Document rxwGO3nrnROEnwoq03qzNMr9J

816 CHAPTER 37 1953-Guide' analysis of probable power requirements,' compressor size; etc.: Further,1 the equations used in analysis1 of a simple saturation cycleTomr the: basis' of the more complex treatments required for compound refrigeration cycles.- For these-reasons a; typical-simple saturation problem \rillr be worked1' in detail. - ` 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,Fl-12 as.the refrigerant. Determine: (a) the cooling effect provided by each pound of refrigerant; (6) .the re frigerant circulating rate; (c) the horsepower required; (d) the'quantity bf 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 bf a double acting cylinder (neglecting the effect of the piston rod) if speed of compressor is 500 revolutions per minute; (p) coefficient of perform ance. f .<- ' 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 Btuiper 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. Pressure-Enthalpy Diagram for Simple Saturation Cycle (6) 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, TV, -- (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 the1 con version factor 42.42, which is the number of Btu per minute corresponding to 1 hp, (hp) = TV, (Ad - A,,) H- 42.42 ' ' . (7) where - 1 hp = horsepower. W, = refrigerant circulating rate in pounds per-minute. ... Ad = enthalpy of vapor at condition of discharge from compressor... Av< = enthalpy of saturated vapor entering compressor. .. - Wr is known from (5) and A.. is the enthalpy of refrigerant as it-enters the com pressor in a saturated vapor state at 52.7 psia;'thus =, 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 stage must have the same entropy at 121 psia. From the table the entropy of vapor superheated 25 deg is - Refrigeration 817 0.17330; so the superheat, possessed by the actual gas discharged from this com- pressor'can be obtained by interpolation as, . '; ; t^d _ 0.16828 -- 0.16608 25 " 0.17330 - 0.16608 from which = 7.6 deg. . As the saturation temperature at 121. psia is .94 F the actual temperature, Id, of the vapor leaving the compressor is, id = 94 + = 94 + 7.6 = 101.6 F, By the same kind of interpolationthe enthalpy of the discharged vapor can be determined from the enthalpies given for vapor superheated 25 F and for saturated vapor,! (Ad - 88.10) (0,16828 - 0.16608) (92.16 - 88.10) " (0.17330 - 0.16608) :' from which, Ad = 89.34 Btu .per pound.' Then substituting in Equation 7, ` (hp) = 26.3 (89.34 -- 82.82) + 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,, Q,, = 53.14 + (89.34 - 82.82) = 53.14 + 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, ..... Qi = 26.3 (89.34 - 29.68) - 1569 Btu per minute. (e) The cooling wafer rate (based on a gallon as 8.34 lb) is 1569 -e (8 'X 8.34) = 23.5 gpm. - (J) 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, M>) -e 4 = 39.4 d = 3.69 in. ' h: : (p)' (CPj = (Av. - A(c) -e (Ad -- A,,) .. = (82.82 - 29.68) -b (89.34 - 82.82) = 8.17. , . where Atc is the specific 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 T, = 501 F (which is 41, F + 460) and To = 554-F (which is 94 F + 460) and, - 501 (CP) = 554 - 501 The actual cycle is therefore 8.17 -5- 9.6 of 85 percent as effective as a Carnot cycle between the same temperature limits. Influence of Suction Pressure Brief consideration of the analytical procedure used in discussion of the simple saturation cycle will bring out the need for maintaining the suction pressure oh any refrigeration system as high as the load will permit. As the suction pressure increases, for fixed discharge pressure, the enthalpy