1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Reactor design

  1. Nov 1, 2014 #1


    User Avatar
    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
    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted