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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4$\begingroup$ Does this answer your question? Intuition behind Bernoulli's equation $\endgroup$– NotMeCommented Jun 4, 2020 at 20:11
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$\begingroup$ 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. $\endgroup$– ShashaankCommented Jun 5, 2020 at 10:13
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$\begingroup$ Everything has some sort of intution or reason,some we can easily understand, some we don't get easily. $\endgroup$– user794763Commented Jun 5, 2020 at 10:19
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$\begingroup$ @user794763 yes someone who couldn't get the intuition has down voted my answer without an explanation $\endgroup$– ShashaankCommented Jun 5, 2020 at 12:46
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$\begingroup$ Very good explanation on atomic scale is in this video: youtube.com/watch?v=TcMgkU3pFBY $\endgroup$– BachorCommented Oct 17, 2020 at 11:11
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3 Answers
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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.
answered Jun 4, 2020 at 18:51
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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... $\endgroup$ Commented Jun 5, 2020 at 12:45
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$\begingroup$ 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?. $\endgroup$ Commented Jun 5, 2020 at 19:33
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$\begingroup$ @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... $\endgroup$ Commented Jun 6, 2020 at 15:19
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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).
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: $$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2$$ $$P_2=P_1-\frac{\rho}{2}(V_2^2-V_1^2)\quad (\because \ \ Z_1=Z_2)$$ $$P_2<P_1\quad (\because \ \ V_2>V_1)$$
answered Jun 4, 2020 at 19:51