When you are buying electrical energy (kilowatt-hours), why should your supplier care about your power factor (PF) and why should he give you a better deal for having a high power factor? The answer: At power factors less than one, your supplier's generators, transmission cables and transformers hit their kVA output limit before his generators have reached their full rated kW output.

At power factors less than one, your supplier's generators, transmission cables and transformers hit their kVA output limit before his generators have reached their full rated kW output. Photo courtesy of Staco Energy

When you are buying electrical energy (kilowatt-hours), why should your supplier care about your power factor (PF) and why should he give you a better deal for having a high power factor?

The answer: At power factors less than one, your supplier's generators, transmission cables and transformers hit their kVA output limit before his generators have reached their full rated kW output. This denies him the extra revenue that delivering full rated kW output to his consumer base would give him.

Figure 1 shows a generator feeding power to an inductive load such as an induction motor.

## How We Usually Describe Power Factor

Two figures can help demonstrate these concepts. Figure 1 shows a generator feeding power to an inductive load -- for example, an induction motor. Figure 2 shows voltage and current waveforms of the inductive load and their vector representation. The current is shown lagging the voltage byfo.

The motor does not get the benefit of all the current (I) shown here. It can only use the component that is in phase with the voltage, i.e., Icosf.

VIcosf is the effective motor input in watts and cosf is called the power factor.

The other component of current (Isin f) lags V by 90o and makes no contribution to motor power. It is sometimes called quadrature or wattless current. It does, however, make an unwanted and costly contribution to the resistance losses and consequent capacity limitation all the way from the generator to the motor terminals.

Another way to define load power factor is watts/volt-amps.

Figure 2 shows the voltage and current waveforms of the inductive load and their vector representation. The current is shown lagging the voltage by fo

## Power Factor Correction

You can eliminate that 90olagging component by connecting capacitors in parallel with your load. Capacitors take a 90oleading current that can reduce or totally cancel out the 90olagging (quadrature) component, thereby minimizing your total current and increasing the power factor. You do not need to connect enough capacitors to reach a power factor of one because at some point, you will spend more on adding capacitors than you save on your electric bill.

A typical bulk tariff bill would have two parts:

• A monthly demand charge based on the largest kVA recorded, averaged over any 15 min period in the month. This penalizes a consumer for excessive demands on the capacity of the system.

• An energy charge based on total consumption. The meter is calibrated to record kVA multiplied by 0.8 instead of recording kWh. This is an incentive for a consumer to maintain a high power factor.

Figure 3 shows an example of phase-angle control with full power at 200 A and half power 141.4 A.

## Loads That Take Nonsinusoidal Currents

Consider a resistive heater R (negligible inductance) controlled by a silicon-controlled rectifier (SCR).

Myth 1. "With a resistive load (no inductance), there is no lagging current; therefore, power factor must equal one." A look closer will show why this is not true. First, I will look at phase-angle control. Figure 3 shows an example of phase-angle control with full power at 200 A and half power at 141.4 A.

At full power, the RMS line voltage is constant at 500 V. The current is 200 A RMS, it is sinusoidal and in phase with the voltage so

kW = kVA = 500 V x 200 A = 100 kW
and Power Factor = kW/kVA = 1

At half power, the current does not appear until the voltage wave is at 90o. The current looks like it is lagging and somewhat out of shape. It cannot be represented by a vector and definitions based on cosf do not apply. So let's calculate the VA and the power factor at half power (50 kW). By removing a symmetrical half of each half cycle of the current wave, the power has been halved to 50 kW.

So heater power I2R has been halved, so I2 has been halved-- that is, multiplied by 0.5. Therefore the RMS value (not the average value) of current I has come down by a factor

?.5 = 0.707
and the reduced current is 141.4 A RMS. So,
kVA = 500 x 141.4 = 70.7 kVA and
Power Factor (kW/kVA) = 0.707

Figure 4 shows the waveforms representing an SCR-controlled resistive load in fast cycle (time-proportioning) mode.
Figure 4 shows the waveforms representing an SCR-controlled resistive load in fast cycle (time-proportioning) mode. The fast cycling waveform shows the current to be present for half the time, resulting in half power (50 kW) into the same load as in the phase-angle case. A damped RMS ammeter would show 141.4 A as before. Multiply this by the line voltage (500) and you get 70.7 kVA as before and a power factor of 0.707.

Myth 2. "With sine waves and resistive loads, power factor must be one." This myth is still circulated even by some manufacturers of SCR controls.

In both SCR firing modes above, suppose you turn down the power by a factor "A" - call it the per unit power instead of percentage of full power. The power factor turns out to be the square root of A.