CHAPTER 5
POROUS MEDIA
1
5.1 Examples of Conduction in Porous Media
porous
shield
coolant
coolant
blade
coolant
porous ring
(a)
(c)
(b)
micro channels
coolant
electronic
component
coolant
porous
material
(e)
(d)
Fig. 5.1
2
5.2 Simplified Heat Transfer Model
• Assume:
At any point the solid and liquid
are at the same temperature
5.2.1 Porosity
• Definition: Porosity
pore volume V f
P

total volume V
V  AL
(5.1)
(a)
3
Model: Pores are straight channels
V f  Af L
(b)
(a) and (b) into eq. (5.1)
P
Af
(c)
A
or
A f  PA
(5.2)
total flow area Af
Fig. 5.2
and solid wall area As
As  (1  P ) A
• Assume: Porosity is constant
(5.3)
4
5.2.2 Governing Equation: Cartesian
Coordinates
One-dimensional transient conduction
A = surface area
m = flow rate
P = porosity
q = energy generation
dx
q
m
• Assumptions:
(1) Constant porosity
(2) Constant flow rate
0
x
Fig. 5.3
(3) Constant properties (solid and fluid)
5
(4) Solid and fluid at same temperature
(5) Negligible changes in kinetic and potential energy
• Conservation of energy for element dx
E in  E g  E out  E
(d)
Use Fourier’s law
T
T

E in   k s (1  P ) A
 k f PA
 m c pf T
x
x
(e)
Subscripts: s = solid, f = fluid
• Define: Conductivity of the solid-fluid matrix k as
6
k  (1  P ) k s  P k f
(5.4)
(e) becomes
T

 c pf T
E in   k A
m
x
(f)
Use (f) to formulate
E out
T
 2T
T
 c pf (T 
 k A
 k A 2 dx  m
dx )
x
x
x
E g  qA dx
(g)
(h)
Rate of energy change within the element E
T
T

E   s c ps (1  P ) A
dx  f c pf PA
dx
t
t
(i)
7
Define: Heat capacity of the solid-fluid matrix  c p as
 c p  (1  P )  s c ps  P  f c pf
(5.5)
(i) becomes
T

E  c p A
dx
t
(j)
(f), (g), (h) and (j) into (d)
m c pf T q 1 T



2
A k  x k  t
x
 2T
(5.6)
 = thermal diffusivity of solid-fluid matrix, defined as
k
 
c p
(5.7)
8
5.2.3 Boundary Conditions
porous wall
m
m
T
To
T (0)
h
reservoir
0
x
Fig. 5.4
(i) Specified temperature
At inlet or outlet
T (0, t )  T1 or T ( L, t )  T2
(5.8)
9
porous wall
m
To
m
T (0)
0
Fig. 5.4
reservoir
T
h
x
(ii) Convection at outlet boundary
T ( L, t )
 ks
 hT ( L, t )  T 
x
(5.9)
(iii) Inlet supply reservoir: Conservation of energy for the
control volume shown
T (0, t )
m c pf To  T (0, t )   k A
(5.10)
x
10
5.2.4 Cylindrical Systems
Conservation of energy and Fourier’s conduction law
applied to the element dr:
r
dr
m

Fig. 5.5
m c pf 1 T q 1 T
 (1 
)
 
2
2 k L r  r k   t
r
 2T
(5.10)
11
5.3 Applications
Example 5.1: Steady State Conduction in a
Porous Plate
porous wall
Plate thickness = L
Heated at x = L by convection: h, T
T
m

m

Coolant reservoir temperature = To T
T(0)
h
o
Design hot side temperature = Td
reservoir 0 x
Determine: m / A
Fig. 5.4
Solution
(1) Observations
• 1-D conduction in porous plate
• T ( L)  Td
12
• T ( x ) depends on m
(2) Origin and Coordinates
See Fig. 5.4.
(3) Formulation
(i) Assumptions
(1) Solid and fluid at same temperature
(2) Constant porosity
(3) Constant properties
(4) Constant flow rate
(5) No energy generation
(6) Steady state
13
(ii) Governing Equation
Eq. (5.6):
d 2T
m
 c pf dT

0
2
Ak d x
dx
or
d 2T
dx
2

 dT
Ldx
0
 is coolant flow parameter defined as
m c pf L

Ak
(a)
(b)
14
(iii) Boundary Conditions
porous wall
Supply reservoir, eq. (5.10), use
m

m

definition of 
T(0)
To

dT (0)
(c)
To  T (0)  
0
reservoir
L
dx
T
h
x
Fig. 5.4
Convection at x  L, eq. (5.9)
dT ( L)
 ks
 h[T ( L)  T ]
dx
(d)
k s = conductivity of the solid material
15
(4) Solution
Integrate (a) twice
T ( x )  C1

L
exp(  x / L)  C 2
(e)
BC (c) and (d) give C1 and C 2
(T  To ) exp(  )
C1 
and C 2  To
L (1 /  )  k s / hL
(f)
Substitute into (e), introduce the Biot number
T ( x )  To
Bi

exp  ( x / L)  1
T  To
Bi  
(g)
16
hL
Bi 
ks
(h)
Determine required m for T  Td : set x = L and T  Td
in (g)
Td  To
Bi

T  To Bi  
Solve for  , use definitions of  and Bi
m
k h (T  Td )

A k s c pf (Td  To )
(i)
17
(5) Checking
Dimensional check
Limiting check:
(i) If h = 0 ( Bi  0), x  L becomes insulated, entire
plate is at To . Setting Bi  0 in (g) gives T ( x )  To .
(ii) If h   ( Bi  ), T (L) should be at T . Setting
Bi   and x  L in (g) gives T ( L)  T .
(iii) If m   (   ), entire plate is at To . Setting
   in (g) gives T ( x )  To .
(iv) If m  0 (   0), BC (c) shows plate is insulated at
x  0, entire plate at T . Setting   0 in (g) gives
T ( x )  T .
18
(6) Comments
(i) Solution (g) shows that increasing m (  ) lowers T
(ii) Solution depends on two parameters, Bi and 
(iii) Alternate solution for the required flow rate:
Conservation of energy for a control volume from
supply reservoir and x  L :
dT ( L)
m c pf To  m c pf Td  k f PA

dx
hA(1  P )(Td  T )
Use (d) to eliminate dT ( L) / dx and rearrange
19
h
m c pf (Td  To )  k f PA (T  Td ) 
ks
h(1  P ) A(T  Td )
or
m h[(1  P )  ( k f / k s ) P ](T  Td )

A
c pf (Td  To )
Using the definition of k
m k h (T  Td )

A k s c pf (Td  To )
20
Example 5.2:Transient Conduction in a Porous
Plate
L
Plate thickness = L
Reservoir temperature = To
Initial temperature = To
T1
m
 /A
Sudden change in surface temperature
T1
to
Determine: Transient temperature
(1) Observations
• One-D transient, porous plate
• At steady stateT ( x, )  T1
x
21
(2) Origin and Coordinates
(3) Formulation
(i) Assumptions
(1) Solid and fluid at same temperatures
(2) Constant porosity
(3) Constant properties
(4) Constant flow rate
(5) No energy generation
(6) Initially flow is established, plate temperature
= To
22
(ii) Governing Equation
m c pf T q 1 T



2
A k  x k  t
x
 2T
(5.6)
(iii) Boundary and Initial Conditions
L
(1) T (0, t )  T1
T ( L, t )
(2)
0
x
(3) T ( x ,0)  To
m
 /A
T1
x
23
(4) Solution
Dimensionless form. Let
T  T1

= dimensionless temperature
To  T1
x
= distance

L

 t
2
L
= time
1 m c pf L

= coolant flow rate parameter
2 Ak
24
Governing equation and boundary and initial conditions
become
 2
 
 2

2
 x 
x
(a)
(1)  (0, )  0
(2)  (1, )  0

(3)  ( ,0)  1
Assume a product solution to (a)
 ( ,  )  X ( )  ( )
(b)
25
(b) into (a), separate variables
d 2Xn
d2
dX n
 2
 2n X n  0
d
(c)
d n
 2n  n  0
d
(d)
Solution to (c) depends on  2 and 2n . Only  2  2n
gives characteristic values. Thus solution is given by
equation (A-6a)
X n  exp(  ) An sin M n  Bn cos M n 
(e)
An and Bn are constant, M n is defined as
M n  2n   2
(f)
26
The solution to (d) is
  C n exp(2n  )
(g)
C n is constant. BC (1) gives
Bn  0
(h)
B.C. (2) gives the characteristic equation
M n tan M n  Bi
(i)
(e) and (g) into (b)

 ( , )   an exp(   2n ) sin M n
(j)
n 1
27
Initial condition
1

 an exp(  ) sin M n
(k)
n 1
Orthogonality gives the constant an .
Comparing (c) with the Sturm-Liouville equation, eq.
(3.7a), shows that the weighting function w ( ) is
w ( )  exp(2  )
(l)
Multiplying both sides of (k) by
w( ) exp(  )[sin M m ] d
28
Integrate from   0 to   1 and apply orthogonality,
eq. (3.9),
1
1
0 exp(  )[sin M n ]d  an 0 exp(  )[sin M n ]d
Evaluate the integrals and solve for an
an 
4M n
( 
2
M n2 )( 2 M n
 sin 2 M n )
(m)
(5) Checking
Dimensional check
Limiting: check
29
At t   , entire plate at T1 . Setting    in (j) gives
 ( ,  )  0 , or T ( x, )  T1 .
(6) Comments
In applications where coolant weight is an important
design factor, weight requirement as determined by a
transient solution is less than that given by a steady state
solution. The saving in weight depends on the length of
the protection period.
30