Estimate the electricity cost of grain storage equipment from fan or motor power, operating hours, load factor, and electricity price. The calculator converts the entered horsepower to approximate electrical energy use and then calculates total cost and cost per ton of stored grain. It is designed for planning aeration, ventilation, conveying, or similar storage operations when detailed meter data is not available. Load factor lets you represent equipment that does not draw full rated power continuously. The estimate is not a substitute for a utility meter or motor-efficiency study: actual electrical input can differ with motor efficiency, power factor, fan static pressure, control strategy, and utility demand charges.
1. Enter rated power
Use the motor or fan nameplate horsepower for the equipment being estimated.
2. Enter operating hours
Use the total run time for the storage period.
3. Set average load factor
Estimate the average percentage of rated power used while operating.
4. Enter electricity price
Use an effective price per kilowatt-hour if possible.
5. Enter stored grain quantity
This allows the result to show energy cost per ton for comparison.
The 0.746 factor converts mechanical horsepower to kilowatts. This simplified model does not separately model motor efficiency.
The main result is the estimated electricity expense for the entered equipment and operating schedule. Cost per ton provides a normalized storage metric.
For billing accuracy, use measured kWh and the actual tariff. Demand charges and fixed service fees are not included unless reflected in the entered effective price.
Given: 25 hp fan, 240 hours, 80% load, $0.15/kWh, and 600 tons stored.
Calculation: Loaded power = 25 × 0.746 × 0.80 = 14.92 kW. Energy = 14.92 × 240 = 3,580.8 kWh. Cost = 3,580.8 × $0.15 = $537.12. Cost per ton = $537.12 ÷ 600 = $0.90.
Result: Estimated electricity cost = $537.12.
Measured input power may be higher because real motors and systems are not perfectly efficient.
Nameplate horsepower is a rated capacity, not necessarily the power used every moment. Load factor provides a simple way to represent partial loading.
No. The 0.746 conversion is a mechanical power conversion. Actual electrical input can be higher because of motor and drive losses.
Yes, and metered energy is usually better for cost accounting. In that case you can multiply metered kWh directly by your effective electricity price.
It normalizes the storage-energy estimate by grain quantity, which can help compare storage units or operating periods of different sizes.
Not explicitly. If demand charges are important, add them separately or use an effective all-in energy price derived from your bill.