1. One Dimensional Steady State Heat Conduction
MODULE 2
One Dimensional Steady State Heat Conduction

2. 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
T(x,y,z)

3. Unidirectional heat conduction (1D)
Area = A x
x+x
Solid bar, insulated on all long sides (1D heat conduction) qx
qx+x A
= Internal heat generation per unit vol. (W/m3)

4. Unidirectional heat conduction (1D)st
gen
out in E
& = + - ) (
First Law (energy balance) t E q x A = D + -
& ) ( t T xc A u x E D = r ) ( x q T kA D + = -

5. Longitudinal conduction
Unidirectional heat conduction
(1D)(contd…) t T c q x k Ac A kA = + ÷ ø ö ç è æ D - r
&
Longitudinal conduction
Thermal inertia
Internal heat generation t T k c q x = + a r 1 2
&
If k is a constant

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.

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).

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 x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction * 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir.
- Heat equation is first order in time. Hence one IC needed

9. 1- Dimensional Heat Conduction
The Plane Wall : … … k
T∞,2
Ts,1
Ts,2
x=0
x=L
Hot fluid
Cold fluid
Const. K; solution is:

10. Thermal resistance (electrical analogy)
OHM’s LAW :Flow of Electricity
V=IR elect
Voltage Drop = Current flow×Resistance

11. Thermal Analogy to Ohm’s Law :
Temp Drop=Heat Flow×Resistance

12. 1 D Heat Conduction through a Plane Wall… … k
T∞,2
Ts,1
Ts,2
x=0
x=L
Hot fluid
Cold fluid
T∞,1
T∞,1
Ts,1
Ts,2
T∞,2 qx A h 2 1 A k L A h 1
(Thermal Resistance )

13. Resistance expressions

14. Composite Walls : T∞,1 T∞,2 A h Overall heat transfer coefficient h1B C
T∞,1
T∞,2 h1 h2
K A
K B
K C
L A
L B
L C
q x
T∞,1
T∞,2 A h 1 2
Overall heat transfer coefficient

15. Overall Heat transfer Coefficient
Contact Resistance : A B TA TB

16. Series-Parallel : A B D C T1 T2
K c
K D
K A
K B
AB+AC=AA=AD
LB=LC

17. Series-Parallel (contd…)
Assumptions :
Face between B and C is insulated.
(2) Uniform temperature at any face normal to X.

18. Example: qx Consider a composite plane wall as shown:
kI = 20 W/mk
AI = 1 m2, L = 1m
kII = 10 W/mk
AII = 1 m2, L = 1m
T1 = 0°C
Tf = 100°C
h = 1000 W/ m2 k qx
Develop an approximate solution for the rate of heat transfer through the wall.

19. 1 D Conduction(Radial conduction in a composite cylinder)h1
T∞,1 k1 r1 r2 k2
T∞,2 r3 h2

20. Critical Insulation Thickness :
Insulation Thickness : r o-r i
Objective : decrease q , increases
Vary r0 ; as r0 increases ,first term increases, second term decreases.

21. Critical Insulation Thickness (contd…)
Maximum – Minimum problem
Set at
Max or Min. ? Take :

22. Critical Insulation Thickness (contd…)
Minimum q at r0 =(k/h)=r c r (critical radius)
good for steam pipes etc.
good for electrical cables
R c r=k/h r0
R t o t

23. 1D Conduction in Sphere r1 r2 k
Ts,1
Ts,2
T∞,1
T∞,2
Inside Solid:

24. Conduction with Thermal Energy Generation
= Energy generation per unit volume V E q
& =
Applications: * current carrying conductors
* chemically reacting systems
* nuclear reactors

25. Conduction with Thermal Energy Generation
The Plane Wall : k
Ts,1
Ts,2
x=0
x=+L
Hot fluid
Cold fluid
T∞,2
T∞,1
x= -L q
&
Assumptions:
1D, steady state, constant k,
uniform q
&

26. Conduction With Thermal Energy Generation (contd…)x k q T
Solution L
cond
Boundary dx d s + - = 2 : , . 1
&

27. Conduction with Thermal Energy Generation (cont..)
Use
boundary
conditions to
find C
and C 1 2 æ ö q
& L 2 x 2 T - T x T + T
Final
solution : T = ç 1 - ÷ + s , 2 s , 1 + s , 2 s , 1 ç ÷ 2 k L 2 è ø 2 L 2
Not linear any more dT
Heat
flux : q = -
Derive the expression and show that it is not independent of x any more k x dx
Hence thermal resistance concept is not correct to use when there is internal heat generation

28. Cylinder with heat sourcero r Ts
T∞ h q
&
Assumptions:
1D, steady state, constant k, uniform q
&
Start with 1D heat equation in cylindrical co-ordinates: k q dr dT r d = + ÷ ø ö ç è æ 1
&

29. Cylinder With Heat Sourcek T
cond ø è = 4 , : . s q
Solution dr dT
Boundary + ÷ ö ç æ - 2 1 ) (
&
Ts may not be known. Instead, T and h may be specified.
Exercise: Eliminate Ts, using T and h.

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.