In the Journeyman's class I'm taking we had the following test question:
A plant has an existing load of 4500KVA with a 0.6 lagging PF. Use an 800HP synchronous motor with 80% efficiency to correct the power factor to 0.9. Calculate the PF for the motor.
so given:
total load 4500 KVA
current 0.6 PF
motor 800 HP
efficiency 80%
Calculations:
Plant KW = KVA * PF = 4500 * .6 = 2700KW
Plant KVAR = KVA - KW = 4500 - 2700 = 1800KVAR
Ugly's .6 to .9 PF = .848
needed KVAR = KW * .848 = 2700 * .848 = 2289KVAR
HP to KW conversion
HP = (V*A*Eff*PF*1.73) / 746
HP*746 / (Eff*PF*1.73) = VA
800*746 / (.8*1.0*1.73) = 432.21KW (1.0 PF)
PF=KW/KVA
PF=432 / 2289.6 = .19PF
But Ideal Electric's web site suggests a Syncro can produce 1.5*KW KVARs which means we are expecting 2289 - (432 * 1.5) = 1642 more KVARs than we can get from the Synchro.
Does this make any sense? Do I have errors in my calc or application of formulas.
If we provide the same amount of KVARs leading as KVARs lagging then don't we correct the PF to unity? Here we have 1800KVARs lagging and supply 2289.6KVARs leading to correct to .9 PF?
The instructor only has a numeric answer but no description of the calculation.
Thanks
