Document DM3eLRg2383GVpdJKxmdK23q4

58 CHAPTER 3 1955 Guido , 20.270 - A, 0.003164. -If, 1 A, - 0.668 = TP, - 0.0006304 .\ ' from Which A, 16.350 and IP, 0.002657. The enthalpy of the final mixture may also be expressed by Equation 28: A, A* -h #*Am Since m by definition is W%/W,, Equation 28 may be rewritten as 16.350 = A. + (0-002657/TP.) X At 56 F the right side of the equation is 16.332, and at 57 F it is 16.582.'' Interpola tion gives as tne final dry-bulb temperature of the mixture 56.07 F. At this tem perature the humidity ratio at saturation is 0.00060 lb of water vapor per lb of dry air. Therefore, the final degree of saturation is #, W0X02657/0.00960 = 0.277 Solution b. From the AJ3.H.AJE. Chart. Equation 34 indicates that the state point of the resulting mixture lies on a straight line connecting the state points of Fig. 10. Illustration of Mixing of Two Steady Flow Streams at Constant Pressure the two streams being mixed, and^diyides this Une into two segments whose'respec tive lengths are inversely proportional to the rates of dry air flow in the correspond ing streams. This is illustrated in Fig. 11. Points 1 and 2 are located and connected by a straight line. The state of the final mixture is set so that........ Gj _ fin _ 1 Gt Dt-l ; 4..--`. Scaling the distances on the .chart, the required solution to Example 5 is 56 F dry-bulb temperature and 0.28 degree of saturation. Addition of Moisture to an Adiabatic Stream Consider a stream of moist air flowing adiabatically. between two sections, 1 and 2, as in Fig. 12. with moisture addition at the' rate (ri(IF, -- JPO and the moisture having the enthalpy Km Btu per pound of moisture. An energy balance yields^. ' \ C,5, + OAWt - W,)h. - G,h, (35) Example 6: Liquid water chilled to 40 F is injected into an air stream initially at Thermodynamics 59 Fig. 11. Solution of Example 5 on A.S.H.V.E. Psychrometric Chart 95 F dry-bulb temperature and 80 F thermodynamic wet-bulb temperature. At what temperature will saturation be reached? now much water must be evaporated to reach saturation? Solution a: From the data of Table 2. The solution of Equation 35 for As yields A, A, + (W, -- TP0A. The initial enthalpy of the moist air A* must be found from Equation 8, A, - A* - (IP* - IPOAw* 22.827 + **40.49 = 43.69 - (0.02233 - 0.03673#*) (48.05) from which #* 0311. Hence, As 22327 + 0311(40.49) = 4332 Btu per lb of dry air: and IP, = 0.03673(0311) 0.01877 lb per lb of dry air. The solution of Equation 35is At = 4332 + (IP, - 0.01877) (839) By trial and error, this equation will be satisfied at the temperature 79.87 F. At this temperature the humidity ratio TP, is 0.02223. The weight of water evaporated is therefore 0.02223 -- 0.01877 0.00346 lb per lb of dry air. AT ENTHALPY h. Fig. 12. Illustration of Addition1 of Moisture to an Adiabatic Stream