Bernoulli's equation:

$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$

There are some conditions to use Bernoulli's equation. But often we neglect them.

  1. The fluid must be incompressible. But we use this equation in many cases where the fluid is air, such as to find the lifting force on the wings of an airplane (image below), to describe how a hurricane rips a roof of a house off, to measure the speed of an air current (using venturi meter), etc. Though as far as I know, air is compressible. 1

  2. Bernoulli's equation is intended for laminar flows. But we apply this to turbulent flows assuming them to be laminar flows too (e.g. wind).

  3. The word 'incompressible' implies that the density is uniform throughout the fluid. Then consider the following image which shows a tank containing two liquids with different densities: 2 It does not matter whether we apply Bernoulli's equation to AB streamline or CD streamline, the result for the exit velocity will be the same$\star$. But according to the conditions, we can apply the equation to a streamline going through one fluid.

If so, why are such conditions imposed?

$\star$ To elaborate 3$^{\text{rd}}$ point
Applying Bernoulli's equation to points A and B: $$\small{\color{red}{(P_{\text{atm}}+h_1 \rho _1 g)} +\color{green}0 +\color{blue}{\rho _2 gh_2} = \color{red}{P_{\text{atm}}} +\color{green}{\frac 12\rho _2 v^2} +\color{blue}0}$$ Applying Bernoulli's equation to points C and D: $$\small{\color{red}{P_{\text{atm}}}+\color{green}0+ \color{blue}{(\rho _1gh_1 +\rho _2gh_2)}= \color{red}{P_{\text{atm}}} +\color{green}{\frac{1}{2}\rho_2 v^2} +\color{blue}0}$$

$\small{\text{$h_1$= height of the upper liquid}}$ $\small{\text{$h_2$= height of the lower liquid measured from the level where liquid exits}}$ $\small{\text{$\rho_1$= density of the upper liquid}}$ $\small{\text{$\rho_2$= density of the lower liquid}}$

Obviously, values for $v$ from these two equations are equal!

  • $\begingroup$ The airfoil/Bernoulli 'theory' is false: see here grc.nasa.gov/www/k-12/airplane/wrong1.html $\endgroup$
    – Gert
    Jul 22, 2021 at 11:11
  • $\begingroup$ @Gert , having a read of the above post, they say 'Equal transit theorem' is wrong, not 'Bernoulli's theorem'. See the last point there. $\endgroup$
    – ACB
    Sep 15, 2021 at 5:16

1 Answer 1


Sorry for my poor english. My native language is french.

You have a too strict conception of the applicability conditions of the theorem. These conditions must be approximately verified.

Incompressible: No fluid is strictly incompressible! The question is whether the relative variations in density are important or not. Very often, it is the changes in pressure that cause changes in density. Very often too, these pressure changes are related to velocity changes. It is therefore not complicated to justify that if the velocity remains much lower than the speed of sound, the relative variations in density are much lower than 1. (Sound waves are waves of compression of the fluid: so, in a sense, the speed of sound is a "measure" of its compressibility).

Turbulence: Bernoulli's theorem is applicable to a perfect fluid: non-viscous. To apply it, we therefore neglect the viscosity. It can be shown that the turbulent terms in the Navier Stokes equation are equivalent to a fictitious viscosity. We neglect this term when we apply Bernoulli's theorem.

In your example with two fluids, you cannot apply Bernoulli's theorem as it is. You have to break it down into two parts.

Finally, don't forget that the flow is also supposed to be permanent, which is not strictly the case here: but in an approximate way, we can sometimes consider that it is!

Edit about the last point :

  • You apply the theorem from C to D by involving an intermediate point A in the middle of the path : the theorem applies in principle between the starting point C and the ending point D. What you are doing is equivalent to applying the theorem twice in its original version.

If you consider the quantity $1/2v^2+gz+P/\mu$ on either side of point A, it is clear that it cannot remain constant when you pass from one side of A to the other side since the height does not vary, the velocity does not vary and the pressure does not vary. But the density varies.

  • Another remark: you suppose that the pressure between C and A varies as in static which is not necessarily true.
  • $\begingroup$ I agree with your other points. But this is a little unclear to me. "In your example with two fluids, you cannot apply Bernoulli's theorem as it is. You have to break it down into two parts." Can you explain it, please? $\endgroup$
    – ACB
    Jul 22, 2021 at 12:35
  • $\begingroup$ The constant in Bernoulli's theorem cannot be the same in the two parts of the flow since the pressure is continuous, the velocity is continuous, the height is continuous: therefore $\mu g h$ cannot be continuous. I am lacking a little time to go into this point, which seems interesting to me. $\endgroup$ Jul 22, 2021 at 12:58
  • $\begingroup$ Sorry, I am still unable to get your point. I have edited the question. Please take a look. $\endgroup$
    – ACB
    Jul 23, 2021 at 5:47
  • $\begingroup$ The usual theorem is $P/\mu+(1/2)v^2+gz=cst$. between the starting point and the ending point. I have the impression that this is not what you write. I will be away this morning. We can talk about it again this afternoon. $\endgroup$ Jul 23, 2021 at 6:48
  • $\begingroup$ Of course your equation and my equation are the same. Divide my equation by $\rho$ and you get the corresponding equation. That's the equation I'm used to. Both of these equations are correct. No problem, focus on this when you can. $\endgroup$
    – ACB
    Jul 23, 2021 at 8:26

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