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Please imagine a situation where we stir cool water. As far as I remember, this process should increase temperature of the water (faster than heat exchange with air around would do). In essence, by stirring we increase the velocity of the water particles and this is what we observe as rise in temperature.

Now, how does this increase in velocity relate to Bernoulli's Law, where increase in velocity should be associated with increase in dynamic pressure. If so, static pressure would drop and so would the temperature (ideal gas law)?

Because Bernoulli's Law is that total pressure is constant, I can imagine the aforementioned scenario works in closed water container (constant volume), but will not hold for a water in open glass, where instead of decreasing temperature we can increase water volume.

I know there is a lot of simplification here, but is that thinking correct? The effect is probably negligible, but please confirm there is one? If the problem is that ideal gas law does not relate to liquids, would that work for the same scenario, but with a gas instead?

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"Dynamic pressure" is a (very inappropriate) name for the quantity $\frac{1}{2}\rho v^2$, where $v$ is velocity of a fluid particle (a better name is "velocity head"). A fluid particle is not a fluid molecule, but is a creature that arises out of treating fluid as a continuum. In real world flows, fluid particle is taken to be an aggregate of several molecules (at a given point in space and time), and velocity of fluid particle is defined as an average of velocity of all those molecules (assuming molecules are identical). If you go down to the level of molecules in a flow, take an aggregate of molecules which are close together, and subtract out their mean velocity (which is the fluid particle velocity), what remains is the random motion of molecules, and thermodynamic concepts such as pressure and temperature are defined based on this random molecular motion. Therefore if you increase the mean velocity of molecules, which we call fluid velocity or flow velocity, while leaving the intensity of random motion of molecules unchanged, then you will not see a change in its thermodynamic properties.

Coming to your example, imparting greater velocity to fluid by stirring, will not by itself change its temperature. However due to fluid viscosity, there is loss of kinetic energy of fluid to heat, which will tend to increase fluid temperature, but in ordinary situations this heating is negligible.

Bernoulli equation in its most unadorned form ($p/\rho+\frac{1}{2} v^2+gh$=constant) applies to flows (more precisely, to flow streamlines) when there is no dissipation. Say in your stirring example, dissipation is negligible, compared to change of other forms of energy (which appear in the Bernoulli equation). If you apply Bernoulli equation to a flow streamline in which change in height is negligible then we have the simpler equation: $p+\frac{1}{2}\rho v^2$=constant. In that case increase in flow velocity $v$ does indeed result in decrease in pressure. Whether this decrease in pressure results in decrease of temperature depends on what other assumption you make (two thermodynamic variables are required to specify the thermodynamic state of system, in the simplest case). If fluid density is assumed constant, then usually decrease in pressure causes a decrease in temperature of fluid.

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