# Deriving Conservation of Mass & Momentum for an Inviscid Compressible Gas

In summary: AUsing the product rule, we can simplify the last two terms to get:\int_V ({\bf u}. \bigtriangledown){\bf u} . {\bf u} dV = \frac{1}{2} \int_V \bigtriangledown ({\bf u}. {\bf u}) dV = \frac{1}{2} \int_S ({\bf u}. {\bf u}) {\bf n} dA = \frac{1}{2} \int_S u^2 {\bf n} dA\int_S \rho {\bf u.n} {\bf u} dA = \int_S \rho u^2 {\bf u.n

## 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.

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}

## 1. What is the definition of an inviscid compressible gas?

An inviscid compressible gas is a type of gas that has no viscosity, meaning it has no internal friction or resistance to flow. It is also compressible, meaning its density can change under the influence of pressure.

## 2. How is conservation of mass applied to an inviscid compressible gas?

Conservation of mass states that the total mass of a closed system remains constant, meaning it cannot be created or destroyed. In the case of an inviscid compressible gas, this means that the total mass of the gas remains the same even if it is compressed or expands.

## 3. What is the equation for conservation of momentum in an inviscid compressible gas?

The equation for conservation of momentum in an inviscid compressible gas is known as the Euler equation. It states that the change in momentum of a gas is equal to the net force acting on it, taking into account any changes in pressure and density.

## 4. How does the compressibility of a gas affect its conservation of mass and momentum?

The compressibility of a gas plays a crucial role in its conservation of mass and momentum. Since a compressible gas can change its density, this affects the overall mass of the gas and can lead to changes in its momentum. As a result, the equations for conservation of mass and momentum must take into account the compressibility of the gas.

## 5. What are some practical applications of deriving conservation of mass and momentum for an inviscid compressible gas?

Understanding the principles of conservation of mass and momentum for an inviscid compressible gas is essential in many fields of science and engineering. It is commonly applied in the study of fluid dynamics, such as in the design of aircraft and rockets, as well as in weather forecasting and understanding the behavior of gases in various industrial processes.

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