Power Factor Capacitors and Reactive Current

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mull982

Senior Member
I've had a situation in my plant recently where we have been blowing fuses on some of the power factor correction capacitors on our large 5kV motors. Our typical power factor for the plant is about 80-90% and we DO NOT pay a power factor penalty with the utility.

Recently when the fuses in these capaciotrs have been blowing we have not had the replacement fuses in stock. We needed to get back up and running so I stated that since we did not pay any sort of PF penalty then to go ahead and run the motor without the Caps.

A production manager noticed that the motor running without the Caps caused the current on the motor to increase (Current monitored by CT's) and became concerned. He took the numerical current increase he saw and performed a calculation and stated that by not having the capacitor this increased current is costing us some $150,000.00 in our power bill.

I tried to explain to him that the increase in current he is seeing is due to Reactive Amps that were coming from the capacitor before but now are coming across the line from our plant power. I explained however that these amps were only Reactive amps and were amps that we did not pay for. I tried to explain the whole concept of power factor and Active (kW) vs Reactive (kVAR) amps with a power triangle but he is still not believing me.

Just so that I am not missing anything, am I correct by stating that the increase in amps he is seeing are reactive amps and that there is no direct corrolation between these amps and an increased electric bill. The only way there would be an increased electric bill is if we paid a PF penalty which we do not, so there is no increase here either.
 

mivey

Senior Member
It does cost you. You will pay for the heat losses caused by pushing the "reactive current" around your plant because the conductors in the plant have resistance.

[PS: I doubt the cost is what he calculated unless he actually made a heat loss calculation (this is the part I doubt)]
 
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mull982

Senior Member
mivey said:
It does cost you. You will pay for the heat losses caused by pushing the "reactive current" around your plant because the conductors in the plant have resistance.

[PS: I doubt the cost is what he calculated unless he actually made a heat loss calculation (this is the part I doubt)]

I agree that there will be heat looses due to the additional reactive current but I dont see these losses amounting to much when were dealing with about 8A or so.

No you are correct he did not perform a heat loss calculation. He took the additional Amps he was seeing which was about 8A and multiplied this by the voltage to come up with a power quantity (which I'm sure he did wrong) he then took this power quantiy and coverted it to a kWH quanity for the entire year. He then took this annual kWH quantiy and multiplied it by what he thinks we are paying per kWH from the utility to come up with some high number around $15,000/yr. I explained to him that this calculation was completly wrong and that it had no relevance for the cost based on the additional current we are seeing. He does not see it however and I'm having a hard time explaining it to him.

Just out of curosity, since we dont pay a PF penalty what are some other reasons to be concerned with our plant PF, if any?
 

mivey

Senior Member
FWIW the annual losses are a function of:
C * resistance/pf^2

Where C is a constant based on a few parameters like load, voltage, # phases, etc.

The point is, you can see resistance and power factor play a role in the net kWh used even if the end load is held constant.
 

mivey

Senior Member
mull982 said:
I agree that there will be heat looses due to the additional reactive current but I dont see these losses amounting to much when were dealing with about 8A or so.

No you are correct he did not perform a heat loss calculation. He took the additional Amps he was seeing which was about 8A and multiplied this by the voltage to come up with a power quantity (which I'm sure he did wrong) he then took this power quantiy and coverted it to a kWH quanity for the entire year. He then took this annual kWH quantiy and multiplied it by what he thinks we are paying per kWH from the utility to come up with some high number around $15,000/yr. I explained to him that this calculation was completly wrong and that it had no relevance for the cost based on the additional current we are seeing. He does not see it however and I'm having a hard time explaining it to him.

Just out of curosity, since we dont pay a PF penalty what are some other reasons to be concerned with our plant PF, if any?
If you make no effort to correct your poor pf, the POCO may impose a penalty on you, even if they have not in the past.

You will need larger system capacity to be able to push the extra current around the plant. Bigger wire, bigger switchgear, etc = higher cost. Along those same lines, you will lose capacity that might otherwise be available for other uses.

There are also concerns over power quality issues like voltage problems and motors overheating, etc.

[Add:
FWIW: It usually does not make sense to correct to a unity power factor with capacitors as there is a knee on the cost/benefit curve that you run into.

It also may not make sense to add correction to a motor that is rarely used, system capacity being a separate issue. This is even more true if you are not paying a pf penalty because with very little kWh being used, your recovery of the cost of the added capacitors is spread out over a very long time.]
 
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mayanees

Senior Member
Location
Westminster, MD
Occupation
Electrical Engineer and Master Electrician
... absolutely no way..

... absolutely no way..

mull982 said:
He took the numerical current increase he saw and performed a calculation and stated that by not having the capacitor this increased current is costing us some $150,000.00 in our power bill.

The first thing I would do is correct the 'orders of magnitude' error in his calculation.
There is no way that the motor you're dealing with is large enough that there would level of cost associated with it.

Please, correct me if I'm wrong!!
 

jghrist

Senior Member
You could also have an increased current because of lower voltage at the motors. This could result in lower efficiency and higher energy use.

Where is the current being measured? The OP said "current on the motor", but the reactive current increase from disconnecting the capacitors would show up in the current that goes to the motor and capacitor combination. The current to the motor alone would not change except as a result of lower voltage.

Have you had any power bills since the capacitors were disconnected? If the plant operation did not change otherwise, a before and after comparison should convince the production manager that the magnitude of the power bill increase is much less than he calculated.
 

mivey

Senior Member
Maybe yes way?

Maybe yes way?

mayanees said:
The first thing I would do is correct the 'orders of magnitude' error in his calculation.
There is no way that the motor you're dealing with is large enough that there would level of cost associated with it.

Please, correct me if I'm wrong!!
OK.
I don't think it has as much to do with the orders of magnitude as it has to do with methodology.

Without information on the system layout, the size of the motors involved, the run hours, location of the meter point, etc, you can't tell if this is a possibility or not.
 

mull982

Senior Member
Thanks for all of the responses. I can see how heating losses and system capacity can be affected by these Reactive amps. I think in this case however when talking about 8A or so these isses are small. The motor we are talking about is a 2300hp motor at 4.16kV.

Can we all agree however that simply taking the increased current and converting it to a power quanity is not the correct way for trying to calculate an increased electrical cost associated with it.

For this motor I can see may a couple of thousand dollars over a years time but nowhere near the number he is coming up with.
 

mull982

Senior Member
jghrist said:
You could also have an increased current because of lower voltage at the motors. This could result in lower efficiency and higher energy use.

Where is the current being measured? The OP said "current on the motor", but the reactive current increase from disconnecting the capacitors would show up in the current that goes to the motor and capacitor combination. The current to the motor alone would not change except as a result of lower voltage.

Have you had any power bills since the capacitors were disconnected? If the plant operation did not change otherwise, a before and after comparison should convince the production manager that the magnitude of the power bill increase is much less than he calculated.

The current is being measured before the capacitor. So with the capacitor in the circuit the CTs do not see the reactive amps from the Caps going to the motor. With the caps out of the circuit the CT's see all of this current on the line thus the visable increase.

This is a recent incident so we do not have any power bills for comparison yet.
 

mivey

Senior Member
mull982 said:
Thanks for all of the responses. I can see how heating losses and system capacity can be affected by these Reactive amps. I think in this case however when talking about 8A or so these isses are small. The motor we are talking about is a 2300hp motor at 4.16kV.

Can we all agree however that simply taking the increased current and converting it to a power quanity is not the correct way for trying to calculate an increased electrical cost associated with it.

For this motor I can see may a couple of thousand dollars over a years time but nowhere near the number he is coming up with.
agreed. I would be surprised if it were a couple of thousand dollars.
 

mivey

Senior Member
At 8 amps, you probably would not save enough to get a cup of coffee.

Are you sure about the 8 amps? Just a rough calc but if you were 95% corrected, then you would be about 92% uncorrected.

Maybe he said 80 amps. This would be more like 72% uncorrected. Now you might can save enough in heat losses for lunch once a year.
 

mull982

Senior Member
mivey said:
At 8 amps, you probably would not save enough to get a cup of coffee.

Are you sure about the 8 amps? Just a rough calc but if you were 95% corrected, then you would be about 92% uncorrected.

Maybe he said 80 amps. This would be more like 72% uncorrected. Now you might can save enough in heat losses for lunch once a year.

The motor is a 2300 hp motor at 4kv and has a motor nameplate of 312A with a PF of 86% at full load. The power factor was corrected from 86% to 95% using a 450kVAR capacitor. The corrected motor FLA is about 285A. So with the cap out of the circuit at full load he probably saw the motor jump from 285A to 312A for an increase of about 27A. I dont see this increase causing much wich respect to heating losses as you mentioned.
 

mivey

Senior Member
mull982 said:
The motor is a 2300 hp motor at 4kv and has a motor nameplate of 312A with a PF of 86% at full load. The power factor was corrected from 86% to 95% using a 450kVAR capacitor. The corrected motor FLA is about 285A. So with the cap out of the circuit at full load he probably saw the motor jump from 285A to 312A for an increase of about 27A. I dont see this increase causing much wich respect to heating losses as you mentioned.
That is correct. Less than $5/yr, I would guess.
 

mivey

Senior Member
mull982 said:
Thanks Mivey

Whats an aproximation for calculating the heat losses for current in an application like this.
As an off-the-cuff number, I just took the change in amps, squared, times the wire resistance, times the # run hours in a year, divided by 1000, times the average cost in cents/kWh, times the # conductors.
 

jghrist

Senior Member
Loss reduction from fully compensating for reactive current is WLR = 3?Ir??R where Ir is the reactive current in amps, R is the total circuit resistance per phase, and WLR is the loss reduction in watts.
 

mivey

Senior Member
jghrist said:
Loss reduction from fully compensating for reactive current is WLR = 3?Ir??R where Ir is the reactive current in amps, R is the total circuit resistance per phase, and WLR is the loss reduction in watts.
For three phase
 
mull982 said:
I've had a situation in my plant recently where we have been blowing fuses on some of the power factor correction capacitors on our large 5kV motors. Our typical power factor for the plant is about 80-90% and we DO NOT pay a power factor penalty with the utility.

Recently when the fuses in these capaciotrs have been blowing we have not had the replacement fuses in stock. We needed to get back up and running so I stated that since we did not pay any sort of PF penalty then to go ahead and run the motor without the Caps.

A production manager noticed that the motor running without the Caps caused the current on the motor to increase (Current monitored by CT's) and became concerned. He took the numerical current increase he saw and performed a calculation and stated that by not having the capacitor this increased current is costing us some $150,000.00 in our power bill.

I tried to explain to him that the increase in current he is seeing is due to Reactive Amps that were coming from the capacitor before but now are coming across the line from our plant power. I explained however that these amps were only Reactive amps and were amps that we did not pay for. I tried to explain the whole concept of power factor and Active (kW) vs Reactive (kVAR) amps with a power triangle but he is still not believing me.

Just so that I am not missing anything, am I correct by stating that the increase in amps he is seeing are reactive amps and that there is no direct corrolation between these amps and an increased electric bill. The only way there would be an increased electric bill is if we paid a PF penalty which we do not, so there is no increase here either.

Tell him that he is full of..............................ignorance. :grin:

If you want to keep your job, then you can ask him if he knows what the power company meters are called. If he does not know then explain it to him that they are called (kilo)watt-hour meters and not (kilo)volt-ampere-hour meters.

kVAhr=SQRT(3)*V*I*Hr

kWHar=SQRT(3)*V*I*pf*hr

p.f.(power factor) - as you probably know - representative of the difference between the voltage and the current that occurs when the load is not purely resistive. In the case of inductive circuits the current will 'lag' behind the voltage, so the sine wave does not peak simultaneously. since the V and I component produces the 'work' together the power factor, that is always less than 1.00, will 'adjust' the calculation to arrive at the actual work produced.

kW represents the actual 'work' that the energy provides for the user. The difference can be thought of that the water company charges you for the cubic yards you are using regardless at what pressure it is delivered. (I know the analogy is less than perfect, but hey, offer yours.)

Since the current component determines the size of the wire, or transformer, some POCO's charge for either kVA Demand or power factor a flat monthly fee to recoup the cots of the extra equipment, wiring they need to install to support the 'current' component. Since power factor capacitors 'lower' the current they are used sometimes to avoid the charges or even in plant to reduce the size of wires need to be installed.
 
mull982 said:
The motor is a 2300 hp motor at 4kv and has a motor nameplate of 312A with a PF of 86% at full load. The power factor was corrected from 86% to 95% using a 450kVAR capacitor. The corrected motor FLA is about 285A. So with the cap out of the circuit at full load he probably saw the motor jump from 285A to 312A for an increase of about 27A. I dont see this increase causing much wich respect to heating losses as you mentioned.

The boss wasn't 'completely' wrong, just ignorant.

If all the 'current' is usable at unity power factor it would have cost $155,635.00 at $0.1/kWhr and 8000 hours.
 
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