ACElec2: Cylindrical capacitor
This is an example of the cylindrical capacitor simulation, performed with QuickField software.
Problem Type:
An axisymmetric problem of AC conduction.
Geometry:
Capacitor consists of ceramic tube with silver electrodes mounted at the surface.
Given:
Relative permittivity of air ε = 1.
Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10-8 S/m;
Voltage U = 10 V.
Frequency f = 1000 Hz.
Problem:
Determine capacitance and dissipation factor of the capacitor.
Solution:
The value of dissipation factor can be calculated as
The value of dissipation factor can be calculated as tg(δ) = PA/PR. Capacitance can be calculated as C = q/U, where U is potential difference between electrodes and q - is a charge on electrodes.
Results:
|
QuickField |
q, C |
2.77·10-11 |
C, F |
2.77·10-12 |
PA, W |
2.45·10-8 |
PR, W |
8.17·10-7 |
tg(δ) |
0.03 |
|
