Chemical Engineering Homework: Azeotropes, Distillation, Particles

This document presents a comprehensive solution to a chemical engineering assignment, addressing key concepts such as azeotropes, distillation columns, and particle analysis. The solution begins by analyzing the azeotrope temperature and composition of a mixture using a T-x-y diagram. It then explores the construction of an equilibrium diagram for a binary system and provides a schematic diagram of a binary distillation column. The assignment further delves into calculating flow rates, compositions, and reflux ratios in distillation processes. Additionally, the solution covers particle analysis, including calculations of surface area, volume, and surface area-to-volume ratios, along with the application of the Reynold's number. The document provides detailed calculations, graphical representations, and explanations to facilitate understanding of these fundamental chemical engineering principles.

Running head: CHEMICAL ENGINEERING 1
Chemical Engineering
Name
Institution

CHEMICAL ENGINEERING 2
Chemical Engineering
Question 1
Part a
The azeotrope temperature of the mixture from the graph is 53.6℃ . The composition of the
azeotrope is 0.65 mole of MVC liquid and 0.65mol of MVC vapor.
From the graph, the feed contains 40mol% of MVC at 54.4 and 55mol% Vapor at the same℃
temperature.
We need to get the temperature at which 75% of the feed evaporates. Meaning, 75 % of 40 mol %
to evaporate. That is, 0.75 × 40 %=30 % mole MVC=0.3 mole MVC
0.3 mol MVC corresponds to a temperature of 58.8 . Hence, 75% of the feed would evaporate℃
at 75 .℃
Part b
We can read the X and Y values at different temperatures from the T-x-y diagram to obtain the X
and y co-ordinates. The results are shown in table 1.
Table 1: X and Y coordinates
Temperature, ℃ Mole fraction in
Liquid, X A
Mole fraction in
Vapor, Y A
64 0.00 0.00
61.6 0.06 0.16
60.0 0.10 0.24
58.4 0.15 0.32
57.2 0.20 0.38
56.4 0.25 0.44

CHEMICAL ENGINEERING 3
55.6 0.30 0.48
54.8 0.35 0.52
54.4 0.40 0.55
54.0 0.45 0.57
53.8 0.50 0.59
53.6 0.55 0.61
53.4 0.65 0.65
53.6 0.75 0.68
53.8 0.80 0.70
54.2 0.85 0.72
54.8 0.90 0.76
57.0 0.96 0.85
60.8 1.00 1.00
Then, we can plot the y-values against the x-values to obtain the equilibrium diagram as shown
in figure 1.

CHEMICAL ENGINEERING 4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
1.2
Equilibrium Curve for the Binary System at 1 atm
Mole Fraction in Liquid
Mole fraction in Vapor,
Figure 1: Equilibrium curve for the binary system at 1 atm
Question 2
Part a

CHEMICAL ENGINEERING 5
Figure 2: Schematic diagram of binary distillation column
Part b
The feed consists of liquid, vapor or a mixture of both. The feed entering the feed section can be
cold feed, superheated vapor, or saturated feed. The feed condition thus determines what leaves
the feed stage. The product leaving at the top of the column is called a distillate or overhead

CHEMICAL ENGINEERING 6
product and can either be solid, liquid or both depending on the condenser applied. On the other
hand, the product leaving at the bottom of the column is called a bottom product.
Question 3
Given the mole fractions of liquid (x), we can get the corresponding mole fractions of vapor (y)
from the graph as follows.
From the graph, it is evident that when x=0.58 , y=0.82∧when x=0.75 , y=0.885
Assumptions to note:

CHEMICAL ENGINEERING 7
We assume that the feed is a saturated liquid and the latent heat of vaporization is equal for both
components A and B.
yn+ 1=( R
R+1 ) xn+ ( 1
R+1 ) xD
0.885=( R
R+1 )0.58+ ( 1
R+1 )0.94
0.885=( R
R+1 ) ( 0.58+0.94 )
( R
R+1 )= 0.885
( 0.58+0.94 ) =0.5822368
R−0.5822368 R=0.5822368
R=1.394
Question 4

CHEMICAL ENGINEERING 8
Flow rate of bottom ¿ Feed flow rate−Distilate flow rate
¿ ( 2500−1358.7 ) kmol/h
¿ 1141.3 kmol /h
Distillate composition , Y D=96 %=0.96
Bottom composition, X D= XW=4 %=0.04
Assuming that the molar overflow rate is constant,
Y m +1 ¿ Y D=0.42
X m=X D=0.04
Reflux ratio ¿ 2 Lm
We know that, Lm=Gm+1 +W
Also, Lm=Gm+1=L
Y m +1=( L
L−W ) Xm− ( W
L−W ) X W
0.42= ( L
L−1141.3 ) 0.3−( 1141.3
L−1141.3 ) 0.04
Solving for L we get , L=3614.12 kmol/h
Gm +1=Lm−W =3614.12−1114.3=2472.82 kmol/h
Reflux ratio= L
D =2

CHEMICAL ENGINEERING 9
L=L+qF , q= L−L
F = 3614.12−2472.82
2500 =0.45652=45.652 %
Fraction of vapor= (100−45.652 ) %=54.348 %
The feed is saturated vapor
The q-value is 0.45652.

CHEMICAL ENGINEERING 10
SECTION B
Part a
Volume of the particle¿ 1
2 s2 sin 60 × thickness= √3
4 s2 × thickness
¿ √3
4 ×22 ×1= √3
Surface area of particle¿ √3
4 ×22 ×2+ ( 2 ×1 ) 2+ ( 2× 1 )=9.461 mm2
Surface area to volume ratio of particle¿ 9.461
√3 =5.4641mm2 /mm3
Surface area of sphere ¿ π d2
Volume of sphere¿ π
6 d3
Surface area to volume ratio¿ π d2
π
6 d3
= 6
d
5.4641mm2
mm3 = 6
d
xsv=d= 6
5.4641 =1.098 mm
Part b
Sphericity= S . A of sphere of same volume
S . A of particle = π
1
3 (6 V P)
2
3
AP
Where V P=volume of particle∧¿ AP=areaof particle

CHEMICAL ENGINEERING 11
Volume of particle= √3
4 ×22 ×1= √3
S . A of particle=9.461
Sphericity= π
1
3 (6 √ 3)
2
3
9.461 =0.7372
Part c
2D projection of the particle will be an equilateral triangle of side length 2mm.
Area of the equilateral triangle ¿ √ 3
4 s2= √ 3
4 ×22=1.732
Area of equivalent circle¿ π d2
4
π d2
4 =1.732
d2=1.732 × 4
π =2.2053
d=1.485mm
Part d
True. Reynold number does always depend on fluid density and particle diameter.
Reynold number , Nℜ= DVρ
μ
Where
D the diameter, ρthe fluid density, μ the fluid viscosity and ν the average velocity. Therefore, it
is evident from the formula that Reynold number depends on fluid density and the particle
diameter.