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Reactor + Separator control outlet concentration

  1. Sep 16, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-9-16_12-28-2.png
    upload_2015-9-16_12-28-19.png

    2. Relevant equations


    3. The attempt at a solution
    Hello,

    I get confused with input and output and state variables. Since we want to control ##C_{Ao}##, I assume that needs to be the output variable in the vector ##\vec {y}##. I think the inputs should be ##C_{Ai}## and ##F_{i}##, since those are not something that we can control. I think mostly the state and output variables should be the same.

    I am also convinced my prof. never did this problem in class, because I have no recollection and I always come to class (and I don't fall asleep!).

    So for third order kinetics, I assume the reaction should be
    $$3A \xrightarrow{k} P$$

    I am unsure where I should do the mass balance, around just the reactor or the whole system (reactor and separator). My balance equations are (##V_{t}## denotes volume of reactor and separator). I wonder if ##V_{t}## should be a constant? Or do I do a separate balance with control volumes being the reactor and separator?

    State ##\vec {x}##, Input ##\vec {u}##, and output ##\vec {y}##

    $$ \vec{x} = \begin{bmatrix} C_{Ao} \\ C_{Po} \\ C_{A} \\ C_{P} \\ V \end{bmatrix} $$
    $$ \vec{u} = \begin{bmatrix} C_{Ai} \\ F_{i} \end{bmatrix} $$
    $$ \vec{y} = \begin{bmatrix} C_{Ao} \\ C_{Po} \\ C_{A} \\ C_{P} \\ V \end{bmatrix} $$

    Balance around the reactor
    $$ \frac {dV}{dt} = F_{i} + F_{r} - F $$
    $$ \frac {d(C_{A}V)}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3} $$
    $$ \frac {d(C_{P}V)}{dt} = F_{r}C_{Pr} - FC_{P} + \frac {kC_{A}^{3}}{3} $$

    Balance around reactor + separator
    $$ \frac {d(C_{Ao}V_{t})}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - F_{o}C_{Ao} - kC_{A}^{3} $$
    $$ \frac {d(C_{Po}V_{t})}{dt} = F_{r}C_{Pr} - F_{o}C_{Po} + \frac {kC_{A}^{3}}{3} $$

    Am I on the right track here?
     
    Last edited: Sep 16, 2015
  2. jcsd
  3. Sep 16, 2015 #2

    Maylis

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    The instructor posted a note (I was right about the problem not being done in class), so now I have some more information to move forward with the problem.
    First off, I don't need to worry about the product, just species A. Second, the separator is acting at steady state. The output is ##C_{Ao}##, and the states are ##C_{A}## and ##V##.So my governing equations and variable vectors are
    $$ \vec{x} = \begin{bmatrix} C_{A} \\ V \end{bmatrix} $$
    $$ \vec{u} = \begin{bmatrix} F_{i} \\ C_{Ai} \\ F_{o} \\ F_{r} \end{bmatrix} $$
    $$ \vec{y} = \begin{bmatrix} C_{Ao} \end{bmatrix} $$

    Reactor
    (1) $$ \frac {dV}{dt} = F_{i} + F_{r} - F $$
    (2) $$ \frac {d(C_{A}V)}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
    Separator (steady state)
    (3) $$ 0 = F - F_{r} - F_{o} $$
    (4) $$ 0 = FC_{A} - F_{r}C_{Ar} - F_{o}C_{Ao} $$

    For equation (2), I expand out
    $$ V \frac {dC_{A}}{dt} + C_{A} \frac {dV}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
    I substitute (1) into ##\frac {dV}{dt}##
    $$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} + F_{r} - F) = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
    The ##FC_{A}## terms cancel
    $$V \frac {dC_{A}}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} -F_{i}C_{A}-F_{r}C_{A} - kC_{A}^{3}V $$
    I know I need to use the separator balance equations to substitute into this equation to cast in the form needed for part (c), but I am unsure which way to substitute.
     
    Last edited: Sep 16, 2015
  4. Sep 16, 2015 #3

    Maylis

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    A new thought came to mind, with equation (2),
    $$ V \frac {dC_{A}}{dt} + C_{A} \frac {dV}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
    I will modify equation (1)
    $$\frac {dV}{dt} = F_{i} + F_{r} - F$$
    Using equation (3)
    $$ 0 = F - F_{r} - F_{o} $$
    Solving for ##F_{o} = F - F_{r}##, I will substitute back into (1)
    $$ \frac {dV}{dt} = F_{i} - F_{o} $$
    Then substitute this into (2)
    $$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
    I will use equation (4)
    $$0 = FC_{A} - F_{r}C_{Ar} - F_{o}C_{Ao}$$
    Solve ##F_{r}C_{Ar} = FC_{A}-F_{o}C_{Ao}##. I substitute this into (2) and now have
    $$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} +FC_{A}-F_{o}C_{Ao} - FC_{A} - kC_{A}^{3}V $$
    The ##FC_{A}## terms cancel, leaving me with
    $$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} -F_{o}C_{Ao} - kC_{A}^{3}V $$
    But I'm still not quite sure how to get the the form they are looking for.
     
  5. Sep 16, 2015 #4

    Maylis

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    Okay, I have some idea now. This is what I got so far.
    Using (3)
    $$ 0 = F - F_{r} - F_{o} $$
    I divide through by ##F##, ##0 = 1 - \frac {F_{r}}{F} - \frac {F_{o}}{F} = 1 - r - \frac {F_{o}}{F}##
    Then I use equation (4)
    $$ 0 = FC_{A} - F_{r}C_{Ar} - F_{o}C_{Ao} $$
    I isolate ##F_{o}C_{Ao} = FC_{A}-F_{r}C_{Ar}##, then I divide through by ##FC_{Ao}##, leaving me with ##\frac {F_{o}}{F}= \frac {C_{A}}{C_{Ao}}-K_{H}r ##. I know that ##\frac {F_{o}}{F}=1-r##, so I put into the equation ##1-r = \frac {C_{A}}{C_{Ao}}-K_{H}r##.
    Next, I isolate
    $$\frac {C_{A}}{C_{Ao}} = 1 - r + K_{H}r = 1-r(1-K_{H}) $$
    I take the inverse
    $$\frac {C_{Ao}}{C_{A}} = \frac{1}{1-r(1-K_{H})}$$

    We will go back to my equation (2) that I had gotten from the previous post
    $$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} -F_{o}C_{Ao} - kC_{A}^{3}V $$
    I move the ##C_{A}(F_{i} - F_{o})## term to the RHS,
    $$ V \frac {dC_{A}}{dt} = F_{i}(C_{Ai}-C_{A}) -F_{o}(C_{Ao}-C_{A}) - kC_{A}^{3}V $$
    I solve for ##C_{Ao}##
    $$C_{Ao} = \frac {C_{A}}{1-r(1-K_{H})}$$
    I substitute this into equation (2)
    $$V \frac {dC_{A}}{dt} = F_{i}(C_{Ai}-C_{A}) -F_{o} \bigg (\frac {C_{A}}{1-r(1-K_{H})} -C_{A} \bigg ) - kC_{A}^{3}V $$
    Pull out ##C_{A}## and divide by ##V##, leaving me with
    $$ \frac {dC_{A}}{dt} = \frac {F_{i}}{V}(C_{Ai}-C_{A}) - \frac {F_{o}C_{A}}{V} \bigg ( \frac {1}{1-r(1-K_{H})}+1 \bigg ) - kC_{A}^{3} $$
    This looks similar to the equation in part (c) (part (c) assumed the volume in the reactor was constant)
    upload_2015-9-16_18-17-38.png
     
    Last edited: Sep 16, 2015
  6. Sep 17, 2015 #5
    In your final equation, the +1 should be a -1. With this change, the final equation is identical to the equation in the problem statement, except for the additional term:$$C_A\frac{(F_0-F_i)}{V}$$

    Chet
     
  7. Sep 17, 2015 #6

    Maylis

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    I noticed that, but my math did not come up with that expression. I wonder if it is correct or if I made a sign error in any of my steps
     
  8. Sep 17, 2015 #7
    You made a sign error in your last step.
     
  9. Sep 17, 2015 #8

    Maylis

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    Okay, I see it now. Thank you
     
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