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4. TUBULAR REACTORS WITH LAMINAR FLOW

Tubular reactors in which homogeneous reactions are conducted can be empty or packed conduits

of various cross-sectional shape. Pipes, i.e tubular vessels of cylindrical shape, dominate. The

flow can be turbulent or laminar. The questions arise as to how to interpret the performance of

tubular reactors and how to measure their departure from plug flow behavior.

We will start by considering a cylindrical pipe with fully developed laminar flow. For a

Newtonian fluid the velocity profile is given by

u = 2u 1r

R

2

(1)

where u =umax2

is the mean velocity. By making a balance on species A, which may be a

reactant or a tracer, we arrive at the following equation:

D2CAz2

uCAz

+D

r rrCAr

rA =

CAt

(2)

For an exercise this equation should be derived by making a balance on an annular cylindrical

region of length z , inner radius r and outer radius r . We should render this equation

dimensionless by defining:

=z

L; =

r

R; =

t

t ;c =

CACAo

where L is the pipe length of interest, R is the pipe radius, t = L / u is the mean residence time,CAo is some reference concentration. Let us assume an n-th order rate form.

The above equation (2) now reads:

D

u L

2c2 2 1

2( )c

+D

u R

L

R

1 c

kCA0

n 1t c n =c

(2a)

We define:

Pea =u L

D=

L2 /D

L /u =

characteristic diffusion time axial( )process time characteristic convection time( )

(3)

= axial Peclet number.

2

Per =u R

D= radial Peclet number (4)

u R2

DL=

R2 / DL /u

=characteristic radial diffusion time

characteristic convection time=

Per xR

L

= radial Peclet number( ) x pipe aspect ratio( )

Dan = k C Aon 1 t =

L / u 1

k CAon 1

=process time

characteristic reaction time(5)

In terms of the above dimensionless groups which involve: characteristic reaction time,

characteristic diffusion time, characteristic convection or process time and the aspect ratio, we

can rewrite the above equations as:

Pea =u L

D=

u dtD

L

dt= Pe

L

dt

(3a)

Per =u R

D=

u dtD

R

dt=

Pe

2(4a)

1

Pe

dtL

1

Pe a6 7 4 8 4 2c2 2 1

2( )c

+1

Pe

4L

dt

1

Per

L

R

6 7 4 8 4

1 c

Danc

n=

c

1

Pe

dtL

1

Pea6 7 4 8 4 2c2 2 1

2( )c

+1

Pe

4L

dt

1

Per

L

R

6 7 4 8 4

1 c

Danc

n=

c(2b)

where Pe =u dtD

= Re Sc is the Peclet number for the flow. In terms of Pea and Per we have:

1

Pea

2c2 2 1

2( )c

+1

Per

L

R

1 c

Danc

n=

c(2c)

The above equation can be simplified if we deal with a steady state reactor problem for whichc= 0 , or if we deal with a nonreactive tracer dynamic response, for which Dan = 0 . In either

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event we need to solve a cumbersome partial differential equation and need two boundary

conditions in axial coordinate and two in radial coordinate .

4.1 Segregated Flow Model

The question arises whether we really need to solve the above equation (2c) numerically at all

times or whether we can find simple solutions which are valid under certain conditions. Since

reactant or tracer dimensionless concentration, c, is a smooth function, based on physical

arguments, the value of the function and its derivatives is of similar order of magnitude except

perhaps at a finite number of points. It can be shown then that when Pea >> 1 and

PerR

L>> 1 the first and third term of eq (2c) can be neglected. For a steady state reactor

problem this results in the following equation:

2 1 2( )c

Dancn= 0 (6)

= 0 ; c =1 (7)

The exit concentration c =1,( ) is a function of radial position, i.e of the stream line on

which the reactant travels. The overall average exit concentration, or mixing cup concentration,

is obtained as

c ex =C AexitCAo

=

2 r u CA z = L,r( ) dro

R

2 ru CAo dro

R =

4 ru 1r

R

2

CA L,r( ) dr

o

R

R2u CAo

where u = 2 u 1r

R

2

. Using dimensionless variables we get:

c ex = 4 12( ) c 1,( ) d

o

1

(8)

For a first order reaction (n = 1) we readily find from eqs. (6) and (7) that

c = 1,( ) = e

Da

2 1 2( )(9)

Using eqn (9) in eqn (8) we obtain the exit mixing cup concentration:

4

c ex = 4 12( ) e

Da

2 1 2( ) do

1

(10)

Change variables to 1

1 2= u ;

2 d

1 2( )2 = du to get

c ex = 41 2( ) 3

21

eDa

2u

du = 2e

Da

2u

u 31

du = 2 E3Da

2

(10a)

where En x( ) =e xu

un1

du is the n-th exponential integral.

In contrast, the cross-sectional average concentration is:

c ex = 2 c 1,( )o

1

d = 2 e

Da

2 1 2( )

o

1

d

=e

Da

2u

u21

du = E2Da

2

(11)

c ex = E2Da

2

= e

Da2 Da

2E1

Da

2

(11a)

Thus, if we measure by an instrument the cross-sectional average concentration, and try to infer

reactant conversion from it, our results may be in error since conversion is only obtainable from

mixing cup (flow averaged) concentration and clearly there is a discrepancy between equation

(11a) and eq (10a).. You should examine the deviation of eq (11a) compared to eq (10a) and plot

it as a function of Da . The needed exponential integral are tabulated by Abramowitz and Stegun

(Handbook of Mathematical Functions, Dover Publ. 1964).

Using the following relationship among exponential integrals

En+1 z( ) =1

ne z z En z( )[ ] (12)

we get the following expression for conversion from eqn (10a)

1 xA = c ex = 1Da

2

e

Da

2 +Da2

4E1

Da

2

(10b)

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where E1 x( ) =e xu

u1

du =e t

tx

dt

We should realize immediately, upon reflection, that by eliminating the diffusion terms in eq (2c)

and in arriving at eq (6-7) we deal with the segregated flow system. Indeed, in absence of

diffusional effects there is no mixing among various stream lines. The reactant that enters on a

particular stream line exits on the same stream line, i.e at the same radial position , and hence

is surrounded by elements of the same age as its own at all times during its journey through the

reactor. Every stream line has a different residence time. The shortest residence time is

experienced by the elements on the center line t / 2( ), the mean residence time t ( ) is experienced

by the fluid traveling on the stream line that has the mean velocity, i.e, at = 1/ 2 = 0.707 ,

while infinite residence time is experienced by the elements on the stream line at the wall ( = 1).

However, since the stream line at the wall receives infinitesimal amount of new fluid the mean

residence time for the system exists and is t . We recall that for the segregated flow condition and

first order reaction the exit concentration can be written as:

c ex = eDa E ( )

o

d = E s( ) s = Da (13)

This means that we have found the Laplace transform of the exit age density function for fully

developed laminar flow of Newtonian fluid in a pipe

L E ( ){ } = E s( ) = 2 E3s

2

= 1

s

2

e

s2 +

s2

4E1

s

2

(14)

However, even the extensive transform pairs of Bateman (Tables of Integral Transform Vol. 1) do

not show this transform.

We can, however, derive the RTD or the F function for laminar flow readily based on physical

arguments. Let us imagine that we have switched from white to red fluid at the inlet at t = 0.

Red fluid will appear at the outlet at z = L only starting at time t /2 . The fraction of the

outflow that is younger than t is given by the fraction of the fluid which is red. This is obtained

by integrating the flow from the center stream line, where the red fluid is present at the outlet

from time t / 2 , to the stream line at position r at which red fluid just at the outlet at time t and

by dividing this flow rate by the total flow rate.

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Recall that

t =L

u=

L

2u 1r

R

2

By definition the F curve is given by:

F t( ) =

2 r' udr'o

r

Q=

4 r' u 1r'

R

2

dr'

o

r

R2u (15)

F t( ) = 4 ' 1 '2( )d 'o

, for tt

2 (15a)

where t =t

2 1 2( )or = 1

t

2t

1/ 2

. Hence,

F t( ) = 42

2

4

4

= 2 2 1

2

2

for t

t

2(16)

F t( ) = 2 1t

2t

11

t

2t2

= 1

t

2t

1 +

t

2t

= 1

t 2

4t2

F t( ) =0

1t 2

4t2

t 1000 and PerR

L> 85 . Recall

that Pea = PeL

dt, Per =

Pe

2.

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4.1.1 Use of Seg