Boiler Efficiency Cost Scenario Calculator

How the boiler efficiency cost scenario calculator works

Both scenarios deliver the same entered useful heat. Fuel input energy is useful heat divided by the output/input efficiency expressed as a fraction. The efficiencies must refer to the same annual or period boundary and the same fuel heating-value convention. They are percentages, not a heat-pump COP.

Fuel input energy is multiplied by one shared price per input kWh. Each fuel charge rounds half-up to cents before its separately entered fixed operating cost is added. The saving is existing rounded cost less alternative rounded cost, and cost verdicts use those displayed amounts. Fixed costs remain even at zero useful demand.

Efficiencies and prices are user assumptions. This does not infer combustion performance from flue-gas measurements, select equipment, calculate capital payback or guarantee a reduction in fuel use. Auxiliary electricity, distribution losses, tariff details and operating changes belong in a separately reviewed boundary.

Worked example

At 8,000 useful kWh, 80% efficiency needs 10,000 fuel-input kWh and 100% needs 8,000. At 0.10 per input kWh with no fixed charges, costs are 1,000 and 800, saving 200. Setting the alternative to 80% makes the costs equal and saving zero.

Sources and assumptions

The source material explains the underlying calculation or pricing structure. Entered prices, costs, schedules and scenario choices are your assumptions; this tool does not obtain supplier quotes or predict actual results.

• U.S. DOE: COP and cooling-efficiency rating definitions
• EIA: energy conversion and fuel heat-content conventions