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62 CHAPTER 3 1957 Guide Combining equations 37 and 38 and solving for the ratio (A, - Ai)/(Wi Wx), th - A, Q h. W, -- Wi G,, (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 55 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 A., + + Aw*(W* - uW.,) = h* & m M i- Thermodynamics 63 the protractor on the chart (Fig. 16). From Equation 39 A -- Ai W,-Wx -- + A. = 7500 Therate of water supply was determined in Solution a, but will be found from the Cu8xt< Xv is ,, ,20,000 <?w = (0.0055 - 0.0017) = 6.28 lb per min Q = Gw(7500 - A.) = 6J28(7500 - 23) =* 46,900 Btu per min. Table 6. Pressure and Temperature for Altitudes in U. S. Standard Atmosphere Fig. 16. Solution op Example 8 on A.S.H'.A.E. Psychrometbic Chart The values of these properties are: A* = 35.39; W* = 0.01668; A,* = 39.61; h,,i = 90.70; W,, = 0.08149; A,, = 28.84. Making the proper substitutions and solving for degree of saturation, ix = 0.0881. 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 is obtained from Equation 38. (Vj (7. = -- (0.005549 - 0.00172) 12.12 = 6.32 lb per min. The heat supplied is obtained from Equation 37. Q = Gi(Aj -- Ai) -- G,,A = (35.02 _ 6.65) - 6.32(28.08) in irt ' 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, and read the value of the ratio (Ai -- ht')/(W, -- Wt) as 7500 from 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 In. 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 I +62.6 +60.8 +59.0 +57.2 +55.4 +41.2 +23.4 + 5.5 -12.3 -30.1 -47.9 -6518 -67.0 -67.0 -67.0 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 the limit of the isothermal atmosphere at 35,332 ft. Thus, T = T0- 0.003566 Z 2. The air is dry. 3. Air is a perfect gas obeying the laws of Charles and Boyle: (40) 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