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Subject
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Date
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Uploaded by DoctorDragonflyMaster198
Mullis, Ashton Rayne
Professor Bowman
ChE-201
20240210
Homework 4
1.
The attached drawing 01-A-005/1 is a process flow diagram (PFD) for one small section of a much larger butane isomerization plant. PFDs are normally printed on 11”x17” sheets of paper. For this problem we will focus on a mass balance around the reboiler E303. a) Determine the mass flows missing in the stream table for streams 21, 22, and 23. b) Calculate the mass fractions of each component in streams 21, 22, and 23. c) Suggest an explanation for why the stream compositions are not the same. Solutions
(A)
System boundary around E-303 and streams 21,22, and 23 in question.
General balance equation, In=Out
S
21
=
S
22
+
S
23
S21, ˙
m
=
¿
820,000 lb/hr
S
21
(
nC
4
)
=
820,000
−
480,000
−
7,700
−
15,000
=
317,300
lb
hr
Round down to 300,000 lb/hr… Balance Equation
S22 (C5+) = S21(C5+) – S23(C5+) 15,000
−
3,100
=
11,900
lb
hr
Round up to 12,000 lb/hr… S22(C5+)
Solve for S22 Total ˙
m
(lb/hr)
S22(
˙
m
) = ∑
product
❑
S22(
˙
m
) = 200,000+320,000+4,000+12,000 = 536,000 lb/hr
round up to 540,000 lb/hr… Solve for S23(iC4)
Balance Equation
S23(iC4) = S21(iC4) – S22(iC4) S23(iC4) = 480,000 – 320,000 S23(iC4) = 160,000 lb/hr
Balance Equation
S23(C1-C3)
S23(C1-C3)
S23(
˙
m
) = ∑
streams
❑
S23(
˙
m
) = 276,800 lb/hr
Round up, 280,000 lb/hr (B)
Mass Fractions of Components (S21-S23)
χ
m
=
m
m
total
S21
(nC4)…
χ
m
=
300,000
820,000
=
0.37
(iC4)…
χ
m
=
480,000
820,000
=
0.59
(C1-C3)…
χ
m
=
7700
820,000
=
0.0094
(C5+)…
χ
m
=
15000
820,000
=
0.018
S22
(nC4)… = 0.37
(iC4)… = 0.59
(C1-C3)… = 0.0074
(C5+)… = 0.022
S23
(nC4)… = 0.39
(iC4)… = 0.57
(C1-C3)… = 0.013
(C5+)… = 0.011
(C) Stream compositions will vary with different boiling points. The boiling points will be higher or lower depending on the components relative mass and density values. Also, since the streams do consist of different chemical varieties, they may be subject to chemical reactions when placed under stress, such as heating, cooling, or mixing. 2.
Brown sugar is a mixture of white sugar and molasses. Unrefined brown sugar contains a large portion of molasses. After one round of crystallization of this unrefined sugar, the brown sugar product contains 15.00 wt% molasses and the remainder white sugar. A second
round of crystallization reduces the molasses content to 1.10 wt%
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molasses. A plant wishes to produce a brown sugar ideal for baking that contains 6.50 wt% molasses. In order to achieve this concentration of molasses, a plant feeds once-crystallized brown sugar
into a recrystallizer. Some of this feed is diverted into a bypass stream that rejoins the sugar that leaves the crystallizer. Pure molasses leaves
the crystallizer as a bypass stream. This process is described by the flow chart below. a) If the plant wishes to produce 350.0 g/min of brown sugar today, what is the feed rate, m
1
? b) What is the flow rate of pure molasses from the crystallizer, m
4
? c) What is the fraction of the feed that bypasses the crystallizer? Solution
(A)
˙
m
1
Sugar, given a feed rate of 350 g/min.
m
1
g
min
=
1.839
total
×
350
g
min
0.8500
g
m
1
=
757.24
g
min
(B)
˙
m
4
Pure Molasses
.
0.1500
g
m
1
(
757
mL
min
)
= ˙
m
4
+
0.0650
g m
6
(
350
g
min
)
113.55
= ˙
m
4
+
22.75
˙
m
4
=
113.55
−
22.75
=
90.8
g
min
(C) Molasses Feed Fraction.
0.1500
90.8
=
1.65
E
−
3
3. Sea water is to be desalinized by reverse osmosis (RO) using the process shown in Figure 1. The process is designed to purify 1500 lb/hr of sea water which is comprised of 96.9 wt% water with the remainder comprised of various salts. Determine: a) The flow rates of fresh water (m
4
) and brine waste (m
6
). b) The percentage of salt rejected by the RO unit. In other words, what
percent of the salt fed to the RO unit in m
3 exits the RO unit in m
5
. c) The percentage of brine exiting the RO unit that is recycled. d) How much salt (in lb/hr) would be rejected if the recycle stream was
removed (as shown in Figure 2)? Note: 99.8% of the salt is rejected by the RO unit. How does this compare to the process in Figure 1? Can you determine the waste brine and purified water flow rates for the process shown in Figure 2?
Solution
(A)
Balance on m
4
∧
m
6
Equation IN=Out
m
1
=
m
4
+
m
6
--
1500
lb
hr
(
0.031
)
=
m
4
(
500
ppm
)
+
m
6
(
0.0525
)
—
46.5 lb
hr
=
m
4
(
0.0005
)
+
m
6
(
0.0525
)
System of two Equations
(1)(0.0525)-(2)
78.75 = 0.0525
m
4
+
0.0525
m
6
-(46.5 = m
4
0.0005
+
m
6
0.0525
¿
32.25 = 0.052
m
4
m
4
=
620.1923077 620.2
lb
hr
1500
lb
hr
=
620.1923077
+
m
6
m
6
=
879.8076923 879.8
lb
hr
(B) Balance
In=out m
1
+
m
2
=
m
3
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∵
m
1
=
1500
lb
hr
∴
1500
=
m
3
−
m
2
—1
Salt Balance
m
1
(
0.031
)
+
m
2
(
0.0525
)
=
m
3
(
0.04
)
46.5
+
m
2
(
0.0525
)
=
m
3
(
0.04
)
46.5
=
m
3
(
0.04
)
−
m
2
(
0.0525
)
–-2
Systems of two equations
(1)(0.0525)-(2)
78.75
=
0.0525
m
3
−
0.0525
m
2
−
¿
)
32.25
=
0.0525
−
0.04
m
3
m
3
=
2580
lb
hr
1500
=
2580
−
m
2
m
2
=
1080
lb
hr
Salt Rejected 4%(
m
3
¿
2580
×
0.04
=
103.2
lbs
Salt Rejected
5.25%
(
m
5
)
m
5
=
m
6
+
m
2
m
5
=
879.8
+
1080
(
lb
hr
)
m
5
=
1959.8
lb
hr
(
0.0525
)
=
102.89
Total Salt Rejected % 102.89
103.2
×
100
=
99.7%
(C.) m
2
m
5
×
100%
1080
1959.8
×
100
=
55.1%
(D)
Salt Rejected 1500 lb
hr
×
0.31
wt
%
salt
=
46.5
lb
hr
RO Recycle Unit Output = 500ppm*100% = 0.05
m
4
=
46.5
0.05
=
930
lb
hr
m
6
=
m
1
−
m
4
m
6
=
1500
−
930
m
6
=
570
lb
hr
4.
Carbon dioxide (along with other gases) has been implicated in global warming. A process is being developed which will separate CO
2 from flu
gas in a coal fired power plant and subsequently sequester the CO
2 underground. A schematic of the proposed process is shown below.
A cleaned and dry flu stream (m
1)
comprised of 26 wt% CO
2 and 74 wt% N
2
, enters into an absorber at a flow rate of 5.0 × 10
4 kg/hr where it contacts liquid-phase amine solvent (MEA) in an absorber. The amine solvent selectively absorbs CO
2
. The flu gas stream exiting the absorber (m
2
) is composed of 98 wt% N
2 and 2.0 wt% CO
2
. The liquid stream exiting the absorber (m
3
) contains only amine and CO
2
. The liquid stream is subsequently fed to a stripper where it is heated and 95% of the available CO
2 is desorbed from the amine. The CO
2 exiting the stripper (m
4
) is subsequently pressurized and injected into subterranean rock formations. The MEA exiting the stripper (m
5
), comprised of 99 wt% amine and 1.0 wt% CO
2
, is recycled back into the absorber. a) What percent of the CO
2 in stream m
1 is removed? b) What is the concentration (in wt%) of CO
2 in stream m
3
? c) What is the flow rate (in kg/hr) of the recycle stream (m
5
)? d) Tests conducted on experimental apparatus reveal that there will be a trace amount
of amine in the pure CO
2 stream. If the concentration of amine in stream m
4 is 25 ppm (not 0 as the figure shows) determine the required flow rate of make-up MEA (in kg/hr) required in the process. For part d) please assume that the total flow rate of m
4 to be 15,000 kg/hr. Solution
m
1
CO
2
=
0.26
∗
5.0E4
kg
hr
=
13,000
kg
m
1
N
2
=
0.74
∗
5.0E4
kg
hr
=
37,000
kg
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m
2
=
m
1
N
2
0.98
m
2
=
37,000
0.98
=
37,755.10204
kg
37,755.1
kg
m
2
C O
2
=
m
2
(
0.02
)
m
2
C O
2
=
37,755.1
×
0.02
=
755.102
kg
C O
2
Removed
=
13,000
−
755.102
=
12244.979
kg
12,245.0
kg
C O
2
%
Removed
=
12,245.0
5E4
×
100
=
24.5%
(B.)
C
%
M
3
m
5
=
0.99%
amine,
0.01%
C O
2
mC O
2
,
0.01
m
5
=
0.05
(
0.01
m
5
+
12244.979
)
kg
/
hr
0.01
m
5
=
0.0005
m
5
+
612.24895
0.0095
m
5
=
612.24896
(
C
)
−−
→m
5
=
64447.3
kg
/hr
m
3
C O
2
=
0.01
(
64447.3
)
+
12244.979
m
3
C O
2
=
12889.5
kg
MEA
0.99
∗
64447.3
=
63802.8
kg (B)-
C
O
2
WT
%
=
12889.5
12889.5
+
63802.8
×
100
=
16.8%
(D)
0.0025 MEA ppm, ˙
m
4
=
15,000
kg
hr
0.0025
MEA×
15000
kg
hr
=
37.5
kg
hr
MEA
m
3
=
m
4
∗
25
ppm
+
m
5
(
0.99
)
m
3
=
15,000
∗
0.0025
+
64447.3
(
0.99
)
m
3
=
63,840.3
k g
/hr
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- E PAP Chem G diagram sh x H PAP Chemi x G Reactions t x K PAP Chemi: x I PAP Chemi x b ar-2903012.agilixbuzz.com/student/135113422/activity/c412d902-b23e-4202-b813-76alf9f4c196 I Student Ap x Session Tir X Mp Rocking WX Mastery Assess It_8 200% + Reset PAP Chemistry-2903012-42100. Enrique Solis All changes saved 3. Reactions that occur in the gaseous state are affected by changing the volume of the container. Which choice best describes what will happen to the reaction if the volume of the container is decreased? The reaction will speed up. The reaction will shift left. The reaction will not change. SAVE & EXIT NEXT The reaction will shift right. 3 of 25 PREVIOUSarrow_forwardGiven that a theoretical yield for isolating Calcium Carbonate in this experiment would be 100%. From that information and based on the results you obtained in this experiment, describe your success in the recovery of calcium carbonate and suggest two possible sources of error that would have caused you to not obtain 100% yield. Results are attached form experimentarrow_forwardprinciple of the chemistry analyser pentra c200, calibration and quality controlarrow_forward
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