Document 50wOpdLwqkqQnKR3Jrbwqz338

54 CHAPTER 3 1946 Guide at this same temperature.. 'This equilibrium temperature is called wetbulb temperature. It is clear that the readings of an actual wet-bulb thermometer cannot be regarded as values of a thermodynamic property of moist air; for these readings are importantly affected by a number of non-thermodynamic factors including design, construction, installation, and technique of using the instrument. Thus, unless the wet-bulb is effectively shielded against radiation from relatively warm surfaces the process will not be strictly adiabatic as tacitly assumed in writing Equation 7. - Also, partial . drying of the wick will prevent the air immediately adjacent to it from reaching complete saturation as assumed in Equation 7. A working theory developed by Arnold 6.enables the calculation of corrections to be applied to the readings of the actual instrument in order to make them agree with the values of temperature calculated from Equation 7. For tunately, and indeed fortuitously, these corrections can be made small, but to emphasize the necessity of making them in accurate experimen. tation, the temperature defined by Equation 7 is called thermodynamic wet-bulb temperature. Example 2. Find the degree of saturation of moist air at 90 F dry-bulb, 63 F thermo dynamic wet-bulb, atmospheric pressure. Solution. The answer is read directly from the Mollier Diagram. Inserting numerical data from Table 1 into Equation 7 gives (21.625 + 34.31 p) + (0.01235 - 0.03118p) X .31.12 = 28.57 The solution of this equation is direct and the final answer is p = 19.67 per cent , Example 3. Find the temperature to which moist air initially saturated at 40 F and at standard atmospheric pressure must be heated in order to have a thermodynamic wet-bulb temperature of 60 F. . Solution. On the Mollier Diagram follow a horizontal line from the saturation curve at 40 F to its intersection with the 60 F thermodynamic wet-bulb line and read the corre: sponding temperature directly. inserting numerical data from Table 1 into Equation 7, this becomes : fta + O.0b5213*as/1F8 = 26.46 - (0.01108 - 0.005213) . X .28.12 = 26.295 . At 85 F the letthand member of this equation has the value 26.147; at 86, F itsvajue is 26.389; by linear interpolation the answer is . . .. 1 . t = 85.61 F ' Dew-Point Temperature. Corresponding to any given state of. moist air there exists another state on the saturation curve having the same humidity ratio W and same pressure p as the given state. , The tempera ture at. this other state on the saturation curve is called the dew-point ;temperature of the given state. Obviously, if moist air is cooled at con stant pressure and constant humidity ratio it will reach saturation when its temperature.falls to.a value equal to its dew-point temperature. . This 'will usually be marked by the first appearance of a coexisting condensed phase. In one type of dew-point apparatus a continuous sample .of air is passed over a mirror which can be cooled by external refrigeration and ;whose temperature can be.accurately measured. The measured ,tem perature at which the intensity of light reflected from the mirror is ab ruptly diminished by condensation is taken to be the dew-point tempera ture' of the air sample/ Examples 4 and.5 illustrate the relation between' ' the dew-point, degree of saturation and dry-bulb temperature. Thermodynamics 55 Example h- . Find the dew-point temperature of moist air at 80 F, 50 per cent 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 and read the temperature at the intersection. From the data in Table 1, the humidity ratio of the air is W = 0.50 X 0.02233 -- 0.01117 lbw/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. Solution. On 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. From the data in Table 1, the humidity ratio of the air must be W = 0.005213. But the humidity ratio at saturation at 90 F is 0.03118; hence the degree of saturation is p = 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, of standard atmospheric pressure. - Heating The process of adding heat to moist air is represented by a horizontalline on the Mollier Diagram. The length of the line between the initial and finalstate 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 ce'nt saturation is heated to 120 F. Find the quantity of heat required to process 20,000 elm of heated air. Solution. From the data in Table 1: the initial humidity ratio is 0.80 X 0.002152 = 0.001722 lbw/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 enthalpy is 28.841 + 0.02113 X 90.70 = 30.757 Btu/lba. ' - ^ may be supposed that the air is heated between two sections of a duct. The quantities of energy convected 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; thiis,- aQb = 30.757 - 6.646 = 24,111 Btu/lba. The final volume is 14.611 + 0.02113 X 1.905 = 14.651 cu ft/lba. Since 20,000 cfmof heated air is to be processed,+he 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 Theproccss 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