Air refrigeration systems

Air refrigeration systems, Carnot refrigeration cycle, Brayton refrigeration or the Bell-Coleman air refrigeration cycle, Aircraft refrigeration system, Simple air cooling system, Simple air evaporative cooling system, Bootstrap air cooling, Bootstrap air evaporative air cooling, Regenerative air cooling, Reduced ambient air cooling, Comparison of different systems

1. Air Refrigeration Systems

2. Air cycle refrigeration systems use  air  as their refrigerant, compressing it and expanding it to create heating and cooling capacity. Heat flows in direction of decreasing temperature i.e. From high temperature to low temperature regions. But the reverse process (i.e. heat transfer from low to high temperature) cannot occur by itself (Claussius Definition of Second Law). This process requires a special device called Refrigerator. Carnot Refrigeration Cycle The Carnot cycle is used to convert the heat into the mechanical work whereas the reverse Carnot cycle is used to absorb the heat from the system and rejects to the surroundings to maintain the system cool. Carnot Refrigeration cycle is a totally reversible cycle which consists of two reversible isothermal processes and two isentropic processes. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump.   12/28/2021 2 Suhail Bhatia

3. The term  heat pump  is usually reserved for a device that can heat a house in winter by using an electric motor that does work W to take heat  Q cold  from the outside at low temperature and delivers heat  Q hot  to the warmer inside of the house. The operating principle of  refrigerators ,  air conditioners , and  heat pumps  is  the same,  and it is just the  reverse  of a  heat engine . In general, a  heat pump  is a device that transfers heat energy from a  heat source  to a “ heat sink ,” but in this case, the transfer occurs in the opposite direction of spontaneous heat transfer by  absorbing heat  from a  cold space  and  releasing it  to a  warmer one. 12/28/2021 3 Suhail Bhatia

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5. We know that; COP of Refrigerator = Heat absorbed / work supplied = Heat absorbed / (Heat rejected – Heat absorbed) = T 2 (S b -S a ) / [T 1 (S b -S a ) – T 2 (S b -S a ) = T 2 / (T 1 -T 2 ) Therefore, COP of Carnot Refrigerator is : where  T H  is the source temperature and  T L  is the sink temperature where heat is rejected (i.e., lake, ambient air, etc.). 12/28/2021 5 Suhail Bhatia

6. Brayton Refrigeration or The Bell Coleman Air Refrigeration System Brayton cycle is a thermodynamic cycle that describes the operation of certain heat engines that have air or some other gas as their working fluid. A  Brayton Refrigeration cycle  that is driven in the reverse direction is known as the  reverse Brayton cycle . Its purpose is to move heat from the colder to the hotter body rather than produce work. This cycle is also known as the gas refrigeration cycle or the Bell Coleman cycle . This type of cycle is widely used in jet aircraft for air conditioning systems using air from the engine compressors. It is also widely used in the LNG industry . 12/28/2021 6 Suhail Bhatia

7. In compliance with the second law of thermodynamics,  heat cannot spontaneously flow  from cold system to hot system without external work being performed on the system. Heat can flow from colder to the hotter body, but  only when forced by external work . This is exactly what refrigerators and heat pumps accomplish. These are driven by electric motors requiring work from their surroundings to operate. One of the possible cycles is a reverse Brayton cycle, which is similar to the ordinary Brayton cycle, but it is driven in reverse via net work input. 12/28/2021 7 Suhail Bhatia

8. Aircraft air refrigeration systems are required due to heat transfer from many external and internal heat sources (like solar radiation and avionics) which increase the cabin air temperature. With the technological developments in high-speed passenger and jet aircrafts, the air refrigeration systems are proving to be most efficient, compact and simple. Despite outside temperature is low refrigeration is required in inside the occupant cabin because : 1. large internal heat is generated due to occupant and equipment inside the cabin. 2.Heat generation due to skin friction caused by the fast moving aircraft. 3. At high altitude, the outside pressure will be sub-automatic. 4.Solar radiation. For low speed aircraft flying at low altitudes, cooling system may not be required, however, for high speed aircraft flying at high altitude, a cooling system is must. Aircraft Refrigeration Systems 12/28/2021 8 Suhail Bhatia

9. Various types of aircraft air refrigeration systems used these days are: Simple air cooling system Simple air evaporative cooling system Boot strap air cooling system Boot strap air evaporative cooling system Reduced ambient air cooling system Regenerative air cooling system. 12/28/2021 9 Suhail Bhatia

10. Simple Air Cooling System Ambient air gets rammed before the main compressor. The air required for refrigeration system is then bled off from this compressor. It is then sent to the heat exchanger where this high pressure and high temperature air is cooled using ram air. Air is further cooled in the cooling turbine due to its expansion. The turbine work drives the cooling fan which draws ram air through the heat exchanger. The cooled air from turbine is then sent to the aircraft cabin. 12/28/2021 10 Suhail Bhatia

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14. Process 1-2= Isentropic ramming of air, Process 2-3i= Isentropic compression in main compressor, Process 2-3= Actual compression in main compressor, Process 3-4= Constant pressure heat rejection in heat exchanger, Process 4-5i= Isentropic expansion in cooling turbine, Process 4-5= Actual expansion in cooling turbine, Process 5-6= Constant pressure heat addition in cabin 12/28/2021 14 Suhail Bhatia

15. If a Q tonnes of refrigeration is the Cooling load in the cabin then the air required for the refrigeration purpose, Power required for the refrigerating system, And C O.P. of the refrigerating system The initial mass of evaporant (Me) required to be carried for the given flight time is given by where Qe = Heat to be removed in evaporation in kJ/min, and Hfg = Latent heat of vaporisation of evaporant in kJ/kg 12/28/2021 15 Suhail Bhatia

16. Problems on aircraft cooling systems - Simple Refrigeration Cycle Example A simple air cooled system is used for an aeroplane having a load of 10 tones. The atmospheric pressure and temperature are 0.9 bar and 10°C respectively. The pressure increases to 1.013 bar due to ramming. The temperature of the air is reduced by 50°C in the heat exchanger. The pressure in the cabin is 1.01 bar and the temperature of air leaving the cabin is 25°C. Determine : 1 Power required to take the load of cooling in the cabin; and 2. C.O.P. of the system. Assume that all the expansions and compressions are isentropic. The pressure of the compressed air is 3.5 bar. Solution Given : Q=10TR; p 1=0.9 bar ; T 1=10° C =10+273=283 K; p 2=1.013 bar ; p 5= p 6=1.01 bar ; T 6=25° C =25+273=298 K ; p 3=3.5bar take λ = 1.4 12/28/2021 16 Suhail Bhatia

17. 1. Power required taking the load of cooling in the cabin First of all, let us find the mass of air ( ma ma) required for the refrigeration purpose. Since the compressions and expansions are isentropic, therefore the various processes on the T-s diagram are as shown in diagram below Let T 2  = Temperature of air at the end of ramming or entering the main compressor, T3 = Temperature of air leaving the main compressor after isentropic compression, T4 = Temperature of air leaving the heat exchanger, and T5 = Temperature of air leaving the cooling turbine 12/28/2021 17 Suhail Bhatia

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19. Simple Air Evaporative Cooling System 12/28/2021 19 Suhail Bhatia

20. It is similar to the simple cooling system except that the addition of an evaporator between the beat exchanger and cooling turbine. The evaporator provides an additional cooling effect through evaporation of a refrigerant such as water. At high altitudes, the evaporative cooling may be obtained by using alcohol or ammonia. The water, alcohol and ammonia have different refrigerating effects at different altitudes. At 20000 metres height for example, water boils at 400 C, alcohol at 90 C and ammonia at - 70° C. The T-s diagram for simple air cycle evaporative cooling system is shown below: 12/28/2021 20 Suhail Bhatia

21. If a Q tonnes of refrigeration is the Cooling load in the cabin then the air required for the refrigeration purpose, Power required for the refrigerating system, And C O.P. of the refrigerating system The initial mass of evaporant (Me) required to be carried for the given flight time is given by where Qe = Heat to be removed in evaporation in kJ/min, and Hfg = Latent heat of vaporisation of evaporant in kJ/kg 12/28/2021 21 Suhail Bhatia

22. Problem on simple air evaporative cooling system Example Simple evaporative air refrigeration system is used for an aeroplane to take 20 tones of refrigeration load. The ambient air conditions are 20°C and 0.9 bar. The ambient air is rammed isentropically to a pressure of 1bar. The air leaving the main compressor at pressure 3.5 bar is first cooled in the heat exchanger having effectiveness of 0.6 and then in the evaporator where its temperature is reduced by 5°C. The air from the evaporator is passed through the cooling turbine and then it is supplied to the cabin which is to be maintained at a temperature of 25°C and at a pressure of 1.05 bar. If the internal efficiency of the compressor is 80% and that of cooling turbine is 75% determine: 1. Mass of air bled off the main compressor; 2. Power required for the refrigerating system; and 3. C.O.P. of the refrigeration system. Solution Given: Q=20TR;  T1=20°C=20+273=293K; p1=0.9bar; p2=1bar; p3=p′3=3.5bar; η H=0.6; T6=25°C=25+273=298K; p6=1.05bar; η C=80%= 0.8 ; η T=75%=0.75 12/28/2021 22 Suhail Bhatia

23. The T-s diagram for the simple evaporative air refrigeration system with the given conditions is shown below: 12/28/2021 23 Suhail Bhatia

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27. Boot-strap Air Cooling 12/28/2021 27 Suhail Bhatia

28. This boot strap air cooling system has two heat exchangers instead of one, and a cooling turbine drives a secondary compressor instead of cooling fan. The air bled from the main compressor is first cooled by the ram air in the first heat exchanger. This cooled air, after compression in the secondary compressor, is led to the second heat exchanger where it is again cooled by the ram air before passing to the cooling turbine. This type of cooling system is mostly used in transport type aircraft. The T-s diagram for a boot-strap air cycle cooling system: 12/28/2021 28 Suhail Bhatia

29. The various processes are as follows: The process 1-2' represents the isentropic ramming of ambient air from pressure P1, and temperature T1 to pressure P2 and temperature T2. The process 1-2' represents the actual ramming process because of internal friction due to irreversibilities. The process 2'-3 represents the isentropic compression of air in the main compressor and the process 2'-3' represents the actual compression of air because of internal friction due to irreversibilities. The process 3'-4 represents the cooling by ram air in the first heat exchanger. The pressure drop in the heat exchanger is neglected. The process 4.5 represents compression of cooled air, from first heat exchanger, in the secondary compressor. The process 4-5' represents the actual compression process because of internal friction due to irreversibilities. The process 5'-6 represents the cooling by ram air in the second heat exchanger. The pressure drop in heat exchanger in neglected. The process 6-7 represents isentropic expansion of the cooled air in the cooling turbine upto the cabin pressure. The process 6-7' represents actual expansion of the cooled air in the cooling turbine. The process 7'-8 represents the heating of air upto the cabin temperature T8. 12/28/2021 29 Suhail Bhatia

30. If Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be Power required for the refrigerating system, and C.O.P. of the refrigerating system 12/28/2021 30 Suhail Bhatia

31. Problem on boot strap cooling system Example A boot-strap cooling system of 10 TR capacity is used in an aeroplane. The ambient air temperature and pressure are 20°C and 0.85 bar respectively. The pressure of air increases from 0.85 bar to 1 bar due to ramming action of air. "The pressure of air discharged from the main compressor is 3 bar. The discharge pressure of air from the auxiliary compressor is 4 bar. The isentropic efficiency of each of the compressor is 80%, while that of turbine is 85%. 50% of the enthalpy of air discharged from the main compressor is removed in the first heat exchanger and 30% of the enthalpy of air discharged from the auxiliary compressor is removed in the second heat exchanger using rammed air. Assuming ramming action to be isentropic, the required cabin pressure of 0.9 bar and temperature of the air leaving the cabin not more than 20° C, find : 1. the power required to operate the system, and 2. the C.O.P. of the system. Draw the schematic and temperature -entropy diagram of the system. Take  γ =1.4 and cp=1 kJ/kg K. 12/28/2021 31 Suhail Bhatia

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36. Boot-strap Evaporative Air Cooling 12/28/2021 36 Suhail Bhatia

37. A boot-strap air cycle evaporative cooling system is shown above. It is similar to the boot-strap air cycle cooling system except that the addition of an evaporator between the second heat exchanger and the cooling turbine. The T-s diagram for a boot-strap air evaporative cooling system: 12/28/2021 37 Suhail Bhatia

38. The various processes of this cycle are same as a simple boot-strap system except the process 5''-6 which represents cooling in the evaporator using any suitable evaporant. If Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be: Power required for the refrigeration system is given by: and C.O.P. of the refrigerating system Note : Since the temperature of air leaving the cooling turbine in boot-strap evaporative system is lower than the simple boot-strap system, therefore mass of air (ma) per tonne of refrigeration will be less in boot- strap evaporative system. 12/28/2021 38 Suhail Bhatia

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40. Problem on boot strap evaporative cooling system Example The following data refer to a boot strap air cycle evaporative refrigeration system used for an aeroplane to take 20 tonnes of refrigeration load: Ambient air temperature = 15°C Ambient air pressure = 0.8bar Mach number of the flight = 1.2 Ram efficiency = 90 Pressure of air bled off the main compressor = 4bar Pressure of air in the secondary compressor = 5bar Isentropic efficiency of the main compressor = 90 Isentropic efficiency of the secondary compressor = 80 Isentropic efficiency of the cooling turbine = 80 Temperature of air leaving the first heat exchanger =? Temperature of air leaving the second heat exchange = ? Temperature of air leaving the evaporator = 100°C Cabin temperature = 25°C Cabin pressure = 1bar Find: 1. Mass of air required to take the cabin load, 2. Power required for the refrigeration system, and 3. C.O.P. of the system. 12/28/2021 40 Suhail Bhatia

41. Solution Given: Q=20 TR ; T 1=15° C =15+237=288 K ; p 1=0.8 bar ; M =1.2; ηR =90 p 3= p ′3= p 4=4 bar ; p 5= p ′5= p 5”= p 6= bar ; ηC 1=90; ηC 2=80 T 6=100° C =100+273=373 K ; T 8=25° C =25+273=298 K ; p 8= p 7= p 7”=1 bar The T-s diagram for the boot-strap air cycle evaporative refrigeration system, with the given conditions, is shown in the figure below: 12/28/2021 41 Suhail Bhatia

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45. Regenerative Air Cooling System 12/28/2021 45 Suhail Bhatia

46. The regenerative air cooling-system is a modification of a simple air cooling system with the addition of a regenerative heat exchanger. The high pressure and high temperature air from the main compressor is first cooled by the ram air in the heat exchanger. This air is further cooled in the regenerative heat exchanger with a portion of the air bled after expansion in the cooling turbine. This type of cooling system is used for supersonic aircrafts and rockets. 12/28/2021 46 Suhail Bhatia

47. The T-s diagram for the regenerative air cooling system is shown above. The various processes are as follows: The process 1-2 represents' isentropic ramming of air and process 1-2' represents actual ramming of air because of internal friction due to irreversibilities. The process 2'-3 represents isentropic compression of air in the main compressor and the process 2'-3' represents actual compression of air because of internal friction due to irreversibilities. The process 3'-4 represents cooling of compressed air by ram air in the heat exchanger. The process 4-5 represents cooling of air in the regenerative heat exchanger. The process 5-6 represents isentropic expansion of air in the cooling turbine up to the cabin pressure and the process 5-6' represents actual expansion of air in the cooling turbine. The process 6'-7 represents heating of air upto the cabin temperature T7. 12/28/2021 47 Suhail Bhatia

48. If Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be For the energy balance of regenerative heat exchanger, we have where T8 = Temperature of air leaving to atmosphere from the regenerative heat exchanger. Power required for the refrigeration system: and C.O.P. of the refrigerating system: 12/28/2021 48 Suhail Bhatia

49. Problem on regenerative cooling system Example A regenerative air cooling system is used for an airplane to take 20 tonnes of refrigeration load. The ambient air at pressure 0.8 bar and temperature 10°C is rammed isentropically till the pressure rises to 1.2 bar. The air bled off the main compressor at 4.5 bar is, cooled by the ram air in the heat exchanger whose effectiveness is 60%. The air from the heat exchanger is further cooled to 60°C in the regenerative heat exchanger with a portion of the air bled after expansion in the cooling turbine. The cabin is to be maintained at a temperature of 25°C and a pressure of I bar. If the isentropic efficiencies of the compressor and turbine are 90% and 80% respectively, find: 1. Mass of the air bled from cooling turbine to be used for regenerative cooling; 2. Power required for maintaining the cabin at the required condition; and 3. C. O. P. of the system. Assume the temperature of air leaving to atmosphere from the regenerative heat exchanger as 1 00° C . Solution Given : Q=20TR; p1=0.8bar; T1=10°C=10+273=283K; p2=1.2bar; p3=p−4=p5=4.5bar ; η H=60%= 0.6; T5=60°C=60+273=333K; T7=25°C=25+273=298K ; p7=p6=p′6=1bar; η c=0.9 ; η T=80 %=0.8;=0.8; T8=100°C=100+273=373K 12/28/2021 49 Suhail Bhatia

50. The T-s diagram for the regenerative air cooling system with the given conditions is 12/28/2021 50 Suhail Bhatia

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54. Reduced Ambient Type System 12/28/2021 54 Suhail Bhatia

55. The reduced ambient air cooling system includes two cooling turbines and one heat exchanger. The air reduced for the refrigeration initially in system is bled off from the main compressor. This high pressure and high temperature air is cooled the heat exchanger. The air for cooling is taken from the cooling turbine which lower the high temperature of rammed air. The cooled air from the heat exchanger is passed through the second cooling turbine from where the air is supplied to the cabin. The work of the cooling turbine is used to drive the cooling fan (through reduction gears) which draws cooling air from the heat exchanger. The reduced ambient air cooling system is used for very high speed aircrafts. 12/28/2021 55 Suhail Bhatia

56. The T-s diagram for the reduced ambient air cycle cooling system is: 12/28/2021 56 Suhail Bhatia

57. The various processes are as follows: The process 1-2 represents isentropic ramming of air and the process 1-2' represents actual ramming of air because of internal friction due to irreversibilities. The process 2'-3 represents isentropic compression in the main compressor and the process 2'-3' represents actual compression of air, because of internal friction due to irreversibilities. The process 3'-4 represents cooling of compressed air by ram air which after passing through the first cooling turbine is led to the heat exchanger. The pressure drop in the heat exchanger is neglected. The process 4-5 represents isentropic expansion of air in the second cooling turbine upto the cabin pressure. The actual expansion of air in the second cooling turbine is represented by the curve 4-5'. The process 5'-6 represents the heating of air upto the cabin temperature T6. 12/28/2021 57 Suhail Bhatia

58. If Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be: Power required for the refrigeration system is given by and C.O.P. of the system 12/28/2021 58 Suhail Bhatia

59. Problems on reduced ambient cooling system Example The reduced ambient air refrigeration system used for an aircraft consists of two cooling turbines, one heat exchanger and one air cooling fan. The speed of aircraft is 1500 km/h. The ambient air conditions are 0.8 bar and 10° C. The ram efficiency may be taken as 90%. The rammed air used for cooling is expanded in the first cooling turbine and leaves it at a pressure of 0.8 bar. The air bled from the main compressor at 6 bar is cooled in the heat exchanger and leaves it at 100° C: The cabin is to be maintained at 20° C and 1 bar. The pressure loss between the second cooling turbine and cabin is 0.1 bar. If the isentropic efficiency for the main compressor and both of the cooling turbines are 85% and 80% respectively, find: 1. mass flow rate of air supplied to cabin to take a cabin load of 10 tonnes of refrigeration ; 2. quantity of air passing through the heat exchanger if the temperature rise of ram air is limited to 80 K ; 3. power used to drive the cooling fan; and 4. C.O.P. of the system . Solution Given : V=1500 km/h=417m/s ;p2=0.6bar; T1=10°C=10+273=283K; η R=90%=0.9 ; p3=p4=6bar; T4=100°C=100+273=373K; T6=20°C=20+273=293K ; p6=1bar; η C=85%=0.85 ; η T1=80%=0.8 ; Q=10TR 12/28/2021 59 Suhail Bhatia

60. The T-s diagram for the reduced ambient air refrigeration system with the given conditions is below: 12/28/2021 60 Suhail Bhatia

61. Let T ′2 T2′=Stagnation temperature of ambient air entering the main compressor, p 2 p2=Pressure of air at the end of isentropic ramming, and p ′2 p2′= Stagnation pressure of air entering the main compressor. 12/28/2021 61 Suhail Bhatia

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63. T-s diagram of first cooling tube 12/28/2021 63 Suhail Bhatia

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66. Comparison of different systems Comparison of Various Air Cooling Systems used for Aircraft Dry Air Rated Temperature (DART):  The concept of Dry Air Rated Temperature is used to compare different aircraft refrigeration cycles. Dry Air Rated Temperature is defined as the temperature of the air at the exit of the cooling turbine in the absence of moisture condensation. For condensation not to occur during expansion in the  turbine , the dew point temperature and hence moisture content of the air should be very low, i.e., the air should be very dry. The aircraft refrigeration systems are rated based on the mass flow rate of air at the design DART. The cooling capacity is then given by: where, m is the mass flow rate of air,  T   DART  and  Ti  are the dry air rated temperature and cabin temperature, respectively. 12/28/2021 66 Suhail Bhatia

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68. The performance curves for the various air cooling systems used for aircrafts are shown above. These curves show the dry air rated turbine discharge temperature (DART) against the Mach number. From the figure, we see that the simple air cooling system gives maximum cooling effect on the ground surface and decreases as the speed of aircraft increases. The boot strap system on the other hand, requires the air plane to be in flight so that the ram air can be used for cooling in the heat exchangers. One method of overcoming this drawback of boot strap system is to use part of work derived from turbine to drive a fan which pulls air over the secondary heat exchanger, thus combining the features of a simple and boot strap system .As the speed of aircraft increases, the temperature of ram cooling air rises and the ram air becomes less effective as a coolant in the heat exchanger. In such cases, a suitable evaporant is used with the ram air so that the cabin temperature does not rise. For high speed aircrafts, the boot strap evaporative or regenerative systems are used because they give lower turbine discharge temperature than the simple cooling system. In some cases, aeroplanes carry an auxiliary gas-,turbine for cabin pressurisation and, air conditioning. From the chart, we see that the turbine. Discharge temperature of the air is variable. Therefore, in order to maintain the content temperature of simply air to the cabin, it requires some control system. 12/28/2021 68 Suhail Bhatia

69. Summary: DART increases monotonically with Mach number for all the systems except the reduced ambient system. The simple system is adequate at low Mach numbers. At high Mach numbers either bootstrap system or regenerative system should be used. Reduced ambient temperature system is best suited for very high Mach number, supersonic aircrafts. 12/28/2021 69 Suhail Bhatia

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