Document MMOdzV7qvnyvjw0D25M879jxz

/' 58 CHAPTER 3 1953 Guide ' 20,270 - A, 0.003164 - W, 1 Aj - 0.668 ~ W, - 0.000630 " 4 from which As = 16.350 and W. = *0.002657. The enthalpy of the final mixture may also be expressed by Equation 28: , As = A. + pA* Since /> by definition is Wt/W,, Equation 28 may be rewritten as 16.350 =. A, + (0.002657/W.) 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 per lb of dry air. Therefore, the final degree of saturation is M = 0.002657/0.00960 = 0.277 Solution b. From the A.S.H.V.E. Chart. Eauation 34 indicates that the state point of the resulting mixture lies on a straight line connecting the state points of Fig. 10. Illustration op Mixing of Two Steady Flow Streams at Constant Pkessube 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 1 and 2 are located and connected by a straight line. The state of the final mixture is set so that O. -D^ l Gt 4 Scaling the distances on the chart, the required solution to Example B 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 <?i(Wi -- Wi) and the moisture having the enthalpy A, Btu per pound of moisture. An energy: balance yields G,A, + G,(W, - W,)hw = GiAj . (35) Example 6: Liquid water chilled to 40 F is injected into an air stream initially at Thermodynamics 59- Fig. 11. Solution op Example 5 on A.S.H.V.E. Psychbometbic 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 A> yields Ai = Ai + (Wi -- Wi)A The initial enthalpy of the moist air Ai must be found from Equation 8, Aj = A* - (W* - Wi)A,,* 22.827 + M40.49 = 43.69 - (0.02233 - 0.03673*.) (48.05) from which p = 0.511. Hence, - A, = 22.827 + 0.511(40.49) = 43.52 Btu per lb of dry air and W, = 0.03673(0.511) = 0.01877 lb per lb of dry air. The solution of Equation 35 is A, = 43.52 + (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 Wt is 0.02223. The weight of water evaporated is therefore 0.02223 - 0.01877 = 0.00346 lb per lb of dry air. Fig. 12. Illustbation op Addition op Moisture to an Adiabatic Stream