Power Factor Correction: Method

Power Factor Correction

The tool sizes shunt capacitors for a target power factor and follows the consequences through the bill, the supply system and the economics. Every relation is a closed form of the power triangle or of a lumped R-L-C circuit; the worksheets and screening rules are those of the Bonneville Power Administration Industrial Power Factor Analysis Guidebook (1995) and of Chapter 8 of the U.S. Department of Energy guide Energy Management for Motor Driven Systems (2000). Quantities owned by a tariff, a code or an equipment standard (rates, clause power factor, discharge time, capacitor duty limits, harmonic currents of the drives) are entered inputs. All inputs are plain numbers in kW, kvar, kVA, volts, amperes, ohms and dollars, so the unit-system switch changes nothing.

Power triangle

Real power P (kW), reactive power Q (kvar) and apparent power S (kVA) form a right triangle: S² = P² + Q², PF = P / S = cos φ, Q = P tan φ. Any two of P, S, Q, PF fix it; with the line current, S = √3 V I / 1000 three-phase (V line to line) or V I / 1000 single-phase. A pasted list of loads is summed row by row: input kW = rated output kW × load factor / efficiency × quantity and kvar = kW × tan(arccos PF), the power factor being the one at the operating load (a lightly loaded induction motor has a much lower power factor than its nameplate value, because its magnetizing kvar hardly changes with load).

Capacitor kvar

A capacitor supplies reactive power at the load so that the source no longer has to. To move from PF₁ to PF₂ at the same kW it must supply Q_c = P (tan φ₁ − tan φ₂) = P [√(1/PF₁² − 1) − √(1/PF₂² − 1)]. The bracket is the kW multiplier tabulated in the handbooks; the page prints a small table of it around the present power factor, and the test script compares the formula with the 763 printed cells of Table 8-3 of the DOE guide (three cells of that printing are misprints: 0.64 to 0.81, 0.64 to 0.99 and 0.86 to 0.99).

A fixed bank is assembled from the entered standard unit ratings: the smallest total at or above the requirement that the ratings can make, with the fewest units for that total (an exact search over the combinations). An automatic bank has equal steps; the fewest steps that reach the target at the demand peak are installed, and at every other load the controller is taken to keep on the fewest steps that reach the target there, one fewer if that would make the bus leading. An entered installed kvar overrides both. Capacitor units rated for a higher voltage than the system deliver only (V / V_rated)² of their rated kvar (600 V units on 480 V: 64 percent), so the rating bought, and its cost, is larger by the inverse.

After correction the source supplies Q₂ = Q₁ − Q_c, S₂ = √(P² + Q₂²); the line current and the kVA fall by 100 (1 − PF₁/PF₂) percent. A negative Q₂ is a leading power factor.

Released capacity

The kVA taken off the source is S₁ − S₂. The guidebook's Figure 12.1 asks a different question: how much more load of the present power factor can be added before the source is back at its present kVA. That is the positive root x of (P + x PF₁)² + (Q₂ + x sin φ₁)² = S₁², which the page solves exactly (1,000 kVA at 0.8 with 300 kvar: 150.8 kVA; the guidebook reads 150 from its chart). The guidebook's straight-line estimate, 100 (1 − PF₁/PF₂) percent, is printed beside it.

Losses

Resistive loss varies with the square of the current, so in every element upstream of the capacitors it falls by 100 [1 − (PF₁/PF₂)²] percent; nothing changes downstream. The page counts two elements: a feeder of entered resistance per conductor (3 I² R three-phase, 2 I² R single-phase two-wire) and the service transformer, whose load loss is (S / S_T)² R S_T with R its per-unit resistance, the form of the guidebook's loss worksheet. A bank at the main bus relieves only the transformer; capacitors at the loads relieve the feeder as well. The demand saving is the loss reduction at the peak. The energy saving is the loss reduction at the average load of the month (monthly kWh over the hours of the month, at the peak power factor, with the bank the controller would have on) times the hours; since loss is convex in load this understates the saving of a strongly varying load. Capacitor losses, a fraction of a watt per kvar, are neglected.

Voltage rise and light load

The source at the capacitor bus is the transformer impedance plus the utility's, 100 S_T / S_sc percent on the transformer kVA (magnitudes added, as in the guidebook's worked case), which gives the short-circuit kVA at the bus, kVA_sc = 100 S_T / Z. Capacitor current flowing through that reactance raises the bus voltage by about ΔV % = 100 kvar / kVA_sc; this is the guidebook's %Q × %Z rule and is identical to kvar × X / (10 kV²) with X in ohms. For a purely reactive source the exact no-load value is 100 [1/(1 − kvar/kVA_sc) − 1], printed beside it. At the entered light load a fixed bank stays connected: the net kvar, the power factor and whether the bus has gone leading are reported. A leading bus means more current than with no capacitors at all, and is the usual reason for switching part of a bank.

Utility bill

Six clause forms are offered, each with entered rates: no clause; demand billed in kVA; kW demand multiplied by the clause power factor over the actual one when the actual is lower (the DOE guide's Example 8-2); a charge per kvar of lagging demand above a free allowance stated as kvar per kW; a charge per lagging kvarh above an allowance per kWh; a percentage added to the demand and energy charges for every 0.01 of power factor below the clause value, with an optional credit above it (the form of the Army TM 5-811-1 example). The power factor is taken at the demand peak; the kvarh of the month are the net lagging kvar at the average load times the hours. Leading kvar is never charged here; some tariffs do, so check the schedule. The bill after correction also carries the kW and kWh of losses saved. Other items of a bill (customer charge, taxes, ratchets, time-of-use blocks) are outside the comparison.

Economic evaluation

Installed cost = cost per kvar × rated kvar + fixed cost. Annual saving = 12 × monthly saving − annual maintenance. Simple payback = cost / annual saving. Present worth of the savings = annual saving × [1 − (1 + i)⁻ⁿ] / i; net present value = present worth − cost; benefit to cost ratio = present worth / cost: the measures of the guidebook's Economic Evaluation Worksheet. The chart and table of saving against target power factor repeat the whole calculation for each target with exactly its required kvar; on a clause that stops at a stated power factor the saving flattens there while the cost keeps rising, which is why the net present value peaks near the clause value.

Capacitor bank data

Three elements in delta each see the line voltage, Q = 3 (2πf) C_Δ V²; in wye each sees V/√3, Q = (2πf) C_Y V², so C_Y = 3 C_Δ. Single-phase, Q = (2πf) C V². The line current is Q / (√3 V) and the reactance per phase V² / Q. A disconnected element holds the voltage at which it was interrupted, at most the crest √2 V_element, and discharges through its resistor as e^−t/RC; reaching the entered residual voltage within the entered time requires R ≤ t / [C ln(√2 V / V_res)]. The default 50 V in 60 s is the usual low-voltage requirement and is an entered value: take the figures from the code that governs the installation. Energizing a single bank through the source inductance L excites the circuit at its natural frequency f √(I_sc/I_c) with a current crest of √2 √(I_sc I_c) (lossless series L-C closed at the voltage crest); switching a step against steps already energized is far more severe and is not computed.

Capacitor at a motor

A capacitor on the load side of the motor contactor is switched with the motor at no extra cost, but after the contactor opens the spinning machine is excited by the capacitor; if the capacitor supplies more than the magnetizing kvar the voltage rises above normal. The guidebook's rule is to keep the capacitor at or below the no-load kVA of the motor, √3 kV I_{no load} (its example: 37 A at 480 V is 31 kvar, so a 50 kvar unit is 19 kvar too large). With the capacitor downstream of the overload relay the relay carries √[(I PF)² + (I sin φ − I_c)²] and its setting is reduced in proportion. The guidebook also lists the cases in which motor and capacitor should not be switched together without the maker's advice: reversing, plugging or jogging duty, restarting while spinning, multispeed and open-transition starters, loads that can drive the motor.

Harmonics screen

The bank and the source inductance are in parallel as seen from a harmonic current source on the bus; their reactances are equal at the order h = √(X_c/X_L) = √(kVA_sc/kvar), where the impedance peaks. The page compares that order, for the whole bank and for every number of steps of an automatic bank, with the 5th, 7th, 11th and 13th harmonics of six-pulse converters and flags a value within the entered band. The guidebook's own case (420 kvar on this source) gives 11.66; the default bank of 475 kvar lands on 10.97, and the page shows what that would mean if drives were added.

The voltage distortion is estimated with the guidebook's harmonic estimation worksheet (Figure 23.1): one bus, the source reactance h X_L in parallel with the capacitor reactance X_c/h, harmonic currents I_h equal to the fundamental current of the drives times an entered fraction at each of the four orders (the worksheet's typical fractions are the defaults), V_h = I_h X_h, THD = √ΣV_h² / V₁ with V₁ the line-to-neutral voltage. From the harmonic voltages follow the capacitor's harmonic currents V_h h / X_c and its duties in percent of rating: rms current √ΣI², rms voltage √ΣV², crest voltage by the arithmetic sum ΣV, and kvar ΣV_hI_h. They are compared with entered limits, defaulted to the figures the guidebook quotes from ANSI/IEEE Std 18-1980 (180, 110, 120 and 135 percent); the distortion limit defaults to 5 percent. The model has no resistance, so close to resonance its numbers are upper bounds that say "study this", not predictions.

A series reactor of p per unit of the capacitor reactance (a detuned bank) tunes the branch to h = 1/√p (7 percent: 3.78) and makes it inductive above that order, so it can no longer resonate with the source at the 5th harmonic and above; the remaining parallel resonance lies lower, at 1/√(p + kvar/kVA_sc). The capacitor units then run at 1/(1 − p) of the bus voltage and the branch delivers Q/(1 − p). The worksheet is evaluated with the branch reactance h p X_c − X_c/h.

Checks

The test script reproduces the DOE fact sheet Reducing Power Factor Cost (100 kW at 0.70 corrected to 0.95 with 67 kvar), Examples 8-1, 8-2 and 8-3 and Table 8-3 of the DOE guide, Figure 5-3 of TM 5-811-1 (29.27 kW of line loss, $2,459 and $2,690 a year), and from the Bonneville guidebook the worksheets of its first case study (0.8023, 0.3286, 0.4737, 475 kvar, $14,250; 2,476 and 3,876 kWh; 0.325; 2.624; $20,435; $6,185; 1.43; 1.83 years), the resonance at 11.66, the 150 kVA of Figure 12.1, the 2.5 percent voltage rise, the 31 kvar motor limit and the 64 and 75 percent voltage deratings. An independent numpy and scipy script recomputes twenty seeded plants with complex power and impedance, root finding, a brute-force bank search, year-by-year discounting and time-domain integration of the inrush.

Limits

Balanced sinusoidal fundamental quantities: the power factor is the displacement power factor, and capacitors do not correct the distortion component of a true power factor. One bus, one transformer, one feeder resistance; the kW at the meter is held constant when the triangle is corrected. The loss energy uses one average load, not a load curve. The resonance screen and the worksheet neglect resistance, cable impedance, other capacitors on the system and utility background distortion; IEEE Std 519 current limits at the point of common coupling are not evaluated. No back-to-back switching transients, fuse or breaker selection, or filter design. Tariffs are represented by their arithmetic only.