Chapter 11. Refrigeration Cycles

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1 Chapter 11 Refrigeration Cycles

2 The vapor compression refrigeration cycle is a common method for transferring heat from a low temperature space to a high temperature space. The figures below show the objectives of refrigerators and heat pumps. 2

3 The purpose of a refrigerator is the removal of heat, called the cooling load or effect, from a low-temperature medium. The purpose of a heat pump is the transfer of heat, called the heating load or effect, to a high-temperature medium. When we are interested in the heat energy removed from a lowtemperature space, the device is called a refrigerator (or air conditioner). When we are interested in the heat energy supplied to the hightemperature space, the device is called a heat pump. In general, the term heat pump is used to describe the cycle as heat energy is removed from the low-temperature space and rejected to the high-temperature space. 3

4 The performance of refrigerators and heat pumps is expressed in terms of coefficient of performance (COP), defined as COP COP R HP Desired output Cooling effect = = = Required input Work input Desired output Heating effect = = = Required input Work input Q W Q W L net, in H net, in Both COP R and COP HP can be larger than 1. Under the same operating conditions, conservation of energy (Q L + W net,in = Q H ) says that the COPs are related by COP HP = COP +1 R 4

5 Refrigerators, air conditioners, and heat pumps are rated with a SEER number or Seasonal Adjusted Energy Efficiency ratio. The SEER is defined as the Btu/hr of heat transferred per watt of work energy input. The Btu is the British thermal unit and is equivalent to 778 ft-lbf of work (1 W = Btu/hr). A SEER of 10 yields a COP of 10/ = Refrigeration systems are also rated in terms of tons of refrigeration. One ton of refrigeration is equivalent to 12,000 Btu/hr or 211 kj/min. 5

6 Reversed Carnot Refrigerator and Heat Pump Shown below are the cyclic refrigeration device operating between two constant temperature reservoirs and the T-s diagram for the working fluid when the reversed Carnot cycle is used. Recall that in the Carnot cycle heat transfers take place at constant temperatures. If our interest is the cooling load, the cycle is called the Carnot refrigerator. If our interest is the heating load, the cycle is called the Carnot heat pump. 6

7 Comparison Between The Carnot-Cycle Heat Engine and Carnot-Cycle Refrigerator Carnot-Cycle Heat Engine Carnot-Cycle Refrigerator 7

8 The standard of comparison for refrigeration cycles is the reversed Carnot cycle. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump, and their COPs are COP COP R, Carnot HP, Carnot 1 TL = TH / TL 1 = TH T 1 TH = = 1 T / T T T L Notice that a turbine is used for the expansion process between the high and low-temperatures. While the work interactions for the cycle are not indicated on the figure, the work produced by the turbine helps supply some of the work required by the compressor from external sources. Why not use the reversed Carnot refrigeration cycle? 1.) Easier to compress vapor only and not liquid-vapor mixture. 2.) Cheaper to have irreversible expansion through an expansion valve. H H L L 8

9 The Ideal Vapor-Compression Refrigeration Cycle The vapor-compression refrigeration cycle has four components: evaporator, compressor, condenser, and expansion (or throttle) valve. The most widely used refrigeration cycle is the vapor-compression refrigeration cycle. In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator pressure and vaporizes as it absorbs heat from the refrigerated space. The ideal vapor-compression refrigeration cycle consists of four processes. Process Description 1-2 Isentropic compression 2-3 Constant pressure heat rejection in the condenser 3-4 Throttling in an expansion valve 4-1 Constant pressure heat addition in the evaporator 9

10 Ideal Vapor-Compression Refrigeration Cycle Process Description 1-2 Isentropic compression 2-3 Constant pressure heat rejection in the condenser 3-4 Throttling in an expansion valve 4-1 Constant pressure heat addition in the evaporator 10

11 Ideal Vapor-Compression Refrigeration Cycle The T-s diagram Note that h 4 = h 3 while s 4 = s 3. The P-h diagram is another convenient diagram often used to illustrate the refrigeration cycle. 11

12 The ordinary household refrigerator is a good example of the application of this cycle. 12

13 Ideal Vapor-Compression Refrigeration Cycle 13

14 Example 11-1 Refrigerant-134a is the working fluid in an ideal compression refrigeration cycle. The refrigerant leaves the evaporator at -20 o C and has a condenser pressure of 0.9 MPa. The mass flow rate is 3 kg/min. Find COP R and COP R, Carnot for the same T max and T min. Using the Refrigerant-134a Tables, we have State 2 State1 kj h Compressor exit kj Compressor inlet = kg h2 s = s o P = P = kpa kg T1 = 20 C kj o s1 = kj T2 s = C x1 = 1.0 kg K s2s = s1 = kg K State3 kj State 4 Condenser exit h3 = x Throttle exit = kg o kj P3 = 900 kpa kj T4 = T1 = 20 C s4 = s3 = kg K x3 = 0.0 kg K h4 = h 3 14

15 Example 11-1 (Continued) h 1 = h o C) = kj/kg and s 1 = s o C) = kj/(kg. K) 15

16 Example 11-1 (Continued) 16

17 Example 11-1 (Continued) h 3 = h kpa) = kj/kg and s 3 = s kpa) = kj/(kg. K) 17

18 Example 11-1 (Continued) 18

19 Example 11-1 (Continued) The COP is The Carnot Result is COP R COP Q& L m& ( h h ) h h = = = W& m& ( h h) h h = = net, in ( ) ( ) 3.44 R, Carnot = T H TL T L kj kg kj kg ( ) K = (43.79 ( 20)) K =

20 Actual Vapor-Compression Refrigeration Cycle 20

21 Multistage compression refrigeration systems 21

22 An Ideal Gas Refrigeration System The power cycles can be used as refrigeration cycles by simply reversing them. Of these, the reversed Brayton cycle, which is also known as the gas refrigeration cycle, is used to cool aircraft and to obtain very low (cryogenic) temperatures after it is modified with regeneration. 22

23 An Ideal Gas Refrigeration System (Continued) The work output of the turbine can be used to reduce the work input requirements to the compressor. Thus, the COP of a gas refrigeration cycle is COP R ql = = w w ql w net, in comp, in turb, out The energy equations (neglecting kinetic and potential energy effects) are as follows: 23

24 An Ideal Gas Refrigeration System (Continued) Then, and, the COP of a gas refrigeration cycle is given by 24

25 Example 11-2 An ideal gas refrigeration cycle (reversed Brayton cycle) using air as the working fluid is to maintain a refrigerated space at 0 o F while rejecting heat to the surrounding medium at 80 o F. The pressure ratio of the compressor is 4. Determine the maximum and minimum temperature in the cycle, the coefficient of performance and the rate of refrigeration for a mass flow rate of 0.1 lbm/s. The T-s diagram is shown on the right. We are given that the refrigerated space is at 0 o F, so that T 1 = 0 o F = 460 R and the surrounding medium is at 80 o F so that T 3 = 80 o F = 540 R. Using Table A-17E, this yields h 1 = Btu/lbm and P r1 = along with h 3 = Btu/lbm and P r3 =

26 Example 11-2 (Continued) 26

27 Example 11-2 (Continued) Since the process from 1 to 2 is isentropic, we must have P r2 /P r1 = P 2 /P 1 = 4 yielding P r2 / = 4, or P r2 = Using Table A-17E and linear interpolation then yields (see the next two slides) T 2 = 223 o F as the maximum temperature and h 2 = Btu/lbm. Since the process from 3 to 4 is isentropic, we must have P r3 /P r4 = P 3 /P 4 = 4 yielding /P r4 = 4, or P r4 = Using Table A-17E and linear interpolation then yields (see the next two slides) T 4 = -97 o F as the minimum temperature and h 4 = 86.7 Btu/lbm. 27

28 Example 11-2 (Continued) 28

29 Example 11-2 (Continued) 29

30 Example 11-2 (Continued) This then leads to the following results. 30

31 Example 11-2 (Continued) The rate of refrigeration is then Extra Assignment Show that an ideal vapor-compression cycle working under similar conditions has a COP value greater than 3. 31

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