Document mm3m2o4dpKNQVrVpOp85pYgek

6 CHAPTER 1 1965 Guide And Data Book jjjexcept that the temperature of the refrigerant during the heat emu* is 520 R, or 20 deg higher than the atmosphere. Solution: " W - 0.5(520 - -100) - 60 Btu (CP) - 200/60 - 3.33 AW - 60 - 50 *=* 10 Btu 1 Q. -- 200 + 60 - 260 Btu AS. = 260/500 - 0.52 Btu/R deg : AS - - 200/400 - - 0.50 Btu/R deg ASi, = 0.52 + (- 0.50) - 0.02 Btu/R deg ToAS**.-- 500(0.02) = 10 Btu Hie flow of energy, available energy and unavailable energy, and their representations on Fig. 5 are: Energy <?< Q. W (Bj)i (Bah (Be). (Eo)* Btu 200 260 60 .-50 250 10 250 Area e a+b+c a+b -b b+c .. a b+c The heat efflux from the' refrigerant (260 Btu) is composed of its available portion (10 Btu) and its unavailable portion (250 Btu). When thin heat efflux is absorbed by the atmosphere, the above available enemy (10 Btu) is degraded by the irreversible process of heat transfer through the 20 deg finite temperature difference, and the heat thus absorbed -by. the-atmosphere is rendered wholly unavailable. It is this degradation of 10-Btu of available energy that has caused the change in the entropy of the universe to be 0.02 Btu per R deg, and T.ASni*, as well as'the additional net work input,- to be 10 Btu. '* Example 3: The 'data are the*same as in Example t, except.that the reversed Carnot cycle has been modifiedothat the adrabatiocompression proces is irreversible, due to the presence of friction, and undergoes an entropy, increase of.0,05 Btu/R deg. This cycle is depicted m Fig. 6. Solution:' 0. - 500 X 0.55 275 Btu ' - *' .-*! W - 275 - 200 - 75 Btu- --............ .- (CP) - 200/75 = 2.67... ,; AW - 75 - 50 - 25 Btu -AS, -- 0.55Btu/Rdeg .* . aSm -- -- 0.50 Btu/R deg A&*, - 0.55 - 0.50 = 0.05 Btu/R deg 7'#&So^*b* 500 X 0.05 a 25 Btu Energy Qi Q. W (Eeh (Bah (Bah (Eeeh Btu 200 275 75 -50 250 0 275 -r 25 . Area ;H-o ' d. - a+b+c+e` --a a+d (a-*e) b-t-c-t-e The irreversibility of the adiabatic process has caused: ;(1) a degradation of available energy {(Eie)A of 25 Btu, (2)an increase in the entropy'of the universe (ASusO-of 0.05 Btu/R deg,'(3) an increase in net work input (AIP) of 25 Btu, and (4) the value of . TaA&^u to be 25 Btu. As there were no heat transfers through finite temperature differences, there was no degradation of energy from that source. Example 4 Tbe data are the as in Example l, except that the reversed Camot cycle has been modified so that the refriger ant, during the process of heat influx, undergoes an irreversible process and has an additional entropy increase, .due- to friction, of 0.05 Btu/R deg. The cycle is presented in fig. 7. Hie evaluation of this cycle is rimilaj- to that in Example 3, with the following differences. In Example 3,' Si -- St -- 0.5 Btu/R deg, and was due solely to the heat influx. In Example 4t Si -- & -- 0.55 Btu/R deg, which is partly due to the influx of heat. (0.5 Btu/R deg), and the remainder (0.05 Btu/R deg) is caused by friction. In Example 3, AW, TjiS^, and Bee (each equal to 25 Btu) were due to the irreversibility of the adiabatic- compression process, while in Example 4 their value is the result of the fluid friction during the process of heat influx. Tbe tabulated values in Example 3 apply exactly to Example 4- Example 5: In this raw, the departures from the idealities of the reversed Carnot cycle of Example 1 are a combination of those in Examples t, 3, ana 4 All four processes are irreversible and have a change in entropy AS/, which is due to friction alone and is equal to 0.05 Btu/R aeg. During the efflux of heat, the temperature of the refrigerant is 520 F (Example f), and during the neat influx, the refrigerant temperature is 380 R (20 deg lower,than the re frigerated space). Tne refrigeration load and the temperatures of the atmosphere and tbe refrigerated space remain the same. The cycle is shown in Fig. 8. Solution: A&a - 200/380 ~ 0.52632 Btu/R deg a5q* -- (Si -- Si) -- AS/ - - 0.67632 - 0.05 - - 0.72632 Btu/R deg Q. - 520 X 0.72632 " 377.69 Btu W - 377.69 - 200 - 177.69 Btu ThemKjdynamics-and Refrigeration Cydes nr _ 1779i- 50 - 127.69.Btu .. fCP) -i 200/177.69 1.1256 % *s, - - 200/400 - - 0.5 Btu/R deg 7s. - 377.69/5000.75538 Btu/R deg aS ,T - 0.75538 - 0.5 - 0.25538 Btu/R deg 7v3it500 X 0.25538.- 127.69 Btu - Tbe energy flows to and from the refrigerant, and their area representations on Fig. 8 are:} Energy Q, u Q. . W (Fx)< (Bah (Beh (Bah (Bee), .. Btu 200 "377.69 177.69 -63.16 263.16 14.53 363.16 100 Area' 0 (a--*n) + p -k k+o (a-*g) (h -*p) (h) + 0 + j) +(l+p)(m+n) Tbe equations for the various energy balances on the refriger ant are given below, and under each term is its numerical value. Qt.+ W -0, ...200 + 177.69 = 377.69 ' Qi - (Eeh +(Eo)i r ' 200--63.16 + 263.16 ` g;:= (Eeh. ' + (Bah . . ." 377.69.- 14.53 + 363.16 - W + (.,)( - (Be). + (Eeeh 177.69 - 63.16 - ,14.53 .100 . 114.53-114.53 (Eu)i + (Ejj), -- (Eu). - 263.16 + 100 - 363.16 The b'eat flow from the refrigerated space to the refrigerant has occurred through & temperature drop of 20 deg. This in itself has caused a degradation of available enemy equal- to 13.16 Btu, which is the decrease in availability of the heat flow (from -- 50 Btu in Example 1 to -63.16 Btu in this example). Also, the process of heat flow from tbe refrigerant to the atmosphere has been accomplished by a 20 deg temperature drop. The available energy degraded here is 14.53 Btu, . which is the value of the available energy rejected by the refrigerant, and which was 7 rendered unavailable when the efflux heat flow was absorbed by the atmosphere. The available energydegraded by tbe refrigerant in its cyclical operation is 100 Btu. Thus, tbe total degradation for the combi nation of refrigerant, refrigerated space, and atmosphere is 127.69 Btu, which equals the additional net work input to the cycle. THE ACTUAL BASIC VAPOR COMPRESSION REFRIGERATION CYCLE Although the idea) basic vapor compression refrigeration cycle shown in Figs. 2 and 3 has been termed ideal, the cycle, in its interaction with the atmosphere, entails two irreversi bilities: (1) the inherent irreversibility in the expansion Valve process, and (2) the process of heat transfer through a finite, temperature difference during the desuperheating of the vapor in the condenser. To remove the first irreversibility, a reversi ble adiabatic expansion engine would be substituted for the expansion valve, and the work output from this engine would supply part of the work input to the cycle. Hie process for tins engine is shown as 3-4' in Fig. 9. To remove the second irreversibility, the compression process l-2 of Fig. 9-wouid be replaced by a reversible two part compression, the isentropic process 1--2' and the isothermal process 2*--2" of Fig. 9. These improvements would produce a reversed Carnot cycle which is equivalent in all respects to the one in Fig. 4, except that, due to the increased entropy change during the heat influx to the refrigerant, the refrigeration load, net work input, and heat rejection would all be increased 4.52 percent. ' The complexity of the equipment needed for these improve ments precludes its application, and the.practical arrangement is th&t.shown in Fig. 1. In; the following three' demonstrations, Example 6 will present a Second-Law analysis of an ideal vapor cycle'as de picted on Fig. 3 and Examples 7 and 8 will analyze vapor cycles that represent the actual irreversible' conditions of frictioo and heat transfer found urpractice. Example 8: The data are tiioae of Example /. The vapor is Re frigerant 12. All processes for tbe vapor are reversible, except that through the expansion valve. Heat'transfers are accomplished with negligible temperature differences,' except for the-desuper heating process in the condenser. The cycle is shown in Fig. 9. fig. 6 . ..Temperature-Entropy, Diagram for. Modified Reversed Camot- Cyde of Example 3 fig. 7 ........Temperature-Entropy Diagram .for Modified Reversed Carnot Cyde' of: Examp/e 4 fig. 8 .... Temperature-Entropy Diagram for Modified Reversed Camot.Cyde of Examp/e 5 fig. 9.;...: Temperature-EnjropyiDiagram.for.ideal/ Basic Vapor,Cyde ofExample 6