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yulyashka [42]
2 years ago
7

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

g power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.
Engineering
1 answer:
Leni [432]2 years ago
6 0

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

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