I have read in several places now that for compressible flow the Bernoulli equation $$ \frac {v^2}{2}+ P\ + \Psi = \text{constant}$$


$$P=\int {dp\over \rho(p)} $$

But I don't get what is meant by that. Anything more, than that the density is a function of pressure? What do people want to express by this?

  • 2
    $\begingroup$ Could you give a source? Incompressible means $\rho =$ constant locally. So you could take that outside the integral if you wished, if you are dealing with an compressible flow then that's a different story. Density as a function of pressure means it's a barotropic flow if you wish to google that. $\endgroup$
    – Novice C
    Nov 23, 2015 at 13:15
  • $\begingroup$ I suppose what they have there is the general case of a barotropic fluid, and it is your freshman Bernoulli equation when $\rho$ = constant, i.e., $1/2 \rho v^2 + P + \rho \Psi =$ constant. The reason why they wanted $\rho(p)$ was because in the derivation of Bernoulli's equation you wanted to get everything in terms of gradients, thus you wanted $\frac{1}{\rho} \nabla p$ to be more like $\nabla \int \frac{\textrm{d}p}{\rho}$, which you need to say it was barotropic to avoid complications from the gradient on non-barotropic densities being integrated over pressure. Source for verification? $\endgroup$
    – Novice C
    Nov 23, 2015 at 13:29
  • $\begingroup$ What does "little p" $p$ mean in your equation? If you have incompressible flow, $\rho$ is not a function of pressure, by definition... $\endgroup$
    – Floris
    Nov 23, 2015 at 13:54
  • $\begingroup$ your're pf course right floris this is the general case. Akward... since this was the entire point of the question.. $\endgroup$
    – Kuhlambo
    Nov 23, 2015 at 15:38

1 Answer 1


To derive the Bernoulli equation for inviscid fluids, the plan is to rewrite the Euler equation in such a way that we have gradients. I'll write the Euler equation with gravity here

$$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u} = -\frac{1}{\rho} \vec{\nabla} p + \vec{g}.$$

Recall $g = - \vec{\nabla} \Psi$, and $\vec{u} \cdot \vec{\nabla} \vec{u} = \vec{\nabla}(\frac{1}{2} \vec{u}^2) - \vec{u} \times (\vec{\nabla} \times \vec{u})$. Now we also consider a steady flow, $\frac{\partial}{\partial t} \rightarrow 0$. Note, you can consider non-steady if you have potential flows, i.e., $\vec{\nabla} \times \vec{u} = 0 \rightarrow \vec{u} = \vec{\nabla}\phi$ (a gradient appears). Now you look at quantities along flow lines, i.e., dot the equation with $\vec{u}$, this returns

$$ \vec{u} \cdot \vec{\nabla} \left(\frac{1}{2} \vec{u}^2 + \Psi\right) = \vec{u} \cdot \left( -\frac{1}{\rho} \vec{\nabla} p\right).$$

Note that $\vec{u} \cdot (\vec{u} \times \vec{A}) = 0$, for any $\vec{A}$, including $\vec{A} = \vec{\nabla} \times \vec{u}$. Now we'd like to put that density inside that gradient so that we can have $\vec{u} \cdot \vec{\nabla}(\textrm{stuff}) = 0$, then ``stuff'' is conserved along $\vec{u}$.

Take $\rho(p,\xi)$, so a function of pressure and perhaps something else. Then let's look at how we might put $\rho$ inside the gradient but be equivalent to $\frac{1}{\rho} \vec{\nabla} p$. Consider $f = \int \frac{\textrm{d} p}{\rho(p,\xi)}$, the reason why will become apparent since we want to use the chain rule and fundamental theorem of calculus. Then

$$\vec{\nabla} f(p,\xi) = \frac{\partial f}{\partial p} \vec{\nabla} p + \frac{\partial f}{\partial \xi} \vec{\nabla} \xi = \frac{1}{\rho} \vec{\nabla} p - \int \frac{1}{\rho^2} \frac{\partial \rho}{\partial \xi} \textrm{d}p \vec{\nabla}\xi = \vec{\nabla} \int \frac{\textrm{d} p}{\rho(p,\xi)}.$$

The first step was the chain rule, and the second was the fundamental theorem of calculus to calculate $\partial_p f$. Therefore we either need that $\vec{\nabla}\xi = 0$, or $\partial_\xi \rho = 0$ so that $\vec{\nabla} f = \frac{1}{\rho} \vec{\nabla}p$. Thus either $\xi$ needs to be constant or pressure never depended on $\xi$. Therefore we see this only works for barotropic fluids, which is to equivalently say by definition, $\rho = \rho(p)$.

In conclusion

$$ \vec{u} \cdot \vec{\nabla} \left(\frac{1}{2} \vec{u}^2 + \Psi + \int \frac{\textrm{d}p}{\rho} \right) = 0.$$

A small tid-bit for anyone interested, I mentioned potential flows earlier being able to talk about time dependent flows too. I should note another nice property is that these quantities are not conserved along flow line alone, but over the entire space! The reason we considered along flow lines was the get rid of the pesky curl term, but in potential flows those drop out naturally and we never had need to consider the quantity along a flow line and the result is much more powerful. However, in a time dependent flow they need not necessarily be conserved in time (still more powerful than having to assume time independent flows).

  • $\begingroup$ should it not be that $ \frac{\partial f}{\partial \xi}=0$ is the result, showing, that everything in the function can only depend on the preassure, excluding the preassure which can depend on $\vec r$? We can shurely not conclude that $\nabla \xi=0$ since this could be anything... $\endgroup$
    – Kuhlambo
    Nov 30, 2015 at 16:09
  • $\begingroup$ And one more question. Is the $\rho$ in the integral? I think it should not be but I am not sure somehow... $\endgroup$
    – Kuhlambo
    Nov 30, 2015 at 16:38
  • $\begingroup$ you say $\partial_{\xi} \rho=0$ which is equivalent here since you set $f=\int {dp \over \rho(p,\xi)}$ $\endgroup$
    – Kuhlambo
    Dec 14, 2015 at 10:19

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