Document 0YRXqbVwLqzZV5emzE64noOb

56 CHAPTER 3 1956 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 W = Wt line into the liquid vapor region. (Tem peratures below 32 F would involve the solid-vapor region). Consider that thefinal temperature is The final enthalpy is then A,; the liquid water formed is (Hr -- Wx), 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 i,. Example 4- How much heat must be removed from 20,000 efm 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 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 . = x (43 -- 34.2) = 12,200 Btu per min. 14.4 ' .; The other method is to use an energy balance, iff, B G[hi -- hx -- A.,(IFi -- IF,)] ( The initial humidity ratio is 0.01831b 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 ... l9, = ^529 x (43 - 34;i - 0.0025 X 38.07) 14.4 = 12,130 Btu per min. W, W5 Flo. 8. Cooling of Aje at Constant Phessobe Shown on A;S.H.A.E. PSYCHBOMETBIC CHABT final temperature is 0.01582 lb of water vapor per lb of dry air; the quantity of liquid foniied is 0.01837 - 0.01582 = 0.00255.lb of water vapor per lb of dry air; Kx at .70 F is 38vll"Btu per lb of water; the initial specific-volume is 13.980 -f 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 Ghi = Ghx + G(iVi -- 1F,)Aw* -f 1Q1 or. i2 = C[A, -- hx -- (IF, -- IF,)A,,,] "lsix (43'072 " 34-09 - -00255 X 38-07>' ; = 12,350 Btu per min. Solution b: From the A.S.H.A.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. loft 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 corre sponding to the final temperature,'70 F. This is shown as point 2 on the diagram. ' en - .iff* -- G(hi -- hx) . The initial enthalpy is 43 Btu per lb of dry air; the initial specifie volume is 14.4 GCwr-w,)hwJ Fig* 9* Illustration of Process of Example 4 Adiabatic Mixing of Two Steady plow Air Streams at Constant Pressure The process is diagrammed in Fig. 10. By applying the principles of the con servation^ maffltmd.energy, .three equations piay be written:: Mass balance for the dry air, Gi + (h~G. ' Energy balanceforthe process, Gihi 4* Gtht = , Mass balance for the water vapor, ; GiWi + GtWt = GtWi : .. .; Eliminating G% and combining the three equations yield the equation, f . hi -- hi W.% -- Wt ^ G\ . (34) hr -- htr Wt -- Wj' Example 6: Outside air at 0 F dry-bulb temperature,andU.80 degree.of saturation is to be mixed adiabatically with recirculated inside air at 70 Fdiy-biilb-temperature and 0.20 degree of saturation, in the ratio of one pound of dry.air in the former to four in the latter. Find the temperature and degree of saturation in the resulting mixture. , Solution a.* From the data of- Table 2. The only unknown properties aTe the 'hqinidity rii'tiO'TPi'and the enthalpy hi of the resulting mixture... TheSe may be determined from Equation 34. Thus,