NCEES 523

[kris7o2]

On question 63 do we ignore the reactive power since the real power has even symmetry with the power factor?


Two Generators Have The Following Characteristics:



thus wouldn't the total power be the following

Pt= Real( 293 kVA ang (-cos^-1(.82)) + 1,131 kVA ang (cos^-1 (.9)) )

 

[no_concentrate]

We don't need to calculate angle and if you want to calculate angle and do the calculation then it will be

Gen 1 Cos phi =0 .82 then angle will be 34.90
Gen2 Cos phi = 0.90 then angle will be 25.84

293kVA* cos (34.90) + 1131 kVA* cos (25.84) = 1258.16kW

They have ask to calculate real power output (P) and they have given the apparent power (S). To calculate real power in Kw we have to multiply apparent power(S) with power factor so P = S*PF or P/S = Cos phi

For Gen 1 P1 = 293Kva*.82 = 240.26kW

For Gen 2 P2 = 1131*.90 = 1017.9kW

Total = 240.26+1017.9 = 1258.16kW

I hope this will help!

 

[kris7o2]

Hi, thanks for this. Also, would 34.9 deg be negative?

 

[no_concentrate]

Hi, thanks for this. Also, would 34.9 deg be negative?

Oh Yes it will be -ve for Gen2. Its my fault! but for cos(+ang) and cos(-ang) remains the same.

 

[kris7o2]

Thanks, appreciate it.