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Reactor design

  1. Nov 1, 2014 #1

    Maylis

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    Gold Member

    1. The problem statement, all variables and given/known data
    The elementary liquid phase reaction
    [tex] A + B \rightarrow C [/tex]
    is carried out in a 500 L reactor. The entering concentrations of streams A and B are both 2 M and the specific reaction rate is 0.01 L/(mol min).

    (a) Calculate the time to reach 90% conversion if the reactor is a batch reactor filled to the brim

    Assuming a stoichiometric feed (10 mol A/min) to a continuous-flow reactor, calculate the reactor volume and space-time to achieve 90% conversion if the reactor is
    (b) a CSTR (Ans.: V = 22,500 L)
    (c) a PFR (Ans: V = 2,250 L)
    (d) redo (a) through (c) assuming the reaction is first order in B and zero order in A with k = 0.01/min.
    (e) Assume the reaction is reversible with ##K_{e} = 2 \frac {L}{mol}##. Calculate the equilibrium conversion and the CSTR and PFR volumes necessary to achieve 98% of the equilibrium conversion.

    2. Relevant equations


    3. The attempt at a solution
    (a)
    I use the equation ##t = N_{A0} \int_0^X \frac {dX}{-r_{A}V} ##
    [tex] t = C_{A0} \int_0^X \frac {dX}{-r_{A}} [/tex]
    I assume this is a first order reaction with respect to A, ##-r_{A} = kC_{A}##, and I know for a liquid phase reaction, ##C_{A} = C_{A0}(1 - X)##. Therefore, ##-r_{A} = kC_{A0}(1-X)##.
    [tex] t = C_{A0} \int_0^X \frac {dX}{kC_{A0}(1-X)} [/tex]
    [tex] t = \frac {1}{k} \int_0^X \frac {dX}{1-X} [/tex]
    [tex] t = - \frac {1}{k} ln(1 - X) = - \frac {1}{0.01} ln(1 - 0.9) = 230.3 \hspace{0.05 in} min [/tex]

    (b) This is where I get stuck, I use the equation ##V = \frac {F_{A0}X}{-r_{A}}##
    [tex] V = \frac {F_{A0}X}{kC_{A0}(1-X)} [/tex]
    [tex] V = \frac {10(0.9)}{0.01(2)(1-0.9)} = 4,500 \hspace{0.05 in} L [/tex]
    But this is not the answer given in the textbook, 22,500 L and I haven't figured out why.
    For the space time,
    [tex] \tau = \frac {V}{v_{0}} [/tex]
    where ##v_{0}## is the initial volumetric flow rate. ##F_{A0} = C_{A0}v_{A0}##, so ##10 = 2v_{A0}##, therefore the volumetric flow rate for A is 5 L/min, and since this is a stoichiometric feed, the total volumetric flow rate is 10 L/min, so the space time is 4,500 L/10 = 450 min.
     
    Last edited: Nov 1, 2014
  2. jcsd
  3. Nov 1, 2014 #2

    Maylis

    User Avatar
    Gold Member

    My problem was my assumption, I should have said that ##-r_{A} = kC_{A}C_{B}##, since this reaction is elementary.
     
    Last edited: Nov 1, 2014
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