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Energy and Fields overview

Topic 6 of 6

Power, losses and efficiency

Energy measures how much transfers. Power measures the rate of transfer. Efficiency compares the useful output with the total input for a stated task.

Average and instantaneous power

Average power = energy transferred / elapsed time
Paverage = ΔE/Δt

The unit is W = J/s. Instantaneous power is the rate at a particular time. Always identify the transfer being described: total input, useful output or the work of a named force.

Over a short interval, a force's work is Fs cos θ. Divide by the interval and use displacement per unit time in the limit of a short interval:

P = Fv cos θ

Here θ is the angle between force and velocity. Mechanical power is the force multiplied by the velocity component in its direction. P = Fv is the aligned special case. An opposing force has negative power; a perpendicular force has zero power.

For the crate in the angled-pull example, consider an instant when its speed is 2.0 m/s right. The 8.0 N pull at 60 degrees supplies 8.0 × 2.0 × cos 60° = 8.0 W. The 2.0 N resistance contributes -4.0 W, giving net mechanical power +4.0 W, the rate of kinetic-energy increase. The perpendicular support contributes zero.

Compare the useful output and total input

Efficiency = useful energy output / total energy input

Efficiency is unitless and is often expressed as a percentage. For powers averaged over the same interval, the same ratio can be calculated as useful output power / total input power.

Worked power and efficiency

The steady lifting motor

A motor raises a 40.0 N load through 5.00 m in 4.00 s at steady speed. Its measured electrical input energy is 320 J. The useful gravitational increase is 40.0 × 5.00 = 200 J, and the remaining 120 J increases internal energy of the device and surroundings.

Useful output power = 200/4.00 = 50.0 W
Input power = 320/4.00 = 80.0 W
Efficiency = 200/320 = 50.0/80.0 = 0.625 = 62.5%

The load's speed is 5.00/4.00 = 1.25 m/s. Lifting-force power is therefore 40.0 × 1.25 = 50.0 W. Use the lifting force, not the zero resultant force. The resultant's zero power accounts for unchanged kinetic energy; it does not mean the motor transfers no energy.

Useful lifting power and input power describe different transfers

A steady lift can transfer energy with zero resultant force

A forty-newton lifting force at 1.25 metres per second supplies fifty wattsA motor pulls a load steadily upward. On the load, the upward lifting force and downward weight are each forty newtons, shown by equal-length purple arrows. A separate blue arrow shows upward velocity 1.25 metres per second. The steady interval covers five metres in four seconds. Although the resultant force is zero, the lifting force supplies fifty watts and gravity's mechanical power on the load is negative fifty watts. The load's kinetic energy does not change, while load-Earth gravitational potential energy increases. Arrangement and velocity-arrow lengths are schematic.MotorLift: 40.0 NWeight: 40.0 N1.25 m/supwardSteady interval: 5.00 m in 4.00 sUseful P = 40.0 × 1.25 = 50.0 W

Use the lifting force when finding its power, not the zero resultant force. Gravity's negative work balances the positive work on the load, so its kinetic energy stays constant while the load-Earth gravitational store increases.

Same energies in half the time

Two supplied runs have equal energy and efficiency but different powersBoth runs receive 320 joules and provide 200 joules usefully. Their input bars are both 288 drawing units wide and useful bars both 180, using one energy scale. Run A takes four seconds and Run B two; their separate time bars are 240 and 120 units wide. Average input and useful powers are eighty and fifty watts for A, and 160 and one hundred watts for B. Both efficiencies are 200 divided by 320, or 62.5 percent. The equal energies are supplied conditions; a real device's efficiency need not remain constant when its speed changes.Run A: 4.00 sInput: 320 JUseful: 200 JTime: 4.00 sInput 80.0 W; useful 50.0 WRun B: 2.00 sInput: 320 JUseful: 200 JTime: 2.00 sInput 160 W; useful 100 WBoth efficiencies: 62.5%

Energy bars use one scale; time bars use another. Halving the time doubles both powers here. Equal input and useful energies keep the efficiency at 62.5% under the supplied conditions.

The lifting-force calculation identifies the useful transfer. In the two supplied runs, equal input and useful energies are transferred over different times, so both powers change while their ratio stays the same.

If a second supplied run transfers the same 320 J input and 200 J useful output in 2.00 s, its input power is 160 W and useful output power 100 W. Both powers double; efficiency remains 62.5%. The equal energies are supplied conditions, not a general claim that changing a real motor's speed leaves efficiency unchanged.

The 120 J internal increase does not lift the load, so the motor needs more than 200 J input. Reducing unwanted mechanical friction can reduce that loss in an otherwise comparable lifting task. All energy is still accounted for; the loss is a reduction in useful output for this task.

A quoted efficiency alone does not give an energy amount: the input is also needed. For a complete account with the stated useful output and total input, efficiency cannot exceed 100%.

Optional check Two supplied motor runs each transfer 320 J in and raise the load-Earth gravitational store by 200 J. One run takes 4.00 s and the other 2.00 s. Which comparison follows from these data?
Two supplied motor runs each transfer 320 J in and raise the load-Earth gravitational store by 200 J. One run takes 4.00 s and the other 2.00 s. Which comparison follows from these data?

Kilowatt-hours measure energy

A kilowatt-hour is the energy transferred at 1 kW for 1 hour. Its symbol is kWh; multiplying power by time gives energy.

1 kWh = 1000 J/s × 3600 s = 3.6 × 106 J

For a supplied constant power of 2.0 kW over 3.0 minutes:

E = Pt = 2.0 × (3.0/60) = 0.10 kWh
= 3.6 × 105 J

The 2.0 kW value is a rate. The 0.10 kWh value is the amount transferred during the stated interval.

Investigate a lift with matching measurements

To determine a lifting device's work, power and efficiency, measure quantities for the same defined interval:

  1. Load force: use a suitable force reading, or measure mass and multiply by the stated local g. Check zero and choose a range that covers the load.
  2. Vertical rise: measure the difference between start and finish heights with a suitable rule or tape. A sloping string length is not the vertical rise. View a pointer perpendicular to the scale to avoid parallax, and retain the same reference.
  3. Elapsed time: identify the same start and finish events used for the displacement and energy readings. Use a steady interval, or account for kinetic-energy changes if the speed changes.
  4. Input energy: use the change in a suitable calibrated energy reading over that interval. A motor's rated power is not a measurement of the energy taken in during a particular run.
  5. Analysis: calculate useful work from force and vertical rise, then divide each measured energy by the same time. Use useful energy / input energy for efficiency.

Choose ranges and resolutions that suit the force, length and time scales. Record raw values with units, repeat comparable runs to assess scatter, and retain guard digits in calculations. An underestimated rise makes the calculated useful work and efficiency too low if the other readings are correct. Repeating the same incorrect height reference will not remove that error.

Load force × speed gives useful mechanical output power here, not electrical input power. An input-energy reading supplies the latter account without assuming that every input joule lifts the load. Friction and warming may explain a real energy difference; they are not automatically faults in the instruments.

Estimate before calculating: a roughly 1 kg load has weight about 10 N near Earth. Raising it about 1 m in about 1 s at steady speed takes roughly 10 J and needs useful power of order 10 W. An inefficient device needs greater input power. The mass, rise, time and local field are the stated assumptions behind this estimate.