Sorry, beginner's question, but this counter-intuition is boggling my mind. Say I have a hose of a certain diameter. If I squeeze the hose at a certain point, then the pressure applied on that point increases by the action of my hand. Consequently (I think), the speed of the water at that point increases. So how come in Bernoulli's model, when there is a constriction, the static pressure decreases? This is counter-intuitive to me and I can't make sense of it.
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$\begingroup$ Yes it is counter-intuitive, as acknowledged in the following web site, but true nonetheless hyperphysics.phy-astr.gsu.edu/hbase/pber.html $\endgroup$– Bob DCommented Dec 23, 2023 at 19:52
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2 Answers
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Your hand is doing a force on a area of the hose. So you are applying a pressure on it. But the consequence for the water flow inside has nothing to do with the stress status of the hose.
There is a continuous flow, so the product $Av$ (cross section times velocity) must be constant. In the squeezed region, the cross section is smaller, so the velocity increases there, or in another words, it accelerates.
From the second law of Newton, there must be a net force to produce an acceleration. It comes exactly from the gradient of pressure. So the pressure before is bigger than inside the squeezed region.
When it goes out of it the effect is the opposite. The water velocity decreases, it decelerates, and the force is counterflow. So the pressure after the squeezed region is bigger that inside it.
answered Dec 23, 2023 at 20:46
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$\begingroup$ You went straight to my point of confusion. Sorry if I'm sounding dumb but I'm a regular Joe, just curious for physics. "Common sense" tells me that pressure is the combined force exerted onto something, in the case of air pressure is the amount of air "pressing" on top of us, then, when I'm squeezing the hose, am I not increasing the pressure on it? This may be childish of me to depict the sense of "applying pressure" onto something like this, and per your comment, if I were to measure the static pressure where I'm squeezing the hose, it would read lower, that is boggling me! 🤣😵 $\endgroup$ Commented Dec 23, 2023 at 21:00
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$\begingroup$ Suppose you apply a pressure on a rigid pipe. If it is not great enough to deform it, you agree that nothing changes for the water inside. $\endgroup$ Commented Dec 23, 2023 at 21:15
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$\begingroup$ Ok, that makes sense, so in practical terms even if I can change the shape of the pipe at a certain point, I'm not changing the pressure inside by doing it, the only thing changing the pressure is the speed flow difference, basically saying that my hand is not really "applying pressure", just doing the same as if the hose was already narrower at that point, is that it? $\endgroup$ Commented Dec 23, 2023 at 21:19
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Your hand is not increasing the pressure at that point. It is decreasing the area available to flow through. Conservation of mass requires that to keep the same flow rate, the flow speed at that point must increase.
Bernoulli's principle then states that as a consequence of the increasing speed, the static pressure at that location must decrease to compensate.
Note that if you had a pressure gauge somewhere upstream of your restriction, the pressure it reads may very well increase from its value before the restriction was introduced. But that is a "secondary" effect of sorts. It happens because you've slowed down the flow at that upstream point, thus increasing the pressure per Bernoulli. But it will never rise higher than the source supply line pressure (about 100 psi in US water systems) unless you are talking about transient spikes from slamming a valve shut.
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$\begingroup$ This all started as me trying to get the airflow on top of a wing, and I wasn't getting it, because it was as if the air was being "compressed" and that was causing me a huge headache! Common sense is a tricky thing, sometimes. $\endgroup$ Commented Dec 23, 2023 at 21:25