Document Qk2zYokJ5enJ5BDO7jJqVjy05

56 CHAPTER 3 1946' Guide Fig. 2. Illustration of Use of Mollier Diagram in Solution of Example 6 the initial and final states is the quantity of heat removed, or refrigeration supplied, 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 deter-. mined. The difference between the initial humidity ratio and this ordi. nate 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 iii Table 1: the initial humidity ratio is 0.50 X 0.03673 = 0.01837 lbw/lba; the initial enthalpy is 22.827 + 0.50 X 40.49 - 43.072 Btu/lba; the humidity ratio at saturation at the final temperature is 0.01582 lbw/lba; the quantity of liquid formed is 0.01837 -- 0.01582 ; 0.00255 lbw/lba; the enthalpy of the final two-phase mixture is 34.09 + 0.00255 X 38.11 = 34.187 Btu/lba. It may be supposed that the air is cooled between two sections of a duct. The; quantities of energy convected across the two sections per pound of dry air crossing them . 15.82 24.70 Fig; 3. Illustration of Use of Mollier Diagram in Solution of Example 7 . Thermodynamics ------------------------------- -------- ------------------------------------ :------- ----------:-------- :-----------:---------------------- the two enthalpies calculated above. Conservation of energy requires that the difference between these two enthalpies be the quantity of heat removed, or refrigeration supplied, between the two sections. Therefore, -a5b = 43,072 - 34.187 = 8.885 Btu/lba The initial volume is 13.980 + 0.50 X 0.822 = 14.391 cu (t/lba. Since 20,000 cfm of air is to be processed, the total refrigeration required is -sQb = 8.885 X 20,000 h- 14.391 = 12,348 Btu per minute On the Mollier 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 adiabatic mixing of two air streams. Referring to Fig. 4, let mi, mi, m3 denote the weights of dry air convected across sections Fi, F2, F3, respect ively, per minute. Then m3Wi, miWi, m3W3 and mihi, mihi, m3h3 will Fig. 4. Adiabatic Mixing of 2 Air Streams denote the weights of water and the quantities of energy similarly con vected. If the mixing is adiabatic, it must be governed by the. three equations, mi + ffi = mi miWi + mtWt = mtWi (8) .mihi -(- mthi = mihi Elimination of m3 'gives, ' hi -- h ._ Wi -- W3 _ nn ' ht -- hi Wi -- Wi mi (9) according to which: on the Mollier Diagram the state point of the resulting mixture lies on the straight line connecting 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. Examples. Outside Air at 0 F and 80 per cent saturation is to be mixed adiabatically 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., .: Solution. The humidity iutio Wi and the enthalpy hi of the resulting mixture must .satisfy Equations 9, namely, 0.003164 JFs ^ 20.270 - hi _ 1 IFS - 0.000630 hi -- 0:668 7 ' i-