4  0.025   60
2

4
7800    0.025   2  103
3
3
 0.01385
T  T
 exp  0.01385
T0  T
T  1161K or 888 C
Rate of cooling = T0 – T = 12℃ /min
11.Derive the general 3- dimensional heat conduction equation cylindrical co-ordinates. Assume the material
as homogeneous isotropic continues.
General Heat Conduction Equation in cylindrical Co-ordinates
The general heat conduction equation in Cartesian co-ordinates derived in the previous section is used
for solid with rectangular boundaries like square, cubes, slabs, etc. But the Cartesian co-ordinates system is not
applicable for the solid, like cylinders, cones, spheres etc. For cylindrical solids a cylindrical co-ordinate system
is used.
Consider a small cylindrical element of sides dr, d∅ and dz shown in fig. 1.2

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Fig. 1.2
The volume of element dv = r d∅ dr dz
Let us assume that thermal conductivity k, Specific heat Cp and density ρ are constant.
The energy balance of this cylindrical element is obtained from first law of thermodynamics.
Set heat 
conducted int o  Heat Heat 
     
    stored 
 element from all   generated within the    
 the co  ordinate  element  in the 
    
element 
directions
 

Net heat conducted into element from all the co-ordinate directions
Heat entering in the element through (r, ∅) plane in time dθ
T
Qz  k  r d dr  d
z
Heat leaving from the element through (r, ∅) plane in time d

Qz  dz  Qz   Qz  dz
z
Net heat conducted into the element through (r, ∅) plane time dθ.

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 Q z  Q z  dz

  Qz  dz
z
   T  
  k  rd.dr  .   d  dz
z   z  
 2T 
 k  2   dr.rd.dz  d
 z 
 2T 
Net heat conducted through  r,   Plane  k  2   dr.rd.dz  d ............ 1.20 
 z 
Heat entering in the element through (∅, z) plane in time dθ.
T
Qr  k  r d dz  d
r
Heat leaving from the element through (∅, z) plane in time dθ.

Qr  dr  Qr   Qr  dr
r
Net heat conducted into teh element through (∅, z) plane in time dθ.
 Q r  Q r  dr

  Qr  dr
r
dr
   T  
  k  rd, dz  ,   d 
r   r  
  T 
 k  dr d, dz  , r. d =
r  r 
  2 T 1 T 
 k  dr.rd.dz   2   d
 r r r 
  2 T 1 T 
Net heat conducted through  , z  Plane  k  dr.rd.dz   2   d .......... 1.21
 r r r 
Heat entering in the element through (z, r) plane in time dθ.
T
Q  k  dr.dz  d
r
Heat leaving from the element through (z, r) plane in time dθ.

Q d  Q 
r
 Q  rd
Net heat conducted into the element through (z, r) plane in time dθ.

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
Q  Q d 
r 2 
 Q  rd
  T 
 k  dr dz  . d rd
r 2   r 
  1 T 
K  dr d dz  d
  r  
 1 2T 
 k  2 2   dr rd dz  d
 r  
 1 2T 
Net heat conducted through  z, r  plane  k  2 2   dr rd dz  d ............. 1.22 
 r  
Net heat conducted into element from all the co-ordinate directions
2T
k  dr rd dz  d
z 2

  2 T 1 T 
k  dr rd dz   2   d
 r r r 
 1 2T 
k  2 2   dr rd dz  d
 r  
[Adding equation 1.20, 1.21 and 1.22]
  2 T  2 T 1 T 1  2 T 
 k  dr rd dz  d  2  2   
 z r r r r 2 2 
  2 T 1 T 1  2 T  2 T 
 k  dr rd dz  d  2    
 r r r r 2 2 z 2 
Net heat conducted int oelement from all the co  ordiante directions
  2 T 1 T 1  2 T  2 T  ......... 1.23
 k  dr rd dz  d  2     
 r r r r 2 2 z 2 
Heat generated within the element
Total heat generated within the element is given by
Q  q  dr rd dz  d ........... 1.24 
Heat stored in the element
The increase in internal energy element is equal to the net heat stored in the element.
Increase in internal energy,
=Net heat stored in the element

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T
   dr rd dz  Cp  d ......... 1.25 

Substituting equation (1.23), (1.24) and (1.25) in (1.19)
  2 T 1 T 1  2 T  2 T 
1.19   k  dr rd dz  d       q  dr rd dz  d
 r
2
r r r 2 2 z 2 
T
   dr rd dz  Cp  d

Dividing by (dr rdϕ dz) dθ
  2 T 1 T 1  2 T  2 T  T
k  2   2 2  2   q  .Cp
  r r r r   z  
  2 T 1 T 1  2 T  2 T  q Cp T
  2      ........ 1.26 
 r r r r 2 2 z 2  k k 
It is a general three dimensional heat conduction equation in cylindrical co-ordinates,
 2 T 1 T 1  2 T  2 T q 1 T
     
r 2 r r r 2 2 z 2 k  
 k 
   
  Cp 
If the flow is steady, one dimensional and no heat generatin, equation (1.26) becomes:
 2 T 1 T
 0 ....... 1.27 
z 2 r r
 OR 
1 d  dT 
 r. 0 .......... 1.28 
r dr  dr 
12. The wall of a cold room is composed of three layer. The outer layer is brick 30cm thick. The middle layer
is cork 20cm thick, the inside layer is cement 15 cm thick. The temperatures of the outside air is 25℃ and
on the inside air is 20℃. The film co-efficient 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 cor5k = 0.05 W/mK
k for cement = 0.28 W/mK
Given:
Thickness of brick, L3 = 30 cm = 0.3 m
Thickness of cork, L2 = 20 cm = 0.2 m
Thickness of cement, L1 = 15cm = 0.15m

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Inside air temperature, Ta = -20℃ + 273 = 253 K
Outside air temperature, Tb = 25℃ + 273. = 298 K
Film co-efficient for outside, hb = 55.4 W/m2K
Kbrik = k3 = 2.5 W/mK
Kcork = k2 = 0.05 W/mK
Kcement = k1 = 0.28 W/mK
To find:
Heat flow rate (Q/A)
Solution:
Heat flow through composite wall is given by
Toverall
Q  From Equation no.1.42or HMT Data book Page no.43and 44
R
Where
∆T = Ta- Tb
1 L L L 1
R  1  2  3 
Q
Ta  Tb 
1 L1 L L 1
  2  3 
Q/A 
Ta  Tb 
1 L1 L 2 L3 1
   
h a k1 k 2 k 3 h b
253  298
Q/A 
1 0.15 0.2 0.3 1
   
17 0.28 0.05 2.5 55.4
Q / A  9.5W / m 2

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The negative sign indicates that the heat flows from the outside into the cold room.
Result:
Heat flow rate, Q/A = -9.5 W/m2.
13. 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) Overall heat transfer coefficient

b) Overall thermal resistance
c) The rate of heat transfer
d) 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
Thickness of Plaster L4 = 1.2 cm = 0.012 m
Interior heat transfer coefficient ha = 5.6 W/m2K
Exterior heat transfer co-efficient hb = 11 W/m2K
Interior room temperature T a = 22C + 273 = 295 K
Outside air temperature Tb = -5C + 273 = 268 K.
Solution:
Heat flow through composite wall is given by
Toverall
Q [From equation (13)] (or) [HMT Data book page No. 34]
R

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Where,  T = Ta– Tb
1 L L L L 1
R  1  2  3  4 
ha A K1 A K 2 A K 3 A K 4 A hb A
Ta  Tb
Q
1 L L L L 1
 1  2  3  4 
ha A K1 A K 2 A K 3 A K 4 A hb A
295  268
Q/ A 
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 (T a – Tb) [From equation (14)]
Where U – overall heat transfer co-efficient
Q
U 
A  (Ta  Tb )
34.56
U 
295  268
Overall heat transfer co - efficient U = 1.28 W/m2 K
We know
Overall Thermal resistance (R)
1 L L L L 1
R  1  2  3  4 
ha A K1 A K 2 A K3 A K 4 A hb A
For unit Area
1 L1 L2 L3 L4 1
R     
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

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Ta  T1 T1  T2 T2  T3 T3  T4 T4  T5 T5  Tb
Q     
Ra R1 R2 R3 R4 Rb
Ta  T1
Q
Ra
295-T1  1 
Q=  R a  
1/ ha A  ha A 
295  T1
Q/ A
1/ ha
295  T1
 34.56 
1/ 5.6
 T1  288.8 K
T1  T2
Q
R1
288.8  T2  L1 
Q  R1  
L1  k1 A 
K1 A
288.8  T2
Q/ A
L1
K1
288.8  T2
 34.56 
0.20
0.66
 T2  278.3 K
T2  T3
Q =
R2
278.3  T3  L2 
Q  R 2  
L2  K2 A 
K2 A
278.3  T3
Q/ A
L2
K2
278.3  T3
 34.56 
0.03
0.6
 T3  276.5 K
Temperature between Mortar and limestone (T 3 is 276.5 K)
14.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 co-efficient for outer side is 8W/m2K. Calculate the heat
lost per square meter area of the furnace wall and also find outside surface temperature.

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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]
Toverall
Heat flow Q  , Where
R
 T = Ta– Tb
1 L L L 1
R  1  2  3 
ha A K1 A K 2 A K 3 A hb A
Ta  Tb
 Q=
1 L1 L L 1
  2  3 
[The term L3 is not given so neglect that term]

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Ta  Tb
 Q=
1 L L L 1
 1  2  3 
The term L3 is not given so neglect that term]
Ta  Tb
Q=
1 L L 1
 1  2 
ha A K1 A K 2 A hb A
923  300
Q/ A 
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
Ta  Tb Ta  T1 T1  T2 T2  T3 T3  Tb
Q     ......( A)
R Ra R1 R2 Rb
T3  Tb
( A)  Q 
Rb
where
1
Rb 
hb A
T3  Tb
Q
1
hb A
T3  Tb
 Q/A =
1
hb
T3  300
 2907.79 
1
8
T3  663.473 K
15. 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

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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 :
Toverall
Heat flow Q  [From equation No.(19) or HMT data book Page No.35]
R
Where  T = T a– T b
1  1 1 r  1 r  1 r  1 
R   In  2   In  3   In  4   
2 L  h a r1 K1  r1  K 2  r2  K 3  r3  hb r4 
Ta  Tb
Q =
1  1 1 r  1 r  1 r  1 
  In  2   In  3   In  4   
2 L  h a r1 K1  r1  K 2  r2  K 3  r3  hb r4 
[The terms K3 and r4 are not given, so neglect that terms]
Ta  Tb
Q =
1  1 1 r  1 r  1 
  In  2   In  3   
2 L  h a r1 K1  r1  K 2  r2  hb r3 
589 - 303
Q =
1  1 1  0.0381  1  0.0631 1 
 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.
16.Derive an expression of Critical Radius of Insulation For A Cylinder.
Consider a cylinder having thermal conductivity K. Let r 1 and r0 inner and outer radii of insulation.
Ti  T
Heat transfer Q  [From equation No.(3)]
r 
In  0 
 r1 
2 KL
Considering h be the outside heat transfer co-efficient.

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Ti  T
Q =
r 
In  0 
 r1   1
2 KL A 0h
Here A 0  2 r0L
Ti  T
Q
r 
In  0 
 r1   1
2 KL 2 r0Lh
To find the critical radius of insulation, differentiate Q with respect to r 0 and equate it to zero.
 1 1 
0  (Ti  T )   2

dQ
  2 KLr0 2 hLr0 
dr0 1 r  1
In  0  
2 KL  r1  2 hLr0
since (Ti  T )  0
1 1
  0
2 KLr0 2 hLr0 2
K
 r0   rc
h
17. 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 :
K
1. Critical radius rc  [From equation No.(21)]
h
0.11
rc   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

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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 
 
 
From HMT data book Page No.35

2 L (Ta  Tb )
=
  0.005  
 In  0.003  
   1

 0.11 25  0.005 
 
2 L (Ta  Tb )
Q1 =
12.64
Heat flow through an insulated wire when critical radius is used is given by
Ta  Tb
Q2  r2  rc 
  rc  
 In   
1   r1  1 

2 L  K1 hbrc 
 
 
2 L (Ta  Tb )
=
 4.4  10 3 
In  
 0.003   1
0.11 25  4.4  10 3
2 L (Ta  Tb )
Q2 =
12.572
 Percentage of increase in heat flow by using
Q2  Q1
Critical radius =  100
Q1
1 1
  100
 12.57 12.64
1
12.64
 0.55%

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Unit –2
CONVECTION
Part – A
1. 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 temperature.
2. Define Reynolds number (Re).
Reynolds number is defined as the ratio of inertia force viscous force
Inertia force
Re 
Viscouse force
3. Define Nusselt number (Nu)
It is defined as the ratio of the heat flow buy 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 meter.
Qconv
Nusselt number (Nu) 
Qcond
4. Define Grashof number (Gr)
It is defined as the ratio of production of inertia force and Buoyancy force to the square of viscous force
Inertia force  Buoyancy force

Gr 
 Viscous force 
2
5. What is meant by non-Newtonian fluids?
The fluid which obey the Newton‘s Law of viscosity are called Newtonian and those which do not
obey are called non-Newtonian thinks.
6. What is meant by Stanton number(St)
Stanton number is the ratio of Nusselt number to the product of Reynolds number and parandtl
number
Nu
St 
Re Pr
7. 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
8. Define boundary layer thickness
The thickness of boundary layer has been defined as the distance from the surface at which the velocity
or temperature reaches 99% of the external velocity or temperature.

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9. What is the from of equation used to calculate heat transfer for flow through cylindrical pipes?
Nu  0.023  Re   Pr 
0.8 n
n  0.4 for heating of fluids

n  0.3for cooling of fluids
10. What is dimensional analysis?
Dimensional analysis is mathematical method which makes us the study of dimension for solving
several engineering problems. This method can be applied to all types of fluid resistance, heat flow problems in
fluid mechanism and thermodynamics.
11. What are all advantage of dimensional analysis?
1. It expressed the functional relationship between the variable in dimensional terms

2. It enables getting up a theoretical solution in a simplified dimensionless form
3. The result of one series of tests can be applied to a large number of other similar problems with the
help of dimensional analysis.
12. What is hydrodynamic boundary layer?
In hydrodynamic boundary layer, velocity of the fluid less than 99% of free steam velocity.
13. What is thermal boundary layer?
In thermal boundary layer, temperature of the fluid is less than 99% of the free stream temperature.
14. What are the dimensionless parameters used in forced convection?
1. Reynolds number (Re)
2. Nusselt number (Nu)
3. Prandtl number (Pr)
15.Indicate the concept or significance of boundary layer.
1. A thin region teh body called the boundary layer where the velocity and the temperature gradients
are large.
2. The region outside the boundary layer where the velocity and the temperature gradients are very
nearly equal to their free stream values.
16. Define displacement thickness.
The displacement thickness is the distance, measured perpendicular to the boundary, by which the free
stream is displaced an account of formation of the boundary layer.
17. Define momentum thickness.
The momentum thickness is defined as the distance through which the total loss of momentum per
second be equal to if it were passing a stationary plate.
18. Define energy thickness.

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It is defined as the distance, measured perpendicular to the boundary of the solid, by which the
boundary should be displaced to compensate for the reduction in kinetic energy of the following fluid on
account of boundary layer.
19. Define Prandtl number (Pr).
Prandtl number is the ratio of the momentum diffusivity of the thermal diffusivity.
Momentum diffusivity
Pr 
Thermal diffusivity
20. Define Stanton number (St).
Stanton number is the ratio of nusselt number to the product of Reynolds number and prandtl number.
Nu
St 
Re Pr
21. What is meant by 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.
1. A long 10 cm diameter steam pipe whose external surface temperature is 110℃ passes through some open
area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit length
when the air is 1 atm and 10℃ and the wind is blowing across the pipe at a velocity of 8 m/s.(8)
Given data:
Diameter of steam pipe, d = 10 cm = 0.1 m
Surface temperature, Tw = 110℃
(Air) fluid temperature, T ∞ = 10℃
Velocity, u = 8 m/s
To find: rate of heat loss from the pipe per unit length
Q  h A  Ts  T 

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Solution:
Film temperature,
Tw  T 110  10 120
Tf     60 C
2 2 2
From HMT data book, page No.34 (Seventh edition) Properties of air at 60℃.
Density, ρ = 1.060 kg/m3
Kinematic viscosity, v = 18.97 × 10-6 m2/s
Prandtl number, Pr = 0.696
Thermal conductivity, K = 0.02896 W/mK
Reynolds number,
uD 8  0.1
Re    0.421105
v 18.96 106
 42194  5 105
Nusselt number, Nu = C (Re) m (Pr)0.333
From HMT data book, page No. 116 (Seventh edition)
Re value is 42,194, corresponding C and values are 0.0266 and 0.805respectively.
Nu= 0.0266 (42194)0.805 (0.696)0.333
Nu = 124.67
We know,
hD
Nu 
K
Nu.K 124.67  0.02896
h 
D 0.1
Heat transfer co-efficient, h = 35.91 W/m2K
(∴ A = π DL)
Heat transfer, Q = h A (Ts - T∞)
= 35.91 × (π× 0.1 × 1) (110 – 10)
Q = 1128.14 W
Result:
Rate of heat loss from the pipe per unit length
Q = 1128.14 W
2.Air at 0℃ flow over a flat plate at a speed of 90 m/s and heated to 100℃. The plate is 60 cm long and
75cm wide. Assuming the transition of boundary layer take place at Re = 5 × 10 5. Calculate the following:

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1. Average friction transfer co efficient.
2. Average heat transfer co-efficient
3. Rate of energy dissipation.
Given: Fluid temperature, T∞ = 0℃
Speed, U = 90m/s
Surface temperature, Tw = 100℃
Length, L = 60 cm = 0.60 m
Wide , W = 75 cm = 0.75 m
To Find:
1. Average friction co-efficient
2. Average heat transfer co-efficient
3. Rate of energy dissipation.
Solution:
Tw  T
Tf 
2
100  0
Film temperature, 
2
Tf  50 C
Properties of air at 50℃:
[From HMT data book, page no. 33 (sixth edition)
Ρ = 1.093 kg/m3
V = 17.95 × 10-6m2/s
Pr = 0.698
K = 0.02826 W/mK
We know,
UL
Re 
v
90  0.60
Reynolds Number, 
17.95  106
Re  3.0  106  5  105
Since Re > 5 × 105, flow is turbulent.
[Note: Transition occurs means flow is combination of laminar and turbulent flow. i.e. the flow is said to be
laminar upto Re value is 5 × 105. After that flow is turbulent]

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For flat plate, Laminar- turbulent flow
[From HMT data book, page No. 114 (sixth edition)]
Average friction 
 CfL  0.074  Re   1742  Re 
0.2 1.0
Co efficient 
0.2 1.0
 CfL  0.074 3.0  106   1742 3.0  106 
CfL  3.16  103
Average friction  3
 CfL  3.16  10
co  efficient 
Average Nusselt 
 Nu   Pr  0.037  Re 0.8  871
0.333
number   
 From HMT data book, page no.114  sixth edition  
0.037  3  106   871

0.333 
  Pr 
0.8
 
  0.698 
0.333 0.037  3  106 0.8  87 
 
Nu  4215
We know,
hL
Nu 
k
Average Nusselt number, h  198.5W / m 2 K
Average heat transfer 
 h  198.5 W / m K
2
co  efficient 
Q  h A  Tw  T 
 h  L  W  Tw  T 
Rate of energy dissipation,
 198.5  0.60  0.75 100  0 
Q  8932.5 K
Result:
1. CfL = 3.16 × 10-3
2. h = 198.5 W/m2K
3.Q = 8932.5 W
3.A 6 m long section of an 8 cm diameter horizontal hot water pipe passes through a large room whose
temperature is 20℃. If the outer surface temperature and emissivity of the pipe are 70℃ and 0.8
respectively, pipe by
(1) Natural Convection
(2) Radiation. (10)
Given data:

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In horizontal hot water pipe (cylinder and internal flow)
Length of hot water pipe = 6m
Diameter of hot water pipe = 8 cm = 0.08 m
Emissivity of the pipe, ε = 0.8
Outer surface temperature, T w = 70℃
Hot (fluid) water temperature, T∞ = 20℃
To find:
Rate of heat loss from the pipe,
(1) Natural convection, Qconv
(2) Radiation (Qrad)
Solution:
Tw  T 70  20
Film temperature, Tf    45 C
2 2
10-5 < GrD. Pr < 1012
2
  
0.167

   
  1.868  106  
 
 0.60  0.387   
0.5625 0.296
    0.559    
  1      
    0.7241    
Nu D  22.89
We know,
hD
Nu 
K
K.Nu 0.02699  22.89
h   7.7225 W / m 2 K
D 0.08
Rate of heat loss from the pipe
Qconv  h As  Ts  T 
 7.7225  1.508   70  20 
Qconv  582.46 W
Area of the surface,
As = π DL
= π × 0.08 × 6
As = 1.508 m2

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Qrad = ε As σ (Ts4 – T4∞)
= 0.8 × 1.508 × 5.67 ×10-8 × (3434 – 2934)
Qrad = 442.65 W
Result:
Rate of heat loss from the pipe by,
(i) Natural convection, Qconv= 582.46 W
(ii) Radiation, Qrad= 442.65 W
4. (i) Explain the concept of hydrodynamic and thermal boundary layers.
(1) Velocity distribution in hydrodynamic boundary layer.
(2) Temperature distribution in thermal boundary layer.
(3) Variation of local heat transfer co-efficient along the flow. (8)
5. Discuss briefly the development of velocity boundary layer for flow through a pipe.
The velocity at any cross section of a pipe varies from zero at wall to a maximum at the centre, and that
there is no well defined free stream. There is a need to define and work in terms of a mean velocity, u m. It is

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defined as that velocity which is multiplied by the fluid density and the cross-sectional area of the tube gives the
rate of mass flow through the tube.
Thus,

m   u m .D2 \
4
The velocity distribution for fully developed, stead laminar flow can be determined by considering the force
equilibrium of a cylindrical fluid element in fig. 1.
Fig. 1. Flow regions in a circular tube
The various forces are,
(i) Shear on the cylindrical surface.
(ii) Normal force due to pressure on the ends.
Since for a fully developed flow,
 u 
Vr  0 and    0
 x 
The axial velocity u,is only a function of r.

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Force balance on a differential element in laminar development flow and the net momentum
flow is zero everywhere.
6. Water at 60℃ and a velocity of 2 cm/s flows over a 5 m long flat plate which is maintained at a
temperature of 20℃. Determine the total drag force and the rate of heat transfer per unit width of the
entire plate.
Given Data: T∞ = 60℃ (fluid temperature)
Velocity, u = 2 cm/s = 0.02 cm/s
X= 5m, Ts = 20℃, L = 5m
To find:
(i) Drag force, FD
(ii) Rate of heat transfer per unit width of the entire plate.
@ Solution: We know that,
Tw  T 60  20
Film temperature, Tf    40 C
2 2
Properties of water @ 40℃
[From HMT data book, page no. 33, 6th edition]
Ρ = 995 kg/m3
V = 0.657 × 10-6m2/s
Pr = 4.34
K = 0.628 W/mK
uL 0.02  5

v 0.657  106
0.1 106
Reynolds‘s number, Re1= 
0.657
Re L  1.522  105  5  105
Since, Re < 5 × 105, ∴ The flow is laminar.
CfL  1.328  Re L 0.5

 1.328  1.522  105 
0.5
Average friction co-efficient,
1.328
  3.4  103
390.128
Drag force, FD = Area × Average shear stress

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 u2
 1 5  CfL
2
995  0.02 
2
 5  3.4  103
2
1.99
 3.4  103 
2
 0.995  3.43  103
FD  3.41 103 N
Local heat transfer co-efficient, hx:
Local Nusselt number, Nux = 0.332 (Re)0.5 (Pr)0.333
 0.332 1.522 105   4.34 

0.5 0.333
 0.332  390.128 1.6303

Nu x  211.168
hxL 211.16  0.628
Nu x    hx
k 5
h x  26.52 W / m 2 K
h  2  h x  2  26.52  53.04 W / m 2 K
(ii) Rate of heat transfer
Q  h A  Ts  T 
 53.04  5   333  293
Q  10608 W
7. Considering a heated vertical plate in a quiescent fluid, draw the velocity and temperature profiles.

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8. A horizontal pipe of 6m length and 8 cm diameter passes through a large room in which the air and
walls are at 18℃. The pipe outer surface is at 70℃. Find the rate of heat loss from the pipe by natural
convection.
Given Data:
Length of the pipe, L = 6m
Diameter of the pipe, d = 8 cm = 0.08 m
Surface temperature, T s = 70℃ + 273 = 343 K
Fluid temperature, T∞ = 18℃ + 273 = 291 K
To find: Q/L
@ Solution:
Ts  T
Tf 
Film temperature, 2
70  18
  44 C  or  317K
2
The properties of air @ 44℃ ≈ 45℃
Ρ = 1.11 kg/m3
V = 17.45 × 10-6 m2/s

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∝ = 25.014 × 10-6 m2/s
K = 0.02791 W/mK
Pr = 0.6985
1 1
   0.00315 K 1
Tf 317
g   T d3
Gr 
v2
9.81 0.00315   343  291 0.08 
3

17.455 10  6 2
8.229  104 1012

  2.7  106
304.677
Gr.Pr  2.7  106  0.6985
GrD Pr  1.88 106
For horizontal cylindrical,
2
  
0.167

   
   
 Gr .Pr 
Nu D  0.6  0.387  D
 
0.5625 0.296
    0.559    
  1      
    Pr    
105  GrD Pr  1012
2
  
0.167

   
  1.88  106  
 
 0.6  0.387  0.296  
    0.559  0.5625
  
  1      
    0.6985    
2
 1.88  106 
0.167

 0.6  0.387   
  1.2058  

 0.6  0.387 1559130.86  
0.167 2
 0.6  0.387 10.819 

2
 0.6  4.186953
2
hd
Nu D  22.916  Nu D 
k
22.916  0.02791
h  7.99
0.08
Q  h A  Ts  T 
 7.99   0.08  6   343  291
Q  626.529 W

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9.Castor oil at 30℃ flows over a flat plate at a velocity of 1.5 m/s. The length of the plate is 4m. The plate
is heated uniformly and maintained at 90℃. Calculate the following.
1. Hydrodynamic boundary layer thickness,
2. Thermal boundary layer thickness,
3. Total track force per unit width on one side of the plate,
4. Heat transfer rate.
90  30
At the mean film temperature Tf   60 C.
2
Properties are taken as follows:
Ρ = 956.8 kg/m3; v = 0.65 × 104 m2/s;
K = 0.213 W/mK; ∝ = 7.2 × 10-8 m2/s.
Given: Fluid temperature, T∞= 30℃
Velocity, U = 1.5 m/s
Length, L = 4m
Plate surface temperature, Tw = 90℃
At Tf = 60℃, ρ = 956.8 kg/m2
k = 0.213 W/mK
v = 0.65 × 10-4 m2/s
∝ = 7.2× 10-8 m2/s
To find:
1. Hydrodynamic boundary layer thickness,
2. Thermal boundary layer thickness,
3. Total drag force per unit width on one side of the place
4. Heat transfer rate.
Solution: We know that,
UL
Re 
v
1.5  4
Reynold‘s Number, 
0.65  104
Re  9.23  104  5  105
Since Re < 5 × 105, flow is laminar.
For flat plate, laminar flow:

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[Refer HMT data book, page No. 112 (Sixth edition)
1. Hydrodynamic boundary layer thickness:
hx  5  x   Re  h  0.5
 5  4   9.23  104 
0.5
 x  L  4m
hx  0.065 m
2. Thermal boundary layer thickness:
Tx   hx   Pr 
0.333
 0.065   902.77 
0.333
 v 0.65 104 
 Pr   8
 902.77 
  7.2  10 
Tx  6.74  103 m
3. Total drag force on one side of the plate:
Average skin friction co-efficient,
CfL  1.328  Re 
0.5
CfL  1.328   9.23 104 

0.5
CfL  4.37 103

CfL 
 U2
2

We know  4.37  103 
956.8  1.5 
2
2
   4.70N / m 2
Average shear stress τ = 4.70N/m2
Drag force, FD = Area × Average shear stress
= (L × W) × 4.70
= (4 × 1) × 4.70 [∵ W = 1m]
Drag force, FD  18.8N
4. Heat transfer rate:
We know that,
Local Nusselt Number
Nux = 0.332 × (Re)0.5 (Pr)0.333

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=0.332 × (9.23 × 104)0.5 × (902.77)0.333
Nu x  972.6
We know,
hxL
Nu x 
k
hx  4
Local Nusselt Number
972.6 
0.213
 h x  51.7W / m 2 K
local heat transfer co  efficient, h x  51.7 W / m 2 K
Average heat transfer co-efficient
h  2 hx
h  2  51.7
h  103.58 W / m 2 K
Heat transfer, Q=h A (Tw - T∞)
= h × L × W (Tw-T∞)
=103.58 × 4 × 1 (90-30)
` Q  24.859 kW
Result:
1. δhx = 0.065 m,
2. δrx = 6.74 × 10-3m,
3. Drag force, FD = 18.8 N,
4. Heat transfer, Q = 24.859 kW.
Example 2: A vertical plate of 0.7m wide and 1.2 m length maintained at a temperature of 90℃ in a
room at 30℃, calculate the convective heat loss.
Given: Wide, W = 0.7m
Height (or) Length, L = 1.2m
Wall temperature, Tw = 90℃
Room temperature, T∞= 30℃
To find: Convective heat loss (Q),
Solution: Velocity (U) is not given, So, this is natural convection type problem.
We know that,

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Tw  T
Film temperature, Tf 
2
90  30

2
Tf  60 C
[From HMT data book, page No. 33 (Sixth Edition)]
ρ = 1.060 kg/m3
v =18.97 × 10-6 m2/s
Pr = 0.696
k = 0.02896 W/mK
we know,
Co  efficient of  1
 
thermal exp ansion  Tf in K
1
   3  103 K 1
60  273
  3  103 K 1
g  L3  T
Grashof Number, Gr 
v2
[From HMT data book, page No. 134 (Sixth Edition)]
9.81 3  103  1.2    90  30 

3

18.97 10 6 2
Gr  8.4  109
Gr  Pr  8.4  109  0.696
Gr Pr  5.9 109
Since Gr Pr > 109, flow is turbulent.
For turbulent flow,
Nusselt Number, Nu = 0.10 (Gr Pr)0.333
[From HMT data book, Page no.135 (Sixth edition)]
0.333
Nu  0.10 5.9 109 
Nu  179.3
We know that,

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hL
Nu 
k
h  1.2
Nusselt number, 179.3 
0.02896
Convective heat 
 h  4.32 W / m K
2
transfer co  efficient 
Heat loss, Q = h A (∆T)
= h × W × L ×(Tw-T∞)
= 4.32 × 0.7 × 1.2 × (90 – 30)
Q  218.16 W
Result: Convective heat loss, Q = 218.16 W.
10. Calculate the heat transfer from a 60 W incandescent bulb at 115℃ to ambient air at 25℃. Assume the
bulb as a sphere of 50 mm diameter. Also find the % of power lost by free convection.
Given Data:
Assume bulb as a sphere, D = 50mm = 0.050 m
Surface temperature, Tw = 115℃ + 273 = 383 K
Ambient air temperature, T∞= 25℃ + 273 = 298 K
Tw  T
Tf 
To find: Film temperature, 2
115  25
  70 C
2
To properties of air at 70℃,
k = 0.02966W/mK
v= 20.02 × 10-6 m2/s
Pr = 0.694
1 1
  
Tf in K 70  273
1
  2.915  103 K 1
343
g D3  Tw  T 
Gr 
v2
Grashof number,
9.81  0.050   2.915 103   383  298 
3

 20.02 10 
6 2
Gr=7.58 × 105

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Gr Pr = 7.58 × 105 × 0.694
=5.26 × 105
[Refer HMT data book , page no. 137]
Nusselt number, Nu = 2 + 0.50 (Gr Pr)0.25
Nu  15.46
hD
  15.46
k
h  9.15W / m 2 K
Q  h A  Ts  T 
Heat transfer, Q  9.15  4  r 2  383  298 
Q  6.10W
Percentage of power lost by free convection
Q
  100
60
6.10
  100  10.18%
60
11. Define teh velocity boundary layer and thermal boundary layer and thermal boundary layer thickness
for flow over a flat plate.
Velocity boundary layer and Thermal boundary layer
The thickness of the boundary layer has been defined as the distance from the surface at which the velocity or
temperature reaches 99% of the external velocity or temperature.
In velocity boundary layer, velocity of the fluid is less than 99% of free stream velocity and in thermal boundary
layer, temperature of the fluid is less than 99% of the free stream.
12.Air at 30℃, at a pressure of 1 bar is flowing over a flat plate at a velocity of 4 m/s. If the plate is
maintained at a uniform temperature of 130℃, calculate the average heat transfer co-efficient over the 1.5
m length of the plate and the air per 1m width of the plate.
Given: Fluid temperature, T∞=30℃
Velocity, U = 4 m/s
Plate temperature, Tw = 130℃

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Length, L = 1.5m
Width, W = 1m
To find: 1. Average heat transfer co-efficient, h
2. Heat transfer, Q.
Tw  T
Tf 
2
130  130
Film temperature, 
2
Tf  80 C
[From HMT data book, Page No. 33 (Sixth Edition)]
ρ= 1 kg/m3
v= 21.09 × 10-6 m2/s
Pr = 0.692
k= 0.03047 W/mK.
UL
Re 
v
4 1.5
Reynolds number, 
21.09  106
Re  2.84  105  5  105
Since Re < 5 × 105, flow is laminar flow,
For flat plate, laminar flow,
Local nusselt 
  0.332  Re   Pr 
0.5 0.333
Number, Nu x 
[From HMT data book, page No. 112 (Sixth edition)]
Nu x  0.332  2.84 105    0.692 

0.5

0.333
Nu x  156.51
We know that,
hx L
Local Nusselt Number, Nu x 
k

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h x 1.5
156.51 
0.03047
 h x  3.179 W / m 2 K
Local heat transfer 
  3.179 W / m K
2
coefficient, h x 
We know that,
Average heat transfer 

  2 hv
coefficient, h 
 2  3.179
h  6.358
We know that,
Heat transfer, Q = h A (Tw - T∞)
= h × W × L (Tw-T∞)
=6.358 × 1 × 1.5× (130 – 30)
Q  953.7 W
Result: 1. h = 6.358 W/m2K
2. Q = 953.7 W
13. A steam pipe 80 mm in diameter is convect with 30mm thick layer of insulation which has a surface
emissivity of 0.94. The insulation surface temperature is 85℃ and the pipe is placed in atmosphere air at
15℃. If the heat lost both by radiation and free convection, find the following:
1. The heat loss from 5m length of the pipe.
2. The overall heat transfer co-efficient.
3. Heat transfer co-efficient due to radiation.
Given: Diameter of pipe =-80 mm
= 0.080 m
Insulation thickness = 30 mm = 0.030 m
Actual diameter of 
  0.080  2  0.030
the pipe.D 
 0.14 m
Emissivity, ε = 0.94
Air surface temperature, T u = 85℃
Air temperature, T∞= 15℃

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To find: 1. Heat transfer from 5m length of pipe, Q
2. Overall heat transfer co-efficient, hr.
3. Heat transfer co-efficient due to radiation, hr.
Tw  T
2
85  15

2
Tf  50 C
[From HMT data book, Page no. 33 (Sixth edition)]
ρ = 1.093 kg/m3
v = 17.95 × 10-6 m2/s
Pr= 0.698
k = 0.02826 W/mK
Co  efficent of thermal  1

exp ansion,   Tf in K
1

50  273
  3.095  103 K 1
We know that,
g  D3  T
Grashof number, Gr 
v2
[From HMT data book, Page No.134 (Sixth edition )]
9.81 3.095 103   0.14   85  15 

3

17.95 10 6 2
Gr  18.10 106
Gr Pr  18.10 106  0.698
Gr Pr  1.263 107

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For horizontal cylinder,
Nusselt number, Nu = C [Gr Pr]m
[From HMT data book, Page No. 137 (Sixth edition0]
C = 1.263 × 107,
Corresponding C = 0.125, and m =-0.333
0.333
 Nu  0.125 1.263 107 
Nu  28.952
hD
We know that, Nu 
k
h  0.14
 28.952 
0.02826
 h  5.84 W / m 2 K
Convective heat transfer coefficient, h c  5.84 W / m 2 K
Heat lost by convection,
Qconv  h A  T 
 h  D  L   Tw  T 
 5.84   0.14  5  85  15 
Qconv  898.88 W
Heat lost by radiation,
Qrad = ε ς A [T4w-T4∞]
Where, ε = Emissivity
A= Area – m2
ς = Stelen Boltzman Constant
=5.67 × 10-8 W/m2 K4
Tu= = Surface temperature, K.
T∞= Fluid temperature, K.
Tw  85  273 T  15  273
Tw  358K T  288K
 Qrad     DL  T 4 w  T 4  

 0.94  5.67  108   0.14  5  3584  2884 
Qrad  1118.90 W

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Total heat loss, Qt = Qconv+ Qrad
= 898.99 + 118.90
Qt  2017.89W
Total heat transfer, Qt = ht A ∆ T
= ht × π DL × (Tw-T∞)
2017.89 = ht × π × 0.14 × 5 × (85 – 15)
 h t  13.108 W / m 2 K
Overall heat coefficient, h t  13.108 W / m 2 K
Radiative heat transfer coefficient,
Hr = ht-hc
=13.108 – 5.84
h r  7.268 W / m2 K
Result:
1. Heat loss from 5m length of pipe
(i) By convection, Qc = 898.99 W
(ii) By radiation, Qr== 118.90 W
2. Overall heat transfer coefficient, ht= 13.108W/m2K
3. Radiative heat transfer co-efficient, hr= 7.268 W/m2K.
14. 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 T w
= 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.
Tw  T
Solution: We know Film temperature Tf  T
2

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300  40
  443 K
2
Tf  170C
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
UL
Reynolds number Re=
v
50  0.8
 6
 1.26  10 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:
Local nusselt number} Nux = 0.0296  (Re)0.8  (Pr)0.333
NUx = 0.0296  (1.26 106)0.8  (0.6815)0.333

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NUx = 1977.57
hx  L
We know NUx 
K
hx  0.8
1977.57 
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
Q2  Q1
2. Percentage error =
Q1
23589.2 - 16390.4
=  100
16390.4
= 43.9%
15. 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

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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
Tmi  Tmo
Mean temperature Tm   65C
2
To find: Heat transfer coefficient (h)
Solution:
UD
Reynolds Number Re =


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   D2  U
4

0.056  1.044   (0.035)2  U
4
 U = 55.7 m/s
UD
(1)  Re =

55.7  0.035
=
7.98  10-7
Re = 2.44  106

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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  106 )0.8  (0.7)0.3
Nu  2661.7
hD
We know that, Nu 
K
h  0.035
2661.7 
0.0298
Heat transfer coefficient h = 2266.2 W/m2K
16. 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:
Tmi  Tmo
Mean temperature Tm =
2

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16  32

2
Tm  24C
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 
Dh 
4A
 4
P  Do  Di 

Do  Di  Do  Di 
(Do  Di )
 Do  Di
= 0.05 – 0.03125
Dh = 0.01875 m
UDh
Reynolds number Re =

30  0.01875

15.9  106
Re = 35.3  10-6
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

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hDh
We know Nu =
K
h  0.01875
 87.19=
26.3  10-3
 h = 122.3 W/m2K
17. 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
UD
Reynolds number Re =

0.8  0.05

7  106
Re = 5714.2
Since Re < 2300 flow is turbulent
L 2
  40
D 0.050
L
10   400
D

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For turbulent flow, (Re < 10000)
0.055
0.8 0.33 D
Nusselt number Nu = 0.036 (Re) (Pr) L
 
0.055
 0.050 
Nu  0.036 (5714.2) 0.8
 (116) 0.33
 
 2 
Nu  142.8
hD
We know Nu =
K
h  0.050
 142.8 =
133.8  10-3
 h = 382.3 W/m2K
18. 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:
Tw  T
Film temperature Tf 
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
1
Coefficient of thermal expansion}  =
Tf in K

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1 1
 
356  273 629
 = 1.58  10-3K 1
g    L3  T
Grashof number Gr =
v2
9.81 2.4  10-3  (4)3  (606  106)
 Gr =
(55.46  10 6 )2
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
 Nu = 0.10 [1.08  1011]0.333

Nu = 471.20
We know that,
hL
Nusselt number Nu 
K
h 4
 472.20 =
49.08  10-3
Heat transfer coefficient h = 5.78 W/m2K
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

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21.
22.

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23.
24.

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25.

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Unit-3
PHASE CHANGE HEAT TRANSFER AND HEAT EXCHANGERS
Part-A
1. what is meant by boiling and condonation?
Ans: The change of phase from liquid to vapour state is known as condensation.
2. Give the appliocations of boiling and condensation?
Ans: Boiling and condensation process finds wide applications as mwntioned below.
1. Thermal and nuclear power plant.

2. Rrfrigerating systems.
3. Process of heating and cooling
4. Air conditioniong systems
3. what is meanrt by pool boiling?
Ans: If heat is added to a liquid from a submerged soliod surface, the boiling process referred to as
pool boiling. In this case the liquid above the hot surface is essentially stagnant and its motion near the
surface is due to free convection and making induced by bubble growth and detachment.
4. What is meant by film wise and Drop wise condenstaion?
Ans: The liquid condensation wets the solid surface, spreads out and forms a continuous film over
the entire surface is known as film wise condensation.
5. Give the merits of drop wise condensation?
Ans: In drop wise condensation, a large portion of the area of the plate is directly exposed to
vapour. The heat transfer rate in drop wise condensation is 10 times higher than in film condensation.
6. What is heat exchanger?
Ans: A heat exchanger is defined as an equipment which transfer the heat from a hot fluid to cold
fluid.
7. What are the types of heat exchangers?
Ans: The types of heat exchangers are as follows:
1. Direct contact that exchangers

2. Indirect contact heat exchangers
3. Surface heat exchangers
4. Parallel flow heat exchangers
5. Counters flow heat exchangers
6. Cross flow heat exchangers
7. Shell and tube heat exchaangers
8. Compact heat exchangers.
8. What is meant by direct heat exchanger (or) open heat exchangers?
Ans: In direcrt contact heat exchanger, the heat exchange takes place by direct miding of hot and cold
fluids.
9. What is meant by Indirect comtact heat exchanger?

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Ans: 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.
10. What is meant by Regenerators?
Ans: In this tyype of heat exchangers, hot and cold fluids flow alternatively through the space.
Examploes: IC engines, gas turbines.
11. what is meant by recuperator (or) Surface heat exchangers?
Ans: 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 seperated by a tube wall or a surfacxe.
12. wehat is meant by parallel flow and counter flow heat exchanger?
Ans: In this type of heat exchanger, hot and cold fluids move in the same directions.
In this type of heat exchanger hot and cold fluids parallel but opposite directions.
13. What is meant by shell and tube heat exchanger?
Ans: 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.
14. What is meant by compact heat exchangers?
Ans: There are many special purpose heat exchanegrs called compact heat exchangers called compact
heat exchangers. They are generally employed when convective heat transfer co-efficient associated with one of
the fluids such smaller than that associated with the other fluid.
15. What is meant by LMTD?
Ans: We know that the temperature difference between the hot and cold fluids in the heat exchanege
varies from Point in addition various modes of heat transfer are involved. Therefore based on concept of
appropriate mean tempoerature 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 co-efficient W/m2K A – Area m2
(∆T)m – Logarithmic mean temperature differnce.
16. What is meant by fouling factor?
Ans: 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 scaliong or deposits. The effect of these deposits the value overall heat
transfer co-efficient. This effect is taken care of by introducing an additional thermal resistance called the
fouling resistance.
17. What is meant by effectiveness?
Ans: The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum
possible heat transfer.
Actual heat transfer Q

Effectiveness   
Maximum possible heat transfer Qmax

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18. Give the application of boiling and condensation?
Ans: 1. Thermal and nuclear power plant
2. Refrigerating systems
3. Process of heat and cooling
4. Air conditionaring systems.
19. What are the method of condensation?
Ans: 1. Filmwise condenstaion
2. Dropwise condenstaion
20. Draw the difference regions of boiling and what is nucleate boiling?
Ans: Nucleate boiling exists in regions ii and iii. The nuclear boiling begins at region ii. As the excess
temperature is further increased, bubbles are formed more rapidly and raoid eveporartion take place. This is
indicated in region iii. Nucleate boiling exists upto T = 50℃.
21. What is meant by counter flow heat exchanger?
Ans: In this types of heat exchanger, hot and cold fluids move in parallel but opposite in direction.

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PART- B
1.A heating element is added with metal is 8 mm diameter and 0 emissivity is 0.92. The element is horizontally
immersed in water bath. The surface temperature of the metal is 260℃ under steady state boiling
conditions. Calculate the power dissipation per unit length of the herater.
Given:
Diameter, D = 8 mm = 8 × 103 m
Emissivity, ∈ = 0.92
Surface temperature, Tw = 260℃.
To find:
Power dissipation
Solution:
We know that, saturation temperature of water is 100℃.

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i.e. Tsat  100 C
T  Tw   Tsat
Excess temperature, T  260  100
T  160 C  50 C
So, this Film pool boiling
Tw  Tsat
2
260  100

2
Tf  180 C
Properties of water Vapour at 180℃. (Saturated Steam)
[From HMT data page no.39 (Sixth edition)]
Ρv = 5.16 kg/m3
Kv = 0.03268 W/mK
Cpv = 2709 J/kg K
Uv = 15.10 × 10-6 Ns/m2
Properties of saturated water at 100℃
[From HMT data book page No. 21 (Sixth edition]
Ρt = 961 kg/m3
From steam table At 100℃
[R.S Khurmi steam table Page No. 4]
h fg  2256.9 kJ / kg
h fg  2256.9 103 J / kg
In film pool boiling, heat is transferred due to both convection and radiation.
Heat transfer co-efficient, h = hconv + 0.75 hrad ……….(1)
 k 3v  v   t  v   g   h fg  0.4  CPv T   
0.25
h conv  0.62   
  v D T 
[From HMT data book page No. [42]
  32.68 103 3  5.16   961  5.16   9.81  2256.9  103   0.4  2709  160 
0.25
h conv  0.62   
 15.10 106  8  103  160 
 

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0.25
 4.10 106 
h conv  0.62  5 
1.93 10 
h conv  421.02 W / m2 K ..........  2 
 T 4  T 4sat 
h rad     w   From HMT data book page no.142
 Tw  Tsat 
  260  2734  100  2734 
h rad  5.67  108  0.92   
  260  273  100  273 
[∴ Stefan boltzman constant, σ = 5.67 × 10-8 W/m2K4]
h rad  20W / m2 K ........... 3
Substitute (2), (3) in (1)
(1) ⟹ h = 421.02 + 0.75(20)
h  436.02 W / m2 K
Heat transferred, Q = h A (T w – Tsat)
= h × π × D × L (Tw – Tsat)
= 436.02 × π × 8 × 10-3 × 1 × (260 – 100)
Q = 1753.34 W/m [∵ L = 1m]
(or)
Power dissipation, P = 1753.34 W/m
Result:
Power dissipation, P = 1753.34 W/m.
2.Hot oil Cp = 2200 J/kg K is to be cooled by water (C p = 4180 J/kg K) in a 2-shell- pass amd 12-tube- pass
heat exchanger. The tubes are thin walled and are made of copper with a diameter of 1.8 cm. The length
of each tube pass in the heat exchanger is 3m, and the overall heat transfer co-efficient is 340 W/m 2K.
Water flows through the tubes at a total rate of 0.1 kg/s, and the oil through the shell at a rate of 0.2 kg/s.
The water and the oil enter at temperatures 18℃ and 160℃ respectively. Determine the rate 0, heat
transfer in the heat exchanger and the outlet temperatures of the water and the oil. (16)
Given Data:
In a 2-shell- pass and 12-tube pass heat exchanger, specific heat capacity of hot oil. Cph= 2200 J/kg K
Specific heat capacity of water, Cpc = 4180 J/kgK
Diameter of copper tube, d = 1.8 cm = 0.018 m

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Length of heat exchanger L = 3m
Overall heat transfer co-efficient, U = 340W/m2K
Mass flow rate of water, mc = 0.1 kg/s
Mass flow rate of oil, mh = 0.2 kg/s
Inlet temperature of cooling water, t1 = 18℃
Inlet temperature of hot oil, T 1= 160℃
To find:
(i) Rate of heat transfer in the heat exchanger, Q
(ii) Outlet temperature of the water. T 2
(iii) Outlet temperature of hot oil, T 2.
Solution:
Capacity rate of hot liquid,
Ch = mh × Cph = 0.2 × 2200
Ch = 440 W/K
Capacity rate of cooling water,
Cc = mc × Cpc = 0.1 × 4180
Cc = 418 W/K
Maximum possible heat transfer,
Qmax = Cmin (T1 – t1)
= 418 (160 – 18)
Qmax= 5.935 × 104 W
Cmin = Cc = 418 W/K
Cmax = Ch = 440 W/K
Cmin 418
  0.95
Cmax 440
Cmin
 0.95
Cmax
Surface Area, As = 12 tube ×As
= 12 × (π D L)

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= 12 × (π × 0.018 × 2)
= 2.04 m2
UA
Number of transfer units, NTU 
Cmin
[From HMT data book, page no. 152]
340  2.04
NTU   1.659
418
To find effectiveness ε, refer HMT data book page No. 165.
(2 shell – 12 tube pass heat exchanger)
From graph,
Xaxis ⟶ NTU ⟶ 1.659
Cmin
Curve   0.95
Cmax
Corresponding Yaxis value is 58%
I.e., ε = 0.58
Q = ε Qmax = 0.58 (5.935 × 104)
= 3.442 × 104 W
Q = mh Cph (T1-T2)
34423 = 0.2 × 2200 (160 – T2)
T2 = 160 – 78.23
T2 = 81.76℃ Effectiveness
Q = mc Cpc (t2 – t1)
34423 = 0.1 × 4180 (t2 – t1)
T2 = 82.35 + 18
T2 = 100.35℃
58%
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NTU updated
1.659 materials from Rejinpaul Network App
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Result:
(i) Rate of heat transfer in the heat exchanger, Q = 34423 bW
(ii) Outlet temperature of the water, t2 = 100 35℃
(iii) Outlet temperature of the hot oil, T 2 = 81.76℃
3.Boiling heat transfer phenomena
Boiling is a convection process involving a change of phase from liquid to vapour state. This is possible only
when the temperature of the surface (T w) exceeds the saturation temperature of liquid (T sat).
According to convection law,
Q h A (Tw – Tsat)
Q h A (∆T)
Where
∆T = (Tw – Tsat) is known as excess temperature.
If heat is added to a liquid from a submerged solid surface the boiling process referred to as pool boiling. In
thios case the liquid above the hot surface is essentially stagnant and its motion near the surface is due to free
convection and mixing induced by bubble growth and detachment.
Fig. 3.1 shows the temperature distribution in saturated pool boiling with a liquid – vapour
interface.
Fig.3.1 Pool Boiling

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The different regions of boiling are indicated in fig. 3.2. This specific curve has been obtained from an
electrically heated platinum wire submerged in a pool of water by varying its surface temperature and measuring
the surface heat flux (q).
Excess Temperature ∆Te = Tw - Tsol
I – Free convection
II – Bubbles Condense in super heated liquid
III – Bubbles raise to surface
IV – Unstable film
V – Stable film
VI – Radiation coming into play
Fig. 3.2. pool Boiling Curve for water
4.What are the different types of fouling in heat exchangers?
The surfaces of a heat exchanger do not remain clean after has been in use for some time. The
surface becomes fouls with scaling or deposits. The effect of these deposits affecting the value of overall heat
transfer co-efficient. This effect is taken care of by introducing an additional thermal resistance called the
fouling resistance.
Types of fouling in heat exchanger:

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(i) Corrosion fouling,
(ii) Chemical fouling, and
(iii) Biological fouling, etc.
5. Hot exhaust gases which enter a cross-flow heat exchanger at 300℃ and leave at 100℃ are
used to heat water at a flow rate of 1 kg/s from 35 to 125℃. The specific heat of the gas is 1000 J/kg. K
and the overall heat transfer co-efficient based on the gas side surface is 100 W/m2. K.
Find the required gas side surface area using the NTU method and LMTD method.
Given Data: In cross flow heat exchanger,
Hot Gas: (Hot fluid)
Inlet temperature, T1 = 300℃
Outlet temperature, T2 = 100℃
Water: (Cold fluid)
Inlet temperature, t1 = 35℃
Outlet temperature, t2 = 125℃
Overall heat transfer co-efficient, u = 100 W/m2K
Specific heat of hot gas, Cph = 1000 J/kgK
Specific heat of water, Cpw = 4186 J/kgK
To find: (i) Area (LMTD method)
(ii) Area (NTU method)
@Solution:
(i) LMTD method:
For cross-flow heat exchanger,
Logarithmic mean temperature difference, LMTD
 T1  t 2    T2  t1 

T t 
ln  1 2 
 T2  t1 
175  65 110
 T LMTD    111.06
 175  0.99
ln  
 65 
To find correction factor, F (both fluids unmixed):
[Refe3r HMT data book, Page No. 162]
From Graph,

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t 2  t1
X  axis value, P 
T1  t 2
125  35 90
   0.34
300  35 265
T1  T2
Curve value, R 
t 2  t1
300  100 200
   2.22
125  35 90
X- axis value is 0.34, curve value is 2.22, corresponding
Y – axis value is 0.87.
i.e, F  0.87
Q = mc Cpc (t2 – t1)
= 1 × 4186 × (125 – 35)
Q = 376.74 kW
We know that, Q = U Ah F (∆T)LMTD
376.74 × 103 = 100 × Ah × 0.87 × (111.06)
 Ah  38.99 m2
(iii)NTU method:
Q = mh Cphg (T1 – T2)
376.74 × 103 = mh × 1000 (300 – 100)
m h  1.883kg / s
mC  1kg / s
Capacity rate of hot liquid, Ch = mh × Cph = 1.883 × 1000
Cmin = Ch = 1883.7 W/K
Capacity rate of cold liquid, Cc = mc × Cpc × = 1 × 4186
Cmax = Cc = 4186 W/K
Cmin 1883.7
  0.45
Cmax 4186
Q max  Cmin  T1  t1 
 1883.7  300  35 
 499180.5W

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Q
Effectiveness,  
Q max
mc Cpc  t 2  t1 
  0.75
499180.5
To find NTU: [Refer HMT data book, page no. 166]
From graph,
Cmin 1883.7
Curve    0.45
Cmax 4186
Y  axis    0.75
Corresponding X-axis value is NTU = 2.1
Uh Ah
NTU 
C min
100  A h
2.1 
1883.7
A h  39.55 m 2
6.Water is to be boiled at atmospheric pressure in a polished copper pan by means of an electric heater.
The diameter of the pan is 0.38 m and is kept at 115℃. Calculate the following
1. Power required to boil the water
2. Rate of evaporation
3. Critical heat flux.
Given:
Diameter, d = 0.38 m;
Surface temperature, Tw = 115℃.
To find:
1. Power required, (P)
2. Rate of evaporation, (m)
Q
3. Critical heat flux,  
A

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Fig. 3.3
Solution:
We know that, saturation temperature of water is 100℃
i.e. Tsat  100 C
Properties of water at 100℃.
[From HMT data book page no. 21, sixth edition]
Density, ρl = 961 kg/m3
Kinematic viscosity, v = 0.293 × 10-6 m2/s
Prandtl Number, Pr = 1.740
Specific heat, Cpl = 4216 J/kg k
Dynamic viscosity, μl = ρl × v = 961 × 0.293 × 10—6
= 281.57 × 10-6 Ns/m2
From steam Table [R.S. Khurmi Steam table page No 4]
At 100℃
Enthalpy of evaporation, hfg = 2256.9 kJ/kg
Hfg = 2256.9 × 103 J/kg
Specific Volume of Vapour, vg = 1.673 m3/ kg
1
Density of vapour, v 
vg
1

1.673
v  0.597 kg / m3
∆T = Excess temperature = Tw – Tsat = 115∘ - 100 = 15℃

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T  15 C  50 C. So this is Nucleate pool boiling process.
1. Power required to boil the water
For Nucleate pool boiling
 g   l   v  
3
 Cpl  T 
0.5
Q
Heat flux,  l  h fg    n 
.......... 1
A     Csf  h fg P r 
[From HMT data book page no. 142 (sixth edition0]
Where, σ = Surface tension for Liquid vapour interface
At 100℃
  0.0588 N / m  From HMT data book page No.144

For water  copper  Csf  Surface fluid cons tan t  0.013
N=1 for water [From HMT data book page No. 143]
Substitute
l , h fg , l , v , , Cpl , T, Csf , h fg , n and Pr values in Equation 1

3
 9.81  961  0.597    
0.5
Q 4216 15
1   281.57 10  2256.9 10  
6 3
  
 0.013  2256.9  103  1.74  
1
A  0.0588 
Q
Heat flux,  4.83105 W / m2
A
⇒ Heat transfer, Q = 4.83 × 105 × A

 4.83  105  d 2
4

 4.83  105   0.38 
2
4
Q  54.7 103 W
 Q  54.7 103  P
 Power  54.7 103 W
2. Rate of evaporation, (m)
We know that,
Heat transferred, Q = m × hfg

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Q
 m
h fg
54.7  103

2256.9 103
m  0.024 kg / s
3. critical heat flux
For Nucleate pool boiling, critical heat flux,
   g   l   v  
0.25
Q
 0.18 h fg  v  
A  2 v 
0.25
 0.0588  9.81  961  0.597  
 0.18  2256.9  10  0.597  
3

 0.597 
2
 
Q
 1.52  106 W / m 2
A
Q
Critical Heat flux, q   0.52  106 W / m 2
A
Result:
1. P = 54.7 × 103 W
2. m = 0.024kg/s
Q
3. = q = 1.52 × 106 W/m2.
A
(OR)
7. Calculate for the following cases, the surface area required for a heat exchanger which is
required to cool 1200 kg hr of bezene (Cp 1.74 kJ/kg℃) from 72℃ ro 42℃. The cooling water (Cpo 4.18
kJ/kg℃) at 15℃ has a flow rate of 200 kg/hr.
(i) Single pass counter-flow
(ii) 1- 4 exchanger (one-shell pas and four-tube passes and
(iii) Cross flow single pass with water mixed and benezene unmixed. Assume all the cases U =
0.28 kW/m2K. (16)
Given:
Hot fluid-Benzene Cold fluid-Water
(T1, T2) (t1, t2)
Mass Flow rate of benzene, mh = 3200 kg/h
= 0.889 kg/s

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Entry temperature of benzene, T 1 = 72℃
Exit temperature of benzene, T2 = 42℃
Specific heat of benzene, Cph = 1.74 kJ/kg℃
= 1.74 × 103 J/kg℃
Specific heat water, Cpc = 4.18 kJ/kg℃
= 4.18 × 103 J/kg℃
Entry temperature of water, t1 = 15℃
Mass flow rate of water, mc = 2200 kg/h i.e., 0.611 kg/s
Overall heat transfer co-efficient, U = 0.28 kW/m2K
= 0.28 × 103 W/m2K
To find:
(i) Surface area for single pass counter flow.
(ii) Surface area for 1 – 4 exchanger.
(iii) Surface area for cross flow single pass with water mixed and benzene unmixed.
Solution:
Case (i):
Heat lost by benzene (Hot fluid)
= Heat gained by water (Cold fluid)
Qh = Qc
Mh Cph (T1 – T2) = mc Cpc (t2- t1)
⇒ 0.889 × 1.74 × 103 (72 – 42) = 0.611 × 4.18 × 103 (t2 - 15∘ )
 t 2  33.2 C
⇒ Q = mh Cph (T1 – T2) or mc Cpc (t2 – t1)
⇒ Q = 0.889 × 1.74 × 103 (72 – 42)
Q  46.405 103 W
We know that,
Heat transfer, Q = U A (∆T)m ……….(1)
[From HMT data book, page no. 151]
Where, (∆T)m = Logarithmic Mean Temperature

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Difference (LMTD)
For counterflow,
 T1  t 2    T2  t1 
 T m 
T  t 
ln  1 2 
 T2  t1 
 72  33.2    42  15

 72  33.2 
ln 
 42  15 
 T m  32.5 C
Substitute (∆T)m and Q values in equation (t).
46.405 103  0.28 103  A  32.5 

(1) ⟹
 A  5.1m 2
Case (ii):
One shell pass and four tube passes,
To find correction factor F, refer HMT data book, page no. 158.
From graph,
t 2  t1
X  axis value, P   0.32
T1  t1
T1  T2
Curve value, R   1.65
t 2  t1
X-axis value is 0.32, curve value is 1.65, corresponding Y-axis value is (From graph) 0.9.
i.e., F = 0.9
⟹ Q = F U A (∆T)m
46.405 × 103 = 0.9 × 0.28 × 103 × A × 32.5
⇒ A  5.66 m2
Case (iii):
Cross flow single pass with water mixed and benzene unmixed. To find correction factor, F, refer HMT
data book, page no. 160.
From graph,

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t 2  t1
X  axis value, P 
T1  t1
 0.32
T  T2
Curve value, R  1
t 2  t1
 1.65
From graph, corresponding Y-axis value is 0.92,
i.e., F = 0.92
⟹ Q = F U A (∆T)m
46.405 × 103 = 0.92 × 0.28 × 103 × A × 32.5
 A  5.54 m2
8. Consider laminar film condensation of a stationary vapour on a vertical flat plate of length L and
width b. Derive an expression for the average heat transfer co-efficient. State the assumptions mode.
Laminar Film wise condensation on a vertical plate:
Nusselt‘s analysis of film condensation makes the following simplifying assumptiuons.
(1) The plate is maintained at a uniform temperarture T s.
(2) Condensate flow is laminar.
(3) Fluid properties are constant.
(4) Shear stress at the liquid vapour interface is negligible.
(5) Acceleration of fluid within the condensate layer is neglected.
(6) Heat transfer is pure conduction and temperature distribution is linear.
The momentum equation is given by
 2 u 1 dp Bx
 
y2  t dx l
dp
Where Bx is the body force in x-direction. The body force within the film is ρ1 g,   v g.
dx

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2u g
    1  v 
y 2
l
u
Integrating twice using u = 0 at y = 0 and  0 at y  
y
g  1  v  2  y 1  y 2 
u  y      
1   2    
The condensate mass flow rate through any x-position of the film is given by
g  1  v  3
mx 
3 l
l  l  v  g  2 d
Also dm 
l
The heat transfer at the wall is given by
k l  Tsat  Ts  dx
dx.qs 

The heat removed by the wall is given by,
k l l  Tsat  Ts 
2 d  dx
g l  l  v  h fg
The local heat transfer co-efficient may now be expressed as
 g 1  l  v  k 3l h fg 
0.25
k
h x  l or h x   
x  4 l  Tsat  Ts  x 
The average value of heat transfer co-efficient is given by

1 4
hL 
L0 h x dx  h L
3
 g 1  l  v  k 3l h fg 
1/ 4
h L  0.943  
 l  Tsat  Ts  L 
9. 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 was scaling or deposits. The effect of these deposits affecting the value of
overall heat transfer co-efficient. This effect is taken care of by introducing an additional thermal
resistance called the fouling resistance .
10.In a cross flow both fluids unmixed heat exchanger, water at 6℃ flowing at the rate of 1.25
kg/s of air that is initially at a temperature of 50℃. Calculated the following
1. Exit temperature of air

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2. Exit temperature of water
Assume overall heat transfer co-efficient is 130 W/m2-K and area in 23 m2.
Given:
Cold fluid-water Hot fluid-air
Inlet temperature of water, tl = 6℃
Mass flow rate of water, mh = 1.2 kg/s
Initial temperature of air, T l = 50℃
Overall heat transfer co-efficient, U = 130 W/m2K
Surface area, A = 23m2
To find:
1. Exit temperature of air, (T2)
2. Exit temperature of water, (t2)
Solution:
We know that,
Specific heat of water, Cpc = 4186 J/kg K
Specific heat of air, C-ph = 1010J/kg K (Constant)
We know
Capacity rate of water
C = mc × Cpc
= 1.25 × 4186
C  5232.5W / k 1
Capacity rate of air
C  m h  Cph
1.2 1010
C  1212 W / K  2
From equation (1) and (2), we know that,
Cmin = 1212 W/K
Cmax = 5232.5 W/K

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Cmin 1212
  0.23
Cmax 5232.5
Cmin
 0.23  3
Cmax
UA
NTU   From HMT data book page no.151
Cmin
130  23
Number of transfer units, 
1212
NTU  2.46  4
To find effectiveness ∈, refer HMT data book page No 165]
(Cross flow, both fluids unmixed)
From graph,
X axis  NTU  2.46

Cmin
Curve   0.23
Cmax
Corresponding Yaxis valueis 0.85
i.e,  0.85
Maximum heat transfer
Q max  Cmin  Tl  t l 
 1212  50  6 
Q max  53.328 W
Cmin
85%  0 C
Cmax
Effectiveness
∈
NTU 2.46%
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Actual heat transfer rate
Q Q max
 0.85  53,328
Q  45,328 W
Heat transfer, Q = mc Cpc (t2 – t1)
45,328 = 1.25 × 4186 (t2 – 6)
⟹ 45,328 = 5232.5 t2- 31,395
⟹ t2 = 14.6℃
outlet temperatureof water, t 2  14.6 C
We know that,
Heat transfer, Q  m h C ph  T1  T2 
45,328  1.2 1010  50  T2 
 45.328  60, 600  1212 T2
 T2  12.6 C
Outlet temperature of air, T2  12.6C
Result:
1. T2 = 12.6℃
2. t2 = 14.6℃
11. In a cross flow heat exchangers, both fluids unmixed, hot fluid with a specific heat of 2300 J/kg K
enters at 380℃ and leaves at 300℃. Cold fluids enters at 25℃ and leaves at 210℃. Calculate the
requirement surface area of heat exchanger. Take overall heat transfer co-efficient is 750 W/m2K. 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℃
Exit temperature of hot fluid, T2 = 300℃
Entry temperature of cold fluid, t1 = 25℃

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Exit temperature of cold fluid, t2 = 210℃
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  F U A  T mcounter flow 1
[From HMT data book page no. 15] (sixth editiuon)
Where
F Correction factor
(∆T)m – Logarithmic mean temperature difference for counter flow
For Counter flow,
 T1  t 2  T2  t1  
 T m  
T t 
ln  1 2 
T2 t1 
 380  210    300  25 

 380  210 
ln 
 300  25 
 T m  218.3 C
Heat transfer, Q  m h Cph  T1  T2 

 Q  1 2300  380  300 
Q  184 103 W
To find correction factor F, refer HMT data book page no 161 (Sixth edition)
[Single pass cross flow heat exchanger – Both fluiods unmixed]
From graph,
t 2  t1 210  25
X axis Value P    0.52
T1  t1 380  25
T1  T2 380  300
Curve value R    0.432
t2  tl 210  25

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Xaxis value is 0.52, curve value is 0.432, corresponding Y axis value is 0.97,
i.e. F  0.97
0.94 0.97
0.9
0.8
R = 0.432
0.7
0.6
0.5
0 P = 0.52
\Fig.3.15
Substitute Q, F, (∆T)m and U value in Equation (1)

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1  Q  F U A  T m
184 103  0.97  750  A  218.3
 A  1.15m 2
Result:
Surface area, A = 1.15 m2
12.) In a refrigerating plant water is cooled from 20℃ to 7℃ by heat solution entering at 2℃ and leaving at 3℃.
The design heat loan is 5500 W and the overall heat transfer co-eficient is 800W/m2k. What area required
when using a shell and tube heat exchange with the water making one shell pass and the brine making is
tube passes.
Given:
Hot fluid- water Cold fluid- brine solution
(T1-T2) (t1 – t2)
Entry temperature of water, T 1 - 20℃
Exit temperature of water, T 2 - 7℃
Entry temperature of brine solution, t1 - 2℃
Exit temperature of brine solution, t2 - 3℃
Heat load, Q = 5500 W
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 flow heat exchanger.
Q  F U A   T mcounter flow ....... 1
[From HMT data book page No. 151]
Where
F – Correction factor
(∆T)m – Logaritmic mean temperature difference for counter flow.
Fpr counter flow,

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 T1  t 2    T2  t1  
 T m  
T  t 
ln  1 2 
 T2  t1 
 20  3   7  2 

 20  3 
ln 
 7  2 
 T m  12.57 C
To find correlation factor refer HMT data book page no. 158
[One shell pass and two tube passes]
t 2  t1 3  2 15
X axis value, P  
T1  T2 20  2 22
P  0.22
T1  T2 20  7 13
curve value, R   
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,
i.e, F  0.94
0.94
0.9
0.8 R = 2.6
F 0.7
0.6
0.5
0 P = 0.22
Substitute (∆T)m, A, U and F value in Equation (1)

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1  Q  F U A  T m
5500  0.94  800  A 12.57
 A  0.58 m 2
Result:
Area of heat exchanger, A = 0.58 m2.

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15. Water is boiling on a horizontal tube whose wall temperature is maintained ct 15C above the
saturation temperature of water. Calculate the nucleate boiling heat transfer coefficient. Assume the
water to be at a pressure of 20 atm. And also find the change in value of heat transfer coefficient when
1. The temperature difference is increased to 30C at a pressure of 10 atm.

2. The pressure is raised to 20 atm at  T = 15C
Given :
Wall temperature is maintained at 15C above the saturation temperature.
Tw  115C.  Tsat  100C Tw  100  15  115C
= p = 10 atm = 10 bar
case (i)

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T  30C; p  10 atm = 10 bar
case (ii)
p = 20 atm = 20 bar; T - 15C
Solution:
We know that for horizontal surface, heat transfer coefficient
h = 5.56 (T)3 From HMT data book Page No.128
h = 5.56 (Tw – Tsat)3
= 5.56 (115 – 100)3
h  18765 w/m2K
Heat transfer coefficient other than atmospheric pressure
hp = hp0.4 From HMT data book Page No.144
= 18765  100.4
Heat transfer coefficient hp  47.13  103 W / m2K
Case (i)
P = 100 bar T = 30C From HMT data book Page No.144
Heat transfer coefficient
h  5.56 ( T)3 = 5.56(30)3

h  150  103 W / m2K

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hp = hp0.4
 150  103 (10)0.4

hp  377  103 W / m2K
Case (ii)
P = 20 bar; T = 15C
Heat transfer coefficient h = 5.56 (T)3 = 5.56 (15)3
h  18765 W/m2K
hp = hp0.4
= 18765 (20)0.4
hp  62.19  103 W/m2K
16. A vertical flat plate in the form of fin is 500m in height and is exposed to steam at atmospheric
pressure. If surface of the plate is maintained at 60C. calculate the following.
1. The film thickness at the trailing edge
2. Overall heat transfer coefficient
3. Heat transfer rate
4. The condensate mass flow rate.
Assume laminar flow conditions and unit width of the plate.
Given :
Height ore length L = 500 mm = 5m

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Surface temperature Tw = 60C
Solution
We know saturation temperature of water is 100C
i.e. Tsat = 100C
(From R.S. Khurmi steam table Page No.4
hfg = 2256.9kj/kg
hfg = 2256.9  103 j/kg
We know
Tw  Tsat
2
60  100

2
Tf  80C
Properties of saturated water at 80C
(From HMT data book Page No.13)
 - 974 kg/m3
v  0.364  106 m2 / s
k = 668.7  10-3 W/mk

 = p  v= 974  0.364  10-6
  354.53  106 Ns / m2
1. Film thickness x

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We know for vertical plate
Film thickness
0.25
 4K  x  (Tsat  Tw ) 
x   
 g  hfg   2
 
Where
X = L = 0.5 m
4  354.53  106  668.7  103  0.5  100  60

x 
9.81 2256.9  103  9742
 x  1.73  104 m
2. Average heat transfer coefficient (h)
For vertical surface Laminar flow
0.25
 k 3   2  g  hfg 
h  0.943  
   L  Tsat  Tw 
The factor 0.943 may be replace by 1.13 for more accurate result as suggested by Mc Adams
0.25
 (668.7  103 )3  (974)2  9.81 2256.9  103 
1.13  
 354.53  106  1.5  100  60 
h  6164.3 W/m k. 2
3. Heat transfer rate Q
We know
Q  hA(Tsat  Tw )
= h  L  W  (Tsat  Tw )
= 6164.3  0.5  1 100-60
Q = 123286 W

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4. Condensate mass flow rate m
We know
Q  m  hfg
Q
m
hfg
1.23.286
m
2256.9  103
m  0.054 kg/s
17. Steam at 0.080 bar is arranged to condense over a 50 cm square vertical plate. The surface
temperature is maintained at 20C. Calculate the following.
a. Film thickness at a distance of 25 cm from the top of the plate.

b. Local heat transfer coefficient at a distance of 25 cm from the top of the plate.
c. Average heat transfer coefficient.
d. Total heat transfer
e. Total steam condensation rate.
f. What would be the heat transfer coefficient if the plate is inclined at 30C with horizontal plane.
Given :
Pressure P = 0.080 bar
Area A = 50 cm  50 cm = 50  050 = 0.25 m2
Surface temperature Tw = 20C
Distance x = 25 cm = .25 m
Solution
Properties of steam at 0.080 bar
(From R.S. Khurmi steam table Page no.7)

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Tsatj / kg  41.53C
hfg  2403.2kj/kg = 2403.2  103 j / kg
We know
Tw  Tsat
2
20+41.53
=
2
Tf  30.76C
Properties of saturated water at 30.76C = 30C
From HMT data book Page No.13
  997 kg/m3
  0.83  10-6 m2 / s
k  612  10-3 W / mK
  p  v  997  0.83  10 6
  827.51 10 6 Ns / m2
a. Film thickness
We know for vertical surfaces
0.25
 4 K  x  (Tsat  Tw ) 
x   
 g  hfg   2
 
(From HMT data book Page No.150)
4  827.51 10 6  612  10 3  .25  (41.53  20)100
x 
9.81 2403.2  103  997 2
 x  1.40  104 m
b. Local heat transfer coefficient hx Assuming Laminar flow

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k
hx 
x
612  10 3
hx 
1.46  10 4
hx  4,191 W/m2K
c. Average heat transfer coefficient h
(Assuming laminar flow)
0.25
 k 3   2  g  hfg 
h  0.943  
   L  Tsat  Tw 
The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc adams
0.25
 k 3  2g hfg 
h  0.943  
   L  Tsat  Tw 
Where L = 50 cm = .5 m
0.25
(612  103 )3  (997)2  9.81 2403.2  103
h  1.13
827.51 106  .5  41.53  20
h  5599.6 W/m2k
d. Heat transfer (Q)
We know
Q = hA(Tsat – Tw)
h  A  (Tsat  Tw )
 5599.6  0.25  (41.53  20
Q  30.139.8 W

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e. Total steam condensation rate (m)
We know
Heat transfer
Q  m  hfg
Q
m
hfg
30.139.8
m
2403.2  103
m  0.0125 kg / s
f. If the plate is inclined at  with horizontal
hinclined  hvertical  sin 1/ 4

hinclined  hvertical  (sin30)1/ 4
 
1/ 4
hinclined  5599.6  1
2
hinclined  4.708.6 W/m2k
Let us check the assumption of laminar film condensation
We know
4m
Reynolds Number R e 
w
where
W = width of the plate = 50cm = .50m
4  .0125
Re 
0.50  827.51 10 6
Re  120.8  1800
So our assumption laminar flow is correct.

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UNIT-4
RADIATION
PART-A
1. Define emissive power [E].
Ans: The emissive power is defined as the total amount of radiation emitted by a body per unit area. It is
expressed in W/m2.
2. What is meant by absorptivity ?
Ans: Absorptivity is defined as the ratio between radiation absorbed and incident radiation.
3. What is black body?
Ans: Black body is an ideal surface having the following properties.
A black body absorps all incident radiation, regardless of wave length and direction. For a prescribed
temperature and wave length, no surface can emit more energy than black body.
4. State planck’s distribution law.
Ans: The relationship between thee 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.
C1 5
E b 
C 
e 2 
 T  1
Where Ebλ = Monochromatic emissive power W/m2
λ = wave length – m
C1 = 0.374 x 10-15 W m2
C2 = 14.4 x 10-3 mK.
5. State Wien’s displacement law.
Ans: The Wien‘s law gives the relationship between temperature and wavelength corresponding to the
maximum spectral emissive power of the black body at that temperature.
λmas T = c3
where c3 = 2.9 x 10-3 [Radiation constant]
 λmas T = 2.9 x 10-3 mK.
6. State Stefan-Boltzmann law.
Ans: The emissive power of a black body is proportional to the fourth power of absolute temperature.
Eb ∞ T 4
Eb = ς T 4

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Where Eb = Emissive power, w/m2
ς = Stefan, Boltzmann constant
= 5.67 x 10-8 W/m2 K4
T = Temperature, K.
7. Define Emissivity.
Ans: It is defined as the ability of the surface of a body to radiate heat. It is also defined as the radio of emissive
power of any body to the emissive power of a black body of equal temperature.
E
Emissivity  
Eb
8. State Kirchoff’s law of radiation.
Ans: This law states that the ratio of total emissive power to the absorptivity is constant for all surfaces which
are in thermal equilibrium with the surroundings. This can be written as:
E1 E 2 E3
 
1  2 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.
α 1 = E1 ; α2 = E2 and so on.
9. Define intensity of radiation (Ib).
Ans: 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.
Eb
In 

10. State Lambert’s cosine law.
Ans: 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 θ.
11. What is the purpose of radiation shield?
Ans: Radiation shields constructed from low emissivity (high reflective) materials. It is used to reduce the neet
radiation transfer between two surface.
12. Define irradiation (G).
Ans: It is defined as the total radiation incident upon a surface per unit time per unit area. It is expressed in
W/m2.
13. What is meant by shape factor?

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Ans: 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 Fn . Other names for
radiation shape factor are view factor, angle factor and configuration factor.
14. What is meant by reflectivity?
Ans: Reflectivity is defined as the ratio of radiation reflected to the incident radiation.
15. What is meant transmissivity?
Ans: Transmissivity is defined as the ratio of radiation transmitted to the incident radiation.
16. What is gray body?
Ans: 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.
17. Define monochromatic emissive power.[Eb2]
Ans: 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.
18. Define emissivity.
Ans: It is defined as the ability of the surface of a body to radiate heat. It is also defined as the ratio of emission
power of any body to the emissive power of a black body of equal temperature. Emissivity, * = E/Eb.
19. Define radiosity (J).
Ans: It is used to indicate the total radiation leaving a surface per unit time per unit area. It is expressed in
W/m2.
20. What is meant by shape factor and mention its physical significance?
Ans: The shape factor is defined as ― The fraction of the radiative energy that is diffused from one surface
element and strikes the other surface directly with no intervening reflections‖.

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PART-B
1) Two very parallel plates are maintained at uniform temperature of T1 = 100 K and T2 = 800 K and
have emissivities of 𝛆1 = 𝛆2 = 0.2 respectively. It is desired to reduce the net rate of radiation heat transfer
between the two plates to one-fifth by placing thin aluminium sheets with an emissivity of 0.15 on both
sides between the plates. Determine the number of sheets that need to be inserted. (10)
ε1 = 0.2; ε2 = 0.2

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T1 = 1000K : T
εs = ε3 = 0.15
To find:
Number of shields required
Solution:
Heat transfer without shield, i.e, n = 0
A  T14  T2 4 
Q 12 
1 1
(no shield)
 1
1  2
1 5.67  108 10004  8004 


1 1
 1
0.2 0.2
3.34  104

9
= 0.37 x 104
Q12  3711.12 W m2 ,
 no shield 
we know that,
1 1
th of Q12   3711.12  742.2 W m 2
5  no shield  5
Heat transfer with n shield is given by.
A  T14  T2 4 
Q12 
1 1  2n 
      n  1
with shield
1  2  s 
1 5.67 108  10004  8004 

742.4 
1 1 2n
    n  1
0.2 0.2 0.15
 2n 
742.4 10    n  1   3.34 104
 0.15 
1484.8n
7424   742.4n  742.4  3.34 10 4
0.15
1113.6 + 1484.8n – 111.36n – 111.36 = 0.15(3.34 x 104)
= 5010
1373.44 + 1002.24 = 5010

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4007.76
n  2.91  3
1373.44
Result :
Number of aluminium sheets = 3
2. Define the following terms:
(1) Monochromatic emissivity
(2) Gray body (3) Shape factor
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.
E
Emissivity,  
Eb
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.
The shape factor is defined as ― The fraction of the radiative energy that is diffused from one surface element
and strikes the other surface directly with no intervening reflection‖. It is represented by F ij. Other names for
radiation shape factor are view factor, angle factor and configuration factor. The shape factor is used in the
analysis of radiative heat exchange between two surfaces.
3) The spectral emissivity function of an opaque surface at 1000 K is approximated as
𝛆𝛌1 = 0.4, 0  𝛌 < 2 m;
𝛆𝛌2 = 0.7, 2 m  𝛌 < 6 m;
𝛆𝛌3 = 0.3, 6 m  𝛌 < ∞
Determine thee average emissivity of the surface and the rate of radiation emission from the surface, in
W/m2.
Given data :
ε1 = 0.4: ε2 = 0.7; ε3 = 0.3
T = 1000K
λ1 = 2 m
λ2 = 6 μm
To find :
(i) Average emissivity of the surface.
(ii) Rate of radiation emission from the surface, in W/m2
Solution:
(i) Average emissivity of the surface,

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E b  0  1T  E b  1   2  Eb  2   
  1  2  3 .... 1
T 4
T 4
T 4
1T = 2 x 1000 = 2000 mK
2 T = 6 x 1000 = 6000 mK
1 T = 2000 m K, corresponding Fractional Emission.
From HMT data book, page. No.83 (seventh edition)
E b  0  1T 
F1   0.066728
T 4
2T = 6000 m K, corresponding Fractional Emission.
From HMT data book, page No.83 (seventh edition)
Eb  0  2T 
F 2   0.737818
T 4
E b  1   2 
i.e,.F 2  F1   0.737818  0.066728
T 4
 0.67109
Eb  2   
 F  F 2  1  F 2 F  1
T 4
1  0.73781
 0.26219
Equation (1) 
  1  F1   2  F 2  F1   3 1  F 2 
  0.4  0.066728   0.7  0.737818  0.066728  0.3 1  0.73781

 0.02669  0.46976  0.07865
  0.575
(ii) Rate of emission =   T 4
= 0.575  5.67 108  1000

4
= 32.6 x 103 W/m2
Result:
(i) Average emissivity of the surface,   0.575
(ii) Rate of emission = 32.6 KW / m2
4) Emissivities of two large parallel plate maintained at 800°C and 300°C are 0.3 and 0.5 respectively.
Find net radiant heat exchange per square metre for these plates. Find the percentage reduction in heat

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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
ε1 = 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 transfer due to radiation shield.
3. Temperature of the shield (T 3).
Solution: Heat exchange between two large parallel plates without radiation shield is given by
Q12    A T14  T24 
[From equation no.(4.28)]
1 1
Where,   
1 1 1 1
 1  1
1  2 0.3 0.5
  0.230

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Q12  0.230   A T14  T24 
 0.230  5.67  108  A  1073   573 

4 4

 
Q12
 15,880.7 W m 2  15.88 kW m 2
A
Heat transfer per square metre without radiation shield
Q12
 15.88kW m ... 1
2
Heat exchange between plate 1 and radiation shield 3 is given by
 Q13    A T14  T34 
1

1 1
 1
1 3
Where,
  A T14  T34 
 Q13  .....  A 
1 1

1 3  1
Heat exchange between radiation shield 3 and plate 2 is given by
Q32    A T34  T24 
Where,
1

1 1
 1
3  2
  A T34  T24 
 Q32  .....  B 
1 1

3  2  1
We know that, Q13 = Q32
A T14  T34  A T34  T24 

 
1 1 1 1
 1  1
1 3 3  2
T14  T34  T34  T24 
 
1 1 1 1
 1  1
0.3 0.06 0.06 0.5
1073   T3  T34   573

4 4 4
 
19 17.6
17.6 1073   T3  
4 4
   573 4
T  4
3  
19

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 T34  0.926 1073   T3     573

4 4 4
 
 T34  0.926  1073  0.926   T3    573
4 4 4
 T3   0.926  T3   1.33 1012

4 4
1.926  T3   1.33 1012

4
 T3   6.90 1011
4
T3  911.5K
Radiation shield temperature,T3  911.5K
Substituting T3 value in equation (A) (or) equation (B),
Heat transfer with radiation shield
5.67 108  A  1073   911.5  

4 4
 Q13   
1 1
 1
0.3 0.06
Q13
 1895.76 W m 2
A
Heat transfer with radiation shield
Q13
  1.89 kW m2 ....  2 
A
Reduction in heat transfer due to radiation shield
Q without shield  Q with shield Q12  Q13

 
Q without shield Q12
15.88  1.89

15.88
 0.88  88%
Result:
1. Heat exchanger per square meter without radiation shield
Q12 = 15.88 kW/m2
2. Percentage reduction in heat transfer = 88%
3. Temperature of radiation shield T 3 = 911.5 K
5) State and prove Kirchhoff’s law of thermal radiation.
KIRCHOFF’S LAW OF RADIATION
The law states that the ratio of total emissive power to the absorptivity is constant for all surfaces which are in
thermal equilibrium with the surroundings. This can be written as

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E1 E 2 E3
  .........
1  2 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.
6) What is a black body? A 20 cm diameter spherical bull at 527° C is suspended in the air. The ball
closely approximates a black body. Determine the total black body emissive power, and spectral black
body emissive power at a wavelength of 3 m
Black body : Refer Page no.4.3, section 4.6.
Given data: In sphere, (Black body)
Diameter of sphere, d = 20cm = 0.2 m
Temperature of spherical ball, T = 527°C + 273 = 800 K
To find:
(i) Total black body emissive power, Eb
(ii) Spectral black body emissive power at a wavelength 3 m
Solution:
(i) Total black body emissive power, Eb
Eb =  A T4 = 5.67 x 10-8 x  x (0.2)2 x (800)4
= 23224.32 x 0.12573
E b  2920 W
(ii) Spectral black body emissive power: @  = 3 m
C1
E b 
 C  
 5 exp  2   1
  T  
0.374 1015

5  103  
 3 106  exp  314.14
6   1
   10  800  
Eb = 3824.3 x 106 W/m2 (or) 3824.3 W/m2 m
7) An oven is approximated as a long equilateral triangular which has a heated surface maintained at a
temperature of 1200 K. The other surface is insulated while third surface is at 500 K. The duct has a
width of a 1 m a side and the heated and insulated surfaces have an emissivity of 0.8. The emissivity of the
third surface is 0.4. Steady state operation find the rate at which energy must applied to the heated side
per unit length of the duct to maintain its temperature at 1200 K. What is the temperature the insulated
surface?
Solution: A1 = A2 = AR

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From electrical network diagram,
1  1 1  0.8
  0.25
1A1 0.8 1
1   R 1  0.8
  0.25
 R A R 0.8 1
1   2 1  0.4
  1.5
 2 A 2 0.4 1
We know that, F11 + F12 + F1R = 1
From symmetric of three surface enclosure, from one surface shares two equal radiations into two different
surfaces.
i.e., F12 = 0.5
But F11 = 0
0 + 0.5 + F1R = 1
F1R  0.5
Similarly, F21 + F22 + F2R = 1
0.5 + 0 + F2R = 1
F2R  0.5
1 1
 2
A1F12 0.5
1 1
 2
A1F1R 0.5
1 1
 2
A 2 F2R 0.5
From Stefan Boltzmann law,
Eb =  T 4
Eb1 =  T14 = 5.67 x 10-8 x (1200)4 = 117573 W/m2

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Eb2 =  T24 = 5.67 x 10-8 x (500)4 = 3544 W/m2
EbR = JR=  TR4
The radiosities J1 and J2 can be calculated by using Kirchoff‘s law.
At node J1:
E b1  J1 J1  J 2 J1  J R
 
1  1 1 1
1A1 A F
1 12 A 1F1R
117573  J1 J1  J 2 J1  J R
 
0.25 2 2
10J1 – J2 – JR = 940584 ...(1)
At node J2 :
E b2  J 2 J 2  J1 J 2  J R
 
1  2 1 1
2 A 2 A 2 F21 A 2 F2R
3544  J 2 J 2  J1 J 2  J R
 
1.5 2 2
-J1 + 3.33 J2 – JR = 4725 .......(2)
At node JR :
E bR  J R J R  J1 J R  J 2
 
1  3 1 1
3 A R A R FR1 A R F2R
J R  J1 J R  J 2
0  E bR  J R 
2 2
-J1 -J2 +2JR = 0 .......(3)
By solving equations (1), (2) and (3), we get
J1 = 1,08,339 W/m2
J2 = 59,093 W/m2
JR = 83,716.33 W/m2
We know that,
EbR = JR =  TR4
1
 83716.33  4
TR   8 
 5.67  10 
TR  1102.31K
8) Derive wien’s displacement law of radiation from planck’s law. (8)
Solution:

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We know that, planck‘s distribution law.
c1 5
 E  b 
c 
exp  2   1
 T 
(E)b becomes maximum (if T remains constant)
d  E  b
0
d
 
When
d  E  b d  c1 5 
   0
d d   c2  
 exp  T   1 
   
  c2   5   c2  c2  1  
exp  T   1  5c1   c1 ex[ T  T   2  
6
        
 2
0
  c2  
exp  T   1
   
c  1 c 
 5c1 6 exp  2   5c1 6  c1c2  5 2 exp  2   0
 T  T  T 
Dividing both sides by 5 c1 -6, we get
c  1 1 c 
 exp  2   1  c2  exp  2   0
 T  5 T  T 
Solving this equation by trial and error method, we get
c2 c
 2  4.965
T  max T
c2
  max T 
4.965
1.439  104
 mK
4.965
= 2898 mK
  max T  2.9 103 mK
9) A black body at 3000 K emits radiation.
Calculate the following :
1. Monochromatic emissive power at 1 m wave length.
2. Wave length at which emission is maximum.
3. Maximum emissive power.
4. Total emissive power,

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5. 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 = 3000 K
To find :
1. Monochromatic emissive power Eb at
 = 1  = 1 x 10-6 m.
2. Maximum wave length, (max).
 Maximum emissive power. (Eb)max
4. Total emissive power, Eb.
5. Emissive power of real surface at  = 0.85.
Solution:
1. Monochromatic Emissive Power:
From Planck‘s distribution law, we know that,
c1 5
E b 
 c2 
e 
 T  1
[From HMT data book, page no.81]
Where c1 = 0.374 x 10-15 W m2
c2 = 14.4 x 10-3 mK
 = 1 x 10-6 m [Given]
5
0.374 1015 1106 
 E b 
 14.4 103 
e 1106  3000 
  1
Eb  3.10 1012 W m2
2. Maximum wave length, (max):
From wien‘s law, we know that,
max T = 2.9 x 10-3 mK
2.9  103
  max 
3000
 max  0.966  106 m

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3. Maximum emissive power (Eb)max :
Maximum emissive power
(Eb)max = 1.307 x 105 I5
= 1.307 x 105 x (3000)5
(Ebl )max  3.17 x 1012 W / m2
4. Total emissive power (Eb) :
From Stefan-Boltzmann law, we know that
Eb =  T 4
[From HMT data book, page no 81]
Where  = Stefan-Boltzman constant
= 5.67 x 10-8 W/m2 K4
 Eb = (5.67 x 10-8) x (3000)4
E b  4.59 106 W m2
5. Total emissive power of a real surface:
(Eb)real =   T4
Where  - Emissivity = 0.85
(Eb)real = 0.85 x 5.67 x 10-8 x (3000)4
 Eb real  3.90 106 W m2
Result:
1. Eb = 3.10 x 1012 W/m2
2. max = 0.966 x 10-6 m
3. (Eb)max = 3.17 x 1012 W/m2
4. Eb = 4.59 x 106 W/m2
5. (Eb)real = 3.90 x 106 W/m2
10)Two parallel plates of size 1 m x 1 m are spaced 0.5 m apart are located in a very large room, the walls
of which are maintained at a temperature of 27°C. One plate is maintained at a temperature of 900°C and
the other at 400°C. Their emissivities are 0.2 and 0.5 respectively. If the plates exchange heat between
themselves and to the room. Consider only the plate surfaces facing each other.
Solution:

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Size of the plates =1m1m
Distance between plates = 0.5 m
Room temperature, T3 = 27°C + 273 = 300 K
First plate temperature, T1 = 900C + 273 = 1173 K
Second plate temperature, T2 = 400C + 273 = 673 K
Emissivity of first plate, 1 = 0.2
Emissivity of second plate, 2 = 0.5
To find: 1. Net heat transfer to each side plate.
2. Net heat transfer to 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.
Fig. 4.62. Electrical network diagram
Area, A1 = 1 x 1 = 1 m2
 A1  A2  1m2
Since the room is large, A3 = 
From electrical network diagram.
1  1 1  0.2
 4
A11 1 0.2
1   2 1  0.5
 1
A 2  2 1 0.5
1  3
0  A3  
A 3 3
1  1 1  2 1  3
Apply  4,  1,  0 valuesin electrical
A11 A2 2 A 3 3

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Fig.4.63 Electrical network diagram
To find shape factor F12, refer HMT data book page no.91 & 92 (sixth edition).
Fig.4.64
L 1
X  2
D 0.5
B 1
Y  2
D 0.5
X value is 2, Y value is 2. From that, we can find corresponding shape factor value is 0.41525.
i.e., F12 = 0.41525
we know that,
F11 + F12 + F13 = 1
But F11 = 0
 F13 = 1 – F12
 F13 = 1 – 0.41525
F13  0.5847
Similarly, F21 + F22 + F23 = 1
We know that, F22 = 0
 F23 = 1 – F21
= 1 – F12
= 1 – 0.41525
F23  0.5847

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1 1
  1.7102
A1F13 1 0.5847
1 1
  1.7102
A 2 F23 1 0.5847
1 1
  2.408
A1F12 1 0.41525
From Stefan-Boltzman law,
Eb =  T 4
Eb1 =  T14
= 5.67 x 10-8[1173]4
Eb1  107.34 103 W m2
= 5.67 x 10-8[300]4
Eb3 = 459.27 W/m2
From electrical network diagram, we know that,
Eb3  J3  459.27 W m2
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 :
E b1  J1 J 2  J1 E b3  J1
  0
4 1 1
A1F12 A1F13
[From electrical network diagram]
107.34  103  J1 J2  J1 459.27  J1

   0
4 2.408 1.7102
J1 J J J1
 26835   2  1  268.54  0
4 2.408 2.408 1.7102
At Node J2:
J1  J 2 E b3  J 2 E b2  J 2
  0
1 1 2
A1F12 A 2 F23
J1  J 2 459.27  J 2 11.63 103  J 2

   0
2.408 1.7102 2

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J1 J J2 11.63 103 J 2
  2  268.54    0
2.408 2.408 1.7102 2 2
0.415J1  0.415J 2  268.54  0.5847J 2  
 0
5.815 103  0.5J 2 

 0.415J1 – 1.499J2 + 6.08103 = 0
 0.415J1 – 1.4997J2 = -6.08  103 ....(2)
Solving equation (1) and (2),
1.2497J1  0.415J 2  27.10 103

0.415J1 1.4997J 2  6.08 103
By solving, J 2  11.06 103 W m 2
J1  25.35 103 W m 2
E b1  J1
Heat lost by plate (1), Q1 
1  1
A11
[From electrical network diagram]
107.34 103  25.35 103


1  0.2
1 0.2
Q1  20.49 103 W
E b2  J 2
Heat lost by plate (1), Q1 
1  2
A2 2
11.63  103  11.06  103


1  0.5
1 0.5
Q 2  570W
Total heat lost by the 

 Q  Q1  Q 2
plates (1) and (2) 
 20.49 103  570
Q  21.06 103 W
Total heat received or  J1  J 3 J 2  J3

Q  1  1
absorbed by the room 
A1F13 A 2 F23

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25.35  103  459.27 11.06 103  459.27

 
1.7102 1.7102
E b3  J 3  459.27 W m 2 
Q  20.752 103 W
[Note: Heat lost by the plates is equal to heat received by the room.]
Result: 1. Net heat lost by each plates
Q1 = 20.49 x 103 W
Q2 = 570 W
2. Net heat transfer to the room
Q = 20.752 x 103 W
11) Distinguish between irradiation and radiosity. (4)
It is defined as the total radiation incident upon a surface per unit time per unit area. It is expressed in W/m 2.
It is used to indicate the total radiation leaving a surface per unit time per unit area. It is expressed in W/m 2.
11 ) ii Consider a cylindrical furnace with outer radius = height = 1m. The top (surface 1) and the base
(surface 2) of the furnace have emissivities 0.8 & 0.4 and are maintained at uniform temperatures of 700
K and 500 K respectively. The side surface closely approximates a black body and is maintained at a
temperature of 400 K. Find the net rate of radiation heat transfer at each surface during steady state
operation. Assume the view factor from the base to the top surface as 0.38. (12)
The view factor F1-3 can be obtained from summation rule.
F1-1 + F1-2 + F1-3 = 1
F1-3 = 1 – 0 – 0.38 = 0.62
Radiation resistance,
1  1 1  0.8
R1    0.0796m 2
1A1 0.8  3.141
1  2 1  0.4
R2    0.4777m 2
 2 A 2 0.4  3.141
1 1
R1 2    0.8381m 2
A 2 F1 2 3.141 0.3
1 1
R13    0.5137m 2
A1F13 3.141 0.62

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Substituting the numerical values,
13614  J1 J 2  J1 1452  J1
   0 ... 1
0.0796 0.8881 0.5137
354  J 2 J1  J 2 1543  J1
   0 ....  2 
0.4777 0.5137 0.8381
Solving these equations
J1  11418W m2 J 2  4562 W m2
Net rate of radiation heat transfer for top surface (1) and base surface (2).
Eb1  J1 13614  11418

Q1    Q1  27588 W
R1 0.0796
Eb2  J 2 3544  4562

Q2    Q2  2131W
R2 0.4777
Net rate of heat transfer at cylindrical surface
J 3  J1 J 3  J 2
Q3  
R1  3 R 2  3
1452  11418 1452  4562
   Q3  25455 W
0.5137 0.5137
12) Considering radiation in gases, derive the exponential decay formula.
Exponential – Decay formula:
Consider absorption of thermal radiation by a gas layer.
I0 is radiation intensity at the left face and Ix propagates in a gas layer is proportional to the thickness dx.
dI(x) = -m I (x) dx
where the proportionality constant m is spectral absorption coefficient of gas. Integrating both sides
I L
dI  x  L
I 

I 0
I  x 
 m  dx or ln  l   m  L
0  I 0 
or IL  I 0 exp  m L 

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The radiation intensity IL decreases exponentially with thickness of gas layer.
13) Two very large parallel plates with emissivities 0.5 exchange heat. Determine the percentage
reduction in the heat transfer rate if a polished aluminium radiation shield of  = 0.04 is placed in
between the plates.
Given:
Emissivity of plate 1, 1 = 0.5
Emissivity of plate 2, 2 = 0.5
Emissivity of radiation shield, 3 = 0.04 = s
Fig.4.23.
To find: Percentage of reduction in heat transfer due to radiation shield.
Solution:
Case 1 : Heat transfer without radiation shield:
Heat exchange between two large parallel plates without radiation shield is given by,
Q12 =   A T14  T24 
1
Where,  
1 1
 1
1  2
1

1 1
 1
0.5 0.5
  0.333
Q12  0.333  A T14  T24 

Qwithout shield  0.333  A T14  T24  ..... 1
Case 2 : Heat transfer with radiation shield :
We know that,

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Heat transfer with n shield,
A  T14  T24 
Q with shield 
1 1 2n
    n  1
1  2 s
Where, s – Emissivity of radiation shield.
n - Number of radiation shield.
A  T14  T24 
Q with shield 
1 1 2 1
   1  1
0.5 0.5 0.04
A  T14  T24 
 
52
Q with shield  0.0192 A  T14  T24  ....  2 
We know that,
Re duction in heat transfer  Qwithout shield  Qwith shield


due to radiation shield  Qwithout shield
0.333A  T14  T24   0.0192 A  T14  T24 

=
0.333A  T14  T24 
0.333  0.0192
=
0.333
= 0.942 = 94.2%
Result : Percentage of reduction in heat transfer rate
14. 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

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 T14  T2 4 
where Q12  [From equation No.(6)]
1  1 1 1 2
 
A11 A1F12 A1 2
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 F 12, refer HMT data book, Page No.76.
Smaller side
X axis =
Distance between planes
2
=
0.5
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
15. 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 =3m2m
Distance between plates =1m
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

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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 takes place between two plates and the room. So this is three surface
problems and the corresponding radiation network is given below. Area A1 = 3  2 = 6 m2
A1  A 2  6m2
Since the room is large A3  
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 [  A 3  ]
3 A3
1 3 1-1 1 2
Apply  0,  0.309,  0.136 values in electrical network diagram.
3 A3 1A1 2A2
To find shape factor F12 refer HMT data book, Page No.78.
b 3
X  3
c 1
a 2
Y   2
c 1
X value is 3, Y value is 2, corresponding shape factor [From table]
F12 = 0.47
F12  0.47
We know that,
F11 + F12 + F13 = 1 But, F11 = 0
 F13  1  F12
 F13  1  0.47
F13  0.53

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Similarly, F21 + F22 + F23 = 1 We know F22 = 0
 F23  1  F21
 F23  1  F12
F13 = 1 - 0.47
F23  0.53
1 1
  0.314 ....(1)
A1F13 6  0.53
1 1
  0.314 ....(2)
A 2F23 6  0.53
1 1
  0.354 ....(3)
A1F12 6  0.47
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)
[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:

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Eb1  J1 J2  J1 Eb3  J1
  0
0.309 1 1
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 510.25 J2 4.24  103 J2
     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
Eb1  J1
Q1 
 1  1 
 
 1A1 
26.01 103  10.73  103

Q1 
1  0.35
0.35  6
Q1  49.36  103 W

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Heat lost by plate 2 is given by
Eb2  J2
Q2 
 1 2 
 
 2A2 
4.24  103  4.73  103

Q2 
1  0.55
6  0.55
Q2  3.59  103 W
Total heat lost by the plates
Q = Q 1 + Q2
= 49.36  103 – 3.59  103
Q  45.76  103 W ......(9)
Heat received by the room
J1  J3 J2  J3
Q 
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.
16. 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 = 2 atm
Temperature T = 927C + 273

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= 1200 K
Mean beam length Lm = 0.3 m
To find: Emissivity of mixture (mix).
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 CO 2.
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 CO 2
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

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H o  0.048
2
To find correction factor for H2o :
PH2o  P 0.1  2
  1.05
2 2
PH2o  P
 1.05,
2
PH2o Lm  0.03 m - atm
From HMT data book, Page No.92 we can find emission of H 20
17. 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
 T14  T2 4 
where Q12 
1  1 1 1 2
 
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 F 12, refer HMT data book, Page No.76.

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Smaller side
X axis =
Distance between planes
2
=
0.5
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
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 0.1
  1.05
PH2o  PCO2 0.1  0.2
PH2o
 0.333
PH2o  PCO2
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 CO 2 and H2o.

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20.

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UNIT 5 MASS TRANSFER
1. What is mass transfer?
Ans: The process of transfer of mass as a result of the species concentration difference in a mixture is known as
mass transfer.
2. Give the examples of mass transfer.
Ans: Some examples of mass transfer.
1) Humidification of air in cooling tower
2) Evaporation of petrol in the caburetor of an IC engine
3) The transfer of water vapour into dry air.
3. What are the modes of mass transfer?
Ans: There are basically two modes of mass transfer.
1) Diffusion mass transfer
2) Convective mass transfer.
4. What is molecular diffusion?
Ans: The transport of water on a microscopic level as a result of diffusion from a region of higher
concentration to a region of lower concentration in a mixture of liquids or gases is known as molecular
diffusion.
5. What is Eddy diffusion?
Ans: When one of the diffusion fluids is in turbulent motion, eddy, diffusion takes place.
6. What is convective mass transfer?
Ans : convective mass transfer is a process of mass transfer that will occur between surface and a fluid medium
when they are at different concentration.
7. State Fick's law of diffusion.
The diffusion rate is given by the Fick's law. which states that molar flux of an element per unit area is directly
proportional to concentration gradient..
ma dCa
 Dab
A dx
ma kg  mole
where,  Molar flux,
A s  m2
Dab Diffusion co-efficient of species a and b, m2/s
dCa
- concentration gradient, kg/m3.
dx
8. What is free convective mass transfer?

Ans: If the fluid motion is produced due to change in density resulting from concentration gradients, the mode
of mass transfer is said to be free or natural convection mass transfer. Example: Evaporation of alcohol.

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9. Define forced convective mass transfer.
Ans: If the fluid motion is artificially created by means of an external force like a blower or fan, that type of
mass transfer is known as convective mass transfer. Example: The evaluation if water from an ocean w hen air
blows over it.
10. Define Schmidt Number.
Ans: It is defined as the ratio of the molecular diffusivity of momentum to the molecular diffusivity of mass.
Molecular diffusivity of momentum

S
Molecular diffusivity of mass
11. Define Scherwood Number.
Ans : It is defined as the ratio of concentration gradients at the boundary
hmx
Sc 
Dab
hm – Mass transfer co – efficient, m/s
Dab – Diffusion co – efficient , m2/s
x - Length, m
12. Give two example convective mass transfer.
Ans:
1) Evaporation of alcohol
2) Evaporation of water from an ocean when air blows over it.
13. Define mass concentration or mass density.
Ans: Mass of a component per unit volume of the mixture lt is expressed in Kg/m3.
14. Define molar concentration or molar density.
Ans: Number of molecules of a component per unit volume of the mixture, it is expressed in Kg mole/m3.
15. Define mass fraction.
Ans: The mass fraction is defined as the ratio of mass concentration of species to the total mass density of the
mixture.
16.. Define mode fraction.
Ans: The mode fraction is defined as the ratio of mole concentration of species to the total molar concentration.

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PART B
1.A 3 –cm diameter Stefan tube is used to measure the binary diffusion coefficient of water vapour in air
at 20C at an elevation of 1600 m where the atmospheric pressure is 83.5 kPa. The tube is partially filled
with water, and the distance from the water surface to the open end of the tube is 40 cm. Dry air blown
over the open end of the tube so that water vapour rising to the top is removed immediately and the
concentration of vapour at the top of the tube is zero. In 15 days of continuous operation at constant
pressure and temperature, the amount of water that has evaporated is measured to be 1.23 g. Determine
the diffusion coefficient of water vapour in air at 20C and 83.5 kPa. (10)
Given data:
Diameter d = 3 cm = 0.03 m
Deep, (x2 – x1) = 40 cm = 0.4 m
Temperature, T = 20C + 273 = 293 K
Atmospheric pressure, P = 83.5 KPa
Dry saturated = 83.5 x 103 N/m2
ma = 1.23 g.
To find :
Diffusion coefficient of water vapour, Dab,
Solution :
We know that, for isothermal evaporation.

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md Dab P  P  Pw 2 
Molar flux,  .ln   ..... 1
A GT  x 2  x1   P  Pw1 
 2 
Area. A  d    9 104 
. 4 4
A  7.068 104 m 2
ma = 1.23  10-3 kg/15 days.
1.23 103
ma  kg sec
15  24  3600
ma = 9.49  10-10 kg/sec
ma
Molar flux, N ax 
M.wt of vapour  Area
9.49 1010

18  7.068 104
= 0.0745  10-6 = 7.45  10-8 Kmol/m2
Where,
J
G – Universal gas constant = 8314
Kg  mole  K
Pw1 – Partial pressure at the bottom of the (water vapour) Stefan tube corresponding to saturation temperature at
20C.
At 20C,
Pw1 = 0.0234 bar (From steam table, page, No.2)
Pw1 = 2.34 kPa
Partial pressure at the top of the Stefan tube.
Pw2 = 0
ma Dab P  83.5  0 
 Nax   ln  
A GT  x1  x 2   83.5  2.34 
Dab = 3.06  10-5 m2/sec
Result:
Diffusion coefficient of water vapour.
Dab = 3.06  10-5 m2/sec
2.State some analogies between heat and mass transfer.
S.No Parameter Heat transfer Mass transfer

1. Driving force Temperature gradient. Concentration gradient.

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2. Proportionality Thermal conductivity Diffusion coefficient

constant. (Fourier‘s Law). (Fick‘s Law).
3. Modes Conduction, convection Conduction (Diffusion)

and Radiation and convection only.
4. Internal heat .Heat generation takes Species generation takes

generation place in systems. place.
5. Convection Heat transfer coefficient Mass transfer coefficient

coefficients. (h). (hm)
6. Dimensionless Depends upon Nusselt Depends upon Sherwood

numbers. number and parallel number and Schmidt
number. number.
3.A thin plastic membrane separate hydrogen from air. The molar concentrations of hydrogen in the
membrane at the inner and outer surfaces are determined to be 0.045 and 0.002 kmol/m 3 , respectively.
The binary diffusion coefficient of hydrogen in plastic at the operation temperature is 5.3  10-10 m2/s.
Determine the mass flow rate of hydrogen by diffusion through the membrane under steady conditions if
the thickness of the membrane is (8)
(1) 2 mm and
(2) 0.5 mm.
Given data :
Case (i)
Thickness of the membrane, I = 2 mm = 0.002 m
Concentration at inner side, Ca1 = 0.045 kmol / m3
Concentration at outer side, Ca2 = 0.002 kmol / m3
Diffusion coefficient of hydrogen in plastic,
Cab = 5.3  10-10 m2 / sec
Case (ii)
Thickness of the membrane, L = 0.5 mm = 0.0005 m
To find :

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(1) The mass flow rate of hydrogen by diffusion through the membrane. (ma)
Case (i) L = 0.002 m
Case (ii) L = 0.0005 m
Solution:
Case(i) L =0.002 m
We know that, for plane membrane
ma D  C  Ca 2 
Molar flux,  N ax  ab a1
A L
5.3 1010   0.045  0.002 

0.002
Nax = 1.139  10-8 kmol/m2.sec
Mass flow rate of hydrogen, m = Nax  (Mol.wt of H2)
= 1.139  10-8  2
m = 2.278  10-8 kg/m2.sec
Case (ii) L = 0.5 mm = 0.0005 m
We know that, for plane membrane.
ma D  C  Ca 2 
Molar flux,  N ax  ab a1
A L
5.3 1010   0.045  0.002 

0.0005
Nax = 4.558  10-8 kmol/m2.sec
Mass flow rate of hydrogen, m = Nax  (Mol.wt of H2)
= 4.558  10-8  2
= 9.116  10-8 (or)
m = 0.9116 10-7
Result : Mass flow rate of hydrogen by diffusion through membrane.
Case (i) L = 0.002 m ; m = 2.278  10-8 kg/m2 sec
Case (ii) L = 0.0005 m; m = 0.9116  10-7 kg/m2 sec.
4.Dry air at 15C and 92 kPa flows over a 2 m long wet surface with a free stream velocity of 4 m/s.
Determine the average mass transfer coefficient. (8)
Given data:
Fluid temperature, T = 15C

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Velocity, U = 4m/sec
Length, X = 2m
Air flow pressure, P2 = 92 kPa
Atmospheric pressure, P1 = 100 kPa
To find :
Average mass transfer coefficient, hm
Solution :
Properties of air at 15C.
[From HMT data book, page No.34 ] (seventh edition)
Kinematic viscosity, v1 = 1.47  10-5 m2/sec
We know that
P1 v 2

P2 v1
100 v2

92 1.47  105
V2 = 2.57  10-5 m2/s
Ux 1.47 105
Reynolds number, Re    544217.6
v 2.57 105
v 1.47 105
Schmidt number, Sc    0.5719
Dab 2.57 105
Sherwood number, sh = 0.664 Re0.5 Sc0.333
= 0.664 (544217.6)0.5 (0.5719)0.333
= 0.664 (737.71) (0.8302)
Sh = 406.6
h m .L sh.Dab
Sherwood number, sh   hm 
Dab L
406.6  2.57 105

hm   0.00522 m sec .
2
Result :
Average mass transfer coefficient = 0.00522 m/sec.
5.Explain equimolal counter diffusion in gases.
STEADY STATE EQUIMOLAR COUNTER DIFFUSION

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Consider two large chambers a and b connected by a passage as shown in Fig.5.3
Na and Nb are the steady state molar diffusion rates of components a and b respectively.
Fig.5.3
Equimolar diffusion is defined as each molecules o, ‗a‘ is replaced by each molecule of ‗b‘ and vice versa.
The total pressure P = Pa – Po is uniform throughout the system.
P = Pa + Pb
Differentiating with respect to x
dp dpa dp b
 
dx dx dx
Since the total pressure of the system remains constant under steady state conditions.
dp dpa dp b
  0
dx dx dx
dpa dp
  b
dx dx
Under steady state conditions, the total molar flux is zero
 Na + N b = 0
Na = -Nb
A dpa A dp b
 Dab   Dba ....  5.5 
GT dx G1 dx
From Fick‘s law,
A dpa
N a  Dab
GT dx
A dp b
N b  D ba
GT dx
We know,
dpb dp
 a [From equations5.4]
dx dx
Substitute in equation (5.5)
A dp a A dpa
(5.5)  Dab   Dba
GT dx GT dx

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 Dab  Dba  D
6.Discuss briefly the Analog between heat and mass transfer.
In a system consisting of two or more components whose concentrations vary from point to point, there is a
natural tendency for species (particles) to be transferred from a region of higher concentration side (higher
density side) to a region of lower concentration side (lower density side).
This process of transfer of mass as a result of the species concentration difference in a mixture is known as
mass transfer.
Some examples of mass transfer are
1. Humidification of air in cooling tower.
2. Evaporation of petrol in the carburetter of an IC engine
3. The transfer of water vapour into dry air.
4. Dissolution of sugar added to a cup of coffee.
7.Define mass transfer coefficient. Air at 1 bar pressure and 25C containing small quantities of iodine
flows with a velocity of 5.2 m/s. Inside a tube having an inner diameter of 3.05 cm. Find the mass transfer
coefficient for iodine transfer from the gas stream to the wall surface. If c m is the mean concentration of
iodine in kg.mol/m3 in the air stream. Find the rate of deposition on the tube surface by assuming the wall
surface is a perfect sink for iodine deposition. Assume D = 0.0834 cm2/s.
Solution:
Mass transfer coefficient :
The mass transfer coefficient is a diffusion rate constant that relates the mass transfer rate, mass transfer area
and concentration gradient as driving force. (Unit is m/s).

mconv
hm 
A  A,s  A, 
Where, hm = Mass transfer coefficient;
A = Area

m = Mass transfer rate
A,s -  Concentration gradient
Mass transfer coefficient determines the rate of mass transfer across a medium in response to a concentration
gradient.
Given data:
Velocity, u = 5.2 m/s
d = 3.05 cm = 0.0305 m

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Diffusion coefficient, Dab = 0.0834 cm2/s
= 0.0834  10-4 m2/s
Dab = 8.34  10-6 m2/s
To find:
(i) Mass transfer coefficient for iodine transfer, h m
(ii) Rate of deposition of tube, N = hm(Cm – CA,)
8.Air at 25C flows over a tray full of water with a velocity of 2.8 m/s. The tray measures 30 cm along the
flow direction and 40 cm wide. The partial pressure of water present in the air is 0.007 bar. The partial
pressure of water present in the air is 0.007 bar. Calculate the evaporation rate of water if the
temperature on the water surface is 15C. Take diffusion co-efficient is 4.2  10-5 m2/s.
Given :
Speed, U = 2.8 m/s
Flow direction is 30 cm side. So, x = 30 cm = 0.30 m
Area, A = 30 cm  40 cm = 0.30  0.40 m2
Partial pressure of water, pw2 = 0.007 bar
pw 2  0.007 105 N m2
Water surface temperature, Tw = 15C
Diffusion co-efficient, Dab = 4.2  10-5 N/m2
To find:
Evaporation rate of water,(mw)
Solution:
We know that,
Tw  T 15  25
Tf  
Film temperature, 2 2
Tf  20C
Properties of air at 20C
Kinematic viscosity, v = 15.06  10-6 m2/s
We know that,

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Ux
Reynolds Number, Re =
v
2.8  0.30
=
15.06  106
Re = 0.557  105 < 5  105
Since, Re < 5  105 , flow is laminar.
For flat plate, Laminar flow :
Sherwood Number, (Sh) = [0.664 (Re) 0.5 (Sc)0.333] .....(1)
Where,
v
Sc – Schmidt Number =
D ab
15.06 106
Sc 
4.2 105
Sc  0.358
Substitute Sc, Re values in equation (1)
(1)  Sh = [0.664 (0.557  105)0.5 (0.358)0.333]
Sh  111.37 We know that,
hm x
Sherwood Number,Sh 
Dab
h m  0.30
111.37 
4.2  105
Mass transfer co  efficient, hm  0.0155 m s
Mass transfer co-efficient based on pressure difference is given by.
hm 0.0155
h mp  
RTw 287  288
Tw  15C  273  288 K, R  287 J kg K 
7
h mp  1.88 10 m s
Saturation pressure of water at 15C
Pw1 = 0.017 bar [From steam table (R.S. Khumi) page no.1]
pw1  0.017 105 N m2
The evaporation rate of water is given by,

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Mw = hmp  A[pw1 – pw2]
= 1.88  10-7  (0.30  0.40)  [0.017  105 – 0.007  105]
mw  2.25 105 kg s
Result:
Evaporation rate of water, mw = 2.25  10-5 kg/s
9.O2 and air experience equimolar counter diffusion in a circular tube whose length and diameter are 1.2
m and 60 mm respectively. The system is at a total pressure of 1 atm and a temperature of 273 K. The
ends of the tube are connected to large chambers. Partial pressure of CO2 at one end is 200 mm of Hg
while at the other end is 90 mm of Hg. Calculate the following
1. Mass transfer rate of Co2 and
2. Mass transfer rate of air
Given :
Diameter, d = 60 mm = 0.060 m
Length. (x2 – x1) = 1.2 m
Total pressure. P = 1 atm = 1 bar
Temperature, T = 273 K
Partial pressure of CO2 at one end
200
Pa1  200 mm of Hg  bar
760
 Pa1 = 0.263 bar [∵ 1 bar = 760 MM OF Hg]
 Pa1  0.263 105 N m2 1bar 105 N m2 
Partial pressure of CO2 at other end
90
Pa2 = 90 mm of Hg = bar
760
 Pa2 = 0.118 bar
 Pa 2  0.118 105 N m2

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To find :
1. Mass transfer rate of CO2
2. Mass transfer rate of air
Solution:
We know that, for equimolar counter diffusion
ma Dab  Pa1  Pa 2 
Molar flux.    .... 1
A GT  x 2  x1 
Where,
Dab = Diffusion co – efficient – m2/s
The diffusion co-efficient for CO2 – Air combination 11.89  10-6 m2/s
[from HMT data book page no.180 (sixth edition)
Dab  11.89 106 m2 s
J
g  universal gas constant  8314
kg  mole  k
 2
A  Area  d
4

 0.060
2
=
4
A  2.82 103 m2
ma 11.89 106  0.263 105  0.118 105 

1  3
  
2.82 10 8314  273  1.2 
kg  mole
Molar transfer rate of CO2 , ma  1.785 1010
s
We know,
Mass transfer rate of CO2 = Molar transfer  Molecular weight
= 1.785  10-10  44.01
[Molecular weight of CO2 = 44.01, refer HMT data, page no.182 (sixth editional)]
Mass transfer rateof CO2  7.85 109 kg s
We know,
kg  mole
Molar transfer rate of air, mb  1.785 1010
s
[∵ ma = - mb]

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Mass transfer rate of air = Molar transfer  Molecular weight of air
= -1.785  10-10  29
Mass transfer rateof air   5.176 109 kg s
Result :
1. Mass transfer rate of CO2 = 7.85  10-9 kg/s
2. Mass transfer rate of air = -5.176  10-9 kg/s
10.Air at 25C flows over a tray full of water with a velocity of 2.8 m/s. The tray measures 30 cm along
the flow direction and 40 cm wide. The partial pressure of water present in the air is 0.007 bar. The
partial pressure of water present in the air is 0.007 bar. Calculate the evaporation rate of water if the
temperature on the water surface is 15C. Take diffusion co-efficient is 4.2  10-5 m2/s.
Given :
Speed, U = 2.8 m/s
Flow direction is 30 cm side. So, x = 30 cm = 0.30 m
Area, A = 30 cm  40 cm = 0.30  0.40 m2
Partial pressure of water, pw2 = 0.007 bar
pw 2  0.007 105 N m2
Water surface temperature, Tw = 15C
Diffusion co-efficient, Dab = 4.2  10-5 N/m2
To find:
Evaporation rate of water,(mw)
Solution:
We know that,
Tw  T 15  25
Tf  
Film temperature, 2 2
Tf  20C
Properties of air at 20C
Kinematic viscosity, v = 15.06  10-6 m2/s
We know that,

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Ux
Reynolds Number, Re =
v
2.8  0.30
=
15.06  106
Re = 0.557  105 < 5  105
Since, Re < 5  105 , flow is laminar.
For flat plate, Laminar flow :
Sherwood Number, (Sh) = [0.664 (Re) 0.5 (Sc)0.333] .....(1)
Where,
v
Sc – Schmidt Number =
D ab
15.06 106
Sc 
4.2 105
Sc  0.358
Substitute Sc, Re values in equation (1)
(1)  Sh = [0.664 (0.557  105)0.5 (0.358)0.333]
Sh  111.37 We know that,
hm x
Sherwood Number,Sh 
Dab
h m  0.30
111.37 
4.2  105
Mass transfer co  efficient, hm  0.0155 m s
Mass transfer co-efficient based on pressure difference is given by.
hm 0.0155
h mp  
RTw 287  288
Tw  15C  273  288 K, R  287 J kg K 
7
h mp  1.88 10 m s
Saturation pressure of water at 15C
Pw1 = 0.017 bar [From steam table (R.S. Khumi) page no.1]
pw1  0.017 105 N m2
The evaporation rate of water is given by,

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Mw = hmp  A[pw1 – pw2]
= 1.88  10-7  (0.30  0.40)  [0.017  105 – 0.007  105]
mw  2.25 105 kg s
Result:
Evaporation rate of water, mw = 2.25  10-5 kg/s
Analogy between heat and mass transfer.
There is similarity among heat and mass transfer. The three basic equations dealing with these are
(i) Newtonian equation of momentum
(ii) Fourier law of heat transfer
(iii) Fick law of mass transfer
∴ The momentum, heat and mass transfer equation can be written as
u v
Continuity equation,  0
dx dy
u v 2u
Momentum Transfer, u v v 2
dx dy y
T T 2T
Heat transfer, u v  2 u
x y y
ca c 2c
Mass transfer, u  v n  D 2a
x y y
11.Evaporation process in the atmosphere.
Ans. Refer Page no.5.34, section 5.16.
12. ISOTHERMAL EVAPORATION OF WATER INTO AIR
Consider the isothermal evaporation of water from a water surface and its diffusion through the stagnant air
layer over it as shown in Fig.5.4. The free surface of the water is exposed to air in the tank.

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Fig.5.4
For the analysis of this type of mass diffusion, following assumptions are made,
1. The system is isothermal and total pressure remains constant.
2. System is in steady state condition.
3. There is slight air movement over the top of the tank to remove the water vapour which diffuses to
that point.
4. Both the air and water vapour behave as ideal gases.
From Fick‘s law of diffusion. We can find
ma Dab p  P  Pw 2 
Molar flux.  ln   ....  5.9 
A GT  x 2  x1   P  Pw1 
Where,
mA kg  mole
 Molar flux 
A s  m2
Dab – Diffusion coefficient – m2/s
J
G  universal gas constant  8314
kg  mole  k
T – Temperature – K
P – Total pressure in bar
Pw1 – Partial pressure of water vapour corresponding to saturation temperature at 1 in N/m 2
Pw2 – Partial pressure of dry air at 2 in N/m2
13.A mixture of O2 and N2 with their partial pressures in the ratio 0.21 to 0.79 is in a container at 25C.
Calculate the molar concentration, the mass density, and the mass fraction of each species for a total
pressure of 1 bar. What would be the average molecular weight of the mixture?
[Dec 2004 & 2005, Anna Univ]

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Given:
Partial pressure of O2, PO2 p = 0.21  Total pressure
= 0.21  1 bar
= 0.21  1  105 N/m2
Partial pressure of N2, PN 2 = 0.79  Total pressure
= 0.79  1 bar
= 0.79  1  105 N/m2
Temperature, T = 25 C + 273
= 298 K
To find :
1. Molar concentrations, Co2 , CN2
2. Mass densities, PO2 , PN2
 
3. Mass fractions, mO2 , m N2
4. Average molecular weight, M
Solution:
We know that,
P
Molar concentration, C 
GT
PO2
 CO2 
GT
0.211105

8314  298
[∵ Universal gas constant, G = 8314 J/Kg – mole – K]
CO2  8.476 103 kg  mole / m3
PN2
 C N2 
GT
0.79 1105

8314  298

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CN2  31.88 103 kg  mole m3
We know that,

Molar concentration, C = F
M
 =CM
O 2  c O 2  M O 2
 8.476 103  32

Molecular weight of O2 is 32
O2  0.271kg m3
N2  c N2  M N2
 31.88 103  28

Molecular weight of N 2 is 28
 N2  0.893kg m3
Overall density   O2   N2
= 0.271 + 0.893
  1.164 kg m3
Mass fractions :
 O 2 0.271
m O2  
 1.164

m O2  0.233
 N 2 0.893
m N2  
 1.164

m N2  0.767
Average Molecular weight
M  PO2 MO2  PN2 MN2
= 0.21  32 + 0.79  28
M = 28.84
Result:

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1. CO2  8.476 103 kg  mole m3

CN2  31.88 103 kg  mole m3
2. O2  0.271kg m3
N2  0.893kg m3

3. mO2  0.233

m N2  0.767
4. M = 28.84
14.Consider air inside a tube of surface area 0.5 m2 and wall thickness 10 mm. The pressure of air drops
from 2.2 bar to 2.18 bar in 6 days. The solubility of air in the rubber is 0.072 m3 of air per m3 rubber at 1
bar. Determine the diffusivity of air in rubber at the operating temperature of 300 K if the volume of air
in the tube is 0.028 m3.
Given:
A = 0.5 m2
L = 10 mm = 0.010 m
Pi = 2.2 bar = 2.2  105 N/m2
Pd = 2.18 bar = 2.18  105 N/m2
S = 0.072 m3
T = 300 K
V = 0.028 m3
To find : Diffusivity of air in rubber [D]
Solution: Initial mass of air in the tube,
Pi V 2.2 105  0.028

mi  
RT 287  300
mi  0.0715 kg
Final mass of air in the tube
Pd V 2.18 105  0.028

md  
RT 287  300
md  0.07089 kg
Man of air escaped = 0.0715 – 0.07089 = 0.00061 kg
The man flux of air escaped is given by

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ma Man of air escaped

Na  
A Time elapsed  Area
0.00061

 6  24  3600   0.5
= 2.35  10-9 kg/s – m2.
The solubility of air should be calculated at the mean operating pressure,
2.2  2.18
i.e  2.19 bar
2
The solubility of air i.e, volume at the mean inside pressure,
S = 0.072  2.19
= 0.1577 m3/m3 of rubber
The air which escapes to atmospheric will be at 1 bar pressure and its solubility will remain at 0.072 m 3 of air
per m3 of rubber.
The corresponding mass concentrations at the inner and outer surfaces of the tube, from characteristic
gas equation, are calculated as;
p1V1 2.19 105  0.1577

Ca1  
RT1 287  300
 0.4011kg m3
P2 V2 1105  0.072
Ca 2  
RT2 287  300
 0.0836 kg m
3
The diffusion flux rate of air through the rubber is given by
m a D  Ca1  Ca 2 
Na  
A  x 2  x1 
D  Ca1  Ca 2 

L
D 0.4011  0.0836
 2.35  10-9 =
0.01
 D = 0.74  10-10 m2/s
Result : The diffusivity of air in rubber
D = 0.74  10-10 m2/s
15. Derive an expression for mass flux in steady state molecular diffusion:
(a) A through non diffusing B
(b) Equimolar Counter Diffusion.

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Ans: (a) Steady state molecular diffusion, A through non-diffusion, A through non – diffusing B
N Ax  CDAB
 y
A 2  y A1 
L
Mass current
QmA 
DAB .AS
L

CA1  CA1 
(b) Steady state equimolar counter diffusion:
NAx   NBx
dy A
N Ax  CD AB
dx
dy B
N Bx  CD BA
dx
dy A dy
 B
dx dx
N Ax 
CD AB

y
 x 2  x1  A1  yA2

D AB  cA1  cA 2 
N Ax 
 x 2  x1 

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16.NH2 gas (A) diffuses through N2(B) under steady state condition with non – diffusing N2 . The total
pressure is 101.325 kPa and temperature is 298 K. The diffusion thickness is 0.15 m the partial pressure
of NH3 at one. Point is 1.5  104 Pa and at the outer point is 5  103 Pa. The DAB for mixture at 1 atm and
298 K is 2.3  10-5 m2/sec. (i) Calculate flux of NH3. (A through non diffusing B) . Calculate flux for
equimodal counter diffusion.(8)
Given : P = 102.325 KPa.
T = 298 k.
L = 0.15 m
Pa1 = 1.5  104 Pa.
Pa2 = 5  103 Pa.
Ans : (a) Equimolar counter diffusion
 ma  ab Pa1  Pa 2
 A   R Tx x
  a 2 1
8314
Ra   489 J kg K.
17
1.052 105 kg gm 2
(b) A through non – diffusing B
ma P  Pb 
 Dab . ln  2 
A R a T  Y2  Y1   Pb1 

Pa1  Pb1  P
1.5  104 + Pb1  101.325  103
Pb1  86325Pa.
Pa2  Pb2  P
5 103  Pb2  101.325 103
Pb2  96325Pa.
ma 2.3 105 101.325 103

 
A  8314   96325 
   298 0.15  LN  
 17   86325 
= 1.168  10-5 kg/gm2.
17.Write a note on the convective mass transfer coefficients for liquids and gases. Ans: Convective
Mass Transfer Co-efficient:
A fluid of species molar concentration, CA, flouring a surface at which species concentration is C AS.

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As long as CAS ≠ CA, mass transfer by convection will occur:
NA = hm (CAS – CA)
NA
hm  m / s.
CAS  CA
ii) Give a brief description on heat, momentum and mass transfer analogies.
Ans : Heat, Momentum and Mass Transfer Analogy:
In has been sent that there is a marked similarly between the laws governing the boundary layer growth of the
three transport phenomena, of momentum, heat and mass. These equations for a laminar boundary layer over a
flat plate are:
(i) Momentum Transfer
u u 2u
u v v 2 ..... 1
x y y
(ii) Heat Transfer
T T 2T
u v  2 ....  2 
x y y
(iii) Mass Transfer
CA C  2 CA
u  v A  DAB ....  3
x y y2
It was shown that the momentum and thermal boundary layers are identical for v =  or when prandtl number
is unity.
v
i.e., Pr  1

Note the similarity between Equations. The velocity and concentration profiles will have the same shape when
v = DAB. The dimensionless ratio v/DAB is called the Schmidt number
v
SC  .....  4 
DAB
The Schmidt number is important in problems involving both momentum and convection mass transfer. It
assumes the same importance in mass transfer as does the Prandtl number to convection heat transfer problems.
Table 14.1 gives the values of the Schmidt for same common gases diffusing into air at 25°C and 1 atmosphere.
Obviously, the temperature and concentration profiles will be similar when  = DAB. The dimensionless ratio
/DAB is called the Lewis number.

Le 
D AB
Table : Schmidt Number for some Gases Diffusing

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Into Air at 1 atm, and 25C
Gas Schmidth Number

Ammonia 0.66
Carbon dioxide 0.94
Hydrogen 0.22
Oxygen 0.75
Water Vapour 0.6
Ethyl ether 1.66
Benzene 1.76
The Lewis number is of significance in problem involving both heat and mass transfer. All the three boundary
layer profiles will become identical when
Pr = Sc = Le = 1
Just like the Nusselt number in convective heat transfer,. We define a non-dimensional parameter called
Sherwood number as:
hm x
Sh  ....  5
DAB
Where x is characteristic length.
Similarly, corresponding to Stanton number,
Nu h
St  
Re.Pr pu  CP
We have a dimensionless number in mass transfer St m, given by:
Sh h
St m   m
Re.Sc 
We have a dimensionless number in mass transfer St m, given by:
Sh h
St m   m
Re.Sc 
As seen, the forced heat transfer correlations are of thee form:
Nu (Re, Pr)
Likewise in forced convection mass transfer, the correlations would be of the form:
Sh = sh(Re, Sc)
The free convection heat transfer correlations have been seen to be the form:
Nu= Nu(Gr, Pr)
2g L3 T
Where Gr  G
2

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We need to define a new ,ass Grashof number, Gr m, because the density variation in mass transfer is
due to concentration difference and not temperature difference., The buoyancy force in mass transfer is given
by:
g
      gm  mA  mAw 

1   
Where m     is a quantity analogous to β.
  m A 
It indicates the variations of density with composition. The mass Grashof number is then defined as :
Grm 

m gL3 m Aw  m Aw  ......  6 

v
gL3    w 
or Grm  ........  7 
v 2 
Where subscript w refers to the wall. The correlations for natural mass transfer can then be written in
the form:
Sh = sh Grm, Pr).

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20.

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21.

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HEAT AND MASS TRANSFER

1. State Fourier’s law of condition

K – Temperature conductivity W/mK

2. Defined Thermal Conductivity

Thermal conductivity is defined as the ability of a substance to conduct heat.

5. State Newton’s law of cooling or conservation law.

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Heat transfer by conservation given by Newton‘s law of cooling

Where A – Area exposed to heat transfer in m2

h – heat transfer co efficient in W/m2K

Ts – Temperature of the surface in K

T∞ - Temperature of the fluid in K

7. Define overall beat transfer co efficient

9. What is critical radius of insulation or critical thickness?

10. Define external surfaces

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11. State applications of fins.

The main application of fins are,

1. Cooling of electronic components

2. Cooling of motor cycle engines

4, Cooling of small capacity compressors.

12. Define Fin efficiency.

13. Define Fin effectiveness.

14. List the mode of heat transfer.

15. List down the types of boundary condition.

2) Prescribed heat flux

3) Convection boundary condition

16. What is meant by lumped heat analysis?

17. What is meant by semi infinite solids?

18. Explain the significance of Fourier number.

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19. What are the factors affecting in thermal conductivity?

20.Explain the significance of thermal diffusivity?

Thermal conductivity of glass

Thermal conductivity of air K2 = 0.026 W/mK

3mm 12mm 3mm

Height of double pane window = 1.2 m

Width of double pane window = 2 m

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Area of double pane window = 1.2 × 2 = 2.4 m2

Interior room temperature, T a = 24° C

Outside air temperature, Tb = -5° c

Thickness of glass, L1 = L3 = 3mm = 0.003m

Thickness of air, L2 = 12mm = 0.012m

Interior convective heat transfer coefficient,

Outer convective heat transfer coefficient,

(a) Rate of heat transfer, Q

(b) Inner surface temperature, T i

Heat flow through composite wall is given by.

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2. Derive the general 3 dimension heat conduction equation in Cartesian coordinates.

Consider a small rectangular element of sides dx, dy and dz shown in figure

Where k – Thermal conductivity, W/mK

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Heat stored in the element

Heat stored  Mass of  Specific  Risein 

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