{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# 8 kmolh the product compositions at minimum reflux

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: roots are probably necessary. These are those between values of α for: (1) iso-butane and n-butane, (2) n-butane and iso-pentane, and (3) iso- Analysis: (continued) Exercise 9.11 (continued) pentane and n-pentane, because iso-butane and n-pentane boil closely to n-butane and isopentane, respectively. Using a spreadsheet, the 3 roots of θ are: (1) 2.7159 between 3.50 and 2.44, (2) 1.02575 between 2.44 and 1.00, and (3) 0.78274 between 1.00 and 0.75. Eq. (9-29) is applied in the following form for each of the three 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. 3.50 (d iC4 ) 0.75 (d nC5 ) 8.56 (2500) 2.44 (594) 1.00 (15) D + Lmin = + + + + 8.56 − 2.7159 350 − 2.7159 2.44 − 2.7159 1.00 − 2.7159 0.75 − 2.7159 . = 36618 + 4.4637(d iC 4 ) − 5253.2 − 8.7418 − 0.3815(d nC5 ) = 4.4637(d iC4 ) − 1600.14 − 0.3815(d nC5 ) . D + Lmin = 0.75 (d nC5 ) 3.50 (d iC4 ) 1.00 (15) 8.56 (2500) 2.44 (594) + + + + 8.56 − 102575 350 − 102575 2.44 − 1.02575 100 − 102575 0.75 − 102575 . . . . . . = 2840.36 + 141457(d iC 4 ) + 1024.83 − 582.52 − 2.71986(d nC5 ) = 141457(d iC4 ) + 3282.67 − 2.71986(d nC5 ) . . D + Lmin = 3.50 (d iC4 ) 0.75 (d nC5 ) 8.56 (2500) 2.44 (594) 1.00 (15) + + + + 8.56 − 0.78274 3.50 − 0.78274 2.44 − 0.78274 100 − 0.78274 0.75 − 0.78274 . = 2751.61 + 1.28806(d iC 4 ) + 874.55 + 69.04 − 22.9078(d nC5 ) = 128806(d iC4 ) + 3695.20 − 22.9078(d nC5 ) . Because the total distillate rate, D, is also an unknown, the following equation is also needed: D = dC3 + diC4 + dnC4 + diC5 + dnC5 = 2500 + diC4 +594 + 15 + dnC5 = 3109 + diC4 + dnC5 Solving these 4 linear equations, the following results are obtained: D = 4712.8 kmol/h, Lmin = 795.4 kmol/h, diC4 = 1593.4 kmol/h, dnC5 = 10.45 kmol/h Because the flow rate of iC4 in the distillate is greater than its feed rate, iC4 does not distribute. Recalculate omitting the first of the four equations that uses θ = 2.7159. Replace diC4 = fiC4 = 400 kmol/h. The equations to be solved now are: D + Lmin + 2.71986 d nC5 = 3848.5 D + Lmin + 22.9078 d nC5 = 4210.4 D − d nC5 = 3509.0 Analysis: (continued) Exercise 9.11 (continued) Solving, D = 3526.9 kmol/h, Lmin = 272.8 kmol/h, and dnC5 = 17.9 kmol/h Because the distillate rate of nC5 is positive and less than its feed rate, it does distribute. If we had assumed that only the two key components distributed, then it is only necessary to solve the following two equations using only θ = 1.02575, the one between the two key components, with dnC5 = 0.0 : D + Lmin = 3848.5 D = 3509.0 Then, Lmin = 3848.5 - 3509.0 = 339.5 kmol/h = internal reflux rate, which is high and incorrect. Assume that the minimum external reflux rate = minimum internal reflux rate = 272.8 kmol/h The product compositions at minimum reflux rate are: Component Propane iso-Butane n-Butane iso-Pent...
View Full Document

{[ snackBarMessage ]}