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I have derived the energy conservation equation:

\begin{equation} \frac{\partial}{\partial t} \left [ \frac{1}{2} \rho v^2 + \varepsilon + \rho \phi \right ] + \frac{\partial}{\partial x_j} \left [ \frac{1}{2} \rho v^2 v_j + \rho h v_j + \rho \phi v_j \right ]=0 \end{equation}

where $\varepsilon$ is internal energy and $\rho \phi$ is the gravitational energy. I was trying to derive the Bernoulli equation from the above equation for time independent flow. It is not clear to me which term is Bernoulli equation term. How can derive the Bernoulli equation?

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what is h and $\epsilon$? – Ron Maimon Aug 27 '12 at 4:17
@RonMaimon h is the enthalpy – Yrogirg Aug 27 '12 at 7:31
actually I hope your energy equation is right, otherwise my Bernoulli's law is wrong :-) – Yrogirg Aug 27 '12 at 7:33

I was trying to derive the Bernoulli equation from the above equation for time independent flow

If you are studying something time independent then you just let $\frac{\partial}{\partial t}$ to be zero:

$$ \frac{\partial}{\partial x_j} \left [ \frac{1}{2} \rho v^2 v_j + \rho h v_j + \rho \phi v_j \right ]=0 $$

Next step is to get rid of $\rho$. Bernoulli's equation doesn't contain $\rho$, does it? We'll need mass balance for the stationary state:

$$\frac{\partial \rho v_j}{\partial x_j} = 0$$

which together with the first equation leads to: $$ \rho v_j \frac{\partial}{\partial x_j} \left [ \frac{1}{2} v^2 + h + \phi \right ]=0 $$

Now one should recall that Bernoulli's law is valid only along streamlines/pathlines, which do coincide for a steady flow. If something called $A$ is constant along the vector field $\boldsymbol v$, then it should suffice the equation

$$v_j \frac{\partial A}{\partial x_j} = 0$$

Actually it is just a derivative of $A$ along the vector field $\boldsymbol v$ --- if the derivative is zero then $A$ is constant along the vector field.

Thus we got the Bernoulli's law:

$$ \frac{1}{2} v^2 + h + \phi = const $$

along the streamline/pathline for the stationary case.

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