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%
50-kW resistance heater
4-pole,60-Hz, 300-hp cylindrical synchronous motor ½ rated load torque angle = -16.4

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 VT and Ia
Calculate Pin to determine Ef
Calculate Ia

10. (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 = 

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

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

13. Adding Components

14. 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 = kW
Fp,sys = cos(5.74) = lagging

15. (d) Adjust power Factor to Unity
Power Triangle for the Synchronous Motor
Psynmot = 116,562.5 W
Qsyn mot = (78, ,231.1) VARS
Ssynmot = 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,
Ef  Φf  If  use Ef
ΔEf = ( – )/( ) x 100%
ΔEf = 9.38%

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