4.4-4.5 - Efficiency and Sankey diagrams

4.4-4.5 - Efficiency and Sankey diagrams

Many devices transfer some input energy in unwanted ways, often to the surroundings. What counts as useful depends on the intended purpose. Efficiency is the percentage of the total energy output that is useful, and Sankey diagrams give a visual way to compare the useful and wasted pathways.

What efficiency means

A device is useful because it makes a wanted energy transfer. A lamp is useful because some electrical energy is transferred by radiation as light. A motor is useful because some input energy is transferred mechanically to a kinetic energy store. A speaker is useful because some energy is transferred by radiation as sound.

The useful output is the part of the output energy that does the job the device is meant to do. The wasted output is the part that is transferred in other ways. Wasted energy is not destroyed; it has usually spread out to the surroundings and is harder to use.

Efficiency

Efficiency is the percentage of the total energy output that is transferred as useful energy output.

The total energy output is all the output pathways added together:

total energy output=useful energy output+wasted energy output\text{total energy output} = \text{useful energy output} + \text{wasted energy output}

When no energy accumulates in the device, conservation of energy means total input equals total output. If its own stores change, include that change in the energy account. Edexcel states the efficiency relationship using useful energy output and total energy output. So if a question gives the total energy input instead, use that value as the total energy output. You must remember this equation: it is not given in the exam.

Efficiency

efficiency=useful energy outputtotal energy output×100%\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy output}} \times 100\%

An efficiency of 100%100\% would mean every joule of output energy is useful. Efficiency cannot exceed 100%, because the useful output is only part of the total output. If a calculation gives more than 100%100\%, check that the useful output is on top and the total output is underneath. A device with unwanted outputs has a lower efficiency; an ideal heater can be treated as 100% efficient if all its output heats the intended space.

Using the equation

The efficiency equation is a ratio, so the two energy values must be in the same unit. If the useful output is in joules, the total output must also be in joules. If the values are both in kilojoules, the units cancel in the same way. Because the units cancel, efficiency has no unit. Without the ×100%\times 100\%, the same calculation gives efficiency as a decimal, such as 0.180.18; both forms are correct, and a decimal efficiency is never more than 11.

Worked example: a lamp transfers 100J100 J of energy in total. 18J18 J is transferred usefully as light. Calculate the efficiency of the lamp.

efficiency=useful energy outputtotal energy output×100%\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy output}} \times 100\% efficiency=18100×100%=18%\text{efficiency} = \frac{18}{100} \times 100\% = 18\%

The remaining 82J82 J is wasted output, mostly transferred as thermal energy to the surroundings. The calculation uses 18J18 J, not 82J82 J, because efficiency is about the useful part.

If efficiency is given and you need the useful output, change the percentage into a decimal multiplier. For example, 65%=0.6565\% = 0.65.

Worked example: a device has a total energy output of 240J240 J and an efficiency of 65%65\%. Find the useful output.

useful output=0.65×240=156 J\text{useful output} = 0.65 \times 240 = 156 \text{ J}

Then the wasted output is:

240−156=84 J240 - 156 = 84 \text{ J}

The answer is sensible because the useful and wasted outputs add back to the total output: 156J+84J=240J156 J + 84 J = 240 J.

If a motor transfers 140 J usefully and 60 J is wasted, first add the outputs: 140+60=200 J140+60=200\,\mathrm{J}. Efficiency is 140/200×100%=70%140/200\times100\%=70\%. Dividing by 60 J would compare useful with wasted energy, not with the total.

To find the total output from the useful output and the efficiency, divide by the efficiency as a decimal. For example, a motor that is 40%40\% efficient gives 300J300 J of useful output:

total output=useful outputefficiency=3000.40=750 J\text{total output} = \frac{\text{useful output}}{\text{efficiency}} = \frac{300}{0.40} = 750 \text{ J}

Reading Sankey diagrams

A Sankey diagram represents energy transfers using arrows. The arrow on the left usually shows the total input energy. Arrows leaving the device show the output energy pathways.

The important feature is arrow width. A wider arrow represents a larger amount of energy. The length of an arrow is not the quantity being measured.

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