Deriving Conservation of Mass & Momentum for an Inviscid Compressible Gas

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Homework Statement

Starting from the equations of conservation of mass and momentum for an inviscid compressible gas,
$$\frac{\partial \rho}{\partial t} + \bigtriangledown . (\rho {\bf u}) = 0$$

$$\rho ( \frac{\partial {\bf u}}{\partial t} + ({\bf u}. \bigtriangledown){\bf u}) = \bigtriangledown p + {\bf F}$$

derive for a fixed volume V enclosed by a surface S:

$$\frac{d}{dt} \int_V \frac{1}{2} \rho u^2 dV + \int_S \frac{1}{2} \rho u^2 {\bf u.n} dA = - \int_S p {\bf u.n} dA + \int_V (p\bigtriangledown. {\bf u} + {\bf F.u}) dV$$.

Homework Equations

Divergence theorem:

$$\int_V \bigtriangledown . {\bf u} dV = \int_S {\bf u.n} dA$$
where A is the surface enclosing V

The Attempt at a Solution

I started from the first equation, multiplied this through by $$\frac{u^2}{2}$$, applied the divergence theorem and looked at what the full derivative given in the question is in terms of the chain rule, and ended up with

$$\frac{d}{dt} \int_V \frac{1}{2} \rho u^2 dV + \int_S \frac{1}{2} \rho u^2{\bf u.n} dA = \int_V \rho {\bf u} . \frac{\partial {\bf u}}{\partial t} dV$$

Which looks like it might be right. However I am a bit wary though because I don't know if I should be considering the del operator acting on the $$\frac{u^2}{2}$$ that I multiplied through by. Anyway, assuming this is right (I have the alternative expression if it's not, and that also looks like it could be right) I tried to proceed by dotting the second fluid equation with $${\bf u}$$, and things were looking quite promising but it didn't quite come out. I seemed to be having particular trouble with the $$p \bigtriangledown.{\bf u}$$ term.

Am I going along the right lines here? Any hints on how I can finish this off?

Thanks.

Edit - Sorry about my lack of vectors, I can't seem to find a command that works on this forum. u and n are vectors through, and u^2 = u.u Also, I guess I should have mentioned (although the question doesn't actually specifiy) that rho, u and p are function sof position and time.

 

[mighty2000]

I can see that you are on the right track with your approach. The key is to use the divergence theorem and the chain rule to manipulate the equations. Here is how I would proceed:

Starting from the first equation, we can write it in the following form:

\frac{\partial \rho}{\partial t} + \bigtriangledown . (\rho {\bf u}) = 0
\Rightarrow \frac{\partial}{\partial t} \int_V \rho dV + \int_S \rho {\bf u.n} dA = 0

Now, let's multiply through by \frac{1}{2}u^2 and use the chain rule to obtain:

\frac{\partial}{\partial t} \int_V \frac{1}{2} \rho u^2 dV + \int_S \frac{1}{2} \rho u^2{\bf u.n} dA = \int_V \rho {\bf u} . \frac{\partial {\bf u}}{\partial t} dV + \int_S \rho {\bf u.n} {\bf u} dA

Next, we can use the second equation to substitute for the term \frac{\partial {\bf u}}{\partial t}:

\rho ( \frac{\partial {\bf u}}{\partial t} + ({\bf u}. \bigtriangledown){\bf u}) = \bigtriangledown p + {\bf F}
\Rightarrow \frac{\partial {\bf u}}{\partial t} = - \frac{1}{\rho} \bigtriangledown p + \frac{1}{\rho} {\bf F} - ({\bf u}. \bigtriangledown){\bf u}

Substituting this into the previous equation, we get:

\frac{\partial}{\partial t} \int_V \frac{1}{2} \rho u^2 dV + \int_S \frac{1}{2} \rho u^2{\bf u.n} dA = - \int_V \bigtriangledown p dV + \int_V \frac{1}{\rho} {\bf F} . {\bf u} dV - \int_V ({\bf u}. \bigtriangledown){\bf u} . {\bf u} dV + \int_S \rho {\bf u.n}