Document zdyvaxByON0X4M2V1pDwmnR6R

56 CHAPTER 3 , 1953. Guide' temperature at this point of intersection is by definition the dew-point temperature for state 1. ]*?1u1> tuvi Ai-AbnUKinliiIjt ^mKtilAufuiitgKllcunAAOBi tin Anluiili<Vkim/uuumm nw*vnutwnco ilnOnOnV/ViAWmUnlpnanUtAtb/1U ktij nWn-U> rUlntnilnMAli' tion. The succession of states for the total system, moist air and liquid water, is represented by a continuation of the W = Wi line into the liquid vapor region. (Tern-, peratures below 32 F would involve the solid-vapor region). Consider that the final temperature is U. The final enthalpy is then As; the liquid water formed is (Wi -- IFi), where point 3 is at the intersection of the isotherin through 2 and the saturation curve; the final humidity ratio of the moist air is W3\ 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 + 0.50(40.49) = 43.072 Btu per lb of dry air; the humidity ratio at saturation at the Thermodynamics - 157 cii ft per lb of dry air; andithe final enthalpy is 34.2 Btu per lb of dry air. The solu tion of the problem is , i5a = 2--0 000 X (43 -- 34.2) = 12,200 Btu per min. - ! ' .14.4 ...... i 1 -: - l"! The other method is to use an energy balance, V >?. = GPl - & - - 1F,)J ' .: The initial humidity ratio is 0.0183 lb of water vapor perlb 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 - iQs = 90 on-o X (43 - 34.1 - 0.0025 X 38.07) : . 14.4 = 12,130 Btu per min. Fig. 8. hU *, Cooling op Aib at Constant Pressure Shown on A.S.H.V.E. PSTCHROMETRIC 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; A.> 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 GA, = Gh, + G(W, - JFS)A + ,? or = (7[Ai -- A -- (Wi -- !Ft)Aw3] 20,000 14.391 X (43.072 - 34.09 - 0.00255 X 38.07) = 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 corre sponding to the final temperature, 70 F. This is shown as point 2 on the diagram. Then, = G(hi -- Aj) The initial enthalpy is 43 Btu per lb of dry air; the initial specific volume is 14.4 Fig. 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 + Gj = Gz Energy balance for the process, Gihi -f- Gzhs -- Gzhz Mass balance for the water vapor, GzWz + GzWz = G,W, Eliminating G and combining the three equations yield the equation, Ai -- A Wt -- Wz Gi A, - A, = Wi-Wr G,' 1^. Example 5: 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 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 are the humidity ratio Wt and the enthalpy A of the resulting mixture. These may be determined from Equation 34. Thus,