Capacitance power factor question

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Discussion Overview

The discussion revolves around calculating the capacitance required to improve the power factor of a 50 kW load from 60% lagging to 90% lagging. The context includes theoretical calculations and the application of formulas related to power factor correction in electrical engineering.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a detailed calculation for determining the required capacitance, including the calculation of reactive power for both the initial and target power factors.
  • Another participant expresses uncertainty about the correctness of the initial calculations and requests verification.
  • A later reply confirms that the initial calculations are correct, but another participant indicates they obtained a different value, suggesting a potential discrepancy.
  • Discussion includes clarification on the use of signed reactance in calculations, with one participant explaining the significance of the negative sign in capacitive reactance.
  • An alternative method using complex impedance is mentioned as a possible approach to retain the sign throughout the calculations.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the calculations, as one participant reports a different value. Some participants agree with the initial calculations, while others express uncertainty or confusion.

Contextual Notes

Participants discuss the implications of using signed versus unsigned reactance in their calculations, indicating a potential area of misunderstanding regarding the formulas used.

oxon88
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Homework Statement



A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

The Attempt at a Solution


for 60% p.f:-

θ1 = Cos-1(0.6) = 53.13 degrees

Q1 = 50,000*tan(53.13) = 66.667 kVAR

for 90% p.f:-

θ2 = Cos-1(0.9) = 25.84 degrees

Q2 = 50,000*tan(25.84) = 24.22 kVAR
Qcapacitor = Q2 - Q1 = -42.45 kVAR
Xc = VRMS /Qc = (104)2 / -42450 = -2356 Ωω = 2.∏.f = 2.∏.60 = 377

C = 1 / ω.Xc = 1 / 377*2356 = 1.126 μF
can anyone check if I'm on the right path here please?
 
Last edited:
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anyone?
 
I posted a response to this about 6 hours ago, saying that's the answer I got.

Not sure where my reply went. https://www.physicsforums.com/images/icons/icon9.gif
 
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Ok that's good. thank you
 
Was this answer correct as i am getting a different value..?
 
yes it is correct.
 
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Last edited:
oxon88 said:

Homework Statement



A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

The Attempt at a Solution


for 60% p.f:-

θ1 = Cos-1(0.6) = 53.13 degrees

Q1 = 50,000*tan(53.13) = 66.667 kVAR

for 90% p.f:-

θ2 = Cos-1(0.9) = 25.84 degrees

Q2 = 50,000*tan(25.84) = 24.22 kVAR
Qcapacitor = Q2 - Q1 = -42.45 kVAR
Xc = VRMS /Qc = (104)2 / -42450 = -2356 Ωω = 2.∏.f = 2.∏.60 = 377

C = 1 / ω.Xc = 1 / 377*2356 = 1.126 μF
can anyone check if I'm on the right path here please?
why isn't it - 2356 in the final formula ?
 
grinder76 said:
why isn't it - 2356 in the final formula ?
oxon88 calculated the signed reactance of the capacitor to be -2356 Ω. The minus sign reflects the fact that it is capacitive reactance. He then used the formula for the magnitude of the reactance, which is unsigned, to find the value of the capacitor.

An alternative approach might have been to use complex impedance which retains the sign throughout, but that would achieve the same end result.
 
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Thanks for the reply i now understand
 

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