This document discusses one-dimensional steady heat conduction problems. It introduces the electrical circuit theory analogy of thermal resistance and describes how to model heat transfer through composite walls and radial systems using thermal resistance networks. The document also discusses how to calculate the critical thickness of insulation on pipes to maximize heat transfer rate while maintaining a safe surface temperature for worker contact.

1. One Dimensional Steady Heat Conduction problems
P M V Subbarao
Associate Professor
Mechanical Engineering Department
IIT Delhi
Simple ideas for complex Problems…

2. Electrical Circuit Theory of Heat Transfer
• Thermal Resistance
• A resistance can be defined as the ratio of a driving
potential to a corresponding transfer rate.
i
V
R


Analogy:
Electrical resistance is to conduction of electricity as thermal
resistance is to conduction of heat.
The analog of Q is current, and the analog of the
temperature difference, T1 - T2, is voltage difference.
From this perspective the slab is a pure resistance to heat
transfer and we can define

3. th
R
T
Q




5. The composite Wall
• The concept of a thermal
resistance circuit allows
ready analysis of problems
such as a composite slab
(composite planar heat
transfer surface).
• In the composite slab, the
heat flux is constant with x.
• The resistances are in series
and sum to R = R1 + R2.
• If TL is the temperature at the
left, and TR is the
temperature at the right, the
heat transfer rate is given by

6. Wall Surfaces with Convection
2
1
1
2
2
0 C
x
C
T
C
dx
dT
dx
T
d
A 





Boundary conditions:
 
1
1
0
)
0
( 



 T
T
h
dx
dT
k
x
 
2
2 )
( 



 T
L
T
h
dx
dT
k
L
x
Rconv,1 Rcond Rconv,2
T1 T2

7. Heat transfer for a wall with dissimilar
materials
• For this situation, the total heat flux Q is made up of the heat flux
in the two parallel paths:
• Q = Q1+ Q2
 with the total resistance given by:

8. Composite Walls
• The overall thermal resistance is given by

9. Desert Housing & Composite Walls

10. One-dimensional Steady Conduction in Radial
Systems
0







dr
dr
dT
kA
d
Homogeneous and constant property material
0







dr
dr
dT
A
d

11. At any radial location the surface are for heat conduction
in a solid cylinder is:
rl
Acylinder 
2

At any radial location the surface are for heat conduction
in a solid sphere is:
2
4 r
Asphere 

The GDE for cylinder:
0







dr
dr
dT
r
d

12. The GDE for sphere:
0
2







dr
dr
dT
r
d
General Solution for Cylinder:
    2
1 ln C
r
C
r
T 

General Solution for Sphere:
 
r
C
C
r
T 1
2 


13. Boundary Conditions
• No solution exists when r = 0.
• Totally solid cylinder or Sphere have no physical relevance!
• Dirichlet Boundary Conditions: The boundary conditions in any heat
transfer simulation are expressed in terms of the temperature at the
boundary.
• Neumann Boundary Conditions: The boundary conditions in any heat
transfer simulation are expressed in terms of the temperature gradient
at the boundary.
• Mixed Boundary Conditions: A mixed boundary condition gives
information about both the values of a temperature and the values of its
derivative on the boundary of the domain.
• Mixed boundary conditions are a combination of Dirichlet boundary
conditions and Neumann boundary conditions.

14. • If A, is increased, Q will increase.
• When insulation is added to a pipe, the outside
surface area of the pipe will increase.
• This would indicate an increased rate of heat
transfer
• The insulation material has a low thermal conductivity, it reduces the
conductive heat transfer lowers the temperature difference between the outer
surface temperature of the insulation and the surrounding bulk fluid
temperature.
• This contradiction indicates that there must be a critical thickness of
insulation.
• The thickness of insulation must be greater than the critical thickness, so
that the rate of heat loss is reduced as desired.
Mean Critical Thickness of Insulation
Heat loss from a pipe:
 


 T
T
hA
Q s
h,T
Ts
ri
ro

15. Electrical analogy:
total
R
T
er
heattransf
of
Rate


o
o
i
o
i
Lh
r
r
r
Lk
T
T
Q

 2
1
ln
2
1










 
As the outside radius, ro, increases, then in the denominator, the first term
increases but the second term decreases.
Thus, there must be a critical radius, rc , that will allow maximum rate of
heat transfer, Q
The critical radius, rc, can be obtained by differentiating and setting the
resulting equation equal to zero.

17. Ti,Tb, k, L, ro, ri are constant terms, therefore:
0
1
2


o
o
o r
h
k
r
When outside radius becomes equal to critical radius, or ro = rc,
we get,

18. Safety of Insulation
• Pipes that are readily accessible by workers are subject to safety
constraints.
• The recommended safe "touch" temperature range is from 54.4 0C to
65.5 0C.
• Insulation calculations should aim to keep the outside temperature of
the insulation around 60 0C.
• An additional tool employed to help meet this goal is aluminum
covering wrapped around the outside of the insulation.
• Aluminum's thermal conductivity of 209 W/m K does not offer much
resistance to heat transfer, but it does act as another resistance while
also holding the insulation in place.
• Typical thickness of aluminum used for this purpose ranges from 0.2
mm to 0.4 mm.
• The addition of aluminum adds another resistance term, when
calculating the total heat loss:

19. Structure of Hot Fluid Piping
Rconv,1 Rpipe
Rconv,2
T1 T2
Rinsulation RAl

20. • However, when considering safety, engineers need a quick way to
calculate the surface temperature that will come into contact with the
workers.
• This can be done with equations or the use of charts.
• We start by looking at diagram:

21. At steady state, the heat transfer rate will be the same for each layer:
Al
insulation
pipe R
T
T
R
T
T
R
T
T
Q 4
3
3
2
2
1 






22. Solving the three expressions for the temperature difference yields:
Each term in the denominator of above Equation is referred to as the
“Thermal resistance" of each layer.
total
Al
insulation
pipe R
T
T
R
T
T
R
T
T
R
T
T
Q 4
1
4
3
3
2
2
1 








23. Design Procedure
• Use the economic thickness of your insulation as a basis for your
calculation.
• After all, if the most affordable layer of insulation is safe, that's the one
you'd want to use.
• Since the heat loss is constant for each layer, calculate Q from the bare
pipe.
• Then solve T4 (surface temperature).
• If the economic thickness results in too high a surface temperature,
repeat the calculation by increasing the insulation thickness by 12 mm
each time until a safe touch temperature is reached.
• Using heat balance equations is certainly a valid means of estimating
surface temperatures, but it may not always be the fastest.
• Charts are available that utilize a characteristic called "equivalent
thickness" to simplify the heat balance equations.
• This correlation also uses the surface resistance of the outer covering
of the pipe.