Document XRN5mn1yVgMYy4xMo2KQ75mDR

62 CHAPTER 3 1952 Guide Thermodynamics 63 Combining equations 37 and 38 and solving for the ratio' (A, -- hi)/{W? -- +i), . the protractor on the chart:(Fig. 16). From Equation 39 I s . ., ;' hr --_ Ai Wr-Wr (39) . A, - A, Q - Wt-Wi ~Gw+*w-7S0 ;V;,! 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 4.802 + 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,j + uh,,r + h,*(W* - w.0 = h*. I I I The^rate^of water supply was determined in Solution a, but. will be found fiqm the 20,000 Gw = 12.1 (0.0055 --; 0.0017) `. '< =.- 6.28 lb per min Q = Gw(7500 - Aw) .=. = 6.28(7500 -- 23) .. = 46.900 Btu per min. Table 6. Pressure and Temperature for Altitudes in U. S. : . Standard Atmosphere The values of these properties are: A* = 35.39; W* = 0.01668; A** = 39.61; A,,i = 90.70; Wa = 0.08149; A., = 28.84. Making the proper substitutions and solving for degree of saturation, M = 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 is obtained from Equation 38. G, = (0.005549 - 0.00172) ll.YZ -- - = 6.82 lb per min. The heat supplied is obtained from Equation 37. Q = Gi(As -- A,) -- G.A, (vi rwt = (35.02 - 6.65) - 6.32(28.08) m = 46,667 Btu per min. . .Solution b: From the A5.H.V.E. Chart. Locate the initial and final states on she 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 (At -- hi)/(W, -- H',) as 7500 from 1 1 i 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. 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 T 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: ., *uere is a linear decrease m temperature 3 tae 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 lawB 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