Efficiency (\(\eta\)) measures how well a device or system converts the energy (or power) supplied to it into useful output. It is expressed as a percentage.
Energy form
\[
\eta (\%) = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%
\]
Power form
\[
\eta (\%) = \frac{\text{useful power output}}{\text{total power input}} \times 100\%
\]
Because power is the rate of energy transfer \((P = \Delta E/\Delta t)\), the two equations give identical results.
| Resource / Device | Typical efficiency (% of input energy that becomes useful output) | Units (if applicable) | Principal reasons for the range |
|---|---|---|---|
| Coal‑fired power station (steam‑turbine) | 30 – 38 % | – | Large heat loss in boiler, flue gases and condenser; limited by Carnot cycle. |
| Natural‑gas combined‑cycle plant | 55 – 60 % | – | Gas turbine (high‑\(T\)) + steam turbine recovers waste heat; higher \(T_{\text{hot}}\) raises Carnot limit. |
| Nuclear power station (steam‑turbine) | 30 – 35 % | – | Thermodynamic cycle similar to coal; heat‑rejection dominates losses. |
| Hydroelectric dam (turbine‑generator) | 80 – 90 % | – | Low mechanical friction; most gravitational potential energy is converted to electricity. |
| Wind turbine (rotor‑generator) | 30 – 45 % | – | Betz limit (59 % theoretical max) plus gearbox, aerodynamic and electrical losses. |
| Solar photovoltaic (PV) panel | 10 – 25 % (commercial); >30 % (research cells) | – | Band‑gap and carrier‑recombination limits; temperature and shading reduce output. |
| Bio‑fuel (ethanol, biodiesel) in an internal‑combustion engine | 20 – 30 % | – | Similar thermodynamic cycle to gasoline; fuel quality and combustion efficiency affect range. |
| Petrol / diesel car (internal‑combustion engine) | 20 – 30 % | – | Most energy lost as heat through exhaust and cooling system. |
| Electric motor (appliances, EVs) | 70 – 95 % | – | DC brushed ≈ 70‑80 %; brush‑less AC or permanent‑magnet motors reach 90‑95 %. |
When drawing a Sankey diagram for a power plant, the width of each arrow represents the amount of energy. The efficiency is the ratio of the width of the “useful output” arrow to the total input width. This visual tool is explicitly required in the syllabus for assessing energy‑resource efficiency.
A 6 kW electric heater raises the temperature of water. In 50 s it delivers 250 kJ of heat to the water.
\[
\eta = \frac{E{\text{useful}}}{E{\text{input}}}\times100
=\frac{250\ \text{kJ}}{(6\ \text{kW}\times50\ \text{s})}\times100
=\frac{250}{300}\times100 = 83\%
\]
\[
P{\text{useful}} = \frac{E{\text{useful}}}{t}= \frac{250\ \text{kJ}}{50\ \text{s}} = 5\ \text{kW}
\]
\[
\eta = \frac{P{\text{useful}}}{P{\text{input}}}\times100
=\frac{5\ \text{kW}}{6\ \text{kW}}\times100 = 83\%
\]
Both approaches give the same result, confirming the equivalence of the two formulas.
Input chemical energy from coal: \(1.0\times10^{9}\ \text{J}\).
Electrical energy generated: \(3.2\times10^{8}\ \text{J}\).
\[
\eta = \frac{3.2\times10^{8}}{1.0\times10^{9}}\times100 = 32\%
\]
Thus 68 % of the coal’s energy is lost as heat in the boiler, flue gases and cooling water – a major source of thermal pollution.
\[
\eta = \frac{12}{60}\times100 = 20\%
\]
Comment on why the efficiency is low (most energy emitted as infrared heat).
| Stage | Energy (×10⁹ J) |
|---|---|
| Potential energy of water (input) | 5.0 |
| Mechanical energy at turbine shaft | 4.5 |
| Electrical energy generated | 4.2 |
| Transmission losses | 0.2 (lost) |
Solution: \(\eta = \dfrac{4.2}{5.0}\times100 = 84\%\). High efficiency reduces fuel‑equivalent cost and limits environmental impact.
Solution sketch (students should draw a bar chart):
Gasoline car: useful work = 100 km × energy per km; primary input = useful work / 0.25.
Electric car: useful work / 0.90 (motor) then divided by 0.60 (generation) → overall η = 0.54. Hence the electric car needs roughly half the primary energy of the gasoline car.
\[
\eta = \frac{2.5}{10}\times100 = 25\%
\]
Compare with a typical diesel engine (≈ 30 %) and comment on why the difference is small.
\[
\eta = \frac{m c \Delta T}{V I t}\times100
\]
Efficiency is a percentage that tells us how much of the supplied energy (or power) is turned into useful output. The two formulas – based on energy or on power – are mathematically identical. Knowing typical efficiency ranges, the physical reasons behind them, and being able to calculate and interpret efficiency are essential skills for the IGCSE 0625 exam and for evaluating real‑world energy systems.
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