Document KR13Q5jrXBJw3v5oD07X441rX

54 CHAPTER 3 _____________ . 1948 Guide Example 4. Find the dew-point temperature of moist air at 80 F, 50 percent satu ration, atmospheric pressure. Solution. On the Mollier Diagram follow a horizontal line from a given state point (80 F, 50 per cent) to the saturation curve arid read the temperature at the intersection. To solve from Table 1: From the data in Table 1, the humidity ratio of the air.is W. - 0.50 X 0.02233 = 0.01117 Ibw/lba. By interpolation this is'found to be the humidity ratio at saturation at 60.22 F which' is therefore the required answer:--'- Example 5. Find' the degree of saturation of moist air at 90 F dry-bulb, 40 F (dew point), atmospheric pressure. SolutionOn the Mollier Diagram follow a horizontal line from 40 F on the saturation curve to the 90 F isotherm (dry-bulb) and read the degree of saturation directly. To solve from Table 1: From the data in Table 1, the humidity ratio of the air riiust be W = 0.005213. But the humidity ratio at saturation at 90 F is 0.03118; hence the degree of saturation is H = 0.005213/0.03118 = 16.72 per cent TYPICAL AIR CONDITIONING PROCESSES The use of Table 1 and the Mollier Diagram in analyzing typical air. conditioning processes is best explained by means of illustrative ex amples. In each of the following, it is to be understood that the process in question takes place at a constant pressure of 29.921 in. Hg,' or standard atmospheric pressure. Heating The process of adding heat to. moist air is represented by a horizontal line on the Mollier Diagram. The length of the line between the initial and final state points is the increase of reduced enthalpy; but, since the humidity ratio is constant, it is also the increase of enthalpy itself and therefore the quantity of heat added per pound of dry air. Example 6. Air initially at 20 F, 80 per cent saturation is heated to 120 F. Find the quantity, of heat required to process 20,000 cfm of heated air. Solution. From the data in Table 1: the initial humidity ratio is 0.80 X 0.002152 = 0.001722 Ibw/lba; the initial enthalpy is 4.804 + 0.80 X 2.302 = 6.646 Btu/lba; the final degree of saturation is 0.001722/0.08149 = 2.113 per cent; the final enthalDV is 28.841 + 0.02113 X 90.70 = 30.757 Btu/lba. , - It may be supposed that the air is heated between two sections of a duct. The quantities of energy converted across the two sections per pound of dry air crossing them are the two enthalpies calculated. Conservation of energy requires that the differ ence between these two enthalpies be the quantity of heat added; thus, ags = 30.757 - 6.646 = 24.ill Btu/lba. The final volume is 14.611 + 0.02113 X 1.905 = 14.651 cu ft/lba. Since 20,000 cfm of heated air is to be processed, the total quantity of heat required is aQb = 24.111 X 20,000/14.651 = 32,914 Btu per minute. On the Mollier Diagram the process is represented by the horizontal line AB, Fig. 2; whose length is the quantity of heat added per pound of dry air. The reduced enthalpy at A is 4.92 while that at B is 29.03, both being read directly from the chart. Since humidity ratio is constant the difference between these reduced enthalpies is also the difference between the enthalpies themselves, namely, 24.11 Btu/lba. Cooling The process of cooling moist air is also represented by a horizontal line on the Mollier Diagram. The line may extend across the saturation curve into the two-phase region, nevertheless, the length of the line between Thermodynamics 55 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 in Table 1: the initial humidity ratio is 0.50 X 0.03673 = 0.01837 ibw/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 Ibw/lba; the quantity of liquid formed is 0.01837 - 0.01582 = 0.00255 lbw/Iba; the enthalpy of the final two-phase mixture is 34.09 -b 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 converted 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