# A Primer on Power Factor Correction

Jun 1, 2004 12:00 PM, By Gonzalo Sandoval and Cesar Chavez Soria, Inelap

Improving power factor can reduce system and conductor losses, boost voltage levels, and free up capacity

Power factor correction is a frequently misunderstood topic. Improper techniques can result in over-correction, under-correction, and/or harmonic resonance, so it can be helpful to understand the process for determining the correct methods of sizing capacitors for various applications. It's also important for calculating the values of system and conductor losses, power factor improvement, voltage boost, and freed-up system capacity (kVA) you can expect to realize from their installation.

**The power triangle and its components.** The term “factor” implies a proportional relationship between two quantities. In this case, power factor is the proportion of active power (P) to apparent power (S). A power triangle is used to represent the proportion and calculate the reactive power, using the Pythagorean Theorem (**Fig. 1**below).

Active power, also known as working power, is the energy converted into useful work. Apparent power, on the other hand, is the total energy consumed by a load or delivered by the utility. So power factor is the proportion of power converted into useful work to the total power consumed by the loads or delivered by the power source. The power not converted into useful work is called reactive power (Q). You need this power to generate the magnetic field in inductors, motors, and transformers.

Suppose you have a load with a power factor of 0.95. What does this mean? Basically, it means that the load consumes 95% of the apparent power and converts it into work. But how much is reactive power? The answer, using the power triangle and the Pythagorean Theorem is calculated as follows:

S^{2}^{2}^{2}

Q^{2}^{2}^{2}

Q = sq rt (S^{2} - P^{2})

Q = sq rt (100^{2} - 95^{2}) = 31.225

If PF is corrected to 1.0, then reactive power=0 and apparent power=SQRT (95^{2}^{2}

Ideal power factor is unity (1.00), which means that the load is using 100% of the power to perform actual work.

**Why reactive power is undesirable.** Although you need reactive power for all the magnetic devices found in any electrical system, it's nevertheless undesirable because it causes a low power factor. A low power factor means a higher apparent power, which translates into excessively high current flows and inefficient use of electrical power. These currents cause elevated losses in transmission lines, excess voltage drop, and poor voltage regulation. Additionally, the installation must have sufficient capacity to conduct both the active and the reactive power. Many utilities apply a penalty to users with a low power factor as a way to get reimbursed for supplying the total apparent power.

The most common method for improving power factor is to add capacitors banks to the system. Capacitors are attractive because they're economical and easy to maintain. Not only that, they have no moving parts, unlike some other devices used for the same purpose.

When you add a capacitor bank to your system, the capacitor supplies the reactive power needed by the load. If you size and select the capacitor bank to compensate to a unity power factor, it can supply all the reactive power needed by the load, and no reactive power is demanded from the utility. If you design the capacitor bank to improve the power factor to a quantity less than 1.0, the reactive power supplied by the bank will be its rated kVARs (or MVARs), while the rest of the reactive power needed by the load will be supplied by the utility.

**How power factor correction releases system capacity.** Consider **Fig. 2**. Prior to the installation of a capacitor bank, all the reactive power of the facility (Q1 in **Fig. 2b**) is supplied by the utility, so the apparent power (S1 in **Fig. 2b**) is high because both the active and the reactive power have to be supplied by the utility. Equipment like transformers, switchgear, and cables must be large enough to handle the total apparent power, but this results in higher equipment costs when the power factor is low.

After installation of power factor correction capacitors, the capacitor bank (Q_{cap}**Fig. 2a**) supplies reactive power to the load, so the facility doesn't have to draw this reactive power from the utility, but rather only the difference (Q1-Q_{cap}

Calculate the release of system capacity by using the following equation:

%I_{r} = (1 - pf_{1}/pf_{2}) x 100

R = P (1/pf_{1} - 1/pf_{2})

where R is the release of system capacity, P is the active power of the load, pf_{1}_{2}

When power factor is low, the potential capacity to be released for the facility is greater than for high power factor.

**Fig. 3** shows the release of power in per-unit of the active power in a facility. It demonstrates that the maximum release occurs in those cases where the existing power factor is low and the target power factor is high, such as near or equal to one.

For example, suppose an industrial facility has a 1MVA transformer and its load has a power factor of 0.82 and 800kW of power consumption. The transformer is therefore loaded at 975kVA (97.5% of its total capacity).

S=P÷pf=800kW÷0.82=975kVA

If you connect a capacitor bank in the secondary of the transformer to improve the power factor to 0.98, the released capacity in the transformer is:

R = 800 (1/0.82 - 1/0.98) = 159kVA

This means you have reduced the transformer load to 816kVA, which is 81.6% of its total capacity.

The reduction of apparent power translates into reduced current because both are directly proportional. Before the use of the capacitor bank, the current has two components, active and reactive. With the addition of the bank, power factor can be corrected to unity and the current will have only the active component.

Calculate the percentage of current reduction by using the following equation:

This equation shows that the reduction obtained is a function of the power factor before and after the improvement.

**Fig. 4** shows the percent of current reduction in the circuit feeders when the power factor is improved. The greatest current reductions occur in those systems with a low power factor when the improvement is near or equal to one.

**How power factor correction reduces conductor losses.**The losses in conductors are equal to the product I2R, so the improvement of the power factor also accomplishes a reduction in the system losses. Transformer losses are also reduced because the current through its windings is reduced by the capacitor bank. The losses are also smaller in the circuit that carries the current from the secondary to the main panel if the capacitor bank is installed close to the load.

The curves in **Fig. 5**show the reduction in feeder losses when you improve power factor. This graph shows that the greatest reduction of losses in feeders occurs when the original power factor in the facility is low and is improved to (or near) unity.

With the reduction of current, voltage drop in the facility also decreases because both are directly proportional. If the capacitor bank has sufficient capacity, the voltage could rise above its nominal value. Calculate the voltage rise in a central compensation scheme, neglecting short circuit resistance of the windings, by using the following equation:

V_{2} = [(S_{T}/Z_{PU})/(S_{T}/Z_{PU} - Q_{cap})] x V'_{1}

where S_{T}_{pu}_{cap}_{1}_{2}

Suppose a facility has a 750kVA transformer with an impedance of 6.3%. Although its nominal secondary voltage is 220V, the voltage with load decreases to 215V. If you install a 75kVAR capacitor bank on the secondary side of the transformer, the voltage will be:

V_{2} = (750/0.063)/(750/0.063^{-75}) x 215 = 216.3V

The corresponding reduction in current means that the apparent power demanded from the utility will be lower than without the capacitor bank. This reduction of apparent power translates directly into energy savings because the power consumption from the utility feed is lower. If the utility penalizes you for a low power factor, then power factor correction will not only eliminate the penalty, it might also help you get a refund if the corrected power factor is greater than the minimum value required by the utility.

**Determining the capacitor kVAR requirement.** Calculate the capacity, in kVAR or MVAR (P must be kW or MW), of the capacitor bank needed to improve power factor from pf_{1}_{2}

Q_{cap} = P x [(sq rt (1-pf_{1}^{2})/pf_{1}) - (sq rt (1-pf_{2}^{2})/pf_{2})]

When active power is constant, you can use this equation to calculate the reactive power of the capacitor bank. But when active power isn't constant, you must consider other factors. You should consider the average value of the active power (P) as well as the average power factor in the facility. Using these two values, you can calculate the capacitor bank for the average operating condition. You should also consider the worst case operating conditions (highest active power and lowest power factor).

**Calculating the kVAR requirement based on maximum active power.** Looking at the preceding equation (hereafter referred to as Equation A), you can see that either of two factors can cause the calculated value of reactive power of the capacitor bank to be less than the value required:

The active power (P) is higher than the average value used in Equation A.

The power factor (pf

) is lower than the average value used in Equation A._{1}

Taking this into account, you need to re-calculate the reactive power requirement of the capacitor bank using the maximum active power in the system and the power factor measured under this operating condition. Equation A can now be expressed as:

Q_{cap} = P_{max} [(sq rt (1 - pf_{1}^{2}_{-Pmax})/pf_{1-Pmax}) - (sq rt (1 - pf_{2}^{2})/pf_{2})]

(hereafter referred to as Equation B) where P_{max}_{1-Pmax}_{max}

If the reactive power requirement for the capacitor bank, as calculated using Equation B, is greater than the average value previously calculated using Equation A, then the capacitor bank sized for the average value won't be sufficient for compensating the reactive power of the load when the active power reaches its maximum value. As a result, the power factor in the facility won't reach the target value. In this case, you should select the capacitor based on the maximum active power and the actual power factor under that operating condition (Equation B).

**Calculating the kVAR requirement based on minimum power factor.** The next consideration is to calculate the capacitor bank needed when the power factor is minimum. Do so by using the following equation:

Q_{cap} = P_{pf1min} [(sq rt (1 - pf_{1}^{2}_{-min})/pf_{1min}) - (sq rt (1 - pf_{2}^{2})/pf_{2})]

(hereafter referred to as Equation C) where pf_{1min}_{pf1min}_{1min}

If the two previously calculated values (average and maximum active power conditions) are less than the value calculated using Equation C, the capacitor bank kVAR previously determined using Equation A or Equation B won't be sufficient to compensate for the reactive power of the load when the power factor reaches its minimum value, and the power factor in the facility won't reach the target value. In this case, you should select the capacitor based on the minimum power factor as calculated in Equation C.

*Note:* The best value of capacitance will be the greater of all calculations above (Equation A, Equation B, and Equation C), because the capacitor bank will have the capacity for compensating for the maximum active power condition as well as minimum power factor condition. Automatic capacitor banks can ensure high power factor under widely varying operating conditions.

Once you've calculated the capacitor bank size in kVAR or MVAR, you should perform a voltage and current spectrum analysis to check for the presence of harmonics in the facility because they can cause severe damage to capacitors. When harmonics are present, you should use only capacitors equipped with capacitor protection reactors.

**Beware of power system resonance.** From the load point of view, the capacitor and the transformer form a parallel resonant circuit, while the same elements form a series resonant circuit from the source point of view. At the resonant frequency, the admittance of a parallel circuit is very low, so a small current in the load side of the circuit results in very large currents in both paths (inductive and capacitive) when the frequency of the current is equal to the resonant frequency. The impedance of a series circuit at its resonant frequency is very low, so a small voltage on the primary side of the transformer results in very large currents in the capacitor bank and the transformer when the frequency of the voltage is equal to the resonant frequency.

To avoid resonance, you should perform an electrical survey prior to installation of the capacitor bank. Harmonic currents on the load side can excite the parallel circuit, while the harmonic voltages in the primary side of the transformer can excite the series circuit. Both situations will result in very high currents in the transformer and the capacitor bank, as well as very distorted voltages in the secondary side of the transformer.

To perform the survey correctly, use a power quality monitor that can record the trend of electrical parameters, such as voltage, current, active, reactive and apparent power, and power factor, with a time stamp. It should also be able to monitor and record the waveform, spectrum analysis, and harmonic distortion levels.

With this kind of instrument, it's relatively easy to determine the reactive power rating of the capacitor bank necessary for power factor improvement. The trend graphs will allow you to calculate the reactive power requirement for maximum active power and minimum power factor conditions, thus helping you to determine the best value of the bank. The waveform and spectrum analysis will provide an indication of any harmonic components present in the voltage and current.

Determining the resonant frequency of the power system will require system modeling and analysis based on an equivalent electrical circuit. You can do this by using simulation software or by doing the calculations by hand. If the system is small and relatively simple, you should be able to do a hand calculation of the resonant point fairly easily. However, you'll need large simulation software on large and complex systems. Once you've developed the equivalent electrical circuit, then you can perform a frequency scan to determine the resonant point of the system.

The power monitor, and its associated software, must be able to calculate total harmonic distortion (THD) for voltage according to the definition in IEEE Std. 519 - 1992 (distortion based on the nominal system voltage) and calculate total demand distortion (TDD) defined in the same standard (distortion based on the rms value of the fundamental current component at maximum demand). These capabilities are a must.

*Sandoval is an electrical engineer and Chavez is manager of electrical engineering at Inelap in Naucalpan, Mexico. John Houdek, electrical engineer and president of Allied Industrial Marketing in Milwaukee, provided editing assistance on this article*.

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