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Pulley systems

  1. Jan 13, 2015 #1
    1. The problem statement, all variables and given/known data
    From the below information provided, I have to calculate the power which has been transfered to the second pully.

    2 pulley system, first pulley has a diameter of 150mm and driven by an electric motor at 250 rev min -1. , second pulley has a diameter of 400mm and is driven by a v-belt, at a load of 200NM. The distance between the two pulleys is 600mm.
    Included angle of the pulley groove is 40°. Coefficient of friction is 0.4 between belt and pulley. The belt has an ultimate strength of 8kN.

    2. Relevant equations
    Power = Torque x Angluar Velocity

    3. The attempt at a solution
    From one of the other threads I found the below -

    "Driver speed = 250 revs min-1 = 25pi/3 rad s-1 = 26.18 rad s-1
    Driven Speed (angular Velocity) = (25pi/3) * (0.075/0.2) = 25pi/8 rad s-1 = 9.817 rad s-1
    Power dissipated at pulley 2 = (25pi/8) * 200 = 625*pi W
    = 9.817 * 200 = 1963.4 W = 1.9634 kW"

    I need to understand how the angular velocity has been calculated, as I am unsure where the figures have come from.
     
  2. jcsd
  3. Jan 13, 2015 #2
    Anyone :)
     
  4. Jan 13, 2015 #3

    haruspex

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    Are you asking for an explanation of this:
    ?
    If the driving pulley is turning at rate ##\omega_1## and has radius r1, what is the rate of travel of the belt?
    If the driven pulley has radius r2, what is its angular rate?
     
  5. Jan 14, 2015 #4
    I'm unsure on how to calculate the above.

    I have found that to calculate the power, the net torque is multiplied by the angular velocity and radius. Is this correct ?

    Motor speed = 250 rev min -1

    Angular velocity = (250 x 2pi) / 60
    = 26.18 rads -1

    Is this correct so far?
     
  6. Jan 14, 2015 #5

    haruspex

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    If a pulley radius r turns through an angle ##\theta## (in radians), how far does the belt move?
    If it does that in time t, what is the average angular velocity (##\omega##) and what is the average belt speed?
    Yes.
     
  7. Jan 15, 2015 #6
    Still struggling to understand.

    If I take the angular velocity calculated above, I need to multiply this by the torque and effective radius to achieve the power transmitted into the second pulley, correct?

    I'm assuming that in the answer previously calculated in the other thread, uses the effective radius of 0.075 (rad 1) / 0.2 (rad 2????) = 0.375, but I need an understanding of why these are used as I cannot find anything through internet sources etc.

    The final answer quoted in my original post was calculated by another member so this is irrelevant, I just need an understanding of the full working out process.
     
  8. Jan 15, 2015 #7
    Can anyone advise?
     
  9. Jan 15, 2015 #8

    haruspex

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    No. Check the dimensions:
    angular velocity dimension = T-1
    torque dimension = force * distance = ML2T-2
    radius dimension = L
    multiply together: ML3T-3
    power dimension = energy/time = ML2T-3
    So what do you think you might have wrong?
    I was trying to explain that to you. Please try to answer my questions:
     
  10. Jan 17, 2015 #9
    I have finally worked it out and now understand it. Can you please review my below answer to find the power transfered if the max tension is halved to 4kN.

    F1 = 4000N
    F1/F2 = e^(μθ/sin⁡α )
    4000/F2 = e^((0.4 x 2.722)/sin⁡20 )
    4000/F2 = e^3.1834
    4000/F2 =24.1296
    F2= 4000/24.1296
    F2= 165.7715

    T= (F1 - F2 ) r
    T= (4000 -165.7715) x 0.075
    T=287.5671 Nm

    P= Tω
    ω= (250 x 2π)/60 = 26.1799

    ∴ using a torque load of 287.5671 Nm the actual power transmitted:
    P= 26.1799 x 287.5671
    P= 7.5285 kW
     
    Last edited: Jan 17, 2015
  11. Jan 17, 2015 #10

    haruspex

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    That's the max tension, but at this stage we don't know if that will be reached. Start with the actual load.
    Please define all your variables, otherwise I have to guess what you are doing. What are these angles?
     
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