In order to meet the requirement of power frequency voltage withstand test for large electrostatic capacitance test , some departments have installed AC series resonant test equipment.
The samples which have the large electrostatic capacitance are usually refers to cable, six sulfur fluoride pipeline, the capacitor and the large capacity generator of more than 300MW.
(1) The capacity of power supply transformer T and voltage regulator AT is small. This is because the voltage Uc=QUs. Us is a high supplied voltage. Since the current flowing in the high voltage circuit is the same, the capacity of the power supply transformer and the voltage regulator is Q times smaller than the required capacity in theory.
(2)The output voltage waveform of the series resonant device is better. This is because only the power frequency resonance, and for other power supply by the high harmonic component, the total impedance of the circuit is very large, so the harmonic component is very weak, the test waveform is better.
(3) If the flashover occurs during the test, the high voltage will disappear immediately because of the absence of resonance condition, so the arc will be extinguished immediately.
(4) The recovery of voltage is longer, and it is easy to control the power trip before the flashover voltage is reached again, so as to avoid repeated breakdown.
(5)The recovery voltage does not appear any overshoot caused by over-voltage.
Because of the characteristics of the above (3) and (4), the burn point formed after the breakdown of the sample is not large, which is beneficial to the study of the breakdown of the test sample. Due to the above characteristics, the device is safe to use, and can not generate large short-circuit current and will not restore over-voltage.
Below we will analyze and explain the above two characteristics about (4) and (5) .
When an AC high voltage series resonant device is used for testing, the recovery voltage of the sample after the first flashover has been eliminated:
U0-1=cos(ωt)-[cos(βt)+αsin(βt)/ β]exp(-at)
β=√1/(LC)-R2/(4L )2, α=R/(2L)
Because Q=ωL/R=40~80,1/(2Q)≈10-2
β=√ω2-[ω2/(4Q2)] =ω√1-1/4(Q2)≈ω
α=R/(2L) =ω√(2Q)
So,
U0-1=cosωt-[cosωt+sinωt/(2Q)]exp[-ωt/(2Q)]
Because 1/(2Q) is small
U0-1≈{1-exp[-ωt/(2Q)]}cosωt
n:The number of cycles counted from 0
T:Period of a Sine Wave
f:Sinusoidal frequency
Then
ωnT=2πfn/f=2πn
So, ωT/(2Q)=πn/Q
If the second times (the flashover voltage in voltage than reignition) first appeared slightly lower values, assuming exponential decay of e to 5%, nπ/Q≈3,n≈Q.
Since the Q value in the resonant device is always larger, the n value is greater. So, before another flashover occurs, it takes dozens of cycles, and it's easy to cut the power off for a long time.
Figure 1: Recovery voltage waveform of test sample after flashover
Figure 2: Full waveform of recovery voltage after flashover test
Figure 1 and Figure 2 draw the recovery voltage waveform of an actual series resonance test device after the test case flashover, the quality factor Q=40. It can be seen from the diagram that the negative overshoot is not occurred after the test case has been flashover. The time interval for recovering the voltage to reach the secondary breakdown value is close to 1s.
The use of series resonant devices to do withstand voltage test is limited, for example, it can not be used for external insulation of wet flashover and pollution flashover test.
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