Figuring required power to spin a reel

Discussion in 'Physics' started by strantor, Apr 10, 2013.

  1. strantor

    Thread Starter AAC Fanatic!

    Oct 3, 2010
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    I'm working with a gearbox manufacturer to upgrade a customer's machine. All the machine does is spin a reel of wire. The diameter of the reel is 3ft and the weight is 1000lbs when full and the diameter is 1ft and the weight is ~150lbs when empty. It needs to pay off wire at 1000 feet per minute. I calculated 300 max rpm of the gearbox, which will occur when the reel is almost empty.

    My customer prefers that the gearbox manufacturer handle the selection of the motor because they want it to mate to the gearbox, out of the box. So this gearbox manufacturer is coming up with out-to-lunch (in my opinion) horsepower figures and trying to convince my customer that they need an expensive motor and drive when they don't. I spoke up about it and tried to back myself up with some math, but I'm not 100% confident that I figured it right. Can someone please look over this and let me know if I got it wrong.

    From the gearbox people:
    My reply:
     
  2. strantor

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    Well, I guess I was close enough. After I sent this email and CC'd my customer, the bearing company sent a quote for a 15HP motor, with no discussion or explanation about difference between 15HP and 40HP.
     
  3. WBahn

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    Mar 31, 2012
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    As you noted, one key factor is how much time is available to change speeds. But another is the steady state friction loading, which are probably small compared to the acceleration power until you can tolerate a more gradual speedup curve. But then there is the external load on the thing. Is this thing being actively controlled (even if manually) so that the payout rate matches the rate at which wire is being consumed, or is there a load on the wire in either direction.
     
  4. THE_RB

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    Feb 11, 2008
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    I didn't see any compensation for losses. :)

    Something that weighs 1000 pounds will have maybe 5-10% bearing losses, and the reduction gearbox that drives it might have 20-30% losses.
     
  5. strantor

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    The external load is actually working with the reel rather than against it. imagine if you had a 1000lb reel of sewing thread and you wanted to be able to grab end of it and take off at a sprint and unwind it. It would break; unless that reel were motorized and programmed to match your speed. It is being controlled with PID signal from a dancer (example).
    That's what the extra 4hp is for ;) I calculated 11HP then rounded up to 15
     
  6. WBahn

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    But it's apples and oranges. The bearing and gearbox losses when you are running at your steady state speed are not acceleration dependent. For instance, if you had said that the acceleration time was ten times as long you would have gotten a number more like 1hp. What would you have rounded that up to?
     
  7. strantor

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    Not sure, maybe 1.5HP. I understand it's apples and oranges, and I was only looking at the apples; seem to me that acceleration time is my limiting factor, so that's all I really cared about. Steady state power is going to be some small fraction of acceleration power, so I really didn't care about it.

    I have seen similar (actually higher rated) applications that used only a 10HP motor, so before doing any math I was already confident that not much more than 10HP would be needed. But I felt that in this situation, corresponding openly in front of my customer with someone who is supposed to be " the expert," like I needed to boost my credibility with some numbers. I just wasn't sure the numbers were 100% correct and I wanted to be prepared and have a rebuttal ready for any attempts to undermine what I'd said. No attempts were made, so it's not really an emergency situation any more.

    That being said, knowing exactly how to calculate this in the future would be nice. The main source of my lack of confidence is the variable K, radius of gyration. From what I can tell, to get the proper variable would require quite a lot of math or empirical testing. I just stuck a 1 in the formula.
     
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