Refrigeration and air conditioning basics

Why it matters

Refrigeration and air conditioning keep machine tools thermally stable, cool hydraulic oil and cutting fluids, preserve food and medicines, and make factories and control rooms habitable. Air conditioning of precision measurement labs and electronics assembly areas is a production requirement, not a luxury. The vapour-compression cycle analysed here is in nearly every chiller, cold store and room air conditioner.

Key ideas

Refrigerator and heat pump. Both move heat from a low-temperature space to a high-temperature one using work input, as the second law requires. A refrigerator is rated by the heat removed (refrigerating effect), a heat pump by the heat delivered. Performance is expressed as the coefficient of performance (COP); the reversed Carnot cycle gives the upper limit between two temperatures.

Unit of refrigeration. One tonne of refrigeration (TR) is the rate of heat removal that freezes one short ton of water at 0 °C in 24 hours: 1 TR = 211 kJ/min ≈ 3.517 kW.

Simple vapour-compression refrigeration (VCR) cycle.

• 1→2 Compressor: saturated (or slightly superheated) vapour is compressed, ideally isentropically, to condenser pressure; its temperature rises above ambient.
• 2→3 Condenser: refrigerant rejects heat at constant pressure, desuperheating then condensing to saturated (or subcooled) liquid.
• 3→4 Expansion (throttle) valve: pressure and temperature drop with no work and no heat, so h₄ = h₃. The process is irreversible; this is why the VCR cycle cannot reach the Carnot COP.
• 4→1 Evaporator: the low-pressure wet mixture absorbs heat from the cold space at constant pressure and evaporates. Each component is a steady-flow device; enthalpies come from refrigerant tables or the p–h chart, which is the standard chart for refrigeration work.

Effects of operating conditions. Lowering the evaporator temperature or raising the condenser temperature lowers the COP and the capacity. Subcooling the liquid after the condenser increases the refrigerating effect. Superheating in the evaporator increases the refrigerating effect slightly but also the compressor work, and protects the compressor from liquid.

Other systems. Vapour-absorption refrigeration replaces the compressor with an absorber, pump and generator driven by heat (useful where waste heat or steam is available; ammonia–water and lithium bromide–water pairs). Gas (air) refrigeration uses a reversed Brayton cycle and is used in aircraft cabin cooling. Thermoelectric (Peltier) coolers are used for small electronic loads.

Refrigerants. Chosen for suitable pressures and boiling points, high latent heat, chemical stability, safety, and environmental impact. CFCs (R-12) were phased out under the Montreal Protocol for ozone depletion; HCFCs (R-22) are being phased out; HFCs (R-134a, R-410A) have zero ODP but high global-warming potential and are being phased down under the Kigali Amendment. Lower-GWP options include R-32, HFOs (R-1234yf), hydrocarbons (R-290 propane, R-600a isobutane), ammonia (R-717) and CO₂ (R-744).

Psychrometry basics (air conditioning). Atmospheric air is a mixture of dry air and water vapour.

• Dry-bulb temperature (DBT): ordinary air temperature. Wet-bulb temperature (WBT): from a thermometer with a wet wick; it is lower than DBT unless the air is saturated. Dew-point temperature (DPT): the temperature at which vapour starts to condense on cooling at constant pressure.
• Humidity ratio (specific humidity) ω: kg of water vapour per kg of dry air.
• Relative humidity φ: actual vapour pressure divided by saturation pressure at the same DBT.
• For saturated air DBT = WBT = DPT.
• Comfort conditions are typically about 22–26 °C DBT and 40–60% RH.
• Common processes: sensible heating or cooling (ω constant), cooling and dehumidification (cool below dew point, then often reheat), heating and humidification, evaporative cooling (WBT roughly constant), and adiabatic mixing of two air streams.

Formulas

COP_R = Q_L/W, COP_HP = Q_H/W = COP_R + 1

• Q_L heat removed, Q_H heat rejected, W work (kJ or kW).

COP_Carnot,R = T_L/(T_H − T_L)

• T in K.

RE = h₁ − h₄, w_c = h₂ − h₁, q_c = h₂ − h₃, h₄ = h₃

• VCR cycle per kg of refrigerant (kJ/kg).

COP = (h₁ − h₄)/(h₂ − h₁)

• Simple VCR cycle.

ṁ = (capacity in kW)/RE, P = ṁ·w_c

• Refrigerant flow (kg/s) and compressor power (kW).

1 TR = 3.517 kW

ω = 0.622·p_v/(p − p_v)

• Humidity ratio (kg/kg dry air); p_v partial pressure of vapour, p total pressure (same units).

φ = p_v/p_sat

• Relative humidity; p_sat at the dry-bulb temperature (from steam tables).

Worked examples

Example 1 (standard): 10 TR vapour-compression plant. Given (from refrigerant tables, given as data): evaporator −10 °C, condenser 40 °C. Enthalpies: h₁ = 395 kJ/kg (saturated vapour at compressor inlet), h₂ = 425 kJ/kg (compressor exit, isentropic), h₃ = 255 kJ/kg (saturated liquid leaving condenser). Capacity 10 TR. Find COP, refrigerant flow, compressor power and condenser load.

1. Throttling: h₄ = h₃ = 255 kJ/kg.
2. RE = h₁ − h₄ = 395 − 255 = 140 kJ/kg; w_c = h₂ − h₁ = 30 kJ/kg.
3. COP = 140/30 = 4.67
4. Capacity = 10 × 3.517 = 35.17 kW; ṁ = 35.17/140 = 0.2512 kg/s.
5. P = 0.2512 × 30 = 7.54 kW
6. Q_cond = ṁ·(h₂ − h₃) = 0.2512 × 170 = 42.71 kW (= 35.17 + 7.54 ✓) Answer: COP ≈ 4.67; ṁ ≈ 0.251 kg/s; P ≈ 7.54 kW; condenser load ≈ 42.7 kW

Example 2 (GATE level): comparison with Carnot. Given: the same temperature limits (−10 °C and 40 °C). Find the Carnot COP and the second-law efficiency of the plant in Example 1.

1. T_L = 263.15 K, T_H = 313.15 K
2. COP_Carnot = 263.15/(313.15 − 263.15) = 263.15/50 = 5.26
3. η_II = COP_actual/COP_Carnot = 4.667/5.263 = 0.887 Answer: COP_Carnot ≈ 5.26; second-law efficiency ≈ 89% (losses mainly from throttling and the superheat horn after compression)

Example 3 (quick): humidity ratio and RH. Air at 101.325 kPa and 30 °C has p_v = 2.0 kPa; p_sat at 30 °C = 4.246 kPa (steam-table value). ω = 0.622 × 2.0/(101.325 − 2.0) = 0.0125 kg/kg dry air; φ = 2.0/4.246 = 0.471, i.e. RH ≈ 47%.

Common mistakes

• Using °C in Carnot COP formulas.
• Treating the expansion valve as isentropic; it is isenthalpic (h₄ = h₃).
• Taking the refrigerating effect as h₁ − h₃ without noting that h₄ = h₃ (same number, but understand why), or as h₂ − h₃ (that is the condenser heat).
• Confusing COP_R and COP_HP.
• Forgetting the factor 0.622 or using total pressure instead of (p − p_v).
• Saying HFCs are environmentally harmless; they have zero ODP but high GWP.

For GATE PI

Expect COP calculations (Carnot and VCR with given enthalpies), refrigerant flow and compressor power for a rated TR, effect of evaporator and condenser temperatures on COP, and basic psychrometric definitions (DBT, WBT, DPT, ω, RH) with simple calculations. Practise converting TR to kW and drawing the cycle on a p–h diagram.

Quick check

1. Convert 5 TR to kW.
2. What property is constant across the expansion valve?
3. A refrigerator has COP 4. What is the COP of the same machine used as a heat pump?
4. For saturated air, how do DBT, WBT and DPT compare?

Answers: 1. 17.6 kW 2. Enthalpy 3. 5 4. All three are equal