Power, energy and runtime calculations

Core idea

Power is the rate at which energy is transferred. Energy equals power multiplied by time. With power in kilowatts and time in hours, the result is kilowatt-hours: energy (kWh) = power (kW) × time (h). Runtime equals energy divided by power.

What to recognise

• Convert watts to kilowatts by dividing by 1000.
• Convert minutes to hours by dividing by 60.
• A 3 kW heater running for 40 minutes uses 3 × 40/60 = 2 kWh.
• Two identical loads running together double power, not time.
• Use actual duty cycle or measured input when the appliance does not run continuously.
• 1 kilowatt-hour equals 3.6 megajoules (3,600,000 joules).
• The watt is one joule each second, so a 2 kW heater converts 2,000 joules, or 2 kJ, every second.
• Efficiency is the useful output divided by the input, given as a percentage: an appliance that takes in 10 kW and delivers 8.5 kW of useful heat is 8.5 ÷ 10 = 85% efficient. The rest is lost, and no appliance can be more than 100% efficient.
• The cost of running something is its energy use in kWh multiplied by the price per kWh, for example 135 kWh at 25p per kWh is £33.75.

Apply it

Write the units beside every value before multiplying. For a 6 kWh energy requirement supplied by a constant 2 kW heater, runtime is 6 ÷ 2 = 3 hours. Sense-check: a larger power should meet the same energy requirement in less time.

Limits

This simple sum assumes the input stays constant. Thermostats, cycling, defrost, standby loads and changing output all affect real use. For a heat pump, electrical input and useful heat output are different, so use the one the question asks for.