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Torque -- units and transfer thereof

  1. Mar 21, 2017 #1
    Have a general question about torque. I have a 102 inch diameter, ring shaped, flywheel that weighs 500 pounds. It is rotating at 25 RPMs. I know that the energy produced is 1304 joules, but you need to know how many Nm that equates to.

    Second question: this flywheel is placed in a direct-drive configuration with an arbor or shank of a generator. If the flywheel is producing 961ftlbs of torque, how much of that torque is being transfered to the 2" diameter cylinder that is turning the generator? See the picture for clarification. The bike tire would be my flywheel and the smaller PMG would be my generator.
     

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    Last edited by a moderator: Mar 21, 2017
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  3. Mar 21, 2017 #2

    CWatters

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    You can't do a direct conversion. The torque depends on how quickly you stop the wheel...

    Torque = Moment of inertia * angular acceleration


    I can't see the picture. Might be the tablet I'm using.
     
  4. Mar 21, 2017 #3

    CWatters

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    Ok can see the picture now.

    The torque is reduced by the ratio of the diameters so it would be 961 * 2/102. The rpm would be increased to 25 * 102/2.
     
  5. Mar 21, 2017 #4

    CWatters

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    PS If the flywheel is turning at a constant velocity (not decelerating) then it's not delivering any power or torque. Any power or torque will be provided by whatever is turning the flywheel.
     
  6. Mar 21, 2017 #5
    Are you saying the driving motor of the flywheel must stop completely? or just idle back enough to let the flywheel do the work?

    Back to the "torque added by the flywheel" question: the rate at which the flywheel slows down will be dependent on the load from the generator. I know the generator needs 20 ftlbs of torque to turn it. Does that help?
     
  7. Mar 21, 2017 #6

    CWatters

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    No they could share the load.

    For the moment let's assume the motor has been disconnected...

    In principle that figure of 20 ftlbs can be used in the equation i gave to calculate how fast the flywheel will slow down and so also calculate how long it will take to come to a halt.

    In practice it's likely to be more complicated. For example if the generator is delivering power to a fixed load then as the rpm falls the torque required to turn it must necessarily increase. So the rate at which the flywheel decelerates will increase.
     
  8. Mar 22, 2017 #7
    Thanks for the input. It is great to have a knowledge base like this to learn from.

    One more question. If the motor is producing 600ftlbs to turn the flywheel, and the flywheel is contributing 961ftlbs, would my total theoretical torque be the sum of these two values? 1561ftlbs?
     
  9. Mar 22, 2017 #8

    CWatters

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    Yes but that would be quite a difficult situation to contrive... The fly wheel and hence the generator would be slowing down at a specific rate (or the flywheel can't contribute that amount of torque). If the flywheel is slowing down the motor isn't really "turning" the flywheel, it's just reducing the rate at which it slows down. If the generator is slowing down then a constant torque of 1561ft translates into a reducing power output. Since the power output is usually dictated by the load the load must be unusual. It must be reducing as well.

    I suspect you are approaching this from the wrong direction. What are you trying to achieve with this set up? Best start by defining the characteristic of the load and working backwards. Why do you even need a flywheel?
     
  10. Mar 22, 2017 #9

    CWatters

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    I've run some numbers...

    For a hoop/ring with all the mass concentrated at the circumference the moment of inertia is given by
    I=MR2

    The energy stored in a flywheel is..
    E = 0.5Iw2
    where w is the angular velocity
    Substitute for I..
    E = 0.5MR2w2

    I work in metric so..
    500lbs = 227kg
    102" = 2.6m, R = 1.3m
    25rpm = 2.6 radians/sec

    E = 0.5 * 227 * 1.32 * 2.62
    = 1297 Joules

    That's close to your figure of 1304 Joules.

    If your generator needs 20ftlbs at 25*102/2 = 1275rpm then the power required can be calculated..

    20ftlbs = 27Nm
    1275rpm = 134 radians/sec

    Power = Torque * angular velocity
    = 27Nm * 134 rads/sec
    = 3618W (3.6kW)

    You can calculate how long the flywheel will turn the generator on it's own....

    Power = energy/time
    so
    Time = Energy/power
    = 1297/3618
    = 0.35 seconds

    Unless I've made a mistake the flywheel will stop really quickly.
     
  11. Mar 23, 2017 #10
    Good morning, sir.

    I have an engine that is producing 600ftlbs of torque at 25-30 RPMs. The engine surges-its inherent to it's design. We want to add a flywheel to smooth out some of the surging and to capture some of the torque as well. So we decided to direct drive (via outside perimeter of the flywheel) the generator off of a flywheel (102" diameter, 500lbs, 25-30rpms).

    We need 20ftlbs of torque at 1200 RPMs to get the Kw we need out of the generator.

    Can you run a smaller flywheel on the generator too? As a separate flywheel ?
     
  12. Mar 23, 2017 #11

    CWatters

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    Ah ok that makes things clearer but I see a bit of a problem...

    The available power (600ftlbs at 25-30rpm) works out at 2.1 to 2.5 kW.
    The power required to turn the generator (20ftlbs at 1200rmp) works out at 3.4kW

    So your engine isn't powerful enough to do the job by >1kW. A flywheel designed to run at a constant speed can't solve this.

    You could use a variable speed flywheel but it would be quite a challenge to make one and potentially dangerous in operation. How long would it need to deliver the missing 1kW for?

    Perhaps you can live with a lower power output?
     
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