# Hydrodynamics: Conservation of momentum violated by a fan in a cylinder?

Consider the following situation: there is a hollow cylinder with both caps removed. The cylinder is submerged into water. Inside the cylinder, we place a fan (or pump) that pumps the water through the cylinder parallel to it's symmetry axis.

The pump-cylinder system pushes the water backwards, which quite intuitively results in the generation of force acting on the system. However, on the other hand, the momentum of the entering water ($v_{water} \frac{dm}{dt}$) is exactly cancelled by the momentum carried out by exiting water. According to this, there is no momentum transfer, thus no induced force. Where is my logic flawed?

Edit: I assume the water has no viscosity (or we can just do the same experiment with a superfluid and use a pump instead of fan). Thus, by Bernoulli's law, the pressures are equal on both ends.

I don't agree with this part:

the momentum of the entering water is exactly cancelled by the momentum carried out by exiting water

The pump is adding momentum to both the water that enters and the water that exits...in the same direction; thus the equal and opposite momentum imparted on the apparatus. As the water goes through the tube, it is gaining momentum the whole time. The pump directly applies a force on the water that it is in contact with, and water pressure transmits that force both forwards and backwards through the water in the tube. There's no "cancellation" taking place. I think it seems a little confusing because it seems like the water is rushing into the tube under its own impulse, but the force that draws the water into the tube comes from the pump.

I found it easier to imagine that instead of water going through a tube, you have a rope being drawn through a tube by a motor. The motor pulls the rope from one pile through the tube to the other pile. It seems clearer that in this case, the rope is just gaining momentum as it gets pulled through the tube, but the situation is quite analogous.

• Water gives the system some momentum. Exactly the same amount of momentum is given to the exiting water by the system. So these two cancel. Indeed, if water would freely (without fans) flow through a segment of hose without viscosity, would the segment (and water) gain momentum? No. Jan 4, 2015 at 20:36
• You're adding momenta of two different objects - momentum given to the pump can't cancel out momentum given to the water. The pump gives the entering and exiting water momentum. Water gives the pump momentum. An inviscid fluid would not start flowing through a tube without a pump. Jan 4, 2015 at 20:44
• In the same way, in a frictionless outer-space environment, a rope could continuously pass through a tube without imparting any momentum on it, and yet if a motor were pulling the rope through, it would. Jan 4, 2015 at 20:46
• Ok, I misunderstood your answer (how is it possible, it was very clearly written?). This is true for a viscous fluid. However for an ideal fluid without viscosity, Bernoulli's law implies that the pressures are equal on both caps. +1 though Jan 4, 2015 at 20:58

Away from the cylinder, water is stationary. It is being accelerated as it approaches the cylinder: once inside, it is conceivable that its momentum doesn't change (incompressible fluid, ignoring vorticity). Water leaving the other end of the cylinder is carrying away momentum.

So looking at the entire body of water in which the cylinder is submerged, there is definitely an increase of momentum going on. If you want the cylinder to be "infinite" you still need to worry about end effects. Where is the water "at infinity" coming from and going to? You can't ignore them - and that's where your argument is going astray.

• But if you create a cylinder of zero thickness, no force can be exerted on the cylinder from the ends (as all surface normals are perpendicular to symmetry-axis). Jan 4, 2015 at 20:43