What measurements would be taken and in what units. If the Power Factor correction capacitor was in an individual electrical panel for a 3 phase 400v centrifigal pump and was rated at 10KVA
2006-07-30 08:43:19 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
Centrifugal pump should be a constant load so
Measure the current with cap in circuit.
Disconnect the cap, some have fuses in them that can be pulled. Power off of course.
Start and run motor again. Current should be higher without the cap in the circuit.
Need to measure the current before the point the cap is connected of course.
Cap does not lower the current in the motor only in the conductors servicing the motor.
2006-07-30 09:00:26 · answer #1 · answered by buderosdad1 2 · 3⤊ 0⤋
In theory, the capacitance/reactance of the corrected circuit is zero, but there are two problems with that.
1) Most folks don't have anything to measure the power factor, I don't.
The capacitor may be there only to correct the condition that occurs at motor startup, when a heavy inductive load is suddenly connected to the 3 phase 400V circuit.
I would not recommend poking at the capacitor unless you are too qualified to ask this question. A 10 KVA capacitor can electrocute you, even if the power is turned off.
2006-07-30 09:16:36 · answer #2 · answered by Computer Guy 7 · 0⤊ 0⤋
Use a Phase angle meter to determine how close to unity the voltage and current are.
The cosine of the phase angle is your power factor.
If you want to determine how well the capacitor is doing its job.
Remove the Capacitor from the circuit measure your power factor, then do the same with it in the circuit. You can then determine the VAR (volt amp reactance ) flow and the savings you are receiving for the investment.
a rough estimate can be made with an clip on Amp meter. Read the motor amperes with the capacitor in the circuit, then disconnect the capacitor and measure amperes without.
The reading should be lower with the capacitor in the circuit.
Crude but it will give you a indication.
Yours: Grumpy
2006-07-30 08:55:04 · answer #3 · answered by Grumpy 6 · 0⤊ 0⤋
I don't know for sure I never worried about it. If the capacitor is sized correctly, installed correctly it should work correctly.
The power factor has to do with the voltage and amperage getting out of phase due to inductive loads like motors.
The utilities have power factors readings on their meters so I'm guessing you could buy a meter to check that. With a low power factor the apparent Volt amps reading is lower than the actual power consumption.
2006-07-30 08:50:11 · answer #4 · answered by Roadkill 6 · 0⤊ 0⤋
well you would want to compare the input phase with the output phase. If you could test it without the caps and with the caps you would know right away.
You are using the caps to reduce the inductance in the system so you get almost all real power and no reactive power.
S=V*Conj(i)
2006-07-30 09:01:29 · answer #5 · answered by DoctaB01 2 · 0⤊ 0⤋
Each and every day. It took some time to understand what was going on. The Holy Spirit was guiding me but I was not following along very well. But once you accept him fully and completely and surrender yourself, you then become a whole new person and listen and understand whats going on. Yea, I feel his power each and every day and it's wonderful!
2016-03-27 07:23:45 · answer #6 · answered by Sandra 4 · 0⤊ 0⤋
Measure the power factor without it. If when it is in, the power factor is closer to one, it is working.
2006-07-30 18:27:23 · answer #7 · answered by blind_chameleon 5 · 0⤊ 0⤋
is the light on.
2006-07-30 08:46:38 · answer #8 · answered by Anonymous · 0⤊ 0⤋
. . . have you tried sticking your finger in . . .
2006-07-30 08:54:06 · answer #9 · answered by Astra 6 · 0⤊ 0⤋