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I need to mount a motor with a variable speed drive(VSD) onto a honey extractor but need to know how much torque I need to start the extractor to start turning.

The diameter is 700mm and the max weight at this diameter will be 50kg.

I believe a 1,1kW motor will do the job, but not sure if the configuration with the VSD will produce enough power for it to start turning automatically.

Could somebody please help me in solving this challenge?

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  • $\begingroup$ Not an easy problem to simulate, let alone solve. Depends on the honeycomb, the temperature, the material, shape, size, and number of vanes on your extractor. $\endgroup$ – Carl Witthoft Jul 1 '14 at 17:12
  • $\begingroup$ Hi Carl. It does not have to be a precise answer. Lets forget about that its a honey extractor. Lets just say that it a 50kg weight to turn at a radius of 350mm (Diameter of 700mm) $\endgroup$ – Riaan Jonker Jul 1 '14 at 17:20
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    $\begingroup$ I suggest you measure the torque required by attaching a handle to the shaft. Then measure the force needed and multiply that by the length of the handle. This is the torque. Add some margin into this when sizing the motor, perhaps 20-30% more. $\endgroup$ – user6972 Jul 1 '14 at 17:21
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    $\begingroup$ The amount of torque you need will depend on its inertia and how quickly you need it to accelerate. $\endgroup$ – Eric Jul 1 '14 at 17:22
  • $\begingroup$ More Phys.SE questions about torque of engine: physics.stackexchange.com/search?q=is%3Aq+torque+engine $\endgroup$ – Qmechanic Jul 1 '14 at 20:48
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Well, since this a physics forum, here is the physical answer:

Moment of Inertia if defined as such: $$ \tau=I \alpha $$

Where $\tau$ is torque, $I$ is moment of inertia, and $\alpha$ is the angular acceleration. The moment of inertia for an evenly distributed "barrel" (A cylindrical honey-extractor, for example) is $$I=mr^2$$

You said $r=350mm, m=50kg$, so $I=6.125 [kg\cdot m^2]$

Meaning, for acceleration of $\alpha=18 \frac{deg}{s^2}$ (which is converted into $\alpha=0.314 \frac{rad}{s^2}$) you'll need $$ \tau=6.125\times 0.314 = 1.924 N\cdot M$$

I've set $\alpha=18 \frac{deg}{s^2}$ in order to reach 15RPM (90 deg/sec) in 5 seconds. So for that angular acceleration, I'll play safe around a 3-4 NM engine :)

Friendly warning: Don't forget to add a clutch, and a brake system!! These are quite powerful engines

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  • $\begingroup$ That's a radically high angular acceleration rate! How about $\pi/100 rad/s^2$ ? That'll get up to a reasonable speed, say 10 rpm, in no time. $\endgroup$ – Carl Witthoft Jul 1 '14 at 20:00
  • $\begingroup$ You're right - Changed the angular accerelation rate to a reach 15RPM in 5 seconds $\endgroup$ – Nitay Jul 2 '14 at 9:58

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