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# Lecture note refrigeration cycle

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### Lecture note refrigeration cycle

1. 1. 11/12/2013 1 Refrigeration Cycle 12/11/2013Refrigeration Cycle 1 Chapter Objective To carry out first law analysis on a refrigeration cycle in which the working fluid undergoes changes of phase as it completes the cycle. 12/11/2013Refrigeration Cycle 2
2. 2. 11/12/2013 2 12/11/2013Refrigeration Cycle 3 What is Refrigerator? A reversed heat engine. Absorb heat QL from a low-temperature medium and reject the heat QH to a high-temperature medium. A working fluid called refrigerant flows through components of the refrigerator, forming a thermodynamic cycle called refrigeration cycle. To perform the heat absorption and rejection processes, the refrigerator requires a work input, W net, in. 12/11/2013Refrigeration Cycle 4
3. 3. 11/12/2013 3 12/11/2013 5 Environment Reverse Heat Engine COLD refrigerated space W net, in = required input QL = desired output QH WQQWQQ LHHL =−→−=− LH LL ref QQ Q W Q COP − == 12/11/2013Refrigeration Cycle 6 Total heat transfer to a cycle = Total work done by the cycle ∑ ∑= WQ Coefficient of performance of a refrigerator, (1)
4. 4. 11/12/2013 4 12/11/2013Refrigeration Cycle 7 The most efficient and ideal refrigerator is the one in which the refrigerant undergoes a reversed Carnot cycle while working between 2 specified temperatures. Q23 condenser engine compressor evaporator 1 3 2 4 W12W34 Q41 Components of a refrigerator working on reversedCarnot cycle 12/11/2013Refrigeration Cycle 8 Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2 T-s diagram for a refrigerator working on reversed Carnot cycle Q23 condenser engine compressor evaporator 1 3 2 4 W12W34 Q41
5. 5. 11/12/2013 5 12/11/2013Refrigeration Cycle 9 The processes shown on the T-s diagram are as follows: 1-2 Wet vapor at state 1 enters the compressor and is compressed isentropically to state 2. The required input is denoted W12. 2-3 The vapor at state 2 and is condensed at constant pressure and temperature to state 3, when it is completely liquid. The heat rejected from the refrigerant is denoted by Q23. Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2 12/11/2013Refrigeration Cycle 10 3-4 The liquid refrigerant at state 3 expands isentropically behind the piston of an engine, doing work of amount denoted by W34. 4-1 The refrigerant at state 4 enters the evaporator where it absorb heat denoted by Q41 from the cold space, until it reaches state 1. Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2
6. 6. 11/12/2013 6 4123 41 QQ Q W Q COP L ref − == ∑ dsTdQ T dQ ds =→= 12/11/2013Refrigeration Cycle 11 Cycle’s Performance The performance of the cycle is measured by the Coefficient of Performance (COP), defined as, Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2 The definition of entropy change is, ( )41141 ssTQ −=∴ ( )32223 ssTQ −= ( ) ( )3241 ssss −=− 12/11/2013Refrigeration Cycle 12 Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2 and and
7. 7. 11/12/2013 7 ( ) ( )( )4112 411 ssTT ssT COPref −− − =∴ ( )12 1 TT T COPref − = 12/11/2013Refrigeration Cycle 13 Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2 i.e. (2) Note: - where T1 = evaporator temperature and T2 = condenser temperature - only valid for reversed Carnot cycle (for refrigerator) - COP value is the maximum value for a cycle A vapor-compression refrigeration system operates on a Carnot cycle, uses R-12 as the working fluid. The refrigerant is dry saturated at entry to the condenser at 28°C, and saturated liquid leaving the condenser. The evaporator temperature is -10°C. Determine, (a) The compressor input work (kJ/kg); (b) The heat transfer in the evaporator (kJ/kg); (c) The COP Show the cycle on a T-s diagram. 12/11/2013Refrigeration Cycle 14
8. 8. 11/12/2013 8 12/11/2013Refrigeration Cycle 15 Q41 Q23 W s T T2 = T3 = 28oC T4 = T1 = -10oC s4 =s3 s1 =s2 1 4 3 2 12/11/2013Refrigeration Cycle 16 Main components of a household refrigerator Evaporator coils Freezer compartment
9. 9. 11/12/2013 9 12/11/2013Refrigeration Cycle 17 Q23 condenser engine compressor evaporator 1 3 2 4 W12W34 Q41 T2 T1 T2’ T1’T2’ = T atmosphere T1’ = T cold space T2’ < T 2 T1’ > T 1 Therefore, Q23 can be transferred from the refrigerant to the atmosphere and Q41 can be transferred from the refrigerated space to the refrigerant. 12/11/2013Refrigeration Cycle 18 T2’ T2’ < T 2 Q41 Q23 s T T2 T1 1 4 3 2 T1’ T1’ > T 1
10. 10. 11/12/2013 10 12/11/2013Refrigeration Cycle 19 Reversed Carnot cycle is not practical since, • All the reversible processes cannot be achieved. • Process 1-2 is bad for compressor – low efficiency and two phase compression is impractical. • It is difficult to stop evaporation process at point 1 and compressed it just to state 2. To get practical cycle, modification to the reversed Carnot cycle is made. Q41 Q23 W s T T2 = T3 T4 = T1 s4 =s3 s1 =s2 1 4 3 2 12/11/2013Refrigeration Cycle 20 Carnot Cycle/ Ideal Replacing the turbine with a throttling Valve (Ideal/actual) Throttling valve + Dry saturated or superheatedvapor at entry to the Compressor (Ideal/actual) Throttling valve + Dry saturated or superheatedvapor at entry to the compressor + Under-cooling at the condenser exit (Ideal/actual) Two or more throttling valves + Flash chamber + 2-stage compression (Ideal/actual)
11. 11. 11/12/2013 11 12/11/2013Refrigeration Cycle 21 Process 3-4 is replaced with a throttle valve. s Replacement of expansion cylinder with a throttle valve q41 q23 T T2 T1 s4 s1 =s2 14 3 2 Q in Wc Expansion valve Evaporator Condenser Compressor Q out Saturated or superheated vapor 1 2 3 4 The expansion of the liquid refrigerant through the throttle valve is represented by a dashed line 3-4 since the process is not reversible. 12/11/2013Refrigeration Cycle 22 s 4-1 Constant Pressure Evaporation Heat from a cold space is absorbed by the refrigerant. As a result, the refrigerant evaporates at a constant evaporator pressure, from state 4 to become a drier liquid-vapor mixture at state 1. q41 q23 T T2 T1 s4 s1 =s2 14 3 2
12. 12. 11/12/2013 12 12/11/2013Refrigeration Cycle 23 s 1-2 Isentropic compression The liquid-vapor mixture is compressed from the evaporator pressure to the condenser pressure, in a reversible adiabatic manner. The refrigerant exits the compressor as a saturated vapor at state 2. q41 q23 T T2 T1 s4 s1 =s2 14 3 2 12/11/2013Refrigeration Cycle 24 s 2-3 Constant Pressure Condensation Heat is rejected from the refrigerant to a warm space. As a result, the refrigerant condenses at a constant condenser pressure until it becomes a saturated liquid at state 3. q41 q23 T T2 T1 s4 s1 =s2 14 3 2
13. 13. 11/12/2013 13 12/11/2013Refrigeration Cycle 25 s 3-4 Constant Enthalpy Expansion The refrigerant expands through the throttle valve adiabatically. As a result, it’s pressure drops from the condenser to the evaporator pressure. The enthalpy is constant during the process, i.e. h3 = h4. Note: The expansion process is highly irreversible, thus making the vapor-compression cycle an irreversible cycle. q41 q23 T T2 T1 s4 s1 =s2 14 3 2 12/11/2013Refrigeration Cycle 26 s Effects of using an expansion valve i) A throttling process is an expansion process with constant enthalpy, (h3 = h4). Therefore, no work done by the cycle, i.e. w34 = 0. ii) The refrigeration effect, q41 decreases. q41 q23 T T2 T1 s4 s1 =s2 14 3 2
14. 14. 11/12/2013 14 12/11/2013Refrigeration Cycle 27 s Analysis of the Cycle The cycle operates as steady flow. So, each component of the vapor- compression refrigeration cycle applies the steady flow energy equation (SFEE) to analyze the energy interaction. The changes in the kinetic and potential energy are ignored. q41 q23 T T2 T1 s4 s1 =s2 14 3 2 1212 hhw −= 4141 hhq −= 12/11/2013Refrigeration Cycle 28 s q41 q23 T T2 T1 s4 s1 =s2 14 3 2 Analysis of the Cycle Evaporation process (4-1): (3) (Refrigerating effect / heat absorbed by the refrigerant) Compression process (1-2): (4) (input work to the compressor)
15. 15. 11/12/2013 15 43 hh = 3223 hhq −= 12/11/2013Refrigeration Cycle 29 s Analysis of the Cycle Condensation process (2-3): (5) (heat rejected from the refrigerant) Expansion process (3-4): (6) (constant enthalpy process) q41 q23 T T2 T1 s4 s1 =s2 14 3 2 12 41 12 41 hh hh w q COPref − − == 12/11/2013Refrigeration Cycle 30 s Performance of the Cycle The performance of the vapor- compression refrigeration cycle is measured by the coefficient of performance of refrigerator, COPref, defined as, (7) q41 q23 T T2 T1 s4 s1 =s2 14 3 2
16. 16. 11/12/2013 16 An ideal vapor-compression refrigeration cycle uses R-134a as the working fluid. The evaporator and condenser operate at –20oC and 1000 kPa respectively. The refrigerant is dry saturated at entry to the compressor and saturated liquid at the condenser outlet. The mass flow of the refrigerant is 3 kg/min. Determine, (a) The COPref; (b) The refrigerating effect (kW); (c) The COPhp; (d) The heat transfer to the cooling water in the condenser (kW). Show the cycle on a T-s diagram. 12/11/2013Refrigeration Cycle 31 12/11/2013Refrigeration Cycle 32 Q41 Q23 T T2 T1 = -20oC s4 s1 =s2 1 4 3 2 P2 = P3 = 1000 kPa s
17. 17. 11/12/2013 17 12/11/2013Refrigeration Cycle 33 To make full use of the specific enthalpy of evaporation of the refrigerant, the evaporation process is continued until the refrigerant becomes a saturated vapor at 1. q41 q23 T T2 T1 1 4 3 2 s The refrigerant exits the evaporator as a saturated vapor 12/11/2013Refrigeration Cycle 34 In actual condition, the refrigerant evaporates until it becomes superheated vapor at state 1. This is to prevent the carry-over of liquid refrigerant into the compressor. However, the amount of superheat is usually kept to minimum. q41 q23 T T2 T1 1 4 3 2 s
18. 18. 11/12/2013 18 12/11/2013Refrigeration Cycle 35 q41 q23 T T2 T1 1 4 3 2 s Effects of This Process 1. The refrigerant becomes superheated vapor as it exits the compressor. 2. More work needs to be supplied to the compressor to compress the refrigerant. 3. The condensation process still occurs at a constant condenser pressure, but not at a constant temperature. 4. More heat needs to be rejected from the refrigerant to the warm region. 12/11/2013Refrigeration Cycle 36 The refrigerant is condensed until it’s temperature is lower than the saturation temperature at the condenser pressure, when it exits the condenser. Under-cooling of the refrigerant in the condenser q41 q23 T Tsat 2 1 4 3 2 s Saturation temperature at condenser pressure P evaporator P condenser Degree of sub-cooling Degree of superheat Tsat 1
19. 19. 11/12/2013 19 12/11/2013Refrigeration Cycle 37 Effects of This Process • The line 3-4 representing the expansion process is shifted to the left, thus line 4-1 representing the evaporation process increases in length. • Therefore, under-cooling of the refrigerant increases the refrigerating effect of the cycle, which is a desirable effect. q41 q23 T Tsat 2 1 4 3 2 s Saturation temperature at condenser pressure P evaporator P condenser Degree of sub-cooling Degree of superheat Tsat 1 ( )41 1 4q h h= − ( )12 2 1w h h= − 12/11/2013Refrigeration Cycle 38 Analysis of the Cycle Evaporation process (4-1): (8) Compression process (1-2): (9) q41 q23 T Tsat 2 1 4 3 2 s Saturation temperature at condenser pressure P evaporator P condenser Degree of sub-cooling Degree of superheat Tsat 1
20. 20. 11/12/2013 20 ( )23 2 3q h h= − 43 hh = 12/11/2013Refrigeration Cycle 39 Analysis of the Cycle Condensation process (2-3): (10) Expansion process (3-4): (11) q41 q23 T Tsat 2 1 4 3 2 s Saturation temperature at condenser pressure P evaporator P condenser Degree of sub-cooling Degree of superheat Tsat 1 ( ) ( ) 1 441 12 2 1 ref h hq COP w h h − = = − 12/11/2013Refrigeration Cycle 40 Performance of the Cycle Performance of the vapor-compression refrigeration cycle is measured by the Coefficient of Performance of refrigerator, COPref, defined as, (12)
21. 21. 11/12/2013 21 12/11/2013Refrigeration Cycle 41 The p-h diagram Another diagram frequently used in the analysis of vapor- compression refrigeration cycles is the P-h diagram. Three of the four processes appear as straight lines. The heat transfer in the condenser and the evaporator is proportional to the lengths of the corresponding process curves. P, (bar) h 1 2 2’3 4 T sat condenser T sat evaporator The air temperature inside a cold room is controlled at a constant value using a vapor compression refrigeration system with R-134a as the refrigerant. The refrigerant leaves the evaporator dry saturated and is compressed to 1 MPa. The refrigerant leaves the condenser at 35oC. The refrigerant is then throttled to the evaporator pressure of 240 kPa. The isentropic efficiency of the compressor is 85%. The refrigeration load is 100 kW. Determine, (a) The temperature of the refrigerant which leaves the compressor (oC); (b) The mass flow rate of the refrigerant (kg/s); (c) The coefficient of performance of the refrigeration system. Show the cycle on a p-h diagram. 12/11/2013Refrigeration Cycle 42
22. 22. 11/12/2013 22 12/11/2013Refrigeration Cycle 43 P, (bar) h 10.84 2.610 Ts = 45oC 1 2 2’3 4 It is proposed to use a heat pump working on the ideal vapor- compression cycle for the purpose of heating the air supply to an incubation room from 25oC to 37oC at a rate of 1.06 kg/s. The supply of heat is taken from a refrigerated room at 7oC. For the air, take cp =1.005 kJ/kgK. The refrigerant is R12 which is dry saturated leaving the evaporator. A temperature difference of 17 K is necessary for the transfer of heat from the refrigerated room to the refrigerant in the evaporator. The delivery pressure of the compressor is 10.84 bar and there is 5 K of under-cooling of refrigerant in the condenser. 12/11/2013Refrigeration Cycle 44
23. 23. 11/12/2013 23 Show the schematic diagram of the components and the cycle on a T-s diagram and p-h diagram. Calculate, (a) The heat load (kW); (b)The mass flow of the refrigerant (kg/s); (c) The refrigerating effect (kW); (d)The motor power required to drive the compressor if the mechanical efficiency is 87%; (e) The COPhp; (f) The COPref. 12/11/2013Refrigeration Cycle 45 12/11/2013Refrigeration Cycle 46 Q41 Q23 T Tsat 14 3 2 s 5K 17K -10oC 10.84 bar P h 4 1 23 2’ 2’
24. 24. 11/12/2013 24 • The most important quantity in refrigeration or freezing application is the total of heat required to be moved from a cold space. • This heat is called refrigerating load. • Measured in `ton’ or kilowatt or Btu/min. • Unit conversion: 1 ton = 200 Btu/min = 3.516 kW. 12/11/2013Refrigeration Cycle 47 ( )414141 hhmQxmQ refref −== ••• ∴ ( ) ( )skg hh Q mref / 41 41 − = • • 12/11/2013Refrigeration Cycle 48 (13) Refrigerating load is, The mass flow rate of the refrigerant is,
25. 25. 11/12/2013 25 12/11/2013Refrigeration Cycle 49 Features of the diagram It is more convenient to represent the vapor- compression refrigeration cycle on a pressure-enthalpy (p-h) diagram, because the enthalpies required for the calculation can be read off directly from the diagram. Enthalpy (kJ/kg) Pressure(bar) 12/11/2013Refrigeration Cycle 50 Figure shows the vapor- compression refrigeration cycle shown on a p-h diagram. Note that the cycle shown is with under- cooling of the refrigerant at the exit of the condenser. Q1 = (h1 – h4) W = (h2 – h1) Vapor-compression cycle on a p-h diagram
26. 26. 11/12/2013 26 12/11/2013Refrigeration Cycle 51 What is Flash Chamber? • Flash chamber is a device used to separate vapor refrigerant from the liquid refrigerant. • When a flash chamber is used, the compression process needs to be carried out in two stages. Schematic diagram of a refrigeration plant with two-stage compression and a flash chamber. Wc Expansion valve 1 Condenser Compressor Q out 3 5 6 Wc Expansion valve 2 Evaporator Flash chamber Compressor 1 2 (1) 8 Direct contact heat exchanger 9 (1) (1) (x) (1-x)27 (1-x) 4 (1-x) Qin 12/11/2013Refrigeration Cycle 52 The refrigerant leaving the expansion valve 1 as a mixture of vapor and liquid (wet vapor). The vapor refrigerant has no more capability to absorb heat in the evaporator. Thus, it is no use to pass the vapor through the evaporator. It is more practical to separate the vapor from the mixture at some intermediate pressure, pi and allow only the liquid refrigerant to flow through the evaporator. The liquid refrigerant has full capability to absorb the heat in the evaporator. Wc Expansion valve 1 Condenser Compressor Q out 3 5 6 Wc Expansion valve 2 Evaporator Flash chamber Compressor 1 2 (1) 8 Direct contact heat exchanger 9 (1) (1) (x) (1-x)27 (1-x) 4 (1-x) Qin
27. 27. 11/12/2013 27 12/11/2013Refrigeration Cycle 53 Process Suppose 1 kg of refrigerant flowing through the condenser. At the flash chamber, x kg of dry saturated vapor at pressure pi and enthalpy hgi is bled off to the inter-sage of the compressor. The remaining mass of (1-x) kg of liquid with enthalpy hfi passes through the throttle valve 2 and then to the evaporator. At the intermediate pressure pi , (1-x) kg of vapor at state 2 is mixed with x kg of flash vapor of enthalpy hgi . The resultant mixture at state 3 is compressed in the second stage compressor, to state 4. Wc Expansion valve 1 Condenser Compressor Q out 3 5 6 Wc Expansion valve 2 Evaporator Flash chamber Compressor 1 2 (1) 8 Direct contact heat exchanger 9 (1) (1) (x) (1-x)27 (1-x) 4 (1-x) Qin 12/11/2013Refrigeration Cycle 54 • 2 compressors, 1 evaporator, 1 condenser, and 1 flash chamber. • Point 2, 3, 6, 7 and 9 are located on the same intermediate pressure pi . pi is chosen so that the compressor work is minimum. pi = (p1 . p4)1/2 • The flash chamber pressure = The intermediate pressure, pi . Vapor-compression refrigeration cycle with two-stage compression and a flash chamber on a p-h diagram P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg
28. 28. 11/12/2013 28 12/11/2013Refrigeration Cycle 55 • Saturated liquid enter the 2nd expansion valve at point 7. • The flash chamber contains mixture of saturated water and vapor. • The dry saturated vapor from the flash chamber at point 9 mix with the superheated vapor from the 1st stage compressor at point 2 to yield superheated vapor at point 3. • The superheated vapor at point 3 is compressed to point 4 through the 2nd stage compressor. P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg fgifi hxhhh )(56 +== 12/11/2013Refrigeration Cycle 56 The amount of dry saturated vapor bled off is given by the dryness fraction x at state 6, at the intermediate pressure pi . Since the enthalpy of the refrigerant at state 6 is equal to the enthalpy at state 5, we have, P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg
29. 29. 11/12/2013 29 fgi fi h hh x − = 5 12/11/2013Refrigeration Cycle 57 Thus, (14) P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg ( )[ ]293 .1..10 hxhxh HHWQ inletexit −+−= −=/−/ ( )9223 . hhxhh −−= 12/11/2013Refrigeration Cycle 58 Thus, (15) 9 (x) kg 2 3 (1-x) kg (1) kg P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg
30. 30. 11/12/2013 30 ∑ += 3412 WWWin ( )( ) ( )34121 hhhhxWin −+−−=∑ 12/11/2013Refrigeration Cycle 59 a) Total work input to the compressors Since the compression process is carried out in two stages, the total work input to the cycle is, P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg i.e. (16) ( )( )8181 1 hhxQ −−= 12/11/2013Refrigeration Cycle 60 b) Refrigerating effects The refrigerating effects of the cycle is (17) P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg
31. 31. 11/12/2013 31 ( )5445 hhQ −= 12/11/2013Refrigeration Cycle 61 c) Heat rejected in the condenser The heat rejected from the refrigerant is (18) P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg ( )( ) ( )( ) ( )3412 81 1 1 hhhhx hhx COPref −+−− −− = ∑ = in ref W Q COP 81 12/11/2013Refrigeration Cycle 62 d)Coefficient of performance The coefficient of performance of the cycle is (19) P h 1 4 3 9 2 67 5 8 (1-x) kg (x) kg (1) kg
32. 32. 11/12/2013 32 • Increase in refrigerating effects of the cycle. • Lower amount of compression work. • Increase in the coefficient of performance. 12/11/2013Refrigeration Cycle 63 • The refrigeration system becomes more complex since there are more equipments and piping system required. • Higher capital costs for setting up the plant • Higher maintenance costs. 12/11/2013Refrigeration Cycle 64
33. 33. 11/12/2013 33 12/11/2013Refrigeration Cycle 65 A two-stage vapor-compression refrigeration system uses Refrigerant- 134a as the working fluid. Saturated vapor enters the first compressor stage at -30oC and saturated liquid leaves the condenser at 11.595 bar. The system is equipped with a flash chamber, which operates at a pressure of 4.1459 bar. If each compressor stage operates isentropically and the refrigerating capacity of the system is 10 tons, determine, (a) the mass flow rate of the refrigerant through the evaporator kg/s; (b) the power input to each compressor (kW); (c) The coefficient of performance. Show the cycle on a T-s diagram. (1 ton refrigeration = 3.516 kW) 12/11/2013Refrigeration Cycle 66 4 T oC s 1 5 3 2 6 9 7 8 (1) (x) (1-x)
34. 34. 11/12/2013 34 12/11/2013Refrigeration Cycle 67 A vapor-compression refrigeration plant uses refrigerant R134a and has a suction saturation temperature of -5oC and a condenser saturation temperature of 45oC. The vapor is dry saturated on entering the compressor and there is no under-cooling of the condensate. The compression is carried out isentropically in two stages and a flash chamber is employed at an intersage saturation temperature of 15oC. Determine: a) the amount of vapor bled off at the flash chamber; b) the state of vapor at inlet to the 2nd compressor stage; c) the refrigerating effect; d) the work input; e) the COPref. A two-stage compression refrigeration system is to operate between the pressure limits of 1 MPa and 0.1 MPa. The working fluid is Refrigerant-134a. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.32 MPa. Part of the refrigerant evaporates leaving the low flashing process, and this vapour mixed with the refrigerant leaving the low-pressure compressor. The mixture is then compressed to the condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves the evaporator as a saturated vapour and both compressors are isentropic, determine: 12/11/2013Refrigeration Cycle 68
35. 35. 11/12/2013 35 (a) The fraction of the refrigerant which evaporates as it throttled to the flash chamber; (b) The amount of heat removed from the refrigerated space (kJ/kg); (c) The compressor work (kJ/kg); (d) The COPref. Illustrated the schematic layout of the refrigeration system showing clearly the respective components and also its p-h diagram. 12/11/2013Refrigeration Cycle 69 12/11/2013Refrigeration Cycle 70 4 T oC s 1 5 3 2 6 9 7 8 (1) (x) (1-x) Wc Expansion valve 1 Condenser Compressor Q out 3 5 6 WcExpansion valve 2 Evaporator Flash chamber Compressor 1 2 (1) 8 Direct contact heat exchanger 9 (1) (1) (x) (1-x)27 (1-x) 4 (1-x) Qin