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Thabiso KaliEnergy engineer · CEM

For engineers and facility managers

Power-factor correction sizing

On a kVA demand tariff you pay for apparent power, not just the kW that does the work. A capacitor bank supplies the reactive power that motors and transformers need, so the same kW arrives with fewer kVA.

Inputs

Pre-filled with an illustrative case. Results update as you type. Jump to results

Load at maximum demand
kW

kW at the time of the kVA maximum demand.

From the bill (kWh ÷ kVAh) or a logger.

0.95 to 0.98 is usual; avoid 1.00.

Capacitor bank
kvar

The bank is rounded up to whole steps.

R/kvar
V
Tariff
R/kVA/month

All charges billed per kVA of monthly maximum demand.

months

Results

375 kvar bank lifts the power factor to 0.975, cutting demand by 155 kVA and saving about R84k a year.

Bank size
375kvar
357.9 kvar needed
kVA reduction
155kVA
15.9% of 976 kVA
Annual saving
R84k
kVA demand charges
Simple payback
2.9years
Bank cost R244k

NoteCheck harmonics before installing: with variable-speed drives or rectifier loads, a detuned (reactor) bank may be needed, which costs more.

Apparent power before and after

kVA at the time of maximum demand: the number the demand charge is billed on.

02004006008001,000Before: 976Before976After correction: 821After correction821 pf 0.975Real power (floor): 800Real power (floor)800

Power triangle

kVA² = kW² + kvar². The bank supplies reactive power locally, so the network carries less.

MeasureBeforeAfter
Real power800 kW800 kW
Reactive power558 kvar183 kvar
Apparent power976 kVA821 kVA
Power factor0.8200.975
Line current at 400 V1,408 A1,185 A
Annual demand chargesR526,829R443,207

Add to a business case

What is this?

Save this result as one measure in a project kept in this browser, then combine it with results from other tools for payback, NPV, IRR and a cost curve.

Saving, year 1
R83,622/yrAssumed
Demand reduced
155 kVACalculated
Capex
R243,750Assumed

Every input above is still the illustrative example, so the result is tagged assumed. Enter your own numbers to make it yours.

Assumed
years

Capacitor banks typically last 10 to 15 years or more with maintenance.

Assumed
R/yr

Inspection, thermal scans and the odd capacitor replacement. Default: 1% of capex a year.

Method

Reactive and apparent power now
Q₁ = P × tan φ₁, S₁ = P ÷ cos φ₁, tan φ = √(1 − cos²φ) ÷ cos φ
Reactive power to add
Q_c = P × (tan φ₁ − tan φ₂)
The bracket is the familiar ‘multiplier table’ value: 0.421 kvar per kW from 0.80 to 0.95.
Bank size
Q_bank = ceil(Q_c ÷ step) × step
After correction
Q₂ = Q₁ − Q_bank, S₂ = √(P² + Q₂²), pf₂ = P ÷ S₂ (leading if Q₂ < 0)
Line current
I = S × 1,000 ÷ (√3 × V_LL)
Annual saving and payback
Saving = (S₁ − S₂) × R/kVA × months, payback = Q_bank × R/kvar ÷ saving

Assumptions

  • The kW and power factor are those at the time of the monthly maximum demand, and stay similar every month.
  • Real power does not change; only reactive power is compensated.
  • The saving counts kVA-based charges only. Some tariffs also charge for reactive energy (kvarh) above a threshold; correction usually removes that too, which this tool does not add.
  • Capacitor losses, harmonics, detuning reactors and maintenance are ignored. A detuned bank costs more per kvar.
  • Cost per kvar is an illustrative installed figure; get quotes.

Sources

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