Heat Engines & Thermal Efficiency

A single expansion can turn heat into work with nothing left over: let an ideal gas expand isothermally and it absorbs heat Q and does work W = Q, every joule of it. The trouble is that it finishes hot, large and at low pressure, and cannot do it again. An engine has to come back. So the working substance is carried around a closed path and returned to the pressure, volume and temperature it started from, ready for the next stroke. Internal energy is a function of state, so returning the state returns U: ΔU_cycle = 0 exactly, whatever route was taken and however violent that route was. Feed this into the first law ΔU = Q − W, with W the work done by the substance, and the books close on their own — over one loop Wₙₑₜ = Qₙₑₜ, with nothing stored and nothing borrowed.

A reservoir is a body large enough that exchanging heat with it does not measurably change its temperature: a furnace, a boiler, the outside air, a river. An engine touches two of them. Call Q_H the heat the working substance absorbs from the hot one and Q_C the heat it exchanges with the cold one. With the usual convention that heat into the substance counts positive, Q_C comes out negative, which is why the magnitude bars appear everywhere. Substituting into Wₙₑₜ = Qₙₑₜ gives W = Q_H − |Q_C|, and with it the energy-flow picture: a stream Q_H arriving, a branch W leaving sideways as work, and |Q_C| continuing to the cold side. Nearly every engine problem in this section is that diagram with two of the three numbers known.

Work done by a gas is ∫p dV, so an expanding branch contributes positive work and a compressing branch negative. Around a closed loop the two branches run between the same pair of volumes, and what survives is W = ∮p dV, the area enclosed. Sense matters. An engine traverses its loop clockwise on a p-V diagram, expanding along the high-pressure branch and being pushed back along the low-pressure one, so the outward work beats the return work and the enclosed area is positive. Run the identical loop counterclockwise and W is negative: work must be supplied, which is a different machine. A rectangular cycle makes the area trivial — expand at 300 kPa from 1.0 L to 4.0 L and return at 100 kPa, and W = (200 kPa)(3.0 L) = 600 J, because 1 kPa × 1 L = 1 J. Tall and thin or long and flat, both enclose little; the work lives in the product.

Thermal efficiency is defined as e = W/Q_H: net work delivered divided by the heat the working substance absorbed. The denominator is deliberately not the net heat and not the fuel's chemical energy — it is the heat that actually entered the substance from the hot side, because that is what a designer is buying. Using W = Q_H − |Q_C| gives the equivalent form e = 1 − |Q_C|/Q_H, which says what the first form hides: the only route to a higher efficiency is a smaller exhaust relative to the intake. For the 600 J rectangle above the gas absorbs 2550 J, counted branch by branch in the next section, so e = 600/2550 = 0.235. That is unremarkable company: a petrol engine reaches roughly 25–35%, a large marine diesel about 50%, a modern steam plant near 40%.

Q_H is not the net heat; it is the sum of only those branches where heat flows in, so the loop must be taken apart before efficiency can be computed. Take the same rectangle with a monatomic ideal gas at p₁ = 100 kPa, p₂ = 300 kPa, V₁ = 1.0 L, V₂ = 4.0 L, and use nRΔT = Δ(pV). The isochoric heating does no work, so its heat is ΔU = (3/2)nRΔT = (3/2)V₁Δp = 1.5 × 1.0 × 200 = 300 J in. The isobaric expansion takes Q = nCₚΔT = (5/2)p₂ΔV = 2.5 × 300 × 3.0 = 2250 J in. Those two branches are the intake: Q_H = 2550 J. The cooling branch rejects (3/2)V₂Δp = 1200 J and the compression rejects (5/2)p₁ΔV = 750 J, so |Q_C| = 1950 J. The audit closes: 2550 − 1950 = 600 J, the area already read off the diagram. Netting the four heats first returns that same 600 J and tells you nothing about the denominator.

Two different things get quoted about engines and they must not be mixed: efficiency, a pure ratio fixed by the cycle, and power, which also depends on how fast the cycle repeats. Convert with P = W f. A cycle delivering 500 J at 24.0 cycles per second is 12.0 kW; the same cycle at 60.0 cycles per second is 30.0 kW at identical efficiency. An audit then has three steps. Check the balance first: does the heat supplied equal the work out plus the heat rejected? Then check e = W/Q_H against 1, since a claim of e > 1 creates energy and dies on the first law. Then check which denominator the claimant used, because work quoted against the absorbed heat, against the fuel's chemical energy, or against the net heat gives three different numbers for one engine, and only the first is thermal efficiency. A claim surviving all three still has the second law to face.