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Gearbox and flywheel

  1. Aug 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A flywheel is attached on the output of the gearbox. The output shaft rotates in the opposite direction to the input shaft at 5 times its speed. The gearbox has an efficiency of 92%.

    If the flywheel is solid, has a mass of 50kg, a diameter of 1.5m and is to accelerate from rest to 300 revs min-1 in 1min:

    a) Calculate the torque required at input T1.

    b) Calculate the magnitude and direction of the torque required to hold the gearbox stationary (holing torque Th). Show the direction of the holding torque applied to the shaft with the aid of a sketch.

    c) Plot a graph of the input power against time when taking the flywheel from rest to 300 revs min-1.

    2. Relevant equations
    α=(ω21)/t
    T=I*α
    I=0.5*m*r2
    η=(-Too)/(Tii)

    3. The attempt at a solution
    a)
    ωi=-0.2ωo
    ω2=300 revs min-1=300*(2π/60)=10π rads-1
    ωo=-10π rads-1

    α=(ω21)/t=(10π-0)/60=π/6 rads-2
    I=0.5*m*r2=0.5*50*0.752=14.062 kgm2
    To=I*α=14.062*(π/6)=7.3628 Nm

    ωi=-0.2*-10π=2π rads-1

    η=(-Too)/(Tii)
    η(Tii)=-Too
    Tii=(-Too)/η

    Ti=(-Too)/(η*ωi)=(-7.3628*-10π)/(0.92*2π)=36.814/0.92=40.0152 Nm

    b)
    Ti+To+Th=0
    Th=-Ti-To=-40.0152-7.3628=-47.378 Nm

    As the holding torque Th has a negative sign the direction of it is clockwise which is opposite direction to the Ti and To.

    20150802_123638.jpg

    The a) and b) seem to be not too difficult (if I got it right), however after spending several hours I can't workout c). Can someone put me in the right direction. I hope a) and b) are correct.
     
  2. jcsd
  3. Aug 2, 2015 #2

    SteamKing

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    1. What's the definition of power if you know the torque on a shaft and its angular velocity?

    2. If you know the relationship between torque, mass moment of inertia, and angular acceleration, can't you calculate the change in angular velocity of the shaft versus time?

    3. If you can calculate the change in angular velocity, can't you also calculate the angular velocity of the shaft at a particular time t = t0 ?

    4. Apply the results of 3.) above with the formula in 1.) and make a table of Power versus time.

    5. Plot results in table of 4.) above.
     
  4. Aug 2, 2015 #3
    Thanks SteamKing for your reply. I will give it a go. Any comment about a) and b)?
     
  5. Aug 2, 2015 #4

    SteamKing

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    Not at this time.
     
  6. Aug 3, 2015 #5
    My attempt:
    1. P=T*ω
    2. T=I*α
    so far easy
    Change in angular velocity of the shaft versus time?
    Is it α=(ω21)/t
    4. T=I*[(ω21)/t]
    as ω1=0
    T=I*(ω2/t)
    therefore P=(I*ω2i)/t
    as I=14.062 kgm-2
    ω2=10π rads-1
    ωi=2π rads-1

    P=(14.062*10π*2π)/t=2775.7275/t

    Just need to now plot a graph of P for t between 0 and 60s.

    Does it make any sense?

    Any luck with a) and b) - sorry to be a pest.
     
  7. Aug 3, 2015 #6

    SteamKing

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    No. There's some mistakes here.

    1. P = T ω, not P = T ω1ω2 . I think you mistyped your equation here, multiplying the omegas instead of subtracting them.

    2. Once you calculate the angular acceleration α of the flywheel, you want to calculate the ω of the shaft for a few intervals between t = 0 and t = 1 min. and the power applied to the shaft. ω = ω0 + α*t . Set up a table of the values of time, the speed of the shaft, and the power. Plot Power versus time.
     
  8. Aug 4, 2015 #7
    SteamKing,

    I think I'm to stupid for this, let's take it a step at a time.

    You are correct, P=T*ω.
    as T=I*α
    and α=(ω21)/t
    then T=[I*(ω21)]/t

    as ω1=0

    so T=(I*ω2)/t

    substituting into power equation

    P=(I*ω2i)/t
     
  9. Aug 4, 2015 #8

    SteamKing

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    OK, I see your derivation now.

    Can you complete answering the question?
     
  10. Aug 4, 2015 #9
    As I=14.062 kgm2
    ω2=300 revmin-1=10π rads-1
    ωi=2π rads-1

    then, P=(14.062*10π*2π)/t=2775.7275/t
     
  11. Aug 6, 2015 #10
    Hi SteamKing, I had another think about it and my above solution is incorrect as I used the moment of inertia from the flywheel where I have to calculate the power input. I believe it should be:
    P=T*ω

    as α=(ω2-ω1)/t
    ω1=0, so
    α=ω2/t
    ω2=α*t

    Substituting this into power equation

    P=T(α*t)
    as T=40.0152 Nm and α=π/6 rads-2

    P=40.0152*(π/6)*t=20.9519*t

    The graph looks like this:

    Book1.jpg

    That is only if I got the question a) right.
     
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