Chapter2.pptx

2.1 INTRODUCTION
The heat flux given in equation (1.4)
(1.4)
can be written in a more general (three
dimensional) form as
(2.1)
or
(2.2)
dx
dT
k
A
Q
q 


.





















z
T
k
y
T
j
x
T
i
k
T
k
q
 
z
y
x kq
jq
iq
q 




2.1 INTRODUCTION…
Comparison of one dimensional and three dimensional heat flow

2.1 INTRODUCTION…
Major objectives in a conduction analysis is
 to determine the temperature field,
 to determine the conduction heat flux,
 for a solid, to ascertain structural integrity through
 determination of thermal stresses,
 expansions, and
 deflections.
 to optimize the thickness of an insulating material.

2.1 INTRODUCTION…
Fig. 2.1 Differential control volume, dx.dy.dz, for conduction analysis in Cartesian
coordinates
Derivation of heat equation in
rectangular coordinate

2.1 INTRODUCTION…
The conservation of energy principle is used to find the
conduction (energy diffusion) equation.
Referring to Fig. 2.1,
dxdydz
t
T
ρc
volume
control
the
in
stored
energy
of
rate
E
dxdydz
q
volume
control
the
in
generated
energy
of
rate
E
volume
control
the
leaving
energy
of
rate
E
volume
control
the
to
entering
energy
of
rate
E
p
st
.
.
gen
.
out
.
in
.









2.1 INTRODUCTION…
Applying the energy balance equation to the control volume,
(2.3)
(2.4)
But from Taylor series approximation,
(2.5)
And from Fourier’s law
(2.6)
st
out
gen
in E
E
E
E
.
.
.
.



dxdydz
t
T
c
dxdy
q
dxdz
q
dydz
q
dxdydz
q
dxdy
q
dxdz
q
dydz
q
p
dz
z
dy
y
dx
x
z
y
x












 
)
(
)
(
.
z
q
q
q
dy
y
q
q
q
dx
x
q
q
q z
z
dz
z
y
y
dy
y
x
x
dx
x











 

 ,
,
z
T
k
q
y
T
k
q
x
T
k
q z
y
x











 ,
,

2.1 INTRODUCTION…
Substituting equations (2.5) and (2.6) in to equation
(2.4) and dividing by dxdydz gives
(2.7)
Where = energy generated per unit volume ( )
ρ = density ( )
=specific heat capacity (J/kg.K)
Equation (2.7) is a general transient three dimensional
diffusion equation in rectangular coordinate system
with energy generation.
t
T
c
z
T
k
z
y
T
k
y
x
T
k
x
q p





































 
.
.
q
p
c
3
/ m
kg
3
/ m
W

2.1 INTRODUCTION…
For materials with constant thermal conductivity k,
equation (2.7) can be written as
is called thermal diffusivity (m2/s ). It is clear
from the above equation that the ability of a material
to let heat pass through it increases with increasing
thermal diffusivity. This can be due to a high thermal
conductivity k or a low heat capacity of the material.
t
T
t
T
k
c
z
T
y
T
x
T
k
q p
















 1
2
2
2
2
2
2
.
p
c
k

 

2.1 INTRODUCTION…
Fig. 2.2 Differential control volume, dr.rdf.dz, for conduction analysis in cylindrical
Derivation of heat equation in
cylindrical coordinate

2.1 INTRODUCTION…
The heat flux in cylindrical coordinates is
(2.8)
Where
(2.9)
Applying the principle of conservation of energy, the
energy diffusion equation can be obtained as
(2.10)





















z
T
k
T
r
j
r
T
i
k
T
k
q
f
1
z
T
k
q
T
r
k
q
r
T
k
q z
r











 ,
,
f
f
t
T
c
z
T
k
z
T
k
r
r
T
kr
r
r
q p





































 
f
f
2
. 1
1

2.1 INTRODUCTION…
For constant thermal conductivity,
(2.11)
Following similar procedure, the heat diffusion equation
can be obtained in spherical coordinate system (Fig. 2.3).
For constant thermal conductivity
(2.12)
t
T
α
1
t
T
k
ρc
z
T
φ
T
r
1
r
T
r
T
r
1
k
q p
2
2
2
2
2
2
2
.


















t
T
α
1
t
T
k
ρc
θ
T
r
1
θ
T
tanθ
r
1
φ
T
θ
sin
r
1
r
T
r
T
r
2
k
q p
2
2
2
2
2
2
2
2
2
2
.






















2.1 INTRODUCTION…
Fig. 2.3 Differential control volume, dr.rsinθdf.rdθ, for
conduction analysis in spherical coordinates
Derivation of heat equation in spherical
coordinate

2.1 INTRODUCTION…
2.1.1 Boundary and Initial Conditions

2.1 INTRODUCTION…
Example 2.1
Passage of an electric current through a long conducting rod
of radius ri and thermal conductivity kr results in a uniform
volumetric heating at rate of . The conducting rod is
wrapped in an electrically non-conducting cladding
material of outer radius ro and thermal conductivity kc, and
convection cooling is provided an adjoining flow. For
steady state conditions, write appropriate forms of the heat
equations for the rod and cladding. Express appropriate
boundary conditions for the solution of these equations.
.
q

2.1 INTRODUCTION…
Fig. Example 2.1

2.1 INTRODUCTION…
Solution
From Equation 2.10, the appropriate forms of the heat equation are:
In the Conducting Rod,
In the Cladding,
Appropriate boundary conditions are:
symmetry
common temperature at interface
heat flux crossing interface area
convection heat transfer at outer surface
0
.








dr
dT
r
dr
d
k
q
0







dr
dT
r
dr
d
0
0


r
r
dr
dT
)
(
)
( i
c
i
r r
T
r
T 
ri
c
c
ri
r
r
dr
dT
k
dr
dT
k 
 


 T
r
T
h
dr
dT
k o
c
ro
c
c )
(

2.2 STEADY STATE CONDUCTION
2.2.1 One Dimensional Steady State Conduction
In "one-dimensional“ system:
 only one coordinate is needed to describe the spatial
variation of the dependent variables.
 temperature gradients exist along only a single
coordinate direction and heat transfer occurs
exclusively in that direction.
Equation (2.7) for one dimensional steady state heat
conduction with no energy generation in rectangular
coordinate system reduces to
(2.14)
0







dx
dT
k
dx
d

2.2 STEADY STATE CONDUCTION…
Integration of equation (2.14) gives us
constant slope (2.15)
and
linear variation (2.16)
The constants C1 and C2 are obtained from boundary conditions.
Similarly for cylindrical coordinate system (equation (2.11)) reduces
to
(2.17)
Integration of equation (2.17) gives
(2.18)
1
C
k
C
dx
dT


2
1
)
( C
x
C
x
T 

0
1







dr
dT
kr
dr
d
r
2
1 )
ln(
)
( C
r
C
r
T 


2.2.2 Thermal Resistance Concept
Consider equation (2.16) applied to the slab
shown in Fig.2.4. Applying boundary
conditions
The heat transfer rate in the slab becomes
(2.19)
Where R is called thermal resistance.
L
T
T
C
dx
dT
and
T
C
L
T
T
C 1
2
1
1
2
1
2
1 ,






kA
L
R
R
T
T
Q
R
T
T
kA
L
T
T
L
T
T
kA
L
T
T
kA
dx
dT
kA
Q















,
2
1
.
2
1
2
1
2
1
1
2
.
Fig. 2.4 Thermal Resistance Concept
L

Defining resistance as the ratio of a driving potential to the corresponding
transfer rate, the conduction thermal resistance can be written as,
(2.20)
For electrical conduction in the same system, Ohm’s law provides an
electrical resistance of the form
(2.21)
We can also write a thermal resistance for convection heat transfer from
Newton’s Law of cooling
And the convection thermal resistance is
(2.22)
kA
L
Q
T
T
Rcond 

 .
2
1
A
L
I
E
E
Relect



 2
1
R
T
T
hA
T
T
Q
or
T
T
hA
Q s
s
conv
s
conv
)
(
1
)
(
)
(
.
.









hA
Q
T
T
R
conv
s
conv
1
.


 

The thermal resistance concept is more appropriate for heat
transfer problems with composite materials in series or parallel
as shown in Fig. 2.5.
Fig. 2.5 Composite wall in series and corresponding thermal circuit

The heat transfer rate for this system may be expressed as
(2.23)
(2.24)
Alternatively,
(2.25)
In terms of an overall heat transfer coefficient,
(2.26)
Where ΔT is the overall temperature difference.
R
T
R
T
T
Q





 
 4
,
1
,
.
A
h
A
k
L
A
k
L
A
k
L
A
h
T
T
Q
C
C
B
B
A
A
4
1
4
,
1
,
.
1
1





 

A
h
T
T
A
k
L
T
T
A
k
L
T
T
A
k
L
T
T
A
h
T
T
Q s
C
C
s
B
B
A
A
s
s
4
4
,
4
,
4
,
3
3
2
2
1
,
1
1
,
1
,
.
1
1

 









T
UA
Q 

.

From equations (2.23) and (2.26) we see that UA = 1 /ΣR. Hence, for the composite wall of
Fig. 2.5,
(2.27)
In general, we may write
(2.28)
The thermal resistance concept can be applied to radial systems too (Fig.2.6).
Assuming heat transfer only in the radial direction, equation (2.18) can be used to obtain an
expression for the thermal resistance.
Introducing the boundary conditions T(r1)=Ts,1 and T(r2)=Ts,2 , the constants C1 and C2
can be obtained and equation (2.18) becomes
(2.29)
4
1
1
1
1
1
h
k
L
k
L
k
L
h
R
A
U
C
C
B
B
A
A







.
1
Q
T
AU
R
Rtot





2
1 )
ln(
)
( C
r
C
r
T 

2
,
2
2
1
2
,
1
,
ln
ln
)
( s
s
s
T
r
r
r
r
T
T
r
T 



















The conduction heat transfer rate in the cylinder will be
obtained from Fourier’s law,
(2.30)
Equation (2.30) shows that conduction thermal resistance in
cylindrical coordinate system is given by
(2.31)

























1
2
2
,
1
,
2
1
2
,
1
,
.
ln
2
ln
2
2
r
r
T
T
L
k
r
r
r
T
T
rL
k
dr
dT
rL
k
dr
dT
kA
Q s
s
s
s



R
T
L
k
r
r
T
T
Q s
s 












2
ln
1
2
2
,
1
,
.
L
k
r
r
R

2
ln
1
2










For the hollow cylinder shown in Fig. 2.6, the total thermal resistance can
be written as
(2.32)
out
Conv
Cond
in
Conv
tot
R
R
R
L
r
h
L
k
r
r
L
r
h
R
,
,
2
2
1
2
1
1 2
1
2
ln
2
1

















Fig. 2.6 Thermal resistance of a hollow cylinder

Example 2.2
Uninsulated thin-walled pipe of 100mm diameter is used
to transport water to equipment that operates
outdoors and uses the water as a coolant. During
particularly harsh winter conditions the pipe wall
achieves a temperature of -150C and a cylindrical layer
of ice forms on the inner surface of the wall. If the
mean water temperature is 30C and a convection
coefficient of 2000W/m2K is maintained at the inner
surface of the ice (k1.94W/mK), which is at 00C, what
is the thickness of the ice layer?

Fig. Example 2.2

Solution
Performing an energy balance for a control surface about
the ice/water interface, it follows that, for a unit length
of pipe,
Dividing both sides of the equation by r2,
The equation is satisfied by r2/r1=1.114, in which case
r1=0.05m/1.114=0.045m, and the ice layer thickness is
k
r
r
o
Ts
i
Ts
T
T
r
h
q
q
i
s
i
i
cond
conv


2
/
)
/
ln(
,
,
)
)(
2
(
1
2
,
.
1





097
.
0
3
15
05
.
0
2000
94
.
1
,
,
)
/
(
)
/
ln(
,
.
2
1
2
1
2






 x
T
T
o
Ts
i
Ts
r
h
k
r
r
r
r
i
s
i
i
mm
m
r
r 5
005
.
0
1
2 





2.2.3 Thermal Contact Resistance
Surfaces of solids are practically rough with numerous picks and
valleys. When two or more such surfaces are pressed together,
the picks form good contact(conductor) and the valleys form
voids filled with air(insulator), Fig. 2.7(b).
(a) Ideal (b) Actual
Fig. 2.7 Thermal Contact Resistance

Thus, an interface offers some resistance to heat transfer, and
this resistance per unit interface area is called the thermal
contact resistance, Rc, given as
(2.33)
It can also be expressed in the form of Newton’s law of cooling as
(2.34)
Where hc= thermal contact conductance
A=apparent interface area
∆Tinterface=effective temperature difference at interface
The thermal contact resistance is given by
(2.35)
gap
contact Q
Q
Q
.
.
.


erface
c T
A
h
Q int
.


A
Q
T
h
R
erface
c
c
/
1
.
int




Table 2.2 Thermal contact conductance of some metal surfaces in air

2.2.4 Critical Thickness of Insulation
When a plane surface is covered with insulation, the rate of heat
transfer always decreases.
However, the addition of insulation to a cylindrical or spherical
surface increases the conduction resistance but reduces the
convection resistance because of the increased surface area.
The critical thickness of insulation corresponds to the condition
when the sum of conduction and convection resistances is a
minimum. The rate of heat transfer from the insulated pipe to
the surrounding air can be expressed as (Fig. 2.8)
(2.36)
L
hr
L
k
r
r
T
T
Q
2
1
2
1
.
2
1
2
ln












 

The variation of with the outer radius of the insulation r2
is plotted in Fig. 2.9. The value of r2 at which reaches a
maximum is determined from the requirement that .
Solving this for r2 yields the critical radius of insulation for
a cylinder to be
(2.37)
Similarly the critical radius of insulation for a sphere is
given by
.
(2.38)
.
Q
.
Q
0
/ 2
.

dr
Q
d
h
k
rcr 
h
k
rcr
2


From Fig. 2.8 it can be seen that insulating the pipe may actually
increase the rate of heat transfer from the pipe instead of decreasing it
when r2<rcr .
Fig. 2.8 Insulated pipe exposed
to convection from external
Fig. 2.9 Variation of heat transfer
rate with insulation thickness

2.2.5 Optimum Thickness of Insulation
 Insulation does not eliminate heat transfer but it merely
reduces it.
 The thicker the insulation, the lower the rate of heat
transfer but the higher the cost of insulation.
 Therefore, there should be an optimum thickness of
insulation corresponding to a minimum combined cost of
insulation and heat lost (Fig. 2.10).
 The total cost, which is the sum of insulation cost and lost
heat cost, decreases first, reaches a minimum, and then
increases.
 The thickness corresponding to the minimum total cost is
the optimum thickness of insulation, and this is the
recommended thickness of insulation to be installed.

Fig. 2.10 Optimum Insulation Thickness

2.3 EXTENDED SURFACES
 The term extended surface is used to describe a system
in which the area of a surface is increased by the
attachment of fins.
 A fin accommodates energy transfer by conduction
within its boundaries, while its exposed surfaces
transfer energy to the surroundings by convection or
radiation or both.
 Fins are commonly used to augment heat transfer from
electronic components, automobile radiators, engine
and compressor cylinders, control devices, and a host
of other applications.

2.3 EXTENDED SURFACES …
Fig. 2.11 Use of fin to enhance heat transfer from a plane wall

To determine the heat transfer rate associated with a fin, we
must first obtain the temperature distribution along the fin.
The following assumptions in determining the temperature
distribution:
 one-dimensional conduction in the x direction,
 Steady-state conditions,
 Constant thermal conductivity,
 Negligible radiation from the surface,
 heat generation effects are absent, and
 Convection heat transfer coefficient h is uniform over the
surface.

Fig. 2.12 Fin element used for analysis

Applying the conservation of energy requirement to the differential
element of Fig.2.12, we obtain
(2.39)
But from Fourier’ Law,
(2.40)
The conduction heat transfer at x+dx can be expressed as
(2.41)
Inserting equation (2.40) in to equation (2.41)
(2.42)
The convection heat transfer will be expressed as
(2.43)
conv
dx
x
x Q
d
Q
Q
.
.
.

 
dx
dT
kA
Q c
x 

.
dx
dx
Q
d
Q
Q x
x
dx
x
.
.
.



dx
dx
)
dx
dT
kA
d(
dx
dT
kA
Q
c
c
dx
x
. 




)
)(
(
.


 T
T
dA
h
Q
d s
conv

Then equation (2.39) becomes,
Or
(2.44)
For fins with constant cross-section Ac, the element
surface area dAs=P dx (where P is fin perimeter) and
equation (2.44) becomes,
(2.45)
)
)(
(
)
(







 T
T
dA
h
dx
dx
dx
dT
kA
d
dx
dT
kA
dx
dT
kA s
c
c
c
0
)
)(
( 









T
T
dx
dA
k
h
dx
dT
A
dx
d s
c
0
)
(
2
2


 
T
T
kA
hP
dx
T
d
c

Let (x)=(T(x)-T∞)then since T∞ is constant. Equation (2.45) becomes,
Or
(2.46)
Where . The general solution of differential equation (2.46) is
(2.47)
The constants C1 and C2 are obtained from boundary conditions.
0
2
2

 

c
kA
hP
dx
d
0
2
2
2

 

m
dx
d
c
kA
hP
m 
mx
mx
e
C
e
C
x 

 2
1
)
(


The boundary conditions used occur at fin base and
tip.
A. Boundary condition at fin base is specified
temperature condition. This temperature is usually
assumed to be known.
(2.48)
From equation (2.47),
(2.49)




 T
T
x b
b

 )
0
(
2
1 C
C
b 



B. The boundary condition at fin tip has three options
I. Infinitely long fin(L→∞,Ttip=T∞)
(x=L)=Ttip-T∞=0=C1emL+C2e-mL
But as L→∞, e-mL=0 and C1=0 and C2=b (from
equation (2.49))
So, for an infinitely long fin equation (2.47) becomes
or
(2.50)
x
kA
hP
b
mx
b
c
e
e
x



 

 )
(
x
kA
hP
b
c
e
x 


 )
(

The heat removed by the fin at base is
(2.51)
0
0
.






x
c
x
c
base
dx
d
kA
dx
dT
kA
Q

c
b
c
b
c
base
kA
hP
kA
m
kA
Q 
 

.
c
b
base hPkA
Q 

.

II. Negligible heat loss from fin tip
(2.52)
Combining equations (2.49) and (2.52), and solving
for the constants C1 and C2,
(2.53)
The heat loss from fin base is
(2.54)










0
L
x
dx
d
  0
2
1 

 

mL
mL
L
x
e
C
e
C
m
dx
d
 
)
cosh(
)
(
cosh
)
(
mL
x
L
m
x
b




)
tanh(
.
mL
hPkA
Q c
b
base 


III. Convection from fin tip
(2.55)
Or
(2.56)
Solving for the constants C1 and C2
(2.57)
The corresponding heat loss from fin base will be,
(2.58)
L
x
c
L
x
c
dx
dT
kA
T
T
hA
Q






 )
(
.
L
x
c
L
x
c
dx
d
kA
hA






     
  )
sinh(
/
)
cosh(
)
(
sinh
/
)
(
cosh
)
(
mL
mk
h
mL
x
L
m
mk
h
x
L
m
x
b 






)
sinh(
)
/
(
)
cosh(
)
cosh(
)
/
(
)
sinh(
.
mL
mk
h
mL
mL
mK
h
mL
hPkA
Q c
b
base




Example 2.3
A brass rod 100mm long and 5mm in diameter extends
horizontally from a casting at 200 c. The rod is in an air
environment with T=20 c and h=30w/m2c. What is
the temperature of the rod 25, 50 and 100mm from the
casting? Take thermal conductivity of brass to be
k=133W/m.K.

Solution
Based on the assumption of convection heat loss from
fin tip, the temperature distribution, from equation
(2.57), has the form
The temperatures at the prescribed location are
tabulated below
x(m)
T(0C)
0.025 156.5
0.05 128.9
0.10 107.0
     
  )
sinh(
/
)
cosh(
)
(
sinh
/
)
(
cosh
)
(
mL
mk
h
mL
x
L
m
mk
h
x
L
m
x b




 


2.3.1 Fin Effectiveness
The performance of fins is judged on the basis of the
enhancement in the heat transfer relative to the no-fin
case. The performance of fins expressed in terms of the fin
effectiveness efin isdefined as (Fig. 2.13)
(2.59)
In any rational design the value of efin should be as large as
possible, and in general, the use of fins may rarely be
justified unless efin 2.
)
T
(T
hA
Q
Q
Q
A
area
of
surface
the
from
rate
transfer
Heat
A
area
base
of
fin
the
from
rate
transfer
Heat
ε
b
b
fin
.
fin
no
.
fin
.
b
b
fin






Fig. 2.13 Fin Effectiveness

2.3.2 Fin Efficiency
 Another measure of fin thermal performance is provided by the fin efficiency,
fin.
 The maximum driving potential for convection is the temperature difference
between the base (x = 0) and the fluid, b=Tb-T∞.
 Hence the maximum rate at which a fin could dissipate energy is the rate that
would exist if the entire fin surface were at the base temperature.
 However, since any fin is characterized by a finite conduction resistance, a
temperature gradient must exist along the fin and the above condition is an
idealization.
 A logical definition of fin efficiency is therefore
(2.60)
Where Afin is the surface area of the fin.
)
T
(T
hA
Q
Q
Q
b
fin
fin
.
.
fin
.
fin




max


2.3.3 Proper Length of Fin
 The temperature of a fin drops along the fin
exponentially and reaches the environment
temperature at some length.
 The part of the fin beyond this length does not
contribute to the heat transfer.
 Designing such an extra long fin results in material
waste, excessive weight and increased size and
cost.

Fig. 2.14 Proper length of fin
Fig. 2.15 Variation of heat
transfer from a fin relative to that
from relatively long fin

To get the sense of the proper length of a fin, we
compare the heat transfer from a fin of finite length to
the heat transfer from an infinitely long fin with the
same conditions.
(2.61)
This ratio becomes unity for mL2.5 as can be seen from
Fig. 2.15. Therefore, gives proper length of a fin
and the designer should make proper compromise
between heat transfer performance and fin size.
)
tanh(
)
(
)
tanh(
)
(
mL
T
T
hpkA
mL
T
T
hpkA
Q
Q
Ratio
Transfer
Heat
b
c
b
c
fin
long
.
fin
.







m
L
5
.
2


2.4 CONDUCTION WITH THERMAL
ENERGY GENERATION
Fig. 2.16 Conduction in plane wall with uniform energy generation

ENERGY GENERATION…
 Consider a one dimensional heat flow through plane wall
of Fig. 2.16.
 For constant thermal conductivity k, equation (2.7) reduces
to
(2.62)
 The general solution is
(2.63)
 Where C1 and C2 are the constants of integration obtained
from boundary conditions.
0
.
2
2


k
q
dx
T
d
2
1
2
.
2
)
( C
x
C
x
k
q
x
T 




 For the prescribed boundary conditions
T(-L)=Ts,1 and T(L)=Ts,2
 Solving for constants
and
 The temperature distribution becomes
(2.64)
 Consider the long solid cylinder of Fig.2.17
L
T
T
C s
s
2
1
,
2
,
1


2
.
1
,
2
,
2
2
2
L
k
q
T
T
C s
s



2
2
1
2
)
( 1
,
2
,
1
,
2
,
2
2
2
.
s
s
s
s T
T
L
x
T
T
L
x
k
L
q
x
T















 For constant thermal conductivity k, equation (2.10)
reduces to
(2.65)
 And the temperature distribution will be
(2.66)
 The constants of integration C1 and C2, are obtained by
applying the boundary conditions
and
0
1
.








dr
dT
r
dr
d
r
k
q
2
1
2
.
)
ln(
4
)
( C
r
C
r
k
q
r
T 


s
o T
r
T 
)
( 0
0


r
dr
dT

Fig. 2.17 Conduction in a solid cylinder with uniform energy generation

 The constants will be
and
 And the temperature distribution becomes
(2.67)
0
1 
C
2
.
2
4
o
s r
k
q
T
C 

s
o
o
T
r
r
k
r
q
r
T 









 2
2
2
.
1
4
)
(

Example 2.4
A nuclear fuel element of thickness 2L is covered with a steel
cladding of thickness b. Heat generated within the nuclear
fuel at a rate removed by a fluid at T∞, which adjoins one
surface and is characterized by a convection coefficient h.
The other surface is well insulated, and the fuel and steel
have thermal conductivities of kf and ks, respectively.
a. Obtain an equation for the temperature distribution T(x) in
the nuclear fuel. Express your results in terms of , kf, L,W
ks, h and T∞.
b. Sketch the temperature distribution T(x) for the entire
system.
.
q
.
q

Fig. Example 2.4

Solution
(a) The heat equation for the fuel is,
And the corresponding temperature distribution is
The insulated wall at x = -(L+b) dictates that the heat
flux at x = - L is zero (for an energy balance applied to
a control volume about the wall, Ein=Eout=0).
)
(
0
.
2
2
L
x
L
k
q
dx
T
d
f





2
1
2
.
2
)
( C
x
C
x
k
q
x
T
f





Hence
(a)
The value of Ts,1 may be determined from the energy
conservation requirement that or on a
unit area basis
f
f k
L
q
C
or
C
L
k
q
dx
dT
.
1
1
.
0
)
( 






2
.
2
.
2
)
( C
x
k
L
q
x
k
q
x
T
f
f




conv
cond
g Q
Q
E
.
.
.


   





 T
T
h
T
T
b
k
L
q s
s
s
s
2
,
2
,
1
,
.
)
2
(

Hence,
(b)
Combining equations (a) and (b),
(c)
This gives
The temperature distribution for (-L≤x≤L) is



 T
h
L
q
k
Lb
q
T
s
s
)
2
(
)
2
(
.
.
1
,
2
2
.
.
.
1
,
2
3
)
2
(
)
2
(
)
( C
k
L
q
T
h
L
q
k
Lb
q
T
L
T
f
s
s 





 












 T
k
L
h
k
b
L
q
C
f
s 2
3
2
2
.
2















 T
k
L
h
k
b
L
q
x
k
L
q
x
k
q
x
T
f
s
f
f 2
3
2
2
2
)
(
.
.
2
.

(b) The temperature distribution is shown in the
following three regions
t
cons
dx
dT
b
L
x
L
x
with
dx
dT
L
x
L
T
T
dx
dT
L
x
b
L
tan
:
:
,
0
:
)
( max

















2.5 TRANSIENT CONDUCTION
 Practical problems in heat transfer, in general, involve
the variation of temperature with
 Position (x,y,z) and
 Time (t).
 Analysis of transient heat conduction is more
complicated than that of steady state conduction and
making simplifying assumptions is more appropriate.
 Lumped capacitance method and
 One dimensional assumptions.

2.5 TRANSIENT CONDUCTION…
2.5.1 Lumped Capacitance System
 The temperature variations within some bodies remain
essentially uniform at a given time during a heat transfer
process (Fig. 2.18).
 Such bodies are said to act like a ‘lump’.
 Consider a hot metal forging that is initially at a uniform
temperature Ti and is quenched by immersing it in a liquid
of lower temperature T∞<Ti (Fig.2.18).
 The essence of the lumped capacitance method is the
assumption that the temperature of the solid is spatially
uniform at any instant during the transient process
(T=f(t)).

Fig. 2.18. Cooling of a hot metal forging

 The variation of temperature of the hot metal with
time can be obtained by applying energy balance
within a time interval of dt.
(2.68)
(2.69)
Rearranging equation (2.69) and noting that dT=d(T-
T∞), since T∞ is constant, we obtain
(2.70)
dt
in
body
the
of
content
energy
the
in
decrease
dt
during
body
the
from
transfer
Heat 
Out
conv E
Q
.
.

dt
dT
mC
T
T
hA p


  )
(
dt
mC
hA
T
T
dT
p


 

 Integration of equation (2.70) gives us
Or
(2.71)
 The temperature of the metal gradually decreases and
eventually equals the ambient air temperature.




 
t
p
T
T
dt
mC
hA
T
T
dT
i 0
t
mC
hA
T
T
T
t
T
p
i














)
(
ln
t
mC
hA
i
p
e
T
T
T
t
T






)
(

 The rate of convection heat transfer is
(2.72)
 Total amount of heat transfer from the hot body to the
surrounding, time 0-t, is
(2.73)
 The maximum amount of heat transfer is
(2.74)
]
)
(
[
.


 T
t
T
hA
Q
 




)
(
]
)
(
[
t
T
T
i
p
p
i
T
t
T
mC
dT
mC
Q
]
[
]
[
max 
 



 T
T
mC
T
T
mC
Q i
p
i
p

Validity of lumped capacitance method
 The lumped capacitance analysis gives us a simple and
convenient way of analyzing transient heat transfer
problems.
 But this method is ideal since it assumes uniform
temperature distribution within a body at an instant
and it is better to know when to use it.
 Before establishing a criterion for the validity of the
method, it is worthy to define the terms known as
characteristic length, Lc, and Biot number, Bi.

(2.75)
(2.76)
 The Biot number can also be expressed as
Or
A
V
Lc 
k
hL
B c
i 
Body
the
within
Transfer
Heat
Conduction
Body
the
of
Surface
the
at
Transfer
Heat
Convection
T
L
k
T
h
B
c
i 



Body
the
of
Surface
the
at
ce
sis
Convection
Body
the
within
ce
sis
Conduction
h
k
L
B c
i
tan
Re
tan
Re
/
1
/



 The lumped capacitance analysis assumes uniform
temperature distribution within the body.
 This is true when the conduction resistance within the
body is zero, which is in turn true only when the Biot
number is zero.
 Therefore, the lumped capacitance analysis is exact
only when Bi=0 and approximate when Bi>0.
 From experience the lumped capacitance method is, in
general, acceptable if Bi≤0.1.

2.5.2 Transient heat conduction in large plane walls,
long cylinders and spheres
 A more realistic assumption than lumped capacitance
method is the case where temperature varies with time
and position in one dimension, T(x, t) or T(r, t), which is
applicable for large plane walls, long cylinders and
spheres (Fig. 2.19).
 If heat transfer takes place between the environment (h
and Ti>T∞) and the large plane wall (initially at Ti ) of
Fig. 2.19(a) by convection, the temperature at the surface
of the wall drops.

(a) Large plane wall (b) Long cylinder (c) Sphere
Fig. 2.19 Bodies where one dimensional temperature variation can be assumed

Fig. 2.20 Transient temperature variation in a large plane wall

 For the large plane wall the diffusion equation (2.7)
becomes,
(2.77)
 The solution of equation (2.77) results in infinite series
which are inconvenient and time consuming to
evaluate.
 Therefore, the solutions are presented in tabular or
graphical form.
t
T
c
x
T
k
x
p














t
T
x
T






1
2
2

 Before presenting the solutions graphically, some
parameters need to be nondimensionalzed to reduce
number of parameters.
 
time
ess
Dimensionl
number
Fourier
t)
coefficien
transfer
heat
less
(Dimension
number
Biot
center
from
distance
ess
Dimensionl
e
temperatur
ess
Dimensionl
)
,
(
)
,
(
2
L
t
k
hL
Bi
L
x
X
T
T
T
t
x
T
t
x
i












 For Fourier number , the infinite series solutions of
equation (2.77) can be approximated by taking only
the first terms of the series. These solutions are given
below,
(2.78)
(2.79)
(2.80)
2
.
0
,
cos
)
,
(
)
,
( 1
1
2
1










 




 

L
x
e
A
T
T
T
t
x
T
t
x
i
wall
2
.
0
,
)
,
(
)
,
( 1
1
2
1












 




 

o
o
i
cyl
r
r
J
e
A
T
T
T
t
r
T
t
r
2
.
0
,
sin
)
,
(
)
,
(
1
1
1
2
1












 





 

o
o
i
sph
r
r
r
r
e
A
T
T
T
t
r
T
t
r

 The constants A1 and 1 are obtained from table 2.3.
 The function J0 is the zeroth-order Bessel function of the
first kind whose value can be determined from Table 2.4.
 The temperature of the body changes from Ti to T∞ at the
end of the transient heat conduction.
 The maximum amount of heat transfer during this process
can be obtained by
(2.81)
)
(
)
(
max i
p
i
p T
T
VC
T
T
mC
Q 


 
 

 The fraction of heat transfer within time t is obtained
by the following equations for the large wall, long
cylinder and sphere.
(2.82)
(2.83)
(2.84)
1
1
1
max
)
sin(
1
2
1















e
A
Q
Q
wall
1
1
1
1
max
)
(
2
1
2
1



 J
e
A
Q
Q
cylinder











3
1
1
1
1
1
max
)
cos(
)
sin(
3
1
2
1





 









 
e
A
Q
Q
wall

Example 2.5
A load of peas at a temperature of 250C is to be cooled
down in a room at constant air temperature of 10C.
(a) How long the peas will require to cool down to 20C when the
surface heat transfer coefficient of the pea is 5.81W/m2K?
(b) What is the temperature of the peas after a lapse of 10
minutes from the start of cooling?
(c) What air temperature must be used if the peas were to be
cooled down to 50C in 30 minutes? The peas are supposed to
have an average diameter of 8 mm, their density is 750kg/m3
and specific heat 3.35 kJ/kgK.

Solution
The problem can be solved by making use of the lumped
capacitance method, neglecting any variation of
temperature within the peas due to its small diameter.
From equation (2.68)
(a) Solving for time t,
t
mC
hA
i
p
e
T
T
T
t
T






)
(
min
6
.
30
1835
1
25
1
2
ln
008
.
0
*
*
81
.
5
3350
*
6
008
.
0
*
*
750
)
(
ln 2
3


































s
T
T
T
t
T
hA
mC
t
i
p



(b) From equation (2.65)
(c)
  C
T
e
T
e
T
T
T
t
T t
mC
hA
i
p 0
600
*
3350
*
6
008
.
0
*
*
750
008
.
0
*
*
81
.
5
48
.
9
600
353
.
0
1
25
1
)
600
(
)
(
3
2









 



















C
T
e
T
T
e
T
T
T
t
T t
mC
hA
i
p 0
1800
*
3350
*
6
008
.
0
*
*
750
008
.
0
*
*
81
.
5
08
.
4
044
.
0
25
5
)
(
3
2


































2.6 NUMERICAL METHODS
Most of the practical problems encountered in
engineering involve:
 Complicated geometries,
 Complex boundary conditions, and
 Variable properties.
Since such problems cannot be solved analytically,
the need for numerical solution methods,
especially in cases of multidimensional problems,
is inevitable.

2.6 NUMERICAL METHODS…
2.6.1 Finite Difference Equation
There are several types of numerical methods. Some
are:
 Finite Difference Method,
 Finite Element Method,
 Boundary Element Method, and
 Control Volume Method.
Because of its ease of application, the finite-
difference method is well suited for an
introductory treatment of numerical techniques.

 A numerical solution enables determination of the
temperature at only discrete points.
 The first step in any numerical analysis must,
therefore, be to select these points.
 This is done by subdividing the medium of interest
into a number of small regions and assigning to each a
reference point that is at its center.
 The reference point is frequently termed as nodal
point (or simply a nod), and the aggregate of points is
termed a nodal network, grid, or mesh.

 The nodal points are designated by a numbering
scheme that, for a two-dimensional system, may take
the form shown in Fig.2.21.
 The x and y locations are designated by the m and n
indices, respectively.
 The temperature of node (m, n) is assumed to be the
average of the surrounding shaded area.
 The accuracy of a numerical analysis is increased by
increasing the number of nodes (fine nodes).
 But the increased number of nodes requires more
computing time and capacity.

Fig. 2.21 Nodal network of two-dimensional conduction

 The finite-difference equation for a node can be
obtained by applying conservation of energy to a
control volume about the nodal region.
 Since the actual direction of heat flow (into or out of
the node) is often unknown, it is convenient to
formulate the energy balance by assuming that all the
heat flow is into the node.
 For steady-state conditions with no generation, the
appropriate form of equation (2.3) is
(2.85)
0
.

 in
E

There are different finite difference equations for interior
and boundary nodes.
1. For interior node (m, n) of Fig. 2.22, the finite
difference equation can be obtained, assuming unit
depth, as
for x=y, the above equation simplifies to
(2.86)
0
)
,
(
)
,
1
(
.
)
,
(
)
,
1
(
.
)
,
(
)
1
,
(
.
)
,
(
)
1
,
(
.



 






 n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m Q
Q
Q
Q
0
1
.
1
.
1
.
1
. )
,
(
)
,
1
(
)
,
(
)
,
1
(
)
,
(
)
1
,
(
)
,
(
)
1
,
(















































 



x
T
T
y
k
x
T
T
y
k
y
T
T
x
k
y
T
T
x
k n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
0
4 )
,
(
)
,
1
(
)
,
1
(
)
1
,
(
)
1
,
( 



 


 n
m
n
m
n
m
n
m
n
m T
T
T
T
T

Fig. 2.22 Interior node for finite difference equation formulation

2. Internal corner node with convection (Fig. 2.23)
(2.87)
0
.
)
,
(
)
,
1
(
.
)
,
(
)
,
1
(
.
)
,
(
)
1
,
(
.
)
,
(
)
1
,
(
.




 






 conv
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m Q
Q
Q
Q
Q
  0
1
.
2
2
1
.
1
.
2
1
.
2
1
.
)
,
(
)
,
(
)
,
1
(
)
,
(
)
,
1
(
)
,
(
)
1
,
(
)
,
(
)
1
,
(







 























































n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
T
T
y
x
h
x
T
T
y
k
x
T
T
y
k
y
T
T
x
k
y
T
T
x
k
0
2
3
2
)
(
)
(
2 )
,
(
)
,
1
(
)
1
,
(
)
,
1
(
)
1
,
( 







 




 



 T
k
xh
T
k
xh
T
T
T
T n
m
n
m
n
m
n
m
n
m

Fig. 2.23 Internal corner node with convection

3. Plane surface node with convection (Fig. 2.24)
(2.88)
0
.
)
,
(
)
,
1
(
.
)
,
(
)
1
,
(
.
)
,
(
)
1
,
(
.



 




 conv
n
m
n
m
n
m
n
m
n
m
n
m Q
Q
Q
Q
  0
1
.
1
.
1
.
2
1
.
2
)
,
(
)
,
(
)
,
1
(
)
,
(
)
1
,
(
)
,
(
)
1
,
(











































n
m
n
m
n
m
n
m
n
m
n
m
n
m
T
T
y
h
x
T
T
y
k
y
T
T
x
k
y
T
T
x
k
0
2
2
2
)
2 )
,
(
)
1
,
(
)
,
1
(
)
1
,
( 







 



 


 T
k
xh
T
k
xh
T
T
T n
m
n
m
n
m
n
m

Fig. 2.24 Plane surface node with convection

4. External corner node with convection (Fig. 2.25)
(2.89)
0
.
)
,
(
)
,
1
(
.
)
,
(
)
1
,
(
.


 


 conv
n
m
n
m
n
m
n
m Q
Q
Q
  0
1
.
2
2
1
.
2
1
.
2
)
,
(
)
,
(
)
,
1
(
)
,
(
)
1
,
(







 





























n
m
n
m
n
m
n
m
n
m
T
T
x
y
h
x
T
T
y
k
y
T
T
x
k
0
2
1
2
) )
,
(
)
1
,
(
)
,
1
( 







 


 

 T
k
xh
T
k
xh
T
T n
m
n
m
n
m

Fig. 2.25 External corner node with convection

5. Plane surface node with heat flux (Fig. 2.26)
(2.90)
0
.
)
,
(
)
,
1
(
.
)
,
(
)
1
,
(
.
)
,
(
)
1
,
(
.



 




 flux
n
m
n
m
n
m
n
m
n
m
n
m Q
Q
Q
Q
0
'
'
1
.
1
.
1
.
2
1
.
2
)
,
(
)
,
1
(
)
,
(
)
1
,
(
)
,
(
)
1
,
(





































 


q
y
h
x
T
T
y
k
y
T
T
x
k
y
T
T
x
k
n
m
n
m
n
m
n
m
n
m
n
m
0
'
'
2
4
2 )
,
(
)
1
,
(
)
,
1
(
)
1
,
( 




 


k
xq
T
T
T
T n
m
n
m
n
m
n
m

Fig. 2.26 Plane surface node with heat flux

2.6.2 Solution of the finite difference equations
The equations obtained for each type of node reduce
the heat transfer problem to solving of system of
linear equations, which can be written in matrix
notation as,
(2.91)
Where [A] is coefficient matrix, {T} is vector of
nodal temperatures and {C} is vector of constants
obtained from boundary conditions.
    
C
T
A 

 





















nn
n
n
n
n
a
a
a
a
a
a
a
a
a
A
....
.
.
.
.
.
.
.
.
.
....
....
2
1
2
22
21
1
12
11
 























n
T
T
T
T
.
.
.
2
1
 























n
C
C
C
C
.
.
.
2
1

 Equation (2.91) can be solved using either the matrix
inversion or the iterative methods. In the matrix
inversion method, the nodal temperatures will be
obtained from
(2.92)
 A good example of iterative methods of solving linear
system of equations is the Gauss-Seidel Iteration
method.
     
C
A
T
1



 Consider the following system of equations for
explanation of the solution procedure.
a) Solve each equation for one of the variables (one with
larger coefficient) in terms of other variables,
13
2
4
5
2
10
3
3
2
1
3
2
1
3
2
1











x
x
x
x
x
x
x
x
x
)
(
)
2
/(
)
13
(
)
(
5
/
)
2
4
(
)
(
3
10
2
1
3
3
1
2
3
2
1
c
x
x
x
b
x
x
x
a
x
x
x












b) Make initial guess for each unknown,
Let x2=0 and x3=0
c) Using equations from step 1, find new values for each
unknown,
Using the initial guess and equation (a) of step 1,
x1=10+3(0)-0=10
Using the updated value x1=-5 and equation (b) of step 1,
x2=(4-2(10)-0)/5=-3.2
From equation (c), x3=(-13+10-(-3.2))/(-2)=-0.1
d) Repeat step 3 until a desired convergence criterion is
satisfied.

Example 2.6
Consider the square channel shown in the sketch operating
under steady-state conditions. The inner surface of the
channel is at a uniform temperature of 600K, while the
outer surface is exposed to convection with a fluid at 300K
and a convection coefficient of 50W/mK.
a) Beginning with properly defined control volume, derive
the finite-difference equations for interior and boundary
nodes. Due to symmetric nature of the problem, take one
eighth of the geometry with Δx=Δy=0.01m. Calculate the
temperatures for all nodes.
b) Calculate the heat loss per unit length from the channel.

Fig. Example 2.6

Solution
(a) Define control volumes about the nodes taking
advantage of symmetry where appropriate and
performing energy balances, , with
Δx=Δy=0.01m. The one eighth geometry of the channel
is meshed in to nine nodes as shown below.
0
.
.

 out
in E
E

 Node 1:
 
 
)
1
(
150
5
.
2
2
0
0
2
2
2
0
1
5
2
1
5
2
1
1
5
1
2
1
1
5
1
2
.
1
5
.
1
2
.












 





























T
T
T
T
k
x
h
T
k
x
h
T
T
T
T
k
x
h
T
T
T
T
T
T
x
h
x
T
T
x
k
x
T
T
x
k
Q
Q
Q Conv

 Node 2:
 
)
2
(
300
5
2
0
2
2
0
2
3
6
1
2
2
3
2
6
2
1
.
2
3
.
2
6
.
2
1
.




























T
T
T
T
gives
equation
this
solving
T
T
x
h
x
T
T
x
k
x
T
T
x
k
x
T
T
x
k
Q
Q
Q
Q Conv

 Node 3: Similar to Node 2,
 Node 4:
)
3
(
300
5
2 3
4
7
2 



 T
T
T
T
 
)
4
(
150
5
.
1
0
2
2
0
4
3
4
4
3
.
4
3
.














T
T
gives
equation
this
solving
T
T
x
h
x
T
T
x
k
Q
Q Conv

 Node 5:
)
5
(
600
4
2
0
2
2
0
5
6
1
5
8
5
6
5
1
5
8
.
5
6
.
5
1
.


















 


T
T
T
gives
equation
this
solving
x
T
T
x
k
x
T
T
x
k
x
T
T
x
k
Q
Q
Q

 Node 6: Interior node
 
)
6
(
600
4
0
4
int
83
.
2
6
7
5
2
6
9
7
5
2










T
T
T
T
T
T
T
T
T
node
erior
for
equation
From

 Node 7:
)
7
(
0
2
0
0
7
6
3
7
6
7
3
7
6
.
7
3
.












 

T
T
T
gives
equation
this
solving
x
T
T
x
k
x
T
T
x
k
Q
Q

 Equations (1) through (7) can be solved simultaneously
using either matrix inversion method or Gauss-Seidel
iteration method. But since the number of equations is
few, the matrix inversion method can be used.
(8)
0
2
600
4
600
4
2
150
5
.
1
300
5
2
300
5
2
150
5
.
2
7
6
3
6
7
5
2
5
6
1
4
3
3
4
7
2
2
3
6
1
1
5
2





























T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T

 Equation (8) can be written in matrix form as
[A]{T}={C} where
 






























2
,
1
,
0
,
0
,
1
,
0
,
0
1
,
4
,
1
,
0
,
0
,
1
,
0
0
,
2
,
4
,
0
,
0
,
0
,
1
0
,
0
,
0
,
5
.
1
,
1
,
0
,
0
2
,
0
,
0
,
1
,
5
,
1
,
0
0
,
2
,
0
,
0
,
1
,
5
,
1
0
,
0
,
1
,
0
,
0
,
1
,
5
.
2
A  























7
6
5
4
3
2
1
T
T
T
T
T
T
T
T  





























0
600
600
150
300
300
150
C
,
,

The temperatures can be obtained using matrix
conversion as, {T}=[A]-1{C}
The solution will be,
 














































112
.
443
092
.
492
571
.
503
755
.
362
133
.
394
684
.
421
102
.
430
7
6
5
4
3
2
1
T
T
T
T
T
T
T
T

b) Referring to Fig. Example 2.6 Solution (a), the heat loss
from the channel is eight times the heat loss from the one
eighth portion shown.
m
3746.946W/
















 



))
300
755
.
362
(
5
.
0
)
300
133
.
394
(
)
300
684
.
421
(
)
300
102
.
430
(
5
.
0
(
03
.
0
50
8
))
(
5
.
0
)
(
)
(
)
(
5
.
0
(
8 4
3
2
1
.
x
x
T
T
T
T
T
T
T
T
hA
Qloss

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