Slide 1 MODULE 2
One Dimensional Steady State Heat Conduction
Slide 2 T(x,y,z)
Objectives of conduction analysis
To determine the temperature field, T(x,y,z,t), in a body
(i.e. how temperature varies with position within the body)
T(x,y,z,t) depends on:
 boundary conditions
 initial condition
 material properties (k, cp, …)
 geometry of the body (shape, size)
Why we need T(x,y,z,t) ?
 to compute heat flux at any location (using Fourier’s eqn.)
 compute thermal stresses, expansion, deflection due to temp. etc.
 design insulation thickness
 chip temperature calculation
 heat treatment of metals
Slide 3 Unidirectional heat conduction (1D)
Area = A
x
x+x
0
x
A
qx
qx+x
= Internal heat generation per unit vol. (W/m3)
Solid bar, insulated on all long sides (1D heat conduction)
Slide 4 &
&
&
&

+
=
(
E
E
)
E
E
in
out
gen
st
¶
E

+
D
=
q
q
A
(
x
)
q
&
+
D
x
x
x
¶
t
=
r
D
¶
E
(
A
x
)
u
T
=

q
kA
x
¶
x
¶
¶
¶
E
u
T
¶
=
r
D
=
r
D
q
A
x
A
xc
=
+
D
x
q
q
x
¶
¶
¶
t
t
t
+
D
x
x
x
¶
x
Unidirectional heat conduction (1D)
First Law (energy balance)
Slide 5 ¶
¶
¶
¶
¶
æ
ö
T
T
T
T

+
+
D
+
D
=
r
D
ç
÷
kA
kA
A
k
x
A
x
q
Ac
x
&
¶
¶
¶
¶
¶
x
x
x
x
t
è
ø
¶
¶
¶
æ
ö
T
T
+
=
r
ç
÷
k
q
c
&
¶
¶
¶
x
x
t
è
ø
Thermal inertia
Internal heat generation
Longitudinal conduction
¶
r
¶
¶
2
T
q
c
T
1
T
&
+
=
=
¶
¶
a
¶
2
x
k
k
t
t
Unidirectional heat conduction
(1D)(contd…)
If k is a constant
Slide 6 Unidirectional heat conduction (1D)(contd…)
 For T to rise, LHS must be positive (heat input is positive)
 For a fixed heat input, T rises faster for higher
 In this special case, heat flow is 1D. If sides were not insulated, heat flow could be 2D, 3D.
Slide 7 Boundary and Initial conditions:
 The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body.
 We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).
Slide 8 Boundary and Initial conditions (contd…)
How many BC’s and IC’s ?
 Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate.
* 1D problem: 2 BC in xdirection * 2D problem: 2 BC in xdirection, 2 in ydirection * 3D problem: 2 in xdir., 2 in ydir., and 2 in zdir.
 Heat equation is first order in time. Hence one IC needed
Slide 9 …. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . . . .
.. .. . . … . . . . . . .. . . . . . . . . . . . . . . . .
Cold fluid
k
Ts,1
.. . . . . . . . . . .. . . . . . . . . . . . . .
Ts,2
T∞,2
Hot fluid
x=L
x=0
1 Dimensional Heat Conduction
The Plane Wall :
Const. K; solution is:
Slide 10 Thermal resistance(electrical analogy)
OHM’s LAW :Flow of Electricity
V=IR elect
Voltage Drop = Current flow×Resistance
Slide 11 Thermal Analogy to Ohm’s Law :
Temp Drop=Heat Flow×Resistance
Slide 12 T∞,1
…. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . . . .
.. .. . . … . . . . . . .. . . . . . . . . . . . . . . . .
Cold fluid
k
Ts,1
.. . . . . . . . . . .. . . . . . . . . . . . . .
Ts,2
T∞,2
Hot fluid
x=L
x=0
1
L
1
A
A
A
h
h
k
1
2
1 D Heat Conduction through a Plane Wall
T∞,1
Ts,1
Ts,2
T∞,2
qx
(Thermal Resistance )
Slide 13 Resistance expressions
Slide 14 T∞,1
A
B
C
h1
K A
K B
K C
h2
T∞,2
T∞,1
T∞,2
L A
L B
L C
q x
1
1
A
A
h
h
1
2
Composite Walls :
Overall heat transfer coefficient
Slide 15 TA
TB
A
B
Overall Heat transfer Coefficient
Contact Resistance :
Slide 16 B
AB+AC=AA=AD
A
D
K B
LB=LC
T1
T2
K D
K A
C
K c
SeriesParallel :
Slide 17 T1
T2
SeriesParallel (contd…)
Assumptions :
 Face between B and C is insulated.
 (2) Uniform temperature at any face normal to X.
Slide 18 kI = 20 W/mk
AI = 1 m2, L = 1m
qx
Tf = 100°C
h = 1000 W/ m2 k
T1 = 0°C
kII = 10 W/mk
AII = 1 m2, L = 1m
Example:
Consider a composite plane wall as shown:
Develop an approximate solution for the rate of heat transfer through the wall.
Slide 19 h1
r1
T∞,1
r2
h2
T∞,2
k1
r3
k2
T∞,1
T∞,2
1 D Conduction(Radial conduction in a composite cylinder)
Slide 20 T∞
r0
ri
h
Ti
Critical Insulation Thickness :
Insulation Thickness : r or i
Objective :decrease q , increases
Vary r0 ; as r0 increases ,first term increases, second term decreases.
Slide 21 Set
Critical Insulation Thickness (contd…)
Maximum – Minimum problem
at
Max or Min. ?Take :
Slide 22 R t o t
good for electrical cables
good for steam pipes etc.
R c r=k/h
r0
Critical Insulation Thickness (contd…)
Minimum q at r0 =(k/h)=r c r (critical radius)
Slide 23 r2
r1
k
T∞,2
Ts,2
Ts,1
T∞,1
1D Conduction in Sphere
Inside Solid:
Slide 24 &
E
=
q
&
V
= Energy generation per unit volume
Conduction with Thermal Energy Generation
Applications: * current carrying conductors
* chemically reacting systems
* nuclear reactors
Slide 25 k
Ts,1
Ts,2
Assumptions:
1D, steady state, constant k,
uniform
q
&
T∞,1
T∞,2
Hot fluid
Cold fluid
q
&
x= L
x=+L
x=0
Conduction with Thermal Energy Generation
The Plane Wall :
Slide 26 2
d
T
q
&
+
=
0
2
dx
k
=

=
Boundary
cond
.
:
x
L
,
T
T
,
1
s
=
+
=
x
L
,
T
T
,
2
s
q
&
=

+
+
2
Solution
:
T
x
C
x
C
1
2
2
k
Conduction With Thermal Energy Generation (contd…)
Slide 27 Conduction with Thermal Energy Generation (cont..)
Use
boundary
conditions
to
find
C
and
C
1
2

+
æ
ö
T
T
T
T
2
2
q
L
x
x
&
ç
÷
=

+
+
,
2
,
1
,
2
,
1
s
s
s
s
Final
solution
:
T
1
ç
÷
2
2
k
L
2
L
2
è
ø
Not linear any more
dT
Derive the expression and show that it is not independent of x any more
¢
¢
=

Heat
flux
:
q
k
x
dx
Hence thermal resistance concept is not correct to use when there is internal heat generation
Slide 28 T∞ h
ro
q
&
r
æ
ö
1
d
dT
q
&
+
=
ç
÷
r
0
r
dr
è
dr
ø
k
Ts
q
&
Cylinder with heat source
Assumptions:
1D, steady state, constant k, uniform
Start with 1D heat equation in cylindrical coordinates:
Slide 29 =
=
Boundary
cond
.
:
r
r
,
T
T
0
s
dT
=
=
r
0
,
0
dr
æ
ö
2
q
r
&
ç
÷
2
=

+
Solution
:
T
(
r
)
r
1
T
ç
÷
0
s
2
4
k
r
è
ø
0
Cylinder With Heat Source
Ts may not be known. Instead, T and h may be specified.
Exercise: Eliminate Ts, using T and h.
Slide 30 Cylinder with heat source(contd…)
Example:
A current of 100A is passed through a stainless steel wire having a thermal conductivity K=25W/mK, diameter 3mm, and electrical resistivity R = 2.0 . The length of the wire is 1m. The wire is submerged in a liquid at 100°C, and the heat transfer coefficient is 10W/m2K. Calculate the centre temperature of the wire at steady state condition.