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58 CHAPTER 3 1954 Guide 20.270 - A, 0.003164 -- W, _ 1 h3 - 0.668 = W, - 0.000630 ~ 4 from which h, = 16:350 and W, = 0.002657. The eDthalpy.ofThe final mixture may also be expressed by Equation 28: h% = hh + /jhu Since /i by definition ia Wz/lf's, Equation 28 may be rewritten 16.350 = K + (0.002657/W*) 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 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 r = 0.002657/0.00960 = 0.277 Solution 6. From the A.S.H.V.E. point of the resulting mixture lies on Chart. E a straight luation 34 indicates that line connecting the state the state points of Fig 10 Illustration of Mixing of Two Steady 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. 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 Gi _ Ds-* _ 1 (?i Da-i 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 adiabatic&lly between two sections, 1 and 2, as in Fig. 12, with moisture addition at the rate -- IFi) and the moisture having the enthalpy h* Btu per pound of moisture. An energy balance yields Gihi + G,(W, - W\)h,, = Giht (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 w . ury-uuio temperature ana 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 *,= *,+ (W, - ff\)AThe initial enthalpy of the moist air ht must be found from Equation 8, h, = h* - {W - Wt)hw* 22.827 + ,,40.49 = 43.69 - (0.02233 - 0.03673,,) (48.05) from which ,, = 0.511. Hence, hi = 22.827 + 0.511(40.49) -- 43.52 Btu per lb of dry air and W, = 0.03673(0.5X1) = 0.01877 lb per lb of dry air. The solution of Equation 35 is h, = 43.52 + (IF, - 0.01877) (8.09) --a-- TV V sauaueu at tne temperature /y.87 At this temperaturethe 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 evaporated is AT ENTHALPY h. Fig. 12. Illustration of Addition of Moisture to an Adiabatic Stream