From fluid energy conservation equation to the continuity equation

[happyparticle]

Homework Statement

Derive the continuity equation from the energy conservation equation

Relevant Equations

$\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$
$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$
$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$
$e = \frac{1}{\gamma -1} \frac{p}{\rho}$

Hey there,

First of all, all energy conservation equations for a fluid I found on google hadn't the $\gamma$ coefficient. What exactly is the difference?

Secondly, by substituting e by $e = \frac{1}{\gamma -1} \frac{p}{\rho}$ in the following equation $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ I got $\frac{D}{Dt}(\frac{1}{\gamma -1} \frac{p}{\rho}) + (\gamma - 1)(\frac{1}{\gamma -1} \frac{p}{\rho}) \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

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

Then, I'm wondering if in this case the pressure is constant to have the right hand side equal to 0?
Anyway, I don't get the expecting result.

 

[vanhees71]

That's the energy-balance and momentum-balance equation for adiabatic processes. So you deal with ideal hydrodynamics. The task is to derive the continuity equation
$$\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{u})=0.$$

 

[happyparticle]

If I understood
$$
\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p \hspace{10pt} (1)
$$
Is the energy-balance equation

and
$$
\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0 \hspace{10pt} (2)
$$
is the momentum-balanced equation

I'm should be able to get (2) from (1) by replacing $e$ in (1).

 

[vanhees71]

No, why?

 

[happyparticle]

No, why?

I have a problem that I have get equation (2) and $\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

From equation (1) by replacing e with $\frac{1}{\gamma -1} \frac{p}{\rho}$

 

[hunt_mat]

Homework Statement: Derive the continuity equation from the energy conservation equation
Relevant Equations: $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$
$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$
$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$
$e = \frac{1}{\gamma -1} \frac{p}{\rho}$

Hey there,

Secondly, by substituting e by $e = \frac{1}{\gamma -1} \frac{p}{\rho}$ in the following equation $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ I got $\frac{D}{Dt}(\frac{1}{\gamma -1} \frac{p}{\rho}) + (\gamma - 1)(\frac{1}{\gamma -1} \frac{p}{\rho}) \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

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

Then, I'm wondering if in this case the pressure is constant to have the right hand side equal to 0?
Anyway, I don't get the expecting result.

For an ideal gas, you just need a constitutive equation, the speed of sound will do. So from:
\frac{dp}{d\rho}=a^{2}=\frac{\gamma p}{\rho}
You write:
dp=\frac{\gamma p}{\rho}d\rho\Rightarrow\frac{Dp}{Dt}=\frac{\gamma p}{\rho}\frac{D\rho}{Dt}

From here, you can manipulate it however you want.