Thermodynamics & Heat Transfer
Study of
Heat and Work transfer
(quantitatively)
Study of
“How heat flows”
Thermodynamics
Heat Transfer
every activity involves
heat transfer
Page  1
Driving Potential & Resistance
Students’
Interest &
Capability
Class Room Ambience
Teacher’s
Interest &
Knowledge
Knowledge
Transfer
Teaching – Learning Process in a Class Room
DRIVING POTENTIAL
Difference between
Teacher’s interest + knowledge &
students’ interest + capability
RESISTANCE FACTORS
1. Time of lecture
2. Nature of subject
3. Ambience (conduciveness)
However, higher potential difference leads to increase in entropy
Page  2
Conduction
The transfer of energy in a solid or fluid via molecular contact
without bulk motion
MODE
Solids > Lattice vibrations
Fluids > Molecular collisions
T
dT 

q    
 dx 
T T0
x
PHYSICAL
PagePHENOMENON
3
MATHEMATICAL
EQUATION
Conduction (contd.)
Fourier Law of Heat Conduction
dT 

q    
 dx 
T 


qx   k

x 

• The heat flux, q is directly proportional to temperature gradient
• The proportionality constant, k, is defined as the thermal
conductivity, a thermo physical property.
Page  4
Conduction (contd.)
Thermal Conductivity, k
Silver = 410 Wm-1K-1
METALS
NON-METALS
k/ksilver
Silver
Gold
Copper
Aluminum
Brass (70% Cu:30% Ni)
Platinum, Lead
Mild steel (0.1% Cu), Cast iron
Bismuth
Mercury
Page  5
1
0.7
0.93
0.86
0.33
0.25
0.12
0.07
0.04
k/ksilver
Air
Water
Granite, Sandstone
Average rock
Limestone
Ice
Glass (crown)
Concrete (1:2:4)
Brick
Snow (fresh or average)
Soil (sandy, dry)
Soil (8% moist)
Wood
0.19
0.0014
0.011
0.012
0.007
0.015
0.0058
0.0042
0.0038
0.005
0.002
0.0033
0.0045
Convection
Convection occurs in liquids and gases.
Energy is carried with fluid motion when convection occurs.
Q  hA(Tw  Ta )
PHYSICAL
PHENOMENON
Page  6
MATHEMATICAL
EQUATION
Convection (contd.)
Newton’s Law of Cooling
Q  hA(Tw  Ta )
•
The quantity h is called the convective heat transfer coefficient (W/m2-K).
•
It is dependent on the type of fluid flowing past the wall and the velocity
distribution.
•
Thus, h is not a thermo physical property.
Page  7
Convection Process
Free convection
Gases
Liquids
50–1000
Forced convection
Gases
Liquids
50–20,000
Convection phase change
h(W/m2-K)
2–25
25–250
2,500–200,000
Convection (contd.)
Convective Processes

Single phase fluids (gases and liquids)
– Forced convection
– Free convection, or natural convection
– Mixed convection (forced plus free)

Convection with phase change
– Boiling
– Condensation
Page  8
Radiation
Energy transfer in the form of electromagnetic waves
A,Ts
E   Ts
PHYSICAL
PHENOMENON
Page  9
4
MATHEMATICAL
EQUATION
Radiation (contd.)
Stefan-Boltzman Law
Eb  T
4
s
The emissive power of a black body over all wave
lengths is proportional to fourth power of temperature
Page  10
One Dimensional Heat Conduction
Heat flow out
k
[Aq]x+Δx
Heat flow in
g
[Aq]x
Net rate of heat
gain by conduction
[Aq]x – [Aq]x+ Δx

Page  11
Rate of energy
generation
+
+
A Δx g
Δx
=
=
Rate of increase
of internal energy
A x  c p
1 [ Aq]x  x  [ Aq]x
T ( x, t )
 g  c p
A
x
t
T ( x, t )
t
One Dimensional Heat Conduction (contd.)
Heat flow out
k
[Aq]x+Δx
Heat flow in
[Aq]x
g
Δx
1 [ Aq]x  x  [ Aq]x
T ( x, t )

 g  c p
A
x
t
As Δx  0, the first term on the LHS, by definition, becomes the derivative of
[Aq] with respect to x
1 
T ( x, t )
 Aq   g  c p

A x
t
Page  12
1  
T 
T ( x, t )
 Ak
  g  c p
A x 
x 
t
One Dimensional Heat Conduction (contd.)
Rectangular Coordinates
  T 
T ( x, t )
 k   g  c p
x  x 
t
n=0
Cylindrical Coordinates
1   T 
T (r , t )
 rk   g  c p
r r  r 
t
n=1
Spherical Coordinates
1   2 T 
T (r , t )
 r k   g  c p
2
r r 
r 
t
A Compact Equation
1   n T 
T (r , t )
 r k   g  c p
n
r r 
r 
t
Page  13
n=2
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Page  14
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
T1
T (x,t) | x=0 = T (0,t) = T1
T2
T (x,t) | x=L = T (L,t) = T2
0
Page  15
L
x
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Heat
Supply
Plate
Conduction
flux
q0  k
W/m2
T
x
Conduction
flux
T
k
x
Page  16
0
T
k
x
x 0
Heat
Supply
 qL
W/m2
xL
L
x
T
k
x
 q0
x 0
 qL
xL
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
T
qa   k
r
k
r a
a
Heat
Supply
b
W/m2
Page  17
T
r
r
 qb
r b
T
k
r
T
k
r
 qa
r a
 qb
r b
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Conduction
Fluid
Flow
T
k
x
Convection
Plate
 h2 (T2  T
xL
T1, h1
Convection
)
Convection heat flux
from the fluid at T1 to
the surface at x = 0
Conduction
T
h1 (T1  T x 0 )  k
x
x 0
Fluid
Flow
T2, h2
Page  18
xL
Conduction heat flux
from the surface at
x= 0 into the plate
T
h1 (T1  T x 0 )  k
x
x 0
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Conduction
Fluid
Flow
k
T
x
Convection
Plate
 h2 (T2  T
xL
T1, h1
Convection
)
Convection heat flux
from the fluid at T2 to
the surface at x = L
Conduction
T
h1 (T1  T x 0 )  k
x
x 0
Fluid
Flow
T2, h2
Page  19
xL
Conduction heat flux
from the surface at
x = L into the plate
T
h2 (T2  T x  L )   k
x
xL
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Fluid
Flow
Hollow Cylinder or
hollow sphere
T1, h1
k
T
r
r b
 h2 (T2  T r b )
b
a
h1 (T1  T r a )   k
Page  20
T
r
Convection heat flux
from the fluid at T1 to
the surface at r = a
Heat
Supply
r a
Fluid
Flow
T2, h2
r
Conduction heat flux
from the surface at
r= a into the plate
T
h1 (T1  T r a )   k
r
r a
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Fluid
Flow
Hollow Cylinder or
hollow sphere
T1, h1
k
T
r
r b
 h2 (T2  T r b )
b
a
h1 (T1  T r a )   k
Page  21
T
r
Convection heat flux
from the fluid at T2 to
the surface at r = b
Heat
Supply
r a
Fluid
Flow
T2, h2
r
Conduction heat flux
from the surface at
r= b into the plate
T
h2 (T2  T r b ) k
r
r b
Steady State One Dimensional Heat Conduction
Rectangular Coordinates
 2T
0
2
x
Governing Equation
T ( x)  c1 x  c2
0
L
T = T1
T = T2
Page  22
x
 T2  T1 
T ( x)  
 x  T1
 L 
K . A.(T1  T2 )
Qx 
L
L
R
K .A
Steady State One Dimensional Heat Conduction
Cylindrical Coordinates (Solid Cylinder)
1 d  dT ( r )  g 0
r

0


r dr 
dr  k
Governing Equation
 dT (r )


0
 dr

T (r )  T2 at
at r  0
b
0
r
Solving,
T = T1
Page  23
T = T2
r b
g0
T (r )  
r  c1 ln r  c2
2k
2

g0
r 
T (r )  
1      T2
4k   b  
dT (r ) g 0 r
q(r )  k

dr
2
Steady State One Dimensional Heat Conduction
Cylindrical Coordinates (Solid Cylinder)
Solved Example
For r=1cm
g0 = 2 x 108 W/m3
k = 20 W/(m.°C)
T2 = 100 °C
b
0
r
What will be the
1. Centre temperature T(0)
2. Heat flux at the boundary surface (r=1cm)
Equations to use (derive)
T = T1
Page  24
T = T2
g
T (r )   0
4k
g0r
q(r ) 
2
  r 2 
1      T2
  b  
Solution
T(0) = 350 °C
q(r) = 106 W/m2
Steady State One Dimensional Heat Conduction
Cylindrical Coordinates (Hollow Cylinder)
Determination of Temperature Distribution
Mathematical formulation of this problem is
d
dr
k
b
T2
a
T1
0
Page  25
r
 dT ( r ) 
r dr   0
in a < r < b
T (r )  c1 ln r  c2
T2  T1
Solving, c1 
ln( b / a)
ln( a)
c2  T1  (T2  T1 )
ln( b / a)
T (r )  T1 ln( r / a )

T2  T1
ln( b / a )
Steady State One Dimensional Heat Conduction
Cylindrical Coordinates (Hollow Cylinder)
Expression for radial heat flow Q over a length H
The heat flow is determined from,
dT ( r )
Q  q ( r ).area   k
2 rH
dr
k
 k 2Hc1
Since,
b
T2
2kH
Q
(T1  T2 )
ln( b / a)
a
T1
0
Page  26
dT (r ) / dr  (1 / r )c1
Rearranging,
r
T  T2
Q 1
R
ln( b / a )
where, R 
2 kH
Steady State One Dimensional Heat Conduction
Cylindrical Coordinates (Hollow Cylinder)
Expression for thermal resistance for length H
ln( b / a )
R
2 kH
k
Above equation can be rearranged as,
ln( b / a) (b  a) ln[ 2bH /( 2aH )]
R

2kH
(b  a)2Hk
b
T2
t
R
kAm
a
T1
0
Page  27
r
where, Am 
A1  A0
ln( A1  A0 )
here, A0 = 2πaH =area of inner surface of cylinder
A1 = 2πbH =area of outer surface of cylinder
Am = logarithmic mean area
t = b – a = thickness of cylinder
Steady State One Dimensional Heat Conduction
Spherical Coordinates (Hollow Sphere)
Expression for temperature distribution
The mathematical formulation is given by,
d  2 dT (r ) 
r
0
dr 
dr 
a
r
0
b
Page  28
in a < r < b
c1
T ( r )    c2
r
ab
(T1  T2 )
where, c1  
ba
bT2  aT1
c2 
ba
a br
b r a
T (r )  .
.T1  .
.T2
r ba
r ba
Steady State One Dimensional Heat Conduction
Spherical Coordinates (Hollow Sphere)
Expression for heat flow rate Q and thermal resistance R
Heat flow rate is determined using the equation,
 dT (r ) 
Q  ( 4 r )   k

dr


c1 
2 
 (4r )  k 2   4kc1
r 

ab
from last slide
c1  
(T1  T2 )
ba
2
a
r
0
using,
b
Page  29
ab
T1  T2
Q  4k
(T1  T2 ) 
ba
R
ba
where, R 
4 kab
Composite Medium
BRICK WALL
REFRACTORY LINING 2
FURNACE
REFRACTORY LINING 1
Example (Furnace Wall)
FURNACE WALL
Page  30
Ambient
Composite Medium
Example (Condenser Water Tube)
Tube Wall
Scale
Cooling Water
Condensing Medium (Steam)
Page  31
Composite Medium
Composite Slab (resistance in series)
L1
L2
Fluid
Flow
Tb, hb
L3
Ta
T0
Q
T1
Q
Tb
T2
Fluid
Flow
T3
Ta, ha
Q
Ta
T1
T0
Ra
R1
R2
R3
Ta  T0 T0  T1 T1  T2 T2  T3 T3  Tb
Q




Ra
R1
R2
R3
Rb
Ra 
Page  32
Tb
T3
T2
L
1
L
L
1
; R1  1 ; R2  2 ; R3  3 ; Rb 
Aha
Ak1
Ak2
Ak3
Ahb
Q
Rb
Q
Ta  Tb
R
W
R  Ra  R1  R2  R3  Rb
Composite Medium
Composite Slab (resistance in parallel)
Insulated
Q
B
T1
A
C
E
T2
Insulated
RA
RB
Rc
Page  33
RD
RE
W
R  RA  Req. p  RE
D
T1
T1  T2
R
T2
1
1
1
1



Req. p RB RC RD
Composite Medium
Composite Cylinder
Q
Ta
Ra
k2
ha
T1
T0
R1
Rb
Q
Ta  T0 T0  T1 T1  T2 T2  T3 T3  Tb
Q




Ra
R1
R2
R3
Rb
k3
k1
hb
Ra 
1
1
r
1
r
; R1 
ln 1 ; R2 
ln 2
2r0 Hha
2Hk1 r0
2Hk 2 r1
R3 
r
1
1
ln 3 ; Rb 
2Hk 3 r2
2r3 Hhb
Ta  Tb
Q
R
Page  34
R3
R2
Tb
T3
T2
W
R  Ra  R1  R2  R3  Rb
Composite Medium
Composite Spheres
Q
Ta
T1
T0
Ra
R1
Rb
Q
Ta  T0 T0  T1 T1  T2 T2  T3 T3  Tb
Q




Ra
R1
R2
R3
Rb
ha
k1
k2
hb
Ra 
1
1 r1  r0
1 r2  r1
;
R

;
R

1
2
4r02 ha
4k1 r1r0
4k 2 r2 r1
R3 
1 r3  r2
1
; Rb 
4k3 r3 r2
4r32 hb
k3
Ta  Tb
Q
R
Page  35
R3
R2
Tb
T3
T2
W
R  Ra  R1  R2  R3  Rb
Composite Medium
H
Convection
into an
ambient at
T∞,h0
r1
T1
ro
Heat Loss, q
Critical Thickness of Insulation
rc
Insulation Radius, r
Page  36
Composite Medium
Critical Thickness of Insulation
The rate of heat loss Q from the tube is given by
Convection
into an
ambient at
T∞,h0
H
r1
T1
ro
Ti  T
Q
Rins  R0
Rins 
r
1
ln 0
2 kH ri
2kH (Ti  T )  1
dQ
k 

0


2 
2 
dr0
ln( r0 / ri )  k /( h0 r0 )  r0 h0r0 
For Cylinder
r0 c 
Page  37
1
R0 
2r0 Hh0
k
h0
For Sphere
r0 c 
2k
h0
Composite Medium
Solved Example (Composite Cylinder)
Calculate,
1. Heat loss from tube for length H=10m
2. Temperature drops resulting in thermal
resistances
7.6 cm
Determination of heat loss
Ta=330°C
ha=400 W/(m2.°C)
Insulation
t =2 cm
K=0.2 W(m.°C)
Ra 
W
1
1

2r0 Hha 2  0.025 10  400
Ra  1.59 10 3 C / W
K = 15 W/(m °C)
Ambient air
Tb=30°C
hb= 60 W/(m2.°C)
Page  38
Ta  Tb
Q
Ra  R1  R2  Rb
R1 
1
r
1
3.8
ln 1 
ln
2Hk1 r0 2 10 15 2.5
R1  0.44 10 3 C / W
Composite Medium
Solved Example (Composite Cylinder)
Calculate,
1. Heat loss from tube for length H=10m
2. Temperature drops resulting in thermal
resistances
7.6 cm
Determination of heat loss
Ta=330°C
ha=400 W/(m2.°C)
R2  33.65 10 3 C / W
Insulation
t =2 cm
K=0.2 W(m.°C)
Rb  4.2110 3 C / W
R  Ra  R1  R2  R3  Rb  39.89 10 3 C / W
K = 15 W/(m °C)
Ambient air
Tb=30°C
hb= 60 W/(m2.°C)
Page  39
Q
330  30
 7521 W
3
39.89 10
Composite Medium
Solved Example (Composite Cylinder)
Calculate,
1. Heat loss from tube for length H=10m
2. Temperature drops resulting in thermal
resistances
7.6 cm
Determination of temperature drops
Ta=330°C
ha=400 W/(m2.°C)
Q
Insulation
t =2 cm
K=0.2 W(m.°C)
Ta  T0 T0  T1 T1  T2 T2  Tb



Ra
R1
R2
Rb
Thotgas  QRa  12.0C
Ttube  QR1  3.3C
Tinsulation  QR2  253.0C
K = 15 W/(m °C)
Ambient air
Tb=30°C
hb= 60 W/(m2.°C)
Page  40
Toutside  QRb  31.7C
Composite Medium
Solved Example (Composite Wall)
GYPSUM SHEATH
2 x 4 wood studs have actual dimensions of
4.13 x 9.21 cm with k = 0.1 W/m.°C
Calculate,
1.Overall heat transfer coefficient
2.R value of the wall
Outside Air Convection
h=15 W/m2 C
Thermal resistance model
COMMON BRICK, k =0.69
k=0.96
8cm
1.9cm
INSULATION, k=0.04
k=0.48
1.9cm
Two parallel heat flow paths are possible
1. Through the studs
2. Through the insulation
40.6 cm
Inside Air Convection
h=7.5 W/m2 C
Rsheath
outside
2x 4 STUDS
Tair
Rconvection
Rconvection
Rsheath
outside
Note: ‘k’ is expressed in W/m °C
inside
inside
outside
Page  41
Rsheath
Rbrick
Tair
outside
Rinsul
Rstud
Rsheath
inside
inside
Composite Medium
Solved Example (Composite Wall)
Calculate,
1. Overall heat transfer coefficient
2. R value of the wall
GYPSUM SHEATH
Outside Air Convection
h=15 W/m2 C
Heat flow through the studs
Area = 0.0413m2/unit depth
COMMON BRICK, k =0.69
k=0.96
8cm
1.9cm
INSULATION, k=0.04
k=0.48
1.9cm
40.6 cm
Inside Air Convection
h=7.5 W/m2 C
Heat flow occurs through 6 thermal resistances
1. Convection Resistance outside of brick
2. Conduction resistance in brick
3. Conduction resistance through outer sheet
4. Conduction resistance through wood stud
5. Conduction resistance through inner sheet
6. Convection resistance on inside
2x 4 STUDS
Recall,
Rconvection  1 / hA
Rconduction  x / kA
Rtotal  1.614  2.807  0.48  22.3  0.96  3.23  31.39C / W
Page  42
Note: ‘k’ is expressed in W/m °C
Composite Medium
Solved Example (Composite Wall)
Calculate,
1. Overall heat transfer coefficient
2. R value of the wall
GYPSUM SHEATH
Outside Air Convection
h=15 W/m2 C
COMMON BRICK, k =0.69
k=0.96
Heat flow through the insulation
8cm
1.9cm
INSULATION, k=0.04
k=0.48
1.9cm
40.6 cm
The five of the materials are same, but the
resistances involve different area terms, i.e.,
40.6 - 4.13 cm instead of 4.13 cm.
Inside Air Convection
h=7.5 W/m2 C
2x 4 STUDS
Thus the total resistance of the insulation
section is given below
Rtotal  7.337 C / W
Page  43
Note: ‘k’ is expressed in W/m °C
Composite Medium
Solved Example (Composite Wall)
Calculate,
1. Overall heat transfer coefficient
2. R value of the wall
GYPSUM SHEATH
Outside Air Convection
h=15 W/m2 C
COMMON BRICK, k =0.69
k=0.96
1. Overall heat transfer coefficient
Overall resistance is obtained by combining
the parallel resistances as calculated earlier.
8cm
1.9cm
INSULATION, k=0.04
k=0.48
1.9cm
Roverall 
1
(1 / 31.39)  (1 / 7.337)
40.6 cm
 5.947C / W
Inside Air Convection
h=7.5 W/m2 C
Overall heat transfer coefficient is found by,
2x 4 STUDS
q  UAT 
U
Page  44
Note: ‘k’ is expressed in W/m °C
T
Roverall
(here, A = 0.406m2)
1
1

 0.414W / m 2 C
RA (5.947)(0.406)
Composite Medium
Solved Example (Composite Wall)
Calculate,
1. Overall heat transfer coefficient
2. R value of the wall
GYPSUM SHEATH
Outside Air Convection
h=15 W/m2 C
COMMON BRICK, k =0.69
k=0.96
2. R Value of the wall
8cm
1.9cm
The resistance of the wall is calculated using
the overall heat transfer coefficient, as given
below:
INSULATION, k=0.04
k=0.48
1.9cm
40.6 cm
Inside Air Convection
h=7.5 W/m2 C
2x 4 STUDS
Page  45
Note: ‘k’ is expressed in W/m °C
1
1
Rvalue  
 2.414C.m 2 / W
U 0.414
Composite Medium
Solved Example (Critical Thickness of Insulation)
Calculate, the critical thickness of rubber and the
maximum heat transfer rate per metre length of
conductor.
The temperature of rubber is not to exceed 65 °C (due
to heat generated within).
Ambient at
30°C, 8.5 W/m2K
Critical thickness
r0c  k / ho  0.155 / 8.5  0.0182 m
r = 5mm
Maximum heat transfer rate
Rins 
r
1
1
 0.0182 
ln 0 
ln 
  1.32 
2kH ri 2  0.155  0.005 
C /W  m
Rubber
k = 0.155 W/mK
1
1
R0 

 1.02 C / W  m
2r0 Hh0 2  0.0182  8.5
Q
Page  46
Ti  T
65  30

 14.89 W / m
Rins  R0 1.32  1.02
Heat Source Systems
Plane wall with heat generation
Expression for mid plane temperature is given
by,
g 2
T0 
L  Tw
2k
x=0
q = heat generated
per unit volume
The temperature distribution can also be written
in alternative form as:
T  Tw
x
 1  
T0  Tw
 L
Tw
x
Tw
L
L
Page  47
2
Conduction-Convection Systems
Fins / Extended Surfaces
• Necessity for fins
• Biot Number
hx ( x / k )

k
1/ h
=
Internal Conductive resistance
Surface Convective resistance
FIN TYPES
Page  48
LONGITUDINAL
RECTANGULAR FIN
RADIAL FIN
Conduction-Convection Systems
Governing Equation (Rectangular Fin)
Net Heat Conducted – Heat Convected = 0
dqconv  hPdx(T  T )
t
base
A
qx
qx+dx
x
dx
L
Page  49
Z
[ Aqx  Aqx dx ]  hPdx(T  T )  0
Aqx  Aqx  dx
 hP(T  T )  0
dx
d
  Aqx   hP (T  T )  0
dx
d 
dT 
   kA   hP(T  T )  0
dx 
dx 
d 2T hP

(T  T )  0
2
dx
kA
d 2
2

m
 0
2
dx
where,
hP
m
kA
&
  T  T
Conduction-Convection Systems
Boundary Conditions
LONG FIN
d 2 ( x)
2

m
 ( x)  0 in x  0
2
dx
 ( x)  T0  T   0 at x  0
 ( x)  0
as x  
SHORT FIN
(end insulated)
d 2 ( x)
2

m
 ( x)  0 in 0  x  L
2
dx
 (x)  T0  T   0
at x  0
d (x)
0
dx
Page  50
SHORT FIN
( end not insulated)
at x  L
d 2 ( x)
 m 2 ( x)  0 in 0  x  L
2
dx
 (x)  T0  T   0
at x  0
k
d (x)
 he ( x)  0 at x  L
dx
Conduction-Convection Systems
Types of Fin Boundaries
Type of FIN
boundary
Long Fin
(TL= T∞)
Short Fin
(end insulated)
Temperature Distribution
Heat transferred by fin
T  T
Tb  T
Q
e-mx
(Tb-T∞)(hPkA)0.5
Cosh m(L-X)
Cosh (mL)
(hPkA)0.5 (Tb-T∞) tanh (mL) *
Cosh[m( L  X )]  (hL / mk ) Sinh[m( L  X )] (T  T ) tanh( mL)  (hL / mk ) (hPkA) 0.5
Short Fin
b

1  (hL / mk ) tanh( mL)
Cosh(mL)  (hL / mk ) Sinh (mL)
(end not insulated)
Specified End
Temperature
At x=L; T=TL
TL  T
Sinh (mx)  Sinh[m( L  x)]
Tb  T
Sinh (mL)
[(Tb  T )  (TL  T )]
* For higher values of mL (i.e., m=4), tanh mL = 0.999 ≈ 1.
Thus Qshort fin  Qlong fin for higher values of mL
Page  51
Cosh(mL)  1
(hPkA) 0.5
Sinh (mL)
Conduction-Convection Systems
Performance Parameters
Fin Efficiency
Q fin
Actual heat trans fer throug h fin


Ideal heat trans fer throug h fin if entire
Qideal
fin surface were at fin base temperatu re , T0
In practical applications, a finned heat transfer surface is composed of the
finned surfaces and the unfinned portion. In such cases total heat transfer is used.
Qtotal
= Qfin + Qunfinned
= η af h θ0 + ( a – af ) h θ0
Where, a = total heat transfer area (i.e., fin surface + unfinned surface)
af = heat transfer area of fins only
Qtotal
= [ηβ+(1-β)] a h θ0 ≡ η‘ a h θ0
Where, η‘ = βη +1 – β = area – weighted fin efficiency
β = af / a
Page  52
Conduction-Convection Systems
Fin Efficiency, η
Performance Parameters
Fin Efficiency
Each curve is specific for
specific fin configuration
L (2h/kt)0.5
Page  53
Fin efficiency curves are available for fins of
various configuration (eg. Axial, circular disk
fins of various length, thickness etc)
Conduction-Convection Systems
Performance Parameters
Fin Effectiveness
Effectiven ess 
Q with fin
Q withoutfin
Although the addition of fins on a surface increases surface area, it also increases
thermal resistance over the portion of the surface where fins are attached. Therefore
there may be situations in which the addition of fins does not improve heat transfer.
Pk / (Ah) > 1
Page  54
(to justify usage of fins)
Conduction-Convection Systems
Solved Example
A steel rod is exposed to ambient air. If one end of the
rod is maintained at a temperature of 120 °C, calculate
the heat loss from the rod
The condition for other end of the rod is not specified
explicitly. By considering L/D ratio, it appears that a
long fin assumption is applicable. Using the simplest
analysis to solve, computing mL:
Diameter = 2cm
Length = 25 cm
k = 50 W / m. °C
Tbase=120°C
Tamb = 20°C
h = 64 W / m2. °C
hP
hD
4h
4  64



kA ( / 4) D 2 k kD 50  0.02
m  16 and mL  16  0.25  4
m2 
Therefore, expression for Qlong fin can be used.


Q  Tb  T  hPkA  Tb  T  (D) D 2 kh
4

 120  20
Page  55

2
(0.02) 3  50  64  25.1 W
Conduction-Convection Systems
Solved Example (Fin Efficiency)
L
t
CIRCULAR DISK FIN
Circular disk fins of constant thickness are attached on a 2.5 cm OD
tube with a spacing of 100 fins per 1m length of tube.
Fin Properties: Aluminium k = 160 W / m.°C, t = 1mm L = 1 cm
Tube wall temperature = 170 °C; Ambient temperature = 30 °C
Heat transfer coeff. of ambient , h = 200 W/m2. °C.
Calculate,
1. Fin Efficiency and area weighted fin efficiency
2. Heat lost to the ambient air per 1m length of tube
3. Heat loss with that if there were no fins on tube
Fin Efficiency, η
Fin Efficiency
Page  56
Fin efficiency is determined using the graph shown aside.
The following parameters are calculated, firstly:
ro/ri
L (2h/kt)0.5
L
2h
2  200
 110  2
 0.5
kt
160 10 3
ro 1.25  1

 1. 8
ri
1.25
The fin efficiency is determined from graph
  0.9
Conduction-Convection Systems
Solved Example (Fin Efficiency)
Calculate,
1. Fin Efficiency and area weighted fin efficiency
2. Heat lost to the ambient air per 1m length of tube
3. Heat loss with that if there were no fins on tube
L
Area Weighted Fin Efficiency
t
CIRCULAR DISK FIN
Tube OD = 2.5 cm
100 fins per 1m tube length
kfin = 160 W/m°C
t = 1mm; L = 1cm
Ttube = 170°C; Tamb = 30°C
hamb = 200 W/m2. °C
Ratio of heat transfer area for fin to the total heat transfer area, β
Fin Surface per cm of tube length = 2π(r02-ri2) = 2π[2.252-1.252]
= 21.99 cm2
Total heat transfer surface per cm of tube length
= 2π (r02-ri2) + 2πri (1 – t)
= 2π[2.252-1.252] + 2π(1.25)(1 – 0.1)
= 29.06 cm2
β = af / a = 21.99 / 29.06 = 0.757
Area Weighted Fin Efficiency, η’ = βη +1 – β = 0.757(0.9) + 0.243
= 0.924
Page  57
Conduction-Convection Systems
Solved Example (Fin Efficiency)
Calculate,
1. Fin Efficiency and area weighted fin efficiency
2. Heat lost to the ambient air per 1m length of tube
3. Heat loss with that if there were no fins on tube
L
Heat lost to ambient per 1m length of tube
t
CIRCULAR DISK FIN
Tube OD = 2.5 cm
100 fins per 1m tube length
kfin = 160 W/m°C
t = 1mm; L = 1cm
Ttube = 170°C; Tamb = 30°C
hamb = 200 W/m2. °C
Total heat transfer surface a per 1m of tube length
a = 29.06 x 100 cm2 = 0.29 m2
Q = η’ahθ0 = 0.924 x 0.29 x 200 (170 – 30) = 7503 W
Heat lost per 1m length of tube with no fins
Qno fin = 2πrihθ0 = 2π x 0.0125 x 200 x (170 – 30) = 2199 W
Clearly, the addition of fins increases the heat dissipation by a
factor of about 3.4
Page  58
Transient Conduction
• If the surface temperature of a solid body is suddenly altered, the
temperature within the body begins to change over time.
• Variation of temperature both with position and time makes determination of
temperature distribution under transient condition more complicated.
• In some situations, variation of temperature with position is negligible under
transient state, hence the temperature is considered to vary only with time.
• The analysis under the above assumption is called lumped system
analysis.
• Biot Number, Bi = (hx) / k
• Lumped System Analysis is applicable only when Bi < 0.1
Page  59
Systems with Negligible Internal Resistance
Q
Lumped Heat Analysis
•
The convective heat loss from the body (shown aside) has its
magnitude equal to decrease in internal energy of solid.
Q  hA(T  T )  pcV
Volume V
Area A
dT
 hA

dt
T  T
pcV
T∞
T
On Integration,
T=T0 at t=0
T0
Page  60
 hA
t  C1
pcV
T  T
 exp  (hA / pcV ).t 
T0  T
1/hA
T∞
ln( T  T ) 
Solving and rearranging,
S
Cth=ρcV
dT
dt
Systems with Negligible Internal Resistance
Biot Number
• It is a non-dimensional parameter used to test the validity of the lumped heat
capacity approach.
Bi 
hL c
internal resistance

convective resistance
k
• The characteristic length (Lc) for some common shapes is given below:
Plane Wall (thickness 2L)
Long cylinder (radius R)
A.2L
Lc 
L
2.A
R 2 .L R
Lc 

2R.L
2
Sphere (radius R)
Cube (side L)
(4 / 3)R 3
R
Lc 

4R 2
3
L3
L
Lc 

2
6L
6
• The lumped heat capacity approach for simple shapes such as plates, cylinders,
spheres and cubes can be used if Bi < 0.1
Page  61
Systems with Negligible Internal Resistance
Response time of a Temperature measuring Instrument
T  T
 exp  (hA / pcV ).t 
T0  T
• For a rapid response of temperature measuring device, the index, (hAt/ρcV) should
be large to make the exponential term reach zero faster.
• This can be achieved by decreasing wire diameter, density and specific heat or by
increasing value of ‘h’.
• The quantity (ρcV/hA) has the units of time and is called ‘time constant’ of system.
Hence at time t=t* (one time constant),
T  T
 e 1  0.368
T0  T
• At the end of time period t* the temperature difference between the body and
ambient would be 0.368 of the initial temperature difference.
• In other words, the temperature difference would be reduced by 63.2 percent.
• This reduction in 63.2 percent of initial temperature difference is called ‘sensitivity’
• Lower the value of time constant, better the response of instrument.
Page  62
Systems with Negligible Surface Resistance
Large Flat Plate with
Negligible Surface Resistance
Ts
T0(x) for
t=0
Ts = T∞ (t>0)
x
L
Page  63
• When convective heat transfer coefficient at the surface is
assumed to be infinite, the surface temperature remains
constant at all the time (t>0) and its value is equal to that of
ambient temperature.
• The systems exhibiting above said conditions are considered
to have ‘negligible surface resistance’
• An important application of this process is in heat treatment of
metals by quenching, viz., the dropping of a metallic sphere
initially at 300 °C into a 20 °C oil bath.
• Mathematical formulation of this case is :
 2T
1 T

0 x L
2
x
 t
T  T0 (x) at t  0 for 0  x  L (initial condition)
T  Ts
at x  0 for t  0
Boundary
Conditions
T  Ts
at x  L for t  0
Heat flow in an Infinitely Thick Plate
Semi-infinite body
• A semi-infinite body is one in which at any instant of time there
is always a point where the effect of heating / cooling at one of
its boundaries is not felt at all.
• At this point the temperature remains unchanged.
• Mathematical formulation is :
Semi-Infinite Plate
Ts
at t=0
Qo
To
x
 2T 1 T

2
x  t
with initial and boundary conditions,
T  T0 at t  0 for all x
T  Ts at x  0 for all t  0
T  T0 as x   for all t  0
Page  64
Systems with Finite Surface and Internal Resistance
Mathematical formulation :
Infinitely Large Flat Plate
of Finite Thickness (2L)
 2T
1 T

2
x
 t
T  T0 at t  0 (for - L  x  L)
h
T
 0 at x  0 (centre line)
x
T∞
 T h
 (T  T ) at x   L
x
k
h
T∞
at t=0
-x
x
x=-L
x=0
x=L
Page  65
Chart Solutions of Transient Heat Conduction Problems
Heisler Charts (by Heisler, 1947)
Infinite Plate
Time History
Mid Plane
hL/k
T(x,t) - T∞
Ti- T∞
0.1
1
0.8
0.9
0.6
0.5
0.4
0.2
Fourier number, ατ/L2
Page  66
Chart Solutions of Transient Heat Conduction Problems
Heisler Charts (by Heisler, 1947)
Infinite Plate
Time History
Any Position, x
1
x/L
0.2
0.4
T(x,t) - T∞
0.6
Ti- T∞
0.8
0.9
1
0
Page  67
0.1
Biot Number, hL/k
100
Chart Solutions of Transient Heat Conduction Problems
Heisler Charts (by Heisler, 1947)
Infinite Plate
Heat Flow
hL/k
0.001
0.01
0.05 0.1
0.5
1
10
20
40
50
Q/Qo
Page  68
h 2
2

Bi
Fo
k2
Where, Qo  cV (To  T )
Lumped System Analysis
Solved Example
Determination of Time required to cool
4
mass
5.5
V  R 3 

 2.037 10 3
Volume
3
density 2700
Aluminium Ball
ρ = 2700 kg/m3
c = 900 J/kg K
k = 205 W/mK
5.5 kg
Tinitial=290°C
Radius
R  3V / 4 
Characteristic
Length
Lc 
1/ 3
 0.0786m
R
 0.0262m
3
  hA  
T  T
.t 
 exp   
T0  T

cV
 
 
T  95C
T  15C
T0  290C
Tfluid = 15°C
h = 58 W / m2. °C
Time required to cool the
aluminium ball to 95°C
Page  69
?
hA
3h
3  58


 9.1 10  4 / s
cV cR 2700  900  0.0786
95  15
80

 exp( 9.1 10  4.t )
290  15 275
3.4375  exp( 9.1 10  4.t )
t  1357 s
Lumped System Analysis
Temperature Measurement by Thermocouples
Solved Example
The temperature of a gas stream is to be measured by
a thermocouple whose junction can be approximated
as a 1mm diameter sphere (shown aside)
Determine how long it will take for the thermocouple to
read 99% of initial temperature difference
Thermocouple Wire
Gas T∞
h=210 W/m2 °C
Junction (Sphere)
D= 1mm
ρ = 8500 kg/m3
k = 35 W/mK
c = 320 J/kg K
How long will it take for the
thermocouple to read 99 %
of Initial Temperature
difference
Page  70
?
Lc
= V/As = (1/6)D = (1/6)x0.001 = 1.67x10-4 m
Bi = hL/k = (210x 1.67x10-4) x 35
= 0.001 < 0.1
Therefore, lumped system analysis is applicable.
In order to read 99% of initial temperature difference
Ti – T∞ between the junction and the gas, we must
have
T  T
 0.01
T0  T
Lumped System Analysis
Temperature Measurement by Thermocouples
Solved Example
The temperature of a gas stream is to be measured by
a thermocouple whose junction can be approximated
as a 1mm diameter sphere (shown aside)
Determine how long it will take for the thermocouple to
read 99% of initial temperature difference
Thermocouple Wire
Time
Gas T∞
h=210 W/m2 °C
Junction (Sphere)
D= 1mm
ρ = 8500 kg/m3
k = 35 W/mK
c = 320 J/kg K
How long will it take for the
thermocouple to read 99 %
of Initial Temperature
difference
Page  71
?
T  T
 0.01  exp  (hA / pcV).t 
T0  T
hA s
h
210
1



0
.
462
s
cV cL c 8500  320 1.67 10 4
exp (0.462).t   0.01
t = 10s
Transient Conduction in
Semi-infinite Solids
Solved Example
A water pipe is to be buried in soil at
sufficient depth from the surface to prevent
freezing in winter.
What minimum depth is required to
prevent the freezing of pipe when soil is at
uniform temperature of Ti = 10 °C, the
surface is subjected to a uniform temperature
of T0 = -15 °C continuously for 50 days.
Also the pipe surface temperature should
Page  72
Transient Conduction in
Semi-infinite Solids
Solved Example
Water Pipe
(to be buried)
Tsurface = -15 °C
?
SOIL
Tsoil = 10 °C
Condition : Tpipe wall should not fall below 0 °C
What burial depth is needed to
prevent freezing of the pipe ?
Page  73
Temperature Distribution in
Semi-infinite Solid
T(x,t) - Tsurface
Tinitial- Tsurface
Error Function,  
Page  74
x
2 t
Determination of Burial depth
T ( x, t )  T0
0  15
 0.6

 ( x, t ) 
10  15
Ti  T0
For  ( x, t )  0.6,   0.6 (from graph)
For   0.2 10 m / s and t  50  24  3600s
-6
2
the error function  is given by,

Page  75
x
2 t

x
2 0.2 10-6  50  24  3600
 0.538 x
Determination of Burial depth
0.538 x  0.6
0.6
x
 1.12m
0.538
The pipe should be buried at least to a depth of
1.12 m to prevent freezing.
Page  76
Application of Heisler Charts
Aluminium Slab
Thickness=10cm
α = 8.4x10-5 m2/s
ρ = 2700 kg/m3
c = 900 J/kg K
k = 215 W/mK
Tinitial=500°C
Tfluid = 100°C
h = 1200 W / m2. °C
Page  77
Mid-plane Temperature and Surface
Temperature after 1 min?
Determination of Mid plane Temperature
2L=10 cm ; L = 5 cm ; t = 1min = 60 s
αt/L2
= (8.4x10-5 x 60) / 0.052 = 2.016
Bi = hL/k = (1200x0.05) x 215
= 0.28
Using above two parameters in Heisler Chart,
T  T
 0.68
T0  T
 T  100  0.68(500  100)  372C
Page  78
Determination of Surface Temperature
For x/L = 1 and Bi = 0.28,
T  T
 0.88
T0  T
 T  100  0.88(372  100)  339.36C
Page  79
Energy Loss
h2αt/k2
= (12002x8.4x10-5 x 60) / 2152 = 0.157
Bi = hL/k = (1200x0.05) x 215
= 0.28
Using above 2 parameters in Heisler Chart for
Heat flow, Q/Q0 = 0.32
Q0
cVT0  T 

 c(2L)(T0  T )
A
A
 2700 900 0.1400 
 97.2 10 J / m
6
Page  80
2
Heat removed per unit surface area
Q
6
 0.32  97.2 10
A
6
2
 31.110 J / m
Page  81