Document 3JRV9j9MzZryzaXrgj3rrG3X0

856- CHAPTER 37 1955 Guide i analysis of. probable power requirements, compressor size, etc. Further,1 f. the equations used in analysis of a simple saturation cycle form the basis1 f 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 suetion and discharge pressure, of 52.7 psia and 121 psia with F-12 as the refrigerant.1 Determine :(a) the cooling effect provided by each pound of refrigerant; (b) 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 gallons1 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 pf 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 re frigerating effect must be 82.82 -- 29.68 = 53.14 Btu per pound. Fig. 4. Pbessure-Enthalpy Diagram ; fob Simple Saturation Cycle (5) The refrigerant circulating rate is equal to the total heat to be picked up in unit time, divided by the pickrup per pound of refrigerant or, W, = (7 ton X 200) + 53.14 = 26.3 lb per minute. (c) 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 version factor 42.42, which is the number of Btu per minute corresponding to 1 bp, (bp) = W, (E, - hr.) + 42.42 (7) where hp = horsepower. Wr = refrigerant circulating rate in pounds per minute. ' Ki = enthalpy of vapor at condition of discharge from compressor. h,, = enthalpy of saturated vapor entering compressor. , Wr is known.from (6) and hr. is the enthalpy of refrigerant as it enters the com pressor in a saturated vapor state at 52.7 psia; thus hr, = 82.82. In order to determine Aa, 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.m Refrigeration 857 0.17330, so the superheat, possessed by the actual gas discharged from this com pressor can be obtained by interpolation as, ' Ua 0.16828 - 0.16608 '-'-/i 25 0.17330 - 0.16608 from which Ua ~ 7.6 .deg. . ' - . ' As the saturation temperature at 121 psia is 94 F the actual temperature, fa, of the vapor leaving the compressor is, U = 94 -1- = 94 + 7.6 = 101.6 F. By the same kind of interpolation the enthalpy of the discharged vapor can be determined from the enthalpies given for vapor superheated 25 F and for saturated vapor, (hj - 88.10) __ (0.16828 - 0.16608) (92.16 --88.10) _ (0.17330 -- 0.16608) . from which, ha = 89.34 Btu.per pound. Then substituting in Equation 7, (hp) = 26.3 (89.34 - 82.82) -i- 42.42 = 4.03. (d) The rate of heat loss from the condenser, Q., must be equal to the sum of the energies picked up by the refrigerant in the evaporator and the compressor, Qa = 53.14 4- (89.34 -- 82.82) = 53.14 4- 6.52 = 59.66 Btu per- pound or 26.3 X 59.66 = 1569 Btu per minute. This same figure can, of course, be determined more directly by subtraction of the enthalpy , of liquid leaving the condenser from the enthalpy of superheated vapor going into it, thus, Qc = 26.3 (89.34 -- 29.68) = 1569 Btu per minute. () The cooling water rate (based on a gallon as 8.34 lb) is 1569 4- (8 X 8.34) = 23.5 gpm. ' if) The compressor size is fixed by the volume of gas which must be drawn into the machine per unit time. Saturated-vapor -at 52.7 psia has a specific volume; from Table 1, of 0.779 cu ft per. pound, hence 26.3 X 0.779 = 20.49. cfm of gas must be handled. Assuming a volumetric efficiency of 90 percent, the compressor must then displace 20.49 -s- 0.9 = 22.8 cfm. The speed is given as 500 rpm and, as the unit is known to be double-acting, the displacement is therefore (22.8 X 1728) + (2 X 500) = 39.4 cu in. If the unit were designed so that bore d and stroke were the same, (vcP) + 4 39.4 d = 3.69 ini - iff) (CP) - (h,, - h,c) + (hi -- hr,) = (82.82 - 29.68) -e (89.34 - 82.82) = 8.17 where hi, is the speoific enthalpy of liquid at discharge.from the condenser. The coefficient of performance of Example 1 may be compared with that of an ideal system operating on the Carnot cycle between the same tempera ture limits. Then TB = 501 F (which is 41 F + 460) and 7V = 554 F (which is 94 F + 460) and, 501 (CP) = 554 - 501 = 9.6 The actual cycle is therefore 8T7 -f- 9.6 or 85 percent as effective as a Carnot cycle between the same temperature limits. Influence of Suction Pressure consideration of the analytical procedure used in discussion of the unple saturation cycle will.bring out the need for maintaining the suction Pressure on any refrigeration system as high as the load will permit. As e Morion pressure increases, for fixed discharge pressure; the enthalpy