mivey
Senior Member
He mentioned capacitors reducing voltage drop so he must have meant a higher voltage at the motor terminals.
Capacitors, Dirty Power, Power Ripple
- Thread starter mivey
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That's what I took it to mean. He also mentioned reduced current as an outcome.
My point was that, if the voltage is low prior to adding PFC, that doesn't mean that it will necessarily take more current. The corollary is that, by adding PFC to "improve" the voltage, that it will necessarily take less current. Hence my point about not generalising.
gar
Senior Member
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- EE
130408-1443 EDT
Experiments with my 1/3 HP single phase induction motor with no mechanical load. The test is at my bench. Using Kill-A-Watt EZ and Fluke 27. Test capacitor is 67.5 mfd or about 39.3 ohms at 60 Hz.
The source impedance at my bench is such that a 10.7 A resistive load (1256 W) causes a voltage change of 4.7 V (123.9-119.2).
Motor only load voltage change is 0.8 V (124.1-123.3) with 5.3 A and 167 and 168 W. Call it 167.5 W.
One can assume for these changes that there is virtually no change in mechanical load or speed.
Capacitor only as load. The voltage change is possibly 0.1 V, but not more. Capacitor current is 3.3 A and Kill-A-Watt reads 2 W, but power is much less than this. Small errors from Kill-A-Watt at 90 deg current shift. The calculated capacitor current is about 3.1 A. The TED system is much worse under these conditions.
Motor plus capacitor is 2.34 A. Not too far from 5.3 - 3.3 = 2.0 A. Using calculated current the result would be 2.2 A. Power is still 167 and 168 W.
.
Experiments with my 1/3 HP single phase induction motor with no mechanical load. The test is at my bench. Using Kill-A-Watt EZ and Fluke 27. Test capacitor is 67.5 mfd or about 39.3 ohms at 60 Hz.
The source impedance at my bench is such that a 10.7 A resistive load (1256 W) causes a voltage change of 4.7 V (123.9-119.2).
Motor only load voltage change is 0.8 V (124.1-123.3) with 5.3 A and 167 and 168 W. Call it 167.5 W.
One can assume for these changes that there is virtually no change in mechanical load or speed.
Capacitor only as load. The voltage change is possibly 0.1 V, but not more. Capacitor current is 3.3 A and Kill-A-Watt reads 2 W, but power is much less than this. Small errors from Kill-A-Watt at 90 deg current shift. The calculated capacitor current is about 3.1 A. The TED system is much worse under these conditions.
Motor plus capacitor is 2.34 A. Not too far from 5.3 - 3.3 = 2.0 A. Using calculated current the result would be 2.2 A. Power is still 167 and 168 W.
.
Girl Engineer
Member
- Location
- Portland, OR
I agree that their wording is weird, and confuses things more than simplifies things.
I have been sizing caps a lot lately for individual motors, so I thought I would chip in. I go to the EC&M Elec. Calc Handbook for the equations and examples for doing these calcs. In one that I did, for a 100hp motor, the capacitor eliminated 17.82kVA of reactive power draw, which is about 21 Amps (FLA on the motor being 124A)
We know that real power is the power that does work in a system and reactive power contributes to the apparent power. If we can reduce the reactive power drawn from the utility, we will reduce the overall apparent power.
This article gives some good info that may help with understand how this works: http://ecmweb.com/power-quality/basics-pf-correction-single-induction-motors
This article also sheds a little more light on this: http://ecmweb.com/power-quality/troubleshooting-power-factor-correction-capacitors
I have been sizing caps a lot lately for individual motors, so I thought I would chip in. I go to the EC&M Elec. Calc Handbook for the equations and examples for doing these calcs. In one that I did, for a 100hp motor, the capacitor eliminated 17.82kVA of reactive power draw, which is about 21 Amps (FLA on the motor being 124A)
We know that real power is the power that does work in a system and reactive power contributes to the apparent power. If we can reduce the reactive power drawn from the utility, we will reduce the overall apparent power.
This article gives some good info that may help with understand how this works: http://ecmweb.com/power-quality/basics-pf-correction-single-induction-motors
This article also sheds a little more light on this: http://ecmweb.com/power-quality/troubleshooting-power-factor-correction-capacitors
jumper
Senior Member
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Nice post.:thumbsup:
mivey
Senior Member
The articles mention that harmonics decrease the power factor and detailed capacitor sizing to correct power factor. What they did not mention is that capacitors correct for displacement power factor caused by inductors, not the distortion power factor caused by harmonics.
Nor, as far as I can tell, did they mention any measures that need to be taken to prevent the capacitors taking excessive currents as a result of the harmonics that you typically see with a "dirty" supply.
The increased proliferation of non-linear loads means that virtually all supplies have harmonic content to some degree.
robbietan
Senior Member
- Location
- Antipolo City
The sellers here dont say "power ripple" , they say "harmonics" and their power savers will take of harmonics.
yeah, right
Wrong, when an induction motor is taking full load: there is around 11% increase in full load current when the voltage is 90% of rated value per page no.36 of
http://www.em-ea.org/Guide Books/book-3/Chapter 3.2 Electical Motors.pdf
mivey
Senior Member
Here is a quote from the PACS's documentation:
They make the claim and get out. No elaboration.
Here's my post #15:
Nope. Not wrong.
My post #15
That is not wrong.
I'll let you ponder/work out why that might be. You might find it instructive.
Nope.
Think about it some more.
I don't doubt that one bit.
I said you simply cannot generalise like that.
Reduced voltage at the motor terminals might make it take more current or less current.
OK. Rather than look at the general case you took the specific case of the motor running at rated load.
That's where you failed to understand the point.
What if the motor was running at half rated load - or even no load?
How then would "improving" the voltage affect the current? Up? Down? Stay the same?
What do you think?
Okay. Do you accept the application capacitor at motor terminals may serve to reduce the voltage drop along the motor circuit in case of low system fault level and motor running at full load ?
Perhaps you'd care to answer the question I posed in post #36 before we move on?
No OT please.
If you don't want to or can't answer a perfectly straightforward question, that's fine.
Perhaps you might bow out gracefully at this point.
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