1

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

2

• 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

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

• 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)

40

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

41

IC: (6) iL TxT )0,( (7) 0)0( ix

SolutionSimilarity method:

Assume

and

tx

s

4 (6.21)

)(ss TT (a)

)(LL TT (b)

42

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)

43

and

BC (2):

DCT LL /erf s (g)

Dt

xATs

if

4erf

Thus

or

txi

tx si 4 (h)

Transformation of BC and IC:

44

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

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)

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

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)

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

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)

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)

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

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 ),(

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

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)

55

Rearrange, separate variables and integrate

ddddTddTd

s

s 00 /

)/(

AddTs lnlnln

ddT

As1ln

BdeAdTs

00

A and B are constants of integration.

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

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.

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)

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

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

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