Power Factor Improvement

Power Factor Improvement

The low power factor is mainly because of a.c. motors which are of induction type having low lagging power factor. Similarly lighting load, arc lamps, electric discharge lamps, industrial induction furnaces etc. also contribute to low power factor operation. To improve the power factor, it is necessary to connect some device which takes leading power to neutralize lagging effect. One of such devices is a capacitor which draws a leading current. Let us see the action of capacitor in a power factor improvement.

Consider a lagging power factor load as shown in the Fig. 7.20.1 (a). The corresponding phasor diagram is shown in the Fig. 7.20.1 (b).

Let I₁ be the current drawn at a lagging power factor angle of ϕ₁

Now to improve the power factor, a capacitor is connected across the load as shown in the Fig. 7.20.2.

Now the capacitor takes a leading current I₂, which leads voltage V by an angle of 90° as shown in the Fig. 7.20.3 (a). Now this leading component of current 12 tries to neutralise the lagging effect of I p Hence the resultant current becomes in the Fig. 7.20.3 (b). 

Now from the Fig. 7.20.3 (b), it can be seen that the efective power factor angle becomes ϕ which is less than ϕ₁ Hence cos ϕ is more than cos ϕ₁ thus there is improvement in the power factor of the system. Thus by connecting a leading power factor device, across the supply power factor can be improved which is advantageous from the supplier as well as consumer point of view.

It can be seen from above phasor diagram that the resultant current I drawn by circuit from supply after connecting capacitor in parallel is less than that of circuit current I₁ without capacitor in circuit.

The active or wattful component remains same after and before power factor improvement because the capacitor only reduces the reactive component. Active power also remains unchanged

I₁cos ϕ = I cos ϕ

VI₁cos ϕ = VI cos ϕ

The reactive component which is lagging reduces after power factor improvement. Its value equal to difference between reactive component of load and capacitor current I₂

I sin ϕ = I₁ sin ϕ - I₂

Multiplying both sides of above equation by V

VI sin ϕ = VI₁ sin ϕ - VI₂

The capacitor current leads the supply voltage by 90°

ϕ_c = 90°

sin ϕ_c = sin 90° = 1 

Leading kVAR supplied by capacitor = VI₂ sin ϕ_c = VI₂ sin 90° = VI₂

, kVAR after power factor improvement = Lagging kVAR before power factor improvement - Leading kVAR supplied by capacitor

1. Importance of Power Factor Improvement

The improvement in power factor is important for both consumers and generating stations.

The industrial and other large consumers pay electricity charges based on maximum demand in kVA in addition to units actually consumed. With improvement in power factor there is corresponding decrease in maximum demand as

Maximum demand (in kVA) =          Peak load ( kW ) / cos ϕ

Due to reduction in maximum demand, the demand charges can be saved by the consumers annually. Thus there is net saving for the consumer if the p.f. is maintained at proper value even though p.f. improvement device requires an extra cost.

Additionally consumers can get incentive or discount in their electricity bills if they maintain power factor closer to unity.

The power factor consideration is important for generating stations also. The alternators in power stations are having rating in kVA while their useful output depends upon kW output supplied.

Now, output in kW = kVA × cos ϕ

The more the p.f., the output of station is more and higher are the units (kWh) supplied by that station to the system. This helps in improving the revenue obtained from that station. Thus improved p.f. enhances earning capacity of generating station.

Example 7.20.1 What should be the kVA rating of a capacitor which would raise the power factor of load of 100 kW from 0.5 lagging to 0.9 lagging

Solution : p.f. before improvement = cos ϕ = 0.5

The capacitor is assumed to operate at p.f. angle of 90° means the phase angle between voltage and current is 90° then sin 90° = 1( ϕ_c = p.f. angle of capacitor = 90°)

We have, kVAR = kVA sin ϕ_c

kVA = kVAR / sin ϕ_c  = 125 / 1 = 125

kVA rating of capacitor = 125

Example 7.20.2 A 3 phase, 50 Hz, 3500 V star connected induction motor develops 250 hp (186.5 kW), the power factor being 0.707 lagging and the efficiency 0.86. Three capacitors in delta are connected across the supply terminals and power factor raised to 0.9 lagging. Calculate i) The kVAR rating of the capacitor bank ii) The capacitance of each unit

Solution :

Example 7.20.3 A 3 phase, 50 Hz, 3000 V motor develops 447.6 kW, the p.f. being 0.75 lagging and the efficiency 0.93. A bank of capacitors is connected in delta across the supply terminals and p.f. raised to 0.95 lagging. Each of the capacitance units is built of five similar 600 V capacitors. Determine the capacitance of each capacitor.

Solution :

Review Questions

1. Explain the concept of power factor improvement.

2. Comment of the imporance of power factor improvement.