Calculate Torque on Bent Pipe | Density 805 kg/m3 | V = 5 m/s | ω = 10 rad/s

[thshen34]

Homework Statement

Consider a steady flow of liquid with a density of 805 kg/m3 through a rotating tube as shown in the sketch. The flow speed is V = 5 m/s. If ω = 10 rad/s, find the torque necessary to rotate the pipe. Assume a uniform velocity distribution at the exit from the pipe, and that the
incoming fluid has no angular momentum.

Homework Equations

I'm assuming we're using conservation of mass and/or conservation of angular momentum.

The Attempt at a Solution

I'd know what to do if the ω was in the plan of the paper, but it seems to be asking the torque to rotate the pipe about the sketched axis. However, the fluid forces from the fluid force points down and to the right, none of which create a moment about that axis, so there's no angular momentum. My other idea is that one of the components of the fluid's force is the centripetal force, which contributes to the ω. Does anyone else have any ideas regarding this problem?

 

[rude man]

I would focus on the outlet. Each second there are x kg of fluid emanating from the outlet, with angular momentum about the indicated axis, which you can compute. So this means there is y amount of rotational kinetic energy associated with each second, and that energy must be supplied by the rotating of the shaft. Knowing that energy, you can compute the torque about the pipe as the energy input to your system.

 

[thshen34]

Hi rude man,

Thanks for the reply. I understand what you are talking about, but from my understanding is that none of the forces from the fluid are normal to the axis. they point down and to the right, as I stated before. For angular momentum, the force would have to come either out of the page or into the page wouldn't it?

Thanks again

 

[rude man]

Hi rude man,

I understand what you are talking about, but from my understanding is that none of the forces from the fluid are normal to the axis. they point down and to the right, as I stated before. For angular momentum, the force would have to come either out of the page or into the page wouldn't it?

Thanks again

Oh, but they are normal to the axis.

The force needed to give the emanating fluid its (rotational) kinetic energy is imparted as it snakes along the bent section. As it proceedsd along the section, more and more angular momentum is imparted until it finally spouts out of the pipe. That force is continuously normal to the axis of rotation.

Think of the liquid spouting from the outlet. You can visualize each bit of fluid as it leaves the pipe and hits the air. It's rotating with the same angular rate as the edge of the pipe oulet, right? And the edge is rotating about the indicated axis with angular rate ω which is given, and surely you can compute I = rotational inertia for each segment of fluid.

 

[rude man]

Hi rude man,

For angular momentum, the force would have to come either out of the page or into the page wouldn't it?

Thanks again

Which it does! The force is circular about the axis. It's directed into & out of the page, and also to the right and to the left above & below the page. It's circular.

 

[thshen34]

Hi rude man,

I appreciate your help, but that doesn't make any sense to me. I'm looking at this as a fluid going through a pipe bend problem, and it's a very easy one. The net force of the fluid on the pipe in the y direction is ρVQ - ρVQcosθ (up) and the force in the x direction is ρVQsinθ (to the right). I don't understand how there is any force normal to the page. The axis drawn goes coaxial through the pipe, but there would only be rotation about the z-axis (in and out of the page), as if the pipe were one half of a sprinkler head.

Basically, the fluid creating a force normal to the page just doesn't make sense to me because I'm thinking of it as a 2D problem about fluid through a pipe bend first, where all the forces are within the plane and not normal to it. Thus, I think that it doesn't create any moment about the drawn axis.

If you could help clear up this concept, I'd appreciate it greatly.

Thanks

 

[rude man]

The z axis is NOT in and out of the page. It's in the PLANE of the page, pointing up. Look at the picture.

The bent part (with the exit port) moves in and out of the page. The input part, which is coaxial with the z axis, rotates but does not move at all.