Start Capacitors Increasing Their Capacitance

[jfrye]

I have two centrifugal 3HP single phase submersible sewage pumps, which use 2 start caps (rated 72-86µF) wired in parallel and 2 run caps (55µF) wired in series to operate. These pumps have been failing out due to overtemp during high flow conditions, so "continuous" running. I believe I need to choke the supply voltage down; the mfr says that when continuously running the voltage variation should be ±5% of 230VAC, line to line the supply is 251VAC which is greater than +5%.

That said, there is still another mystery to me with the start caps, as I said before they are labeled 72-86µF, but when metered with a multi-meter set for capacitance they all read ~130µF, this is with all the wiring removed, there is only a discharge resistor(which is soldered on) attached to the start cap terminals. I personally installed the new capacitor kits a month or two ago, and checked everything which at the time was nameplate normal. So has anyone ever seen the capacitance of a capacitor change and not explode?

 

[ScottyUK]

The resistor is probably skewing the measurement because test current will pass through it in addition to the expected path through the capacitance.

 

[crshears]

Spot on, ScottyUK!

If the OP wanted to confirm the resistor was causing the problem, is there a way [other than removing one of the soldered leads] that could be used to measure the phase angle of the applied current through first the resistor then the cap indivually so the capacitance value could be calculated? [I don't get to much testing in my job...]

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[jfrye]

This was confirmed using a new capacitor same nameplate and a removable resistor. Rated at 72-86µF with no resistor read 78µF, with resistor read 135µF.

So I am guessing the multi-meter takes its reading based on the RC time constant, so by adding impedance it changes the "capacitance".

Thanks all.

 

[ScottyUK]

I think most types charge the cap for a defined period of time using a known constant current, then calculate capacitance based on the change in voltage at the end of the defined period.

charge = current * time
capacitance = charge / voltage

 

[itsmoked]

Yes, so since a bunch of current was leaking off the result would be a larger assumed capacitance. The resistor in this case was bleeding about half the current off to give twice the capacitance.

Keith Cress
kcress -

http://www.flaminsystems.com