We often hear the term 'Power Factor' in electrical power systems. But, how do we define 'Power Factor' in an electrical world?
In this article, we will be discussing power factor in detail including its significance, adversity,
calculations and a few methods for improving power.
Power Factor can be defined as the ratio between Real Power (Watts) and Apparent Power (VA). In simplerwords, it tells how effectively your device utilizes electricity. We already know that the apparent power is the combination of Real power (kW) and Reactive power (kVAR).
Formula for calculation of PF:
PF = REAL POWER / APPARENT POWER
PF = REAL POWER / (REAL POWER+REACTIVE POWER)
PF = KW / KVA OR KW / (KW+KVAR)
A simple diagram can be illustrated to understand Power Factor considering the following analogy. The combination of drinkable beer(kW) and foam (kVAR) inside the mug represents the Apparent Power (kVA). That said, Power Factor is simply the ratio of Real Power (kW) to Apparent Power (kVA) and is represented by the following formula: PF = kW / kVA. Using our beer analogy, you can write the formula like this:
PF = Beer / Drinkable Beer + Foam
As Apparent Power is made up of two parts, the resistive power (the in-phase power in watts) and the reactive power (the out-of-phase power in volt-amperes), we can show the vector addition of these two power components in the form of a power triangle.
A power triangle has four parts: P, Q, S and θ.
The three elements which make up power in an AC circuit can be represented graphically by the three sides of a right-angled triangle as the previous impedance triangle. The horizontal (adjacent) side represents the circuits real power (P), the vertical (opposite) side represents the circuits reactive power (Q) and the hypotenuse represents the resulting apparent power (S) of the power triangle as shown below.
Power Triangle of an AC Circuit:
The power factor is calculated as the ratio of the Real Power to the Apparent Power as the ratio equals cos(Φ).
Low Power Factor is majorly caused by huge inductive load which causes the reactive power to increase in magnitude, ultimately dropping the power factor. As discussed above, lower power factor doesn't allow effective usage of electric power. From an industrial point of view, this low power factor is resulted by the following causes.
The main cause of low Power factor is Inductive Load. Current lags 90° from voltage in pure inductive circuit. This huge difference of phase angle between current and voltage causes zero power factor.
All circuits having capacitance or inductance have a power factor due to a difference of phase angle (θ) between current and voltage. An exception to this rule are resonance circuits (also called tuned circuits), where inductive reactance is equal to capacitive reactance (XL = Xc), so the circuit becomes a resistive circuit.
Following are the causes of low Power factor:
Full load, Pf = 0.8 -0.9
Small load, Pf = 0.2 -0.3
No Load, Pf may drop to Zero (0)
2. Varying Load in Power System (when the power system is lightly loaded, the ratio of real power to reactive power is reduced, resulting in a decreased power factor).
3. Industrial heating furnaces.
4. Electrical discharge lamps (High intensity discharge lighting) Arc lamps (which operate at a very low power factor).
5. Transformers.
6. Harmonic Currents.
Since power factor is the ratio of active power and apparent power, it has no unit. That said, it is the quantitative measure of how much effective power is being used, without a unit.
It is imperative to know that power factor is only calculated for AC currents and circuits both for single phase and threephase. Following are some of the helpful formulae which can help in calculating the power factor of single and three phase circuit.
| Single Phase | Three Phase L-L | Three Phase L-N |
Main Calculation | PF = |cos φ| = 1000 × P_{(kW)} / (V_{(V)} × I_{(A)}) | PF = |cos φ| = 1000 × P_{(kW)} / (√3 × V_{L-L}_{(V)} × I_{(A)}) | PF = |cos φ| = 1000 × P_{(kW)} / (3 × V_{L-N(V)} × I_{(A)}) |
Apparent Power | |S_{(kVA)}| = V_{(V)} × I_{(A)} / 1000 | |S_{(kVA)}| = √3 × V_{L-L(V)} × I_{(A)} / 1000 | |S_{(kVA)}| = 3 × V_{L-N(V)} × I_{(A)} / 1000 |
Real Power | Q_{(}_{kVAR}_{)} = √(|S_{(kVA)}|^{2} - P_{(kW)}^{2}) | Q_{(}_{kVAR)}_{ }= √(|S_{(kVA)}|^{2} - P_{(kW)}^{}^{2}) | Q_{(}_{kVAR}_{)} = √(|S_{(kVA)}|^{2} - P_{(kW)}^{2}) |
Capacitor Capacitance | S_{corrected} _{(kVA)} = P_{(kW)} / PF_{corrected} Q_{corrected} _{(}_{kVAR}) = √(S_{corrected} _{(kVA)}^{2} - P_{(kW)}^{2}) Q_{c (}_{kVAR}_{)} = Q_{(}_{kVAR}_{)} - Q_{corrected} _{(}_{kVAR}_{)} C_{(F)} = 1000 × Qc _{(}_{kVAR}_{)} / (2πf_{(Hz)}×V_{(V)}^{2}) | Qc _{(}_{kVAR}_{)} = Q_{(}_{kVAR}_{)} - Q_{corrected} _{(}_{kVAR}_{)} C_{(F)} = 1000 × Qc _{(}_{kVAR}_{)} / (2πf_{(Hz)}×V_{L-L(V)}^{2}) | Q_{c} _{(}_{kVAR}_{)} = Q_{(}_{kVAR}_{)} - Q_{corrected}_{ (kVAR}_{)} C_{(F)} = 1000 × Q_{c (}_{kVAR}_{)} / (3×2πf_{(Hz)}×V_{L-N(V)}^{2}) |
Power factor correction can be defined as the method of improving power factor value to make it reach unity or nearby unity value,such that the angle between voltage and current reduces. The following are the factors to be considered when carrying out power factor correction in three main conditions:
Low Power Factor can be a problem for both consumers and generating utilities/stations, thus its improvement is vital for both:
The useful output of a power station is the kW output delivered by it to the supply system. Sometimes, a power station is required to deliver more kW to meet the increase in power demand. This can be achieved by either of the following two methods:
Knowing about power factor is very crucial for any electrical power system as it tells the amount of power wasted (Reactive Power) and consumed (Real Power) by it. Taking corrective measures will result in reduced power losses, increased voltage stability and eventually result in lowering the electric utility bills.