Power factor correction and capacitor sizing on the PE Power exam

Power factor is the cosine of the angle between voltage and current, which equals real power divided by apparent power. Correction problems on the PE Power exam start from a lagging load and ask for the capacitor bank that reaches a target, or for what an installed bank does to current, loss or transformer loading. The NCEES handbook holds the sizing and loss relations under Power Factor Correction and Voltage Support, the power triangle under 3-Phase Circuits and AC Circuits, and the capacitance relation under Capacitors. Installation rules are in NEC Article 460.

What you actually need to own

• The sizing relation. Q_bank = P × (tan Φ1 − tan Φ2), where each Φ is the inverse cosine of a power factor. Real power is the fixed quantity. If the problem gives kVA, convert first with P = S × pf. With round numbers, 800 kW moved from 0.75 to 0.95 lagging needs 800 × (0.882 − 0.329), about 443 kvar.
• The power triangle at a bus. Q = P × tan Φ for each load. Real powers add, reactive powers add, and a capacitor's kvar subtracts. Then S = √(P² + Q²) and pf = P/S. If the bank exceeds the load's reactive demand, the bus is leading.
• What correction buys upstream. At constant real power, apparent power falls from P/pf1 to P/pf2, and line current falls in the same ratio. Loss goes as current squared, so the handbook's loss reduction is 100 × [1 − (pf1/pf2)²] percent. A capacitor relieves only the conductors and transformers between itself and the source.
• The bank as a fixed capacitance. Output is Q = 2πf × C × V², per phase with the voltage across that phase. A bank running below its rated voltage delivers less than nameplate, by the square of the voltage ratio. Three capacitors in wye deliver V_LL²/X_C in total. The same three in delta deliver three times that. Rated line current is Q/(√3 × V_LL).
• Fixed against switched. A bank's kvar does not follow the load. A fixed bank sized for full load over-corrects at light load and the power factor goes leading. A load that swings widely calls for switched steps.
• The Article 460 rules. Capacitor circuit conductors need an ampacity of at least 135 percent of the capacitor's rated current. With a capacitor on the load side of a motor overload device, the overload rating follows the improved current, while the motor conductors are sized as if the capacitor were absent. The article also sets how fast stored charge must drain after disconnection, with separate allowances by voltage class.

Where people lose points

• Multiplying the tangent difference by kVA. The relation is written on real power.
• Using sine in place of tangent. Sine goes with apparent power, tangent with real power.
• Correcting to unity when the target is lower. That returns the load's whole reactive demand.
• Reporting the drop in kVA as the bank size. P/pf1 − P/pf2 is capacity released upstream. The bank kvar is larger, because P and Q add at right angles.
• Leaving the ratio unsquared. 1 − pf1/pf2 is the fractional fall in current, and loss takes the square of the ratio.
• Calling a leading bus lagging. Compare the bank with the load's Q first.
• Treating nameplate kvar as the output at any voltage, or scaling it with voltage to the first power.
• A stray √3. In a wye bank each capacitor sees V_LL/√3. In delta it sees V_LL.

A note on the codes

The exam supplies its codes, including the 2020 NEC and the 2017 NESC, and scores code answers against those editions. For this topic the chapter to know is the one holding Article 460. Learn where its conductor, overload and discharge rules sit before exam day. See the PE Power handbook and code books.

How to study it

Take one plant load and correct it to three targets. For each, write the bank kvar, the new kVA, the new line current and the percent loss reduction, then recompute the power factor from P and the residual Q. Repeat with the load cut to a quarter and the bank left in, and name the sense. Then tie it to the feeder: lower current is lower voltage drop. Capacitors at motor terminals lead into motor circuits.

Part of the Transmission and Distribution Analysis area of the PE Power exam. → Practice it on CastorPrep