Question

. A 3-phase, 400 V synchronous motor having a power consumption of 50 kW is connected in parallel with an induction motor which takes 200 kW at a power factor of 0·8 lagging. (i) Calculate the current drawn from the mains when the power factor of the synchronous motor is unity.​

Answer 1

Answer:

Explanation:

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)Read more on Sarthaks.com - https://www.sarthaks.com/494697/a-synchronous-motor-absorbing-50-kw-is-connected-in-parallel-with-a-factory-load-of-200-kw

Answer 2

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