Vapour Cycles: Carnot Vapour Cycle, Ideal Rankine Cycle and Vapour-Compression Refrigeration

This chapter opens Section XE4.7 of XE4 Thermodynamics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. It answers the first two lines of the heading “Thermodynamic Cycles”: “Carnot vapor cycle, ideal Rankine cycle; Simple vapor-compression refrigeration cycle”. It teaches why the Carnot vapour cycle is not built, the state-by-state analysis of the ideal Rankine cycle (efficiency, work ratio, the effect of boiler pressure, condenser pressure and superheat, reheat and regeneration in outline) and the simple vapour-compression cycle (refrigeration effect, COP and the effect of evaporator pressure, condenser pressure, superheating and subcooling). The numericals are cycle efficiency, steam flow, refrigeration power and COP, every one with its enthalpy data given in the stem. Sign convention throughout: heat added to the working fluid and work done by it are positive; w_net = Σq_in − Σq_out.

1. The Carnot vapour cycle and why it is impractical

Run a Carnot cycle inside the vapour dome. 1→2 isentropic compression of a wet mixture, 2→3 isothermal heat addition in the boiler (liquid fully boiled to saturated vapour, at T_H), 3→4 isentropic expansion in a turbine to a wet state, 4→1 isothermal heat rejection in the condenser (partial condensation, at T_L). Its efficiency is η = 1 − T_L/T_H, the highest possible between the two temperatures, and it is a rectangle on the T-s diagram. Yet no steam plant uses it.

  • Process 4→1 stops part-way, so the compressor must handle a liquid-vapour mixture. Pumping a two-phase mixture is hard (cavitation, poor efficiency); a pump can compress only liquid, a compressor only vapour.
  • Heat addition at constant temperature is possible only in the wet region, so T_H cannot exceed the critical temperature (about 374 °C for water). Superheating raises T but then heat addition is no longer isothermal, and the cycle is no longer Carnot.
  • The wet turbine exhaust has low dryness, and the compressor work is large, so the work ratio is poor.
⚠️ Carnot is a limit, not a design
The fix that gives the Rankine cycle is to condense completely (state 1 saturated liquid) and pump a liquid. Pump work becomes tiny, because v is small. The price: heat is no longer added entirely at T_H, so η falls below the Carnot value.

2. The ideal Rankine cycle, state by state

Four steady-flow devices, all ideal: pump and turbine isentropic, boiler and condenser at constant pressure with no pressure drop. 1→2 pump: saturated liquid at condenser pressure to boiler pressure. 2→3 boiler: heat added until the steam leaves as saturated or superheated vapour. 3→4 turbine: isentropic expansion to condenser pressure (s₄ = s₃). 4→1 condenser: heat rejected until saturated liquid. On the T-s diagram the heating line 2→3 runs along the liquid curve, then the flat boiling line, then the superheat.

Steady-flow energy equation per kg, neglecting kinetic and potential energy, gives every term:

  • pump: w_p = h₂ − h₁ = v₁(p₂ − p₁) (liquid incompressible, v₁ = v_f at condenser pressure)
  • boiler: q_in = h₃ − h₂
  • turbine: w_t = h₃ − h₄
  • condenser: q_out = h₄ − h₁
  • η = w_net/q_in = [(h₃ − h₄) − (h₂ − h₁)]/(h₃ − h₂) = 1 − q_out/q_in
  • back work ratio = w_p/w_t (tiny, about 0.4 % here); work ratio = w_net/w_t = 1 − bwr
  • specific steam consumption = 3600/w_net kg/kWh (w_net in kJ/kg).
Worked example (data given: v₁ = 0.00101 m³/kg, p₁ = 10 kPa, p₂ = 4 MPa)
Stateh (kJ/kg)How
1 (sat. liquid, 10 kPa)192given
2 (pump exit)196.03w_p = 0.00101 × 3990 = 4.03; h₂ = 192 + 4.03
3 (turbine inlet)3214given
4 (turbine exit, s₄ = s₃)2150given

Then q_in = 3214 − 196.03 = 3017.97, w_t = 1064, w_net = 1064 − 4.03 = 1059.97 kJ/kg, so η = 1059.97/3017.97 = 35.1 %, bwr = 4.03/1064 = 0.38 %, and steam consumption = 3600/1059.97 = 3.40 kg/kWh. Check by the other route: q_out = 2150 − 192 = 1958, and 1 − 1958/3017.97 = 35.1 %.

3. What raises Rankine efficiency; reheat and regeneration in outline

The cycle behaves like a Carnot cycle at the mean temperature of heat addition T_m,in and the heat-rejection temperature T_L: roughly η ≈ 1 − T_L/T_m,in. Every change that raises T_m,in without raising T_L helps. Each change below is for an otherwise fixed cycle.

Effect of the main variables
ChangeEfficiencyTurbine exit dryness x₄Limit
Lower condenser pressurerises (T_L falls)falls (wetter)cooling-water temperature; air leakage; blade erosion
Higher boiler pressure (same T₃)rises (T_m,in rises)falls (wetter)strength and cost of boiler; moisture at exit
Superheat (same pressure)risesrises (drier)metallurgical limit on T₃

The practical limit on moisture is roughly x₄ ≳ 0.88 to 0.90, because water droplets erode the last turbine blades and lower turbine efficiency. This is why high boiler pressure is paired with superheat or reheat.

Reheat (outline): expand steam in a high-pressure turbine to an intermediate pressure, return it to the boiler, reheat it (often to the same temperature as T₃) and expand it in the low-pressure turbine. Reheat mainly raises exit dryness; the efficiency gain is small unless the reheat pressure is well chosen.

Regeneration (outline): bleed some steam from the turbine to heat the feedwater (open or closed feedwater heaters) so that the water enters the boiler hotter. The mean temperature of heat addition rises, so η rises; the work per kg of boiler steam falls, and less steam reaches the condenser.

⚠️ Higher efficiency is not always drier steam
Lowering condenser pressure and raising boiler pressure both increase efficiency and decrease dryness. Only superheat (and reheat) raises both. A question that offers “η up and x₄ up together” for a pressure change is a trap.

4. The simple vapour-compression refrigeration cycle

The reversed Rankine idea: move heat from a cold space at T_e to the surroundings at T_c using work. Four states: 1 saturated vapour leaving the evaporator at p_e; 1→2 isentropic compression to the condenser pressure p_c (state 2 superheated); 2→3 condenser: desuperheating, condensation, leaving as saturated liquid at p_c; 3→4 throttling in an expansion valve, an irreversible isenthalpic process with h₄ = h₃, ending as a cold wet mixture; 4→1 evaporator, where heat is absorbed from the cold space. A throttle replaces the turbine because a small turbine on a liquid-rich stream recovers almost no work and is costly.

Per kg of refrigerant:

  • refrigeration effect q_L = h₁ − h₄
  • compressor work w = h₂ − h₁
  • heat rejected q_H = h₂ − h₃ = q_L + w
  • COP = q_L/w = (h₁ − h₄)/(h₂ − h₁), and the Carnot limit is COP_C = T_e/(T_c − T_e) (absolute temperatures)
  • mass flow ṁ = refrigeration capacity/q_L, compressor power = ṁw.

On the p-h diagram the cycle is a lopsided loop: the compression line slants up, the condenser is horizontal at p_c, throttling is a vertical line (constant h), the evaporator is horizontal at p_e. On T-s the throttle is a dashed irreversible path to the right.

Worked example: 10 kW capacity, h₁ = 392, h₂ = 428, h₃ = h₄ = 245 kJ/kg. q_L = 147, w = 36, COP = 147/36 = 4.08, ṁ = 10/147 = 0.0680 kg/s, power = 0.0680 × 36 = 2.45 kW. Heat rejected = 0.0680 × (428 − 245) = 12.45 kW = 10 + 2.45. For a Carnot machine between 270 K and 310 K, COP_C = 270/40 = 6.75, and the actual 4.08 is below it.

5. Effect of evaporator and condenser pressure, superheating and subcooling

Changes to the simple cycle (all else fixed)
ChangeRefrigeration effectCompressor workCOP
Lower evaporator pressurefalls (h₄ unchanged, h₁ lower)rises (pressure ratio up)falls
Higher condenser pressurefalls (h₃, h₄ rise)risesfalls
Subcooling the liquid below saturationrises (h₄ = h₃ lower)unchangedrises
Superheating the vapour before compressionrises only if the heat comes from the cold spacerisesmay rise or fall; depends on the refrigerant

Why subcooling is the clean win: it shifts state 3 (and so state 4) left on the p-h diagram, adding refrigeration effect without touching the compressor line. Why lowering T_e hurts: the Carnot COP T_e/(T_c − T_e) falls as the lift T_c − T_e grows, and the real cycle follows. The designer therefore keeps the evaporator as warm, and the condenser as cool, as the application and the cooling medium allow.

Worked example, subcooling: in the 10 kW example, subcool the liquid so that h₃ = h₄ = 235 kJ/kg, with the same compressor work 36 kJ/kg. Then q_L = 392 − 235 = 157 and COP = 157/36 = 4.36, up from 4.08, a 6.8 % gain for no extra work.

🧠 Reading the p-h diagram fast
Horizontal lengths are enthalpy differences: evaporator length = q_L, compressor length = w, condenser length = q_H. COP is simply (evaporator length)/(compressor length), and a throttle is always a straight drop at constant h.

6. Non-ideal turbine and pump, and Rankine versus Carnot

Real machines are irreversible, so the exit state is at higher entropy. With isentropic efficiencies, η_t = (h₃ − h₄)/(h₃ − h₄s) and η_p = (h₂s − h₁)/(h₂ − h₁), the actual turbine work is η_t(h₃ − h₄s) and the actual pump work is v₁Δp/η_p. A less efficient turbine leaves the exhaust hotter and drier (h₄ > h₄s), which is the one case in which a loss makes the steam look better; it still lowers w_net and η.

Worked example: in the 4 MPa, 10 kPa cycle above take h₄s = 2150, η_t = 0.85 and η_p = 0.80. Turbine work = 0.85 × 1064 = 904.4 kJ/kg. Pump work = 4.03/0.80 = 5.04, so h₂ = 197.04 and q_in = 3214 − 197.04 = 3016.96. Then w_net = 904.4 − 5.04 = 899.4 kJ/kg and η = 899.4/3016.96 = 29.8 %, against 35.1 % for the ideal cycle.

Compare with a Carnot cycle between the boiling and condensing temperatures of this cycle. If steam is boiled at 523 K and condensed at 319 K, η_Carnot = 1 − 319/523 = 39.0 %, above the Rankine value, because in the Rankine cycle most of the heat is not added at the top temperature: the liquid is heated from about 319 K up to 523 K, which pulls T_m,in down. Regeneration attacks exactly this low-temperature heating, which is why a regenerative cycle approaches the Carnot value.

Key takeaways

  • The Carnot vapour cycle needs a compressor for a wet mixture and caps T_H at the critical temperature, so it is not built; the Rankine cycle condenses fully and pumps a liquid.
  • Ideal Rankine: w_p = v₁(p₂ − p₁), q_in = h₃ − h₂, w_t = h₃ − h₄, η = w_net/q_in; the back work ratio is tiny and 3600/w_net is the steam consumption in kg/kWh.
  • Efficiency rises with a higher mean temperature of heat addition and a lower condenser temperature; lower condenser pressure and higher boiler pressure also make the exhaust wetter, while superheat and reheat make it drier.
  • Regeneration (bled steam heating the feedwater) raises the mean heat-addition temperature and the efficiency; reheat mainly protects the last blades from moisture.
  • Vapour-compression: q_L = h₁ − h₄, w = h₂ − h₁, COP = q_L/w, q_H = q_L + w, with h₄ = h₃ across the throttle; the COP is always below T_e/(T_c − T_e).
  • Subcooling raises COP at no extra work; a lower evaporator pressure or higher condenser pressure lowers COP.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The main reason the Carnot cycle is not used with steam as a practical power cycle is that

    1. its efficiency is lower than that of the Rankine cycle between the same temperatures
    2. compression starts from a liquid-vapour mixture no machine handles well
    3. heat cannot be added to boiling water at constant temperature at any pressure
    4. the turbine would have to run on dry saturated vapour only, and such a vapour does not exist
    Show answer

    Answer: B — compression starts from a liquid-vapour mixture no machine handles well

    State 1 lies inside the dome, so the compressor must handle a two-phase mixture, and the maximum temperature is also capped at the critical temperature. The Carnot efficiency between two given temperatures is the highest possible, so the claim of a lower efficiency is false; isothermal heat addition is easy in the wet region (boiling at constant p and T).

  2. In the ideal Rankine cycle the four processes, in order starting from the condenser exit, are

    1. isentropic pumping, isobaric heating, isentropic expansion, isobaric heat rejection
    2. isothermal pumping, constant-volume heating, isentropic expansion, constant-pressure heat rejection
    3. isentropic pumping, constant-volume heating, isothermal expansion, constant-volume heat rejection
    4. throttling, constant-pressure heating, isentropic expansion, isothermal heat rejection
    Show answer

    Answer: A — isentropic pumping, isobaric heating, isentropic expansion, isobaric heat rejection

    The boiler and the condenser are constant-pressure heat exchangers, and the pump and turbine are ideal, i.e. isentropic. Throttling belongs to the refrigeration cycle, and no process of the Rankine cycle is isothermal except the boiling and condensing sections, which are isobaric as well.

  3. In an ideal Rankine cycle, other things fixed, lowering the condenser pressure

    1. raises the efficiency and makes the turbine exhaust drier
    2. lowers the efficiency and makes the turbine exhaust wetter
    3. raises the efficiency and makes the turbine exhaust wetter
    4. leaves the efficiency unchanged because the pump work is negligible
    Show answer

    Answer: C — raises the efficiency and makes the turbine exhaust wetter

    A lower condenser pressure means a lower heat-rejection temperature, so η rises; the isentropic expansion then ends at a lower pressure and a lower x₄, so the exhaust is wetter. Pump work being small does not make the efficiency insensitive to T_L; the turbine work and q_out change a lot.

  4. For the ideal Rankine cycle, the pump work per kg is best estimated by

    1. h₃ − h₄, the enthalpy difference between boiler exit and turbine exit steam
    2. v_g(p₂ − p₁), using the specific volume of the saturated vapour
    3. c_p(T₂ − T₁), treating the pump fluid as a perfect gas with constant specific heat
    4. v_f(p₂ − p₁), with v_f of the saturated liquid at condenser pressure
    Show answer

    Answer: D — v_f(p₂ − p₁), with v_f of the saturated liquid at condenser pressure

    The pump handles saturated liquid, which is practically incompressible, so ∫v dp ≈ v_f(p₂ − p₁) for an isentropic process. Using v_g would overstate the pump work by about three orders of magnitude, which is exactly why a vapour compressor is far more costly.

  5. At a fixed boiler pressure and condenser pressure, adding superheat to the Rankine cycle raises the efficiency mainly because

    1. the pump work falls to zero when the steam leaving the boiler is superheated
    2. the heat rejection temperature falls as the turbine exhaust becomes drier
    3. heat is added above saturation temperature, raising the mean heat-addition temperature
    4. the turbine isentropic efficiency rises to 100 % once the steam is superheated
    Show answer

    Answer: C — heat is added above saturation temperature, raising the mean heat-addition temperature

    The extra heat is added above the saturation temperature, so T_m,in rises while T_L, fixed by the condenser pressure, does not change; hence η rises. A drier exhaust does not change T_L, and the ideal turbine is already 100 % isentropic by assumption.

  6. In the simple vapour-compression refrigeration cycle, the throttling process through the expansion valve is

    1. isentropic, with the refrigerant entropy remaining constant
    2. isenthalpic and irreversible, so h₄ = h₃ and the entropy increases
    3. isothermal, with the refrigerant temperature remaining constant
    4. isochoric, with no change in the specific volume
    Show answer

    Answer: B — isenthalpic and irreversible, so h₄ = h₃ and the entropy increases

    A valve is an adiabatic device with no work and negligible kinetic-energy change, so h₄ = h₃ from the steady-flow energy equation; the pressure drop is dissipative, so the entropy rises. The refrigerant temperature falls (it flashes to a cold wet mixture), so it is not isothermal, and the specific volume increases greatly.

  7. A vapour-compression refrigerator runs with saturated vapour entering the compressor and saturated liquid leaving the condenser. If the condenser pressure is raised (the evaporator pressure unchanged), the refrigeration effect and the COP change as

    1. refrigeration effect increases and COP increases
    2. refrigeration effect decreases and COP decreases
    3. refrigeration effect decreases and COP increases
    4. refrigeration effect is unchanged and COP increases
    Show answer

    Answer: B — refrigeration effect decreases and COP decreases

    A higher p_c raises the saturated-liquid enthalpy h₃ = h₄, so h₁ − h₄ shrinks, and the compressor work h₂ − h₁ grows with the pressure ratio; both push COP down. The refrigeration effect cannot stay unchanged because h₄ moves.

  8. The main purpose of reheating steam between a high-pressure and a low-pressure turbine in a Rankine plant is to

    1. reduce the pump work required to feed the boiler at its high pressure
    2. lower the condenser pressure below what the cooling-water temperature allows
    3. raise the exhaust dryness and so limit moisture damage to the last blades
    4. convert the cycle into a Carnot cycle by making heat addition isothermal
    Show answer

    Answer: C — raise the exhaust dryness and so limit moisture damage to the last blades

    Without reheat the expansion from a high pressure ends deep in the wet region. Reheating to a high temperature at an intermediate pressure moves the last expansion into a drier state. The condenser pressure and pump work are not affected, and the cycle remains a Rankine one.

  9. For a vapour-compression refrigerator and a reversed Carnot refrigerator working between the same evaporator and condenser temperatures, the actual vapour-compression COP is

    1. always lower, owing to throttling and finite temperature differences
    2. equal, since both use the same two temperatures
    3. higher whenever the refrigerant has a large latent heat of vaporisation
    4. higher, since the isenthalpic throttling recovers some useful work
    Show answer

    Answer: A — always lower, owing to throttling and finite temperature differences

    No cycle between two reservoirs can exceed the reversed-Carnot COP T_e/(T_c − T_e). The throttle is a pure loss (it produces no work), and part of the heat is rejected over a temperature range above T_c, so the real COP is below the Carnot value for any refrigerant.

  10. Which of the following are genuine reasons the Carnot vapour cycle is not adopted for steam power plants?

    1. Compression would have to start from a liquid-vapour mixture
    2. The constant-temperature heat addition restricts the maximum temperature to below the critical temperature
    3. The cycle efficiency would exceed the Carnot limit and violate the second law
    4. Superheating the steam while keeping the heat addition isothermal is not possible, so the cycle cannot use the high-temperature superheat region
    Show answer

    Answer: A — Compression would have to start from a liquid-vapour mixture; B — The constant-temperature heat addition restricts the maximum temperature to below the critical temperature; D — Superheating the steam while keeping the heat addition isothermal is not possible, so the cycle cannot use the high-temperature superheat region

    The mixture compression, the critical-temperature cap and the isothermal-superheat problem are all real limitations. Nothing can exceed the Carnot efficiency between the same two temperatures; the cycle attains exactly that value, so there is no violation of the second law.

  11. In a regenerative Rankine cycle in which part of the steam is bled from the turbine to heat the feedwater, compared with the simple cycle with the same boiler and condenser conditions

    1. the mean temperature of heat addition is higher
    2. the steam flow reaching the condenser, per kg leaving the boiler, is smaller
    3. the heat rejected in the condenser per kg of boiler steam is larger
    4. the turbine inlet temperature must be raised to keep the same efficiency
    Show answer

    Answer: A — the mean temperature of heat addition is higher; B — the steam flow reaching the condenser, per kg leaving the boiler, is smaller

    Bled steam heats the water that would otherwise be warmed in the boiler at low temperature, so T_m,in rises; and the bled fraction no longer expands to the condenser, so less steam arrives there and the condenser heat per kg of boiler steam falls. No change of turbine inlet temperature is needed for the efficiency to improve.

  12. For the simple vapour-compression cycle (saturated vapour leaving the evaporator and saturated liquid leaving the condenser as the base case), which of the following are correct?

    1. Subcooling the liquid raises the refrigeration effect without changing the compressor work
    2. Lowering the evaporator pressure, with the condenser pressure fixed, lowers the COP
    3. The heat rejected in the condenser equals the refrigeration effect plus the compressor work
    4. The expansion valve delivers useful work that reduces the compressor input
    Show answer

    Answer: A — Subcooling the liquid raises the refrigeration effect without changing the compressor work; B — Lowering the evaporator pressure, with the condenser pressure fixed, lowers the COP; C — The heat rejected in the condenser equals the refrigeration effect plus the compressor work

    The subcooling, evaporator-pressure and heat-balance statements follow from the p-h diagram and the energy balance q_H = q_L + w for a cycle. The expansion valve is a throttle with no work output, since h₄ = h₃ and nothing is extracted.

  13. An ideal Rankine cycle has these data (all in kJ/kg): enthalpy of saturated liquid leaving the condenser h₁ = 192, turbine inlet h₃ = 3214, turbine exit (isentropic expansion, given) h₄ = 2150. The pump raises the pressure from 10 kPa to 4 MPa, v₁ = 0.00101 m³/kg, incompressible liquid. Heat added is positive, work done by the fluid is positive. The thermal efficiency is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 35.1

    w_p = v₁Δp = 0.00101 × (4000 − 10) = 4.03 kJ/kg, h₂ = 196.03. q_in = 3214 − 196.03 = 3017.97; w_t = 3214 − 2150 = 1064; w_net = 1059.97. η = 1059.97/3017.97 = 0.3512 = 35.1 %. Check: 1 − (2150 − 192)/3017.97 = 35.1 %. Neglecting the pump gives 1064/3022 = 35.2 %, a near miss.

  14. For the same ideal Rankine cycle (h₁ = 192, h₃ = 3214, h₄ = 2150 kJ/kg; v₁ = 0.00101 m³/kg; pump from 10 kPa to 4 MPa), the plant must deliver a net power of 100 MW. The steam mass flow rate required is ____ kg/s (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 94.3

    w_net = (3214 − 2150) − 0.00101 × 3990 = 1064 − 4.03 = 1059.97 kJ/kg. ṁ = 100 000 kW/1059.97 = 94.34 kg/s. Using w_t alone gives 94.0, which is the pump-neglect error.

  15. A refrigerator works on the simple vapour-compression cycle with a refrigeration capacity of 10 kW. The refrigerant enthalpies (kJ/kg) are: evaporator exit (saturated vapour) h₁ = 392, compressor exit h₂ = 428, condenser exit (saturated liquid) h₃ = 245. The expansion is isenthalpic. The compressor power input is ____ kW (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.45

    h₄ = h₃ = 245, so q_L = 392 − 245 = 147 kJ/kg and ṁ = 10/147 = 0.06803 kg/s. w = 428 − 392 = 36 kJ/kg, so P = 0.06803 × 36 = 2.449 = 2.45 kW (COP = 4.08). Using h₂ − h₃ as the work would give 12.45, the heat-rejection error.

  16. In the refrigerator above (h₁ = 392 and h₂ = 428 kJ/kg, compressor work unchanged), the liquid leaving the condenser is subcooled so that its enthalpy, and hence the enthalpy after the expansion valve, becomes 235 kJ/kg. The COP is now ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 4.36

    q_L = h₁ − h₄ = 392 − 235 = 157 kJ/kg; w = 428 − 392 = 36 kJ/kg. COP = 157/36 = 4.361 = 4.36, against 4.08 before subcooling.

  17. An ideal reversed Carnot refrigerator extracts heat from a cold space at 270 K and rejects it to the surroundings at 310 K. Its coefficient of performance is ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 6.75

    COP = T_e/(T_c − T_e) = 270/(310 − 270) = 270/40 = 6.75. Using T_c/(T_c − T_e) = 7.75 is the heat-pump COP, the usual slip.