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
Bus Type - Known Parameter - Unknown Parameter Load Bus -P, Q - V, phase angle Generator Bus - P, V (magnitude) - Q, Voltage phase angle Slack Bus Voltage - magnitude and phase angle - P, Q
Current flow through the circuit, I = V/Z = 400/10 = 41.5 A Power factor = 8/10 = 0.8 After connecting a capacitor bank, the reactive power component of current changes but the active power component of current is unchanged. I1*cosφ1 = I2*cosφ2 41.5*0.8 = I2*0.9 I2 = 36.88 A Q1 (before connecting capacitor) = √3*VL*IL*sinφ1 = √3*415*41.5*sin(36.86) = 26.847 kVAR Q2 (after connecting capacitor) = √3*VL*IL*sinφ2 = √3*415*36.88*sin(25.84) = 15.867 kVAR kVAR supplied by capacitor bank = Q1 - Q2 = 26.847 - 15.867 = 10.98 kVAR
Let, E→EMF in volt φ→Flux in weber A→Parallel path (for wave winding = 2) Z→No. of armature conductors N→speed in rpm P = Number of poles E = PφNZ/(60A) E = 8*0.020*625/(60*2) E = 500 V
EXAMIANS