Power and Efficiency

Spec mapping: OCR H556 Module 3.3 — Work, energy and power (definition of power as rate of energy transfer; $P = W/t$ and $P = Fv$ ; efficiency = useful output / total input as a ratio or percentage; Sankey diagrams). Refer to the official OCR H556 specification document for exact wording.

Work and energy tell you how much energy is transferred; power tells you how fast. A marathon runner and a sprinter can do the same total work by climbing a single flight of stairs, but the sprinter does it in a tenth of the time — so the sprinter's power output is ten times higher. This rate-of-energy-transfer is the central quantity in engineering, environmental physics and everyday electrical billing. Efficiency, meanwhile, captures how much of the input energy actually becomes useful output, and how much is dissipated as heat, sound or friction. The second law of thermodynamics imposes hard upper limits on efficiency; no device, however clever, can exceed $100\%$ , and most fall well short.

Key Definition: The power of a device is the rate at which it transfers energy: $P = W/t = \Delta E / \Delta t$ . SI unit: watt (W), equal to one joule per second. The efficiency of a device is the dimensionless ratio (or percentage) of useful energy output to total energy input.

Definition of Power

For a process in which work $W$ is done in time $t$ , the average power is:

$\overline{P} = \frac{W}{t} = \frac{\Delta E}{\Delta t}$

The instantaneous power at a given moment is the time-derivative:

$P = \frac{dW}{dt} = \frac{dE}{dt}$

For steady-state processes (motor running at constant load, kettle boiling water at constant temperature) average and instantaneous power are equal. For accelerating systems (rocket lift-off, lift starting from rest, car launching from a traffic light) they differ — instantaneous power varies through the motion.

SI unit: watt (W), defined as $1\,\text{W} = 1\,\text{J s}^{-1}$ . The unit honours James Watt, the engineer whose improvements to the steam engine triggered the industrial revolution.

Scale	Typical power
Human basal metabolism (at rest)	~ 80 W
Bright incandescent light bulb	100 W
Person cycling hard, sustained	300 W
Domestic kettle	2 kW
Sports car, peak	150 kW
Large wind turbine	2–3 MW
Nuclear reactor unit	1–3 GW
Total human civilisation	$\sim 1.9 \times 10^{13}$ W
Sun's total luminosity	$3.83 \times 10^{26}$ W

Two non-SI units appear regularly in UK contexts:

Horsepower (hp): $1\,\text{hp} \approx 746\,\text{W}$ (imperial mechanical horsepower). Car-engine peak output is still routinely quoted in hp.
Kilowatt-hour (kWh): a unit of energy, not power. $1\,\text{kWh} = 1\,\text{kW} \times 1\,\text{h} = 1000 \times 3600 = 3.6 \times 10^6$ J $= 3.6$ MJ. UK domestic electricity meters bill in kWh.

Common Exam Mistake: Confusing kWh (energy) with kW (power). A $2$ kW kettle run for $30$ minutes consumes $2 \times 0.5 = 1$ kWh of energy. The kettle's power is constant; the energy used scales with time.

Power in Terms of Force and Velocity

For a constant force $F$ acting along the direction of motion, over time $t$ the body moves $s = vt$ . Work done is $W = Fs = Fvt$ , so:

$P = \frac{W}{t} = \frac{F v t}{t} = F v$

So:

$\boxed{P = F\,v}$

This is the most useful form for motors, engines and pumps maintaining constant speed against a resistive force. It generalises (with $\vec{F}\cdot\vec{v}$ ) to non-aligned force and velocity.

Two notes on interpretation:

For a constant-power engine (the typical idealisation of an electric motor running on a fixed supply), $P = Fv$ means that the driving force $F = P/v$ decreases as the vehicle speeds up. This is why a car has its maximum acceleration at low speeds and rapidly diminishing acceleration as it approaches its top speed.
For a constant-force engine (e.g. a steady cable tension), $P = Fv$ means the power grows with speed. This is unusual mechanically but is the standard assumption for SUVAT-style problems.

Worked Example 1 — Car at Constant Speed

A car travels at a steady $30$ m s $^{-1}$ along a level road against a total resistive force of $700$ N. Find the useful power output of the engine.

At constant velocity, Newton I requires the driving force to balance the resistive force: $F_{\text{drive}} = 700$ N.

$P = F v = 700 \times 30 = 21\,000\;\text{W} = 21\;\text{kW}$

This is the useful mechanical power delivered at the wheels — not the fuel power consumed. A petrol engine is roughly $25\%$ efficient, so the fuel input power is about $21/0.25 = 84$ kW, the remaining $\sim 63$ kW being dumped to the radiator and exhaust as heat.

Worked Example 2 — Cyclist Climbing a Hill

A cyclist plus bike has total mass $75$ kg and rides up a $5°$ slope at a steady $5.0$ m s $^{-1}$ . Ignore drag. Find the mechanical power output.

Component of weight along the slope: $mg\sin\theta = 75 \times 9.81 \times \sin 5° = 735.8 \times 0.0872 = 64.1$ N. To climb at constant speed the cyclist's driving force must equal this.

$P = F v = 64.1 \times 5.0 = 321\;\text{W}$

This is sustainable for several minutes by a fit amateur; Tour de France riders maintain $\sim 400$ W for hours and can hit $\sim 1000$ W in short sprints. The world-record one-hour cycling distance (Wout van Aert and others) corresponds to an output around $440$ – $460$ W sustained for the full hour.

Worked Example 3 — Lift Motor (Constant Velocity vs Accelerating)

A lift of total mass $800$ kg ascends a shaft. Ignore friction in the cables and pulley.

(a) Steady ascent at $2.5$ m s $^{-1}$ : at constant velocity the cable tension equals the weight: $T = mg = 800 \times 9.81 = 7\,848$ N.

$P = T v = 7\,848 \times 2.5 = 19\,620\;\text{W} \approx 19.6\;\text{kW}$

(b) Lift now accelerating upward at $1.5$ m s $^{-2}$ . At the instant when $v = 2.0$ m s $^{-1}$ , find the power.

Net upward force required: $F_{\text{net}} = ma = 800 \times 1.5 = 1\,200$ N. Cable tension: $T = mg + ma = 7\,848 + 1\,200 = 9\,048$ N. Power: $P = T v = 9\,048 \times 2.0 = 18\,100$ W $\approx 18.1$ kW.

Note (b) is less than (a) instantaneously, because the lift is moving more slowly even though the cable tension is higher. As the lift accelerates and $v$ rises, the power required will increase rapidly — and once cruising at the same $2.5$ m s $^{-1}$ but no longer accelerating, the tension drops back to $7\,848$ N and power back to $19.6$ kW. Power is a time-rate quantity and depends explicitly on the instantaneous velocity.

Worked Example 4 — Pump Lifting Water

A pump raises $15$ kg of water per second from a well to a tank $8.0$ m above. Find the useful power output of the pump.

The water gains GPE at rate (mass-per-second) × $g$ × $h$ :

$P_{\text{useful}} = \dot{m}\,g\,h = 15 \times 9.81 \times 8.0 = 1\,177\;\text{W} \approx 1.18\;\text{kW}$

If the pump's electrical input is $1.5$ kW, then its efficiency is $1\,177 / 1\,500 = 78.5\%$ . The remaining $21.5\%$ is lost as heat in motor windings, friction in bearings and turbulent dissipation in the pipework.

Efficiency

No real energy-transfer process is perfectly efficient. Some fraction of the input always ends up in unwanted forms — typically heat (always) and sometimes sound, vibration or radiation. Efficiency is defined as:

$\eta = \frac{\text{useful energy output}}{\text{total energy input}}$

equivalently, for a steady-state device, as the ratio of useful output power to total input power:

$\eta = \frac{P_{\text{useful}}}{P_{\text{input}}}$

Often quoted as a percentage:

$\eta\,(\%) = \frac{P_{\text{useful}}}{P_{\text{input}}} \times 100$

Efficiency is dimensionless and always satisfies $0 \leq \eta \leq 1$ (or $0\% \leq \eta \leq 100\%$ ). The upper limit $\eta = 1$ is forbidden for any heat engine by the second law of thermodynamics; the maximum possible efficiency for a heat engine operating between hot reservoir $T_H$ and cold reservoir $T_C$ is the Carnot efficiency $\eta_{\text{Carnot}} = 1 - T_C/T_H$ , attained only by an idealised reversible cycle. Real engines fall short.

Typical Efficiencies

Device	Efficiency
Incandescent light bulb (tungsten filament)	$\sim 5\%$ (95% wasted as heat)
Compact fluorescent (CFL)	15–20%
LED bulb	30–40%
Petrol internal-combustion engine	25–30%
Diesel engine	35–40%
Coal-fired steam power station	35–40%
Combined-cycle gas turbine (CCGT)	$\sim 60\%$
Industrial electric motor	85–95%
Hydroelectric turbine	up to 90%
Silicon photovoltaic panel (commercial)	15–22%
Human muscle (cycling)	$\sim 25\%$

Values are typical, not exact, and depend strongly on operating conditions. The pattern is unmistakable: electrical devices (motors, generators) are highly efficient because the transformation is direct; thermal devices (engines, power stations) are limited by Carnot.

Worked Example 5 — Crane Efficiency

A crane motor draws $15$ kW of electrical power. It lifts a $600$ kg load vertically at a steady $2.0$ m s $^{-1}$ . Find the efficiency.

$P_{\text{useful}} = F v = m g v = 600 \times 9.81 \times 2.0 = 11\,772\;\text{W}$

$\eta = \frac{P_{\text{useful}}}{P_{\text{input}}} = \frac{11\,772}{15\,000} = 0.785 = 78.5\%$

So $21.5\%$ of the electrical input is lost — primarily as heat in motor windings (Joule heating $I^2R$ ), friction in the cable drum and gearbox, and a small amount of acoustic energy.

Worked Example 6 — Power Station Efficiency

A gas-fired CCGT power station burns natural gas at a rate that releases chemical energy at $1.2 \times 10^9$ J s $^{-1}$ ( $1.2$ GW). The useful electrical output is $720$ MW. Find the efficiency.

$\eta = \frac{720 \times 10^6}{1.2 \times 10^9} = 0.60 = 60\%$

Modern CCGT plants achieve $\sim 60\%$ efficiency — close to the theoretical limit set by Carnot for the gas-turbine operating temperatures ( $\sim 1500$ °C at the turbine inlet, $\sim 25$ °C at the condenser). Older coal-fired plants achieve $35$ – $40\%$ . The $40\%$ "lost" leaves the plant as waste heat to the cooling water, and is one driver of the move toward combined heat-and-power (CHP) plants that use that low-grade heat for district heating.

Sankey Diagrams — Visualising Energy Flow

A Sankey diagram depicts energy flow through a device with arrows whose widths are proportional to the energy carried. Useful output emerges in one arrow; each form of waste in another. OCR exam questions frequently give a Sankey diagram (or partial Sankey) and ask candidates to calculate the missing quantity.

flowchart LR
  A[Chemical fuel energy: 100 J] --> B[Useful mechanical work: 25 J]
  A --> C[Exhaust heat: 50 J]
  A --> D[Engine-block heat: 15 J]
  A --> E[Friction and acoustic: 10 J]

For the petrol engine sketched: efficiency = $25/100 = 25\%$ . Of the $75$ J wasted, $50$ J ( $67\%$ of waste) is exhaust heat — which is why cars need radiators and exhaust pipes to dump it. The energy bookkeeping in a Sankey must balance: useful + waste = input, and Sankey arrows summed (by width) must equal the input arrow.

Worked Example 7 — Reading a Sankey Diagram

A petrol engine delivers $25$ J of useful work for every $100$ J of fuel chemical energy. $50$ J is lost as exhaust heat, $15$ J as engine-block heat, $10$ J as friction (final losses, ending up as heat).

(a) Efficiency: $25/100 = 25\%$ .

(b) Total waste energy: $75$ J. Breakdown by fraction of waste: exhaust $50/75 = 67\%$ , block $15/75 = 20\%$ , friction $10/75 = 13\%$ .

(c) If the engine produces $25$ kW of useful output, then the fuel input must be $25/0.25 = 100$ kW. At a fuel calorific value of $\sim 44$ MJ kg $^{-1}$ , the petrol consumption is $100\,000 / 44 \times 10^6 = 2.27 \times 10^{-3}$ kg s $^{-1}$ = $8.2$ kg h $^{-1}$ $\approx 11.0$ L h $^{-1}$ (density $\sim 0.75$ kg L $^{-1}$ ). At $70$ mph ( $112$ km h $^{-1}$ ) this is about $9.8$ L per $100$ km — typical motorway consumption.

Power and Efficiency

Power and Efficiency

Definition of Power

Power in Terms of Force and Velocity

Worked Example 1 — Car at Constant Speed

Worked Example 2 — Cyclist Climbing a Hill

Worked Example 3 — Lift Motor (Constant Velocity vs Accelerating)

Worked Example 4 — Pump Lifting Water

Efficiency

Typical Efficiencies

Worked Example 5 — Crane Efficiency

Worked Example 6 — Power Station Efficiency

Sankey Diagrams — Visualising Energy Flow

Worked Example 7 — Reading a Sankey Diagram

The Kilowatt-Hour and Electricity Bills

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