Energy & electrical
Fuel input from useful output
- where
- fuel input (kW)
- useful heat / power rate (kW)
- efficiency on stated energy basis (fraction; any unit you pick is converted)
Explain this equation
Works out the fuel input needed to deliver a useful heat output at a given efficiency. Use it to find gas input from a heat demand.
P is the useful heat needed. η is the efficiency, as a fraction. P_in is the fuel input required.
It is the efficiency equation turned round: useful output divided by efficiency gives input. A lower efficiency means more input for the same output.
Sense check: 24 kW out at 90% needs about 26.7 kW of input. At 80% it would need 30 kW.
Watch out: use an efficiency on the same gross or net basis as the input you want. The calculator caps efficiency at 100%.
Enter your values to see the result.
Good to know
Enter a compatible efficiency basis. Use a factor above 100% only in a separately justified net-CV method; this calculator limits the entered efficiency to 100%.
Sources
Results are arithmetic only. Worked examples are not recommended design settings. This library does not provide automatic gas pipe selection, live gas tightness tests, purging procedures, combustion acceptance limits, installation approval or commissioning certification. Manufacturer data and the applicable current method remain essential.