# Need some help -- Power Factor Correction for a Motor

• Kasim
In summary, a 12.5 kW motor needs to have an operating efficiency of 85% and a power factor of 0.86 lagging. The power utility requires that the power factor be 0.96 lagging or better, so adding a capacitor to the motor would achieve this. The current in the capacitor leads the supply voltage by 90°, and the capacitance must be equal to 0.96 sin(16.26°), or 80.44 μF.f

#### Kasim

< Mentor Note -- thread moved to HH from the technical engineering forums, so no HH Template is shown >

Hi there. I am wondering if someone can help me.. Below is a question and answer, I need help with IV . Where does the 0.51 and 0.28 come from? I've tried cos and sin and thitha but doesn't give me correct number

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Seriously? You ask people for help by posting a tiny unreadable image of something

Cant you enlarge it by opening the actual picture? Sorry but I am new here..

Cant you enlarge it by opening the actual picture? Sorry but I am new here..
The "enlargement" is tiny

A 12.5 kW motor is to be powered from a 440 V, 50 Hz supply. Data from

factory testing indicates the motor has an operating efficiency of 85% and

that the power factor is 0.86 lagging. The power utility feeding the factory

requires that the motor’s power factor is 0.96 lagging or better.

(i) Explain how the addition of a capacitor to the motor would give the required power factor.

(ii) Determine the motor current in the original configuration.

(iii) Determine the supply current with the capacitor in place.

(iv) Determine the current flowing in the capacitor.

(v) Calculate the value of the capacitance which must be added to give the

desired performance

(vi) Determine the ‘VAr’ rating of the additional component.

(ii) Efficiency of motor = Power output/power input,

ð 0.85 = 12500/power input

ð power input = 14706 W

Power input = VImcosФ = (440)(I)(0.86)

ð I am = 14706/(440)(0.78) =

ð I am = 42.85 A

The motor current I am is 42.85 A

(iii) The current in the parallel capacitor leads the supply voltage by 90°.

The phasor sum of the original current I am and the capacitor current Ic must be such that the supply current Is lags by 0.96 or better.

Resistive component of motor current = I am cosФ = (42.85)(0.86) = 36.85 A

Resistive component of new supply current = Is cosФ’ = Is (0.96) = 0.96 Is A

As resistive component of current does not change => 36.85 = 0.96 Is

=> New supply current, Is = 38.39 A

(iv) Reactive component of motor current = I am sinФ = (42.85)(0.51) = 21.87 A

Reactive component of new supply current = Is sinФ’ = (38.39)(0.28) = 10.75 A

As reactive component of current in capacitor is balance of current => Ic = 21.87 – 10.75

=> Capacitor current, Ic = 11.12 A

(v) Capacitor current, Ic = V/Xc => Xc = V/Ic

ð 1/(2πfC) = V/Ic

ð C = Ic /(2πfV) = 11.12/(2π(50)(440))

ð C = 1.26 x 10-4 = 80.44 μF

(vi) The var rating of the capacitor = VIc = (440)(11.12) = 4.89 kvar

Old PF is cosФ = 0.86 = arccos(0.86) = 30.68° so the sin of 30.68° degrees is sin (30.68°) = 0.51
Since the new PF must be equal to cosФ = 0.96 = arccos(0.96) = 16.26° so the sin of 16.26° is 0.28.

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Thank you