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Differential equation of gradually varied flow

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data
    I have no idea how the third formula of dy/dx is derived ...

    2. Relevant equations


    3. The attempt at a solution
    I know that the Q = (1/n)(A)(R^2/3) [(s)(^0.5)] ,
    Q = K [(s)(^0.5)]
    , so , K= (1/n)(A)(R^2/3)

    i know that for very wide channel , y = R
    A = by
    K= (1/n)(A)(R^2/3)
    = (1/n)(by)(y^2/3)
    = (1/n)(b)(y^5/3)
    Thus , (K^2) = [[ (1/n)(b)]^2 ](y^10/3)
    but , it seems that the author just got (y^10/3) ,

    So , is the author wrong ?
     

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  3. Feb 7, 2017 #2

    BvU

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    On first sight: no. Consider ##{K_0\over K}## instead of just ##K##.
    There is no second sight because your use of the template is utterly unenlightening...
     
  4. Feb 7, 2017 #3
    sorry , i mean $$K_0$$ corresponds to $$y_0$$
    and $$K$$ corresponds to $$y$$
     
  5. Feb 7, 2017 #4

    BvU

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    Yes, I understand. Don't the n's and the b's cancel ?
     
  6. Feb 7, 2017 #5
    in my working , i cant cancel the n and b ... Is there anything wrong with my working ?
     
  7. Feb 7, 2017 #6

    BvU

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    Not from what I can see :smile: because what I can see is nothing ...(refer to post #2). I have no idea what this is about.

    I see you write ##K^2 = (b/n)^2\; y^{10/3}## and I add ##K_0^2 = (b/n) ^2\; y_0^{10/3}## leading to ##\left ( {K_0\over K}\right ) ^2 = \left ({y_0\over y}\right )^{10/3}##. That simple.
     
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