Power Factor Capacitors and Reactive Current

[mivey]

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.

It is kvarh, kWh, and kVA. I prefer to think of kW as speed. kWh is the energy, kW is how fast you pull the energy off of the system.

The faster you pull energy, the bigger the equipment you must have (like a bigger hose and pump to get the water faster). Bigger kW = bigger kVA

 

[mull982]

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.

Weressl

Can you please explain this statement? I do not follow?

 

[mull982]

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.

I agree that this reactive current effects the size of wire, transformer etc.. However with this equipment already in place there is no additional direct cost associated with this extra current other that kw losses due to heating.

If a power company does not charge a pf penalty or any other penalty how does it recoup its losses for giving away these reactive amps?

 

[mivey]

I agree that this reactive current effects the size of wire, transformer etc.. However with this equipment already in place there is no additional direct cost associated with this extra current other that kw losses due to heating.

If a power company does not charge a pf penalty or any other penalty how does it recoup its losses for giving away these reactive amps?

You had kWh losses, not kW losses. You lost energy, not power.

You are correct that other than heat, there are essentially no other costs, other than those you may bring on yourself.

The power companies usually reserve the right to charge you for excess kvar, usually below about 90-95% pf. Some will start charging you a kVA charge instead of a kW charge. The power companies have to manage their var load. If you "abuse" the power system, they have a right to charge you.

Your company may also have contract obligations with the POCO concerning power factor that they have to abide by.

The power company gives nothing away. They either have power plants that generate the var needed or they have capacitor banks.

Most of these costs are embedded in the rates plus they add penalties for excessive var load. Transmission level customers will usually receive reactive control as an ancillary service.

 

[mivey]

Weressl

Can you please explain this statement? I do not follow?

He was just saying your boss used the following:
kWh = #phases*volts*amps/phase*pf*hours
and kWh * rate = $

so: 3*2.3_kV*27_A*1.0*8000=1,490,400_kWh
and 1,490,400_kWh*$0.1/kWh=$149,040

I don't know how he got to $155,635.00

Remember, the real cost is:
#phases * (reactive amps per phase)^2 * resistance * hours / 1000 * $/kWh

For example (using 27 amps/ph, 0.004 ohms, 4500 hours, $0.10/kWh):
3*27^2*0.004*4500/1000*$0.10 = $3.94/yr

 

[wirenut1980]

From a power quality perspective, I would want the cap online. They can be used to keep voltage levels higher during heavily loaded periods (like in the summer), which can drag down the utility voltage. As discussed, it does not really hurt you financially to keep the cap offline, but my next step would be to find out why the cap fuses are blowing, solve that problem, buy the fuses and put the cap back online.

 

[weressl]

He was just saying your boss used the following:
kWh = #phases*volts*amps/phase*pf*hours
and kWh * rate = $

so: 3*2.3_kV*27_A*1.0*8000=1,490,400_kWh
and 1,490,400_kWh*$0.1/kWh=$149,040

I don't know how he got to $155,635.00

Remember, the real cost is:
#phases * (reactive amps per phase)^2 * resistance * hours / 1000 * $/kWh

For example (using 27 amps/ph, 0.004 ohms, 4500 hours, $0.10/kWh):
3*27^2*0.004*4500/1000*$0.10 = $3.94/yr

(((sqrt)3*4160V*27A)/1000)*0.1$*800hrs=$155635.16

 

[weressl]

From a power quality perspective, I would want the cap online. They can be used to keep voltage levels higher during heavily loaded periods (like in the summer), which can drag down the utility voltage. As discussed, it does not really hurt you financially to keep the cap offline, but my next step would be to find out why the cap fuses are blowing, solve that problem, buy the fuses and put the cap back online.

The caps -with age, about 10 years - get 'tired' and short out. THAT'S when they usually blow the fuse. This is usually because the electrolyte dries out. So buy new caps and you're back in business.

We had discussed capacitors age and failure in relationship to UPS's and ASD's.

 

[jim dungar]

(((sqrt)3*4160V*27A)/1000)*0.1$*800hrs=$155635.16

4160V
27A
800hrs/yr
$0.10/KWH

(1.73*4160*27)/1000=194.3136kW
194.3436*800=155,450.88KWH
155,450.88*.1=$15,545

 

[Cold Fusion]

I'm pretty new at this. Perhaps someone could check my math and reasoning.

It seems one might consider overall system losses (copper, transfomer) at 3 - 5%.

So, (8A x 4160V x 5% x 8736hrs/year)/1000 = 14,500kwh

At $0.10/kwh, that's ~ $1450

for 3% losses and $0.05kwh, that's ~ $450

Perhaps you could explain it this way (Yes, most all of this has been said):
The current is reactive, the motor magnetic field stores the energy on one half cycle and sends it back to the generating plant on the other half. With the caps on-line, this energy is exchanged from the motor mag field to the cap electric feild. With out the caps, the additional losses are the copper losses from the motor to the service point.

cf

 

[weressl]

4160V
27A
800hrs/yr
$0.10/KWH

(1.73*4160*27)/1000=194.3136kW
194.3436*800=155,450.88KWH
155,450.88*.1=$15,545

8000hrs/yr.
$155,635.16

(SQRT)3= 1.732051

 

[mivey]

2.3kV?

2.3kV?

8000hrs/yr.
$155,635.16

(SQRT)3= 1.732051

Just noticed I've been sticking 2.3 kV in there :roll: and should have had 2.4 kV so:

3*2.4_kV*27_A*1.0*8000=1,555,200_kWh
and 1,555,200_kWh*$0.1/kWh=$155,520
which would be the same number you would get if you used 2400*sqrt(3) or 4156.92 V

 

[mull982]

Just noticed I've been sticking 2.3 kV in there :roll: and should have had 2.4 kV so:

3*2.4_kV*27_A*1.0*8000=1,555,200_kWh
and 1,555,200_kWh*$0.1/kWh=$155,520
which would be the same number you would get if you used 2400*sqrt(3) or 4156.92 V

I agree with both of these calculations. This is probalby the calculation that my colleuge used to come up with his number. However we all agree that this calculation is incorrect for this application. These 27A are reactive amps and do not factor into the kW number therefore do not directly corrolate to an increased charge as the above calculation would indicate. If we had a unitity power factor as wressel implied then these amps would be real amps and this calculation would apply.

Agreed?

 

[mivey]

I'm pretty new at this. Perhaps someone could check my math and reasoning.

It seems one might consider overall system losses (copper, transfomer) at 3 - 5%.

So, (8A x 4160V x 5% x 8736hrs/year)/1000 = 14,500kwh

At $0.10/kwh, that's ~ $1450

for 3% losses and $0.05kwh, that's ~ $450

Perhaps you could explain it this way (Yes, most all of this has been said):
The current is reactive, the motor magnetic field stores the energy on one half cycle and sends it back to the generating plant on the other half. With the caps on-line, this energy is exchanged from the motor mag field to the cap electric feild. With out the caps, the additional losses are the copper losses from the motor to the service point.

cf

Part 1) No can do. The % is wrong as well as the formula. What is the basis for the 3-5% loss assumption?
The total kWh losses are: #phases * (amps per phase)^2 * resistance * hours / 1000

The reactive kWh losses are: #phases * (reactive amps per phase)^2 * resistance * hours / 1000

Part 2) That is a simple way to put it, like the longer version here:
http://forums.mikeholt.com/showpost.php?p=827136&postcount=6

 

[mivey]

I agree with both of these calculations. This is probalby the calculation that my colleuge used to come up with his number. However we all agree that this calculation is incorrect for this application. These 27A are reactive amps and do not factor into the kW number therefore do not directly corrolate to an increased charge as the above calculation would indicate. If we had a unitity power factor as wressel implied then these amps would be real amps and this calculation would apply.

Agreed?

This formula would never apply for calculating the heat loss.

If you could actually reduce the real load, then you could estimate the load portion of the savings this way. You would still have a separate calculation for reduced heat loss (call it feeder savings).

 

[weressl]

I agree with both of these calculations. This is probalby the calculation that my colleuge used to come up with his number. However we all agree that this calculation is incorrect for this application. These 27A are reactive amps and do not factor into the kW number therefore do not directly corrolate to an increased charge as the above calculation would indicate. If we had a unitity power factor as wressel implied then these amps would be real amps and this calculation would apply.

Agreed?

OP data: "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."

Read my post on WHY this calculations are not appropriate under post:

http://forums.mikeholt.com/showthread.php?p=875611#poststop

The only correct formula is for three phase calculations is:

((SQRT(3)*4160V*27A)/1000)*0.1$/kWhr*8000hrs

Where 4160V is the line-to-line nominal voltage in order to arrive the effective voltage factor in the power calculation of a 3 phase circuit is to calculate the momentary value of the three phases spaced at 120* to each other at any given time is by multiplying it by SQRT(3).

As you can see the formula is not complete since we are multiplying kVA with $/kWhr, so kVA and kW do NOT cancel each other out.

If you go back to the original numbers and use the kW it will result in the following:

With pf corrective capacitors:
1.732051*4kV*285A*.9451pf=1859.027kW or kW*.923eff/.746=2300.11HP

Without power factor capacitors:
1.732051*4kV*312A*.86pf=1858.975kW or kW*.923eff/.746=2300.046HP

HP is the mechanical work produced on the shaft and kW is the electrical energy that is consumed by the motor. As you can see the caonversion from electrical to mechanical energy has ony one factor that is variable and that is the efficiency. Efficiency changes with the actual load. Power factor has insignificant impact on efficency. Voltage has a slightly greater impact, but it still remains meaningless in real life to worry about if that voltage change has to do with voltage drop change diue to the current decrease. If you have 3% voltage drop calculated and you drop less than 10%(27A) then the voltage drop will increase by 0.3%. You start to pay attention to efficency when your voltage drop is about 5% or greater. Voltage inballance has a greater effect on efficiency, but in this case it would be the same for both before and after cap's so, again, you don't care.

 

[jim dungar]

8000hrs/yr.
$155,635.16

(SQRT)3= 1.732051

Your formula showed only 800 hours.

In any case because this is reactive current the result should be kVARH not KWH.

FWIW, Many (if not most) POCO's in the Wisconsin area stopped applying PF penalties more than 15 years ago. Medium voltage customers often negotiate their KWH rates and service KVA capacity using formulas that include PF multipliers that stay unchanged until the next rate negotiation.

 

[weressl]

Your formula showed only 800 hours.

In any case because this is reactive current the result should be kVARH not KWH.

FWIW, Many (if not most) POCO's in the Wisconsin area stopped applying PF penalties more than 15 years ago. Medium voltage customers often negotiate their KWH rates and service KVA capacity using formulas that include PF multipliers that stay unchanged until the next rate negotiation.

Absolutely. I just demonstrated how his boss could have got the, erroneous, results.

 

[mivey]

The only correct formula is for three phase calculations is:

((SQRT(3)*4160V*27A)/1000)*0.1$/kWhr*8000hrs

or:
(3*2400V*27A)/1000*0.1$/kWhr*8000hrs

Where 2400V is the line-to-ground nominal voltage.

 

[mivey]

FWIW, Many (if not most) POCO's in the Wisconsin area stopped applying PF penalties more than 15 years ago. Medium voltage customers often negotiate their KWH rates and service KVA capacity using formulas that include PF multipliers that stay unchanged until the next rate negotiation.

We still charge a var penalty in Georgia. We also have some contract agreements, some using kVA charges, and others just have pf clauses that differ from the regular rates.