PFR reactor, concentration profile

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dRic2

When I set the differential equation (mass transfer) to find the the concentration profile of the i-spice I also need boundary condition to solve it. Inside the pipe (along its axis of symmetry) the concentration has a maximum (first derivate has to be 0), while on the internal surface... should I put the concentration egual to 0 (like i think) or should I force the function to have a minimum?
 
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When I set the differential equation (mass transfer) to find the the concentration profile of the i-spice I also need boundary condition to solve it. Inside the pipe (along its axis of symmetry) the concentration has a maximum (first derivate has to be 0), while on the internal surface... should I put the concentration egual to 0 (like i think) or should I force the function to have a minimum?
Did you really expect us to figure out what you are describing here.? Please provide a precise statement of the problem or at least show the model equations that you are trying to solve.
 
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dRic2

Sorry, I thought it was a "general" question, I don't have a problem to solve, I was just thinking about it.

here are my assmption:

A PFR reactor = A pipe.

cylindrical coordinates.

Reagent "i" comes in and flow through the pipe.

Reaction ##i→ products## takes place. (equilibrium not reached)

Turbulent flow ## → \frac {\partial v} {\partial r} = 0##

Also ## \frac {\partial v} {\partial z} = 0##

My thinking:

Since the velocity, even if it can be considered as constant in the section of the pipe, has to be 0 when ##r = R_{pipe}## (boundary layer) so technically the time of residence of ##i## at the internal surface of the pipe has to be infinite ##→## the concentration of i at ##r=R_{pipe}## has to be zero.

My question:

Is it right or should I say, instead, that the concentration of i at ##r=R_{pipe}## is minumum?
 
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The PFR model is only an idealization, featuring a square wave residence time distribution, so it doesn't really pay to worry about what's happening at the wall. However, even in real reactors where the velocity is profiled and is zero at the wall, the system doesn't behave as if the reaction has an infinite time to occur for the material near the wall. There is still diffusion of chemical species that allows reactants to diffuse from the bulk toward the wall and products to diffuse from the wall to the bulk. So the species concentrations at the wall are not what they would be if there were no diffusion. Typically, in modeling laminar flow reactors, we include radial diffusion.
 
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dRic2

So, let's say I write the equation of mass transfer for i specie like this:
$$ v_z \frac {\partial c_i} {\partial z} = -D_{diff} \left[ \frac 1 r \frac {\partial} {\partial r} \left(r \frac {\partial c_i} {\partial r} \right) + \frac {\partial ^2 c_i} {\partial z^2}\right] + kc_i$$

what boundary condition should I apply?

I thought:

## lim_{z→∞} c_i(r, z) = 0 ##
## \frac {\partial c_i} {\partial r} |_0 = 0 ##
## \frac {\partial c_i} {\partial r} |_R = 0 ## ?? Not sure at all

and maybe the fourth could be:

## v_zc_i^0 = v_zc_i(r, 0) - D_{diff} \frac {\partial c(0, z)} {\partial z} ## (comes from ##\dot m_{in} = \dot m_{out}##)
 
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So, let's say I write the equation of mass transfer for i specie like this:
$$ v_z \frac {\partial c_i} {\partial z} = -D_{diff} \left[ \frac 1 r \frac {\partial} {\partial r} \left(r \frac {\partial c_i} {\partial r} \right) + \frac {\partial ^2 c_i} {\partial z^2}\right] + kc_i$$
Both signs on the right hand side of this equation are incorrect, and need to be flipped. Also, the diffusion in the axial direction is virtually always negligible compared to the radial diffusion (and is virtually always neglected).

what boundary condition should I apply?

I thought:

## lim_{z→∞} c_i(r, z) = 0 ##
This boundary condition is unnecessary if diffusion in the axial direction is neglected.
## \frac {\partial c_i} {\partial r} |_0 = 0 ##
## \frac {\partial c_i} {\partial r} |_R = 0 ## ?? Not sure at all
Both are correct.
and maybe the fourth could be:

## v_zc_i^0 = v_zc_i(r, 0) - D_{diff} \frac {\partial c(0, z)} {\partial z} ## (comes from ##\dot m_{in} = \dot m_{out}##)
If axial diffusion is neglected, this becomes: ## c_i(r, 0)=c_i^0##. Usually, the inlet concentration is assumed to be radially uniform.
 
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dRic2

Thank you very much!

This boundary condition is unnecessary if diffusion in the axial direction is neglected.
but if I do not intend to neglect it, would it be correct?
 
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Thank you very much!



but if I do not intend to neglect it, would it be correct?
But then it has to be solved in conjunction with what is happening upstream and downstream (i.e., be matched up with), rather than just standing alone. At the upstream end, you still use ##c=c_i##. The downstream end is problematic, and depends on the details of what is happening after the pipe exit (unless you really do have an infinite pipe).
 

BvU

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Fogler discusses these boundary conditions in a reasonable textbook format
 
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Fogler discusses these boundary conditions in a reasonable textbook format
Over 50 years ago, Scott Fogler was on my preliminary exam committee at the Univ. of Mich. He was a new young professor and I was a 1st year graduate student. He is the one remaining faculty member from back then who is still active in the department.
 
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dRic2

thank you very much
 
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thank you very much
Considering your boundary conditions on r, c is not going to be a function of r. So your equation reduces to an ODE. If axial diffusion is neglected, it reduces to a first order ODE which can immediately be integrated analytically.
 
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dRic2

Considering your boundary conditions on r, c is not going to be a function of r.
Why?

Anyway i think it is still possible to integrate it analytically using the separation of variables without any further assumption
 
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Why?

Anyway i think it is still possible to integrate it analytically using the separation of variables without any further assumption
Does c not being a function of r satisfy the differential equation and boundary conditions?
 
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dRic2

sorry if you read the previous comment. I take it back. You're right, I need to think about it
 

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