Generator Rated Power Factor and Load Power Factor

Thanks for ur reply.

my connected load is near about 400KW so no worries about 600kw capacity.I found the following written documents from a person of catterpillar:

"The industry standard for generators is for them to be rated for up to a 0.8 power factor. If your loads are likely to exceed this power factor, an oversized generator can solve this problem without having to oversize the entire generator set."----------dont get his idea coz when my PF is 1 :

KW= 600 KW

KVA= 600/PF =600/1 = 600 KVA;WHERE my generator rating is 750kva(still not reach the limit);why I should make the generator size higher not engine size?Any issue is there related to generator exciting current?

need more detail please

On the basis of this additional detail, it is likely that the selected generator will be adequate. If it is a package as suspected, then the manufacturer's on-board controls will sort out all the other issues, and the warranty provisions on the equipment will provide technical and commercial safeguards. Ensure that the vendor's commissioning personnel attend site and leave the equipment in a warranted condition before they leave.

Dear Mr.mzkrazib,

Even though you can maintain a P.F near Unity i.e P.F.1, the Generator is to be designed for a P.F of 0.8, to take care of the Heat Generated on account of Inductive Load, which will lead to RISE IN WINDING TEMPERATURE, As You know the Inductive Power results in HEAT GENERATION only.

Though you may plan to run at Unity P.F., at times the P.F will be lower and at that time the Line current ( for the same Power) will be more, and Rise in Temp. for the windings of Alternator CANNOT BE AVOIDED.

Then the kick load comes in to picture - where the H.P of the motor, Type of the Starter etc. comes in to picture.

Further, in your post no.2 you have stated "Any issue is there related to Generator Exciting Current?"

The Answer is "Yes." The P.F. will have an IMPACT on the EXCITATION CURRENT. Higher the P.F, Lesser the Excitation Current and Viseversa.

Hence you should take care while arriving at the KVA required, based on the above 2 factors.

Thanks,

DHAYANANDHAN.S

Thanks for ur reply.Please consider the following situation:

Genset=generator+engine

1.Genset output @ Generator rated PF=0.8 and Load power factor=0.8

KVA=750;KW=600;Amp= 900

2.Genset output @ Generator rated PF=0.8 and Load power factor=0.9

KVA=600;KW=600(cannot get 675kw as its beyond engine HP);Amp= 720

3.Genset output @ Generator rated PF=0.8 and Load power factor=1

KVA=600;KW=600;Amp= 720 *NO CHANGE!! then wht is the difference bet load pf=0.9 and load PF=1;Will it l cause any operational problem if I keep load power factor more than engine rated?Will it overheat the generator while generator still nt reach at 100%KVA of rated KVA.

On the other hand regarding KVA overload,any load power factor below than rated 0.8 will result the 750kva at lower KW(example: @pf0.75 KW=562 and KVA=750)

In addition to #1, #4 & #5 and the fact that you are loading only 400kW @ 1 pf:

for eaxample, If, for any reason, the load pf drops to 0.65 instead of remaining 1, while the kW load remains at 400 kW, the kVA will be 615. And for pf = 0.53, the kVA will become ~750, the maximum allowed.

Therefore, you must make sure that, if you reach a load near the 600 kW, with pf=1, you do not allow the pf to drop to a level where the equivalent kVA will suddenly exceed the rated 750 kVA for longer than the time stated in the specs. This will overload the prime mover (Diesel Engine or else).

That is why, thinking that you were loading the genset to the 600kW @pf=1, they recommended you to get a bigger size engine. With ~400 kW, you will be OK as long as pf is < 0.53, and you are advised to be near or slightly better than 0.85 at worst if possible. If without the PF correction, you can maintain > 0.7, then fine for 400 kW.

"Therefore, you must make sure that, if you reach a load near the 600 kW, with pf=1, you do not allow the pf to drop to a level where the equivalent kVA will suddenly exceed the rated 750 kVA for longer than the time stated in the specs. This will overload the prime mover (Diesel Engine or else)."

At a load of 600KW, if the power factor drops from unity to a level where KVA exceeds value of 750, it will the stator and the rotor windings of the generator that will get over loaded, not the prime mover. The prime mover supplies only the active power (KW power) which does not change if power factor changes.

Correct! Slipped. I should have said that the Alternator will be overloaded with a higher amps running through it.

But I would point out that the KVA is the total apparent power and that the reactive power is absorbed by the load (hence the p.f. effect). therefore it has to be supplied by the prime mover. It is not going to do any active power at the load end, but it is still generated at the prime mover. If the resultant kVA is higher than the 750 specified, the current will be higher in the alternator for sure, but in an inefficient way since only a portion of it will be for active power. The inefficient portion (in a sense) is producing the reative power necessary to the load as demanded by it.

Thus the prime mover will supply the KVA value as mechanical power while the alternator will generate the kW value as active and the kVAr as a reactive, non mechanical.

Sorry, prime movers only supply kW, they know nothing about kVArs. It's the excitation system and the iron in the alternator that produce or absorb kVArs. Still not convinced, check out the Synchronous Condenser, a device that draws only enough kW to overcome its windage and bearing losses (both mechanical) but can provide or absorb large quantities of reactive power. It doesn't even have a prime mover!

I understand ur logic.But the point is they didnt recommend higher size engine but only higher size alternator.My confusion is on this point.

1) Rated Amps = 750,000/(480*√3) = 902 A~

2) loaded to 600kW the amps = 600,000/(480*√3 *0.8) = 902 A~

3) Loaded to 600 kW at P.F. = 1, Amps = 600,000/(480*√3 *1) = 722 A~ equivalent to 600 kVA

4) You can load the alternator to 750 kW at P.F.=1, which makes 750 kVA, and the amps will stil be 902 A~. The engine will give that power output. But the norms are specified at P.F. 0.8 so that a fair comparison is done between all the offered units on the market.

The Prime Mover has nothing to do with the P.F. It provides the whole power required by the Alternator. Depending on the load circuit configuration, you will have Active power converted to mechanical power or else, and the reactive power necessary due to the load configuration to enable the conversion.

If the Reactive power creates more heat in the winding of the alternator while, at the same time the active power is being supplied, surely you can see that the extra energy has to come from somewhere! It comes from the prime mover, as all the energy generated on the Alternator side. Therefore, the kVA rating is not to be exceeded for the sake of the prime mover, if that unit is closely matched to the alternator. Since the manufacturers can use the same engine for different alternator sizes (with some tweeking...), they can recommend to oversize the alternator in some cases, avoiding the overheating problem due to the extra kVArs ...

If you are arguing that the additional heat load caused by the additional I squared R losses from the reactive current place an additional kW load on the prime mover, then technically you are correct. But in real life this should be in the 1-3% range and not enough to overload the prime mover.

I did not say that the prime mover produces kVArs. It is true that it only produces kW.

When you are drawing 600 kW at 0.8 p.f., on a gen set rated 750 kVA, the kVAr is = 450. To be able to produce the 450 kVAr, the alternator will need some power from the prime mover, and that will be 150 kW. Therefore, the prime mover will be running as if producing 750 kW, out of which, 600 is active kW and the balance is transformed into 450 kVAr (90 deg out of phase).

A Synchronous converter will consume power otherwise it will not rotate and produce reactive power for the grid. The fact is that it does not consume the value of kVAr it produces, but the required kW to operate as required.

The details of the theory are more elaborate that what is in the wikipedia article.

I don't wish to argue this in a public forum but you are still claiming that the reactive power in your scenario "requires" 150kW (which can only come from the prime mover) to produce, it doesn't. KWs come from fuel, kVArs come from the excitation system.

Next time you are in front of a control panel ask the operator to increase the kVArs while holding the kWs constant, the voltage will rise but the fuel consumption will not, all other things (frequency, load, etc.) being equal.

I, also, do not wish to argue this issue here and I still claim that you need, in the scenario I presented (the OP), the 150 kW from the prime mover.

Your proposition to check it on the panel of a generating set, by increasing the kVArs (outputed by the set) does not stand unless you have 2 sets running in sync (paralleled). Increasing the kVARs by manipulating the excitation field, on an island unit (Single set-grid) is not a simple forward action.

Please remember that the kVAR demand depends on the load configuration. You cannot just send kVARs down acircuit not requiring them (same as kW: you need to increase/decrease you load).

Now, if you have 2 sets in parallel, increasing the kVARs of one set is possible because it will reduce the kVArs produced by the other set. In this case, an increase of 10 - 11% of KVArs produced, will increase the total kVA by ~ 4% only. This is already in the region of your previous 3%~ statement. I think it will be a difficult measurement to make in the normal environment (not impossible but delicate process), and probably will reflect into an 8% increase in fuel consumption.

In the same 2 sets in //: You normally increase the Voltage of the intended set (increase excitation) to send more of the kVARs down the load. This will drop your p.f. for this set. The other set's p.f. will improve. The whole mechanism of this operation is quite complicated. Instead of sending me to experiment the result, try to work it theoretically, starting from the point of view of the prime mover.

I don't think that I will go down that road again since it is more than 40 years ago...

I should have thought of this earlier:

"..Usually, the exciter's rating varies from 2.0 to 3.5 kW/MVA of a generator's rating...", from Kundur, P. Power System Stability and Control. - McGraw-Hill, 1994.

Since this is for a DC exciter feeding the field through brushes, it is a good proxy for the amount of power consumed by the field of any alternator. Knowing nothing about the internals of a 750 kVA machine, we can use that ratio to estimate that the field requires approximately 2.5 kW of power to reach its rated kVArs, a far cry from 150 kW.

No hard feelings intended, but I am being pedantic about this because this because I consider CR4 to be a learning site, and there tends to be much confusion on the web about VARs, where they come from, how they're controlled, and their function in AC networks. I hope this helps to keep the record straight.