Statement: An electrical load withdraws 60 kW at PF of 0.8. It is desired to improve the PF from 0.8 to 0.9. Find the reactive power which should be provided.

Solution: Here Cos 𝜽 = 0.8, or 𝜽 = 36.86,

From Power Factor triangle:

At 0.8 the reactive power is: Q = tan (36.86) * 60 kW = 44.98 kvar

At 0.8 the reactive power is: Q = tan (36.86) * 60 kW = 44.98 kvar

At PF of 0.9, Cos 𝜽 = 0.9, or 𝜽 = 25.84

& Q = tan (25.84) * 60 kW = 29.056 kvar

The reactive power which is supplied by capacitor = 44.98 - 29.056 = 15.9

## No comments:

## Post a Comment