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Combustion mass conservation integral

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data
    integrate [(rs^2)*(rhos)*(us)*(db/dr)=(d/dr)*(r^2)*(rho)*(D)*(db/dr))]

    rs=radius at surface
    rhos=density at surface
    us=velocity at surface
    r = radius
    rho = density
    D= diffusivity
    b=spalding non-dimensional parameter


    2. Relevant equations



    3. The attempt at a solution
    This is the solution after one integration based on the assumptions that r, rho, D, u are not dependent

    (rs^2)*(rhos)*(us)*(b)=(r^2)*(rho)*(D)*(db/dr)+(c1)

    I think this involves separation of variables to solve and integrate but I dont see the steps clearly. Can someone please show me?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 6, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    First off you mean that rs is a constant right, not r? Secondly the right hand side seems a bit awkward. The d/dr operator does it act on everything that comes after it or?

    If so, ignoring the constants, is this the differential equation you're trying to solve?:

    [tex]
    \frac{db}{dr}=\frac{d}{dr}\left(r^2\frac{db}{dr}\right)
    [/tex]

    Edit: Looking at your attempt at a solution this must be the case.

    So after one integration you're left with [itex]b=r^2 b'+c \Rightarrow b'/(b-c)=1/r^2[/itex]. You should be able to integrate this expression.
     
    Last edited: Jun 6, 2009
  4. Jun 6, 2009 #3
    Yes you are correct that is the basic equation I am trying to solve. I am getting hung up on the first integral though. I dont understand the steps from solving this:
    db/dr=d/dr*(r^2*(db/dr))

    I am thinking of this as separation of variables. IE move the dr on the left over to the right side. then the left side just becomes b, but the right side becomes more complicated. What are the steps for solving the original equation?
     
  5. Jun 6, 2009 #4
    not the original equation i wrote but the original equation cyosis wrote.
    Thanks
     
  6. Jun 6, 2009 #5

    Cyosis

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    Homework Helper

    The first integral is pretty easy. Note that both sides are just "terms" that are getting differentiated with respect to r. Therefore integrating with respect to r will cancel out the differentiation operation bar a constant.

    [tex]
    \begin{align*}
    \int \frac{db}{dr} dr &=\int \frac{d}{dr}\left(r^2\frac{db}{dr}\right)dr
    \\
    b &=r^2\frac{db}{dr}+c
    \end{align*}
    [/tex]
     
    Last edited: Jun 6, 2009
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