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

# 33 946 gpm 5970 cm3s from eq 6 54 cl 0362 0317

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Unformatted text preview: ued) Exercise 7.41 (continued) (b) From Eq. (7-42), Eo = 13.3 - 66.7 log µ Take the viscosity as that of the feed = 0.34 cP Eo = 13.3 - 66.7 log (0.34) = 44.6% This is poor agreement with the performance data. (c) From Eq. (7-43), Eo = 50.3(αµ)-0.226 At the feed composition, x = 0.36 and y = 0.71. Therefore, from Eqs (2-19) and (2-21 combined, the relative volatility is, α = ( y/x)/[(1 - y)/(1 - x)] = (0.71/0.36)/(0.29/0.64) = 4.4 Exercise 7.41 (continued) Analysis: Eo = 50.3[(4.4)(0.34)]-0.226 = 45.9% Now correct for length of liquid path from Fig. 7.5. Column diameter = 6 ft. Assume length of liquid path = 70% of column diameter = 0.7(6) = 4.2 ft. From Fig. 7.5, correction to be added = 10%. Therefore corrected Eo = 45.9 + 10 = 55.9% This also appears to be low. (d) From Eq. (6-56), NOG = - ln (1 - EOV). Therefore, EOV = 1 - exp(-NOG) Use Eqs. (6-62, (6-64), (6-66), and (6-67) as in Example 6.7. Carry out the calculations at the bottom tray based on methanol diffusion. Conditions are: Molar flow rate, kmol/h Molecular weight Density, kg/m3 Gas 630 18.5 0.657 Liquid 1,192 18.0 940 Estimate liquid diffusivity of methanol in water at 212oF = 373 K. From Perry's Handbook, DL = 1.6 x 10-5 cm2/s at 25oC. The viscosity of water at 212oF = 0.25 cP. Using Eq. (3-39) to correct for temperature and viscosity, DL = 1.6 x 10-5 (373/298)(1/0.25) = 8 x 10-5 cm2/s Estimate gas diffusivity of methanol in water vapor from Eq. (3-36). T = 373 K MAB = 2/[(1/32) + (1/18)] = 23. Using Table 3.1, = 15.9 + 2.31(4) + 611 = 31.3, . = 131 . V V methanol water 0.00143(373)1.75 DV = = 0.30 cm2/s 1/ 2 1/ 3 1/ 3 2 (151 / 14.7)23 [31.3 + 131 ] . . Required tray dimensions: DT = 6 ft, A = 3.14(6)2/4 = 28.3 ft2 = 2.63 m2 Aa = 0.91 A = 0.91(2.63) = 2.39 m2 = 23,900 cm2, Lw = 42.5 in. = 1.08 m Tray conditions: φε = 1 - 0.617 = 0.383, qL = 47,300/(60)(8.33) = 94.6 gpm = 5,970 cm3/s From Eq. (6-54), Cl = 0.362 + 0.317 exp[-3.5(2)] = 0.362 From Eq. (6-51), f = 0.40, 94.6 hl = 0.383 2.0 + 0.362 (42.5)(0.383) From Eq. (6-64), 2/3 = 1.21 in. = 3.08 cm2 t L = (3.08)(23,900)/5,970 = 12.3 s Analysis: (d) (continued) The continuity equation is, mV = U a Aa ρV , so U a , ft/s = From Eq. (6-65), tG = From below Eq. (6-67), Exercise 7.41 (continued) mV 630(18.5)(2.205) / 3600 = = 6.8 ft/s = 2.07 m/s Aa ρV 28.3(0.91)(0.657 /16.02) (0.617)(3.08) = 0.024 s (0.383)(6.8)(2.54)(12) 0.5 F=Ua ρV = 2.07(0.657)0.5 = 1.68 (kg/m)0.5/s From Eq. (6-67), k L a = 78.8(8 × 10 −5 ) 0.5 (168 + 0.425) = 1.48 s-1 . From Eq. (6-66), k G a = 1,030(0.30)1/ 2 0.40 − 0.842(0.40) 2 3.081/ 2 = 85.3 s-1 From Eq. (6-63), N L = k L at L = 148(12.3) = 18.2 . From Eq. (6-62), N G = k G at G = 85.3(0.024) = 2.05 From the vapor-liquid equilibrium data at the bottom of the column, Kmethanol = 0.156/0.0246 = 6.34 Absorption factor = KV/L = (6.34)(630)/1,192 = 3.35 1 1 ( KV / L) 1 3.35 = + = + = 0.488 + 0.184 = 0.672 N OG N G NL 2.05 18.2 NOG = 1/0.672 = 1.488 From Eq. (6-61), Thus, the gas-phase resistance is more importa...
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