-
A 3 – φ induction motor takes 120 kW at 415 V, 50 Hz and a power factor of 0.8 lagging. What is the capacitance per phase, if its power factor is improve to 0.9 and capacitor bank is connected in star ?
-
- 587 µF
- 440 µF
- 560 µF
- 600 µF
Correct Option: A
Power per phase,
| PP = | = 40 kW | |
| 3 |
cos φ1 = 0.8
⇒ φ1 = 36.86°
∴ tan φ1 = 0.749
Final power factor,
cos φ2 = 0.9
⇒ φ2 = 25.84°
∴ tan φ2 = 0.4843
Now QCP = PP (tan φ1 – tan φ2)
= 40 (0.749 – 0.4843)
= 10.588 KVAr/phase.
| VP = | = | = 239.6 volts | ||
| √3 | √3 |
| IC = | = | = 44.19 Ampere | ||
| VP | 239.6 |
Capacitance per phase,
| CY = | = | ||
| ω . VP | 2πƒ . VP |
| = | = 587.0 × 10-6 F | |
| 2π × 50 × 239.6 |
= 587 µF
