CourseWiki/Engineering Fundamentals/Force and Power

Power and Efficiency

Power is work per unit time - three equivalent ways to calculate it, the horsepower conversions, and what an efficiency figure can and cannot be.

Power is how fast work gets done

Two hoists that lift the same load to the same height do the same work. The one that does it in half the time needs twice the power.

P = W / t — average power equals work divided by elapsed time.

The two power units and the factors between them.
System	Unit	Symbol	Definition and conversion
SI	watt	`W`	`1 W = 1 J/s`; `1 W = 0.001341022 hp`
Foot–slug–second	horsepower	`hp`	`1 hp = 550 ft·lbf/s`; `1 hp = 745.6999 W`

The symbol W does double duty: upright W after a number is the watt, while W standing in a formula for a quantity is usually work. Context separates them: P = W/t uses W for work, while 1 W = 1 J/s uses it for the watt.

Worked example: powering the lift

On Work and Energy, lifting 2,000 lbf through 10 ft took 20,000 ft·lbf, or 27.1 kJ. How much power does doing it in 4 s require?

In U.S. units: 20,000 ft·lbf ÷ 4 s = 5,000 ft·lbf/s, and 5,000 ÷ 550 = 9.09 hp.
In SI: 27,116 J ÷ 4 s = 6,779 W, or 6.78 kW.
Cross-check the two: 9.0909 hp × 745.6999 W/hp = 6,779 W. They agree, which is the point of doing both.

That is the power delivered to the load. A motor sized for this job needs more than that, because no drive is perfectly efficient.

The same power, three ways to reach it; Watch out

The same power, three ways to reach it

Which form you use depends on what you have been given, not on which is more correct. All three produce watts in SI. In P = Fv, force and velocity must be parallel; otherwise use the force component along the velocity.

Three routes to a power, in each system.
Form	SI	Foot–pound
Work over time `P = W/t`	`J/s = W`	`(ft·lbf/s) ÷ 550 = hp`
Force times speed `P = Fv`	`N × (m/s) = W`	`(lbf × ft/s) ÷ 550 = hp`
Torque times angular velocity `P = τω`	`(N·m) × (rad/s) = W`	`(lbf·ft × rad/s) ÷ 550 = hp`

In the third form the angular velocity must be in rad/s. An rpm figure substituted directly is wrong by a factor of about 9.55. Convert it first — Angular Motion.

The divisor 550 is not a fudge factor. One horsepower is defined as 550 pound-force feet of work per second, so dividing by 550 is simply converting the unit.

Two shortcut formulas, and the conditions attached to them

These two save a lot of arithmetic in practice. Both are unit conversions with an assumption hidden inside.

From volts and amps

hp = (volts × amps) ÷ 746

The 746 is exact: NIST defines a separate unit, horsepower (electric), as 746 W. So the formula is really two steps — find the electrical power in watts, then convert watts to horsepower.

The hidden condition is in the first step. Volts times amps gives real power in watts for a direct-current circuit or a purely resistive load. For single-phase AC, RMS volts times RMS amps gives apparent power, in volt-amperes; the real power in watts is that multiplied by the power factor, the ratio of real power to apparent power. Use the shortcut on an AC motor without its power factor and the answer comes out high. See Current, Voltage, and Resistance.

From torque and rpm

hp = (torque in lbf·in × speed in rpm) ÷ 63,025

This one is the P = τω row above, with the rpm-to-rad/s conversion and the inch-to-foot conversion already folded into the constant. It only works with the torque in pound-force inches and the speed in rpm; feed it lbf·ft and the answer is twelve times too small.

Where 63,025 comes from

One horsepower is 550 ft·lbf/s, which is 550 × 12 = 6,600 lbf·in per second. One revolution per minute is 2π/60 rad/s. Putting a torque τ in lbf·in and a speed N in rpm into P = τω gives τN(2π/60) lbf·in per second, and dividing by 6,600 converts that to horsepower:

hp = τN × (2π/60) ÷ 6,600 = τN ÷ (6,600 × 60 ÷ 2π) = τN ÷ 63,025

Every constant in the shortcut is a unit conversion. Nothing about the physics is buried in it, which is why the same relationship in coherent SI units needs no constant at all.

Efficiency is the fraction that makes it through; Watch out

Efficiency is the fraction that makes it through

The efficiency of a machine is the ratio of useful output to total input. Its symbol is η (eta), and it can be written with either power or work as long as both sides of the ratio use the same one:

η = P_out / P_in · η = W_out / W_in

It is a ratio of two like quantities, so the units cancel: efficiency is a plain number, usually quoted as a percentage.

For a real machine in steady operation, η < 1; the ideal lossless limit is 1. You cannot get more power out than you put in, and some of every input always leaves as heat, noise, or friction rather than as useful output. A calculated efficiency above 1 means an error in the numbers, not a remarkable machine.

Worked example: an electric hoist motor

A motor consumes 500 W and applies 1,000 N of force to lift an object at 0.45 m/s. What is the efficiency of the system?

Useful output, using P = Fv: 1,000 N × 0.45 m/s = 450 W.
η = 450 W ÷ 500 W = 0.90, or 90%.

The missing 50 W has not vanished. It is power lost rather than delivered to the load, typically as heat in the motor and drive.

Worked example: follow power through two stages

A hypothetical steady system receives 100 W. Stage 1 is given as 80% efficient, and Stage 2 is given as 90% efficient. Each percentage applies to the power entering that stage.

A 100 W input passes through an 80 percent stage, leaving 80 W useful and 20 W lost; that 80 W passes through a 90 percent stage, leaving 72 W useful and 8 W lost. — Useful power continues across the top. Each loss branches at the boundary where it occurs.

Stage 1 output: 100 W × 0.80 = 80 W. Stage 1 loss: 100 W − 80 W = 20 W.
Stage 2 receives 80 W, not the original 100 W. Its output is 80 W × 0.90 = 72 W. Its loss is 80 W − 72 W = 8 W.
Overall efficiency: 72 W / 100 W = 0.72, or 72%. Equivalently, 0.80 × 0.90 = 0.72.
Total accounted loss: 20 W + 8 W = 28 W. Check the boundary: 72 W useful + 28 W loss = 100 W input.

Do not add 20% + 10% and call the system 70% efficient. The 20% loss is measured from 100 W, while the 10% loss is measured from the 80 W that reaches Stage 2. Watts may be added here because 20 W and 8 W are amounts of power accounted across the same overall boundary.

Work backward when the required output is known: Stage 2 needs 72 W / 0.90 = 80 W at its input, and Stage 1 needs 80 W / 0.80 = 100 W. The same result comes from 72 W / 0.72 = 100 W. Use Gear Ratios to explore the same multiplication idea for stated per-mesh losses.

These are given, illustrative constant efficiencies at steady operation with no changing stored energy. They are not a motor performance prediction. The accounted losses leave the useful-flow path rather than disappearing; this example does not predict their temperature or physical mechanism.

Efficiency figures are only comparable when the boundary is the same. A motor efficiency, a drive efficiency, and a system efficiency measured from the wall socket to the load are three different numbers for the same machine, and the largest of them is the one a datasheet prefers to print. Say which one you mean.

Sources

Where these values were checked

NIST SP 811, Appendix B lists horsepower (550 ft·lbf/s) to watt as 7.456 999 E+02 and horsepower (electric) to watt as 7.46 E+02 — two separately defined units, which is why 745.6999 and 746 both appear above. The 0.001341022 hp in one watt is the reciprocal of the first. Foot pound-force to joule is 1.355 818 E+00 in the same table.

NIST SP 811, Chapter 4, Table 3, gives the watt as W, expressed as J/s and in base units as m²·kg·s⁻³.

Power factor is defined by Fluke as “the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).” The 63,025 constant is derived above from the horsepower definition and the rpm-to-rad/s conversion, not quoted.

Appears in

DET 403 DET Work, Power, and Efficiency