Transcript Power Factor Correction

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
P
indmot
P
indmtr
(400hp)(0.75)(746W / hp)

0.958
 233, 611.7W
System Active Power (continued)
• Heater
P
heater
 50, 000W
System Active Power (continued)
• Synchronous Motor
P
synmot
P
synmot
(300hp )(0.5)(746W / hp)

0.96
 116, 562.5W
System Active Power (continued)
P
P
P
 (233, 611.7  50, 000  116, 562.5)W
P
 400,174.2W  400.2kW
system
system
system
indmot
P
heater
P
synmot
(b) Power Factor of the Synchronous Motor
• Determine the angle between VT and Ia
• Calculate Pin to determine Ef
• Calculate Ia
V
T
P
460

 265.58V
3
V E
3
sin 
X
PX

3V sin 
(116, 562.5)(0.667)

 345.615V
(3)(265.58)(sin  16.4)
 345.615  16.4V
T
in
f
s
E
in
f
T
E
f
E
f
s
E  V  I jX
f
T
a
s
V E
(265.58 0)(345.615  16.4)
I 

jX
(0.667 90)
T
f
a
s
I  175.41 34.06
a
  (   )  (0  34.06)  34.06
v
i
F  cos( 34.06)  0.828leading
p
(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
S
Qindmtr = Ptanθ
θ = 27
Qindmot = 119,031.1 VARS
Pindmot = 233,611.7 W
Power Triangles (continued)
• Synchronous Motor
Psynmot = 116,562.5 W
θ = -34.06
Qsynmot = Ptanθ
Qsynmot = -78,800 VARS
• Heater
Pheater = 50,000 W
Adding Components
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
S
166, 589.98  45.6
S 

3
3
S  55, 532.99  45.6  V I
3
1
*
1
T
a
55, 532.99  45.6
I 
 209.1  45.6
265.58 0
*
a
Rotor Circuit for one phase
V  E  I jX
s
E  V  I jX
s
T
f
f
T
a
a
E  265.58 0  (209.1 45.6)(0.667 90)
f
E  378.04
f
 14.6
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