Mixing
Heat Model Example for liquid and solid change in state
$begingroup$
If i have a region $x>0$ that is initially liquid at constant temp $T_0$ above freezing temp $T_f$ (so $T_0 > T_f$). The surface at $x=0$ stays at $T_1$ below freezing temp (so $T_1<T_f<T_0$)
And $T_s(x,t)$ and $ T_L(x,t)$ satisfy:
$frac{partial {T_s}}{partial{t}} = D_sfrac{partial^2{T_s}}{partial x^2}$ and $frac{partial {T_L}}{partial{t}} = D_Lfrac{partial^2{T_L}}{partial x^2}$
For constants $D_s$ and $D_L$, with $T_s(x,t)=T_1$ and $T_L(x,0)=T_0$.
The boundary between liquid and solid is $x=s(t)$ with $s(0)=0$
So the boundary conditions i found were:
At $x=0$ $T=T_1$ and is maintained at $T_1$ until it starts to change to solid at $x=s(t)$ where $T=T_f$
Using boundary conditions That $T_s$ and $T_L$ must satisfy show the solution can be written as:
$T_s(x,t) = T_1+frac{T_f-T_1}{erf(m)}erf(frac{x}{sqrt{4D_Lt}})$
$T_L(x,t) = T_0-frac{T_0-T_f}{1-erf(msqrt{frac{D_s}{D_L}})}(1-erf(frac{x}{sqrt{4D_Lt}}))$
That took flipping ages to write... anyway i understand how to find the boundary conditions, that's the easy part however i have zero clue how to go about showing these solutions are true, any help would be absolutely amazing! Thanks in advance!
pde mathematical-modeling heat-equation
asked Jan 13 at 18:59
L GL G
248
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add a comment |
$begingroup$
If i have a region $x>0$ that is initially liquid at constant temp $T_0$ above freezing temp $T_f$ (so $T_0 > T_f$). The surface at $x=0$ stays at $T_1$ below freezing temp (so $T_1<T_f<T_0$)
And $T_s(x,t)$ and $ T_L(x,t)$ satisfy:
$frac{partial {T_s}}{partial{t}} = D_sfrac{partial^2{T_s}}{partial x^2}$ and $frac{partial {T_L}}{partial{t}} = D_Lfrac{partial^2{T_L}}{partial x^2}$
For constants $D_s$ and $D_L$, with $T_s(x,t)=T_1$ and $T_L(x,0)=T_0$.
The boundary between liquid and solid is $x=s(t)$ with $s(0)=0$
So the boundary conditions i found were:
At $x=0$ $T=T_1$ and is maintained at $T_1$ until it starts to change to solid at $x=s(t)$ where $T=T_f$
Using boundary conditions That $T_s$ and $T_L$ must satisfy show the solution can be written as:
$T_s(x,t) = T_1+frac{T_f-T_1}{erf(m)}erf(frac{x}{sqrt{4D_Lt}})$
$T_L(x,t) = T_0-frac{T_0-T_f}{1-erf(msqrt{frac{D_s}{D_L}})}(1-erf(frac{x}{sqrt{4D_Lt}}))$
That took flipping ages to write... anyway i understand how to find the boundary conditions, that's the easy part however i have zero clue how to go about showing these solutions are true, any help would be absolutely amazing! Thanks in advance!
pde mathematical-modeling heat-equation
asked Jan 13 at 18:59
L GL G
248
$endgroup$
1
$begingroup$
Can you at least write the boundary conditions down?
$endgroup$
– Ian
Jan 13 at 19:01
$begingroup$
@Ian Sorry didn't even think that that would obviously help, i'll add them now
$endgroup$
– L G
Jan 13 at 19:09
add a comment |
$begingroup$
If i have a region $x>0$ that is initially liquid at constant temp $T_0$ above freezing temp $T_f$ (so $T_0 > T_f$). The surface at $x=0$ stays at $T_1$ below freezing temp (so $T_1<T_f<T_0$)
And $T_s(x,t)$ and $ T_L(x,t)$ satisfy:
$frac{partial {T_s}}{partial{t}} = D_sfrac{partial^2{T_s}}{partial x^2}$ and $frac{partial {T_L}}{partial{t}} = D_Lfrac{partial^2{T_L}}{partial x^2}$
For constants $D_s$ and $D_L$, with $T_s(x,t)=T_1$ and $T_L(x,0)=T_0$.
The boundary between liquid and solid is $x=s(t)$ with $s(0)=0$
So the boundary conditions i found were:
At $x=0$ $T=T_1$ and is maintained at $T_1$ until it starts to change to solid at $x=s(t)$ where $T=T_f$
Using boundary conditions That $T_s$ and $T_L$ must satisfy show the solution can be written as:
$T_s(x,t) = T_1+frac{T_f-T_1}{erf(m)}erf(frac{x}{sqrt{4D_Lt}})$
$T_L(x,t) = T_0-frac{T_0-T_f}{1-erf(msqrt{frac{D_s}{D_L}})}(1-erf(frac{x}{sqrt{4D_Lt}}))$
That took flipping ages to write... anyway i understand how to find the boundary conditions, that's the easy part however i have zero clue how to go about showing these solutions are true, any help would be absolutely amazing! Thanks in advance!
pde mathematical-modeling heat-equation
asked Jan 13 at 18:59
L GL G
248
$endgroup$
If i have a region $x>0$ that is initially liquid at constant temp $T_0$ above freezing temp $T_f$ (so $T_0 > T_f$). The surface at $x=0$ stays at $T_1$ below freezing temp (so $T_1<T_f<T_0$)
And $T_s(x,t)$ and $ T_L(x,t)$ satisfy:
$frac{partial {T_s}}{partial{t}} = D_sfrac{partial^2{T_s}}{partial x^2}$ and $frac{partial {T_L}}{partial{t}} = D_Lfrac{partial^2{T_L}}{partial x^2}$
For constants $D_s$ and $D_L$, with $T_s(x,t)=T_1$ and $T_L(x,0)=T_0$.
The boundary between liquid and solid is $x=s(t)$ with $s(0)=0$
So the boundary conditions i found were:
At $x=0$ $T=T_1$ and is maintained at $T_1$ until it starts to change to solid at $x=s(t)$ where $T=T_f$
Using boundary conditions That $T_s$ and $T_L$ must satisfy show the solution can be written as:
$T_s(x,t) = T_1+frac{T_f-T_1}{erf(m)}erf(frac{x}{sqrt{4D_Lt}})$
$T_L(x,t) = T_0-frac{T_0-T_f}{1-erf(msqrt{frac{D_s}{D_L}})}(1-erf(frac{x}{sqrt{4D_Lt}}))$
That took flipping ages to write... anyway i understand how to find the boundary conditions, that's the easy part however i have zero clue how to go about showing these solutions are true, any help would be absolutely amazing! Thanks in advance!
pde mathematical-modeling heat-equation
pde mathematical-modeling heat-equation
asked Jan 13 at 18:59
L GL G
248
asked Jan 13 at 18:59
L GL G
248
edited Jan 13 at 19:13
L G
asked Jan 13 at 18:59
L GL G
248
asked Jan 13 at 18:59
L GL G
248
asked Jan 13 at 18:59
L GL G
248
248
1
$begingroup$
Can you at least write the boundary conditions down?
$endgroup$
– Ian
Jan 13 at 19:01
$begingroup$
@Ian Sorry didn't even think that that would obviously help, i'll add them now
$endgroup$
– L G
Jan 13 at 19:09
add a comment |
1
$begingroup$
Can you at least write the boundary conditions down?
$endgroup$
– Ian
Jan 13 at 19:01
$begingroup$
@Ian Sorry didn't even think that that would obviously help, i'll add them now
$endgroup$
– L G
Jan 13 at 19:09
1
1
$begingroup$
Can you at least write the boundary conditions down?
$endgroup$
– Ian
Jan 13 at 19:01
$begingroup$
Can you at least write the boundary conditions down?
$endgroup$
– Ian
Jan 13 at 19:01
$begingroup$
@Ian Sorry didn't even think that that would obviously help, i'll add them now
$endgroup$
– L G
Jan 13 at 19:09
$begingroup$
@Ian Sorry didn't even think that that would obviously help, i'll add them now
$endgroup$
– L G
Jan 13 at 19:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is how I managed to work it out, if anyone notices an inconsistency please point it out, I'm definitely no expert.
Boundary Conditions
$$T_S(0,t)=T_1$$
$$T_S(s(t),t)=T_L(s(t),t)=T_f$$
$$T_L(x,0)=T_0$$
Solution will be in the form of
$$T(x,t)=c_1mbox{erf}(frac{x}{sqrt{4Dt}})+c_2$$
Therefore
$$T_S(x,t)=c_1mbox{erf}(frac{x}{sqrt{4D_St}})+c_2$$
$$T_L(x,t)=c_3mbox{erf}(frac{x}{sqrt{4D_Lt}})+c_4$$
Using the first boundary condition
$$T_S(0,t)=c_1mbox{erf}(frac{0}{sqrt{4D_St}})+c_2
=c_1mbox{erf}(0)+c_2
=c_2=T_1$$
Using the second boundary condition
$$T_S(s(t),t)=c_1mbox{erf}(frac{s(t)}{sqrt{4D_St}})+T_1
=T_f$$
In this case, $c_1$ can only be a constant if $$s(t)=msqrt{4D_St}$$
Therefore
$$T_S(s(t),t)=c_1mbox{erf}(m)=T_f-T_1$$
$$c_1=frac{T_f-T_1}{mbox{erf}(m)}$$
Therefore
$$mbox{erf}(frac{x}{sqrt{4D_St}})$$
$$T_S(x,t)=(frac{T_f-T_1}{mbox{erf}(m)})mbox{erf}(frac{x}{sqrt{4D_St}})+T_1$$
Which is the first part of the answer rearranged but I'll leave it as is for continuity's sake
Using the third boundary condition
$$T_L(x,0)=c_3mbox{erf}(infty)+c_4=T_0$$
$$c_3+c_4=T_0$$
For ease of use later, let's rearrange it to:
$$c_4=T_0-c_3$$
Using the second boundary condition again
$$T_L(s(t),t)=c_3mbox{erf}(frac{s(t)}{sqrt{4D_Lt}})+T_4=T_f$$
$$T_f=c_3mbox{erf}(frac{msqrt{4D_St}}{sqrt{4D_Lt}})+(T_0-C_3)$$
$$T_f-T_0=c_3mbox{erf}(msqrt{frac{D_S}{D_L}})-c_3$$
$$T_f-T_0=-c_3(1-mbox{erf}(msqrt{frac{D_S}{D_L}}))$$
$$c_3=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Therefore
$$c_4=T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Put both results for the liquid state together
$$T_L(x,t)=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}mbox{erf}(frac{x}{sqrt{4D_Lt}})+T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Which simplifies
$$T_L(x,t)=T_0-(frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})})(1-mbox{erf}(frac{x}{sqrt{4D_Lt}})$$
At this point, we should have both equations, again please comment any problems you can find.
edited Jan 14 at 17:39
postmortes
1,95121117
answered Jan 14 at 17:07
user634729user634729
1
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is how I managed to work it out, if anyone notices an inconsistency please point it out, I'm definitely no expert.
Boundary Conditions
$$T_S(0,t)=T_1$$
$$T_S(s(t),t)=T_L(s(t),t)=T_f$$
$$T_L(x,0)=T_0$$
Solution will be in the form of
$$T(x,t)=c_1mbox{erf}(frac{x}{sqrt{4Dt}})+c_2$$
Therefore
$$T_S(x,t)=c_1mbox{erf}(frac{x}{sqrt{4D_St}})+c_2$$
$$T_L(x,t)=c_3mbox{erf}(frac{x}{sqrt{4D_Lt}})+c_4$$
Using the first boundary condition
$$T_S(0,t)=c_1mbox{erf}(frac{0}{sqrt{4D_St}})+c_2
=c_1mbox{erf}(0)+c_2
=c_2=T_1$$
Using the second boundary condition
$$T_S(s(t),t)=c_1mbox{erf}(frac{s(t)}{sqrt{4D_St}})+T_1
=T_f$$
In this case, $c_1$ can only be a constant if $$s(t)=msqrt{4D_St}$$
Therefore
$$T_S(s(t),t)=c_1mbox{erf}(m)=T_f-T_1$$
$$c_1=frac{T_f-T_1}{mbox{erf}(m)}$$
Therefore
$$mbox{erf}(frac{x}{sqrt{4D_St}})$$
$$T_S(x,t)=(frac{T_f-T_1}{mbox{erf}(m)})mbox{erf}(frac{x}{sqrt{4D_St}})+T_1$$
Which is the first part of the answer rearranged but I'll leave it as is for continuity's sake
Using the third boundary condition
$$T_L(x,0)=c_3mbox{erf}(infty)+c_4=T_0$$
$$c_3+c_4=T_0$$
For ease of use later, let's rearrange it to:
$$c_4=T_0-c_3$$
Using the second boundary condition again
$$T_L(s(t),t)=c_3mbox{erf}(frac{s(t)}{sqrt{4D_Lt}})+T_4=T_f$$
$$T_f=c_3mbox{erf}(frac{msqrt{4D_St}}{sqrt{4D_Lt}})+(T_0-C_3)$$
$$T_f-T_0=c_3mbox{erf}(msqrt{frac{D_S}{D_L}})-c_3$$
$$T_f-T_0=-c_3(1-mbox{erf}(msqrt{frac{D_S}{D_L}}))$$
$$c_3=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Therefore
$$c_4=T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Put both results for the liquid state together
$$T_L(x,t)=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}mbox{erf}(frac{x}{sqrt{4D_Lt}})+T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Which simplifies
$$T_L(x,t)=T_0-(frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})})(1-mbox{erf}(frac{x}{sqrt{4D_Lt}})$$
At this point, we should have both equations, again please comment any problems you can find.
edited Jan 14 at 17:39
postmortes
1,95121117
answered Jan 14 at 17:07
user634729user634729
1
$endgroup$
add a comment |
$begingroup$
This is how I managed to work it out, if anyone notices an inconsistency please point it out, I'm definitely no expert.
Boundary Conditions
$$T_S(0,t)=T_1$$
$$T_S(s(t),t)=T_L(s(t),t)=T_f$$
$$T_L(x,0)=T_0$$
Solution will be in the form of
$$T(x,t)=c_1mbox{erf}(frac{x}{sqrt{4Dt}})+c_2$$
Therefore
$$T_S(x,t)=c_1mbox{erf}(frac{x}{sqrt{4D_St}})+c_2$$
$$T_L(x,t)=c_3mbox{erf}(frac{x}{sqrt{4D_Lt}})+c_4$$
Using the first boundary condition
$$T_S(0,t)=c_1mbox{erf}(frac{0}{sqrt{4D_St}})+c_2
=c_1mbox{erf}(0)+c_2
=c_2=T_1$$
Using the second boundary condition
$$T_S(s(t),t)=c_1mbox{erf}(frac{s(t)}{sqrt{4D_St}})+T_1
=T_f$$
In this case, $c_1$ can only be a constant if $$s(t)=msqrt{4D_St}$$
Therefore
$$T_S(s(t),t)=c_1mbox{erf}(m)=T_f-T_1$$
$$c_1=frac{T_f-T_1}{mbox{erf}(m)}$$
Therefore
$$mbox{erf}(frac{x}{sqrt{4D_St}})$$
$$T_S(x,t)=(frac{T_f-T_1}{mbox{erf}(m)})mbox{erf}(frac{x}{sqrt{4D_St}})+T_1$$
Which is the first part of the answer rearranged but I'll leave it as is for continuity's sake
Using the third boundary condition
$$T_L(x,0)=c_3mbox{erf}(infty)+c_4=T_0$$
$$c_3+c_4=T_0$$
For ease of use later, let's rearrange it to:
$$c_4=T_0-c_3$$
Using the second boundary condition again
$$T_L(s(t),t)=c_3mbox{erf}(frac{s(t)}{sqrt{4D_Lt}})+T_4=T_f$$
$$T_f=c_3mbox{erf}(frac{msqrt{4D_St}}{sqrt{4D_Lt}})+(T_0-C_3)$$
$$T_f-T_0=c_3mbox{erf}(msqrt{frac{D_S}{D_L}})-c_3$$
$$T_f-T_0=-c_3(1-mbox{erf}(msqrt{frac{D_S}{D_L}}))$$
$$c_3=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Therefore
$$c_4=T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Put both results for the liquid state together
$$T_L(x,t)=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}mbox{erf}(frac{x}{sqrt{4D_Lt}})+T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Which simplifies
$$T_L(x,t)=T_0-(frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})})(1-mbox{erf}(frac{x}{sqrt{4D_Lt}})$$
At this point, we should have both equations, again please comment any problems you can find.
edited Jan 14 at 17:39
postmortes
1,95121117
answered Jan 14 at 17:07
user634729user634729
1
$endgroup$
add a comment |
$begingroup$
This is how I managed to work it out, if anyone notices an inconsistency please point it out, I'm definitely no expert.
Boundary Conditions
$$T_S(0,t)=T_1$$
$$T_S(s(t),t)=T_L(s(t),t)=T_f$$
$$T_L(x,0)=T_0$$
Solution will be in the form of
$$T(x,t)=c_1mbox{erf}(frac{x}{sqrt{4Dt}})+c_2$$
Therefore
$$T_S(x,t)=c_1mbox{erf}(frac{x}{sqrt{4D_St}})+c_2$$
$$T_L(x,t)=c_3mbox{erf}(frac{x}{sqrt{4D_Lt}})+c_4$$
Using the first boundary condition
$$T_S(0,t)=c_1mbox{erf}(frac{0}{sqrt{4D_St}})+c_2
=c_1mbox{erf}(0)+c_2
=c_2=T_1$$
Using the second boundary condition
$$T_S(s(t),t)=c_1mbox{erf}(frac{s(t)}{sqrt{4D_St}})+T_1
=T_f$$
In this case, $c_1$ can only be a constant if $$s(t)=msqrt{4D_St}$$
Therefore
$$T_S(s(t),t)=c_1mbox{erf}(m)=T_f-T_1$$
$$c_1=frac{T_f-T_1}{mbox{erf}(m)}$$
Therefore
$$mbox{erf}(frac{x}{sqrt{4D_St}})$$
$$T_S(x,t)=(frac{T_f-T_1}{mbox{erf}(m)})mbox{erf}(frac{x}{sqrt{4D_St}})+T_1$$
Which is the first part of the answer rearranged but I'll leave it as is for continuity's sake
Using the third boundary condition
$$T_L(x,0)=c_3mbox{erf}(infty)+c_4=T_0$$
$$c_3+c_4=T_0$$
For ease of use later, let's rearrange it to:
$$c_4=T_0-c_3$$
Using the second boundary condition again
$$T_L(s(t),t)=c_3mbox{erf}(frac{s(t)}{sqrt{4D_Lt}})+T_4=T_f$$
$$T_f=c_3mbox{erf}(frac{msqrt{4D_St}}{sqrt{4D_Lt}})+(T_0-C_3)$$
$$T_f-T_0=c_3mbox{erf}(msqrt{frac{D_S}{D_L}})-c_3$$
$$T_f-T_0=-c_3(1-mbox{erf}(msqrt{frac{D_S}{D_L}}))$$
$$c_3=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Therefore
$$c_4=T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Put both results for the liquid state together
$$T_L(x,t)=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}mbox{erf}(frac{x}{sqrt{4D_Lt}})+T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Which simplifies
$$T_L(x,t)=T_0-(frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})})(1-mbox{erf}(frac{x}{sqrt{4D_Lt}})$$
At this point, we should have both equations, again please comment any problems you can find.
edited Jan 14 at 17:39
postmortes
1,95121117
answered Jan 14 at 17:07
user634729user634729
1
$endgroup$
This is how I managed to work it out, if anyone notices an inconsistency please point it out, I'm definitely no expert.
Boundary Conditions
$$T_S(0,t)=T_1$$
$$T_S(s(t),t)=T_L(s(t),t)=T_f$$
$$T_L(x,0)=T_0$$
Solution will be in the form of
$$T(x,t)=c_1mbox{erf}(frac{x}{sqrt{4Dt}})+c_2$$
Therefore
$$T_S(x,t)=c_1mbox{erf}(frac{x}{sqrt{4D_St}})+c_2$$
$$T_L(x,t)=c_3mbox{erf}(frac{x}{sqrt{4D_Lt}})+c_4$$
Using the first boundary condition
$$T_S(0,t)=c_1mbox{erf}(frac{0}{sqrt{4D_St}})+c_2
=c_1mbox{erf}(0)+c_2
=c_2=T_1$$
Using the second boundary condition
$$T_S(s(t),t)=c_1mbox{erf}(frac{s(t)}{sqrt{4D_St}})+T_1
=T_f$$
In this case, $c_1$ can only be a constant if $$s(t)=msqrt{4D_St}$$
Therefore
$$T_S(s(t),t)=c_1mbox{erf}(m)=T_f-T_1$$
$$c_1=frac{T_f-T_1}{mbox{erf}(m)}$$
Therefore
$$mbox{erf}(frac{x}{sqrt{4D_St}})$$
$$T_S(x,t)=(frac{T_f-T_1}{mbox{erf}(m)})mbox{erf}(frac{x}{sqrt{4D_St}})+T_1$$
Which is the first part of the answer rearranged but I'll leave it as is for continuity's sake
Using the third boundary condition
$$T_L(x,0)=c_3mbox{erf}(infty)+c_4=T_0$$
$$c_3+c_4=T_0$$
For ease of use later, let's rearrange it to:
$$c_4=T_0-c_3$$
Using the second boundary condition again
$$T_L(s(t),t)=c_3mbox{erf}(frac{s(t)}{sqrt{4D_Lt}})+T_4=T_f$$
$$T_f=c_3mbox{erf}(frac{msqrt{4D_St}}{sqrt{4D_Lt}})+(T_0-C_3)$$
$$T_f-T_0=c_3mbox{erf}(msqrt{frac{D_S}{D_L}})-c_3$$
$$T_f-T_0=-c_3(1-mbox{erf}(msqrt{frac{D_S}{D_L}}))$$
$$c_3=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Therefore
$$c_4=T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Put both results for the liquid state together
$$T_L(x,t)=frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}mbox{erf}(frac{x}{sqrt{4D_Lt}})+T_0-frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})}$$
Which simplifies
$$T_L(x,t)=T_0-(frac{T_0-T_f}{1-mbox{erf}(msqrt{frac{D_S}{D_L}})})(1-mbox{erf}(frac{x}{sqrt{4D_Lt}})$$
At this point, we should have both equations, again please comment any problems you can find.
edited Jan 14 at 17:39
postmortes
1,95121117
answered Jan 14 at 17:07
user634729user634729
1
edited Jan 14 at 17:39
postmortes
1,95121117
edited Jan 14 at 17:39
postmortes
1,95121117
edited Jan 14 at 17:39
postmortes
1,95121117
1,95121117
answered Jan 14 at 17:07
user634729user634729
1
answered Jan 14 at 17:07
user634729user634729
1
answered Jan 14 at 17:07
user634729user634729
1
1
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$begingroup$
Can you at least write the boundary conditions down?
$endgroup$
– Ian
Jan 13 at 19:01
$begingroup$
@Ian Sorry didn't even think that that would obviously help, i'll add them now
$endgroup$
– L G
Jan 13 at 19:09