che656 2012 homework1 solutions

8/10/2019 CHE656 2012 Homework1 Solutions

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CHE 656

Computer Applications for Chemical Practice

Homework Set #1 Solutions

Class-16

Prepared by

Dr. Hong-ming Ku

King Mongkuts University of Technology ThonburiChemical Engineering Department

Chemical Engineering Practice School

May 2012 Use with Permission of the Author Only


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1. Mass Balances and Constraints with Elementary Modules, I

Consider the following process which consists of a pump, a reactor, an absorber, and adistillation column. The feed consists of 2 chemical components, A and B, which are fed intothe reactor in which the following reactions take place:

Reaction 1: A + B CReaction 2: A D + 2EReaction 3: 2C D + F

Pure F Pure E S7

S4 S5

Components S2A and B

S1 Reactor S6

Pump S3S8

Absorber

Distillation Column

Pure liquid component E is fed into the absorber, in which all of component F in the absorbervapor feed S3 leaves the overhead while the rest of the components in stream S3 are absorbed(assume 100% absorption). Components A, B, C, D, and E are then separated by distillation(relative volatility is: C, D, A, B, E in decreasing order).

(a) Express the flowsheet above in terms of elementary modules. Note that except for mixers,only 1 inlet stream is allowed in all elementary modules.

(b) The following information is available about the process:

Reactor:Outlet stream S3 contains 21.6 lbmol/hr of component FThe overall conversion of Reactions 1 and 2 based on A is 80%

Column specifications:Light key = A Heavy key = BMole-recovery of B in bottom = 98%

Process Feed:


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Molar flow rate of A in feed = 100 lbmol/hrMolar flow rate of B in feed = 150 lbmol/hr

Other information:Mole fraction of A in column overhead = 24.06%Total molar flow ratio of stream S5 to S4 = 9.26Molar ratio of component B flow rate in stream S8 to that in stream S1 = 0.5096

Is this problem constrained? If so, how many constraints are there, and what are they?Using the degree of freedom analysis, is this problem under-specified, fully specified, orover-specified?

(c) Perform a mass balance to determine the flow rates and compositions of every stream inyour elementary-module flowsheet.

Solution:

Selectivity of Reaction 1: A + B C = _____0.90______

Fractional conversion of reaction 3: 2C D + F = ____0.60______

Pure component E feed flow rate into absorber = ____200.0_____ lbmol/hr

Mole-recovery of light key in column overhead = ____0.95_____

Total molar flow rate of streams:

S4: ____21.6_____ lbmol/hr

S7: ____78.96____ lbmol/hr

S8: ____293.44___ lbmol/hr


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2. Fractional conversion of Reaction 3: 2C ----> D + F

3. Pure component E feed flow rate into absorber

4. Mole-recovery of light key (component A) in column overhead

Therefore, the problem is fully specified!

(c) Perform mass balance:

Define variables:

x1 = reaction 1 selectivity

x2 = reaction 2 fractional conversion

x3 = component E molar flow rate into absorber

x4 = mole-recovery of light key

Stream S2: Answers

AS2 = 100 (0.8)(100) = 20 lbmol/hr 20 lbmol/hr

BS2 = 150 0.8x 1 (100) = 150 80x 1 78

CS2 = 80x 1 72

DS2 = 0.8(1 x 1)(100) = 80(1 x 1) 8

ES2 = 2.80(1 x 1) = 160(1 x 1) 16

Stream S3: Answers

AS3 = 20 20 lbmol/hr

BS3 = 150 80x 1 78

CS3 = 80x 1(1 x 1) 28.8

DS3 = 80(1 x 1) + 40x 1x2 29.6


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ES3 = 160(1 x 1) 16

FS3 = 40x 1x2 = 21.6 lbmol/hr 21.6

Stream S5: Answers

ES5 = x 3 200 lbmol/hr

Stream S9: Answers

AS9 = 20 20 lbmol/hr

BS9 = 150 80x 1 78

CS9 = 80x 1(1 x 2) 28.8

DS9 = 80(1 x 1) + 40x 1x2 = 80(1 x 1) + 21.6 29.6

ES9 = 160(1 x 1) + x 3 216

FS9 = 21.6 21.6

Stream S4:

FS4 = 40x 1x2 = 21.6 lbmol/hr

Stream S6: Answers

AS6 = 20 20

BS6 = 150 80x 1 78

CS6 = 80x 1(1 x 1) 28.8

DS6 = 80(1 x 1) + 21.6 29.6

ES6 = x 3 + 160(1 x 1) = 200 + 160(1 x 1) 216

But 26.96.21

3

6

3 == x

F

x

S

x3 = 200 lbmol/hr

Stream S7: Answers

CS7 = 80x 1(1 x 2) 28.8 lbmol/hr


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DS7 = 80(1 x 1) + 21.6 29.6

AS7 = 20x 4 19

BS7 = (0.02)(150 80x 1) = 3 1.6x 1 1.56

Stream S8: Answers

AS8 = (1 x 4)20 = 20(1 x 4) 1 lbmol/hr

BS8 = (0.98)(150 80x 1) = 147 78.4x 1 76.44

ES8 = 200 + 160(1 x 1) 216

From the last constraint:

5096.0150

4.78147 11

9 =

= x

B

B

S

S

x1 = 0.9 x2 =)9.0)(40(

6.21 = 0.60

From the first constraint:

4

4

2056.16.298.2820

x

x

+++ = 0.2406

83.1255 x 4 = 20x 4 + 59.6

x4 = 0.95


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2. Mass Balances and Constraints with Elementary Modules, II

Consider the following process, which consists of a reactor and a flash vessel. Two feedsenter the reactor, each containing pure component A and pure component C, in which thefollowing reactions take place:

Reaction 1: 2A BReaction 2: 2B + C D

Pure A

Flash VesselPure C

Reactor

(a) Express the flowsheet above in terms of elementary modules. Note that except formixers, only 1 inlet stream is allowed in all elementary modules.

(b) The following information is available about the process:

Reactor: Reaction 2 fractional conversion (based on one mole of component B) is 1.2 times

higher than that of Reaction 1 (based on one mole of component A).

Flash Vessel: 90 mole% of component A and 40 mole% of component D in the feed to the flash

drum are known to go to the overhead vapor stream. The mole fractions of component A and component B in the overhead stream are

0.5907 and 0.0067, respectively. The mole fraction of component A in the bottom liquid stream is 0.1214.

Process Feeds:Molar flow rate of A in reactor feed = 100 lbmol/hrMolar flow rate of C in reactor feed = 40 lbmol/hr

How many constraints are there, and what are they? Using the degree of freedom analysis,is this problem under-specified, fully specified, or over-specified?

(c) Perform a mass balance to determine the molar flow rates of all species in every stream inyour elementary-module flowsheet.


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Solution

(a) Elementary modules

4

Reactor Reactor

A 1 2 3 Separator

C 2A ---> B 2B + C ---> D

5

(b) Degree of freedom analysis:

First, there are 8 standard input specifications required.

FA, FC, f 1, (Rxn 1 conversion), f 2 (Rxn 2 conversion), t A, tB, tC, tD

Component split fractionsin overhead

There are 4 constraints:

1) f 2 = 1.2f 1 = 1.2f

2) x A,4 = 0.5907

3) x B,4 = 0.0067

4) x A,5 = 0.1214

Four standard input specs are given:

FA = 100 F C = 40 t A = 0.90 t D = 0.40

# of missing input = # of constraints

Therefore, this problem is fully specified!

Mixer Rxn1 Rxn2


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(c) Mixer:

A1 = 100 C 1 = 40 lbmol/hr Answers

Reactor 1:

A2 = (1 f)100 35.0 lbmol/hr

B2 = 0.5f(100) = 50f 32.5 lbmol/hr

C2 = 40 40.0 lbmol/hr

Reactor 2 :

A3 = (1 f)100 35.0 lbmol/hr

B3 = (1 1.2f)(50f) = 50f(1 1.2f) 7.15

C3 = 40 0.5(1.2f) (50f) = 40 3f 2 27.325

D3

= 0.5(1.2f)(50f) = 30f 2 12.675

Separator:

A4 = 0.9 A 3 = 90 (1 f)

A = 0.5907 ; A = 0.1214

stream 4 stream 5

A3 = A 4 + A 5 = (1 f)100 A5 = (1 f)100 A 4

B3 = B 4 + B 5 = 50(1 1.2f)f B5 = 50(1 1.2f)f B 4

C3 = C 4 + C 5 = 40 30f 2 C5 = 40 30f 2 C 4

D3 = D 4 + D 5 = 30f 2 D5 = 30f 2 D 4


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( 1 f )100 90 (1 f ) = 0.1214

100 - 100f + 50f 60 f + 40 30f +30f - stream 4

90(1 f )

0.5907

10 (1 f ) = 0.1214

-12.3616 + 102.3616 60f 2

10 10f = -1.5007 + 12.4267f 7.284f 2

7.284f 2 22.4267f + 11.5007 = 0

f 1 = f = 22.4267 12.9566 = 0.65 2 (7.284)

f 2 = 0.78

A1 = 31.5 lbmol/hr stream 4 = 53.3266

A5 = 3.5 lbmol/hr stream 5 = 28.8303

B4 = 0.0067 B 4 = 0.3573 lbmol/hr

stream 4

B5 = 6.7927 t = 0.05


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31.5 + 0.3573 + C 4 + D 4 = 53.3266 C4 + D 4 = 21.4693

3.5 + 6.7927 + C 5 + D 5 = 28.8303 C5 + D 5 = 18.537

D4 = 0.4 D4 = 0.3573 lbmol/hr

12.675

D5 = 7.605

C 4 = 16.3993

C 5 = 10.9326

t C = 0.60


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Component A

Component B

S1

S2

S3 S4 S5

S6

Reactor 1 Reactor 2

Separator 1

Separator 2S7

S8S9

S10

S11

7. Degree of Freedom Analysis Consider the following flowsheet which consists of one mixer, two reactors in series, twoseparators, and a splitter, all elementary modules. Feed stream S1 contains pure component A,while feed stream S2 contains pure component B which react in the first reactor according tothe following two parallel reactions.

A + B ----> C Reaction 1 A ----> D + E Reaction 2

The following reaction takes place in the second reactor:

C + D ----> E Reaction 3

The overall conversion of the two parallel reactions in Reactor 1 based on component A is80%, while the conversion of Reaction 3 based on component C is 70%. The first separatorremoves 100% of component A in the Reactor 2 outlet as overhead and removes 100% ofcomponent E in Reactor 2 outlet as the bottom. In the second separator, 95% of component Bin S7 is recovered as overhead while 99% of component E in S7 is recovered as the bottom.

The following additional information is available about the process.

1. The ratio of molar flow rate between component A in S1 and component B in S2 is2 to 1.

2. The flow rate of S2 is 100 lbmol/hr.3. The flow rate of component C in stream S4 is 60 lbmol/hr.4. The total flow rate of S8 is 10 lbmol/hr.5. The product purity of component E in S11 is 99.9 mole%.6. S7 contains 50 mole% C and 20 mole% D

Perform a degree of freedom analysis on this process. Is the problem fully specified, under-specified, or over-specified? If the problem is over-specified, how many pieces of data are


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8. Determination of Tear Streams and Computation Order I Consider the flowsheet given below.(a) Find all irreducible groups in the flowsheet using the method by Sargent and Westerberg .

Write down a valid precedence order (calculation sequence) of this flowsheet comprisingthese irreducible groups .

(b) For group (s) with more than 1 unit in Part (a), find the minimum number of tear streams andtheir locations by B & M method . Also, write down a complete calculation sequence for theentire flowsheet .

3

2 18

4

1 9

5 814 10

6

11 12

1315

1916

717

Solution:

(a) The irreducible groups are: ___B1B2B4B3B5B10, B9, and B6B7B8B11___

The precedence order or computational order is ___B1B2B4B3B5B10 B9

B6B7B8B11___

(b) For each irreducible group, the minimum number of tear streams and their locations:

___For B1B2B4B3B5B10, the tear streams are 3, 4, and 18 (so minimum is 3)_________

B1 B2 B3 B4 B5

B6 B7 B11B10B9B8


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1 3

2 3

3 2 , 18 3 self loop

4 1 , 5 3 4

5 4

6 2 , 18 3

9 4

10 6

18 9 , 10 4 6

Node Precursors

4 4 solf loop

6 18

18 4 , 6 18

Node Precursors

18 18 self loop

The tear streams are: 3, 4, and 18

Irreducible group: B6 B7 B8 B11


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Node Precursors

11 14

12 11 , 13 14

14 12 , 17 , 19 15 14 self-loop

15 12 , 17 , 19 15 14

17 15

19 14

Node Precursors

12 13

15 12 , 15 self-loop

Node Precursors

12 13

The tear stream are: 14, 15

A complete computational sequence is:

Tear 3, 4, 18 B1 B4 B2 B3 B10

B5 Update tears 3, 14, 18 B9 Tear 14 , 15

B11 B6 B7 B8 Update tear 14, 15


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5 2, 15

6 5

7 5

8 3, 11, 16 14

9 3, 11, 16 14

10 6, 9, 19 5 12

11 10, 20 5

12 10, 20 5

13 8

14 13, 18 8 12

15 13, 18 8 12

Drawing this flowsheet will show that there is one (and only one) two-way edge pair betweenNodes 5, 2, and 4 with Node 2 being the common node.

So eliminate Node 2 and Stream 2 is the first tear stream we found.

Then we do another round of graph reduction as follows:

Streams Precursors

3 15

4 15

25 4


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5 15

8 3, 11, 14 15

9 3, 11, 14 15

10 5, 9, 12 15

11 5, 10 15

14 8, 12

15 8, 12

Drawing this flowsheet reveals another two-way edge pair between Nodes 14, 8, and 15 withNode 8 being the common node.

So eliminate Node 8 and Stream 8 is the second tear stream we found.Then we do another round of graph reduction as follows:

Streams Precursors

9 11, 14, 15 12

10 9, 12, 15 12

11 10, 15 12

12 10, 15 12

14 12

15 12 Elimination:

Streams Precursors

8 1514


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9 11, 12

10 9, 12

11 10, 12

12 10, 12 Self-loop

So Node 12 is another tear stream (the third one).

Graph Reduction:

Streams Precursors

9 11

10 9 11

11 10 11 Another self-loop

So Node 11 is the last tear stream (the fourth one).

The minimum number of tear streams is 4, and they are: Streams 2, 8, 11, and 12.

The complete computation order is:

Tear 2, 8, 11, and 12 B1 B10 B4 B5 B2 B7 B1 B6 B3 B8

B9 Update tear streams

ASPEN PLUS also found four tear streams, namely Streams 2, 9, 12, and 13 with acomputation sequence of:

Block $OLVER01 ( Method: WEGSTEIN) has been defined to con e!"e st!ea#s: 2 9 13 12

OM%&T'TION OR(ER OR T*E LOWS*EET: $OLVER01 B10 B5 B2 B7 B6 B8 B9 B3 B4 B1(RET&RN $OLVER01)


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Othe! alid tea! st!ea# sets incl+de:

S,- S.- S/0- and S/1

S,- S2- S/0- and S/1


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11. Finding Tear Streams and Computational Sequence Using ASPEN PLUS (a) Consider the flowsheet below. Use ASPEN PLUS to find the minimum number of tearstreams, their locations, and a computation order for the flowsheet. You must use the sameIDs for streams and blocks as given when you draw your flowsheet in ASPEN PLUS. Submita copy of your flowsheet from ASPEN PLUS.

Hint: Pick a simple (e.g. requiring the least input) unit operation module or modules tosimulate blocks A to P in the flowsheet. Your model needs not converge since we are onlyinterested in the flowsheet analysis, although it is very easy to pick modules that give youconvergence.

2

1 3 4 5 6 8

719 18 17 16 11 10 9

13

20 21 15 12

22 23 24 25 14

29 28 27 26

3132

30 3334

38 36 35

37

(b) For the irreducible group that contains the block D, we are interested in knowing if thisgroup contains any exclusive tear sets. Use Forder-Hutchisons loop analysis to determineif there are any exclusive tear sets in this irreducible group? If so, write down 3 such tearsets.

Solution:

(a) (Not assigned but good to know) Output from the Run Control Panel

A B C D E

F G H I J

K ML

O

N

P


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FSPLIT

A

FSPLIT

C

MIXER

B

FSPLIT

D

FSPLIT

E

MIXER

F

FSPLIT

G

FSPLIT

H

FSPLIT

I

FSPLIT

J

FSPLIT

K

FSPLIT

L

MIXER

M

MIXER

N

MIXER

O

MIXER

P

1

3

19

2

45

6

7

8

910

11

1213

14

15

161718

20

21

22 2324 25

2627

28

29

30

31

32

33 34

3536

37

38

->Processing input specifications ...

Flowsheet Analysis :

Block $OLVER01 (Method: WEGSTEIN) has been defined to convergestreams: 29 34 36

Block $OLVER02 (Method: WEGSTEIN) has been defined to convergestreams: 6 12

COMPUTATION ORDER FOR THE FLOWSHEET:A$OLVER01 N O K F G B C H M L P(RETURN $OLVER01)$OLVER02 I D J E(RETURN $OLVER02)

Note: Some of you may get other sets of tear streams such as Streams 24, 31, and 28 in the firstloop from ASPEN PLUS.


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The minimum number of tear streams is: 5

The tear streams are: Streams 29, 34, 36, 6, and 12

The computational order is: A Tear streams 29, 34, and 36 N O K F G B C H M L P Update streams 29, 34, and 36 Tear streams 6 and

12 I D J E Update streams 6 and 12

(b) Use Forder-Hutchison loop finder:

i) D S7 E S6 D A loop

ii) D S10 J S9 E S6 D A loop

ii) D S10 J S12 I S11 D A loop

iv) D S10 J S12 I S14 J A loop, but this loop doesnt containBlock D

So there are altogether 3 loops.

The incidence matrix looks as follows:

S7 S6 S10 S9 S12 S11 S14

Loop 1 X XLoop 2 X X XLoop 3 X X X

Yes, there are exclusive tear sets, namely {S6, S12}, {S6, S11}, and {S7, S10}.

che656 2012 homework1 solutions

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