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3 phase AC question

  • Thread starter phillies09
  • Start date
  • #1
The system with a balanced 3 phase load of 100kW at a 0.75 power factor, the system voltage is 12500V line to line, grounded wye.

What size of capacitor can bring the power factor to 0.90 lagging, then what will be the real power?
 

Answers and Replies

  • #2
This was my work.
Would anyone please verify my approach is correct?

Initial 3 phase load of Ptotal =520kW, Pf1=0.85, and Vline=31540volts
3*Pf=Ptotal
Pf=520kW/3=173.33kW

For grounded wye,
Vf=Vline/SQRT3=18209volts

Pf1=0.85,
Pf1=Sf1 * (cosf1), where Sf1= Pf1/(cosf1)
Sf1=173.33W/0.85=203.92kVA
Qf1=SQRT(Sf12-Pf12)=107kVAR


Pf2=0.96,
Sf2=Pf2/(cosf2) = 173.33kW/0.96=180.55kVA
Qf2=SQRT(Sf22-Pf22)=50.54kVAR

Xf(reactive) = Vf2/Qf=182092/Qf=331.567E6/ Qf
Xf(before) = 331.567E6/107k=3086Ω
Qfcap= Qf1-Qf1=107k-50.54k=56460VAR
Xfcap= Vfcap2/Qfcap= 331.567E6/56460= 5872 Ω

a) The size of the capacitor per phase C is defined as
C=1/(2*p*f* Xfcap)=1/(376.99*5872)=0.416mF since it is assumed f=60Hz

b) Real power should remain same since we are not changing anything about the resistance part of the impedance so
Ptotal = 3* Pf =520kW
Pf=173.33kW per phase

C) Qtotal = 3*Qf
So Qtotal before = 3*107kVAR=321kVAR and Qtotal after = 3*50.54kVAR=151.62kVAR
 

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