Four hundred and fifty lbmol/h (204 kmol/h) of a mixture of 60 mol% benzene (LK) and 40 mol% toluene (HK) is to be separated into a liquid distillate and a liquid bottoms product of 95 mol% and 5 mol % benzene, respectively. The feed enters the column with a molar per- vaporization equal to the distillate-to-feed ratio. Use the McCabe-Thiele method to compute, at 1 atm (101.3 kPa): (a) Nmin, and the optimal feed-stage location. Also, compare the results with (b) Amin, and (c) number of equilibrium stages N, for R/Rmin: 1.3, those from a process simulator. and cent
Four hundred and fifty lbmol/h (204 kmol/h) of a mixture of 60 mol% benzene (LK) and 40 mol% toluene (HK) is to be separated into a liquid distillate and a liquid bottoms product of 95 mol% and 5 mol % benzene, respectively. The feed enters the column with a molar per- vaporization equal to the distillate-to-feed ratio. Use the McCabe-Thiele method to compute, at 1 atm (101.3 kPa): (a) Nmin, and the optimal feed-stage location. Also, compare the results with (b) Amin, and (c) number of equilibrium stages N, for R/Rmin: 1.3, those from a process simulator. and cent
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Question
Redo Example 7.1, keeping everything the same but assuming a relative volatility of 1.7.
![**Example 7.1: Distillation of a Mixture of Benzene and Toluene**
Four hundred and fifty kmol/h (204 kmol/h) of a mixture of 60 mol% benzene (LK) and 40 mol% toluene (HK) is to be separated into a distillate and a liquid bottoms product of 95 mol% and 5 mol% benzene, respectively. The feed enters the column with a molar percent vaporization equal to the distillate-to-feed ratio. Use the McCabe-Thiele method to compute:
a. Minimum number of stages at 1 atm (101.3 kPa).
b. Number of equilibrium stages \( N \), for \( R/R_{\text{min}} = 1.3 \), and the optimum feed-stage location. Compare the results with those from a process simulator.
**Solution**
First, calculate \( D \) and \( B \). An overall material balance on benzene gives:
\[ 0.60(450) = 0.95D + 0.05B \tag{1} \]
A total balance gives:
\[ 450 = D + B \tag{2} \]
Combining (1) and (2) and solving, \( D = 275 \) kmol/h, \( B = 175 \) kmol/h, and \( D/F = 0.611 \). Thus, the molar vaporization of the feed is 61.1%.
Calculate the slope of the q-line:
\[ V/F = D/F = 0.611 \]
and for a partially vaporized feed is
\[ \frac{L_f}{F} = \frac{F - V_f}{F} = 1 - \frac{V_f}{F} = 0.389 \]
From (7-26), the slope of the q-line is:
\[ \frac{q}{q-1} = \frac{0.389}{0.389 - 1} = -0.637 \]
1. **Figure 7.13:** Determination of Minimum Stages for Example 7.1
The graph illustrates the mole fraction of benzene in the liquid versus the mole fraction in the vapor for benzene-toluene at 1 atm. The equilibrium curve and 45° line are also shown. Stages are stepped off between these two curves.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F403d5df5-c9a3-47af-a06d-d9770154fd48%2F64725e5e-9183-4678-9eb7-51cd109bd63e%2F3uq1msp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example 7.1: Distillation of a Mixture of Benzene and Toluene**
Four hundred and fifty kmol/h (204 kmol/h) of a mixture of 60 mol% benzene (LK) and 40 mol% toluene (HK) is to be separated into a distillate and a liquid bottoms product of 95 mol% and 5 mol% benzene, respectively. The feed enters the column with a molar percent vaporization equal to the distillate-to-feed ratio. Use the McCabe-Thiele method to compute:
a. Minimum number of stages at 1 atm (101.3 kPa).
b. Number of equilibrium stages \( N \), for \( R/R_{\text{min}} = 1.3 \), and the optimum feed-stage location. Compare the results with those from a process simulator.
**Solution**
First, calculate \( D \) and \( B \). An overall material balance on benzene gives:
\[ 0.60(450) = 0.95D + 0.05B \tag{1} \]
A total balance gives:
\[ 450 = D + B \tag{2} \]
Combining (1) and (2) and solving, \( D = 275 \) kmol/h, \( B = 175 \) kmol/h, and \( D/F = 0.611 \). Thus, the molar vaporization of the feed is 61.1%.
Calculate the slope of the q-line:
\[ V/F = D/F = 0.611 \]
and for a partially vaporized feed is
\[ \frac{L_f}{F} = \frac{F - V_f}{F} = 1 - \frac{V_f}{F} = 0.389 \]
From (7-26), the slope of the q-line is:
\[ \frac{q}{q-1} = \frac{0.389}{0.389 - 1} = -0.637 \]
1. **Figure 7.13:** Determination of Minimum Stages for Example 7.1
The graph illustrates the mole fraction of benzene in the liquid versus the mole fraction in the vapor for benzene-toluene at 1 atm. The equilibrium curve and 45° line are also shown. Stages are stepped off between these two curves.
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