adv math project

8/12/2019 Adv Math Project

1/19

Concentration Profile in

a Tubular Reactor

Pinili, Jeanne Kamille E.

Rejuso, Mark Arthur V.


2/19

Description of the Problem

A catalytic tubular reactor is shown below. Adilute solution of solute A in a solvent S is in a

fully developed, laminar flow in the region z < 0.

When it encounters the catalytic wall in theregion 0 z L, solute A is instantaneously andirreversibly rearranged to an isomer B. Write

the diffusion equation appropriate for this

problem, and find the solution for shortdistances into the reactor. Assume that the flow

is isothermal and neglect the presence of B.


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Figure 1. Boundary Conditionsfor a tubular reactor


4/19


The flowing liquid will always be nearly pure

solvent S and neither A nor B will be presentin any large quantity. The product D

ASis

considered constant and the diffusion of A in

S is described by the steady-state version of

the equation of continuity in terms of molar

concentration. The presence of a smallamount of the reaction product B is ignored.


5/19

Figure 2. A control volume for tubular

coordinates of component A

Development of the Mathematical Model(Continuity of A)


6/19

the velocity field will be described as

the volume of the differential control volume is

the number of moles of component A in the control volume is

the rate of change of mole or accumulation the control volume is

(1)

(2)

(3)

(4)



7/19

the net flow in the r direction is

, 1

+

the net flow in the direction is

,

1

the net flow in the z direction is

,

combining ( 3 ) to (6), the continuity equation simplifies to

+

+

+

+

=0

dividing by dV and rearranging the r components of the velocity

+

( )

+

()

+()

=0

(4)

(5)

(6)

(7)



8/19

continuity equation(7) expressed in molar

flux components , where =

+

(

)

+

(

)

+

()

=0

the components of the molar flux are given for a binary system by

Ficks law of diffusion in the form

equation (8) combined with equations (9) through (11) simplifies to

+

( )

+

()

+

()

=

+

2

22 +

2

22

(8)

(9)

(10)

(11)

(12)

Development of the Mathematical Model


9/19

From equation 12,

The diffusion of A in S can be described as steadystate and it is one -dimensional, the flowing liquid moves along the z-axis.

Continuity of A:

We can make an assumption that axial diffusion can be neglected with respect

to axial convection, and therefore delete the second term

2

21

z

C

r

Cr

rrz

C AAAz ASD


+

( )

+

()

+

()

=

+

222

+

222

r

Cr

rrz

C AAz

1ASD


10/19

Development of the Mathematical Model(Equation of Motion)

Equation of Motion for a Newtonian Fluid with Constant and (App. B.6)

Cylindrical coordinates (r,,z):

Equation of Continuity

zzzzz

zzz

rz

zrrr

rrzzrrtg

P

2

2

2

2

2

11

011

zr

zv

rr

rrt

r

rr

r

r

z

rr

r

r v

rzrr

rrrrrzrrtg

P

22

2

2

2

2

2 211

g

P

rr

zr

v

rzrr

rrrrrzrrt 22

2

2

2

2

2111


11/19

Considering a steady-state, laminar flow of a fluid of constant density in a

tubular reactor of length L and radius R, the equation of motion will be

simplified. For small flow rates, the viscous forces will prevent continued

acceleration of the liquid through the tube so that will become independent of

z in a short distance. Therefore, it is reasonable to postulate that

there is no radial flow , no tangential flow , and further that .Consequently, we can discard many terms from the previous equations, leaving

equation of continuity (1)

requation of motion (2)

equation of motion (3)

zequation of motion (4)The first equation indicates that depends only on r; hence the partial

derivatives in the second term on the right side of equation 4 can be replaced

by ordinary derivatives . Equations 2 and 3 show that Pis a function of z

alone, and the partial derivative in the first term of equation 4 may be replaced

by an ordinary derivative.


0zz

dr

dz

rzz

0r 0 zpp

r

P

0

P

0

rrrrz

1

0 z

P


12/19


The equation now becomes:

Motion: (5)

The only way that we can have a function of r plus a function of z equal to zero

is for each term individually to be a constant, C0so that equation 5 reduces to

The P equation can be easily integrated. The vzequation can be integrated by

double integration. This gives

Boundary Conditions:1. z=0, P=P0

2. z=L, P=PL

3. r=R. vz=0

4. r=0, vz=finite

dr

dr

dr

d

rdz

dz

1

0 P

dz

d

Cdr

d

rdr

d

r

z P

0

1

10 CzCP

32

20 ln

4

CrCrC

z


13/19

Applying B.C.1

Motion:

Applying B.C.2 Velocity distribution for laminar,

Incompressible flow of a Newtonian

fluid in a tube is parabolic. The

maximum velocity ,occurs at r=0,

and is

Applying B.C.3 and B.C.4

Since at r=0 (middle of the pipe), the Substituting this result to the

Velocity profile must be finite. C2must be equation

0.


00 PzCP

0C

L

PP Lo

L

zPPPP L

00

10 CP

00

PLCPL

L

RLz

4

2

max,

PP0

L

R

C 4

2

3

L0 PP

22

14 R

r

L

RLz

PP0

2

max, 1 R

r

r zz

r

Cr

rrz

C

R

r AAS

Az

11

2

max, D


14/19

The new P.D.E.

Boundary Conditions:

1. z=0, CA=CA0

2. r=R. CA=0

3. r=0, CA=finite


r

Cr

rrz

C

R

r AAS

Az

11

2

max, D


15/19

For short distances z into the reactor, the concentration CAdiffers from

CA0only near the wall, where the velocity profile is practically linear. Hence we

can introduce the variable , neglect curvature terms, and replace B.C.3

by a fictitious boundary condition at .

The reformulated problem statement is then

with the boundary conditions1. Z=0, CA = CA0

2. y=0, CA = 03. ,CA= CA0

Solution of the Mathematical Model

2

2

max,2y

C

z

C

R

y AAz

AS

D

rRy y

y


16/19

This problem can be solved by the method of combination of independent

variables by seeking a solution of the form , where .

One thus obtains the ordinary differential equation .

The appendix C.1-9 gives the solution to this general form of ODE:

as

where z is just a dummy variable of integration.

The system that we have in this problem is analogously

fC

C

A

A 0

3

12

max,

9

2

z

R

R

y z

ASD

03 '2'' ff


03 22

2

dx

dyx

dx

yd

x

CdzzCy0

2

3

1 exp

03 2

2

2

d

df

d

fd

1

00

f

f


17/19


where

Applying the boundary condition to get

Applying the boundary condition to get

fC

C

A

A 0

3

12

max,

9

2

z

R

R

y z

ASD

00 f

0

0 2

3

1 exp0 CdzzC

2

2

0

00

C

C

1f

0

3

1 exp1 dzzC

3

4

exp

1

0

31

dzzC


18/19

When everything is put back together,

34

3

0

30

3

0

0

exp

exp

exp

d

d

d

C

Cf

A

A



19/19

Interpretation of the Solution

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