Heat engine efficiency and the environment

Can an engine be improved so that all the heat received in every cycle becomes work? Follow the energy from the hot source to the shaft and surroundings.

For gas to do work repeatedly, it must receive energy and return to its initial state. In the heat engine model, a hot reservoir supplies heat Q₁ to the working substance, such as gas. The working substance does work and transfers some energy Q₂ to a colder body, the cold reservoir. This name describes its role in heat exchange: it may be air or water, rather than a household refrigerator.

Choose one complete cycle. The working substance returns to its initial state, so its final internal energy equals its initial value. The heat received therefore does not remain as extra stored energy: Q₁ = A + Q₂. Here Q₁ and Q₂ are the positive magnitudes of heat received and rejected, and A is the net work over a cycle. Increase Q₂ at unchanged Q₁ and less energy remains for work. This checks energy conservation.

The energy balance alone would allow Q₂ = 0 and A = Q₁. But observations of thermal processes impose another restriction: a cyclic engine cannot have as its sole result the complete conversion of heat from one hot reservoir into work. It must reject energy to a colder environment. This is not caused by poorly designed gears: eliminating friction would still not give such an engine 100% efficiency. Real friction and heat losses reduce the output further.

Compare two engines receiving the same Q₁: the one producing more work A is more efficient. Thus efficiency is η = A/Q₁. Use energies for the same cycle in matching units; the units cancel, and multiply the ratio by 100% for a percentage. At fixed Q₁, doubling A doubles η; if A = 0, efficiency is zero. The balance also gives η = 1 − Q₂/Q₁: a larger fraction sent to the cold reservoir leaves a smaller fraction for work.

Where should the engine boundary be drawn? Fuel may release qm, but some energy can bypass the working substance during heating, making the Q₁ it receives less than qm. Engine efficiency based on heat received by the working substance is A/Q₁; efficiency of the whole installation based on fuel is A/(qm). Both ratios are meaningful but answer different questions. Do not substitute one denominator for the other: identify the energy input first, then compare systems with the same definition.

Imagine two installations lifting the same load through the same height. The useful work is equal. From the definition of efficiency, input energy equals useful work divided by efficiency. Higher efficiency needs less input energy and, for the same fuel, usually less fuel mass. Unused energy does not disappear: it enters the surroundings. Saving energy therefore means using less input to achieve the same result.

Distinguish two effects of a heat installation. Heat rejection warms the surroundings, while fuel combustion produces reaction products whose composition depends on the fuel and process. Higher efficiency reduces the input needed for a given work output; emission treatment changes what enters the air; choosing a different source changes how energy is obtained. One improvement does not automatically remove every other impact.

An engine receives 500 J and rejects 350 J to the cold reservoir per cycle. Neglect other losses. Find the work and efficiency.

Over a cycle, the working substance's internal energy does not change. Of the 500 J received, 350 J goes to the cold reservoir, so A = 500 − 350 = 150 Дж.

Find the fraction of incoming heat converted into work: η = 150/500 = 0,30 = 30%.

The remaining fraction is 350/500 = 70%. Together, 30% and 70% give 100%, so energy balances.

Work is 150 J and efficiency is 30%. In this operating mode, of every 100 J of heat received, 30 J becomes work and 70 J is transferred to the cold reservoir.

Over a cycle, the working substance stores no net energy: Q₁ is divided between work and heat rejection Q₂.

Reducing friction alone cannot remove the need for a cold reservoir.

Efficiency is meaningful only when the input energy is clearly specified and the calculation boundary is consistent.

Next we begin electricity: why objects acquire charge and interact at a distance.