Refrigeration, Air Conditioning and Psychrometry
Vapour compression refrigeration cycle and COP, refrigerants, air refrigeration, psychrometric properties (humidity ratio, relative humidity, dew point, wet bulb) and air-conditioning processes.
Refrigerators and air conditioners move heat from cold to hot using work. Air conditioning also controls humidity, which needs psychrometry: the study of moist air.
Vapour compression refrigeration
| Process | Device |
|---|---|
| 1 → 2: isentropic compression of vapour | Compressor |
| 2 → 3: condensation at constant pressure | Condenser |
| 3 → 4: throttling (h₄ = h₃) | Expansion valve |
| 4 → 1: evaporation at constant pressure | Evaporator |
- Refrigerating effect = h₁ − h₄. Compressor work = h₂ − h₁.
- COP = (h₁ − h₄)/(h₂ − h₁).
- 1 tonne of refrigeration (TR) = 3.5167 kW (about 211 kJ/min).
- Subcooling the liquid and lowering the condenser temperature raise COP; a lower evaporator temperature reduces it.
Refrigerants: older CFCs (R-12) damage ozone; HFCs (R-134a) do not but have high global warming potential; ammonia (R-717) and CO₂ (R-744) are natural refrigerants.
Air refrigeration (Bell–Coleman / reversed Brayton) is used in aircraft: COP = 1/(r_p^((γ−1)/γ) − 1).
Psychrometry
Moist air = dry air + water vapour. With total pressure p and vapour partial pressure p_v:
- Humidity ratio (specific humidity) ω = 0.622 p_v/(p − p_v) kg vapour per kg dry air.
- Relative humidity φ = p_v/p_sat (at the dry-bulb temperature).
- Dew point: the temperature at which the vapour starts to condense when cooled at constant ω.
- Wet-bulb temperature: read by a thermometer with a wet wick; it lies between dew point and dry bulb (all three are equal at φ = 100%).
- Enthalpy of moist air ≈ 1.005 t + ω(2500 + 1.88 t) kJ/kg dry air (t in °C).
Air-conditioning processes
| Process | ω | Dry bulb |
|---|---|---|
| Sensible heating / cooling | Constant | Rises / falls |
| Cooling and dehumidification | Falls | Falls |
| Heating and humidification | Rises | Rises |
| Evaporative cooling | Rises | Falls (wet bulb constant) |
| Adiabatic mixing of two streams | Mass-weighted average | Approximately mass-weighted |
Bypass factor of a cooling coil = (t_out − t_coil)/(t_in − t_coil).
Worked example
A refrigerator has h₁ = 400, h₂ = 430 and h₃ = h₄ = 250 kJ/kg.
- Refrigerating effect = 150 kJ/kg; work = 30 kJ/kg; COP = 5.
- For 1 TR: ṁ = 3.5167/150 = 0.0234 kg/s.
Practice
A refrigerator has refrigerating effect 120 kJ/kg and compressor work 40 kJ/kg. What is its COP?
Show answer and explanation
Answer: 3COP = 120/40 = 3.
Air at 101.325 kPa has a vapour pressure of 2 kPa. What is its humidity ratio in g/kg dry air? (Give 2 decimal places.)
Show answer and explanation
Answer: 12.52ω = 0.622 × 2/99.325 = 0.01252 kg/kg = 12.52 g/kg.
A plant needs 10 TR of cooling. What is that in kW? (Give 2 decimal places.)
Show answer and explanation
Answer: 35.171 TR = 3.5167 kW, so 10 TR = 35.17 kW.
During evaporative cooling of air, which stays approximately constant?
Show answer and explanation
Answer: CEvaporative (adiabatic) saturation follows a line of constant wet-bulb temperature: dry bulb falls and humidity rises.
A cooling coil at 10 °C cools air from 30 °C to 15 °C. What is the bypass factor?
Show answer and explanation
Answer: 0.25BF = (t_out − t_coil)/(t_in − t_coil) = 5/20 = 0.25.