Topic 5 of 6
Efficiency
Efficiency is the fraction of the total input that becomes the useful output for a stated purpose.
Conservation accounts for all the energy. Efficiency identifies which part serves the intended task. Use the same process and time interval for input and output.
For a lifting motor, the intended increase in gravitational energy is useful. Energy increasing the motor's or surroundings' internal stores does not perform that lifting task. For a heater, an increase in the intended object's internal energy can be the useful output instead.
Useful lifting accounts for 300 J of the 400 J input
The intended result is an increase in the gravitational energy store of the load and Earth. The remaining 100 J increases internal stores of the motor and surroundings.
300 J useful: gravitational store increase
100 J: internal store increases
Efficiency = 300 / 400 = 0.75 = 75%.
Worked example
Useful lifting output from a 400 J input
- State the useful output: the intended gravitational increase is 300 J.
- Divide by the total input: efficiency = 300 / 400 = 0.75.
- Convert to a percentage: 0.75 x 100% = 75%.
- Account for the remainder: 400 - 300 = 100 J increases internal stores, corresponding to 25% of the input.
The 100 J has not been destroyed. It has a destination, but it is not useful for the stated lifting purpose.
Use the fraction when finding an unknown
Useful output = efficiency x total input, with efficiency written as a fraction. A device that is 0.75 efficient and receives 600 J gives 0.75 x 600 = 450 J of useful output.
Alternatively, total input = useful output / efficiency. Supplying 300 J of useful output at 0.75 efficiency requires 300 / 0.75 = 400 J. Substituting 75 instead of 0.75 would confuse a percentage with a fraction.
For a complete account with no uncounted initial store supplying the process, useful output cannot exceed total input, so the efficiency lies between 0 and 1, or 0% and 100%. An apparent result above 100% indicates a reversed ratio or an incomplete or inconsistent account.
Efficiency and power answer different questions
A powerful motor transfers energy rapidly. An efficient motor directs a large fraction of its input to the intended output. A motor could have high power but low efficiency, so one property does not establish the other.
Reducing friction in a lifting mechanism can reduce the transfer to internal stores, leaving a larger useful fraction for the same total input. The appropriate improvement depends on where the unwanted transfers occur.
Define useful before calculating. Compare energies over the same interval and use the full input. "Useful" depends on the task; it is not the name of a special energy store.