I know the Bernoulli equation that states that the energy along a streamline is constant:

$$ \frac{v^{2}}{2}+gh+\frac{p}{\rho}=C $$

However I can't seem to figure out what exactly $p$ is in this equation. In every book I look and also on Wikipedia it only says "$p$ is the (static) pressure". I always considered $p$ to be the internal pressure of the liquid, i.e. the pressure exerted by the fluid on its surroundings (that includes covering fluid). But this raises a problem that I wasn't sure of how to solve.

enter image description here

Lets assume that the flow in the picture is an incompressible flow, like water. When the water flows from the wide to the narrow part of the tube, the speed $v$ increases, lowering the pressure $p$ (as $gh$ and $\rho$ remain constant). As this is an incompressible fluid, the volume cannot change. But if the pressure is lowered and the volume does not change, doesn't that mean that the temperature is lowered too? This just seems weird to me.

What goes wrong in this thought process?

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    $\begingroup$ Why does it seem weird the temperature would change? Have you ever used compressed air cans that get cold when used, or let the air out of a tire and felt how cold the air was when moving? $\endgroup$ – tpg2114 May 28 '14 at 11:10
  • $\begingroup$ @tpg2114 I always though that was due to the speed of the air, like when using a ventilator. Is it actually true that the temperature changes in this example? That would solve my difficulties. And if so, what relation is there between pressure, temperature and volume in fluids (like $\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ when dealing with ideal gasses)? Thanks for your reply :-) $\endgroup$ – Jori May 28 '14 at 11:14
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    $\begingroup$ It's all related -- air is fast because it's expanding (so pressure decreasing) and temperature decreases. If you use a bike pump you'll notice the hose gets hot from the air compressing. And if you want to use the ideal gas law for water you certainly can (and for this type of problem that's probably what they want, unless you've been using other equations of state). $\endgroup$ – tpg2114 May 28 '14 at 11:19

Bernouilli is all about conservation of energy. The drop in pressure is necessary so that work can be done on the incompressible fluid - because when it flows faster, it has more kinetic energy and that had to come from somewhere.

There is no change in temperature. This is not an ideal gas - you said yourself it is an incompressible fluid. So $PV=nRT$ does not apply.

  • $\begingroup$ What is weird is that pressure can change in an incompressible fluid. How can that be visualized? $\endgroup$ – hyportnex May 28 '14 at 15:27
  • $\begingroup$ Think of a the hydraulics in your car. When you press the brake pedal, you increase the pressure of the fluid, but it doesn't (really) compress - it just transmits the pressure from the master cylinder to the slave cylinder. At the microscopic level, imagine tightly packed balls. If I push on the top ball, the pressure is transmitted to the ball at the bottom. If the balls can jiggle, they get closer; but if they are already as close as possible, all they can do is transmit the force/pressure. $\endgroup$ – Floris May 28 '14 at 15:39
  • $\begingroup$ My problem is exactly with $your$ picture, namely that $I$ cannot imagine microscopic particles with empty space between them forming an incompressible fluid; and once I imagine it is incompressible $and$ continuous I cannot see how internal pressure can change, sure it will transmit the force form one end to another but how can that be felt $internally$. $\endgroup$ – hyportnex May 28 '14 at 15:59
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    $\begingroup$ @user31748: of course nothing is incompressible-it would violate General Relativity. But when we talk of an incompressible liquid, we mean the compression can be ignored. For water, the compressibility is around $4-5 \cdot 10^{-10}\ Pa^{-1}$, so going from 0 to 1 atm results in a change in volume of around $4-5 \cdot 10^{-5}$. If you can ignore that in your calculation, water is incompressible. $\endgroup$ – Ross Millikan May 28 '14 at 21:26
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    $\begingroup$ And yes, the speed of sound under the incompressible approximation is, in fact, infinite. $\endgroup$ – tpg2114 May 28 '14 at 22:19

If we think of the forces $(F=PA)$ acting on the fluid as it flows along the pressure gradient the problem becomes clearer.

An incompressible fluid flowing from a region of higher pressure to a region of lower pressure feels a net force in the direction of the pressure gradient. This net force causes the fluid to accelerate (bulk motion).

Thus there is an increase in kinetic energy. On the assumption that the water is incompressible there will not be a change in temperature.


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