5.2.2 - Mixing in a Pipeline

7/25/2019 5.2.2 - Mixing in a Pipeline

1/4

1

5.2 Mixing in a Pipeline

When instead of a reactor problem we deal with problems of mixing of one material that flows

after another as both are being pumped through the same pipe line, we use the dispersion or

Taylor diffusion model (depending whether the flow is turbulent or laminar) to describe thespreading of the material in the axial direction.

The governing equations in the coordinate system that moves at the mean velocity of flow can be

written according to the diffusion equation:

c

' =

1

Pe

2c

2 (122)

where Pe =u L

Dapp

and Dapp is either the dispersion coefficient, Ez , or Taylor diffusivity.

If we try to describe an impulse response in a very long pipe, then all the material was

concentrated originally at the plane = 0 . At the same time

' = 0; c = 0 except at = 0 (123a)

' > 0; c 0 (123b)

no material can reach axial position of infinity at a finite time which is expressed by the second

condition above. Finally, since the mass mtof the tracer material injected at time zero at the

plane at zero axial position must be conserved we have the last condition:

c d =mt

A CoL

(124)

where A = R2 is the cross-sectional area of the system, L is the length with respect to whichwe have dimensionalized the axial coordinate, Co is a normalizing concentration.

The solution to the above problem can be obtained by either

a) taking the Laplace transform of the PDE, and B.C., solving the resulting ODE and inverting


2/4

2

the solution, or

b) by similarity transform i.e. by introducing new variables

=

'and u = c '

The solution is

c ,'( )= mtR2LCo

1

2 '

Pe

e

2

4'

Pe (125)

If we consider c Co = C actual concentration and turn back to the fixed coordinate system

=z

L= + ; = '=

t

t=

u t

L (126)

C ,( )= mtR2L

Pe

4e

Pe ( )2

4 (127)

C z,t( )=mt

R2L

LPe

4u tePe zu t( )2

4u Lt (128)

or taking Pe =u L

Dapp

C z,t( )=mt

R2

1

4Dappte zu t( )

2

4Dappt (128a)

The impulse response at z = L, G (t) (not necessarily the age density function because the

system may be open) is now given by

G t( ) =R2u C

mt=

u2

4Dappte

Lu t( )2

4Dappt (129)

and

G ( ) = t G t( ) =L2

4DappL

u

e

L2 1( )2

4DappL

u

(130a)


3/4

3

G ( ) =Pe

4e

Pe 1( )2

4 (130)

Pe =u L

Dapp

L

Those that had probability theory will recall that the Gaussian density function is given by:

f ( ) =1

D 2e

( )2

2D2

(131)

It can readily be shown that in long beds G ( ) given by eq (130) is small for all exceptthose in the vicinity of = 1 . In the first approximation one can represent the G ( ) by a

Gaussian density function

G ( ) 1

D 2e 1( )2

2 D2

(132)

where

D2 =

2

Pe=

2Dapp

u L (133)

This shows that given the flow velocity profile u, then the mean velocity u and the apparent

axial dispersion coefficient

u 2R2

48D is fixed, if Taylor diffusion model is applicable.

The relative spread of the curve around the mean then is reduced as the length L between the

injection and monitoring station is increased.

The absolute spread, however,

2 =D

2t

2 =2

Pe

L2

u2=

2DappL

u3 (134)

increases as the length of the conduit is increased. If the dispersion model holds, the increase ofthe spread measured by = 2 is proportional toL 1./2.

Step response is now given by:

Fs ( ) =1

D 2e

1( )2

2 D2

d

(135)


4/4

4

x = 1

2 D, d= 2D dx, =1 + 2D x

Fs

( ) =1

e

x 2dx =

1

ex 2

dx + e x2

dxo

1

2D

o

1

2D

=1

2

2

ex 2

dx +2

ex2

dx

o

1

2D

o

=1

21+ erf

12D

=1

21 erf

1 2D

=1

2erfc

12D

with D =

2

Pe

which is the normal distribution.

Fs ( ) =

1

21 erf

1 2D

for< 1

1

21+ erf

1

2D

for> 1

(136)

This formula can be used to

a) determine after switching at = 0 at positionz=0 from fluidA to fluidB what fraction of

fluidB appears at the outflow and how long a time it takes until the outflow contains 95% or

more of fluidB.

b) determine the length over which one has a mixture between p % of fluidA and p % of fluidB.

c) other problems of similar type.

5.2.2 - Mixing in a Pipeline

Documents

Transcript of 5.2.2 - Mixing in a Pipeline