Solutions of the Conduction Equation 2.ppt

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1. The Conduction Equation
),(''. trgq
t
H



  ),(.. trgTk
t
T
C
p 



Incorporation of the constitutive equation into the energy
equation above yields:
Dividing both sides by C
p
and introducing the thermal
diffusivity of the material given by
s
m
m
s
m
C
k
p

2



2. Thermal Diffusivity
•Thermal diffusivity includes the effects of properties like
mass density, thermal conductivity and specific heat
capacity.
•Thermal diffusivity, which is involved in all unsteady heat-
conduction problems, is a property of the solid object.
•The time rate of change of temperature depends on its
numerical value.
•The physical significance of thermal diffusivity is
associated with the diffusion of heat into the medium
during changes of temperature with time.
•The higher thermal diffusivity coefficient signifies the
faster penetration of the heat into the medium and the less
time required to remove the heat from the solid.

3. pp C
trg
T
C
k
t
T

),(
.. 













This is often called the heat equation.

p
C
trg
T
t
T


),(
.. 


For a homogeneous material:
p
C
txg
T
t
T


),(
2




4. This is a general form of heat conduction equation.
Valid for all geometries.
Selection of geometry depends on nature of application.

5. General conduction equation based on Cartesian
Coordinates
x
q
xx
q

yy
q

yq
zz
q

zq

6. ),(. txgTk
t
T
C
p




For an isotropic and homogeneous material:
),(
2
txgTk
t
T
C
p




):,,(
2
2
2
2
2
2
tzyxg
z
T
y
T
x
T
k
t
T
C
p 



















7. General conduction equation based on Polar
Cylindrical Coordinates
):,,(
1
2
2
2
2
2
tzrg
z
TT
rr
T
r
r
k
t
T
C
p 

 

























8. General conduction equation based on Polar
Spherical Coordinates
):,,(
sin
1
sin
sin
11
2
2
222
2
2
trg
T
r
T
rr
T
r
rr
k
t
T
C
p




 































X
Y

9. Thermal Conductivity of Brick Masonry Walls

10. Thermally Heterogeneous Materials
 zyxkk ,,
),(. txgTk
t
T
C
p 



),,,( tzyxg
z
z
T
k
y
y
T
k
x
x
T
k
t
T
C
p 







































11. ),,,(
2
2
2
2
2
2
tzyxg
z
T
k
z
T
z
k
y
T
k
y
T
y
k
x
T
k
x
T
x
k
t
T
C
p 



























More service to humankind than heat transfer rate calculations

12. Satellite Imaging : Remote Sensing

13. Thermal Imaging of Brain

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

17. Desert Housing & Composite Walls

18. Steady-State One-Dimensional Conduction
•Assume a homogeneous medium with invariant thermal
conductivity ( k = constant) :
•For conduction through a large
wall the heat equation reduces to:
),,,(
2
2
tzyxg
x
T
k
x
T
x
k
t
T
C
p












),,,(
2
2
tzyxg
x
T
k
t
T
C
p 






One dimensional Transient conduction with heat generation.

19. Steady Heat transfer through a plane slab
0
2
2

dx
Td
A
0),,,(
2
2



tzyxg
x
T
k
No heat generation
211
CxCTC
dx
dT


20. Isothermal Wall Surfaces
Apply boundary conditions to solve for
constants: T(0)=T
s1
; T(L)=T
s2
211
CxCTC
dx
dT

The resulting temperature distribution is:
and varies linearly with x.

21. Applying Fourier’s law:
heat transfer rate:
heat flux:
Therefore, both the heat transfer rate and
heat flux are independent of x.

22. Wall Surfaces with Convection
2112
2
0 CxCTC
dx
dT
dx
Td
A 
Boundary conditions:
 
11
0
)0(


 TTh
dx
dT
k
x
 
22
)(


 TLTh
dx
dT
k
Lx

23. Wall with isothermal Surface and Convection Wall
2112
2
0 CxCTC
dx
dT
dx
Td
A 
Boundary conditions:
1
)0( TxT 
 
22
)(


 TLTh
dx
dT
k
Lx

24. 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

25. q
T
R
R
T
q
th
th




WK
mW
Km
m
kA
L
L
TT
kA
TT
q
T
R
ss
ss
cond
th /
1.
2
12
21












 
WK
mW
Km
hATThA
TT
q
T
R
s
s
conv
th /
1.1
2
2









26.  
WK
mW
Km
AhTTAh
TT
q
T
R
rsurrsr
surrs
rad
th /
1.1
2
2







27. 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
th
= R
th1
+ R
th2
.
•If T
L
is the temperature at
the left, and T
R is the
temperature at the right, the
heat transfer rate is given by
21 thth
RL
th
RR
TT
R
T
q






28. Wall Surfaces with Convection
2112
2
0 CxCTC
dx
dT
dx
Td
A 
Boundary conditions:
 
11
0
)0(


 TTh
dx
dT
k
x
 
22 )(


 TLTh
dx
dT
k
Lx
R
conv,1
R
condR
conv,2
T
1 T
2

29. 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 = Q
1
+ Q
2

with the total resistance given by:

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

31. Desert Housing & Composite Walls