Download Power Factor Correction Most domestic loads (such

Power Factor Correction
The process of increasing the power factor without
altering the voltage or current to the original load is
known as power factor correction
Most domestic loads (such as washing machines, air
conditioners, and refrigerator) and industrial loads (such as
induction motors) are inductive and operate at a low lagging
power factor. Although the inductive nature of the load can not
be changed, but the power factor can be increased.
Most loads are inductive, as shown in figure below (a) which has
a power factor of cos  1 , a load’s power factor is improved or
corrected by installing a capacitor in parallel with load, as shown
in Figure below (b) which has a power factor of cos  2.
I
IL
V
IL
Inductive
load
V
Inductive
load
(b)
(a)
Figure Power factor correction: (a) original inductive load,
(b) inductive load with improved power factor
IC
C
The effect of adding the capacitor can be illustrated using power
triangle or the phasor diagram of the current involved.
IC
1
2
V
IC
I
IL
The original load has power factor pf 1  cos  1
and the inductive
load with improved power factor has pf 2  cos 2
The power factor correction can be looked from another perspective.
Consider the power triangle in Figure below.
QC
S1
Q1
S2
Q2
1
2
P
If the original inductive load has apparent power S1, then
P  S1 cos 1
Q1  S1 sin  1  P tan 1
If desired to increase the power factor from pf  cos to pf  cos
without altering the real power (i.e P  S 2 cos 2 ), then the new reactive
power is
1
1
2
2
Q2  P tan  2
The reduction in the reactive power is caused by the shunt
capacitor, that is
QC  Q1  Q2  P(tan  1  tan  2 )
;
QC
2
Vrms
2

  CVrms
XC
The value of the required shunt capacitance of capacitor C is
determined by
C
QC
P(tan  1  tan  2 )

2
2
Vrms
Vrms
Although the most command situation in practice is that of an
inductive load, it is also possible that the load is capacitive, that is, the
load is operating at a leading power factor. In this case, an inductor
should be connected across the load for power factor correction. The
required shunt inductance L can be calculated from
2
2
Vrms
Vrms
QL 

XL
L

2
Vrms
L
 QL
Where QL  Q1  Q2 , the difference between the new and old reactive
powers.
Example 1
When connected to a 120 V (rms), 60 Hz power line, a load absorbs 4 kW
at a lagging power factor of 0.8. Find the value of capacitance necessary
to rise the pf to 0.95.
Solution
Before improving power factor :
If the pf=0.8, than
cos 1  0.8   1  36.87 0
The apparent power from the real power
S1 
P
cos 

1
400
 500 VA
0.8
After improving power factor :
The pf is raised to 0.95
cos 2  0.95   2  18.19 0
The real power has not changed, but the apparent power has changed,
its new value :
S2 
P
4000

 4210.5 VA
cos  2
0.95
The new reactive power is
Q2  S 2 sin  2  4210.5 sin( 18.19 0 )  1314.4 VAR
The reactive power due to installation of capacitor is
QC  Q1  Q2  3000  1314.4  1685.6 VAR
and
C
QC
1685.6

 310.5  F
2
Vrms
2 x60 x120 2
Example 2
The pf correction capacitor in a 240 V, 50 Hz fluorescent light unit has broken
down and needs replacing. A test on the unit shows that, without the capacitor,
the supply current is 0.86 A at pf of 0.5 lagging. The value quoted on the
original capacitor have faded and the only other information is that the working
pf of the unit should be 0.95. Determine the value of the capacitor needed
(look in Figure below).
Choke and lamp
0.86 A
Pf=0.5
240 V
50 Hz
Three phase system
The total average power P :
P  3V I  cos   3VL I L cos 
The total reactive power Q :
Q  3V I  sin   3VL I L sin 
The total apparent power S :
S  P  jQ  3VL I L 
For a Y-connected load :
VL  3V
, but
I L  I
For a - connected load :
I L  3I
V L , V , I L and I 
, but
V L  V
are all rms value and  is the angle between the phase voltage
and the phase current
Example 3
A three-phase motor can be regarded as a balanced Y-load. A threephase motor draws 5.6 kW when the line voltage is 220 V and the line
current is 18.2 A. Determine the power factor of motor.
Solution
The power factor is :
3V L I L cos 
P
pf  cos 


P
3V L I L
5600
3 x 220 x18.2
 0.8075
Example 4
A balanced three phase load draws 50 kW at a lagging power factor of 0.85
from a 480 V supply.
a. Find the apparent power S
b. Find the line current I L
c. Find the reactive power Q
Example 5
A three phase industrial load consists of 136 kW at 0.8 power factor
lagging. What size capacitor bank is required to correct the power factor
to 0.95 lagging?