Document YrZky6D5xM2ydeppGqpqpEgrk

754 CHAPTER 39 1949 Guide . :The most common and least complicated type of refrigeration , cycle is called 'the simple saturation cycle and is shown diagrammatically in Fig. 3 and plotted upon pressure-enthalpy coordinates in Fig. 4. For this system saturated vapor flows without gain or loss of heat from the evaporator to the suction of the compressor. During passage through the compressor the energy added as shaft work goes entirely to increase the enthalpy of the refrigerant, and the compression process, which is assumed to occur irreversibly and without external heat transfer, is characterized by constant entropy. Thus the state of the superheated vapor leaving the compressor can be determined from the tables of thermodynamic properties by noting the discharge pressure and fixing also the entropy of the saturated vapor at entrance to the compressor.' : Table 4. Properties of Monofluorotrichloromethane (F-ll) Sat. FTbrp. Abb. Press. . La-fsb Sain. Liquid Vapor Ebtbalpt abb Entbopt Taken Fbox --40 F Enthalpy liquid ` Vapor Entropy 25 F Superheat SO F Superheat' liquid Vapor Enthalpy Entropy Enthalpy Entropy 0 5 10 15 20 1 25 2.59 0.01020 13.700 2.96 0.01024 12.100 3.38 0.01028 10.700 3.85 0.01032 9.530 4.36 0.01036 8.490 4.94 0.01040 7,580 7.81 8.81 9.82 10.80 11.90 12.90 90.4 0.0178 0.1975 91.2 0.0200 0.1974 92.0 0.0222 0.1973 92.8 0.0243 0.1971 93.7 0.0264 0.1970 94.5 0.0286 0.1969 93.9 0.2049 97.4 0.2120 94.7 0.2047 98;2 0.2117 95.5 0.2045 99.0 0.2114 96.3 0.2043 99.8 0.2111 97.2 0.2041 100.7 0.2109 98.0 0.2039 101.5 0.2107 30 5.57 0.01045 6.770 13.90 95.3 0.0307 0.1969 98.8 0.2038 102.3 0.2105 35 6.27 0.01049 6.080 14.90 96.1 0.0328 0.1968 99.6 0.2037 103.1 0.2103 40 7.03 0.01053 5.460 16.00 96.8 0.0349 0.1968 100.3 0.2036 103.8 012101 45 7.88 0.01057 4.920 17.00 97.6 0.0370 0.1967 101.1 0.2035 104.6 0.2099 50 8.79 0.01062 4.440 18.10 98.4 0.0391 0.1967 101.9 0.2034 105.4 0.2098 55 9.80 0.01066 4.020 19.10 99.2 0.0412 0.1967 102.7 0.2033 106.2 0.2097 60 10.90 0.01071 3.640 20.20 100.0 0.0432 0.1967 103.5 0.2033 107.0 0.2096 65 12.10 0,01076 3.300 21.30 100.8 0.0453 0.1967 104.3 0.2032 107.8 0.2094 70 13.40 0.01081 3.000 22.40 101.5 0.0473 0.1967 105.0 0.2032 108.5 0.2093 75 14.80 0.01086 2.740 23.50 102.2 0.0493 0.1967 105.7 0.2031 .109.2 0.2092 80 16.30 0.01091 2.500 24.50 102.9 0.0513 0.1966 106.4 0.2030 109.9 0.2090 85 17.90 0.01096 2.280 25.60 103.6 0.0533 0.1966 107.1 0.2029 110.6 0.2089 90 19.70 0.01101 2.090 26.70 104.4 0.0553 0.1966 107.9 0.2028 111.4 0.2088 . 95 21.60 o:ouo6 1.918 27.80 105.1 0.0573 0.1966 108.6 0.2028 112.1 0,2087 100 23.60 0.01111 1.761 28.90 105.7 0.0593 0.1965 109.2 0.2027 112.7 0.2085 105 25.90 0.01116 1.620 .30.10 106.4 0.0613 0.1965 109.9 0.2026 113.4: 0.2084, \ -Superheated vapor from the compressor flows to the condenser where de-superheating and condensation take place: From the condenser the re frigerant flows to the expansion valve, undergoes a constant-enthalpy pres sure reduction and'returns to the evaporator where it again removes a quantity of undesired heat. When the evaporator is arranged to permit direct cooling of room air by the refrigerant, the system is said to be of the direct expansion type, while a system in which the evaporating refrigerant 'cools water or brine, which in turn cools the air, is said to be indirect. Although many differences exist between most actual systems and that of the' simple saturation cycle this latter is nonetheless of great value in that it provides an extremely simple method of rapidly achieving an approximate analysis of probable power requirements, compressor size, etc. Further, the equations used in analysis of a simple saturation cycle form the basis Refrigeration 755 of the more complex treatments required for compound refrigeration cycles: For these reasons a typical simple saturation problem will be worked in detail. Example 1. A simple saturation cycle carries a 7 ton load when operating between suction and discharge pressures of 52.7 psia and 121 psia with F-12 as the.refrigerant.. Determine: (a) the cooling effect provided by each pound of refrigerant, (5)'"the.re frigerant circulating rate, (c) the horsepower required, (d) the quantity of heat to be dissipated from the condenser, (e) the required condenser cooling water, in gallons per minute, if temperature rise of water passing through the condenser is 8 deg,; (/) the bore and stroke of a double acting cylinder (neglecting the effect of the piston rod) if speed of compressor is 500 revolutions per minute, (g) coefficient of perform ance. Solution, (a) Saturated liquid F-12 at 121 psia leaves the condenser and enters the expansion valve. The enthalpy of this material (from Table 1) is 29.68 Btu per.. pound and this must also be its enthalpy at entrance to the evaporator. Leaving the' evaporator as a saturated vapor at 52.7 psia, its enthalpy is 82.82 so the refrigerating effect must be 82.82 -- 29.68 = 53.14 Btu per pound: (5) The refrigerant circulating rate is equal to the total heat to be picked up in unit : time divided by the pick-up per pound of refrigerant or, W, = (7 ton.X 200) T- 53.14 ^ 26.3 lb per minute. (c) The horse power required is equal to the increase in energy of the refrigerant, passing through the compressor (expressed in Btu per minute) divided by the con-' version factor 42.42, which, is the number of Btu per minute corresponding to 1 hp, ` (hp) = W, (hi - fcv.) -i- 42.42 (7) where hp = horsepower. Ir, = refrigerant circulating rate in pounds per minute. ) hi = enthalpy of vapor at condition of discharge from compressor, ftv. = enthalpy of saturated vapor entering compressor. Wr is known from (6) and Avs is the enthalpy of refrigerant as it enters the com pressor in a saturated vapor state at 52.7 psia; thus hi, = 82.'82: " In order to determine Ad; the state of the refrigerant must first be determined at the compressor discharge. At the known suction state the entropy (from Table 1 for saturated vapor, at 52.7 psia) is 0.16828 and, since the compression-is'assumed'to occur, isentropically, it therefore follows that the discharge state must have the Bame. entropy at 121 psia. ' From the table the entropy of vapor superheated'25' deg is'