This preview shows page 1. Sign up to view the full content.
Unformatted text preview: hanol boils at 78.4oC. From Fig. 2.4, the vapor pressure of ethanol at 64.7oC = 8.2 psia.
The vapor pressure of methanol at 78.4oC = 25 psia. For the Raoult's law Kvalue, Eq. (244)
applies. Combining this equation with the definition of the relative volatility of Eq. (221),
gives, for methanol with respect to the less volatile ethanol, αΜ,Ε = (Ps)M/ (Ps)E . At 64.7oC, αΜ,Ε
= 14.696/8.2 = 1.79. At 78.4oC, αΜ,Ε = 25/14.696 = 1.70. Since these values are close (within
about 5%), use an equilibrium curve based on a constant α = (1.70 + 1.79)/2 = 1.745. From Eq.
(73), the curve is computed from,
y = αx/[1 + x(α1)] = 1.745x/(1+0.745x)
(3)
The McCabeThiele diagram on the next page shows the operating lines for determining the
minimum reflux ratio. It assumes that the pinch region occurs at the feed stage and not at the
sidestream stage. From Eq. (718), q for 25 mol% vaporized = 0.75. From Eq. (726), the slope
of the qline = q/(q1) = 0.75/(0.751) = 3. The upper section operating line intersects the qline
and equilibrium curve at y=0.823 and x=0.727. Thus, slope of the upper section operating line =
L/V = (0.960.823)/(0.960.727) = 0.588. From Eq. (727), Rmin = (L/V)/[1  (L/V)] = 0.588/(10.588) = 1.427. Therefore, L = 1.427(74.45) = 106.3 kmol/h and V = L + D = 106.3 + 74.45 =
180.75 kmol/h. In the middle section between the feed stage and sidestream stage, for 25 mol%
vaporization, L' = L + 0.75(100) = 106.3 + 75 = 181.3 kmol/h and V' = V  0.25(100) = 180.75 25 = 155.75 kmol/h. Thus, L'/V' = 181.3/155.75 = 1.164. The middle section operating line has
this slope and passes through y=0.823 and x=0.727. In the lower section operating line below the
sidestream, for a liquid sidestream flow rate of 15 kmol/h, L" = L'  S = 181.3  15 = 166.3
kmol/h and V" = V' = 155.75 kmol/h. Thus, L"/V" = 166.3/155.75 = 1.068. This line passes
through the point y=0.05 and x = 0.05 with the slope of 1.068. It also intersects the middle
section operating line at the sidestream composition, xs = 0.20. These three operating lines are
drawn in the McCabeThiele diagram, showing that the middle section operating line lies below Analysis: (continued) Exercise 7.36 (continued) the equilibrium curve. Therefore, the assumption that the minimum reflux is controlled by the
feed stage is verified and the minimum reflux ratio is 1.427. For actual operation, reflux ratio = 1.2Rmin = 1.2(1.427) = 1.712. Reflux rate in upper
section = L = 1.2Lmin = 1.2(106.3) = 127.6 kmol/h. Vapor rate = V = L + D = 127.6 + 74.45 =
202.05 kmol/h. Slope of upper section operating line = L/V = 127.6/202.05 = 0.632. For
constant molar overflow, the flow rates in the other sections are:
Middle section:
L' = 202.6 kmol/h
V' = 177.05 kmol/h L'/V' = 1.144
Lower section:
L" = 187.6 kmol/h
V" = 177.05 kmol/h L"/V" = 1.060 Analysis: (continued) Exercise 7.36 (continued) In the McCabeThiele diagrams below, upper, middle, and lower operating lines are
based on these values start...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details