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One Dimensional Steady Heat Conduction problems
P M V Subbarao
Associate Professor
Mechanical Engineering Department
IIT Delhi
Simple ideas for complex Problems…
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
th
R
T
Q



Mel242 6
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
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
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:
Composite Walls
• The overall thermal resistance is given by
Desert Housing & Composite Walls
One-dimensional Steady Conduction in Radial
Systems
0







dr
dr
dT
kA
d
Homogeneous and constant property material
0







dr
dr
dT
A
d
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
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 

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.
• 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
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.
Mel242 6
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,
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:
Structure of Hot Fluid Piping
Rconv,1 Rpipe
Rconv,2
T1 T2
Rinsulation RAl
• 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:
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 





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 







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.

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Mel242 6

  • 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
  • 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.