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Evgenii Konkin
Evgenii Konkin

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Why Your Utility Bills kVA but Your Equipment Runs in kW

#electrical #engineering #energy #power

Most commercial electricity bills have two charges:

Energy charge:  kWh consumed × rate
Demand charge:  peak kVA (or kW) × rate
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The demand charge is where power factor costs real money.

The kVA vs kW gap 

kW  = real power (performs work)
kVA = apparent power (what the utility supplies)
kW  = kVA × PF
kVA = kW / PF
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A facility with 500 kW peak demand at power factor 0.85:

kVA = 500 / 0.85 = 588 kVA
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The utility meter reads 588 kVA. The equipment uses 500 kW.

What that gap costs

If the demand rate is $12/kVA per month:

Billed in kW:   500 × $12 = $6,000/month
Billed in kVA:  588 × $12 = $7,059/month

Difference: $1,059/month = $12,706/year
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Same equipment. Same energy. Same hours. $12,706 more per year because the utility bills apparent power, not real power.

Power factor penalty (alternative billing)

Some utilities bill in kW but apply a penalty when PF drops below 0.90:

Adjusted demand = actual kW × (0.90 / actual PF)
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Example at PF = 0.78:

Adjusted = 500 × (0.90 / 0.78) = 577 kW billed
Penalty:  77 kW × $12 = $924/month
Annual:   $11,088/year
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The facility did not use 577 kW. It used 500. But low power factor inflated the bill.

What causes low power factor 

Load type              Typical PF
──────────────────────────────────
Resistive (heaters)    1.0
LED lighting           0.90–0.98
Motors (full load)     0.85–0.90
Motors (half load)     0.70–0.80
Motors (quarter load)  0.50–0.60
Mixed commercial       0.80–0.85
Heavy industrial       0.75–0.85
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Induction motors are the primary cause. They draw reactive current for their magnetic field. That current does no work but occupies utility capacity.

Lightly loaded motors are the worst. Reactive current stays roughly constant regardless of load.

The capacitor fix

Power factor correction capacitors generate reactive current locally:

Before correction (PF 0.78):
  kVA = 500 / 0.78 = 641 kVA

After correction (PF 0.95):
  kVA = 500 / 0.95 = 526 kVA

Reduction: 115 kVA
Monthly savings: 115 × $12 = $1,380
Annual savings: $16,560
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Typical capacitor bank cost: $8,000–$15,000 installed.

Payback: $15,000 / $16,560 = 10.9 months
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First step

Check your utility bill:

1. Find demand charge (kVA or kW)
2. If in kVA: PF = kW / kVA
3. If PF < 0.90: you are paying a penalty
4. Calculate correction savings before ordering capacitors
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