AIR REFRIGERATION CYCLE.ENGINEERING NOTES FOR MECHANICAL ENGINEERSpptx

ENGINEERING NOTES FOR MECHANICAL ENGINEERS

1. NUMERICALS /Systems

2. INTRODUCTION OF AIR REFRIGERATION CYCLE In advent of high-speed passenger aircraft, jet aircraft and missiles has introduced the need for compact and simple refrigeration systems, capable of high capacity with minimum reduction of pay Load. The air cycle systems usually prove to be most Efficient. Type of Air-craft Tonne Require An Ordinary passenger Aircraft 08TR Fighter Jet at 950Km/h speed 10 to 20 TR Missiles & other High-Speed Jet 03 TR Advantages of Air Refrigeration System: - Easily available no cost for Refrigerant. Non-Toxic and Non-Inflammable. Leakage of air is tolerable. Main compressor used no extra require. Air is light in weight per TR. No any Requirement of separator for air. Fabrication and Maintenance is Simple. Disadvantages of Air Refrigeration System: - Lower COP as compared to others Running cost is high. Mass of air required is very high.

3. Open and closed air refrigeration cycles: - Open Air Refrigeration Cycle: - In an open refrigeration system, the air is directly passed over the space is to be cooled, and allowed to circulate through the cooler. The pressure of open refrigeration cycle is limited to the atmospheric pressure. A simple diagram of the open-air Refrigeration system is given below. Closed Air Refrigeration Cycle: - In closed or dense air refrigeration cycle, air refrigerant is contained within pipes and component part of the system at all time. The circulated air does not have to direct contact with the space to be cooled. The air is used to cool another fluid (brine), and this fluid is circulated into the space to be cooled.

4. Bell Coleman or Reversed Joule air refrigeration cycle: The Bell-Coleman Cycle, also known as the Air Refrigeration Cycle, is a thermodynamic cycle used in air refrigeration systems to provide cooling. It was developed by Alexander Graham Bell and Charles E . Coleman in the late 19th and early 20th centuries. The Bell-Coleman cycle is a reversed Brayton cycle that comprises two isentropic processes and two constant-pressure (isobaric) processes. It efficiently absorbs heat from a low-temperature cabin and transfers it to a high-temperature cabin. This versatile cycle is utilised for both heating and cooling purposes, employing air as the refrigerant or working fluid to achieve the desired thermal effects. The four processes in the Bell-Coleman cycle are: 1-2: - Isentropic compression from P1 to P2. 2-3: - Constant pressure heat rejection. 3-4: - Isentropic expansion from P3 to P4. 4-1: - Constant pressure heat addition process .

5. Bell Coleman or Reversed Joule air refrigeration cycle: COMP. 1 2 3 4 VOLUME (V) PRESSURE (P) COOLER 1 2 3 4 Isen . Comp. Isen . Exp. Constt . Pres. Constt . Pres. Q R Q A EXP. REFRIGERATOR 1-2: Isentropic Compression The first step of the Bell-Coleman Cycle involves drawing air from the refrigerator into the compressor cylinder, where it undergoes isentropic compression. During this process, no  heat  is transferred. The air is compressed, resulting in a decrease in volume from V1 to V2, an increase in pressure from P1 to P2, and an increase in temperature from T1 to T2, while entropy remains constant. 2-3: Constant pressure heat rejection Following isentropic compression, the warm air is passed through a cooler, where it is cooled at constant pressure. The cooler can be air-cooled or water-cooled, with the latter being commonly used. In this cooling process, pressure remains constant while temperature, volume, and  entropy  decrease. 3-4: Isentropic Expansion In this step, the cooled air is expanded isentropically , resulting in an increase in volume and a decrease in pressure. Since the process is isentropic, entropy remains constant while temperature decreases. 4-1: Constant pressure heat addition The last step of the Bell-Coleman Cycle involves the cooled and expanded air absorbing heat from the refrigerator. Heat is transferred from the refrigerator to the air during this process. As a result, the air expands from volume V4 to V1, temperature increases from T4 to T1, and entropy also increases from S4 to S1. (Heat extracted from the refrigerator or the refrigerating effect produced) during constant pressure expansion per kg of air is We know that work done during the cycle per kg of air = Heat rejected - Heat absorbed = =   ⸫ Coefficient of performance,  

6. COP CALCULATION IN DETAILS Work done during compression (W comp)   Work done during expansion (W exp)   Theoretical Coefficient of Performance (COP)  

7. NUMERICALS A refrigerator working on Bell-Coleman cycle operates between pressure limits of 1.10bar and 9.5bar. Air is drawn from the cold chamber at 11 C, compressed and then it is cooled to 31 C before entering the expansion cylinder. The expansion and compression follow the law PV 1.3 =constant. Determine the theoretical C.O.P of the system P 1 =P 4 =1.10 bar P 2 =P 3 =9.50 bar Substituting the given values: So, the temperature after compression (T 2 ​) is 491.03 K . T 1 =284K T 2 =491.03K Step 1: Temperature after compression (T₂) Step 2: Work done during compression (W comp) The work done during the compression process is given by:  

8. P 1 =P 4 =1.10 bar P 2 =P 3 =9.50 bar T 2 =491.03K T 1 =284K Step 3: Temperature after expansion (T₄) For a polytropic expansion process , the relation is similar to compression. Using the polytropic relation: Substituting the given values: So, the temperature after expansion (T 4 ) is 234.6 K . T 3 =304K T 4 =234.6K Step 4: Work done during expansion (W exp) The work done during expansion is given by:   Step 5: Refrigeration effect (Q in) The refrigeration effect is the heat absorbed from the cold chamber, which happens during the constant-pressure process from T 1 to T 4 ​ Step 6: Theoretical Coefficient of Performance (COP) The COP of the Bell-Coleman cycle is given by the ratio of the refrigeration effect to the net work done:  

9. Methods of Air Refrigeration Systems The various methods of air refrigeration systems used for aircrafts these days are as following 1. Simple air-cooling system, 2. Simple air evaporative cooling system, 3. Boot strap air cooling system, 4. Boot strap air evaporative cooling system, 5. Reduced ambient air-cooling system, and 6. Regenerative air-cooling system. 1. Simple Air-Cooling System: - A simple air-cooling system for aircrafts is shown in Fig. The main components of this system are the main compressor driven by a gas turbine. a heat exchanger, a cooling turbine and a cooling air fan. The air required for refrigeration system is bled off from the main compressor. This high pressure and high temperature air is cooled initially in the heat exchanger where ram air is used for cooling. It is further cooled in the cooling turbine by the process of expansion. The work of this turbine is used to drive the cooling fan which draws cooling air through the heat exchanger. This system is good for ground surface cooling and for low flight speeds. The T-s diagram for a simple air-cooling system is shown in Fig.

10. 1. Ramming process . Let the pressure and temperature of ambient air is P 1 and T 1 respectively. The ambient air is rammed isentropically from pressure P 1 and temperature T 1 to the pressure P 2 and temperature T 2. This ideal ramming action is shown by the vertical line l-2 in fig. If V is the aircraft velocity or the velocity of air relative to the aircraft in metres per second. Then kinetic energy of outside air relative to aircraft, From the energy equation, we know that   The temperature T 2 = T 2 ’ is called the Stagnation Temperature of the ambient air entering the compressor. The stagnation pressure after isentropic compression (P 2 ) is given by   Due to the irreversible compression in the ram, the air reaches point 2' instead of point 2 at the same stagnation temperature but at a reduced stagnation pressure P2'. The pressure P2' may be obtained from the expression of ram efficiency ( η R ) which is given as

11. 2. Compression Process . The isentropic compression of air in the main compressor is represented by the line 2’- 3. In actual practice, because of internal friction, due to irreversibility’s, the actual compression curve is represented by 2’- 3’. The work done during compression process is given by. Where m a =mass of air bled from the main compressor for refrigeration purposes.   3. Cooling Process. The compressed air is cooled by the ram air in the heat exchanger this process is shown by the curve 3’- 4 in fig. the heat rejected in the heat exchanger during the cooling process.   4. Expansion Process. The cooled air is now expanded isentropically in the cooling turbine as shown by the curve 4 – 5.In actual practice, the actual expansion in the cooling turbine is shown by the curve 4 – 5’. Then the work done by the cooling turbine will be given by the equation. The work of this turbine is used to drive the cooling air fan which draws cooling air from the heat exchanger.  

12. 5. Refrigeration Process. The air from the cooling turbine is sent to the cabin and the cock pit where it gets heated by the heat of equipment and occupancy. This process is shown by the curve 5’ – 6. The refrigerating effect produced or heat absorbed is given by Where, T 6 =Inside temp. of cabin T 5 ’=Exit temp. of cooling turbine We know that C.O.P. of air Cycle is If Q tonnes of refrigeration is the cooling load in the cabin, then the air required for the refrigeration purpose, Power required for the refrigeration system is and C.O.P. of the refrigerating system  

13. ENTROPY 3 2 2 3 4 5 Cooled air to cabin Ambient Air Turbine Free Comp. Main Comp. COOLER 1 1 4 5 P3=P4 1.25 bar TEMPERATURE 1 bar .2 bar 1. An air conditioning system unit of a pressurised aircraft receives its air from the jet engine compressor at a pressure of 1.25 bar. The ambient pressure and temperature are 0.25 bar and 237 K respectively. The air-conditioning unit consists of a freewheeling compressor and turbine mounted on one shaft. The work produced by the turbine is sufficient to drive the compressor. The compressed air is then cooled in the cooler at constant pressure and then expanded in the turbine to the cabin pressure of 1 bar and temperature of 280 K. Calculate the compressor discharge pressure and the cooler exit temperature. Given Data: P 2 = 1.25 bar ; P 1 = 0.2 bar; T 1 = 237 K ; P 5 = 1 bar ; T 5 = 280 K

14. Given Data: P 2 = 1.25 bar ; P 1 = 0.2 bar; T 1 = 237 K ; P 5 = 1 bar ; T 5 = 280 K Since the turbine drives the compressor, the work produced by the turbine equals the work consumed by the compressor:  

15. 2. Simple air evaporative cooling system, A simple air evaporative cooling system is shown in Fig. It is similar to the s imple cooling system except that the addition of an evaporator between the heat 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 , water boils at 40°C, alcohol at 9°C and ammonia at - 70°C. The T-S diagram for a simple air cycle evaporative cooling system is shown in Fig. If Q tonnes of refrigeration is the cooling load in the cabin, then the air required for the refrigeration purpose,   Power required for the refrigeration system is   The initial mass of evaporant ( ) required to be carried for the given flight time is given     = Flight time in Minutes and. = Latent Heat of vaporization of evaporant in kJ/kg