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

# 36 09248 236 09248 188 09248 100 09248 084 09248

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Unformatted text preview: 5 20 3.600 nC 5 35 0.000 Total: 100 46.725 Bottoms 0.000 0.000 1.875 16.400 35.000 53.275 The results with K-values and relative volatilities are as follows, using the SRK equation of state for K-values and the bubble-point equation, (4-12): Component C3 iC 4 nC 4 iC 5 nC 5 K, distillate, 140oF 2.100 1.056 0.818 0.401 0.329 αi,iC5 , 140oF K, bottoms, 230oF αi,iC5 , 230oF 5.24 2.63 2.04 1.00 0.82 3.880 2.260 1.863 1.068 0.922 3.63 2.12 1.74 1.00 0.86 αi,iC5 , geometric mean 4.36 2.36 1.88 1.00 0.84 Exercise 9.17 (continued) Analysis: (continued) (a) Using the Fenske equation (9-12), with nC4 as the LK and iC5 as the HK, log N min = d nC4 biC5 d iC5 bnC4 log log ( α nC4,iC5 )avg = ( 23.125 ) (16.4 ) ( 3.6 ) (1.875 ) log1.88 = 6.38 (b) To compute the distribution of nonkey components at total reflux, use for the lighter than light key components, LLK, Eq. (9-15), and use Eq. (9-16) for the heavier than heavy key, HHK, with iC5 as the reference component, r, and the above value of Nmin. fi fi = (1) Thus, for the LLK, i, bi = d iC5 N min 1 + 3.6 α 6.38 i,iC5 1+ α i,iC5 16.4 b iC5 fi di = For the HHK, i, 1+ d iC 5 biC5 d iC5 biC5 α iN min ,iC 5 3.6 6.38 α i ,iC5 16.4 = 3.6 6.38 1+ α i ,iC5 16.4 fi α iN min ,iC5 (2) Using the above geometric mean values of αi, iC5 , followed by use of Eqs. (1) and (2) with the material balance, fi = di + bi , the following results are obtained: lbmol/h: Component Feed Distillate Bottoms C3 5 4.998 0.002 iC 4 15 14.720 0.280 nC 4 25 23.125 1.875 iC 5 20 3.600 16.400 nC 5 35 2.356 32.644 Total: 100 48.799 51.201 (c) Because the split of the LK and HK components is not sharp and because iC4 boils close to nC4, and iC5 boils close to nC5, it is possible that both iC4 and nC5 will distribute at minimum reflux. It is questionable that C3 will distribute. Nevertheless, first assume a Class 1 Underwood minimum reflux and use the following rearrangement of Eq. (9-21) for a feed at the bubble point: F Lmin = d nC 4 f nC 4 − α nC 4 ,iC5 α nC4 ,iC5 − 1 d iC5 f iC5 100 = 23125 . 3.6 − 188 . 25 20 188 − 1 . = 66.7 l Exercise 9.17 (continued) Analyze: (c) (continued) Now, use the Class 2 Underwood equation, Eq. (9-28),which for a bubble-point feed is: α i ,iC5 zi , F . . 4.36(0.05) 2.36(015) 188(0.25) 1.00(0.20) 0.84(0.35) = 1− q = 0 = + + + + . 4.36 − θ 2.36 − θ 188 − θ 1.00 − θ 0.84 − θ i α i ,iC5 − θ (3) There are four roots to Eq. (3), which from a spreadsheet calculation are: θ = 0.9248, 1.317, 2.173, 3.996 Assume that the C3 does not distribute. Then, the 3.996 root is not needed and dC3 = 5 lbmol/h. Eq. (9-29) is applied in the following form for each of the three remaining values of θ, in terms of the three unknowns, diC4 , dnC5 , and Lmin. The form used is obtained by multiplying Eq. (9-29) by the total distillate rate, D, which converts distillate mole fractions to component distillate rates, and the reflux ratio, R to L. 2.36 (d iC 4 ) 1.88 (23125) 0.84 (d nC5 ) 4.36 (5) . 1.00 (3.6) D + Lmin = + + + + 4.36 − 0.9248 2.36 − 0.9248 188 − 0.9248 1.00 − 0.9248 0.84 − 0.9248 . = 6.346 + 1644(d iC4 ) + 45514 +...
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