Question

The power consumption of an industry is \(500\;kVA,\;at\;0.8\;p.f.\) lagging. A synchronous motor is added to raise the power factor of the industry to unity. If the power intake of the motor is \(100\;kW\), the \(p.f.\) of the motor is _____________

Answer 1

Given,

Apparent power S = 500 kVA

Power factor cos ϕ = 0.8 

sin ϕ = √(1 - cos² ϕ) = 0.6

Reactive power of motor Q = S sin ϕ = 500 × 0.6

Q = 300 kVAr

Given real power of motor P = 100 kW 

Total power  = √(P² + Q²)

Total power \(= \sqrt{{{100}^2} + {{300}^2}} = 316.22 kVA\) 

⇒ Power factor = \(cos \phi = \frac{P}{S}\)

\(cos \phi= \frac{{100}}{{\sqrt {{{100}^2} + {{300}^2}} }} = 0.316\) leading