# Expansion 1D Euler Eq.?

1. Oct 27, 2011

### fury902

Expansion 1D Euler Eq.??

Trying to figure out an expansion step for 1D Euler Equations for unsteady gas flow.
Continuity:
$\frac{\partial(\rho F)}{\partial t}$+$\frac{\partial (\rho uF)}{\partial x}$=0

After Expansion:
$\frac{\partial(\rho)}{\partial t}$+$\frac{\partial (\rho u)}{\partial x}$+$\frac{\rho u}{F}\frac{dF}{dx}$=0

I understand to go from step 1 to step 2, you divide by F; however, for Step 2 where is the third term coming from and what does it mean?

I have the same question for the momentum equation:
Momentum:
$\frac{\partial (\rho uF)}{\partial t}$+$\frac{\partial (\rho u^{2}+p)F}{\partial x}$-$p\frac{dF}{dx}$+$\frac{1}{2}\rho u^{2}f\pi$ D=0

After Expansion:
$\frac{\partial (\rho u)}{\partial t}$+$\frac{\partial (\rho u^{2}+p)}{\partial x}$+$\frac{\rho u^{2}}{F}\frac{dF}{dx}$+$\frac{1}{2}\rho u^{2}f\pi$ D=0

Again, where is the $\frac{\rho u^{2}}{F}\frac{dF}{dx}$ term coming from and what does it mean?

2. Oct 27, 2011

### LCKurtz

Re: Expansion 1D Euler Eq.??

It would be helpful to us who aren't familiar with gas flow problems and notation if you would tell us what variables ρ, u, and F depend on. I'm going to assume, from what I see, that F doesn't depend on t but does depend on x.
$$\frac{\partial(\rho F)}{\partial t} = \frac {\partial \rho}{\partial t}F$$
because F doesn't depend on t. But $\frac{\partial (\rho uF)}{\partial x}$ requires the product rule since $\rho u$ and F depend on x:
$$\frac{\partial (\rho uF)}{\partial x}=F\frac{\partial (\rho u)}{\partial x} +\rho u\frac{\partial (F)}{\partial x}$$

So if you differentiate the equation you get those three terms on the right above and you divide the equation by F. I didn't check your second question but I'm guessing it might be the same problem.