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Math Help - Separable Equation Help

  1. #1
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    Separable Equation Help

    I was assigned a problem that stated pressure, p, and density, ρ, are related by the equation dp = -gρ dy. It said it follow the adiabatic equation p=p0(ρ/ρ0)^γ, where γ is not equal to 0, and p0 is pressure at earths surface and ρ0 is density at earths surface. It then wants me to show that

    p = p0(1-((γ-1)/γ)*((ρ0gy)/p0))^(γ/(γ-1))

    I've tried for an hour or more, and cannot figure out this problem. If anyone has any insight or even knows where to start, it would be greatly appreciated.
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  2. #2
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    hi there rickyticki, I have a suggestion.

    Have a go at solving this

    Quote Originally Posted by rickyticki View Post
    I was assigned a problem that stated pressure, p, and density, ρ, are related by the equation dp = -gρ dy
    After this you will have an equation for p, for this subtitute in

    Quote Originally Posted by rickyticki View Post
    p=p0(ρ/ρ0)^γ,
    and if all goes well it might end up being

    Quote Originally Posted by rickyticki View Post

    p = p0(1-((γ-1)/γ)*((ρ0gy)/p0))^(γ/(γ-1))
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  3. #3
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    Thanks for the suggestion.

    When I do that, I end up with -gρy = p0(ρ/ρ0)^(γ).

    Is that how it should be? The problem is I do not have a p variable to solve for now.
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  4. #4
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    I see, I was confusing your \rho and p
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  5. #5
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    yes, and the \gamma and y are different too.

    didnt know how to use math symbols
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  6. #6
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    Quote Originally Posted by rickyticki View Post
    I was assigned a problem that stated pressure, p, and density, ρ, are related by the equation dp = -gρ dy. It said it follow the adiabatic equation p=p0(ρ/ρ0)^γ, where γ is not equal to 0, and p0 is pressure at earths surface and ρ0 is density at earths surface. It then wants me to show that

    p = p0(1-((γ-1)/γ)*((ρ0gy)/p0))^(γ/(γ-1))

    I've tried for an hour or more, and cannot figure out this problem. If anyone has any insight or even knows where to start, it would be greatly appreciated.
    This is a somewhat confusing question. The first thing to do is to obtain the derivative of the adiabatic equation and substitute this in the given DE. This is:

    dp=p_0 \gamma \left( \frac{\rho}{\rho_0}\right) ^{\gamma-1}\frac{d \rho}{\rho_0}

    and so:

    \frac{p_0 \gamma}{\rho_0^{\gamma}} \rho^{\gamma-2}d \rho = -gdy

    Integrating this gives:

    \frac{p_0}{\rho_0^{\gamma}}\frac{\gamma}{\gamma-1} \rho^{\gamma-1}= -gy + K

    The constant of integration can be found by the boundary condition that at y=0, the density [imath]\rho=\rho_0[/imath]. This gives:

    K=\frac{p_0}{\rho_0}\frac{\gamma}{\gamma-1}

    The solution is now -after some algebra- the one given in the post as:

    p=p_0 \left[1-\left( \frac{\gamma-1}{\gamma} \right) \left(\frac{gy \rho_0}{p_0} \right) \right]^{\frac{\gamma}{\gamma-1}}

    coomast
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