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Feeding reactant into side to keep concentration high

  1. Nov 12, 2014 #1

    Maylis

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    Hello, I am studying the maximizing of selectivity of a particular product with two competing reactions

    ##A \xrightarrow {k_{D}} D##, ##r_{D} = k_{D}C_{A}^{\alpha_{1}}##
    ##A \xrightarrow {k_{U}} U##, ##r_{U} = k_{U}C_{A}^{\alpha_{2}}##

    where D is the desirable species, and U is the undesirable species. The selectivity is ##S_{DU} = \frac {k_{D}}{K_{U}}C_{A}^{a}##, where ##a = \alpha_{1} - \alpha_{2}##. In this case, ##a > 0##, so to maximize selectivity, I wish to run at a high concentration.

    Now I was reading a section about feeding into the side. I am confused about their explanation for why it is not beneficial to feed through the side. It seems like injecting high concentration into the reacting mixture will keep the concentration higher throughout the PFR. How is stream 2 diluted, as it is just be separated from stream 1 and fed into the side. Just separating a stream won't dilute it.
     

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    Last edited: Nov 12, 2014
  2. jcsd
  3. Nov 14, 2014 #2
    When streams 1 and 2 are intimately mixed at the cross section where stream 2 is introduced, the concentration of the new overall stream will be a weighted average of the concentrations of streams 1 and 2. Streams 1 and 2 do not maintain their individual integritiesbeyond the location at which stream 2 is introduced.

    Chet
     
  4. Nov 14, 2014 #3

    Maylis

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    That is confusing me. Isn't stream 2 just being siphoned off a stream 1? I mean, it's like saying having red dye mixed in a cup of water. If I take half the water and pour it into a new cup, they both have the same concentration of red dye, no?
     
  5. Nov 15, 2014 #4
    I think I'm beginning to get the idea of what you are saying. The way they are describing this system is that reaction can not take place until a stream "officially" enters the reactor. Before streams 1 and 2 enter the reactor, no reaction is occurring. They failed to mention that in their underlying assumptions.

    But, once a fluid parcel is "officially" inside the reactor, the reaction can proceed according to the prescribed rate law. How something like this could be done in practice would be to introduce catalyst just before the stream enters the reactor, or heat the stream up to reaction temperature just before it enters. More typically, the residence time in the inlet piping to the reactor is much shorter than the residence time in the reactor, so that the amount of reaction taking place in the inlet piping (including the distribution manifold for stream 2 is negligible.

    Chet
     
    Last edited: Nov 15, 2014
  6. Nov 16, 2014 #5

    Maylis

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    I see, somehow I totally missed what they were trying to say. Would the concept of siphoning off reactant and feeding it alongside the reactor be helpful though? The stream that enters down the reactor will have a smaller residence time, but due to its higher concentration, it makes the reaction throughout the reactor faster. I guess it would require further numerical analysis to determine if it would be worth pursuing.
     
  7. Nov 16, 2014 #6
    I don't exactly follow. But, if it seemed complicated to reason through, I would just do what you suggested and try some modelling calculations.

    Chet
     
  8. Nov 16, 2014 #7

    Maylis

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    What I mean is this. Suppose the reaction rate increases with a higher concentration of the reactant.

    If you siphon off reactant and put a parcel of fluid halfway through the PFR, the center will have a higher concentration of reactant than it would otherwise have without a side feed.

    However, since the reactant was inserted halfway into the reactor, that particular parcel of fluid will have spent less time in the reactor since it was introduced halfway through.
     
  9. Nov 16, 2014 #8
    What I would do with this situation would be to do some modeling to help me understand what's involved. I could specify the problem that I would start with, but I wouldn't want to deprive you of this learning experience.

    Chet
     
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