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

# 4 fp v 65 00752 therefore uo 95 fts the fraction of

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Unformatted text preview: ripping of water. Find: (a) Minimum water rate. (b) Column diameter and height for water rate of 2 times minimum and 50% of flooding gas velocity if 38-mm ceramic Berl saddles are used. (c) Same as Part (b) for 50-mm Pall rings. (d) Packing recommendation. Analysis: First compute the ammonia material balance. Flow rate of NH3-free air = 2,000/29 = 69 lbmol/h Flow rate of NH3 in the entering gas = 12(69)/(760-12) = 1.107 lbmol/h Flow rate of NH3 in the exiting liquid = 0.996(1.107) = 1.1026 lbmol/h Flow rate of NH3 in the leaving gas = 1.107 - 1.1026 = 0.0044 lbmol/h (a) Operation in dilute region, where we can apply Henry's law. From Fig. 6.49, using results from Exercise 6.33, when x=0.0108, y=0.007. Therefore, the equilibrium equation is y = 0.65x. Therefore, K = 0.65. Entering gas rate = V = 69 + 1.107 = 70.1 lbmol/h From Eq. (6-11), Lmin = VK(fraction absorbed) = 45.4 lbmol/h (b) L = 2Lmin = 2(45.4) = 90.8 lbmol/h Compute column diameter for 50% of flooding, using Fig. 6.36. Take properties of gas and liquid as those of air and water, respectively. From the ideal gas law, ρG = PM/RT = (1)(29)/(0.7302)(528) = 0.0752 lb/ft3 LM L ρV In Fig. 6.36(a), X = VM V ρ L 1/ 2 (45.4)(18) 0.0752 = (70.1)(29) 62.4 1/ 2 = 0.014 Take f{ρL} and f{µL} =1.0. From Fig. 6.36(a), at flooding, Y = 2 uo FP ρV = 0.22 g ρH 2O (1) For 38-mm Berl saddles, the packing factor, FP, is not included in Table 6.8. However, it can be found in Perry's Chemical Engineers' Handbook, which gives FP = 65 ft2/ft3. Yg ρH2 O (0.22)(32.2) 62.4 2 Solving Eq. (1), uo = = = 90.4 FP ρV 65 0.0752 Therefore, uo = 9.5 ft/s. The fraction of flooding = f = 0.5. From Eq. (6-103), the column diameter is, 4VM V DT = fuo πρV 1/ 2 4(70.1/ 3, 600)(29) = (0.5)(9.5)(3.14)(0.0752) 1/ 2 = 1.42 ft and S=πDT2/4 =3.14(1.42)2/4=1.58 ft2 Exercise 6.34 (continued) Analysis: (b) (continued) For column height, use Eq. (6-89), because mass-transfer is controlled by the gas phase. For NOG , use Eq. (6-88) with xin = 0. Absorption factor = A = L/KV = (90.8)/(0.65)(70.1) = 2 From Eq. (6-93), using yin = 1.107/70.1 = 0.0158, yout = 0.0044/69 = 0.0000638 N OG = ln ( A − 1) / A ( yin ) / ( yout ) + (1 / A) ( A − 1) / A = ln (2 − 1) / 2 0.0158 / 0.0000638 + (1 / 2) (2 − 1) / 2 = 9.65 For HOG, literature data can be used, as found on page 16-38 of Perry's Chemical Engineers' Handbook, 6th edition or on page 663 of &quot;Equilibrium-Stage Separation Operations in Chemical Engineering&quot; by Henley and Seader, as shown below. To use this plot, L&quot; = 90.9(18)/1.58 = 1,040 lb/h-ft2 and V”=G&quot; = 70.1(29)/1.58 = 1,290 lb/h-ft2 Then, HOG = 1.9 ft and column height = lT = NOGHOG = (9.65)(1.9) = 18.3 ft. Air-Ammonia-Water system with 1.5-inch Berl Saddles, AE units Exercise 6.34 (continued) Analysis: (c) (continued) (c) Assume the 50-mm Pall rings are metal. From Table 6.8, FP = 27 ft2/ft3. 2 Solving Eq. (1), uo = Yg ρH 2O (0.22)(32.2) 62.4 = = 235 FP ρV 27 0.0752 Therefore, uo = 15.3 ft/s. The fraction of flooding = f = 0.5. From Eq. (6-103), the column d...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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