1. 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 .

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

3. (a) System Active Power Induction Motor

4. System Active Power (continued) Heater

5. System Active Power (continued) Synchronous Motor

6. System Active Power (continued)

7. (b) Power Factor of the Synchronous Motor Determine the angle between V T and I a Calculate P in to determine E f aCalculate I a

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

11. Look at the Power Triangles Induction Motor P indmot = 233,611.7 W Q indmtr = Ptanθ Q indmot = 119,031.1 VARS S θ = 27 

12. Power Triangles (continued) Synchronous Motor Heater P synmot = 116,562.5 W Q synmot = Ptanθ Q synmot = -78,800 VARS P heater = 50,000 W θ = -34.06 

13. Adding Components

14. Resultant System Power Triangle P system = 400.2 kW Q system = 40,231.1 VARS θ = 5.74  F p,sys = cos(5.74  ) = 0.995 lagging θ = tan -1 (40,231.1/400,200) θ = 5.74 

15. (d) Adjust power Factor to Unity Power Triangle for the Synchronous Motor P synmot = 116,562.5 W Q syn mot = (78,800 + 40,231.1) VARS S synmot = 166,598.98  -45.6  additional VARS provided by the synchronous motor

16. For one phase

17. Rotor Circuit for one phase

18. Rotor Circuit for one phase (cont) Neglecting saturation, –E f  Φ f  I f  use E f –ΔE f = (378.04 – 345.615)/(345.615) x 100% –ΔE f = 9.38%

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