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

# Separation Process Principles 2n Seader& Henley Solutions Manual

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: VM V ρ L 2 From Fig. 6.36, Y = 0.18. Therefore, uo = 1/ 2 (514 + 18.5)(18) 0.0734 = (312)(28.3) 62.4 1/ 2 Yg ρH 2O (0.18)(32.2) 62.4 = = 88 FP ρV 56 0.0734 Therefore, flooding velocity = uo = 9.4 ft/s For 50% of flooding, f = 0.5 4GM G From Eq. (6-103), DT = fuo πρG 1/ 2 = 0.0373 4(312 / 3, 600)(28.3) = (0.5)(9.4)(3.14)(0.0734) 1/ 2 = 3.0 ft. Exercise 6.38 (continued) Analysis: (continued) Assume gas phase controls mass transfer rate. Then, from Eq. (6-89), lT = HOGNOG From Eq. (6-93), for xin = 0, yin = 0.06, and yout = 0.00065, ln 92.3 N OG = A −1 1 + A A A −1 A (1) For the absorption factor, A = L/KV. Take L = 514 lbmol/h. Take V = 312 lbmol/h. From Fig. 6.49, K = 0.82. Therefore, A = (514)/(0.82)((312) = 2.0 From Eq. (1), 2 −1 1 ln 92.3 + 2 2 = 7.7 N OG = 2 −1 2 Mass-transfer data for 2-inch metal Pall rings for the absorption of NH3 from air into water are presented in Figs. 6.42 and 6.43. In the solutions to Exercise 6.34, these data are fitted to the equation, (3) kGa = 43.4 F0.7 uL0.45 , s-1 where the gas capacity factor = F = uV(ρV)0.5 = f uo (ρV)0.5 in kg1/2-s-1-m-1/2 and from the continuity equation, uL = m/SρL in m/s or m3/m2-s. Here, S = 3.14(3)2/4 = 7.07 ft2 For our case, F = (0.5)(9.4)(0.0734)0.5 = 1.27 lb1/2-s-1-ft-1/2 or 3.10 kg1/2-s-1-m-1/2 and uL = [(532.5)(18)/3,600)]/(7.07)(62.4) =0.006 ft/s = 0.00182 m3/m2-s Substitution of these values into Eq. (3) gives, kGa = 43.4(3.10)0.7(0.00182)0.45 = 5.6 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, HOG = HG = VMV/kGaSρV = (312/3,600)(28.3)/(5.6)(7.07)(0.0734) = 0.84 ft Now, we must correct this value to 1-inch metal Pall rings. From Table 6.8, using interpolation when necessary, we have the following characteristics: Packing 1.0-inch Pall rings 2.0-inch Pall rings a, m2/m3 223.5 112.6 ε 0.954 0.951 Ch 0.719 0.784 CV 0.336 0.410 Analysis: (continued) Exercise 6.38 (continued) From Eqs. (6-136) to (6-140), aPh/a is proportional to ε1/2/a . Therefore, aPh / a for 1.0-inch Rings 0.954 = aPh / a for 2.0-inch Rings 0.951 1/ 2 112.6 = 0.505 223.5 From Eq. (6-133), if we ignore holdup in the term (ε - hL), then, H G is proportional to Therefore, 1 a 1.25 CV a aPh 1 0.954 H G for 1.0-inch Rings 0.336 223.51.25 = 1 0.951 H G for 2.0-inch Rings 0.410 112.61.25 This ratio should be about the same for HOG . Therefore, for the 1.5-inch rings, HOG = 0.84(1.03) = 0.87 ft. From Eq. (2), column height = 7.7 (0.87) = 6.7 ft. 1 = 1.03 0.505 Exercise 6.39 Subject: Absorption of SO2 from air by water in a packed column. Given: Mass velocity of entering gas = 6.90 lbmol/h-ft2 on SO2-free basis, containing 80 mol% air and 20 mol% SO2. Water enters at a mass velocity of 364 lbmol/h-ft2. Gas exits with 0.5 mol% SO2. Tower operates at 30oC and 2 atm. Equilibrium y-x equation given for SO2. Tower is packed with Montz B1-200 metal structured packing. Assumptions: No stripping of water. No absorption of air. Find: (a) Molar material balance operating line. (b) NOG Analysis: (a) A material balance for the solute, SO2, from the top of the column to an intermediate point gives: Yn +1V ′ + X in L′ = YoutV ′ + X n L′ (1) Solving Eq. (1), the operating line is given by, Y= L′ L′ X + Yout − X in V′ V′ (2) From the given data, L'/V' = 364/6.90 = 52.75. Xin = 0.0 and Yout = 0.005/0.995 = 0.005025 Rearrange Eq. (2), letting Y = y/(1-y) and X = x/(1-x), which is approximately x be...
View Full Document

## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online