From Bernoulli's principle we know for an incompressible fluid (constant density $\rho$) in a gravitational potential $\psi=gz$, that we can state the equation along a streamline from point 1 to point 2:

$$\int_1^2\dfrac{\partial u}{\partial t}ds+\frac{u_2^2}{2}+\frac{p_2}{\rho}+gz_2=\frac{u_1^2}{2}+\frac{p_1}{\rho}+gz_1.$$

Where $u_i$ is the velocity at point $i$, $p_i$ is the pressure at point $i$ and $z_i$ ist the height at point $i$.

Now lets consider an airfoil which is has no curvature at the bottom (lower camber) and has convex curvature at the top (upper camber). If we only consider stationary flow past the airfoil, we can eliminate the integral. Additionaly we assume that the change in height is small. As reference point (point 1) we take a point upstream far away from the airfoil where $p=p_{\infty}$ and $u=u_{\infty}$ (these values will be approximately the same for other streamlines near to that point).


If we compare the flow above the airfoil ($2$) and below the airfoil ($2'$). We notice that the flow above has to accelerate, hence $u_2>u_{2'}$. Using Bernoulli's principle we can conclude $p_2<p_{2'}$.

My Question: Is there an intuitive explanation to the relationship between faster airflow and lower pressure for this situation? This intuitive explanation should not use Bernoulli's principle or energy conservation (using pressure energy). It is not the aim to explain the generation of lift, it is to explain intuitively the relationship between velocity and pressure.

EDIT: I think I came up with an explanation. Lets imagine two particles with the same total amount of kinetic energy $T=\frac{1}{2}mv^2=\frac{1}{2}mv_t^2+\frac{1}{2}mv_n^2$. If a particle has higher tangential velocity $v_t$ this implies a lower normal velocity $v_n$. But normal velocity is a measure for pressure, because at the molecular level pressure is generated by collisions of the particles with the surface. If there is less normal velocity this implies less pressure. Does this explanation make sense? Is there any contradiction?

  • $\begingroup$ This isn't exactly the explanation of lift (a result of pressure differences). Try looking up the theory of flow turning over an airfoil to see why the speed is higher on the upper surface. $\endgroup$ May 2, 2016 at 9:21
  • $\begingroup$ That was not the question. I don't want to know why the speed is higher on the upper surface. I want to know why a higher velocity will result in a lower pressure. $\endgroup$
    – MrYouMath
    May 2, 2016 at 9:24
  • 1
    $\begingroup$ Oh, my bad! Well, the question needs an edit. The title says "higher pressure above airfoil", which isn't what happens; it's normally lower than freestream pressure. $\endgroup$ May 2, 2016 at 9:37
  • $\begingroup$ The problem with all of this is that energy is not even conserved. Drag will change the total energy in the flow. $\endgroup$
    – CuriousOne
    May 2, 2016 at 20:26
  • $\begingroup$ Do you mean drag or friction? $\endgroup$
    – MrYouMath
    May 2, 2016 at 20:28

2 Answers 2


The other way around is more intuitive; if the pressure is lower on the right, the fluid would feel a net positive force in that direction and accelerate toward right. hence it will have higher velocity there. So, lower pressure will result in higher velocity.

you can rephrase the above in a way that it sounds as what you may like but is not scientifically accurate: When an element of the gas starts to accelerate toward right, it wont transfer the pressure it feels from left to the element in its right. it uses a part of that to accelerate and transfers what is remained. so the pressure will drop as you move from left to right; hence higher velocity will result in lower pressure.

  • $\begingroup$ I had the same idea with spreading out, but that somehow contradicts incompressiblity of the flow in some way, or doesn't it? $\endgroup$
    – MrYouMath
    May 2, 2016 at 10:41
  • $\begingroup$ @MrYouMath: I agree with this explanation, but I would put it in simpler terms. It's just a consequence of $F=ma$. There is a velocity difference if and only if there is a pressure difference. One implies the other. As far as incompressibility is concerned, all that means is we're dealing with speeds well below the speed of sound. It does not mean the fluid is stiff. $\endgroup$ May 2, 2016 at 21:03
  • $\begingroup$ Can you guys check my explanation (Edit section) for consistency? $\endgroup$
    – MrYouMath
    May 2, 2016 at 21:54
  • $\begingroup$ About incompressibility there is no contradiction, the element will be elongated in in one direction and shrinks in another in such a way that its volume remains constant. $\endgroup$
    – Seyed
    May 3, 2016 at 15:10
  • $\begingroup$ I think the mean density does not change, but the local density changes with this explanation. $\endgroup$
    – MrYouMath
    May 3, 2016 at 15:33

Bernouli does not explain wing lift. You can measure an older light plane with a "plank" wing, factor in the wing area, distance over the upper and lower surfaces, cruise speed, and air density, and come up with a total lift figure of about 25% of the aircraft weight. Bernouli equations were published in an aviation text decades ago and the error propagated through the literature ever since. Accurate projections of wing lift are modeled numerically using the Navier-Stokes equations.

Disregarding your wing lift example, and applying the Bernoulli equations to a venturi, I think that your normal velocity vs tangential velocity visualization has merit. For me, "energy is conserved" always proved sufficient. Force a flow of air through a smooth area reduction in a (mostly) adiabatic regime and it has to accelerate. Energy is conserved, so it has to come out in the pressure. Visualizing that the aggregate of the molecule's velocities shifts from homogeneous to biased toward the direction of travel, thereby reducing normal pressure on surfaces, adds a nice level of visualization.

  • $\begingroup$ When you say "Bernoulli does not explain wing lift" are you by any chance confusing Bernoulli with the "equal transit time fallacy"? The Bernoulli principle isn't wrong - what's wrong is the way it is typically used to explain wing lift, assuming air parcels are re-united at the trailing edge. I've found this to be the clearest explanation. $\endgroup$ May 3, 2016 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.