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Homework Help: Constants in chemcial reaction

  1. Dec 28, 2008 #1
    1. The problem statement, all variables and given/known data
    This is an example from a book. I am having trouble in substituting the numbers to get the given result.

    2. Relevant equations
    The bar notation is to indicate the operating point. The subscript A is for the reactant A (which reacts and becomes product B in a A->B reaction). The subscript i stands for the input. So [tex]c_{A_i}[/tex] means the concentration of A entering the reactor.

    fc is the flow rate of the coolant
    Tci is the temperature of the coolant going in
    Tc is the temperature of the coolant going out

    f is the flow rate of the reactant going in
    Ti is the temperature of the reactant going in
    cAi is the concentration of the reactant going in
    T is the temperature of the reactant going out
    cA is the concentration of the reactant going out

    3. The attempt at a solution
    My problem is that when I plug in the values, I don't get 2.07 for [tex]\tau_1[/tex]. I first assumed that [tex]\bar{f} = f= 1.3364[/tex] and [tex]\bar{T}_{\bar{c}_A} = 678.9[/tex]. But I got [tex]\tau_1 = 0.5148[/tex].

    I assumed that my [tex]\bar{f}[/tex] was correct, so I set [tex]\tau_1=2.07[/tex] and rearranged the equation to solve for [tex]\bar{T}_{\bar{c}_A}[/tex].

    [tex]\bar{T}_{\bar{c}_A} = -\frac{E}{R}(ln(\frac{1-\frac{\tau_1 \bar{f}}{V}}{2\tau_1 k_o}))^{-1}[/tex] . . . (2)

    When I do that I got [tex]\bar{T}_{\bar{c}_A} = 631[/tex], which doesn't match any of the given temperature. So I thought [tex]\bar{f}[/tex] might be 0.8771. And I got [tex]\bar{T}_{\bar{c}_A} = 633[/tex], which still doesn't match any number.

    I have tried some other equations but I couldn't get any result to match. Since I don't have a background in studying chemical reactions, I was suspecting that there is something fundamental that I am missing here.

    Do you know how the author got 2.07 for [tex]\tau_1[/tex]?

    - Thank you.
  2. jcsd
  3. Dec 28, 2008 #2


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    Science Advisor
    Homework Helper

    Hi Saltine! :smile:

    Look at the dimensions …

    it isn't e(-E/RTcA)

    it's (e(-E/RT))cA :wink:
  4. Dec 28, 2008 #3
    That is right!

    A second problem:


    Now I am trying to verify T2. Here, Cv is the heat capacity at constant volume of the reaction mixture (Unit of Cv: Btu/lb-°R). However, the book did not give its value. This is the same for other examples where it seems that every other value is listed, except the Cv. Is it directly derivable from the other values?

    U is the overall heat transfer coefficient, assumed to be constant, Btu/ft²-°R-min
    A is heat transfer area, ft²
    ∆Hr is the heat of reaction, Btu/lbmole of A reacted
    rA is the rate of reaction

    I first computed [tex]\bar{r}_A = 0.03938[/tex]. Since Cv is just alone in the numerator, I first evaluated T2 without using Cv and got 0.2833. Therefore,

    Cv = -7.96 / 0.2833 = -28.10 Btu/lb-°R

    There is nowhere else in this particular example where Cv is used, so I couldn't check Cv in another way. Does it make sense that Cv is negative? This is an exothermic reaction. The problem didn't state this, but I suppose it is since coolant is used.
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