Power Factor Correction Example 8.3 page 324 of the text by Hubert A three-phase (y-connected) 60-Hz, 460-V system supplies the following loads: 6-pole, 60-Hz, 400-hp induction motor at ¾ load with an efficiency of 95.8% and power factor of 89.1% 50-kW delta-connected resistance heater 300-hp, 60-Hz, 4-pole, synchronous motor at ½ load with a torque angle of -16.4.

Power Factor Correction 6-pole, 60-Hz, 400-hp induction motor ¾ rated load efficiency = 95.8% power factor = 89.1% 50-kW resistance heater 4-pole,60-Hz, 300-hp cylindrical synchronous motor ½ rated load torque angle = -16.4

(a) System Active Power Induction Motor

System Active Power (continued) Heater

System Active Power (continued) Synchronous Motor

System Active Power (continued)

(b) Power Factor of the Synchronous Motor Determine the angle between VT and Ia Calculate Pin to determine Ef Calculate Ia

(c) System Power Factor Look at each load Induction Motor Fp = 0.891 θ = cos-1(0.891) = 27 Heater θ = 0 Synchronous Motor θ = -34.06

Look at the Power Triangles Induction Motor Qindmtr = Ptanθ Qindmot = 119,031.1 VARS S θ = 27 Pindmot = 233,611.7 W

Power Triangles (continued) Synchronous Motor Heater Psynmot = 116,562.5 W θ = -34.06 Qsynmot = Ptanθ Qsynmot = -78,800 VARS Pheater = 50,000 W

Adding Components

Fp,sys = cos(5.74) = 0.995 lagging Resultant System Power Triangle θ = tan-1(40,231.1/400,200) θ = 5.74 Qsystem = 40,231.1 VARS θ = 5.74 Psystem = 400.2 kW Fp,sys = cos(5.74) = 0.995 lagging

(d) Adjust power Factor to Unity Power Triangle for the Synchronous Motor Psynmot = 116,562.5 W Qsyn mot = (78,800 + 40,231.1) VARS Ssynmot = 166,598.98-45.6 additional VARS provided by the synchronous motor

For one phase

Rotor Circuit for one phase

Rotor Circuit for one phase (cont) Neglecting saturation, Ef  Φf  If  use Ef ΔEf = (378.04 – 345.615)/(345.615) x 100% ΔEf = 9.38%

(e) The power angle for unity power factor δ = -14.96