Question

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.

Answer 1

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

Step-by-step 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`