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Induction motor improving power factor

Short answer

Motor supplies 302.9 kVAR leading. Its kVA rating is 319 kVA, operating at 0.313 leading power factor.

An induction motor improves the power factor of a load of 500 KW from 0.707 lagging to 0.95 lagging. Simultaneously the motor carries a load of 100 KW. Find: a) The leading KVAR supplied by motor b) KVA rating of the motor and c) Power factor at which the motor operates

Note: an induction motor always draws lagging kVAR, so it cannot improve the power factor. The machine that does this while also driving a mechanical load is an over-excited synchronous motor; the solution below treats the motor as such.

Given: load P1 = 500 kW at 0.707 lag; motor load Pm = 100 kW; final pf = 0.95 lag.

Reactive power of the original load:

tan(cos⁻¹0.707) = 1.0003
Q1 = 500 × 1.0003 = 500.15 kVAR (lagging)

Combined load after adding motor:

P  = 500 + 100 = 600 kW
tan(cos⁻¹0.95)  = 0.3287
Q  = 600 × 0.3287 = 197.21 kVAR (lagging)

a) Leading kVAR supplied by motor

Qm = Q1 − Q = 500.15 − 197.21 = 302.94 kVAR (leading)

b) kVA rating of the motor

Sm = √(Pm² + Qm²) = √(100² + 302.94²)
   = √(10 000 + 91 773) = 319.02 kVA

c) Power factor of the motor

cos φm = Pm / Sm = 100 / 319.02 = 0.313 (leading)

Phasor (power triangle) picture:

         Q (kVAR)
 load  ↑ 500.15 lag
         |
 motor   ↓ 302.94 lead
 -----------------------
 total   197.21 lag  → pf 0.95 at 600 kW

Answer: (a) 302.9 kVAR leading, (b) about 319 kVA, (c) about 0.31 leading.

✓ Answer checked by Claude Opus 5.5 (high effort)

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