Question

A 3-$\varphi$, 50 Hz, 400 V motor develops 80 kW. The power factor being 0.75 lagging and motor has efficiency of 95%. A bank of capacitors is connected in delta across the supply terminals to improve power factor to 0.95 lagging. Then the KVAR rating of capacitor bank is_______.

• A. 46.58

Answer 1

The correct option is A 46.58
Input to motor $=\frac{Motoroutput}{\eta }=\frac{80kW}{0.95}=84.21kW$

Initial power factor, $cos{\varphi }_1=0.75$ (lagging)

Power factor after improvement,
$cos{\varphi }_2=0.95$ (lagging)

KVAR rating of capacitor bank $=P(tan{\varphi }_1-tan{\varphi }_2)$

$=84.21\left[tan\right(co{s}^{-1}\left(0.75\right)-tan\left(co{s}^{-1}\right(0.95\left)\right)]$

$=46.58$ kVAR