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

So2 free inlet air rate 00621 0016 0061 kmols so2

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Unformatted text preview: Eq. (6-52), EOV = 1 − exp(− N OG ) = 1 − exp(−0.066) = 0.064 or 6.4% (e) From Example 6.2, the separation requires 3 equilibrium stages. If we assume the liquid on a tray is well mixed. Then, from Eq. (6-31), EMV = EOV = 0.064. From Eq. (6-21), Na =Nt /Eo = 3/0.064 = 47 trays. If we assume plug flow of liquid on a tray, then, if we take, below Eq. (6.33), λ=KV/L=9.89, from Eq. (6-37), log 1 + E MV ( λ − 1) log 1 + 0.064(9.89 − 1) = = 0197 . log λ log 9.89 Then we need from Eq. (6-21), Na =Nt /Eo = 3/0.197 = 15 trays. Sufficient information is not given to establish the partial mixing prediction. Therefore, the number of trays required ranges widely from 15 to 47. Eo = Exercise 6.24 Subject: Absorption of SO2 from air into water in an existing packed column. Given: Feed gas flow rate of 0.062 kmol/s containing 1.6 mol% SO2. Absorbent is 2.2 kmol/s of pure water. Packed column is 1.5 m2 in cross sectional area and packed with No. 2 plastic super Intalox saddles to a 3.5-m height. Exit gas contains an SO2 mole fraction of 0.004. Operating pressure is 1 atm. At operating temperature, equilibrium curve for SO2 is y = Kx = 40x Assumptions: No stripping of water. No absorption of air. Find: (a) (b) (c) (d) L/Lmin NOG and Nt HOG and HETP KGa Analysis: Compute material balance. SO2-free inlet air rate = 0.062(1-0.016) = 0.061 kmol/s SO2 inlet rate in feed gas = 0.062(0.016) = 0.001 kmol/s = V' SO2 outlet rate in gas = 0.061(0.004/0.996) = 0.00025 kmol/s SO2 rate in outlet water = 0.001 - 0.00025 = 0.00075 kmol/s Fraction absorbed = 0.00075/0.001 = 0.75 or 75% (a) From Eq. (6-11), L'min = V'K(fraction absorbed) = 0.061(40)(0.75) = 1.83 kmol/s Therefore, L/L'min = 2.2/1.83 = 1.20 0.887 N +1 − 0.887 (b) Take A = L/KV = 2.2/[(40)(0.062)] = 0.887, From Eq. (6-13), 0.75 = 0.887 N +1 − 1 Solving, Nt = 4. For NOG, use Eq. (6-89) with yin = 0.016, yout = 0.004, xin =0.0, ln N OG = 0.887 − 1 0.016 1 + 0.887 0.004 0.887 = 3.78 (0.887 − 1) / 0.887 (c) Given height of packing = 3.5 m = lT From Eq. (6-73), HETP = lT /Nt = 3.5/4 = 0.875 m From Eq. (6-89), HOG = lT /NOG = 3.5/3.78 = 0.926 m V K G aPS 3 = 0.045 kmol/s-m -atm (d) From Table 6.7, HOG = Therefore, K G a = V H OG PS = 0.062 (0.926)(1)(1.5) Exercise 6.25 Subject: Operating data for absorption of SO2 from air into water in a packed column. Given: Column operates at 1 atm (760 torr) and 20oC. Solute-free water enters at 1,000 lb/h. Mole ratio of water to air is 25. Liquid leaves with 0.6 lb SO2/100 lb of solute-free water. Partial pressure of SO2 in exit gas is 23 torr (0.0303 atm). Equilibrium data are given as partial pressures of SO2 in air as a function of lb SO2 dissolved/ 100 lb H2O. Assumptions: No stripping of water. No absorption of air. Density of liquid taken as water. Find: (a) % of SO2 absorbed. (b) Concentration of SO2 in the liquid at the gas-liquid interface in lbmol/ft3 at a point where the bulk liquid concentration is 0.001 lbmol SO2/lbmol of water and: kL = 1.3 ft/h kp = 0.195 lbmol/h-ft2-atm Analysis: (a) Inlet water rate = 1,000/18.02 = 55.5 lbmol/h Inlet air rate = water rate/25 = 55.5/25 = 2.22 lbmol/h Partial pressure of air in exit gas = 760 - 23 = 737 torr By partial pressure ratio, SO...
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