raback wrote: ↑04 Jul 2022, 16:35
Hi Dustin,
Electric field here has 6 components.
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Electric field[Electric field re:3 Electric field im:3]
So in your expression tx(0) is real field in x-direction and tx(5) is imaginary in z-direction. These are not treated as complex variables so it does not look so pretty as you have treat it by its components.
(It would make sense to add the few lines of code in the CalcFields routine so that it would directly compute the losses.)
I think the heat source in Elmer is per mass (historical choice). You might want to check that also.
-Peter
Dear Peter,
Thank you so much for your advice.The heat source is working.So I ran a bunch of models yesterday.And then I have a new problem.
Potato is the heated material.When I change its heat conductivity from 0.1 to 450, my potato don't get heated.
According to my simulations, the higher the heat conductivity, the less accurate the temperature.
Here are the Elmer and COMSOl temperature results for heat conductivity of 0.1 and 450, respectively.The size of the potato is 0.03m*0.03m*0.02m, and the maximum mesh of the potato is set to 0.0008m.
- Elmer and COMSOl temperature results for heat conductivity of 0.1 and 450.png (123.17 KiB) Viewed 1194 times
I don't know what the problem is.Here's what I did with your advice.
I added
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Field Variable = Electric field E[Electric field re E:3 Electric field im E:3]
to CalcFields.
The heat source looks like this.The heat source divided by the density becomes per mass.
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Heat Source = Variable "Electric Field E"
Real MATC "pi*8.8542e-12*2.45e9*20*((tx(0))^2+(tx(1))^2+(tx(2))^2+(tx(3))^2+(tx(4))^2+(tx(5))^2)/1050"
Is that right?I don't understand your sentence "so that it would directly compute the losses".I don't know what code to add to calculate the losses directly in CalcFields.
Dustin
The 0.0008m grid file is too big, I uploaded a 0.0016m grid.
- 0016.7z
- (851.64 KiB) Downloaded 64 times