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920 CHAPTER 36 1958 Guide de-superheating and condensation take place. -From the condenser the re frigerant flows to the expansion valve, undergoesa.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 of the more complex treatments required for compound refrigeration cycles. For these reasons a typical simple saturation problem will be worked in detail. Heat of Compression Added to Gas Refrigeration 921' where hp = horsepower. . W, -- refrigerant circulating rate in pounds per minute. hd = enthalpy of vapor at condition of discharge from compressor. kvs -- enthalpy of saturated vapor entering compressor. Wj is known from (5) and his the enthalpy of refrigerant as it enters the com pressor in a saturated vapor state at 52.7 psia; thus h,, = 82.82. In order to determine hi, 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 stage must have the same entropy at 121 psia. From the table the entropy of vapor superheated- 25 deg is 0.17330, so the superheat, possessed by the actual .gas discharged from this com pressor can be obtained by interpolation as, from which 1*i -- 7.6 deg. t.i~,__ 0.16828 - 0.16608 25 ~ 0.17330 - 0.16608 Example 1: A simple saturation cycle carries a 7 ton load when.operating between suction and discharge pressure of 52.7 psia and 121 psia with dichlorodifluoromethane, CCljFj, as the refrigerant. Determine: (a) the cooling effect provided by each pound of refrigerant; (6) the refrigerant circulating rate; (c) the horsepower required; (d) the quantity of heat to be dissipated from the condenser; (e) the required conden ser 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. performance. Solution: (a) Saturated liquid CCltFj 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 re- , frigerating effect must be 82.82 -- 29.68 = 53.14 Btu per pound. '.- (b) 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) + 53.14 = 26.3 lb per minute. (e) The horsepower required is equal to the increase in energy of the refrigerant . passing through the compressor (expressed in Btu per minute) divided by the con-., j version factor 42.42, which is the number of Btu per minute corresponding to 1 hP; - i (hp) = W, (Ad - fcv.) + 42.42 <7) i Fig. 4. Pressure-Enthalpy Diagram for Simple Saturation Cycle As the saturation temperature at 121 psia is 94 F the actual temperature, ti, of the vapor leaving the compressor is, U = 94 + = 94 + 7.6 = 101.6 F. By the same Jund of interpolation the enthalpy of the discharged vapor can be determined from tne enthalpies given for vapor superheated 25 F and for saturated vapor, (Aj-88.10) (0.16828 - 0.16608) (92.16 - 88.10) " (0.17330 - 0.16608) from which, Ai = 89.34 Btu per pound. Then substituting in Equation 7, (hp) = 26.3 (89.34 - 82.82) + 42.42 = 4.03. r.a*'e f heat loss from the condenser, Qi, must be equal to the sum of the gies picked up by the refrigerant in the evaporator and the compressor, 59 66 = fHV" (89 34 .- 82-82) = 53.14 + 6.52 = 59.66 Btu per pound or 26.3 X dirpptl k 9 iu Per. minute. This same figure can, of course, be determined more pn*Li^ b subtraction of the enthalpy of liquid leaving the condenser from the ipy of superheated vapor going into it, thus, Q = 26.3 (89.34 -- 29.68) = 1569 Btu per minute. 23 5Cgp^e cohng water, rate (based on a gallon as 8.34 lb) is 1569 -5- (8 X 8.34) = maSii^p 6 comP(esspr size is fixed by the volume of gas which must be drawn into the Tahlp i P,r,,unit time. Saturated vapor at 52:7 psia has a specific volume, from of 0-779 cu ft per pound, hence 26.3 X 0.779 = 20.49 cfm of gas must be