Answer to Question #193163 in Mechanical Engineering for Joel

Question

A station supplies 250KVA at a lagging power factor of 0.8. A synchronious motor is connected in parallel with the load. If the combined load is 250KW with a lagging P.F of 0.9, determine: 

1. The leading KVAR taken by the motor

2. The KVA rating of the motor

3. P.F at which the motor operates

Answer 1

Load = - 250 kVA,

0.8 lagging power $\phi$ = 36.86'

P_L = (250 x 0.8) kW = 200 kW

Q_L = (250 x sin 36.86) = 150 kVAR Combined Load = - 250 kW,

0.9 lagging $\phi$ = 25.84;

Q_Total = P x tan $\phi$ = (250 x tan 25.84 ) kVAR = 121 kVAR

Q_Total = (Q_Load + Q_motor)

121 = 150 + (-Q_motor)

Q_motor = 29 kVAR

P_Total = (P_Load + P_motor)

250 = 200 + P_motor

P_motor = 50 kW

s = (P² + Q²)^1/2 kVA = (50² x 29²)^1/2

s = 57.80 kVA

cos $\phi$_m = P/s = 50/57.80

cos $\phi$_m = 0.865