1 CHAPTER 6 CONDUCTION WITH PHASE CHANGE: MOVING BOUNDARY PROBLEMS 6.1 Introduction • Applications: Melting and freezing Casting Ablation Cryosurgery Soldering Permafrost Food processing L k L liquid ps c s k s ) , ( t x T s pf c ) , ( t x T L moving interface Fig. 6.1 i x x dt dx i solid T solid liquid
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CONDUCTION WITH PHASE CHANGE: MOVING BOUNDARY PROBLEMS
• Features: 6.2 The Heat Equations • Assumptions: Moving interface Properties change with phase changes Density change ® liquid phase motion Temperature discontinuity at interface Transient conduction Non-linear problems 6.2 The Heat Equations Two heat equations are needed • Assumptions: One-dimensional
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Properties change with phase changes Density change liquid phase motion Temperature discontinuity at interface Transient conduction Non-linear problems
6.2 The Heat EquationsTwo heat equations are needed• Assumptions: One-dimensional
3
Uniform properties of each phase One-dimensional
Negligible liquid phase motion No energy generation
t
T
x
T s
s
s
1
2
2
ixx 0 (6.1)
tT
x
T L
L
L
1
2
2
ixx (6.2)
6.3 Moving Interface Boundary Conditions(1) Continuity of temperature
4
fiLis TtxTtxT ),(),( (6.3)
fusion (freezing or melting) temperaturefT
interface location at time t )(txx ii
(2) Energy equation
During dt element changes tosolid Density change volume change
Lqsq
ix
Ldxsi dxdx
at t at t+dt
liquid
at t
Fig. 6.2
element
interface interfaceLiquid element LdxMass m
5
Conservation of energy:
sL dxdx
si dxdx
EEE outin (a)
dtqE sin (b)
dV = change in volume
pdVdtqE Lout (c)
p = pressuremuuE Ls )ˆˆ( (d)
6
(b), (c) and (d) into (a)
Fourier’s law
energy per unit massu
pdVmuudtqq LsLs )ˆˆ()( (e)
,),(
x
txTAkq is
ss
x
txTAkq iL
LL
),( (f)
is Adxm (g)
mdVLs
11 (h)
7
(f), (g) and (h) into (e) and assume p = constant
x
txTkx
txTk iLL
iss
),(),(
dt
dxpupu i
LL
sss
ˆˆ (i)
(j)
sL
ss
LL hhpupu ˆˆˆˆ
L
8
h enthalpy per unit massL = latent heat of fusion
(j) into (i)
where
Eq. (6.4) is the interface energy equation for solidification.
For melting Ls
(6.4)siL
Lis
s xtxTk
xtxTk
),(),(dt
dxiL
9
(3) Convection at the interface
+ solidification, - melting
6.5 Non-linearity of Interface Energy Equation
But dtdxi / depends on temperature gradient:Total derivative of sT in eq. (6.3)
(6.5)sfis
s TThx
txTk
)(),(
dtdxiL
(6.4)siL
Lis
s xtxTk
xtxTk
),(),(dt
dxiL
10
(6.6) into (6.5)
(6.7) has two non-linear terms
0),(),(
dt
t
txTdx
x
txT isis
xtxTdttxT
dtdx
is
isi
/),(/),( (6.6)
(6.7)t
txT iL
),(s
isiLL
iss x
txTx
txTkx
txTk
),(),(),( 2
L
11
6.5 Non-dimensional Form of the GoverningEquations: Governing Parameters
Ste = Stefan number
(6.1), (6.2) and (6.4)
,of
fss TT
TT
,
of
fL
s
LL TT
TT
kk
,
Lx
tL
Ste s2
(6.8)
ss Ste2
2
(6.10)
)( ofps TTcSte
(6.9)
L
12
• NOTE:
LL
s
L Ste2
2
(6.11)
iiLis dtt ),(),( (6.12)
Two governing parameters: and Ste sL /
(1) Parameter is eliminated due to the definition of
)/( sL kk
L
(2) Biot number appears in convection BC(3) The Stefan number: Ratio of the sensible heat to the
latent heat
13
Sensible heat )( ofps TTc
Latent heat = L
Stefan number for liquid phase:
6.6 Simplified Model: Quasi-Steady Approximation• Significance of small Stefan number:
sensible heat << latent heat
)( fopf TTcSte
(6.13)L
)( fopf TTc L
14
Ste = 0Temperature can be changed instantaneouslyby transferring an infinitesimal amount of heat
Limiting case: specific heat = zero, 0Ste
Small Stefan number: Interface moves slowly
Ste = 0stationary interface
• Quasi-steady state model: neglect sensible heat for set in (6.10) and (6.11),1.0Ste 0Ste
Alternate limiting case: 0, SteL
15
02
2
s (6.14)
02
2
L (6.15)
(1) Interface energy equation (6.12) is unchanged• NOTE:
(2) Temperature distribution and are time dependent
)(txi
16
Thickness = L
Find: Time to solidify
T
Fig. 6.3
L
fT
oT0
ixx
liquidsolid
Example 6.1: Solidification of a Slab at the Fusion Temperature fT
Initially liquid fT
is suddenly maintained 0xfo TT
is kept at Lx fT
Solution
(1) Observations• Liquid phase remains at fT
17
(2) Origin and Coordinates
(3) Formulation(i) Assumptions:
(1) One-dimensional conduction(2) Constant properties of liquid and solid(3) No changes in fusion temperature
• Solidification starts at time 0t• Solidification time Ltxt oio )(:
(4) Quasi-steady state, 1.0Ste
(ii) Governing Equations
18
BC
02
2
x
Ts (a)
02
2
x
TL (b)
T
Fig. 6.3
L
fT
oT0
ixx
liquidsolid (1) os TtT ),0(
(2) fis TtxT ),(
(3) fiL TtxT ),(
(4) fL TtLT ),(
19
Interface energy condition
Initial conditions (6) fL TxT )0,(
(7) 0)0( ix
T
Fig. 6.3
L
fT
oT0
ixx
liquidsolid
(4) Solution
Integrate (a) and (b)
BAxtxTs ),( (c)
(5) siL
Lis
s xtxTk
xtxTk
),(),(L
dtdxi
20
and
A, B, C and D are constants. They can be functions oftime.
BC (1) to (4):
DCxtxTL ),( (d)
)()(),(
txxTTTtxTi
ofos (e)
fL TtxT ),( (f)
• NOTE: Liquid remains at fT
Interface location: (e) and (f) into BC (5)
21
Integrate
IC (6) gives 01 C
12 )(2
CtTTk
xs
ofsi
L
tTTk
txs
ofsi
)(2)(
(6.16a)
L
dtTTk
dxxs
ofsii
)(
L
si
ofs x
TTk
0 L
dtdxi
22
Dimensionless form: Use (6.8)
2i (6.16b)
Set Time to solidify :ot Lxi in (6.16a)
Or at ,o ,1i (6.16b) gives
2/1o (6.17b)
(5) Checking
Dimensional check
)(2 ofso TTk
t
(6.17a)s L L2
23
(6) Comments
(i) Initial condition (6) is not used. However, it is satisfied
(ii) Fluid motion plays no role in the solution
Limiting check:(i) If no solidification, thus Setting
in (6.17a) gives
,fo TT .ot
fo TT .ot
then Setting in (6.17a) gives(ii) If ,L .ot L.ot
24
Example 6.2: Melting of Slab with Time Dependent Surface Temperature
Thickness = L
Find: )(txi and Melting time
ix
liquid
Fig. 6.4
L
solid
x0fT
T
tTo exp
Initial solid temperature fT
is at0x
tTtT oL exp),0(
Lx is at fT
25
Solution(1) Observations
• Solid phase remains at fT
(2) Origin and Coordinates
(3) Formulation (i) Assumptions:
(1) One-dimensional(2) Constant properties (liquid and solid)(3) No changes in fusion temperature
• Melting starts at time 0t
• Melting time Ltxt oio )(:
26
(4) Quasi-steady state, 1.0teS
(5) Neglect motion of the liquid phase
(ii) Governing Equation
02
2
x
TL (a)
No heat transfer to the solid:
fs TtxT ),( (b)
BC:
ix
liquid
Fig. 6.4
L
solid
x0fT
T
tTo exp
(1) tTtT oL exp),0(
(2) fiL TtxT ),(
27
Interface energy condition
IC:
(4) Solution
Integrate (a)
ix
liquid
Fig. 6.4
L
solid
x0fT
T
tTo exp
(4) 0)0( ix
BAxtxTL ),(
BC (1) and (2)
tTxx
tTTtxT o
i
ofL
exp
exp),(
(c)
(3) LiL
L xtxTk
),( L
dtdxi
28
(c) into condition (3)
Integrate, use IC (4)
iiofL
L dxxdttTTL
k )exp(
(d)
ix
ii
t
ofL
L dxxdttTTL
k
00
)exp(
2
)/()exp()/((2
ioof
L
L xTtTtTL
k
Solving for )(txi
29
Solve for ot by trial and error.
Melt time Set :ot Lxi in (6.18)
(5) Checking
Dimensional checkLimiting check: Special case:
)/()exp()/(2)( ofoL
i TtTtTktx (6.18)L L
)/()exp()/(2 oofooL TtTtTkL (6.19)
L L
30
This result agrees with eq. (6.16a)
Constant temperature at Can not set in (6.18). Expanded for small values of
,0x .0 0)exp( ot t
Set 0
tTTktx foL
i )(2)( (6.20)L L
)/(...)!1/)(1()/(2)( ofoL
i TtTtTktx L L
31
(6) Comments
(i) No liquid exists at time t = 0, no initial condition isneeded.
(ii) Quasi-steady state model is suitable for timedependent boundary conditions.
6.7 Exact Solutions6.7.1 Stefan’s Solution
• Semi-infinite liquid region• Initially at fT
• Boundary at is suddenly maintained at 0x fo TT
• Liquid remains at fT
32
Determine: Temperature distribution and interface location
Liquid solution:
Solid phase:ix
Fig. 6.5
x
fT
oT
T
dtdxi
solid liquidfL TtxT ),( (a)
tT
xT s
s
s
1
2
2 (b)
BC:
(a) into (6.4)
(1) os TtT ),0(
(2) fis TtxT ),(
(3) sis
s xtxTk
),( L
dtdxi
33
Solution Use similarity method
Assume
(b) transforms to
IC: (4) 0)0( ix
tx
s
4 (6.21)
)(ss TT (c)
022
2
ddT
dTd ss (d)
34
Solution to (d)
BC (2) and (6.21)
This requires that
Let
BATs erf (6.22)
Bt
xATs
if
4erf (e)
txi
tx si 4 (f)
is a constant
• NOTE:
35
BC (1):
BC (2):
(f) satisfies initial condition (4)
oTB (g)
erfof TT
A
(h)
(g) and (h) into (6.22)
erferf
),( ofos
TTTtxT
(6.23)
Interface energy condition (3) determines (f) and (6.23) into BC (3)
.
36
or
• NOTE:(1) (6.24) does not give explicitly
depends on the material as well as temperature at (2) 0x
• Special Case: Small
sof cTT )(
erf)(exp 2 (6.24)
L
sxx
ofs
idxd
ddTT
k
)erf(erf
Lt
s 42
37
and
Substituting into eq. (6.24) gives
Small corresponds to small Evaluate eq. (6.24) for small First expand and
.ix. 2exp erf
1...!2!1
1exp42
2
2...103
2!)12(
)1(2erf53
0
12
nn
nn
for small 2
( )ofs TTc (6.25)
L
38
Examine the definition of Stefan number
Substitute (6.25) into (f)
Eq. (6.26) is identical to eq. (6.16a) of the quasi-steadystate model.
Therefore a small corresponds to a small Stefan number.
tTTk
x ofsi
)(2 for small (6.26)(small Ste)
s L
2)( ofs TTc
Ste
(6.13)L
39
6.7.2 Neumann’s Solution: Solidification of Semi- Infinite Region