Generator Rating
Generator Rating
(OP)
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 !
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 !






RE: Generator Rating
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
RE: Generator Rating
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)
RE: Generator Rating
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
RE: Generator Rating
"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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RE: Generator Rating
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
RE: Generator Rating
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.
RE: Generator Rating
Calculate your min and max kVAR using actual load in kW and .715 pf.
RE: Generator Rating
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.
RE: Generator Rating
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RE: Generator Rating
Thanks
RE: Generator Rating
What do the your last two posts have to do with your orignal question?