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Power transmission - Pulleys

  1. Mar 7, 2015 #1
    OK, struggling a little here to grasp this one. Any help guiding me in the correct direction would be welcome.

    150mm diameter pulley is directly driven at 250 rev per min. A V-belt is used to transmit power from this electrically driven pulley to another, which is 400mm in diameter. The centre point distances between the two pulley's is 600mm.

    The included angle of the pulley groove is 40 degrees.
    The coefficient of friction is 0.4
    The ultimate strength of the belt is 8kN

    What I need to find is the actual power transmitted to the second pulley.
     
  2. jcsd
  3. Mar 7, 2015 #2
    Here's what I have thus far and the relevant formula's

    I should hasten to add that we're working against a load from the second pulley of 200 Nm.

    Belt tension for V belts: F1/F2=e^((μθ)/sinα))

    F1 = Tension in the tight side of the belt (N)
    F2 = Tension in the slack side of the belt (N)
    e = The constant 2.718
    θ = Angle of lap of belt around the pulley (Rads)
    μ = Friction coefficient
    2α = included pulley groove angle

    So from the info I'm given, I know that 2α = 40° α = 20°

    I then need to find the angle of lap from each pulley.

    Pulley A = 0.075m Radius
    Pulley B = 0.2m Radius
    The Radial difference of the two = 0.125m
    ω = 250revs min^-1 = 250x((2π)/60) = 25/3 π Rad
    Distance between the pulley's = 0.6m

    Sin^-1α = (0.125/0.6) = 12.02°
    ∴ angle of lap on Pulley A = 180 - (2x12.02) = 155.96 = 2.72 rad
    ∴ angle of lap on Pulley B = 180 + (2x12.02) = 204.04 = 3.56 rad

    I guess this is where I start to become unstuck, I need to use the tension formula previously stated, but am I assuming that F1 = 8kN max?
     
  4. Mar 7, 2015 #3
    Update: Thinking about it, I don't/shouldn't need to use the tension formula to calculate power transmitted (I think that's going to be used for the second part of this problem). I need to know the angular velocity of the second pulley so I can use P= Tω

    So, 250 revs min^-1 = 250(2π)/60 = 25/3 π rad s^-1

    ω2 = 25/3 π rad s^-1 x (0.075/0.2) = 25/8 π

    P= Tω and T = 200Nm
    ∴ P = 200 x (25/8 π) = 1963.5
    Power transmitted to second pulley = 1963.5 W or 1.9kW
     
  5. Mar 7, 2015 #4

    billy_joule

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    What you've done looks correct.

    Yes. Once you have F2 you can find the torque on either pulley when the belt is about to break.
    Then you should be able to answer a question like "What is the greatest possible load on the second pulley?"
     
  6. Mar 9, 2015 #5
    Thanks for feedback. That's pretty much the question to the next part.
     
  7. Mar 11, 2015 #6
    Ok, so following on from this, to calculate the max load on the second pulley when the ultimate strength of the belt is halved!

    If I use the V-Belt tension formula, using my new value (4kN) as F1, do I calculate for F2 using the larger angle of lap or the smaller? Large angle being the second pulley, smaller being pulley 1.
     
  8. Mar 11, 2015 #7

    billy_joule

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    You should always use the smaller pulley. The smaller pulley will always slip before the larger pulley (assuming alpha and mu are constant)
    If you do both ways and compare you'll see that using the larger pulley lap angle will give a lower F2 value: in which case the smaller pulley would be slipping.
     
  9. Mar 12, 2015 #8
    Thanks Billy, One final note, once I've calculate my new value of F2 and apply it to P = Tw, am I also using the smaller pulley's rotational speed for this formula?
     
  10. Mar 12, 2015 #9

    billy_joule

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    Once you have the net tangential force (F1-F2) on both pulleys you can use that to find information about either pulley by using the appropriate radius.
    In your case you want the torque load on pulley B. You don't need to use power or rotational speed, just good old T=Fr.
     
  11. Mar 12, 2015 #10
    Sorry, I've misled you somewhat due to my own mis-information. I'm trying to find the power transmitted once the belt strength is halved. So I've found what power is transmitted in the initial question, I need to find what it'll be after halving the original 8kN Max ultimate strength.

    So, as I've already calculated the angle of lap, I can easily find out F2 using F1/F2=e^((μθ)/sinα)) and as you've pointed out, using the smaller pulley's angle of lap.

    Once I have my new torque value, is it simply adding that new value to P= Tw to find new value of power transmitted (using rotational speed of second pulley) ?

    Thanks again
     
  12. Mar 12, 2015 #11

    billy_joule

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    Yep. Good work.
     
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