A plate column as show in Figure Q4 is used to recover solute A from a dilute gas mixture using recycle water as solvent. The solvent entering flow rate is 280 kgmol/h and consists of 99.9 mol% of water and 0.1 mol% of A. The gas mixture enter the column with 6 mol% of solute A and flow rate of 150 kgmol/h. The required concentration of solute A in leaving stream, xA is 0.03. Assume the operating temperature and pressure are constant with the equilibrium relation for this system is у- 1.25х. L- 280 kgmol/h X= 0.001 L = 280 kgmol/h X = 0.03 V = 150 kgmol/h YA=0.06 Figure Q4: Gas absorption System (a) Determine the amount of streams leaving the column, L, and Vz. (b) Determine the number of ideal stages analytically. (c) Based on the minimum solvent flow rate, comment on the actual solvent flow rate for this system.

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
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Question
A plate column as show in Figure Q4 is used to recover solute A from a dilute gas
mixture using recycle water as solvent. The solvent entering flow rate is 280 kgmol/h
and consists of 99.9 mol% of water and 0.1 mol% of A. The gas mixture enter the
column with 6 mol% of solute A and flow rate of 150 kgmol/h. The required
concentration of solute A in leaving stream, xA is 0.03. Assume the operating
temperature and pressure are constant with the equilibrium relation for this system is
у- 1.25х.
L- 280 kgmol/h
X= 0.001
L = 280 kgmol/h
X = 0.03
V = 150 kgmol/h
YA=0.06
Figure Q4: Gas absorption System
(a)
Determine the amount of streams leaving the column, L, and Vz.
(b)
Determine the number of ideal stages analytically.
(c)
Based on the minimum solvent flow rate, comment on the actual solvent flow
rate for this system.
Transcribed Image Text:A plate column as show in Figure Q4 is used to recover solute A from a dilute gas mixture using recycle water as solvent. The solvent entering flow rate is 280 kgmol/h and consists of 99.9 mol% of water and 0.1 mol% of A. The gas mixture enter the column with 6 mol% of solute A and flow rate of 150 kgmol/h. The required concentration of solute A in leaving stream, xA is 0.03. Assume the operating temperature and pressure are constant with the equilibrium relation for this system is у- 1.25х. L- 280 kgmol/h X= 0.001 L = 280 kgmol/h X = 0.03 V = 150 kgmol/h YA=0.06 Figure Q4: Gas absorption System (a) Determine the amount of streams leaving the column, L, and Vz. (b) Determine the number of ideal stages analytically. (c) Based on the minimum solvent flow rate, comment on the actual solvent flow rate for this system.
Expert Solution
Step 1

Plate column is used to recover solute from the dilute gas mixture using water as a solvent.

molar flowrate of entering solvent, L2=280 Kmol/h

mole fraction of water in entering solvent, xwL2=0.999

mole fraction of solute A, xAL2=0.001

molar flowrate of entering gas mixture, V1=150 Kmol/h

mole fraction of A in entering gas, xAV1=0.06

mole fraction of A in existing solvent, xAL1=0.03

The equilibrium relationship is given as,

y=1.25x                                                                        ...... (1)

Step 2

(a)

The Schematic of the process is given as,

Chemical Engineering homework question answer, step 2, image 1

Write the overall material balance on the extraction column,

V1+L2=V2+L1                                                                               ...... (2)

Substitute V1=150 Kmol/h, L2=280 Kmol/h, L1=280 Kmol/h in equation (2)

150+280=V2+280V2=150 Kmol/h

Write the material balance of A in the column,

V1×yAV1+L2×xAL2=L1×xAL1+V2×yAV2                                               ...... (3)

Substitute V1=150 Kmol/hyAV1=0.06, L2=280 Kmol/h, xAL2=0.001, L1=280 Kmol/h, xAL1=0.03, V2=150 Kmol/h in equation (3)

150×yAV2+280×0.03=280×0.001+150×0.06yAV2=0.00586

Therefore, A in L1= xAL1×L1=0.03×280=8.4 Kmol/h

A in V2yAV2×V2=0.00586×150=0.879 Kmol/h

 

 

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