Generator Power Factor

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Dark Sparky

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Scope: the USA.

When performing load calculations involving generators, when does one use the (industry standard) 0.8 power factor (pf), and when does one use the load's power factor (let's assume it's 0.9)?

Case in point - if determining the # of generators (500 kW each) required to feed a given load (load is 500 kW or 555 kVA at 0.9 pf), how many generators are needed?

Alternately, if one has a 500 kW/625 kVA genset - will that supply a 500 kW/555 kVA load (with 0.9 pdf)?

Thanks in advance for helping me wrap my head around this.
 
Generators supply power to the loads. It is the load power factor that determines how big a generator you'll need of how many small generators you need. You got it backwards. E.g. if your load is 1000 kW at 0.85 pf, you'll need a total of 1000/0.85 = 1177 kVA generator's.
 
jonathankpa said:
Scope: the USA.

When performing load calculations involving generators, when does one use the (industry standard) 0.8 power factor (pf), and when does one use the load's power factor (let's assume it's 0.9)?

Case in point - if determining the # of generators (500 kW each) required to feed a given load (load is 500 kW or 555 kVA at 0.9 pf), how many generators are needed?

Alternately, if one has a 500 kW/625 kVA genset - will that supply a 500 kW/555 kVA load (with 0.9 pdf)?

Thanks in advance for helping me wrap my head around this.
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500 kW is 500 kW to the generator prime mover. Reactive current because of power factor adds current and additional losses in the lines and source windings, but is not something directly seen by the prime mover, the additional heat losses in the lines are a real load however and do add to the kW.

If the genset is rated 500kW/625 kVA ... 500/625 = .80, all that means is the thing was designed to operate full kW at power factor as low as .80 before excess heating begins to occur.
 
kwired,

the reactive power is seen by the prime mover. it is temporary energy that is not permanently absorbed by the load. this energy is exchanged back and forth from load and generator during the cycle and the generator must be sized to handle it in addition to the load watts.
 
mivey said:
kwired,

the reactive power is seen by the prime mover. it is temporary energy that is not permanently absorbed by the load. this energy is exchanged back and forth from load and generator during the cycle and the generator must be sized to handle it in addition to the load watts.
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Watt losses resulting from the reactive power is real power that is typically lost in the circuit as heat. AFAIK the prime mover only sees the real power (kW) plus those watt losses and any mechanical losses between it and the generator.
 
The way I think of it is there are two ratings on the generator you cannot exceed, the kVA and kW. Whichever you reach first with the load you apply, you stop.
If your load is 500kW/555kVA, then in theory you need a 500kW/625kVA generator because you will hit the 500kW rating first before you hit the kVA rating of that same machine with adding more load.
I say in theory, because most gensets will be standby rated which you will want to keep to 70% of the rating if this is important equipment.
ISO-8528-1 limits the 24-hour average load factor on most standby generator sets to 70 percent of nameplate capacity. For utility outages lasting a few minutes or a few hours, one or two times a year, standby generator sets are designed to be loaded to 100 percent of nameplate capacity for the duration of the outage. However, if an outage lasts days instead of hours and the standby power system is loaded to 100 percent of its nameplate capacity, it is likely that the 24-hour average load will exceed the power system's design parameters.
 
kwired said:
Watt losses resulting from the reactive power is real power that is typically lost in the circuit as heat. AFAIK the prime mover only sees the real power (kW) plus those watt losses and any mechanical losses between it and the generator.
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The prime mover supplies the load's real power, the real power due to delivery losses, and also sees and must provide the capacity needed to exchange the reactive power back and forth within each cycle.

So the generator must generate extra energy to create the fields inside the load and load path associated with reactive power. When these fields collapse and reverse within the cycle, the generator must be able to re-absorb or convert this temporary energy and then send extra energy back to charge the load and path fields in the opposite direction.

The prime mover not only generates the used energy and loss energy but also has to handle the initial generation and subsequent exchanging of this extra temporary field energy, including associated makeup losses.

So the prime mover sees the load energy, the loss energy, and the exhanged reactive energy. It sees all three and must be sized to handle all three.
 
mivey said:
The prime mover supplies the load's real power, the real power due to delivery losses, and also sees and must provide the capacity needed to exchange the reactive power back and forth within each cycle.

So the generator must generate extra energy to create the fields inside the load and load path associated with reactive power. When these fields collapse and reverse within the cycle, the generator must be able to re-absorb or convert this temporary energy and then send extra energy back to charge the load and path fields in the opposite direction.

The prime mover not only generates the used energy and loss energy but also has to handle the initial generation and subsequent exchanging of this extra temporary field energy, including associated makeup losses.

So the prime mover sees the load energy, the loss energy, and the exhanged reactive energy. It sees all three and must be sized to handle all three.
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Supply a VFD from the generator - the motor driven by the VFD has reactive power between it and the drive but supply side power factor is for the most part negligible and all the generator sees is watts. Any line losses because of extra reactive current are real watts to the supply, but are in addition to the motor load. Generator only sees motor watts plus those losses, it doesn't see the reactive current.

Same thing if you use power factor capacitors, reactive current flows between motor and capacitors but not to source (presuming 100% correction)
Again generator must produce any watts that are given up as line loss in addition to the motor load.

Line watt losses are not equal to kVAR though.
 
kwired said:
Supply a VFD from the generator - the motor driven by the VFD has reactive power between it and the drive but supply side power factor is for the most part negligible and all the generator sees is watts. Any line losses because of extra reactive current are real watts to the supply, but are in addition to the motor load. Generator only sees motor watts plus those losses, it doesn't see the reactive current.

Same thing if you use power factor capacitors, reactive current flows between motor and capacitors but not to source (presuming 100% correction)
Again generator must produce any watts that are given up as line loss in addition to the motor load.

Line watt losses are not equal to kVAR though.
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yes, but that makes them a near unity power factor load to the generator. other than providing the initial field charge, the generator sees a load with kVA essentially equal to kW. So kVA and kW needs are both met. any residual var load will still be seen and handled by the generator.

the op was about a non-unity load presented to the generator.

high power factor loads are either resistive in nature or have internal power factor correction. keep in mind that capacitors only correct for reactive vars not distortion vars. both type vars contribute to low power factor and are on 90 degree axis relative to each other and to the real axis (real on the x-axis, reactive on the y-axis and distortion on the z-axis).

to correct for distortion and keep the pf high, you need a distortion filter as capacitors work in a different plane.
 
Belated - but thank you all for the responses. For practical generator sizing purposes it seems like I need to consider both the kW and the kVA of a load. And all the discussion on losses (heat & otherwise) helps me grasp the concept as a whole.

To my original question:
"Alternately, if one has a 500 kW/625 kVA genset - will that supply a 500 kW/555 kVA load (with 0.9 pdf)? "

I assume the answer is "YES, a 500 kW/625 kVA genset can supply a 500 kW/555 kVA load" - because the genset exceeds the load's kW and also exceeds the load's kVA?

Thanks again!
 
A little background theory that might make sense:
In the case of a motor, KW or HP is what is measured at the shaft of the motor. Due to the power factor, induction, magnetic field or whatever you like to call it, the electricity to make this power at the shaft is called KVA. This number will always be higher and is directly related to the power factor.

Now lets relate that to a generator.
Lets arbitrarily say that a 500KW (670HP) diesel motor turns the generator. Let's pick a power factor of .8. (500KW /.8= 625KVA) This will now produce 625KVA of electrical power.

Let's now say we have a 500KW (670HP) electric motor that we need to run off this generator that also has a power factor of .8. This motor just happens to need 625KVA of electrical power to run. Conveniently, the generator in the paragraph above this happens to make 625KVA. This 500KW (670HP) electric motors runs just dandy off that generator and produces 500KW (600HP) at the shaft. Isn't mother nature beautiful?
There may be some fluff factors in there, but hoping the general exercise of theory is helpful.
 
Thanks powerpete69 - that helps too. I love how you can do math 10 different ways, but get the same answer.
 
powerpete69 said:
A little background theory that might make sense:
In the case of a motor, KW or HP is what is measured at the shaft of the motor. Due to the power factor, induction, magnetic field or whatever you like to call it, the electricity to make this power at the shaft is called KVA. This number will always be higher and is directly related to the power factor.

Now lets relate that to a generator.
Lets arbitrarily say that a 500KW (670HP) diesel motor turns the generator. Let's pick a power factor of .8. (500KW /.8= 625KVA) This will now produce 625KVA of electrical power.

Let's now say we have a 500KW (670HP) electric motor that we need to run off this generator that also has a power factor of .8. This motor just happens to need 625KVA of electrical power to run. Conveniently, the generator in the paragraph above this happens to make 625KVA. This 500KW (670HP) electric motors runs just dandy off that generator and produces 500KW (600HP) at the shaft. Isn't mother nature beautiful?
There may be some fluff factors in there, but hoping the general exercise of theory is helpful.
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Which follows what I have been trying to say, the generator prime mover still only sees real power (kW). Resistance losses because of increased reactive power are for the most part an inefficiency just like any friction in the generator bearings is, the reactive current itself otherwise is current that only flows between the motor and the source, or some PF correction capacitors if installed. If you have current flow you have some resistance and some loss in the conductors as a result.

Even if generator is supplying 1.0 PF load there is still going to be some inefficiencies that mean the prime mover sees the real load plus the losses from inefficiency.
 
mivey said:
Ever run a 60% pf 10 kW load using a 10 kW 80% pf supply? What happened?
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It sounds like you are saying what happens if you have a load of 16.6 KVA and try to feed it with a supply of 12.5 KVA.
I'm gonna go out on a limb and say the load doesn't receive enough power and doesn't work.
 
powerpete69 said:
It sounds like you are saying what happens if you have a load of 16.6 KVA and try to feed it with a supply of 12.5 KVA.
I'm gonna go out on a limb and say the load doesn't receive enough power and doesn't work.
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Haven't tried it, suspect there is a chance it will run the load, but may have excess heating in the generator, proper overcurrent protection should still open the circuit before it does it for too long.
 
powerpete69 said:
It sounds like you are saying what happens if you have a load of 16.6 KVA and try to feed it with a supply of 12.5 KVA.
I'm gonna go out on a limb and say the load doesn't receive enough power and doesn't work.
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Another way to state this assuming this is a 480V three phase load:
What happens if you have a load of 20 amps and your supply is only 15 amps?
 
powerpete69 said:
Another way to state this assuming this is a 480V three phase load:
What happens if you have a load of 20 amps and your supply is only 15 amps?
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The supply attempts to deliver the demand, if supply impedance is too high, you may suffer some extreme voltage drop. Supply sees heating beyond it's design at very least, marginally sized prime mover slows down, volts and frequency also drop, it still attempts to deliver the demand in many cases.
 
kwired said:
The supply attempts to deliver the demand, if supply impedance is too high, you may suffer some extreme voltage drop. Supply sees heating beyond it's design at very least, marginally sized prime mover slows down, volts and frequency also drop, it still attempts to deliver the demand in many cases.
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The math suggests it will not work, however I certainly never tried it in a lab or real life.
Perhaps a real life simulation will shed light on the absolute truth.
 
powerpete69 said:
The math suggests it will not work, however I certainly never tried it in a lab or real life.
Perhaps a real life simulation will shed light on the absolute truth.
Click to expand...
I see it often with marginally sized supply, in particular POCO transformers - add a heavy load and you get voltage drop problems. A genarator has lesser power available from prime mover and would have even greater effects.

Run any typical household standby generator and compare volt changes when a motor starts up to what it looks like when running same loads on utility power.

Not a thorough investigation, but certainly a good look at the fact the smaller source can't deliver the same performance as the utility source.
 
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