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Bernoulli's principle states that air molecules moving with high velocity create an area of low pressure which in turn causes an object to fly, but how does high velocity cause low pressure?

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    $\begingroup$ Does this answer your question? Intuition behind Bernoulli's equation $\endgroup$ – Semoi Jun 4 at 20:11
  • $\begingroup$ I would like to add that none of the question there in answer why a higher velocity causes lower pressure. Rather they talk about the reverse. Also they say the velocity increases in lower pressure area as the fluid has been "forced" to flow from a higher pressure area and is thus accelerated. But if it is accelerated horizontally the force on it will also be horizontal. But they give NO INTUITION as to why the fluid which is flowing fast should GIVE RISE to a lower pressure in the 1st place. Applying bernouli etc is just maths not intuition. $\endgroup$ – Shashaank Jun 5 at 10:13
  • $\begingroup$ Everything has some sort of intution or reason,some we can easily understand, some we don't get easily. $\endgroup$ – user794763 Jun 5 at 10:19
  • $\begingroup$ @user794763 yes someone who couldn't get the intuition has down voted my answer without an explanation $\endgroup$ – Shashaank Jun 5 at 12:46
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Assume air to be incompressible ($M<0.33$) and non viscous flowing under steady condition then according to Bernoulli's principle the sum of pressure energy, kinetic energy and potential energy remain constant. Thus when velocity increases, the pressure decreases at that section to keep the total energy (head) constant.

Consider a steady flow of an incompressible and non-viscous fluid with constant density $\rho$ through a horizontal pipe with converging from $A_1$ at section X-X to $A_2$ at section Y-Y (As shown in figure).

enter image description here

From continuity equation $$A_1V_1=A_2V_2\implies V_2=\frac{A_1}{A_2}V_1$$ $$V_2>V_1\quad (\because \ \ A_1>A_2)$$

From Bernoulli's equation: $$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2$$ $$P_2=P_1-\frac{\rho}{2}(V_2^2-V_1^2)\quad (\because \ \ Z_1=Z_2)$$ $$P_2<P_1\quad (\because \ \ V_2>V_1)$$

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It is more usual to think that low pressure causes high velocity because a pressure differential implies a force which accelerates the particles of a fluid into the region of low pressure.

OTOH if we know that there is a high velocity, then this must have been caused by a pressure differential, so the logical implication works both ways.

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That is a wonderful question. See if there is low pressure in some region then in surrounding areas there will be a high pressure. Imagine a pipe which is constricted in the middle. The high pressure in surrounding areas will cause the molecules to flow into the constricted area with a larger horizontal velocity and less vertical velocity. Essentially the pressure comes due to vertical velocity, when the molecules bounce off the surface below, a lower vertical velocity implies a lower pressure.

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  • $\begingroup$ Would the down voter like to explain the reason for the down vote. Down voting an answer without an explanation is ridiculous... $\endgroup$ – Shashaank Jun 5 at 12:45
  • $\begingroup$ Re. "The high pressure in surrounding areas will cause the molecules to flow into the constricted area with a larger horizontal velocity and less vertical velocity". Is that due to conservation of momentum?. $\endgroup$ – user45664 Jun 5 at 19:33
  • $\begingroup$ @user45664 No, the different molecules of water will have different velocities. Only those molecules with a large horizontal velocity ( and less vertical velocity) will be able to move into the constricted area and move parallely through it ( imagine this) with the highest probability. But then their vertical velocity will be less than those which are present in the broader zone. So they will exert less pressure on the walls since pressure is nothing but rate of change of vertical momentum per unit area... $\endgroup$ – Shashaank Jun 6 at 15:19
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I think you doubt is the reason of the Bernoulli principle that fluid velocity causes a decrease of pressure.

For air, that is your question, it is related to the molecular source of pressure and temperature.

It is a good approximation to say that the energy of a gas comes from the translational kinetic energy of its molecules. And the macroscopical variable related to the energy is temperature.

So, at a given temperature, the energy of the gas is:

$$\sum \frac{1}{2}mv^2 = \sum \frac{p^2}{2m} = \frac{1}{2m}\sum ({p_x^2} + {p_y^2} + {p_z^2}) $$ over all molecules, where $p$ is the momentum.

If the air is not moving, the situation is symmetric, and the sum of momentum for each direction at each instant is the same. But if the air is flowing in the $x$ direction for example it is not more the case.

If temperature is the same, that sum (energy) is also the same. So for a greater $p_x$ => $p_y$ and $p_z$ must be smaller.

Pressure is the result of the number of collisions of molecules in a given area and of the force of each collision. That force is $F=\frac{dp}{dt}$.

For any given time, the force of the collisions of the molecules with any surface perpendicular to $y$ or $z$ will decrease due to decrease in $p_y$ and $p_z$. So the pressure decreases.

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