Carnot cycle, reversibility and thermodynamic temperature scale

Why it matters

The Carnot cycle tells you the best any engine, refrigerator or heat pump could possibly do between two temperatures. That limit is used every day to check whether a design claim is believable, to see why gas-turbine designers push turbine inlet temperatures higher, and to rate how far a real machine is from ideal. It also defines temperature itself, independent of any thermometer fluid.

Key ideas

Reversible process. A process that can be reversed so that both the system and the surroundings return exactly to their initial states. It must be quasi-static and free of dissipation. All real processes are irreversible; reversible ones are the limiting ideal.

Sources of irreversibility. Friction; heat transfer across a finite temperature difference; unrestrained (free) expansion; mixing of different substances or states; electrical resistance; inelastic deformation; chemical reaction and combustion. Internally reversible means no irreversibility inside the system boundary; externally reversible means none in the surroundings (heat exchanged with a reservoir at the system's own temperature).

Carnot cycle (reversible heat engine, e.g. a gas in a cylinder):

1. 1→2 Reversible isothermal expansion at T_H, absorbing Q_H from the source.
2. 2→3 Reversible adiabatic (isentropic) expansion, temperature falling from T_H to T_L.
3. 3→4 Reversible isothermal compression at T_L, rejecting Q_L to the sink.
4. 4→1 Reversible adiabatic compression back to T_H.

On a T–s diagram it is a rectangle; on a p–v diagram it is bounded by two isotherms and two steeper adiabats. Run backwards it becomes the Carnot refrigerator or heat pump.

Carnot principles.

• No heat engine operating between two reservoirs can be more efficient than a reversible engine between the same reservoirs.
• All reversible engines operating between the same two reservoirs have the same efficiency, regardless of working fluid or mechanism.

Both are proved by showing that violating them would violate the Kelvin–Planck statement.

Thermodynamic (Kelvin) temperature scale. Because the reversible efficiency depends only on the reservoir temperatures, Kelvin defined temperature through Q_H/Q_L = T_H/T_L for a reversible engine. This scale is independent of any substance and coincides with the ideal-gas scale. It is fixed by assigning 273.16 K to the triple point of water.

Why Carnot is not built. Isothermal heat transfer needs infinitesimal temperature differences, so it would be infinitely slow (zero power). With a gas, the cycle's net work is small compared with the gross work in the four strokes, so friction would wipe it out. Practical cycles (Otto, Diesel, Brayton, Rankine) approximate it in different ways.

Two Carnot engines in series (first rejects to an intermediate reservoir at T_m, second takes from it):

• Equal work: T_m = (T_H + T_L)/2
• Equal efficiency: T_m = √(T_H·T_L)

Formulas

η_Carnot = 1 − T_L / T_H Q_H / Q_L = T_H / T_L (reversible devices only) COP_R,Carnot = T_L / (T_H − T_L) COP_HP,Carnot = T_H / (T_H − T_L)

• T_H, T_L: absolute temperatures of the hot and cold reservoirs, K
• Q_H, Q_L: magnitudes of heat exchanged with these reservoirs, kJ or kW
• η: thermal efficiency (fraction); COP: dimensionless

Use these as the upper limit for any real device; a claimed η or COP above them is impossible.

Worked examples

Example 1 (standard). (a) An inventor claims an engine that receives 1000 kJ at 1000 K, rejects heat at 300 K and produces 750 kJ of work. Is it possible? (b) A Carnot heat pump keeps a building at 295 K when it is 270 K outside, delivering 20 kW. Find the minimum power input.

1. (a) η_max = 1 − T_L/T_H = 1 − 300/1000 = 0.70
2. Claimed η = 750/1000 = 0.75 > 0.70 → impossible (it violates the second law).
3. (b) COP_HP = T_H/(T_H − T_L) = 295/25 = 11.8
4. W = Q_H/COP_HP = 20/11.8 = 1.69 kW

Example 2 (GATE level). Two reversible engines operate in series between a source at 1000 K and a sink at 300 K. Engine A receives 500 kW from the source and rejects heat to an intermediate reservoir at T_m, from which engine B takes all of it. Both engines produce the same power. Find T_m, each engine's power and each efficiency.

1. Equal work means the drop T_H − T_m equals T_m − T_L, so T_m = (T_H + T_L)/2 = 650 K. (Proof: for reversible engines W = Q·ΔT/T, and A and B handle heats proportional to their source temperatures.)
2. η_A = 1 − T_m/T_H = 1 − 650/1000 = 0.35, so W_A = 0.35 × 500 = 175 kW
3. Heat to the intermediate reservoir = 500 − 175 = 325 kW
4. η_B = 1 − T_L/T_m = 1 − 300/650 = 0.5385, so W_B = 0.5385 × 325 = 175 kW ✓
5. Overall: 350 kW from 500 kW = 0.70 = 1 − 300/1000 ✓ (series reversible engines behave like one Carnot engine).

Answer: T_m = 650 K, W_A = W_B = 175 kW, η_A = 35%, η_B = 53.8%

Common mistakes

• Using °C in 1 − T_L/T_H. Always kelvin.
• Applying Q_H/Q_L = T_H/T_L to an irreversible device — it holds only for reversible ones.
• Thinking a lower T_L and a higher T_H give the same benefit. For efficiency, lowering T_L by ΔT helps more than raising T_H by ΔT (∂η/∂T_L = −1/T_H, ∂η/∂T_H = T_L/T_H²).
• Confusing COP_R and COP_HP formulas (numerator T_L versus T_H).
• Assuming reversible adiabatic means any adiabatic; an adiabatic process with friction is not isentropic.

For GATE AE

Expect quick numericals on Carnot efficiency and COP, feasibility checks of claimed devices, engines in series (equal work or equal efficiency), and a Carnot engine driving a Carnot refrigerator or heat pump. Conceptual questions list processes and ask which are irreversible, or ask about the Carnot principles. Practise building combined systems with energy balances at each reservoir.

Quick check

1. Carnot engine between 600 K and 400 K absorbs 1000 J. Find the work.
2. Carnot refrigerator between 250 K and 350 K removes 500 J. Find the work input.
3. Name three sources of irreversibility.
4. Which processes in the Carnot cycle involve heat transfer?
5. Two reversible engines in series between 1000 K and 400 K have equal efficiencies. Find the intermediate temperature.

Answers: 1. 333.3 J. 2. 200 J. 3. Any three of: friction, heat transfer across finite ΔT, free expansion, mixing, electrical resistance, combustion. 4. The two isothermal processes. 5. About 632.5 K.