Document LJbXVBRdJDaY2Vyro4g1j64rq

56 CHAPTER 3 1950 Guide Thermodynamics 57 It may be supposed that the air is cooled between two sections of a duct The tqhueamntaitrieesthoef etwneorgeyncthoanlvpeiecstecdaalccurolastsetdh.e tCwoonsseecrtvioantisonpeorfpeonuenrgdyorfedaruyiraesir tchro*stsJinhgf difference between these two enthalpies be the quantity of heat removed, ofrefrigera tion supplied,, between the two sections. Therefore, *eingera -a?b = 43,072 - 34.187 = 8.885 Btu/lb. The initial volume is 13.980 + 0.50 X 0.822 = 14.391 cu ft/lb. Since 20,000 cfm of air is to be processed, the total refrigeration required is L-aQb = 8.885 X 20,000 + 14.391 = 12,348 Btu per min On the Goff Diagram the process is represented by the horizontal line AB, Fig. 3, whose length is the quantity of refrigeration required per pound of dry air. ' Adiabatic Mixing of Two Air Streams A typical air conditioning process requiring special analysis is the adi abatic mixing of two air streams. Referring to Fig. 4, let mt, ms, m3 denote 3 the weights, of dry air convected across sections Fi, Fs, Fj, respectively, The process of cooling moist air is also represented by a horizontal line on the Goff Diagram. The line may extend across the saturation curve into the two-phase region, nevertheless, the length of the line between the initial and final states is the quantity of heat removed, or refrigeration sup plied, per pound of dry air. By following the final isotherm downward to the. right to-the saturation curve and reading the ordinate there, the weight' of water vapor per pound of dry air in the vapor phase is determined.' The difference between the initial humidity ratio and this ordinate is the; weight of condensed phase per pound of dry air in the final state. Example 7. Air at 95 F and 50 per cent saturation is cooled to 70 F. Find the refrigeration required to process 20,000 cfm of uncooled air. Solution. From the data in Table 1: the initial humidity ratio is 0.50 X 0.03673 = 0.01837 lbw/lb.; the initial enthalpy is 22.827 + 0.50 X 40.49 = 43.072 Btu/lb.; the humidity ratio at saturation at the final temperature is 0.01582 lbw/lb.; the quantity, of liquid formed is 0.01837 -- 0.01582 = 0.00255 lbw/lb,; the enthalpy of the final two-' phase mixture is 34.09 + 0.00255 X 38.11 = 34.187 Btu/lb,. . '' ) ' Fig. 4. Adiabatic Mixing of 2 Ain Streams per minute., Then niiWi, m3Wi, m3Wz and mhi, mji2, mjii will denote the weights of water and the quantities of energy similarly.convected. If the mixing is-adiabaticrit must be governed by the three equations, 77il + Hit m\Wi + Hlihi -j- itithi Itltht (8) Elimination of m* gives, ht -- hi W* -- Wt mi h, - A,= W -- W, = m> ...; (9) according to which: on Ike Goff. Diagram the state point of the resulting mixture lies on the straight line contacting the state points of the two streams being mixed and divides the line into'two segments which are in the same ratio as are the weights of dry air in the two 'streams. Example 8. Outside Air at 0 F and 80 per cent saturation is to be mixed adiabatic- ally with recirculated Inside Air at 70 F and 20 per cent saturation in the ratio of one pound of dry air in the former to seven in(the latter. Find the temperature and degree of saturation of the resulting mixture. j Solution. - The humidity ratio Wt and the enthalpy h, of the resulting mixture must satisfy Equation 9, namely, . - 0.003164 - Wt = 20.270 - h. = 1 Wt - 0.000630 " ' h, - 0.668 " 7 ` '