Energy consumption is power multiplied by time:
kWh = kW × hours
Cost is energy multiplied by rate:
Cost = kWh × rate per kWh
The challenge is getting the power number right. For single-phase loads, power in watts is:
P = V × I × PF
For three-phase loads:
P = √3 × V × I × PF
P = 1.732 × V × I × PF
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
Step 1 — calculate power:
P = 1.732 × 480 × 100 × 0.85
P = 70,668 W
P = 70.7 kW
Step 2 — daily energy:
kWh/day = 70.7 × 8
kWh/day = 565.3
Step 3 — monthly energy:
kWh/month = 565.3 × 30.44
kWh/month = 17,207
Step 4 — monthly cost:
Cost/month = 17,207 × $0.15
Cost/month = $2,581
Step 5 — annual cost:
Cost/year = $2,581 × 12
Cost/year = $30,972
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
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
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
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
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
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
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
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.
