Shaft and Torque Calculations with Mass

[Mechaman]

Suppose I have a motor driving a shaft that runs 1.5m long with a screw attached, the screw is mixing a product.

Power = Torque x Speed
Torque = Power/Speed

To find my Torque I can use my Shear Max = Tr/J
And J for a shaft is Tmax = (π / 16) τmax D3

My question is, where does the mass of the shaft itself come into the formula? And How would I calculate the force the product is causing on the screw attached to the shaft?

 

[Baluncore]

My question is, where does the mass of the shaft itself come into the formula?

Once the shaft and screw mass is rotating it has stored kinetic energy. It takes no energy to keep it rotating at that fixed speed. So mass is not important once it is running, only during acceleration.

How would I calculate the force the product is causing on the screw attached to the shaft?

That will depend on the flow of the product. Mixing usually involves lifting material and dropping it again, or causing a circular flow with eddies that causes the mixing. In either case energy is needed to lift mass or circulate a viscous fluid. Once you know the energy flow needed to carry out the mixing process, you can calculate the power and so the torque on the shaft. The forces on the screw will be dependent on screw geometry. But you know the total energy flow rate = power and so can distribute that energy flow over the area of screw in contact with the product.

 

[Mechaman]

Once the shaft and screw mass is rotating it has stored kinetic energy. It takes no energy to keep it rotating at that fixed speed. So mass is not important once it is running, only during acceleration.That will depend on the flow of the product. Mixing usually involves lifting material and dropping it again, or causing a circular flow with eddies that causes the mixing. In either case energy is needed to lift mass or circulate a viscous fluid. Once you know the energy flow needed to carry out the mixing process, you can calculate the power and so the torque on the shaft. The forces on the screw will be dependent on screw geometry. But you know the total energy flow rate = power and so can distribute that energy flow over the area of screw in contact with the product.

Thanks for clearing that up, makes a lot of sense about the shaft already running. I suppose getting it to that running speed shouldn't be a problem in most situations.

 

[JBA]

That depends on whether your product is a Newtonian or Non-Newtonian fluid.

 

[Mechaman]

That depends on whether your product is a Newtonian or Non-Newtonian fluid.

It's actually a grain that I'm looking to move with an auger, would an approach like, how much weight grain is per meter and then how much flighting on the screw is in contact with the grain?

 

[Baluncore]

Thank goodness you are not mixing a non-Newtonian cornflower custard. I was thinking mixing material in an open chamber, but if you are moving a flowing material like grain with an auger, then it will be the mass flow and the change in height that will dictate the energy input requirement. Will the transfer be horizontal ?

There will also be a matter of grain friction against the rotating auger screw. You might consider the bottom half of the space between two turns of the auger as a pocket that is filled with grain. There will be a boundary layer of grain between the auger and auger casing. Steeper augers need closer pitch turns as only a small cylindrical wedge of grain can be moved in each pocket.

 

[CWatters]

If possible I would try measuring the torque required.

Is the grain discharged at the top of the auger or does it continue up a pipe?

 

[Guidestone]

Hi, sorry for getting in the middle of the topic. I just found it interesting. And I have a question.

Once the shaft and screw mass is rotating it has stored kinetic energy. It takes no energy to keep it rotating at that fixed speed. So mass is not important once it is running, only during acceleration.

Why doesn't it require energy to keep the screw rotating if the screw is displacing material in a circular motion and therefore doing work?

 

[Baluncore]

Guidestone; welcome to the thread.

Why doesn't it require energy to keep the screw rotating if the screw is displacing material in a circular motion and therefore doing work?

Because the OP question was in two distinct parts and the answer was in two parts. You have only read one.

My question is, where does the mass of the shaft itself come into the formula? And How would I calculate the force the product is causing on the screw attached to the shaft?

Without material to mix or move, no energy is needed to keep the shaft and screw rotating.
The energy needed to mix or move the material is dependent on many other unspecified material and flow rate parameters.

 

[Rx7man]

OK, so here's what I'm imagining you're working with, but instead of transferring grain from one place to another, you're putting it back into the same container, eventually mixing it.

The way I'd approach the problem is in 2 parts.. One is how high you're lifting the grain in total, The length of the auger is of little relevance here, but it's displacement per turn is.. Let's say it moves 10kg per revolution and moves it 1m vertically (though it may take 30 turns to get a particular kernel to the top), you can find the work it takes (10 * 9.8 * 1 = 98NM per turn).
The second part is a lot harder.. Friction will be a function of how much grain is on top of the auger (downward pressure) and the total area of the auger exposed to it.. I'm not quite sure how you'd go about the details of that though other than empirical measurements.