Expansion 1D Euler Eq.?

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Expansion 1D Euler Eq.??

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

After Expansion:
[itex]\frac{\partial(\rho)}{\partial t}[/itex]+[itex]\frac{\partial (\rho u)}{\partial x}[/itex]+[itex]\frac{\rho u}{F}\frac{dF}{dx}[/itex]=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:
[itex]\frac{\partial (\rho uF)}{\partial t}[/itex]+[itex]\frac{\partial (\rho u^{2}+p)F}{\partial x}[/itex]-[itex]p\frac{dF}{dx}[/itex]+[itex]\frac{1}{2}\rho u^{2}f\pi[/itex] D=0

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

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

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

After Expansion:
[itex]\frac{\partial(\rho)}{\partial t}[/itex]+[itex]\frac{\partial (\rho u)}{\partial x}[/itex]+[itex]\frac{\rho u}{F}\frac{dF}{dx}[/itex]=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?
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.
[tex]\frac{\partial(\rho F)}{\partial t} = \frac {\partial \rho}{\partial t}F[/tex]
because F doesn't depend on t. But [itex]\frac{\partial (\rho uF)}{\partial x}[/itex] requires the product rule since [itex]\rho u[/itex] and F depend on x:
[tex]\frac{\partial (\rho uF)}{\partial x}=F\frac{\partial (\rho u)}{\partial x}
+\rho u\frac{\partial (F)}{\partial x}[/tex]

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.
 

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