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I'm trying to understand a derivation of Bernoilli's Equation and I'm having a hard time understanding the math behind this indentity

$$ \frac{1}{\rho} \nabla p = \nabla \int \frac{\textrm{d}p}{\rho}.$$

I should say that this is for a barotropic flow so $p=p(\rho).$ I'd also appreciate a quick comment on how seeing isentropic fluids are barotropic flows (as lead to me believing by wikipedia, but I do not believe that work both ways).

I've tried showing the LHS is equal by doing differentiation under the integral but I'm a bit lost, from the notation and I suppose math. My guess is the bounds are meant to be from 0 to $p$. Then I get the first term back, but now I'm stuck with an integral I wish to say was zero, but don't know how.

$$ \nabla \int_0^p \frac{\textrm{d}p}{\rho} = \frac{1}{\rho} \nabla p + \int_0^p \nabla \left(\frac{1}{\rho}\right) \textrm{d}p.$$

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  • $\begingroup$ Bernoulli equation is valid for incompressible flow so $\rho$ is a constant and that term is indeed zero. $\endgroup$ – nluigi Nov 5 '15 at 11:10
  • $\begingroup$ Very true, but I also want to consider a compressible flow. If I can show this identity, in that case, I can reach agreement with what is listed on wikipedia for this subject. en.wikipedia.org/wiki/… $\endgroup$ – Novice C Nov 6 '15 at 4:10
  • $\begingroup$ Here is a believed solution I came up with, using chain rule+ vanilla Fundamental Theorem of Calculus (noting the integral is only a function of pressure, so a partial is equal to a total in this case) instead. $$\nabla \int \frac{\textrm{d}p}{\rho} = \frac{\partial}{\partial p} \left(\int \frac{\textrm{d}p}{\rho}\right) \nabla p = \frac{1}{\rho} \nabla p.$$ Which raises an interesting question if this shows the integral in my answer is zero, and what the means physically. But perhaps I just don't understand a subtlety of calculus and either attempt flawed—wouldn't be the first time. $\endgroup$ – Novice C Nov 6 '15 at 4:11
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I think your solution in the comment is right. Now to add some physical explanation to it.

Bernoulli's constant is as such $$H = \frac{1}{2}u^2 + \Psi + \int \frac{dp}{\rho} $$

The first term is the specific kinetic energy. The second is the specific gravitational potential energy. The third is what some people call the pressure potential.

You can think of the pressure potential as a cleverly hacked up internal energy term of the fluid element: In (monoatomic) ideal gases, internal energy refers to the thermal motion of the particles. For van der waals gases, internal energy also includes long range attraction and short range repulsion interaction energy. In MHD fluids, internal energy may even include magnetic energy density.

These internal energies have very different microscopic origins, but they all share one property: they contribute to the pressure of the fluid. Said another way, if we isolate a fluid element and remove all external fluid around it, these internal energies have their own little forces that pushes the fluid element to expand.

But how do we quantify the internal energy from the pressure? We now waddle into the tricky concept of enthalpy vs internal energy. I think we can't quantify internal energy (wrong natural variables). But we can quantify the enthalpy (which fluid engineers seem to do a lot).

Briefly, the enthalpy (with a tilde to distinguish it from Bernoulli's constant) is defined as $$ d\tilde{H} = TdS + VdP $$ .

Specific enthalpy: $$d\tilde{h} = \frac{dH}{m} = T\frac{dS}{m} + \frac{V}{m}dP = Tds + \frac{dP}{\rho}$$

For an isoentropic fluid, no heat transfers between fluid elements so specific entropy $s$ doesnt change $$d\tilde{h} = \frac{dP}{\rho}$$

Now, looking at this, we can interpret $\tilde{h}$ as a potential on its own right if the fluid is barotropic.

$$ \tilde{h} = \int \frac{dP}{\rho} $$

And if we do that, the pressure potential $\tilde{h}$ is a field on its own like the gravitational potential $\Psi$. It depends on position implicitly through the density (or pressure) field. We now reinterpret the equation in the question as: \begin{align*} d\tilde{h} &= \frac{dP}{\rho} \\ (\mathbf{\nabla }\tilde{h}) \cdot d\mathbf{r} &= \frac{(\mathbf{\nabla }P) \cdot d\mathbf{r}}{\rho} \\ \mathbf{\nabla }\tilde{h} &= \frac{\mathbf{\nabla }P }{\rho} \end{align*}

References: http://www.astro.yale.edu/vdbosch/astro320_summary11.pdf https://www.grc.nasa.gov/www/k-12/airplane/enthalpy.html

Added (21 May 2017):

To give some feel of numbers, the pressure potential (or specific enthalpy) $\tilde{h}$ is larger than the specific internal energy. Assuming an equation of state of $ p = K \rho^\gamma$, where $\gamma = \frac{5}{3}$ for monoatomic gas

$$ \tilde{h} = \int \frac{dP}{\rho} = \frac{\gamma}{\gamma -1} \frac{P}{\rho} = \frac{5}{2} \frac{P}{\rho} $$

whereas specific internal energy (per mass), derived by assuming (Tds = 0), gives $$ d\epsilon = \frac{-PdV}{m} = \frac{-(K\rho^{\gamma})d(m/\rho)}{m} = d\left(\frac{P}{\rho(\gamma - 1)} \right)$$

$$ \epsilon = \frac{1}{\gamma - 1} \frac{P}{\rho} = \frac{3}{2}\frac{P}{\rho}$$

This reminds us that specific enthalpy is $\tilde{h} = \epsilon + \frac{P}{\rho}$

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