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%

4-pole,60-Hz, 300-hp cylindrical synchronous motor ½ rated load torque angle = -16.4

50-kW resistance heater

(a) System Active Power

• Induction Motor

(400 )(0.75)(746 / )0.958

233,611.7

indmot

indmtr

hp W hpP

P W

System Active Power (continued)

• Heater

50,000heaterP W

System Active Power (continued)

• Synchronous Motor

(300 )(0.5)(746 / )0.96

116,562.5

synmot

synmot

hp W hpP

P W

System Active Power (continued)

(233,611.7 50,000 116,562.5)

400,174.2 400.2

system indmot heater synmot

system

system

P P P P

P W

P W kW

(b) Power Factor of the Synchronous Motor

• Determine the angle between VT and Ia

• Calculate Pin to determine Ef

• Calculate Iaa

460265.58

3

3 sin

3 sin

(116,562.5)(0.667)345.615

(3)(265.58)(sin 16.4 )345.615 16.4

T

T f

in

s

in s

f

T

f

f

V V

V EP

XP X

EV

E V

E V

(265.58 0 )(345.615 16.4 )(0.667 90 )

175.41 34.06

( ) (0 34.06 ) 34.06

cos( 34.06 ) 0.828

f T a s

T f

a

s

a

v i

p

E V I jX

V EI

jX

I

F leading

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(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

Pindmot = 233,611.7 W

Qindmtr = Ptanθ

Qindmot = 119,031.1 VARS

S

θ = 27

Power Triangles (continued)

• Synchronous Motor

• Heater

Psynmot = 116,562.5 W

Qsynmot = Ptanθ

Qsynmot = -78,800 VARS

Pheater = 50,000 W

θ = -34.06

Adding Components

Resultant

• System Power Triangle

Psystem = 400.2 kW

Qsystem = 40,231.1 VARS

θ = 5.74

Fp,sys = cos(5.74) = 0.995 lagging

θ = tan-1(40,231.1/400,200) θ = 5.74

(d) Adjust power Factor to Unity

• Power Triangle for the Synchronous Motor

Psynmot = 116,562.5 W

Qsyn mot = (78,800 + 40,231.1) VARSSsynmot = 166,598.98-45.6

additional VARS provided by the synchronous

motor

For one phase

3

1

*

1

*

166,589.98 45.63 355,532.99 45.6

55,532.99 45.6209.1 45.6

265.58 0

T a

a

SS

S V I

I

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Rotor Circuit for one phase

265.58 0 (209.1 45.6 )(0.667 90 )

378.04 14.6

T f a s

f T a s

f

f

V E I jX

E V I jX

E

E

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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