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Modeling a tubular reactor

  1. Oct 26, 2015 #1
    Good evening, PF! I am supposed to model the following system. I will be using the same notation as in BSL.
    reactor.png
    Fluid A enters the reaction zone at z = 0 at a concentration CA0. A reacts to form B in the first order reaction A → B at a rate of [itex]R_A = k_1''' C_A[/itex]. We assume the whole mixture to be incompressible, with constant global C and ρ, so the mass average velocity equals the molar average velocity. Diffusion effects can not be neglected. I have to consider two situations, plug flow and laminar flow. However, the velocity profile in laminar flow only depends on the radial direction, so I only performed one mass balance for species A. The shell mass balance for species A in the volume element πR2Δz is
    [tex]\pi R^2 N_{Az} |_z - \pi R^2 N_{Az} |_{z+\Delta z} - \pi R^2 \Delta z R_A = 0[/tex]
    Then, dividing by the whole volume and taking the limit as Δz approaches zero
    [tex]\frac{d N_{Az}}{dz} + R_A = 0[/tex]
    The molar flux in the z direction is given by
    [tex]N_{Az} = -D_{AB} \frac{dC_A}{dz} + C_A v_z[/tex]
    Substituting this expression in the first differential equation, we arrive at the equation which must be solved to find the concentration profile
    [tex]D_{AB} \frac{d^2 C_A}{dz^2} - v_z \frac{dC_A}{dz} - k_1''' C_A = 0[/tex]
    I solved this equation by proposing [itex]C_A = c_1 e^{m_1 z} + c_2 e^{m_2 z}[/itex] as a solution and using the auxiliary equation of the form
    [tex]D_{AB} m^2 - v_z m - k_1''' = 0[/tex]
    Solving the above equation, I got
    [tex]m_1 = \frac{v_z + \sqrt{v_z^2 + 4D_{AB} k_1'''}}{D_{AB}^2}[/tex]
    [tex]m_2 = \frac{v_z - \sqrt{v_z^2 + 4D_{AB} k_1'''}}{D_{AB}^2}[/tex]
    Using the following boundary conditions in order to find the constants of integration
    B.C. 1: z = 0; CA = CA0
    B.C. 2: z = L; CA = CA1
    I arrived at the concentration profile of A
    [tex]C_A = \left( \frac{C_{A0} e^{m_2 L} - C_{A1}}{e^{m_2 L} - e^{m_1 L}} \right) e^{m_1 z} + \left( \frac{C_{A1} - C_{A0} e^{m_1 L}}{e^{m_2 L} - e^{m_1 L}} \right) e^{m_2 z}[/tex]
    Now, in the plug flow case, vz is a constant and we have nothing to worry about when using the model. In the laminar flow case, vz is a function of r, so, does that mean that CA becomes a function of r and z? We could also use the average velocity <vz>, which is also a constant.
    Also, I have questions about the relationship between NA and NB. Stoichiometry tells us that [itex]N_A = - N_B[/itex], however, given the geometric nature of molar fluxes, that would mean that B flows in the negative z direction, and this is not the case. That would also mean [itex]x_A(N_A + N_B) = 0[/itex], thus no convective term in the modified form of Fick's first law [itex]N_{Az} = -D_{AB} \frac{dC_A}{dz} + x_A(N_A + N_B)[/itex]. Right now, my guess is that [itex]N_A = N_B[/itex], but I'm not sure. I'm still struggling to comprehend the physical interpretation of the mathematics of mass transfer.

    This is it, for the moment. Thanks in advance for any kind of input!
     
    Last edited: Oct 26, 2015
  2. jcsd
  3. Oct 27, 2015 #2

    BvU

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    Hi Mex,
    Just reacting to your last paragraph: Stoechiometry doesn't tell us ##N_A = - N_B## but ##N_A + N_B = N_{A, z = 0}## so ##dN_A = - dN_B##.

    Read around in BSL some more and read up on plug flow reactors with backmixing in e.g. Levenspiel (page 312, 346).
     
  4. Oct 27, 2015 #3
    When they said that diffusion is not negligible, what they almost certainly meant was that radial diffusion is not negligible. In my judgement, they expected you to neglect axial diffusion. So, for the laminar flow case, you are going need to do shell balances to include radial radial diffusion.

    Chet
     
  5. Oct 27, 2015 #4
    I assumed this was the case, as it was done previously in BSL. In section 18.3, the reaction 2A → B is studied. The authors say the relationship between the molar fluxes is [itex]\frac{1}{2} N_A = -N_B[/itex]. Then, in example 19.4-3, we work with the reaction O2 + 2CO → 2CO2, and the relationship between fluxes is [itex]N_{O_2} = \frac{1}{2} N_{CO} = - \frac{1}{2} N_{CO_2}[/itex]. These relationships resemble the relationships between reaction rates in chemical kinetics. I should mention, however, that in both cases only diffusion is happening, no convection. I did check chapters 13 and 14 of Levenspiel's, but found no mention of the relationship between molar fluxes, he uses different terminology from the one used in BSL.
    What is the reasoning behind this relationship? How can I derive it?
    That makes sense. I'm going to keep considering axial diffusion in the plug flow case, and I will rework the laminar case considering radial diffusion. My intuition tells me that the radial parabolic velocity profile induces a radial concentration gradient, is this right?. I will return later with the new model.
    One more thing, is there a dimensionless number that can be used to determine whether a flow is plug or not?

    Thank you both!
     
    Last edited: Oct 27, 2015
  6. Oct 27, 2015 #5
    Yes. As in heat transfer, the radial "conduction" is going to dominate over the axial "conduction."
    Plug flow is not a physical reality because of the no-slip boundary condition. But it provides a very valuable limiting case for comparison and, under certain physical circumstances, plug flow can be approached. For example, in a shear-thinning fluid, the velocity profile over most of the cross section is flater than a parabolic profile.
     
  7. Oct 28, 2015 #6
    Oops, I actually had that in mind, that plug flow is just an approximation to simplify things. However, I was looking for some kind of relation, such as when one calculates the Biot number in heat transfer in order to determine if it is appropiate to use a lumped system analysis.

    As for the laminar flow reactor model, here we go. We know the radial velocity profile is given by
    [tex]v_z = v_{z,max} \left(1- \left( \frac{r}{R} \right)^2 \right)[/tex]
    Now, performing a mass balance for species A in a cylindrical shell of volume 2πrΔrΔz, we arrive at the following PDE
    [tex]\frac{\partial }{\partial r} (rN_{Ar}) + \frac{\partial }{\partial z} (rN_{Az}) + k_1''' C_A = 0[/tex]
    The components of the molar flux vector are
    [tex]N_{Ar} = -D_{AB} \frac{\partial C_A}{\partial r}[/tex]
    [tex]N_{Az} = C_A v_z[/tex]
    Substituting these expressions in the original PDE gives
    [tex]\frac{\partial }{\partial r} \left( -D_{AB} \frac{\partial C_A}{\partial r} \right) + \frac{\partial }{\partial z} (r C_A v_z) + k_1''' C_A = 0[/tex]
    [tex]D_{AB} \frac{\partial^2 C_A}{\partial r^2} = v_{z,max} \ r \left[1- \left( \frac{r}{R} \right)^2 \right] \frac{\partial C_A}{\partial z} + k_1''' C_A[/tex]
    This is the partial differential equation governing the reaction zone. It's way over my math level, though I'm not expected to solve it. Still, a very similar (if not the same) equation is solved in chapter 19 of BSL, so I might give it a try.
     
  8. Oct 28, 2015 #7
    You have the right idea, but you need to re-do your shell balance. The equation immediately after "performing a mass balance for species A in a cylindrical shell of volume 2πrΔrΔz, we arrive at the following PDE" is not set up correctly in terms of the balance on the shell.

    Chet
     
  9. Oct 28, 2015 #8
    I see. Seems like I forgot to multiply the reaction term by the shell volume. The correct equation is
    [tex]\frac{D_{AB}}{r} \frac{\partial^2 C_A}{\partial r^2} = v_{z,max} \left[1- \left( \frac{r}{R} \right)^2 \right] \frac{\partial C_A}{\partial z} + k_1''' C_A[/tex]
    By any chance, do you happen to have any comments about the physical interpretation of [itex]N_{Az} |_{z=0} = N_{Az} + N_{Bz}[/itex]

    Thanks!
     
  10. Oct 28, 2015 #9
    The first term is incorrect. It is not even consistent dimensionally with the other two terms. It should read:
    $$\frac{D_{AB}}{r}\frac{\partial}{\partial r}\left(r\frac{\partial C_A}{\partial r}\right)$$
    Well, for plug flow,
    [tex]N_{Az} + N_{Bz}=cv=const[/tex]
    where c is the total molar density (constant for this problem). Any decrease in the molar density (concentration) of species A is compensated by an increase in the molar density (concentration) of species B, because each mole of A that reacts forms one mole of B. So the overall molar density c remains constant, as does the overall axially transport of A + B.

    Chet
     
  11. Oct 29, 2015 #10
    Yikes, I just got rid of the r inside the parentheses when substituting the expression for NAr, and for no apparent reason. I didn't notice. Thanks for pointing that out.
    Great explanation. In other words, mass is conserved. I guess this was intuitive, however, I'm still not very comfortable playing around with mass/molar fluxes, given their geometric (physical) nature. I have noticed BSL doesn't go into full detail when introducing the most fundamental concepts such as fluxes, maybe I need a complementary textbook, especially for the mass transfer/diffusion sections, any suggestions? Still, I think BSL is one of the best textbooks ever written for any field, in my somewhat limited opinion. The sheer amount of engineering wisdom within those pages still amazes me.

    Thanks for all your insights!
     
  12. Oct 29, 2015 #11
    The mass transfer portion of BSL is a little hard to work with because there are so many different mass- and molar fluxes that are used, and mass and molar average velocities (and combinations thereof). During my career, whenever I encountered a system that I had to analyze with this formalism, I always needed to go back and review the first couple of chapters.

    chet
     
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