CONDUCTION WITH PHASE CHANGE: MOVING BOUNDARY PROBLEMS
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CHAPTER 6CONDUCTION WITH PHASE CHANGE:
MOVING BOUNDARY PROBLEMS
6.1 Introduction• Applications: Melting and freezing Casting Ablation Cryosurgery Soldering Permafrost Food processing
Lk
L
liquid
pscsk
s
),( txTs
pfc
),( txTL
moving interfaceFig. 6.1
ixx
dtdxi
solid
T
solid liquid
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• Features: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 EquationsTwo heat equations are needed• Assumptions: One-dimensional
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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
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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
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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)
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(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)
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(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
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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
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(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
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(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
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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
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• 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
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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
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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
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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
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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
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(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
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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 ),(
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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
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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)
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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
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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
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(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
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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
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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 )(:
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(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 ),(
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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
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(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
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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
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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
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(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
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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
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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)
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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:
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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)
.
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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
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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
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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
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6.7.2 Neumann’s Solution: Solidification of Semi- Infinite Region
• Stefan’s problem: fTxT )0,(
ixxoT
T
dtdxi
0
solid liquid
Fig. 6.6• Assumptions: One-dimensional Constant properties Stationary
• Neumann’s problem:fi TTxT )0,(
• Differential equations:
tT
xT s
s
s
1
2
2ixx 0 (6.1)
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tT
xT L
L
L
1
2
2 xxi (6.2)
ixxoT
T
dtdxi
0
solid liquid
Fig. 6.6
BC:
(1) os TtT ),0( (2) fis TtxT ),(
(3) fiL TtxT ),( (4) iL TtT ),(
Interface energy equation:
(5) siL
Lis
s xtxTk
xtxTk
),(),(L dt
dxi
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IC: (6) iL TxT )0,( (7) 0)0( ix
SolutionSimilarity method:
Assume
and
tx
s
4 (6.21)
)(ss TT (a)
)(LL TT (b)
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Equations (6.1) and (6.2) transform to
,022
2
ddT
dTd ss
i 0 (c)
,022
2
ddT
dTd L
L
sL i (d)
where
tx
s
ii
4
(e)
Solutions to (c) and (d) areBATs erf (f)
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and
BC (2):
DCT LL /erf s (g)
Dt
xATs
if
4erf
Thus
or
txi
tx si 4 (h)
Transformation of BC and IC:
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liquid
Fig. 6.7
solid0
oT
T
iTfT
(1) os TT )0(
(2) fs TT )(
(3) fL TT )(
(4) iL TT )(
(6) iL TT )(
(7) 0)0( ix
(5)
ssL
Ls
s ddTk
ddTk 2)()(
L
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45
• NOTE: • (h) satisfies IC (7)
• Conditions (4) and (6) are identical
• BC (1) - (4) give A, B, C and D. Solutions (f) and (g) become
• Interface appears stationary at
txTT
TtxTs
ofos 4
erferf
)(),(
(6.27)
and
txTT
TtxTLL
ofiL 4
erf-1)/(erf-1
)(),(
s(6.28)
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46
Determine substitute (6.27) and (6.28) into the energycondition (5)
:
• NOTE:in Neumann’s equations (6.27), (6.28) and Set fi TT
(6.29) to obtain Stefan’s solution.
).(
)()/erf)/exp(
erf)exp(
296
(1 s
22
ofpsL
Ls
of
fi
s
L
L
s
TTcTTTT
kk
L
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47
• Follow the same procedure used in solving the solidification problem
6.7.3 Neumann’s Solution: Melting of Semi-infinite Region
Solution
• Modify energy condition (5): Ls
tx Li 4 (6.30)
txTT
TtxTL
ofoL 4
erferf
)(),(
(6.31)
txTT
TtxTss
ifis 4
erf-1/erf-1
)(),(
L(6.32)
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48
where is given by
6.8 Effect of Density Change on the Liquid Phase Change
Changes in density cause theliquid phase to move.
Heat equation becomes: ix
Fig. 6.8
x0
dtdxi
)(tusolid liquid
).(
)()/erf)/exp(
erf)exp(
336
(1
22
foLsL
sL
fo
if
L
s
s
L
TTcTTTT
kk
L
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49
Determine u first
Ldx = liquid element
sdx = element in solid phase
xTu
tT
xTa LLL
L
2
2(6.34)
idx = displacement of interface during solidification
Ldxdx i = displacement of liquid
si dxdx
interfaceat t at t+dt
interface
liquid at t
element
Ldx
9.6.Fig
si dxdx (a)
dtdxdxu Li
(b)
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50
Conservation of mass for the element
or
(c) into (b)
isssLL dxdxdx
iL
sL dxdx
(c)
dtdxu i
L
s
1 (6.35)
Eq. (6.35) into eq. (6.34)dtdxi / is time dependent. Thus • )(tuu
tT
xT
dtdx
xT LLi
L
sLL
12
2(6.36)
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51
Solution: By similarity method
6.9 Radial Conduction with Phase Change
• Limitations: Infinite domain
Example: Solidification due to a line heat sink
Heat is removed at a rate oQ perunit length
Liquid is initially at fi TT
Fig. 6.10
solidliquid
interface
fTiT
r
rir
liquid
solid
sink line
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52
Assume:(1) Constant properties in each phase
(2) Neglect effect of liquid motion
tT
rT
rrT s
s
ss
11
2
2)(0 trr i
(6.37)
tT
rT
rrT L
L
LL
11
2
2 rtri )((6.38)
BC:
(1) os
sr
Qr
Trk
2lim
0
(2) fis TtrT ),(
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53
Interface energy equationFig. 6.10
solidliquid
interface
source line
fTiT
r
rir
liquid
solid (3) fiL TtrT ),(
(4) iL TtT ),(
IC: (6) iL TrT )0,(
(7) 0)0( ir
(5) siL
Lis
s rtrTk
rtrTk
),(),(dtdriL
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54
Equations (6.37) and (6.38) transform to
SolutionSimilarity method:
tr
s
4
2 (6.39)
0112
2
ddT
dxTd ss (6.40)
Separate variables and integrate eq. (6.40)
012
2
ddT
dxTd L
L
sL (6.41)
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55
Rearrange, separate variables and integrate
ddddTddTd
s
s 00 /
)/(
AddTs lnlnln
ddT
As1ln
BdeAdTs
00
A and B are constants of integration.
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56
Similarly, solution to eq. (6.41) is
BC (2):
DdeCTi
Ls
L
)/( (b)
is value of at the interface. It is determined bysetting in eq. (6.39)
i
irr
tr
s
ii
4
2 (6.42)
BdeATi
f
0
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57
Thus
constant, therefore is independent of time.fT i
tri 2
• NOTE:
Eq. (6.43) into (6.42):
Eq. (a) and BC (1) give A
tr si 42 (6.43)
(6.43) satisfies IC (7). ir
i (c)
where constant.
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58
Solving for A
(a), (c) and B.C. (2) give B
trerAk
rddTrk
sss
422lim2lim
00
os QerAk
4
s
ok
QA4
(d)
0
)()(),( BdeATTTtrT fsisis
de
kQTB
s
of
04 (e)
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59
(b), (c) and BC (3) give D
Similarly, BC (4) gives
DDdeCTi
Ls
f
0)/(
fTD (f)
de
TTC
Ls
fi)/(
(g)
Interface energy equation (5) gives
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60
• Examine integrals in the solution:The exponential integral function Ei(x) is defined as
dvv
exEix
v
)( (6.45)
Values of this function at and are0x x
,)0( Ei 0)( Ei (6.46)
Use the definition of )(xEi in (6.45) and the constantsin (d)-(e), equations (a), (b) and (6.44), become
(6.44)ss
fiL
s
o Ls
Ls
e
de
TTke
kQ
)/()/(
)(4 L
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61
6.10 Phase Change in Finite Regions
(6.48))4/()/(
2 trEiEi
TTTT s
Ls
ifiL
)4/()(4
),( 2 trEiEik
QTtrT ss
ofs
(6.47)
(6.49)ssLs
ifL
s
o LseEi
TTke
kQ
)/(
)/()(
4 L