Document gyM1Xoea0pLE406k8oZ48rge

56 CHAPTER 3 1955 Guide temperature at this point of intersection is by definition the dew-point temperature for state 1. Further cooling through successive equilibrium states is accompanied by condensa tion. The succession of states for the total system, moist air and liquid water, is represented by a continuation of the IV = W, line into the liquid vapor region. (Tem peratures below 32 F. would involve the solid-vapor region). Consider that the final temperature is li- The final enthalpy is then k3; the liquid water formed is (IVi -- JFi), where point 3 is at the intersection of the isotherm through 2 and the saturation curve; the final humidity ratio of the moist air is-IF*; and this final moist air has dew-point, wet-bulb and dry-bulb temperatures all equal to ij. Example 4: How much heat must be removed from 20,000 cfm of air at 95 F dry-bulb temperature and 0.50 degree of saturation to cool the air to 70 F, saturated? . Solution a: From the data of Table 2. The initial humidity ratio is 0.50(0.03673) = 0.01837 lb of water vapor per lb of dry air; the initial enthalpy is 22.827 + 0.50(40.49) = 43.072 Btu per lb of dry air; the humidity ratio at saturation at the Fig. 8. Cooling op Air at Constant Pressure Shown on A.S.H.V.E. PSYCHROMETRIC CHART final temperature is 0.01582 lb of water vapor per lb of dry air; the quantity of liquid formed is,0.01837 -- 0.01582 = 0,00255 lb of water vapor per lb of dry air; hwt at 70 F is 38:11 Btu per lb of water; the initial specific volume is 13.980 + 0.50(0.822) = 14.391 cu ft per lb of dry air. Fig. 9 illustrates the process diagrammatically. The energy equation for the process is Gh, = Gh, + G(W, - Wt)hn + i or igj = G[hi -- As -- (Wi -- PFs)A.j] = 20 000 x (43.072 - 34.09 - 0.00255 X 38.07) 14.391 = 12,350 Btu per min. Solution b: From the A.S.H.V.E. Chart. Two methods may be used to solve the problem by use of the psychrometric chart. The simpler is to use the region to the left of the saturation line (Fig. 8). From point 1 draw a horizontal line on the chart until it intersects the constant temperature line in the liquid-vapor region corresponding to the final temperature, 70 F. This is shown as point 2 on the diagram, Then, | i?i = G(ht -- hi) The initial enthalpy is 43 Btu per lb of dry air; the initial specific volume is 14.4 i Thermodynamics 57 cu ft per lb of dry air; and the final enthalpy is 34.2 Btu per lb of dry air. The solu tion of the problem is 20,000 i<b = ' ^ X (43 -- 34.2) = 12,200 Btu. per min. The other method is to use an energy balance, = GTA, - hi - Awj(H,l - W,)] , The initial humidity ratio is 0.0183 lb of water vapor per lb of dry air, and the final humidity ratio is 0.0158 lb of water vapor per lb of dry air. Therefore, the heat to be removed is 9j = -1r4r.--4 X (43 - 34.1 - 0.0025 X 38.07) = 12,130 Btu per min. Q -W VV--j GG ///// 777777 Z/S, 'f l3 c(wi-w3)bwJ Fio. 9. Illustration op Process of Example 4 Adiabatic Mixing of Two Steady Flow Air Streams at Constant Pressure The process is diagrammed in Fig. 10. By applying the principles of the con servation of mass and energy, three equations may be written: Mass balance for the dry air, Gi 4- G% = Gt Energy balance for the process, Gihi + Gtht -- Gihi Mass balance for the water vapor, GiWi -f GtWt - GzWt Eliminating Gt and combining the three equations yield the equation, ' ht -- kt W\ -- Wi Gi . ht- h " Wt^Wt ~ Gt; m is Outside air at 0 F dry-bulb temperature and 0.80 degree of saturation and n <^diabatically with recirculated inside air at 70 F dry-bulb temperature in thft l *+ reen- saturation, in the ratio of one pound of dry air in the former to four e latter. Find the temperature and degree of saturation in the resulting mixture. a-` From the data of Table 2. The only unknown properties are the the enthalpy ht of the resulting mixture. These may be uetermined from Equation 34. Thus,