What book is this problem from?
The answer is 667MVA. Either the generator is operating in overload, or It appears that it was written without realizing that 0.8pu of real output power at 0.6 lagging power at an output rating of 500MVA results in an output that is greater than the rating of the generator.
The first step is to calculate the real power output from the given apparent power rating of the machine, using it as base, and the real power output in PU:
Pout = Pout_pu•Sb
Pout = 0.8•500MVA
Pout = 400MW
It's a little confusing because the base is in units of MVA and real power output is in the units of MW, the key here is remembering that base values are magnitudes only and can be multiplied by the real or imaginary component of a complex number like this:
if Sout = Pout +jQout,
and: Sout_pu = Pout_pu + jQout_pu,
then: Sout = Sout_pu•|Sb|,
Sout = (Pout_pu + jQout_pu)•|Sb|
Sout = Pout_pu•|Sb| + jQout_pu•|Sb|
Pout = Pout_pu•|Sb| and Qout = Qout_pu•|Sb|
In this problem, we are given Pout_pu, |Sb|, and Power factor. We are solving for |Sout|.
Now that we know Pout = 400MW, we can find |Sout| using the given power factor. Remember apparent power is always the largest power quantity when power factor is not unity, so to make Pout larger to solve for |Sout|, we need to divide Pout by PF:
|Sout| = Pout/PF
|Sout| = 400MW/0.6
|Sout| = 667MVA
The answer is 667MVA, assuming that the apparent power output of the machine is consumed by the load neglecting any increase in VA due to the reactive power absorbed in the synchronous reactance of the machine itself.
Thanks Zach for your insight. Someone found this problem on chegg.com.