2. Basis: 4500 kg/h Fresh Feed. The flowchart for the system without recycle appears as shown below. 4500 kg/h 0.333 kg K/kg 0.667 kg W/kg m[kg W(v)/h] EVAPORATOR m₂(kg/h) 0.494 kg K/kg 0.506 kg W/kg CRYSTALLIZER AND FILTER ms(kg/h) m3[kg K(s)/h] m4(kg soln/h) 0.364 kg K/kg soln 0.636 kg W/kg soln 0.364 kg K/kg 0.636 kg W/kg
2. Basis: 4500 kg/h Fresh Feed. The flowchart for the system without recycle appears as shown below. 4500 kg/h 0.333 kg K/kg 0.667 kg W/kg m[kg W(v)/h] EVAPORATOR m₂(kg/h) 0.494 kg K/kg 0.506 kg W/kg CRYSTALLIZER AND FILTER ms(kg/h) m3[kg K(s)/h] m4(kg soln/h) 0.364 kg K/kg soln 0.636 kg W/kg soln 0.364 kg K/kg 0.636 kg W/kg
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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I need help understanding the answers for m3 and m5. Be thorough to the fullest
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4500 kg/h
0.333 kg K/kg
0.667 kg W/kg
2. Basis: 4500 kg/h Fresh Feed.
The flowchart for the system without recycle appears as shown below.
Q Search
T Add text
EVAPORATOR
m₁[kg W(v)/h]
■
Draw
m₂(kg/h)
0.494 kg K/kg
0.506 kg W/kg
CRYSTALLIZER
AND FILTER
m5(kg/h)
Highlight
m3[kg K(s)/h]
m4(kg soln/h)
0.364 kg K/kg soln
0.636 kg W/kg soln
0.364 kg K/kg
0.636 kg W/kg
Erase
1
{9
50
We will not go through the detailed solution but will simply summarize. The degree-of-freedom analysis
4.6 Chemical Reaction Stoichiometry 129
V
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ENG
X
:
o
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1:15 PM
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Transcribed Image Text:=
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CD Page view A Read aloud
+
4500 kg/h
0.333 kg K/kg
0.667 kg W/kg
2. Basis: 4500 kg/h Fresh Feed.
The flowchart for the system without recycle appears as shown below.
Q Search
T Add text
EVAPORATOR
m₁[kg W(v)/h]
■
Draw
m₂(kg/h)
0.494 kg K/kg
0.506 kg W/kg
CRYSTALLIZER
AND FILTER
m5(kg/h)
Highlight
m3[kg K(s)/h]
m4(kg soln/h)
0.364 kg K/kg soln
0.636 kg W/kg soln
0.364 kg K/kg
0.636 kg W/kg
Erase
1
{9
50
We will not go through the detailed solution but will simply summarize. The degree-of-freedom analysis
4.6 Chemical Reaction Stoichiometry 129
V
{"
ENG
X
:
o
r
6
+
→→
1:15 PM
2/1/2023
![=
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[T] Add text
Draw
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J
60
We will not go through the detailed solution but will simply summarize. The degree-of-freedom analysis
leads to the results that the overall system has one degree of freedom, the evaporator has zero, and the
crystallizer-filter has one. (Verify these statements.) The strategy is therefore to begin with the evaporator and
solve the balance equations for m₁ and m₂. Once m2 is known, the crystallizer has zero degrees of freedom
and its three equations may be solved for m3, m4, and m5. The rate of production of crystals is
m3 = 622 kg K(s)/h
With recycle it was 1470 kg/h, a dramatic difference. The mass flow rate of the filtrate is
4.6 Chemical Reaction Stoichiometry 129
m5 = 2380 kg/h
The filtrate (which is discarded) contains 0.364 × 2380 = 866 kg/h of potassium chromate, more than the
filter cake contains. Recycling the filtrate enables us to recover most of this salt. The obvious benefit of
recycling is the revenue from selling the additional potassium chromate. The costs include the purchase and
{"
D
ENG
X
:
♂
r
6
+
1:17 PM
2/1/2023](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ba88ecd-654b-4f6b-9783-e93c33ca6a16%2F7c8580f9-87d9-4b86-a00a-e431e0c99a79%2Fgsd4das_processed.png&w=3840&q=75)
Transcribed Image Text:=
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CD Page view A Read aloud
[T] Add text
Draw
Q Search
Highlight
Erase
J
60
We will not go through the detailed solution but will simply summarize. The degree-of-freedom analysis
leads to the results that the overall system has one degree of freedom, the evaporator has zero, and the
crystallizer-filter has one. (Verify these statements.) The strategy is therefore to begin with the evaporator and
solve the balance equations for m₁ and m₂. Once m2 is known, the crystallizer has zero degrees of freedom
and its three equations may be solved for m3, m4, and m5. The rate of production of crystals is
m3 = 622 kg K(s)/h
With recycle it was 1470 kg/h, a dramatic difference. The mass flow rate of the filtrate is
4.6 Chemical Reaction Stoichiometry 129
m5 = 2380 kg/h
The filtrate (which is discarded) contains 0.364 × 2380 = 866 kg/h of potassium chromate, more than the
filter cake contains. Recycling the filtrate enables us to recover most of this salt. The obvious benefit of
recycling is the revenue from selling the additional potassium chromate. The costs include the purchase and
{"
D
ENG
X
:
♂
r
6
+
1:17 PM
2/1/2023
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