Refrigeration and liquefaction cycles

Why it matters

Refrigeration runs crystallisers, chillers for reactor cooling, gas drying and every cold box in an air-separation or LNG plant. Liquefaction makes oxygen, nitrogen and natural gas transportable. Compressor power is the main operating cost of these units, so estimating COP and liquid yield correctly decides whether a design makes money.

Key ideas

Refrigerator and heat pump. Both move heat Q_C from a cold region to a hot one at temperature T_H using work W, rejecting Q_H = Q_C + W. A refrigerator is judged by the heat removed per unit work (COP_R = Q_C/W); a heat pump by the heat delivered (COP_HP = Q_H/W = COP_R + 1). Unlike efficiency, COP is usually greater than 1.

Carnot refrigerator. The reversed Carnot cycle sets the upper limit: COP = T_C/(T_H − T_C). COP falls as the temperature lift grows, which is why deep refrigeration is expensive and why multistage or cascade systems are used for large lifts.

Why not run a Carnot cycle in practice? Isentropic compression of a wet vapour damages compressors, and an expansion engine handling mostly liquid gives very little work for its cost. So the practical cycle compresses dry vapour and replaces the turbine with a throttle valve.

Vapour-compression cycle (four steady-flow steps). 1→2 Compressor: saturated (or slightly superheated) vapour compressed, ideally isentropically, to condenser pressure. 2→3 Condenser: heat rejected at roughly constant pressure; vapour desuperheats and condenses to saturated liquid. 3→4 Throttle valve: isenthalpic expansion, h₄ = h₃; the fluid partly flashes and cools. 4→1 Evaporator: the cold liquid-vapour mixture absorbs the refrigeration load and evaporates. Each step is analysed with the steady-flow energy balance, using enthalpies from the refrigerant's P-h chart or tables (take these from your data book).

Refrigerant choice. Suitable vapour pressures (above atmospheric in the evaporator to avoid air leaks, moderate in the condenser), large latent heat, low toxicity and flammability, and low ozone-depletion and global-warming potential. Ammonia is common in industry; HFCs and HFOs in commercial units.

Absorption refrigeration. The compressor is replaced by an absorber, pump and generator (for example ammonia-water or LiBr-water), driven mainly by heat. Useful where waste heat is cheap; its COP is lower than vapour compression.

Liquefaction. A gas is liquefied by cooling it below its critical temperature at a pressure where it condenses. Processes:

• Linde (Joule-Thomson) process: high-pressure gas is cooled in a counter-current heat exchanger by returning cold gas, then throttled. Part of it liquefies; the rest returns. It works only if the gas cools on throttling, i.e. it is below its inversion temperature (hydrogen and helium need precooling).
• Claude process: part of the high-pressure gas is expanded in a turbine (nearly isentropic, giving much more cooling than throttling and recovering work); the expanded cold gas cools the remaining stream before its throttle. Higher liquid yield than Linde. The liquid fraction follows from an energy balance on the cold section (heat exchanger, valve and separator) treated as adiabatic with no work.

Formulas

COP_R = Q_C / W, COP_HP = Q_H / W = COP_R + 1 Q_H = Q_C + W COP_Carnot,R = T_C / (T_H − T_C) q_evap = h₁ − h₄, w_comp = h₂ − h₁, q_cond = h₂ − h₃, h₄ = h₃ (vapour compression, per kg) COP_R = (h₁ − h₄) / (h₂ − h₁) ṁ = Q̇_C / (h₁ − h₄) y = (h_LP − h_HP) / (h_LP − h_f) (Linde, adiabatic cold section)

Symbols: Q_C heat removed from the cold space (J or W); Q_H heat rejected (J or W); W work input (J or W); T_C, T_H absolute temperatures (K); h specific enthalpy (kJ/kg) at the states numbered above; ṁ refrigerant mass flow (kg/s); y fraction of compressed gas that is liquefied; h_HP enthalpy of high-pressure gas entering the cold section, h_LP enthalpy of low-pressure gas leaving it (both at the warm end), h_f saturated-liquid enthalpy at the separator pressure.

Worked examples

Example 1 (standard): vapour-compression refrigerator. Given: a 10 kW refrigerator with evaporator at −10 °C and condenser at 40 °C. From the refrigerant chart: h₁ = 395 kJ/kg (saturated vapour leaving the evaporator), h₂ = 430 kJ/kg (after isentropic compression), h₃ = 250 kJ/kg (saturated liquid leaving the condenser). Find COP, refrigerant flow, compressor power, condenser duty, and compare with Carnot.

1. Throttle: h₄ = h₃ = 250 kJ/kg.
2. Refrigerating effect: q = h₁ − h₄ = 395 − 250 = 145 kJ/kg.
3. Compressor work: w = h₂ − h₁ = 430 − 395 = 35 kJ/kg.
4. COP = q/w = 145/35 = 4.14.
5. ṁ = Q̇_C/q = 10/145 = 0.0690 kg/s.
6. Power = ṁ·w = 0.0690 × 35 = 2.41 kW. Condenser duty = ṁ(h₂ − h₃) = 0.0690 × 180 = 12.41 kW = 10 + 2.41 kW. Consistent.
7. Carnot: T_C = 263.15 K, T_H = 313.15 K, COP = 263.15/50 = 5.26. The real cycle reaches 4.14/5.26 = 79% of it. Answer: COP = 4.14, ṁ = 0.069 kg/s, power = 2.41 kW, condenser duty = 12.4 kW

Example 2 (GATE level): Linde liquid yield. Given: air enters the cold section of a Linde liquefier at 300 K and 200 bar with h_HP = 423 kJ/kg; the unliquefied gas leaves the warm end of the exchanger at 300 K and 1 bar with h_LP = 461 kJ/kg; saturated liquid at 1 bar has h_f = 29 kJ/kg (all values read from an air property chart on one reference). Find the fraction liquefied and the liquid produced per 1000 kg of compressed air.

1. Energy balance on the adiabatic, work-free cold section, per kg of feed: h_HP = y·h_f + (1 − y)·h_LP.
2. Rearranged: y = (h_LP − h_HP)/(h_LP − h_f) = (461 − 423)/(461 − 29) = 38/432 = 0.0880.
3. Liquid per 1000 kg of feed = 0.0880 × 1000 = 88 kg. Answer: y = 0.088, about 88 kg of liquid air per tonne compressed Note that the yield depends entirely on the enthalpy drop of the gas on isothermal compression (h_LP − h_HP), a real-gas effect; an ideal gas would give y = 0.

Common mistakes

• Using °C in the Carnot COP; T must be in kelvin.
• Writing h₄ ≠ h₃; the throttle is isenthalpic.
• Taking COP = Q_H/W for a refrigerator (that is the heat-pump COP).
• Calling the compressor work h₁ − h₂ and getting a negative COP.
• In Linde problems, using the liquid's enthalpy at the high pressure or forgetting that the returning gas leaves at the warm end.
• Assuming any gas cools on throttling; above the inversion temperature it warms.

For GATE CH

• NAT problems on Carnot COP and on vapour-compression cycles with enthalpies given: COP, mass flow, compressor power, condenser load.
• Linde or Claude liquid-yield energy balances with property values given.
• Conceptual MCQs on why a throttle replaces the expander, the effect of evaporator and condenser temperatures on COP, and COP_HP = COP_R + 1.
• Practise drawing the cycle on T-s and P-h diagrams and labelling each state.

Quick check

1. A Carnot refrigerator operates between −23 °C and 27 °C. What is its COP?
2. Which process in the vapour-compression cycle is isenthalpic?
3. If COP_R = 3, what is COP_HP between the same temperatures and the same cycle?
4. Why does the Linde process fail for hydrogen at room temperature without precooling?
5. What is the main advantage of the Claude process over Linde?

Answers: 1. 250.15/50 ≈ 5.0. 2. The throttling (expansion valve) step. 3. 4. 4. Room temperature is above hydrogen's inversion temperature, so throttling warms it. 5. The expansion turbine gives much more cooling (and recovers work), so the liquid yield is higher.