Cyclic processes, heat engines & efficiency

Internal energy is a function of state, so dU is an exact differential and its integral around any closed path vanishes: ∮dU = 0. Heat and work are not state functions — đQ and đW depend on the path — so their loop integrals need not vanish, and the first law ΔU = Q − W applied to one complete cycle gives Qₙₑₜ = Wₙₑₜ. That single line is the reason an engine can run indefinitely: nothing accumulates in the working substance, so whatever heat it does not return to the surroundings must have left as work, cycle after cycle. Declare the sign convention once: Q > 0 into the gas, W > 0 by the gas. A gas that absorbs 1800 J and rejects 1400 J in one cycle has done 400 J of work, and its internal energy at the end of the cycle is exactly what it was at the start. For an ideal gas you can check that directly, because U depends on pV alone — (3/2)pV for a monatomic gas — and the loop lands on the same corner it left.

Along any leg the gas does W = ∫p dV, the area under that leg on the p-V diagram. Around a loop the expansion legs contribute the area under the upper path and the compression legs subtract the area under the lower path, so what survives is the area between them: Wₙₑₜ = ∮p dV = the enclosed area. Its sign is fixed by the sense of traversal. Clockwise — expanding along the high-pressure path and returning along the low-pressure path — gives positive net work: an engine. Anticlockwise gives the same magnitude with a minus sign: the surroundings drive the gas round, and you are looking at a refrigerator, the subject of a later topic. Check the units: 1 kPa × 1 L = 10³ Pa × 10⁻³ m³ = 1 J, which is why a rectangle from 100 kPa to 300 kPa and from 1.0 L to 3.0 L encloses 200 kPa × 2.0 L = 400 J. A curved leg needs the integral — an isotherm contributes nRT ln(V₂/V₁) — but the closure rule does not change: Wₙₑₜ is the area, and the arrows set its sign.

Efficiency needs the heat that was bought, Q_H, so each leg must be audited for its heat and sorted by sign, not by which leg looks hottest. Take the 100-to-300 kPa, 1.0-to-3.0 L rectangle traversed clockwise with a monatomic ideal gas, for which U = (3/2)pV. On an isochoric leg W = 0 and Q = ΔU = (3/2)V Δp; on an isobaric leg Q = ΔU + p ΔV = (5/2)p ΔV. Leg A → B, isochoric at 1.0 L, 100 → 300 kPa: Q = (3/2)(1.0 L)(200 kPa) = +300 J. Leg B → C, isobaric at 300 kPa, 1.0 → 3.0 L: Q = (5/2)(300 kPa)(2.0 L) = +1500 J. Leg C → D, isochoric at 3.0 L, 300 → 100 kPa: Q = (3/2)(3.0 L)(−200 kPa) = −900 J. Leg D → A, isobaric at 100 kPa, 3.0 → 1.0 L: Q = (5/2)(100 kPa)(−2.0 L) = −500 J. Two legs absorb: Q_H = 300 + 1500 = 1800 J. Two reject: Q_C = 900 + 500 = 1400 J. Their difference, 400 J, is the enclosed area, so the audit closes. Miss the 300 J of isochoric heating and you would report Q_H = 1500 J and an efficiency a fifth too high.

η = Wₙₑₜ/Q_H = 1 − Q_C/Q_H. The denominator is Q_H alone, because that is what the fuel pays for; the rejected heat is not a cost you recover, and dividing by Q_H + Q_C or by Q_C answers a question nobody asked. For the rectangle, η = 400/1800 = 0.222: 22% of the heat became work and 78% went out to the cold side. Write the general form as η = Δp ΔV / [(3/2)V₁ Δp + (5/2)p₂ ΔV] and every term is a pressure times a volume, so the ratio is dimensionless as it must be. The limiting cases behave: shrink the pressure swing Δp to zero and Wₙₑₜ → 0 while Q_H → (5/2)p₂ ΔV stays finite, so η → 0 — heat in, all of it out, no work; shrink ΔV and the same happens. η = 1 would need Q_C = 0, a cycle that rejects nothing, which is exactly what the second law forbids. The next topic sets the actual ceiling for reservoirs at the cycle's extreme temperatures; with pV running from 100 J to 900 J here, Tₘₐₓ/Tₘᵢₙ = 9 and that ceiling is 0.889. A rectangle sits far below it, and that gap — not the 22% — is the number worth thinking about.

The two-box picture — heat from a hot reservoir, heat to a cold reservoir — assumes each reservoir's temperature is unchanged by what it gives or takes. That is an idealisation with a checkable condition: a body of heat capacity Cᵣₑₛ that exchanges heat Q shifts by ΔTᵣₑₛ = |Q|/Cᵣₑₛ, and 'reservoir' means this is negligible on the timescale that matters. A 1000 kg water tank (Cᵣₑₛ = 4.18 × 10⁶ J K⁻¹) receiving the rectangle's 1400 J per cycle warms by 1400/(4.18 × 10⁶) = 3.3 × 10⁻⁴ K per cycle — invisible for one cycle, but after 10⁴ cycles it is 3.3 K and the 'cold' side has drifted. A furnace fed with fuel or the open atmosphere qualify because they are replenished, not merely because they are large. The idealisation also hides something about the rectangle itself: the gas absorbs heat while its pV runs from 100 J to 900 J, a ninefold range of temperature, so no single hot reservoir can feed it without a temperature difference across most of the leg. The two-box diagram lumps every absorbing leg into Q_H; the entropy topics later price what that lumping costs.

A cycle's numbers are per revolution; an engine's are per second. With f cycles per second, P = f Wₙₑₜ, the heat input rate is f Q_H and the cooling load is f Q_C. The 400 J rectangle turning over at 25 Hz delivers 10 kW, draws 45 kW of heat and must dump 35 kW into its cold side continuously — which is why the radiator on a real engine is a larger component than 22% suggests, and why the reservoir check has to be done per second, not per cycle: a 1000 kg tank absorbing 35 kW warms at 35 000/(4.18 × 10⁶) = 8.4 × 10⁻³ K s⁻¹, half a kelvin a minute. Efficiency is untouched by f, since it is a ratio of two per-cycle quantities, so a faster engine is a more powerful one, not a more efficient one. Check the limiting case: f → 0 gives P → 0 while η stays at 0.22, the correct statement that a slower engine turns the same fraction of its heat into work, only fewer joules of it per second.