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Overload Relay Sizing when using PF capacitors

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DFordPE

Electrical
Aug 18, 2014
20
I recently heard that the OLR must be "resized" if a PF capacitor is connected on the load side of the OLR. Ref: NEC 430.22 and 460.9

Has anyone done this? I can't find a way to calculate. However, I have always placed the capacitor ahead of the OLR.
 
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If the cap is connected to the terminals of the motor and the Over Load Protection is in a motor Control Assembly some where else, I see the following:-

45kW, 380v, 4 pole, IE2 has a full load current of 81.6A power factor of 0.86 at full load. The required kVAr to get it to unity is 26.7kVAr. You don't get these so you add a 20kVAr unit. The pf goes to 0.99 and the current now drops to 69.13A (Calculated) If your overload is still set at 81.6A the motor can now add another 12.47A or another 15% before the overload can react to the current which it assumes is FLC but has been corrected by the cap. The motor has a chance to fail as it can sit as a constant overload of 14% and the overload will never trip as it has never reached the overload condition at the protection device.

This is the problem. I wold start the motor and see what it runs at and set the overload accordingly. As the cap looses efficency it may trip. I don't like this method.

 
From John McGraw, P.Eng. These USA references will make more sense to you.

Re: The sizing of the motor Overload Protection when using Power Factor Correction Capacitors
Ref. NEC Article 460.9
Rating or Setting of Motor Overload Device
Where a motor installation includes a capacitor connected on the load side of the motor overload device, the rating or setting of the motor overload device shall be based on the improved power factor of the motor circuit.
The effect of the capacitor shall be disregarded in determining the motor circuit conductor rating in accordance with 430.2
Where a capacitor is connected on the load side of the overload relays, as shown in Exhibit 460.1, diagram (a), consideration must be given when selecting the rating or setting of the motor overload device because the line current will reduced due to an improved power factor. A value lower than that indicated in 430.32
should be used for proper protection of the motor.

An article by John Paschal, P.E., Bechtel Corp. in the Jun 1, 2002
of EC&M magazine entitled Correcting Power Factor at Your Motor Terminals covers the topic very well and also refers to the NEC article I mentioned earlier. You can go to it by typing in the following in your favorite web browser.

 
Yes, I read that article, but I still am somewhat confused.

Using Squeeky's example:

I have a motor with FLA of 81.6.

I add PF correction and the current is now 69.13.

How could going back to 81.6a be considered an overload?

It seems to me that by reducing the imaginary current and resultant heat, I should get an net improvement in my load capacity.

What am I missing?
 
I made a rude presentation showing what it looks like in the Re/Im current plane, using the Ossanna diagram.

See
The reduction in line current (that is what the protection reacts to) is usually quite small and definitively smaller than all other uncertainties like ambient temperature, dust covering cooling finns and the tolerance in the protection device itself.

The last picture shows that PFC reduces starting current. But the the reduction is quite small, so that blink at start is reduced, but not at all eliminated. To eliminate blink, one may need rather high capacitor values.

The problem with self-excitation and overvoltage is not covered. But it is a reality.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
"I made a rude presentation"

No offence meant - read "I made a crude presentation"

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
I believe that what I've researched says that you subtract capacitor current from motor FLA and set (I prefer IEC OLR) that number. That's where I'm stuck. It appears that I'm de-rating the motor after I improved the system.

Next, the chart Skogsgurra drew shows the PF worsening under an overload. Does that really happen? Not that it matters a lot but I'm curious.
 
Yes, that's exactly what you do. You allow less power when you "adjust" the overload setting. It is all very clear in my diagram and the comments. I don't find this very useful or important since you have a lot more uncertainties in many other factors that decide the temperature of the motor. It may be an interesting "conversation piece" to think about how PFC influences the setting of the thermal protection but I have never cared for the "allowance" that motor terminal connected PFC causes. But, then again, the motor terminal PFC practice was abandoned by me and my customers many, many years ago.


"shows the PF worsening under an overload. Does that really happen?"

Of course it does! Overload means more torque out from the motor. More torque and only a slight decrease in speed (increasing slip) means more kW - or in your parlance, HP. It IS the HP that is the load.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
Sorry, my second answer was about the wrong thing.

Yes, PF Changes when motor load changes. From a very low PF when idling to something like 0.8 at rated load and then getting worse as load increases. Locked rotor, the dashed Arrow, has a very low PF, worse than idling.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
Wherever you read that is wrong. It's not a straight subtraction. You have to do some vector math.

FLA x PF = real current.
reactive current = sqrt(FLA^2 - real current^2)
Corrected reactive current = inductive current - capacitor current
new motor current = sqrt(real current^2 + corrected reactive current^2)

Example;
FLA = 100A
PF = 87% or 0.87
capacitor rated current = 35A

real current = 100 x 0.87 = 87A
reactive current = sqrt(100^2 - 87^2) = 49A
corrected current = 49A - 35A = 14A
overload setting = sqrt(87A^2 + 14A^2) = 88A

Just subtracting the capacitor current puts you at 65A or 23A low.


 
[bigsmile]I like what Skogsgurra did with his down load. Wish I know how, but you Yanks do everything strange. You drive on the wrong side of the road, use inches and miles and your electrical drawings have the inductive phasor in the wrong direction. But it still works and I understand it. The self excitation bit still leaves me thinking. I just avoid this and only use pfc at the load on lighting. Thanks for the responses.
 
A. Skogsgurra is not a yank and lives in the perfectly normal IEC world :)

B. The inductive phasor is in the correct direction. It is lagging the active, and that's how these things work.
 
I stand corrected. Our "in phase" goes to the right. May be I was absent on that day and everyone I deal with is too polite to correct me. :). Others just jump right in and tell me..... :)
 
Thanks for all the input. I found the simple answer to be that the nameplate current is "apparent" current not "real" current, so the OLR is simply adjusted to account for that.
 
It's not about what I "thought was being done;" I never thought about it at all. Adding PF capacitors is just something you do with motors, for overall plant PF improvement. It was not necessary to think about what was going on in the motor, you just did it.

I've worked for a couple of motor control manufacturers and we always taught that you always use motor FLA for OLR selection/setting (IEC). There was never any mention of re-calculating after PF caps, so I just never thought about it.

Now I've thought about it.


 
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