4. TUBULAR REACTORS WITH LAMINAR FLOW · PDF filetubular reactors and how to measure their...
Transcript of 4. TUBULAR REACTORS WITH LAMINAR FLOW · PDF filetubular reactors and how to measure their...
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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:
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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