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Intgral of dp^2?

  1. Aug 25, 2011 #1
    intgral of dp^2??

    Hi,

    I have the following equation:

    [itex]\dot{m} = c \frac{dp^{2}}{dx}[/itex]

    Integration gives:

    [itex]\dot{m} \int dx = c \int dp^{2}[/itex]

    How can I solve [itex]\int dp^{2}[/itex]??

    The lower boundary is Ps and the upper boundary is P0 for the integral!
     
  2. jcsd
  3. Aug 25, 2011 #2

    tiny-tim

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    hi hermano! :smile:

    do you mean dq/dx where q = p2 ?

    just integrate over q from q = (ps)2 to q = (p0)2 :wink:
     
  4. Aug 25, 2011 #3
    Re: intgral of dp^2??

    Hi tiny-tim ;-)

    I want to integrate dp^2!! So you take some parameter q and say that this parameter q is equal to p^2 and then solve the integral of dq instead of dp^2. This gives q again and thus p^2! Then fill in the upper and lower boundary for p: P0^2 - Ps^2

    Am I right?
     
  5. Aug 25, 2011 #4

    tiny-tim

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    if i'm correctly understanding your original equation, yes :smile:
     
  6. Aug 25, 2011 #5
    Re: intgral of dp^2??

    My original question is how to solve this equation:

    [itex]\dot{m} \int dx = c \int dp^{2}[/itex]

    My problem is solving the right hand side of the equation namely:

    [itex]c \int dp^{2}[/itex]

    Your answer is: replace [itex] p^{2}[/itex] by q and solve this 'normal' integral [itex]c \int dq[/itex]! This gives c q, then replace q by [itex] p^{2}[/itex] again so the solution is c [itex] p^{2}[/itex]. Then I only have to fill in the boundaries for p. Right?
     
  7. Aug 25, 2011 #6

    tiny-tim

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    I think so, yes. :smile:
     
  8. Aug 25, 2011 #7

    K^2

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    Re: intgral of dp^2??

    dp²/dx is an infinitesimal quantity. On the left side of equation, you have a finite quantity. Something is wrong.

    You either have (dp/dx)² or d²p/dx². In either case, review your equation. Where do you get it from, anyways?
     
  9. Aug 26, 2011 #8
    Re: intgral of dp^2??

    Hi K^2,

    This equation comes from Darcy's law namely [itex]\dot{m} = \frac{p Q}{R T}[/itex]

    Fill Q (volumetric flow rate form Darcy's law) in in the above equation gives:

    [itex]\dot{m} = c p \frac{dp}{dx}[/itex]

    With [itex] c [/itex] all the constants in the equation.

    [itex] p \frac{dp}{dx}[/itex] is equal to [itex] \frac{1}{2} \frac{dp^2}{dx}[/itex]

    And thus finally I got [itex] \dot{m} = \frac{c}{2} \frac{dp^2}{dx}[/itex]

    Solving this:

    [itex]\dot{m} \int dx = \frac{c}{2} \int dp^{2} [/itex]

    Therefore my question: How to solve [itex] \int dp^{2} [/itex] ??
     
  10. Aug 26, 2011 #9

    tiny-tim

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    ah, now i see where it comes from! :rolleyes:

    yes, the whole point of changing pdp/dx to 1/2 d(p2)/dx

    (and btw, it's better to write it that way, with brackets)

    is because that is a perfect derivative, and you can immediately integrate it to pf2 - pi2. :smile:

    Incidentally, the same trick enables us to go from F = ma (in mechanics) via a = vdv/dx to ∆(1/2 mv2) :wink:
     
  11. Aug 26, 2011 #10
    Re: intgral of dp^2??

    Tiny-tim,

    Do I make a fault against the mathematics by saying that [itex]\dot{m} = \frac{c}{2} \frac{dp^2}{dx}[/itex] can be solved through [itex]\dot{m} \int dx = \frac{c}{2} \int dp^{2}[/itex]??

    Then still my question is: how to solve the right hand site of this equation namely the integral of [itex]\int dp^{2}[/itex] ??
     
  12. Aug 26, 2011 #11

    K^2

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    Re: intgral of dp^2??

    Yeah, you shouldn't write it as dp²/dx, because that means (dp)²/dx, and that just isn't right. d(p²)/dx, on the other hand, does make sense, and d(p²) can be integrated over using variable substitution.

    [tex]\int f(u)du = \int f(x)\frac{du}{dx}dx[/tex]

    Which in this case, turns integral over d(p²) into integral over 2p dp.
     
  13. Aug 26, 2011 #12

    tiny-tim

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    hi hermano! :smile:

    (btw, we say "a fault in the mathematics" … though you can say "a crime against mathematics"! :biggrin:)
    No, we often treat a derivative exactly like a fraction, and multiply both sides by dx (and then integrate) …

    that's fine :smile:
    ∫ d(p2) = [p2] :wink:
     
  14. Aug 26, 2011 #13
    Re: intgral of dp^2??

    Thanks, now it is clear for me!

    Maybe you both can help me with another small issue I'm facing with this equation! For the integral of my pressure, I have to fill in the upper and lower boundary for p^2! However, I doubt of the input pressure is my upper or lower boundary, and also the same for the output pressure. What tells me which pressure (input - output) is the upper and lower boundary for my integral?

    The equation gives the mass flow through a pipe (Poiseuille) from the high input pressure to the lower ouput pressure.
     
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