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Shaft and Torque Calculations with Mass

  1. Dec 15, 2016 #1
    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?
     
  2. jcsd
  3. Dec 16, 2016 #2

    Baluncore

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    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.
     
  4. Dec 18, 2016 #3
    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.
     
  5. Dec 18, 2016 #4

    JBA

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    That depends on whether your product is a Newtonian or Non-Newtonian fluid.
     
  6. Dec 19, 2016 #5
    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?
     
  7. Dec 19, 2016 #6

    Baluncore

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    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.
     
  8. Dec 19, 2016 #7

    CWatters

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    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?
     
  9. Dec 20, 2016 #8
    Hi, sorry for getting in the middle of the topic. I just found it interesting. And I have a question.

    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?
     
  10. Dec 20, 2016 #9

    Baluncore

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    Guidestone; welcome to the thread.
    Because the OP question was in two distinct parts and the answer was in two parts. You have only read one.
    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.
     
  11. Dec 21, 2016 #10
    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.
    SP-Auger-570x380-pixels.jpg

    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.. Lets 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.
     
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