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Evgenii Konkin
Evgenii Konkin

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The Energy Cost Formula Most Engineers Use Wrong for Three-Phase Loads

#electrical #engineering #energy #tutorial

Energy consumption is power multiplied by time:

kWh = kW × hours
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Cost is energy multiplied by rate:

Cost = kWh × rate per kWh
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The challenge is getting the power number right. For single-phase loads, power in watts is:

P = V × I × PF
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For three-phase loads:

P = √3 × V × I × PF
P = 1.732 × V × I × PF
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The 1.732 is the square root of 3. It accounts for the phase relationship in a three-phase system. Leaving it out underestimates three-phase power by 42 percent.

Example: one motor, one year

A commercial HVAC chiller motor:

Voltage:      480V line-to-line
Current:      100A
Power factor: 0.85
Runtime:      8 hours/day
Rate:         $0.15/kWh
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Step 1 — calculate power:

P = 1.732 × 480 × 100 × 0.85
P = 70,668 W
P = 70.7 kW
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Step 2 — daily energy:

kWh/day = 70.7 × 8
kWh/day = 565.3
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Step 3 — monthly energy:

kWh/month = 565.3 × 30.44
kWh/month = 17,207
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Step 4 — monthly cost:

Cost/month = 17,207 × $0.15
Cost/month = $2,581
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Step 5 — annual cost:

Cost/year = $2,581 × 12
Cost/year = $30,972
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One motor. Thirty thousand dollars a year.

The three mistakes that break the calculation

Mistake 1: using nameplate power for cycling loads

A 5 kW compressor with a 60 percent duty cycle does not consume 5 kW continuously.

Nameplate calculation: 5 kW × 8 hr = 40 kWh/day
Actual calculation:    5 kW × 8 hr × 0.60 = 24 kWh/day
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The nameplate method overstates consumption by 67 percent. Use measured average power or apply the duty cycle factor.

Mistake 2: ignoring power factor

Calculating power as voltage times current without power factor gives apparent power in kVA, not real power in kW.

Without PF: P = 1.732 × 480 × 100 = 83,138 VA = 83.1 kVA
With PF:    P = 1.732 × 480 × 100 × 0.85 = 70,668 W = 70.7 kW
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The electricity meter measures real power (kW), not apparent power (kVA). Omitting power factor overstates energy consumption by 18 percent for a motor at PF 0.85.

Mistake 3: entering rate as whole cents 

Wrong: rate = 15    → cost = 17,207 × 15 = $258,105/month
Right: rate = 0.15  → cost = 17,207 × 0.15 = $2,581/month
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One hundred times too high. This happens in spreadsheet calculations more often than anyone admits.

Why it matters for equipment decisions

When you know the per-equipment energy cost, upgrade decisions become math:

Motor annual cost:           $30,972
VFD savings (30% reduction): $9,292/year
VFD installed cost:          $5,000
Payback period:              6.5 months
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Without the per-equipment calculation, this decision is a guess. With it, the payback is obvious.

What the formula does not cover 

It does not account for:
  Demand charges (kW peak billing)
  Tiered rate structures
  Time-of-use pricing
  Power factor penalties
  Tax and distribution fees
  Motor efficiency losses
  Seasonal load variation
  Actual metered data
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The kWh cost calculation is a screening estimate. It tells you the approximate operating cost of a piece of equipment at a flat rate. For precise energy accounting, use metered data and your actual utility rate schedule.

Quick reference

For any piece of equipment, you need three numbers:

1. Power in kW
2. Daily runtime in hours
3. Electricity rate in $/kWh
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Multiply all three for daily cost. Multiply by 30.4 for monthly. By 365 for annual.

For three-phase equipment where you only know voltage and current:

kW = 1.732 × V × I × PF
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Use PF = 0.85 for motors, 0.95 for electronic loads, 1.0 for heaters.

For quick energy cost estimates with single-phase, three-phase, and direct kWh entry, there is an energy consumption calculator on CalcEngineer.

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