Refrigeration and Air-Conditioning

What to remember

• Refrigeration removes heat from a cold space and rejects it to a warmer one using work (vapour compression) or heat (vapour absorption). COP = refrigerating effect ÷ work input. 1 ton of refrigeration (TR) = 3.517 kW.
• Vapour-compression cycle: compressor, condenser, expansion (throttle) valve, evaporator. COP = (h₁ − h₄)/(h₂ − h₁) where 1 is the evaporator exit, 2 the compressor exit, 4 the evaporator inlet.
• Air-conditioning controls temperature, humidity, air motion and cleanliness. Comfort is about 22 to 26 °C and 40 to 60% relative humidity. Psychrometric processes (sensible cooling, dehumidification) decide the cooling load.

1. Basic terms

• Refrigeration effect: heat absorbed in the evaporator per kg of refrigerant (kJ/kg).
• Ton of refrigeration: heat removal rate to freeze one tonne of water at 0 °C into ice at 0 °C in 24 hours. 1 TR = 211 kJ/min = 3.517 kW. Example: a 10 kW cooling duty = 10/3.517 = 2.84 TR.
• COP = refrigeration effect ÷ work input. Carnot (reversed) COP = T_L/(T_H − T_L). Example: evaporator at 250 K and condenser at 300 K gives COP = 250/50 = 5. Real cycles are lower.
• Heat pump COP = COP of refrigerator + 1.
• Relative COP = actual COP ÷ Carnot COP between the same temperatures.

2. Vapour-compression cycle

Four processes on the p–h chart:

• 1. Compression in compressor (1 → 2): dry saturated vapour is compressed isentropically to high pressure and superheated.
• 2. Condensation (2 → 3): heat rejected in the condenser at constant pressure; leaves as saturated liquid.
• 3. Expansion (3 → 4): throttling at constant enthalpy (h₃ = h₄). Pressure and temperature fall. Some liquid flashes to vapour.
• 4. Evaporation (4 → 1): heat absorbed at constant pressure and temperature.

Formulas: refrigerating effect = h₁ − h₄; compressor work = h₂ − h₁; heat rejected = h₂ − h₃; COP = (h₁ − h₄)/(h₂ − h₁).

Worked example: h₁ = 250, h₂ = 290, h₃ = h₄ = 100 kJ/kg. COP = (250 − 100)/(290 − 250) = 3.75. If the cooling load is 7 kW and the refrigerating effect is 140 kJ/kg, the mass flow is 7/140 = 0.05 kg/s.

Effect of operating changes:

• Wet compression (liquid droplets in suction vapour) damages the compressor. So dry compression is preferred.
• Throttling uses an expansion valve (thermostatic or float) or a capillary tube in domestic units. An expander would recover work but is not used.
• Compressor types: reciprocating (small to medium), rotary and scroll (small), screw (medium), centrifugal (large chillers).

3. Refrigerants

• Desirable: high latent heat, low freezing point, moderate pressures, non-toxic, non-flammable, chemically stable, low cost, compatible with oil, high thermal conductivity, low ozone depletion potential (ODP) and low global warming potential (GWP).
• Designation: R-12 (CCl₂F₂) is a CFC; R-22 is an HCFC; R-134a is an HFC (zero ODP, but significant GWP); R-717 is ammonia; R-718 is water; R-744 is carbon dioxide; R-290 is propane; R-600a is isobutane.
• Classes: primary refrigerants (circulate and absorb heat directly in the evaporator); secondary refrigerants (brines such as chilled water or calcium chloride solution that carry cold from a chiller to the load).
• Montreal Protocol phased out CFCs because they damage the ozone layer. HCFCs are being phased out, and HFCs are being phased down for climate reasons.
• Ammonia: excellent thermodynamic properties, used in large industrial and cold-storage plants, toxic and corrosive to copper. Freon-type refrigerants are used in small units.

4. Vapour-absorption system

• Uses heat instead of mechanical work. Components: absorber, solution pump, generator, condenser, expansion valve, evaporator (and rectifier and heat exchanger in ammonia plants).
• Working pairs: ammonia (refrigerant) with water (absorbent); water (refrigerant) with lithium bromide (absorbent). Lithium bromide systems give chilled water above 0 °C for air conditioning.
• The absorber takes refrigerant vapour into the weak solution. The pump raises its pressure with very little work. The generator heats the strong solution to drive off the refrigerant vapour at high pressure.
• COP = refrigeration effect ÷ heat supplied to generator (the pump work is small). It is usually below 1. Advantages: waste heat or solar heat can run it; quiet; few moving parts.
• Electrolux refrigerator uses ammonia, water and hydrogen with no pump; heat alone drives circulation.

5. Air refrigeration and other systems

• Bell-Coleman (reversed Brayton) cycle: compress air, cool it, expand it in a turbine to get cold air. COP = 1/(r_p^((γ−1)/γ) − 1). Example: pressure ratio 4 and γ = 1.4 gives COP = 1/(1.486 − 1) = 2.06.
• Air cycle is used in aircraft cooling because the working fluid is cheap and light, and plenty of pressurised air is available.
• Steam jet refrigeration uses water as the refrigerant and a steam ejector to produce vacuum. It suits large chilled water duties where steam is available.
• Thermoelectric (Peltier) cooling has no moving parts, used in small special devices.
• Dry ice is solid carbon dioxide that sublimates at about −78 °C.

6. Air-conditioning

• Comfort air-conditioning: 22 to 26 °C, 40 to 60% RH, low air velocity (below about 0.25 m/s) and clean air. Industrial air-conditioning targets the process.
• Cooling load = sensible heat + latent heat. Sources: heat from walls, roof, glass (solar gain), people, lights, equipment, and fresh (ventilation) air.
• Room sensible heat factor RSHF = room sensible heat/(room sensible + room latent heat). It fixes the slope of the supply-air line on the psychrometric chart.
• Cooling and dehumidifying coil: apparatus dew point (ADP) is the effective surface temperature of the coil. Bypass factor BPF = (t_out − t_ADP)/(t_in − t_ADP). Contact factor = 1 − BPF. A deeper coil (more rows) lowers BPF.
• Summer systems cool and dehumidify; winter systems heat and humidify. Unitary systems: window, split and packaged. Central systems use chilled-water plants with air handling units.
• Air distribution: ducts sized by equal friction or velocity methods; air-changes per hour and fresh-air quantity are set to the occupancy. Filters remove dust. Cooling towers reject condenser heat in water-cooled plants.

7. Components and practical points

• Evaporators: flooded (liquid refrigerant covers the surface; high heat transfer) and dry expansion (small liquid amount; superheated exit). Finned coils increase air-side area.
• Condensers: air-cooled (small units), water-cooled (shell-and-tube, with cooling tower or river water), and evaporative (water spray plus air). A higher condensing temperature means higher compressor power.
• Thermostatic expansion valve (TEV) senses superheat at the evaporator exit and adjusts the refrigerant flow. Float valves keep a constant liquid level in flooded evaporators.
• Accumulator, receiver, drier and sight glass are common accessories. A liquid-suction heat exchanger sub-cools the liquid and superheats the suction vapour.
• Defrosting removes frost from the evaporator coil, because frost insulates the surface and reduces heat transfer.
• Cold storage uses insulated walls (such as polyurethane or cork) and a vapour barrier. Moisture in insulation destroys its value.
• Ventilation is needed to supply fresh air and remove carbon dioxide and odours. Unitary units and chilled-water plants both need regular filter cleaning and refrigerant leak checks.

Exam traps

• COP of a refrigerator can be more than 1; efficiency of a heat engine cannot.
• Throttling in the expansion valve is constant enthalpy, not isentropic.
• In the p–h chart, compression is almost along an isentropic line, with enthalpy increasing.
• R-12 is a CFC, R-22 is an HCFC, R-134a is an HFC.
• The vapour-absorption system replaces the compressor by absorber, pump and generator; the condenser, expansion valve and evaporator stay.
• Ammonia is the refrigerant in the NH₃-water pair; in the LiBr system water is the refrigerant.
• Latent heat loads change moisture; sensible loads change dry-bulb temperature.
• Lower bypass factor means a more efficient coil.

One-liners

• 1. 1 TR = 3.517 kW (210 to 211 kJ/min).
• 2. Refrigerant in a domestic refrigerator is commonly an HFC or a hydrocarbon such as isobutane.
• 3. Ammonia is R-717; carbon dioxide is R-744.
• 4. COP of heat pump = COP of refrigerator + 1.
• 5. Condenser rejects heat to the surroundings.
• 6. The capillary tube replaces the expansion valve in small units.
• 7. LiBr–water absorption system serves air-conditioning.
• 8. Ammonia–water absorption system can reach below 0 °C.
• 9. Bell-Coleman cycle is a reversed Brayton cycle.
• 10. Comfort RH is about 40 to 60%.
• 11. Contact factor = 1 − bypass factor.
• 12. The Montreal Protocol deals with ozone-depleting substances.

Practice questions

1. One ton of refrigeration is equal to about

   1. 1.0 kW
   2. 3.517 kW
   3. 35.17 kW
   4. 0.3517 kW
   Answer

   B. 3.517 kW

   1 TR = 211 kJ/min = 3.517 kW.

2. A 5 TR air-conditioning unit has a cooling capacity of about

   1. 5.0 kW
   2. 7.0 kW
   3. 35.2 kW
   4. 17.6 kW
   Answer

   D. 17.6 kW

   5 × 3.517 = 17.585 kW.

3. The coefficient of performance of a refrigerator is the ratio of

   1. refrigerating effect to work input
   2. heat rejected to work input minus heat rejected
   3. work input to refrigerating effect
   4. work output to heat input
   Answer

   A. refrigerating effect to work input

   COP = Q_evaporator/W.

4. A reversed Carnot refrigerator works between an evaporator at 250 K and a condenser at 300 K. Its COP is

   1. 0.2
   2. 1.2
   3. 6
   4. 5
   Answer

   D. 5

   COP = T_L/(T_H − T_L) = 250/50.

5. A refrigerator of COP 3.5 operated as a heat pump between the same limits has COP

   1. 2.5
   2. 4.5
   3. 3.5
   4. 0.29
   Answer

   B. 4.5

   COP_HP = COP_ref + 1.

6. In the ideal vapour-compression cycle the expansion through the throttle valve is a process of constant

   1. pressure
   2. temperature
   3. enthalpy
   4. entropy
   Answer

   C. enthalpy

   Adiabatic throttling with no work: h₃ = h₄.

7. In the condenser of a vapour-compression system, heat is rejected at

   1. constant pressure
   2. constant enthalpy
   3. constant volume
   4. constant entropy
   Answer

   A. constant pressure

   Desuperheating and condensation occur at high constant pressure.

8. In a cycle h₁ = 250, h₂ = 290 and h₃ = h₄ = 100 kJ/kg. The COP is

   1. 0.27
   2. 6.25
   3. 3.75
   4. 4.75
   Answer

   C. 3.75

   COP = (250 − 100)/(290 − 250) = 150/40.

9. A cooling load of 7 kW is met with a refrigerating effect of 140 kJ/kg. The refrigerant mass flow is

   1. 0.5 kg/s
   2. 0.05 kg/s
   3. 20 kg/s
   4. 0.02 kg/s
   Answer

   B. 0.05 kg/s

   m = 7/140 = 0.05 kg/s.

10. With m = 0.05 kg/s and h₂ − h₁ = 40 kJ/kg, the compressor power is

    1. 8 kW
    2. 20 kW
    3. 0.8 kW
    4. 2 kW
    Answer

    D. 2 kW

    P = 0.05 × 40 = 2 kW.

11. With h₂ = 290 and h₃ = 100 kJ/kg, the heat rejected in the condenser per kg is

    1. 190 kJ/kg
    2. 390 kJ/kg
    3. 40 kJ/kg
    4. 150 kJ/kg
    Answer

    A. 190 kJ/kg

    q = h₂ − h₃.

12. Lowering the evaporator pressure while keeping the condenser pressure constant generally

    1. does not change the COP
    2. reduces compressor work to zero
    3. reduces the COP
    4. raises the COP
    Answer

    C. reduces the COP

    Refrigerating effect falls and compression work rises.

13. Subcooling the liquid refrigerant leaving the condenser

    1. increases compressor work greatly
    2. has no effect on enthalpy at expansion valve exit
    3. decreases the refrigerating effect
    4. increases the refrigerating effect
    Answer

    D. increases the refrigerating effect

    Lower h₃ = h₄ gives bigger h₁ − h₄.

14. Wet compression (liquid droplets entering the compressor) is avoided because it

    1. raises the COP
    2. can damage the compressor
    3. reduces condenser load to zero
    4. is needed for lubrication
    Answer

    B. can damage the compressor

    Liquid is incompressible and causes hammering.

15. In small domestic refrigerators the expansion device is usually a

    1. turbine
    2. capillary tube
    3. float of a boiler
    4. compressor valve
    Answer

    B. capillary tube

    A long narrow tube gives the required pressure drop.

16. R-12 (dichlorodifluoromethane) is classified as a

    1. HCFC
    2. hydrocarbon
    3. HFC
    4. CFC
    Answer

    D. CFC

    It contains chlorine, fluorine and carbon only.

17. Statements: 1. R-134a has zero ozone depletion potential. 2. R-12 is a chlorofluorocarbon. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    R-134a has no chlorine; R-12 is a CFC.

18. The refrigerant number for ammonia is

    1. R-717
    2. R-290
    3. R-744
    4. R-718
    Answer

    A. R-717

    700 series are inorganic refrigerants: 717 ammonia, 718 water, 744 CO₂.

19. Carbon dioxide as a refrigerant is designated

    1. R-134a
    2. R-600a
    3. R-717
    4. R-744
    Answer

    D. R-744

    R-744 is CO₂ (molecular mass 44).

20. A secondary refrigerant such as brine is used to

    1. carry cold from the chiller to the cooled space
    2. compress the vapour
    3. produce vacuum in the generator
    4. replace the condenser
    Answer

    A. carry cold from the chiller to the cooled space

    It transfers heat without evaporating itself.

21. The Montreal Protocol is concerned with

    1. noise from compressors
    2. cost of electricity
    3. substances that deplete the ozone layer
    4. water purity in cooling towers
    Answer

    C. substances that deplete the ozone layer

    CFCs were phased out under it.

22. A vapour-absorption refrigeration system is driven mainly by

    1. a large electric motor driving a compressor
    2. heat energy
    3. high-speed turbine work
    4. a centrifugal fan only
    Answer

    B. heat energy

    Heat is supplied to the generator.

23. In an ammonia-water absorption system, the absorbent is

    1. ammonia
    2. lithium bromide
    3. hydrogen
    4. water
    Answer

    D. water

    Water absorbs ammonia vapour; ammonia is the refrigerant.

24. In a lithium bromide absorption chiller, the refrigerant is

    1. ammonia
    2. lithium bromide
    3. water
    4. R-22
    Answer

    C. water

    LiBr is the absorbent; water vapour is the refrigerant.

25. Statements on an absorption system: 1. The solution pump needs very little work compared with a vapour compressor. 2. The COP is based on the heat supplied to the generator. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Pumping a liquid is cheap; heat input is the cost.

26. In an absorption system, the refrigerant vapour is driven out of the strong solution in the

    1. generator
    2. condenser
    3. absorber
    4. evaporator
    Answer

    A. generator

    Heat in the generator liberates the vapour at high pressure.

27. The Electrolux refrigerator differs from other absorption systems because it

    1. uses a screw compressor
    2. has no pump and uses hydrogen to balance pressure
    3. needs lithium bromide
    4. uses steam jets
    Answer

    B. has no pump and uses hydrogen to balance pressure

    Heat alone drives it; hydrogen equalises total pressure.

28. The Bell-Coleman air refrigeration cycle is the

    1. reversed Diesel cycle
    2. reversed Rankine cycle
    3. reversed Otto cycle
    4. reversed Brayton (Joule) cycle
    Answer

    D. reversed Brayton (Joule) cycle

    Gas is compressed, cooled at constant pressure and expanded to give cold air.

29. A Bell-Coleman cycle works with pressure ratio 4 and γ = 1.4 (4^0.2857 ≈ 1.486). The ideal COP is about

    1. 2.06
    2. 1.49
    3. 0.49
    4. 4.0
    Answer

    A. 2.06

    COP = 1/(r^((γ−1)/γ) − 1) = 1/0.486.

30. Air-cycle refrigeration is common for

    1. large cold storages
    2. aircraft cooling
    3. household water coolers
    4. domestic refrigerators
    Answer

    B. aircraft cooling

    Light, safe working fluid and bleed air are available on aircraft.

31. The refrigerant in a steam jet refrigeration system is

    1. ammonia
    2. R-12
    3. mercury vapour
    4. water
    Answer

    D. water

    A steam ejector keeps a vacuum so water boils at low temperature.

32. Solid carbon dioxide (dry ice) changes directly to gas at atmospheric pressure at about

    1. −78 °C
    2. 0 °C
    3. −150 °C
    4. −20 °C
    Answer

    A. −78 °C

    It sublimes at about −78 °C.

33. For human comfort the usual relative humidity range is about

    1. 80 to 100%
    2. 40 to 60%
    3. 10 to 20%
    4. 0 to 5%
    Answer

    B. 40 to 60%

    Along with 22 to 26 °C.

34. The room sensible heat factor is

    1. total heat divided by fresh-air load
    2. coil load divided by compressor power
    3. room sensible heat divided by (sensible plus latent) room heat
    4. room latent heat divided by sensible heat
    Answer

    C. room sensible heat divided by (sensible plus latent) room heat

    RSHF = RSH/(RSH + RLH).

35. The total cooling load of a room is the sum of

    1. only sensible loads
    2. compressor and condenser work
    3. sensible and latent heat loads
    4. pump and fan work only
    Answer

    C. sensible and latent heat loads

    Both temperature and moisture are controlled.

36. The apparatus dew point (ADP) of a cooling coil is

    1. the compressor discharge temperature
    2. the temperature of room air
    3. the dew point of outdoor air
    4. the effective surface temperature of the coil
    Answer

    D. the effective surface temperature of the coil

    Air leaving the coil lies on the line from inlet state toward ADP.

37. Air enters a cooling coil at 30 °C and leaves at 14 °C. The coil ADP is 10 °C. The bypass factor is

    1. 0.2
    2. 0.4
    3. 0.8
    4. 0.25
    Answer

    A. 0.2

    BPF = (14 − 10)/(30 − 10) = 0.2.

38. Statements: 1. Contact factor = 1 − bypass factor. 2. Adding more rows to a cooling coil lowers the bypass factor. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    More surface contact means less air bypasses.

39. One ton of refrigeration expressed in kJ/min is about

    1. 21.1
    2. 211
    3. 35
    4. 3517
    Answer

    B. 211

    1 TR = 211 kJ/min.

40. A refrigerator has actual COP 3 while the Carnot COP between the same temperatures is 5. The relative COP is

    1. 1.67
    2. 0.6
    3. 15
    4. 2.0
    Answer

    B. 0.6

    Relative COP = 3/5.

41. A 1 TR load is handled by a unit with COP 3.517. The power input is

    1. 1 kW
    2. 12.4 kW
    3. 3.517 kW
    4. 0.28 kW
    Answer

    A. 1 kW

    P = 3.517/3.517 = 1 kW.

42. A heat pump with COP 4 supplies 12 kW of heating. The work input is

    1. 8 kW
    2. 16 kW
    3. 48 kW
    4. 3 kW
    Answer

    D. 3 kW

    W = Q/COP = 12/4.

43. For large chilled-water plants, the type of compressor most commonly used is

    1. small rotary vane
    2. hermetic scroll only
    3. centrifugal
    4. hand-operated
    Answer

    C. centrifugal

    Large volume flows suit centrifugal compressors.

44. The refrigerant leaving the expansion valve is

    1. a superheated vapour at high pressure
    2. a dry vapour at condenser pressure
    3. a sub-cooled liquid at high pressure
    4. a low-pressure liquid-vapour mixture
    Answer

    D. a low-pressure liquid-vapour mixture

    Throttling flashes part of the liquid to vapour.

45. Which common refrigerant is toxic and attacks copper and its alloys?

    1. Ammonia
    2. Isobutane
    3. Carbon dioxide
    4. R-134a
    Answer

    A. Ammonia

    Ammonia needs steel piping; it is used in large industrial plants.