2
5.1 Examples of Conduction in Porous Media
porous ring
coolant
(a)
blade
coolant
(c)
componentelectronic
micro channels
coolant
(d)
coolant
porous material
(e)
Fig. 5.1
(b)
porous shield
coolant
3
5.2 Simplified Heat Transfer Model
• Assume:
5.2.1 Porosity
• Definition: Porosity
(a)ALV
V
V
volumetotal
volumeporeP f (5.1)
At any point the solid and liquid are at the same temperature
4
Fig. 5.2
area flow total fA
Model: Pores are straight channels
(a) and (b) into eq. (5.1)
or
LAV ff (b)
A
AP f (c)
PAA f (5.2)
and solid wall area s
A
APAs )1( (5.3)
• Assume: Porosity is constant
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5.2.2 Governing Equation: Cartesian Coordinates
One-dimensional transient conduction
= surface areaA
= flow ratem
Fig. 5.3
0 x
q dx
m= porosityP
= energy generationq
• Assumptions:
(1) Constant porosity
(2) Constant flow rate
(3) Constant properties (solid and fluid)
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(4) Solid and fluid at same temperature
(5) Negligible changes in kinetic and potential energy
• Conservation of energy for element dx
Use Fourier’s law
Subscripts: s = solid, f = fluid
EEEE outgin (d)
TcmxT
PAkxT
APkE pffsin
)1( (e)
• Define: Conductivity of the solid-fluid matrix ask
7
fs kPkPk )1( (5.4)
(e) becomes
Use (f) to formulate
Tcmx
TAkE pfin
(f)
)(2
2
dxx
TTcmdx
x
TAk
x
TAkE pfout
(g)
dxAqEg (h)
Rate of energy change within the element E
dxt
TPAcdx
t
TAPcE pffpss
)1( (i)
8
Define: Heat capacity of the solid-fluid matrix aspc
pffpssp cPcPc )1( (5.5)
(i) becomes
(f), (g), (h) and (j) into (d)
dxt
TAcE p
(j)
t
T
k
q
x
T
kA
cm
x
T pf
1
2
2 (5.6)
= thermal diffusivity of solid-fluid matrix, defined as
pc
k
(5.7)
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5.2.3 Boundary Conditions
(i) Specified temperature
Fig. 5.4
0
wallporous
reservoir x
m
oT (0)Tm T
h
1),0( TtT or 2),( TtLT (5.8)
At inlet or outlet
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(ii) Convection at outlet boundary
Fig. 5.40
wallporous
reservoir x
m
oT (0)Tm T
h
TtLTh
x
tLTks ),(
),( (5.9)
(iii) Inlet supply reservoir: Conservation of energy for the control volume shown
x
tTAktTTcm opf
),0(),0( (5.10)
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5.2.4 Cylindrical Systems
Conservation of energy and Fourier’s conduction lawapplied to the element dr:
m
rdr
Fig. 5.5
tT
kq
rT
rLk
cm
r
T pf
11
21 )(
2
2 (5.10)
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5.3 Applications
Example 5.1: Steady State Conduction in a Porous Plate
Plate thickness = L
Solution(1) Observations • 1-D conduction in porous plate
Heated at x = L by convection: h, TCoolant reservoir temperature = oT
Design hot side temperature = dT
Determine: Am /
• dTLT )(
Fig. 5.40
wallporous
reservoir x
moT (0)Tm T
h
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• depends on)(xT 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
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(ii) Governing Equation
Eq. (5.6):
or
02
2
xd
dT
Lxd
Td (a)
is coolant flow parameter defined as
02
2
xd
dT
kA
cm
xd
Td pf
kA
Lcm pf (b)
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(iii) Boundary Conditions
xd
dTTT
L o)0(
)0(
(c)
Convection at eq. (5.9),Lx
])([)(
TLThxd
LdTks (d)
sk = conductivity of the solid material
Supply reservoir, eq. (5.10), use definition of
Fig. 5.40
wallporous
reservoir x
moT (0)Tm T
h
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(4) Solution Integrate (a) twice
Substitute into (e), introduce the Biot number
21 )/exp()( CLxL
CxT (e)
and (f) hLkL
TTC
s
o
/)/1(
)exp()(1
oTC 2
(g) 1)/(exp)(
Lx
Bi
Bi
TT
TxT
o
o
BC (c) and (d) give and 2C1C
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sk
hLBi (h)
Determine required for : set x = L and in (g)
m dTT dTT
Bi
Bi
TT
TT
o
od
Solve for , use definitions of and Bi
)(
)(
odpf
d
s TTc
TTh
k
k
A
m
(i)
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Dimensional check
Limiting check:
(5) Checking
(i) If h = 0 becomes insulated, entire Setting in (g) gives
),0( Bi Lx plate is at .oT 0Bi .)(
oTxT
should be at Settingand in (g) gives
(ii) If h ),( Bi )(LT .TBi Lx .)( TLT
entire plate is at Setting (iii) If m ),( .oT in (g) gives .)(
oTxT
BC (c) shows plate is insulated at entire plate at Setting in (g) gives
(iv) If 0m ),0( ,0x .T 0
.)( TxT
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(6) Commentsm (i) Solution (g) shows that increasing lowers T)(
(ii) Solution depends on two parameters, andBi
(iii) Alternate solution for the required flow rate: Conservation of energy for a control volume from supply reservoir and :Lx
))(1(
)(
TTPhAdx
LdTPAkTcmTcm
d
fdpfopf
Use (d) to eliminate and rearrangedxLdT /)(
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or)()1(
)()(
d
ds
fodpf
TTAPh
TTk
hPAkTTcm
)(
)]()/()1[(
odpf
dsf
TTc
TTPkkPh
A
m
Using the definition of k
)(
)(
odpf
d
s TTc
TTh
k
k
A
m
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Example 5.2:Transient Conduction in a Porous Plate
Plate thickness = L L
Am/
x
1T
Determine: Transient temperature
Reservoir temperature = oT
Initial temperature = oT
(1) Observations• One-D transient, porous plate
• At steady state 1),( TxT
Sudden change in surface temperatureto T1
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(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
= oT
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(ii) Governing Equation
(iii) Boundary and Initial Conditions
t
T
k
q
x
T
kA
cm
x
T pf
1
2
2 (5.6)
L
Am/
x
1T
(1) 1),0( TtT
(2) 0),(
x
tLT
(3) oTxT )0,(
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(4) SolutionDimensionless form. Let
= dimensionless temperature1
1
TT
TT
o
= distanceL
x
= time2L
t
= coolant flow rate parameterkA
Lcm pf2
1
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Governing equation and boundary and initial conditionsbecome
Assume a product solution to (a)
xx2
2
2
(a)
(1) 0),0(
(2) 0),1(
(3) 1)0,(
)()(),( X (b)
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(b) into (a), separate variables
02 22
2
nnnn X
d
dX
d
Xd
(c)
02 nnn
d
d
(d)
Solution to (c) depends on and . Only
gives characteristic values. Thus solution is given by
equation (A-6a)
2 2n 22
n
nnnnn MBMAX cossin)exp( (e)
and are constant, is defined asnA nB nM
22 nnM (f)
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The solution to (d) is
)exp( 2 nnC (g)
is constant. BC (1) givesnC
0nB (h)
B.C. (2) gives the characteristic equation
(e) and (g) into (b)
BiMM nn tan (i)
1
2 sin)exp(),(n
nnn Ma (j)
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Initial condition
1
sin)exp(1n
nn Ma (k)
Orthogonality gives the constant . na
)2exp()( w (l)
Multiplying both sides of (k) by
dMw m ])[sinexp()(
is(3.7a), shows that the weighting function )(w
Comparing (c) with the Sturm-Liouville equation, eq.
29
Integrate from to and apply orthogonality,
eq. (3.9),
0 1
1
0
1
0])[sinexp(])[sinexp( dMadM nnn
Evaluate the integrals and solve for na
)2sin2)((
422
nnn
nn
MMM
Ma
(m)
(5) Checking
Dimensional check
Limiting check:
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(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 statesolution.The saving in weight depends on the length of the protection period.
At , entire plate at . Setting in (j) gives
, or .
t 1T 0),( 1),( TxT