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Unformatted text preview: roots are probably necessary. These are those
between values of α for: (1) isobutane and nbutane, (2) nbutane and isopentane, and (3) iso Analysis: (continued) Exercise 9.11 (continued) pentane and npentane, because isobutane and npentane boil closely to nbutane 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. (929) 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. (929) 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
isoButane
nButane
isoPent...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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