Impeller Design Question

I have a simple Impeller or Turbine that is inside of a tube. The impeller will be used to measure the volume of liquid passed through the tube. The tube ID is very close to the impeller OD. It is a four blade turbine with a 1/4 twist (the end of the blade is offset 90 degrees from the beginning). How can I theoretically estimate the volume of water that it takes to turn the turbine one full revolution?

I know the volume of the turbine and how much water can fit around the turbine inside of the tube.

Any help is appreciated....

[SailDesign]

Your best bet with that is empirical testing. Honestly.
If that is not allowed, then you'll have to resort to CFD codes, and create a 3D model of the beast - but empirical is simpler

I will work with that some more... But I am looking for a baseline to compare that data to.. I have a 3D model and can get a volume from it... Would it just be the amount of water that would travel the length of the blades times four to get a complete revolution? Disregarding any frictional losses of course....

[SailDesign]

Would it just be the amount of water that would travel the length of the blades times four to get a complete revolution? Disregarding any frictional losses of course....

Not quite... You have to take intyo account things like blade efficiency, rotational resistance, etc., as well as frictional losses onthe walls, blades, and so on.

It is NOT as simple as it looks.

Steve

[SailDesign]

<blush>
Obviously, punctuation and spelling are not as simple as they look at this time of night either...
</blush>

[tspeer]

The blades are twisted 90 degrees? That would make the tips flat against the flow. Or do you mean 90 degrees from tip to tip, resulting in 45 degrees of twist to the blades? Either way, it sounds excessive.

Oh - I get it. The tips are aligned with the flow and the hub is at 90 degrees to the flow. That's got to be a lousy way to make an impeller for measuring flow rate - it's a huge obstruction! A much better shape would be a piece of metal lined up edge-on to the flow and twisted like a propeller. At the center, the linear velocity of such an impeller would be zero and the flow would pass by it freely. At the tip, the rotation of the impeller would also line up the blade with the local flow and present minimal resistance.

For this alternate design, I would look at the advance ratio corresponding to blade pitch and and back out the rpm's from that. Since there's negligible torque on the blades, they ought to rotate fast enough to have a small angle of attack at the blade tips. So if the blade twist is theta (with theta=zero meaning the blade is flat and aligned with the flow) and the flow through the tube is moving with a speed, U (in linear units per second), the rotational speed of the impeller should be approximately

tan(theta)=R*(2*pi*rpm/60) / U

So the flow speed is approximately

U = [(R*2*pi/60) / tan(theta)] * rpm

Once you know the flow speed, the volume flow rate, Q, is the cross sectional area times the flow rate:

Q = U * pi * R^2

Q = [(R^3 * 2*pi^2/60) / tan(theta)] * rpm

The quantity in square brackets is a constant defined by the design of your impeller and tube. The more twist, the faster it will spin at a given flow rate.

If I'm right about your baseline design, there's no way you're going to be able to estimate its characteristics before testing. The middle of the blade will be driving the impleller, the tip will be resisting the rotation and the center will be separated and stalled. Which part is going to win out is anyone's guess. It may even depend on the density, viscoscity and speed of your fluid.

[CDBarry]

Look at a good reference on turbomachinery design. This is a pretty straightforward problem, but you will need the design charts.