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Power Transmission Problem

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data

    I have this self study question, can anyone provide some guidance?...

    A pulley 150 mm diameter is driven directly by an electric motor at 250 revs min–1. A V-belt is used to transmit power from this pulley to a second pulley 400 mm diameter against a load of 200 Nm.

    The distance between the centre of the pulleys is 600 mm, the included angle of the pulley groove = 40°, the coefficient of friction between the belt and pulley is 0.4 and the ultimate strength of the belt is 8 kN.


    (a) Calculate the actual power transmitted to the second pulley.
    (b) Calculate the power which can be transmitted if the maximum tension in the belt is limited to half of the ultimate strength of the belt.
    (c) What would be the effect of the following factors on the maximum power which can be transmitted (give reasons for your answer):
    (i) increasing the coefficient of friction
    (ii) increasing the included angle of the pulley groove.
    (d) What would be the effect on the following if the load torque is increased and the speed maintains constant (give reasons for your answer):
    (i) the tension in the tight side of the belt
    (ii) the tension in the slack side of the belt
    (iii) the power transmitted.

    2. Relevant equations

    P(t) = τ.ω (power is the product of the torque τ and angular velocity ω)


    3. The attempt at a solution

    a)

    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
     
  2. jcsd
  3. Apr 25, 2014 #2
    Part a) looks right.
     
  4. Apr 25, 2014 #3

    adjacent

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    Gold Member

    Where is part b?
     
  5. May 7, 2014 #4
    can anyone advise on how i should solve part b?
     
  6. May 7, 2014 #5

    adjacent

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    Gold Member

    You should try it first.
     
  7. Jul 10, 2014 #6
    really struggling with this.

    do i need to equate F1 to 4000n?
     
  8. Jul 10, 2014 #7
    ok so how does this look for part B?


    F1 equated to 4000n

    F1 / F2 = e^((μ*θ/(sin(α))

    4000 / F2 = e^((0.4*2.722) / (sin(20))

    4000 / F2 = 24.13

    F2 = 165.77

    T = (F1 - F2) * r

    T = (4000 - 165.77) * 0.075m

    T = 287.56724


    P = 387.56725 * 8.333*∏ = 7524.5 W
     
  9. Jul 10, 2014 #8
    Where did you get this formula?
     
  10. Jul 11, 2014 #9
    its one of the given formula in my learning material.
     
  11. Jul 14, 2014 #10
    does my answer look reasonable?
     
  12. Jul 14, 2014 #11
    How did you know to apply that formula? What is θ?
     
  13. Aug 4, 2014 #12
    its the angle of lap of belt round pulley (radians)
     
  14. Jan 4, 2015 #13
    Trying to attempt part d) ...anybody got any pointers?
     
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