Different forms of fluid energy conservation equations

[happyparticle]

Homework Statement

Derive the different forms of the continuity equation

Relevant Equations

$\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

$e = \frac{1}{\gamma -1} \frac{p}{\rho}$

I asked this question about one year ago, but at that time I didn't really understand what I was doing.

After spending a lot of time in this problem, I still fail to get the asked answer.

Starting with $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ I have to derive the following expressions.

$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

I don't know what I'm doing wrong or what I'm missing, but I couldn't get those expressions.

Here is what I did.
$\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ (1)

$\frac{De}{Dt} = \frac{\partial}{\partial t} (\frac{P}{(\gamma - 1) \rho}) + (\vec{u} \cdot \nabla) (\frac{P}{(\gamma -1) \rho})$ (2)

Replacing (2) into (1) and multiplying both side by $(\gamma - 1) \rho$
$\frac{\partial p}{\partial t} - \frac{p \partial \rho}{\rho \partial t} + \vec{u} \cdot \nabla p - \frac{p}{\rho} \vec{u} \cdot \nabla \rho + (\gamma -1)p \nabla \cdot \vec{u} = - (\gamma -1) \vec{u} \cdot \nabla p$

Then using the definition of the material derivative.
$\frac{Dc}{Dt} = \frac{\partial c}{\partial t} + \vec{u} \cdot \nabla c$

$\frac{Dp}{Dt} - \frac{P}{\rho} \frac{D\rho}{Dt} + (\gamma -1)p \nabla \cdot \vec{u} = - (\gamma -1) \vec{u} \cdot \nabla p$

So far there are too much terms on the left hand side. For instance, there are two terms with $\gamma$ and in the final answer there is only one.

 

[pasmith]

The continuity equation is $$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u}) = 0.$$ This can be rewritten as $$\frac{D\rho}{Dt} + \rho \nabla \cdot \vec{u} = 0.$$ Your first relevant equation for $De/Dt$ is the conservation of internal energy, which is an independent result.

From $$(\gamma -1 )e\rho = p$$ you can take the material derivative of both sides and use the product rule on the left to get an expression for $Dp/Dt$ in terms of $De/Dt$ and $D\rho/Dt$, both of which you already know. Then you can use the quotient rule on $$\frac{D}{Dt}\left(\frac{p}{\rho^\gamma}\right).$$

 

[Chestermiller]

$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

This equation is the expression for adiabatic reversible expansion or compression of an ideal gas.

$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

This can be derived from the previous equation using the continuity equation.