Calculating Power Factor for 3-Phase AC System of 100kW @ 12500V

[phillies09]

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?

 

[phillies09]

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

 

[Mene]

I would first like to clarify that power factor is a measure of the efficiency of an electrical system and is defined as the ratio of real power (measured in watts) to apparent power (measured in volt-amperes). A power factor of 1 indicates a perfectly efficient system, while a power factor less than 1 indicates a less efficient system.

Given the information provided, we can calculate the apparent power of the 3-phase AC system using the equation: Apparent Power = (Voltage * Current * √3), where voltage is the line-to-line voltage of 12500V and current is the total current drawn by the load which can be calculated as (100kW/0.75)/(√3) = 115.47 amps.

Therefore, the apparent power of the system is (12500V * 115.47A * √3) = 199.89 kVA.

To bring the power factor to 0.90 lagging, we need to introduce a capacitive load in the system. The size of the capacitor needed can be calculated using the equation: Capacitance = (Apparent Power * tan θ)/(2π * Frequency * Voltage^2), where θ is the angle between the voltage and current in the system.

Assuming a frequency of 60 Hz, the required capacitance would be (199.89kVA * tan 25.84°)/(2π * 60Hz * 12500V^2) = 64.82 microfarads.

Finally, to determine the real power in the system with a power factor of 0.90, we can use the equation: Real Power = Apparent Power * power factor = 199.89kVA * 0.90 = 179.90 kW.

In conclusion, in order to bring the power factor of the 3-phase AC system to 0.90 lagging, a capacitor with a capacitance of 64.82 microfarads would be needed. This would result in a real power of 179.90 kW in the system.