Document YD637LndBd2yzZRZoRnYqDNa0

64 CHAPTER 3 1951- Guide 20.270 - A, = 0003164 - W, 1 k, - 0.668 " W, - 0.000630 " 4 from which A, = 16.350 and IF, = 0.002657. The enthalpy of the final mixture may also be expressed by Equation 28: A, *= A, + Mm Since a by definition is Wt/W,, Equation 28 may be rewritten as 16.350 = A, + (0.002657/IE,) X Am 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 a = 0.002657/0.00960 = 0.277 Solution b. From the A.S.H.V.E. 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 op Mixing op Two Stuart Flow Streams At Constant 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. ThisisiUustratedinFig.il. Points 1 and 2 are located and connected by a straight line. The state of the final mixture is set so that ft i ft=I>A4 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. ,P M> Addition of Moisture to an Adiabatic Stream Consider a stream of moist air flowing adiabaticaUy between two sections, 1 and 2, as in Fig. 12, with moisture addition at the rate ft(IF, -- Wt) and the moisture having the enthalpy A. Btu per pound of moisture. An energy balance yields . , G1A1+ ft(Wt -- IEi)A* = Giht (35) Example 6. Liquid water chUIed to 40 F is injected into an airstream initially at M J Thermodynamics 1. \ 65 w2 w? w, 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? 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, = A2 + (W, - Wi)hw The initial enthalpy of the moist air Ai must be found from Equation 8, A2 = A* - (IF* - IF2)A,,* 22.827 + M0.49 = 43.69 - (0.02233 - 0.03673a) (48.05) from.which a =' 0.511. Hence, A, = 22.827 + 0.511(40.49) = 43.52 Btu per lb of dry air and Wi = 0.03673(0511) = 0.01877 lb per lb of dry air. The solution of Equation 35 is A, = 43.62 + (IF, - 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 Wi is 0.02223. The weight of water evaporated is therefore 0.02223 -- 0.01877 = 0.00346 lb per lb of dry air. . Fig. 12. ATENTHALPY h,, .......................... . ......... Illustration of Aodition'of Moisture'to an Adiabatic Stream