| Author | Message | ||
     Danny-Boy (dannyboy) New member Username: dannyboy Post Number: 1 Registered: 11-2005 |
I have a student license and am trying to model heat flow in a bottle-like container. The container is relatively simple and has 4 layers. I keep getting the error "Initial Mesh Size (2257) Exceeds Node Limit!", which also suggests trying a smaller NGRID among other things. I have tried NGRID as low as possible but with no success. My script is still very much a work in progress and is included below. Can anyone suggest how to remedy this problem? (Unfortunately, I cannot afford the professional license...) COORDINATES Cartesian3 VARIABLES Temp SELECT NGRID=1 DEFINITIONS KGlass =1 KLiquid=1 K=0 RBody = 1 RNeck=0.5 Glass=0.25 TAmbient=20 TIce=0 INITIAL VALUES Temp=TAmbient EQUATIONS dt(Temp)=K*(dxx(Temp)+dyy(Temp)+dzz(Temp)) EXTRUSION SURFACE "Bottom" Z=0 LAYER "Base" SURFACE "Base-Body Interface" Z=Glass LAYER "Body" SURFACE "Lower Neck-Body Interface" Z=1 LAYER "Neck-Bottle Interface" SURFACE "Upper Neck-Body Interface" Z=1+Glass LAYER "Neck" SURFACE "Top" Z=2 BOUNDARIES SURFACE "Bottom" VALUE(Temp)=TIce SURFACE "Top" VALUE(Temp)=TIce REGION 1 LAYER "Neck" VOID LAYER "Neck-Bottle Interface" VOID LAYER "Body" VOID LAYER "Base" K=KGlass START (RBody,0) VALUE(Temp)=TIce ARC(CENTER=0,0) ANGLE=360 TO CLOSE REGION 2 LAYER "Neck" VOID LAYER "Neck-Bottle Interface" VOID LAYER "Body" K=KLiquid START (RBody,0) VALUE(Temp)=TIce ARC(CENTER=0,0) ANGLE=360 TO CLOSE START (RBody-Glass,0) VALUE(Temp)=TIce ARC(CENTER=0,0) ANGLE=360 TO CLOSE REGION 3 LAYER "Neck" VOID LAYER "Neck-Bottle Interface" K=KLiquid START (RBody,0) VALUE(Temp)=TIce ARC(CENTER=0,0) ANGLE=360 TO CLOSE START (RBody-RNeck,0) VALUE(Temp)=TIce ARC(CENTER=0,0) ANGLE=360 TO CLOSE LIMITED REGION 4 LAYER "Neck" K=KGlass START (RNeck,0) VALUE(Temp)=TIce ARC(CENTER=0,0) ANGLE=360 TO CLOSE START (RNeck-Glass,0) ARC(CENTER=0,0) ANGLE=360 TO CLOSE TIME 0 TO 1 PLOTS FOR T=0 BY 1 TO 1 GRID(x,y,z) END | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 494 Registered: 06-2003 |
In order to make it feasible for us to distribute a free student version, we have had to place limitations on the capabilities of the free version. Unfortunately, your problem appears to exceed these restrictions. There may be some things you can do: 1) I see no azimuthal variations that require a 3D model. Why not use 2D cylindrical coordinates? Write your equation as dt(Temp) = div(K*grad(Temp)) and FlexPDE will expand the divergence for you. 2) If you really need azimuthal variation, perhaps you could cut the cylinder in half on a symmetry plane and compute only one half. Use Natural(Temp)=0 on the cut face. Incidentally, you have inconsistent initial and boundary values. This means a discontinuous initial temperature, which will require dense gridding at the surface to track the steep initial temperature wave, far beyond the limitations of the student version. You will instead see serious overshoot in the solution in the outer cells as FlexPDE tries and fails to approximate a step function with a parabola. You might ameliorate this condition if you use a convective boundary condition Natural(Temp) = c*(Temp-TIce), with c a coupling constant characteristic of the thermal resistance between the cylinder and the reservoir. |