Work out efficiency both ways, avoid the inverted fraction, and see why controls, balancing and return temperature decide real efficiency.
Gross efficiency compares useful output with gross fuel input. The worked figures below show the arithmetic; real-world performance also depends on the system around the boiler.
The calculation
Efficiency (%) = useful output ÷ input × 100. Both figures must use the same units, usually kW, and refer to the same operating condition and basis.
- With 30 kW entering as gross fuel input and 24 kW reaching the water, 24 ÷ 30 = 0.80, or 80%. The other 6 kW is not delivered as useful heat.
- 27 kW from a 30 kW gross input: 27 ÷ 30 = 0.90, and 0.90 × 100 = 90%.
- 20 kW from 25 kW: 20 ÷ 25 × 100 = 80%.
To find output, convert the percentage to a decimal and multiply: 30 kW at 88% gives 30 × 0.88 = 26.4 kW. To find input from a known output, divide the output by the decimal efficiency.
Write the fraction with labels before entering numbers, with useful output on top. Putting input on top, 30 ÷ 27, gives about 111%, which is wrong for these figures.
Gross and net
Gross calorific value includes the heat available from condensing the water vapour formed by combustion; net calorific value excludes that latent heat. A condensing boiler can exceed 100% on a net basis because it recovers some of that heat. It does not create energy. On a gross basis the answer must come out below 100%. Label every input and efficiency figure gross or net, and never compare figures on different conventions.
COP, used for heat pumps, is a heat-transfer performance ratio and can exceed 1. It is not boiler percentage efficiency.
Why real efficiency differs from the sum
A calculation or a flue-gas test captures one operating condition. Comfort and running cost also depend on the system around the boiler.
- Seasonal performance. A real system spends the year modulating, cycling and responding to controls under varying loads, so seasonal performance can differ noticeably from a one-off test figure.
- Controls decide how often the boiler fires and at what temperatures it runs.
- Return temperature. Natural-gas flue products have a dew point around 55 °C, varying with combustion conditions. Cool return water helps the heat exchanger recover latent heat. High flow settings, uneven circulation or excess bypass flow can warm the return, but measure the actual operating temperatures under load before judging the effect. Retain any bypass required to maintain the boiler’s specified minimum flow.
- Balancing. Set each emitter’s flow to meet its design duty. Uneven distribution can leave rooms too warm or too cool, prompting extra demand or higher settings; it does not add a fixed amount to the boiler’s running time.
- Sizing. Too little output can leave rooms below design temperature. Frequent cycling can occur when the lowest available boiler output exceeds the active load, so assess its modulation range as well as the maximum rating.
- System water and distribution. A boiler can give an excellent combustion reading while the system around it performs badly.
Common mistakes
- Putting input on top of the fraction.
- Multiplying by 88 instead of 0.88.
- Comparing a net-basis figure with a gross-basis one.