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I'm reading a book "A Mathematical Introduction to Fluid Mechanics" by Alexandre J. Chorin, and I came across the derivation of Euler's equations for isentropic flow. Page 15, the author goes from

$$\frac{d}{dt}\int_{W_t} (\frac{1}{2} \rho ||\vec{u}^2|| + \rho \epsilon ) dV = -\int_{\partial W_t} p \vec{u}\cdot \vec{n} dA + \int_{W_t}\rho \vec{u}\cdot \vec{b}dV $$

to

$$\frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla)\vec{u} = -\nabla \omega + \vec{b}$$

Now, this is supposed to be a compressible flow, so $\nabla \cdot \vec{u}$ is not necessarily equal to 0, and change in internal energy $\epsilon$ is not necessarily zero either.

The author writes

This follows from the balance of momentum using our earlier expressions for $(d/dt)E_{kinetic}$, the transport theorem, and $p = \rho^2 \frac{\partial \epsilon}{\partial \rho}$`

This is what I believe to be the earlier expressions for the $(d/dt)E_{kinetic}$

$$d/dt E_{kinetic} = \frac{d}{dt}\int_{W_t} (\frac{1}{2} \rho ||\vec{u}^2||)dV = \int _{W_t} \rho ( \vec{u}\cdot(\frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla)\vec{u}))dV$$

When I tried to reach the result myself, I get stuck at:

$$ \frac{d}{dt}\int_{W_t} (\frac{1}{2} \rho ||\vec{u}^2|| + \rho \epsilon ) dV = -\int_{\partial W_t} p \vec{u}\cdot \vec{n} dA + \int_{W_t}\rho \vec{u}\cdot \vec{b}dV\\ \int_{W_t} (\rho(\vec{u}\cdot \frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot ((\vec{u} \cdot \nabla)\vec{u})) + \rho \frac{D}{Dt}\epsilon ) dV = \int_{W_t} (- \nabla \cdot (p \vec{u}) + \rho\vec{u}\cdot \vec{b}) dV\\ \rho(\vec{u}\cdot \frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot ((\vec{u} \cdot \nabla)\vec{u})) + \rho \frac{D}{Dt}\epsilon = - (\vec{u}\cdot(\nabla p) + p\nabla\cdot \vec{u}) + \rho\vec{u}\cdot \vec{b} \\ \rho\vec{u}\cdot(\frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla)\vec{u}) + \rho \frac{\partial \epsilon}{\partial t} + \rho \nabla \cdot (\epsilon \vec{u})= - \vec{u}\cdot(\rho \nabla \omega) - p\nabla\cdot \vec{u} + \rho\vec{u}\cdot \vec{b} \\ $$

which doesn't seem to be reducible any further. UNLESS I presume it's incompressible, that is; $(D/Dt) \epsilon = 0$ and $\nabla \cdot \vec{u} = 0$. When I do, I can then do: $$\rho\vec{u}\cdot(\frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla)\vec{u}) + \rho \frac{\partial \epsilon}{\partial t} + \rho \nabla \cdot (\epsilon \vec{u})= - \vec{u}\cdot(\rho \nabla \omega) - p\nabla\cdot \vec{u} + \rho\vec{u}\cdot \vec{b} \\ \rho\vec{u}\cdot(\frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla)\vec{u})= - \vec{u}\cdot(\rho \nabla \omega) + \rho\vec{u}\cdot \vec{b} \\ \frac{\partial \vec{u}}{\partial t} + (\vec{u} \cdot \nabla)\vec{u} = -\nabla \omega + \vec{b}$$ which is exactly the answer the book claims. But this equation is supposed to describe (together with equation of conservation of mass and boundary condition for trapped volume $\vec{u}\cdot \vec{n} = 0$) compressible isentropic flow. How do I get there?

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Let's start from

$$ \frac{\mathrm{d}}{\mathrm{d}t} \int_{W_t}\left(\tfrac{1}{2}\rho\rvert\rvert\mathbf{u}\rvert\rvert^2+\rho\epsilon\right)\mathrm{d}V = -\int_{W_t}p\mathbf{u}\cdot\mathbf{n}\mathrm{d}A + \int_{W_t}\rho\mathbf{u}\cdot\mathbf{b}\mathrm{d}V, $$

and use the transport theorem and divergence theorem to obtan

$$ \int_{W_t}\left(\rho\mathbf{u}\cdot\frac{D\mathbf{u}}{Dt} + \rho\frac{D\epsilon}{Dt}\right)\mathrm{d}V = \int_{W_t}\left(-\nabla\cdot(p\mathbf{u}) + \rho\mathbf{u}\cdot\mathbf{b}\right)\mathrm{d}V $$

$$ \Longrightarrow\;\,\rho\mathbf{u}\cdot\frac{D\mathbf{u}}{Dt} + \rho\frac{D\epsilon}{Dt} = -\nabla\cdot(p\mathbf{u}) + \rho\mathbf{u}\cdot\mathbf{b}. $$

Now, we divide through by $\rho$ and use $\nabla w = \nabla p/\rho$ to obtain

$$ \mathbf{u}\cdot\frac{D\mathbf{u}}{Dt} + \frac{\partial\epsilon}{\partial\rho}\frac{D\rho}{Dt} = -\mathbf{u}\cdot\nabla w - \frac{p}{\rho}\nabla\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{b}. $$

Finally, using $D\rho/Dt=-\rho\nabla\cdot\mathbf{u}$ and $p/\rho = \rho\partial\epsilon/\partial\rho$, we find

$$ \mathbf{u}\cdot\frac{D\mathbf{u}}{Dt} - \frac{p}{\rho}\nabla\cdot\mathbf{u} = -\mathbf{u}\cdot\nabla w - \frac{p}{\rho}\nabla\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{b} $$

$$ \Longrightarrow\;\, \frac{D\mathbf{u}}{Dt} = -\nabla w + \mathbf{b}. $$

P.S. Note that $D\epsilon/Dt = \partial\epsilon/\partial t + \mathbf{u}\cdot\nabla\epsilon$, which is different from $\partial\epsilon/\partial t + \nabla(\epsilon\mathbf{u})$.

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  • $\begingroup$ Thank you very much. $D\rho / Dt = (-) \rho \nabla \cdot \vec{u}$. I tried to edit your post, but it won't let me do edits that are less than 6 characters. $\endgroup$ – Aiman Al-Eryani Feb 18 '18 at 9:47
  • $\begingroup$ No problem! And you're right, I've edited it. $\endgroup$ – Elbers Feb 18 '18 at 13:34

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