Document pvvVVO6n3583aeYr10brGY8d

62 CHAPTER 3 1956 Guide Combining equations 37 and 38 and solving for the ratio (A, -- h\)/(Wt -- \Pi), ' hi -- hi Wt-Wi (39) Example 8: Moist air at 20 F dry-bulb temperature and 0.80 degree of saturation is heated and humidified until it is at 120 F dry-bulb temperature and 71.5 F thermo dynamic wet-bulb temperature. Water at 65 F is supplied. If the air flow rate is 20,000 cfm at the initial conditions, how much heat is required? Solution a: From the data of Table 2. The initial humidity ratio is 0.80(0.002152) = 0.00172; the initial enthalpy is 4.804 + 0.80(2.302) = 6.6456; the initial specific volume is 12.084 + 0.80(0.042) = 12.118. The degree of saturation at the final state may be determined from Equation 8 which may be rewritten as h., + phut + h.'(W' - uW.i) = A* Fio. 16. Solution of Example 8 on A.S.H.A.E. Pstcurometric Chabt The values of these properties are: A* -- 35.39; W* = 0.01668; A.* = 39.61; AmS = 90.70; = 0.08149; A., = 28.84. , Making the proper substitutions and solving for degree of saturation, u = 0.0681. The final humidity ratio is therefore 0.0681(0.08149) = 0.005549; the final enthalpy is 28.84 + 0.0681(90.70) - 35.02 Btu per lb dry air. The rate of water addition iB obtained from Equation 38. G. = (O.OOSS49 - 0.00172) 6.32 lb perlmin... The heat supplied is obtained from Equation 37. Q = <7i(A* -- hi) -- Gwhw = (35.02 - 6.65) - 6.32(28.08) *= 46,667 Btu per min. Solution b: From the A.S.H.A.E. ,Chart. . Locate the initial and final states on the chart and connect them with a. straight line. Through the reference point on the chart, draw a line parallel to the line connecting the initial and final state points, the condition line, ana read the value of the ratio (5i -- h\)/:(W* -- Wt) as 7500 from Thermodynamics 65 the protractor on the chart (Fig. 16). From Equation 39 - Ai - hi Wt-Wi 7500 The rate of water supply was determined in Solution a, but will be found from the; chart. It is . ' Gw = (0.0055 - 0.0017) = 6-28 lb per min Q - <7.(7500 - h.) = 6.28(7500 -- 23) => 46,900 Btu per min. Table 6. Pressure and Temperature pob Altitudes in U. S. Standard Atmosphere Altitude Feet Z - 1,000 - 500 0 + 500 + 1,000 + 5,000 10,000, 15,000 20,000 25,000 30,000 35,000 40,000 45,000 50,000 Pressure Ik. op Hg P 31.02 30.47 29.921 29.38 28,86 24.89 , 20.58 16.88 13.75 11.10 8.88 7.04 5.54 4.36 3.436 - Temp F t v +62.6 +60.8 +59.0 .. , +57.2 +55.4 +41.2 +23.4 + 5.5 -12.3 -30.1 --47.9 -85.8 -67.0 -67.0 -67.0 U. S. STANDARD ATMOSPHERE The definition of the U. S= Standard Atmosphere is important to .the air conditioning engineer as an essential standard of reference. The basic assumptions in defining, the Standard Atmosphere are: 1. There is a linear decrease in temperature T with altitude up to theJimit of the isothermal atmosphere at 35,332 ft. Thus, T = To - 0.003566 Z (40) 2. The air is dry. 3. Air is a perfect gas obeying the laws of Charles and Boyle: " /. PV - RT 4. Gravity is constant at all altitudes with the standard value.' 5. The temperature of the isothermal atmosphere is --66 F. Standard values at sea level, which are part of the definition of the Standard Atmosphere, are: Pressure Temperature 29.921 in. Hg 59 F