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Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# From page 14 59 of perrys chemical engineers handbook

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Unformatted text preview: iameter is, 4VM V DT = fuo πρV 1/ 2 4(70.1/ 3, 600)(29) = (0.5)(15.3)(3.14)(0.0752) 1/ 2 = 1.12 ft and S = 3.14(1.12)2/4 = 0.985 ft2 The value of NOG = 9.65, the same as in Part (b) For HOG , can use Figs. 6.42 and 6.43 for 50 mm Pall rings. Gas capacity factor = F = uG(ρV)0.5 = f uo(ρV)0.5 = (0.5)(15.3)(0.0752)0.5 = 2.1 lb1/2-s-1-ft-1/2 or 5.12 kg1/2-s-1-m-1/2 By the continuity equation, liquid load = QL/S = uL = m/SρL = (90.8/3600)(18)/(0.985)(62.4) =0.0074 ft/s or 0.00225 m/s or 2.25 x 10-3 m3/m2-s In Fig. 6.42, from the slope of the correlating line, kGa is proportional to F0.7 In Fig. 6.43, from the slope of the correlating line, kGa is proportional to uL0.45 Using the value of kGa = 3.1 s-1 at uL = 2.25 x 10-3 m3/m2-s and F = 1.16, then the two plots are represented by the equation, kGa = 43.4 F0.7 uL0.45 For our case, kGa = 43.4 (5.120.7)(0.002250.45) = 8.8 s-1 Note that this kGa is in concentration units for the driving force. Therefore, from Table 6.7 by analogy to the liquid phase case, HG = VMV/kGaSρV = (70.1/3,600)(29)/(8.8)(0.985)(0.0752) = 0.87 ft The column height is lT = 9.65(0.87) = 8.4 ft. Exercise 6.34 (continued) Analysis: (d) Now compare the two packings on the basis of flooding and mass transfer: Packing 38-mm ceramic Berl saddles 50-mm metal Pall rings Column diameter, ft 1.42 1.12 Column height, ft 18.3 8.4 Clearly the 50-mm metal Pall rings are superior in performance. However, the selection would have to be made on the basis of cost. Because of the much smaller column and volume of packing, it is likely that the 50-mm Pall rings would be favored. From page 14-59 of Perry's Chemical Engineers' Handbook, 7th edition (1997), the cost of 38mm ceramic Berl saddles is approximately \$21/ft3. The cost of 50-mm Pall rings depends on whether they can be made of carbon steel or must be made of stainless steel, with the former \$19.9/ft3 and the latter \$99.0/ft3. So the material of construction is a very important factor. Exercise 6.35 Subject: Absorption of Cl2 from air with water in a packed column. Given: 100 kmol/h of feed gas containing 20 mol% Cl2. Column operation at 20oC and 1 atm. Table of y-x data for Cl2. Exit gas to contain 1 mol% Cl2. Assumptions: No stripping of water. No absorption of air. Find: (a) Minimum water rate in kg/h. (b) NOG for a water rate of 2 times minimum. Analysis: Solve this problem using mole ratios. Feed gas is 80 kmol/h of air = G' , and 20 kmol/h of Cl2 with Yin = 20/80 = 0.25. Exit gas has Yout = 1/99 = 0.0101. Xin = 0.0. Highest value of Y in the table of equilibrium data is 0.06/0.94 = 0.0638. Therefore, data for higher values of Y are needed. Can obtain this from Perry's Handbook, page 3-102 in the 6th edition and page 2-126 in the 7th edition. The expanded table of equilibrium data with conversion to mole ratios includes the following values added from Perry's Handbook, using the conversions: y = pCl2, torr/760 torr x = [(grams Cl2/L)/71]/[1,000/18 + (grams Cl2/L/71)] pCl2, torr 100 150 200 gra...
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