CHAPTER X PHASE-CHANGE PROBLEMS

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Chapter X Phae-Change Problem December 3, 18 917 CHAPER X PHASE-CHANGE PROBLEMS X.1 Introducton Clacal Stefan Problem Geometry of Phae Change Problem Interface Condton X. Analytcal Soluton for Soldfcaton n Half Space (Neumann Soluton) X.3 Analytcal Soluton for Meltng n Half Space (Eerce)

918 Chapter X Phae-Change Problem December 3, 18 X.1 Inroducton ranent heat tranfer problem nvolvng evaporaton, meltng, condenaton or oldfcaton wth movng nterface between phae. Clacal Stefan Problem heat tranfer problem wth lqud-old phae-change. Problem non-lnear due to nterface condton., ubcrpt referred to old and lqud phae, repectvely. m meltng or freezng temperature (temperature at whch lqud oldfe or old melt) or equlbrum phae change temperature (temperature at whch both old and lqud phae can tay together n thermodynamc equlbrum). temperature n the old phae. ( ) temperature n the lqud phae. S( t ) S( t) equaton of the phae-change boundary. ρ denty (t aumed n ome problem, that ρ ρ ρ ) Phae-change problem n the em-nfnte regon: Soldfcaton old lqud heat flu m Meltng lqud S( t ) old heat flu m S( t)

Chapter X Phae-Change Problem December 3, 18 919 Interface Condton: conder the control urface contanng the nterface boundary S( t ) Soldfcaton: ds t A m ρ q h dt h A A freezng ρh ds t dt rate of heat releaed durng oldfcaton per unt area of nterface kg J m 3 m kg A S V A k k S v t ds t dt v ( t) S( t ) Energy balance: ds t dt k k ρh ds dt k k ρh Meltng: ρh ds t dt rate of heat aborbed durng meltng per unt area of nterface k k S( t) Energy balance: ds t + dt k k ρh ds Same condton both for oldfcaton and meltng. dt k k ρh

9 Chapter X Phae-Change Problem December 3, 18 X. Analytcal oluton for oldfcaton n half pace (Neumann Soluton): old lqud m Mathematcal Model: S( t ) Heat Equaton: 1 t 1 t < < S( t) t > (8.1) S( t) < < t > (8.) Boundary Condton:,t t > (8.3) Interface Condton:,t t > (8.4) S t,t S t,t m contnuty t ds ( t ) ρ S( t) dt k k h Intal Condton: > (8.5) t > (8.6), > (8.7) Soluton: Recallng the oluton of the Heat Equaton n the doman > (Secton IX.1, p., and Eerce), note that the functon erf t, erfc, any contant C cont t (where, 1 1 a ) erf ( ) and any ther lnear combnaton are oluton of the homogeneou Heat Equaton (8.1-). hen look for the oluton of the equaton (8.1) n the form 1 erfc ( ) (,t ) + A erf A cont (8.8) t whch atfe the Heat Equaton (8.1) and the boundary condton (8.3). And look for the oluton of the equaton (8.) n the form + B erfc B cont (8.9) t whch atfe the Heat Equaton (8.), condton (8.4) and the ntal condton (8.7). hen ue the nterface condton (8.5-6) to determne the coeffcent A, B and S( t ).

Chapter X Phae-Change Problem December 3, 18 91 Subttute the tral oluton (8.8-9) nto the contnuty condton (8.5): S( t) S t + A erf + B erfc m t t S( t) S t + A erf + B erfc m t t + A erf ( ) + B erfc m (8.1) S( t) where S( t) t (8.11) t Equaton (8.1) mple (becaue both erf and erfc are monotonc functon) that S t cont t Solve Equaton (8.1) for coeffcent A and B : m A erf m B erfc and ubttute them nto equaton (8.8-9): (,t ) m + erf erf ( ) t m + erfc t erfc (8.1) (8.13) he remanng contant ha to be determned from the nterface condton (8.6): 1 ep t π t m erf ( ) 1 π erf t ( ) t m ep 1 m π ep t t erfc Dfferentate equaton (8.11): ds ( t) dt d t dt 1 t t 1 m ep π t t erfc

9 Chapter X Phae-Change Problem December 3, 18 Apply the nterface condton (8.6): k 1 m S ( t) S ep ( t 1 m ) k ep π erf t ( ) t π t t erfc ρh ds t dt k S( t 1 1 m ) ep π erf t ( ) t S m ( t 1 1 ) k ep π t t erfc ρh t k 1 S m ( t) ep π erf ( ) t 1 S m ( t) k ep π t erfc ρh k 1 m π erf ( ) 1 m ep k ep π erfc ρh e e k ( m ) ( m ) π ρ erf ( ) ρc p k k h erfc Equaton for contant (ha to be olved numercally): e k e π h ( m ) ( m ) erf ( ) k c p erfc (8.14) After found, the oluton of the phae-change problem for oldfcaton gven by the equaton (8.11-13): S( t ) t (8.11) erf t (,t ) + ( m ) S( t) erf ( ) (8.1) + ( ) m erfc t erfc S t (8.13) where a potve root of Equaton (8.14)

Chapter X Phae-Change Problem December 3, 18 93 Eample: Conder oldfcaton of the materal under the followng condton: 1 1 m k 1. ρ 1 cp 5 ρcp k. h 3 k 5. ρ 1 p c 1 k.5 ρ c p Solve equaton (8.14) for contant :.547 Phae-change boundary: S( t) ranent temperature profle: t 5 t t 5 m

94 Chapter X Phae-Change Problem December 3, 18 Maple Soluton: 1 PHASE-CHANGE SOLIDIFICAION > retart; > wth(plot): > :-1;:1;m:; > k:1.;kl:5.; > r:1;rl:1; > cp:5;cpl:1; > a:k/r/cp;al:kl/rl/cpl; > h:3; : -1 : 1 m : k : 1. kl : 5. r : 1 rl : 1 cp : 5 cpl : 1 a :. al :.5 h : 3 > W(v):(m-)*ep(-v^)/erf(v)-kl/k*qrt(a/al)*(m-)*ep(- v^*a/al)/erfc(v*qrt(a/al))-qrt(p)*h/cp*v; 1 e ( v ) erf( v) > plot(w(v),v.1..1,y-5..1); 31.67766 e (.4v) erfc(.6345553 v) W( v ) : + 6 π v > v:folve(w(v),v.1..1); (correpond to contant ) Phae-Change Boundary: v :.5466644957 > S(t):*v*qrt(a*t); > plot(s(t),t..5); S( t ) :.1546687 t

Chapter X Phae-Change Problem December 3, 18 95 emperature Profle: > S(,t):+(m-)*erf(//qrt(a*t))/erf(v); S (, t ) : 1 + 17.843546 erf 35.3553397 t > L(,t):+(m-)*erfc(//qrt(al*t))/erfc(v*qrt(a/al)); L (, t ) : 1 16.317578 erfc.3667978 t > SH:S(,t)*(Heavde()-Heavde(-S(t))); > LH:L(,t)*Heavde(-S(t)); SH : 1 + 17.843546 erf 35.3553397 t ( Heavde( ) Heavde (.1546687 t )) LH : 1 16.317578 erfc.3667978 Heavde (.1546687 t ) t > anmate({sh,lh,,m,},..15,t..1,frame,aeboed); > S1:ub(t5,SH):L1:ub(t5,LH): > S:ub(t,SH):L:ub(t,LH): > S3:ub(t5,SH):L3:ub(t5,LH): > plot({s1,l1,s,l,s3,l3,m,,},..15,colorblack,aeboed);

96 Chapter X Phae-Change Problem December 3, 18 X.3 Eerce: Analytcal Soluton for Meltng n Half Space lqud old m S( t) Mathematcal Model: Heat equaton: 1 t 1 t < < S( t) t > (8.1) S( t) < < t > (8.) Boundary Condton:,t t > (8.3) Interface Condton: ( ) ( ),t t > (8.4) S t,t S t,t Intal Condton: m contnuty t ds ( t ) ρ S( t) k k h dt > (8.5) t > (8.6), > (8.7) Soluton: Look for the oluton of the equaton (8.1) n the form + A erf A cont (8.8) t whch atfe the Heat Equaton (8.1) and the boundary condton (8.3). And look for the oluton of the equaton (8.) n the form (,t ) + B erfc B cont (8.9) t whch atfe the Heat Equaton (8.), condton (8.4) and condton (8.7). hen ue the nterface condton to determne the coeffcent A, B and S( t ).

Chapter X Phae-Change Problem December 3, 18 97

98 Chapter X Phae-Change Problem December 3, 18