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A synchronous motor absorbing 50 kW is connected in parallel with a factory load of 200 kW having a lagging power factor of 0.8. If the combination has a power factor of 0.9 lagging, find the kVAR supplied by the motor and its power factor. |
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Answer» Load kVA = 200/0.8 = 250 Load kVAR = 250 × 0.6 = 150 (lag) [cos φ = 0.8 sin φ = 0.6] Total combined load = 50 + 200 = 250 kW kVA of combined load = 250/0.9 = 277.8 Combined kVAR = 277.8 × 0.4356 = 121 (inductive) (combined cos φ = 0.9, sin φ = 0.4356) Hence, leading kVAR supplied by synch, motor = 150 – 121 = 29 (capacitive) kVA of motor alone = √(kW2 + kVAR2) = √(502 + 292) = 57.8 p.f. of motor = kW/kVA = 50/57.8 = 0.865 (leading) |
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