Air pressure in closed (circular) tube I have learned that for fluids and gases $P_\text{total}=P_\text{dynamic} + P_\text{static}$.
Suppose we have a closed circular tube in a form of ring filled with air under some pressure.  In this case I believe the following is true.
$P_t = P_s$ as $P_d=0$.
No suppose some kind of propeller starts to move the air, so it circulates within the tube. Now what will happen?


*

*$P_t$ will stay constant $P_s$ will go down and $P_d$ will go up.

*$P_s$ will stay constant and $P_t$ will go up due to increase in $P_d$.

 A: Assuming the system with the propeller is isentropic (which is a pretty bad assumption), then $P_t$ is conserved which means that $P_s$ must decrease and $P_d$ must increase.
If the propeller is not isentropic, then $P_t$ may change and then you can't really say what happens to $P_s$ or $P_d$ without more information. 
So the answer depends on whether Bernoulli's equation is a good assumption or not. That's up to you to decide.
A: I have pondered the exact same question and believe that even if the propellor is Isentropic then the conversion between $P_s$ and $P_d$ according to Bernoullis cannot maintain an Adiabatic process within the system and as a result $P_t$ will increase
Reasoning
Still assuming the ideal frictionless fluid, as the propeller converts more $P_s$ to $P_d$ then the energy available to be shared (via collisions with the walls) with the pipes surroundings has reduced and the pipe is now no longer in thermal equilibrium with its surroundings and a net flow of energy will enter the fluid.
Once that fluid is brought to rest its total energy is increased $Final P_t$ is greater than $Initial P_t$ until thermondynamic equilibrium is again achieved.
