Power factor is the ratio of the real power to apparent power and is either less than or equal to unity. In an AC circuit, current can either lag behind or lead to the voltage depending on the circuit elements and accordingly the power factor is termed as lagging or leading.
Reasons for Low Power Factor
Various causes can be attributed for low power factor:-
- Inductive loads, particularly lightly loaded induction motors and transformers. In these load current can be resolved in two parts: working current Iw in phase with voltage and magnetising current Im perpendicular to voltage. Since Im is constant, lightly loaded induction motor means total current I lags the voltage by more angle thus decreasing the power factor.
- Induction furnaces.
- Arc lamps and arc furnaces with reactors.
- Fault limiting reactors.
- High voltages, as it increases the magnetising current, whreas the working current remains constant.
Table of Contents
Disadvantages of low power factor:
- For same power requirement low power factor means higher current, which implies higher line loss and lower efficiency. For example current drawn at 0.5 PF will be twice the current at unity pf. So for low power factors, the transformers, switchgear, cables etc have to be oversized.
- It adversaly affects generators, transformers and transmission lines as higher than rated current causes higher temperature rise.
- Low power factor gives rise to higher voltage regulation, which is undesirable. As voltage regulation = IR cos ϕ + IX sin ϕ, and X >> R, sin ϕ > cos ϕ at incresed angle ϕ, so voltage regulation increases for low power factor.
- Prime mover of alternator is underloaded. This is because alternator is rated with reference to current, wheras prime mover is rated with reference to its mechanical output.
- High billing cost on account of more kVA for same consumption in kW. Also electricity supply companies levy penalty for low power factor (normally, if lower than 0.85). Also some supply companies gives discount if power factor is beyond certain limits.
Power Factor Improvement
Various methods of improving power factor are:-
- By Phase invertor.
- By Synchronous motor in overexcited mode.
- Fully loaded induction motors and transformers: If the loading is continually low, replace with lower rating machine to improve loading.
Use of high rpm machines, as torque T ∝ Φ I, where Φ is the flux and power P ∝ n T, n is speed in rpm, so for fixed P, T ∝ 1/n,
So 1/n ∝ ΦI, n ∝ 1/(Φ I) = 1/(B A I) = 1/ (μ H A I) = (f × lg ) / (μ N Iμ I),
thus n ∝ 1/ Iμ, where Iμ is magnetising current, lg is air gap length.
So for higher rpm machines, magnetising current reduces and power factor improves.
- By capacitors: The capacitors are connected in parallel with the supply and are made ON/OFF depending on the power factor.
Selection of Capacitors
The ideal place to connect the capacitor bank is the receiving end i.e. at nearer to load point as possible so that all the cables, wires and associated switchgears are releived of the excess current due to poor p.f.. In fact big induction motors should always be provided with a capacitor bank of its own, the capacity of which depends on the magnetising and working current component. A table for the selection of capacitor for motors is shown in Table 2. But generally in a distribution system the capacitor banks are provided on the LT feeder through power factor supervisor relay which makes or brakes certain banks automatically as per the requirements.
The capacity of capacitors depends upon the existing power factor and the desired power factor. Also the selection of the capacitors depends on the cost effectiveness i.e. how much money can be saved and the cost of capacitor bank. Capacitors in a bank can be grouped either in series or in parallel depending on the voltage and current rating and in star or delta for three phase systems.
Examlpe: A factory takes a load of 1000 kVA at 0.7 lag p.f.. if the power factor is improved with the help of capacitor bank to 0.9 lag, determine (i) The value of additional load which can be taken from the mains (ii) The rating of the capacitor banks.
kW at 0.7 p.f = 1000 × 0.7 = 700 kW.
So at 0.9 p.f. kVA = 700/0.9 = 778 kVA
Hence additional 1000 - 779 = 222 kVA can be taken.
KVAR1 = 700 tan ϕ1, KVAR2 = 700 tan ϕ2
So rating of capacitor = KVAR1 – KVAR2 = 700 (tan ϕ1 - tan ϕ2) = 375 kvar.
If we assume constant kVA rating then
Rating of capacitor = 1000 (sin ϕ1 – sin ϕ2) = 278 kvar.
A table for selection of power factor is givel below in Table 1. Here multiplier factors are shown to improve the power factor. To obtain the kvar rating of the capacitor, the load in kW can be multiplied by the multiplying factor.
|Initial Power Factor||Proposed Power Factor|
|Motor HP||Motor Speed (rpm)|
|KVA Rating||Required Capacitor|