Xing,

I have not studied the physics of your problem in detail so I can't say if your equations are correct. However, I ran bot L = 0.005 and L=0.01 and I see that the extra thickness pulls more heat from the boundary -- this may be reasonable since you are warming that extra material and your boundary condition on the far end is perfectly insulated. I would expect the extra material to draw more heat. To better visualize your problem you can see how it changes over time:

Manipulate[
 Plot3D[Evaluate[T[t, x, z] /. dd /. t -> tt], {x, 0, M}, {z, 0, L}, 
  ImageSize -> {500, 400}, 
  PlotRange -> {{0, M}, {0, L}, {200, 1500}}], {tt, 0, tend}]

This will show you how the heat moves through the material.

You can verify your equation (since it is a simple heat transfer PDE with a textbook. If you still have some doubts, my next step would be to try a simplified, lumped parameter model to see if it also behaves that same way. If your two models converge than you have a second check on your PDE. I hope that helps.

That makes so much sense. And I indeed need the instantaneous temperature for the BC.

Do you have any idea why the temperature changes so much with z length (the material thickness) changing? It could stay at room temperature if I changed L to 0.05. Physically speaking, the temperature shouldn't change much with the material thickness increasing.

As general debugging advice:

1. If you do any numerical solve or numerical plot, make sure the expression is ALL numerical. If it has one variable left in it (like you had cp above) it will fail. I always do this test on complicated expressions. For your equation there should be nothing but T[] and its derivatives in the PDE (this is how I found the two problems -- I just evaluated the expression and looked for extra variables)

2. Evaluate your problem at low resolution so it solves quickly, make sure everything works and then increase to your required resolution and you can let it run for a long time.

Regards

It worked for me. I deleted the

"MinPoints" -> 2000, "MaxPoints" -> 9999

Because it was taking a long time but you can slowly add more resolution until it brings your computer to its knees.

Regards

Thanks, Neil. I used T to replace T1. And changed a little bit about the BCs since the NDsolve shows BCs are inconsistent. The errors didn't show anymore, but one weird thing is that there is no figure shown.

p = 2931.5;
c = 335;
k = 1.45;
\[Alpha] = 0.64;
\[Beta] = 1/(260*10^-6);
w = 0.0018;
L = 0.005;
tend = 10;
P = 20;
M = 0.005;
dd = NDSolve[{c*p*\!\(
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(T[t, x, z]\)\) == k*(\!\(
\*SubscriptBox[\(\[PartialD]\), \({z, 2}\)]\(T[t, x, z]\)\) + \!\(
\*SubscriptBox[\(\[PartialD]\), \({x, 2}\)]\(T[t, x, 
           z]\)\)) + \[Beta]*\[Alpha]*(2*P)/(Pi*w^2)*
       Exp[-((2*(x - 0.0025)^2)/w^2)]*Exp[-\[Beta]*z], 
    T[0, x, z] == 293.13, 
\!\(\*SuperscriptBox[\(T\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, x, 0] == 0, 
\!\(\*SuperscriptBox[\(T\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, x, L] == 0, 
\!\(\*SuperscriptBox[\(T\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, 0, z] == 0, 
\!\(\*SuperscriptBox[\(T\), 
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[t, M, z] == 0}, 
   T[t, x, z], {t, 0, tend}, {z, 0, L}, {x, 0, M}, 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"TensorProductGrid", 
       "MinPoints" -> 2000, "MaxPoints" -> 9999}}];
Plot3D[Evaluate[T[t, x, z] /. dd /. t -> 2], {x, 0, M}, {z, 0, L}]