MODULE 2. One Dimensional Steady State Heat Conduction . 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  PowerPoint PPT Presentation
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MODULE 2
One Dimensional Steady State Heat Conduction
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
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)
&
&
&
&

+
=
(
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
First Law (energy balance)
¶
¶
¶
¶
¶
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
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
…. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . . . .
.. .. . . … . . . . . . .. . . . . . . . . . . . . . . . .
Cold fluid
k
Ts,1
.. . . . . . . . . . .. . . . . . . . . . . . . .
Ts,2
T∞,2
Hot fluid
x=L
x=0
The Plane Wall :
Const. K; solution is:
OHM’s LAW :Flow of Electricity
V=IR elect
Voltage Drop = Current flow×Resistance
Temp Drop=Heat Flow×Resistance
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
T∞,1
Ts,1
Ts,2
T∞,2
qx
(Thermal Resistance )
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
Overall heat transfer coefficient
TA
TB
A
B
Contact Resistance :
B
AB+AC=AA=AD
A
D
K B
LB=LC
T1
T2
K D
K A
C
K c
SeriesParallel :
T1
T2
Assumptions :
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.
h1
r1
T∞,1
r2
h2
T∞,2
k1
r3
k2
T∞,1
T∞,2
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.
Set
Critical Insulation Thickness (contd…)
Maximum – Minimum problem
at
Max or Min. ?Take :
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)
r2
r1
k
T∞,2
Ts,2
Ts,1
T∞,1
Inside Solid:
&
E
=
q
&
V
= Energy generation per unit volume
Conduction with Thermal Energy Generation
Applications: * current carrying conductors
* chemically reacting systems
* nuclear reactors
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
The Plane Wall :
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
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
T∞ h
ro
q
&
r
æ
ö
1
d
dT
q
&
+
=
ç
÷
r
0
r
dr
è
dr
ø
k
Ts
q
&
Assumptions:
1D, steady state, constant k, uniform
Start with 1D heat equation in cylindrical coordinates:
=
=
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
Ts may not be known. Instead, T and h may be specified.
Exercise: Eliminate Ts, using T and h.
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.