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 This problem shows the application of FlexPDE to the melting of metal.


 We choose as our system variables the temperature, "temp", and the

 fraction of material which is solid at any point, "solid".


 The temperature is given by the heat equation,


       rho*cp*dt(temp) - div(lambda*grad(temp)) = Source


 where cp is the heat capacity, rho the density and lambda the conductivity.


 The latent heat, Qm, is an amount of heat released as "Solid" changes from

 zero to one.  We have Qm = integral[0,1]((dH/dSolid)*dSolid), or assuming

 dH/dSolid is constant, dH/dSolid = Qm.

 Then heat source from freezing is


       dH/dt = (dH/dSolid)*(dSolid/dt) = Qm*dt(Solid).


 We assume that the solid fraction can be represented by a linear ramp from

 one down to zero as the temperature passes from (Tm-T0/2) to (Tm+T0/2).


       solid = 1                   when  temp  <  Tm-T0

               (Tm+T0/2-temp)/T0   when  Tm-T0 <= temp <= Tm+T0

               0                   when  temp  >  Tm+T0


 where Tm is the melting temperature, and T0 is a temperature range over

 which the melting transition occurs.  Since there are no spatial derivatives

 in this equation, we introduce a diffusion term with small coefficient to act

 as a noise filter.


 The particular problem addressed here is a disk of cold solid material

 immersed in a bath of liquid.  The initial temperatures are such that material

 first freezes onto the disk, but after  equilibrium is reached all the material

 is liquid.    The outer boundary is insulated.


 Since the initial condition is a discontinuous distribution, we use a separate

REGION to define the cold initial disk, so that the grid lines will follow the

 shape.  We also add a FEATURE bounding the disk  to help the gridder

 define the abrupt transition. SELECT SMOOTHINIT helps minimize

 interpolator overshoots.




'Melting Metal'










 Qm= 225000       { latent heat }

 Tm=1850           { Melting temperature }

 T0= 20           { Melting interval +- T0 }

 temp_liq=2000     { initial liquid temperature }

 temp_sol=400     { initial solid temperature }


 R_inf = 0.7       { Domain Radius m}

{ plate }


 dd = d/5         { a defining layer around discontinuity }


 K = 30+4.5e-5*(temp-1350)^2 { Conductivity }

 rho=2500                     { Density kg/m3 }

 cp = 700                     { heat capacity }



 solid =  0.5*erfc((tinit-Tm)/T0)


 temp:  rho*cp*dt(temp) - div(K*grad(temp)) = Qm*dt(solid)

 solid: solid - 1e-6*div(grad(solid)) = RAMP((temp-Tm), 1, 0, T0)    


region 'Outer'

    Tinit = temp_liq

    start 'outer' (0,-R_inf)

      value(temp)= temp_liq   arc(center=0,0) angle 180

      natural(temp)=0         line to close

region 'Plate'

    Tinit = temp_sol



      line to (R_Plate,0) to (R_Plate,d) to (0,d) to close

TIME  0 by 1e-5 to 600


for cycle=10

  grid(r,z) zoom (0,-0.1,0.25,0.25)

  elevation(temp) from(0.1,-0.1) to (0.1,0.15) range=(0,2000)

  elevation(solid) from(0.1,-0.1) to (0.1,0.15) range=(0,1)


  for t= 0 1e-4 1e-3 1e-2 0.1 1 10 by 10 to 100 by 100 to 300 by 300 to endtime

  contour(temp)   range=(0,2000)

  contour(temp)   zoom (0,-0.2,0.45,0.45) range=(0,2000)

  elevation(temp) from(0.1,-0.1) to (0.1,0.15) range=(0,2000)

  contour(solid)   range=(0,1)

  contour(solid)   zoom (0,-0.2,0.45,0.45) range=(0,1)

  surface(solid)   zoom (0,-0.2,0.45,0.45) range=(0,1) viewpoint(1,-1,30)

  elevation(solid) from(0.1,-0.1) to (0.1,0.15) range=(0,1)


  history(temp) at (0.051,d/2) (0.075,d/2) (R_plate,d/2)

  history(temp) at (0.051,d)   (0.075,d)   (R_plate,d)

  history(solid) at (0.051,d/2) (0.075,d/2) (R_plate,d/2)

  history(solid) at (0.051,d)   (0.075,d)   (R_plate,d)

  history(integral(cp*temp+Qm*(1-solid))) as "Total Energy"