Coupl5: Temperature distribution in the conducting sheet
The voltage is applied to the sides of conducting sheet placed vertically and surrounded by the still air. The flowing current heats the sheet due to resistive losses. The front and back surfaces of the sheet are cooled by the air (natural convection).
Problem Type:
A plane-parallel problem of electro-thermal coupling.
Geometry:
Given:
Data for magnetic analysis:
Sheet thickness d = 1 mm;
Material resistance ρ = 10-7 Ohm/m;
Voltage applied U = 0.02 V;
Material heat conductivity λ = 380 W/K·m;
Convection coefficient α = 10 W/K·m2;
Ambient air temperature T0 = 20°C.
Problem:
Calculate the current and temperature distribution in a conducting sheet.
Solution:
The resistive losses are calculated in the DC conduction problem. Then these losses are transferred to the linked heat transfer problem.
The side surfaces area is much smaller then the sum of face surface areas. We can ignore the convection from side surfaces. For simplicity, let's assume that the convection coefficient is constant across the vertically positioned flat sheet.
The convection from the faces is modelled by the heat sink Q(T) = -λ·T, where λ - convection coefficient. We should take into account the convection from the back face of the sheet. Both front and back vertical sides are washed by the air and subjected to the same cooling conditions. Thus we should multiply the convection coefficient by two.
The losses in the sheet are calculated per 1 meter of depth, which is 1000 times greater than the losses in the real sheet. Therefore we should increase the convection coefficient l by the 1000.
Results:
Current distribution in the conducting sheet |
Temperature distribution in the conducting sheet |
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The Coupl5CF.pbm is the problem of calculating the current distribution in the sheet, and Coupl5HT.pbm analyzes temperature field.
Student's version |
Professional version |
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Coupl5.zip (2488 nodes) |
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