chapter x phase-change problemsvps/me505/iem/10 00.pdf · 918 chapter x phase-change problems...

Chapter X Phase-Change Problems December 3, 2018 917

CHAPTER X

PHASE-CHANGE PROBLEMS X.1 Introduction – Classical Stefan Problem Geometry of Phase Change Problems Interface Conditions X.2 Analytical Solution for Solidification in Half Space (Neumann’s Solution) X.3 Analytical Solution for Melting in Half Space (Exercise)


X.1 Inroduction Transient heat transfer problems involving evaporation, melting, condensation or solidification with moving interface between phases.

“Classical Stefan Problem” – heat transfer problems with liquid-solid

phase-change. Problem is non-linear due to interface conditions. s, subscripts referred to solid and liquid phases, respectively.

mT melting or freezing temperature (temperature at which liquid solidifies or solid melts) or equilibrium phase change temperature (temperature at which both solid and liquid phases can stay together in thermodynamic equilibrium) .

( )sT x temperature in the solid phase.

( )T x

temperature in the liquid phase.

( )S t ( )x S t= equation of the phase-change boundary . ρ density (it is assumed in some problems, that sρ ρ ρ= =

) Phase-change problems in the semi-infinite region: Solidification Melting

( )S t →

x0

( )T x,t

( )sT x,t

solid

0T

liquid

iT

mT

( ) S t←

x0

( )sT x,t

( )T x,t

liquid

0T

solid

iT

mT

heat flux

heat flux


Interface Conditions: consider the control surface containing the interface boundary ( )S t Solidification:

Energy balance: ( )s

s fg

dS tT Tk k h 0x x dt

ρ∂ ∂ − − − − = ∂ ∂

ss fg

T T dSk k hx x dt

ρ∂ ∂

− =∂ ∂

Melting:

Energy balance: ( )s

s fg

dS tT Tk k h 0x x dt

ρ∂ ∂ − − − + = ∂ ∂

ss fg

T T dSk k hx x dt

ρ∂ ∂

− =∂ ∂

Same condition both for solidification and melting.

( )S t →

Tkx

∂−

∂

( )fg

dS th

dtρ

ss

Tk

x∂

−∂

rate of heat releasedduring solidification per unit area of interface

( ) S t←

Tkx

∂−

∂

( )fg

dS th

dtρ

Tkx

∂−

∂

rate of heat absorbedduring melting per unit area of interface

3

kg J mkg sm

⋅ ⋅

( )

freezing fg fg

dS tAm dtq h h

A A

ρ′′ = =

( ) ( )x

dS tv t

dt=

A

xS v t∆ ∆= ⋅

A S V∆ =


X.2 Analytical solution for solidification in half space (Neumann’s Solution): Mathematical Model:

Heat Equation: 2

s s2

s

T T1tx α

∂ ∂=

∂∂ ( )0 x S t< < t 0> (8.1)

2

2

T T1tx α

∂ ∂=

∂∂

( )S t x< < ∞ t 0> (8.2)

Boundary Conditions: ( )s 0T 0,t T= t 0> (8.3) ( ) iT x,t T=

x → ∞ t 0> (8.4) Interface Conditions: ( ) ( )s mT S t ,t T S t ,t T= =

continuity t 0> (8.5)

( )ss fg

dS tT Tk k hx x dt

ρ∂ ∂

− =∂ ∂

( )x S t= t 0> (8.6)

Initial Condition: ( ) iT x,0 T=

x 0> (8.7) Solution: Recalling the solutions of the Heat Equation in the domain x 0> (Section IX.1, p., and

Exercise), note that the functions

xerf2 tα

, xerfc2 tα

, any constant C const= (where, 2 1aα

= )

and any their linear combination are solutions of the homogeneous Heat Equation (8.1-2). Then look for the solution of the equation (8.1) in the form

( )sT x,t 0s

xT A erf2 tα

= + ⋅

A const= (8.8)

which satisfies the Heat Equation (8.1) and the boundary condition (8.3). And look for the solution of the equation (8.2) in the form

( )T x,t

ixT B erfc

2 tα

= + ⋅

B const= (8.9)

which satisfies the Heat Equation (8.2), condition (8.4) and the initial condition (8.7). Then use the interface conditions (8.5-6) to determine the coefficients A , B and ( )S t .

( )S t →

x0

( )T x,t

( )sT x,t

solid

0T

liquid

iT

mT

1

( )erfc x

1

( )erf x


Substitute the trial solutions (8.8-9) into the continuity condition (8.5):

( ) ( )

0 i ms

S t S tT A erf T B erfc T

2 t 2 tα α

+ ⋅ = + ⋅ =

( ) ( ) s

0 i ms s

S t S tT A erf T B erfc T

2 t 2 tααα α

+ ⋅ = + ⋅ =

( ) s0 i mT A erf T B erfc T

αλ λ

α

+ ⋅ = + ⋅ =

(8.10)

where ( )

s

S t2 t

λα

= ⇒ ( ) sS t 2 tλ α= (8.11)

Equation (8.10) implies (because both erf and erfc are monotonic functions) that

( )s

S tconst

2 tλ

α= =

Solve Equation (8.10) for coefficients A and B :

( )

m 0T TA

erf λ−

=

m i

s

T TB

erfc αλ

α

−=

and substitute them into equations (8.8-9):

( )sT x,t ( )

m 00

s

T T xT erferf 2 tλ α

−= + ⋅

(8.12)

( )T x,t

m ii

s

T T xT erfc2 t

erfcαα

λα

−= + ⋅

(8.13)

The remaining constant λ has to be determined from the interface condition (8.6):

( )sT x,tx

∂∂

( )

2

m 0

s s

T T1 2 xexperf2 t 2 tλα π α

− = ⋅ −

( )

2

m 0

s s

T T1 xexperft 2 tλπα α

− = ⋅ −

( )T x,tx

∂∂

2

m i

s

T T2 1 xexp2 t 2 t

erfcπ α αα

λα

− = − ⋅ −

2

m i s

ss

T T1 xexpt 2 t

erfc

ααπα αα

λα

− = ⋅ −

Differentiate equation (8.11):

( )dS tdt

sd 2 tdt

λ α = s

s

122 t

αλ

α= s

stα

λα

=


Apply the interface condition (8.6):

( )( )

2

m 0s

s s

S tT T1k experft 2 tλπα α

− ⋅ −

( )2

m i s

ss

S tT T1k expt 2 t

erfc

ααπα αα

λα

− − ⋅ −

( )fg

dS th

dtρ=

( )( )

2

m 0s

s s

S tT T1 1k experft 2 tλπ α α

− ⋅ −

( )2

s m i s

s ss

S tT T1 1k expt 2 t

erfc

α αα απ α αα

λα

− − ⋅ −

sfg

s

ht

αρ λ

α=

( )( )

2

m 0s

s

S tT T1k experf 2 t

λ

λπ α

− ⋅ −

( )2

s m i s

ss

S tT T1k exp2 t

erfcλ

α αα απ αα

λα

− − ⋅ −

s fghα ρ λ=

( )2m 0

sT T1k experf

λλπ

− ⋅ − 2s m i s

s

T T1k experfc

α αλ

α απ αλ

α

−− ⋅ −

s fghα ρ λ=

( ) ( ) ( )2 s

2

s ss m 0 m i fg

pss

ke ek T T k T T herf c

erfc

αλ

αλ απ ρ λ

λ α ραλ

α

−−

− − − =

Equation for constant λ (has to be solved numerically):

( ) ( ) ( )2 s

2

fgsm 0 m i

s pss

hke eT T T Terf k c

erfc

αλ

αλ παλ

λ α αλ

α

−−

− − − =

(8.14)

After λ is found, the solution of the phase-change problem for solidification is given by the equations (8.11-13):

( )S t s2 tλ α= (8.11)

( )sT x,t ( ) ( )s

0 m 0

xerf2 t

T T Terf

αλ

= + − ( )0 x S t≤ ≤ (8.12)

( )T x,t

( )i m is

xerfc2 t

T T Terfc

α

αλ

α

= + −

( )x S t≥ (8.13)

where λ is a positive root of Equation (8.14)


Example: Consider solidification of the material under the following conditions: 0T 10= − iT 10= mT 0=

sk 1.0= s 100ρ = psc 50= ss

s ps

k0.0002

cα

ρ= = fgh 3000=

k 5.0=

100ρ =

pc 100=

p

k 0.0005c

αρ

= =

Solve equation (8.14) for constant :λ Phase-change boundary: Transient temperature profiles:

( )S t

( )T x,t

( )sT x,t

( )T x,t

iT

mT

0T

t 5000=t 20000=

t 50000=

0.547λ =


Maple Solution: 12 PHASE-CHANGE SOLIDIFICATION > restart; > with(plots): > T0:=-10;Ti:=10;Tm:=0;

:= T0 -10

:= Ti 10

:= Tm 0

> ks:=1.0;kl:=5.0; := ks 1.0

:= kl 5.0

> rs:=100;rl:=100; := rs 100

:= rl 100

> cps:=50;cpl:=100; := cps 50

:= cpl 100

> as:=ks/rs/cps;al:=kl/rl/cpl; := as 0.0002000000000

:= al 0.0005000000000

> hfg:=3000; := hfg 3000

> W(v):=(Tm-T0)*exp(-v^2)/erf(v)-kl/ks*sqrt(as/al)*(Tm-Ti)*exp(-v^2*as/al)/erfc(v*sqrt(as/al))-sqrt(Pi)*hfg/cps*v;

:= ( )W v + − 10 e

( )−v2

( )erf v31.62277660 e

( )−0.4000000000v2

( )erfc 0.6324555320 v 60 π v

> plot(W(v),v=0.01..1,y=-50..100);

> v:=fsolve(W(v)=0,v=0.01..1); (corresponds to constant λ )

:= v 0.5466644957

Phase-Change Boundary: > S(t):=2*v*sqrt(as*t);

:= ( )S t 0.01546200687 t

> plot(S(t),t=0..5000);


Temperature Profiles: > TS(x,t):=T0+(Tm-T0)*erf(x/2/sqrt(as*t))/erf(v);

:= ( )TS ,x t − + 10 17.84003546

erf 35.35533907 x

t

> TL(x,t):=Ti+(Tm-Ti)*erfc(x/2/sqrt(al*t))/erfc(v*sqrt(as/al));

:= ( )TL ,x t − 10 16.00317578

erfc 22.36067978 x

t

> TSH:=TS(x,t)*(Heaviside(x)-Heaviside(x-S(t)));

TSH

− + 10 17.84003546

erf 35.35533907 x

t :=

( ) − ( )Heaviside x ( )Heaviside − x 0.01546200687 t

> TLH:=TL(x,t)*Heaviside(x-S(t));

:= TLH

− 10 16.00317578

erfc 22.36067978 x

t( )Heaviside − x 0.01546200687 t

> animate({TSH,TLH,Ti,Tm,T0},x=0..15,t=0..100000,frames=200,axes=boxed);

> S1:=subs(t=5000,TSH):L1:=subs(t=5000,TLH): > S2:=subs(t=20000,TSH):L2:=subs(t=20000,TLH): > S3:=subs(t=50000,TSH):L3:=subs(t=50000,TLH): > plot({S1,L1,S2,L2,S3,L3,Tm,Ti,T0},x=0..15,color=black,axes=boxed);


X.3 Exercise: Analytical Solution for Melting in Half Space Mathematical Model:

Heat equation: 2

2

T T1tx α

∂ ∂=

∂∂

( )0 x S t< < t 0> (8.1)

2

s s2

s

T T1tx α

∂ ∂=

∂∂ ( )S t x< < ∞ t 0> (8.2)

Boundary Conditions: ( ) 0T 0,t T=

t 0> (8.3) ( )s iT x,t T= x → ∞ t 0> (8.4) Interface Conditions: ( )( ) ( )( )s mT S t ,t T S t ,t T= =

continuity t 0> (8.5)

( )ss fg

dS tT Tk k hx x dt

ρ∂ ∂

− =∂ ∂

( )x S t= t 0> (8.6)

Initial Condition: ( )s iT x,0 T= x 0> (8.7) Solution: Look for the solution of the equation (8.1) in the form

( )T x,t

0xT A erf

2 tα

= + ⋅

A const= (8.8)

which satisfies the Heat Equation (8.1) and the boundary condition (8.3). And look for the solution of the equation (8.2) in the form

( )sT x,t is

xT B erfc2 tα

= + ⋅

B const= (8.9)

which satisfies the Heat Equation (8.2), condition (8.4) and condition (8.7). Then use the interface conditions to determine the coefficients A , B and ( )S t .

( )S t

x0

( )sT x,t

( )T x,t

liquid

0T

solid

iT

mT

chapter x phase-change problemsvps/me505/iem/10 00.pdf · 918 chapter x phase-change problems...

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