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Power on gear train

  1. Jun 17, 2015 #1
    1. The problem statement, all variables and given/known data
    a) Determine the input power required at shaft 1.
    b) Specify the efficiency of the gear train as a percentage.
    c) Determine an equation for the efficiency of the gear train in terms of the load (torque) on shaft 2 (all other factors remaining constant).

    Knows:
    4 gear train lie on a line.
    NA=80, rA=80mm, ωA=10π/3 rads-1 (input shaft)
    NB=50, rB=50mm, ωB=-16π/3 rads-1
    NC=50, rC=50mm, ωC=16π/3 rads-1
    ND=20, rD=20mm, ωD=-40π/3 rads-1 (output shaft)

    All shafts have a friction resistance of Tfr=5Nm
    Load on shaft 2 TL=200Nm

    2. Relevant equations


    3. The attempt at a solution
    a)
    Total torque TT=TL+2*Tfr=200+10=210Nm
    P=TTD=210*(-40π/3)=-2800π=-8.7965kW

    Required input power at shaft 1 due to load and friction is 8.7965kW.

    b)
    gear train efficiency η=(-power at output/power at input)*100%
    TA={[ωD*(TL+Tfr)]/ωA}-Tfr=815Nm

    input power=TAA=815*10π/3=2716.6667π=8534.6600W=8.5347kW

    Something not right here as input should be greater then output?

    C)
    Will need some help with it, not sure where to start.
     
  2. jcsd
  3. Jun 17, 2015 #2

    CWatters

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    Can you explain how you get 2*Tfr ? All four shafts have friction and have different ratios between where that friction occurs and where the total Torque is calculated.
     
  4. Jun 17, 2015 #3
    I thought when it says shafts it means the input and output shafts, not all 4 gears. That is way I've 2*Tfr

    I've got no example in my txt book on how to calculate torque when friction is present so I'm not entirely sure if it's correct.
     
  5. Jun 18, 2015 #4

    CWatters

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    It says all shafts and I'd assume that includes the shafts for the two middle gears.

    You could work out the torque at the input that is attributable to friction in each gear but there is a slightly easier way. The question doesn't actually ask you to calculate the torque at the input just the power at the input. Have you considered applying conservation of energy to the gearbox? Have a think about that. If you can't work out what I mean let us know and I'll give another hint.

    When do you have to submit this homework?
     
  6. Jun 18, 2015 #5
    I have done some research and I might found a way, however it seams a bit simple, maybe to simple.

    a)
    Pin=Pout+Ploss

    Pout=TLD=200*40π/3=8000π/3=8377.5804W

    Power losses due to friction on each gear
    Gear A
    P=TfrA=5*10π/3=50π/3=52.3599W
    Gear B & C (ωBC)
    P=TfrB/C=5*16π/3=80π/3=83.7758W
    Gear D
    P=TfrD=5*40π/3=200π/3=209.4395W

    Total power loss=2*83.7758+52.3599+209.4395=429.351W

    Pin=8377.5804+429.351=8806.9314W=8.8069kW

    b)
    η=(Pout/Pin)*100%=(8377.5804/8806.9314)*100%=09512*100%=95.12%

    Is a) & b) correct now?

    I don't know where to start with c).

    I don't have a deadline for the homework, however I've been on it for some time now and I need to move on.
     
  7. Jun 18, 2015 #6
    My attempt for c)

    η=output power/input power

    where input power=output power + power loss

    therefore

    η=(TLD)/(TLD+power loss)=(40π/3*TL)/(40π/3*TL+429.351)=(41.8879*TL)/41.8879*TL+429.351)

    How's that look like??
     
  8. Jun 18, 2015 #7

    CWatters

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    Post #5 looks good to me. Exactly the approach I was hinting at.

    As for c). Yes the approach looks right but try and avoid substituting the value for the friction power loss. Then pi will probably cancel.
     
  9. Jun 18, 2015 #8
    Thanks for the help on that. As of c) - I was thinking about simplifying the term too, however as there is sum in the denominator I don't think it's possible.
     
  10. Jun 18, 2015 #9

    CWatters

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    η=(TL*ωD)/(TL*ωD + power loss)

    Should simplify because all terms contain angular velocity and all are specified in terms of Pi.

    Edit: Actually all contain Pi/3
     
    Last edited: Jun 18, 2015
  11. Jun 18, 2015 #10

    CWatters

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    For what it's worth the simplest I could get was..

    Power Loss = 410Pi/3

    η=(TL*ωD)/(TL*ωD + power loss)

    = (TL*40Pi/3) / ((TL*40Pi/3 + 410Pi/3)

    Pi/3 cancels

    = (TL*40) / (TL*40 + 410)

    = TL/(TL + 10.25)
     
  12. Jun 18, 2015 #11
    Forgot that I can show the power loss in terms of π. You learn something new everyday. Thanks for your help.
     
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