Ethylene Glycol IBI 0.9 0.1 0.8/ 0.2 0.7 0.3 0.4 Ey boundan Plait 0.5 Point P. 05 0.6 0.3 Tie ine 0.7 0.2 0.8 0.1 0.9 Furfural (C) Water (A) D 0.9 0.8 0.5 0.4 0.1 G 0.7 0.6 0.3 0.2 Mass fraction furfural ethylene gyol as fract ion witer Mass fraction %3B Etrect Reffinate 10 0.9 0.8 0.7 0.6 45" Line 0.5 0.4 0.3 0.2 0.1 o 0.1 0.2 03 0.4 O5 0.6 0.7 0.8 09 1.0 Mass fraction glyool in raffinate Mass fraction glycol inextmct
Ethylene Glycol IBI 0.9 0.1 0.8/ 0.2 0.7 0.3 0.4 Ey boundan Plait 0.5 Point P. 05 0.6 0.3 Tie ine 0.7 0.2 0.8 0.1 0.9 Furfural (C) Water (A) D 0.9 0.8 0.5 0.4 0.1 G 0.7 0.6 0.3 0.2 Mass fraction furfural ethylene gyol as fract ion witer Mass fraction %3B Etrect Reffinate 10 0.9 0.8 0.7 0.6 45" Line 0.5 0.4 0.3 0.2 0.1 o 0.1 0.2 03 0.4 O5 0.6 0.7 0.8 09 1.0 Mass fraction glyool in raffinate Mass fraction glycol inextmct
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
Forty-five kg of a solution of 30 wt% ethylene glycol in water is to be extracted with furfural. Using Figures, calculate the: (a) minimum kg of solvent; (b) maximum kg of solvent; (c) kg of solvent-free extract and raffinate for 45 kg solvent, and the percentage glycol extracted; and (d) maximum purity of glycol in the extract and the maximum purity of water in the raffinate for one stage.
Transcribed Image Text:Ethylene Glycol
IBI
0.9
0.1
0.8/
0.2
0.7
0.3
0.4
Ey boundan
Plait
0.5
Point
P.
05
0.6
0.3
Tie ine
0.7
0.2
0.8
0.1
0.9
Furfural
(C)
Water
(A)
D 0.9
0.8
0.5
0.4
0.1 G
0.7
0.6
0.3
0.2
Mass fraction furfural
ethylene gyol
as fract ion witer
Mass fraction
%3B
Etrect
Reffinate
Transcribed Image Text:10
0.9
0.8
0.7
0.6
45" Line
0.5
0.4
0.3
0.2
0.1
o 0.1 0.2 03 0.4 O5 0.6 0.7 0.8 09 1.0
Mass fraction glyool in raffinate
Mass fraction glycol inextmct
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Step 1: a) Draw a line from to Solvent to Feed, Place M on Raffinate curve
VIEWStep 2: a) Minimum kg of solvent=Maximum solubility
VIEWStep 3:b)Draw a line from to Solvent to Feed, Place M on Extract curve
VIEWStep 4: Maximum kg solvent = Minimum solubility
VIEWStep 5: c) % of glycol extracted
VIEWStep 6: % of glycol extracted & purity of glycol in the extract and raffinate
VIEW
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