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

# 38 subject absorption of nh3 from air into water with

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Unformatted text preview: = 1.75, can apply Eq. (5-29), which was derived for extraction but applies to absorption using: Yout = Yin 1 N En where E = L′ 18.23 = = 1.356 K ′G ′ (175)(7.679) . n=0 Therefore, Yout 0.000912 = = 0.0295 = Yin 0.03093 1 N n =0 1356 . Solving, N = 7.5 equilibrium stages n Exercise 6.37 (continued) Analysis: (c) (continued) Could also solve this using Kremser's equation or by a graphical construction like Fig. 6.12. (d) Because operating line and equilibrium line are straight, use Eq. (6-95) with A = E = 1.356 N OG = N t ln(1/ A) ln(1/1.356) = 7.5 = 8 .7 (1 − A) / A (1 − 1.356) /1.356 (e) From Eq. (6-141), lT = HOGNOG From Table 6.7, H OG = V′ 7.679 0.64 = = , where S = cross sectional area of tower in ft2 KY aS (12.0) S S The allowable gas superficial velocity = uV = 2.4 ft/s. Take this at the bottom of the tower where gas rate is the highest. From the continuity equation, Q = uV S = gas volumetric rate Therefore, S = Q/uV = (50/60)/2.4 = 0.347 ft2. Therefore, HOG = 0.64/0.347 = 1.84 ft Packing height = (1.84)(8.7) = 16 ft. (f) To obtain packed height as a function of L'/V' with V' and HOG constant, the only change will be the value of NOG . Use Eq. (6-93), which for xin = 0 and yin/yout = 0.03/0.000911 = 32.9, is: ln 32.9 N OG = A −1 1 + A A A −1 A (2) where for a dilute system, we can take A = L'/K'V' = (L'/V')/K' = (L'/V')/1.75 The results obtained from Eq. (2) are as follows: L'/V' 1.75 2.0 2.5 3.0 5.0 10.0 A 1.00 1.143 1.429 1.714 2.857 5.714 1/ A 1.00 0.875 0.700 0.583 0.350 0.175 NO'G Height, ft 12.9 7.86 6.38 4.74 4.01 23.6 14.5 11.7 8.72 7.38 ∞ ∞ Exercise 6.38 Subject: Absorption of NH3 from air into water with a packed column. Given: Column operating at 68oF and 1 atm. Gas enters at 2,000 ft3/min at 68oF and 1 atm, containing 6 mol% NH3 in air. Entering water rate is twice the minimum. Absorb 99% of the NH3. Gas velocity at 50% of flooding velocity. Assumptions: No stripping of water. No absorption of air. Find: Tower diameter and packed height. Analysis: Entering gas rate is 2,000 60 + 460 60 = 312 lbmol /h 379 68 + 460 Compute material balance. NH3 in entering gas = 0.06(312) = 18.72 lbmol/h NH3 in exiting gas = 0.01(18.72) = 0.19 lbmol/h Air in entering and exiting gas = V' = 312 - 18.72 = 293.28 lbmol/h NH3 in exiting water = 18.72 - 0.19 = 18.53 lbmol/h Therefore, Xin = 0.0 Yin = 18.72/293.28 = 0.06383 Yout = 0.19/293.28 = 0.000648 At the minimum water rate, the exiting liquid is in equilibrium with the entering gas. From Fig. 6.49, X*out = 0.072 for Yin = 0.06383. Therefore, L'min = 18.53/0.072 = 257 lbmol/h Operating liquid rate = 2 times minimum = 2(257) = 514 lbmol/h Now, compute column diameter using Fig. 6.36. Corrections for liquid density and viscosity are neglected. From the ideal gas law on the entering gas, with an average molecular weight = 28.3, ρV =PM/RT = (1)(28.3)/(0.7302)(68+460) = 0.0734 lb/ft3 From Table 6.8, packing factor for 1-inch metal Pall rings = 56 ft2/ft3. LM L ρV The abscissa in Fig. 6.36 is, X =...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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