Reactive power taken, Q = √3*(W1 - W2)
Q = √3*(375 + 50)
Q = √3*425 kVAR
Related Question
For lagging power factor = (IR*cosφ + Ixsinφ)/Vr
For leading power factor = (IR*cosφ - Ixsinφ)/Vr
Zero voltage regulation occurs on leading power factor only.
For single phase induction motor,
P = VIcosφ
P = 400*30*0.7
P = 8400 W
Q1 = P*tan(φ) = 8400*1.020
Q = 8568.82 VAR
Now power factor change to 0.9
So, φ' = 25.84 degree
Q = 8400*tan(25.84)
Q = 4067.95
Qnet = Q - Q'
Qnet = 8568.82 - 4067.95
Qnet = 4500.8
C = V²/(2ᴨf)
C = (400*400)/(2ᴨ*50)
C = 8.95*10⁻⁵
C = 89.5 μF
The ratio of full load current to short circuit current = 1/5
Xsc = j/(1/5)
External reactance required = j*((1/5) - 0.05))
= j*0.15 pu
Full load current = (30*1000)/(√3*11)
= 1574.6 A
Per unit reactance = j*0.15 = (I*Xr)/V
j*0.15 = (1574.6*Xb)/((11/√3)*1000))
= 0.60 ohm
The methods often employed in practice to improve system stability are:
1. Increasing System Voltage
2. Reduction in Transfer Reactance
3. Using High Speed Circuit Breaker
4. Automatic Reclosing
5. Transient Stability
6. Turbine Fast Valving
7. Application of Braking Resistors
8. Single Pole Switching: