Generator Rating

[ccamts]

Hi Guys,
I got induction generator ratings from the manufacturer,
A little confusing needs help: It says KW=298.4 *** V=575 v
400 HP

LOAD 100% 75% 50%
EFFICIENCY% 94.00 93.80 92.70
POWER FACTOR % 71.50 66.20 54.60
AMPS 419.10 339.50 274.40
RPM 405.00 403.75 402.50
My questin is what is the actual power factor, should I say if the HP is 400 then KVA =298.4 but that means it also says KW=298.4 it is operating at unity power factor, but what about the stuff at the top that says P.F. 71.50
Please explain.
cheers !

 

[ccamts]

Hi Every body
I forgot to add some thing here
I have Locked rotor current= 1920 A
%FLC 458 referred to full load current 100 % load
code letter E,
I need to calculate the 3 ph fault current and single line to ground fault current, as it is an induction generator, am I supposed to model it like synch generator like a voltage source in series with reactance, and what that reactance will be as induction motor model is completely differnt.
Thanks

 

[rcw retired EE]

1 HP = 0.746 kW, not KVA. 400 Hp x 0.746 = 298.4 kW.

At 71.5% pf that is 417 kVA.

For Short cirucit, modle it as an induction motor, that is what it is. It will deliver 1920 A fault current (Locked Rotor Current)

 

[ccamts]

Thanks
but what does 71.5 % pf means cos(fi)=.715 fi=44.35
so how can 400 Hp= 417 KVA be in this case.
Please advice

 

[electricpete]

rcwilson is right on all aspects.

"so how can 400 Hp= 417 KVA:
It is not strictly true with the equal sign your wrote. The left represents real power P and right represents apparent power S.

P = S * pf
400hp * (0.746 kw/hp) = 417 kva * (0.71 kw/kva)
Both sides evaluate to approximately 278kw






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[ccamts]

I got it now, so if I elaborate that means

if I have to calcualte to rating of capacitor for this induction generator and the utility has asked that it should opeate between .9 lag to .98 leading. So what
capacitor rating I need ?

This is what I think it should happen,
cos(fi)=.715
sin(fi)=.699

so KVAR= S*sin(fi)=417*.699= 291.48
So this should be the rating of capacitor bank and it will still fulfill .9 lag condition ?
Thanks

 

[rcw retired EE]

The induction generator when operating at full load will put out 298 KW at 71.5% power factor leading = 417 KVA. (The generator is consuming Vars and producing watts, so it is leading power factor). It is consuming 291.5 KVARS.

Adding a 300 kVAR bank at the generator terminals will bring the output to unity power factor at full load, negelecting any KW and KVARs consumed in your facility.

At 50% load the generator needs 229 KVAR, based on its .546 power factor [50% x 298 kW x tan(inv cos(.546))= 229 kVAR].
So at 50% load there will be 300 - 229 = 71 kVAr left for the utility. [cos (inv tan( 71 KVAR / 149 kW))= 0.9 pf.

So if you leave the 300 kVar bank connected you will run between 0.9 pf lagging at 50% load to unity at 100% load.

Find out how you are going to keep form going overvoltage when the generator drops below 50% load.

check these figures and make sure I understand what you are trying to do.

 

[rbulsara]

While your thought process is correct for the kVARs, the caps should be sized based on the actual load the generator will see. The caps shall be rated so that at minumum load the pf does not become any more leading than .98 (that is .95 leading is not acceptable) and at the maximum load the pf does not become any worse than .9 lag.

Calculate your min and max kVAR using actual load in kW and .715 pf.

 

[Cooda76]

I found this discussion interesting, so thought should participate,
so the parameters mentioned below
Rotor Resistance (Rr): p.u.
Rotor Reactance (Xr): p.u.
Stator Resistance (Rr): p.u.
Stator Reactance (Xr): p.u.
Magnetizing Reactance (Xm): p.u.
Short Circuit Reactance (Xd”): p.u.

will be provided by the manufacturer, that means only having a locked rotor current will not help. am I correct in assuming that.

 

[electricpete]

will not help what?

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[Cooda76]

I mean if we have FLA and LRC, can we find all the values given above or manufacturer has to provide us with that.
Thanks

 

[rbulsara]

cooda:

What do the your last two posts have to do with your orignal question?