Vapour compression and vapour absorption refrigeration

The ideal and actual vapour-compression cycle on the p–h chart, refrigerating effect, COP and tonnage, subcooling and superheat, refrigerant choice, and NH₃–water and LiBr–water absorption systems with their COP limit.

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Why it matters

Refrigeration and air conditioning use a large share of the electricity in Indian buildings, cold chains and process industry, and the same cycle runs heat pumps and battery or electronics chillers. The vapour-compression cycle on a p–h chart is how these machines are sized and diagnosed. The absorption cycle matters wherever waste heat, steam or solar heat is cheaper than electricity — hotels, hospitals, cogeneration plants.

Key ideas

Ideal vapour-compression refrigeration (VCR) cycle, usually drawn on a p–h diagram: 1→2 isentropic compression of saturated (or slightly superheated) vapour from evaporator pressure to condenser pressure; 2→3 constant-pressure heat rejection in the condenser: desuperheating, then condensation to saturated liquid; 3→4 throttling in the expansion valve or capillary tube, isenthalpic (h₄ = h₃), giving a cold, low-quality wet mixture; 4→1 constant-pressure evaporation in the evaporator, absorbing heat from the cold space.

  • Throttling replaces an expansion engine because the work recoverable from a liquid is small and the device is simple. It is the main irreversibility of the ideal cycle, which is why COP is below the reversed-Carnot value even for ideal components.
  • Compression of wet vapour is avoided (slugging damages reciprocating compressors), so the cycle compresses dry vapour.

Refrigerating effect and capacity. Refrigerating effect per kg = h₁ − h₄. Capacity is often given in tonnes of refrigeration: 1 TR = 211 kJ/min ≈ 3.517 kW (the rate of freezing one US ton of water at 0 °C in 24 h).

Effects of operating changes.

  • Lower evaporator temperature or higher condenser temperature: higher pressure ratio, more compressor work, lower COP and lower volumetric efficiency.
  • Subcooling the liquid leaving the condenser lowers h₃ = h₄, increasing refrigerating effect with no extra work — COP rises.
  • Superheating in the evaporator (useful superheat) raises the refrigerating effect but also compressor work; superheating in the suction line outside the cold space is a loss. A thermostatic expansion valve maintains a few kelvin of superheat to protect the compressor.

Actual cycle. Pressure drops in lines and heat exchangers, non-isentropic compression (isentropic efficiency 0.7–0.85), suction superheat and liquid subcooling.

Refrigerants. Chosen for suitable pressures at working temperatures, high latent heat, low specific volume at suction, chemical stability, safety and environmental impact. CFCs (R-12) were phased out for ozone depletion (ODP); HFCs such as R-134a and R-410A have zero ODP but high global-warming potential (GWP), and are being phased down. Low-GWP options: R-32, R-1234yf, hydrocarbons R-290 (propane) and R-600a (isobutane) in domestic units, ammonia (R-717) in industry, CO₂ (R-744) in transcritical systems.

Vapour absorption refrigeration (VAR). The compressor is replaced by a "thermal compressor": absorber, pump, generator and (usually) a solution heat exchanger.

  • Low-pressure refrigerant vapour from the evaporator is absorbed into a weak solution in the absorber, which rejects heat. The pump raises the liquid solution to condenser pressure — far less work than compressing vapour. In the generator, heat drives the refrigerant back out of solution; the vapour goes to the condenser, the expansion valve and the evaporator as in a VCR system.
  • NH₃–water: ammonia is the refrigerant; it can reach sub-zero temperatures, but needs a rectifier/analyser to strip water vapour.
  • LiBr–water: water is the refrigerant, so evaporator temperatures stay above about 4 °C; it is used for air-conditioning chillers; watch for crystallisation.
  • COP (refrigeration ÷ generator heat) is typically 0.6–0.8 for single-effect and 1.0–1.3 for double-effect units — lower than VCR COP, but the input is low-grade heat.

Ideal limit for an absorption machine. It is a reversible heat engine (heat from the generator at T_G, rejected at T₀) driving a reversible refrigerator (between T_E and T₀).

Formulas

RE = h₁ − h₄ — refrigerating effect (kJ/kg); h₄ = h₃ across the throttle. w_c = h₂ − h₁ — isentropic compressor work (kJ/kg); actual w_c = (h₂s − h₁)/η_c. q_c = h₂ − h₃ — heat rejected in the condenser (kJ/kg). COP_R = (h₁ − h₄) / (h₂ − h₁) COP_HP = (h₂ − h₃) / (h₂ − h₁) = COP_R + 1 COP_Carnot,R = T_E / (T_C − T_E) — T in K. ṁ = Q_E / (h₁ − h₄) — refrigerant flow (kg/s); Q_E in kW. P = ṁ w_c — compressor power (kW). 1 TR = 3.517 kW = 211 kJ/min x₄ = (h₄ − h_f,E) / h_fg,E — quality entering the evaporator. COP_VAR = Q_E / (Q_G + W_pump) ≈ Q_E / Q_G COP_VAR,max = (1 − T₀ / T_G) · T_E / (T₀ − T_E) — T_G: generator, T₀: heat-rejection (ambient), T_E: evaporator temperature (all K).

Worked examples

Enthalpies below are from R-134a tables using the IIR reference state (h = 200 kJ/kg for saturated liquid at 0 °C). Tables using another reference give values shifted by a constant; differences are the same.

Example 1 (standard, ideal VCR cycle). An ideal R-134a cycle has the evaporator at −10 °C (p = 200.6 kPa) and condenser at 40 °C (p = 1016.6 kPa). Data: h₁ = 392.7 kJ/kg (saturated vapour at −10 °C), h₂ = 426.5 kJ/kg (after isentropic compression to 1016.6 kPa, about 46 °C), h₃ = 256.4 kJ/kg (saturated liquid at 40 °C). Find RE, compressor work, COP and compare with Carnot.

  1. h₄ = h₃ = 256.4 kJ/kg.
  2. RE = h₁ − h₄ = 392.7 − 256.4 = 136.3 kJ/kg.
  3. w_c = h₂ − h₁ = 426.5 − 392.7 = 33.8 kJ/kg.
  4. COP = RE / w_c = 136.3/33.8 = 4.03.
  5. COP_Carnot = T_E/(T_C − T_E) = 263.15/50 = 5.26; the ideal VCR cycle reaches 77 % of it, the gap coming mostly from throttling and desuperheating.

RE = 136.3 kJ/kg, w_c = 33.8 kJ/kg, COP ≈ 4.03

Example 2 (GATE level, plant sizing). The cycle of Example 1 must provide 10 TR, and the compressor's isentropic efficiency is 0.80. Find the refrigerant flow rate, compressor power, actual COP and condenser heat rejection.

  1. Q_E = 10 × 3.517 = 35.17 kW.
  2. ṁ = Q_E/(h₁ − h₄) = 35.17/136.3 = 0.2580 kg/s.
  3. Actual work per kg = 33.8/0.80 = 42.25 kJ/kg; power P = 0.2580 × 42.25 = 10.90 kW.
  4. COP = 35.17/10.90 = 3.23.
  5. Condenser heat = Q_E + P = 35.17 + 10.90 = 46.07 kW.

ṁ ≈ 0.258 kg/s, P ≈ 10.9 kW, COP ≈ 3.23, Q_cond ≈ 46.1 kW

Example 3 (absorption limit). A LiBr–water chiller has its generator at 120 °C, rejects heat at 30 °C and evaporates at 5 °C. Find its maximum possible COP.

  1. COP_max = (1 − T₀/T_G) · T_E/(T₀ − T_E) = (1 − 303.15/393.15) × 278.15/25.
  2. = 0.2289 × 11.126 = 2.55.

COP_max ≈ 2.55 (real single-effect units achieve about 0.7)

Common mistakes

  • Taking compressor work as h₁ − h₄ or refrigerating effect as h₂ − h₃.
  • Writing h₄ = h_f at evaporator pressure. Throttling keeps h₄ = h₃; the refrigerant enters the evaporator as a wet mixture.
  • Forgetting to convert TR to kW (3.517 kW per TR).
  • Applying a compressor isentropic efficiency to the refrigerating effect.
  • Comparing VAR COP directly with VCR COP without noting that one uses heat and the other electrical work.
  • Using LiBr–water for sub-zero applications; water as refrigerant would freeze.

For GATE ME

Expect p–h-based VCR problems with supplied enthalpies: refrigerating effect, COP, mass flow per TR, compressor power, and effects of subcooling and superheating; Carnot COP limits; and absorption-system concepts (component roles, refrigerant–absorbent pairs, maximum COP from three temperatures). Practise reading the four states quickly and checking that COP_HP = COP_R + 1.

Quick check

  1. What property is conserved across the expansion valve?
  2. Convert 5 TR to kW.
  3. Does subcooling the condensate increase or decrease COP?
  4. In an LiBr–water system, which substance is the refrigerant?
  5. An ideal VCR cycle has RE = 150 kJ/kg and w_c = 30 kJ/kg. Find COP_R and COP_HP.

Answers: 1. Enthalpy. 2. 17.6 kW. 3. Increase. 4. Water. 5. 5 and 6.

Interview questions

All Thermal Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is vapour compression refrigeration?Concept

    Vapour compression refrigeration is a process that uses a refrigerant in a closed loop to absorb heat from a low-temperature space and reject it at a higher temperature. It involves four main components: a compressor, a condenser, an expansion valve, and an evaporator. The refrigerant undergoes phase changes from liquid to vapour and back to liquid, allowing it to absorb and release heat efficiently.

  2. 2.Explain the working principle of vapour absorption refrigeration.Concept

    Vapour absorption refrigeration uses a refrigerant and an absorbent pair to achieve cooling. The refrigerant, typically ammonia, is absorbed by a liquid absorbent, such as water, in the absorber. The solution is then heated in the generator to release the refrigerant vapour, which is condensed and expanded to produce cooling in the evaporator. This system uses heat energy instead of mechanical energy to drive the refrigeration cycle.

  3. 3.What are the main differences between vapour compression and vapour absorption refrigeration systems?Concept

    The main differences are in the energy source and components used. Vapour compression systems use mechanical energy from a compressor, while vapour absorption systems use thermal energy, often from waste heat or solar energy. Vapour compression systems are generally more efficient but require electricity, whereas absorption systems can operate on low-grade heat sources. Additionally, absorption systems are quieter and have fewer moving parts.

  4. 4.Why is ammonia commonly used as a refrigerant in absorption refrigeration systems?Application

    Ammonia is commonly used because it has a high latent heat of vaporization, which makes it efficient for heat absorption. It is also readily available and relatively inexpensive. Ammonia's thermodynamic properties are well-suited for absorption systems, and it can be easily absorbed by water, which is a common absorbent in these systems.

  5. 5.What happens if the condenser of a vapour-compression system cannot reject heat properly (for example, its fan fails or it is fouled)?Application

    The condensing temperature and pressure (head pressure) rise until the condenser can reject the heat again. The higher pressure ratio increases compressor work and discharge temperature, and the liquid may leave the condenser not fully condensed, so flash gas enters the expansion device and the refrigerating effect falls. COP and capacity drop, and if the pressure keeps rising the high-pressure cut-out trips the compressor to protect it.

  6. 6.How does the coefficient of performance differ between vapour-compression and vapour-absorption systems?Application

    Vapour-compression systems typically have COP of about 2.5–5, because the input is high-grade mechanical work. Absorption systems are rated as cooling ÷ generator heat: about 0.6–0.8 for single-effect and 1.0–1.3 for double-effect machines. The numbers are not directly comparable, since one input is work and the other is low-grade heat; an absorption chiller makes sense when that heat is waste heat, steam or solar energy that would otherwise be unused.

  7. 7.Calculate the COP of a vapour compression refrigeration system if the refrigerant absorbs 200 kJ of heat in the evaporator and the work input to the compressor is 50 kJ.Numerical

    COP = Qc / W = 200 kJ / 50 kJ = 4. The coefficient of performance (COP) is the ratio of the heat absorbed by the refrigerant in the evaporator (Qc) to the work input to the compressor (W). In this case, the COP is 4, indicating that the system provides 4 units of cooling for every unit of work input.

  8. 8.In a vapour-absorption system, what is the effect of increasing the generator temperature?Application

    A hotter generator drives more refrigerant out of the solution, so the solution circulation needed per kW of cooling falls and capacity and COP rise. The ideal limit, COP_max = (1 − T₀/T_G)·T_E/(T₀ − T_E), also rises with T_G. The gain flattens beyond an optimum, because losses in the solution heat exchanger and rectifier grow; in LiBr–water machines an excessive generator temperature also risks crystallisation and corrosion.

  9. 9.What role does the expansion valve play in a vapour compression refrigeration system?Concept

    The expansion valve reduces the pressure of the refrigerant, allowing it to expand and cool before entering the evaporator. This pressure drop is essential for the refrigerant to absorb heat efficiently from the surroundings in the evaporator. The expansion valve helps control the flow of refrigerant, ensuring the system operates efficiently and maintains the desired cooling effect.

  10. 10.A vapour absorption refrigeration system uses water as the absorbent and ammonia as the refrigerant. If the absorber temperature is too high, what impact does it have on the system?Application

    If the absorber temperature is too high, the ability of the absorbent (water) to absorb the refrigerant (ammonia) decreases. This can lead to reduced system efficiency and cooling capacity, as less refrigerant is absorbed and circulated through the system. Maintaining an optimal absorber temperature is crucial for effective operation of the absorption refrigeration system.

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