Document bBQwdOJX59XqELQXkXRw5peay

62 CHAPTER 3 1951 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 cur 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 the final temperature is U. The final enthalpy is then h*; the liquid water formed is (JFj -- Wi), 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 Wi\ and this final moist air has dew-point, wet-bulb and dry-bulb temperatures all equal to l<. , 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(0.03763) -- 0.01837 lb of water vapor per lb of dry air; the initial enthalpy is 22.827 4- 0.50(40.49) = 43.072 Btu per Id of dry air; the humidity ratio at saturation at the Thermodynamics 63 1K o-v ai,.. and the final enthalpy is 34.2 Btu per lb of dry air. The solucu ft per m oi ***j tion of the problem is = 99^99 X (43 - 34.2). = 12,200 Btu per min.. * 14.4 The other method is to use an energy balance, i3 " G[A. -- hi -- h^tiWi -- Wi)] Tti^hbeein'reitmiaol hveudm'ii?dity1ra58tio*is0f.0W1*8t3*lb^of wat^er viabpofr p^er lb of dTrhveareirfoaren^dththe*,,tl = ?9i999 X (43 - 34.1 - 0.0025 X 38.07) 14.4 = 12,130 Btu per min. Fig. 8. PSYCHBOMETRIC CHABT 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; 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 Ghi = G&* + G(Wi -- W-^)h*+ ig* igi G[hi -- ht -- (Wi -- _ 20,000 7 14.391 X (43.072 - 34.09 - 0.00255 X 38.07) *= 12,350 Btu per min. Solution ))', : From th^A.S.H.-ViE.. Chart. Two methods may be used to solve the problem by^use of the'psyeh/6nietric`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 corre sponding to the final temperature, 70 F. This is shown as point 2 on the'diagram. Then * ig- = f?(h$'.x-- ... r The initial enthalpy is 43 Btu per lb of dry air; the initial specific volume is 14.4" Fio. 9. Illustration op Process op 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 + <7, = Gt Energy balance for the process, GiAi + Gift, =* GJit Mass balance for the water vapor, GiWi + G.IF, - G,W, Eliminating G and combining the three equations yield the equation, h% -- hi '. Wt -- Wi Gi A, - h, = IF, - IF, " G, * '' . Example S.. Outside air at 0 F dry-bulb temperature and 0.80 degree of saturation is to be mixed adiabatically with recirculated inside air at 70 F dry-bulb temperature and 0.20 degree of saturation, in the ratio of one pound of dry air in the former to four mjhe latter. Find the temperature and degree of saturation in the resulting mixture. ' Solution a. From the, data of Table 2. The only unknown properties are the huandity ratio IF. and the.enthalpy h. of the resulting mixture. These may be 'determined from Equation" 34.' Thus,