- #1

happyparticle

- 426

- 20

- 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.

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.

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