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

Sagot :

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Answer:

[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]

Explanation:

From the question we are told that:

Voltage [tex]V=480/0 \textdegree V[/tex]

Power [tex]P=120kW[/tex]

Initial Power factor [tex]p.f_1=0.77 lagging[/tex]

Final Power factor [tex]p.f_2=0.9 lagging[/tex]

Generally the equation for Reactive Power is mathematically given by

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

Since

[tex]p.f_1=0.77[/tex]

[tex]cos \theta_1 =0.77[/tex]

[tex]\theta_1=cos^{-1}0.77[/tex]

[tex]\theta_1=39.65 \textdegree[/tex]

And

[tex]p.f_2=0.9[/tex]

[tex]cos \theta_2 =0.9[/tex]

[tex]\theta_2=cos^{-1}0.9[/tex]

[tex]\theta_2=25.84 \textdegree[/tex]

Therefore

[tex]Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]

[tex]Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]

[tex]Q=-41.33VAR[/tex]

Therefore

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

[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]

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