**1. Power factor**Active power can directly do work, such as making a resistive heater generate heat or making a motor generate mechanical energy to drive a machine. Reactive power can create a magnetic field in a coil or an electric field in a capacitor; energy is stored for part of a cycle and returned to the circuit for another part of the same cycle. The combination of active power and reactive power is apparent power.

The ratio of active power to apparent power is called power factor. Power factor is defined as: PF=active power/apparent power, or PF=watts/volt-amps. This relationship can be organized as

P_{VA}=P_{TRUE}/PF and P_{TRUE}=PF×P_{VA}

The value of P_{TRUE}/P_{VA} is between 0 and 1 with no unit. A power factor of 0 means that the active power is zero, or the circuit is purely reactive, and the apparent power is equal to the reactive power. A power factor of 1 means that the circuit is purely resistive, with apparent power equal to active power. Figure 1 shows the phase and power factor of the circuit when the power factor is 1 and 0, respectively.

The power company usually requires the user’s power factor to be greater than 0.9, that is, the active power is greater than 90%. The reason why utilities want loads with larger power factor is that low power factor loads (reactive loads) draw more current than high power factor loads. Larger currents require thicker cables and larger transformers, which will increase the cost of generating and transmitting electricity.

Let’s see why lower power factor loads require more current. If a load requires 5kW of active power and the power factor is 1, the power system only needs to transmit 5kVA of apparent power, that is, P_{VA}=P_{TRUE}/PF_{ }=5kW/1 = 5kW; if the power factor is 0.75, the power system needs to transmit The apparent power of 6.66kVA, that is, P_{VA}=P_{TRUE}/PF =5kW/0.75 = 6.66kW is used to supply the load with only 5kW active power.

If the working voltage of the load is 120V, PF=1, then the current is I=P/E=5kW/120=41.6A, if PF=0.75, then the current is I=P/E=6.66kW/120=55.5A, To transmit the required 5kW of active power, the current is increased by 33.4%.

The power company delivers apparent power to your home or business, but the kilowatt-hour (kW·h) measured by the meter in your home is active power, and you are only billed for active power, not reactive power. Therefore, for low power factor factory users, the electricity price is usually higher.

example 1

A wind turbine powers a load, as shown in Figure 2. Calculate its power factor.

Active power is

PTRUE =E²/R=240²/2kW=28.8kw

Apparent power is

P_{VA}=E²/2=240²/0.82kW=70.244kW

The power factor is

PF=P_{TRUE}/P_{VA}=28.8kW/70.24kW=0.41

This is a low power factor, so the power transfer from the wind turbine to the load is not efficient. Thicker cables must be used to reduce the line voltage drop from the wind turbine to the load.

Example 2

A photovoltaic array supplies power to a DC-AC inverter, the rated power of the inverter is 2.4kW, the rated output voltage is AC120V, and PF=0.92. When the inverter provides power with rated power for the load, calculate the supply Apparent power and total current of the load.

PF=PTRUE/PVA, the formula can be transformed into

PVA=PTRUE/PF=2.4kW/0.92=2.608kW

The total current is

I_{T}=PVA/E=2608/120A=21.73A

(If PF is 1, then I_{T}=2400/120 =20A.)

Figure 3 shows the power curve comparison between PF=1 and PF=0.

When PF=1, the power curve is always positive (above the zero axis), and all power is active power, which is consumed by the load.

When PF=0, the parts of the power curve above and below the zero axis are equal. The energy is stored by the circuit and then returned to the circuit without producing active power.

**2. Power factor correction**The purpose of power factor correction is to make the load more resistive, so that the power factor approaches 1. For this purpose, capacitors can be added to inductive loads or inductors can be added to capacitive loads to reduce the original inductive or capacitive effect. Figure 4 is a vector diagram illustrating the effect of this reduction in reactive power (VAR).

In Figure 4a, P=1000W, Px_{L}, =1200VAR, PxC=0VAR. Figure 4b adds PxC=600VAR to the circuit. Figure 4c shows the effect of adding 600W of capacitive power, which makes

Px_{L} drops to 600W ((1200-600)W=600W). The PVA values of Figures 4a and 7-33c are PVA=√(P²+P²x_{L})=√(1000² +1200²)W=√2440000w = 1562W

PVA=√[P²+(Px_{L}-Px_{C})]-=√(1000² + 600 ²)W=√1360000 =1166 W

The resistance in Figure 4c is smaller, with a PF closer to 1.

There are many methods of power factor correction. Industrial power users often have many motors that are inductive due to their internal field windings. In practice, capacitors are often installed at the load end where the power is connected. The power factor can be kept above the set value by connecting capacitors in or out of the load circuit. As shown in Figure 5, if a capacitor is connected in parallel with the load, the XC of the capacitor will weaken the X_{L} of the inductor, which increases the resistance of the load and improves the power factor.

Example 3

Using the circuit shown in Figure 5 in this example, the PF can be corrected from 0.75 to 0.90. Before any correction, the total current and active power need to be measured to determine the power factor. The steps to calculate the capacitance value to correct the power factor from 0.75 to 0.90 are described below.

(1) Calculate the apparent power according to the measured value:

P_{VA}=E×I_{T}=120×10W= 1200W

(2) Calculate the power factor according to the measured value:

P_{F}=P_{TRUE}/P_{VA}=W/2VA=900/1200 =0.75

(3) Calculate the PVA required to raise the power factor to 0.90

P_{VA}=P_{TRUE}/P_{F}=W/P_{F}=900/0.90W=1000W

(4) Calculate the inductive VAR before circuit correction (active power is unchanged):

P_{VA}=√(P²+VAR²_{L})

Arranging the functional formula, we have

VAR_{L}=√(P²_{VA}-P²)=√(1200²-900²)W=√1359000 W= 1165.76W

(5) Calculate the new inductive VAR value with P_{VA}=1000W (active power remains unchanged):

VAR_{L}=√(P²_{VA}-P²)=√(1200²-900²) W =√190000 =435.89W

(6) By adding capacitive VAR, the original inductive VAR is reduced to 435.89W:

VAR_{C}=VAR_{L}-435.89W=(1165.76-435.89)W=729.87w

(7) Calculate the capacitive reactance Xc required to correct the power factor:

X_{c}=E²/VAR_{C}

Xc=120²/729.87Ω=19.73Ω

(7) Calculate the capacitance required to correct the power factor:

X_{c}=1/(2πfC)

C=1/(2πfX_{c})=1/(2×3.14×60×19.73)F=0.0001345F = 134.5μF

The PF correction circuit is shown in Figure 6. The PF correction circuit reduces the reactive power supplied by the power supply from 1165.76W to 435.89W