Document QX3gO7RwREedY9eZDe3414va5

60 CHAPTER 3 1946 Guide the weight of water evaporated is 0.01894 -- 0.01559 = 0.00335 lb per pound of dry air. On the Mollier Diagram, Fig. 7, a line is drawn through the initial state point (90 F, 50 per cent saturation) in the direction given by the protractor for a specific enthalpy of 3.06 Btu/lbw. 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 11.1 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 above remarks the solution of this example reduces to the determination-of the thermodynamic wet-bulb temperature of the air. Its humidity Fig. 8. Illustration of Use of Mollier Diagram in Solution of Example 11 - 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 As* - (Wf - 0.01129) Aw* = 30.372 At 65 F the value of the lefthand 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 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 Mollier 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 ordi nates 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load for an air conditioned space usually reduces to the determination of the quantity of inside air that must be withdrawn .and the condition to which it must- be brought by suitable processing so that its return to the conditioned space-will have the net effect of removing given amounts of energy and water from 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 Let M denote the weight of dry'air withdrawn with inside air per hour. With it will be withdrawn energy of amount Mhi and water of.amount .Thermodynamics --------------------------------------- ' '---------------------------- ----------------------------------------------------------- MWx per hour, where hi and Wi 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.. LetAQ and A W denote the given amounts of energy and water to be removed from the conditioned space ner hour; then Mh = Mhi - AQ MW= MW, - A IE . Eliminating M and letting q denote the ratio of energy removed to water removed, that is, q = AQ/AIT, h -- h\ W - Wi " 2 (ID according to which: all possible states for the Conditioned Air lie on astraight line on the Mollier 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 appa ratus dew-point for the given problem. The protractor on the Mollier Diagram facilitates the drawing of the condition 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 12. 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 estimated 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 ventilation requirement is 30,000 cu ft per hour. The number of occupants is 50. Find the ap paratus dew-point. Solution. The properties of inside air arid outside air are readily calculated from the data in Table 1, see especially Example 2. ti h W v Inside Air = . 0.5024 = 31.514 = , 0.01122 . = ........... Outside Air 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 energyof amount 2098.5 X.31.514 = 66,132 Btu per houpand water of amount 2098.5 X 0.01122 = 23.541 lb per hour. Each occupant may be regarded as a normal person standing at rest and therefore evaporating 0.198 lb of water per hour at about 79 F (Table 4, Chapter 12).' 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, convectiori, and radiation, making a total for 50 persons of 11,300 Btu per hour. ., The net energy gain is 16,000 4- 48,000 + 13,900 + 80,595 -- 66,132 -f* 10,847 + 11,300 = 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 -* 16.018 = 7148.8 Btu per pound of water., , * On the Mollier Diagram, Fig. 9, the direction of the' condition linc U^yemby t>.e pro tractor for a specific enthalpy of water added of* 7148.8 Btu per- pound.- The line itself