Document zdg5ojqboEvNNDGGv2XXB0V9a

58 CHAPTER 3 1957 Guide 20.270 - A, 0.003164 - Wi A, - 0.668 W, - 0.000630 4 from which h, = 16.350 and W, = 0.002657. The enthalpy of the final mixture may also be expressed by Equation 28: A* = Aa 4- phu Since m by definition is Equation 28 may be rewritten as * JB.V M 16.350 = A. + (0.002657/1FJ X A,, At 56 F the right side of the equation is 16.332, and at 57 F it is 16.582. Interpola tion gives as the final dry-bulb temperature of the mixture 56.07 F. At this tem perature the humidity ratio at saturation is 0.00960 lb of water vapor ner lb of dry air. Therefore, the final degree of saturation is M = 0.002657/0.00960 = 0.277 Solution b. From the AJ3.HA.E. Chart. Equation 34 indicates that the state point of the resulting mixture lies on a straight line connecting the state points of flU. 1U. iLiiUOJOAxxuir vx _ .. .. Pressure the two streams being mixed, and divides this line 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.l and 2 are located and connected by a straight line. The state of the final mixture is set so that G, _ D,, _ 1 G, ftj 4 Scaling the distances on the chart, the required solution to Example 6 is 66 F dry-bulb temperature and 0428 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 G,(Wi -- IFi) and the moisture having the enthalpy K, Btu per pound of moisture. An energy balance yields <5iAi + GtiWt - W,)hw = Gib, (35) Example 8: Liquid water chilled to 40 F is injected into an air stream initially at Thermodynamics 59 Fig. IX. Solution op Example 5 on ASHAE Psycheometkic Chart 95 F dry-bulb temperature and 80 F thermodynamic wet-bulb temperature. At what temperature will saturation be reached? How much water must be evaporated to reach saturation? Solution a: From the data of Table 2. The solution of Equation 35 for hi yields A = Ai + (W, -- IFi)A,, The initial enthalpy of the moist air hi must be found from Equation 8, A, = A* -- (XT* - IPi)A,,* 22.827 + ^40.49 = 43.69 - (0.02233 - 0.03673m) (48.05) from which u = 0.511. Hence, Ai = 22.827 + 0.511(40.49) <= 43.52 Btu per lb of dry air and Wi = 0.03673(0.511) = 0.01877 lb per lb of dry air. The solution of Equation 35 is A, = 43.62 + (W, - 0.01877) (8.09) By trial and error, this equation will be satisfied at the temperature 79.87 F. At this temperature the humidity ratio W, is 0.02223. The weight of water evaporated is therefore 0.02223 - 0.01877 = 0.00346 lb per lb of dry air. G(wj-Wi) AT ENTHALPY h. 'I. 12. Illustbation op Audition op Moisture to an Adiabatic Stream