Document 1Y6Gm9oLQpNpeRzN1RnvVg75

62 CHAPTER 3 v 1955 Guide Combining equations 37 and 38 and solving for the ratio (A, - hJ/W, Wi), ~ = -- + A, W, - Wi ff. (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) u= 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 Btate may be determined from Equation 8 which may be rewritten as AaI + i + hv*(W* - nWn) = A* I Fig. 16. Solution op Example 8 on A.S.H.V.E. Psychrometric Chart The values of these properties are: A* = 35.39; W* = 0.01668; Aw* = 39.61; A,,* =* 90.70; W%t =* 0.08149; Aal = 28.84. Making the proper substitutions and solving for degree of saturation, p = 0.0681. The final humidity ratio is therefore 0.0681(0.08149) = 0.005549; the final enthalpy i> 28.84 + 0.0681(90.70) = 35.02 Btu per lb dry air. The rate of water addition is obtained from Equation 38. 20000 Gy, = (0.005549 - 0.00172) = 6.32 lb per min. The heat supplied is obtained from Equation 37. Q = Gifai -- Ai) Gy/hy, = ^000 12.12 m _ 6 65) _ 6 32(28.08) = 46,667 Btu per min. Solution b: From the A.S.H.V.E. Chart. Locate the initial and final'states on the chart and connect them with a straight line. Through the reference point on J 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 (At -- hi)/(Wi -- Wt) as 7500 from | .Thermodynamics 63 the protractor on the chart (Fig. 16). From Equation 39 At-A. Wi - W, + A. 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 = G,(7500 - A.) = 6.26(7500 - 23) = 46,900 Btu per min. Table 6. Pressure and Temperature for Altitudes in U. S. Standard Atmosphere Altitude Feet Z - 1,000 - 500 0 -1- 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. of 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 +62.6 +60.8 +59.0 +57.2 +55.4 +41.2 +23.4 + 5.5 -12.3 -30.1 -47.9 -65.8 -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 the limit of the isothermal atmosphere at 35,332 ft. Thus, T = To - 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