1. Filling an initially porous tube under a constant head imposed at x =0
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1 Notes on Moving Bounary problems, Voller U o M, volle00@umn.eu. Filling an initially porous tube uner a constant hea impose at x =0 Governing equation is base on calculating the water volume lux by the arcy law, i.e. q h = K Such that a volume (lux) balance at a point in the wet portion gives the ollowing governing equation h = 0, 0 x s( t) () with bounary conitions h( 0) = ho ; h( s) = 0 () An the aitional volume balance at the wetting ront s (t) which states that the spee o this ront is equal to the spee o the water low at that point, i.e., k h s = s, s(0) = 0 (3) where k is a hyraulic conuctivity ajuste to account or porosity. The Solution o () that satisies () is 0 ( x h = h ) s This provies the linear hea proile in the wet raction o the tube. When is use in (3)it gives the ollowing OE or movement o the wetting ront s (t) k h 0 s = s, s(0) = 0 The solution s = k h0t Gives the movement o the wetting ront
2 . One- melting o ice initially at temperature T = Tm Governing equation (assuming heat conuction T = α, 0 x s( t) () Where α = K / ρc is the thermal iusivity. The main bounary conitions are T ( 0) = T0, T ( s) = () T m the initial conition is T ( x,0) = (3) T m The moving bounary conition (the Stean conition) expressing the heat balance at the ront (see ull erivation below) is K x= s s = ρ H Where H is the latent heat o usion. We seek a similarity solution using the similarity variable x ξ = (5) An assuming the melt ront moves as s = λ (6) Where λis a constant. With (5) an (6) it is observe that = T 4 ξ t, =, = 3 =, T 4 x s = λ (7) With (7) equation () an () becomes
3 T + ξ = 0, 0 ξ λ (8) T ( 0) = T0, T ( λ ) = (9) T m An the moving bounary conition becomes c H = λ λ (0) The solution o (8) an (9) can be shown to be Tm T0 T = er ( ξ ) + T er ( λ) 0 () Where the error unction, its erivative an secon erivative are given by ξ 4 β ξ ξ er ( ξ ) = β, ( ξ ) =, ( ξ ) = ξ ξ π e er e er e () π ξ π 0 Substitution o () into (0) provies an equation or calculating the unknown value o λ λ π λe er ( λ) = S t (3) where S t c( T Tm = 0 ) H Is the imensionless Stean number
4 3. erivation o the one-omain Stean problem with constant properties Consier a 3- omain with surace, initially soli at the constant phase change temperature T = 0. At time t 0 melting is inuce by setting an ixing part o the omain surace to a temperaturet > 0. Such that at some later time the omain is segmente into a liqui region an a ix soli region. The liqui region has a temperature T ix T 0 varying in space an time, while the soli region remains at the initial constant temperature o T = 0. These regions are separate by a melt interace continuously moving into the soli. To arrive at a governing equation, at a given instant in time, we consier an arbitrary close, ully liqui omain where part o the omains surace coincies with the melt ront. The heat content o this omain is given by H Where H = ρ ct + ρ H V () is the latent heat o usion, cis the speciic heat, an ρ, assume constant in both the soli an liqui phase, is the ensity. The surace o out omain is mae o two segments one entirely in the liqui phase an the other on the moving melt interace. In consiering the melt interace it is oten helpul to consier separately the surace o material points on the liqui sie o this interace, rom the surace mae up o material points on the soli o this interace +. In particular it is note that on the liqui sie the temperature graient T > 0, whereas on the soli sie the graient is T = 0 (ue to the constant soli temperture); a iscontinuity in temperature graient require to provie the latent heat o usion or the soli to change phase. From the einition o heat in () a transient heat balance on our arbitrary volume can be written as + ρ ct + ρ H V = K T. n A + K T. n A () Or since we have zero graient ρ ct + ρ H V = K T. n A (3) Where the unit outwar normal points out rom the volume.on using the Reynols transport ormula an noting that T = 0on the melt interace, this becomes ρ c V + H v n A K T n A t ρ. =. Where, without loss o exactness, the secon term on the let is evaluate on material points on the liqui sie o the melt interace an vis the velocity o the material points on this interace. On aing to both sies the lux on the liqui sie o the melt ace (3) is reorme as
5 ρ c V + ρ H v. n + K T. n A = K T n A t. (5) + On recognizing that the term on the let is over a suraces that encloses our arbitrary volume an that the term K T is continuous in this volume we can use the ivergence theorem to arrive at ρ c + K T V = 0 ρ H v n K T n A (6) t But since our volume choice is arbitrary both arguments in the above integral must be ientically zero at every point in the omain o integration, i.e, ρ c = K T, in, (7) the appropriate governing equation an K T = ρ H v, on (8) which is seen to be the Stean conition on the moving bounary. Note in one imension (7) an (8) become as expecte the previously use one-phase Stean problem ormulization T = α, 0 x s( t) T (0) = T, T ( s) = 0 ix (9) s K = ρ H (0)
6 4. An Enthalpy Formulization We irst exten the previous melting problem in a way that assumes that the transition rom liqui to ull soli takes place over a narrow an arbitrarily small temperature range ε T 0. Within this setting we eine the enthalpy (total heat) at any point (liqui or soli) in H = ct + H () to be given by where is a liqui raction that changes smoothly an monotonically rom a value o = or temperatures T ε to a value o = 0or temperatures T 0. In this way, the latent heat is evolve smoothly across a temperature range an as such no temperature graient iscontinuities are require to rive the phase change. This type o phase change system is reerre to as a iusive phase change system to inicate that the phase transition takes place over a spatial region o inite thickness as oppose to the sharp transition eine in the Stean problem. The assumption is mae (which can be supporte by rigorous analysis) that as the region o transition ecreases in thickness i.e., as the temperature range ε 0 this iusive interace moel becomes an increasingly more accurate representation o the sharp interace Stean moel. With () an the iuse interace assumption the transient heat balance o the entire omain can be orme as ρ H V = K T. n A () An since temperature graients are continuous in the omain we can use the ivergence theorem an arbitrary volume observation to arrive at the governing equation H ρ = K T, in (3) An equation which is applicable in the entire omain (liqui an soli). In a one-phase one-imensional system this equation becomes H T ρ = K, x 0 Where is seeking a numerical solution we can assume an arbitrarily thin temperature range over which the phase can occurs such that with no loss o accuracy we can essentially assume a sharp interace that sets
7 H H, c T = 0 H H H < H (5) An example problem to ollow
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