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Homework Help: Reactor Design and Kinetics of Reactions

  1. Jul 6, 2017 #1
    1. The problem statement, all variables and given/known data

    My problem deals with understanding why we substitute in the reaction kinetics equivalent into mass conservation equation instead of dealing with differentials

    2. Relevant equations
    From the conservation of mass law about an envelope:
    $$\frac{dm}{d t} = \sum m_{in} - \sum m_{out} + Generation - Consumption$$

    Assuming there is a consumption of some compound, then:
    $$Consumption = -\frac{dc}{dt} = kC^n$$

    3. The attempt at a solution

    Why is that when we go to solve the mass balance equation, usually for the unknown concentration that varies with time (in this case for a complete mix reactor), we make the following substitution:

    $$\frac{dm}{d t} = \sum m_{in} - \sum m_{out} - kVC^n$$

    Why can't we do this:

    $$\frac{dm}{d t} = \sum m_{in} - \sum m_{out} + -\frac{dc}{dt} \cdot V$$
    $$ V \cdot \frac{dc}{dt} = \sum m_{in} - \sum m_{out} -\frac{dc}{dt} \cdot V$$

    and then isolate for the differential terms, and integrate to get the final concentration??

    Thank you.
  2. jcsd
  3. Jul 7, 2017 #2


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    1. check dimensions. In your first equation you don't have ##m_{\rm in}## but ##{dm_{\rm in}\over dt}## (usually written as ##\,\dot m_{\rm in}## ).
    2. Consumption is not -dc/dt but dcV/dt -- as you use in your third equation
    3. In your last equation you seem to take V as constant. It's not.
  4. Jul 7, 2017 #3
    The problem is with this equation: $$Consumption = -\frac{dc}{dt} = kC^n$$
    It is only valid for a batch reactor. I know it is often common to write down an expression like this, but it always struck me as just plain stupid when referring to flow reactor. So, in the case of a flow reactor, dc/dt is not equal to ##-kC^n##. In a flow reactor, the rate of consumption of a species is equal to the reactor volume times the reaction rate: ##VkC^n##.
  5. Jul 7, 2017 #4
    Yes I agree @Chester.

    Also, I forgot to state my assumption, sorry:

    Assume that we are dealing with a CSTR ( complete mixing ) and that the volume of the tank is constant.

    But, in the mass balance equation, why do we explicitly equate consumption to


    but not simply use
    $$-V\cdot \frac{dc}{dt}$$

    in the equation?

    I know they are equal, but I have an urge to keep the differential form of consumption in the equation and solve it.
  6. Jul 7, 2017 #5
    In a cstr, they are not equal. Read my post again.
  7. Jul 7, 2017 #6
    @Chestermiller, that's so weird, in my textbook, they say they are equal and transition from dc/dt to kC

    Attached Files:

  8. Jul 8, 2017 #7
    This is why, in my first post, I said that it is "just plain stupid." The author of your textbook (and many others in their textbooks) has done you a disservice by writing the equation in this way and confusing you. Unfortunately, this is not the only thing in textbooks that confuse students. Have you had a course in thermodynamics yet? Thermodynamics texts are chock full of confusing and incorrect statements.
  9. Jul 8, 2017 #8
    Oh wow, I've been cheated :(. No , I will have not and am not required to take any thermodynamics courses (Civil Engineering Student). My original question arose from a Waste Water Process Engineering course I am currently taking.

    Can you please recommend a reference that I can read to help learn about more this?
  10. Jul 8, 2017 #9
    I like Chemical Reaction Engineering by Octave Levenspiel. This book is available as a pdf on line. However, even Levenspiel makes this same boo boo with the time derivative. In his defense, he no longer continues to do this when he gets to sections of the book where he describes how to analyze CSTRs and plug flow reactors.
  11. Jul 8, 2017 #10
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