'Basic' reactor engineering problems on your pocket computer

C 815468 BIBLIOTHEEK TU Delft P 1727 5444

1111111111111

e.M.

van den Bleek

A.W.

Gerritsen

VSSD 1983

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'BASIC'

Reactor Engineering

Problems

Introduction

If you were looking fo r a set of ready-to-run reactor engineering programs, you bought the wr on g book.

This is a study book, so you have to do most of the work yourself. The book is meant to help you to master some BASIC reactor problems, which, if extended to the pr a c t i c al situation, can only be solved numerically. Up to now this cou l d on l y be done on the large machine of your institute. This is of ten raising a barrier to master the techniques of reactor design yourself. The cheap, easy to handle pocket computers bring the numerical approach at your doorstep intro-ducing a new and bet ter insight in design and operation of chemical

reactors.

The problems in this booklet are mainly taken from a course in advanced reactor engineering which we offer at the Delft University of Technology to graduate st ude n t s in Chemical Engineering. It started in 1980 using the only fifty progrannnabie "keystroke language" steps of the Texas Instruments TIS7 which turned out to be somewhat limited in its possibilitie s . A year later we switched to the in BASIC programmabie, user friendly 1.5 k Sharp PCI211 or the equivalent TR580-PC1.

If you want to "ea t this pudd ing " YOll ar e suppos ed to be a graduate student being familiar with the principles of ideal chemical reactors. Moreover we often suppose that the symbo l s used in figures and formulas are self-evident.

All problems involve differenti al equations which, if at all, can only be solved analytically for first order reaction kinetics. In case of non-isothermal behaviour or other kinetics the equations become non-linear and must be solved numerically.

Several methods to solve these equations, e.g. according to Euler, Heun and Runge-Kutta, are shown in detail. Attention is given to the fact that most equations are second order and involve a boundary value problem. Again several methods (garden-hose-, bisection- and finite difference- or matrix method) are explained and used in a program. As the PCI211 has limited capacities (memory size and computing time:) the programs do not offer the sophistication and complexity of those used by engineering companies for their reactor design . They do offer you however achallenge to combine your knowledge of mathematics and

3

In the second part of this booklet comprehensive answers to the greater half of the exercises are given. Most of the ready-to-run programs offered can be found among them. All numerical values asked for in the exercises are given.

The computer programs given here are specially written for a PC1211, PC1212 or TRSSD-PCI pocket computer. The BASIC dialect used can be run on other machines with minor corrections; attention should be given to some special features:

both H • Q and HQ are ways to express a multiplication of memories Hand Q on the PC1211: the lat ter way consumes less memory, but will cause errors on computers which accept two-or more-Iettered variables. - instead of a line number, a name (placed between quotes) can be used

in GOTO or GOSUB statements. If these names are placed in the beginning of a program the computation speed can be increased considerably. Furthermore translation to other programming languages is made easy as flowcharts are provided for all programs.

We like to thank Mr. J.J.B. van Holst for drawing most of the figures. We extend our special acknowledgement to Ms. C. Monna for the, sometimes tedious, typing of the manuscript.

C.M. van den Bleek A.W. Gerritsen

January 1983.

For PCI211 or EPSON HX-2D users a cassette with all programs can be purchased from the authors.

Contents Introduction Exercise Exercise 2 Exercise 3 Exercise 4 Exer::ise 5 Exercise 6 Exercise 7 Exercise 8 Exercise 9 Exercise 10 Exercis e 11 Answers 2 Numerical approach to tubular reactors 5 Temperature profile in an isolated tubular reactor 15 F-curve of a series of two ideal tank reactors 21 Tubular reactor with axial dispersion 25 Mass transfer and chemical reaction model for one tray 33 Chlorination of benzene in a tray-reactor 43 Chlorination of benzene in a tray-reactor 51

Interfac e variable 55

Bubble col umn re a ctor 55

Autothermic fixed bed reactor 59

The combustion of coal particles 71

Heterogeneous catalysis with porous particles 75 Influence of model and reaction kinetics

111 1111 UIlI 111111 111111 tllIJI 111 111 111 111 111 111 111 111 111 111 111 111 111111111 111111111 111111111

Exercise 1: Numerical approach to tubular reactors

to recall some methods to solve a simple or d i n a r y differential equation

numerically,

_ to get some feeling about the number of st e p s required to approximate

the analytical solution and

_ to get an impression of the time, consumed by you r pocke t co mp ute r for

this purpose. This exercise is meant

You are asked to calculat e the conversion in an isothermal co n t i n uo u s ideal tubular reactor ("plug flow" reactor) where the reactant A is con-verted according to the reaction

A _products

From a molar balance of Aove r an infinitely thin "slice" of the reactor

(fig. 1) you can derive:

~v

,

\ cA

I

cAO

i

x+dx I I 1 0 x x+dx - x L

Fig. 1. Plug flow reactor.

OUT IN - CONVERSION - HOLD UP

4>vCA\ = 4>vcA\ - (-rA)Fdx- 0

x+dx x d c 50 4> A dx - (- r A) F dx v dx or -(-r_{A 4>v})~ (1)

with the initial condition:

(2) cAoat x o

.-

-- -- -- -- -- -- -- -- -- - - -

-7

If the ra te equation is known and relatively simple. the concentration profile along the reactor length and 50 the concentration at the end of the reactor can be calculated analytically from equations (I) and (2). In many cases. including most of those of practical interest. these equations have to be solved numerically. Therefore we will recall short-ly the finite difference methods of Euler. Heun and Runge-Kutta.

For all thes e methods it is convenient to transform the equations to be solved into a dimensionless form. Introducing the dimensionless reactor length Z

=

x/L and the degree of conversion ~

=

(c

Ao- cA'/cAo equations (1) and (2) become:

~

dz T5= f(~) (3)

with the initial condition:

o at Z = 0 (4)

Z fig. 2. Moreover. the traject of the independent variable (in our case 0 ~ Z ~ I) along which va lues of the dependent variable (in our case: ~) have to he calculated is divided into N equally spaced subtrajects ("steps") by introducing the gridpoints zO' zl' z2· •••• zk. zk+I •.•· zn as shown in fig. 2.

I

Zo

zl z2 zk zk+1

I f h denotes the spacing then z - z

h N 0 I

N N

so , given the value of ~ at Z zk (= k .. h) we are looking for a goed approximation of the value of ~ at Z = zk+l' at a distance h from zk'

- The Euler Method.

Here the val ue of ~ at zk+1 is approximated from the slope of the solution cur v e at Z zk' This is shown in fig. 3.

!;;k+ 1

f

o

!;;k+1 %!;;k + h

~

I

I

---1

I

I

I

I

h

I

Fig. 3. Euler's method

~equation(3) = !;;k + h f(!;;k)

_ Z

(5)

Starting with the initial value !;; = 0 at z = 0, !;;k at every grid-point zk along the reactorlength can be approximated by means of equation (S), k ranging from 0, I, 2 ... to N-I. It can be shown

(I, 2, 3, 4) that the approximate solution according to Euler's method converges to the exact solution if h tends to zero; the ra te of this convergence is proportional to h. Therefore Euler's method is called first order and the error is Oeh).

- The Beun method.

This method introduces a correction to the Euler method based on the following idea:

The general solution of equation (3) will consist of a family of curves, each curve having a particular value of the constant of integration. Curve "a" (fig. 4) is one member of this family passing through the point (zk' !;;k)' According to Euler's method !;;k+1 is approximated from the tangent to the curve at z = zk by means of

!;;k+l,Euler = !;;k +

It is obvious from figure 4 that this is a poor approximation to the desired answer, unless the curve "a" happens to be a straight line.

-

-

- ~

E,

t

E,k+l,Euler E,k+l, HeUn

Fig. 4. Heun's method.

Through the point (zk+l' E,k+l , Eu l e r ) passes another curve of the solution family, shown as curve "b" in figure 4. Instead of the slope to the curve "a" at zk (line

Q)

in fig. 4), we also could have used the slope to the curve "b" at zk+l (line

@

in fig. 4) in the point (zk' E,k ) to calculate E,k+ l at zk+l by Euler's method (parallel displacement of line

@

to line

GD).

This would give, however, a more or less equally poor approximation of the desired answer at zk+l. It was Heun who showed (1, 2, 3, 4) that using the average of bath slopes (line

@

in fig. 4) in the point (zk,é;k) will give a better approximation af the solution value at zk+l' So:

9

é;k+l,Heun +

_!!S.

dz

_]

z=zk+l

10

eq . (3)

!

~

k ~

!!.2

~

_[

(~

k)

~

f

«

_{k+ l , Eu l er}

)

]

(6)

It can be shown (1, 2, 3, 4) th a t the Heun method is a seco nd order me thod and has an error O(h2) ; this means roughly that the er ro r tends to zero proportional to h2.

- The Runge-Kutta Method.

This method can be seen as an extension of Heun's method. You like to-know the slope of the chord to the solution curve between zk and zk+l

(line

<D

in figure 5), because this would allow you to calculate the value ~k+l exactly, given the value of ~k at zk. The question is, what is the best estimation of that slope? If you apply Euler's method to get a first ap p r o x i mati o n of ~k+l (and thus of the slope of the desired chordl), yo u can obtain a second and much better estimation of th e slope of the chord from the slope of the tangent (line

0,

fig . 5) to the membe r (cu rv e "b") of the family of so l ut ion curves whic h passes through the poi n t (zk + ., h , ~k + .,hf(!;k» .

~k+l,E

uler

-(!)points where the slope of the chord is estimated

I

I

- - - î

I

I

I

o

h Fig. 5. The Runge-Kutt a method . - Z

~

-- -- ~ -

-11 A third estimation of the slope can be found as follows:

Draw a line

(CD

in fig. 5) through the point (zk'~k) parallel to line

@.

At Z = Zk + '2 h this line will intersect member "c" of the solution family. The slope of the tangent (line

@))

to this member at this intersection point will give an estimate of the slope of the chord which is more or less as good as the second estimate.

So now we have got one estimate at zk and two, better ones at zk + '2 h. To balance a fourth can be found at zk+l from the tangent

(CD,

fig. 5) to the member ("d", fig. 5) of the solution curves which passes through the intersection of Z = zk+l and the tangent

@).

The quality of this last estimate is of the order of the first one but if the first one is an overestimate then is the last one an underestimate and vice versa. Averaging the four estimates and weighing the second and the third twice as much as the others results in:

+ 2 + 2 ~ dz z=zk+ '2h ~=~k+i:lhf(~k)

d~

I

dz z=zk+ '2 h ~=~k+ '2h f{~k'+'2h f{~k)} denoting f(~k) f{~k+'2hf(~k)}= f{~k+'2hko)

ft~k

+ '2

hf{~k

+'2 h

f{~k)}}

=

f(~k

_{+ '2 hk l)}

fl~k

+h f

{~k

+'2 h

f{~k

+'2 h

f(~k)}

}]=

f{~k

_{+h k2)} (7) (8) (9) (10) ~k+l,R-Kcan be calculated as

~

_x-i

_.a-x"

E+hd~1

K dZchord

=

~k

+

~

_{{ko + 2 k l + 2 k 2 + k3}} (ll)

Back to our plug flow reactor.

To calculate the conversion in such a reactor numerically you are now asked to write a computer program, consisting of some independent modules. The overall relation between these modules in the program is given in the flow scheme in figure 6.

number of steps N boundaries

Zo

and zN initial values subroutines "EULER" GOSUD "DKSI/ DZ" , RETURN calculation of ste p si ze h initialization of ~ and z to be used _ ~ YES "HEUN" , GOSUB"DKSI/DZ" RET URN IlRUNGE-KUTTA" GOSUB"DKSI/DZ" RETURN "DKSI/DZ " RETURN

-- -- - ---~-~~~=======~~-- --- --~~

-13

Be f o r e wr i t ing down the program first make a flowscheme for each of the me tho d s to be used.

The "OKSI/OZ" subroutine con tai ns the reac t o r diffe rential equation or in other words equation (3 ) in which th e re l e v a n t rate equat ion is sub

-stituted.

In this exercise we only wi l l use first order ki ne tics , so -rA: k CAo (I -~ ) .

Because in that case th e analytical solution is known i t is pos sible to

test your program and to compare the methods of Eu ler , He un and Runge - Ku t t a as far as it concerns the number of steps re qui r e d to approx i ma te that

analytical solution and the time consumed by your pocket co mpu te r. Ca lc ulat e the convers ion at the end of the reactor for kT : I.

s

Fi l l up the next table and use the numb e r of steps as as k e d for.

N

~EULER

t ime _~HEUN time _{~R . K} time

PC PC PC [sj [sj [sj I 2 5

l

a

20 50 cc

Mak e a graph of ~ ver s u s N for the diffe ren t methods. Which method do you

pr efer, consider i ng the accuracy wi th respect to the analytical solution

and the time to reach that accuracy?

To make the comparison with Our answers more comfortable better use the

fol l o wi n g symbols when programming the PCI211 or TRSSO-PCI: memorie s

symbol

Z

o'

starting value of independent variable z

~ at z :

Z

o

z, the di me ns ion les s reactorlength PC1211 or TR5S0-PCI A B F H N w x y Z N,

w,

~,

final value of independent variable z tangent to the ~ vs. z curve

stepsize number of steps stepcounter

Literature

1. A. Schuitman,

Lecture notes Wiskunde lIl,

1978 , Delft Univers ity of Technology. 2. J.M. Ortega , W.G. Poole jr.,

An introduction to numerical methods for differ en tia l equations , Pitman Publishing Inc., 1981, Massachusetts.

3. J.M. Coulson, J.F . Richardson,

Chemica 1 Engineering, Volume three, 2nd edition , Pergamon Press Ltd., 1979, Oxford.

4. R.L. Johnston,

Numerical methad, a software approach, J. Wiley &Sans, 1982, New York.

---

-

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Ex e r c i s e 2: Temperature profile in an iso lated tubular reactor

In th i s exerc ise the temperature and concentration profiles of an isolated ideal tubular re a cto r are calcu l a ted (see fig. I).

dx

~)

:t)cj--

(j-I

I

I

0 __ x x x+dx L

fig. 1

The reaction considered: A - - B has the following ki ne t i c s :

whe r e

spec ifi c conversion rate of A

reaction ra te constant

activation energy

gas constant

T absol ute temperature

cA concentrati on of A 90 4.104 8 mol m-3s-I -I s kJ kmol-1 kJ kmOl- 1K- 1 K mol m-3 3 -1

The reaction is exothermic: 6H

r = - 6 10 kJ kmol A ,the vo l umet ri c

-3 -I

heat capacity of the reaction mixture, pCp' equals 12 kJ m K and is

independent of temperature and composition of the reaction mixture .

In exercise 1 you have seen that the molar balance of A over a "slice"

Fdx of the reactor results in

~

= kTs(I-E; 1

dz (I)

where z is the dime nsio nl e s s reactor l e ng th x/Land E; the dimens ionles s

conversion (c A - cl/CAO·

o

A

Be c au s e the react o r is not opera ted is oth ermal l y, you need to consider the te mp erature de p e nd e nc y hidden in the reaction ra te constant. Intro

-17 ducing the adiabatic temperature rise

(-I\H lC Ar 0 pc

p

and the dimensionless temperature

8

the molar balance (1) becomes:

~

dz

with the initial condition: z

o

o

and a

o

a. In a similar way the heat balance over the "slice" Fdx results in an expression for da . Determine this expression, assuming that all

dz

heat produced is used to raise the temperature of the reaction mixture (the heat capacity of the reactor can be neglected).

The concentration and temperature profiles ca n now be obtained by

simultaneously solving both balances, starting with the initial condition

z = 0 : !; = 0 and a =

o.

Each of the three methods, mentioned before in exercise 1 can be

easily extended to solve this set of differential equations.For the Runge-Kutta method this is done as follows:

In exercise I the Runge-Kutta equations (7) to (1 1 ) were given to

advance the integration from ~k to ~k+l. For two simultaneous differential equations a similar set can he derived. It can he demonstrated that for the set differential equations

and ~

dz f(z , ~, a)

da

dz g(z, C al

the Runge-Kutta equations to advance the integration from ~k to ~k+1

as weIl as from a

ka f(Zk ' E;k'

E

V

ma g(Zk ' E;k ' 8k) k

l f(Zk +!:Ih ,E;k + !:I h ka, 8k+ !:I hmO) mi g(zk +!:Ih,E;k+!:I h ka, 8

k+ !:I hmO)

k

2 f(zk+!:Ih, E;k+!:Ihkl, 8k + !:I hml) m

2 g(zk+ !:Ih ,E;k '!:Ihk l ' 8k+ !:I h mi) k 3 f (zk +,h, E;k + h k2 , 8 k + h m2) m₃ g(zk+ h, E;k + h k_2,8_k+ h m2) so and 8 k,RK

b. Use the method of Runge-Kutta to calculate the conve rs ion of A and the temperature of the reaction mixture at ten equidistant places along the reactor axis, using T

s = 25 s and cAO= I kmol.m- 3 for two inlet temperatures Ta: 500 K and 600 K. Do not forget to make a flowscheme first. Make a graph of the results.

c. If the inlet temperature is chosen to be soa K the execution of your program will result in nonsense. What goes wrong? How can thi s be ove rco me and what does this mean for your results under b.?

Us e the following memories for the variables indicated: memories PC1211 or TRSSO-PC2 A B D E F G H K N T

v

symbol

ZO' starting value of independent variable Z zN' final value of independent variable Z ÓTad' adiabatic temperature rise

E_A, ac:tivation energy

*'

tangent to the E; vs. Z curve

dfl.. , tangent to the 8 vs. Zcurve dz h, stepsize ka Ts N, number of steps 8, dimensionless temperature (T- To )/óT ad Ta, temperature at Z = Zo

w

x y Z w, stepcounter ~, degree of conversion ~o' conversion at

Z

=

Zo

z, dimensionless reactorlength 19

3mnnJn!33] 33D33333331333 3n3J3333Ulm mn mn '33n mn mu m .. nmnn ,nnmn mnnnn nu, nn

'"

'"

''''

u " , 3nJJ33DU333 3JJJl33J3Ul] nnnunn

Exerc1se 3: F-curve of a serie s of two ideal tank reactors C 0 0 - t _Cl C 2

0

_v ; _{5 m}3/s Co = 4 kmol/m3

eb

eb

v> 2.5 m3 v

All ideal plug flow reactorca n be considered as a series of an

infinite

number of infinitely small ideal tank reactors. A real plug flow reactor can therefore be described as a

finite

series of N ideal tank reactors, each of them with a volume of Vreal/N.

Anticipating on this model you are asked to calculate the F(t) curve of a series of two continuous ideal tankreactors.

a. Suppose that a stepwise perturbation is set on the entrance of the first tank by raising the concentration of a tracer from zero to Co (see fig.). Make a molar balance of the tracer over each of the two tanks from t ~ 0 onwards. I'.ake these equations dimensionless by introducing

B , C t

1

s

Solve them numerically using the methad of Runge Kutta. Calculate F as a function of t (e.g. every half a second)until F > 0.99. Start with a value of 6t which equals the space time; af ter that dec rea se it until more or less (say 1 ~oo) the same value of F is found if 6t is halved.

First make a flow diagram.

Make a graph of the calculated F(t) curve afterwards. Use the next memories.

memories PCI211 Ol TRSSO-PCI B C F G H J M T w parameter

C/CO' dimensionless concentration in first tank C2/CO' dimensionless concentrat i on in second ta n k dB derivate of B T dT ve r s u s de. , derivate of C ve r s u s T dT dimensionless time ste p

cou n t e r of time steps between two print-outs #ST/PR, number of timesteps/print instruct ion

t/T

s ' dimensionless time

counter of time steps for calculation of ti me which matches the F-curve

23

b. If a pulse of mmoles of tracer was injected on t~O in the ent rance

of the first tank, determine the E(t) curve analytically. By

int e g r a t i n g this function you can find the F-curve. Do this and co mp a r e the results with the numerically on e s ob t a i n e d.

.«

....

...

...

...

...

...

44... 444

...

...

...

....

...

... ...

...

-...4<W44444444 ...4+44+444444

...

...

...

Exer cis e 'I: Tub ular reactor wi th axial di spers ion

In a tubular reactor whicn can be described by the axial dispers ion

model, A is converted into B according to th e reac t ion A~B. Th e product

B slows down th e reaction ra te as can be se e n from the kine t ics :

-rA

=

k 1cA/(l+k 2CB) wi t h kl

=

1 min -1 and k 2

=

1 m 3/kmo l .

In this exerc ise you are asked to in v e stiga t e th e way in which the con

-centration at the end of the reacto r is influenc e d by the va lu e s of the mo d e l parameters and the bou ndary condi t ions .

4a. Because there is no dispersion in the co n n e c t i o n tubes the reactor

ca n be describ e d by th" so-called "c l o sed -c l o s e d" system.

(j-I

o

~x x x +dx L

Introduce the dimen s i onless groups:

an d Ba v.L

o

where v is the superficial ve l o c i t y [m.s-l] and 0 the dispersion

coefficient [m2.s-l].

Determine the (dimensionless) differential equation that describes the molar balance of A for a "slice" with volume F.dx.

Which boundary conditions must be used?

4b. The second order differential equation obtained in 4a. can be replaced by an equivalent pair of two first order differential equations: dy

_r

dZy dz f ry ) ~ dz2 dJ" f('() dz

27

Yo u are asked to solve this set of differential eq ua t i o ns using the by now known method of Runge-Kutta (see exercise 2) .

However, the two boundary conditions needed to solve the problem are

here not known at the same value of z, but one is given at z

=

0,

while the othe r is known at z

=

1. This means that no matter at which

side of the reactor you start your calculations, you have to make a

gues s fo r the missing boundary value at that side to be able to

ca l c u l a t e the first slice (note that mathematically it does not matter whether you ca l c u l a t e from entrance to outlet or the other way around:).

If the guess was right, the value of the second boundary condition is

reached exa c t l y at the last slice. If the first guess was wrong (which

of course always will happen) you will have to make a new guess for the

first slice, based on the difference between the value

aalaulated

for

the last slice and the condition value

needed

there. The calculation

of all slices has to be repeated until the second boundary condition more or less coincides with the value obtained (translate "more or less" here with "better than 9'00 " ) .

The algorithm for exercise 4 is given in fig. 1. Of the program itself

only the Runge-Kutta part is given (fig. 2). You have to complete the program, based on the algorithm and information given. Start your

calculations at z

=

1, the reactor outlet, because in that way

con-ve r g e nc e is often faster. Read the program description.

4c. Calculate with your program the out let of the reactor with T

s

following va lues for Be: 0.001,

concentrations halfway and at the

2 min, c 0

=

1 kmol/m3 and the

A,

0.01, 1,10 and 20.

(Divide the reactor in 10 slices or more; if the number of slices is too small, you can get nonsense as a result: e.g. negative con-centrations or positive derivatives.)

- What does strike you for small values of Be?

- How might you calculate the concentration in the reactor for small

values of Bo instead of using the axial dispersion model? Do this

and compare the results.

4d. An apparently more obvious set of boundary conditions for the

Itclosed-closed" system is:

z

=

0,

Z :;;::" 1,

y

!ti.

Repeat part c. of this exercise for Bo adjusting your program to this set.

0.01, 1 and 20 af ter

29

The program IlEXC4"

Th e program "EXC4 " is meant to calculate a tubular reactor which can be described by the axial dispersion model. Ba s e d on the differen t ial

eq u a t ion , wh i ch describes the molar ba l a n c e of A, and the Run ge - Kutta me t h o d you can ca lc u l a te (f o r given values of the Bodenstein numbe r Bo, the in l e t concent r a t i o n cA,O and the space time T

s) :

Y1: the dimen s i o n le s s concentration of A at z = 1

f

O: th e dime n s i o n l e ss de r i v a t i v e

~

at z = 0

f

1: the dimensionless derivative ~ at z

Th e flowsc heme of the program is given in figure 1, part of the listin g

is giv e n in fi g u re 2. The numbers indicated in the flowscheme correspo nd wi t h the line numbers of the program.

SHORT DESCRI PTION --4---line 5 lines 10- 14 lin e s 20-56 line 70 lines 100-120 lin es 12 0- 2 0 0

states the name of the program, defines the print format

and jumps over the subroutines lip"I 1lRK2"and nOUT"..

The subroutine "F", where the two first-order differential equations are stated on the lines 10 and 12. As here on e

se con d-order equation is solved, the first equation is

known: Q= U which stands for

dy =

r

dz

The other differential equation must be given in line 12

written as

v f {(U), P, Z} where

V dL

=

~Y2-'

U =

f,

P

=

Y and Z

=

z

dz dz

YOU HAVE TO COMPLETE LINE 12.

The subroutine "RK2" to solve two first-order differential equations simultaneously. Because the values of y and f

at z = 0, 0.5 and 1 are asked for, three jumps (lines 24 , 51 and 54) to the output subroutine "OUT" are present. The subroutine "OUT" to print the results at the boundaries and at z = 0.5.

These lines are meant

- to input Be, cA,O and T

s

- to input the number of steps and to calculate the

stepsize

- to input the boundary conditions

YOU HAVE TO PROGRAM THE LINES 100-120

Th e s e lines are meant to print some co mm e n t s or values of

Be, numbe r of steps etc. to id e n t i fy the different results. YOUHAVE TO PROG~1 THE LINES 120-2 00.

1 ines 200-270 line 27 0 MEMORIES USED A B H N P Q R S T u V X Z A(27) A(28) A(29) A(30)

Th e subprogram "BVCOR", which stdIlds for "~oundary Va1ue and ~ection".

In the first part (lines 200-250) z,

y

and

r

have to be initialized and, dep end ing on the correc tio n method used to adjust your gu ess of the second bounda ry, sa me paramete rs of that meth a d have to get thei r in i t ial value.

In 1ine 250 the Runge-Ku tta methad is cal led for and the reactor will be calcu l ated in N st eps.

In the lines 251-2 70 yo u ha ve to ve r i f y if yo u r gu e s s of the value of the second bo u n d a ry was righ t. If not you ha v e to correct that gues s and repeat the procedur e . YOU HAVE TOWRITE THE SUBPROGRAM "BVCOR".

The end of the program.

CAO

Ba, Boden s tei n number h, steps i ze of z

N, number of steps in Runge-Kutta subroutine y, va ri ab le

!!i.

va riable dz' yi '

Y

at z = 1

r

₁, ~at z = 1 dz T s

r

.

va r i a ble

{ff

,

variabl e counter of step Óz z

intermediate value of y during step intermediate value of

r

during step an auxiliary value of y

5

\I

EXC4"

100-110

'bo •

CAo'

7::

S

#

STEPS

BoutJ'l).

COtJ'I). CALCULflilON STE~\"ZE

z=z

+

62

2-C;OSU~"F· 120-200 CoMt1El\IT5

lSI-'lio

ADJUSTt1UlT

or

GUESS oF Y OZ=1 fig. 1. 200·1~ "g\lC.OR."

IWITI Ç\LI"2.p.n ot.) 2 ,

O'

r

IIJITIRLI2.~TIOlil

?ARRM. ColtlL METHO'!)

2S0

CALCUUITION

'REIKTOR ~

CALLltJ~"RK2"

Listing program "EXC4" (MEM 918/114)

5: "EXC4":USING

".ttttttU

A " :

GOTD

100

10: "F": Q=U

12:'.1=

14: RETURN

20: "RK2"

.

22:GûSUB "F"

24:BEEP l:PRINT

"START VALUE

S:":GOSUB 70

26: FOR X=!TO

t~ 28:A(2~)=P:A(30 )=u

30:A(27)=P+HQ/6

:A(28)=U+HV/

32:P=A(29)+.5HQ

:U=A(30H.5H

V:Z=Z+.5H

34:GOSUB "F"

36:A(27)=A(27)+

HQ"'3: A(28)=A

(28)+HV/3

38:P=A(29)+.5HQ

:U=A(30H.5H

V

40:GOSUB "F"

42:A(27)=A(27)+

H(!/3: A(28)=A

(28)+HV/3

44:P=A(29)+HQ:U

=A(30)+HV:Z=

Z+.5H

46:GOSUB "F"

48: P=A(27)+HQ/t,

:U=A(28)+HV/

51: IF ::<=N/2

GOSUB 54

52:

t~E~n

x

54: BEEP 1: PF.:ltH

-z

(t~)

I,,.'ALUES

: " : C;O:3UB 70

56: RETURt

·!

70:

"OUT":

PRUn

-z

=";Z:

PRun

"GN Z=

";p:PRun

"c;

Gt'lZ="

i

U:

RETURN

100:

lt~PUT

120:BEEP l:PRINT

200:

11

BVCCiR"

250:GOSUB IRK2"

270:PRINT "KLAAR

=READ'r':STiJP"

: END

fig. 2.

Exercise 5: Mass transfer and chem ica ! reactio n model for one tray gas 0₀ I I liquid

•

•

I

•

"7

0b filmI bulk inte r fa ce a m2/ m3 liquid

at interface equilibrium c

• • A D

th1c kn e s s f1 1m = u =~

L n

reacti o n rate = -rA = knc

volumet r ic flow rate

0

_v

c*

o ;

c

v b

In the numerical determinatio n of molar flows

0

₀ (from gas to !iquid )

and

0

b (from film to bulk) the film model is of ten use d. This mod el

results in the second order differential equat ion:

-r

A bou ndary conditions x

=

0

x = (',

c = c"

If on ly the film thickness (',(= ~) and c" are known thi s equa tion kL

relates c

b to

0

0 as weil as to

0

b·

The mol a r balance for the bulk of the liquid giv es an o t her relation

between ~b and cb:

Combination of film model and molar balance of bulk thus gives

0

₀, ~b

and c

b as a function of kinetics (ci n' c*, a, kL, D, VL and ~v are

supposed to be known).

5a. Generally the differential equation and the molar balance are made

dimensionless by the substitution of

y c

.

c

and _z = "F,'x

This leads to the dimensionless group ~:

where k 2

n + 1

k (c*) n - l

35

Prove:

4>2

kc·

Sb. The molar balance of the bulk ca n als o be written in a dimensionless form:

(-rA)b 2

- ---

.

4>

.

Hl

kc· where

Yb cb/c* dimensionless concentration in bulk liquid

Yin cin/c· dimensionless concentration in liquid flow that enters the reactor

f_b (!tr.) _{derivative of Y at bulk-side of film, z}

₌

dz b

T VL/'/J_v residence time of liquid on tray

Prove the equation given and remember that (l-at.)/at. = (kL/a .D)- I Hl I - at. I - a

~

t.!< ak L 2 I ) 1 - .

4>

(=

_H_

k I+HI) 5-:: . The program "EXC5" can be used to ca l cu l a t e Yb' f

b and fo (the derivative of Y at the gas liquid interfa ce ).

Do this with the following set of data:

k 1·c D 2.10- 9 2 -I

-rA m .s

k₁ 3.362 10- 2 s-I c· 0.7 kmol.m-3 k_L 8.2 10-6 m.s-1 cin/c

.

=

0

a 100 m2.m- 3 T 450 s

For time-sake use 2 steps in the "BVCOR" subroutine (this equals about 8 "Euler-steps", or division in 8 slices). If you have more time use 10 steps, the analytic solution is th e n practically reached.

Complete the lines 12 and 230 with the co r re c t statements and determine Yb' f_b an d f_{o '}

5d. Determine the relation between th e enhanc ement factor E, Yb and f

o' (This relation does not depend upon ki net i c s ! )

Se. Ca l c u l a te E fo r the ex a mp le of Sc. and compa r e the res u lts with the

ana l y t ic solution.

Sf. Comp l e te the foll owi n g tabIe.

T(s) Yi n(-) Yb(-} fb(- }

r

0(-) E (-) 450 0 450 0. 2 450 0.5 450 0.8 450 1.0 4.5 0.5 0.045 0.5 0.045 0.8

Why has Yi n such a small ef f e c t on the resu lts in the firs t 5 line s

of this tabIe?

- Ex pla i n the strange value of f_b at T

=

0.045, Yin 0.8

Sg. Assume se cond-order kinetics: -rA k 2

2.c using the same value for k

(con s equ e n t ly k

1 ~ k2~)

Change lines 12 and 23 0 correspondingly.

Determine again the values of Yb ' f

b, fo and E for th e condi t ion s

37

The program "ExeS"

The program "ExeS" calculates Yb'

r

0 and

r

b in the reactor model for mass transfer with chemical reaction.

The va lues of k, D, k

L, c*, Cin' a and T are considered to be known.

SHORT DESeR1PT10N line S

lines 100-1SO

lines 200-270

states the name of the program and jumps over the

sub-routines "F"; "RK2" and "OUT" (see below)

is the section where the input of parameters takes place. 1f desired these parameters can also be printed out. Moreover, the dimensionless groups ~, a and Hl are cal-culated and printed.

form the sub-p r og r am "BVeOR" (which stands for Boun da r y Value, including CORrection) , and is the heart of "EXeS". 1t uses the method of Runge-Kutta, defined as a set of subroutines in line 10-70, to solve the second order differential equation:

!!l:.

f(dz' Y, z )

err,

Y, z)

A problem is that the boundary conditions are given at different values of z:

At z = 1,

r

_b is a function of Yb defined by the molar balance of the bulk. One of the two has to be "guessed" (here Yb).

At z 0, the value of Y

o must be one. However, the value of

r

o depends on Yb' and is not known on forehand. The method us e d to find th e boundary value s is popularly known as the garden-ho se-method: Starting with some value of Yb at z = 1* in line 240 (the correspondi ng va l ue of

r

_b was calc ulated in line 230) th e Runge-K u tta method is used to calculated Y_o and its derivative

r

o at the boundary z ;:: o.

1f the value of Y

o differs toa much from1 (here any value between 0.99 and 1.01 is considered as "goed") a

*1n this case the first value of Y cannot be taken equal to zero because then the correction of this value in

lines \0-\2 lines 20-54 line 70 MEMORIES USED A B c D F G H J K

correction of Yb must be made. This is done by considering Y

o and Yb of the preceding ca l c u l a t i o n (in the first calculation bath are assumed zero) and assuming a linear _{dependence of Yo on Yb'} Thus a new estimation of Yb is obtained in line 260; the calculation is repeated by returning to line 230. In cas e of first orde r kinetics (linear differential equation:) one it e r a t i o n cycle suffices.

Here the two first-order differential equatio n s are stated.

As here one ~-order equat i on is solved the first equati on is known : Q= U which st a nd s for

~

r.

dz

Th e othe r diffe r e n t i a l equation mu st be gi ven in line 12 in the fo rm: v f(U), p. Zwhere V df

Ó

dz dz2 U

r

p _Y Z = z

The Runge - Kut t a met hod to solv e two fir s t-o r d e r diffe r-ential eq ua t i on s, simultaneousl y .

Is a subroutine to print the resul t s at the bounda ries.

a a Hl D = c.

I

c ·

Yin r n h , stepsizeofz

value of _Yo in preceding calculatio n value o_{f Yb} in preceding calculat i on

k, reaction rate constant = (2/ (n +l )) .k (c*)n-l n

L M N

0$

p Q R S T U V X Z A(27) A(28) Á129) A(30)

kL, mass transfer coefficient

c*, equilibrium concentration

number of steps in Runge-Kutta subroutine

+ or -, determines printing of parameters

y. variabie

dy/dz, variabie

Yb and (Yb)start which MUST differ from 0

r

_b

T, the space-time of the liquid in the reactor

r, variabie dr/dz, variabie counter of steps 6z

z, variable which changes from 1 to 0

an intermediate value of y during step

an intermedia te value of r during step an auxiliary value of y

an auxiliary value of r

UF-t:.z.

T

"F

li

z-z

+

~z.

GOSUB "F"

GOSUB "F"

z=z

+

GO.suB

GOSU~

cW/d

Zo=

f

(z,l,

r)

f

I

'RETURN

250

GOSuB

"RK2"

40

Listing program

"

EXeS "

(MEM 532/66)

41 5: "E:-:C5":USH IG

" • 1tUnu

·

"

"

:

GOTO 100 100 : HIPUT "K="j K , "D=" j D, "KL= "jL,"C*="j~l, "GAt1t1A I t·l="j

G,

UA=";

B,

11TA U="jT, "U:3TEP S="jt·l 110: HIPUT "PF.:. "iA RI ABLE::;?+OR -?:"j

0$: I F 0$

="-"GOTO 140 120:BEEP l:PPIHl II~::

=

11;

K: PRltH "D " jD: PRUil

"t<

L ="jL: PF.:ItH "Ct' " jt'l:PF:HlT "G HI="jG 130: PRItH "A "jH:PRIHT "T AU =";T 140: F=.f< KD)",'L:A= BLT/(l+KT>:C =L···BD-l 150: BEEP 1: PF:I Hl "PHI ="jF: PF.:ItH "ALFA= "jA:PRItlT "H L =II;C 200:"BVCOF.:" 210 : PP I NT "USTP= "j t·l:H=-l...~1 220:BEEP l:INPUT "GAt1t'lA

B

:::TA RT=";P: I=O:} =0 230: "BUUO:":S= 240 : Z= 1: P=F.::U=S 250: GO::;UB "F:K2" 260: IF ABS (l-P) >. 0 1LET ::;=p: R=P+( t-P >* AH::; « F.:- J>....0:. P-I» : I=F':J= :3: GCiTO 230 27ü: PP I NT "KL AAR =READ ....·:::;TOF'" :H lIi 10:"F":Q=U 12:\0'= 14:RETURN

20:

1I·~:K2" 22:GCiSUB

"F"

24: BEEP 1:PRUlT "HULK VALUES : ":GOSUB 70 26: FCiR

)-;=

1

TO ~l 2:;:::A(29)=P:A(30 )=U 30: A(27)=P+HC·...6 : A(2:;::)=U+H\o'/

6

32: P=A(29)+.5HG! :U=A( 30) + . 5H I,':Z=Z+.5H 34: GO:::UB

"F"

36:A(27)=A (27)+ 3::::P=A(29)+.5HO :U=A(30)+.5H V 40: GO::;UB

"F"

42:A(27 )=A (27 )+ HO/3:A (2B )=A (28) +H\·'/ 3 44:P=A(29 )+HQ:U =A(30 ;'+H...·': Z= Z+.5H 46:GOSUB "F" 4:::: P=A(27)+HO/6 : U=A (2:::)+HV/

6

50: GO::;UB "F" 52: NE::<T

>(

54:HEEP l:PPItH "PE::;ULT AT 1 tHF. :u :GO~;UB 70 56:RETURtl 70: "OUT": PRItn PF.:Itn "G Z= "j P: PRItH "F L Z=";U: RETUPN

...

...

...

...

...

...

...

....

_....

...

" ' ' ' H ' f f ' ( ( fff"'f'fUUI

...

....

....

...

...

...

...

...

_...

_...

«UfSf"I""

...

...

r:

-44

Exercise 6: Chlorination of benzene in a tray-reactor

pure Chlorine

I

I

r-,

1 2

-

f - - 3

lb

V

n-2~ ?"" ~ ~ n-1 n

+

pure Benzene c o ... +J

'"

'"'

a

<IJ Ol _{chlorine +} product

...

,.

Benzene can be converted to monochlor benzene according to the reaction:

Bz + Cl

2- . . . Bz - Cl + HCI

Suppose the reactor system sketched above is used and the conversion per pass of benzene is limited to 20\ to minimize consecutive reactions.

Some data (rough estimations~).:

concentration benzene (pure) diffusivity of chlorine in benzene mass transfer coefficient

equilibrium concentration Cl

2 in Bz

specific area interface space time liquid on each tray rate of conversion of chlorine

reaction rate constant

c* a T = 10 = 2.10-9 = 7.10-6 = 0.72 = 70 = 450 kmol.m- 3 2 -1 m.5 -1 m.s kmol.m-3 m2/(m3 liquid) 5 -3 -1 kmol.m s

Neglect at first the fact that the change in benzene concentration in-fluences the reaction rate, and replace k.c

BZ.cCI 2 throughout the reactor

by k.cBZ,O.cCI_2'

With program "EXC6" the convers ion of benzene can be calculated.

45

6a. Check how this 1S done with the algorithm and corresponding listing.

Determine the statement to be given at line 40: Y ;

6b. Determine the number of trays needed to obtain a benzene con-version of 20%.

6c. The program "EXC6" can also be used to study the influence of

different parameters. Do this by changing each of the parameters

successively a factor 10 smaller and bigger (combine 6e with 6d

and 6e).

Check why the number of trays is more or less proportional with some parameters and much less dependant upon others.

Table 6c + 6d + 6e

variabie factor HL cp a #trays B or F c i c ·

out changed

-

1 k 10 0.1 D 10 0.1 k L 10 0.1 c· 10 0.1 cBz,o 10 0.1 a 10 0.1 T 10 0.1

6d. Show in the table where most (use here 80% or more ) of the reaction

takes place on the last tr a y : wri t e for example a B if i t is in the

bulk or an'F if i t is in the film. Do not report anything if bulk

and film do not differ much.

6e. How much chlorine leaves the reactor with the liquid?

6f. Write subroutine 300, comp l e te with algo rit hm.

This subroutine is needed if the concentra t i o n of the sec ond co m-ponent that reacts (benzene) ch a nges to o much to be assume d co nstant on all trays. Suppose now that on l y 1%diff er enc e is al lowed between the co ncen t r a t i o n originally used to calc u late th e co n v e rs i o n on a certain tray, and the concentration obtai n e d after calc u l a t i o n of th e benzene conversion on that speci fi c tray .

6g . Determine the number of trays needed to obt a i n a conVe rs ion of 20%, using the new made subroutine 300.

6h . Is i t allowed to assume that the reaction rate is only influenced by the concentration profile of chlorine?

Must the benzene concentration profile be taken into account? Is this done by using subroutine 300?

47

The program "EXC6"

The program ca l c u l a t e s the convers ion of benzene in a chlorination reactor co ns i s t i n g of a number of trays.

It is assumed that

the film model can be used to determine the enhancement factor the reaction ra te is only influenced by the concentration profile of chlorine on the tray

the eq uil i b r i um concentration of chlorine does not change in the reactor. SHORT DESCRIPTION

----

---lines 100-190 lines 200-27 0 lines 10-2 0 lines 30-50 lines 60-70 lines 300-MEMORIES USED

---A B C D E

control the input of the variables. If desired the variables can be printed out, and the convers ion followed tray by tray. Also a decision has to be given whether the change in benzene concentration along the reactor must be taken into account, or that the rate constant can be assumed constant and equal to k.cBZ,O

Calculation of the conversion tray by tray. Printout if des ired.

Subroutine to calculate some important dimensionless groups like $ and a.

Subroutine to calculate cb/c* and molar flux transferred,

0

_Ö

'

of each tray. The analytic solution of the model is used.

IN LINE 40 THE STATEMENT TO DETERMINE THE CHANGE IN BENZENE CONVERSION HAS TO BE GIVEN YET.

Subroutine to print the variables, if desired.

Subroutine that RAS TO BE MADE BY YOURSELF. It calculates the change in benzene concentration and verifies whether the concentration used in that calculation and the final concentration differ more than 1\. If so the values of $ and a are corrected and the calculation repeated.

a

a, specific area interface

cb/c*, dimensionless bulk concentration

D, diffusion coefficient

F G H I J K L M N

o

p Q R S T u v w

x

y

z

;4(27) ;4 (28) A$(29)

c*, equilibrium concentration chlorine

CBZ,O' concentration pure benzene c

Bz' concentration benzene on tray

cBz,calc' concentration benzen e used conversion drop

to calculate

k, reaction constant

k

L, transfer coefficient

Hl, Hinterland number

=

Vb/Vfilm tray number exp $ exp-$ sinh $ cosh $ 1-0$2

" space time liquid on tray l/(l+k.CBz,calc·')

~Ö' molar flux transferred

number of trays needed & conversion before last tray

~BZ ' conversion benzene

fi~BZ' change in conversion benzene ~ , final conversion benzene needed

Bz,end

+ or - determines printout of variables + or - determines printout tray by tray + or - determines whether k.c

49

RETURN

~

I I

t

I I I I

cp

=

({k·Cf,Z.Q\c.

.I!)

')/kl..

u,=

I/Cl-+-

k:.Cs~.CóI'C.rt ) 0(:a~..l:u, :S" 1-CUf1 s·,,,h :GO~!)q>

G./e*

=

O('f+(c;.,/C*).u..&iohlp

s.~i"hijl

+

crll'co~h

Cf

4>""

k

.c~

.

1ll'P*,hlf' +5- (Ci"/c

·)u.~h.,

50

Listing program "EXc6" (MEM 578/72)

5: "EXC6" : USHlG ".1UUUlA " : GOTO 100 10: "PHI":F=·f(KJ D)/L: U=l/(l+ KJn: A=BL TU: S=1-AFF 20: O=EXP F: P=1/ G:Q=(G-P)/2: R=(û+P)/2: RETURN 30: "DELTA":C=(A F+EUQ)/(SQ+A

HD:

I!=LGF*(A FO/R+S-EU.·'R) /(AF+SQ/R) 40:\,= 50:RETURH 60:"OUT":EEEP 1 : USIt~G ".

nnu

n

··'

,":

PRINT "I<

=";K:

PRItH "D = "jI!:PRINT "I< L ="jL: PRINT "C* II;G 70: PRHlT "CBZO= "jH:PRINT "A

="

j

E:

PRItH "TAU = "j T: PRItH

":x:

Et·m="jz: RETURtl 100: HlPUT "K=" j I< , "I!="jD,"I<L= ti ;L, 11C:t:=IJ ;C;, 11

CI:20=

11 ;

H, "A

="jB, "TAU=";

T

110: 1t·1PUT "CCit·j',,.' EZ EHD="jZ 120: U1PUT "PF.: ',,.'A RIABLE~;?+GF.:­

?:";A$(27)

130:IHPUT "PF.: TR AY I:'r' TRA'r'·~'+ GF.:-·~':"jAt(2:::: ) 140: HlPUT "RATE Ot-1L\' F(CL2)? +GR-?:" jA$(2

9)

150:IF A$(27)="+ "GOSUB "OUT" 160: H=O: E=O:;<=0 : I=H:J=H: GCJ9JB "PHI" 170: f'l=L/BD-1: BEEP 1: PF:Un "HL ="jf'l H::O: IF A$(2'3)="+ "Pf<:UlT "PHI

="

jF: PF:ItH " ALFA="jA 190:IF A$(2E:)="+ "PF:IHT " TF.:A YU CCit·1\"" 200:l'l=>:::GO:;UB "I! ELTA" 210:IF A$(29)= "-"GO:;UB 300 220:X=X+Y:E=C:N= t-l+1 230:IF A$(28)= " + "BEEP 1: PRItH U:;HlG "ttttUtt"j~1j USWG "nUttnU

tt.

ttUtttt"j

i<

24ü:IF X<ZGOTG 2 00 250: IF A$(2::::)= "-"BEEP 1: PF.:ItH "LA::::T CCi~lV CALe:"; USIt-1G "UU.ttU UU"j>:: 260:W=(N-l )+(Z-W ).····C' :- l,D :BEEP 1: PF.: I ur "UTF.: A'(:::; FOF.: ::'::EHD

: " jUSI

t·Ki

"utt UU. UU"jl~ 270: PF.:ItH "KLAAR

=F.:EAD....·: :3TCiP"

:Et-lIi

ft"""ti"",

"11""""'"

IlIft'IIlii""

ToT TTn TTn TTn Tm TTn Tm m m m m m m m m m m m

52

Exercise 7: Chlorination of benzene in a tray-reactor Interface variable.

In this exercise the reactor system sketched in exercise 6 is used again. However, here the change in benzene concentration along the reactor is not t.ken into account (the reaction rate is assumed to

be k .c . c ev erywhere), but the interface area changes from

Bz, O C1

2

tray to tray. It is assumed that the following relation holds:

a. l. where constant .

_o

1 . . C 2,.10,1 ai constant

interface area of tray i from top [m2/(m3 liquid on tray)] apparatus constant [m2s/(kmOl.m3 liquid)]

molar flow of chlorine which enters tray i [kmol/s]

Other data needed are:

interface area top tray

volumetrie flow of liquid in reactor (assumed independent of conversion)

molar flow of chlorine leaving top tray 10-4 kmol/s

The program to calculate the benzene conversion in the reactor: "EXC7" is very similar to "EXC6", they differ slightly in the lines 5, 10, 40, 70, 100, 110, 140, 160, 170, 180 and 210 (see listings). No separate description of "EXC7" will be given.

The new memories are:

A(32) A(33) A(34) A(3s)

0

_v il_C1 2,out,1

0

C1 2 , o u t , i a c a l c

molar chlorine flow leaving tray i from top

area used in subroutines "phi" and "deltall

7a. Check the changes made.

The program "EXC7" will only run properly, that is with a variable inter-face area, if subroutine 400 is added, in which this area is calculated.

7b. Write subroutine 40 0 using the following 12 statements once. (These statements we re shuffled and regrouped in a more or less

"alphabetic" order:) 53 A(34) A(36) A(38) A(38) A(37) * A(38)

VT* A(35) * A(32) + A(34) GOSUB "DELTA" GOSUB "PHI" GOTO 400 GOTO 440 IF ABS (A (36) - A (35»

IA

(35» > .01 IF N= 0 LET A(35) LET A(37) RETURN where A(36) B/A(38) A(36) A(37) A(38)

ai = area tray 1 from top

constant in relation interface area vs. chlorine flow molar chlorine flow entering tray i from top

7c. How many trays are needed to obtain a benzene conversion of 20 %7 7d. Complete the following table (al 70 m2

Im ,

3 !;BZ 20%)

l'lV,BZ l'lC1 #trays 2,out, 1 m3/s kmolis 10- 4 10-4 10-4 10-3 10-3 10-4 10-3 10- 3

Can you expla1n the th1rd line of th1s table where (in comparison to the first line) 2x less trays are needed to convert a 10x larger benzene flow?

Listing program

...5:"E:~C7 " :USING ".ttttttttA " : GGTO 100 -.10: "PHI":F=.[(KJ D)/L:U=V(1+ I(Jn: A=L TU*A (35):S=1-AFF 21]:O=EXP F:P=l/ (1:Q=«J-P)/2: R=(I)+P ) /2: RETLIRt'l 30: "DELTA": C=(A F+EUQ)/(:::;Q+A FR):'./=LGF* (A FQ/R+S-EU/F:) /( AF+SQ/ R) ...40:','= ('·IT * A( 3 5 ) -CG+EG)/H 50: RETUF:t·l 60: "OUT":BEEP 1 : US

I

t·lG ". uutt ttA":PF:HlT "K =";K: PRItH

"11

"jD:PRHlT "K L ="jL: PF:IHT "C'" = 11;(; ...70:PRIHT "CBZO= "jH:PRINT

"A

TOP=" j B: PF:ItH "TAU = "; T: PRItH "j:: Et·m="jz: PRItH

"LIQ

=

";A

(32 ):

PPItH "CL20= RETURtl -.100: nWUT "1':="jK ,"D="jD, "KL=

"

; L,

11(::+:=11; G~ 11

CE:ZO=

11 ;H~ "A TOF' TPA\,=";

B,

"TAU="; T

-+110:

ItlPUT "CON'.,,' BZ Et·m="j

2, "

LW FLCir·j=": A (3 2) , " CL 2 FL CH,j OUT=";14(": 3) - -120:HlPUT "PF.: VA RfAB L E:::: ~' +O F:-­ .:':";~~$(2"7'" "EXC7" (MEM 456 / 5 7 ) 130: HjPUT "PF: TR AY BY TRA'ï'?+ GR-?: "jA$0::28 ) ...140: HWUT "ItHEF: F.CONSTANT?+ OR-?:"jA$( 29 ) 150:IF A$(27)="+

"GOSUB "OUT

"

~lE.I]:N=O:E=O: :":=0: I=H:J=H:A(3 4 )=A(33 ):A(35 )=B:GO:3UB "P Hl"

-+170:

BEEP 1: PPI

m

"PHI .= "jF -+180:IF A$(29)="+ "LET t'l=L/BD-1: PF:nIT "HL =" j t'l: F'RltH "ALFA="jA

1'30:

IF A$(28)="+ "PF:ItH " TPA Ylt COt·r·... " 200:~.j=>(:G03UI: "D ELTA" ...210 : IF A$(29 )="-"GCi::::UB 400 220:X=X+Y:E=C: N= tH

1

230:IF A$(28)= "+ "BEEP

1:

PRHIT

U:;:

;

nlG "ttttttU" ;

I-l

j USWG "uuuuu

u.

uttttU"jjO: 240:IF X<ZGOTO 2 00 250:IF A$(28)=" -"BEEP 1: PRItH "LAST Cm'l',,1 CALC:"; USWG "UU.UU ltlt"j ;': 260:W=(N-1)+(Z-N

)

.···

·

C":-W,:

BEEF'

1:

PRltH "UTR

i~'ï'::; FOP

>:

:E

t

·

.!IJ

: "jU:::HlCï "UU UU. UU";~'J

270 : PP l t-n "KLAAF' =F:EAD....·::::T':W..

---

..

--

-- .

IS B

-

1J88

--

I a 111

-.

--

_{- -}

_{- -}

_-

. ,

...

I

_--

-

I!

--Exercise 8: Bubble column reactor

At the Delft University of Technology sometimes the H2S con t e n t of a gas

stream is determined by leading it through a small bubble co l umn reactor

filled with a MeS04-s01ution. The following reaction takes pl a ce.

+ Me2+ _ MeS' + 2H+

(H2S)dissolved T

(t h e solubility of the sulfide is so low that this reac t i on can be assumed irreversible)

The pH of the solution is kept at a constant value (pH = 4) by ad d i ng 0.1 M

Na OH. From the flow of base and the gasflow the concentration of H2S in the

gas can be calculated. In this calculation it is assumed that all H2S present

in the gas is converted. This can be verified mathematically, by using the

following model for the reactor.

- the kinetic equation for the reaction is:

reactor

the liquid phase is an ideal semi-batch

- the gas phase is an ideal tubular reactor - between both phases only H2S is transferred - temperatures of gas and liquid are the

same everywhere. C 2+ where k' = 250 0 . m3/(kmOl.s) Me

L

= kt

t

G

the diffusion coefficients of D = 10-9 m2.s- 1

Moreover the following assumptions are made:

- the reactor is filled with 80 mi of a 0.1 M MeS04 solution. This solution

2+

is refreshed so often that the Me concentration in the bulk of the liquid does not vary with time

- the interface has a specific area of 1.5 m

2/(m 31iqUid)

- in the gas phase no pressure drop takes place as a result of mass transfer. - in the liquid phase the mass transfer coefficient, k , equals

2.10-5 _mIs

L 3

- the Henri-coefficient of H2S in the solution is 10 kPa. m/kmo l . - the temperature is 27 oe, the pressure 100 kPa (= 1 atm)

the gas constant R is 8,3 kPa. m3/(kmOl.K)

2+ H

57

8a. Make a sketch of the concentration profiles

of liquid at the battom of the rea ctor, fo r

2+ of H

2S and Me in the film the case that the nitregen-gas flows into the reactor with a rate of 100 mi/min and has a concentration of 10 ppm{vol) of H

2S.

Bb, Set up the molar balance of H

2S over a "slice" of the tubular reactor perpendic-ular to the interface. The mass flux transferred should be described with an expression that includes k

L, the specific area a and the enhancement factorE.

Bc. Determine how much of the incoming H

2S leaves the reactor at the conditions mentioned in Ba.

Hd. In Sc i t will be found that the H

2S is absorbed for more than 95% {there the inlet concentration of H

2S was only 10 ppm).Can this absorption also be reached if the concentration of H

2S raises to 1 (vol)%?

Be . Which of the following measures can be taken to obtain an absorption of 95% or more if the.inlet concentration of the reactor is 1 (vol)% ?: - increase stirring speed until k

L becomes 5x higher. (the specific surface area does not change)

- increase temperature to 40

°C.

The reaction rate constant k' increases then a factor of 10, the Henri-coefficient a factor 2 and bath

diffusion coefficients a factor 1.05. The mass transfer coefficient k L does not change.

- increase the MeS0

...

...

...

... ...

.... ....

...

...

...

...

...

... ...

_...

nUS:SSHhH

...

....

....

...

...

...

...

...

...

...

produetion

Exercise 9: Autothermic fixed bed reactor

In case of a stirred tank reactor the heat balance terms can oe

presented graphically (f i g. 1). The sigmoidal curve shows the heat production due to the ch emi c a l reaction and the other curve the com-bined effect of heat loss by exchange, sensible heat etc.

•

T

fig. 1.

The intersection points are solutions of the total heat balance. SI and S2 are stable solutions, S3 is unstable and has the practical meaning

of initiation temperature. A reactor with this behaviour is cal led

autothermic. The principal reason for this behaviour is the effect of feedback: the mass entering the reactor attains immediately the reaction temperature. It "feels" the effect of heat of reaction.

Such a stabilized situation can also be obtained in a catalytic fixed bed reactor. There are several ways to realize the feedback. It can be done by preheating the feed in an external heat exchanger with the reactor outlet stream.

Another way of making the reactor autothermic is to design the tubular reactor itself as a heat exchanger. The feed cools the reactor tubes

on the outside before entering the mass of catalyst inside (see fig. 2) .

Such a solution is, for instanee, realized in same ammonia synthesis

reactors. An outline of the temperature gradient in such a reactor is given in the right part of fig. 2, where

61

z=o

z

=1

~

0

z

1

T

-fig. 2.

Tin temperature of reactor feed

Ta temperature at the inlet of the catalyst bed T = temperature at the reactor outlet

out

T temperature in catalyst bed TI _{temperature in preheat section}

To calculate this behaviour not only the molar balance of the key com-ponent and the heat balance in the catalyst bed are needed, but also the heat balance of the heat exchanger part of the reactor has to be taken into account.

The purpose of this exercise is to calculate the temperature and con-vers ion profile in the tube side and the temperature profile in the shell side of such a reactor and to get an impression of the stability of the system.

The reaction to be considered is exothermal, first order and irreversible so

E

-rA ka exp(-

:;>

cA where -1

The reactor contain s n tubes with radius Rand le ng t h L. The heat exc hangi ng capa c ity of the tubes is

8 _HTU1 ₀ UA

v pCp

0.5

while 8 is thought inde p e nd e nt of temperature and mi xt ur e compos i t i o n .

In this formula A is the total ex ch a ngi ng area of the n reactor tubes .

Th e space time T

s of the reactor (tube side) is 100 s and the adiaba tic temperat ure ri se

250 K

9a. You are asked to set up both heat balances and the molar balance

of A. Make them dimensionless by introducing

= x z L ~ cAD- cA cAD T -T. 0 ~n lÏT a d T' -T. 0'= ~n lÏT a d

, the dime ns ionl ess reactor length

I the dime nsionl ess conversion

, the dimensionless temperature, tube side

, the dimensionless temperature, shell si de

Write down the boundary conditions necessary to solve these bal a n ces.

The problem in solving the equations is the fact that at

b

ot h

sides of the reactor boundary values are missing, just as was for instance the case in the exercises 4 and 5. By now you are familiar with the proc edure

to treat such a problem: make a first guess of the missing boundary

va l ue at one side of the reactor, then calculate the reactor from that side to the other and check if the boundary condition at the other side

is met. If not, adjust your first guess and repeat the procedure etc.

However, you cannot use the garden-hose methode. Though in principl e

only suitable to linear problems, it does weil in a lot of non-li near

problems if the non-linearity is not tOD high (that is how we managed it in exercises 4 and 5). But here our equations ar e extremely non-linear

due to the exponential dependency of the reaction rate on the temperature.

Next to that they are very sensitive to small disturbances just because

we make our calculations in the stability/instability regions.

To master the non-linearities we would have to use the rather slow bisect i on method to ad j us t our guesses. To prevent the program to run

63

away due to some odd values we also would have to build in some safety routines which rapidly increase the number of calculation steps in that case and restart the procedure.

Running sueh an "explosion proof" program would be time consuming. Though you will find some details for sueh a program in the answer of this

exercise, we shall calculate the profiles based on another idea.

9b. Instead of starting with a given feed temp e r a t u r e (Tin)' choose a fixed temperature at the in let of the catalyst bed (T

O)' Because now at z = 0 all boundary values are known, you can calculate

the profiles straightforward from z

=

0 to z

=

1. That will give you the temperature the feedstream should have in order to make your choice of T

Opossible. Or in other words, if you had chosen that feed temperature then the calculated profiles would be valid.

Calculate in this way both the temperature profiles and the con-version profile using T

O= 540 K and 580 K.

Use the program "EXC9" given at the end of this exercise. First read the program description.

Start with T

O= 580 K and choose the number of steps for the

Runge-Kutta procedure 20 , 10 and 5 respectively. Make a graph of the temperature T vs . z anu the sensiti vity of the calculations wi l l be demonstrated. Use only 20 steps to calculate the profiles for

T

O= 540 K.

Present your results as weil tabulated as in graphs showing the profiles of ~, Tand T' for ten points of the dimensionless reactorlength.

To investigate the stability, thus the autothermic character of the system we will have to look for multiple steady states as is the case in the stirred tank reactor, mentioned before. If there are multiple steady states there has to be more than one set of temperature and conversion profiles for one value of feedtemperature Tin' Or in other words, for one value of the feedtemperature (Tin) there shoul d be more than one value of T

Opossible.

9c. Use the same program "EXC9" and calculate for the set of bed entrance temperatures T

O= 440, 480, 520 ... 640 K the value of Tin needed for each entrance temperature.Choose the number of steps 20, but to save time, do not calculate the profiles (so choose #ST/PR 20 too). Make a graph of Tin versus T

64

forget the points you have already got from exc. 9b.).

Is there any multiplicity?

Remark:

Using Tin in 9b. and 9c. to define the dimensionless temperature 8 is now just as arbitrary as any other reference temperature; the only advantage is that the extension of the program "EXC9" to a program which can calculate the profiles starting with a fixed feed temperature

65

The program "EXC9"

The program "EXC9" is meant to calculate the temperature and conversion profiles in an autothermic tubular reactor. It solves the set of three differential equations which describe the molar balance of A, the heat balance in the reaction tubes and the heat balance in the preheat

section and is based upon the Runge-Kutta method.

Before you can run the program you have to comp lete it, as fa r as it concerns the se t of dimensionless different ial equations and the initial

values .

The progr am expects you to key in respecti vely the va l ue s of #STEPS #ST/PR DTADIAB EA KOTS BETA TIN TO

, th e number of steps int o which you like to divi de

the reactor

, the number of steps af ter which (intermediatel results will be printed

the adiabati c tempe r a tu re rise , the activation energy

, kOT s

, the he a t exchange capacity

, the feed temperature Tin (in Kl

, the temperature at the entrance of the catalys t

bed (i n K)

Af ter #ST/PR steps the program returns the values of ~, T (i n Kl and T' (in K).

The flowscheme of the program is given in fig. 3, the listing in fig. 4. The numbe r s in the flowscheme correspond with the line numbers of

the program.

line 5

lines 10-14

lin e s 20-42

states the name of the program, defines the print format and jumps over the subroutines ItF311 and "RK3"

The subroutine "F3" , where the three first-order different1al equations are stated on the lines 11, 12 and 13. They have to be wr1tten as

F f(X, T, U), the molar balance

G g(X, T, U, F), the tubeside heat balance

H h(X, T, U, F, Gl, the shellside heat balance

~ d0 d0'

where F = dz' G= ~, H= ~,X ~,T = 0 and U 0' YOU HAVE TO COMPLETE THESE LINES: