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: McGraw-Hill Education
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question and answer attached.
could someone show clearer working and reasoning behind solotion
Transcribed Image Text:Tutorial 7 - Question Q1
Extraction of acetone from trichloroethane
1000 kg/h of a 45 wt% solution of acetone (A) in water (W) is to be extracted with
pure 1,1,2-trichloroethane (TCE) in a continuous counter-current system. Using the
equilibrium data below:
a) Construct the equilibrium curve for the liquid-liquid region in the ternary system.
b) Identify the solute, carrier and solvent. Briefly explain your choice.
c) Determine the minimum flow rate of solvent to achieve phase split.
Concentrations in phase 1 (wt%) Concentrations in phase 2 (wt%)
Xw
Xw
13
35
4
3
2
1.5
1
ХА
60
50
40
30
20
10
XTCE
27
46
57
68
78.5
89
W
Tutorial 7 - Question Q1
ХА
55
50
40
30
20
10
Extraction of acetone from trichloroethane
W
35
Extraction of acetone from trichloroethane
12
13.
A
30-
x
+
43
57
68
79
89.5
Tutorial 7 - Question Q1a Solution
A
X
A
V
4
K
X
60
A
55
XTCE
10
X
7
3
2
1
0.5
A
University of
Strathclyde
Engineering
Tie line data
Raffinate
Extract
X₁ (wt%) X₁ (wt%)
44
56
29
40
12
18
TCE
TCE
University of
Strathclyde
☆
University
Strathclyde
Engineering
Transcribed Image Text:Tutorial 7 - Question Q1a Solution
Extraction of acetone from trichloroethane
W
Carrier
W
Tutorial 7 - Question Q1b Solution
Extraction of acetone from trichloroethane
d
4
W
Solute A
Feed F
XAF 0.45
.
.
Extraction of acetone from trichloroethane
The minimum concentration of
solvent to achieve phase split
is given by the green dot (M):
XAM = 0.44
XW.M= 0.52
XTCEM = 0.04
A
(.)
.
A
• Solute: Acetone
Tutorial 7 - Question Q1c Solution
TCE
. Solvent: Trichloroethane
This is the compound we want to
extract
> Phase split at low concentrations
• Carrier: Water
➤ Originally carries the solute
Solvent
TCE
The slope of the tie lines increases.
from the carrier to the solvent
Overall balance:
M=F+S
Component balance:
XAM M=XAFF+XAS S
Carathclyde
University of
University of
Strathclyde
Solvent S
✓✓XAS=0
TCE
S=F* (XAF-XAM)/XAM
S = 1000 (0.45-0.44) / 0.44
S = 23 kg/h
University of
Strathclyde
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could you touch further on the last slide, where does the mixing point come from and also tie lines ?
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could you touch further on the last slide, where does the mixing point come from and also tie lines ?
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