Lesson 5 · 30 min

Power and Efficiency

Lifting a crate \(8\ \text{m}\) takes the same work whether a crane does it in four seconds or in a minute. What differs is the power: the rate of doing work. Motors, engines and pumps are rated by power, and every real machine wastes some of what it takes in, which is its efficiency.

Learning objectives

Define power, \(P = dU/dt = \Fvec\cdot\vvec\), in watts, and tell instantaneous from average power.
Calculate the power needed to hoist, to climb a grade and to overcome resistance at constant speed, and while accelerating.
Use mechanical efficiency, \(\varepsilon = P_\text{out}/P_\text{in}\), including several machines in series.
Read the work done as the area under a power–time graph.

Power

Power is the rate at which a force does work. Since \(dU = \Fvec\cdot d\rvec\) and \(d\rvec/dt = \vvec\):

Power of a force

\[ \colP{P} = \frac{dU}{dt} = \Fvec\cdot\vvec = F v\cos\theta \]

The force and the velocity at the same instant. Average power over an interval is \(P_\text{avg} = U/\Delta t\).

The unit is the watt, \(1\ \text{W} = 1\ \text{J/s} = 1\ \text{N}\cdot\text{m/s}\). Engines are often rated in horsepower: \(1\ \text{hp} = 550\ \text{ft}\cdot\text{lb/s} = 745.7\ \text{W}\). A kilowatt-hour, on an electricity bill, is energy, not power: \(1\ \text{kW}\cdot\text{h} = 3.6\ \text{MJ}\).

A force that pushes a particle at constant speed against resistance delivers \(P = Fv\): more force or more speed, more power. That is why a car needs far more power at highway speed than in town, even on the level.

Efficiency

No machine delivers all the power it takes in: friction in bearings and gears, electrical resistance and air turbulence turn part of it into heat. The mechanical efficiency is

Mechanical efficiency

\[ \varepsilon = \frac{\text{power output}}{\text{power input}} = \frac{P_\text{out}}{P_\text{in}} \lt 1 \]

Over a whole cycle, the same ratio holds for energy: \(\varepsilon = E_\text{out}/E_\text{in}\). For machines in series, \(\varepsilon = \varepsilon_1\,\varepsilon_2\,\varepsilon_3\cdots\).

Example 5.1 — An electric hoist

An electric hoist lifts a \(300\ \text{kg}\) crate on a vertical cable. The hoist, motor and drum together, is \(80\%\) efficient. Find the electrical power it draws (a) while the crate rises at a constant \(2\ \text{m/s}\), and (b) at an instant when the crate rises at \(2\ \text{m/s}\) and is accelerating upward at \(0.5\ \text{m/s}^2\).

Show solution

(a) At constant speed the cable force balances the weight: \(F = mg = 2943\ \text{N}\). The cable force and the velocity both point up:

\[ P_\text{out} = Fv = (2943)(2) = 5886\ \text{W}, \qquad \colP{P_\text{in}} = \frac{5886}{0.80} = 7.36\ \text{kW} \ (9.9\ \text{hp}) \]

(b) Now \(\sum F_y = ma\): \(F - mg = (300)(0.5)\), so \(F = 300(9.81 + 0.5) = 3093\ \text{N}\).

\[ P_\text{out} = (3093)(2) = 6186\ \text{W}, \qquad \colP{P_\text{in}} = \frac{6186}{0.80} = 7.73\ \text{kW} \]

Power is instantaneous: it takes the force at that instant, and while the crate accelerates, Newton's second law supplies that force.

Figure 5.1 The hoist of Example 5.1 on a full lift: it speeds the crate up to \(2\ \text{m/s}\) in the first second, runs steadily, then slows to a stop. The output power \(\colP{P = Fv}\) peaks at the end of the speed-up, when the cable force is largest at full speed; the area under the curve is the work done, here \(mgh\) for a lift of \(h = 8\ \text{m}\). Slide through the lift, or change how quickly it speeds up.

A motor must be sized for the peak, not the average. In Figure 5.1 the steady part needs \(5.9\ \text{kW}\), but the end of the speed-up needs \(7.1\ \text{kW}\) at the shaft, and more still from the supply once the efficiency is included.

Power to move a vehicle

At constant speed on a grade, the driving force at the wheels balances the component of the weight along the road, the rolling resistance of the tires and the air drag. The power at the wheels is that force times the speed, and the engine must supply that divided by the efficiency of the drivetrain.

Figure 5.2 The power a car needs at the wheels to hold a steady speed: climbing \(\colVg{mg\sin\theta\,v}\), rolling resistance \(\colF{C_{rr}\,mg\cos\theta\,v}\) and air drag \(\colF{\tfrac12\rho C_D A\,v^3}\), all added in the amber total. Drag grows with the cube of the speed, so above about \(100\ \text{km/h}\) it takes most of the power on the level. Values: \(C_{rr} = 0.012\), \(C_D A = 0.65\ \text{m}^2\), \(\rho = 1.225\ \text{kg/m}^3\).

Example 5.2 — A car on a grade

A \(1500\ \text{kg}\) car climbs a \(5\%\) grade (a rise of \(5\ \text{m}\) for every \(100\ \text{m}\) horizontally) at a constant \(25\ \text{m/s}\). Rolling resistance and air drag together total \(600\ \text{N}\), and the drivetrain is \(85\%\) efficient. What power must the engine deliver?

Show solution

The grade angle is \(\theta = \tan^{-1}0.05 = 2.862^\circ\), so the weight's component along the road is \(mg\sin\theta = (1500)(9.81)(0.04994) = 734.8\ \text{N}\). At constant speed the driving force balances it and the resistance: \(F = 734.8 + 600 = 1335\ \text{N}\).

\[ P_\text{wheels} = Fv = (1335)(25) = 33.4\ \text{kW}, \qquad \colP{P_\text{engine}} = \frac{33.4}{0.85} = 39.3\ \text{kW} \ (52.6\ \text{hp}) \]

Example 5.3 — Efficiencies in series

An electric motor (\(\varepsilon = 0.92\)) drives a gearbox (\(0.95\)) that drives a pump (\(0.75\)). The pump lifts \(5\ \text{L/s}\) of water through a height of \(20\ \text{m}\). What electrical power does the motor draw?

Show solution

The useful output is the work done on the water per second, \(\dot m\,g\,h\) with \(\dot m = 5\ \text{kg/s}\):

\[ P_\text{out} = (5)(9.81)(20) = 981\ \text{W}, \qquad \varepsilon = (0.92)(0.95)(0.75) = 0.6555, \qquad P_\text{in} = \frac{981}{0.6555} = 1.50\ \text{kW} \]

A third of the electrical power ends up as heat, most of it in the pump.

Power and the principle of work and energy

Divide the principle of work and energy by \(dt\): the total power of all the forces on a particle is the rate of change of its kinetic energy,

\[ \textstyle\sum P = \dfrac{dT}{dt} \qquad\text{so}\qquad T_2 - T_1 = \displaystyle\int_{t_1}^{t_2}\textstyle\sum P\,dt \]

This is how power brings time back into energy methods: if you know the power, you know how fast the kinetic energy grows.

Example 5.4 — Constant-power acceleration

A \(1500\ \text{kg}\) car accelerates from rest on a level road with a constant \(60\ \text{kW}\) delivered at the wheels. Ignoring resistance, how long does it take to reach \(100\ \text{km/h}\)?

Show solution

With constant power, \(T_2 - T_1 = P\,t\). At \(100\ \text{km/h} = 27.78\ \text{m/s}\):

\[ t = \frac{\tfrac12 mv_2^2}{P} = \frac{\tfrac12(1500)(27.78)^2}{60\,000} = 9.65\ \text{s} \]

Real cars cannot use full power at the start: \(F = P/v\) would be enormous at low speed, and the tires can only push as hard as their grip allows. So the first part of a real launch is limited by traction, not by power.

Check your understanding

Key takeaways

\(P = dU/dt = \Fvec\cdot\vvec\), in watts. It uses the force and the velocity at the same instant.
At constant speed, the driving force balances the resistances and \(P = Fv\). While accelerating, find the force from \(\sum\Fvec = m\avec\).
\(\varepsilon = P_\text{out}/P_\text{in} \lt 1\); for machines in series, multiply the efficiencies.
The area under a power–time graph is the work; \(\sum P = dT/dt\).
Next, Lesson 6 stores the work of weights and springs as potential energy.