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

Apply this equation to the heavy component o obtain

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Unformatted text preview: ve volatility as (1.167 + 1.159)/2 = 1.163. From Eq. (7-26), the slope of the q-line = q/(q - 1) = 0.9/(0.9 - 1) = -9. Feed is 10 mol% vaporized. Apply the Fenske-Underwood-Gilliland method to obtain an initial estimate of reflux and stage requirements. Can use the SHOR model in CHEMCAD. The results are: Nmin = 48, Rmin = 9.37 R = 1.15Rmin = 10.77 N = 103 (102 + reboiler) N of feed = stage 50 from the top To construct the equilibrium curve for the McCabe-Thiele method, use Eq. (7-3), yP = α P-O xP 1163xP . = 1+ xP α P-O − 1 1 + 0163xP . From Eq. (7-9), the slope of the rectifying section operating line is, L/V = R/(R+1) = 10.77/(10.77 + 1) = 0.9150 (1) Exercise 7.39 (continued) Analysis: (continued) This line passes through x = 0.98 on the 45o line. The equation for this line is given by Eq. (7-9), L 1 1 (2) x+ x D = 0.915x + (0.98) = 0.915x + 0.08326 V R +1 11.77 + 1 Below the feed stage, with 10 mol% vaporization of the feed, L / V = 1.0535 . From a rearrangement of Eq. (7-12), the boilup ratio, VB, is 18.6916. The equation of the stripping section operating line is given by Eq. 7-14), L 1 1 y= x− x D = 10535x − . (0.04) = 10535x − 0.00214 . (3) V VB 18.6916 y= The equation for the q-line is given by Eq. (7-26), y= q 1 0.9 1 x− zF = x− (0.62) = −9 x + 6.2 q −1 q −1 0.9 − 1 0.9 − 1 (4) Based on Eqs. (1) to (4), the McCabe-Thiele diagram in terms of P, the more volatile component, is drawn below for three regions: (2) x = 0.2 to 0.4, (3) x = 0.4 to 0.6, and (4) x = 0.6 to 0.8, in order to gain accuracy. In these three regions, 28 stages are stepped off in the rectifying section up to x = 0.8, and 34.3 stages are stepped off in the stripping section down to x = 0.2. Let region (1) extend from x = 0.8 to 0.98 (i.e. xD). Apply the Kremser equation, Eq. (7-39) to this region. Apply this equation to the heavy component, O. Obtain the K-value for O from the top α of 1.167, taking the K-value for P = 1.00. Therefore, KO = 1/1.167 = 0.857. Therefore, the absorption factor, A = L/KV = 0.915/0.857 = 1.067. Other quantities needed in Eq. (7-39) are: xo = 1 - (xD)P = 1 - 0.98 = 0.02 y1 = xo = 0.02 From Eq. (2), for xN = 0.8, yN+1 for P = 0.8153. Therefore, for O, yN+1 = 1 - 0.8153 = 0.1847. log NR = 1 1 + 1− A A logA yN +1 − xo K y1 − xo K log = 1 1 + 1− 1.067 1067 . 0.1847 − 0.02(0.857) 0.02 − 0.02(0.857) log1.067 = 23.6 Let region (5) extend from x = 0.04 (i.e. xB) to 0.20. Apply the Kremser equation, Eq. (7-40) to this region. Apply this equation to the light component, P. Obtain the K-value for P from the bottom α of 1.159, taking the K-value for O = 1.00. Therefore, KP = 1.159. Therefore, the absorption factor in the stripping section is A = L / KV = 1.0535 / 1.159 = 0.9085 . Other values needed in Eq. (7-40) are x1 = xB = 0.04 and xN+1 = 0.20. Therefore, x −x /K 0.20 − 0.04 / 1159 . log A + 1 − A N +1 1 log 0.9085 + 1 − 0.9085 x1 − x1 / K 0.04 − 0.04 / 1159 . NS = = = 135 . 1 1 log log A 0.0985 Analysis: (continu...
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