Genset load calc check please

[Bribyk]

I've got some resistive load bank (PF =1?) test results and am trying to estimate the flywheel engine horsepower, assuming a 90% efficiency for this old syncronous generator.

Gen specs: 600 kW, 750 kVA, 3 phase, 600 VAC

Test results: 537/537/538 A, 600.0 VAC

My calcs:
kVA = 1.732*600V*573.3A*1/1000 = 558 kVA
ekW = 558 ekW
eHP = 558*1.341 = 749 eHP
BHP = 749/0.9 = 832 BHP @ flywheel.

Does this sound right? I'm confusing myself on this one mostly due to power factors.

Brian Bobyk - Hoerbiger Canada

 

[Bribyk]

Oops. 573.3A should be posted as 537.3A in the kVA calc.

 

[alehman]

I didn't check the math, but the method looks OK. kW=KVA for pf=1.

Alan
----
"It’s always fun to do the impossible." - Walt Disney

 

[edison123]

In your first equation, KVA should really be KW.

KW = Sqrt(3) x Voltage X Current X PF/1000

KVA = Sqrt (3) x Voltage X Current/1000

Since PF = 1 for pure resistive load, KW = KVA (only in this purely resistive case)

90% efficiency for generator is reasonable.

 

[Bribyk]

You're right, Edison. I mis-copied the formula off my reference sheet. I think the generator's vs. load power factor was confusing me. Thanks for the sanity check.

Brian Bobyk - Hoerbiger Canada