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

36 09248 236 09248 188 09248 100 09248 084 09248

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 +...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern