Solution Manual Essentials of Chemical Reaction Engineering

Solutions for Chapter 1 – Mole Balances

P1-1.

This problem helps the student understand the course goals and objectives. Part (d) gives hints on how to solve problems when they get stuck.

P1-2.

Encourages students to get in the habit of writing down what they learned from each chapter. It also gives tips on problem solving.

P1-3.

Helps the student understand critical thinking and creative thinking, which are two major goals of the course.

P1-4.

Requires the student to at least look at the wide and wonderful resources available on the CDROM and the Web.

P1-5.

The ICMs have been found to be a great motivation for this material.

P1-6.

Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives the student an idea of things to come in terms of sizing reactors in chapter 4. An alternative to P1-15.

P1-7.

Straight forward modification of Example 1-1.

P1-8.

Helps the student review and member assumption for each design equation.

P1-9.

The results of this problem will appear in later chapters. Straight forward application of chapter 1 principles.

P1-10.

Straight forward modification of the mole balance. Assigned for those who emphasize bioreaction problems.

P1-11.

Will be useful when the table is completed and the students can refer back to it in later chapters. Answers to this problem can be found on Professor Susan Montgomery’s equipment module on the CD-ROM. See P1-14.

P1-12. Many students like this straight forward problem because they see how CRE principles can be applied to an everyday example. It is often assigned as an in-class problem where parts (a) through (f) are printed out from the web. Part (g) is usually omitted. P1-13. Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with P1-6. P1-14. I always assign this problem so that the students will learn how to use POLYMATH/MATLAB before needing it for chemical reaction engineering problems. P1-15 and P1-16. Help develop critical thinking and analysis.

CDP1-A

Similar to problems 3, 4, and 10. 1-2

Summary Assigned

Difficulty

Time (min)

AA

SF

60

P1-2

I

SF

30

 P1-3

O

SF

30

P1-4

O

SF

30

P1-5

AA

SF

30

P1-6

AA

SF

15

P1-7

I

SF

15

 P1-8

S

SF

15

P1-9

S

SF

15

P1-10

O

FSF

15

P1-11

I

SF

1

P1-12

O

FSF

30

P1-13

O

SF

60

 P1-14

AA

SF

60

P1-15

O

--

30

P1-16

O

FSF

15

CDP1-A

AA

FSF

30

 P1-1

Alternates

1-13

Assigned  = Always assigned, AA = Always assign one from the group of alternates, O = Often, I = Infrequently, S = Seldom, G = Graduate level

1-3

Alternates In problems that have a dot in conjunction with AA means that one of the problem, either the problem with a dot or any one of the alternates are always assigned. Time Approximate time in minutes it would take a B/B+ student to solve the problem. Difficulty SF = Straight forward reinforcement of principles (plug and chug) FSF = Fairly straight forward (requires some manipulation of equations or an intermediate calculation). IC = Intermediate calculation required M = More difficult OE = Some parts open-ended. *

Note the letter problems are found on the CD-ROM. For example A

CDP1-A.

Summary Table Ch-1 Review of Definitions and Assumptions

1,5,6,7,8,9

Introduction to the CD-ROM

1,2,3,4

Make a calculation

6

Open-ended

8

1-4

P1-1 Individualized solution. P1-2 (b)

The negative rate of formation of a species indicates that its concentration is decreasing as the reaction proceeds ie. the species is being consumed in the course of the reaction. A positive number indicates production of the particular compound.

(c)

The general equation for a CSTR is:

FA 0

V

FA rA

Here rA is the rate of a first order reaction given by: rA = - kCA Given : CA = 0.1CA0 , k = 0.23 min-1, v0 = 10dm3 min-1

Substituting in the above equation we get:

V

CA0 v0 C Av0 kC A

C A0 v0 (1 0.1) 0.1kC A0

V = 391.304 m3 (d)

k = 0.23 min-1 From mole balance:

dNA

Rate law:

rA

rA

rA V

dt

k

k CA

NA V

Combine:

dNA dt

k NA

1-5

(10dm3 / min)(0.9) (0.23 min 1 )(0.1)

at t

0 , NAO = 100 mol and t

t=

N 1 ln A0 k NA

1 ln 100 0.23

t , NA = (0.01)NAO

min

t = 20 min

P1-3 Individualized solution.



P1-6 Individualized solution

P1-7 (a) The assumptions made in deriving the design equation of a batch reactor are: -

Closed system: no streams carrying mass enter or leave the system.

-

Well mixed, no spatial variation in system properties

-

Constant Volume or constant pressure.

1-6

P1- 7 (b) The assumptions made in deriving the design equation of CSTR, are: -

Steady state.

-

No spatial variation in concentration, temperature, or reaction rate throughout the vessel.

P1-7(c) The assumptions made in deriving the design equation of PFR are: -

Steady state.

-

No radial variation in properties of the system.

P1-7 (d) The assumptions made in deriving the design equation of PBR are: -

Steady state.

-

No radial variation in properties of the system.

P1-7 (e) For a reaction, A B -rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=] moles/ (dm3.s). -rA’ is the rate of disappearance of species A per unit mass (or area) of catalyst *=+ moles/ (time. mass of catalyst). rA’ is the rate of formation (generation) of species A per unit mass (or area) of catalyst *=+ moles/ (time. mass catalyst). -rA is an intensive property, that is, it is a function of concentration, temperature, pressure, and the type of catalyst (if any), and is defined at any point (location) within the system. It is independent of amount. On the other hand, an extensive property is obtained by summing up the properties of individual subsystems within the total system; in this sense, rA is independent of the ‘extent’ of the system. P 1-8 Rate of homogenous reaction rA is defined as the mole of A formed per unit volume of the reactor per second. It is an Intensive property and the concentration, temperature and hence the rate varies with spatial coordinates. 1-7

rA' on the other hand is defined as g mol of A reacted per gm. of the catalyst per second. Here mass of catalyst is the basis as this is what is important in catalyst reactions and not the reactor volume. Applying general mole balance we get:

dN j dt

Fj0

Fj

r j dV

No accumulation and no spatial variation implies

0 F j0

Fj

r j dV

Also rj = ρb rj` and W = Vρb where ρb is the bulk density of the bed. =>

0 ( Fj 0

Fj )

rj' ( b dV )

Hence the above equation becomes

Fj 0

W

Fj r

' j

We can also just apply the general mole balance as

dN j dt

( Fj 0

Fj )

rj' (dW )

Assuming no accumulation and no spatial variation in rate, we get the same form

Fj 0

W

as above:

Fj rj'

P1-9 Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 9), so there is no cell growth and the nutrients are used in making product. Let’s do part c first.

1-8

[Flowrate In (moles/time)] penicillin

penicillin

+ [generation rate (moles/time)]penicillin – [ Flowrate Out(moles/time)]

= [rate of accumulation (moles/time)]penicillin

dNp dt

Fp,in + Gp – Fp,out =

Fp,in = 0 (because no penicillin inflow) V

Gp =

rp .dV

Therefore, V

rp .dV - Fp,out =

dNp dt

Assuming steady state for the rate of production of penicillin in the cells stationary state,

dNp =0 dt And no variations

V

Fp ,in

Fp ,out rp

Or,

V

Fp ,out rp

Similarly, for Corn Steep Liquor with FC = 0

V

FC 0

FC rC

FC 0 rC

Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed. P1-10 Given

A

2 * 1010 ft 2

TSTP

V

4 * 1013 ft 3

T = 534.7 R

R

0.7302

atm ft 3 lbmol R

491.69 R

H

2000 ft

PO = 1atm

CS

yA = 0.02 1-9

2.04 *10

10

lbmol ft 3

C = 4*105 cars FS = CO in Santa Ana winds

vA

3000

FA = CO emission from autos

ft 3 per car at STP hr

P1-10 (a) Total number of lb moles gas in the system:

N

N=

PV 0 RT

1atm (4 1013 ft 3 ) = 1.025 x 1011 lb mol atm. ft 3 0.73 534.69 R lbmol.R

P1-10 (b) Molar flowrate of CO into L.A. Basin by cars.

FA

y A FT

FT

3000 ft 3 hr car

yA

v A C P0 R TSTP 1lbmol 400000 cars 359 ft 3

(See appendix B)

FA = 6.685 x 104 lb mol/hr P1-10 (c) Wind speed through corridor is v = 15mph W = 20 miles The volumetric flowrate in the corridor is vO = v.W.H = (15x5280)(20x5280)(2000) ft3/hr = 1.673 x 1013 ft3/hr P1-10 (d) Molar flowrate of CO into basin from Sant Ana wind. FS

v0 CS

= 1.673 x 1013 ft3/hr 2.04 10

10

lbmol/ft3

= 3.412 x 103lbmol/hr

P1-10 (e) 1-10

Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO =

FA

FS

vo C co

V

dC co dt

(V=constant, N co

dN CO dt

C coV )

P1-10 (f) t = 0 , C co

C coO Cco

t

dt

V

0

CcoO

dCco FS voCco

FA

F FS vo C coO V ln A vo FA FS vo C co

t

P1-10 (g) Time for concentration to reach 8 ppm.

CCO 0

2.04 10

8

lbmol , CCO ft 3

2.04 lbmol 10 8 4 ft 3

From (f),

F FS vO .CCO 0 V ln A vo FA FS vO .CCO

t

3 lbmol lbmol 3 lbmol 13 ft 3.4 10 1.673 10 2.04 10 8 3 4 ft hr hr hr ft 3 ln ft 3 lbmol lbmol ft 3 lbmol 1.673 1013 6.7 104 3.4 103 1.673 1013 0.51 10 8 hr hr hr hr ft 3

6.7 104

t = 6.92 hr P1-10 (h) (1)

to

=

0

tf = 72 hrs

C co = 2.00E-10 lbmol/ft3

vo Fs

a

a = 3.50E+04 lbmol/hr

= 1.67E+12 ft3 /hr

b = 3.00E+04 lbmol/hr

= 341.23 lbmol/hr

V = 4.0E+13 ft3

b sin

t 6

Fs

vo C co

V

dC co dt

Now solving this equation using POLYMATH we get plot between Cco vs t 1-11

See Polymath program P1-10-h-1.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t

0

0

72

C

2.0E-10

v0

1.67E+12

a

3.5E+04

3.5E+04

3.5E+04

3.5E+04

b

3.0E+04

3.0E+04

3.0E+04

3.0E+04

F

341.23

V

4.0E+13

2.0E-10 1.67E+12

341.23 4.0E+13

72 2.134E-08

1.877E-08

1.67E+12

341.23

1.67E+12

341.23

4.0E+13

4.0E+13

ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+b*sin(3.14*t/6)+F-v0*C)/V

Explicit equations as entered by the user [1] v0 = 1.67*10^12 [2] a = 35000 [3] b = 30000 [4] F = 341.23 [5] V = 4*10^13

1-12

(2) tf = 48 hrs

Fs = 0

a

t 6

b sin

vo C co

V

Now solving this equation using POLYMATH we get plot between Cco vs t See Polymath program P1-10-h-2.pol.

POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t

0

0

48

C

2.0E-10

v0

1.67E+12

a

3.5E+04

3.5E+04

3.5E+04

3.5E+04

b

3.0E+04

3.0E+04

3.0E+04

3.0E+04

V

4.0E+13

4.0E+13

4.0E+13

4.0E+13

2.0E-10 1.67E+12

48 1.904E-08

1.693E-08

1.67E+12

1.67E+12

ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+b*sin(3.14*t/6)-v0*C)/V

Explicit equations as entered by the user [1] v0 = 1.67*10^12 [2] a = 35000 [3] b = 30000 [4] V = 4*10^13

1-13

dC co dt

(3) Changing a  Increasing ‘a’ reduces the amplitude of ripples in graph. It reduces the effect of the sine function by adding to the baseline.

Changing b  The amplitude of ripples is directly proportional to ‘b’. As b decreases amplitude decreases and graph becomes smooth.

Changing v0  As the value of v0 is increased the graph changes to a “shifted sin-curve”. And as v0 is decreased graph changes to a smooth increasing curve.

P1-11 (a) – rA = k with k = 0.05 mol/h dm3

CSTR: The general equation is

FA 0

V

FA rA

Here CA = 0.01CA0 , v0 = 10 dm3/min, FA = 5.0 mol/hr Also we know that FA = CAv0 and FA0 = CA0v0, CA0 = FA0/ v0 = 0.5 mol/dm3 Substituting the values in the above equation we get,

V

C A0v0

C A v0 k

(0.5)10 0.01(0.5)10 0.05 1-14

 V = 99 dm3 FR: The general equation is

dFA dV

rA

k , Now FA = CAv0 and FA0 = CA0v0 =>

dC A v0 dV

k

Integrating the above equation we get

v0 C A dC A k CA0

V

dV

=> V

0

v0 (C A0 k

CA )

Hence V = 99 dm3 Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of concentration. P1-11 (b) - rA = kCA with k = 0.0001 s-1 CSTR: We have already derived that

V

C A0v0

C A v0

v0 C A0 (1 0.01) kC A

rA

k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1 V

(10dm 3 / hr )(0.5mol / dm 3 )(0.99) (0.36hr 1 )(0.01 * 0.5mol / dm 3 )

PFR: From above we already know that for a PFR

dC Av0 dV

rA

kC A

Integrating

v0 C A dC A k CA0 C A

v0 C A0 ln k CA

V

dV 0

V

Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1

1-15

=> V = 2750 dm3

Substituing the values in above equation we get V = 127.9 dm3 P1-11 (c) - rA = kCA2 with k = 3 dm3/mol.hr CSTR:

CA0 v 0 CA v 0 rA

V

v 0CA0 (1 0.01) kCA 2

Substituting all the values we get

(10dm 3 /hr)(0.5mol /dm 3 )(0.99) (3dm 3 /hr)(0.01* 0.5mol /dm 3 ) 2

V

=> V = 66000 dm3

PFR:

dCA v 0 dV

rA

kCA 2

Integrating C

v 0 A dCA k C CA 2 A0

=> V

V

dV => 0

v0 1 ( k CA

10dm 3 /hr 1 ( 3 3dm /mol.hr 0.01CA0

1 ) V CA 0

1 ) = 660 dm3 CA0

P1-11 (d) CA = .001CA0

t

dN rAV

NA0 NA

Constant Volume V=V0

t

CA0 CA

dC A rA

Zero order:

t

1 CA0 0.001CA0 k

.999CAo 0.05

9.99 h

First order:

1-16

t

1 CA0 ln k CA

1 1 ln 0.001 .001

6908 s

Second order:

t

1 1 k CA

1 CA0

1 1 3 0.0005

1 0.5

666 h

P1-12 (a) Initial number of rabbits, x(0) = 500 Initial number of foxes, y(0) = 200 Number of days = 500

dx dt

k1 x k2 xy …………………………….(1)

dy dt

k3 xy k4 y ……………………………..(2)

Given, 1

k1

0.02day

k2

0.00004 /(day

k3

0.0004 /(day rabbits )

k4

0.04day

foxes )

1

See Polymath program P1-12-a.pol.

POLYMATH Results Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value t

0

0

500

x

500

2.9626929

519.40024

4.2199691

y

200

1.1285722

4099.517

117.62928

k1

0.02

k2

4.0E-05

0.02 4.0E-05

500

0.02

0.02

4.0E-05

4.0E-05

1-17

k3

4.0E-04

k4

0.04

4.0E-04 0.04

4.0E-04 0.04

4.0E-04

0.04

ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(t) = (k1*x)-(k2*x*y) [2] d(y)/d(t) = (k3*x*y)-(k4*y)

Explicit equations as entered by the user [1] k1 = 0.02 [2] k2 = 0.00004 [3] k3 = 0.0004 [4] k4 = 0.04

When, tfinal = 800 and k3

0.00004 /(day rabbits )

1-18

Plotting rabbits Vs. foxes

P1-12 (b) POLYMATH Results See Polymath program P1-12-b.pol. POLYMATH Results NLES Solution Variable

Value

f(x)

Ini Guess

x

2.3850387 2.53E-11

2

y

3.7970279 1.72E-12

2

NLES Report (safenewt) Nonlinear equations [1] f(x) = x^3*y-4*y^2+3*x-1 = 0 [2] f(y) = 6*y^2-9*x*y-5 = 0

P1-13 Enrico Fermi Problem – no definite solution

P1-14 Mole Balance:

V=

FA0

FA rA 1-19

Rate Law :

rA

kC A2

Combine:

V=

FA0 FA

FA0 FA kC A2

v0C A v0C A

dm3 2molA 3 . s dm3 3

dm3 0.1molA . s dm3

mol s V= dm3 mol (0.03 )(0.1 3 ) 2 mol.s dm

6molA s 0.3molA s

(6 0.3)

19, 000dm3

1-20

The authors and the publisher have taken care in the preparation of this book but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions. No liability is assumed for the incidental or consequential damage in connection with or arising out of the use of the information or programs contained herein.

Visit us on the Web : www.prenhallprofessional.com

Copyright © 2011 Pearson Education,Inc .

This work is protected by United States copyright laws and is provided solely for the use of the instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including the World Wide Web ) will destroy the integrity of the work and is not permitted . The work and the materials from it should never be made available to the students except by the instructors using the accompanying texts in the classes. All the recipient of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of the other instructors who rely on these materials .

Solutions for Chapter 2 - Conversion and Reactor Sizing

P2-1.

This problem will keep students thinking about writing down what they learned every chapter.

P2-2.

This “forces” the students to determine their learning style so they can better use the resources in the text and on the CDROM and the web.

P2-3.

ICMs have been found to motivate the students learning.

P2-4.

Introduces one of the new concepts of the 4th edition whereby the students “play” with the example problems before going on to other solutions.

P2-5.

This is a reasonably challenging problem that reinforces Levenspiels plots.

P2-6.

Straight forward problem alternative to problems 7, 8, and 11.

P2-7.

To be used in those courses emphasizing bio reaction engineering.

P2-8.

The answer gives ridiculously large reactor volume. The point is to encourage the student to question their numerical answers.

P2-9.

Helps the students get a feel of real reactor sizes.

P2-10.

Great motivating problem. Students remember this problem long after the course is over.

P2-11.

Alternative problem to P2-6 and P2-8.

P2-12.

Novel application of Levenspiel plots from an article by Professor Alice Gast at Massachusetts Institute of Technology in CEE.

CDP2-A

Similar to 2-8

CDP2-B

Good problem to get groups started working together (e.g. cooperative learning).

CDP2-C

Similar to problems 2-7, 2-8, 2-11.

CDP2-D

Similar to problems 2-7, 2-8, 2-11. Summary

P2-1  P2-2  P2-3

Assigned O A A

Alternates

Difficulty

Time (min) 15 30 30

P2-4 P2-5  P2-6 P2-7 P2-8 P2-9  P2-10 P2-11 P2-12 CDP2-A CDP2-B CDP2-C CDP2-D

O O AA S AA S AA AA S O O O O

7,8,11 6,8,11

6,7,8 8,B,C,D 8,B,C,D 8,B,C,D 8,B,C,D

M FSF FSF SF SF SF SF M FSF FSF FSF FSF

75 75 45 45 45 15 1 60 60 5 30 30 45

Assigned  = Always assigned, AA = Always assign one from the group of alternates, O = Often, I = Infrequently, S = Seldom, G = Graduate level Alternates In problems that have a dot in conjunction with AA means that one of the problems, either the problem with a dot or any one of the alternates are always assigned. Time Approximate time in minutes it would take a B/B+ student to solve the problem. Difficulty SF = Straight forward reinforcement of principles (plug and chug) FSF = Fairly straight forward (requires some manipulation of equations or an intermediate calculation). IC = Intermediate calculation required M = More difficult OE = Some parts open-ended. ____________ * Note the letter problems are found on the CD-ROM. For example A CDP1-A. Summary Table Ch-2 Straight forward 1,2,3,4,9 Fairly straight forward 6,8,11 More difficult 5,7, 12 Open-ended 12 Comprehensive 4,5,6,7,8,11,12 Critical thinking P2-8

P2-1 Individualized solution. P2-2 (a) Example 2-1 through 2-3 If flow rate FAO is cut in half.

v1 = v/2 , F1= FAO/2 and CAO will remain same. Therefore, volume of CSTR in example 2-3,

1 FA0 X 2 rA

F1 X rA

V1

1 6.4 2

3.2

If the flow rate is doubled, F2 = 2FAO and CAO will remain same, Volume of CSTR in example 2-3, V2 = F2X/-rA = 12.8 m3 P2-2 (b) Example 2-4

Levenspiel Plot 4.5 4 3.5

Fao/-ra

3 2.5 2 1.5 1 0.5 0 0

0.2

0.4

Now, FAO = 0.4/2 = 0.2 mol/s, Table: Divide each term X [FAO/-rA](m3)

0.82

FA 0 rA

0.8

FA 0 in Table 2-3 by 2. rA

0 0.445

Reactor 1 V1 = 0.82m3 V = (FAO/-rA)X

0.6

Conversion

0.1 0.545

0.2 0.665

0.4 1.025

0.6 1.77

0.7 2.53

0.8 4

Reactor 2 V2 = 3.2 m3

X1

3.2

X1

By trial and error we get: X1 = 0.546 and X2 = 0.8 Overall conversion XOverall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673 P2-2 (c) Example 2-5 (1) For first CSTR, at X=0 ;

FA 0 rA

X2 X2

1

FA0 rA

1.28m3

at X=0.2 ;

FA0 rA

.94 m3

From previous example; V1 ( volume of first CSTR) = .188 m3 Also the next reactor is PFR, Its volume is calculated as follows 0.5

V2 0.2

FAO dX rA

0.247 m3 For next CSTR, X3 = 0.65,

FAO rA

2 m 3 , V3 =

FAO ( X 3 X 2 ) rA

(2) Now the sequence of the reactors remain unchanged. But all reactors have same volume. First CSTR remains unchanged Vcstr = .1 = (FA0/-rA )*X1 => X1 = .088 Now For PFR: X2

V 0.088

FAO dX rA

,

By estimation using the levenspiel plot X2 = .183 For CSTR,

.3m3

VCSTR2 =

FAO X 3 X 2 rA

0.1m3

=> X3 = .316 (3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller CSTR.

Conversion X1 = 0.20 X2 = 0.60 X3 = 0.65

Original Reactor Volumes V1 = 0.188 (CSTR) V2 = 0.38 (PFR) V3 = 0.10 (CSTR)

Worst Arrangement V1 = 0.23 (PFR) V2 = 0.53 (CSTR) V3 = 0.10 (CSTR)

For PFR, X1 = 0.2 X1

V1 0

FAO dX rA

Using trapezoidal rule, XO = 0.1, X1 = 0.1

X1

V1

XO rA

f XO

f X1

0.2 1.28 0.98 m3 2 0.23m3 For CSTR, For X2 = 0.6,

FAO 1.32m3 , rA

V2 =

FAO X2 rA

X 1 = 1.32(0.6 – 0.2) = 0.53 m3

For 2nd CSTR, For X3 = 0.65,

FAO rA

2m3 ,

V3 = 0.1 m3

P2-3 Individualized solution. P2-4 Solution is in the decoding algorithm given with the modules. P2-5

X 0 0.1 0.2 0.4 0.6 0.7 0.8 3 FAO/-rA (m ) 0.89 1.08 1.33 2.05 3.54 5.06 8.0 V = 1.6 m3 P2-5 (a) Two CSTRs in series For first CSTR, V = (FAo/-rAX1) X => X1 = 0.53 For second CSTR, V = (FAo/-rAX2) (X2 – X1) => X2 = 0.76

P2-5 (b) Two PFRs in series X1

V 0

FAo dX rA

X2

X1

FAo dX rA

By extrapolating and solving, we get X1 = 0.62 X2 = 0.84

P2-5 (c) Two CSTRs in parallel with the feed, FAO, divided equally between two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 V = (0.5FAO/-rAX1) X1 Solving we get, Xout = 0.68 P2-5 (d) Two PFRs in parallel with the feed equally divided between the two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 By extrapolating and solving as part (b), we get Xout = 0.88 P2-5 (e)

A CSTR and a PFR are in parallel with flow equally divided Since the flow is divided equally between the two reactors, the overall conversion is the average of the CSTR conversion (part C) and the PFR conversion (part D) Xo = (0.60 + 0.74) / 2 = 0.67 P2-5 (f) A PFR followed by a CSTR, XPFR = 0.50 (using part(b)) V = (FAo/-rA-XCSTR) (XCSTR – XPFR) Solving we get, XCSTR = 0.70 P2-5 (g) A CSTR followed by a PFR, XCSTR = 0.44 (using part(a)) X PFR

V X CSTR

FAO dX rA

By extrapolating and solving, we get

XPFR = 0.72

P2-5 (h) A 1 m3 PFR followed by two 0.5 m3 CSTRs, For PFR, XPFR = 0.50 (using part(b)) CSTR1: V = (FAo/-rA-XCSTR) (XCSTR – XPFR) = 0.5 m3 XCSTR = 0.63 CSTR2: V = (FAo/-rA-XCSTR2) (XCSTR2 – XCSTR1) = 0.5 m3 XCSTR2 = 0.72 P2-6 Exothermic reaction: A  B + C X 0 0.20 0.40 0.45 0.50 0.60 0.80 0.90 P2-6 (a)

r(mol/dm3.min) 1 1.67 5 5 5 5 1.25 0.91

1/-r(dm3.min/mol) 1 0.6 0.2 0.2 0.2 0.2 0.8 1.1

To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as given below. CSTR: Mole balance:

VCSTR

FA0 X rA

300mol / min 0.4 => 5mol / dm 3 . min

=>VCSTR = 24 dm3 PFR: X

V PFR

FA 0 0

Mole balance:

dX rA

= 300(area under the curve) VPFR = 72 dm3

P2-6 (b) For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is constant over this conversion range. .6

VPFR .4

FA 0 dX rA

FA0 .6 dX rA .4

P2-6 (c) VCSTR = 105 dm3 Mole balance: VCSTR

FA0 X rA

FA 0 X rA

.6 .4

105dm 3 300mol / min

X rA

0.35dm 3 min/ mol

Use trial and error to find maximum conversion. At X = 0.70,

1/-rA = 0.5, and X/-rA = 0.35 dm3.min/mol

Maximum conversion = 0.70 P2-6 (d) From part (a) we know that X1 = 0.40. Use trial and error to find X2.

FA 0 X 2 rA

Mole balance: V

X1 X2

Rearranging, we get

X2

0.40 rA

X2

At X2 = 0.64,

V FA 0

X2

0.008

0.40 rA

0.008

X2

Conversion = 0.64 P2-6 (e)

From part (a), we know that X1 = 0.40. Use trial and error to find X2.

X2

Mole balance: V PFR

72

FA 0 0.40

dX rA

X2

300

At X2 = 0.908, V = 300 x (area under the curve) => V = 300(0.24) = 72dm3 Conversion = 0.908.

0.40

dX rA

P2-6 (f) See Polymath program P2-6-f.pol.

P2-7 (a)

V

FS 0 X rS

FS0 = 1000 g/hr At a conversion of 40% Therefore V

1 rS

0.15

0.15 (1000)(0.40)

dm 3 hr g 60 dm 3

P2-7 (b) At a conversion of 80%,

1 rS

0.8

dm 3 hr g

FS0 = 1000 g/hr Therefore V

0.8 (1000)(0.80)

640 dm 3

P2-7 (c) X

VPFR

FS 0 0

dX rS

From the plot of 1/-rS Calculate the area under the curve such that the area is equal to V/FS0 = 80 / 1000 = 0.08 X = 12% For the 80 dm3 CSTR, V

80 dm 3

FS 0 X rS

X/-rs = 0.08. From guess and check we get X = 55%

P2-7 (d) To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to have a CSTR with a volume to achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of 1/-rs. Next is a PFR with the necessary volume to achieve the 80% conversion following the CSTR. This arrangement has the smallest reactor volume to achieve 80% conversion. For two CSTR’s in series, the optimum arrangement would still include a CSTR with the volume to achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of 1/-rs, first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR. P2-7 (e)

rs

kC S CC and C C K M CS

rs

kC S 0.1 C S 0 C S K M CS

1 rs

K M CS kC S 0.1 C S 0 C S

0.1 C S 0

CS

0.001

0.001

0.001

Let us first consider when CS is small.

CS0 is a constant and if we group together the constants and simplify then

1 rs

KM k1C S2

CS k 2 Cs

since CS < KM

1 rs

KM which is consistent with the shape of the graph when X is large (if CS is small X is k1C k 2 Cs 2 S

large and as CS grows X decreases). Now consider when CS is large (X is small) As CS gets larger CC approaches 0:

CC

0.1 C S 0

CS

0.001 and C S

CS 0

If

kC S CC then K M CS

rs

1 rs

K M CS kC S CC

As CS grows larger, CS >> KM

And

1 rs

CS kC S CC

1 kCC

And since CC is becoming very small and approaching 0 at X = 0, 1/-rs should be increasing with CS (or decreasing X). This is what is observed at small values of X. At intermediate levels of CS and X, these driving forces are competing and why the curve of 1/-rS has a minimum. P2-8 Irreversible gas phase reaction 2A + B  2C See Polymath program P2-8.pol. P2-8 (a) PFR volume necessary to achieve 50% conversion Mole Balance X2

V

FA0

dX X 1 ( rA )

Volume = Geometric area under the curve of (FA0/-rA) vs X)

V

1 400000 0.5 2

100000 0.5

V = 150000 m3 . P2-8 (b) CSTR Volume to achieve 50% conversion Mole Balance

V

V

FA 0 X ( rA ) 0.5 100000

V = 50000m3

P2-8 (c) Volume of second CSTR added in series to achieve 80% conversion

V2 V2

FA0 ( X 2 X 1 ) ( rA ) 500000 (0.8 0.5)

V2 = 150000m3 P2-8 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion

VPFR

(

1 400000 0.3) (100000 0.3) 2

VPFR = 90000m3

P2-8 (e) For CSTR, V = 60000 m3 (CSTR) Mole Balance

V

FA0 X ( rA )

60000

( 800000 X

500000 ) X

X = 0.463 For PFR, V = 60000 m3 (PFR) Mole balance X

V

FA0

dX ( rA ) 0 X

60000

( 800000 X 100000)dX 0

X = 0.134

P2-8(f) Real rates would not give that shape. The reactor volumes are absurdly large.

P2-9 Problem 2-9 involves estimating the volume of three reactors from a picture. The door on the side of the building was used as a reference. It was assumed to be 8 ft high. The following estimates were made: CSTR h = 56ft

d = 9 ft

V = πr2h = π(4.5 ft)2(56 ft) = 3562 ft3 = 100,865 L PFR Length of one segment = 23 ft Length of entire reactor = (23 ft)(12)(11) = 3036 ft D = 1 ft V = πr2h = π(0.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L Answers will vary slightly for each individual. P2-10 No solution necessary. P2-11 (a) The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in series and containing equal amounts of catalyst can be calculated from the figure below.

The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR. This figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR. See Polymath program P2-11.pol. P2-11 (b) Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining the area of the shaded region in the figure below.

The area of the rectangle is approximately 23.2 kg of catalyst. P2-11 (c) The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area of the shaded rectangle shown in the figure below.

The area of the rectangle is approximately 7.6 kg of catalyst.

P2-11 (d) The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the shaded region in the figure below.

The necessary catalyst weight is approximately 22 kg. P2-11 (e) The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from calculating the area of the shaded region in the graph below.

The necessary catalyst weight is approximately 13 kg.

P2-11 (f)

P2-11 (g) For different (-rA) vs. (X) curves, reactors should be arranged so that the smallest amount of catalyst is needed to give the maximum conversion. One useful heuristic is that for curves with a negative slope, it is generally better to use a CSTR. Similarly, when a curve has a positive slope, it is generally better to use a PBR. P2-12 (a) Individualized Solution P2-12 (b) 1) In order to find the age of the baby hippo, we need to know the volume of the stomach. The metabolic rate, -rA, is the same for mother and baby, so if the baby hippo eats one half of what the mother eats then Fao (baby) = ½ Fao (mother). The Levenspiel Plot is shown: Autocatalytic Reaction 5

4.5 4

mao/-raM1

3.5 3

Mother 2.5

Baby

2 1.5

1 0.5

0 0

0.2

0.4

Conversion

0.6

0.8

Vbaby

FaoX rA

1.36 *0.34 0.23m3 2

Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the baby’s stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.

Age

4.5 years 2

2.25 years

2) If Vmax and mao are both one half of the mother’s then

1 mA0mother 2 rAM 2

mAo rAM 2 and since

rAM 2

rAM 2baby mAo rAM 2

vmax C A KM CA

then

1 vmax C A 2 KM CA

baby

1 rAM 2mother 2

1 mAo 2 1 rAM 2 2

mAo rAM 2

mother

mother

mAo will be identical for both the baby and mother. rAM 2 Assuming that like the stomach the intestine volume is proportional to age then the volume of the intestine would be 0.75 m3 and the final conversion would be 0.40 P2-12 (c) Vstomach = 0.2 m3 From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:

mA0 = 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499 rAM 1

And since Vstomach =

mA0 X, rAM 1

solve V= 127X5 - 172.36X4 + 100.18X3 - 28.354X2 + 4.499X = 0.2 m3 Xstomach = .067. For the intestine, the Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178. Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive.

P2-12 (d) PFR CSTR PFR: Outlet conversion of PFR = 0.111

CSTR: We must solve V = 0.46 = (X-0.111)(127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499) X=0.42 Since the hippo gets a conversion over 30% it will survive. P2-13 For a CSTR we have : V=X

FA0 rA | X X f

So the area under the

FA 0 versus X curve for a CSTR is a rectangle but the height of rectangle rA

corresponds to the value of

FA 0 at X= Xf rA

But in this case the value of

FA 0 is taken at X= Xi and the area is calculated. rA

Hence the proposed solution is wrong.

CDP2-A (a) Over what range of conversions are the plug-flow reactor and CSTR volumes identical? We first plot the inverse of the reaction rate versus conversion.

Mole balance equations for a CSTR and a PFR:

CSTR: V

FA 0 X rA

X

PFR: V 0

dX rA

Until the conversion (X) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical. 0.5

i.e. VPFR 0

dX rA

FA0 0.5 dX rA 0

FA0 X rA

VCSTR

CDP2-A (b) What conversion will be achieved in a CSTR that has a volume of 90 L? For now, we will assume that conversion (X) will be less that 0.5. CSTR mole balance:

V

FA0 X rA

v0 C A 0 X X rA

V v0 C A 0 rA

3

5

m s

0.09m 3 3 mol 8 m .s 200 3 3 10 mol m

3 10

CDP2-A (c) This problem will be divided into two parts, as seen below:



The PFR volume required in reaching X=0.5 (reaction rate is independent of conversion).

V1 

FA0 X rA

v0 C A 0 X rA

1.5 1011 m 3

The PFR volume required to go from X=0.5 to X=0.7 (reaction rate depends on conversion).

13

Finally, we add V2 to V1 and get: Vtot = V1 + V2 = 2.3 x1011 m3 CDP2-A (d) What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to raise the conversion to 90 %

We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new value into our CSTR mole balance equation:

VCSTR

FA0 X rA

v0 C A 0 X rA

1.4 1011 m 3

CDP2-A (e) If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to the reactor, what length of time is necessary to achieve 40% conversion? Since there is no flow into or out of the system, mole balance can be written as: Mole Balance: rAV

dN A dt

Stoichiometry:

Combine: rAV

NA

N A0 (1 X )

N A0

dX dt

From the stoichiometry of the reaction we know that V = Vo(1+eX) and e is 1. We insert this into our mole balance equation and solve for time (t):

rA

V0 (1 X ) N A0

t

X

dt 0

C A0 0

dX dt

dX rA (1 X )

After integration, we have:

t

1 C A0 ln(1 X ) rA

Inserting the values for our variables: t = 2.02 x 1010 s That is 640 years. CDP2-A (f) Plot the rate of reaction and conversion as a function of PFR volume. The following graph plots the reaction rate (-rA) versus the PFR volume:

Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the conversion exceeds 50%.

The volume required for 99% conversion exceeds 4*1011 m3. CDP2-A (g) Critique the answers to this problem. The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the desired conversion, it would require a reactor of geological proportions (a CSTR or PFR approximately the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time.

CDP2-B Individualized solution

CDP2-C (a)

For an intermediate conversion of 0.3, Figure above shows that a PFR yields the smallest volume, since for the PFR we use the area under the curve. A minimum volume is also achieved by following the PFR with a CSTR. In this case the area considered would be the rectangle bounded by X =0.3 and X = 0.7 with a height equal to the CA0/-rA value at X = 0.7, which is less than the area under the curve.

CDP2-C (b)

CDP2-C (c)

CDP2-C (d)

For the PFR,

CDP2-C (e)

CDP2-D

CDP2-D (a)

CDP2-D (b)

CDP2-D (c)

CDP2-D (d)

CDP2-D (e)

CDP2-D (f)

CDP2-D (g)

CDP2-D (h)

CDP2-E

CDP2-F (a) Find the conversion for the CSTR and PFR connected in series. X -rA 0 0.2 0.1 0.0167 0.4 0.00488 0.7 0.00286 0.9 0.00204

1/(-rA) 5 59.9 204.9 349.65 490.19

CDP2-F (b)

CDP2-F (c)

CDP2-F (d)

CDP2-F (e)

CDP2-F (f)

CDP2-F (g) Individualized solution





3-1

Solutions for Chapter 3 – Rate Laws


P3-2 (a) Example 3-1 For, E = 60kJ/mol

k 1.32 1016 exp

T (K)

For, E = 240kJ/mol

60000 J RT

k (1/sec) 1/T

k 1.32 1016 exp

ln(k)

T (K)

240000 J RT

k (1/sec) 1/T

ln(k)

310 1023100 0.003226 13.83918

310 4.78E-25 0.003226 -56.0003

315 1480488 0.003175

315

14.2087

2.1E-24 0.003175 -54.5222

320 2117757 0.003125 14.56667

320 8.77E-24 0.003125 -53.0903

325 2996152 0.003077 14.91363

325 3.51E-23 0.003077 -51.7025

330 4194548

330 1.35E-22

0.00303 15.25008

335 5813595 0.002985 15.57648

0.00303 -50.3567

335 4.98E-22 0.002985 -49.0511

3-2

P3-2 (b) No solution will be given P3-2 (c) A+

1 1 B C 2 2

Rate law: rA

rA 1 kC

rB 1/2 kB

2

k A CA CB and kA

rC 1/2

1 dm 3 12.5 s mol

1 dm 3 25 s mol 2

25CA CB

2

k B CA 2CB 1/2

kC CA 2CB 1/2

2

P3-3(a) Refer to Fig 3-3 The fraction of molecular collisions having energies less than or equal to 50 Kcal is given by the area under the curve, f(E,T)dE from EA = 0 to 50 Kcal.

P3-3(b) The fraction of molecular collisions having energies between 10 and 20 Kcal is given by the area under the curve f(E,T) from EA = 10 to 20 Kcal. 3-3

P3-3(c) The fraction of molecular collisions having energies greater than the activation energy EA= 30 Kcal is given by the area under the curve f(E,T) from EA =30 to 50 Kcal.

P3-4 (a) Note: This problem can have many solutions as data fitting can be done in many ways. Using Arrhenius Equation For Fire flies: T(in K) 294

1/T 0.003401

Flashes/ min 9

ln(flashe s/min) 2.197

298

0.003356

12.16

2.498

303

0.003300

16.2

2.785

Plotting ln (flashes/min) vs. 1/T, We get a straight line.

See Polymath program P3-4-fireflies.pol. For Crickets: T(in K) 287.2

1/T 3 x10 3.482

chirps/ min 80

ln(chirps/ min) 4.382

293.3 300

3.409 3.333

126 200

4.836 5.298

Plotting ln (chirps/min) Vs 1/T, We get a straight line. Both, Fireflies and Crickets data follow the Arrhenius Model. ln y = A + B/T , and have the similar activation energy.

See Polymath program P3-4-crickets.pol.

P3-4 (b) For Honeybee: T(in K)

V(cm/s)

ln(V)

298

1/T 3 x10 3.356

0.7

-0.357

303

3.300

1.8

0.588

308

3.247

3

1.098

Plotting ln (V) vs. 1/T, almost straight line. 3-4

ln (V) = 44.6 – 1.33E4/T At T = 40oC (313K) At T = -5oC (268K)

V = 6.4cm/s V = 0.005cm/s(But bee would not be alive at this temperature)

See Polymath program P3-4-bees.pol.

P3-4 (c) For Ants: T(in K)

1/T x10

283

3

V(cm/s)

ln(V)

3.53

0.5

-0.69

293

3.41

2

0.69

303

3.30

3.4

1.22

311

3.21

6.5

1.87

Plotting ln (V) vs. 1/T, We get almost a straight line.

See Polymath program P3-4-ants.pol. So activity of bees, ants, crickets and fireflies follow Arrhenius model. So activity increases with an increase in temperature. Activation energies for fireflies and crickets are almost the same. Insect Cricket Firefly Ant Honeybee

Activation Energy 52150 54800 95570 141800

P3-4 (d) There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which would be helpful is the maximum and the minimum temperature that these insects can endure before death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it could be useless.

P3-5 There are two competing effects that bring about the maximum in the corrosion rate: Temperature and HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the column and consequently the rate of corrosion should increase. However, the HCN concentrations (and the HCN-H2SO4 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in the bottom of the column. These two opposing factors results in the maximum of the corrosion rate somewhere around the middle of the column.

3-5

P3-6 Antidote did not dissolve from glass at low temperatures.

P3-7 (a) If a reaction rate doubles for an increase in 10°C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1. Then with k = Ae-E/RT in general, k1

k2 k1

e

E 1 1 R T2 T1

ln

E R

or

Ae

E / RT1

k2 k1

1 T2

and k2

Ae

E / RT2

, or

k2 k1 T1 T2 T1T2

ln

1 T1

Therefore:

ln E

R

k2 k1

T1 T1 10 R

T2 T1

T1 T1 10

ln 2 T1 T1 10 10

10 E R ln 2

which can be approximated by T

10 E 0.5 R ln 2

P3-7 (b) Equation 3-18 is k

Ae

E RT

From the data, at T1 = 0°C, k1

k Dividing gives 2 k1

E

R ln

k2 k1

1 T2

1 T1

1.99 E

e

E 1 1 R T2 T1

Ae

E / RT1

, and at T2 = 100°C, k2

, or

RT1T2 k ln 2 T1 T2 k1

cal 273 K 373 K mol K .050 ln 100 K .001

7960

3-6

cal mol

Ae

E / RT2

A k1e

E RT1

7960 10 3 min 1 exp 1.99

cal mol

cal mol K

2100 min

1

273 K

P3-7 (c) Individualized solution P3-8 From the given data k= rA(dm3/mol.s) T(K) rA/(4*1.5) 0.002 300 0.00033333 0.046 320 0.00766667 0.72 340 0.12 8.33 360 1.38833333

1/T ln (k) 0.003333 -8.00637 0.003125 -4.87087 0.002941 -2.12026 0.002778 0.328104

Plotting ln(k) vs (1/T), we have a straight line:

Since, ln k = ln A - (

, therefore ln A = 41.99 and E/R = 14999 3-7

(a) Activation energy (E),

E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol

(b) Frequency Factor (A), ln A = 41.99 (c) k = 1.72X1018 exp(-

A = 1.72X1018

(1)

Given T0 = 300K Therefore, putting T = 300K in (1), we get k(T0) = 3.33X10-4 Hence, k(T) = k(T0)exp[

P3-9 (a) From the web module we know that

dX dt

k (1 x) and that k is a function of temperature, but not a

linear function. Therefore doubling the temperature will not necessarily double the reaction rate, and therefore halve the cooking time.

P3-9 (b) When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can only be 100°C. When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more than double that of boiling water.

P3-10 1) C2H6 → C2H4 + H2

Rate law: -rA = kC C2 H 6

2) C2H4 + 1/2O2 → C2H4O

1/ 2 Rate law: -rA = kCC2 H 4 C 02

3) (CH3)3COOC(CH3)3 → C2H6 + 2CH3COCH3 A → B + 2C Rate law: -rA = k[CA – CBCC2/KC] 4) n-C4H10 ↔ I- C4H10

Rate law: -rA = k[ C nC4 H10 – C iC4 H10 /Kc]

5) CH3COOC2H5 + C4H9OH ↔ CH3COOC4H9 + C2H5OH A + B ↔ C + D Rate law: -rA = k[CACB – CCCD/KC]

3-8

P3-11 (a) 2A + B → C (1) -rA = kCACB2 (2) -rA = kCB (3) -rA = k (4) -rA = kCACB-1

P3-11 (b) (1) H2 + Br2 → 2HBr

(2) H2 + I2 → 2HI

Rate law: -rHBr

1/ 2 k1C H 2 C Br 2 = C HBr k2 C Br2

Rate law: -rA = k1CH 2 CI2

P3-12 a)

we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition. If we assume the rate law for the reverse reaction (B->A) is =

then:

From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet

0, so:

Solving for KC gives:

3-9

b)

If we assume the rate law for the reverse reaction

is

then:


0, so:


c)

If we assume the rate law for the reverse reaction

is

then: 3-10


0, so:


P3-13 The rate at which the beetle can push a ball of dung is directly proportional to its rate constant, therefore -rA =c*k, where c is a constant related to the mass of the beetle and the dung and k is the rate constant k = A exp( From the data given -rA 6.5 13 18

T(K) 300 310 313

1/T 0.003333 0.003226 0.003195

ln k 1.871802 2.564949 2.890372

3-11

Refer to P3-8 (similar procedure) Therefore, A = 1.299X1011 E = 59195.68 J/mol k = 1.299X1011 exp(-7120/T) Now at T = 41.5 C = 314.5 K k = 19.12 cm/s Therefore, beetle can push dung at 19.12 cm/s at 41.5 C

P3-14 Since the reaction is homogeneous, which means it involves only one phase. Therefore,

So, option (4) is correct.

P3-15 Assuming the reactions to be elementary: 2 Anthracene -> Dimer



rA

2 k (C Anthracene

CDimer ) KC

where, Kc = k+/k-

Similarly for the second reaction:

3-12

Norbornadiene  Quadricyclane



rNorbornadiene

k (CNorbornadiene

CQuadricyclane KC

)

where, Kc = k+/k-

P3-16 The mistakes are as follows: 1. For any reaction, the rate law cannot be written on the basis of the stoichiometric equation. It can only be found out using experimental data. 2. In the evaluation of the specific reaction rate constant at 100o C the gas constant that should have been used was 8.314 J K-1 mol -1. 3. In the same equation, the temperatures used should have been in K rather than oC. 4. The units for calculated k(at 100oC) are incorrect. 5. The dimension of the reaction rate obtained is incorrect. This is due to the fact that the rate law that has been taken is wrong.

CDP3-A Polanyi equation: E = C – α(-ΔHR) We have to calculate E for the reaction CH3• + RBr  CH3Br + R• Given: ΔHR = - 6 kcal/mol From the given data table, we get 6.8 = C – α (17.5) And, 6.0 = C – α (20) => C = 12.4 KJ/mol and α = 0.32 Using these values, and ΔHR = - 6 kcal/mol, we get E = 10.48 KJ/mol

3-13





4-1

Solutions for Chapter 4 –Stoichiometry

P4-1

Individualized solution

P4-2 (a) Example 4-1 Yes, water is already considered inert. P4-2 (b) Example 4-2 For 20% Conversion CD = same as example & CB = CAo (ѲB – XA/3) = 10 (

-

) = 2.33 mol/dm3

–

) = 0 mol/dm3

For 90 % Conversion CD = same as example & CB = CAo (ѲB – XA/3) = 10 (

The final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore 90 % of caustic soda is possible.

P4-2 (c) Example 4-3 For the concentration of N2 to be constant, the volume of reactor must be constant. V = VO. Plot:

1 rA

0.5(1 0.14 X )2 (1 X )(0.54 0.5 X ) 1/(-ra) vs X 180

160

140

1/(-ra)

120

100

80

60

40

20

0 0

0.2

0.4

0.6

0.8

1

1.2

X

The rate of reaction decreases drastically with increase in conversion at higher conversions.

4-2

P4-2 (d) Example 4-4 POLYMATH Report

No Title 05-May-2009

Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

epsilon

-0.14

-0.14

-0.14

-0.14

2

FA0

1000.

1000.

1000.

1000.

3

k

9.7

9.7

9.7

9.7

4

K

930.

930.

930.

930.

5

KO2

38.5

38.5

38.5

38.5

6

KSO2

42.5

42.5

42.5

42.5

7

PO2

2.214

0.19639

2.214

0.1963901

8

PSO2

4.1

0.0115348

4.1

0.011535

9

PSO20

4.1

4.1

4.1

4.1

10 PSO3

0

0

4.754029

4.754029

11 rA

-0.0017387

-0.0063092

5.915E-08

-1.659E-09

12 w

0

0

5.0E+05

5.0E+05

13 x

0

0

0.9975796

0.9975795

Differential equations 1 d(x)/d(w) = -rA/FA0 Explicit equations 1

FA0 = 1000 mol.hr-1

2

PSO20 = 4.1

3

K = 930 atm-1/2

4

KSO2 = 42.5

5

KO2 = 38.5 atm-1

6

k = 9.7

7

epsilon = -0.14

8

PSO3 = PSO20*x/(1+epsilon*x)

9

PO2 = PSO20*(1.08-x)/2/(1+epsilon*x)

10 PSO2 = PSO20*(1-x)/(1+epsilon*x) 11 rA = -k*(PSO2*PO2^0.5 - PSO3/K)/(1+(PO2*KO2)^0.5+PSO2*KSO2)^2 mol/hr.g.cat

General

4-3

Total number of equations

12

Number of differential equations 1 Number of explicit equations

11

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

4-4

Therefore, catalyst weight required for 30% conversion is 1.611x105 g = 161.1 kg

4-5



P4-2 (e) Example 4-5 For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor. Therefore the reverse reaction decreases. CT0 = constant and inerts are varied. N 2O4 2NO2 A ↔ 2B Equilibrium rate constant is given by: KC Stoichiometry:

y A0

y A0 (2 1)

CB,e 2 CA,e

y A0

Constant volume Batch:

CA

N A0 (1 X) V0

CA0 (1 X)

and

CB

2N A 0 X V0

2CA 0 X

Plug flow reactor:

CA

FA0 (1 X) v 0 (1 X)

CA0 (1 X) and CB (1 X)

2FA0 X v 0 (1 X) 4-6

2CA0 X (1 X)

CAO

y AO PO RTO

y AO 0.07176 mol /dm 3

Combining: For constant volume batch:

KC

CB,e 2 CA,e

2 4CAo X2 CAO (1 X)

Xe

KC (1 X e ) 4CA0

Xe

K C (1 X e )(1 4C A0

For flow reactor:

KC

C B ,e

2

C A,e

2 4C Ao X2 C AO (1 X )(1

X)

See Polymath program P4-2-e.pol. POLYMATH Results NLES Report (safenewt) Nonlinear equations [1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao))^0.5 = 0 [2] f(Xef) = Xef - (kc*(1-Xef)*(1+eps*Xef)/(4*Cao))^0.5 = 0

Explicit equations [1] [2] [3] [4]

yao = 1 kc = 0.1 Cao = 0.07174*yao eps = yao

4-7

Xe)

Yinert

Yao 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95 0.956

Xeb

Xef

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.044

0.44 0.458 0.4777 0.5 0.525 0.556 0.5944 0.6435 0.71 0.8112 0.887 0.893

0.508 0.5217 0.537 0.5547 0.576 0.601 0.633 0.6743 0.732 0.8212 0.89 0.896

P4-3 Solution is in the decoding algorithm available separately from the author.

P4-4 A y A0 (a) C B0 (b) C A

CC

C A0

C A0 C A0

0.1

X

X 2 1

1 X 2 1 X

1 (c) C B

(d) C B

C A0

0.1

0.1

1 0.25 1 1 X 2

0.25 2 0.1 1 1 X 2

X

1 2

mol dm 3

1 X 1

1 1 B C 2 2 1 1 1 1 2 2 2

0.875

0.1 0.125 0.875

C A0 1 1

0.1 0.75

1 X 2

1 X 2

mol dm 3 4-8

C A0

0.086

0.0143

0.1

mol dm3

mol dm 3

mol dm 3

rA 4

(e)

rC , rA 2

4 mol dm 3 min

2rC

P4-5 (a) Liquid phase reaction, O

CH2--OH

CH2 - CH2 + H2O → CH2--OH A + B → C CAO = 16.13mol/dm3

CBO = 55.5 mol/dm3

Stoichiometric Table: Species Ethylene oxide Water Glycol

Rate law:

Symbol A

Initial CAO=16.13 mol/dm3

Change - CAOX

B

CBO= 55.5 mol/dm3, B =3.47

Remaining CA= CAO(1-X) = (1-X) lbmol/ft3

-CAOX

CB = CAO(

0

CAOX

=(3.47-X) lbmol/ft3 CC = CAOX = X lbmol/ft3

C

-rA = kCACB

Therefore,

2 -rA = k C AO (1-X) (

At 300K

E = 12500 cal/mol, k = 0.1dm3/mol.s

CSTR

=

B

-X) = k (16.13)2(1-X) (3.47-X) X = 0.9,

16.13 0.9 C AO X = 2 rA 0.1 16.13 1 0.9 3.47 0.9

2.186sec

and, V = τ X FA0 = 2.186 X 200 liters = 437.2 liters At 350K, k2 = k exp((E/R)(1/T-1/T2))= 0.1exp((12500/1.987)(1/300-1/350)) = 19.99 dm3/mol.s Therefore, CSTR

=

16.13 0.9 C AO X = 2 rA 19.99 16.13 1 0.9 3.47 0.9

and, V = τ X FA0 = 0.109 X 200 liters = 21.8 liters

4-9

0.109sec ,

B

-X)

P4-5 (b) Isothermal, isobaric gas-phase pyrolysis, C2H6 C2H4 + H2 A  B + C Stoichiometric table: Species C2H6 C2H4 H2

symbol A B C

Entering FAO 0 0 FTO=FAO

Change -FAOX +FAOX +FAOX

Leaving FA=FAO(1-X) FB=FAOX FC=FAOX FT=FAO(1+X)

= yao = 1(1+1-1) = 1 v = vo(1+ X) => v = vo(1+X)

P RT 1 6atm

CAO = yAO CTO = yAO =

= 0.067 kmol/m3 = 0.067 mol/dm3

3

0.082

m atm K .kmol

1100 K 1 X 1 X

CA =

FA v

FAO (1 X ) vO (1 X )

CB =

FB v

FAO ( X ) vO (1 X )

C AO

X mol/dm3 1 X

CC =

FC v

FAO ( X ) vO (1 X )

C AO

X mol/dm3 1 X

C AO

mol/dm3

Rate law: -rA = kCA= kCAO

1 X 1 X

=0.067 k

1 X 1 X

If the reaction is carried out in a constant volume batch reactor, =>( = 0) CA = CAO(1-X) mol/dm3 CB = CAO X mol/dm3 CC = CAO X mol/dm3

P4-5 (c) Isothermal, isobaric, catalytic gas phase oxidation,

1 O2  C2H4O 2 1 + B  C 2

C2H4 + A

4-10

Stoichiometric table:

B

=

FBO FAO y AO

C AO

CA

1 FAO 2 FAO

CB

FB v

CC

FC v

Symbol A B

Entering FAO FBO

Change -FAOX

C2H4O

C

0

+FAOX

1 2

2 1 1 1 3 2

y AO CTO

FA v

Species C2H4 O2

y AO

P RT

FAO 1 X vO 1 X FAO

B

vO 1

X 2 X

FAO X vO 1 X

y AO

FAO FTO

-

Leaving FA=FAO(1-X)

1 FB=FAO( FAOX 2

FAO FAO FBO

B

-X/2)

FC=FAOX

2 3

0.33

2 3

6atm 0.082

atm.dm mol.K

CAO 1 X 1 0.33 X

0.092

3

533K

mol dm3

0.092 1 X 1 0.33 X

0.046 1 X 1 0.33 X

0.092 X 1 0.33 X

If the reaction follow elementary rate law Rate law: rA

kC AC

0.5 B

rA

0.092 1 X k 1 0.33 X

0.046 1 X 1 0.33 X

P4-5 (d) Isothermal, isobaric, catalytic gas-phase reaction in a PBR C6H6 + 2H2 → C6H10 A + 2B → C Given: 0

50 dm 3 / min

Stoichiometric Table: Species C6H6 H2 C6H10

Symbol A B C

Entering FA0 FB0=2FA0 0

Change -FA0X -2FA0X FA0X

4-11

Leaving FA=FA0(1-X) FB=FA0(θB-2X) Fc=FA0X

0.5

B

C A0

CA CB CC

FB 0 FA 0

2 FA 0 FA 0

2

P 1 RT 3

CT 0 y A0

FA 0 FA0 FB 0

1 3

6atm 1 0.0821 * (443K ) 3

FA0 (1 X ) X) 0 (1

FB

FA0 ( B 2 X ) X) 0 (1

C A0

FA0 X X) 0 (1

1 (1 2 1) 3

2 3

mol dm3

(1 X ) (1 23 X )

C A0

C A0

y A0

0.055

dm3 atm mol K

FA

FC

FA 0 FT 0

y A0

(2 2 X ) (1 23 X )

2 * C A0

(1 X ) (1 23 X )

X 2 3

(1

X)

Rate Law: NOTE: For gas-phase reactions, rate laws are sometimes written in terms of partial pressures instead of concentrations. The units of the rate constant, k, will differ depending on whether partial pressure or concentration units are used. See below for an example.

rA ' kPA PB

mol kgcat min

2

mol * atm * atm 2 3 kgcat min atm

Notice that if you use concentrations in this rate law, the units will not work out.

PA

y A0 P

rA

( y A0 C ) * RT

kPA PB

2

2

C A RT

kC AC B ( RT ) 3

4kC A0

3

(1 X ) 3 ( RT ) 3 (1 23 X ) 3

Design Equation for a fluidized CSTR:

W

W W

FA0 X rA '

FA0 X (1

2 3

X )3

3

4kC A0 (1 X ) 3 ( RT ) 3 0

X (1

2 3

X )3

2

4kC A0 (1 X ) 3 ( RT ) 3

Evaluating the constants:

k

53

mol at 300 K kgcat min atm3

At 170°C (443K),

4-12

k 443

k 300

EA 1 R T300

1 T443

J 1 mol 53 J 300 K 8.314 mol K 80000

1 443K

1663000

mol kgcat min atm

Plugging in all the constants into the design equation: X = 0.8 3

W

3 2 50 dm min 0.8 (1 3 0.8) 3 mol mol 2 atm 4 1663000 kgat min (0.055 dm 0.8) 3 (0.0821 dm 443K ) 3 3 ) (1 mol K atm3

5.25 10 7 kgcat

At 270°C (543K),

k 543

k 300

EA 1 R T300

1 T543

J 1 mol 53 J 300 K 8.314 mol K 80000

1 543K

9079000

mol Plu kgcat min atm

gging in all the constants into the design equation: X = 0.8 3

W

3 2 50 dm min 0.8 (1 3 0.8) 3 3 mol mol 2 atm 4 9079000 kgat min (0.055 dm 0.8) 3 (0.0821 dm 3 ) (1 mol K 543K ) atm3

5.22 10 8 kgcat

P4-6 (a) Let A = ONCB B = NH3

A + 2B -rA

C = Nibroanaline D = Ammonium Chloride

C+D

kC ACB

P4-6 (b) Species A B C D

Entering FA0 FB0 = ΘBFA0 =6.6/1.8 FA0 0 0

Change - FA0X -2 FA0X FA0X FA0X

P4-6 (c) For batch system, CA=NA/V

-rA = kNANB/V2

P4-6 (d) 4-13

Leaving FA0(1-X) FB= FA0(ΘB – 2X) FC= FA0X FD=FA0X

-rA

kC ACB

FA

NA V

NA V0

N A0 1 X V0

FB

NB V

NB V0

N A0 V0

kC A2 0 1 X

rA B

C A0

rA

CB 0 C A0

6.6 1.8

1.8

kmol m3 2

k 1.8

B

B

FA v

FA v0

C A0 1 X

2 X , CB

FB v0

C A0

C A0 1 X , C A

2X

C A0

B

B

2X

3.67

1 X

3.67 2 X

P4-6 (e) 1) At X = 0 and T = 188°C = 461 K

rA0 rA0

kC

2 A0

0.0202

m3 kmol 0.0017 1.8 3 kmol min m

B

2

3.67

kmol m3 min

2) At X = 0 and T = 25C = 298K

k

k O exp

E 1 R TO

1 T

cal m 1 mol k 0.0017 exp cal kmol. min 461 1.987 mol.K m3 2.03 10 6 kmol. min 11273

3

1 298

-rAO = kCAOCBO = 2.41 X 10-5 kmol/m3min 3)

k

k0 exp

E 1 R T0

1 T

4-14

0.0202

kmol m 3 min

2X

k

cal m 1 mol 0.0017 exp cal kmol min 461 K 1.987 mol K

k

0.0152

11273

3

rA0

1 561 K

m3 kmol min

kC A0CB 0

rA

0.0152

m3 kmol 1.8 3 kmol min m

rA

0.1806

kmol m3 min

6.6

kmol m3

P4-6 (f) rA = kCAO2(1-X)(θB-2X) At X = 0.90 and T = 188C = 461K 1) at T = 188 C = 461 K

rA

m3 0.0017 kmol. min

0.00103

kmol 1.8 3 m

2

1 0.9 3.67 2 0.9

kmol m 3 min

2) at X = 0.90 and T = 25C = 298K

rA

2.03 10

1.23 10

6

6

m3 kmol. min

kmol 1.8 3 m

2

1 0.9 3.67 2 0.9

kmol m 3 min

3) at X = 0.90 and T = 288C = 561K

rA

m3 0.0152 kmol. min

0.0092

kmol 1.8 3 m

2

1 0.9 3.67 2 0.9

kmol m 3 min

P4-6 (g) FAO = 2 mol/min 1) For CSTR at 25oC -rA

1.23 10

6

kmol m 3 min

4-15

FAO 1 X rA X 0.9

V

2mol / min 0.1 mol 1.23 10 3 3 m min 2)At 288oC, -rA

V

0.0092

162.60m 3

kmol m 3 min

FAO 1 X rA X 0.9 2mol / min 0.1 mol 0.0092 3 m min

21.739m 3

P4-7 C6H12O6 + aO2 + bNH3 → c(C4.4H7.3N0.86O1.2) + dH2O + eCO2 To calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients a, b, c, d, and e. This can be done with mass balances on each element involved in the reaction. Once all the coefficients are found, you can then calculate the yield coefficients by simply assuming the reaction proceeds to completion and calculating the ending mass of the cells.

P4-7 (a) Apply mass balance For C 6 = 4.4c + e For N b = 0.86c

For O For H

6 + 2a = 1.2c + d + 2e 12 + 3b = 7.3c + 2d

Also for C, 6(2/3) = 4.4c which gives c = 0.909 Next we solve for e using the other carbon balance 6 = 4.4 (0.909) + e e=2 We can solve for b using the nitrogen balance b = 0.86c = 0.86* (0.909) b = 0.78 Next we use the hydrogen balance to solve for d 12 + 3b = 7.3c + 2d 12 + 3(0.78) = 7.3(0.909) + 2d 4-16

d = 3.85 Finally we solve for a using the oxygen balance 6 + 2a = 1.2c + d + 2e 6 + 2a = 1.2(0.909) + 3.85 + 2(2) a = 1.47

P4-7 (b) Assume 1 mole of glucose (180 g) reacts: Yc/s= mass of cells / mass of glucose = mass of cells / 180 g mass of cells = c*(molecular weight) = 0.909 mol* (91.34g/mol) mass of cells = 83.12 g Yc/s = 83.12 g / 180 g Yc/s = 0.46 Yc/o2 = mass of cells / mass of O2 If we assume 1 mole of glucose reacted, then 1.47 moles of O2 are needed and 83.12 g of cells are produced. mass of O2 = 1.47 mol * (32 g/mol) mass of O2 = 47.04 g Yc/o2 = 83.12 g /47.04 g Yc/o2 =1.77

P4-8 (a) Isothermal gas phase reaction.

1 N2 2

3 H2 2

NH 3

Making H2 as the basis of calculation:

H2 A

1 N2 3 1 B 3

2 NH 3 3 2 C 3

Stoichiometric table: Species H2 N2

Symbol A B

Initial FAO

NH3

C

0

FBO=

B

FAO

4-17

change -FAOX -FAOX/3

Leaving FA=FAO(1-X)

+2FAOX/3

FC=(2/3)FAOX

FB=FAO(

B

-X/3)

P4-8 (b) 2 1 1 3 3 y AO C AO

2 3 2 3

0.5

16.4atm

0.5 0.082

CH 2

CNH3

CA

CC

1 3

atm.dm mol.K

C AO 1 X 1 X

= 0.2 mol/dm3

3

500 K 0.2 1 X X 1 3

2 C AO X 3 1 X

0.1mol / dm3

0.2 X X 1 3

2 3

0.1mol / dm3

P4-8 (c) kN2 = 40 dm3/mol.s (1) For Flow system: 1

rN 2

k N2 CN2

2

3

CH 2 1

40 C AO

X 1 3 X 1 3

2

3

rN 2

1.6

2 3

2

2

1 X X 1 3

2

1 X X 1 3

(2) For batch system, constant volume.

4-18

1

rN2

k N2 CN2

CA

NA V

CA

C A0 1 X

CB

NB V

3

2

CH 2

2

N A0 1 X V0

NA V0

N A0

B2

X 3

V0 X 3

C H 2O 1

1

rN2

40 C A0

X 3

1.6 1

X 1 3

2

1

2

1 X

2

3

1 X 3

2

2

P4-9 (a) Liquid phase reaction  assume constant volume Rate Law (reversible reaction):

rA

k C AC B

CC KC

Stoichiometry:

CA

C A 0 1 X , CB

C A0 1 X , CC

C A0 X

To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe.

C AC B K C

CC

C A2 0 1 X e

X e2

2

Xe

0.80

2

KC

C A0 X e

1 Xe 1 0 C A0 K C

To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric relations.

CA

C A0 1 X

CB

C A0 1 X

mol 1 0.80 dm3 mol 2 3 1 0.80 dm 2

mol dm3 mol 0.4 3 dm 0.4

4-19

CA

C A0 X

2

mol mol *0.80 1.6 3 3 dm dm

P4-9 (b) Stoichiometry:

y A0

1 3 1

2 and

N A0 1 X

CA

NA V

V0 1

CC

NC V

3 N A0 X V0 1 X

C A0

X

C A0

0

C

1 X 1 2X

3X 1 2X

Combine and solve for Xe.

K C C A0

3

1 Xe

C A0

1 2Xe

KC 1 X e 1 2 X e

4 Xe

2

3X e 1 2Xe 27CA2 0 X e3

27C A2 0 X e3 3 X e 1 0 KC 0.58

Equilibrium concentrations:

C A0

P0 RT0

CA

0.305

CC

10 atm dm3 atm 400 K 0.082 mol K

1 0.58 1 2 0.58

3 0.58 0.305 1 2 0.58

0.059 0.246

0.305

mol dm3

mol dm3

mol dm3

P4-9 (c) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction. Stoichiometry:

CA

NA V

N A0 1 X V0

C A0 1 X 4-20

CC

NC V

3 N A0 X V0

3C A0 X

Combine and solve for Xe

KC C A0 1 X e Xe

3

3CA0 X e

0.39

Equilibrium concentrations

CA

0.305 1 0.39

CC

0.305 0.39

mol dm3 mol 0.36 3 dm 0.19

P4-9 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction):

rA

CC3 KC

k CA

Stoichiometry:

y A0

1 3 1

2 and

N A0 1 X

CA

NA V

V0 1

CC

NC V

3 N A0 X V0 1 X

1 X

C A0

X

0

C

1 2X

3C A0

X 1 2X

Combine and solve for Xe:

K C C A0 1 X e 1 2Xe Xe

3C A0 X e 1 2Xe

3

0.58

Equilibrium concentrations:

CA

0.305 1 0.58 1 2 0.58

0.059

mol dm3 4-21

3 0.305 0.58

CC

0.246

1 2 0.58

mol dm3

P4-10 Given: Gas phase reaction A + B  8C in a batch reactor fitted with a piston such that V = 0.1P0

k 1.0

ft 3

2

lb mol 2 sec

kC A2 CB

rA

NA0 = NB0 at t = 0 V0 = 0.15 ft3 T = 140°C = 600°R = Constant

P4-10 (a) N A0 N A0 N B 0

y A0

0.5

8 1 1 6 y A0 3 Now

V

Therefore

NA

rA

V0 P0 T P T0

1

10V02 1 10V

V

N A0 1 X

2 A

kC CB

kN A2 N B V3

X

and

X

T T0

1 , P0

or

V2

NB

N A0

kN A3 0 1 X V03 1

X

B

10V0 , and P 10V

V02 1

X

X

B

3

N A0

3 2

Therefore

4-22

N B0 N A0

y A0 P0 V0 RT

1

1 X

y P k A0 0 RT

rA

rA

5.03*10

1

X

1 X

9

1 3X

3 3 2

3 3 2

lb mol ft 3 sec

P4-10 (b) V2

V02 1

0.22

X rA

X

0.152 1

X

0.259 8.63*10

10

lb mol ft 3 sec

P4-11 (1) For any reaction ,we cannot write the rate law on the basis of the stochiometric equation. The rate law is to be obtained from the experimental data. It has been mentioned as an elementary reaction in the problem statement but in the proposed solution the rate law is based on the reaction equation that has been divided by stoichiometric coefficient of A. (2) The value of calculated is incorrect. = yA0δ = 0.6 (3+5-2-3) = 1.8 is the correct value. (3) The expression for CA and CB will therefore be, C A0 1 X CA 1 X C A0 1 CB

1

2 X 3 X

(4) According to the system of units being used in the calculations, R = 0.0821 atm. Liter/ mol & Temperature = 700 K should be used.

4-23





5-1

Solutions for Chapter 5: Isothermal Reactor Design- Conversion


P5-2 (a) Example 5-1 There would be no error! The initial liquid phase concentration remains the same.

P5-2 (b) Example 5-2: For n equal sized CSTRs (volume Vi each) with equal feed FAi0 = FA0/n CSTR mole balance on any reactor:

Since FAi0 = FA0/n

rA= kCA= kCAO(1-X) (1)

For single CSTR with volume nVI and feed rate FA0

CSTR mole balance:

rA= kCA= kCAO(1-X) (2)

From equation (1) and (2), Xi = X. Therefore, we can generalize that the conversion achieved in any one of the parallel reactors with volume V i and feed rate of FA0/n will be identical to the conversion achieved in one large CSTR of volume nVi and feed rate FA0.

P5-2 (c) Example 5-3 For 50% conversion, X = 0.5 and k = 3.07 sec-1 at 1100 K (from Example 5-3) 5-2

FB = 200X106 / (365 X 24 X 3600 X 28) lbmol/sec = 0.226 lbmol/sec F 0.026 FAO = B 0.452lbmol / sec Also, C AO 0.00415lbmol / ft 3 X 0.5 Now, we have from the example

V

FA0 kC A0

1

ln

1 1 X

X

0.452lbmol / sec 3.07sec 1 X 0.00415lbmol / ft 3

1 1 ln

1 1 0.5

1X 0.5

= 35.47 X 0.886 ft3 = 31.44 ft3 Now, n = 31.44 ft3/0.0205 ft2 X40 ft = 38.33 So, we see that for lower conversion and required flow rate the volume of the reactor is reduced.

P5-2 (d) Example 5-4 New Dp = 3D0/4 Because the flow is turbulent o

1 Dp

1

2

y

Now, 1

2

1

o

L

Po

Dp 2 D p1

2 L Po

0.0775

1 2

1 0.75

0.1033

atm 2 X 0.103 X 60 ft ft 1 10atm

1 2

0.24

1 2

0 , so too much pressure drop P = 0 and the flow stops.

P5-2 (e) Example 5-5 For P = 60atm, CAO = 0.0415 lbmol/ft3

( C AO

y AO PO RTO

60 ) 0.73 1980

5-3

Using equation E-4-3.6, for X = 0.8 We see that the only thing that changes is CA0 and it increases by a factor of 10, therby decreasing the volume by a factor of 10.

1 P

V

P5-2 (f) Example 5-6 For turbulent flow

1 and Dp

2

1

DP1 DP 2

1 P0

P01 P02

1

1 1 5

1 5

1

Therefore there is no change.

P5-2 (g) Example 5-3, Using ASPEN, we get (Refer to Aspen Program P5-2g from polymath CD) (1) At 1000K, for the same PFR volume we get only 6.2% conversion. While at 1200K, we get a conversion of nearly 100%. This is because the value of reaction constant ‘k’ varies rapidly with reaction temperature. (2) Earlier for an activation energy of 82 kcal/mol we got approx. 81% conversion. For activation energy of 74 kcal/mol keeping the PFR volume the same we get a conversion of 71.1%. While for an activation energy of 90 kcal/mol we get a conversion of 89.93%. (3) On doubling both flow rate and pressure we find that the conversion remains the same.

P5-2 (h) Individualized solution. P5-2 (i) Individualized solution.

P5-3 Solution is in the decoding algorithm given with the modules.

5-4

P5-4 We have to find the time required to cook spaghetti in Cuzco, Peru. Location Ann Arbor Boulder Cuzco

Elevation (km) 0.21 1.63 3.416

Pressure (mm Hg) 739 625 493

Boiling Point (°C) 99.2 94.6 88.3

Time (min) 15 17 ?

Assume reaction is zero order with respect to spaghetti conversion: E

rA

dC A dt

Ae RT

k

so that CA

CC

Ae

E RT

t

A0

For complete conversion (i.e.: well cooked) CA = 0 at time t.

Therefore

C A0 C A0 A

ln

tAe te

C A0 A

E RT

E RT

ln k

ln t

E 1 R Tb

Now, plot the natural log of the cooking time versus 1/Tb and get a linear relationship. Extrapolation to Tb = 88.3°C = 361.45 K yields t = 21 minutes.

5-5

P5-5 (a) The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal. So, we get 0.57 as conversion instead of 0.5.

5-6

P5-5 (b) A CSTR is been created at the bend due to back mixing, so the effective arrangement is a PFR is in series with a CSTR.

CSTR zone due to back mixing

A→B k = 5 min-1 vo = 5 dm3/min. Xexpected = 0.632 Xactual = 0.618 X

V 0

FAO dX rA

vo 1 5 1 = ln ln k 1 X Expected 5 1 .632

1.0

Now, For PFR,

VP

For CSTR, Also,

ln

FAO ( X actual rA

VC

VP V

1 ………………………………….(1) 1 X1

VC V

X1 )

X actual X 1 ……………(2) 1 X actual

1 ………………………………………..(3)

Solving 1, 2 and 3 by using polymath,

See Polymath program P5-5-b.pol. Calculated values of NLE variables Variable Value f(x) Initial Guess 1 Vc

0.3001282 -2.831E-15 1. 5-7

2 X1

0.503351 7.864E-11 0

Variable Value 1 V

1.

2 Vp

0.6998718

3 X2

0.618

Nonlinear equations 1 f(X1) = ln(1/(1-X1)) -Vp = 0 2 f(Vc) = (X2-X1)/(1-X2) - Vc = 0 Explicit equations 1 V=1 2 X2 = .618 3 Vp = V - Vc General Settings Total number of equations

5

Number of implicit equations 2 Number of explicit equations 3 Elapsed time

0.0000 sec

Solution method

SAFENEWT

Max iterations

150

Tolerance F

0.0000001

Tolerance X

0.0000001

Tolerance min

VC = 0.3 dm

0.0000001 3

; VP = 0.7 dm3 ;

X1 = 0.5

P5-5 (c) CAO = 2 mol/dm3 A B Assuming 1st order reaction,

C AO X rA

For CSTR,

-rA = kCA = kCAO(1-X) =>

X 1 X

k

X

For PFR,

V

FA0 0

=> X PFR

0.4 0.6

0.67

dX , k kC A0 1 X

1 exp(

X

dX 1 X 0

k ) =1-exp(-0.67) = 0.486

Now assuming 2nd order reaction, 5-8

For CSTR,

Now, assuming 2nd order reaction,

C AO X rA

For CSTR,

rA

2 kC A2 = kC AO 1 X

X

=> kC AO

1 X

1 kC AO

For PFR,

=> X

1

1

X

0

0.4 0.62

2

dX 1 X 1 kC AO

2

1.111

1 X kC AO 1 X

2

1

1 .526 2.111

So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second order reaction.

P5-5 (d) A graph between conversion and particle size is as follows: Originally we are at point A in graph, when particle size is decreased by 15%, we move to point C, which have same conversion as particle size at A. But when we decrease the particle size by 20%, we reach at point D, so a decrease in conversion is noticed. Also when we increase the particle size from position A, we reach at point B, again there is a decrease in the conversion.

5-9

P5-6 (a)

(3) 50% (Liquid phase reactions do not depend on pressure)

(b)

(2) <50% (Low concentration if we lower the pressure) 2

kC A0 1 X

rA

(c)

1

X

2

, C A02

2

1 C A01 , low rate 2

(1) >0.234 2

rA

kC A0 1 X

rA

ky 2

y2

y2

1

k true y 2 k true

2

k 0.234 0.234 0.234 y2

P5-7 (a)

Ans. (1) X > 0.5

~

1 D

Increase D, decrease , increase X. (b) Ans. (3) Xe = 0.75

Xe

(c)

KC 1 KC

3 1 3

Ans. (3) remain the same

5-10

0.75

KC

CA C Ae

C A0 1 X e y 1 X e C A0 X e y Xe

Xe is not a function of . (d) Ans. (3) remain the same Xe is not a function of . (e)

Ans. (4) insufficient information to tell Xe dp

P5-8 (a)

X

V kC A2 0 1 X

2

10

5-11

dX dV

X 1 X

rA FA0

2 kC AO (1 X )2 FA0

kC A2 0V FA0

X=0.85 So, considering the above results, we will choose a CSTR.

P5-8 (b)

P5-8 (c)

P5-8 (d) 1) CSTR and PFR are connected in series:

5-12

(200dm3 )(0.07dm3 / mol.min)(1mol / dm3 ) 2 (1 X ) 2 10mol / min

X CSTR

Solving the quadratic equation, XCSTR = 0.44 For PFR,

FA0

dX dV

rA

(0.07dm3 / mol / min)C AO (1 X )(1 X )2 dV 10 mole / min

dX

X

(0.07dm3 / mol / min)(1mol / dm3 )2 (800dm3 ) 10mole / min

dX (1 X )3 0.44

X

0.736

2) when CSTR and PFR are connected in parallel,

X CSTR

(200dm3 )(0.07dm3 / mol.min)(1mol / dm3 ) 2 (1 X ) 2 5mol / min .

XCSTR = 0.56 For PFR, X

0

dX (1 X )2

(0.07dm3 / mol.min)(1mol / dm3 )2 (800dm3 ) 5mol / min

XPFR = 0.92 Hence, final conversion X =

0.56 0.92 =0.74 2

P5-8 (e) To process the same amount of species A, the batch reactor must handle

5dm3 2M min

60 min hr

24h day

14400

mol day

If the reactants are in the same concentrations as in the flow reactors, then

V

14400

mol day

1dm3 mol

14400

dm3 day

So the batch reactor must be able to process 14400 dm3 every 24 hours.

5-13

Now we find the time required to reach 90% conversion. Assume the reaction temperature is 300K.

rAV N A0

dX dt

kC A2 0 1 X V N A0

tR

N N A0 X , and since A0 2 V VkC A0 1 X

tR

1 X kC A0 1 X

C A0

1

0.9 dm mol 0.1 4.2 *1 mol hr dm3 3

2.14hr

Assume that it takes three hours to fill, empty, and heat to the reaction temperature. tf = 3 hours ttotal = tR + tf ttotal = 2.14hours + 3 hours = 5.14 hours. Therefore, we can run 4 batches in a day and the necessary reactor volume is

14400dm3 4

3600dm3

Referring to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be approximately $85,000 for the reactor.

P5-8 (f) The points of the problem are: 1) To note the significant differences in processing times at different temperature (i.e. compare part (b) and (c)). 2) That the reaction is so fast at 77°C that a batch reactor is not appropriate. One minute to react and 180 to fill and empty. 3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black.

P5-9 PFR

dX dV

rA FA0

kCA0 1 X C A0 0 1 X 5-14

k 1 X X 0 1

X

1

X

0

1 X 1

k

dX

1

ln

1

y A0

k

1

2

2s

11 1 1

1

1

1 1 ln

0.8

1 0.8

2 1.61 0.8

2

X

1 X

5 dm 3 s 1

k

0

10dm 3

k

dV

k

1.209 at 300K

CSTR

FA0 X

V

0C A0 X

rA

0X

1 X

kC A0

1

1 2

1

X

k1 X

X

1 1 1 0.5

k

1 X1

X

1 X

1 0.8 1 0.5 0.8 2

0.2

2.8

ln

ln

k2

E 1

1

k1

R T1

T2

1

1

E

2.8

E

1.21

R 300

E R

300 320 20

ln

320 2.8 1.21

E 8010 5-15

20

R 320 300 , R 1.987 cal

mol

cal mol K

P5-10 A

kC A0 1 X X 0C A0 1

dX dV kV

3B

1

ln

0

1 X

1 3 1 2

y A0 kV V0

1

1 1 ln

FA0 X 2 X1 rA2

V

X

0C A0

V

kC A0

1

X 2 X1 1 1 X2

k

1 0.5 1 0.5

X 22

k 0.886

X2 1

X 2 X1 1 X2 1 X2

X2

X 2 X1 1 X 2 1 X2

k X1

X1

k

0

X 22 1.386X 2 1.386 0 X2

1.386X 2

1.386 2

4

1.386

2 X2

0.673

P5-11 (a) POLYMATH Report Ordinary Differential Equations

04-May-2009

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 w

0

0

2000.

2000.

2 x

0

0

0.9978192

0.9978192

3 FBuO

50.

50.

50.

50.

4 PBuO

10.

10.

10.

10.

5 PBu

10.

0.0109159

10.

0.0109159

6 k

0.054

0.054

0.054

0.054

7 K

0.32

0.32

0.32

0.32

8 rBu

-0.0306122

-0.0421872

-0.0005854

-0.0005854

5-16

Differential equations 1 d(x)/d(w) = -rBu / FBuO Kgcat-1

Explicit equations 1 FBuO = 50 Kmol.hr-1

2 PBuO = 10 atm

3 PBu = PBuO * (1 - x) / (1 + x) atm

4 k = 0.054 kmol.kgcat-1.hr-1.atm-1

5 K = 0.32 atm-1

6 rBu = -k * PBu / (1 + K * PBu) ^ 2 Kmol.kgcat-1.hr-1

General Total number of equations

7


6

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001


5-17

5-18

Hence, weight of catalyst required for 80% conversion is 1054.1 kg .

P5-11 (b) Differential Mole Balance:

FBuO

= -rBuʹ

PBu = PBuO Rate equation

:

For Fluidized CSTR

:

Here,

-rBuʹ = K*PBu

5-19

Therefore,

w = FBuo*X Given, X = 0.8 FBuO = 50 Kmol.hr-1 KBu = 0.32 atm-1 PBuO = 10 atm Therefore, putting the values in the above equation

w = 50*0.8

=

1225 Kg

Therefore, 1225 Kg of fluidized CSTR catalyst weight is required to achieve 80% conversion.

P5-11 (c)

With pressure drop parameter alpha = 0.0006 kg-1

POLYMATH Report Ordinary Differential Equations

04-May-2009


w

0

0

1100.

1100.

2

x

0

0

0.7750365

0.7750365

3

y

1.

0.2689385

1.

0.2689385

4

FBuO

50.

50.

50.

50.

5

f

1.

1.

6.600157

6.600157

6

PBuO

10.

10.

10.

10.

7

PBu

10.

0.3408456

10.

0.3408456

8

k

0.054

0.054

0.054

0.054

9

K

0.32

0.32

0.32

0.32

10 rBu

-0.0306122

-0.0421871

-0.0149635

-0.0149635

11 alpha

0.0006

0.0006

0.0006

0.0006

Differential equations 1 d(x)/d(w) = -rBu / FBuO

5-20

Kgcat-1

2 d(y)/d(w) = -alpha * (1 + x) / 2 / y Kgcat-1

Explicit equations 1 FBuO = 50 Kmol.hr-1

2 f = (1 + x) / y 3 PBuO = 10 atm

4 PBu = PBuO * (1 - x) / (1 + x) * y atm

5 k = 0.054 kmol.kgcat-1.hr-1.atm-1

6 K = 0.32 atm-1

7 rBu = -k * PBu / (1 + K * PBu) ^ 2 Kmol.kgcat-1.hr-1

8 alpha = 0.0006 Kg-1

General Total number of equations

10


8

Elapsed time

1.157 sec

Solution method

RKF_45

Step size guess. h

0.000001


5-21

So, we see the maximum rate in case with pressure drop is at catalyst weight equal to around 600 Kg.

5-22

To achieve 70% conversion, catalyst weight required is 932.3 kg .

In case of (a), 915.5 kg of catalyst is required to achieve 70% conversion.

5-23

P5-11 (d)


P5-11 (e)


P5-12 (a)

P5-12 (b)

P5-12 (c) For α = 0.001dm-3

See Polymath program P5-12-c.pol.

5-24

POLYMATH Results Calculated values of the DEQ variables Variable v x y Co esp alfa C k r fo

initial value 0 0 1 0.3 2 0.001 0.3 0.044 -0.0132 2.5

minimal value 0 0 0.1721111 0.3 2 0.001 0.0077768 0.044 -0.0132 2.5

maximal value 500 0.656431 1 0.3 2 0.001 0.3 0.044 -3.422E-04 2.5

Differential equations as entered by the user [1] d(x)/d(v) = -r/fo [2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y) Explicit equations as entered by the user [1] Co = 0.3 [2] esp = 2 [3] alfa = 0.001 [4] C = Co*(1-x)*y/(1+esp*x) [5] k = 0.044 [6] r = -k*C [7] fo = 2.5

At V = 500, x = 0.66, y = 0.17

P5-12 (d) Individualized solution P5-12 (e) A

B 2C

5-25

final value 500 0.656431 0.1721111 0.3 2 0.001 0.0077768 0.044 -3.422E-04 2.5

Using these equations in Polymath we get the volume to be 290 dm3.

For a batch reactor.

5-26

P5-12 (f)

A

B 2C

Using Polymath to solve the differential equation gives a volume of 290 dm3

See Polymath program P5-12-f.pol. PFR with pressure drop: Alter the Polymath equations from part (c).

See Polymath program P5-12-f-pressure.pol. POLYMATH Results

5-27

Calculated values of the DEQ variables Variable v x y Kc alfa Cao k esp fo r

initial value 0 0 1 0.025 0.001 0.3 0.044 2 2.5 -0.0132

minimal value 0 0 0.3181585 0.025 0.001 0.3 0.044 2 2.5 -0.0132

maximal value 500 0.5077714 1 0.025 0.001 0.3 0.044 2 2.5 -1.846E-04

ODE Report (STIFF) Differential equations as entered by the user [1] d(x)/d(v) = -r/fo [2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y) Explicit equations as entered by the user [1] Kc = .025 [2] alfa = 0.001 [3] Cao = 0.3 [4] k = 0.044 [5] esp = 2 [6] fo = 2.5 [7] r = -(k*Cao/(1+esp*x))*(4*Cao^2*x^3/((1+esp*x)^2*Kc))

5-28

final value 500 0.5077714 0.3181585 0.025 0.001 0.3 0.044 2 2.5 -1.846E-04

At V = 500 dm3 X = 0.507 and y = 0.381

P5-13 A To = 300K

KCO (300K)= 3.0 V = 1000gal = 3785.4 dm3

FAO X rA

V

Mole balance:

rA

Rate law: Stoichiometry: C A

V

Now using:

where

B

kO C

and

C AO 1 X

FAO 2 kO C AO

1 X

2902.2dm3 , Z

KC

k kO

2

V 1 X

Z

z

C BO X

X2 KC

2

FAO kOC AO 2

Z z

CB

X

Z

V

C B2 KC

2 A

X2 KC

3785.4 0.62 0.4

X k ko

exp

X 1 X

2

E 1 R TO

exp

K CO exp

2

X KC

1 T

H RX 1 R TO

E 1 R T0

1 T

f X

0 V

and

1 T

Solving using polymath to get a table of values of X Vs T.

See Polymath program P5-13.pol. POLYMATH Results NLE Solution Variable X To

Value 0.4229453 300

f(x) 3.638E-12

Ini Guess 0.5

5-29

0.42 3

Z z

X 1 X

2

X2 KC

T z V E R y Kco Hrx Kc

305.5 2902.2 3785.4 1.5E+04 2 1.5684405 3 -2.5E+04 1.4169064

NLE Report (safenewt) Nonlinear equations

0.43 0.42

[1] f(X) = (z/y)*X/((1-X)^2 - X^2/Kc) -V = 0

Explicit equations

0.41

[1] To = 300 [2] T = 305.5 [3] z = 2902.2 [4] V = 3785.4 [5] E = 15000 [6] R = 2 [7] y = exp(E/R*(1/To-1/T)) [8] Kco = 3 [9] Hrx = -25000 [10] Kc = Kco*exp(Hrx/R*(1/To-1/T))

X

0.4 0.39 0.38 0.37 0.36 295

300

305

310

Temperature

T(in K)

X

300

0.40

301

0.4075

303

0.4182

304

0.4213

305

0.4228

305.5

0.4229

305.9

0.4227

307

0.421

310

0.4072

315

0.3635

We get maximum X = 0.4229 at T = 305.5 K.

P5-14 CH 3CO 2 O H 2O A B 5-30

2CH 3COOH 2C

315

320

Part 1 CSTR

A B

3.3 10 3 dm3 s

0

C A0 1M, C B 51.2M

3

V 1 dm

X= ?

V

(1)Mole Balance:

FA0 X rA

rA

(2)Rate Law: (3)Stoichiometry:

k CACB

Liquid

0

Stoichiometric Table

A B C

FA0 FB0

FA0 X

BFA0

FA

FA0 X

0

FB

2FA0 X

FA0 1 X FA0

FC

Concentration

CA CB

FA

FA0 1 X

0

FB

FA0

0

51.2M B

1M

C A0 1 X

0 B 0

X

C A0

B

51.2

Substitute the value of X into CB. 5-31

X

B

2FA0 X

X

CB

CA0 51.2 X

CB0

(4)Combine k

rA

k C AC B0

k C B0 C A0 1 X

k k C B0 1.95 10 FA0 X

V

rA

dm3

4

mol s

0C A0 X

0

kC A0 1 X

kC A0 1 X

51.2

mol dm

3

0.01 s

1

X

k 1 X

X k1 X k

X

1

k

(5)Evaluate

1 dm3

V

3.3 10 3 dm3 s

0

k X

303 s

3.03

3.03 1 3.03

0.75

Calculate the Conversion in a

0

3

3

3.3 10 dm s,

CA0 1M CB 0

V

51.2M

PFR

5-32

0.311 dm3

X= ?

(1)Mole Balance:

(2)Rate Law: (3)Stoichiometry:

dX

rA

dV

FA0

rA k CACB Liquid

0

CA

C A0 1 X

CB

C B0 k

rA

k C B0 C A0 1 X

kC A0 1 X

(4)Combine

dX

rA

dV

FA0

kC A0 1 X 0C A0

Separate variables and Integrate with limits V = 0 then X = 0 and V = V then X = X X 0

ln

dX 1 X

0

1

kV

1 X

k

V

dV

0

k

0

X 1 e

k

(5)Evaluate

0.311 dm3 3.3 10 3 dm3 s k

0.94

X

0.61

5-33

94 s

P5-15 Gaseous reactant in a tubular reactor: A  B

rA

kC A

k

0.0015min 1 at 80 F

E

25, 000

MWA

cal g mol

MWB

58

lb lb mol

P 132 psig 146.7 psia

lb hr lb 58 lb mol 1000

FB

X

0.90

MB

Dt

1 inch (I.D.)

L 10 ft

T

260 F

nt

lb mol 17.21 hr

720 R

FA0

FB X

1000

lb hr

number of tubes

lb mol hr 0.9

17.21

19.1

For a plug flow reactor: 0.9

nt Dt2 L 4

V

FA0 0

dX rA

1 1 0

rA

yA

1.0

kC A0 1 X 0.9

V

yA

FA0 0

C A0 0.9

dX rA

FA0 0

PA RT

FA0 1 ln kC A0 1 0.9

dX kC A0 1 X

0

P RT

FA0 RT ln10 kP

At T2 = 260°F = 720°R, with k1 = 0.0015 min-1 at T1 = 80°F = 540°R,

E 1 R T1

k2

k1 exp

k2

53.6 min

V

1

3219hr

FA0 RT ln10 kP

V

0.72 ft 3

V

nt Dt2 L 4

1 T2

0.0015exp

25000 1 1.104 540

1 720

ft 3 psia lb mol R

720 R

53.6 min

1

19.1

lb mol hr

10.73

3219 hr

1

146.7 psia

5-34

ln10

1

lb mol hr

nt

4V Dt2 L

4 0.72 ft 3 1 ft 12

2

13.2

10 ft

Therefore 14 pipes are necessary.

P5-16(a) A → B/2

Combining 2 VPFR kCA0 FA0

X

1

X

2

0

1 X

2

dX

2 1

ln 1 X

2

2

X

1 X 1 X

(for the integration, refer to Appendix A) from the Ideal Gas assumption.

C AO

y AO CTO

Substituting Eqn. (5), X = 0.8 and = –1/4 to Eqn. (4) yields, 2 2 VPFR ky AO CTO FAO

2( 1/ 4)(1 1/ 4) ln(1 0.8) ( 1/ 4)2 0.8

5-35

(1 1/ 4)2 0.8 1 0.8

2.9 ……….(6)

2 VPFR ky '2AO C 'TO F '2AO

2 2 CTO 8 VPFR ky AO 9 FAO

8 2.9 9

2.58

P5-16(b) Individualized solution.

P5-17(a) A

Given: The metal catalyzed isomerization

rA

B liquid phase reaction

CB with Keq = 5.8 K eq

k1 C A

For a plug flow reactor with yA = 1.0, X1 = 0.55 Case 1: an identical plug flow reactor connected in series with the original reactor.

Since yA = 1.0,

rA

kC A0

B

= 0. For a liquid phase reaction CA

X K eq

1 X

For the first reactor, X1

V1

FA0 0

dX rA

X1

FA0 0

dX kC A0 1 X

or

X K eq 5-36

CA0 1 X and C B

C A0 X

X1

kC A0V1 FA0

X1

dX

0

kC A0V1 FA0

1

1 X K eq

1

1 1 1 K eq

1

ln 1

1

1

1 K eq

ln 1

1

1 X K eq 0

1 X1 K eq

0.853ln .355

0.883

Take advantage of the fact that two PFR’s in series is the same as one PFR with the volume of the two combined. VF = V1 + V2 = 2V1 and at VF , X = X2 XF XF

kC A0VF FA0

0

kC A0VF FA0 2

2

kC A0V1 FA0

1.766

dX 1

1

kC A0V1 FA0

2 0.883

1 1 1 5.8

ln 1

1

1 X K eq 1 1

1 K eq

1 K eq

1

ln 1

1

ln 1

1

1 X K eq 0

1 X2 K eq

1.766

1

1 X2 5.8

X2 = 0.74

P5-17(b) Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,

5-37

The analysis for the first reactor is the same as for case 1.

kC A0V1 FA0

1 1 K eq

1

ln 1

1 X1 K eq

1

By performing a material balance on the separator, FA0,2 = FA0(1-X1) Since pure A enters both the first and second reactor CA0,2 = CA0, CB0,2 = 0,

CA = CA0 1 - X

B

=0

C B = C A0 X for the second reactor. X2

V2

FA0,2 0

kC A0V2 FA0 1 X 1

FA0 1 X

dX rA

kC A0

1 1

1 K eq

X2

0

ln 1

dX 1 X

1

X K eq

1 X2 K eq

and since V1 = V2

kC A0V2 FA0

kC A0V1 FA0

or

1 1

1 K eq

ln 1

1

1

1

1 X1 ln 1 1 1 K eq

1 X1 K eq

1 X2 K eq

1

5-38

1

1 X1 K eq

1

1 1 X1

1 X2 K eq

1

1

X2

1

1 X1 K eq

1 1 X1

1

1 1 K eq

0.356

1 0.45

0.766

1.174

Overall conversion for this scheme:

X

X

FA0

FA0,2 1 X 2

FA0

FA0 1 X1 1 X 2

FA0

FA0

0.895

P5-18 Given: Meta- to ortho- and para- isomerization of xylene.

M

k1

P

M

k2

O

O

P (neglect)

Pressure = 300 psig T = 750°F V = 1000 ft3 cat. Assume that the reactions are irreversible and first order. Then:

rM

k

k1CM

k 2 CM

kCM

k1 k2 0

Check to see what type of reactor is being used. Case 1:

v0

2500

gal hr

X

0.37

gal hr

X

0.50

Case 2:

v0

1667

Assume plug flow reactor conditions:

FM 0 dX

rM dV or 5-39

1

1 X1 1 X 2

X

V

FM 0 0

X

dX rM X

CM 0 v0 dX rM

V 0

v0 0

v0 ln 1 X k

dX k 1 X

CM0, k, and V should be the same for Case 1 and Case 2. Therefore,

kV kV

Case1

v0

Case1

ln 1 X Case1

Case 2

v0

Case1

ln 1 X Case 2

gal ln 1 0.37 hr gal 1667 ln 1 0.50 hr

2500

gal hr gal 1155 hr

1155

The reactor appears to be plug flow since (kV)Case 1 = (kV)Case 2 As a check, assume the reactor is a CSTR.

FM 0 X

V

CM 0 v0 X

CM 0 X v0 rM

rM V

v0 X k 1 X

or kV

v0 X 1 X

Again kV should be the same for both Case 1 and Case 2.

kV

kV

v0 Case1

1 X Case1

v0 Case 2

Case 2

gal 0.37 hr 1 0.37

1468

gal 0.50 hr 1 0.50

1667

2500

X Case1 Case1

X Case 2

gal hr

1667

1 X Case 2

gal hr

kV is not the same for Case 1 and Case 2 using the CSTR assumption, therefore the reactor must be modeled as a plug flow reactor.

kV

k

gal hr gal 1155 gal hr 1.55 3 1000 ft cat. hr ft 3 cat 1155

For the new plant, with v0 = 5500 gal / hr, XF = 0.46, the required catalyst volume is:

V

v0 ln 1 X F k

gal hr ln 1 0.46 gal 1.155 hr ft 3 cat 5500

2931 ft 3 cat

This assumes that the same hydrodynamic conditions are present in the new reactor as in the old. 5-40

P5-19 A→ B in a tubular reactor

Tube dimensions: L = 40 ft, D = 0.75 in. nt = 50

FA0

FA0 0

rA

1

FA0 0

40 6.14 ft 3

6.86

kC A0 1 X

X

dX rA

X

FA0 0

dX kC A0 1 X

y A0 P RT FA0 RT 1 ln or k ky A0 P 1 X

with C A0

lb mol hr

dX rA

kC A0 1 X X

V

lb hr lb 73 lb mol

2

500

mA MWA X

V

V

50

nt D 2 L 4

V

0.75 12 4

FA0 1 ln kC A0 1 X

P RT

FA0 RT 1 ln Vy A0 P 1 X

Assume Arrhenius equation applies to the rate constant. At T1 = 600°R, k1 = 0.00152 At T2 = 760°R, k2 = 0.0740

Ae Ae

E RT1

E RT2

5-41

k2 k1

ln

E 1 R T2

exp

k2 k1

E 1 R T2

1 T1

E T2 T1 R T1T2

1 T1

660 760

TT k 1 2 ln 2 T1 T2 k1

E R

0.740 19,500 R 0.00152

E RT1

A k1 exp so k

100

ln

E 1 R T

k1 exp

1 T1

From above we have


k so


E 1 R T

k1 exp

1 T1

Dividing both sides by T gives:

FA0 R 1 ln Vy A0 P 1 X

6.86 .00152

sec

E 1 R T

k1 exp

lb mol hr

1 T1

T

10.73

sec 3600 hr

psia ft 3 lb mol R

6.14 ft

3

exp ln 5

114.7 psia

Evaluating and simplifying gives:

exp 0.0308 R

1 T

19500

1

1 660 R

T

Solving for T gives: T = 738°R = 278°F

5-42

1 T

19500 T

1 660 R

P5-20 Rate law:

-rA= kCACB

(1)

Molar flow rate: FA0 = FB0 = Since FA0= FB0, therefore equimolar mixture. CA0 = CB0 = FA0/ = 1.51/5.33 = 0.283 lbmol/ft3 From (1), -rA= kCACB = kCA0(1-X)CB0(1-X) = kCA02 (1-X)2

(2)

Now, for PFR V = FA0

(3)

Putting -rA from (2) in (3) and then integrating, we get VkCA02/ FA0 = Now given, 50% conversion, therefore X = 0.5 So, VkCA02/ FA0 =

=1

5.33*k*0.2832/1.51 = 1 k = 3.54 Now we have CSTR in series V = 100 gal = 13.37 ft3 Now, CA20 = CA0/2 = 0.1425 lbmol/ft3 , since we have 50% conversion if PFR FA20 = FA0/2 = 1.51/2 = 0.755 CSTR mole balance: V = FA20 So, 13.37 = 0.755

= FA20 =

13.37 = => X = 0.42 So, CA = 0.1425(1-0.42) = 0.083 So, overall conversion X = (CA0- CA)/CA0 = (0.283 – 0.083)/(0.283) = .70 5-43

X = 0.7

P5-21 Five things wrong with the solution are as follows : (i) The rate law is wrongly written (ii) The integration is wrongly done (iii) Value of Ɛ is wrongly calculated (iv)The equation for calculation of k has been reaaranged wrongly. (v) calculated value of k is therefore wrong correct solution For a first order reaction ; -rA = -

1 dN A V dt

kC A

By transforming into conversion units we have; -rA =

C A0 dX A 1 A X A dt

kC A0 (1 X A ) 1 AXA

Separating and integrating we obtain ; ln(1 X A )

V ) AV0

ln(1

kt

Now in this case ; Ɛ = yA0δ = 1(3-1) = 2 Substituting

V V0

4 2 1 2

we have; -ln(1-0.5) = ln2 = kt => k = ln2/t = 0.693/2hr = 0.3465 hr-1

5-44

P5-22 Reversible isomerization reaction m-Xylene → p-Xylene Xe is the equilibrium conversion.

rm

Rate law:

k Cm

Cp ke

At equilibrium, -rm = 0 =>

Cm

Cp ke

Cmo 1 X e

Xe 1 Xe

Ke

1 Ke

1

rm

1

1 Xe Xe

kC A0 1

Cmo X e Ke

Xe 1 Xe Xe

1 Xe

X Xe

P5-22 (a) For batch reactor, Mole balance:

dX dt

rmV N mO

kC A0 X 1 C A0 Xe

Xe Xe ln k Xe X

V

FAO vo k

For PFR, PFR

PFR

dX rm dX

1

1

1 k

1 X Ke dX

1

1

1 X Ke

Xe Xe ln k Xe X 5-45

P5-22 (b) For CSTR,

V

Fmo X rm X

CSTR

k 1

1

1 X Ke

Putting the value of Ke,

X k

CSTR

Xe Xe

X

P5-22 (c) Volume efficiency =

PFR V CSTR

1 V

Xe Xe ln k Xe X

ln

Xe X k Xe X

Xe Xe

X

Xe X

X Xe

X

X Xe 1 ln X X 1 Xe Xe

Following is the plot of volume efficiency as a function of the ratio (X/Xe),

See Polymath program P5-22-c.pol.

5-46

X

ln

Xe Xe

X

P5-23(a)

5-47

P5-23 (b)

P5-23 (c)

P5-23 (d) For turbulent flow 2

1

DP1 DP 2

Ac1 Ac 2

3

0.018kg

2 1

1

12 1.52

3

0.00316 kg

1

Pexit = P0(1- αW)0.5 = (20atm)(1-0.00316kg-1(50kg))1/2 = 18.35 atm

ln X2

1 0.5 1 X2

10 3 2 20atm atm kg 3 0.00316kg

1

1 1 0.00316kg

0.81

P5-24 Production of phosgene in a microreactor. CO + Cl2 COCl2 (Gas phase reaction) A + B C The euqtions that need to be solved are as follows : d(X)/d(W) = -rA/FA0 d(y)/d(W) = -α*(1+Ɛ *X)/(2*y)

( from 4.30)

5-48

1

50kg

3/ 2

FB0 = FA0; Fb = FB0-FA0*X; Fc = FA0*X

See Polymath program P5-24.pol. POLYMATH Results Calculated values of the DEQ variables Variable W X y e FA0 FB0 Fa Fb v0 v Fc Ca Cb a k rA Cc

initial value 0 0 1 -0.5 2.0E-05 2.0E-05 2.0E-05 2.0E-05 2.83E-07 2.83E-07 0 70.671378 70.671378 3.55E+05 0.004 -19.977775 0

minimal value 0 0 0.3649802 -0.5 2.0E-05 2.0E-05 4.32E-06 4.32E-06 2.83E-07 2.444E-07 0 9.1638708 9.1638708 3.55E+05 0.004 -19.977775 0

maximal value 3.5E-06 0.7839904 1 -0.5 2.0E-05 2.0E-05 2.0E-05 2.0E-05 2.83E-07 4.714E-07 1.568E-05 70.671378 70.671378 3.55E+05 0.004 -0.3359061 53.532416

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -rA/FA0 [2] d(y)/d(W) = -a*(1+e*X)/(2*y) Explicit equations as entered by the user [1] e = -.5 [3] FB0 = FA0 [5] Fb = FB0-FA0*X [7] v = v0*(1+e*X)/y [9] Ca = Fa/v [11] a = 3.55e5 [13] rA = -k*Ca*Cb

[2] FA0 = 2e-5 [4] Fa = FA0*(1-X) [6] v0 = 2.83e-7 [8] Fc = FA0*X [10] Cb = Fb/v [12] k = .004 [14] Cc = Fc/v

5-49

final value 3.5E-06 0.7839904 0.3649802 -0.5 2.0E-05 2.0E-05 4.32E-06 4.32E-06 2.83E-07 4.714E-07 1.568E-05 9.1638708 9.1638708 3.55E+05 0.004 -0.3359061 33.259571

P5-24 (a)

P5-24 (b) The outlet conversion of the reactor is 0.784 The yield is then MW*FA*X = 99 g/mol * 2 e-5 mol/s * 0.784 = .00155 g/s = 48.95 g/ year. Therefore 10,000 kg/year / 48.95 kg/ year = 204 reactors are needed.

P5-24 (c) Assuming laminar flow, α ~ Dp-2, therefore 2

DP21 1 DP2 2

3.55 105 kg

1

4 14.2 105 kg

1

P5-24 (d) A lower conversion is reached due to equilibrium. Also, the reverse reaction begins to overtake the forward reaction near the exit of the reactor.

P5-24 (e) Individualized solution 5-50

P5-24 (f) Individualized solution P5-24 (g) Individualized solution

P5-25

No solution necessary

P5-26

A B

C D

P0

P

1. Mole Balance – Use Differential Form

dX dW

rA FA 0

2. Rate Law – Psuedo Zero Order in B

rA 3. Stoichiometry – Gas:

kC A C 0B 0

1

kC A X

T P0 , Isothermal, therefore T T0 P

5-51

T0

FA0 1 X P X P0 0 1

FA

CA

CA

1

X

y , y

P P0

0

C A0 1 X y

, where W

2y

1 X

1 1 1 1 1 2

y A0

dy dW

C A0

0 then y 1 and X 0

Integrating

y part (a)

1

W

12

Now since P/P0 >0.1 .

So y= .1 for minimum value of W Now (1 – αW)1/2 = .1 => W = 99 kg

4. Combine

rA

kC A

kC A0 1 X y dX dW

kC A0 1 X 1

kC A0 1 X 1 FA0

Separate Variables dX 1 X

k

1

W

12

dW

0

when W = 0, X = 0

ln

1 1 X

2k 1 3 0

1

W

32

5-52

W

12

W

12

2 1.2 dm 3 g min

2k 3

0

ln

3 0.01 g 25 dm 3 min 1

3.2 1 1

1 X y

1

W 0

0

W

3.2

32

12

P0 P 1 y

5-53

W

y

1/y

X

P

0

1

1

0

10 atm

1

1

1

0.47

2.5

0.99

1.01

0.11

5

0.975

1.02

0.21

15

0.92

1.08

0.5

25

0.87

1.16

0.67

50

0.71

1.41

0.87

75

0.5

2.0

0.94

90

0.32

3.16

0.955

99

0.1

10

0.96

X = 0.9

2.3 1 3.2 1

1

W

32

W .429

W = 57g for 5% W = 1.05g for last 5% (85 to 90%) W = 57 – 45 = 12 g Ratio =

12g = 11.4 1.05g

Polymath

dX dW

rA prime FA0

dy dW

alpha 2 y

rA prime

CA

k CA

C A0

1 Xy

k 1.2 dm 3 g min alpha 0.01g

1

5-54

1 atm

5-55

5-56

P5-27 0, y

0 1

W

1

2

Mole Balance/Design Equation

FA 0

dX dW

rA

Rate Law

rA

kC A

2

5-57

Stoichiometry

CA

C A0 1 X y

Combining

FA 0

dX dW

FA 0

2

kC A0 1 X dX

2

kC A0 1 X FA 0

X

1

2

W

2

kC A0 1 X 10 k 300 K

0.1

k 300 K

2000 101

k 400 K

k 300 K exp

y2

W dW

2

W2

99 2

100

2

2

19.8

101 2

dm 6 mol 2 min

E 1 R 300

1 400

19.8 exp

10,000 1 1.987 300

1311.179

dm 6 mol 2 min

1 400

P5-28 A rA

B C kC A

Case1: 10 dm3 PFR with 80% conversion at 300K.

dV FA0

dX A rA

C A0

dX A rA

Here,

= 10 dm3/5 dm3/sec = 2 sec &

C A0

kC A0

dX A 1 XA 1 XA

y A0

1(1 1 1) 1

dX A 1 XA kC A0 1 XA 5-58

0.8

dX A 1 XA 1 XA

2k 0

Integrating, we get k = 1.21 sec-1 Case2: 10 dm3 CSTR with 80% conversion at 320K.

V FA0

XA rA

C A0

XA rA

Here,

= 10 dm3/5 dm3/sec = 2 sec &

C A0

XA 1 XA 1 XA

kC A0

kC A0

y A0

1(1 1 1) 1

XA 1 XA 1 XA

XA 1 XA 1 XA

2k

Putting XA=0.8, we get k = 2.8 sec-1 Now,

k (T2 ) k (T1 )

exp

k (T k (T

Ea 1 ( R T1

320 K ) 300 K )

1 ) T2

exp

Ea 1 1 ( ) 1.986 300 320

Ea 2.8 1 1 exp ( ) 1.21 1.986 300 320 Ea

8000cal / mol

P5-29 For the turbulent flow

0, y

0 1

W

1

2

5-59

Mole Balance/Design Equation

FA 0

dX dW

rA

Rate Law

rA

kC A

2

Stoichiometry

CA

C A0 1 X y

Combining

FA0

dX dW

kC A0 2 1 X

FA0 dX 2 kC A0 1 X FA0 X 2 kC A0 1 X

1

2

W

2

y2

W dW

2

W2

Combining

~

2

DP1 1 DP 2

X 1 X

1

2

1 DP

9.975 10 3 kg 2

13.47 0.244 10

2

4.99 10 3 kg

4.99 10 3 100 100 2

0.0802 100 24.95

X

1

2

6.02

0.86


5-60

1

CDP5-B a)

Polymath solution (Ans CDP5-B-a) b)

5-61

Polymath solution(Ans CDP5-B-b) c)

Polymath solution(Ans CDP5-B-c) d) This part is almost same as part(b) with minor changes: V = 15000000 – 10000t vo = 80000 and vout = 70000 The reason the graph looks so different from(a) is that pure water is evaporated, but water with atrazine is coming in. Polymath code:

5-62

d(ca)/dt=cao*vo/v)-(ca(kv+vout)/v) ca(0)=4.5 # vo = 80000 # v = 15000000 - 10000*t k = 0.0025 # cao = 4.5 # vout = 70000 # t(0)=0 # t(f)=1000

Polymath solution(Ans CDP5-B-d)

CDP5-C

5-63

CD5RD-1 No solution

5-64

CD5RD-2

5-65





6-1

Solutions for Chapter 6 : Isothermal Reactor Design- Molar Flow rates P6-1 Individualized solution. P6-2 (a) Example 6-1 For pressure doubled and temperature decrease CTO = 2*Po/RT and T = 688K

See Polymath program P6-2-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fc E T Cto Ft Ca k ra Fao rb vo rc X Tau rateA

initial value 0 2.26E-04 0 0 2.4E+04 688 0.573773 2.26E-04 0.573773 213.40078 -70.254837 2.26E-04 70.254837 3.939E-04 35.127419 0 0 70.254837

minimal value 0 1.363E-05 0 0 2.4E+04 688 0.573773 2.26E-04 0.0236075 213.40078 -70.254837 2.26E-04 0.1189309 3.939E-04 0.0594654 0 0 0.1189309

maximal value 1.0E-04 2.26E-04 2.124E-04 1.062E-04 2.4E+04 688 0.573773 3.322E-04 0.573773 213.40078 -0.1189309 2.26E-04 70.254837 3.939E-04 35.127419 0.9395277 0.253044 70.254837

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc

6-2

final value 1.0E-04 1.363E-05 2.124E-04 1.062E-04 2.4E+04 688 0.573773 3.322E-04 0.0236075 213.40078 -0.1189309 2.26E-04 0.1189309 3.939E-04 0.0594654 0.9395277 0.253044 0.1189309

Explicit equations as entered by the user [1] E = 24000 [2] T = 688 [3] Cto = 2*1641/8.314/T [4] Ft = Fa+Fb+Fc [5] Ca = Cto*Fa/Ft [6] k = 0.29*exp(E/1.987*(1/500-1/T)) [7] ra = -k*Ca^2 [8] Fao = 0.000226 [9] rb = -ra [10] vo = Fao/Cto [11] rc = -ra/2 [12] X = 1-Fa/Fao [13] Tau = V/vo [14] rateA = -ra

P6-2 (b) Example 6-2 Individualized solution. P6-2 (c) Example 6-3 Using trial and error, we get maximum feed rate of B = 0.0251dm3/s to keep concentration of B 0.01mol/dm3.

See Polymath program P6-2-c.pol. POLYMATH Results Calculated values of the DEQ variables Variable t ca cb cc cd k v00 cb0

initial value 0 0.05 0 0 0 0.22 0.0251 0.025

minimal value 0 0.0063485 0 0 0 0.22 0.0251 0.025

maximal value 500 0.05 0.009981 0.0078965 0.0078965 0.22 0.0251 0.025

6-3

final value 500 0.0063485 0.009981 0.0078965 0.0078965 0.22 0.0251 0.025

v0 5 5 ca0 0.05 0.05 rate 0 0 v 5 5 x 0 0 Differential equations as entered by the user [1] d(ca)/d(t) = -k*ca*cb-v00*ca/v [2] d(cb)/d(t) = -k*ca*cb+v00*(cb0-cb)/v [3] d(cc)/d(t) = k*ca*cb-v00*cc/v [4] d(cd)/d(t) = k*ca*cb-v00*cd/v

5 0.05 3.91E-05 17.55 0.5543321

Explicit equations as entered by the user [1] k = .22 [2] v00 = 0.0251 [3] cb0 = 0.025 [4] v0 = 5 [5] ca0 = 0.05 [6] rate = k*ca*cb [7] v = v0+v00*t [8] x = (ca0*v0-ca*v)/(ca0*v0)

If the concentration of A is tripled the maximum feed rate becomes 0.064 dm3/s

P6-2 (d through g) Individualized solution.

P6-3 Solution is in the decoding algorithm given with the modules. P6-4 (a) A

B 2C

CA= CA0 CA0 = PA0/RT = k = 10-4exp[ Next we express the rate law as:

6-4

5 0.05 1.394E-05 17.55 0.5543321

rA

k CA

CBCC2 KC

At equilibrium,

C F k T0 A FT

CT 0 FT

3

FB FC2 KC

(1)

= 0,

Therefore, from (1), k[CA-

]=0

CA0(1-Xe) – CA0Xe*(2CA0Xe)2/Kc = 0 (1-Xe) - Xe*22* CA02* Xe2/ Kc = 0 (1-Xe) - Xe*22* 0.32* Xe2/ 0.025 = 0 (1-Xe) – 14.4Xe3 = 0 => Xe = 0.355 So, the equilibrium conversion obtained in a PFR is 35.5 %.

P6-4 (b)

dFB dV dFC dV

rB

RB

rC

Transport out the sides of the reactor: RB = kcCB =

kB CT 0 FB FT

Stoichiometery: -rA = rB =1/2 rC

6-5

POLYMATH Report


Ordinary Differential Equations


Co

0.3

0.3

0.3

0.3

2

Fa

2.5

0.0323558

2.5

0.0323558

3

Fb

0

0

0.0197934

0.0002897

4

Fc

0

0

4.935288

4.935288

5

Ft

2.5

2.5

4.967934

4.967934

6

K

0.044

0.044

0.044

0.044

7

Kc

0.025

0.025

0.025

0.025

8

kc

4.8

4.8

4.8

4.8

9

ra

-0.0132

-0.0132

-8.324E-05

-8.324E-05

10 Rb

0

0

0.0099094

8.398E-05

11 V

0

0

1500.

1500.

12 X

0

0

0.9870577

0.9870577

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = -ra - Rb 3 d(Fc)/d(V) = -2*ra Explicit equations 1 Ft = Fa+Fb+Fc 2 K = 0.044 3 kc = 4.8 min-1

4 Kc = 0.025 5 Co = 0.3 6 X = (2.5-Fa)/2.5 7 Rb = kc*Co*Fb/Ft 8 ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2)) General Total number of equations

11


8

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001


6-6

6-7

So, if the reactor volume is 800 dm3, exit conversion will be 0.92.

6-8

We see that volume required for 80% conversion is 467 dm3, therefore plotting the flow rates till 467 dm3

6-9

P6-5 (a) Assume constant volume batch reactor Mole balance: C A0

dX dt

rA

Rate law and stoichiometry:

rA

kC A

kC A0 1 X 6-10

Specific reaction rate: k 25 C

0.0022 weeks

1

Combine: X

t

dX kC A0 1 X

C A0 0

1 ln 1 X 52.2 weeks k

1 ln 1 X 0.0022 weeks 1

X

0.108

CA

C A0 1 X but since volume and molecular weight are constant the equation can be written as:

mA

mA0 1 X

6500 IU mA0

mA0 1 0.108

7287 IU

C A0 C A *100 CA

%OU

7287 6500 *100 12.1% 6500

P6-5 (b) 10,000,000 lbs/yr = 4.58 * 109 g/yr of cereal Serving size = 30g Number of servings per year = 4.58 * 109 / 30 =1.51 * 109 servings/yr Each serving uses an excess of 787 IU = 4.62 * 10-4 = 1.02 * 10-6 lb Total excess per year = (1.51 * 108 servings/yr) * (1.02 * 10-6 lbs/serving) = 154.11 lb/yr Total overuse cost = $100/lb * 154.11 lb/yr = $15411 / yr (trivial cost)

P6-5 (c) If the nutrients are too expensive, it could be more economical to store the cereal at lower temperatures where nutrients degrade more slowly, therefore lowering the amount of overuse. The cost of this storage could prove to be the more expensive alternative. A cost analysis needs to be done to determine which situation would be optimal.

P6-5 (d) k 40 C X

t

C A0 0

0.0048 weeks

dX kC A0 1 X

26weeks

1

6 months = 26 weeks

1 ln 1 X k

1 ln 1 X 0.0048 weeks 1

X

0.12

CA

C A0 1 X but since volume and molecular weight are constant the equation can be written as: 6-11

mA

mA0 1 X

6500 IU mA0

%OU

mA0 1 0.12

7386 IU

C A0 C A *100 CA

7386 6500 *100 13.6% 6500

P6-6 3

Suppose the volumetric flow rate could be increased to as much as 6,000 dm /h (9,000 mol/h) and the total time to fill, heat, empty and clean is 4.5 hours. What is the maximum number of moles of ethylene glycol (CH2OH)2 you can make in one 24 hour period? The feed rate of ethylene cholorhydrin will be adjusted so that the volume of fluid at the end of the reaction time will be 2500 dm3. Now suppose CO2 leaves the reactor as fast as it is formed.

A B

C D CO2

Mole Balance

dN A rAV dt dN B FB0 dt dNC rC V dt N D NC

rBV

Overall Mass Balance Accumulation = In – Out

dm ÝCO 2 0 m dt m V Assume constant density mÝCO 2 dV dt

0

The rate of formation of CO2 is equal to the rate of formation of ethylene glycol (C).

mÝCO 2 dV dt CO 2

FCO

2

0

MWCO2

FCO

2

FCO

2

MWCO2 0

CO 2

MWCO2 6-12

Rate Law and Relative Rates

rA

kCA CB

rB rA rC rA rCO 2 rA Stoichiometry NA V NB V

CA CB

dC A dt dC B dt

dCC dt CD CC dN CO 2 dt FCO 2

0

rA

CO 2

CA

V

0C B0

0

V

CO 2

V 0

CO 2

CB

rA

CC

V 0

FCO 2

rAV

rAV

6-13

P6-6(c) FA0

0.15 mol min

vo Fao /Ca 3

9 mol hr

9 mol hr / 1.5 mol dm 3

6 dm 3 hr 3

1000 dm is needed to fill the reactor. At 6 dm /hr it will take 166.67 hours Now solving using the code form part (a) with the changed equations: See Polymath program

6-14

P6-7 C B0

0.25

mol dm3 V

C A0

0.2

mol

V0

dm3

VT VT

D 2L

1.5m

4 4.42m

2

2.5m

4 3

4, 420 dm 3

We don’t know V0 or v0. First try equal number of moles of A and B added to react. NA C A0 V0 V0

NB

C B0 V C B0 VT VT

4420

C A0

1

C B0

0.8 1.0

V0 2456

V 1964 V 0

k0

tR

3

0.000052 dm mol s

0.187 dm3 mol h

For 1 batch tR = 24 – 3 = 21 VT 0

V0 21

(1) For one batch we see that only 198 moles of C are made so one batch will not work. (2) For two batches, we have a down time of 2 3 = 6 h and therefore each batch has a reaction time of 18h/2 = 9 h. We see that 106 moles of C are made in one batch therefore 2 106 = 212 moles/day are made in two batches. 6-15

P6-7

6-16

P6-7

P6-8 (a) A

B 2C

To plot the flow rates down the reactor we need the differential mole balance for the three species, noting that BOTH A and B diffuse through the membrane

dFA dV dFB dV dFC dV

rA

RA

rB

RB

rC

Next we express the rate law: First-order reversible reaction

rA

k CA

CBCC2 KC

C F k T0 A FT

CT 0 FT

3

FB FC2 KC 6-17

Transport out the sides of the reactor: RA = kACA =

k ACT 0 FA FT

RB = kBCB =

kB CT 0 FB FT

Stoichiometery: -rA = rB =1/2 rC Combine and solve in Polymath code:

See Polymath program P6-9-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable v Fa Fb Fc Kc Ft Co K Kb ra Ka Ra Rb Fao X

initial value 0 100 0 0 0.01 100 1 10 40 -10 1 1 0 100 0

minimal value 0 57.210025 0 0 0.01 100 1 10 40 -10 1 0.472568 0 100 0

maximal value 20 100 9.0599877 61.916043 0.01 122.2435 1 10 40 -0.542836 1 1 2.9904791 100 0.4278998

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(v) = ra - Ra [2] d(Fb)/d(v) = -ra - Rb [3] d(Fc)/d(v) = -2*ra

6-18

final value 20 57.210025 1.935926 61.916043 0.01 121.06199 1 10 40 -0.542836 1 0.472568 0.6396478 100 0.4278998

Explicit equations as entered by the user [1] Kc = 0.01 [2] Ft = Fa+ Fb+ Fc [3] Co = 1 [4] K = 10 [5] Kb = 40 [6] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2)) [7] Ka = 1 [8] Ra = Ka*Co*Fa/Ft [9] Rb = Kb*Co*Fb/Ft

P6-8 (b) The setup is the same as in part (a) except there is no transport out the sides of the reactor.

See Polymath program P6-9-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable v Fa Fb Fc Kc Ft Co K ra Fao X

initial value 0 100 0 0 0.01 100 1 10 -10 100 0

minimal value 0 84.652698 0 0 0.01 100 1 10 -10 100 0

maximal value 20 100 15.347302 30.694604 0.01 130.6946 1 10 -3.598E-09 100 0.153473

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(v) = ra [2] d(Fb)/d(v) = -ra [3] d(Fc)/d(v) = -2*ra

6-19

final value 20 84.652698 15.347302 30.694604 0.01 130.6946 1 10 -3.598E-09 100 0.153473

Explicit equations as entered by the user [1] Kc = 0.01 [2] Ft = Fa+ Fb+ Fc [3] Co = 1 [4] K = 10 [5] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2))

P6-8 (c) Conversion would be greater if C were diffusing out. P6-8 (d) Individualized solution

P6-9 (a)

CO + H 2O

CO2 + H 2

A + B

C

+ D

Assuming catalyst distributed uniformly over the whole volume Mole balance:

dFA dW

Rate law:

r RH 2

rA

dFB dW

r rB

rC

r rD

dFC dW k C AC B

K H2 CD 6-20

r CC CD K eq

dFD dW

r

RH 2

Stoichiometry: C A

FT

CTO

FA FT

FA

FB

CB FC

CTO

FB FT

CC

CTO

FC FT

FD

Solving in polymath:

See Polymath program P6-10.pol. POLYMATH Results Calculated values of the DEQ variables Variable W Fa Fb Fc Fd Keq Ft Cto Ca Cb Kh Cc Cd Rh k r

initial value 0 2 2 0 0 1.44 4 0.4 0.2 0.2 0.1 0 0 0 1.37 -0.0548

minimal value 0 0.7750721 0.7750721 0 0 1.44 3.3287437 0.4 0.0931369 0.0931369 0.1 0 0 0 1.37 -0.0548

maximal value 100 2 2 1.2249279 0.7429617 1.44 4 0.4 0.2 0.2 0.1 0.147194 0.0796999 0.00797 1.37 -0.002567

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(W) = r [2] d(Fb)/d(W) = r [3] d(Fc)/d(W) = -r [4] d(Fd)/d(W) = - r -Rh Explicit equations as entered by the user [1] Keq = 1.44 [2] Ft = Fa+Fb+Fc+Fd [3] Cto = 0.4 [4] Ca = Cto*Fa/Ft [5] Cb = Cto*Fb/Ft [6] Kh = 0.1 [7] Cc = Cto*Fc/Ft [8] Cd = Cto*Fd/Ft [9] Rh = Kh*Cd [10] k = 1.37 [11] r = -k*(Ca*Cb-Cc*Cd/Keq)

For 85% conversion, W = weight of catalyst = 430 kg

6-21

final value 100 0.7750721 0.7750721 1.2249279 0.5536716 1.44 3.3287437 0.4 0.0931369 0.0931369 0.1 0.147194 0.0665322 0.0066532 1.37 -0.002567

CD

CTO

FD FT

(b) In a PFR no hydrogen escapes and the equilibrium conversion is reached.

K eq

CC CD C AC B

C A2 0 X 2 C A2 0 (1 X ) 2

X2 1 X

2

1.44

solve this for X, X = .5454 This is the maximum conversion that can be achieved in a normal PFR. (c) If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of .459

P6-10


P6-11 (a) At equilibrium, r = 0 =>

C AC B

CC C D KC

V = V0 + vot

CA

N AO 1 X V

C AO 1 X V VO

C AO 1 X 1

vo t VO

6-22

C BO CB

vO t C AO X VO 1

CC

vo t VO C AO X

CD

1

C AO

t

vo t VO

v 1 X C BO O t C AO X VO

VO C AO X2 vO C BO K C 1 X

2

X

200 * 7.72 X2 0.05 *10.93 1.08 1 X

t

C AO X KC

X

Solving in Polymath POLYMATH Report Nonlinear Equation

06-May-2009

Calculated values of NLE variables Variable Value f(x) Initial Guess 1 x

0

0

0.495 ( 0 < x < 0.99 )

Nonlinear equations 1 f(x) = 2825.25*(x^2/1.08/(1-x)+x) = 0 General Settings Total number of equations

1

Number of implicit equations 1 Number of explicit equations 0 Elapsed time

0.0000 sec

Solution method

SAFENEWT

Max iterations

150

Tolerance F

0.0000001

Tolerance X

0.0000001

Tolerance min

0.0000001

6-23

If we solve in Excel X 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 0.99

t (sec)

0 148.1466374 311.591358 493.0338235 695.8486111 924.3101852 1183.914286 1481.84765 1827.692593 2234.515909 2720.611111 3312.40216 4049.525 4994.264683 6250.42963 8004.875 10631.31111 15001.7287 23732.1 49902.28611 259188.435

6-24

P6-11 (b) See Polymath program P6-12-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb Cc Cd Kc k

initial value 0 7.72 10.93 0 0 1.08 9.0E-05

minimal value 0 0.2074331 7.6422086 0 0 1.08 9.0E-05

maximal value 1.5E+04 7.72 10.93 3.2877914 3.2877914 1.08 9.0E-05

6-25

final value 1.5E+04 0.2074331 9.51217 1.41783 1.41783 1.08 9.0E-05

ra vo Vo V X

-0.0075942 0.05 200 200 0

-0.0075942 0.05 200 200 0

-1.006E-05 0.05 200 950 0.9731304

-1.006E-05 0.05 200 950 0.9731304

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*vo/V [2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93) [3] d(Cc)/d(t) = -ra - vo*Cc/V [4] d(Cd)/d(t) = -ra - vo*Cd/V Explicit equations as entered by the user [1] Kc = 1.08 [2] k = 0.00009 [3] ra = -k*(Ca*Cb - Cc*Cd/Kc) [4] vo = 0.05 [5] Vo = 200 [6] V = Vo + vo*t [7] X = 1 - Ca/7.72

Polymath solution

P6-11 (c) Change the value of vo and CAO in the Polymath program to see the changes.

P6-11 (d) As ethanol evaporates as fast as it forms: Now using part (b) remaining equations, Polymath code:

CD = 0

See Polymath program P6-12-d.pol. 6-26

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k ra vo Vo V X

initial value 0 7.72 10.93 9.0E-05 -0.0075942 0.05 200 200 0

minimal value 0 0.0519348 6.9932872 9.0E-05 -0.0075942 0.05 200 200 0

maximal value 6000 7.72 10.93 9.0E-05 -3.69E-05 0.05 200 500 0.9932727

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*vo/V [2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93) Explicit equations as entered by the user [1] k = 0.00009 [2] ra = -k*Ca*Cb [3] vo = 0.05 [4] Vo = 200 [5] V = Vo + vo*t [6] X = 1 - Ca/7.72

P6-11 (e)


P6-11 (f)

Individualized solution 6-27

final value 6000 0.0519348 7.8939348 9.0E-05 -3.69E-05 0.05 200 500 0.9932727

P6-12

CA0,

CA0,

CSTR A

B

CA0(1-X), (rejected)

-rA = k[ CA – CB/Ke) -rA = k[ CA0(1-X)- CA0X/Ke) Given : k = 0.4 h-1 Ke = 4 V = 60 m3 CA0= 100 kg/m3

6-28

= 12 m3/hr FA0= 100*12 = 1200 kg/hr FA20= FA0 + CA0 = FA0 (1+ f) = 1200(1+ f) CSTR mole balance: =

= Putting the values, we have =

(1)

Solving we get, X=

Profit = Value or products – Operating Cost Profit = $2*CA0*X (1+f) - $50 (1+f) Profit = $ 600*(1+f)*(4X-1)

(2)

Now, solving equation (1) in excel and corresponding finding out the vale or profit from equation (2), we have f X Profit($) 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55

0.571429 0.56338 0.555556 0.547945 0.540541 0.533333 0.526316 0.519481 0.512821 0.506329 0.5 0.493827

771.4286 789.7183 806.6667 822.3288 836.7568 850 862.1053 873.1169 883.0769 892.0253 900 907.037 6-29

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0.487805 0.481928 0.47619 0.470588 0.465116 0.45977 0.454545 0.449438 0.444444

913.1707 918.4337 922.8571 926.4706 929.3023 931.3793 932.7273 933.3708 933.3333

0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1

0.454545 0.453515 0.452489 0.451467 0.45045 0.449438 0.44843 0.447427 0.446429 0.445434 0.444444

932.7273 932.9116 933.0679 933.1964 933.2973 933.3708 933.417 933.4362 933.4286 933.3942 933.3333

0.96 0.961 0.962 0.963 0.964 0.965 0.966 0.967 0.968 0.969 0.97 0.971 0.972

0.44843 0.44833 0.448229 0.448129 0.448029 0.447928 0.447828 0.447728 0.447628 0.447527 0.447427 0.447327 0.447227 0.447127 0.447027 0.446927 0.446828 0.446728 0.446628 0.446528 0.446429

933.417 933.4202 933.423 933.4256 933.428 933.43 933.4318 933.4333 933.4346 933.4355 933.4362 933.4367 933.4369 933.4368 933.4364 933.4358 933.4349 933.4337 933.4322 933.4305 933.4286

0.973 0.974 0.975 0.976 0.977 0.978 0.979 0.98

6-30

So we see the maximum profit ($933.4369) occurs for f=0.972 Corresponding conversion of A, X = 0.4472

P6-13 A + B -> C

6-31

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 t

0

0

200.

200.

2 Cc

2.

2.

3.893617

3.893617

3 Cb

2.

0.106383

2.

0.106383

4 Ca

2.

0.106383

2.

0.106383

5 k

0.0445

0.0445

0.0445

0.0445

6 Cao

2.

2.

2.

2.

7 ra

-0.178

-0.178

-0.0005036

-0.0005036

8 rb

-0.178

-0.178

-0.0005036

-0.0005036

9 X

0

0

0.9468085

0.9468085

10 rc

0.178

0.0005036

0.178

0.0005036

Differential equations 1 d(Cc)/d(t) = rc 2 d(Cb)/d(t) = rb 3 d(Ca)/d(t) = ra Explicit equations 1 k = 0.0445 6-32

2 Cao = 2 3 ra = -k * Ca * Cb 4 rb = ra 5 X = (Cao - Ca) / Cao 6 rc = -ra

P6-14 The 4 mistakes are as follows: 1) rb = - k * Ca2 2) CA0 = 0.4 and not CTO 3)

CTO = 0.8

4) The 5th equation should have been

alpha FT 2 y FT 0

dy dW

CDGA 6-1 (a) At equilibrium, r = 0 =>

C AC B

CC C D KC

V = V0 + vot

CA

N AO 1 X V

C BO CB

C AO 1 X V VO

C AO 1 X 1

vo t VO

vO t C AO X VO 1

vo t VO

6-33

CC

CD

C AO X 1

C AO

t

vo t VO

v 1 X C BO O t C AO X VO

VO C AO X2 vO C BO K C 1 X

C AO X KC

2

X

Now solving in polymath,

See Polymath program CDGA6-1-a.pol.

CDGA 6-1 (b) See Polymath program CDGA6-1-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb Cc Cd Kc k ra vo Vo V X

initial value 0 7.72 10.93 0 0 1.08 9.0E-05 -0.0075942 0.05 200 200 0

minimal value 0 0.2074331 7.6422086 0 0 1.08 9.0E-05 -0.0075942 0.05 200 200 0

maximal value 1.5E+04 7.72 10.93 3.2877914 3.2877914 1.08 9.0E-05 -1.006E-05 0.05 200 950 0.9731304

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*vo/V [2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93) [3] d(Cc)/d(t) = -ra - vo*Cc/V [4] d(Cd)/d(t) = -ra - vo*Cd/V Explicit equations as entered by the user [1] Kc = 1.08 [2] k = 0.00009 [3] ra = -k*(Ca*Cb - Cc*Cd/Kc) [4] vo = 0.05 [5] Vo = 200

6-34

final value 1.5E+04 0.2074331 9.51217 1.41783 1.41783 1.08 9.0E-05 -1.006E-05 0.05 200 950 0.9731304

[6] V = Vo + vo*t [7] X = 1 - Ca/7.72

Polymath solution

CDGA 6-1 (c) Change the value of vo and CAO in the Polymath program to see the changes.

CDGA 6-1 (d) As ethanol evaporates as fast as it forms: Now using part (b) remaining equations, Polymath code:

CD = 0

See Polymath program CDGA6-1-d.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k ra vo Vo V X

initial value 0 7.72 10.93 9.0E-05 -0.0075942 0.05 200 200 0

minimal value 0 0.0519348 6.9932872 9.0E-05 -0.0075942 0.05 200 200 0

maximal value 6000 7.72 10.93 9.0E-05 -3.69E-05 0.05 200 500 0.9932727

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*vo/V [2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93) Explicit equations as entered by the user

6-35

final value 6000 0.0519348 7.8939348 9.0E-05 -3.69E-05 0.05 200 500 0.9932727

[1] [2] [3] [4] [5] [6]

k = 0.00009 ra = -k*Ca*Cb vo = 0.05 Vo = 200 V = Vo + vo*t X = 1 - Ca/7.72

CDGA 6-1 (e) CDGA 6-1 (f)

Individualized solution Individualized solution

CDGA 6-2 (a) Mole balance on reactor 1:

C A0 v A0 C A1v rA1V C A0 v0 C A1v rA1V 2

dN A1 with v A0 dt dN A1 dt

1 v0 2

Liquid phase reaction so V and v are constant.

C A0 2

C A1

rA1

dC A1 dt

Mole balance on reactor 2:

C A1v0 C A 2 v0 C A1

C A2

rA 2

rA 2V

dN A 2 dt

dC A 2 dt 6-36

Mole balance for reactor 3 is similar to reactor 2:

C A 2 v0 C A3v0 C A2

C A3

rA3V

rA3

dN A3 dt

dC A3 dt

Rate law:

rAi

kC Ai CBi

Stoichiometry For parts a, b, and c CAi = CBi so that

rAi

kC Ai2

Combine:

C A0 2 C A1

C A1

kC A21

C A2

kC A2 2

C A2

C A3

kC A2 3

dC A1 dt dC A 2 dt dC A3 dt

See Polymath program CDGA6-2-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca1 Ca2 Ca3 k Cao tau X

initial value 0 0 0 0 0.025 2 10 1

minimal value 0 0 0 0 0.025 2 10 0.3890413

maximal value 100 0.8284264 0.7043757 0.6109587 0.025 2 10 1

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1^2 [2] d(Ca2)/d(t) = (Ca1 - Ca2)/tau -k*Ca2^2 [3] d(Ca3)/d(t) = (Ca2 - Ca3)/tau -k*Ca3^2

6-37

final value 100 0.8284264 0.7043757 0.6109587 0.025 2 10 0.3890413

Explicit equations as entered by the user [1] k = 0.025 [2] Cao = 2 [3] tau = 10 [4] X = 1 - 2*Ca3/Cao

From Polymath, the steady state conversion of A is approximately 0.39

CDGA 6-2 (b) 99% of the steady state concentration of A (the concentration of A leaving the third reactor) is: (0.99)(0.611) = 0.605 This occurs at t =

CDGA 6-2 (c) The plot was generated from the Polymath program given above.

CDGA 6-2 (d) We must reexamine the mole balance used in parts a-c. The flow rates have changed and so the mole balance on species A will change slightly. Because species B is added to two different reactors we will also need a mole balance for species B. Mole balance on reactor 1 species A:

dN A1 with v A0 dt dN A1 C A1v rA1V dt

C A0 v A0 C A1v rA1V 2C A0 v0 3

2 v0 and v0 3

Liquid phase reaction so V and v are constant.

6-38

200 15

C A0 2

C A1

rA1

dC A1 dt

Mole balance on reactor 1 species B:

dN B1 and vB 0 dt

CB 0 vB 0 CB1v rB1V

1 v0 3

Stoichiometry has not changed so that –rAi = -rBi and it is a liquid phase reaction with V and v constant.

CB 0 3

CB1

rA1

dCB1 dt

Mole balance on reactor 2 species A: We are adding more of the feed of species B into this reactor such that v2 = v0 + vB0 = 20

C A1v0 C A 2 v2

C A1 1

C A2

rA2

2

rA 2V

dN A 2 dt

dC A2 where dt

1

V and v0

2

V v2

Mole balance on reactor 2 species B:

dN B 2 dt

CB1v0 CB 0 vB 0 CB 2 v rB 2V

CB1 1

C B 0 vB 0 V

CB 2

rA2

2

dCB 2 dt

Mole balance for reactor 3 species A:

C A 2 v2 C A3v2

C A2 2

C A3

rA3

2

rA3V

dN A3 dt

dC A3 dt

Mole balance for reactor 3 species B:

CB 2 v2 CB 3v2

CB 2 2

CB 3 2

rA3

rA3V

dN B 3 dt

dCB 3 dt 6-39

Rate law:

rAi

kC Ai CBi

See Polymath program CDGA6-2-d.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca1 Ca2 Ca3 Cb1 Cb2 Cb3 k Cao tau X tau2 V vbo

initial value 0 0 0 0 0 0 0 0.025 2 13.333333 1 10 200 5

minimal value 0 0 0 0 0 0 0 0.025 2 13.333333 0.3721856 10 200 5

maximal value 100 1.1484734 0.7281523 0.6278144 0.4843801 0.7349863 0.6390576 0.025 2 13.333333 1 10 200 5

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca1)/d(t) = (2*Cao/3 -Ca1)/tau -k*Ca1*Cb1 [2] d(Ca2)/d(t) = Ca1/tau - Ca2/tau2 -k*Ca2*Cb2 [3] d(Ca3)/d(t) = (Ca2 - Ca3)/tau2 -k*Ca3*Cb3 [4] d(Cb1)/d(t) = (1*Cao/3-Cb1)/tau-k*Ca1*Cb1 [5] d(Cb2)/d(t) = Cb1/tau+Cao*vbo/V-Cb2/tau2-k*Ca2*Cb2 [6] d(Cb3)/d(t) = (Cb2-Cb3)/tau2-k*Ca3*Cb3 Explicit equations as entered by the user [1] k = 0.025 [2] Cao = 2 [3] tau = 200/15 [4] X = 1 - 2*Ca3/Cao [5] tau2 = 10 [6] V = 200 [7] vbo = 5

Equilibrium conversion is 0.372. This conversion is reached at t = 85.3 minutes.

6-40

final value 100 1.1484734 0.7281523 0.6278144 0.4821755 0.7291677 0.6309679 0.025 2 13.333333 0.3721856 10 200 5

CDGA 6-2 (e) Individualized solution

6-41





7-1

Solutions for Chapter 7 – Collection and Analysis of Rate Data P7-1 (a) Individualized solution P7-1 (b) Individualized solution P7-1 (c) Individualized solution P7-1 (d) Individualized solution

P7-2 (a) In Example 7-1, we have assumed CB as constant, but CB is varying and the error is due to the term

and the maximum error occurs when this ratio is minimum.

Error = 1 CB = CA0(Θ - X) Θ = 0.5/0.05 = 10 CB = CA0(10 - X) So, we have t(min) 0 50 100 150 200 250 300

CA 0.05 0.038 0.0306 0. 256 0.0222 0.0195 0.0174

X 0 0.024 0.0388 0.0488 0.0556 0.061 0.0652

CB 0.5 0.4988 0.49806 0.49756 0.49722 0.49695 0.49674

error(%) 0 0.24 0.388 0.488 0.556 0.61 0.652

So, we have maximum error of 0.652%, hence k = 0.24

.652*0.24/100

k = 0.24

0.0015648

P7-2 (b) Example 7-3 because when α is set equal to 2, the best value of k must be found. P7-2 (c) Example 7-4   rate law: rCH 4  kPCO PH 2

7-2

Regressing the data r’(gmolCH4/gcat.min) 5.2e-3 13.2e-3 30e-3 4.95e-3 7.42e-3 5.25e-3

PCO (atm) 1 1.8 4.08 1 1 1

PH2 (atm) 1 1 1 0.1 0.5 4

See Polymath program P7-2-c.pol. POLYMATH Results Nonlinear regression (L-M) Model: r = k*(PCOâlfa)*(PH2^beta) Variable Ini guess Value k 0.1 0.0060979 alfa 1 1.1381455 beta 1 0.0103839 Precision R^2 = 0.9869709 R^2adj = 0.9782849 Rmsd = 4.176E-04 Variance = 2.093E-06

95% confidence 6.449E-04 0.0850634 0.1951217

Therefore order of reaction = 1.14 Again regressing the above data putting   1 POLYMATH Results Nonlinear regression (L-M) Model: r = k*(PCO^0.14)*(PH2) Variable Ini guess Value k 0.1 0.0040792 Precision R^2 = -0.8194508 R^2adj = -0.8194508 Rmsd = 0.0049354 Variance = 1.754E-04

95% confidence 0.0076284

Therefore, k = 0.004 (gmolCH4/(gcat.min.atm1.14))

P7-3

Solution is in the decoding algorithm given with the modules.

P7-4


7-3

P7-5 (a)

P7-5 (b) First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)

See Polymath program P7-5-b.pol. POLYMATH Results Polynomial Regression Report Model: X = a0 + a1*z + a2*z^2 + a3*z^3 + a4*z^4 + a5*z^5 + a6*z^6 Variable

Value

95% confidence

7-4

a0 a1 a2 a3 a4 a5 a6

2.918E-14 0.0040267 -6.14E-05 7.767E-06 -5.0E-07 1.467E-08 -1.6E-10

0 0 0 0 0 0 0

General Order of polynomial = 6 Regression including free parameter Number of observations = 7 Statistics R^2 = R^2adj = Rmsd = Variance =

1 0 1.669E-10 1.0E+99

Next we differentiate our expression of X(z) to find dX/dz and knowing that

 dX  ln    ln a  n ln 1  X   dz  n kC Ao Ac where, a  FAo  dX   as a function of ln 1  X  gives us similar vaules of slope and intercept  dz 

Linear regression of ln 

as in the finite differences. POLYMATH Results Linear Regression Report Model: ln(dxdz) = a0 + a1*ln(1-X) Variable a0 a1

Value -5.531947 1.2824279

95% confidence 0.0241574 0.3446187

General Regression including free parameter Number of observations = 7 Statistics R^2 = R^2adj = Rmsd = Variance =

0.9482059 0.9378471 0.0044015 1.899E-04

n = 1.28 ln a = -5.53, a = 0.00396 7-5

k

FAO a 45.7 106 moles / s  3.96 103 cm  4.0s 1 n 6 3 2  CAO AC 2.3 10 moles / cm  0.0196cm

Hence rate law is,

 rA  4.0 C 1.28 A

mol dm3 s

P7-6 (a) Liquid phase irreversible reaction: A  B + C ; CAO = 2 mole/dm3

C AO  C A

 kC A

  C  CA  ln  AO   ln k   ln C A    Space time (  )min. CA(mol/dm3) ln(CA) 15 38 100 300 1200

1.5 1.25 1.0 0.75 0.5

0.40546511 0.22314355 0 -0.28768207 -0.69314718

ln((CAO-CA)/  ) -3.4011974 -3.9252682 -4.6051702 -5.4806389 -6.6846117

By using linear regression in polymath:

See Polymath program P7-6-a.pol.

POLYMATH Results Linear Regression Report Model: y = a0 + a1*lnCa

 C  CA  ln  AO   ln k   ln C A    Variable

Value

95% confidence

7-6

a0 -4.6080579 0.0162119 a1 2.9998151 0.0411145 Statistics R^2 = 0.9999443 R^2adj = 0.9999258 Rmsd = 0.003883 Variance = 1.256E-04

Hence,

  slope  3 ln(k) = intercept = -4.6 therefore, k = 0.01 mole-2min-1. Rate law:



dC A  0.01C A3 mol / dm3 min dt

P7-6 (b) Individualized solution P7-6 (c) Individualized solution

P7-7 (a) Constant voume batch reactor: Mole balance:



A B +C

dC A  kC A dt

Integrating with initial condition when t = 0 and CA = CAO for   1.0

t

(1 )  C A(1 ) 1 (2)(1 )  C A(1 ) 1 C AO  ……………substituting for initial concentration CAO = 2 mol/dm3. k (1   ) k (1   )

t (min.) 0 5 9 15 22 30 40 60

CA (mol/dm3) 2 1.6 1.35 1.1 0.87 0.70 0.53 0.35

See Polymath program P7-7-a.pol. POLYMATH Results Nonlinear regression (L-M) Model: t = (1/k)*((2^(1-alfa))-(Ca^(1-alfa)))/(1-alfa) Variable Ini guess Value 95% confidence k 0.1 0.0329798 3.628E-04

7-7

alfa 2 Precision R^2 = 0.9997773 R^2adj = 0.9997327 Rmsd = 0.1007934 Variance = 0.0995612

1.5151242

0.0433727

K= 0.03 (mol/dm3)-0.5.s-1 and   1.5 Hence , rate law is



dC A 3  0.03C 1.5 A mol / dm .s dt

P7-7 (b) Individualized solution P7-7 (c) Individualized solution P7-7 (d) Individualized solution

P7-8 (a) Liquid phase reaction of methanol and triphenyl in a batch rector. CH3OH + (C6H5)3CCl  (C6H5)3COCH3 + HCl A + B  C + D Using second set of data, when CAO = 0.01 mol/dm3 and CBO = 0.1 mol/dm3 CA(mol/dm3) 0.1 0.0847 0.0735 0.0526 0.0357

t (h) 0 1 2 5 10

Rate law:

 rA  kC Am C Bn n For table 2 data: CAO  CBO =>  rA  k ' C Am where k '  kCBO

(1 m )  C A(1 m ) 1 (0.01)(1m )  C A(1m ) 1 C AO  k' (1  m) k' (1  m) See Polymath program P7-8-a-1.pol.

Using eqn E7-3.1, t 

POLYMATH Results Nonlinear regression (L-M) Model: t = (1/k)*((0.1^(1-m))-(Ca^(1-m)))/(1-m) Variable k m

Ini guess 1 2

Value 1.815656 2.0027694

95% confidence 0.0109025 0.0021115

7-8

Nonlinear regression settings Max # iterations = 64 Precision R^2 R^2adj Rmsd Variance

= = = =

1 0.9999999 3.268E-04 8.902E-07

Therefore, m = 2 For first set of data, equal molar feed => CA = CB Hence, rate law becomes rA  kC A2CBn  kC A(2 n ) Observation table 2: for CA0 =0.01 and CB0 = 0.1 t (h) 0 0.278 1.389 2.78 8.33 16.66

t

CA(mol/dm3) 1.0 0.95 0.816 0.707 0.50 0.37

(1 (2  n ))  C A(1(2 n )) 1 (0.1)( 1n )  C A( 1n ) 1 C AO  k (1  (2  n)) k (1  n))

See Polymath program P7-8-a-2.pol. POLYMATH Results Nonlinear regression (L-M) Model: t = (1^(-1-n)-Ca^(-1-n))/(k*(-1-n)) Variable n k

Ini guess 3 2

Value 0.8319298 0.1695108

95% confidence 0.0913065 0.0092096


= = = =

0.9999078 0.9998848 0.0233151 0.0048923

Therefore, n = 0.8

7-9

Hence rate law is:

 rA  0.17 C A2 CB0.8

mol dm3 h

P7-8 (b) Individualized solution

P7-9 (a) At t = 0, there is only (CH3)2O. At t = ∞, there is no(CH3)2O. Since for every mole of (CH3)2O consumed there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure. P(∞) = 3P0 931 = 3P0 P0 ≈ 310 mm Hg

P7-9 (b) Constant volume reactor at T = 504°C = 777 K Data for the decomposition of dimethylether in a gas phase: Time PT(mm Hg)

0 312

390 408

777 488

1195 562

(CH 3 ) 2 O  CH 4  H 2  CO y A0  1

  3 1  2    y A0  2 P  V  V0  0  1   X   V0 because the volume is constant. P

P  P0 1   X  at t = ∞, X = XAF = 1

N dX 1 dN A   A0  rA V dt V0 dt Assume  rA  kC A (i.e. 1st order)

C A  C A0 1  X  (V is constant) dX  kC A0 1  X  dt P  P0 and X   P0 Then: C A0

7-10

3155 799



931

Therefore:

dX 1 dP  dt  P0 dt

 P  P0  1 dP k  k 1  1   P0  P    P0 dt  P0   P0  dP  k 1   P0  P  or dt P

t

dP P 1   P0  P  0 kdt 0 

  2 P0   P0  624   kt   ln    ln   936  P   3P0  P   1    P0  P 

Integrating gives: ln 

Therefore, if a plot of ln

624 versus time is linear, the reaction is first order. From the figure below, 936  P

we can see that the plot is linear with a slope of 0.00048. Therefore the rate law is:

 rA  0.00048C A 1.6 y = 0.00048x - 0.02907

1.2 0.8 0.4 0 -0.4 0

1000

2000

3000

4000

P7-9 (c) Individualized solution P7-9 (d) The rate constant would increase with an increase in temperature.

This would result in the

pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is true for colder temperatures.

7-11

P7-10 (a) Photochemical decay of bromine in bright sunlight: t (min) CA (ppm)

10 2.45

20 1.74

30 1.23

40 0.88

50 0.62

60 0.44

Mole balance: constant V

dC A  rA  kC A dt

 dC  ln   A   ln k    ln C A   dt  Differentiation T (min) Δt (min) CA (ppm) ΔCA (ppm)

C A  ppm    t  min 

10

20 10

2.45

30 10

1.74

40 10

1.23

50 10

0.88

60 10

0.62

-0.71

-0.51

-0.35

-0.26

-0.18

-0.071

-0.051

-0.035

-0.026

-0.018

7-12

10 0.44

After plotting and differentiating by equal area -dCA/dt ln(-dCA/dt) ln CA

0.082 -2.501 0.896

0.061 -2.797 0.554

0.042 -3.170 0.207

0.030 -3.507 -0.128

0.0215 -3.840 -0.478

0.014 -4.269 -0.821

Using linear regression: α = 1.0 ln k = -3.3864 k = 0.0344 min-1

P7-10 (b) dN A  VrA  FB  0 dt ppm mg at CA = 1 ppm rA  0.0344  0.0344 min l min 

 mg   min   1 g  3.7851l  1lbs  lbs FB  25000 gal  0.0344   60     0.426  l min   hr   1000 mg  1 gal  453.6 g  hr 

P7-10 (c) Individualized solution

P7-11 For the reaction,

Oz + wall  loss of O3 Oz + alkene  products

rO z 

Rate law:

dCO z dt

k1 k2

m n  k 1CO z  k 2CO C z Bu

7-13



Using polymath nonlinear regression we can find the values of m, n, k1 and k2

See Polymath program P7-11.pol.

7-14

P7-12 Given: Plot of percent decomposition of NO2 vs V/FA0

% Decomposition of NO 2 100 Assume that  rA  kC An X=

For a CSTR V  or

FA0 X rA

V X X   FA0 rA kC An

with n = 0, X  k

V FA0

X has a linear relationship with

V as FA0

shown in the figure. 7-15

Therefore the reaction is zero order.

P7-13 SiO2  6 HF  H 2 SiF6  2 H 2O

NS = moles of SiO 2 =

A C ρS δ MWS

AC = cross-sectional area ρS = silicon dioxide density MWS = molecular weight of silicon dioxide = 60.0 δ = depth of Si

N F = moles of HF=

w V 100 MWF

w = weight percentage of HF in solution ρ = density of solution V = volume of solution MWF = molecular weight of HF = 20.0 Assume the rate law is rS  kCF Mole balance:

dN S  rSV dt 

  A  d wV  C S k  V MWS dt  100V MWF  kMWS d   dt 100 AC  S



      Vw  MWF 

kMWS d    w where   dt 100 AC  S



     V  MWF 

 d  ln     ln    ln w  dt   d  ln     dt  ln w

-16.629 2.079

-15.425 2.996

-14.32 3.497

-13.816 3.689

7-16

-13.479 3.871

m  d   is in min  dt 

where  

 d   and ln w we have:  dt 

From linear regression between ln  

slope = α = 1.775 intercept = ln β = -20.462 or β = 1.2986 * 10-9

1.775

      MWF 

kMWS  1001.775 AC  S



V



AC  10*106 m 10m 2sides 1000 wafers   0.2 m2

MWS  60

 S  2.32

g gmol

g g  2.32*106 3 ml m

(Handbook of Chemistry and Physics, 57th ed., p.B-155)

g g  106 3 ml m g MWF  20 gmol

 1

V  0.5 dm3  0.0005 m3

  1.2986*109  6 g  10 3 9 m 1.2986*10    20 g  gmol 

1.775

     

 g  k  60   gmol  0.0005 m3 g  1.775  0.2 m2  2.3*106 3  100  m  







7-17



 m3  k  3.224*10    gmol 

0.775

7

Final concentration of HF 

min 1 5  2.316 0.2   0.107 5

weight fraction = 10.7%

Initial concentration of HF = 0.2 (given)

Mole balance for HF:

dN dN F 6 S dt dt 

  dw w   6k   V 100MWF dt  100 MWF 

V

10.7





20

weight fraction = 20%

  dw   6k   1.775 w  100MWF  10.7

where α = 1.775

0.775 t

 dt 0

1  1  106  7   6 3.224*10     0.775  w0.775  20  20*100 





0.775

t

1  1 1   0.775   2.389*104 t  0.775 0.775  10.7 20  t  331 min

P7-14

See Polymath program P7-14.pol

7-18

P7-15 The following are the mistakes in the given solution, followed by the corrected solution:-

AB+C For a batch Reactor,

dC A  rA  kC A dt dC  A  kC A dt Taking ln both sides,

ln(

dC A )  ln k   ln C A dt

7-19

(1) The value of △CA/△t cannot be calculated for t=0. It can only be calculated for t=1 onwards. (2) The values of △CA/△t calculated for t=1, 2, 3 etc. have been incorrectly calculated. t

Ca

ln Ca

Delta t

delta Ca/ delta t ln ( -delta Ca/ delta t)

0

1

0

1

0.5

-0.7

1

-0.5

-0.7

2

0.3

-1.2

1

-0.2

-1.61

3

0.2

-1.61

1

-0.1

-2.3

4

0.15

-1.9

1

-0.05

-3

(3) The graph has to be plotted for ln (-△CA/△t) vs. ln (CA) and not ‘t’.

(4) So, the value of α from the graph = 1.88, not 0.5. (5) ln k = 0.64 from the graph, which gives k = 1.89 Therefore, we get

dC A  rA  1.89C1.88 A dt 7-20





8-1

Solutions for Chapter 8 – Multiple Reactions P8-1 Individualized solution

P8-2 (a) Example 8-1 For PFR (gas phase with no pressure drop or liquid phase),

dC A d dC B d

k1

k 2C A

dC X d dCY d

k 3C A2

k 2C A

k1 k 3 C A2

In PFR with V = 1566 dm3 we get τ = V/v0 = 783 X = 0.958 , SB/XY (instanteneous) = 0.244, and SB/XY (overall) = 0.624 also at τ = 350, SB/XY (instanteneous) is at its maximum value of 0.84 See polymath problem P8-2-1.pol

Calculated values of the DEQ variables Variable initial value tau 0 Ca 0.4 Cx 1.0E-07 Cb 0 Cy 1.0E-06 Cao 0.4 X 0 k1 1.0E-04 k2 0.0015 k3 0.008 Sbxy_over 0 Sbxy_inst 0.4347826

minimal value 0 0.0166165 1.0E-07 0 1.0E-06 0.4 0 1.0E-04 0.0015 0.008 0 0.2438609

maximal value 783 0.4 0.0783001 0.1472919 0.1577926 0.4 0.9584587 1.0E-04 0.0015 0.008 0.6437001 0.8385161

8-2

final value 783 0.0166165 0.0783001 0.1472919 0.1577926 0.4 0.9584587 1.0E-04 0.0015 0.008 0.6238731 0.2438609

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(tau) = -k1-k2*Ca-k3*Ca^2 [2] d(Cx)/d(tau) = k1 [3] d(Cb)/d(tau) = k2*Ca [4] d(Cy)/d(tau) = k3*Ca^2 Explicit equations as entered by the user [1] Cao = 0.4 [2] X = 1-Ca/Cao [3] k1 = 0.0001 [4] k2 = 0.0015 [5] k3 = 0.008 [6] Sbxy_overall = Cb/(Cx+Cy) [7] Sbxy_instant = k2*Ca/(k1+k3*Ca^2) Independent variable variable name : tau initial value : 0 final value : 783

(2) PFR - Pressure increased by a factor of 100. (a) Liquid phase: No change, as pressure does not change the liquid volume appreciably. (b) Gas Phase: Now CA0 = P/RT = 100 (P0_initial)/RT = 0.4 × 100 = 40 mol/dm3 Running with CA0 = 40 mol/dm3, we get: X = 0.998 , SB/XY (instanteneous) = 0.681, and SB/XY (overall) = 0.024

As a practice, students could also try similar problem with the CSTR.

P8-2 (b) Example 8-2 (a) CSTR: intense agitation is needed, good temperature control.

8-3

(b) PFR: High conversion attainable, temperature control is hard – non-exothermic reactions, selectivity not an issue (c) Batch: High conversion required, expensive products (d) and (e) Semibatch: Highly exothermic reactions, selectivity i.e. to keep a reactant concentration low, to control the conversion of a reactant. (f) and (g) Tubular with side streams: selectivity i.e. to keep a reactant concentration high, to achieve higher conversion of a reactant. (h) Series of CSTR’s: To keep a reactant concentration high, easier temperature control than single CSTR. (i) PFR with recycle: Low conversion to reuse reactants, gas reactants, for highly exothermic reactions (j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid reactants , for highly exothermic reactions (k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product , used for thermodynamically limited reactions where the equilibrium shifts on left side. (l) Reactive Distillation: when one product is volatile and the other is not ; for thermodynamically limited gas phase reactions.

P8-2 (c) Example 8-3 (1) For k1 = k2, we get

CA

C AO exp

'

k1

and

d

d C B exp k1 '

dCB d '

Wopt

CB 0

' opt

vo k1

k1C B

k1C AO exp

k1

'

k1C AO

0 , CB = 0

Optimum yield:

'

'

'

d Now at

dC B

k1C AO ' exp

k1C AO exp

k1

'

k1

'

k1 ' exp

k1

'

1 k1 and

X opt

1 exp

k1

'

1 e 1 =0.632

P8-2 (d) Example 8-4 According to the plot of CB versus T the maximum temperature is T = 310.52 0C.

8-4

CB

k2

k1C A0 1 k1 1

POLYMATH Results See polymath problem P8-2-1.pol Calculated values of the DEQ variables Variable t T Cao tau k1o k2o E1 E2 R k1 k2 Cb

initial value 0 300 5 0.5 0.4 0.01 10 20 0.001987 0.4 0.4 0.6944444

minimal value 0 300 5 0.5 0.4 0.01 10 20 0.001987 0.4 0.4 0.0052852

maximal value 100 400 5 0.5 0.4 0.01 10 20 0.001987 26.513034 1757.3524 0.8040226

ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 1 Explicit equations as entered by the user [1] Cao = 5 [2] tau = .5 [3] k1o = .4 [4] k2o = .01 [5] E1 = 10 [6] E2 = 20 [7] R = .001987 [8] k1 = k1o*exp((-E1/R)*(1/T-1/300))

8-5

final value 100 400 5 0.5 0.4 0.01 10 20 0.001987 26.513034 1757.3524 0.0052852

[9] k2 = k1o*exp((-E2/R)*(1/T-1/300)) [10] Cb = (tau*k1*Cao)/(tau*k2+1)/(tau*k1+1)

P8-2 (e) Individualized solution P8-2 (f) Individualized solution P8-2 (g) Individualized solution P8-2 (h) Example 8-8 (1) Membrane Reactor PFR

SD/U Original Problem 2.58 0.666

SD/U P8-2 h 1.01 0.208

Doubling the incoming flow rate of species B lowers the selectivity. (2) The selectivity becomes 6.52 when the first reaction is changed to A+2B  D

P8-2 (i) CDROM Example Original Case – CDROM example

P8-2 i

The reaction does not go as far to completion when the changes are made. The exiting concentration of D, E, and F are lower, and A, B, and C are higher. See Polymath program P8-2-i.pol.

P8-2 (j) For equal molar feed in hydrogen and mesitylene. CHO = yHOCTO = (0.5)(0.032)lbmol/ft3 =0.016 lbmol/ft3 CMO = 0.016 lbmol/ft3 Using equations from example, solving in Polymath, we get

8-6

opt

0.38 hr. At

0.5 hr all of the H2 is reacted and only the decomposition of X takes place. XH CH CM CX ~

S X /T

CD-ROM example 0.50 0.0105 0.0027 0.00507 0.2hr 0.596

This question 0.99 0.00016 0.0042 0.0077 0.38hr 1.865

See Polymath program P8-2-j.pol. POLYMATH Results Calculated values of the DEQ variables Variable tau CH CM CX k1 k2 r1M r2T r1H r2H r1X r2X

initial value 0 0.016 0.016 0 55.2 30.2 -0.1117169 0 -0.1117169 0 0.1117169 0

minimal value 0 1.64E-06 0.0041405 0 55.2 30.2 -0.1117169 0 -0.1117169 -0.0159818 2.927E-04 -0.0159818

maximal value 0.43 0.016 0.016 0.0077216 55.2 30.2 -2.927E-04 0.0159818 -2.927E-04 0 0.1117169 0

ODE Report (RKF45) Differential equations as entered by the user [1] d(CH)/d(tau) = r1H+r2H [2] d(CM)/d(tau) = r1M [3] d(CX)/d(tau) = r1X+r2X Explicit equations as entered by the user [1] k1 = 55.2 [2] k2 = 30.2 [3] r1M = -k1*CM*(CH^.5) [4] r2T = k2*CX*(CH^.5) [5] r1H = r1M [6] r2H = -r2T [7] r1X = -r1M [8] r2X = -r2T ~

Increasing θH decreases τopt and S X/T.

P8-2 (k)

8-7

final value 0.43 1.64E-06 0.0041405 0.0077207 55.2 30.2 -2.927E-04 2.986E-04 -2.927E-04 -2.986E-04 2.927E-04 -2.986E-04

From CD-ROM example Polymath code: See Polymath program P8-2-k.pol. POLYMATH Results NLES Solution Variable CH CM CX tau K1 K2 CHo CMo

Value 4.783E-05 0.0134353 0.0023222 0.5 55.2 30.2 0.016 0.016

f(x) -4.889E-11 -1.047E-11 -9.771E-12

Ini Guess 1.0E-04 0.013 0.002

NLES Report (safenewt)

Nonlinear equations [1] f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0 [2] f(CM) = CM-CMo+K1*CM*CH^.5*tau = 0 [3] f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tau-CX = 0

Explicit equations [1] [2] [3] [4] [5]

tau = 0.5 K1 = 55.2 K2 = 30.2 CHo = 0.016 CMo = 0.016

A plot using different values of is given. For =0.5, the exit concentration are CH = 4.8 ×10-5 lbmol/ft3 CM =0.0134 lbmol/ft3 CX =0.00232 lbmol/ft3 The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for

YMX

FX FMO

CX FM

C MO

CM

0.00232 0.016 0.0134

0.89mole xylene produced mole mesitylene reacted

The overall selectivity of xylene relative to toluene is:

FX FT

~ S X /T CH CM CX

YMX

8.3mole xylene produced mole toluene produced CD-ROM example 0.0089 0.0029 0.0033 0.5 0.41

This Question 4.8 x 10-5 0.0134 0.00232 0.5 0.89 8-8

=0.5:

SX/T

0.7

8.3

P8-2 (l)

At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As the two components are created, the reactant concentration drops and equilibrium forces the production to slow. At the same time the reactions that consume the two components begin to accelerate and the concentration of TF-VIIa and TF-VIIaX decrease. As those reactions reach equilibrium, the reactions that are still producing the two components are still going and the concentration rises again. Finally the reactions that consume the two components lower the concentration as the products of those reactions are used up in other reactions.

P8-2 (m) Individualized solution

P8-3 Solution is in the decoding algorithm given with the modules (ICM problem )

P8-4 (a) Assume that all the bites will deliver the standard volume of venom. This means that the initial concentration increases by 5e-9 M for every bite. After 11 bites, no amount of antivenom can keep the number of free sites above 66.7% of total sites. This means that the initial concentration of venom would be 5.5e-8 M. The best result occurs when a dose of antivenom such that the initial concentration of antivenom in the body is 5.7e-8 M, will result in a minimum of 66.48% free sites, which is below the allowable minimum.

P8-4 (b) 8-9

The victim was bitten by a harmless snake and antivenom was injected. This means that the initial concentration of venom is 0. From the program below, we see that if an amount of antivenom such that the initial concentration in the blood is 7e-9 M, the patient will die. See Polymath program P8-4-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable t fsv fs Cv Ca fsa Cp kv ksv ka kia Cso ksa kp kov koa kop g h m j

initial value 0 0 1 0 7.0E-09 0 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -2.1E-09

minimal value 0 0 0.6655661 0 4.503E-09 0 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -2.1E-09

maximal value 0.5 0 1 0 7.0E-09 0.3344339 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -1.351E-09

ODE Report (STIFF) Differential equations as entered by the user [1] d(fsv)/d(t) = kv * fs * Cv - ksv * fsv * Ca [2] d(fs)/d(t) = -kv*fs*Cv - ka * fs * Ca + kia * fsa + g [3] d(Cv)/d(t) = Cso * (-kv * fs * Cv - ksa * fsa * Cv) + h [4] d(Ca)/d(t) = Cso*(-ka * fs * Ca + kia * fsa) + j [5] d(fsa)/d(t) = ka * fs * Ca - kia * fsa - ksa * fsa * Cv [6] d(Cp)/d(t) = Cso * (ksv * fsv * Ca + ksa * fsa * Cv) + m

Explicit equations as entered by the user [1] kv = 2e8 [2] ksv = 6e8 [3] ka = 2e8 [4] kia = 1 [5] Cso = 5e-9 [6] ksa = 6e8

8-10

final value 0.5 0 0.6655661 0 4.503E-09 0.3344339 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -1.351E-09

[7] kp = 1.2e9 [8] kov = 0 [9] koa = 0.3 [10] kop = 0.3 [11] g = ksa * fsa * Cv + ksv * fsv * Ca [12] h = -kp * Cv * Ca - kov * Cv [13] m = kp * Cv * Ca - kop * Cp [14] j = -Cso * ksv * fsv * Ca - kp * Cv * Ca - koa * Ca

P8-4 (c) The latest time after being bitten that antivenom can successfully be administerd is 27.49 minutes. See the cobra web module on the CDROM/website for a more detailed solution to this problem

P8-4 (d) Individualized Solution

P8-5 (a) Plot of CA, CD and CU as a function of time (t): See Polymath program P8-5-a.pol. POLYMATH Results Calculated values of the DEQ variables

8-11

Variable t Ca Cd Cu k1 k2 K1a K2a Cao X

initial value 0 1 0 0 1 100 10 1.5 1 0

minimal value 0 0.0801802 0 0 1 100 10 1.5 1 0

maximal value 15 1 0.7995475 0.5302179 1 100 10 1.5 1 0.9198198

final value 15 0.0801802 0.7995475 0.1202723 1 100 10 1.5 1 0.9198198

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a)) [2] d(Cd)/d(t) = k1*(Ca-Cd/K1a) [3] d(Cu)/d(t) = k2*(Ca-Cu/K2a) Explicit equations as entered by the user [1] k1 = 1.0 [2] k2 = 100 [3] K1a = 10 [4] K2a = 1.5 [5] Cao = 1 [6] X = 1-Ca/Cao

To maximize CD stop the reaction after a long time. The concentration of D only increases with time

P8-5 (b) Conc. Of U is maximum at t = 0.31 min.(CA = 0.53)

P8-5 (c) Equilibrium concentrations: CAe = 0.08 mol/dm3 CDe = 0.8 mol/dm3 CUe = 0.12 mol/dm3 8-12

P8-5 (d) See Polymath program P8-5-d.pol. POLYMATH Results NLES Solution Variable Ca Cd Cu Ca0 k1 k2 K1a K2a t

Value 0.0862762 0.7843289 0.1293949 1 1 100 10 1.5 100

f(x) -3.844E-14 -2.631E-14 6.478E-14

Ini Guess 1 0 0


Nonlinear equations [1] f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca = 0 [2] f(Cd) = t*k1*(Ca-Cd/K1a)-Cd = 0 [3] f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0

1 min

10 min

100min

CAexit

0.295

0.133

0.0862

CDexit

0.2684

0.666

0.784

CUexit

0.436

0.199

0.129

X

0.705

0.867

0.914

Explicit equations [1] [2] [3] [4] [5] [6] [7]

Ca0 = 1 k1 = 1 k2 = 100 K1a = 10 K2a = 1.5 t = 100 X = 1-Ca/Ca0

P8-6 (a)

A

X

rX

1/ 2 k1C A

A

B

rB

k 2C A

A

Y

rY

k 3C A

2

1/ 2

k1

mol 0.004 dm 3

k2

0.3 min

k3

dm 3 0.25 mol. min

Sketch SBX, SBY and SB/XY as a function of CA

8-13

1

min

See Polymath program P8-6-a.pol. It shows the table of SBX, SBY and SB/XY in correspondence with values of CA .

1)

SB/X =

rB rX

2)

SB/Y =

rB rY

3)

SB/XY =

k 2C A k1C A1/ 2

k 2C A k 3C A 2

rB rX

k2 C A1/ 2 k1

k2 k 3C A

k 2C A rY

k1C A1/ 2

k 3C A 2

8-14

P8-6 (b) Volume of first reactor can be found as follows We have to maximize SB/XY From the graph above, maximum value of SBXY = 10 occurs at CA* = 0.040 mol/dm3 So, a CSTR should be used with exit concentration CA* Also, CA0 = PA/RT = 0.162 mol/dm3 And

=> V

rA

rX

rB

rY

v0 (C A0 C A* ) rA

(k1C A1/ 2

k 3C A 2 )

k 2C A

v0 (C A0 * 1/ 2

(k1 (C A )

C A* )

k 2C A

*

* 2

k 3 (C A ) )

P8-6 (c) Effluent concentrations:

CB rB

We know, τ = 9.24 min => Similarly: C X

*

0.007

mol dm 3

and

CB k 2C A CY *

CB* 0.0037

P8-6 (d) Conversion of A in the first reactor:

C A0

CA

C A0 X

X

0.74

P8-6 (e) A CSTR followed by a PFR should be used. Required conversion = 0.99 => For PFR, Mole balance:

dV dX

F A0 rA

8-15

0.11 mol dm 3

mol dm 3

92.4dm 3

0.99

V

10 0.162

dX

1/ 2 0.74 ( k1C A

k 2C A

2

92.8dm 3

k 3C A )

P8-6 (f) If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of B, decrease as temperature drops. So we have to compromise between high selectivity and production. To do this we need expressions for k1, k2, and k3 in terms of temperature. From the given data we know:

ki

Ei 1.98T

Ai exp

Since we have the constants given at T = 300 K, we can solve for Ai.

A1

A2

A3

.004

1.49e12

20000 exp 1.98 300

.3

5.79e6

10000 exp 1.98 300 .25

1.798e21

30000 exp 1.98 300

Now we use a mole balance on species A

V

V

FA0

FA rA

v C A0 C A rA C A0 C A rA

C A0 C A k1C k2C A k3C A2 0.5 A

A mole balance on the other species gives us:

Fi

vCi

Ci

ri

rV i

8-16

Cb versus temp 0.072 0.07 0.068 0.066 Cb 0.064 0.062 0.06 285

290

295

300

305

310

315

Using these equations we can make a Polymath program and by varying the temperature, we can find a maximum value for CB at T = 306 K. At this temperature the selectivity is only 5.9. This may result in too much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal temperature will depend on the price of B and the cost of removing X and Y, but without actual data, we can only state for certain that the optimal temperature will be equal to or less than 306 K. See Polymath program P8-6-f.pol. POLYMATH Results NLE Solution Variable Ca T R k1 k2 Cao Cb k3 tau Cx Cy Sbxy

Value 0.0170239 306 1.987 0.0077215 0.4168076 0.1 0.070957 0.6707505 10 0.0100747 0.0019439 5.9039386

f(x) 3.663E-10

Ini Guess 0.05

NLE Report (safenewt)

Nonlinear equations [1] f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0

Explicit equations [1] T = 306

8-17

[2] R = 1.987 [3] k1 = 1.49e12*exp(-20000/R/T) [4] k2 = 5790000*exp(-10000/R/T) [5] Cao = .1 [6] Cb = 10*k2*Ca [7] k3 = 1.798e21*exp(-30000/R/T) [8] tau = 10 [9] Cx = tau*k1*Ca^.5 [10] Cy = tau*k3*Ca^2 [11] Sbxy = Cb/(Cx+Cy)

P8-6 (g) Concentration is proportional to pressure in a gas-phase system. Therefore:

PA

S B / XY ~

PA

PA2

which would suggest that a low pressure would be ideal. But as before the

tradeoff is lower production of B. A moderate pressure would probably be best. We know that : SB/XY =

rB rX

k 2C A rY

k1C A1/ 2

k 3C A 2

Substituting PA = CA R T ; in the above co – relation we get

rB SB/XY = rX

rY

k2 ( PA / RT ) k1 ( PA / RT )1/2 k3 ( PA / RT ) 2

See Polymath program P8-6-g.pol.

Thus we find that the maximum is obtained at P= 1 atm . 8-18

P8-7 US legal limit: 0.8 g/l Sweden legal limit: 0.5 g/l

A

k1

B

k2

C

Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stream

dC A k1C A dt dCB k1C A k2 dt k1 10 hr 1

k2

0.192

g L hr

Two tall martinis = 80 g of ethanol Body fluid = 40 L

C A0

80 g 40 L

2

g L

Now we can put the equations into Polymath. See Polymath program P8-7.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0 2 0 10 0.192

minimal value 0 7.131E-44 0 10 0.192

maximal value 10 2 1.8901533 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

P8-7 (a) In the US the legal limit it 0.8 g/L. 8-19

final value 10 7.131E-44 0.08 10 0.192

This occurs at t = 6.3 hours..

P8-7 (b) In Sweden CB = 0.5 g/l , t = 7.8 hrs.

P8-7 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs P8-7 (d) For this situation we will use the original Polymath code and change the initial concentration of A to 1 g/L. Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the second martini is ingested. This means that 1 g/l will be added to the final concentration of A after a half an hour. At a half an hour CA = 0.00674 g/L and CB = 0.897 g/L. The Polymath code for after the second drink is shown below. See Polymath program P8-7-d.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0.5 1.0067379 0.8972621 10 0.192

minimal value 0.5 5.394E-42 0.08 10 0.192

maximal value 10 1.0067379 1.8069769 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

for the US t = 6.2 hours Sweden: t = 7.8 hours Russia: t =10.3 hours.

P8-7 (e)

8-20

final value 10 5.394E-42 0.08 10 0.192

The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g of ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the rate of change in concentration due to the incoming ethanol is 2 g/L/hr. For the first hour the differential equation for CA becomes:

dC A dt

k1C A

2t after that it reverts back to the original equations.

See Polymath program P8-7-e.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0 0 0 10 0.192

minimal value 0 0 -1.1120027 10 0.192

maximal value 11 0.1785514 0.7458176 10 0.192

final value 11 6.217E-45 -1.1120027 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = if(t<1)then(-k1*Ca+2*t)else(-k1*Ca) [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

US: CB never rises above 0.8 g/L so there is no time that it would be illegal. Sweden: t = 2.6 hours Russia: t = 5.2 hours

P8-7 (f) 60 g of ethanol immediately  CA = 1.5 g/L CB = 0.8 g/L at 0.0785 hours or 4.71 minutes. So the person has about 4 minutes and 40 seconds to get to their destination.

P8-7 (g) A heavy person will have more body fluid and so the initial concentration of CA would be lower. This means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They will take longer to reach the legal limit, as their initial concentration will be higher. 8-21

P8-8 (a)

See Polymath program P8-8-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2 k4 k3

initial value 0 6.25 0 0.15 0.6 0.2 0.1

minimal value 0 0.3111692 0 0.15 0.6 0.2 0.1

maximal value 4 6.25 0.5977495 0.15 0.6 0.2 0.1

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca-k2*Ca [2] d(Cb)/d(t) = k1*Ca-k3-k4*Cb Explicit equations as entered by the user [1] k1 = 0.15 [2] k2 = 0.6 [3] k4 = 0.2 [4] k3 = if(Cb<0)then (k1*Ca-k4*Cb) else (0.1)

P8-8 (b)

8-22

final value 4 0.3111692 0.4057018 0.15 0.6 0.2 0.1

8-23

P8-8 (c) If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first six hours. Doing so will result in build up of the drug in the bloodstream that can cause harmful effects.

P8-8 (d) If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also drag the drug to the intestines and may limit its effectiveness. This effect can be seen in the adsorption constant k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is slow and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows the concentration profiles for k1 = 0.10 h-1 and k2 = 0.8 h-1. Note that the maximum amount of the drug in the bloodstream is reduced by two. Concentration Profiles Ca (mg/dm3) Cb (mg/dm3)

0.6

CB (mg/dm3)

CA (mg/dm3)

8 6

0.4 4

0.2 2 0

0 0

2

4

6

8

10

12

Time (h)

Concentration profile for Tarzlon in the stomach (A) and bloodstream (B). The maximum amount of Tarzlon in the bloodstream is 0.3 mg/dm3.

P8-9 (a) Reactor selection

A B

D

rD

r1 A

r1 A 10 exp( 8000 K / T )C AC B

A B

U

rU

r2 A

r2 A

SDU =

rD rU

10 exp( 8000 K / T )C A C B 1/ 2

100 exp( 1000 K / T )C A C B

1/ 2

100 exp( 1000 K / T )C A C B

exp( 8000 K / T )C A 3/ 2

10 exp( 1000 K / T )C B

At T = 300K 1/ 2

k1 = 2.62 x 10-11 &

k2 = 3.57

SD/U

1/ 2

7.35 10 12 C A 1/ 2 CB

At T = 1000K

8-24

1/ 2

3/ 2

-3

k1 = 3.35 x 10 &

k2 =36.78

SD/U

9.2 10 5 C A CB

1/ 2

1/ 2

Hence In order to maximize SDU, use higher concentrations of A and lower concentrations of B. This can be achieved using: 1) A semibatch reactor in which B is fed slowly into a large amount of A 2) A tubular reactor with side streams of B continually fed into the reactor 3) A series of small CSTR’s with A fed only to the first reactor and small amounts of B fed to each reactor.

Also, since ED > EU, so the specific reaction rate for D increases much more rapidly with temperature. Consequently, the reaction system should be operated at highest possible temperature to maximize SDU. 1

CB Note that the selectivity is extremely low, and the only way to increase it is to keep CA

2

10

6

and

add B drop by drop.

P8-9 (b) A B

D

A B

U

SDU =

rD rU

and

r1 A 100 exp( 1000 K / T )C AC B

and

r2 A 106 exp( 8000K / T )C ACB

100 exp( 1000 K / T )C AC B 10 6 exp( 8000 K / T )C AC B

exp( 1000 K / T ) 10 4 exp( 8000 K / T )

At T = 300K k1 = 3.57 & k2 = 2.623 SDU = 1.14 x106 At T = 1000K k1 = 36.78 & k2 = 3354.6 SDU = 0.103 Hence temperature should be kept as low as possible to maximize SDU, but at the same time care should be taken to have a significant reaction extent.

P8-9 (c) A B D

and

r1 A

B D

and

r2 A

U

10 exp( 8000 K / T )C ACB 109 exp( 10, 000 K / T )CB CD

8-25

r1 A r2 A

S DU

10 exp

8000 K / T C ACB

109 exp

10000 K / T CBCD

exp

S DU

8000 / T C A

8

10 exp

10000 / T CD

Therefore the reaction should be run at a low temperature to maximize SDU, but not too low to limit the production of desired product. The reaction should also take place in high concentration of A and the concentration of D should be limited by removing through a membrane or reactive distillation.

P8-9 (d) A D D U1 A

and

r1 A

4280 exp( 12000 K / T )C A

and

r2 D

10,100 exp( 15000 K / T )C D

U 2 and

r3 A

rD

S D / U 1U 2

rU 1

26 exp( 10800 K / T )C A

4280 exp( 12000 K / T )C A 10,100 exp( 15000 K / T )C D 10,100 exp( 15000 K / T )C D 26 exp( 10800 K / T )C A

rU 2

At T = 300K k1 = 1.18 x 10-14 &

k2 = 1.94 X 10-18 & k3 = 6.03 x 10-15

If we keep CA > 1000CD

S D / U 1U 2

1.18 10 1.94 10

At T = 1000K k1 = 0.026 &

14

C A 1.94 10 18 C D 18 C D 6.03 10 15 C A

1.18 1.96 .603

k2 = 3.1 X 10-3 & k3 = 5.3 x 10-4

If we keep CA > 1000CD

S D / U 1U 2

0.026C A 3.1 10 3 C D 3.1 10 3 C D 5.3 10 4 C A

.026 .00053

49

Here, in order to lower U1 use low temperature and high concentration of A But low temperature and high concentration of A favours U2 So, it’s a optimization problem with the temperature and concentration of A as the variables . Membrane reactor in which D is diffusing out can be used.

P8-9 (e) A B

D

and

r1 A

10 9 exp( 10000 K / T )C A C B

8-26

D A B A B U S D /U

rD rU

and

r2 D

20 exp( 2000 K / T )C D

and

r3 A

103 exp( 3000K / T )C AC B

10 9 exp( 10000 K / T )C AC B 20 exp( 2000 K / T )C D 10 3 exp( 3000 K / T )C A C B

At T = 300K k1 = 3.34 x 10-6 &

k2 = 0.025

& k3 = 0.045

The desired reaction lies very far to the left and CD is probably present at very low concentrations so that:

S D /U

3.34 10 6 C A C B 0.025C D 0.045C A C B

S D /U

0

At T = 1000K k1 = 45399.9 &

0.000334 4.5

0.000074

k2 = 2.7 & k3 = 49.7

If we assume that CACB > 0.001CD then,

S D /U

45399.9C A C B 2.7C D 49.7C A C B

45399 49.7

913

Here we need a high temperature for a lower reverse reaction of D and lower formation of U Also we need to remove D as soon as it is formed so as to avoid the decomposition.

P8-9 (f) A B

D

and

r1 A

10 exp( 8000 K / T )C ACB

A

U

and

r2 A

26 exp( 10,800 K / T )C A

U

A

and

r3U

1000 exp( 15, 000 K / T )CU

SD /U

rD rU

10exp( 8000 K / T )C ACB 26exp( 10,800 K / T )C A 1000exp 15, 000 K / T CU

We want high concentrations of B and U in the reactor. Also low temperatures will help keep the selectivity high. If we use pseudo equilibrium and set –rA = 0.

rA 10 exp

8000 C AC B T

26 exp

10800 C A 1000 exp T

8-27

15000 CU T

0

10 exp

CA CU

CA CU

8000 CB T

26 exp

1000 exp

15000 T

10800 T

1 7000 26 4200 exp CB exp 100 T 1000 T

P8-9 (g) A B

D

and

r1 A

800 exp( 8000 K / T )C A0.5CB

A B

U1

and

r2 B

10 exp( 300 K / T )C ACB

D B

U2

and

r3 D

106 exp( 8000 K / T )CD CB

(1)

S D / U1

800 exp 10 exp

8000 / T C A0.5CB

80 exp

300 / T C ACB

exp

8000 / T

300 / T C A0.5

At T = 300

S D / U1

2.098*10 10 0.368C A0.5

At T = 1000

S D / U1

29.43 0.7408C A0.5

To keep this selectivity high, low concentrations of A, and high temperatures should be used.

S D /U 2

800 exp

8000 / T C A0.5CB

106 exp

8000 / T CD CB

800C A0.5 106 CD

To keep this selectivity high, high concentrations of A and low concentrations of D should be used. Try to remove D with a membrane reactor or reactive distillation. The selectivity is not dependant on temperature. To keep optimize the reaction, run it at a low temperature to maximized SD/U1 in a membrane reactor that allows only D to diffuse out. (2)

S D / U 1U 2 S D / U 1U 2

800 exp 10 exp

300 / T C ACB 106 exp 800 exp

10 exp

8000 / T C A0.5CB 8000 / T CD CB

8000 / T C A0.5

300 / T C A 106 exp

8000 / T CD

8-28

At T = 300

S D /U 1U 2

2.09*10 9 C A0.5 3.67C A 2.62*10 6 CD

0

At T = 1000 and very low concentrations of D

S D /U 1U 2

0.268C A0.5 7.408C A 335CD

.03617 C A0.5

If temperature is the only parameter that can be varied, then the highest temperature possible will result in the highest selectivity. Also removing D will help keep selectivity high.

P8-10 (a) Species A : dCa dt

rA

;

-rA = k1 * Ca Species B: dCb dt

rb

rB = k1 * Ca – k2 * Cb Species C: dCc dt

rc

rC = k2*Cb See polymath program P8-10-a.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca

1.6

6.797E-18

1.6

6.797E-18

2 Cb

0

0

1.455486

0.6036996

3 Cc

0

0

0.9963004

0.9963004

4 k1

0.4

0.4

0.4

0.4

5 k2

0.01

0.01

0.01

0.01 8-29

6 ra

-0.64

-0.64

-2.719E-18

-2.719E-18

7 rb

0.64

-0.0132415

0.64

-0.006037

8 rc

0

0

0.0145549

0.006037

9 t

0

0

100.

100.

Differential equations 1 d(Cc)/d(t) = rc 2 d(Cb)/d(t) = rb 3 d(Ca)/d(t) = ra Explicit equations 1 k2 = 0.01 2 rc = k2*Cb 3 k1 = 0.4 4 ra = - k1*Ca 5 rb = k1*Ca - k2*Cb

P8-10 (b) Now the rate laws will change ra = k1-1Cb – k1*Ca rb =k1*Ca – k1-1Cb – k2*Cb rc = k2*Cb See polymath program P8-10-b.pol Calculated values of DEQ variables 8-30

Variable Initial value Minimal value Maximal value Final value 1

Ca

1.6

0.3949584

1.6

0.3949584

2

Cb

0

0

0.8736951

0.5191343

3

Cc

0

0

0.6859073

0.6859073

4

k1

0.4

0.4

0.4

0.4

5

k1b

0.3

0.3

0.3

0.3

6

k2

0.01

0.01

0.01

0.01

7

ra

-0.64

-0.64

-0.0022431

-0.0022431

8

rb

0.64

-0.0047669

0.64

-0.0029483

9

rc

0

0

0.008737

0.0051913

0

0

100.

100.

10 t

Differential equations 1 d(Ca)/d(t) = ra 2 d(Cb)/d(t) = rb 3 d(Cc)/d(t) = rc Explicit equations 1 k1 = 0.4 2 k2 = 0.01 3 k1b = 0.3 4 ra = k1b*Cb - k1*Ca 5 rb = k1*Ca - k1b*Cb - k2*Cb 6 rc = k2*Cb

8-31

P8-10 (c) rC = k2*CB – k-2*Cc See polymath program P8-10-c.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

1.6

0.4500092

1.6

0.4500092

2

Cb

0

0

0.8742432

0.5953978

3

Cc

0

0

0.554593

0.554593

4

k1

0.4

0.4

0.4

0.4

5

k1b

0.3

0.3

0.3

0.3

6

k2

0.01

0.01

0.01

0.01

7

k2c

0.005

0.005

0.005

0.005

8

ra

-0.64

-0.64

-0.0013843

-0.0013843

9

rb

0.64

-0.0044688

0.64

-0.0017967

10 rc

0

0

0.0085186

0.003181

11 t

0

0

100.

100.

Differential equations 1 d(Ca)/d(t) = ra 2 d(Cb)/d(t) = rb 3 d(Cc)/d(t) = rc Explicit equations 1 k1 = 0.4 2 k2 = 0.01 3 k1b = 0.3 4 k2c = 0.005 5 ra = k1b*Cb - k1*Ca 6 rb = k1*Ca - k1b*Cb - k2*Cb + k2c*Cc 7 rc = k2*Cb - k2c*Cc

8-32

P8-10 (d) When k1 > 100 and k2 < 0.1 the concentration of B immediately shoots up to 1.5 and then slowly comes back down , while CA drops off immediately and falls to zero . This is because the first reaction is very fast and the second reaction is slower with no reverse reactions. When k2 = 1 then the concentration of B spikes again and remains high, while very little of C is formed. This is because after B is formed it will not further get converted to C because the reverse reaction is fast. When k-2 = 0.25 , B shoots up , but does not stay as high because for the second reaction in the reverse direction is a slightly slower than seen before , but still faster than the reaction in the forward direction.

P8-11 (a) Intermediates (primary K-phthalates) are formed from the dissociation of K-benzoate with a CdCl2 catalyst reacted with K-terephthalate in an autocatalytic reaction step:

A k1 R S

C AO

P RT

R

k2

Series

S

k3

2S 110kPa kPa.dm3 8.314 683K mol.K

Autocatalytic

0.02mol / dm3

Maximum in R occurs at t = 880 sec. See Polymath program P8-11-a.pol. POLYMATH Results Variable t A

initial value 0 0.02

minimal value 0 0.003958

maximal value 1500 0.02

8-33

final value 1500 0.003958

R S k1 k2 k3 0.00159

0 0 0.00108 0.00119 0.00159 0.00159

0 0 0.00108 0.00119 0.00159

0.0069892 0.0100382 0.00108 0.00119

0.005868 0.0100382 0.00108 0.00119

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 1.08e-3 [2] k2 = 1.19e-3 [3] k3 = 1.59e-3

P8-11 (b) 1) T = 703 K CAO = 0.019 mol/dm3

k1'

k1'

k1 exp

E1 1 R T

1 T'

(1.08 10 3 s 1 ) exp

(42600cal / mol ) 1 (1.987cal / mol.K ) 683K

1 703K

2.64 10 3 s

Similarly,

k2'

3.3 10 3 s

And, k3'

1

3.1 10 3 dm3 / mol.s

Maxima in R occurs at around t =320 sec. See Polymath program P8-11-b1.pol. POLYMATH Results Calculated values of the DEQ variables Variable t A R S k1 k2 k3

initial value 0 0.019 0 0 0.00264 0.0033 0.0031

minimal value 0 3.622E-04 0 0 0.00264 0.0033 0.0031

maximal value 1500 0.019 0.0062169 0.0174625 0.00264 0.0033 0.0031

8-34

final value 1500 3.622E-04 8.856E-04 0.0174625 0.00264 0.0033 0.0031

1

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 2.64e-3 [2] k2 = 3.3e-3 [3] k3 = 3.1e-3

2) T = 663 K CAO = 0.19 mol/dm3

(42600cal / mol ) 1 (1.987cal / mol.K ) 683K

k1'

(1.08 10 3 s 1 ) exp

k2'

0.4 10 3 s

k3'

0.78 10 3 dm3 / mol.s

1 663K

0.42 10 3 s

1

1

See Polymath program P8-11-b2.pol. POLYMATH Results Calculated values of the DEQ variables Variable t A R S k1 k2 k3

initial value 0 0.019 0 0 4.2E-04 4.0E-04 7.8E-04

minimal value 0 2.849E-04 0 0 4.2E-04 4.0E-04 7.8E-04

maximal value 10000 0.019 0.0071414 0.016889 4.2E-04 4.0E-04 7.8E-04

final value 10000 2.849E-04 0.0012573 0.016889 4.2E-04 4.0E-04 7.8E-04

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 0.42e-3 [2] k2 = 0.4e-3 [3] k3 = 0.78e-3 Independent variable variable name : t initial value : 0 final value : 10000

Maxima in R occurs around t = 2500 sec.

P8-11 (c) Use the Polymath program from part (a) and change the limits of integration to 0 to 1200. We get: CAexit = 0.0055 mol/dm3 CRexit = 0.0066 mol/dm3

8-35

CSexit = 0.0078 mol/dm3

P8-12 (a)

P8-12 (b)

P8-12 (c)

P8-12 (d)

P8-12 (e)

P8-12 (f)

P8-12 (g)

P8-12 (h) Mole balance: C AO

CA

rA

CC

rC

CD

rD

8-36

Rate law:

Solving in polymath:

rA

k1 A C A

1 k 2 D C A CC2 3

rC

1 k1 A C A 3

2 k 2 D C A CC2 3

rD

k 2 D C A CC2

CA

0.0069 M C B

k 3 E CC C D

4 k 3 E CC C D 3

0.96M CC

SB/D = rB/rD = 247

SB/C = 1.88

See Polymath program P8-12-h.pol. POLYMATH Results NLES Solution Variable Ca Cb Cc Cd Ce kd ka rb ra ke rc rd re tau Cao

Value 0.0068715 0.9620058 0.5097027 0.0038925 0.2380808 3 7 0.0160334 -0.0498855 2 0.008495 6.488E-05 0.003968 60 3

f(x) -2.904E-10 -1.332E-15 -1.67E-08 -2.391E-08 1.728E-08

Ini Guess 3 0 0 0 0


Nonlinear equations [1] [2] [3] [4] [5]

f(Ca) = Cao-Ca+ra*tau = 0 f(Cb) = Cb - rb*tau = 0 f(Cc) = Cc-rc*tau = 0 f(Cd) = Cd-rd*tau = 0 f(Ce) = Ce - re*tau = 0

Explicit equations [1] [2] [3] [4] [5] [6] [7]

0.51M C D

kd = 3 ka = 7 rb = ka*Ca/3 ra = -(ka*Ca+kd/3*Ca*Cc^2) ke = 2 rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc

8-37

0.004M

[8] re = ke*Cd*Cc [9] tau = 60 [10] Cao = 3

P8-12 (i) For PFR and gas phase: Mole balance:

dFA dV

rA

Rate law:

rA

k1 A C A

rB

1 k1 A C A 3

rC

1 k1 A C A 3

rD

k 2 D C A CC2

rE

k 3E CC C D

Stoichiometry: C A

FT

dy dV

dFC dV

dFB rB dV 1 k 2 D C A CC2 3

CTO

FA y FT

FA

FB

2 k 2 D C A CC2 3

rC

dFD dV

rD

CD

CTO

dFE dV

K 3 E CC C D

4 k 3 E CC C D 3

CC FC

FD

CTO

FC y FT

FD y FT

FE

FT 2 y FTO

Plot of CB and CC are overlapping. See Polymath program P8-12-i.pol. POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fc Fd Fe y Ft Cto Cc ka kd ke Ca

initial value 0 20 0 0 0 0 1 20 0.2 0 7 3 2 0.2

minimal value 0 9.147E-04 0 0 0 0 0.9964621 13.330407 0.2 0 7 3 2 1.367E-05

maximal value 100 20 6.6638171 6.6442656 0.0201258 0.0043322 1 20 0.2 0.0993605 7 3 2 0.2

8-38

final value 100 9.147E-04 6.6638171 6.6442167 0.0171261 0.0043322 0.9964621 13.330407 0.2 0.0993325 7 3 2 1.367E-05

rE

rb ra Cd Fto rc rd re alfa X

0.4666667 -1.4 0 20 0.4666667 0 0 1.0E-04 0

3.191E-05 -1.4 0 20 -1.923E-05 -7.012E-05 0 1.0E-04 0

0.4666667 -9.586E-05 3.0E-04 20 0.4666667 8.653E-04 5.908E-05 1.0E-04 0.9999543

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc [4] d(Fd)/d(V) = rd [5] d(Fe)/d(V) = re [6] d(y)/d(V) = -alfa*Ft/(2*y*Fto) Explicit equations as entered by the user [1] Ft = Fa+Fb+Fc+Fd+Fe [2] Cto = 0.2 [3] Cc = Cto*Fc/Ft*y [4] ka = 7 [5] kd = 3 [6] ke = 2 [7] Ca = Cto*Fa/Ft*y [8] rb = ka*Ca/3 [9] ra = -(ka*Ca+kd/3*Ca*Cc^2) [10] Cd = Cto*Fd/Ft*y [11] Fto = 0.2*100 [12] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc [13] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc [14] re = ke*Cd*Cc [15] alfa = 0.0001 [16] X = 1-Fa/20

8-39

3.191E-05 -9.586E-05 2.56E-04 20 -1.923E-05 -6.742E-05 5.087E-05 1.0E-04 0.9999543

P8-12 (j) Changes in equation from part (i): dFC dV

rC

RC

RC

k diffuse C C

k diffuse

2 min

See Polymath program P8-12-j.pol.

P8-13(a)

8-40

1

See Polymath program P8-13-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

1.5

0.3100061

1.5

0.3100061

2

Cb

2.

0.3384816

2.

0.3384816

3

Cc

0

0

0.1970387

0.105508

4

Cd

0

0

0.6751281

0.6516263

5

Ce

0

0

0.1801018

0.1801018

6

Cf

0

0

0.3621412

0.3621412

7

kd1

0.25

0.25

0.25

0.25

8

ke2

0.1

0.1

0.1

0.1

9

kf3

5.

5.

5.

5.

10 ra

-1.5

-1.5

-0.0694818

-0.0694818

11 rb

-3.

-3.

-0.0365984

-0.0365984

12 rc

1.5

-0.0490138

1.5

-0.0085994

13 rd

1.5

-0.0128609

1.5

-0.0126825

14 rd1

1.5

0.0088793

1.5

0.0088793

15 re

0

0

0.0523329

0.0202008

16 re2

0

0

0.0523329

0.0202008

17 rf

0

0

0.2571468

0.0188398

18 rf3

0

0

0.2571468

0.0188398

19 V

0

0

50.

50.

20 vo

10.

10.

10.

10.

Differential equations 1 d(Ca)/d(V) = ra/vo 2 d(Cb)/d(V) = rb/vo 3 d(Cc)/d(V) = rc/vo 4 d(Cd)/d(V) = rd/vo 5 d(Ce)/d(V) = re/vo 6 d(Cf)/d(V) = rf/vo Explicit equations 1

vo = 10

2

kf3 = 5 8-41

3

ke2 = .1

4

kd1 = 0.25

5

rf3 = kf3*Cb*Cc^2

6

rd1 = kd1*Ca*Cb^2

7

re2 = ke2*Ca*Cd

8

rf = rf3

9

re = re2

10 rd = rd1-2*re2+rf3 11 ra = -rd1-3*re2 12 rb = -2*rd1-rf3 13 rc = rd1+re2-2*rf3

P8-13 (b)

8-42

P8-13 (c)

8-43

8-44

8-45

P8-13 (d) As θB increases the outlet concentration of species D and F increase, while the outlet concentrations of species A, C, and E decrease. When θB is large, reactions 1 and 3 are favored and when it is small the rate of reaction 2 will increase.

P8-13 (e) When the appropriate changes to the Polymath code from part (a) are made we get the following. POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fc Fd Fe Ff vo Ft Cto kd1 ke2 kf3 Cc Cd

initial value 0 20 20 0 0 0 0 100 40 0.4 0.25 0.1 5 0 0

minimal value 0 18.946536 18.145647 0 0 0 0 100 38.931546 0.4 0.25 0.1 5 0 0

maximal value 500 20 20 0.9342961 0.8454829 0.0445942 0.0149897 100 40 0.4 0.25 0.1 5 0.0095994 0.0086869

8-46

final value 500 18.946536 18.145647 0.9342961 0.8454829 0.0445942 0.0149897 100 38.931546 0.4 0.25 0.1 5 0.0095994 0.0086869

Cb Ca rd1 re2 rf3 re rf rd ra rb rc Scd Sef

0.2 0.2 0.002 0 0 0 0 0.002 -0.002 -0.004 0.002 1 0

0.1864364 0.1946651 0.0016916 0 0 0 0 0.0014393 -0.0021989 -0.004 0.0016889 1 0

0.2 0.2 0.002 1.691E-04 8.59E-05 1.691E-04 8.59E-05 0.002 -0.002 -0.003469 0.002 1.1734311 83.266916

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc [4] d(Fd)/d(V) = rd [5] d(Fe)/d(V) = re [6] d(Ff)/d(V) = rf Explicit equations as entered by the user [1] vo = 100 [2] Ft = Fa+Fb+Fc+Fd+Fe+Ff [3] Cto = .4 [4] kd1 = 0.25 [5] ke2 = .1 [6] kf3 = 5 [7] Cc = Cto*Fc/Ft [8] Cd = Cto*Fd/Ft [9] Cb = Cto*Fb/Ft [10] Ca = Cto*Fa/Ft [11] rd1 = kd1*Ca*Cb^2 [12] re2 = ke2*Ca*Cd [13] rf3 = kf3*Cb*Cc^2 [14] re = re2 [15] rf = rf3 [16] rd = rd1-2*re2+rf3 [17] ra = -rd1-3*re2 [18] rb = -2*rd1-rf3 [19] rc = rd1+re2-2*rf3 [20] Scd = rc/(rd+.0000000001) [21] Sef = re/(rf+.00000000001)

8-47

0.1864364 0.1946651 0.0016916 1.691E-04 8.59E-05 1.691E-04 8.59E-05 0.0014393 -0.0021989 -0.003469 0.0016889 1.1734311 1.9686327

P8-13 (f) The only change from part (e) is:

dFD dV

rD

kcD CD

dFB dV

rB

FB 0 where VT = 500 dm3 and FB0 = 20 mol/min VT

P8-13 (g) The only change from part (e) is:

8-48

P8-14 (a) Equations:

Rate law: -rA = k1CACB1/2 + k2CA2 rB = rA rC= k1CACB1/2 - k3CC + k4CD rD = k2CA2 – k4CD rE = k3CC rG = k3CC rG = k4CD Mole Balance:

See polymath problem P8-14-a.pol Maximum values can be found out from this table : Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 8-49

1

Ca

0.098

0.0064314

0.098

0.0064314

2

Cb

0.049

0.0032157

0.049

0.0032157

3

Cc

0

0

0.0138769

0.0004265

4

Cd

0

0

0.0001148

6.459E-07

5

Ce

0

0

0.0642751

0.0642751

6

Cg

0

0

0.0083756

0.0083756

7

Cto

0.147

0.147

0.147

0.147

8

Cw

0

0

0.0642751

0.0642751

9

Fa

10.

0.9041349

10.

0.9041349

10 Fb

5.

0.4520674

5.

0.4520674

11 Fc

0

0

1.463177

0.0599575

12 Fd

0

0

0.0116183

9.08E-05

13 Fe

0

0

9.035817

9.035817

14 Fg

0

0

1.177438

1.177438

15 Ft

15.

14.87618

20.66532

20.66532

16 Fw

0

0

9.035817

9.035817

17 k1

0.014

0.014

0.014

0.014

18 k2

0.007

0.007

0.007

0.007

19 k3

0.014

0.014

0.014

0.014

20 k4

0.45

0.45

0.45

0.45

21 ra

-0.0003709

-0.0003709

-5.395E-06

-5.395E-06

22 rb

-0.0001855

-0.0001855

-2.698E-06

-2.698E-06

23 rc

0.0003037

-3.328E-05

0.0003037

-5.744E-07

24 rd

6.723E-05

-8.355E-07

6.723E-05

-1.1E-09

25 re

0

0

0.0001943

5.971E-06

26 rg

0

0

5.166E-05

2.906E-07

27 rw

0

0

0.0001943

5.971E-06

28 Sae

0

0

25.89133

0.1000612

29 Scd

0

0

660.3424

660.3424

30 Sdg

0

0

0.0501992

7.711E-05

31 V

0

0

2.0E+05

2.0E+05

32 Yc

0

0

0.2299208

0.0663148

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 8-50

6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg Explicit equations 1

Cto = 0.147

2

k1 = 0.014

3

Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg

4

k2 = 0.007

5

Cb = Cto*(Fb/Ft)

6

k3 = 0.014

7

k4 = 0.45

8

Ca = Cto*(Fa/Ft)

9

Cc = Cto*(Fc/Ft)

10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2) 11 rb = ra/2 12 re = k3*Cc 13 Cd = Cto*(Fd/Ft) 14 Ce = Cto*(Fe/Ft) 15 Cg = Cto*(Fg/Ft) 16 Cw = Cto*(Fw/Ft) 17 rg = k4*Cd 18 rw = k3*Cc 19 rd = k2*(Ca)^2-k4*Cd 20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 21 Yc = Fc/Fa 22 Scd = if (V>0.0001)then (Fc/Fd) else (0) 23 Sae = if (V>0.0001)then (Fa/Fe) else (0) 24 Sdg = if (V>0.0001)then (Fd/Fg) else (0)

8-51

8-52

8-53

8-54

P8-14(b) Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.098

0.0951293

0.098

0.0951293

2

Cb

0.049

0.0475647

0.049

0.0475647

3

Cc

0

0

0.0032152

0.0032152

4

Cd

0

0

0.0001412

0.0001411

5

Ce

0

0

0.0002229

0.0002229

6

Cg

0

0

0.0005038

0.0005038

7

Cto

0.147

0.147

0.147

0.147

8

Cw

0

0

0.0002229

0.0002229

9

Fa

10.

9.63739

10.

9.63739

10 Fb

5.

4.818695

5.

4.818695

11 Fc

0

0

0.3257277

0.3257277

12 Fd

0

0

0.0143026

0.014299

13 Fe

0

0

0.0225834

0.0225834

14 Fg

0

0

0.0510394

0.0510394

15 Ft

15.

14.89232

15.

14.89232

16 Fw

0

0

0.0225834

0.0225834

17 k1

0.014

0.014

0.014

0.014

18 k2

0.007

0.007

0.007

0.007

19 k3

0.014

0.014

0.014

0.014

20 k4

0.45

0.45

0.45

0.45

21 ra

-0.0003709

-0.0003709

-0.0003538

-0.0003538

22 rb

-0.0001855

-0.0001855

-0.0001769

-0.0001769

23 rc

0.0003037

0.0003037

0.000336

0.000309

8-55

24 rd

6.723E-05

-1.674E-07

6.723E-05

-1.674E-07

25 re

0

0

4.501E-05

4.501E-05

26 rg

0

0

6.352E-05

6.351E-05

27 rw

0

0

4.501E-05

4.501E-05

28 Sae

0

0

7.875E+05

426.7466

29 Scd

0

0

22.77975

22.77975

30 Sce

0

0

594.3536

14.42332

31 Sdg

0

0

18.10358

0.2801563

32 V

0

0

1000.

1000.

33 Yc

0

0

0.0337983

0.0337983

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg Explicit equations 1

Cto = 0.147

2

k1 = 0.014

3


4

k2 = 0.007

5

Cb = Cto*(Fb/Ft)

6

k3 = 0.014

7

k4 = 0.45

8

Ca = Cto*(Fa/Ft)

9

Cc = Cto*(Fc/Ft)

10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2) 11 rb = ra/2 12 re = k3*Cc 13 Cd = Cto*(Fd/Ft) 14 Ce = Cto*(Fe/Ft) 15 Cg = Cto*(Fg/Ft) 16 Cw = Cto*(Fw/Ft) 17 rg = k4*Cd 18 rw = k3*Cc

8-56

19 rd = k2*(Ca)^2-k4*Cd 20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 21 Yc = Fc/Fa 22 Scd = if (V>0.0001)then (Fc/Fd) else (0) 23 Sae = if (V>0.0001)then (Fa/Fe) else (0) 24 Sdg = if (V>0.0001)then (Fd/Fg) else (0) 25 Sce = if (V>0.0001)then (Fc/Fe) else (0)

Then taking different values of FA0 and FB0, we find the following data: FA0

FB0 Θ= FA0/FB0

0.01 1 5 10 100 1000

5 5 5 5 5 5

0.002 0.2 1 2 20 200

YC

SC/E

SD/G

SC/D

0.139038

4.424789

0.070678

3.48E+04

0.111343

5.379092

0.088876

321.5455

0.057955

9.342066

0.168248

52.61217

0.033798

14.42332

0.280156

22.77975

0.001968

107.1563

2.87001

1.931864

5.99E-05

1001.268

30.05269

0.413009

100

150

200

250

0.16 0.14 0.12 0.1 0.08 Yc

0.06 0.04 0.02 0

-50

-0.02 0

50

ΘO2

8-57

1200 1000 800 SC/E

600 400 200 0

-50

0

50

100

150

200

250

150

200

250

ΘO2

35 30 25 20 SD/G

15 10 5 0

-50

-5

0

50

100 ΘO2

8-58

25 20 15 SD/G

10 5 0

-500

0

500

-5

1000

1500

2000

2500

ΘO2

P8-14(c) Now we have a pressure drop parameter α = 0.002, so we modify our polymath program as follows: Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

alpha

0.002

0.002

0.002

0.002

2

Ca

0.098

0.0041647

0.098

0.0041647

3

Cb

0.049

0.0020823

0.049

0.0020823

4

Cc

0

0

0.0004581

3.917E-05

5

Cd

0

0

5.56E-05

3.278E-06

6

Ce

0

0

8.655E-06

9.533E-07

7

Cg

0

0

3.75E-05

3.755E-06

8

Cto

0.147

0.147

0.147

0.147

9

Cw

0

0

8.655E-06

9.533E-07

10 Fa

10.

9.896873

10.

9.896873

11 Fb

5.

4.948437

5.

4.948437

12 Fc

0

0

0.0930717

0.0930717

13 Fd

0

0

0.0082669

0.0077896

14 Fe

0

0

0.0022655

0.0022655

15 Fg

0

0

0.0089238

0.0089238

16 Ft

15.

14.95962

15.

14.95963

17 Ft0

15.

15.

15.

15.

18 Fw

0

0

0.0022655

0.0022655

19 k1

0.014

0.014

0.014

0.014

20 k2

0.007

0.007

0.007

0.007

21 k3

0.014

0.014

0.014

0.014

8-59

22 k4

0.45

0.45

0.45

0.45

23 ra

-0.0003709

-0.0003709

-2.782E-06

-2.782E-06

24 rb

-0.0001855

-0.0001855

-1.391E-06

-1.391E-06

25 rc

0.0003037

3.587E-06

0.0003037

3.587E-06

26 rd

6.723E-05

-4.638E-06

6.723E-05

-1.354E-06

27 re

0

0

6.413E-06

5.483E-07

28 rg

0

0

2.502E-05

1.475E-06

29 rw

0

0

6.413E-06

5.483E-07

30 Sae

0

0

3.344E+06

4368.521

31 Scd

0

0

11.94813

11.94813

32 Sce

0

0

1218.255

41.08224

33 Sdg

0

0

37.46977

0.8729093

34 V

0

0

500.

500.

35 y

1.

0.0428242

1.

0.0428242

36 Yc

0

0

0.0094042

0.0094042

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg 8 d(y)/d(V) = -alpha/2/y*(Ft/Ft0) Explicit equations 1

alpha = 0.002

2

Ft0 = 15

3

Cto = 0.147

4

k1 = 0.014

5


6

k2 = 0.007

7

Cb = Cto*(Fb/Ft)*y

8

k3 = 0.014

9

k4 = 0.45

10 Ca = Cto*(Fa/Ft)*y 11 Cc = Cto*(Fc/Ft)*y 12 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)

8-60

13 rb = ra/2 14 re = k3*Cc 15 Cd = Cto*(Fd/Ft)*y 16 Ce = Cto*(Fe/Ft)*y 17 Cg = Cto*(Fg/Ft)*y 18 Cw = Cto*(Fw/Ft)*y 19 rg = k4*Cd 20 rw = k3*Cc 21 rd = k2*(Ca)^2-k4*Cd 22 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 23 Yc = Fc/Fa 24 Scd = if (V>0.0001)then (Fc/Fd) else (0) 25 Sae = if (V>0.0001)then (Fa/Fe) else (0) 26 Sdg = if (V>0.0001)then (Fd/Fg) else (0) 27 Sce = if (V>0.0001)then (Fc/Fe) else (0)

8-61

8-62

8-63

P8-14(d) Since C(Formic acid) is our desired product, so temperature corresponding to the maximum yield of C should be recommended.

8-64


Ca

0.098

3.689E-08

0.098

3.689E-08

2

Cb

0.049

1.844E-08

0.049

1.844E-08

3

Cc

0

0

0.0150519

1.497E-12

4

Cd

0

0

0.0089776

1.018E-10

5

Ce

0

0

0.0496431

0.0496431

6

Cg

0

0

0.0477138

0.0477138

7

Cto

0.147

0.147

0.147

0.147

8

Cw

0

0

0.0496431

0.0496431

9

Fa

10.

7.43E-06

10.

7.43E-06

10 Fb

5.

3.715E-06

5.

3.715E-06

11 Fc

0

0

1.854

3.016E-10

12 Fd

0

0

1.030757

2.05E-08

13 Fe

0

0

9.999993

9.999993

14 Fg

0

0

9.611354

9.611354

15 Ft

15.

14.91758

29.61135

29.61135

16 Fw

0

0

9.999993

9.999993

17 k1

0.014

0.014

30.64368

30.64368

18 k10

0.014

0.014

0.014

0.014

19 k2

0.007

0.007

7.341E+07

7.341E+07

20 k20

0.007

0.007

0.007

0.007

21 k3

0.014

0.014

6.707E+04

6.707E+04

22 k30

0.014

0.014

0.014

0.014

23 k4

0.45

0.45

984.9755

984.9755

24 k40

0.45

0.45

0.45

0.45

25 ra

-0.0003709

-0.0625305

-1.0E-07

-1.0E-07

26 rb

-0.0001855

-0.0312653

-5.002E-08

-5.002E-08

27 rc

0.0003037

-0.0238556

0.0236439

-5.93E-12

28 rd

6.723E-05

-0.0093009

0.0120515

-3.804E-10

29 re

0

0

0.0773459

1.004E-07

30 rg

0

0

0.0627128

1.003E-07

31 rw

0

0

0.0773459

1.004E-07

32 Sae

0

0

5.205E+05

7.43E-07

33 Scd

0

0

3.394658

0.0147083

34 Sdg

0

0

18.42613

2.133E-09

35 T

300.

300.

550.

550.

36 V

0

0

1000.

1000.

37 Yc

0

0

0.4483122

4.059E-05

8-65

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg 8 d(T)/d(V) = 0.25 Explicit equations 1

k20 = 0.007

2

k10 = 0.014

3

k1 = k10*exp((10000/1.97)*(1/300-1/T))

4

k2 = k20*exp((30000/1.97)*(1/300-1/T))

5

k30 = 0.014

6

k3 = k30*exp((20000/1.97)*(1/300-1/T))

7

k40 = 0.45

8


9

Cto = 0.147

10 Ca = Cto*(Fa/Ft) 11 Cb = Cto*(Fb/Ft) 12 Cc = Cto*(Fc/Ft) 13 Cd = Cto*(Fd/Ft) 14 Ce = Cto*(Fe/Ft) 15 Cg = Cto*(Fg/Ft) 16 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2) 17 rb = ra/2 18 re = k3*Cc 19 k4 = k40*exp((10000/1.97)*(1/300-1/T)) 20 Cw = Cto*(Fw/Ft) 21 rg = k4*Cd 22 rw = k3*Cc 23 rd = k2*(Ca)^2-k4*Cd 24 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 25 Yc = Fc/Fa 26 Scd = if (V>0.0001)then (Fc/Fd) else (0) 27 Sae = if (V>0.0001)then (Fa/Fe) else (0)

8-66

28 Sdg = if (V>0.0001)then (Fd/Fg) else (0)

Temperature corresponding to maximum yield = 367.8 K

P8-15 (1)

C2H4 + 1/2O2 → C2H4O E + 1/2O → D

(2) E

C2H4 + 3O2 → 2CO2 + 2H2O + 3O → 2U1 + 2U2

8-67

FIO

0.82 FTO

0.007626

P8-15 (a) Selectivity of D over CO2

S

FD FU 1

See Polymath program P8-15-a.pol. POLYMATH Results Variable W Fe Fo Fd Fu1 Fu2 Finert Ft K1 K2 Pto Pe Po k1 k2 X S r1e r2e

initial value 0 5.58E-04 0.001116 0 1.0E-07 0 0.007626 0.0093001 6.5 4.33 2 0.1199987 0.2399974 0.15 0.088 0 0 -0.0024829 -0.0029803

minimal value 0 1.752E-10 4.066E-05 0 1.0E-07 0 0.007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0 0 -0.0024829 -0.0029803

maximal value 2 5.58E-04 0.001116 2.395E-04 6.372E-04 6.371E-04 0.007626 0.0093001 6.5 4.33 2 0.1199987 0.2399974 0.15 0.088 0.9999997 0.4101512 -3.692E-10 -8.136E-10

Differential equations as entered by the user [1] d(Fe)/d(W) = r1e+r2e [2] d(Fo)/d(W) = 1/2*r1e + 3*r2e [3] d(Fd)/d(W) = -r1e [4] d(Fu1)/d(W) = -2*r2e [5] d(Fu2)/d(W) = -2*r2e Explicit equations as entered by the user [1] Finert = 0.007626 [2] Ft = Fe+Fo+Fd+Fu1+Fu2+Finert [3] K1 = 6.5 [4] K2 = 4.33 [5] Pto = 2 [6] Pe = Pto*Fe/Ft [7] Po = (Pto*Fo/Ft) [8] k1 = 0.15 [9] k2 = 0.088 [10] X = 1 - Fe/0.000558 [11] S = Fd/Fu1 [12] r1e = -k1*Pe*Po^0.58/(1+K1*Pe)^2 [13] r2e = -k2*Pe*Po^0.3/(1+K2*Pe)^2

8-68

final value 2 1.752E-10 4.066E-05 2.395E-04 6.372E-04 6.371E-04 0.007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0.9999997 0.3758225 -3.692E-10 -8.136E-10

X = 0.999 and S = 0.376(mol of ethylene oxide)/(mole of carbon dioxide)

P8-15(b) Changes in equation from part (a):

dFO dW RO

1 r1E 3r 2 E RO and FO o 2 0.12 0.0093 0.001116 mol W 2 kg.s

From Polymath program:

0

X = 0.71 S = 0.04 (mol of ethylene oxide)/(mole of carbon dioxide)

P8-15 (c) Changes in equation from part (a):

dFE dW

r1E r 2 E RE

RE

0.06 0.0093 W

From Polymath program:

and FE o

0

0.000558 mol 2 kg.s

X = 0.96 S = 0.41(mol of ethylene oxide)/(mole of carbon dioxide)

P8-15 (d) No solution will be given. P8-16 The reactions are 1. C + W  6 H2 + 6CO 2. L + W  13 H2 + 10 CO Rate laws are : - r1C = k1C CC CW -r2L = k2L CL Cw2 CT0 = P0/RT = ( 1 atm )/(0.082)(1473) = 0.00828 mol/dm3 FC0 = 0.00411 mol/s FL0 = 0.0185 mol/s FW0 = 0.02 mol/s

P8-16 (a) See polymath program P 8-16(a) Calculated values of DEQ variables

8-69

Variable Initial value Minimal value Maximal value Final value 1

ConvC

0

0

0.8337198

0.8337198

2

ConvL

0

0

0.7379696

0.7379696

3

Ct0

0.0082791

0.0082791

0.0082791

0.0082791

4

Fc

0.0041152

0.0006843

0.0041152

0.0006843

5

Fc0

0.0041152

0.0041152

0.0041152

0.0041152

6

Fco

0

0

0.0342518

0.0342518

7

Fh

0

0

0.0383516

0.0383516

8

Fl

0.0018519

0.0004852

0.0018519

0.0004852

9

Fl0

0.0018519

0.0018519

0.0018519

0.0018519

10 Ft

0.0259671

0.0259671

0.0807757

0.0807757

11 Fw

0.02

0.0070028

0.02

0.0070028

12 k1c

3.0E+04

3.0E+04

3.0E+04

3.0E+04

13 k2l

1.4E+07

1.4E+07

1.4E+07

1.4E+07

14 Mc

0.6666673

0.1108536

0.6666673

0.1108536

15 Mco

0

0

0.9590499

0.9590499

16 Mh

0

0

0.0767032

0.0767032

17 Ml

0.333333

0.0873434

0.333333

0.0873434

18 Mw

0.36

0.1260502

0.36

0.1260502

19 P

1.

1.

1.

1.

20 R

0.082

0.082

0.082

0.082

21 r1c

-0.2509955

-0.2509955

-0.0015102

-0.0015102

22 r2l

-0.3361048

-0.3361048

-0.0003587

-0.0003587

23 T

1473.

1473.

1473.

1473.

24 V

0

0

0.417

0.417

Differential equations 1 d(Fc)/d(V) = r1c 2 d(Fl)/d(V) = r2l 3 d(Fw)/d(V) = r1c+7*r2l 4 d(Fh)/d(V) = -6*r1c-13*r2l 5 d(Fco)/d(V) = -6*r1c-10*r2l Explicit equations 1

T = 1473

2

R = 0.082

3

P=1

4

k2l = 14000000

5

k1c = 30000

8-70

6

Ft = Fc+Fl+Fw+Fh+Fco

7

Ct0 = P/(R*T)

8

Fc0 = 0.00411523

9

Fl0 = 0.00185185

10 Mc = Fc*162 11 Ml = Fl*180 12 Mh = Fh*2 13 Mco = Fco*28 14 Mw = Fw*18 15 r1c = -k1c*(Ct0*(Fc)/(Ft))*(Ct0*(Fw)/(Ft)) 16 r2l = -k2l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2) 17 ConvC = (Fc0-Fc)/Fc0 18 ConvL = (Fl0-Fl)/Fl0

Plot of Fc vs V

8-71

Plot of Fl vs V

Plot of Fw vs V

8-72

Plot of Fh vs V

plot of Fco vs V (b) W is the key reactant YC = (-dCC/-dCW) = -r1C/(-r1C -7*r2L) 8-73

The modification is made in the polymath code for Prob 8.16(a) The plot of YC versus V is as follows ;

For poltting YW versus volume ; since W is our key reactant then, Yw = -dW/(-dW) = 1 Now YL = -dL/(-dW) = -r2L/(-r1C – 7*r2L) The plot is as follows :

8-74

Overall selectivity SCO / H 2 = NCO = exit molar flow rate of CO/exit molar flow rate of H2 = N H2 CCO |exit /CH2 | exit at any point in the reactor.

P8-16 (c) Individualized solution P8-17 (a) The reactions are (i) L + 3 W  3 H2 + 3 CO + char (ii) Ch + 4 W  10 H2 + 7CO 8-75

rate laws at 1200 0C are : -r1L = k1LCLCw2 -r2CH = k2ChCchCw2

; k1L = 3721 (dm3/mol)2/sec ; k2CH = 1000 ( dm3/mol)2/sec

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 ConL

0

0

0.99

0.99

2 Ct0

0.2

0.2

0.2

0.2

3 Fch

0

0

0.0076175

0.0047293

4 Fco

0

0

0.089117

0.089117

5 Fh

0

0

0.1115957

0.1115957

6 Fl

0.0123457 0.0001235

0.0123457

0.0001235

7 Fl0

0.0123457 0.0123457

0.0123457

0.0123457

8 Ft

0.1240226 0.1240226

0.2506041

0.2506041

9 Fw

0.1116769 0.0450387

0.1116769

0.0450387

10 k1l

3721.73

3721.73

3721.73

3721.73

11 k2ch

1000.

1000.

1000.

1000.

12 r1l

-2.403113

-2.403113

-0.0004738

-0.0004738

13 r2ch

0

-0.0992252

0

-0.0048763

14 V

0

0

0.417002

0.417002

Differential equations 1 d(Fl)/d(V) = r1l 2 d(Fch)/d(V) = r2ch-r1l 3 d(Fw)/d(V) = 3*r1l+4*r2ch 4 d(Fh)/d(V) = -3*r1l-10*r2ch 5 d(Fco)/d(V) = -3*r1l-7*r2ch Explicit equations 1 k2ch = 1000 2 k1l = 3721.73 3 Ft = Fch+Fl+Fw+Fh+Fco 4 Ct0 = 0.2 5 r1l = -k1l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2) 8-76

6 r2ch = -k2ch*(Ct0*(Fch)/(Ft))*((Ct0*(Fw)/(Ft))^2) 7 Fl0 = 0.012345679 8 ConL = (Fl0-Fl)/(Fl0)

8-77

P8-17 (b)

(c) SCO/Ch Let S1 = S~CO/Ch = Fco/Fch The plot is as follows-

8-78

Considering W as the principle reacting species; Yw = 1 and Yl = (rl)/(rw) = (r1l)/(3*r1l + 4*r2ch) the plot is as follows:

(d) The molar flow rate for char is maximum at about V= 0.02085 dm3.

8-79

P8-18 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst. A

r1'C

B C

k1C C A

AD

r2' D

k2 D C A

2C + D  2E

r3' E

k3 E CC2 CD

C AO

P RT

CB CC K1C

24.6atm 0.082dm atm / mol.K (500 K ) 3

0.6mol / dm3

Fa(0) = 10 mol/min Fb(0) = Fc(0) = Fd(0) = Fe(0) = 0 Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 alfa

0.008

0.008

0.008

0.008

2 Ca

0.6

0.0029024

0.6

0.0029024

3 Cb

0

0

0.1525032

0.0075281

4 Cc

0

0

0.1425644

0.0541764

5 Cd

0

0

0.0254033

0.0036956

6 Ce

0

0

0.274725

0.235079

7 Cto

0.6

0.6

0.6

0.6

8 Fa

10.

0.0702103

10.

0.0702103

9 Fb

0

0

3.391481

0.1821054

10 Fc

0

0

3.16936

1.310531

11 Fd

0

0

0.5564071

0.0893965

12 Fe

0

0

5.686575

5.686575

13 Ft

10.

7.338818

12.8179

7.338818

14 Fto

10.

10.

10.

10.

15 k1c

2.

2.

2.

2.

16 K1c

0.2

0.2

0.2

0.2

17 k2d

0.4

0.4

0.4

0.4

18 k3e

400.

400.

400.

400.

8-80

19 kb

1.

1.

1.

1.

20 r1c

1.2

0.0017264

1.2

0.0017264

21 r2d

0.24

0.001161

0.24

0.001161

22 r3e

0

0

0.200243

0.0043387

23 ra

-1.44

-1.44

-0.0028874

-0.0028874

24 rb

1.2

0.0017264

1.2

0.0017264

25 rc

1.2

-0.0492057

1.2

-0.0026123

26 rd

0.24

-0.0164816

0.24

-0.0010084

27 re

0

0

0.200243

0.0043387

28 W

0

0

100.

100.

29 y

1.

0.5056359

1.

0.5056359

Differential equations 1 d(Fa)/d(W) = ra 2 d(Fb)/d(W) = rb-(kb*Cb) 3 d(Fc)/d(W) = rc 4 d(Fd)/d(W) = rd 5 d(Fe)/d(W) = re 6 d(y)/d(W) = -alfa*Ft/(2*Fto*y) Explicit equations 1 k2d = 0.4 2 K1c = 0.2 3 Ft = Fa+Fb+Fc+Fd+Fe 4 Cto = 0.6 5 Cb = Cto*(Fb/Ft)*y 6 Ca = Cto*(Fa/Ft)*y 7 Cd = Cto*(Fd/Ft)*y 8 Cc = Cto*(Fc/Ft)*y 9 kb = 1 10 k1c = 2 8-81

11 r2d = k2d*Ca 12 k3e = 400 13 r1c = k1c*(Ca-(Cb*Cc/K1c)) 14 ra = -r1c-r2d 15 r3e = k3e*(Cc^2)*Cd 16 rd = r2d-(r3e/2) 17 rb = r1c 18 rc = r1c-r3e 19 re = r3e 20 Ce = Cto*(Fe/Ft)*y 21 alfa = 0.008 22 Fto = 10

P8-18 (b) Species B, C, D, and E all go though a maximum. The concentration of species are affected by two factors: reaction and pressure drop. We can look at species B for example: Species B: 8-82

Production by reaction 1 => tends to increase concentration of B Pressure drop => tends to decrease concentration of B At W<10, concentration of B rises because the production of B by reaction outweighs the dilution of B by pressure drop. Further down the reactor, the concentration of B drops because the effect of pressure drop outweighs the production of B by reaction. Species E: The reasoning for species E is similar to species B. Species C and D: Species C and D are produced in reactions 1 and 2, respectively. However, species C and D are also reactants in reaction 3; this contributes to their decline at later times (in addition to the pressure drop). P8-18 (c) Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 alfa

0.008

0.008

0.008

0.008

2 Ca

0.6

0.0024106

0.6

0.0024106

3 Cb

0

0

0.2447833

0.0593839

4 Cc

0

0

0.1181101

0.0028718

5 Cd

0

0

0.0408517

0.0119119

6 Ce

0

0

0.1034882

0.0252849

7 Cto

0.6

0.6

0.6

0.6

8 Fa

10.

0.2791706

10.

0.2791706

9 Fb

0

0

6.877205

6.877205

10 Fc

0

0

2.623093

0.3325842

11 Fd

0

0

1.379513

1.379513

12 Fe

0

0

2.928221

2.928221

13 Ft

10.

10.

13.00145

11.79669

14 Fto

10.

10.

10.

10.

15 k1c

2.

2.

2.

2.

16 K1c

0.2

0.2

0.2

0.2

8-83

17 k2d

0.4

0.4

0.4

0.4

18 k3e

400.

400.

400.

400.

19 kb

1.

1.

1.

1.

20 r1c

1.2

0.0031158

1.2

0.0031158

21 r2d

0.24

0.0009642

0.24

0.0009642

22 r3e

0

0

0.1548711

3.93E-05

23 ra

-1.44

-1.44

-0.0040801

-0.0040801

24 rb

1.2

0.0031158

1.2

0.0031158

25 rc

1.2

0.001223

1.2

0.0030765

26 rd

0.24

0.0009446

0.24

0.0009446

27 re

0

0

0.1548711

3.93E-05

28 W

0

0

100.

100.

29 y

1.

0.169772

1.

0.169772

Differential equations 1 d(Fa)/d(W) = ra 2 d(Fb)/d(W) = rb 3 d(Fc)/d(W) = rc-(kb*Cc) 4 d(Fd)/d(W) = rd 5 d(Fe)/d(W) = re 6 d(y)/d(W) = -alfa*Ft/(2*Fto*y) Explicit equations 1 k2d = 0.4 2 K1c = 0.2 3 Ft = Fa+Fb+Fc+Fd+Fe 4 Cto = 0.6 5 Cb = Cto*(Fb/Ft)*y 6 Ca = Cto*(Fa/Ft)*y 7 Cd = Cto*(Fd/Ft)*y 8 Cc = Cto*(Fc/Ft)*y 8-84

9 kb = 1 10 k1c = 2 11 r2d = k2d*Ca 12 k3e = 400 13 r1c = k1c*(Ca-(Cb*Cc/K1c)) 14 ra = -r1c-r2d 15 r3e = k3e*(Cc^2)*Cd 16 rd = r2d-(r3e/2) 17 rb = r1c 18 rc = r1c-r3e 19 re = r3e 20 Ce = Cto*(Fe/Ft)*y 21 alfa = 0.008 22 Fto = 10

Major Differences:

8-85

(1) When C diffuses out instead of B, the concentration of C is lower and the concentration of B is higher. (2) The concentration of D is higher because reaction 3 is slower (C is a reactant in reaction 3) (3) The concentration of E is lower because of the same reasoning as (2). P8-18 (d) Individualized Solution P8-20 The reactions are : k3

2A A

k1

D B C

Volume of PFR = 100 dm3 flow rate =υ = 10 dm3/min C0 = 3 mol / m3 K1 = 0.05 min K3 = 0.015 (dm3/mol)/min Kc = 0.5 dm3 /mol The mistakes in the solution are as follows : k1

(i)

k1 is the rate constant for the reaction for A

B C . But the rate constants for

(ii) (iii)

the reactions have been reversed. The molar flow rates have not been used when rate laws are written in the code. The rate law expression for various species (A,B,C,D) is incorrect . The mistakes are corrected hereby; The rate laws are written as follows; rC = rB = k1(CA – (1/KC )CCCB) = 0.05 *(CA – CB*CC/0.5) rA = - k3*CA – k1(CA – (1/KC )CCCB) = - 0.05 *(CA – CA*CC/0.5) - 0.015*CA rD = 1/2 *k3*CA = 0.0075 Now for plotting the concentrations with the volume we need the performance equation for PFR; Thus

8-86

-rA *dV = dFA dFA/dV = -rA dCa/dV = -1/υ *rA This being a gaseous reaction the volumetric flow rate will change with volume as we proceed in the PFR, Assuming the extent of reactions to be XA1 and XA2 for the reactions 1 and 2 ; then υ =υ0 *(1 + εA1XA1 +εA2XA2 ) the performance equations will be ; dXa/dV =

CA0 * 0 (1

1 A1 X A1

A2 X A2 )

* Ca0*(k 3*(1 Xa) k1((1 Xa) (1/ KC )* B * C *(1 Xa1)*(1 Xa2))

Here , Xa = Xa1 + Xa2 Similarly performance equations for other species in the reactor can be written and the equations will have to be solved simultaneously for getting the plots.

8-87

CDGA8-1

8-88

8-89

8-90

8-91





9-1

Solutions for Chapter 9 – Reaction Mechanisms, Pathways, Bioreactions and Bioreactors P9-1 Individualized Solution P9-2 (a) Example 9-1 The graph of Io/I will remain same if CS2 concentration changes. If concentration of M increases the slope of line will decrease.

P9-2 (b) Example 9-2 The inhibitor shows competitive inhibition.

P9-2 (c) Example 9-3 1)Now Curea = 0.001mol/dm3 and t = 10 min = 600 sec.

t

kM 1 ln VMAX 1 X

600 sec

Curea X VMAX

0.0266mol / dm 3 1 ln 3 1 X 0.000266mol / s.dm

0.001mol / dm 3 X 0.000266mol / s.dm 3

Solving, we get X = 0.9974. t 461.7 sec 2) For CSTR,

C urea X rurea rurea

rurea

Curea X

9-2

VMAX Curea K M Curea

0.000266mol / s.dm 3 0.1mol / dm 3 1 X 0.0266mol / dm 3 0.1mol / dm 3 1 X Solving, we get X = 0.675

See Polymath program P9-2-c.pol. POLYMATH Results NLE Solution Variable Value f(x) Ini Guess X 0.6751896 -8.062E-10 0.5

NLE Report (safenewt)

Nonlinear equations [1] f(X) = 0.0000266*(1-X)/(0.0266+0.1*(1-X))-0.1*X/461.7 = 0

3) For PFR, CO

dC urea r Cures

and C urea

kM 1 ln VMAX 1 X

Curea X VMAX

C urea0 1 X

Same as batch reactor, but t replaced by

X = 0.8

P9-2 (d) Example 9-4 CS CP

YS / P

1

YP / S

YS / C

238.7 245 5.03 2.14 1 2.18

YS / P

1 P

YC

P/S

2.18 g / g

0.46 g / g

1 0.075 0.46

1.87 g / g

Yes there is disparity as substrate is also used in maintenance.

P9-2 (e) Example 9-5 See Polymath program P9-2-e.pol. Part (1) Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 24 24 Cc 2.0E-04 1.276E-04 0.002742 0.002742

9-3

0.1mol / dm 3 X 461.7 sec

Cs Cp rd Ysc Ypc Ks m umax rsm kobs rg vo Cso Vo V mp

5.0E-04 5.0E-04 4.5794959 4.5794959 0 0 0.0159354 0.0159354 2.0E-06 1.277E-06 2.742E-05 2.742E-05 12.5 12.5 12.5 12.5 5.6 5.6 5.6 5.6 1.7 1.7 1.7 1.7 0.03 0.03 0.03 0.03 0.33 0.33 0.33 0.33 6.0E-06 3.83E-06 8.226E-05 8.226E-05 0.33 0.3299706 0.33 0.3299706 1.941E-08 1.941E-08 6.598E-04 6.598E-04 0.5 0.5 0.5 0.5 5 5 5 5 1 1 1 1 1 1 13 13 0 0 0.20716 0.20716

ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg-rd-vo*Cc/V [2] d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V [3] d(Cp)/d(t) = rg*Ypc-vo*Cp/V Explicit equations as entered by the user [1] rd = Cc*.01 [2] Ysc = 1/.08 [3] Ypc = 5.6 [4] Ks = 1.7 [5] m = .03 [6] umax = .33 [7] rsm = m*Cc [8] kobs = (umax*(1-Cp/93)^.52) [9] rg = kobs*Cc*Cs/(Ks+Cs) [10] vo = 0.5 [11] Cso = 5 [12] Vo = 1 [13] V = Vo+vo*t [14] mp = Cp*V Comments [2] d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V Cso is concentration of substrate in feed stream, NOT initial concentration of substrate in reactor [17] mp = Cp*V mass of product in grams

9-4

Part (2) Change the cell growth rate law:

rg

CC C S

k obs

2

KS

CS

CS KI

All other equations are the same as in part (1).

9-5

Notice that uncompetitive inhibition by the substrate causes the concentration of cells to decrease with time, whereas without uncompetitive inhibition, the concentration of cells increased with time. Consequently, the concentration of product is very low compared to the case without uncompetitive inhibition. Part (3) Change the observed reaction rate constant:

k obs

max

1

CP 10000

0.52

All other equations are the same as in part (1).

CP Since CP is so small, the factor 1 * CP

0.52

1 whether CP* = 93 g/dm3 or CP* = 10,000 g/dm3. Thus

the plots for this part are approximately the same as the plots in part (1).

P9-2 (f) Example on CDROM For t = 0 to t = 0.35 sec, PSSH is not valid as steady state not reached. And at low temperature PSSH results show greatest disparity.

See Polymath program P9-2-f.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 12 12 C1 0.1 2.109E-04 0.1 2.109E-04 C2 0 0 1.311E-09 1.311E-09 C6 0 0 3.602E-09 3.602E-09 C4 0 0 2.665E-07 1.276E-08 C7 0 0 0.0979179 0.0979179 C3 0 0 0.0012475 0.0012475

9-6

C5 C8 CP5 CP1 k5 T k1 k2 k4 k3

0 0 0.0979179 0.0979179 0 0 6.237E-04 6.237E-04 0 0 0.0979123 0.0979123 0.1 2.166E-04 0.1 2.166E-04 3.98E+09 3.98E+09 3.98E+09 3.98E+09 1000 1000 1000 1000 0.0014964 0.0014964 0.0014964 0.0014964 2.283E+06 2.283E+06 2.283E+06 2.283E+06 9.53E+08 9.53E+08 9.53E+08 9.53E+08 5.71E+04 5.71E+04 5.71E+04 5.71E+04

ODE Report (STIFF) Differential equations as entered by the user [1] d(C1)/d(t) = -k1*C1-k2*C1*C2-k4*C1*C6 [2] d(C2)/d(t) = 2*k1*C1-k2*C1*C2 [3] d(C6)/d(t) = k3*C4-k4*C6*C1 [4] d(C4)/d(t) = k2*C1*C2-k3*C4+k4*C6*C1-k5*C4^2 [5] d(C7)/d(t) = k4*C1*C6 [6] d(C3)/d(t) = k2*C1*C2 [7] d(C5)/d(t) = k3*C4 [8] d(C8)/d(t) = 0.5*k5*C4^2 [9] d(CP5)/d(t) = k3*(2*k1/k5)^0.5*CP1^0.5 [10] d(CP1)/d(t) = -k1*CP1-2*k1*CP1(k3*(2*k1/k5)^0.5)*(CP1^0.5)

P9-2 (g) Individualized solution P9-3 Burning:

9-7

9-8

9-9

P9-4 k1

CH 3 CHO

CH 3 k2

CH 3

CH 3CHO

CHO

CH 3CHO

2CH 3 rAC

k4

r1

r1

CHO

CH 3

CO CH 4

r2

k 2 C AC CCH3

CH 3

2CO H 2

r3

k 3CCHO C AC

k3

r4

C2 H 6 r2

r3

k1C AC

k 2 C AC CCH3

C AC k1 k 2 CCH3

k 3C AC CCHO

k 3CCHO

Active intermediates: CH 3 , CHO

rCHO

r1

r3

k1C AC C AC

k1

CCHO

rCH3

0 k 3C AC CCHO k 3CCHO

0 0

k1 k3

r1 r2 r2 r3 2r4

k1C AC

k1C AC

k3C AC CCHO

0

2k4CCH3

2

0

Plugging in expression for C CHO : 9-10

k 4 CCH3

2

k1C AC

k1C AC

2k4CCH3

2

0

k1C AC k4

CCH3

Now, substitute expressions for CH 3 and CHO into equation for

rAC

C AC k1

k 2 CCH3

k 3CCHO

C AC k1 k2

k1C AC k4

k1C AC 2 k2

C AC k1k4

k1

P9-4 (b) When k2

C AC k1k4

1,

rAC

k2

3 k1 C AC 2 k4

kC AC

3

2

9-11

rAC :

P9-4 (c) CH3CHO k1 k4 C2H6

CH3•

CHO•

k2

k3

CH4

CO

H2

P9-5 (a)

9-12

P9-5 (b) k1

Cl2

2Cl *

k2 k3

Cl * CO

COCl *

k4 k5

COCl * Cl2

rCOCl*

COCl 2 Cl *

0 k3 Cl * CO k4 k5 Cl2

k5 k3 CO Cl * Cl2

k5 COCl * Cl2

rCl

k5 COCl * Cl2

k3 Cl * CO

COCl *

rCOCl2

k4 COCl *

0 k1 Cl2

k2 Cl *

k4 k5 Cl2 2

k3 Cl * CO

add rCOCl to rCl

rCl

rCOCl

Cl * Cl *

2

0 0 k1 Cl2

k2 Cl

2

k1 Cl2 k2 k1 Cl2 k2

9-13

k4 COCl *

k5 COCl * Cl2

rCOCl2 k4

0.5

k1 k5 k3 CO Cl2 k2 k4 k5 Cl2

Cl2

k5 k3

k1 k2 k4

1/ 2

CO Cl2

3 2

k5 Cl2

k5 Cl2

rCOCl2

k11/ 2 k3 k5 CO Cl2 k21/ 2 k4

3 2

P9-5 (c) The proposed mechanism for this reaction is:

Through this mechanism, it may be deduced that the net rate of formation of HBr (product) is given by:

where the concentrations of the intermediates may be determined by invoking the steady state approximation, i.e.:

and:

i.e. from (ii) + (i) we obtain: 9-14

i.e.:

which substituted back in (i) gives:

i.e.:

i.e.:

or: dividing 'top' and 'bottom' by k`b :

Thus, the net rate of formation of HBr may be written solely in terms of reactants,

9-15

i.e.: which simplifies to:

as predicted by the empirical rate law:

where:

9-16

P9-6(a)

9-17

P9-6(b) Low temperatures with anti-oxidant

9-18

P9-6(c) If the radicals are formed at a constant rate, then the differential equation for the concentration of the radicals becomes:

d I dt

k0

and I

ki I

RH

0

k0 k i RH

The substitution in the differential equation for R· also changes. Now the equation is:

d R dt

ki I

RH

k P1 R

O2

and solving and substituting gives: R

k P 2 RO 2

k0

RH

k P 2 RO 2 k P1 O2

0

RH

Now we have to look at the balance for RO2·.

d RO 2 dt

k P1 R

O2

k P 2 RO 2

RH

k t RO 2

and if we substitute in our expression for *R·+ we get

0

k0

RO 2

k t RO 2

2

0 which we can solve for [RO2·+.

k0 kt

Now we are ready to look at the equation for the motor oil. 9-19

2

0

d RH dt

ki I

RH

k P 2 RO 2

RH

and making the necessary substitutions, the rate law for the degradation of the motor oil is:

d RH dt

r RH

k0

kP 2

k0 RH kt

P9-6(d) Without antioxidants

With antioxidants

P9-6 (e) Individualized solution P9-7 (a)

9-20

P9-7 (b)

P9-7 (c)

P9-7 (d) See Polymath program P9-7-d.pol.

9-21

Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is much slower than the rate at which a healthy person becomes ill. Eventually everyone is ill and people start dying.

P9-7 (e) Individualized solution P9-7 (f) Individualized solution

P9-8 (a) E S E.S E.S P E

k1 k2 k3 k4

E.S E S P E E.S

By applying PSSH for the complex [E.S], we have

k1 [ E ][S ] k 2 [ E.S ] k 3 [ E.S ] k 4 [ P][E ]

k1 [ S ] k 4 [ P ] [E] k 2 k3

[ E.S ] rP

0

rS

k 3 [ E.S ] k 4 [ P][E ]

k 3 k1 [ E ][S ] k 3 k 4 [ P][E ] k 4 [ P][E ] k 2 k3 k 3 k1 [ E ][S ] k 3 k 4 [ P][E ] k 2 k 4 [ P][E ] k 3 k 4 [ P][E ] k 2 k3 k 3 k1 [ S ] k 2 k 4 [ P ] [E] k 2 k3

9-22

k1 k 3 [ s ]

k2

[ P] [E] k1 k 2 k2k4 k3

Since E is not consumed, we have where [ ET ] is a constant

[ ET ] [ E ] [ E.S ]

k1 [ S ] k 4 [ P ] [E] k 2 k3

[ ET ] [ E ] k2

k3

k1 [ S ] k 4 [ P ] [E] k 2 k3

So,

k1 k 3 [ S ] rP

k2

k3

[ P] [ ET ] ke

k1k 3 k2k4

where k e

k1 [ S ] k 4 [ P ]

P9-8 (b) E

k1

S

E S

E P

E S

k3

E P

E P

k5

P

rE

k1 E S

S

k4 E P

k2 E S

E S

k5 E P

k3 E S

since E is not consumed: ET

E

k4

S

k3 E S k 4 k5

E P

rE

E

E

0 k3 E S

S

k2

E S

ET

k3

E S

k4

k5

E E S

k4 E P

E S E P or E E

ET

E S

ET 1

E S E P k3

k4

k5

Insert this into the equation for rE·S and solve for the concentration of the intermediate:

E S

k1 S E T 1

rP

k5 E P

k3 k4

k1 S

k2

k3

k 4 k3 k 4 k5

k5

k 3 k5 E S k4 k5 9-23

rP

k1k3 k5 S ET k4 k5 k1 S k2 k4

k3

k3k5

P9-8 (c) k1

E S1

E S1

k2

k3

E S1 S2

E S1S2

k4

k5

E S1S2 (1) rE

S1

(2) rE

S1S2

P E

0 k1 E S1

k2 E S

0 k3 E S S2

k3 E S S2

k4 E S1S2

k4 E S1S2

k5 E S1S2

If we add these two rates we get: (3) rE

rE

S1

0 k1 E S

S1S2

k2 E S

k5 E S1S2

k3 E S S2

From equation (2) we get E S1S2

k4 k5

Plug this into equation 3 and we get:

k1 E S1 k3 S 2 k2 k 4 k5

E S1

E S1S2

k3 S2 k1 E S1 k4 k5

rP

k5 E S1S2

ET

E

ET

E 1

E S

k1k3 E S1 S2

k1k2 k5 S1 S2 k2 k4

k2 k4 k5 k2 k3 S2

k3 S2 k4 k5

k2

E

k2 k5 k3 S2

E S1S2

k1 S1 k3 S 2 k2 k 4 k5

k1k3 S1 S 2 k2 k4

k5 k 2

k3 S 2

9-24

k1k2 k5 S1 S2

rP

ET

k1 S1 k3 S2 k2 k4 k5

k 2 k 4 k 2 k5 k3 S 2 1

k1k3 S1 S2 k2 k4 k5 k2 k3 S2

P9-8 (d)

k4

0 k1 E S

S

k1 S

rP

k5

k3 E S

k4 E P

k1k3 S E k1k3 S k2

k5 E P

k4 E P

k2 k3

k4 E P

E k3

E

ET

E

ET

E 1

E S k1 S

E P E

k4 E P

k2 k3 k1 S k2

k5 k4 P

k3

k5

k1k3 S ET

rP

k1 S k2 k3

k2 k3 1

rP

k5 E P

k4 E P

ET

rP

k3 E S

k3

0 k4 E P

E P

rP

k2 E S

E

k2

P

rP

P

E P

k5

E S rE

E S

k2

E P

rE

k3

k1

E S

k4 P k5

k3 S ET k2 k3 k1

S

k4 k2 k3 k1k5

P

Vmax S KM

S

KP P 9-25

P9-8 (e) No solution will be given P9-8 (f) No solution will be given

P9-9 (a) The enzyme catalyzed reaction of the decomposition of hydrogen peroxide. For a batch reactor:

9-26

P9-9 (b)

P9-9 (c) Individualized solution P9-9 (d) Individualized solution

9-27

P9-10 (a)

P9-10 (b)

9-28

9-29

X

CS 0 CS CS 0

50 2.1 .958 50

P9-10 (c) rS

kCS ET 1 K1Cs K 2CS2

If ET is reduced by 33%, -rS will also decrease by 33%. From the original plot, we see that if the curve –rS is decreased by 33%, the straight line from the CSTR calculation will cross the curve only once at approximately CS = 40 mmol/L

X

CS 0 CS CS 0

50 40 50

0.2

P9-10 (d) Individualized solution P9-10 (e) Individualized solution

P9-11 (a)

9-30

P9-11 (b)

9-31

P9-11 (c) Individualized solution P9-11 (d) Individualized solution P9-12 For No Inhibition, using regression, Equation model:

1 rs a0 = 0.008

a0 a1

1 S

a1 = 0.0266

For Maltose, Equation model:

1 rs

a0 a1

1 S

9-32

a0 = 0.0098

a1 = 0.33

For α-dextran,

1 rs

Equation model:

a0 = 0.008

a0 a1

1 S

a1 = 0.0377

Maltose show non-competitive inhibition as slope and intercept, both changing compared to no inhibition case. α-dextran show competitive inhibition as intercept same but slope increases compared to no inhibition case.

P9-13 (a) rS

rEHS

rP

k EHS

EHS rP

K M EH

kK M EH

EH 2 EH

E

ET

ET

rP

S

S

K2 H K1 H

EH E

EH K1 H E

EH

EH K1 H

EH 2

EH

K2 H

EH

ET

EH 1

1 K1 H

K2 H

Now plug the value of (EH) into rP

9-33

rS

rP

kK M EH

kK M ET K1 H

S

1 K1 H

S

K1 K 2 H

2

At very low concentrations of H+ (high pH) rS approaches 0 and at very high concentrations of H+ (low pH) rS also approaches 0. Only at moderate concentrations of H+ (and therefore pH) is the rate much greater than zero. This explains the shape of the figure.

P9-13 (b) Individualized solution P9-13 (c) Individualized solution P9-14 (a) Case1: Without drug inhibition:Reactions: 1. GLP-1(7-36)  Stimulates insulin releases ( Rate constant k1) (a) 2. GLP-1(7-36) + DPP-4  GLP-1(9-36) (Rate constant k2) (a)

(b)

Rates of reaction (assuming 1st order) are given by: -rGLP-1 =

-

rDPP-4 =

dCa = k1*Ca+ k2*Ca*Cb dt

Sub to: - (Assumed values) at t=0, Ca0 = Cb0 =1 gmol/liter

dCb = k2*Ca*Cb dt

Now, For a 1st order reaction, half life period (t 1/2) = ln (2)/ rate constant Therefore, assuming the half life period of GLP-1(7-36) to be very short (i.e. 10 sec), we get  k1 = ln (2)/ 10 (sec-1)  k1 = 0.0693 sec-1

9-34

Similarly, given that half life of GLP-1(9-36) is 0.5 to 1 min (i.e. average 0.75 min or 45 sec), we get  k2 = ln (2)/ 45 (sec-1)  k2 = 0.0154 sec-1 Solving the differential equations using POLYMATH, we get See Polymath program P9-14-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca

1.

0.0002638

1.

0.0002638

2 Cb

1.

0.831363

1.

0.831363

3 k1

0.0693

0.0693

0.0693

0.0693

4 k2

0.0154

0.0154

0.0154

0.0154

5 t

0

0

100.

100.

6 X

0.0847

2.166E-05

0.0847

2.166E-05

Differential equations 1 d(Cb)/d(t) = -k2*Ca*Cb 2 d(Ca)/d(t) = -k1*Ca-k2*Ca*Cb

Explicit equations 1 k2 = 0.0154 2 k1 = 0.0693 3 X = k1*Ca+k2*Ca*Cb

9-35

Here, we get the rate of reaction (-rGLP-1) value at t = 10 sec to be 0.035973 (gmol/ liter sec). Case2: With drug inhibition Reactions: 1. GLP-1(7-36) + Drug  Stimulates insulin releases (Rate constant k1) (a) (e) 2. GLP-1(7-36) + DPP-4  GLP-1(9-36) (Rate constant k2) (a) (b) 3. Drug + DPP-4  E.DPP-4 (Inactive) (Rate constant k3) (e) (b)

Rates of reaction are given by: -

Sub to: - (Assumed values)

9-36

-rGLP-1 =

dCa = k1*Ca*Ce + k2*Ca*Cb dt

-

rDPP-4 =

dCb = k2*Ca*Cb + k3*Ce*Cb dt

-

rDrug =

dCe = k1*Ca*Ce + k3*Ce*Cb dt

at t=0, Ca0 = Cb0 =Ce0=1 gmol/liter

Now, as derived earlier k1 = 0.0693 sec-1 k2 = 0.0154 sec-1 k3 = rate constant for drug inhibition (to be varied) Taking, k3 = 0.0693, 0.03465, 0.0231, 0.0173, 0.0154, 0.01390, 0.01155, 0.0099, 0.00866, 0.0077, 0.00693 respectively and solving the 3 differential equations using POLYMATH: Example: For k3 = 0.03465 Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca

1.

0.1451749

1.

0.1451749

2 Cb

1.

0.3512064

1.

0.3512064

3 Ce

1.

0.0232788

1.

0.0232788

4 k1

0.0693

0.0693

0.0693

0.0693

5 k2

0.0154

0.0154

0.0154

0.0154

6 k3

0.03465

0.03465

0.03465

0.03465

7 t

0

0

100.

100.

8 X

0.0847

0.0010194

0.0847

0.0010194

Differential equations

9-37

1 d(Cb)/d(t) = -k3*Ce*Cb-k2*Ca*Cb 2 d(Ce)/d(t) = -k1*Ca*Ce-k3*Ce*Cb 3 d(Ca)/d(t) = -k1*Ca*Ce-k2*Ca*Cb Explicit equations 1 k3 = 0.03465 2 k2 = 0.0154 3 k1 = 0.0693 4 X = k1*Ca*Ce+k2*Ca*Cb

Here, we get the rate of reaction (-rGLP-1) value at t = 10 sec to be 0.0234211 (gmol/ liter sec).

Similarly, we obtain the rate of reaction (-rGLP-1) value at t = 10 sec, for all values of k3: 9-38

Rate of reaction without Rate constant for inhibition (-rGLP-1) at t=10 Rate of reaction(-rGLP-1) at inhibition( k3) sec t=10sec Ratio ( t= 10 sec) 0.0693 0.035973 0.0198914 1.808469992 0.03465 0.035973 0.0234211 1.535922736 0.0231 0.035973 0.0249 1.444698795 0.0173 0.035973 0.0257217 1.398546752 0.0154 0.035973 0.026 1.383576923 0.0139 0.035973 0.0262283 1.371533801 0.01155 0.035973 0.0265907 1.352841407 0.0099 0.035973 0.0268515 1.339701693 0.00866 0.035973 0.0270509 1.329826364 0.0077 0.035973 0.0272075 1.322172195 0.00693 0.035973 0.0273344 1.316034008

Plotting the ratio of reaction of –rGLP (without inhibition) to the rate – rGLP (with inhibition) as a function of drug inhibition constant k3, we get

9-39

P9-14 (b) The performance equation for mixed reactor is given by:-

V vCAo

XA rA

(1)

Where, V = Volume of the reactor, v = volumetric flow rate of the reactant XA = Conversion of the reactant, -rA = rate of reaction in term of conversion CAo = Initial concentration of the reactant, τ = Residence time = (V/ v) As derived earlier, in case of drug inhibition the rate of disappearance of GLP-1(7-36) is given by:-rA = k1*Ca*Ce + k2* Ca* Cb Let the conversion be XA, Therefore we get, -rA = k1*CAo (1- XA)* CAo (M1 - XA) + k2* CAo (1- XA)* CAo (M2 - XA) --------- (2) Where, M1 = (CEo / CAo) & M2 = (CBo / CAo) Putting, equation (2) in (1), we get



V vCAo



V v

=



τ

=

XA =

k1CAo 2 1

XA

M1

XA

k2 CAo 2 1

XA

M2

XA k1CAo 1

XA

M1

XA

k2 C Ao 1

X A M2

XA

k2 C Ao 1

X A M2

XA

XA k1CAo 1

XA

M1

XA

P9-15 -rs = (μmax*Cs*Cc)/(Km+Cs)

9-40

XA

μmax= 1hr-1 Km = 0.25 gm/dm3 Yc/s = 0.5g/g (a) Cc0 = 0.1g/dm3 Cs0 = 20 g/dm3 Cc=Cc0 + Yc/s(Cs0-Cs) See polymath problem P9-15-a.pol Calculated values of DEQ variables Variable Initial value Minimal value

Maximal value

Final value

1

Cc

0.1

0.1

10.1

10.1

2

Cc0

0.1

0.1

0.1

0.1

3

Cs

20.

6.301E-11

20.

6.301E-11

4

Cs0

20.

20.

20.

20.

5

Km

0.25

0.25

0.25

0.25

6

rc

0.0493827

1.273E-09

4.043487

1.273E-09

7

rs

-0.0987654

-8.086974

-2.546E-09

-2.546E-09

8

t

0

0

10.

10.

9

umax

1.

1.

1.

1.

0.5

0.5

0.5

0.5

10 Ycs

Differential equations 1 d(Cs)/d(t) = rs Explicit equations 1 Cc0 = 0.1 2 Ycs = 0.5 3 Cs0 = 20 4 umax = 1 5 Cc = Cc0+Ycs*(Cs0-Cs) 6 Km = 0.25 7 rs = -umax*Cs*Cc/(Km+Cs) 8 rc = -rs*Ycs

9-41

Plot of Cc and Cs versus time

Plot of rs and rc with time

(b) Change the polymath code to include rg = μmax*(1-Cc/C∞)*Cc C∞= 1 g/dm3

9-42

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Cc

0.1

0.1

0.5751209

0.5751209

2 Cc0

0.1

0.1

0.1

0.1

3 Cinf

1.

1.

1.

1.

4 Cs

20.

19.04976

20.

19.04976

5 Cs0

20.

20.

20.

20.

6 Km

0.25

0.25

0.25

0.25

7 rc

0.0225

0.0225

0.0624999

0.0610892

8 rg

0.09

0.09

0.2499995

0.2443569

9 rs

-0.045

-0.1249997

-0.045

-0.1221784

10 t

0

0

10.

10.

11 umax

1.

1.

1.

1.

12 Ycs

0.5

0.5

0.5

0.5

Differential equations 1 d(Cs)/d(t) = rs Explicit equations 1 Cc0 = 0.1 2 Ycs = 0.5 3 Cs0 = 20 4 umax = 1 5 Cinf = 1 6 Cc = Cc0+Ycs*(Cs0-Cs) 7 Km = 0.25 8 rg = umax * (1-Cc/Cinf) * Cc 9 rs = -Ycs*rg 10

rc = -rs*Ycs

9-43

Plot of Cc with time

(c)

Cs0 = 20g/dm3

Cc0 = 0 The dilution rate at which wash-out occurs will be by setting Cc=0 in equation; Cc = Ycs*(Cs0 – (DKs)/(umax – D)) Dmax =  Dmax =

= 0.987 hr

-1 -1

Thus dilution rate at which washout occurs is 0.987hr .

(d) DC C

rg , D C SO

m CC vO

rgV

rS

,

rg

CC

CCV

Divide by CCV,

D

CS

max

KmD / (

CS

max

KM

Cs Cs

D)

9-44

Cs Km Cs max

,

Now

rS

DC C

CC

YS / C rg

DYC / S C SO

d DC C dD

 Dmax,prod = 0.88 hr

CS

CS

 DCc = D YC/S (CS0 - KmD / (

Now, for Dmax . prod ,

YC / S C SO

max

D) )

0

-1

Using this value of D we can find the value of Cs Cs = CS

KmD / (

D)

max 3

= 1.83 g/dm

CC

YC / S C SO

CS

= 9.085g/dm

3 3

rs = D(Cs0 – Cs) = 15.98 g/dm /hr

(e) Cell death cannot be neglected. Kd =0.02 hr-1 DCc = rg –rd And D(CS0 –Cs)= rS For steady state operation to obtain mass flow rate of cells out of the system, Fc FC =CCv0 = (rg-rd)V= (µ-kd) CCV After dividing by CcV; D=µ-kd Now since maintenance is neglected. Substituting for µ in terms of substrate concentration; Cs =

( D kd )k s ( D kd ) max

The stoichiometry equation can be written as : -rs = rg YS/C CC = Yc/s

CS 0

( D kd )ks ( D kd ) max

9-45

Now the dilution rate at can be found by substituting Cc =0; Thus Dmax =

max

KS

CS 0 CS 0

kd = 0.96 hr-1

Similarly the expression for dilution rate for maximum production is given by solving the equation ; (D+kd) Cc = (D+kd ) YC/S (CS0 - Km( D

Now for Dmax . prod ,

d DC C dD

kd ) / (

max

( D kd )) )

0

Thus we obtain Dmax,prod = 0.86 hr-1

(f) In this case the maintenance cannot be neglected. m = 0.2 g/hr/dm3 The correlation for steady substrate concentration will remain the same. Cs =

DK S max

D

But the cell maintenance cannot be neglected. Thus the stoichiometry equation will be changed. The equation will be -rs = Ys/Crg + mCc => -rs = rg/YS/C + mCc t

=> (Cs0 – Cs) = Cc/YC/S +

mCc dt 0

Also we know that by mass balance -

dCc / dt

(

D )CC

Using this relation, the stoichiometric equation for substrate consumption changes to -

(CS 0 CS ) CC (1/ YC / S  Cc =

m/(

D))

D(CS 0 DK S / ( max D)) 1/ YC / S m / ( max D))

Now for finding the dilution rate at which wash out occurs, Cc =0;

9-46

So Dwashout = 0.98 hr

-1

Similarly to calculate Dmax . prod ,

d DC C dD

0

Thus we obtain Dmax,prod = 0.74 hr-1

(g) Individualised soultion (h) Individualised solution

P9-16 Tessier Equation,

rg

max

1

e

CS / k

CC

Redoing P9-15 part (a) For batch reaction,

dC S dt CC

rS , C CO

rS YC / S C SO

rg

YS / C rC , CS

See Polymath program P9-16-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 7 7 Cs 20 0.0852675 20 0.0852675 Cco 0.1 0.1 0.1 0.1 Ycs 0.5 0.5 0.5 0.5 Cso 20 20 20 20 Cc 0.1 0.1 10.057366 10.057366 k 8 8 8 8 umax 1 1 1 1 Ysc 2 2 2 2 rg 0.0917915 0.0917915 3.8563479 0.1066265 rs -0.183583 -7.7126957 -0.183583 -0.213253 Rates 0.183583 0.183583 7.7126957 0.213253

ODE Report (RKF45) Differential equations as entered by the user [1] d(Cs)/d(t) = rs

9-47

max

1 e

CS / k

CC

Explicit equations as entered by the user [1] Cco = 0.1 [2] Ycs = 0.5 [3] Cso = 20 [4] Cc = Cco+Ycs*(Cso-Cs) [5] k = 8 [6] umax = 1 [7] Ysc = 2 [8] rg = umax*(1-exp(-Cs/k))*Cc [9] rs = -Ysc*rg [10] RateS = -rs

Redoing P9-15 part (b) D C SO DC C rg m

C C vO

Divide by CCV,

CS

rg V

D

k ln 1

CS

rS

rg

CC

max

1 e

CS / k

CCV max

1 e

CS / k

D max

Now

rS

CC

YS / C rg

YC / S C SO

For dilution rate at which wash out occur,

k ln 1

CSO = CS

CS CC = 0

D max

DMAX

max

1 e

CS 0 / k

1hr 1 1 e

20 g / dm3 / 8 g / dm#

Redoing P9-15 part (c) DC C

DYC / S C SO

CS

CS

k ln 1

D max

9-48

0.918hr

1

DCC

DYC / S C SO

k ln 1

D max

Now

d DC C dD

For Dmax . prod ,

0

Dmax . prod = 0.628 hr-1


P9-17 (a) rg =

max

CC

For CSTR,

KM DC C CS

CS CS D C SO

rg

YC / S C SO

DC C

CS

0.8hr

1

v0CC

dCC dt

V

dCC dt

rg

Substrate Balance:

rg

FC

dCS dt

v0CS 0 v0CS

YS rg C

CS CC K M CS max

9-49

YS / C rg

D

4.5 g / dm

1g / dm 3 4.5 g / dm 3 4 1 g / dm 3

6250dm 3

rgV

3

C S CC K M CS

Flow of cells out = Flow of cells in

Cell Balance: FC

1g / dm 3

0.5(10 1) g / dm

P9-17 (b) FC

rS

max

rg

vO 4.5 g / dm 3 V V

rS

10 g / dm 3 1 0.9

C SO 1 X

CC

CS

3

vO

V

This would result in the Cell concentration growing exponentially. This is not realistic as at some point there will be too many cells to fit into a finite sized reactor. Either a cell death rate must be included or the cells cannot be recycled.

P9-17 (c) Two CSTR’s For 1st CSTR, V = 5000dm3 , DC C

rS

rg

D C SO

YS / C rg

rg =

CS max

KM

CS CC CS

See Polymath program P9-17-c-1cstr.pol. POLYMATH Results NLES Solution Variable Value f(x) Ini Guess Cc 4.3333333 9.878E-12 4 Cs 1.3333333 1.976E-11 5 umax 0.8 Km 4 Csoo 10 Cso 10 Ysc 2 rg 0.8666667 rs -1.7333333 V 5000 vo 1000 D 0.2 X 0.8666667 Cco 4.33


Nonlinear equations [1] f(Cc) = D*(Cc)-rg = 0 [2] f(Cs) = D*(Cso-Cs)+rs = 0

Explicit equations [1] umax = 0.8 [2] Km = 4 [3] Csoo = 10 [4] Cso = 10 [5] Ysc = 2 [6] rg = umax*Cs*Cc/(Km+Cs)

9-50

rS

D

vO

V

[7] rs = -Ysc*rg [8] V = 5000 [9] vo = 1000 [10] D = vo/V [11] X = 1-Cs/Csoo [12] Cco = 4.33

CC1 = 4.33 g/dm3 CS1 = 1.33 g/dm3 For 2nd CSTR, D C C 2

X = 0.867 CP1 = YP/CCC1 =0.866 g/dm3

C C1

rg

D C S1

See Polymath program P9-17-c-2cstr.pol. POLYMATH Results NLES Solution Variable Value f(x) Ini Guess Cc 4.9334151 3.004E-10 4 Cs 0.1261699 6.008E-10 5 umax 0.8 Km 4 Csoo 10 Cs1 1.333 Ysc 2 rg 0.120683 rs -0.241366 V 5000 vo 1000 D 0.2 X 0.987383 Cc1 4.33


Nonlinear equations [1] f(Cc) = D*(Cc-Cc1)-rg = 0 [2] f(Cs) = D*(Cs1-Cs)+rs = 0

Explicit equations [1] umax = 0.8 [2] Km = 4 [3] Csoo = 10 [4] Cs1 = 1.333 [5] Ysc = 2 [6] rg = umax*Cs*Cc/(Km+Cs) [7] rs = -Ysc*rg [8] V = 5000 [9] vo = 1000

9-51

CS

rS

[10] D = vo/V [11] X = 1-Cs/Csoo [12] Cc1 = 4.33

CC2 = 4.933 g/dm3 CS2 = 1.26 g/dm3

X = 0.987 CP1 = YP/CCC1 =0.9866 g/dm3

P9-17 (d) For washout dilution rate,

Dmax

DMAX .PROD

max

1

CC = 0 max

KM

C SO C SO

0.8hr 1 10 g / dm 3 4 g / dm 3 10 g / dm 3

KM K M C SO

0.8hr

1

0.57hr

1

4 g / dm 3 4 g / dm 3 10 g / dm 3

1

0.37 hr

Production rate = CCvO(24hr) = 4.85 g / dm 3 x1000dm3/hrx24hr = 116472.56g/day

P9-17 (e) For batch reactor, V = 500dm3,

dC C dt

rg

dC S dt

rS

CCO = 0.5 g/dm3 CSO = 10g/dm3

rS

YS / C rg

rg =

max

KM

See Polymath program P9-17-e.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 6 6 Cc 0.5 0.5 5.4291422 5.4291422 Cs 10 0.1417155 10 0.1417155 Km 4 4 4 4 Ysc 2 2 2 2 umax 0.8 0.8 0.8 0.8 rg 0.2857143 0.1486135 1.403203 0.1486135 rs -0.5714286 -2.8064061 -0.2972271 -0.2972271

ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg [2] d(Cs)/d(t) = rs

9-52

CS CC CS

1

Explicit equations as entered by the user [1] Km = 4 [2] Ysc = 2 [3] umax = 0.8 [4] rg = umax*Cs*Cc/(Km+Cs) [5] rs = -Ysc*rg

For t = 6hrs, CC = 5.43g/dm3. So we will have 3 cycle of (6+2)hrs each in 2 batch reactors of V = 500dm3. Product rate = CC x no. of cycle x no. of reactors x V = 5.43 g/dm3 x 3 x 2 x 500dm3 = 16290g/day.

P9-17 (f) Individualized solution P9-17 (g) Individualized solution P9-18 (a) max

D CS D

KS max

CS KS max

CS KS

CS max

is the intercept and

1

is the slope

max

9-53

CS gm /dm2 D(day-1) CS/D

1 1 1

Slope = 0.5 day

max

3 1.5 2

4 1.6 2.5

intercept = 0.5 gm/dm3

= 2 KS = 0.5*2 = 1

P9-18 (b) CC DK S

YC S CSO

max

D

Inserting values from dataset 4

4 1.8 1.0 2.0 1.8

YC S 50

YS C

1 YC / S

4 50 9.0

10.25

P9-18 (c) Individualized solution P9-18 (d) Individualized solution P9-19 (a) See Polymath program P9-19-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Cc

1.

1.

1.774022

1.774022

2 f

0

0

0.99985

0

3 t

0

0

48.

48.

Differential equations 1 d(Cc)/d(t) = f* Cc * 0.9 / 24 Explicit equations

9-54

10 1.8 5.6

1 f = (If (((t > 0) And (t < 12)) Or ((t > 24) And (t < 36))) Then (sin(3.14 * t / 12)) Else (0))

(b) Given the initial concentration as Co = 0.5 mg/litre we have, = ln

=

Cc

=

exp

= 0.9 day -1 = 0.0375 hr -1 From 6am to 6pm , t= 12 hrs  Cc

=

exp

 Cc

=

exp

 Cc

=

exp

 Cc =

} } *0.0375*

}

1.332

Thus the concentration progresses as = 1.332(conc. of previous day)

9-55

Therefore the time taken to reach the concentration of 200 mg/dm 3 using MS Excel is 22 days. (d) From part (b) Concentration at the end of 1st day = 1.33Co Concentration at the start of 2nd day =1.33Co/2 + 100

(mg/lit)

Concentration at the end of 2rd day= 1.33(1.33Co/2 + 100) And so forth. Thus the concentration progresses as = 1.332(conc. of previous day)+100 Solving for the progression on MS Excel we find the concentration to converge to 299.4011mg/lit Thus, at the end of 365 days the conc. becomes = 299.4011mg/lit (e) Assuming that dilution and removal of algae is done at the end of the day after the growth period is over. Then, Rate of decrease of conc. =

= - k Cc

where, k = 1 (day-1)

At t=0, Co= 200mg/lit (i.e. maximum concentration allowed for maximum productivity at the end of the day) So, removal per day:Cc= Co exp (-kt) = 200 exp (-1 day-1 x 1 day) = 200 exp (-1)  Cc= 73.5 mg/dm3/day  Volume of the pond = 5000 gallons or 18940.5 dm3  Total mass flow rate of algae = 1392.1 g/day 9-56

(f) Now, since CO2 will affect the rate of growth of algae too, then let the reaction be:CO2 + Algae  more algae Hence, A

+C



2C Where, A: CO2 C: algae

Now, rate of growth of algae should be rc = k sin( CaCc

Earlier, Ca = 1.69gm/kg of water = 1.69 gm/dm3 of water (density of water = 1kg/lit)

We have rate law as rc = sin ( Cc

( CO2 concentration was assumed to be a constant = 1.69g /dm3

of water ) Hence, = k 1.69 gm / dm3 k = (.9 per day)

(1.69 gm / dm3)

k = 0.5325 dm3/g day or 0.022 dm3/g hr A At t=0

+

Ca0

C Cc0



2C 0

Let conversion with respect to C is X Then, At t=t C

Ca0 – XCc0

Cc0 (1-X)

2Cc0X

Cc= Cc0 (1+ X) Ca= Ca0 – XCc0 = Cc0 (M – X)

where, M= Ca0/Cc0

Hence, = k sin (

CaCc

9-57

thus, total number of moles of

=>

= k sin (

=>

= Cc0 k sin (

=>

(1+X) (M-X)

=

=>

ln

= (M+1)k Cc0 {1 -

=> =>

Cc0 (1+X) Cc0 (M-X)

}

= exp [(M+1) k Cc0 {1 X (1 + exp [k (M+1) Cc0 {1 -

{1 -

}

} }

]

]/M) =

exp [k (M+1) Cc0

]-1

=> X = Now, X= 1Hence

,

=1-

We have, Ca0 = Ks = 2g/dm3 & Cc0 = 1mg/ dm3 (assuming initial seeding value of algae from (c)) M= Ca0/Cc0 = 2000 Substituting the values we have the concentration profile i.e. (g)

Let A: the species of unwanted algae C: the species of desired algae Then, Rate of growth of A = 2  

= 2 for A = 2 x

for C

Now overall density of the medium at any time t 9-58

Vs. t

= Ca + Cc Hence, from given conditions, Ca=0.5(Ca + Cc) So, at that time Ca=Cc Now, 

=

Cc

 Cc = Cc0 exp( 

=

Ca

 Ca = Ca0 exp (2 Assuming, Cco= initial seed concentration of desired algae= 1 mg/litre & Cao = 0.1 mg/litre (given). Now, since the concentration is very less assuming there is no constraint of sunlight.  Ca = Cc  Cc0 exp(

= Ca0 exp (2

 ln (Cco/Cao) = exp (

where, = 0.9 per day

Putting the values, we get

t = 2.56 days or 61.4 hrs.

Since the number of days is coming less than 4.347 days which was calculated in part (c), so the effect of daylight can be neglected. Hence, the initial assumption is verified.

P9-20 The following errors are present in this solution1. For the Eadie-Hofstee model, we plot –rS as a function of [-rS/(S)] and for the HanesWoolf model, we plot [(S)/-rS] as a function of (S). Since the y-axis plot parameter here is [(S)/-rS], it is a Hanes-Woolf plot and not the Eadie-Hofstee plot.\ 2. For Competitive Inhibition, Hanes-Woolf form is:

For Uncompetitive Inhibition, Hanes-Woolf form is:

9-59

Since the intercept is fixed and slope is changing in the given plot, this is uncompetitive inhibition rather than being competitive one. 3. The x-axis variable is written wrong as (S + S2). It should be (S) for Hanes-Woolf form. 4. The expression for intercept is correct but the slope is given wrong. The correct expressions are -

5. As concentration of inhibitor (I) increases, slope increases. In the given plot, slope for line 1 is more as compared to line 2 in spite of having lower concentrations. This implies that the concentration values are switched. Also the numerically calculated slope values are wrong. The correct values are as – Line 1:

(I) = 0.05 M

Slope 1 = 35

Line 2:

(I) = 0.02 M

Slope 2 = 15.5

6. Calculations on the basis of correct values – ;

;

Solving these above equations, we get kinetic law parameters

Vmax = 0.4 ;

KM = 0.04

KI = 3.85e-3

;

9-60





10-1

Solutions for Chapter 10 – Catalysis and Catalytic Reactors


P10-2 (a) Example 10-1 (1)

CT .S CB . s KB

Cv KT PT Cv K B PB 1.39

KT PT 0 (1 X ) K B PT 0 X KT

KT (1 X ) KB X

1.038

CT .S Therefore, C = f(X) can be plotted. B.s (2)

Cv Ct

1 1 KT PT K B PB

1 1 KT PT 0 (1 X ) K B PT 0 X

Here, PT0 = yA0PTotal = 0.3X40 atm = 12 atm & K B

1.39 KT

1.038

Cv Therefore, C = f(X) can be plotted. t (3)

CB . S Ct

Cv K B PB Cv (1 KT PT K B PB )

K B PT 0 X 1 KT PT 0 (1 X ) K B PT 0 X

CB .S Therefore, C = f(X) can be plotted. t

P10-2 (b) Example 10-2 (1) See polymath problem P10-2-b.pol Increasing the pressure will increase the conversion for the same catalyst weight. At 40 atm, we had 68.2% conversion for a catalyst weight of 10000kg. While, at 80 atm we have 85.34% conversion for the same weight of the catalyst.

10-2

At 1 atm, we get 0.95% conversion for 10000 kg of catalyst. However, it’s not practically possible to operate at inlet pressure of 1 atm, because there will be no flow of feed into the PFR, due to absence of pressure difference. (2) If the flow rate is decreased the conversion will increase for two reasons: (1) smaller pressure drop and (2) reactants spend more time in the reactor. (3) From figure E10-3.1 we see that when X = 0.6, W = 5800 kg.

P10-2 (c) Example 10-3 With the new data, model (a) best fits the data (a) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Kea*Pea+Ke*Pe) Variable Ini guess Value 95% confidence k 3 3.5798145 0.0026691 Kea 0.1 0.1176376 0.0014744 Ke 2 2.3630934 0.0024526 Precision R^2 R^2adj Rmsd Variance

= = = =

0.9969101 0.9960273 0.0259656 0.0096316

(b) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe) Variable Ini guess Value 95% confidence k 3 2.9497646 0.0058793 Ke 2 1.9118572 0.0054165 Precision R^2 R^2adj Rmsd Variance

= = = =

0.9735965 0.9702961 0.0759032 0.0720163

(c) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/((1+Ke*Pe)^2) Variable Ini guess Value 95% confidence

10-3

k Ke Precision R^2 R^2adj Rmsd Variance

3 2

= = = =

1.9496445 0.3508639

0.319098 0.0756992

0.9620735 0.9573327 0.0909706 0.1034455

(d) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Peâ*Ph^b Variable Ini guess Value 95% confidence k 3 0.7574196 0.2495415 a 1 0.2874239 0.0955031 b 1 1.1747643 0.2404971 Precision R^2 R^2adj Rmsd Variance

= = = =

0.965477 0.9556133 0.0867928 0.107614

Model (e) at first appears to work well but not as well as model (a). However, the 95% confidence interval is larger than the actual value, which leads to a possible negative value for Ka. This is not possible and the model should be discarded. Model (f) is the worst model of all. In fact it should be thrown out as a possible model due to the negative R^2 values. (e) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/((1+Ka*Pea+Ke*Pe)^2) Variable Ini guess Value 95% confidence k 3 2.113121 0.2375775 Ka 1 0.0245 0.030918 Ke 1 0.3713644 0.0489399 Precision R^2 R^2adj Rmsd Variance

= = = =

0.9787138 0.9726321 0.0681519 0.0663527

(f) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Ka*Pea)

10-4

Variable k Ka Precision R^2 R^2adj Rmsd Variance

Ini guess 3 1

Value 95% confidence 44.117481 7.1763989 101.99791 16.763192

= -0.343853 = -0.5118346 = 0.5415086 = 3.6653942

P10-2 (d) Individualized solution

P10-3 Solution is in the decoding algorithm given with the modules

P10-4

P10-4 (a)

10-5

P10-4 (b) Adsorption of isobutene limited

P10-4 (c)

10-6

P10-4 (d)

P10-4 (e) Individualized solution P10-5 (a) H H2 E

Ethylene

A Ethane H2

C 2H 4

H E

cat

C 2H 6

A

Because neither H2 or C2H6 are in the rate law they are either not adsorbed or weakly adsorbed. Assume H2 in the gas phase reacts with C2H6 adsorbed on the surface and ethane goes directly into the gas phase. Then check to see if this mechanism agrees with the rate law Eley Rideal

E S

E S H

E S

A S

rE

k AD PE CV

rS

k SC E SPH 2

CE S KE

10-7

Assume surface reaction

CE

S

rS

k S C E SPH

CT

K E PE C V

CV

rA

CE

k SK E C T

S

PE PH 1 K E PE


P10-6 O2

2S

2O S

C3 H 6 O S

rS

C3 H 5OH

k A PA2 CV2

rAD kA

0

CA S rB

rC

k3 PBCV K A PA

k D CC

S

rC S kD

0

CC

K C PC CV

CT

rB

S

C A2 S KA

CV K A PA rS

S

C S

S

k3 PB C A S

rAD

CV

rS

CA

S

PC CV KD

CC

S

k D CC

CV 1

S

K C PC CV

K A PA

K C PC

k3CT PB K A PA 1

K A PA

2S

2A S

B A S

C3 H 5OH S

C3 H 6OH S

rB

A2

KC PC

10-8

C S

C S

P10-7 (a)

P10-7 (b)

10-9

P10-7 (c)

10-10

P10-7 (d) Individualized solution P10-7 (e) Individualized solution

P10-8

10-11

P10-9

P10-9 (a)

10-12

P10-9 (b)

P10-10

10-13

P10-10 (a)

10-14

P10-10 (b)

10-15

Substituting the expressions for CV and CA·S into the equation for –r’A

P10-10 (c) Individualized solution P10-10 (d) First we need to calculate the rate constants involved in the equation for –r’A in part (a). We can rearrange the equation to give the following

10-16

Thus from the slope and intercept data

See Polymath program P10-10-d.pol. POLYMATH Results Calculated values of the DEQ variables Variable W X e

initial value 0 0 1

minimal value 0 0 1

maximal value 23 0.9991499 1

10-17

final value 23 0.9991499 1

Pao Pa k1 k2 Fao ra rate

10 10 560 2.04 600 -12.228142 12.228142

10 0.0042521 560 2.04 600 -68.5622 2.3403948

10 10 560 2.04 600 -2.3403948 68.5622

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [1] e = 1 [2] Pao = 10 [3] Pa = Pao*(1-X)/(1+e*X) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [7] ra = -k1*Pa/((1+(k2*Pa))^2) [8] rate = -ra

P10-10 (e) Individualized solution P10-10 (f)

10-18

10 0.0042521 560 2.04 600 -2.3403948 2.3403948

Use these new equations in the Polymath program from part (d). See Polymath program P10-10-f.pol. POLYMATH Results Calculated values of the DEQ variables Variable W X y e Pao Pa k1 k2 Fao ra rate alpha

initial value 0 0 1 1 10 10 560 2.04 600 -12.228142 12.228142 0.03

minimal value 0 0 0.0746953 1 10 7.771E-05 560 2.04 600 -68.584462 0.0435044 0.03

maximal value 23 0.9997919 1 1 10 10 560 2.04 600 -0.0435044 68.584462 0.03

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao [2] d(y)/d(W) = -alpha*(1+X)/2/y Explicit equations as entered by the user [1] e = 1 [2] Pao = 10 [3] Pa = y*Pao*(1-X)/(1+e*X) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [7] ra = -k1*Pa/((1+(k2*Pa))^2) [8] rate = -ra [9] alpha = .03

10-19

final value 23 0.9997919 0.0746953 1 10 7.771E-05 560 2.04 600 -0.0435044 0.0435044 0.03

P10-11(a)

10-20

P10-11 (b)

P10-11 (c) The estimates of the rate law parameters were given to simplify the search techniques to make sure that it converged on a false minimum. In real life, one should make a number of guesses of the rate law parameters and they should include a large range of possibilities

P10-11 (d) A ra PA

B H kPA 1 K A PA

K B PB

PA0 (1 X ) , PB (1 X )

K H PH

2

PA0 X , PH (1 X )

PA0 X here, (1 X )

10-21

y A0

1(2 1) 1

Where, k 0.00137 K A 4.77 KB

0.259

KH

0.424

PA0

15atm

FA0 10mol / sec

Now, the design equation for the PFR, rA dX dW FA0 Solving the above equations using Polymath (Refer to P10-11-d.pol) Calculated values of DEQ variables Variable

Initial value

Minimal value

Maximal value

Final value

1

Fao

10.

10.

10.

10.

2

k

0.00137

0.00137

0.00137

0.00137

3

Ka

4.77

4.77

4.77

4.77

4

Kb

0.262

0.262

0.262

0.262

5

Kh

0.423

0.423

0.423

0.423

6

Pa

15.

3.064E-09

15.

3.064E-09

7

Pao

15.

15.

15.

15.

8

Pb

0

0

7.5

7.5

9

Ph

0

0

7.5

7.5

10 ra

-3.904E-06

-1.26E-05

-1.114E-13

-1.114E-13

11 W

0

0

2.0E+06

2.0E+06

12 X

0

0

1.

1.

10-22

Differential equations 1 d(X)/d(W) = -ra/Fao Explicit equations 1

Pao = 15

2

k = 0.00137

3

Ka = 4.77

4

Kb = 0.262

5

Kh = 0.423

6

Fao = 10

7

Ph = Pao*X/(1+X)

8

Pa = Pao*(1-X)/(1+X)

9

Pb = Pao*X/(1+X)

10 ra = -k*Pa/((1+Ka*Pa+Kb*Pb+Kh*Ph)^2)

From the above graph, we get the weight of catalyst required for 85% conversion is 1.2083x106 kg. The required catalyst weight is so high because of very low value of reaction constant ’k’.

10-23

P10-12(a) Proposed Single site Mechanism: S ' H 2O

S ' H 2O

S ' H 2O

S ' H2

S ' H2 S'

1 O2 2

S ' H2

1 O2 2

S

Rate of adsorption: rAD

Rate of surface reaction: rs Rate of desorption: rD

CS '

k AD (Cv ' PH 2O k s CS '

k D (CS '

K H 2O

H 2O

H2

H 2O

)

---------- (1)

K H 2 PH 2 Cv ' )

Assuming, surface reaction to be rate limiting, we get rAD rD 0 k AD k D CS '

H 2O

K H 2OCv ' PH 2O and CS '

H2

K H 2 PH 2 Cv '

From total site balance, we get Ct Cv ' 1 K H 2O PH 2O K H 2 PH 2 Putting all the values in equation (1), we get k s K H 2O Ct PH 2O rs 1 K H 2O PH 2O K H 2 PH 2 rs

kPH 2O 1 K H 2O PH 2O

K H 2 PH 2

Where, k

ks K H 2OCt

P10-12(b) Proposed Mechanism: S ' H 2O

S ' H 2O

S ' H 2O

S H2

S H2

S

H2


k D (Cv ' PH 2O

CS '

H 2O

K H 2O

) 10-24

Rate of surface reaction: rs Rate of desorption: rD

ks (CS '

k D (CS

H 2O

CS

H2

) ---------- (1)

K H 2 PH 2 Cv )

H2

Assuming, surface reaction to be rate limiting, we get rAD rD 0 k AD k D CS '

K H 2OCv ' PH 2O and CS

H 2O

H2

K H 2 PH 2 Cv

From total site balance, we get Ct ' Cv ' 1 K H 2O PH 2O

Cv

Ct 1 K H 2 PH 2

Putting all the values in equation (1), we get K H O PH O Ct ' K H 2 PH 2 Ct rs k s ( 2 2 ) 1 K H 2O PH 2O 1 K H 2 PH 2

P10-12(c) Proposed Mechanism: Step (1)

S h

S ' O2

S ' O2

S ' O2

S ' H 2O

Step (2) S ' H 2O S H2

S ' H 2O S H2 S

H2

For Step (1) Rate of adsorption of O2: rAD O 2 Rate of Reaction: rSO 2

kD O 2 (Cv

kSO 2 (CS ' O2

CS ' O2 KS

)

KO2 PO2 Cv ' )

Rate of surface reaction of water is rate controlling, we get rAD O 2 rS O 2 0 kD O 2 kS O 2 CS ' O2

Cv K S ----------- (2) 10-25

CS ' O2

KO2 PO2 Cv ' ------- (3)

Equating (2) and (3), we get

Cv

KO2 PO2 Cv ' KS

-------- (4)

Now, here the site balance gets modified because oxygen is also getting adsorbed. Ct

Cv Cv ' CS ' O2

CS '

H 2O

CS

H2

Putting, the values and replacing Cv using equation (4), we get Cv ' (1 K O2 PO2

And, Cv

rs

K H 2O PH 2O

KO2 PO2 Cv ' KS

Ct K O2 PO2

K O2 K H 2 PO2 PH 2

KS

KS

= K S (1 KO2 PO2

)

KO2 PO2 Ct KO2 PO2 K H 2O PH 2O KS

Putting all the values in equation (1), we get K H 2O PH 2OCt ks ( KO2 PO2 K O2 K H 2 PO2 PH 2 (1 KO2 PO2 K H 2O PH 2O ) KS KS

P10-13 (a)

10-26

K O2 K H 2 PO2 PH 2 KS

K S (1 KO2 PO2

)

K H 2 PH 2 K O2 PO2 Ct K O2 PO2 K H 2O PH 2O KS

K O2 K H 2 PO2 PH 2 KS

) )

P10-13 (b)

P10-14 Assume the rate law is of the form rDep At high temperatures K

rDep rDep 2 PVTIPO

Run 1

k 0.028 0.45 0.2

Run 5

2 as T and therefore KPVTIPO

2 kPVTIPO

0.05

Run 2

2 kPVTIPO 2 1 KPVTIPO

11.2

2

11.28

2

11.25

7.2 0.8

2

At low temperature and low pressure

rDep

2 kPVTIPO

10-27

1

rDep

k

2 PVTIPO

Run 1

0.004 0.1

Run 2

2

0.015 0.2

2

0.4 0.375

These fit the low pressure data 2 At high pressure KPVTIPO

2 kPVTIPO 2 KPVTIPO

rDep

1

k K

This fits the high pressure data At PVTIPO = 1.5, r = 0.095 and at PVTIPO = 2, r = 0.1 Now find the activation energy At low pressure and high temperature k = 11.2 At low pressure and low temperature k = 0.4

ln

k2 k1

ln

11.2 0.4

E R

E 1 R T1 E R

1 T2

E T2 T1 R T1T2

473 393 473 393

7743

E 15330.65

cal mol

10-28

P10-15

P10-16 (a) Using Polymath non-linear regression few can find the parameters for all models: See Polymath program P10-16.pol. (1) POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PMâ*PH2^b Variable k a b

Ini guess 1 0.1 0.1

Value 1.1481487 0.1843053 -0.0308691

95% confidence 0.1078106 0.0873668 0.1311507

10-29

Precision R^2 R^2adj Rmsd Variance

= = = =

0.7852809 0.7375655 0.0372861 0.0222441

α = 0.184 β = -0.031 k = 1.148 (2) POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM/(1+KM*PM) Variable k KM Precision R^2 R^2adj Rmsd Variance

Ini guess 1 2

= = = =

Value 12.256274 9.0251862

95% confidence 2.1574162 1.8060287

0.9800096 0.9780106 0.0113769 0.0018638

k = 12.26 KM = 9.025

(3) POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM*PH2/((1+KM*PM)^2) Variable k KM Precision R^2 R^2adj Rmsd Variance

Ini guess 1 2

Value 8.4090333 2.8306038

95% confidence 18.516752 4.2577098

= -4.3638352 = -4.9002187 = 0.1863588 = 0.5001061

k = 8.409 KM = 2.83 (4) 10-30

POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM*PH2/(1+KM*PM+KH2*PH2) Variable k KM KH2

Ini guess 1 2 2

Value 101.99929 83.608282 67.213622

95% confidence 4.614109 7.1561591 5.9343217


= -3.2021716 = -4.1359875 = 0.1649487 = 0.4353294

k = 102 KM = 83.6

KH2 = 67.21

P10-16 (b) We can see from the precision results from the Polymath regressions that rate law (2) best describes the data.

P10-16 (c) Individualized solution. P10-16 (d) We have chosen rate law (2) Proposed Mechanism M

S

M S

M S

H 2( g ) T( g )


k AD (Cv PM

Rate of surface reaction: rs

CM S ) KM

k s CM S ---------- (1)

Assuming, surface reaction to be rate limiting: rAD 0 k AD CM

S

Cv PM K M

10-31

Putting, the value of CM S in Equation (1) rs

ks Cv PM K M ----------------- (2)

Now, applying site balance Ct Cv CM S Ct

Cv Cv PM K M

Cv

Ct 1 PM K M

Putting in Equation (2), we get ks K M Ct PM rs 1 K M PM

rs

kPM Where, k = ks K M Ct 1 K M PM

P10-17 Mistakes in the solution: 1) In the CC.S expression, the constant should be KA2 instead of KDC 2) The overall site balance should include the product, as it too is getting adsorbed. 3) The final rate law expression derived is wrong. The correct expression has been derived below:

Assuming, desorption to be rate controlling, we get:

rS1 kS1

rAD k AD

A S C A.S A.S

rS 2 kS 2

0

A.S K A PACV A.S

C A2 .S C A2 .S

A2 .S S K S C A2.s / CV K S K A 2 PA 2CV

10-32

A2 .S B ( g ) rS2

C.S CC .S ] K A2

k S2 [ PB C A 2 .S

CC .S

K A2 PB C A2 .S C A2. S

CC .S

K A2 PB K S

CC .S

K A2 PB K S K A 2 PA 2Cv

CV

now, Ct

CV

C A.S

C A2 .S

CC .S

Putting, all the values in equation (1), we get Ct

CV

Ct

CV (1 K A PA

K S C A2.S CV

K A PACV

K A2 PB K S

K S K A2 PA 2

C A2 . S 2

CV

K A2 K S K A2 PB PA2 )

now, rC rC rC

kc CC .S

kC K A2 PB K S

C A2.S CV

kC K A2 K S K A2 PB PA2Cv

kC K A2 K S K A2 PB PA2 (1 K A PA

K S K A2 PA 2

(1 K A PA

kPB PA2 K S K A2 PA 2 K A2 K S K A2 PB PA2 )

Where, k

K A2 K S K A2 PB PA2 )

kC K A2 K S K A2

10-33

CD10GA-1

10-34

10-35

CD10GA-2

10-36

CD10GA-3

10-37

10-38

10-39





11-1

Solutions for Chapter 11: Non-isothermal Reactor Design-The Steady-state energy balance and adiabatic PFR applications P11-1 Individualized solution P11-2 (a) Example 11-1 For CSTR

V X

FA0 X rA

X k 1 X

k

Ae E RT 1 Ae E RT

1

k

T

T0

0

One equation, two unknowns Adiabatic energy balance H Rx X CPA

In two equations and two unknowns In Polymath form the solution Ae E RT X 1 Ae E RT

f X f T

H Rx X CPA

T0

Enter X, A, E, R, C PA , T0 and HRx to find τ and from that you can find V.

P11-2 (b) Example 11-2 Helium would have no effect on calculation

%Error

CP T TR H Rx

CP T TR

1270 100 5.47% 23,210

11-2

P11-2 (c) Example 11-3 Set Vfinal = 0.8 m3

POLYMATH Results Calculated values of the DEQ variables Variable V X Ca0 Fa0 T Kc k Xe ra rate

initial value 0 0 9.3 146.7 330 3.099466 4.2238337 0.7560658 -39.281653 39.281653

minimal value 0 0 9.3 146.7 330 2.852278 4.2238337 0.7404133 -56.196156 39.281653

maximal value 0.8 0.2603491 9.3 146.7 341.27312 3.099466 9.3196276 0.7560658 -39.281653 56.196156

final value 0.8 0.2603491 9.3 146.7 341.27312 2.852278 9.3196276 0.7404133 -56.196156 56.196156

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra/Fa0 Explicit equations as entered by the user [1] Ca0 = 9.3 [2] Fa0 = .9*163 [3] T = 330+43.3*X [4] Kc = 3.03*exp(-830.3*((T-333)/(T*333))) [5] k = 31.1*exp(7906*(T-360)/(T*360)) [6] Xe = Kc/(1+Kc) [7] ra = -k*Ca0*(1-(1+1/Kc)*X) [8] rate = -ra

Change the entering temperature T0 in the polymath code above (highlighted in yellow), and record the exiting conversion: T X

330 0.26

340 0.54

350 0.68

370 0.66

390 0.65

X

11-3

420 0.62

450 0.59

500 0.55

600 0.48

We see that there is a maximum in the exit conversion corresponding to T0 = 350K. Therefore, the inlet temperature we would recommend is 350K.

P11-2(d) Aspen Problem

P 11-2(e) (1)

(2) At high To, the graph becomes asymptotic to the X-axis, that is the conversion approaches 0. At low To, the conversion approaches 1. (3) On addition of equal molar inerts, the energy balance reduces to:

X EB

CPA T T0

CPI (T T0 ) H Rx

CPA CPI 11-4

 X EB

2CPA T T0 H Rx

(4) It is observed that at the same inlet temperature, the equilibrium conversion on the addition of inerts is greater. This also follows from the Le Chatelier’s principle.

P11-2(f) .

.

Q

m CP T

.

Q

220kcal / s 220 103 m 18(400 270) 854.7 mol / s .

For the2nd case: Hot Stream: 460K  350K Cold Stream: 270K  400K LMTD = 69.52 oC

220 103 A 100 69.52 31.64m2 11-5

P11-3 A B

C

3

Since the feed is equimolar, CA0 = CB0 = .1 mols/dm

CA = CA0(1-X) CB = CB0(1-X) Adiabatic:

T

X[

H R (T0 )] X CP i CPi

T0 CP

C pC C pB C pA

H R (T ) i

T

Ci

HC

C pA

HB B

C pB

30 15 15 0

H A = -41000-(-15000-(-20000) = -6000 cal/mol A cal 15 15 30 mol K

6000 X 300 200 X 30 k C A2 0 (1 X ) 2 .01 k (1 X )2

300 rA

P11-3 (a) VPFR

FA0

VCSTR

dX rA

FA0 X rA

For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mols/dm3 See Polymath program P11-3-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 X

0

0

0.85

0.85

2 V

0

0

308.2917

308.2917

3 Ca0

0.1

0.1

0.1

0.1

4 Fa0

0.2

0.2

0.2

0.2

5 T

300.

300.

470.

470.

6 k

0.01

0.01

4.150375

4.150375

7 ra

-0.0001

-0.0018941

-0.0001

-0.0009338

Differential equations 1 d(V)/d(X) = -Fa0 / ra Explicit equations 1 Ca0 = .1 2 Fa0 = .2

11-6

3 T = 300 + 200 * X 4 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T)) 5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)

V = 308.2917dm3 For the CSTR, X = .85, T = 300+(200)(85) = 470 K. k = 4.31 (Using T = 470K in the formula). -rA = .000971 mol/dm3/s

FA0 X

V

rA

.1 2 .85 175 dm3 -4 9.71 10

The reason for this difference is that the temperature and hence the rate of reaction remains constant throughout the entire CSTR (equal to the outlet conditions), while for a PFR, the rate increases gradually with temperature from the inlet to the outlet, so the rate of increases with length.

P11-3 (b) T

T0

X[

HR ] i CPi

For boiling temp of 550 k, 550 = T0 + 200 T0 = 350K

P11-3 (c)

P11-3 (d) FA0 X rA

VCSTR

VCSTR ( rA ) FA0

X

For V = 500 dm3, FA0=.2

rA

T

k C A2 0 (1 X ) 2 300 200 X

.01 k (1 X )2

Now use Polymath to solve the non-linear equations. 11-7

See Polymath program P11-3-d-1.pol. Calculated values of NLE variables Variable Value

f(x)

Initial Guess

0

480.

1 T

484.4136

2 X

0.9220681 -2.041E-09 0.9

Variable Value 1 k

6.072856

2 ra

0.0003688

Nonlinear equations 1 f(T) = 300 + 200 * X - T = 0 2 f(X) = 500 - .2 * X / ra = 0 Explicit equations 1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T)) 2 ra = 0.01 * k * (1 - X) ^ 2

Hence, X = .922 and T = 484.41 K For the conversion in two CSTR’s of 250 dm3 each, For the first CSTR, using the earlier program and V = 250 dm3, Calculated values of NLE variables Variable Value

f(x)

Initial Guess

1 T

476.482 1.137E-13 480.

2 X

0.88241 -5.803E-09 0.9

Variable Value 1 k

5.105278

2 ra

0.0007059

Nonlinear equations 1 f(T) = 300 + 200 * X - T = 0 2 f(X) = 250 - .2 * X / ra = 0 Explicit equations 1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T)) 2 ra = 0.01 * k * (1 - X) ^ 2

T = 476.48 ad X = .8824 Hence, in the second reactor,

11-8

VCSTR X

T

FA0 ( X X 1 ) rA VCSTR ( rA ) X 1 FA0

Tout ,CSTR1 200( X

X1 )

See Polymath program P11-3-d-2.pol. Calculated values of NLE variables Variable Value

f(x)

Initial Guess

0

480.

1 T

493.8738

2 X

0.9693688 -1.359E-09 0.8824

Variable Value 1 k

7.415252

2 ra

6.958E-05

3 X1

0.8824

Nonlinear equations 1 f(T) = 476.48 + 200 * (X - X1) - T = 0 2 f(X) = 250 - .2 * (X - X1) / ra = 0 Explicit equations 1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T)) 2 ra = 0.01 * k * (1 - X) ^ 2 3 X1 = .8824

Hence, final X = .9694

P11-3 (e) Individualized solution P11-3 (f) Individualized solution

11-9

P11-4 (a)

11-10

P11-4 (b)

P11-4 (c) Individualized solution P11-4 (d)

11-11

11-12

11-13

P11-5 (a) A

B C

CA

CT

I

FA FT

FI FA CT

FT

C A CI FA

C A01

FI C A 0 CI 0

FA0 FA0 FI 0

C A 0 CI 0 1 I

C A01

P11-5 (b) Mole balance:

dX dV

rA FA0

Rate law:

rA

kC A

Stoichiometry: C A

C A01

1 X T0 1 X T

y A0 1 1 1 1 11-14

FA0 FT 0 1

y A0

1

T

FA0 FA0

1 Fi 0

1

i

i

X H RX

CPA

CPA

i

i

CPi T0

CPi

Enter these equations into Polymath See Polymath program P11-5-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable V X Cao Cio theta Fao Cao1 e To dHrx Cpa Cpi T k ra

initial value 0 0 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1100 25.256686 -0.0110894

minimal value 0 0 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1098.3458 24.100568 -0.0110894

maximal value 500 0.417064 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1100 25.256686 -0.0061524

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [1] Cao = 2/(.082*1100) [2] Cio = Cao [3] theta = 100 [4] Fao = 10 [5] Cao1 = (Cao+Cio)/(theta+1) [6] e = 1/(1+theta) [7] To = 1100 [8] dHrx = 80000 [9] Cpa = 170 [10] Cpi = 200 [11] T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi) [12] k = exp(34.34-34222/T) [13] ra = -k*Cao1*(1-X)*To/(1+e*X)/T

11-15

final value 500 0.417064 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1098.3458 24.100568 -0.0061524

P11-5 (c) There is a maximum at θ = 8. This is because when θ is small, adding inerts keeps the temperature low to favor the endothermic reaction. As θ increases beyond 8, there is so much more inert than reactants that the rate law becomes the limiting factor.

P11-5 (d) The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The new code is not shown, but the plots are below. See Polymath program P11-5-d.pol.

11-16

The maximum conversion occurs at low values of theta (θ < 8) because the reaction is now exothermic. This means heat is generated during the reaction and there is no advantage to adding inerts as there was in the endothermic case.

P11-5 (e) We need to alter the equations from part (c) such that rA kC A2 and CA0 = 1 A plot of conversion versus theta shows a maximum at about θ = 5. See Polymath program P11-5-e.pol.

P11-5 (f) We need to alter the equations from part (c) such that We already know that C A

C A0

rA

k CA

CB CC KC

1 X T0 . Now weed need expressions for CB and CC. From 1 X T

stoichiometry we can see that CB = CC. In terms of CA0 we find that:

CB

CC

C A0

X 1

T0 X T

We also need an equation for KC: K C

KC1 exp

H RX 1 R T1

1 T

2 exp

80000 1 1 8.314 1100 T

When we enter these into Polymath we find that the maximum conversion is achieved at approximately θ = 8. See Polymath program P11-5-f.pol.

11-17

P11-5 (g) See Polymath program P11-5-g.pol.

P11-6 (a) Mole Balance

Adiabatic (1)

dX dW W dX dV

Rate Law

Stoichiometry

rA FA0 bV

rA B FA0 k CA

rA FA0 CB KC

(2)

rA

(3)

k

(4)

KC

KC2 exp

(5)

CA

CA0 1 X y T0 T

(6)

CB

CA0Xy T0 T

k 1 exp

E 1 R T1

11-18

1 T H Rx 1 R T2

1 T

dy FT T dW y FT 0 T0 W V dy b T dV 2y T0 Parameters

(7) – (15)

T 2y T0

FA0, k1, E, R, T1, KC2,

H Rx , T2, CA0 , T0 , ,

Energy Balance

CP

Adiabatic and

(16A)

0

Additional Parameters (17A) & (17B) Heat Exchange

(16B)

H Rx X iC Pi

T T0

T0 ,

dT dV

rA

FiCPi

FA0

dT dV

rA

i C Pi

C PA

I C PI

H Rx Ua T Ta FiCPi iCPi

H Rx FA0

11-19

CP X , if CP

Ua T Ta iC Pi

0 then

b

(a) Adiabatic

11-20

Adiabatic

11-21

P11-7 (a) Mole Balance (1)

dX dW

rA

rA FA 0

k C AC B

Rate Law

(2)

Rate Law

(3)

k

Rate Law

(4)

k 1 0.0002

Rate Law

(5)

T1

Rate Law

(6)

E 25,000

k 1 exp

E 1 R T1

CC2 KC

1 T

310

11-22

KC

K C 2 exp

H Rx 1 R T2

1 T

Rate Law

(7)

Rate Law

(8)

Rate Law

(9)

T2

Rate Law

(10)

K C2 1,000 KC 1000exp

H Rx

20,000

305 20,000 1 1.987 305

1 T

2 1 1 0

y A0

Stoichiometry

(11)

CA

0.5 2 1 1

CA0 1 X y

B

Stoichiometry

(12)

Stoichiometry

(13)

Stoichiometry

(14)

Stoichiometry

(15)

2

C B C A0 2 X

CC

2CA 0X

dy dW

11-23

T0 y T

T0 y T

T 2y T0

= 0.00015

(b)

T0 T

0

rA

k C A0 1 X y

rA

kC 2A0

T0 T

C A0 2 X y

T

4X 2

1 X 2 X

2C A0 Xy

T0

KC

T0 T

KC

y

T0

2

T

(c)

Equilibrium Conversion

KC

4C 2A0 X 2e 1 Xe 2 Xe

C 2A0

X 2e 1 Xe 2 Xe

KC 4 2

KC 4

3K C Xe 4

KC 1 X 2e 4

Equilibrium Conversion (19)

Xe

3KC 4

3KC 4 2

Xe

X 2e

3K C K Xe 2 C 4 4 2

KC 4

Xe

K C X 2e 4

2KC

KC 4

0

1

1

T

X W

W

Section 1.01 Case 1 Adiabatic

11-24

2

(d) Energy Balance

(16)

Energy Balance

(17)

T

H Rx X

T0

iC Pi

T0 = 325

H Rx CP

2C PC

HR C PA

C P T TR

C PB CP

i C Pi

C PA 20 2 20

Energy Balance

(e)

iC Pi

(18)

T T0

H Rx X i C Pi

11-25

2 20

20 20 0

0

B C PB

C C PC

0 20

I C PI

Everything that enters

1 40

100 cal mol K

400

20,000X 100

400 200X

(f) Adiabatic

11-26

0

Adiabatic

Gas Phase Adiabatic

11-27

Gas Phase Adiabatic

11-28

P11-8 1 Xe

2

KC

Xe

T

X e2

CC CD C AC B

KC

1

T0

KC

HR X CPA CPB

30000 X 25 25

300

300 600 X

See Polymath program P11-8.pol. Calculated values of NLE variables Variable Value 1 Xe

f(x)

Initial Guess

0.9997644 3.518E-11 0.5 ( 0 < Xe < 1. )

Variable Value 1 T

300.

2 Kc

1.8E+07

Nonlinear equations 1 f(Xe) = Xe - (1 - Xe) * Kc ^ 0.5 = 0 Explicit equations 1 T = 300 2 Kc = 500000 * exp(-30000 / 1.987 * (1 / 323 - 1 / T))

T

X

300

1

320

0.999

340

0.995

360

0.984

380

0.935

400

0.887

420

0.76

440

0.585

460

0.4

480

0.26

500

0.1529

520

0.091

540

0.035

560

0.035

11-29

Xe

1

Xe (Eqm Conversion)

0.9 0.8 0.7 0.6 Xe

0.5 0.4 0.3 0.2 0.1 0 300

350

400

450

500

550

Temp (K)

P11-9 For first reactor,

KC

X e1 or X e1 1 X e1

KC 1 KC

For second reactor

KC

X e2 orX e 2 1 X e2 B2

KC B2 1 KC

For 3rd reactor

KC

X e3 orX e3 1 X e3 B3

KC B3 1 KC

1st reactor: in first reaction Xe = 0.3 So, FB = FA01(.3) 2nd reactor: Moles of A entering the 2nd reactor: FA02 = 2FA01 - FA01(.3) = 1.7FA01

B2

.2 FA01 1.7 FA0

FA02 CPi X

C pA

i

.12 T T0 B

FA02 X

HR

0

C pB T T 0 HR

Slope is now negative 3rd reactor:

11-30

X e2 = 0.3 Say

B

(.2 FA01 ) .3FA02

FA01 (.2 .3 1.8)

FA03

FA01 FA02 (1 X e 2 )

FA03

2.26 FA01

FA01 1.8 FA01 (1 X e 2 ) 1 1.8(1 .3) FA01

Feed to the reactor 3:

(2 FA01 ) .3FA02 B3

FA01 (.2 .3 1.8) 0.7FA01

.74 2.26

Feed Temperature to the reactor 2 is (520+450)/2 = 485 K Feed Temperature to reactor 3 is 480 K Xfinal = .4 Moles of B = .2FA01 + .3FAo2 + .4FA03 = FA01(.2 + .54 + (.4)(2.26)) = 1.64 FA01 X = FB/3FA01 = .54

P11-9 (b) The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is now 450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2 the temperature is (520+450)/2 = 485 K Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3 would be (510+510+450)/3 = 490 K For any reactor j,

FA0 j CPi X

C pA

i

T T0 B

FA0 j X

HR

0

C pB T T 0 HR

and θB for reactor 1 = 0. For reactor 2, θB > 0. This means that the slope of the conversion line from the energy balance is larger for reactor 2 than reactor 1. And similarly θB for reactor 3 > θB for reactor 2. So the line for conversion in reactor 3 will be steeper than that of reactor 2. The mass balance equations are the same as in part (b) and so the plot of equilibrium conversion will decrease from reactor 1 to reactor 2, and, likewise, from reactor 2 to reactor 3.

11-31

P11-10 1) The standard heat of reaction should have been -10000 cal/mol instead of -5000 cal/mol 2) While calculating

CP

1 CP 2 B

CP for the reaction, the following expression should have been employed:

1 CP CPA 2 C

The Cp of inerts should not have appeared in the

CP .

3) In the equation for the variation of T with conversion terms , the following should have been used,

T

0 X ( H Rx (TR )) X CPTR i

 T  T

CPi

i

CPi T

X CP

X (10000) X ( 9)(298) 33*300 33 X ( 9) 7318 X 9900 33 9 X

11-32





12-1

Solutions for Chapter 12: Steady State Non-isothermal Reactor Design- Flow Reactors with Heat Exchange

P12-1 Individualized solution P12-2(a) (1) Part (a) Co-current See polymath program P12-2-a(1).pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca0

9.3

9.3

9.3

9.3

2 Cp0

159.

159.

159.

159.

3 Cpc

28.

28.

28.

28.

4 deltaH

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

5 Fa0

14.67

14.67

14.67

14.67

6 k

0.5927441 0.5927441

654.8375

3.449562

7 Kc

1817.59

215.5325

1817.59

1063.552

8 m

500.

500.

500.

500.

9 Qg

1.103E+05 0.1990791

1.448E+06

0.1990791

10 Qr

-2.5E+04

-2.5E+04

5.301E+05

5.445171

11 ra

-5.51252

-72.4152

-9.954E-06

-9.954E-06

12 rate

5.51252

9.954E-06

72.4152

9.954E-06

13 T

305.

305.

417.9988

327.2341

14 Ta

310.

309.9439

327.233

327.233

15 Ua

5000.

5000.

5000.

5000.

16 V

0

0

5.

5.

17 X

0

0

0.9990603

0.9990603

18 Xe

0.9994501 0.9953818

0.9994501

0.9990606

Differential equations 1 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc 2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0 3 d(X)/d(V) = -ra/Fa0 12-2

Explicit equations 1 Cpc = 28 2 Ua = 5000 3 k = 31.1*exp(7906*(T-360)/(T*360)) 4 Cp0 = 159 5 deltaH = -20000 6 Ca0 = 9.3 7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330))) 8 ra = -k*Ca0*(1-(1+1/Kc)*X) 9 Xe = Kc/(1+Kc) 10 Fa0 = .9*163*.1 11 rate = -ra 12 m = 500 13 Qg = ra*deltaH 14 Qr = Ua*(T-Ta)

Plot of X versus V

Plot of Ta versus V 12-3

Plot of T versus V

Part (b) Counter current First multiply the right hand side of equation (E12-1.2) with by minus one to obtain; d(Ta)/d(V) = - Ua*(T-Ta)/m/Cpc

12-4

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca0

9.3

9.3

9.3

9.3

2 Cp0

159.

159.

159.

159.

3 Cpc

28.

28.

28.

28.

4 deltaH

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

5 Fa0

14.67

14.67

14.67

14.67

6 k

0.5927441 0.1050316

640.6661

0.1050316

7 Kc

1817.59

216.9721

3077.31

3077.31

8 m

500.

500.

500.

500.

9 ra

-5.51252

-73.92857

-0.000591

-0.000591

10 rate

5.51252

0.000591

73.92857

0.000591

11 T

305.

285.9125

417.5158

285.9125

12 Ta

310.

285.8822

310.0614

285.8822

13 Ua

5000.

5000.

5000.

5000.

14 V

0

0

5.

5.

15 X

0

0

0.9990703

0.9990703

16 Xe

0.9994501 0.9954123

0.9996751

0.9996751

Differential equations 1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc 2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0 3 d(X)/d(V) = -ra/Fa0 Explicit equations 1 Cpc = 28 2 Ua = 5000 3 k = 31.1*exp(7906*(T-360)/(T*360)) 4 Cp0 = 159 5 deltaH = -20000 6 Ca0 = 9.3 7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330))) 8 ra = -k*Ca0*(1-(1+1/Kc)*X) 9 Xe = Kc/(1+Kc) 10 Fa0 = .9*163*.1 12-5

11 rate = -ra 12 m = 500 Plot of X versus V

Plot of T versus V

Plot of Ta versus V

12-6

Part (c) Constant Ta For constant Ta, we have to use Polymath program (a) but multiply the right hand side of E12-1.1 by 0 in the program; i.e d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc *0 Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca0

9.3

9.3

9.3

9.3

2 Cp0

159.

159.

159.

159.

3 Cpc

28.

28.

28.

28.

4 deltaH

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

5 Fa0

14.67

14.67

14.67

14.67

6 k

0.5927441 0.5927441

648.9771

0.9009882

7 Kc

1817.59

216.1228

1817.59

1600.16

8 m

500.

500.

500.

500.

9 ra

-5.51252

-72.24709

-0.000498

-0.000498

10 rate

5.51252

0.000498

72.24709

0.000498

11 T

305.

305.

417.8002

310.0078

12 Ta

310.

310.

310.

310.

13 Ua

5000.

5000.

5000.

5000. 12-7

14 V

0

0

5.

5.

15 X

0

0

0.9993161

0.9993161

16 Xe

0.9994501 0.9953943

0.9994501

0.9993755

Differential equations 1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc *0 2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0 3 d(X)/d(V) = -ra/Fa0 Explicit equations 1 Cpc = 28 2 Ua = 5000 3 k = 31.1*exp(7906*(T-360)/(T*360)) 4 Cp0 = 159 5 deltaH = -20000 6 Ca0 = 9.3 7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330))) 8 ra = -k*Ca0*(1-(1+1/Kc)*X) 9 Xe = Kc/(1+Kc) 10 Fa0 = .9*163*.1 11 rate = -ra 12 m = 500 Plot of X versus V

12-8

Plot of T versus V

Plot of Ta versus V Ta will remain constant with V Part (d) Adiabatic Using the polymath program of part (a) and making the parameter Ua =0; we have; 12-9


9.3

9.3

9.3

9.3

2 Cp0

159.

159.

159.

159.

3 Cpc

28.

28.

28.

28.

4 deltaH

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

5 Fa0

14.67

14.67

14.67

14.67

6 k

0.5927441 0.5927441

1115.107

1115.105

7 Kc

1817.59

183.3032

1817.59

183.3033

8 m

500.

500.

500.

500.

9 ra

-5.51252

-265.5745

0.0023739

-4.937E-05

10 rate

5.51252

-0.0023739

265.5745

4.937E-05

11 T

305.

305.

430.1037

430.1037

12 Ta

310.

310.

310.

310.

13 Ua

0

0

0

0

14 V

0

0

5.

5.

15 X

0

0

0.9945744

0.9945742

16 Xe

0.9994501 0.9945742

0.9994501

0.9945742

Differential equations 1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc *0 2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0 3 d(X)/d(V) = -ra/Fa0 Explicit equations 1 Cpc = 28 2 Ua = 5000*0 3 k = 31.1*exp(7906*(T-360)/(T*360)) 4 Cp0 = 159 5 deltaH = -20000 6 Ca0 = 9.3 7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330))) 8 ra = -k*Ca0*(1-(1+1/Kc)*X) 9 Xe = Kc/(1+Kc) 10 Fa0 = .9*163*.1 12-10

11 rate = -ra 12

m = 500

Plot of X versus V;

Plot of T versus V

Plot of Ta versus V; 12-11

Ta remains a constant (2) Since the temperature reaches a maximum which is above 400K in all of the above systems at some point during the operation thus they cannot prevent reaching the 400 K temperatures which will start of another exothermic reaction. Hence the heat exchanger needs to be changed in order that the heat removed by the exchanger is such that the maximum temperature in the reactor does not go above 400K at any point during the operation. (3) Making the necessary changes in the polymath code and adding variables Qg and Qr as asked for in the problem we plot them for a co-current exchanger as follows:

12-12

For counter current heat exchanger we have

For constant Ta;

For adiabatic operation;

12-13

(4) Varying the values of m (i) when m =0; a plot of Qg and Qr versus volume is as follows, this seems to be an unsafe operating condition because generated heat is not taken out at the same time.

(ii) when m =2000; 12-14

In this case the conditions are safer than for m=500 and the heat generated is simultaneously removed. Variation of entering feed temperature: (i) Keeping other conditions constant we change value of T0 only now. Let T0 = 273K then,

12-15

Plot of Qg and Qr versus V is as follows;

(ii) T0 = 315K; Plot of Qg and Qr versus V is as follows;

Since again the heat generation term, Qg is much larger than Qr thus an unsafe operating condition can result in this condition.

12-16

(5) An unsafe condition will result if the entering coolant temperature is made Ta(0) = 300K in a concurrent heat exchanger ;resulting in great amount of heat generation compared to the rate of heat removal. The plot of Qg and Qr with V is as shown;

Similarly other unsafe operating conditions can be found out keeping in view that the heat generated during the course of the reaction, is not removed efficiently by the exchanger such that there is a wide difference between the two terms at any position in the reactor. P12-2(b) (1) The terms Qg = ra*△HRx and Qr = Ua*(T-Ta) are added in the polymath code given in example 12.2. See polymath program P12.2-b(1) Case1: adiabatic operation Qr=0 in this case

12-17

Case 2 : For constant heat exchange conditions

Case 3: Co-current heat exchange

12-18

Case 4 : counter current heat exchange : We need to guess a value of Ta such that at exit Ta = 1250K If we take Ta(0) = 995.15K then this can be done.

(2) Now, V = 0.5 m3 T0 = 1050 K Ta0 = 1250K Case 1: adiabatic operation; Substitute; Ua = 0 in the polymath code 12-19

(3) Substitute, the volume V= 5 m3 in the polymath code and get the results. (4) Plot of Qg for all 4 cases against volume

Plot of Qr versus volume for all cases

Plot of -ra (rate) versus V 12-20

It can be seen that the rate for constant heat exchange fluid temperature Ta , is higher than the rest of cases because the difference beween heat generated and heat removed in this case is highest. The rate of reaction for all cases is decreasing because the temperature of the system is decreasing with volume. The rate of reaction for counter-current heat exchanger system is a U shaped curve plotted against volume. At the front of the reactor, the reaction takes place very rapidly, drawing energy from the sensible heat of the gas causing the gas temperature to drop because the heat exchanger cannot supply energy at an equal or greater rate. This drop in temperature, coupled with the consumption of reactants, slows the reaction rate as we move down the reactor. (5) Introduction of inert will introduce a change in energy balance equation and the value of Ѳ1 as well. (6) (i) ѲI = 0 All the plots will remain as for part(1) ( ii ) ѲI = 1.5 Instead of the energy equation which was used previously

12-21

d (T ) = (Ua*(Ta-T)+ra* H)/(Fa0*(CpA+X*delCp)) the equation will change. d (V )

Now we have to change the value of

i

Cpi

Now the value will be ∑Cp = ѲI*CpI + CpA = 1.5×50 + 1×163 = 238 Value of Ѳ will change as well

Ѳ=

FX

0

FX

FX

0

1

=

(1.5C A0 2C A0 ) (1.5C A0 C A0 ) (1.5C A0 C A0 )

1 2.5

0.4

(iii) Ѳ I = 3 ∑Cp = ѲI*CpI + CpA = 3×50 + 1×163 = 313J/molA/K

FX Ѳ=

0

FX

FX

0

1

=

(3CA0 2C A0 ) (3C A0 C A0 ) 1 0.25 (3CA0 C A0 ) 4

Incorporating these changes in the code and plotting X versus V for different cases. See polymath code P12-2-b(6). The analysis is as follows: Case I: adiabatic operation:

12-22

Case2: Constant heat exchange fluid temperature Ta

Case 3: Co-current heat exchange

The polymath program for reference is for the case of co-current heat exchange with ѲI =3. By changing values of ∑Cp and ε variables as shown we can change the program for various cases and sub cases.

12-23

Case 4: Countercurrent heat exchange

Remarks: Thus we can see that for all cases when inert gas concentration is more then the reaction proceeds faster but then the overall yield is less as well. In the case of adiabatic operation this phenomenon is very significant . In case of constant heat exchange fluid temperature the effect of inert gas is negligible. (7) Here we will change the polymath program as entered in P12-2(b) part1. The Ta value will be changed and the program will be tested for following values of Ta. 1000K, 1175K, 1350K Case 1: adiabatic conditions

12-24

Case 2: Constant heat exchange fluid temperature Ta

Case 3: Co-current conditions

12-25

Case 4: Countercurrent conditions We need to enter Ta (V =0) values such that at V=Vf, Taf = 1000K, 1175K and 1350K respectively Ta ( V=0) (K) 983.75 992 999

Ta( V=V f) (K) 1000 1175 1350

(d) Example 12-3 (1)

CA0 will decrease but this will have no effect 12-26

(2)

Will decrease

401.1 ft 3 466.1 ft 3 s (3)

In the energy balance the slope of the energy balance of X vs. T will be greater i CPC

35

18.65 18

4 501

1.67 19.5

35 335.7 130.2

BTU kmol R XM B XM, T

Base case

Change QM

12-27

Less Conversions

P12-2 (e) Example 12-4 Change CP = 29 and –ΔH = 38700 POLYMATH Results NLES Solution Variable Value f(x) Ini Guess X 0.7109354 2.444E-11 0.367 T 593.6885 1.2E-09 564 tau 0.1229 A 1.696E+13 E 3.24E+04 R 1.987 k 20.01167 NLES Report (safenewt) Nonlinear equations [1] f(X) = X-(397.3*(T-535)+92.9*(T-545))/(38700+7*(T-528)) = 0 [2] f(T) = X-tau*k/(1+tau*k) = 0 Explicit equations [1] tau = 0.1229 [2] A = 16.96*10^12 [3] E = 32400 [4] R = 1.987 [5] k = A*exp(-E/(R*T)) Vary the heat exchanger area to find the effect on conversion. 12-28

P12-2 (f) (1) Example 12-5 See Polymath program P12-2-f.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 alpha

1.05

1.05

1.05

1.05

2 Ca

0.1

1.306E-07

0.1

1.306E-07

3 Cb

0

0

0.0208092

8.93E-07

4 Cc

0

0

0.0038445

1.739E-07

5 Cto

0.1

0.1

0.1

0.1

6 Fa

100.

9.521981

100.

9.521981

7 Fb

0

0

65.11707

65.11707

8 Fc

0

0

12.68047

12.68047

9 Ft

100.

87.31953

100.

87.31953

10 Fto

100.

100.

100.

100.

11 k1a

482.8247

482.8247

1.753E+04

1.706E+04

12 k2a

553.0557

553.0557

1.79E+06

1.683E+06

13 r1a

-48.28247

-136.5345

-0.0022276

-0.0022276

14 r2a

-5.530557

-90.93151

-2.87E-08

-2.87E-08

15 T

423.

423.

682.1122

678.9481

16 To

423.

423.

423.

423.

17 V

0

0

0.8

0.8

18 y

1.

1.922E-05

1.

1.922E-05

Differential equations 1 d(Fa)/d(V) = r1a+r2a 2 d(Fb)/d(V) = -r1a 3 d(Fc)/d(V) = -r2a/2 4 d(T)/d(V) = (4000*(373-T)+(-r1a)*20000+(-r2a)*60000)/(90*Fa+90*Fb+180*Fc) 5 d(y)/d(V) = -alpha/(2*y)*(Ft/Fto)*(T/To) Explicit equations 1 k1a = 10*exp(4000*(1/300-1/T)) 2 k2a = 0.09*exp(9000*(1/300-1/T)) 3 Cto = 0.1 4 Ft = Fa+Fb+Fc 5 To = 423 6 Ca = Cto*(Fa/Ft)*(To/T)*y 7 Cb = Cto*(Fb/Ft)*(To/T)*y 8 Cc = Cto*(Fc/Ft)*(To/T)*y 12-29

9 r1a = -k1a*Ca 10 r2a = -k2a*Ca^2 11 Fto = 100 12 alpha = 1.05

(f) (2) In the polymath program of part (f) (1), rate equation will be changed like – r1a = -k1a*(Ca-Cb/Kc) with Kc= 10*exp(20000*(1/450-1/T)) 12-30

P12-2(g) See Polymath program P12-2-g.pol. Entering Temperature (T0 in K)

Reactor temperatures (Ts in K)

1.

200

250

2.

220

3.

244

275, 563, 644

4.

261

292, 409, 571, 661

265, 625

12-31

5.

283

310, 363, 449, 558, 677

6.

292

373, 463, 562, 674

7.

304

475, 551, 686

8.

338

9.

350

518, 713 720

Plot of steady state reactor temperature; Ts(Y axis) versus T0 (X axis) The extinction temperature is 220 K and the ignition temperature is 338K. Vary UA UA 4000 J m3 s K No steady state occurs in this case and G (T) is always more than R (T). UA 40, 000 J m3 s K five steady states will exist here T = 310, 363K, 449K ,558K and 677 depending how you read the intersection on the graph. UA 4, 00, 000 J m3 s K Only one steady states occurs at around T = 350K (about).

Vary 0.001 Only the lower steady state exists at about T = 316 K 0.01 ; A number of steady states occur; 5 steady states occur 12-32

0.1, Only an upper steady state occurs at 678K.

Thus we see that if we go on increasing the value of residence time, firstly the number of steady states will increase and then it will start decreasing. P12-2(h) Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 V

0

0

10.

10.

2 Fa

5.

0.3455356

5.

0.3455356

3 Fb

10.

0.7169561

10.

0.7169561

4 Fc

0

0

4.62858

4.62858

5 Fd

0

0

0.0258848

0.0258848

6 T

300.

300.

914.7896

518.2818

7 Ta

325.

322.7616

506.8902

506.8902

8 E2

1.2E+04

1.2E+04

1.2E+04

1.2E+04

9 y

1.

1.

1.

1.

10 R

1.987

1.987

1.987

1.987

11 Ft

15.

5.716956

15.

5.716956

12 To

310.

310.

310.

310.

13 k2c

2.

2.

1.511E+06

9619.448

14 E1

8000.

8000.

8000.

8000.

15 Cto

0.2

0.2

0.2

0.2

16 Ca

0.0688889 0.0072303

0.0688889

0.0072303

17 Cc

0

0

0.096852

0.096852

18 r2c

0

-0.0081056

0

-0.0004569

19 Cpco

10.

10.

10.

10.

20 m

50.

50.

50.

50.

21 Cb

0.1377778 0.0150021

0.1377778

0.0150021

22 k1a

40.

3.318E+05

1.14E+04

23 r1a

-0.0523079 -3.727241

-0.0185467

-0.0185467

24 r1b

-0.1046159 -7.454481

-0.0370935

-0.0370935

25 rb

-0.1046159 -7.454481

-0.0370935

-0.0370935

26 r2a

0

-0.0081056

0

-0.0004569

27 DH1b

-1.5E+04

-1.5E+04

-1.5E+04

-1.5E+04

28 DH2a

-10000.

-10000.

-10000.

-10000.

40.

12-33

29 r1c

0.0523079 0.0185467

3.727241

0.0185467

30 Ta55

325.

325.

325.

325.

31 Cpd

16.

16.

16.

16.

32 Cpa

10.

10.

10.

10.

33 Cpb

12.

12.

12.

12.

34 Cpc

14.

14.

14.

14.

77.2731

170.

77.2731

35 sumFiCpi 170. 36 rc

0.0523079 0.0180899

3.725968

0.0180899

37 Ua

80.

80.

80.

80.

38 r2d

0

0

0.0162112

0.0009137

39 ra

-0.0523079 -3.728513

-0.0190036

-0.0190036

40 rd

0

0

0.0162112

0.0009137

41 Qg

1569.238

560.9706

1.118E+05

560.9706

42 Qr

-2000.

-2000.

4.139E+04

911.3246

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(T)/d(V) = (Qg - Qr) / sumFiCpi 6 d(Ta)/d(V) = Ua * (T - Ta) / m / Cpco Explicit equations 1 E2 = 12000 2 y=1 3 R = 1.987 4 Ft = Fa + Fb + Fc + Fd 5 To = 310 6 k2c = 2 * exp((E2 / R) * (1 / 300 - 1 / T)) 7 E1 = 8000 8 Cto = 0.2 9 Ca = Cto * (Fa / Ft) * (To / T) * y 10 Cc = Cto * (Fc / Ft) * (To / T) * y 11 r2c = -k2c * Ca ^ 2 * Cc ^ 3 12-34

12 Cpco = 10 13 m = 50 14 Cb = Cto * (Fb / Ft) * (To / T) * y 15 k1a = 40 * exp((E1 / R) * (1 / 300 - 1 / T)) 16 r1a = -k1a * Ca * Cb ^ 2 17 r1b = 2 * r1a 18 rb = r1b 19 r2a = r2c 20 DH1b = -15000 21 DH2a = -10000 22 r1c = -r1a 23 Ta55 = 325 24 Cpd = 16 25 Cpa = 10 26 Cpb = 12 27 Cpc = 14 28 sumFiCpi = Cpa * Fa + Cpb * Fb + Cpc * Fc + Cpd * Fd 29 rc = r1c + r2c 30 Ua = 80 31 r2d = -2 * r2c 32 ra = r1a + r2a 33 rd = r2d 34 Qg = r1b * DH1b + r2a * DH2a 35 Qr = Ua * (T - Ta)

12-35

Addition of Inerts: The adding of inerts will decrease the peak in the reactor temperature. By trial and error, an inert flow rate of 6.2 mol/min is seen to be sufficient to keep the reactor temperature below 700K. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 V

0

0

10.

10.

2 Fa

5.

0.4822244

5.

0.4822244

3 Fb

10.

0.9811024

10.

0.9811024

4 Fc

0

0

4.501122

4.501122

5 Fd

0

0

0.0166535

0.0166535

6 T

300.

300.

697.1199

570.593

7 Ta

325.

322.0417

477.6044

477.6044

8 E2

1.2E+04

1.2E+04

1.2E+04

1.2E+04

9 y

1.

1.

1.

1.

10 R

1.987

1.987

1.987

1.987

11 Ft

15.

5.981102

15.

5.981102

12 To

310.

310.

310.

310. 12-36

13 k2c

2.

2.

1.914E+05

2.8E+04

14 E1

8000.

8000.

8000.

8000.

15 Cto

0.2

0.2

0.2

0.2

16 Ca

0.0688889 0.0087606

0.0688889

0.0087606

17 Cc

0

0

0.081772

0.081772

18 r2c

0

-0.0040086

0

-0.0011748

19 Cpco

10.

10.

10.

10.

20 m

50.

50.

50.

50.

21 Cb

0.1377778 0.0178237

0.1377778

0.0178237

22 k1a

40.

8.369E+04

2.323E+04

23 r1a

-0.0523079 -1.876998

-0.0523079

-0.0646601

24 r1b

-0.1046159 -3.753997

-0.1046159

-0.1293201

25 rb

-0.1046159 -3.753997

-0.1046159

-0.1293201

26 r2a

0

-0.0040086

0

-0.0011748

27 DH1b

-1.5E+04

-1.5E+04

-1.5E+04

-1.5E+04

28 DH2a

-10000.

-10000.

-10000.

-10000.

29 r1c

0.0523079 0.0523079

1.876998

0.0646601

30 Ta55

325.

325.

325.

325.

31 Fi

6.2

6.2

6.2

6.2

32 Cpa

10.

10.

10.

10.

33 Cpb

12.

12.

12.

12.

34 Cpc

14.

14.

14.

14.

35 Cpd

16.

16.

16.

16.

36 rc

0.0523079 0.0523079

1.876239

0.0634852

37 Ua

80.

80.

80.

80.

38 r2d

0

0

0.0080172

0.0023497

39 ra

-0.0523079 -1.877758

-0.0523079

-0.0658349

40 rd

0

0

0.0080172

0.0023497

41 Qg

1569.238

1569.238

5.632E+04

1951.55

42 Qr

-2000.

-2000.

2.358E+04

7439.087

43 Cpi

10.

10.

10.

10.

141.8776

232.

141.8776

44 sumFiCpi 232.

40.

12-37

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(T)/d(V) = (Qg - Qr) / sumFiCpi 6 d(Ta)/d(V) = Ua * (T - Ta) / m / Cpco

Explicit equations 1 E2 = 12000 2 y=1 3 R = 1.987 4 Ft = Fa + Fb + Fc + Fd 5 To = 310 6 k2c = 2 * exp((E2 / R) * (1 / 300 - 1 / T)) 7 E1 = 8000 8 Cto = 0.2 9 Ca = Cto * (Fa / Ft) * (To / T) * y 10 Cc = Cto * (Fc / Ft) * (To / T) * y 11 r2c = -k2c * Ca ^ 2 * Cc ^ 3 12 Cpco = 10 13 m = 50 14 Cb = Cto * (Fb / Ft) * (To / T) * y 15 k1a = 40 * exp((E1 / R) * (1 / 300 - 1 / T)) 16 r1a = -k1a * Ca * Cb ^ 2 17 r1b = 2 * r1a 18 rb = r1b 19 r2a = r2c 20 DH1b = -15000 21 DH2a = -10000 22 r1c = -r1a 23 Ta55 = 325 24 Fi = 6.2 12-38

25 Cpa = 10 26 Cpb = 12 27 Cpc = 14 28 Cpd = 16 29 rc = r1c + r2c 30 Ua = 80 31 r2d = -2 * r2c 32 ra = r1a + r2a 33 rd = r2d 34 Qg = r1b * DH1b + r2a * DH2a 35 Qr = Ua * (T - Ta) 36 Cpi = 10 37 sumFiCpi = Cpa * Fa + Cpb * Fb + Cpc * Fc + Cpd * Fd + Cpi * Fi

P12-2 (i) CD Example CDR12.4-1 ~

2

1

dP1 P01 dP2 P0 2

1 dP P0 1 1 12 2

1 12-39

No effect for turbulent flow if both dP and P changed at the same time. P12-3

T (K)

(a) 390 380 370 360 350 340 330 320 310

Fao = 8 mol/s Fao = 5 mol/s Fao = 1 mol/s

0

1000

2000

3000

4000

5000

X

W (g)

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0


0

1000

2000

3000

4000

5000

W (g)

1

Xe

0.8 Fao = 8 mol/s

0.6

Fao = 5 mol/s

0.4

Fao = 1 mol/s

0.2 0 0

1000

2000

3000

4000

5000

W (g)

When the flowrate of A is large (high Fao), the reactants spend less time in the reactor. That is, the residence time is short, so the exit conversion is small. At small Fao, the residence time is long, so the exit conversion is large. Notice that at low Fao, the conversion reaches equilibrium conversion, while at high Fao, the conversion does not reach equilibrium conversion.

12-40

(b) 420

T (K)

400 380

thetaI = 4

360

thetaI = 1

340

thetaI = 0.5

320 300 0

1000

2000

3000

4000

5000

W (g)

0.6

X

0.5 0.4

thetaI = 4

0.3

thetaI = 1

0.2

thetaI = 0.5

0.1 0 0

1000

2000

3000

4000

5000

W (g)

1

Xe

0.8 thetaI = 4

0.6

thetaI = 1

0.4

thetaI = 0.5

0.2 0 0

1000

2000

3000

4000

5000

W (g)

As the fraction of inerts in the entering stream increases, the concentration of A in the entering stream decreases. Low concentration of A slower reaction rate lower conversion and less heat evolved by the reaction (that is, lower temperature in the reactor). (c)

12-41

T (K)

400 380

Ua/rho = 0.1 cal/kg/s/K

360 340


320


300 0

1000

2000

3000

4000

5000

W (g)

0.6 0.5


X

0.4


0.3 0.2


0.1 0 0

1000

2000

3000

4000

5000

W (g)

Xe

1 0.8


0.6 0.4


0.2


0 0

1000

2000

3000

4000

5000

W (g)

A higher heat transfer coefficient means there is a faster rate of heat transfer from the reactor to the coolant. Thus, at high Ua the reactor temperature is low. The lower reactor temperature causes reaction rate to be slow, and thus lower conversion is observed.

12-42

(d) 420

T (K)

400 380

To = 310K

360

To = 330K

340

To = 350K

320 300 0

1000

2000

3000

4000

5000

W (g)

0.6

X

0.5 0.4

To = 310K

0.3

To = 330K

0.2

To = 350K

0.1 0 0

1000

2000

3000

4000

5000

W (g)

1 0.8 To = 310K

0.6

To = 330K 0.4

To = 350K

0.2 0 0

1000

2000

3000

4000

5000

A higher entrance temperature causes reaction rate to be fast, thus high conversion is observed. Conversely, a low entrance temperature causes reaction rate to be slow, thus a lower conversion is observed.

12-43

(e) 400

T (K)

380 Tao = 300K

360

Tao = 320K 340

Tao = 340K

320 300 0

1000

2000

3000

4000

5000

W (g)

0.6

X

0.5 0.4

Tao = 300K

0.3

Tao = 320K

0.2

Tao = 340K

0.1 0 0

1000

2000

3000

4000

5000

W (g)

1

Xe

0.8 Tao = 300K

0.6

Tao = 320K 0.4

Tao = 340K

0.2 0 0

1000

2000

3000

4000

5000

W (g)

A lower coolant temperature results in lower reactor temperatures. Lower reactor temperatures result in lower reaction rate, thus lower conversion. (f) T0 = 350K

12-44

420 400

T (K)

380 mc = 50 g/s

360

mc = 1000 g/s

340 320 300 0

1000

2000

3000

4000

5000

W (g)

0.6 0.5

X

0.4 mc = 50 g/s

0.3

mc = 1000 g/s

0.2 0.1 0 0

1000

2000

3000

4000

5000

W (g)

1

Xe

0.8 0.6

mc = 50 g/s

0.4

mc = 1000 g/s

0.2 0 0

1000

2000

3000

4000

5000

W (g)

With a higher coolant flowrate, the coolant can remove heat from the reactor at a faster rate. Thus, the temperature of the reactor will be lower with a higher coolant flowrate. Again, a lower reactor temperature causes slower reaction rate, which results in smaller conversion.

(g) T0 = 350K 12-45

420 400 mc = 50 g/s (countercurrent)

T (K)

380 360

mc = 1000 g/s (countercurrent)

340 320 300 0

2000

4000

6000

W (g)

0.6 0.5 mc = 50 g/s (countercurrent)

X

0.4 0.3


0.2 0.1 0 0

2000

4000

6000

W (g)

1

Xe

0.8 0.6


0.4


0.2 0 0

2000

4000

6000

W (g)

The counter-current case is similar to the co-current case, except when mc=50 g/s, the temperature of the reactor is lower in the counter-current case. This is because, in general, counter-current heat exchange is more efficient than cocurrent heat exchange. (h) Fluidized bed CSTR Parameters: Weight of the bed : 5000 kg UA : 500 cal/s.k Ta : 320 K Density of bed : 2 Kg/m3 Rest all parameters are as specified in the base case. 12-46

Mole Balance: X MB W FAo rA'

rA'

CC2 ) KC

k (C A C B

CA

CB

C A0 1 X MB

CC

2C A0 X MB

T0 T

T0 T

W k C A0 1 X MB

2

T0 T

2C Ao X MB KC

2

X MB

FA0

Solving this implicit equation will give X MB T Energy Balance: C P 0 and B

I

mc C Pc (Ta

1,

C

T ) 1 exp

X EB T

0

UA mc C Pc 0 H Rx

So we have X EB (T ) Equilibrium Conversion: At equilibrium, rA'

C AC B

CC2 KC

C A0 1 X e

Xe

0

2

2C A0 X e KC

2

K C0.5 K C0.5

2

12-47

FA 0

i

C Pi (T Ti 0 )

1.2 1

X

0.8 XMB

0.6

XEB 0.4 0.2

39 3

38 9

38 5

38 1

37 7

37 3

36 9

36 5

36 1

35 7

35 3

34 9

0

T

Xe 0.7 0.6

X

0.5 0.4 Xe

0.3 0.2 0.1

39 3

38 9

38 5

38 1

37 7

37 3

36 9

36 5

36 1

35 7

35 3

34 9

0

T

The intersection of mole balance and energy balance curve gives the conversion achieved in the reactor. Conversion achieved X = 0.438 Temperature = 369.7 K (Xe = 0.441)

T (K)

(i) Endothermic, varying Fao 332 330 328 326 324 322 320 318 316


0

1000

2000

3000

4000

5000

W (g)

12-48

0.12

X

0.1 0.08

Fao = 8 mol/s

0.06

Fao = 5 mol/s

0.04

Fao = 1 mol/s

0.02 0 0

1000

2000

3000

4000

5000

W (g)

0.2 0.15

Xe

Fao = 8 mol/s 0.1

Fao = 5 mol/s Fao = 1 mol/s

0.05 0 0

1000

2000

3000

4000

5000

W (g)

When the flowrate of A is large (high Fao), the reactants spend less time in the reactor. That is, the residence time is short, so the exit conversion is small. At small Fao, the residence time is long, so the exit conversion is large. Notice that at low Fao, the conversion reaches equilibrium conversion, while at high Fao, the conversion does not reach equilibrium conversion. Endothermic, varying thetaI 332 330 328 326

thetaI = 4

324 322

thetaI = 1 thetaI = 0.5

320 318 316 0

1000

2000

3000

4000

5000

12-49

0.08 0.07 0.06 0.05

thetaI = 4

0.04 0.03

thetaI = 1 thetaI = 0.5

0.02 0.01 0 0

1000

2000

3000

4000

5000

0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0

thetaI = 4 thetaI = 1 thetaI = 0.5

0

1000

2000

3000

4000

5000

As the fraction of inerts in the entering stream increases, the concentration of A in the entering stream decreases. Low concentration of A slower reaction rate lower conversion and less heat consumed by the reaction (that is, higher temperature in the reactor). Endothermic, varying To 355 350 345 340 335 330 325 320 315 310 305

To = 310K To = 330K To = 350K

0

1000

2000

3000

4000

5000

12-50

0.12

X

0.1 0.08

To = 310K

0.06

To = 330K

0.04

To = 350K

0.02 0 0

1000

2000

3000

4000

5000

Xe

W (g)

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

To = 310K To = 330K To = 350K

0

1000

2000

3000

4000

5000

W (g)

A higher entrance temperature causes reaction rate to be fast, thus high conversion is observed. Conversely, a low entrance temperature causes reaction rate to be slow, thus a lower conversion is observed. OVERALL: Exothermic reactions: -equilibrium conversion varies inversely with reactor temperature Exothermic reactions: -equilibrium conversion varies with reactor temperature Conversion approaches equilibrium conversion with: -a faster reaction rate (higher T or higher amount of reactants) -Notice in all the plots, a higher T causes X to approach Xe -a longer residence time P12-4 Solution is in the decoding algorithm available in the beginning of this manual

12-51

P12-5 (a) Individualized solution P12-5 (b)

NH 4 NO3

2H 2O g

A

2W g

N 2O g

Bg

From Rate Data

ln

k2 k1

E T2 T1 R T2T1

ln E R

50 E R 970 1020

2.912 0.307 44518

E 1 1 R 970 T

k 0.307 exp Mole Balance

FA0 X rA

V

k

rAV FA 0

X MB

M V V FA 0

kM FA0

Energy Balance

FA0 H A0 FA0 H A0

FA0

W

FW0 H W0 FA H A g

H W0 FA0 1 X H A g H A g, T H Rx

FW H W g FA0

HA

2H W g

W

FB H B g

2FA0 X H W g

,T

H Vap

HB g

HA

0 FA0 XH B g

0

H Rx

H A0 H A g

W

CPA T 660

HA

,T

H W0 H W g

2H W g

HB g

HA

H Vap X

0

0

H A0 XE

1 X

H Vap

CP T 660

W

W

HS 500 F HS 500 F

H W 200 F H W 200 F

H Rx 12-52

C PS T 500 C PS T 500

H Rx X

W

0.17 0.83

FW FA

18 80

0.9103

80lb BTU 30.4 mol lbmol R 18lb BTU C PS 8.46 mol lbmol R 80lb BTU H Rx 26,880 mol lbmol BTU BTU H g 200 F 168 2,916 lb lbmol BTU BTU H W 500 F 1, 202 21,636 lb lbmol C PA

BTU lb R BTU 0.47 lb R BTU 336 lb 0.38

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Cpa

30.4

30.4

30.4

30.4

2 Cps

8.46

8.46

8.46

8.46

3 delH

-2.688E+04

-2.688E+04

-2.688E+04

-2.688E+04

4 Fao

257.3

257.3

257.3

257.3

5 Hg

2916.

2916.

2916.

2916.

6 Hw

2.164E+04

2.164E+04

2.164E+04

2.164E+04

7 k

0.307

0.307

0.7758943

0.7758943

8 M

500.

500.

500.

500.

9 Qg

1.604E+04

1.604E+04

4.053E+04

4.053E+04

10 Qr

2.654E+04

2.654E+04

2.73E+04

2.73E+04

11 t

0

0

20.

20.

12 T

970.

970.

990.

990.

13 thetaw

0.9103

0.9103

0.9103

0.9103

14 Xe

0.9874192

0.9874192

1.015768

1.015768

15 Xm

0.5965799

0.5965799

1.507762

1.507762

Differential equations 1 d(T)/d(t) = 1 Explicit equations 1 thetaw = 0.9103 2 Cpa = 30.4 3 Cps = 8.46 4 delH = -26880 5 k = 0.307*exp(44518*(1/970-1/T)) 6 M = 500 12-53

7 Fao = 310*0.83 8 Hw = 21636 9 Hg = 2916 10 Xm = k*M/Fao 11 Xe = Cpa*(T-660)/(-delH)+thetaw*(Hw-Hg+Cps*(T-960))/(-delH) 12 Qg = -delH*Xm 13 Qr = -delH*Xe

The temperature prior to shutdown is 981.10 0R

12-54

P12-5 (c) Individualized solution P12-5 (d) Individualized solution P12-5 (e) Individualized solution P12-6 A B

2C

A

B

C

10

10

0.0

80

80

-

Btu lb mole F

51

44

47.5

lb lb mol

128

94

222

63

67.2

65

lb mole hr Tio(F) Fio

~

CPio MW ,

i,

lb ft 3

HR

20, 000

Btu , lb mol A

Energy balance with work term included is: Q WS X A HR FA0 A

1,

B

FB 0 FA0

i

CPi T To

10 1, X AF 10

1

Q UA(Ts T ) Substituting into energy balance, UA(TS T ) WS FA0 H R X AF UA(TS T ) WS T

T0

FA0 H R

FA0 C pA C pB T T0 FA0 C pA C pB

UA(TS T ) WS FA0 H R FA0 C pA C pB UA

Ws

63525

T

199 F

Btu hr

12-55

UA T T0

P12-7 (a) A B

C

Since the feed is equimolar, CA0 = CB0 = 0 .1 mol/dm3 CA = CA0(1-X) CB = CB0(1-X) Adiabatic:

T

X[

H R (T0 )] X CP i CPi

T0 CP

C pC C pB C pA

H R (T ) i

T

Ci

HC

C pA

HB B

C pB

30 15 15 0

H A = - 41000 - (-15000 ) - (-20000) = -6000 cal/mol A

15 15 30

cal mol K

6000 X 300 200 X 30 k C A2 0 (1 X ) 2 .01 k (1 X )2

300 rA

VPFR

FA0

dX rA

FA0 X rA FA0 = CA0v0 = (.1)(2) = 0.2 mols/dm3 VCSTR

k = .01*exp((10000 / 2) * (1 / 300 - 1 / T)) See Polymath program P12-7-a.pol. Calculated values of NLE variables Variable Value f(x) Initial Guess 1T

483.8314 -1.421E-13 700.

2X

0.919157 -1.516E-10 0.99

Variable Value 1 Ca0

0.1

2 Fa0

0.2 12-56

3k

5.625546

4 ra

-0.0003677

5V

500.

Nonlinear equations 1 f(T) = T- 300 - 200 * X = 0 2 f(X) = X + V*ra/Fa0 = 0 Explicit equations 1 V = 500 2 k = 0.01*exp((10000 / 2) * (1 / 300 - 1 / T)) 3 Fa0 = 0.2 4 Ca0 = 0.1 5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2) For 500 dm3 CSTR, X = 0.92 For two 250 dm3 CSTR in series, X = 0.967 P12-7 (b) Constant heat exchanger temperature Ta When heat exchanger is added, the energy balance can be written as H Rxn (T ) dT Ua(Ta T ) ( rA ) dV F ( C Cˆ ) A0

So with Cˆ P = 0,

i

C pi

i

pi

P

30 , H Rxn = -6000 cal/mol

dT dV

Ua(Ta T ) ( rA ) 6000 FA0 (30)

Where Ua = 20 cal/m3/s/K, Ta = 450 K

See Polymath program P12-7-b.pol.

12-57

Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.3634806 0.3634806 T 300 300 455.47973 450.35437 k 0.01 0.01 3.068312 2.7061663 Kc 286.49665 9.2252861 286.49665 9.9473377 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0221893 -1.0E-04 -0.0010758 Xe 0.8298116 0.3682217 0.8298116 0.3810642 DH -6000 -6000 -6000 -6000 Ua 20 20 20 20 Ta 450 450 450 450 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp) Explicit equations as entered by the user [1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = -6000 [8] Ua = 20 [9] Ta = 450 [10] Fao = 0.2 [11] sumcp = 30

12-58

500

450

T [K]

400

350

300

250 0

1

2

3

4

5

6

7

8

9

10

6

7

8

9

10

3

V [m ] 0.4

0.35

0.3

X

0.25

0.2

0.15

0.1

0.05

0 0

1

2

3

4

5

V [m3] 0.9

0.8

0.7

0.6

Xe

0.5

0.4

0.3

0.2

0.1

0 0

1

2

3

4

5

6

7

3

V [m ]

12-59

8

9

10

P12-7 (c) For a co-current heat exchanger, CpC = 1cal/g/K, Ta1=450 K, m 50

g sec

See Polymath program P12-7-c.pol. Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.3611538 0.3611538 T 300 300 442.15965 442.15965 Ta 450 434.90618 450 441.60853 k 0.01 0.01 2.1999223 2.1999223 Kc 286.49665 11.263546 286.49665 11.263546 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0160802 -1.0E-04 -0.0019246 Xe 0.8298116 0.4023362 0.8298116 0.4023362 DH -6000 -6000 -6000 -6000 Ua 20 20 20 20 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 mc 50 50 50 50 Cpc 1 1 1 1 12-60

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp) [3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc Explicit equations as entered by the user [1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = -6000 [8] Ua = 20 [9] Fao = 0.2 [10] sumcp = 30 [11] mc = 50 [12] Cpc = 1 500

Ta

450

400

350

T 300

250 0

1

2

3

4

5

6

7

8

V [m3]

12-61

9

10

0.9

0.8

0.7

0.6

Xe 0.5

0.4

0.3

X 0.2

0.1

0 0

1

2

3

4

5

6

7

8

9

10

3

V [m ]

Next increase the coolant flow rate and run the same program to compare results. P12-7 (d) For counter-current flow, swap (T – Ta) with (Ta-T) in dTa/dV equation in the previous polymath program.

See Polymath program P12-7-d.pol. Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 12-62

X 0 0 0.3647241 0.3647241 T 300 300 463.44558 450.37724 Ta 440.71 440.71 457.98124 450.00189 k 0.01 0.01 3.7132516 2.7077022 Kc 286.49665 8.2274817 286.49665 9.9439517 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0256436 -1.0E-04 -9.963E-04 Xe 0.8298116 0.3488462 0.8298116 0.381006 DH -6000 -6000 -6000 -6000 Ua 20 20 20 20 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 mc 50 50 50 50 Cpc 1 1 1 1 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp) [3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc Explicit equations as entered by the user [1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = -6000 [8] Ua = 20 [9] Fao = 0.2 [10] sumcp = 30 [11] mc = 50 [12] Cpc = 1

12-63

500

Ta 450

T 400

350

300

250 0

1

2

3

4

5

6

7

8

9

10

6

7

8

9

10

V [m3] 0.9

0.8

0.7

Xe 0.6

0.5

0.4

0.3

0.2

0.1

X

0 0

1

2

3

4

5

V [m3]

P12-7 (e) Adiabatic See Polymath program P12-7-e.pol. 12-64


0.1

0.1

0.1

0.1

2 Cpc

1.

1.

1.

1.

3 DH

-6000.

-6000.

-6000.

-6000.

4 Fa0

0.2

0.2

0.2

0.2

5 k

0.01

0.01

0.010587

0.010587

6 Kc

286.4967

276.8576

286.4967

276.8576

7 mc

50.

50.

50.

50.

8 Qg

0.6

0.6

0.6286162

0.6286162

9 Qr

0

0

0

0

10 ra

-0.0001

-0.0001048

-0.0001

-0.0001048

11 sumCp

30.

30.

30.

30.

12 T

300.

300.

301.0235

301.0235

13 Ta

450.

450.

450.

450.

14 Ua

0

0

0

0

15 V

0

0

10.

10.

16 X

0

0

0.0051176

0.0051176

17 Xe

0.8298116 0.827152

0.8298116

0.827152

Differential equations 1 d(X)/d(V) = -ra/Fa0 2 d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fa0*sumCp) 3 d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc Explicit equations 1 Kc = 10*exp(-6000/1.987*(1/450-1/T)) 2 k = 0.01*exp((10000/1.987)*(1/300-1/T)) 3 Fa0 = 0.2 4 Ca0 = 0.1 5 ra = -k*(Ca0^2)*((1-X)^2-X/Ca0/Kc) 6 Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 7 DH = -6000 8 Ua = 20*0 9 sumCp = 30 12-65

10 mc = 50 11 Cpc = 1 12 Qg = ra*DH 13 Qr = Ua*(T-Ta)

12-66

P12-7 (f) We see that it is better to use a counter-current coolant flow as in this case we achieve the maximum equilibrium conversion using a lesser volume of the PFR.

12-67

P12-8 Refer to solution P11-6

rA

dT dV

Heat Exchange

FiCPi (16B)

FA0 rA

dT dV

H Rx Ua T Ta FiCPi iCPi

H Rx FA0

CP X , if CP

Ua T Ta iC Pi

A. Constant Ta (17B) Ta = 300 Additional Parameters (18B – (20B): Ta,

i C Pi ,

Ua,

B. Variable Ta Co-Current (17C)

dTa dV

Ua T Ta , V 0 ÝCPcool m

Ta

Tao

C. Variable Ta Counter Current (18C)

dTa dV

Ua Ta T ÝCPcool m

V 0

Guess Ta at V = 0 to match Ta = Tao at exit, i.e., V = Vf (a) Variable Ta Co-Current

12-68

Ta

?

0 then

(b) Gas Phase Counter Current Heat Exchange Vf = 20 dm3

Matches 12-69

(c)

Constant Ta

12-70

----------------------------------------------------------------------------------------------------------------------------(d) refer to problem P11-6 (e) Endothermic PFR A

B

dX dV

k1

1

1 X KC 0

, Xe

KC 1 KC

Xe X XEB

X EB

i C Pi

T T0

H Rx

T0

12-71

C PA

I C PI

H Rx

T T0

T T0

H Rx X I C PI

CPA

PFR Adiabatic FA0 FT

1. Irreversible A A.

B Liquid Phase, Keep FA0 Constant

First order

dX dV

rA FA0

kCA FA0

k

FA0 1 X FA0

k 1 X kCA0 1 X FA0

Constant density liquid volumetric flow rate without inert

0

0

FA0 FI FA0

0

1

I

k1 X 0 1 I

dX dV First Order Irreversible I

ISO

I

1

ISO

k

T

1

V

V

12-72

I

I

k 1 yI X

k 1

I

I OP T

I

Endothermic Adiabatic

12-73

I OP T

I

12-74

12-75

-----------------------------------------------------------------------------------------------------------------------------

P12-9 Refer to solution P11-7 For Heat Exchanger A Energy Balance

terms

(20),

CP

15

Same as adiabatic

0 Ua

Energy Balance

1

dT

b

dW

FA0

Ta

T

rA H Rx

rA H Rx

Ua

T Ta

b i C Pi

FA0

CP X

Ua

320

b

1400

12-76

0.23

i C Pi

(21)

dT

0.23 Ta

T

dW

500

Ta

A Constant Ta (22A)

300 K Ua

dTa

B Co-Current Exchange (22B)

Ýc C PCool m

Ua

C Counter Current (22C)

T Ta

b

dW

dTa dW

rA H Rx

Ta

T

b

mcCPCool

(a) Co-Current Heat Exchange Gas Phase Co-current Variable Ta

12-77

, mc = 018, C PCool

18

(b) Counter Current Ta

Matches

12-78

Gas Phase Counter Current Variable Ta

12-79



12-80

(c) Constant Ta

Constant Ta

12-81

Gas Phase Constant Ta

12-82

Gas Phase Constant Ta

(d) Refer to solution P11-7 for comparison

P12-10 (a) For reversible reaction, the rate law becomes rA

T

300 200 X

k

k (300) exp

KC

k C AC B

E 1 1 R 300 T

K C (450) exp

H Rxn 1 1 R 450 T

Stoichiometry: CC

C A0 X

CA

C A0 (1 X )

CB

C A0 (1 X ) 12-83

CC KC

See Polymath program P12-10-a.pol. POLYMATH Results No Title 03-21-2006, Rev5.1.233 Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.0051176 0.0051176 T 300 300 301.02352 301.02352 k 0.01 0.01 0.010587 0.010587 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 Kc 286.49665 276.85758 286.49665 276.85758 ra -1.0E-04 -1.048E-04 -1.0E-04 -1.048E-04 Xe 0.8298116 0.827152 0.8298116 0.827152 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 Explicit equations as entered by the user [1] T = 300+200*X [2] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] Kc = 10 * exp(-6000 /1.987 * (1 / 450 - 1 / T)) [6] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [7] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2

12-84

301.2

0.006

301

0.005

300.8

X

T [K]

0.004 300.6 0.003

300.4 0.002 300.2

0.001

300

299.8

0 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

3

4

5

V [m ]

0.83

0.8295

Xe

0.829

0.8285

0.828

0.8275

0.827 0

1

2

3

4

5

6

7

8

9

10

V [m3]

P12-10 (b) When heat exchanger is added, the energy balance can be written as H Rxn (T ) dT Ua (Ta T ) ( rA ) dV FA0 ( CöP ) i C pi C 30 , H = -6000 cal/mol So with Cö = 0, P

i

pi

6 3

V [m ]

Rxn

dT dV

Ua(Ta T ) ( rA ) 6000 FA0 (30)

Where Ua = 20 cal/m3/s/K, Ta = 450 K See Polymath program P12-10-b.pol.

12-85

7

8

9

10

Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.3634806 0.3634806 T 300 300 455.47973 450.35437 k 0.01 0.01 3.068312 2.7061663 Kc 286.49665 9.2252861 286.49665 9.9473377 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0221893 -1.0E-04 -0.0010758 Xe 0.8298116 0.3682217 0.8298116 0.3810642 DH -6000 -6000 -6000 -6000 Ua 20 20 20 20 Ta 450 450 450 450 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp) Explicit equations as entered by the user [1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = -6000 [8] Ua = 20 [9] Ta = 450 [10] Fao = 0.2 [11] sumcp = 30

12-86

500

0.4

0.35 450 0.3

0.25

X

T [K]

400

350

0.2

0.15

0.1 300 0.05

250

0 0

1

2

3

4

5

6

7

8

9

10

0

3

0.9

0.8

0.7

0.6

Xe

0.5

0.4

0.3

0.2

0.1

0 0

1

2

3

4

5

1

2

3

4

5

V [m3]

V [m ]

6

7

8

9

10

V [m3]

P12-10 (c) For a co-current heat exchanger,

g sec See Polymath program P12-10-c.pol. CpC = 1cal/g/K, Ta1=450 K, m 50

12-87

6

7

8

9

10

Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.3611538 0.3611538 T 300 300 442.15965 442.15965 Ta 450 434.90618 450 441.60853 k 0.01 0.01 2.1999223 2.1999223 Kc 286.49665 11.263546 286.49665 11.263546 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0160802 -1.0E-04 -0.0019246 Xe 0.8298116 0.4023362 0.8298116 0.4023362 DH -6000 -6000 -6000 -6000 Ua 20 20 20 20 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 mc 50 50 50 50 Cpc 1 1 1 1 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp) [3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc Explicit equations as entered by the user [1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = -6000 [8] Ua = 20 [9] Fao = 0.2 [10] sumcp = 30 [11] mc = 50 [12] Cpc = 1 12-88

500

0.9

0.8

Ta

450

0.7

0.6

Xe

400 0.5

0.4 350 0.3

T

X 0.2

300

0.1

250

0 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

3

5

6

7

8

3

V [m ]

V [m ]

Next increase the coolant flow rate and run the same program to compare results. P12-10(d) For counter-current flow, See Polymath program P12-10-d.pol. Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.3647241 0.3647241 T 300 300 463.44558 450.37724 Ta 440.71 440.71 457.98124 450.00189 k 0.01 0.01 3.7132516 2.7077022 Kc 286.49665 8.2274817 286.49665 9.9439517 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0256436 -1.0E-04 -9.963E-04 Xe 0.8298116 0.3488462 0.8298116 0.381006 DH -6000 -6000 -6000 -6000 Ua 20 20 20 20 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 mc 50 50 50 50 Cpc 1 1 1 1 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp) [3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc 12-89

9

10

Explicit equations as entered by the user [1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = -6000 [8] Ua = 20 [9] Fao = 0.2 [10] sumcp = 30 [11] mc = 50 [12] Cpc = 1 500

0.9

0.8

Ta 450

0.7

Xe

T 0.6 400 0.5

0.4 350 0.3

0.2

300

0.1

250

X

0 0

1

2

3

4

5

6

7

8

9

10

0

V [m3]

1

2

3

4

5

6

7

8

9

10

V [m3]

P12-10 (e) We see that it is better to use a counter-current coolant flow as in this case we achieve the maximum equilibrium conversion using a lesser volume of the PFR. P12-10 (f) If the reaction is irreversible but endothermic, we have rA k C A2 0 (1 X )2 .01 k (1 X )2 as obtained in the earlier problem. H Rxn 6000cal / mol See Polymath program P12-10-f-co.pol. we use 8- 7f cocurrent.pol

12-90

Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.4016888 0.4016888 T 300 300 428.84625 424.16715 Ta 450 425.45941 450 425.45941 k 0.01 0.01 1.4951869 1.314808 Ca0 0.1 0.1 0.1 0.1 Fa0 0.2 0.2 0.2 0.2 ra -1.0E-04 -0.0132694 -1.0E-04 -0.0047067 DH 6000 6000 6000 6000 Ua 20 20 20 20 Fao 0.2 0.2 0.2 0.2 sumcp 30 30 30 30 mc 50 50 50 50 Cpc 1 1 1 1 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp) [3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc Explicit equations as entered by the user [1] k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T)) [2] Ca0 = 0.1 [3] Fa0 = 0.2 [4] ra = -k * (Ca0 ^ 2) *(1 - X) ^ 2 [5] DH = 6000 [6] Ua = 20 [7] Fao = 0.2 [8] sumcp = 30 [9] mc = 50 [10] Cpc = 1

12-91

For counter-current flow, See Polymath program P12-10-f-counter.pol. Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 10 10 X 0 0 0.3458817 0.3458817 T 300 300 449.27319 449.27319 Ta 423.8 423.8 450.01394 450.01394 k 0.01 0.01 2.5406259 2.5406259 Kc 0.3567399 0.3567399 9.8927301 9.8927301 Fa0 0.2 0.2 0.2 0.2 Ca0 0.1 0.1 0.1 0.1 ra -1.0E-04 -0.0141209 -1.0E-04 -0.0019877 Xe 0.0333352 0.0333352 0.3801242 0.3801242 DH 6000 6000 6000 6000 12-92

Ua Fao sumcp mc Cpc

20 0.2 30 50 1

20 0.2 30 50 1

20 0.2 30 50 1

20 0.2 30 50 1

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra / Fa0 [2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp) [3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc Explicit equations as entered by the user [1] k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T)) [2] Kc = 10 * exp(6000 / 2 * (1 / 450 - 1 / T)) [3] Fa0 = 0.2 [4] Ca0 = 0.1 [5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc) [6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2 [7] DH = 6000 [8] Ua = 20 [9] Fao = 0.2 [10] sumcp = 30 [11] mc = 50 [12] Cpc = 1

12-93

P12-11 (a) Constant Ta, PBR Part (1) Calculated values of the DEQ variables Variable initial value minimal value maximal value final value W 0 0 50 50 X 0 0 0.8987768 0.8987768 T 450 450 893.55511 893.55511 To 450 450 450 450 vo 20 20 20 20 Cao 0.2706726 0.2706726 0.2706726 0.2706726 Ca 0.2706726 0.0072668 0.2706726 0.0072668 k 0.133 0.133 8.5739542 8.5739542 ra -0.0359995 -0.1539472 -0.0359995 -0.062305 Ta 323 323 323 323 Cpa 40 40 40 40 delH -2.0E+04 -2.0E+04 -2.0E+04 -2.0E+04 Uarho 0.08 0.08 0.08 0.08 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/(Cao*vo) [2] d(T)/d(W) = (Uarho*(Ta-T)+ra*delH)/(Cao*vo*Cpa) Explicit equations as entered by the user 12-94

[1] To = 450 [2] vo = 20 [3] Cao = 10/0.0821/450 [4] Ca = Cao*(1-X)/(1+X)*To/T [5] k = 0.133*exp(31400/8.314*(1/450-1/T)) [6] ra = -k*Ca [7] Ta = 323 [8] Cpa = 40 [9] delH = -20000 [10] Uarho = 0.08 Ua rho

0.08

J s kgcat K

Part (2)

Ua rho

240

J s kgcat K

12-95

In part (1), Ua/rho is so small that the reactor behaves as if there is no heat exchange occurring. In part (2), Ua/rho is 3000 times larger, causing the heat exchange term Ua/rho*(T-Ta) to be much larger. Therefore, in part (2), we can see the effect of heat exchange in the temperature drop, which causes the reaction to be slower, hence the lower conversion X. Part (3) Keep

Ua rho

240

J s kgcat K

Add differential equation

dy dW

2y

* (1

X )*

T , with To

And include pressure drop in expression for Ca: C a

C ao

0.019 kg -1

(1 X ) To y (1 X ) T

If there is pressure drop in the PBR, we observe a slight decrease in the conversion X. The pressure drop causes Ca to decrease (y 1 always), which decreases reaction rate slightly, which decreases conversion slightly. P12-11 (b) Co-current and Counter-current heat exchange, PBR For co-current heat exchange, add equation:

dTa dW

For counter-current heat exchange, add equation:

with mc

0.2

kg and C p ,coolant s

5000

J kg K

Part (1) 12-96

Ua rho

(T Ta ) mcoolantC p , coolant dTa dW

Ua rho

(Ta T ) , mc C p , coolant

Ua rho

0.08

J s kgcat K

Co-current

Counter-current

Again, Ua/rho is so small that the reactor behaves as if there is no heat exchange. Therefore, it doesn’t matter whether we use a counter-current or co-current heat exchanger; the temperature and conversion profiles will be similar. Part (2) Ua rho

240

J s kgcat K

Co-current

12-97

Counter-current

The heat exchange is significant in this part, so we will see a difference in the temperature and conversion profiles depending on whether a co-current or counter-current heat exchange is used. The counter-current heat exchanger is generally better at heat transfer than the cocurrent heat exchanger. Therefore, in the counter-current case, the temperature of the reactor is almost constant at 323K. Notice that the temperature and conversion profiles are similar to the constant Ta case (P8-9a, part 2). The co-current heat exchanger, on the other hand, is not able to keep the reactor temperature as low as the counter-current. Therefore, the higher reactor temperature results in a higher conversion in the co-current case. Part (3) Keep

Ua rho

240

J s kgcat K

Add differential equation

dy dW

2y

* (1

X )*

T , with To

12-98

0.019 kg -1

And include pressure drop in expression for Ca: C a

C ao

(1 X ) To y (1 X ) T

Co-current

Counter-current

The pressure drop causes Ca to decrease (y 1), which decreases reaction rate slightly, which decreases conversion slightly. P12-11 (c) Step 1: Find XMB = f(T): FA0 X W rA '

rA '

kC A

CA

C A0

(1 X ) T0 (1 X ) T

Combining: FA0 X W (1 X ) T0 kC A0 (1 X ) T 12-99

V

Note

W

0

k T T0

b

(1

X2

b 0

X)

X (1

( TT0

k

T

T0 k

X MB Where k

k1 exp

b

)X

X ) TT0 k T2

b

E 1 R T1

b

0

6TT0 k 2T

b

T0

2 2

k2

2 b

1 T

Step 2: Find XEB = f(T): When C P 0 and pure A enters the reactor,

X EB

UA(T Ta ) C PA (T T0 ) FA0 H RXN 0.16 0.14 0.12

X

0.1 Xmb

0.08

Xeb

0.06 0.04 0.02 0 365

370

375

380

385

T(K)

The intersection of the mass balance and energy balance lines correspond to the conditions in the reactor. Therefore, the reactor operates at T=376.7K and X=0.101. Parameters:

12-100

T0

450K 4000s 0.001

b

k

kg dm 3 E

0.0133* exp

Ta

1

R 450

1

dm 3

T

kg s

323K

UA 500

J s K

FA0

CA 0

CPA

40

H RXN

10 0

mol

0.0821* 450 dm

3

* 20

dm 3 s

05.41

mol s

J mol K 20000

J mol

P12-11 (d) For a reversible reaction, we have all the previous equations, but the rate law is modified as: rA k f C A kr CB CC

X T0 1 X T Plugging the equation for kr, and solving using POLYMATH program, we get the plots. Only the co-current program and plots are shown. CB

CC

C A0

See Polymath program P12-11-d.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 W

0

0

80.

80.

2 X

0

0

0.057593

0.057593

3 T

450.

420.7523

450.

420.7523

4 Ta

323.

323.

426.1627

420.7565

5 T0

450.

450.

450.

450.

6 v0

20.

20.

20.

20. 12-101

7 k

0.133

0.0742131

0.133

0.0742131

8 Uarho

240.

240.

240.

240.

9 P0

1.013E+06 1.013E+06

1.013E+06

1.013E+06

10 CA0

270.8283

270.8283

270.8283

270.8283

11 CA

270.8283

258.1071

270.8283

258.1071

12 kr

0.2

0.076962

0.2

0.076962

13 CC

0

0

15.77362

15.77362

14 CB

0

0

15.77362

15.77362

15 rA

-36.02017

-36.02017

-0.0062421

-0.0062421

Differential equations 1 d(X)/d(W) = -rA / v0 / CA0 2 d(T)/d(W) = (Uarho * (Ta - T) + rA * 20000) / v0 / CA0 / 40 3 d(Ta)/d(W) = Uarho * (T - Ta) / .2 / 5000 + sign = cocurrent, -ve sign = countercurrent in RHS of eqn. Explicit equations 1 T0 = 450 2 v0 = 20 3 k = 0.133 * exp(31400 / 8.314 * (1 / T0 - 1 / T)) 4 Uarho = .08 * 3000 5 P0 = 1013250 6 CA0 = P0 / 8.314 / T0 7 CA = CA0 * (1 - X) / (1 + X) * T0 / T 8 kr = 0.2 * exp(51400 / 8.314 * (1 / T0 - 1 / T)) 9 CC = CA0 * X / (1 + X) * T0 / T 10 CB = CA0 * X / (1 + X) * T0 / T 11 rA = -(k * CA - kr * CB * CC)

12-102

P12-11 (e) Individualized solution

P12-12 (a) Start with the complete energy balance:

dEˆ dt

Q WS

Fi H i

in

Fi H i

out

Fi RV i

The following simplifications can be made: It is steady state. In part (a), there is no heat taken away or added There is no shaft work That leaves us with

Fi H i in Fi H i out Evaluating energy terms:

Fi RV i

out

H A0 FA0 H B 0 FB 0 H C 0 FC 0 Out: H A FA H B FB H C ( FC RCV ) In:

Now we evaluate Fi FA FA0 FA0 X FB FB 0 FA0 X FC FC 0 FA0 X FB0 = FC0 = 0 12-103

out

Inserting these into our equation gives:

H A FA0

H A FA0 X

H B FA0 X

H C FA0 X

H C RCV

H A0 FA0

0

Combining and substituting terms gives:

FA0 H A H A0

FA0 X H RX

H C RC

0

FA0CPA T T0

FA0 X H RX

H C (T0 ) CPC T T0

RC

0

RC

0

Differentiating with respect to V with ΔCP = 0

FA0C p

dT dV

dX dV

FA0

(rA )

dT dV

H Rx T

H Rx T

H C (T0 ) CPC T T0

H C (T0 ) CPC T T0 FA0

i

C pi

Combine that with the mole balance and rate law:

rA

rB

dFA dV CA

rC

k CA

CB CC KC

dF dFB rB ; C rC RC dV dV FA T0 F T CB CT 0 B 0 CC FT T FT T

rA ; CT 0

CT 0

FC T0 FT T

See Polymath program P12-11-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca

0.2710027 0.1839444

0.2710027

0.1839444

2 Cb

0

0

0.0098751

0.0098751

3 Cc

0

0

0.0065147

0.0029402

4 Cpa

40.

40.

40.

40.

5 Cpb

25.

25.

25.

25.

6 Cpc

15.

15.

15.

15.

7 Ct0

0.2710027 0.2710027

0.2710027

0.2710027

8 dHrx

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

12-104

RC

9 Fa

5.42

5.143851

5.42

5.143851

10 Fa0

5.420054

5.420054

5.420054

5.420054

11 Fb

0

0

0.2761489

0.2761489

12 Fc

0

0

0.1641532

0.0822207

13 Ft

5.42

5.42

5.584153

5.502221

14 Hc

-3.735E+04 -3.735E+04

-3.48E+04

-3.48E+04

15 k

0.133

0.133

1.325591

1.325591

16 kc

1.5

1.5

1.5

1.5

17 Kc

0.0006905 0.0001596

0.0006905

0.0001596

18 ra

-0.0360434 -0.0589786

-0.0027523

-0.0027523

19 Rc

0

0

0.009772

0.0044103

20 T

450.

450.

619.7977

619.7977

21 To

450.

450.

450.

450.

22 V

0

0

30.

30.

23 Xe

0.0504144 0.0284733

0.0504144

0.0284733

Differential equations 1 d(Fc)/d(V) = -ra-Rc 2 d(Fb)/d(V) = -ra 3 d(Fa)/d(V) = ra 4 d(T)/d(V) = (Fa0*ra*dHrx - Rc*Hc)/(Fa0*Cpa) Explicit equations 1 k = .133*exp((31400/8.314)*(1/450-1/T)) 2 To = 450 3 Ct0 = 10/0.082/450 4 Ft = Fa + Fb + Fc 5 Cc = Ct0*Fc/Ft*To/T 6 Fa0 = Ct0*20 7 Ca = Ct0*Fa/Ft*To/T 8 dHrx = -20000 9 Cb = Ct0*Fb/Ft*To/T 10 Kc = 0.01*exp((dHrx/8.314)*(1/300-1/T)) 11 Cpc = 15 12 ra = -k*(Ca-Cb*Cc/Kc) 12-105

13 kc = 1.5 14 Rc = kc*Cc 15 Xe = ((Kc*T/(Ct0*To))/(1+(Kc*T/(Ct0*To))))^0.5 16 Cpa = 40 17 Cpb = 25 18 Hc = -40000+Cpc*(T-273)

P12-12 (b) Now, the heat balance equation needs to be modified.

dT dV

Ua(Ta T ) (rA )

H Rx T FA0

H C (T0 ) CPC T T0 i

C pi

Temperature reaches a peak value of 628 K before again dropping down.

12-106

RC

P12-13 (a) Substrate  More cells + Product SC+P G(T) = X*-∆HRX To solve for G(T) we need X as a function of temperature, which we get by solving the mass balance equation. rg FA0 X FS 0 X V and since rS then, rA rS Yc / s

V

rg

Yc / s FS 0 X rg

CC and

(T )

CS KS

CS

6700 T 48000 1 exp 153 T

0.0038* T exp 21.6 where

(T )

1max

if we combine these equations we get: Yc / s FS 0 X V (T )CC CS K S CS 12-107

V

CS V

Yc / s FS 0 X K S

CS

(T )CC CS

CS 0 1 X and CC Yc / s FS 0 X K S

CS 0Yc / s X

CS 0 1 X

(T )CS 0Yc / s XCS 0 1 X

Canceling and combining gives: V

FS 0 K S (T )C

CS 0 1 X 2 S0

1 X

Now solve this expression for X: FS 0 K S X 1 (T )VCS20 FS 0CS 0

It can be observed from graph that if the reactor operates at a steady state temperature of 294.4316.79, Conversion achieved is 0. Now that we have X as a function of T, we can plot G(T). To get R(T) we must calculate the heat removed which is the sum of the heat absorbed by reactants to get to the reaction temperature and the heat removed from any heat exchangers. The heat gained by the reactants = CP 0 T T0 The heat removed by the heat exchanger = UA(T-Ta)/FS0 UA R(T ) CPS T T0 T Ta FS 0 12-108

Now enter the equations into polymath and specify all other constants. The adiabatic case is shown below. The non-adiabatic case would be with explicit equation [12] as A = 1.1. See Polymath program P12-13-a.pol. Differential equations as entered by the user [1] d(T)/d(t) = 1 Explicit equations as entered by the user [1] mumax = .5 [2] Ycs = .8 [3] vo = 1 [4] Ta = 290 [5] mu = mumax*(.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T))) [6] Ks = 5 [7] V = 6 [8] Cso = 100 [9] Fso = vo*Cso [10] Cps = 5 [11] dH = -20000 [12] UA = 0*300 [13] kappa = UA/(Cps*Fso) [14] To = 280 [15] X = 1-((Fso*Ks)/((mu*V*(Cso^2))-Fso*Cso)) [16] Gt = if T<294.4 or T>316.7 then 0 else X*(-dH)*Ycs [17] Rt = Cps*(T-To)+UA*(T-Ta)/Fso Independent variable variable name : t initial value : 0 T(0): 280 final value : 45 Adiabatic case

12-109

Non adiabatic case

12-110

P12-13 (b) To maximize the exiting cell concentration, we want to maximize the conversion of substrate. If we look at G(T) from part A, we see that it is at a maximum at about 310 K. This corresponds to the highest conversion that can be achieved. By changing the values of UA and m c we can change the slope of the R(T). What we are looking to do is get R(T) to intersect with G(T) at 310 K. Since we now have a limited coolant flow rate we will use a different value for Q. Q

mC C PC T

Ta 1 exp

UA mC C PC

and so, R(T )

mC C PC T

Ta 1 exp

UA mC C PC

C PS T

T0

Now we set R(T) equal to the maximum value of G(T) which is 15600 J/h G (T ) 15600 FS 0

mC C PC T

Ta 1 exp

UA mC C PC

FS 0 C PS T

T0

And now plug in the known values. Assume the maximum coolant flow rate and that will give the minimum heat exchange area. 12-111

15600

g J 100 g hr

60000

g hr

1560000

UA

4.2

J hr

J 310 K gK

5040000

290 K 1 exp

J hr

1 exp

UA g 60000 hr

J 4.2 gK

UA 252000

J hr K

J hr K

A=1.85 m2 Plot of G(T) ,R(T) Vs T

12-112

100

11000

g h

J hr

5

J 310 K gK

288 K

P12-13 (c) From the plot of G(T) vs R(T) for (1) Aiabatic case : There are two steady states possible- 281K (stable and X=0), 294.2K (unstable) and 316.76K (stable and X=0.175) (2) Non adiabatic case with large coolant flow rate : There are three steady states –289K (stable and X=0), 294.2K (unstable) and 315.7K (stable and X=0.89) (3) Non adiabatic case with limited coolant flow rate : There are three steady states –289K (Stable and X=0), 294.3K (unstable) and 309K (stable and X=0.975) P12-13 (d) (1) Adiabatic case : Varying To

From the graph, it can be observed that increasing To will increase the steady state temperature.

12-113

(2) Non adiabatic case with large coolant flow rate : Varying Ta

Increasing Ta will increase steady state temperature. (3) Non adiabatic case with limited coolant flow rate Varying coolant flow rate

Increasing coolant flow rate will decrease steady state temperature achieved. 12-114

P12-14 Assume Adiabatic (a) -12,000 cal/mol (b) Ignition = 260°C Extinction = 190°C (c) Ignition, T = 300°C, 475°C Extinction, T = 215°C, 375°C (d) X (Ignition) = 0.281 X (Extinction) = 0.86

P12-15 (a)

G(T) X

HR X k

1

k

,k 6.6 10 3 exp

R(T) C p 0 1

T Tc

C p0

50

i C pi

UA C p0 FA 0 Tc

Ta T0 1

8000 50 80

E 1 R 350

1 T

2

350K

To find the steady state T, we must set G(T) = R(T). This can be done either graphically or solving the equations. We find that for T0 = 450 K, steady state temperature 399.94 K.

12-115

POLYMATH Results 02-22-2006, Rev5.1.233 Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 1000 1000 T 350 350 450 450 RT 0 0 1.5E+04 1.5E+04 EoR 2.013E+04 2.013E+04 2.013E+04 2.013E+04 k 0.0066 0.0066 2346.7972 2346.7972 tau 100 100 100 100 X 0.3975904 0.3975904 0.9999957 0.9999957 GT 2981.9277 2981.9277 7499.968 7499.968 at 2981.9277 -7500.032 4336.6841 -7500.032

12-116

ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 0.1 Explicit equations as entered by the user [1] RT = 150*(T-350) [2] EoR = 40000/1.987 [3] k = 6.6*0.001*exp(EoR*(1/350-1/T)) [4] tau = 100 [5] X = tau*k/(1+tau*k) [6] GT = 7500*X [7] at = GT-RT P12-15 (b) First, we must plot G(T) and R(T) for many different T0’s on the same plot. From this we must generate data that we use to plot Ts vs To. Calculated values of NLE variables Variable Value f(x) Initial Guess 1T

399.9425 1.337E-10 450. ( 300. < T < 600. )

Variable Value 1 EoR

2.013E+04

2 GT

7491.375

3k

8.685615

4 RT

7491.375

5 tau

100.

6X

0.99885

Nonlinear equations 1 f(T) = RT - GT = 0 Explicit equations 1 RT = 150*(T-350) 2 EoR = 40000/1.987 3 k = 6.6*0.001*exp(EoR*(1/350-1/T)) 4 tau = 100 12-117

5 X = tau*k/(1+tau*k) 6 GT = 7500*X

400 390 380 370

T

360 350 340 330 320 310 300 310

330

350

370

390

410

430

To P12-15 (c) For high conversion, the feed stream must be pre-heated to at least 404 K. At this temperature, X = .991 and T = 384.2 K in the CSTR. Any feed temperature above this point will provide for higher conversions. P12-15 (d) For a temperature of 369.2 K, the conversion is 0.935 P12-15 (e) The extinction temperature is 363.3 K (90oC). P12-16 (a)

Mol Balance : FA0 X v0C A0 X V rA k C A CB / K e V v0

X k 1 X

X / Ke 12-118

X 1 X

k (1 1/ K e )

k

k 1

k (1 1/ K e )

G (T )

H Rx X

80000 X

k 1min 1 10 min K e 100 10 X .901 1 10(1.01) G (400) 72080 cal / mol

P12-16 (b) UA FA0C pA R(T ) C pA (1

3600 10* 40

9

)(T TC )

400(T TC )

T0 Ta 310 1 R(T ) 400(T 310) The following plot gives us the steady state temperatures of 310, 377.5 and 418.5 K TC

See Polymath program P12-16-b.pol.

P12-16 (c)

12-119

P12-16 (d)

P12-16 (e) The plot below shows Ta varied.

The next plot shows how to find the ignition and extinction temperatures. The ignition temperature is 358 K and the extinction temperature is 208 K.

12-120

P12-16 (f)

P12-16 (g) At the maximum conversion G(t) will also be at its maximal value. This occurs at approximately T = 404 K. G(404 K) = 73520 cal. For there top be a steady state at this temperature, R(T) = G(T). See Polymath program P12-16-g.pol.

UA FA0C pA R(T ) C pA (1

)(T TC ) 73520

T0 Ta 1 If we plug in the values and solve for UA, we get: UA = 7421 cal/min/K where TC

P12-16 (h) Individualized solution P12-16 (i) The adiabatic blowout flow rate occurs at V 0.0041s v0 v0 V 0.0041*10 v0

0.041

0.0041s

dm3 min

12-121

See Polymath program P12-16-i.pol.

P12-16 (j) Lowing T0 or Ta or increasing UA will help keep the reaction running at the lower steady state. P12-17 Given the first order, irreversible, liquid phase reaction: A B Ta 100 C UA 1.0 cal / min/ C

Pure A Feed Cp

0.5g mol / min

C pA

Design Eqn : V Rate Law :

C pB

Stoichiometry : C A

C pB

HR

0

FA0 X rA rA kC A

C pA

2cal / g mol / C

100cal / g

v0 k (1 X )

C A0 (1 X )

Energy Balance : UA(T TA ) FA0 X H R

FA0

i

C pi (T T0 )

Simplifying, k 1 X 1 k 1 1 k HR 1 1 k The equation for heat removal curve is: UA R(T ) C pA 1 T Tc , , Tc C pA FA0 G (T )

HR X

Ta To 1 Plot this along with the heat generation curve for various T0.

12-122

mol / C

P12-17 (a) (a)

Tc=156oC

R T C PS

CP 0 1

T TC

C PA

ICP I

UA FA0C PA R T Tc

C PA

2

1cal min K mol 2cal 0.5 min mol K

2 1 1 T TC 1 Ta T0 1 1

Ta

1

4 T Tc T0

2

100 T0 2

Draw R(T) with slope 4such that it is tangent to G(T) curve. The point of the tangency is at G(T) = R(T) = 27.5 and T 163°C. We could solve for Tc or read it off the graph by extrapolating to R(T) = 0 to find Tc = 156.

R T

Ta

4 T

T0

4T 200 2T0

2

27.5cal mol A

From the above Figure at the point of tangency R(T) = 27.5 and T = 163oC solving for T0,

T 0 212 C

or

Tc

Ta 1

T0

156

100 T0 , T0 2

212

R(T) slope is 4. Move

27.5 4 163 Tc Tc 156 100 T0 156 2 T0

312 100 212oC 12-123

P12-17 (b) (b)

Ts=180oC

Ts=163oC

We see there are two intersections at which the reactor can operate. At upper intersection T = 180. T = 163oC and 180oC from Figure P12-17.

P12-17 (c) Slope of R(T) curve is 4 if we move to left we see that there are 3 intersections. Because we heated the RcTr above the extraction temperature it will operate at the upper steady state at the upper steady state G(T) = R(T) = 90 cal/mol and T = 180°C. Extrapolating G(T) to large T, X = 1 so G = (– HRx) we find – HRx = 100 cal/mol).

GT

X H Rx

X 100

X 0.9 90

4 180 Tc

100 T0 2 Solving for T Tc 157

or get Tc at R T T0

214

12-124

0, Tc 157

(c)

P12-17 (d) (d)

Tc=148oC

At the point of tangency R(T) = G(T) = 80 and T = 167

R T

80

Tc 147 Tc T0

Ta 1 194

4 167 Tc 100 T0 2 T0

*Note all the temperatures in parts (a) – (d) are 2°C due to accuracy of reading the graph

12-125

P12-17 (e) Individualized solution

P12-18 TC = Ta = T0 = 330 K τ = V/υ0 = 1.2 h V = FA0X/-rA = CA0υ0X/ -rA -rA = k(CA – CB/KC) = kCA0(1-X – X/KC) k = 0.001exp(30000/1.987(1/300-1/T) X = τk/(1+τK+τK/KC) G(T) = (-∆HRx)X G(T) is plotted as a function of T. See Polymath program P12-18.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Cpo

250.

250.

250.

250.

2 DHrx

-4.2E+04

-4.2E+04

-4.2E+04

-4.2E+04

3 E

3.0E+04

3.0E+04

3.0E+04

3.0E+04

4 GT

50.33959

0.0001053

3.61E+04

0.0001053

5 k

0.001

0.001

8.48E+07

8.48E+07

6 Kc

5.0E+06

2.507E-09

5.0E+06

2.507E-09

7 t

0

0

300.

300.

8 T

300.

300.

600.

600.

9 tau

1.2

1.2

1.2

1.2

0.8596127

2.507E-09

10 X 0.0011986 2.507E-09 Differential equations 1 d(T)/d(t) = 1 Explicit equations 1 E = 30000

12-126

2 k = 0.001*exp((E/1.987)*(1/300-1/T)) 3 DHrx = -42000 4 Kc = 5000000*exp((DHrx/1.987)*(1/300-1/T)) 5 Cpo = 250 6 tau = 1.2 7 X = tau*k/(1+tau*k + tau*k/Kc) 8 GT = X*(-DHrx)

From the plot, the maximum conversion achieved, Xmax = 0.86 At Xmax, T = 367.6 K and G(T) = 36100 cal/mol R(T) = G(T) 12-127

κ = 2.84 Since Therefore, UA = (2.84)(10 mol/h)(250 cal/mol/K) = 7,100 cal/h/K

P12-19 Adiabatic POLYMATH Report Ordinary Differential Equations


Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 alpha

0.019

0.019

0.019

0.019

2 Ca

1.9

0.7510709

1.902453

0.7510709

3 Ca0

1.9

1.9

1.9

1.9

4 Cc

0

0

0.1974831

0.1113793

5 Cp0

40.

40.

40.

40.

6 Cpc

4200.

4200.

4200.

4200.

7 deltaCp -30.

-30.

-30.

-30.

8 DH

-2.0E+04

-2.0E+04

-2.0E+04

9 epsilon -1.

-1.

-1.

-1.

10 Fa0

5.

5.

5.

5.

11 k

0.01

0.01

0.0261675

0.0261675

12 Kc

10000.

902.809

10000.

902.809

13 m

0.05

0.05

0.05

0.05

14 ra

-0.0361

-0.0627617

-0.014758

-0.014758

15 T

450.

450.

817.9727

817.9727

16 T0

450.

450.

450.

450.

17 Ta

500.

500.

500.

500.

18 Ua

0

0

0

0

-2.0E+04

12-128

19 W

0

0

50.

50.

20 X

0

0

0.4949381

0.4949381

21 y

1.

0.2174708

1.

0.2174708

Differential equations 1 d(X)/d(W) = -ra/Fa0 2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0 3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0 4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc Explicit equations 1 Cp0 = 40 2 DH = -20000 3 epsilon = -1 4 T0 = 450 5 Ca0 = 1.9 6 Fa0 = 5 7 alpha = 0.019 8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y 9 Ua = 0.8*0 10 deltaCp = 1/2*20-40 11 Kc = 10000*exp(DH/8.314*(1/450-1/T)) 12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y 13 m = 0.05 14 Cpc = 4200 15 k = 0.01*exp(8000/8.314*(1/450-1/T)) 16 ra = -k*(Ca^2-Cc/Kc) General Total number of equations 20 Number of differential equations 4 Number of explicit equations 16 Elapsed time 0.000 sec Solution method RKF_45 Step size guess. h 0.000001 Truncation error tolerance. eps 0.000001 12-129

Now, constant temperature Ta = 300K POLYMATH Report Ordinary Differential Equations



0.019

0.019

0.019

0.019

2 Ca

1.9

0.7615715

1.906858

0.7615715

3 Ca0

1.9

1.9

1.9

1.9

4 Cc

0

0

0.1991675

0.1180575

5 Cp0

40.

40.

40.

40.

6 Cpc

4200.

4200.

4200.

4200.

7 deltaCp -30.

-30.

-30.

-30.

8 DH

-2.0E+04

-2.0E+04

-2.0E+04

9 epsilon -1.

-1.

-1.

-1.

10 Fa0

5.

5.

5.

5.

11 k

0.01

0.01

0.0248669

0.0248669

12 Kc

10000.

1025.524

10000.

1025.524

13 m

0.05

0.05

0.05

0.05

-2.0E+04

12-130

14 ra

-0.0361

-0.0610755

-0.0144197

-0.0144197

15 T

450.

450.

783.9972

783.9972

16 T0

450.

450.

450.

450.

17 Ta

500.

500.

500.

500.

18 Ua

0.8

0.8

0.8

0.8

19 W

0

0

50.

50.

20 X

0

0

0.4848172

0.4848172

21 y

1.

0.2300674

1.

0.2300674

Differential equations 1 d(X)/d(W) = -ra/Fa0 2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0 3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0 4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc *0 Explicit equations 1 Cp0 = 40 2 DH = -20000 3 epsilon = -1 4 T0 = 450 5 Ca0 = 1.9 6 Fa0 = 5 7 alpha = 0.019 8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y 9 Ua = 0.8 10 deltaCp = 1/2*20-40 11 Kc = 10000*exp(DH/8.314*(1/450-1/T)) 12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y 13 m = 0.05 14 Cpc = 4200 15 k = 0.01*exp(8000/8.314*(1/450-1/T)) 16 ra = -k*(Ca^2-Cc/Kc) General Total number of equations

20 12-131

Number of differential equations 4 Number of explicit equations 16 Elapsed time 0.000 sec Solution method RKF_45 Step size guess. h 0.000001 Truncation error tolerance. eps 0.000001

Now, Co-current heat exchanger POLYMATH Report Ordinary Differential Equations



0.019

0.019

0.019

0.019

2 Ca

1.9

0.7614153

1.906781

0.7614153

3 Ca0

1.9

1.9

1.9

1.9

4 Cc

0

0

0.1990505

0.1177523

5 Cp0

40.

40.

40.

40.

6 Cpc

4200.

4200.

4200.

4200.

-30.

-30.

-30.

7 deltaCp -30.

12-132

8 DH

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

9 epsilon -1.

-1.

-1.

-1.

10 Fa0

5.

5.

5.

5.

11 k

0.01

0.01

0.0249108

0.0249108

12 Kc

10000.

1021.01

10000.

1021.01

13 m

0.05

0.05

0.05

0.05

14 ra

-0.0361

-0.0610733

-0.0144393

-0.0144393

15 T

450.

450.

785.126

785.126

16 T0

450.

450.

450.

450.

17 Ta

500.

499.0741

521.1534

521.1534

18 Ua

0.8

0.8

0.8

0.8

19 W

0

0

50.

50.

20 X

0

0

0.4849408

0.4849408

21 y

1.

0.2296895

1.

0.2296895

Differential equations 1 d(X)/d(W) = -ra/Fa0 2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0 3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0 4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc Explicit equations 1 Cp0 = 40 2 DH = -20000 3 epsilon = -1 4 T0 = 450 5 Ca0 = 1.9 6 Fa0 = 5 7 alpha = 0.019 8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y 9 Ua = 0.8 10 deltaCp = 1/2*20-40 11 Kc = 10000*exp(DH/8.314*(1/450-1/T)) 12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y 13 m = 0.05 12-133

14 Cpc = 4200 15 k = 0.01*exp(8000/8.314*(1/450-1/T)) 16 ra = -k*(Ca^2-Cc/Kc) General Total number of equations 20 Number of differential equations 4 Number of explicit equations 16 Elapsed time 1.157 sec Solution method RKF_45 Step size guess. h 0.000001 Truncation error tolerance. eps 0.000001

12-134

Now, counter – current heat exchanger POLYMATH Report Ordinary Differential Equations



0.019

0.019

0.019

0.019

2 Ca

1.9

0.7671508

1.910238

0.7671508

3 Ca0

1.9

1.9

1.9

1.9

4 Cc

0

0

0.2023329

0.1203647

5 Cp0

40.

40.

40.

40.

6 Cpc

4200.

4200.

4200.

4200.

7 deltaCp -30.

-30.

-30.

-30.

8 DH

-2.0E+04

-2.0E+04

-2.0E+04

9 epsilon -1.

-1.

-1.

-1.

10 Fa0

5.

5.

5.

5.

11 k

0.01

0.01

0.025358

0.025358

12 Kc

10000.

976.5939

10000.

976.5939

13 m

0.05

0.05

0.05

0.05

-2.0E+04

12-135

14 ra

-0.0361

-0.062846

-0.0149206

-0.0149206

15 T

450.

450.

796.6909

796.6909

16 T0

450.

450.

450.

450.

17 Ta

520.

500.0379

521.8302

500.0379

18 Ua

0.8

0.8

0.8

0.8

19 W

0

0

50.

50.

20 X

0

0

0.4958573

0.4958573

21 y

1.

0.2280604

1.

0.2280604

Differential equations 1 d(X)/d(W) = -ra/Fa0 2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0 3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0 4 d(Ta)/d(W) = - Ua*(T-Ta)/m/Cpc Explicit equations 1 Cp0 = 40 2 DH = -20000 3 epsilon = -1 4 T0 = 450 5 Ca0 = 1.9 6 Fa0 = 5 7 alpha = 0.019 8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y 9 Ua = 0.8 10 deltaCp = 1/2*20-40 11 Kc = 10000*exp(DH/8.314*(1/450-1/T)) 12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y 13 m = 0.05 14 Cpc = 4200 15 k = 0.01*exp(8000/8.314*(1/450-1/T)) 16 ra = -k*(Ca^2-Cc/Kc) General Total number of equations

20 12-136

Number of differential equations 4 Number of explicit equations 16 Elapsed time 0.000 sec Solution method RKF_45 Step size guess. h 0.000001 Truncation error tolerance. eps 0.000001

12-137

P12-20 (a) Liquid Phase : A B

dX dV rA k

C

rA FA0 kC ACB E 1 R T

.01*exp

1 300

C A C A0 (1 X ) CB The energy Balance: dT Ua(Ta T ) ( rA )( dV FA0 C pA C pB 4 D Assume D

HR )

a

U

4

5J 1cal 1m 2 m 2 Ks 4.184 J 10dm 2

.0120

cal dm 2 Ks

See Polymath program P12-20-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 V

0

0

1000.

1000.

2 T

300.

300.

434.7779

348.2031

3 X

0

0

0.9620102

0.9620102

4 Ta

300.

300.

300.

300.

5 R

1.988

1.988

1.988

1.988

6 E

10000.

10000.

10000.

10000.

7 cao

0.1

0.1

0.1

0.1

8 ca

0.1

0.003799

0.1

0.003799

9 cb

0.1

0.003799

0.1

0.003799

10 k

0.01

0.01

1.808628

0.1018746

11 ra

-0.0001

-0.0010427

-1.47E-06

-1.47E-06

12 cpb

15.

15.

15.

15.

13 cpa

15.

15.

15.

15.

14 fao

0.2

0.2

0.2

0.2 12-138

15 Dhr1

-6000.

-6000.

-6000.

-6000.

16 a

1.

1.

1.

1.

17 U

0.012

0.012

0.012

0.012

Differential equations 1 d(T)/d(V) = (U * a * (Ta - T) + (-ra) * (-Dhr1)) / (fao * (cpa + cpb)) 2 d(X)/d(V) = -ra / fao Explicit equations 1 Ta = 300 2 R = 1.988 3 E = 10000 4 cao = .1 5 ca = cao * (1 - X) 6 cb = cao * (1 - X) 7 k = .01 * exp(-E / R * (1 / T - 1 / 300)) 8 ra = -k * ca * cb 9 cpb = 15 10 cpa = 15 11 fao = .2 12 Dhr1 = -6000 13 a = 1 14 U = .0120

12-139

P12-20 (b) Gas Phase: A dX rA dW FA0

rA

k f CA

CB k

CC

B C

k r CB CC

CA0

X T0 1 X T

0.133 exp

kr

0.2exp

E 1 R T

Er 1 R 450

1 450 1 , Er T

The energy Balance: dT Ua(Ta T ) ( rA )( dW FA0C pA HR U

51.4 kJ mol

HR )

40 50 70=-20 kJ/mol

5

See Polymath program P12-20-b.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 W

0

0

50.

50.

2 X

0

0

0.0560855

0.0560855

3 T

400.

371.902

400.

371.902

4 T0

400.

400.

400.

400.

5 k

0.133

0.0651696

0.133

0.0651696

6 v0

20.

20.

20.

20.

7 kr

0.2

0.0622149

0.2

0.0622149

8 Uarho

5.

5.

5.

5.

9 Ta

323.

323.

323.

323.

10 P0

1.013E+06 1.013E+06

1.013E+06

1.013E+06

11 CA0

304.6819

304.6819

304.6819

304.6819

12-140

12 CA

304.6819

292.8948

304.6819

292.8948

13 CC

0

0

17.40323

17.40323

14 CB

0

0

17.40323

17.40323

15 rA

-40.52269

-40.52269

-0.2446624

-0.2446624

Differential equations 1 d(X)/d(W) = -rA / v0 / CA0 2 d(T)/d(W) = (Uarho * (Ta - T) + rA * 20000) / v0 / CA0 / 40 Explicit equations 1 T0 = 400 2 k = 0.133 * exp(31400 / 8.314 * (1 / T0 - 1 / T)) 3 v0 = 20 4 kr = 0.2 * exp(51400 / 8.314 * (1 / T0 - 1 / T)) 5 Uarho = 5 6 Ta = 323 7 P0 = 1013250 8 CA0 = P0 / 8.314 / T0 9 CA = CA0 * (1 - X) / (1 + X) * T0 / T 10 CC = CA0 * X / (1 + X) * T0 / T 11 CB = CA0 * X / (1 + X) * T0 / T 12 rA = -(k * CA - kr * CB * CC)

12-141

P12-21 First note that ΔCP = 0 for both reactions. This means that ΔHRx(T) = ΔHRx° for both reactions. Now start with the differential energy balance for a PFR: rij ( H Rxij ) Ua (Ta T ) r1 A ( H 1 A ) r2 B ( H Rx 2 B ) dT Ua (Ta T ) dV F j C Pj F j C Pj If we evaluate this differential equation at its maximum we get dT 0 and therefore, Ua (Ta T ) r1 A ( H Rx1C ) r2 B ( H Rx 2 B ) dV We can then solve for r1A from this information. Ua(Ta T ) r2 B ( H Rx 2 B ) r1 A ( H Rx1 A ) Ua(Ta T ) 2k 2 D C B CC ( H Rx 2 B ) r1 A ( H Rx1 A ) 10(325 500) 2(0.4)(0.2)(0.5)(5000) r1A 0.043 50000 1 1 r1 A 0.043 2 k1C C A C B 2 k1C (0.1)( 0.2)

k1C

4 .3

k1C (500 ) 4.3

k1C ( 400 ) exp

0.043 exp

E 1 R 400

E 1 1.987 400

1 500 1 500

cal mol Alternate Solution: E 18300

12-142

0

E = 76575 J /mol

P12-22 T(K)

CC

800

0.92

700

1.06

600

1.0755

500

0.78

650

1.1025

625

1.099

675

1.088

Ans: In the range between 300K to 600K, answer is 600K 600K

12-143

P12-23 Mole balance: dFA rA dW Rate Laws: rA r2 B

dFB dW

rB

dFC dW

rC

r1 A r3 A

rB

r1 A r2 B

rC

r3 A

r1 A

k1C A

r2 B

k B CB

r3 A k3CC Stoichiometry: FT C A CT A 0 FT T FBT0 FT T Energy balance: dT Ua(Ta T ) ( r1 A )( H R1 A ) ( rR 2 B ) ( r3 A )( dW FAC pA FBC pB FC C pC CB

dT dW

CT

H R3 A )

16(500 T ) ( r1 A )1800 ( rR 2 B )1800 ( r3 A )1100 100( FA FB FC )

k1

0.5exp 2(1 320 / T )

k2

k1 KC 12-144

k3

0.005exp 4.6(1 460 / T )

KC

10 exp 4.8(430 / T 1.5)

See Polymath program P12-23.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 w

0

0

100.

100.

2 fb

1.

0.9261241

1.368476

0.9261241

3 fa

1.

0.6296429

1.

0.8694625

4 fc

0

0

0.2044134

0.2044134

5 T

330.

330.

416.3069

416.3069

6 Ua

16.

16.

16.

16.

7 Ta

500.

500.

500.

500.

8 Dhr1a

-1800.

-1800.

-1800.

-1800.

9 Dhr3a

-1100.

-1100.

-1100.

-1100.

10 cpa

100.

100.

100.

100.

11 cpb

100.

100.

100.

100.

12 cpc

100.

100.

100.

100.

13 k1

0.5312401 0.5312401

0.7941566

0.7941566

14 k3

0.0008165 0.0008165

0.0030853

0.0030853

15 ct

2.

2.

2.

2.

16 ft

2.

2.

2.

2.

17 To

330.

330.

330.

330.

18 Kc

3.885029

1.062332

3.885029

1.062332

19 k2

0.1367403 0.1367403

0.74756

0.74756

20 ca

1.

0.5682599

1.

0.6892094

21 cb

1.

0.7341242

1.253213

0.7341242

22 r1a

-0.5312401 -0.5748799

-0.362406

-0.5473402

23 r3a

-0.0008165 -0.002138

-0.0007594

-0.0021264

24 rc

0.0008165 0.0007594

0.002138

0.0021264

25 r2b

-0.1367403 -0.5770243

-0.1367403

-0.5488019

12-145

26 rb

0.3944998 -0.0522707

0.3944998

-0.0014617

27 ra

-0.3953164 -0.3953164

0.0510521

-0.0006647

Differential equations 1 d(fb)/d(w) = rb 2 d(fa)/d(w) = ra 3 d(fc)/d(w) = rc 4

d(T)/d(w) = (Ua * (Ta - T) + (-r1a) * (-Dhr1a) + (-r2b) * (Dhr1a) * (-r3a) * (-Dhr3a)) / (fa * cpa + fb * cpb + fc * cpc)

Explicit equations 1 Ua = 16 2 Ta = 500 3 Dhr1a = -1800 4 Dhr3a = -1100 5 cpa = 100 6 cpb = 100 7 cpc = 100 8 k1 = .5 * exp(2 * (1 - 320 / T)) 9 k3 = .005 * exp(4.6 * (1 - (460 / T))) 10 ct = 2 11 ft = 2 12 To = 330 13 Kc = 10 * exp(4.8 * (430 / T - 1.5)) 14 k2 = k1 / Kc 15 ca = ct * fa / ft * To / T 16 cb = ct * fb / ft * To / T 17 r1a = -k1 * ca 18 r3a = -k3 * ca 19 rc = -r3a 20 r2b = -k2 * cb 12-146

21 rb = -r1a + r2b 22 ra = -r2b + r1a + r3a P12-23 (a) As seen in the above table, the lowest concentration of o-xylene (A) = .568 mol/dm3 P12-23 (b) The maximum concentration of m-xylene (B) = 1.253 mol/dm3 P12-23 (c) The maximum concentration of o-xylene = 1 mol/dm3 P12-23 (d) The same equations are used except that FB0 = 0. The lowest concentration of o-xylene = 0.638 mol/dm3. The highest concentration of m-xylene = 1.09 mol/dm3. The maximum concentration of o-xylene = 2 mol/dm3. P12-23 (e) Decreasing the heat of reaction of reaction 1 slightly decreases the amount of E formed. Decreasing the heat of reaction of reaction 3 causes more of C to be formed. Increasing the feed temperature causes less of A to react and increases formation of C. Increasing the ambient temperature causes a lot of C to be formed. P12-23 (f) Individualized solution

P12-24 (a) A

B C

A D E A C F G We want the exiting flow rates B, D and F Start with the mole balance in PFR: dFC dFA dFB rA rB dV dV dV dFE dFF rE rF dV dV

Rate Laws: rA

r1s

r2 B

rB

r1s

rC

r1s r3T

rD

r2 B

rE

r2 B

rF

r3T

r3T

12-147

rC

dFD dV

dFG dV

rG

rD

r1s

(1

r2 B

(1

r3T

(1

10925 PA T 25000 ) exp 13.2392 PA T 11000 ) exp 0.2961 PA PC T

) exp

0.08359

PB PC K

Stoichiometry:

PA

FA PT 0 FT

PB

FB PT 0 FT

PC

FC PT 0 FT

FT

FA

FI

steamratio .0034

FB

FC

FD

FE

FF

FG

FI

Energy Balance:

dT dV

r1s H R1 A r2 B H R 2 A r3T H R 3 A FA *299 FB *283 FC *30 FD *201 FE *90 FF *249 FG *68 FI *40

K p1

exp b1

b2 T

b3 ln(T )

(b4T b5 )T b6 T

See Polymath program P12-24.pol. For T0 = 800K Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 v

0

0

10.

10.

2 fa

0.00344

0.002496

0.00344

0.002496

3 fb

0

0

0.0008974

0.0008974

4 fc

0

0

0.0008615

0.0008615

5 fd

0

0

1.078E-05

1.078E-05

6 fe

0

0

1.078E-05

1.078E-05

7 ff

0

0

3.588E-05

3.588E-05

12-148

8 fg

0

0

3.588E-05

3.588E-05

9 T

800.

765.237

800.

765.237

10 Hla

1.18E+05

1.18E+05

1.18E+05

1.18E+05

11 H2a

1.052E+05 1.052E+05

1.052E+05

1.052E+05

12 H3a

-5.39E+04

-5.39E+04

-5.39E+04

-5.39E+04

13 p

2137.

2137.

2137.

2137.

14 phi

0.4

0.4

0.4

0.4

15 Kl

0.0459123 0.0196554

0.0459123

0.0196554

16 sr

14.5

14.5

14.5

14.5

17 fi

0.04988

0.04988

0.04988

0.04988

18 ft

0.05332

0.05332

0.0542282

0.0542282

19 Pa

0.1548387 0.1104652

0.1548387

0.1104652

20 Pb

0

0

0.0397155

0.0397155

21 Pc

0

0

0.0381277

0.0381277

22 r2b

2.991E-06

5.16E-07

2.991E-06

5.16E-07

23 rd

2.991E-06

5.16E-07

2.991E-06

5.16E-07

24 re

2.991E-06

5.16E-07

2.991E-06

5.16E-07

25 r3t

0

0

4.196E-06

4.151E-06

26 rf

0

0

4.196E-06

4.151E-06

27 rg

0

0

4.196E-06

4.151E-06

28 rls

0.0002138 2.481E-05

0.0002138

2.481E-05

29 rb

0.0002138 2.481E-05

0.0002138

2.481E-05

30 rc

0.0002138 2.066E-05

0.0002138

2.066E-05

31 ra

-0.0002167 -0.0002167

-2.948E-05

-2.948E-05

Differential equations 1 d(fa)/d(v) = ra 2 d(fb)/d(v) = rb 3 d(fc)/d(v) = rc 4 d(fd)/d(v) = rd 5 d(fe)/d(v) = re 12-149

6 d(ff)/d(v) = rf 7 d(fg)/d(v) = rg 8

d(T)/d(v) = -(rls * Hla + r2b * H2a + r3t * H3a) / (fa * 299 + fb * 273 + fc * 30 + fd * 201 + fe * 90 + ff * 68 + fi * 40)

Explicit equations 1 Hla = 118000 2 H2a = 105200 3 H3a = -53900 4 p = 2137 5 phi = .4 6

Kl = exp(-17.34 - 1.302e4 / T + 5.051 * ln(T) + ((-2.314e-10 * T + 1.302e-6) * T + -0.004931) * T)

7 sr = 14.5 8 fi = sr * .00344 9 ft = fa + fb + fc + fd + fe + ff + fg + fi 10 Pa = fa / ft * 2.4 11 Pb = fb / ft * 2.4 12 Pc = fc / ft * 2.4 13 r2b = p * (1 - phi) * exp(13.2392 - 25000 / T) * Pa 14 rd = r2b 15 re = r2b 16 r3t = p * (1 - phi) * exp(.2961 - 11000 / T) * Pa * Pc 17 rf = r3t 18 rg = r3t 19 rls = p * (1 - phi) * exp(-0.08539 - 10925 / T) * (Pa - Pb * Pc / Kl) 20 rb = rls 21 rc = rls - r3t 22 ra = -rls - r2b - r3t Fstyrene = 0.0008974 Fbenzene = 1.078E-05 12-150

Ftoluene = 3.588E-05 SS/BT = 19.2 P12-24 (b) T0 = 930K Fstyrene = 0.0019349 Fbenzene = 0.0002164 Ftoluene = 0.0002034 SS/BT = 4.6 P12-24 (c) T0 = 1100 K Fstyrene = 0.0016543 Fbenzene = 0.0016067 Ftoluene = 0.0001275 SS/BT = 0.95 P12-24 (d) Plotting the production of styrene as a function of To gives the following graph. The temperature that is ideal is 995K

P12-24 (e) Plotting the production of styrene as a function of the steam gives the following graph and the ratio that is the ideal is 25:1

12-151

P12-24 (f) When we add a heat exchanger to the reactor, the energy balance becomes: Ua Ta T r1s H R1 A r2 B H R 2 A r3T H R 3 A dT

dV FA *299 FB *283 FC *30 FD *201 FE *90 FF *249 FG *68 FI *40 With Ta = 1000 K Ua = 100 kJ/min/K = 1.67 kJ/s/K The recommended entering temperature would be T0 = 440 K. This gives the highest outlet flow rate of styrene.

P12-24 (g) Individualized solution P12-24 (h) Individualized solution

P12-25 Let a=A b = A2 c = A4 PART 1 POLYMATH Report Ordinary Differential Equations

21-Jun-2010

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 12-152

1 Ca

2.

0.155599

2.

0.155599

2 Cb

0

0

0.4963834

0.1995088

3 Cpa

25.

25.

25.

25.

4 Cpb

50.

50.

50.

50.

5 Cpc

100.

100.

100.

100.

6 DH1a

-3.25E+04

-3.25E+04

-3.25E+04

-3.25E+04

7 DH2b

-2.75E+04

-2.75E+04

-2.75E+04

-2.75E+04

8 E1

4000.

4000.

4000.

4000.

9 E2

5000.

5000.

5000.

5000.

10 Fa

100.

7.779951

100.

7.779951

11 Fb

0

0

24.81917

9.975439

12 Fc

0

0

18.06729

18.06729

13 k1

0.6

0.6

0.6

0.6

14 k1a

0.6

0.6

76.97387

6.133163

15 k2

0.35

0.35

0.35

0.35

16 k2b

0.2072016 0.2072016

89.46089

3.787125

17 Qg

7.8E+04

8971.335

1.569E+06

8971.335

18 Qr

-1.5E+04

-1.5E+04

7.696E+05

1.44E+05

19 r1a

-2.4

-42.01417

-0.1484903

-0.1484903

20 r1b

1.2

0.0742452

21.00708

0.0742452

21 r2b

0

-18.03538

0

-0.1507418

22 ra

-2.4

-42.01417

-0.1484903

-0.1484903

23 rb

1.2

-8.261933

17.21274

-0.0764966

24 rc

0

0

9.017691

0.0753709

25 sumCp

2500.

2500.

2500.

2500.

26 T

300.

300.

1084.641

459.0062

27 T1

300.

300.

300.

300.

28 T2

320.

320.

320.

320.

29 Ta

315.

315.

315.

315.

30 Ua

1000.

1000.

1000.

1000.

31 V

0

0

10.

10.

32 vo

50.

50.

50.

50.

Differential equations 12-153

1 d(T)/d(V) = (Qg-Qr)/sumCp 2 d(Fc)/d(V) = rc 3 d(Fb)/d(V) = rb 4 d(Fa)/d(V) = ra Explicit equations 1 Cpc = 100 2 Cpb = 50 3 Cpa = 25 4 k2 = .35 5 Ta = 315 6 Ua = 1000 7 DH2b = -27500 8 DH1a = - 32500 9 T1 = 300 10 E1 = 4000 11 k1 = 0.6 12 T2 = 320 13 E2 = 5000 14 k2b = k2*exp((E2/1.987)*(1/T2-1/T)) 15 vo = 50 16 Cb = Fb/vo 17 Ca = Fa/vo 18 sumCp = (Fa*Cpa+Fb*Cpb+Fc*Cpc) 19 k1a = k1*exp((E1/1.987)*(1/T1-1/T)) 20 r2b = -k2b*Cb^2 21 rc = -r2b/2 22 r1a = -k1a*Ca^2 23 r1b = -r1a/2 24 rb = r1b + r2b 25 ra = r1a 26 Qg = r1a*DH1a+r2b*DH2b 27 Qr = Ua*(T-Ta)

12-154

(b) The required reactor volume = 3.5 dm3 (c)

12-155

Part 2

12-156

12-157

P12-26 Let A = a A3 = b, A6 = c k1 = k1A and k2 = k2A3 Thus the values of quantities exiting the reactor are as follows: FA = 23.15 mol/sec FA3 = 11.906 mol/sec FA6 = 6.85 mol/sec T = 589.73 K

POLYMATH Report Nonlinear Equations

21-Jun-2010

Calculated values of NLE variables Variable Value

f(x)

Initial Guess

12-158

1 Fa

23.15856 -4.263E-14 30.

2 Fb

11.90629 1.243E-14 30.

3 Fc

6.85376 8.882E-16 2.

4T

589.7334 -2.328E-10 1000.

Variable Value 1 Ca

0.5620827

2 Cb

0.2889783

3 Cpa

25.

4 Cpb

75.

5 Cpc

150.

6 Cto

2.

7 DH1a

-8.0E+04

8 DH2b

-1.0E+05

9 E1

4000.

10 E2

5000.

11 Fao

100.

12 Ft

41.91861

13 k1

0.9

14 k1a

24.32177

15 k2

0.45 12-159

16 k2b

16.41453

17 Qg

7.518E+05

18 Qr

7.518E+05

19 r1a

-7.684144

20 r1b

2.561381

21 r2b

-1.370752

22 ra

-7.684144

23 rb

1.190629

24 rc

0.685376

25 T1

300.

26 T2

320.

27 Ta

315.

28 To

300.

29 UA

100.

30 V

10.

Nonlinear equations 1 f(T) = (Qr - Qg) = 0 2 f(Fc) = 0 -Fc+rc*V = 0 3 f(Fb) = 0 - Fb + rb*V = 0 4 f(Fa) = Fao-Fa+ra*V = 0

12-160

Explicit equations 1 Cpc = 150 2 Cpb = 75 3 Cpa = 25 4 k2 = 0.45 5 Ta = 315 6 UA = 100 7 DH1a = -80000 8 DH2b = -100000 9 T1 = 300 10 E1 = 4000 11 k1 = 0.9 12 T2 = 320 13 E2 = 5000 14 k2b = k2*exp((E2/1.987)*(1/T2-1/T)) 15 Cto = 2 16 Ft = Fa + Fb + Fc 17 To = 300 18 Cb = Fb/Ft*Cto*To/T 19 Ca = Fa/Ft*Cto*To/T

12-161

20 k1a = k1*exp((E1/1.987)*(1/T1-1/T)) 21 r2b = - k2b*Cb^2 22 rc = -r2b/2 23 r1a = -k1a*Ca^2 24 r1b = -r1a/3 25 rb = r1b + r2b 26 ra = r1a 27 Qg = r1a*DH1a+r2b*DH2b 28 Fao = 100 29 Qr = UA*(T-Ta)+Fao*Cpa*(T-To) 30 V = 10

12-162

CD12GA1

12-163

12-164

12-165

12-166

See Polymath program CD12GA-1.pol.

12-167

POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 1.0E+08 1.0E+08 X 0 0 0.6495463 0.6495463 Fao 1200 1200 1200 1200 T 550 225.22683 550 225.22683 Cao 0.064 0.064 0.064 0.064 To 550 550 550 550 k 1.949E+05 5.047E-05 1.949E+05 5.047E-05 Kc 1.01E+06 1841.4832 1.01E+06 1841.4832 f 1 1 2.4419826 2.4419826 Ca 0.064 0.0547713 0.064 0.0547713 Cb 0.064 0.0547713 0.064 0.0547713 Cc 0 0 0.2030311 0.2030311 ra -798.17344 -798.17344 -1.503E-07 -1.503E-07 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [1] Fao = 1200 [2] T = -500*X+550 [3] Cao = .064 [4] To = 550 [5] k = .035*exp(8419.5*(1/273-1/T)) [6] Kc = 25000*exp(2405.6*(1/298-1/T)) [7] f = To/T [8] Ca = Cao*(1-X)*f [9] Cb = Cao*(1-X)*f [10] Cc = 2*Cao*X*f [11] ra = -k*(Ca*Cb-Cc^2/Kc)

12-168

CD12GA2

12-169

See Polymath program CD12GA-2.pol. POLYMATH Results NLES Solution Variable Value f(x) Ini Guess Ca 0.0016782 -1.472E-16 0.0017 Cb 0.0016782 -1.472E-16 0.0017 Cd 0.0071436 -1.154E-16 0.0072 Cu 0.0011782 -2.017E-17 0.0012 V 3794.94 1.018E-09 3794 Cao 0.01 Cbo 0.01 vo 6000 k1 6.73 k2 1.11 tau 0.63249 rd 0.0112944 ru 0.0018628 ra -0.0131572 rb -0.0131572 NLES Report (safenewt) Nonlinear equations [1] f(Ca) = tau*ra+Cao-Ca = 0 [2] f(Cb) = tau*rb+Cbo-Cb = 0 [3] f(Cd) = Cd-tau*rd = 0 [4] f(Cu) = Cu-tau*ru = 0 [5] f(V) = 171000/(20190*Ca+6660*Cb)-V = 0 Explicit equations [1] Cao = .01 [2] Cbo = .01 [3] vo = 6000 [4] k1 = 6.73 12-170

[5] k2 = 1.11 [6] tau = V/vo [7] rd = k1*Ca [8] ru = k2*Cb [9] ra = -k1*Ca-k2*Cb [10] rb = ra CD12GA 2 (a)

CD12GA 2 (b)

CD12GA 2 (c) Individualized solution ______________________________________________________________________________

12-171





13-1

Solutions for Chapter 13 – Unsteady State Non-isothermal Reactor Design

P13-1


P13-2 (a) Example 13-1 (1) If the heat of mixing had been neglected, the shape of the graphs would have been as follows:

(2) The new T0 of 20 ˚F (497 ˚R) gives a new

HRn and T. With T=497+89.8X the polymath program of

example 9-1 gives t= 8920 s for 90 % conversion.

13-2

(3) Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 t

0

0

9000.

9000.

2 X

0

0

1.

1.

3 T

480.

480.

894.193

894.193

4 k

8.307E-06

8.307E-06

56.65781

56.65781

5 Ta

896.4

896.4

896.4

896.4

6 Ua

0.22

0.22

0.22

0.22

7 Qr

91.608

0.4855311

91.608

0.4855311

8 V

1.2

1.2

1.2

1.2

9 Nao

1.

1.

1.

1.

10 NiCpi

403.

403.

403.

403.

11 Ca

0.8333333 -5.468E-07

0.8333333

-1.392E-09

12 ra

-6.923E-06 -0.0038571

8.639E-06

7.887E-08

13 Qg

-0.3016323 -168.055

0.3763917

0.0034364

Differential equations 1 d(X)/d(t) = k * (1 - X) 2 d(T)/d(t) = (Qr - (-36309) * (-ra) * V) / NiCpi Explicit equations 1 k = 0.000273 * exp(16306 * ((1 / 535) - (1 / T))) 2 Ta = 896.4 3 Ua = .22 4 Qr = Ua * (Ta - T) 5 V = 1.2 6 Nao = 1 7 NiCpi = 403 8 Ca = (Nao/V) * (1 - X) 9 ra = -k * Ca 10 Qg = (-36309) * (-ra) * V

13-3

P13-2 (b) Example 13-2 To show that no explosion occurred without cooling failure.

13-4

P13-2 (c) Example 13-3 Decreasing the coolant rate to 10 kg/s gives a weak cooling effect and the maximum temperature in the reactor becomes 315 K. An increase of the coolant rate to 1000 kg/s gives a Tmax of 312 K. A big change to the coolant rate has, in this case, only a small effect on the temperature, and because the temperature does not change significantly the conversion will be kept about the same.

P13-2 (d) Example 13-4

13-5

13-6

P13-2 (e) Example 13-5 Using the code from Example 9-5, we could produce the following graphs either by changing TO and finding the steady state conversion or changing the coolant flow rate and finding the steady state conversion and temperature. These are the graphs of those:

13-7

P13-2 (f) Example 13-6 (1) Individualised solution (2) The transition to runaway occurs over a very small range of Ua . If U a value is changed such that Ua is taken from 2.1 ×104 to 1.9 ×104 J/hr/K the runaway occurs. Thus it occurred over a very narrow range of Ua values . (3) The modelling is done and the changes are incorporated in the polymath code for Example 13.6 A new switch type variable is introduced .

13-8

Sw2 such that Sw2 = 1

if 300
0

and a new term is added in the energy equation such that the equation becomes : d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T – 373.15))/SumNCp) + SW1*Sw2*4/60 Case 1: When Ua = 2.77 × 106 J/hr/K But the cooling starts only for T>455K Thus Ua = 2.77 × 106 J/hr/K , T> 455K = 0 for T< 455K See polymath Program for part 3 Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

A1A

4.0E+14

4.0E+14

4.0E+14

4.0E+14

2

A2S

1.0E+84

1.0E+84

1.0E+84

1.0E+84

3

CA

4.3

0.0457129

4.3

0.0457129

4

CB

5.1

0.8457129

5.1

0.8457129

5

CS

3.

2.999997

3.

2.999997

6

Cv1

3360.

3360.

3360.

3360.

7

Cv2

5.36E+04

5.36E+04

5.36E+04

5.36E+04

8

DHRx1A -4.54E+04

-4.54E+04

-4.54E+04

-4.54E+04

9

DHRx2S

-3.2E+05

-3.2E+05

-3.2E+05

-3.2E+05

10 E1A

1.28E+05

1.28E+05

1.28E+05

1.28E+05

11 E2S

8.0E+05

8.0E+05

8.0E+05

8.0E+05

12 FD

2467.445

61.27745

8874.034

61.27745

13 Fvent

2467.445

61.27745

8874.034

61.27745

14 k1A

0.0562573

0.0562573

1.82787

0.7925113

15 k2S

8.428E-16

8.428E-16

2.367E-06

1.276E-08

16 P

4.4

4.4

4.4

4.4

17 r1A

-1.233723

-4.437017

-0.0306385

-0.0306385

18 r2S

-2.529E-15

-7.102E-06

-2.529E-15

-3.829E-08

19 SumNCp 1.26E+07

1.26E+07

1.26E+07

1.26E+07

20 SW1

1.

1.

1.

1.

21 Sw2

0

0

0

0

13-9

22 T

422.

422.

466.4882

454.9731

23 t

0

0

4.

4.

24 UA

0

0

2.77E+06

0

25 V0

4000.

4000.

4000.

4000.

26 VH

5000.

5000.

5000.

5000.

Differential equations 1 d(CA)/d(t) = SW1*r1A mol/dm3/hr 2 d(CB)/d(t) = SW1*r1A change in concentration of cyclomethylpentadiene 3 d(CS)/d(t) = SW1*r2S change in concentration of diglyme 4 d(P)/d(t) = SW1*((FD-Fvent)*0.082*T/VH) 5 d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T-373.15))/SumNCp) + SW1*Sw2*4/60 Explicit equations 1 V0 = 4000 dm3 2 VH = 5000 dm3 3 DHRx1A = -45400 J/mol Na 4 DHRx2S = -3.2E5 J/mol of Diglyme 5 SumNCp = 1.26E7 J/K 6 A1A = 4E14 per hour 7 E1A = 128000 J/kmol/K 8 k1A = A1A*exp(-E1A/(8.31*T)) rate constant reaction 1 9 A2S = 1E84 per hour 10 E2S = 800000 13-10

J/kmol/K 11 k2S = A2S*exp(-E2S/(8.31*T)) rate constant reaction 2 12 SW1 = if (T>600 or P>45) then (0) else (1) 13 r1A = -k1A*CA*CB mol/dm3/hour (first order in sodium and cyclomethylpentadiene) 14 r2S = -k2S*CS mol/dm3/hour (first order in diglyme) 15 FD = (-0.5*r1A-3*r2S)*V0 16 Cv2 = 53600 17 Cv1 = 3360 18 Fvent = if (FD<11400) then (FD) else(if (P<28.2) then ((P-1)*Cv1) else ( (P-1)*(Cv1 +Cv2))) 19 UA = if T> 455 then (2.77e6) else (0) no cooling 20 Sw2 = if ((T>300) and (T<422)) then (1) else (0)

13-11

(4) Solving the equations with Ua = 0 for the entirety of the process, we get that the temperature reaches at 0.94 hours while the system reaches runaway at 3.649 hours. This implies the maximum time in minutes that the cooling can be lost is 163 minutes.

13-12

P13-2 (g) Example RE13-1

P13-2 (h) Example RE13-2

13-13

P13-2 (i) No solution will be given

P13-3 After the feed was shut off,

See Polymath program P13-3.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Cpa

0.38

0.38

0.38

0.38

2 dH

-336.

-336.

-336.

-336.

3 k

0.0081711

0.0081711

4.285E+17

4.285E+17

4 T

980.

980.

3.302E+20

3.302E+20

5 t

0

0

4.

4.

Differential equations 1 d(T)/d(t) = -dH * k / Cpa Explicit equations 1 dH = -336 2 Cpa = .38 3 k = (0.307/60)* exp(44498 * (1 / 970 - 1 / T))

13-14

As can be seen from the above plots, temperature shoots at time t = 3.16 mins Since NA doesn’t come in the temperature equation, there will not be any effect of feed present in the reactor. 0 If T0 = 100 F, Temperature will remain constant for next 4 minutes, which means no explosion. 0 If T0 = 500 F, there will not be any explosion and the temperature profile will look like

13-15

P13-4 (a)

See Polymath program P13-4-a.pol

13-16


Cbo

10.

10.

10.

10.

2

Fbo

10.

0

10.

0

3

k

0.0089352

0.0089352

10.08511

10.08511

4

Na

500.

0.1396707

500.

0.1396707

5

Nao

500.

500.

500.

500.

6

Nb

0

0

42.43357

0.1397745

7

Nc

0

0

499.8603

499.8603

8

Qg

0

0

1.24E+05

11.81314

9

Qr

0

0

0

0

10 ra

0

-0.3460808

0

-1.969E-05

11 T

298.

298.

510.4441

510.4441

12 t

0

0

120.

120.

13 V

50.

50.

100.

100.

14 vo

1.

0

1.

0

15 X

0

0

0.9997207

0.9997207

Differential equations 1 d(Na)/d(t) = ra*V 2 d(Nb)/d(t) = ra*V+Fbo 3 d(Nc)/d(t) = -ra*V 4 d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323)))/(15*Na+15*Nb+30*Nc) 5 d(X)/d(t) = -ra*V/Nao Explicit equations 1 Fbo = if(t<50)then(10)else(0) 2 Nao = 500 3 Cbo = 10 4 k = .01*exp((10000/1.987)*(1/300-1/T)) 5 vo = Fbo/Cbo 6 V = if(t<50)then(50+(vo*t))else(100) 7 ra = -k*Na*Nb/(V^2) 8 Qg = -6000*ra*V 9 Qr = 0

13-17

P13-4 (b) This is the same as part (a) except the energy balance. Energy balance:

13-18

See Polymath program P13-4-b.pol

P13-4 (c) This is the same as part (b) except the reaction is now reversible.

13-19

See Polymath program P13-4-c.pol

13-20

P13-5 (a)

P13-5 (b)

13-21

P13-6

13-22

See Polymath program P13-6.pol POLYMATH Results Calculated values of the DEQ variables Variable t X T k1 Ca Cb k2 Cc ra

initial value 0 0 373 0.002 0.1 0.125 3.0E-05 0 -2.236E-04

minimal value 0 0 373 0.002 0.0749517 0.0999517 3.0E-05 0 -0.1344598

maximal value 10 0.2504829 562.91803 106.13627 0.1 0.125 366.75159 0.0250483 8.644E-06

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(t) = -ra/.1 [2] d(T)/d(t) = ((40000+(10*(T-298)))*(-ra)*(1/.1))/(56.25-(10*X)) Explicit equations as entered by the user [1] k1 = .002*exp((100000/8.314)*(1/373-1/T)) [2] Ca = .1*(1-X) [3] Cb = .1*(1.25-X) [4] k2 = .00003*exp((150000/8.314)*(1/373-1/T)) [5] Cc = .1*X [6] ra = -((k1*(Ca^.5)*(Cb^.5))-(k2*Cc))

13-23

final value 10 0.2504829 562.91802 106.13612 0.0749517 0.0999517 366.75083 0.0250483 1.3E-07

P13-7 (a) Use Polymath to solve the differential equations developed from the unsteady state heat and mass balances. Refer to P13-7-a.pol POLYMATH Results Calculated values of the DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Cc

0.1

0.1

14.25628

14.25628

2

Cpc

4.2

4.2

4.2

4.2

3

Cps

5.

5.

5.

5.

4

Cs

300.

282.3047

300.

282.3047

5

Hrxn

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

6

Iprime

0.0866522

0.0004366

0.3898631

0.0004366

7

Km

5.

5.

5.

5.

8

mu

0.0426159

0.0002145

0.1916769

0.0002145

9

mu1max 0.5

0.5

0.5

0.5

10 neg_rs

0.005327

0.0038221

1.083901

0.0038221

11 Q

0

0

0

0

13-24

12 rg

0.0042616

0.0030577

0.8671209

0.0030577

13 rho

1000.

1000.

1000.

1000.

14 rs

-0.005327

-1.083901

-0.0038221

-0.0038221

15 t

0

0

300.

300.

16 T

278.

278.

334.6251

334.6251

17 V

25.

25.

25.

25.

18 Ycs

0.8

0.8

0.8

0.8

ODE Report (STIFF) Differential equations as entered by the user [1] d(Cc)/d(t) = rg [2] d(Cs)/d(t) = -rg/Ycs [3] d(T)/d(t) = (Q+(-Hrxn)*(rg))/(rho*Cps) Explicit equations as entered by the user [1] Iprime = (0.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T)) [2] Km = 5.0 [3] mu1max = 0.5 [4] Ycs = 0.8 [5] mu = mu1max*Iprime*(Cs/(Km+Cs)) [6] Q = 0 [7] Cps = 5 [8] Hrxn = -20000 [9] V = 25 [10] rho = 1000 [11] rg = mu*Cc [12] Cpc = 4.2 [13] rs = -rg/Ycs

13-25

P13-7 (b) When we change the initial temperature we find that the outlet concentration of species C has a maximum at T0 = 300 K. Refer to P13-7-b.pol

P13-7 (c)

13-26

Cc can be maximized with respect to T0 (inlet temp), Ta (coolant/heating temperature), and heat exchanger area. Therefore, if we are to find the optimal heat exchanger area the inlet temperature and coolant/heating temperature needs to be specified. If we take T0 = 310 and Ta = 290 we find that the optimal heat exchanger area is 0.46 m2.The concentration of cells at the end of 24 hrs under these conditions is 21.55 g/dm3 . Refer to P13-7-c.pol. The plot of Cc Vs A:

P13-8 (a)

13-27

See Polymath program P13-8-a.pol POLYMATH Results Calculated values of the DEQ variables Variable t Na Nb Nc X T vb k V Fbo Ca Cb Cc ra

initial value 0 50 0 0 0 300 1.5 5.0E-04 10 6 5 0 0 0

minimal value 0 0.3324469 0 0 0 283.60209 1.5 2.3E-04 10 6 2.202E-04 0 0 -0.0028611

maximal value 1000 50 5900.6649 49.667553 0.9933511 300 1.5 5.0E-04 1510 6 5 3.9077251 0.0718765 0

final value 1000 0.3324469 5900.6649 49.667553 0.9933511 298.76383 1.5 4.73E-04 1510 6 2.202E-04 3.9077251 0.0328924 -1.59E-06

ODE Report (RKF45) Differential equations as entered by the user [1] d(Na)/d(t) = ra*V [2] d(Nb)/d(t) = 2*ra*V+Fbo [3] d(Nc)/d(t) = -ra*V [4] d(X)/d(t) = -ra*V/50 [5] d(T)/d(t) = ((250*(290-T))-(80*vb*(T-325))+(-55000*(-ra*V)))/(35*Na+20*Nb+75*Nc)

13-28

Explicit equations as entered by the user [1] vb = 1.5 [2] k = .0005*exp((8000/1.987)*(1/300-1/T)) [3] V = 10+(vb*t) [4] Fbo = 4*vb [5] Ca = Na/V [6] Cb = Nb/V [7] Cc = Nc/V [8] ra = -k*Ca*Cb^2

13-29

P13-8 (b)

P13-9

13-30

P13-9 (a)

See Polymath program P13-9-a.pol POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cc Cb T UA V k1 k2 ra1 rb2 rb1

initial value 0 0.3 0 0 283 0 10 1.1172964 4.081E-09 -0.3351889 0 0.3351889

minimal value 0 1.34E-65 0 -2.02E-44 283 0 10 1.1172964 4.081E-09 -35.016552 -27.963241 -6.345E-60

maximal value 0.2 0.3 0.3 0.2895784 915.5 0 10 2.141E+05 1.041E+06 6.345E-60 2.103E-38 35.016552

ODE Report (STIFF) Differential equations as entered by the user [1] d(Ca)/d(t) = ra1 [2] d(Cc)/d(t) = -rb2

13-31

final value 0.2 1.34E-65 0.3 -1.864E-65 915.5 0 10 2.141E+05 1.041E+06 6.345E-60 -3.974E-59 -6.345E-60

[3] d(Cb)/d(t) = rb1+rb2 [4] d(T)/d(t) = ((UA*(330-T))+(55000*(-ra1*V))+(71500*(-rb2*V)))/(200*V*(Ca+Cb+Cc)) Explicit equations as entered by the user [1] UA = 0 [2] V = 10 [3] k1 = 3.03*exp((9900/1.987)*(1/300-1/T)) [4] k2 = 4.58*exp((27000/1.987)*(1/500-1/T)) [5] ra1 = -k1*Ca [6] rb2 = -k2*Cb [7] rb1 = -ra1

P13-9 (b)

13-32

P13-9 (c)

13-33

P13-10 Semi batch with parallel reactions dNa dt dNb dt dNd dt dNu dt

r1 r2 V Fbo

r1 r2 V

r1 V r2 V

Rate laws: r1 k1 Ca r2 k2.Cb k1 10 exp( 2000 / T) k2 20 exp( 3000 / T)

Stoichiometry: Na V Nd Cd V V V0 v b t Ca

Cb Cu

Fbo Cbo Cao 5mol / dm 3

Nb V Nu V

vb

Cbo 1mol / dm 3

13-34

dT dt

3000( r1 V) [5000 10(T 300)]( r2 V) 30 Fbo (T Tbo) 20Na 30Nb 50Nd 40Nu

See polymath program P13-10.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

5.

0.0402932

5.

0.0402932

2

Cao

5.

5.

5.

5.

3

Cb

1.

0.1879238

1.

0.3260075

4

Cbo

1.

1.

1.

1.

5

Cd

0

0

0.7888994

0.6235848

6

Cu

0

0

0.0504077

0.0504077

7

Fbo

10.

10.

10.

10.

8

k1

0.0204583

0.0204583

0.0464803

0.0460806

9

k2

0.0018507

0.0018507

0.0063377

0.0062561

10 Na

500.

28.20523

500.

28.20523

11 Nb

100.

68.89056

228.2052

228.2052

12 Nd

0

0

436.5094

436.5094

13 Nu

0

0

35.28541

35.28541

14 r1

-0.1022916

-0.1022916

-0.0018567

-0.0018567

15 r2

-0.0018507

-0.0020395

-0.001066

-0.0020395

16 T

323.

323.

372.3484

371.7507

17 t

0

0

60.

60.

18 Tao

373.

373.

373.

373.

19 Tbo

323.

323.

323.

323.

20 V

100.

100.

700.

700.

21 vb

10.

10.

10.

10.

Differential equations 1 d(Na)/d(t) = (r1+r2)*V 2 d(Nb)/d(t) = Fbo + (r1+r2)*V 3 d(Nd)/d(t) = -r1*V

13-35

4 d(Nu)/d(t) = -r2*V 5

d(T)/d(t) = (-30*Fbo*(T-Tbo) + 3000*(-r1*V) + (5000 + 10*(T-300))*(r2*V))/(20*Na+30*Nb+50*Nd+40*Nu)

Explicit equations 1

Cbo = 1

2

Fbo = 10

3

k1 = 10*exp(-2000/T)

4

k2 = 20*exp(-3000/T)

5

vb = Fbo/Cbo

6

V = 100 + vb*t

7

Ca = Na/V

8

Cb = Nb/V

9

r1 = -k1*Ca

10 r2 = -k2*Cb 11 Cao = 5 12 Cd = Nd/V 13 Tao = 373 14 Tbo = 323 15 Cu = Nu/V

13-36

13-37

Solution Manual Essentials of Chemical Reaction Engineering

Recommend Documents