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.

The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction

LI

V

LI

VInductiveload

Inductiveload C

CI

I

)(a )(bFigure Power factor correction: (a) original inductive load, (b) inductive load with improved power factor

Most loads are inductive, as shown in figure below (a) which has a power factor of cos , 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 .

1

2

The effect of adding the capacitor can be illustrated using power triangle or the phasor diagram of the current involved.

CI

1 2

LI

CI

V

I

The original load has power factor and the inductive load with improved power factor has

11 cospf

22 cospf

The power factor correction can be looked from another perspective. Consider the power triangle in Figure below.

1S

2S

1 2

1Q

2Q

CQ

P

If the original inductive load has apparent power S1, then

11 cosSP

1111 tansin PSQ

If desired to increase the power factor from to without altering the real power (i.e ), then the new reactive power is

11 cospf 22 cospf

22 cosSP

22 tanPQ

The reduction in the reactive power is caused by the shunt capacitor, that is

)tan(tan 2121 PQQQC

The value of the required shunt capacitance of capacitor C is determined by

; 22

rmsC

rmsC CV

X

VQ

221

2rmsrms

C

V

P

V

QC

)tan(tan

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

L

rmsrms

L

rmsL Q

VL

L

V

X

VQ

222

Where , the difference between the new and old reactive powers.

21 QQQL

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

011 873680 ..cos

The apparent power from the real power

VAP

S 50080

400

11

.cos

After improving power factor :

The pf is raised to 0.95

022 1918950 ..cos

The real power has not changed, but the apparent power has changed, its new value :

VAP

S 54210950

4000

22 .

.cos

The new reactive power is

VARSQ 41314191854210 0222 .).sin(.sin

The reactive power due to installation of capacitor is

VARQQQC 6168541314300021 ..

and

FxxV

QC

rms

C

5310120602

6168522

..

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).

Pf=0.5

240 V50 Hz

Choke and lamp

0.86 A

Three phase system

The total average power P :

coscos LL IVIVP 33

The total reactive power Q :

sinsin LL IVIVQ 33

The total apparent power S :

LL IVjQPS 3

For a Y-connected load :

VVL 3 II L , but

For a - connected load :

II L 3 VVL , but

IandIVV LL ,, 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 three-phase 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 :

807502182203

5600

3

3

..

cos

cos

xx

IV

P

pf

IVP

LL

LL

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

c. Find the reactive power Q

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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?