Document Rpd12Y2wm4mxYRzVob1313GME

60 CHAPTER 3 1950 Guide' 50 per cent saturation) in the direction given by the protractor for a specific enthalpy of 3.06 Btu/lb.. If air is saturated adiabatically with spray water which is recirculated, the water will ultimately assume a temperature such that the air is brought to saturation at exactly the same temperature; that is, the water will assume the thermodynamic wet-bulb temperature of the air. Example 11. Air at 75 F and 60 per cent saturation is saturated adiabatically with recirculated spray water. Find the resulting temperature and the weight of water added per pound of dry air. Solution. In view of the foregoing remarks the solution of this example reduces to the determination of the thermodynamic wet-bulb temperature of the air. Its humidity ratio is 0.60 X 0.01882 = 0.01129; its enthalpy is 18.018 + 0.60 X 20.59 = 30.372; Equation 7 defining thermodynamic wet-bulb temperature becomes h.* - (Wm* - 0.01129)Aw* = 30.372 At 65 F the value of the left-hand member is 29.995; at 66 F its value is 30.746; by inter polation the thermodynamic wet-bulb temperature is 65.51 F where the humidity Fig. 8. Illustration of Use of Goff Diagram in Solution of Example 11 ratio at saturation is 0.01350; consequently the weight of water added is 0.01350 -- 0.01129 = 0.00221 lb per pound of dry air. On the Goff Diagram, Fig. 8, the process is represented by the line AB which is a segment of the 65.51 F thermodynamic wet-bulb line. The `difference between the ordinates at B and at A is the weight of water added per pound of dry air. . Cooling Load ' The problem of calculating the cooling load for an air conditioned space usually reduces to the determination. of the quantity of, inside air that must be.withdrawn and. the condition to which it must be brought by suitable processing so that its return to the conditioned space will haveltbe net effect of removing given amounts of energy and water front the. space. Let M denote the weight of dry air withdrawn with inside air per hour. With it will be withdrawn energy of amount Mhi and water of amount MWi per hour, where hi and W\ denote the enthalpy and humidity ratio of the inside air, respectively. The weight of dry air returned: with; the conditioned air will necessarily be the same as that withdrawn with the inside .air, but with it must be returned a smaller .quantity of energy Mh and a' smaller quantity of water MW.' Let AQ and AW denote the given Thermodynamics 61 amounts of energy and water to be removed from the conditioned space per hour; then Mh = Mht - AQ MW = MWi - AW Eliminating M and letting q denote the ratio of energy removed to water removed, that is, q -- AQ/AW, according to which: all possible states for the conditioned air lie on a straight line on the Goff Diagram passing through the state point of the inside air in the direction specified by the numerical value of the ratio q. This line is called the condition line for the given problem. If the condition line crosses the saturation curve, the point of intersection is called the apparatus dew point for the given problem. The protractor of the Goff Diagram facilitates the drawing of the condi tion line and the locating of the apparatus dew-point. For this purpose the numerical value of the ratio q is to be regarded as a value of the specific enthalpy of water added, Btu per pound. Example IS. A condition of 80 F dry-bulb, and 67 F thermodynamic wet-bulb, is to be maintained in a clothing store, outside conditions being 95 F dry-bulb, and 75 F thermodynamic wet-bulb. The energy gain from normal heat transmission is esti mated at 16,000 Btu per hour, that from solar radiation at 48,000 Btu per hour. The energy generated by lights, fans, etc., is estimated at 13,900 Btu per hour. The venti lation requirement is 30,000 cu ft per hour. The number of occupants is 50. Find the apparatus dew-point. Solution. The properties of inside air and outside air are readily calculated from the data in Table 1, see especially Example 2. Inside Air OursiDn Air p= h= W= v= 0.5024 31.514 0.01122 -- 0.3848 38.408 0.01413 14.296 The weight of dry air entering with the ventilating air is 30,000/14.296 = 2098.5 lb per hour which brings with it energy of amount 2098.5 X 38.408 *= 80.595 Btu per hour and water of amount 2098.5 X 0.01413 = 29.659 lb per hour. The weight of dry air displaced from the store by the ventilating air is 2098.5 lb per hour which takes with it energy of amount 2098.5 X 31.514 = 66,132 Btu per hour and water of amount 2098.5 X 0.01122 = 23.541 lb per hour. Bach occupant may be regarded as a normal person standing, at rest and evaporat ing (1386 grains) 0.198 lb of water per hour (value obtained by interpolation between Curves D and C Fig. 7, Chapter 6) at about 79 F. From this source there is water of amount 50 X 0.198 = 9.90 lb per hour and energy of amount 9.90 X 1095.7 = 10,847 Btu per hour added to the conditioned space. In addition each occupant loses 225 Btu per hour by conduction, convection, and radiation, making a total for 50 persons of 11,300 Btu per hour. The net energy gain is 16,000 4- 48,000 4- 13,900 4- 80,595 -- 66,132 4- 10,847 411,300 = 114,510 Btu per hour. The net water gain is 29.659 -- 23.541 4- 9.90 = 16.018 lb per hour. Accordingly the direction of the condition line is fixed by the ratio, q => 114,510 -f- 16.018 = 7148.8 Btu per pound of water. On the Goff Diagram, Fig. 9, the direction of the condition line is given by the pro tractor for a specific enthalpy of water added of 7148.8 Btu per pound. The line itself passes through the state point of the inside air and intersects the saturation curve at the apparatus dew-point. According to Equation 11 the enthalpy h and"humidity ratio Wi at the apparatus \ dew-point must satisfy the equation '' : '.'...I',-- 7148.8IF. -- A, - 7148.8 X 0.01122 - 31.514-48.68,1