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Mini Auger torque

  1. Feb 11, 2016 #1
    Hi guys,

    I've been a member for a wile and never really posted anything so here goes. I'm trying to determine the torque requirements to turn an Auger. Radius of the core is 15mm and the outside diameter is 28mm. I've read a lot of literature and thesis regarding this subject and most of what I have read goes over my head as the research I've read looks into individual forces on the flight surfaces etc. I've been asked by my supervisor to negate any forces from the powder that is being conveyed by the auger and to simplify the problem by using only the solid core diameter of the shaft. If I use I=mr^2 for the moment of inertia and T = (I/(theta)(((omega2)^2-(omega1)^2)/2). I have specified that although it isn't possible but If I had 100% efficiency from the auger negating windage etc created by the rotating shaft that I need to rotate at 2500RPM or 262 rad/s for the mass flow rate of material needed of specific density.

    If I substitute value of mass of 0.423Kg, and radius of 7.5mm, I end up with a torque requirement of 1.08 Ncm. Does this sound viable? How can I add the contribution of 2 roller bearings to the equation at each end of the shaft?

    Any help would be greatly appreciated!

    Thanks Kris.
     
  2. jcsd
  3. Feb 11, 2016 #2

    Nidum

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    Please supply a better description of auger and tell us what it is being used for . Pictures would be good .
     
  4. Feb 11, 2016 #3
    The torque calculated (1.08 Ncm) is the acceleration torque of the auger system, is that correct? I would think the torque to move the material would be much higher than the calculated empty acceleration torque. Assuming an electric motor will be used to turn the system, an electric motor may be capable of accelerating the rotating mass depending on the type of motor, as some electric motor types develop increased torque at less than operating speed. The torque and speed number listed 1.08 Ncm and 2500 rpm equal 3 watts. This seems to be a very small size motor. The bearing torques are a small number if properly installed and will be the least of the torque demands.

    My opinion of the torque ranges in order of percentage of total auger system torque are:
    1) Friction of the moving material ( (60-100)
    2) Jamming friction of any material between the auger flights and the ID of the transport tube. (20-100)
    3) Accelerating the mass of material from the feed point to the discharge point (1-10)
    4) Misalignment of the mechanical assembly due to manufacturing and assembly tolerance ( ?)
    5) Any seal friction ( 1-5)
    6) Acceleration torque during start up of an empty auger system (0.1)
    7) The roller bearings unless over loaded due to miss alignment. (.0001)

    A service factor on the electric motor of 1.15 is a minimum standard, so the motor is not operating at 100% . Additional safety factors can be added to the motor sizing selection process.

    Of course I could be totally wrong.

    Good Luck
     
  5. Feb 11, 2016 #4

    CWatters

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    I think I would ask your supervisor to clarify the problem , especially what you should include and what you should ignore.

    As it stands it appears no power is required (if the auger is rotating at constant Velocity.)
     
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