Document ZB7wOdR5Ga8MVLgnMwVj48NZL

American Society of Heating and Ventilating Engineers Guide, 1936 dt, the weight of saturated vapor mixed with 1 lb of dry air, Wt, (at a relative humidity of 100 per cent and a barometric pressure, B, of 29.92 in. of mercury), the specific volume of dry air, and the volume of an air-vapor mixture containing 1 lb of dry air (at a relative humidity of 100 per cent and a pressure of 29.92 in. of mercury). The preceding equations or the data from Table 5 may be conveniently used in solving the following typical problems: Example 8. Humidifying and Healing. Air is to be maintained at 70 F with a relative humidity of 40 per cent (4> = 0.4) when the outside air is at 0 F and 70 per cent relative humidity (<1> = 0.7) and a barometric pressure, B, of 29.92 in. of mercury. Find the weight of water vapor added to each pound of dry air and the dew-point temperature of the humidified air. Solution. From Equation 5a and Table 5, Wi = 0.622 ( gg 92^_? 00264 ) = 0-000548 lb per pound of dry air. Wt = 0.622 ( 29^)2^.-* 0^295 ) = 0.00618 lb per pound of dry air. The water vapor added per pound of dry air must be (W, -- W,) or 0.005632 lb. By inspection of Table 5, Wt = 0.00618 at* 44.5 F, so this is the dew-point temperature of the humidified air. .' i An approximation of the same result from Table 5 is W, = 0.7 X 0.0007852 = 0.00054964 lb per pound of dry air. W -- 0.4 X 0.01574 = 0.006296 lb per pound of dry air. The water vapor added per pound of dry air is approximately 0.00574636 lb and the dew-point temperature is approximately 45 F. The degree of approximation is evident. Example S. Dehumidifying and Cooling. Air with a dry-bulb temperature of 84 F, a wet-bulb of 70 F, or a relative humidity of 50 per cent (4> = 0.5), and a barometric pressure, B, of 29.92 in. of mercury is to be cooled to 54 F. Find the dew-point tem perature of the entering air and the weight of vapor condensed per pound of dry air. . Solution. From Equation 5a and Table 5, Wi = o.622 (wW-iilsk) = 001248 lb Per pund of dry air- Wt = 0.622 ( 29 92 ^-^42003 ) = '00887 lb Per Pund of drY air. Since W, = Wt when t = 63.4 F, this is the dew-point temperature of the entering air. The weight of vapor condensed is (Wt -- W3) or 0.00361 lb per pound of dry air: An approximate result is Wi = 0.5 X 0.02543 = 0.012715 lb per pound of dry air. Wt = 1 X 0.008856 = 0.008856 lb per pound of dry air, since the exit air is saturated:. Since Wi = Wt at / = 64 F, this is the dew-point temperature of the entering air.; The weight of vapor condensed is 0.003859 lb per pound of dry air. The degree of appfoxi-: mation is again evident. ADIABATIC SATURATION OF AIR The process :of adiabatic saturation of air is df considerable importance , in air conditioning. Suppose that 1 lb of dry air, initially unsaturated but' carrying W lb of Water vapor with a dry-bulb temperature, t, and a wet-; 20 -X Chapter 1 -Fundamentals of Heating and Air Conditioning bulb temperature, l1,'be made to pass through a tunnel containing an exposed water surface. Further assume the tunnel to tie completely in^ sulated, thermally, so that the only heat transfer possible is that between the air and water. As the air passes over the water surface, it will gradu ally pick up water vapor and will approach saturation at the initial wetbulb temperature of the air, if the water be supplied at this wet-bulb tem perature. During the process of adiabatic saturation, then, the dry-bulb temperature of the air drops to the wet-bulb temperature as a limit, the wet-bulb temperature remains substantially constant, and the weight of water vapor associated with each pound of dry air increases to Wt., as a limit, where Wt< is the weight of saturated vapor per pound of dry air for saturation at the wet-bulb temperature. Example 4. If air with a dry-bulb of 85 F and a wet-bulb of 70 F be saturated adia- batically by spraying with recirculated water, what will be the final temperature and the vapor content of the air? Solution. The final temperature will be equal to the initial wet-bulb temperature or 70 F, and since the air is saturated at this temperature, from Table 5, W = 0.01574 lb per pound of dry air. In the adiabatic saturation process, since the heat given up by the dry air and associated vapor in cooling to the wet-bulb temperature is utilized in evaporation, of water at the wet-bulb temperature, W. H. Carrier has pointed out* that the equation for the process of adiabatic saturation, and hence for a process of constant wet-bulb temperature, is: Vis (Wt. -W)=Cpa(t- l1) + cPsW (t - V) (9a) and using cPa = 0.24 and cPa = 0.45 - where *'fg (Wt. - W) = (0.24 + 0.45BO (t - <') ... (9b). h'ig = latent heat of vaporization at t', Btu per pound. (Wv -- W) = increase in vapor associated with 1 lb of dry air when it is saturated. adiabatically from an initial dry-bulb temperature, /, and an initial vapor content, W, pounds. Knowing any two of the three primary variables, t, f, or W, the third may be found from this equation for any process of adiabatic saturation. TOTAL HEAT AND HEAT CONTENT The total heat of a mixture of dry air and water vapor was originally defined by W. H. Carrier as . 2 = Cpa (t - ) + W [A'fg + Cps (t - f')] (10) where . ........................ S = total heat of the mixture, Btii per pound of dry air. ~ Cpa = mean specific heat at constant pressure of dry air. .. Cpg = mean specific heat at constant pressure of water vapor. / = dry-bulb temperature, degrees-Fahrenheit. v: . t' = wet-bulb temperature, degrees Fahrenheit. ................... Transactions, Vol. 33, 1911, p. 1005. 21: