# Separable Equation Help

• Jan 14th 2010, 11:56 AM
rickyticki
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.
• Jan 14th 2010, 12:05 PM
pickslides
hi there rickyticki, I have a suggestion.

Have a go at solving this

Quote:

Originally Posted by rickyticki
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
p=p0(ρ/ρ0)^γ,

and if all goes well it might end up being

Quote:

Originally Posted by rickyticki

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

• Jan 14th 2010, 01:58 PM
rickyticki
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.
• Jan 14th 2010, 02:22 PM
pickslides
I see, I was confusing your $\rho$ and $p$
• Jan 14th 2010, 02:27 PM
rickyticki
yes, and the $\gamma$ and $y$ are different too.

didnt know how to use math symbols
• Jan 15th 2010, 10:06 AM
Coomast
Quote:

Originally Posted by rickyticki
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