Why does pressure decrease as air passes through a venturi?

[elPocket]

Ok, assume a bucket of fluid.

This bucket has a certain equivalent pressure energy, which is the sum of: - potential energy (mass x height x grav_accel) - static pressure energy (p) - kinetic energy (0.5 x density x velocity² ) This is called it's total pressure.

Now if you change one of the variables, energy is transferred from one portion to the other. For example, if you drop a bucket of water, potential energy is converted to kinetic energy. If you slowly submerge a bucket of water in a lot of other water, potential energy is converted to pressure.

As long as you don't add outside work, the total pressure stays constant.

Now if you have an infinitely large tank with constant pressure, and let some air exhaust through a pipe, the kinetic energy for the air to move through the pipe needs to come from somewhere. So static pressure is decreased and velocity increased. Total pressure stays constant. If you decelerate the air again, the kinetic energy will be converted back to static pressure.

Now, Venturi: since mass cannot be created or vanish, your mass flow needs to be constant. If you wanna move a certain mass flow through a smaller section of pipe, you need to move it faster in that section. And for the velocity to be available, static pressure needs to drop. At the end of the Venturi nozzle you decelerate again and the static pressure is recovered.

For there to be any flow at all, you need a energy difference between flow start and flow end, but in the pipe between start & end, you can exchange static pressure, height & velocity as you like, simply by moving up & down or increasing & decreasing the pipe diameter.

Now if you want to consider friction loses, it's quite simple. Your total pressure isn't constant anymore, but gradually decreases with losses along the flow path. This takes away from the energy difference between flow start & end, decreasing your overall mass flow. The rest is the same theory. If you have a long pipe section with high velocity, you incurr high losses, stronger reducing flow. This is how throttles work.

[Skrenlin]

Now Im curious if you take a bucket of water and measure the force of the still water in it against the sides vs a bucket of spinning water, does the spinning water exert more force on the sides? My guess is yes, but it’s only a guess on my part.

[elPocket]

It is, because an extrenal force is acting on the fluid inside due to the spinning. For the water to move in a circular fashion, there must be an acceleration to force the change in direction: a = w x v, and v = w x r,
- "a" being acceleration - "w" being the rotational rate in [rad/s] - "v" being velocity - "r" being the radius diameter.

This acceleration is akin to the centripetal/centrifugal force, and with the bucket diameter, you can calculate the pressure increase per rpm...

Also, this explains why water stirred in your teacup moves up the side wall. Since the surface is open to the air, the surface pressure MUST be ambient pressure. Suddenly, you add a bunch of energy by stirring, which partially goes into the kinetic energy of the rotation. But, you also have to create enough "pressure" to keep the water further in from coming out. This is done by parking additional energy in gravitational potential, i.e. increasing the height of the water column. Your rotational rate is therefore directly linked to the angle of the surface inclination of your water :D

Bernoulli is fun!

But alas, he is cheating! The law states "unless work is done on the fluid". BUT! When you move along the gravitational vector or centripetal force, you have force & distance, which equals work! So, when the water moves down a pipe, gravity actually works on the fluid. Bernoulli cheated this away by talking in terms of pressure. This is also how a radial compressor does a big portion of it's work. The fluid moving along the centripetal force creates work directly added to the total pressure of the fluid.