Boiler efficiency from input and output (worked example)

Efficiency tells you how much of the gas energy a boiler turns into useful heat: efficiency (%) = useful heat output ÷ energy input × 100. Both figures must be in the same units, usually kW, and on the same gross or net basis.

Example 1: find the efficiency

1. A boiler has a gross heat input of 24 kW and delivers a useful output of 21.6 kW into the water.
2. Efficiency = 21.6 ÷ 24 × 100 = 0.9 × 100 = 90% gross.
3. The remaining 2.4 kW (10%) is not delivered as useful heat, mostly leaving in the flue gases, with some lost from the casing. For this gross-basis comparison, output is below input.

Example 2: find the output

1. Start with a 20 kW gross input and an 85% efficiency for this boiler.
2. Convert the percentage to a decimal: 85% = 0.85.
3. Useful output = input × efficiency = 20 × 0.85 = 17 kW.

Do not divide 20 by 0.85. That answers a different question: the input needed for a known output.

Gross and net

Gross calorific value includes the latent heat available when combustion water vapour condenses; net calorific value leaves it out. Recovering some of that heat can put a condensing boiler above 100% when efficiency is calculated from net input. On a gross basis, useful output remains below input. Always label the basis before comparing figures.

Using the output figure

• Compare the useful output with the building's calculated heat loss and its hot-water demand. A boiler serving both must cover both.
• Do not estimate the load from room size alone. Insulation, glazing, ventilation and exposure set the real heat loss.
• A quoted output is valid only at the operating conditions it was measured under, such as particular flow and room temperatures.
• Too little useful output shows as rooms falling short of temperature in the coldest weather, exactly when heating matters most.

Use the manufacturer's data for real figures.