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Simple Gear Train Efficiency and Torque Help

  1. Jan 5, 2015 #1
    1. The problem statement, all variables and given/known data

    I'm on the very last section of a question for the design of a simple gear train. I've managed to design the gear train with all criteria fulfilled and I am confident that it is correct. I just need clarification that my calculations thus far are correct and also a fresh pair of eyes on the final section, as I'm struggling to interpret it properly. Thanks in advance!

    So I have four spur gears in a straight line:

    Gear A = 100 rpm 80 Teeth (Unknown Input Power)
    Gear B = 160 rpm 50 Teeth (Idler Gear)
    Gear C = 160 rpm 50 Teeth (Idler Gear)
    Gear D = 400 rpm 20 Teeth (Output against a load of 200 Nm)
    All shafts carry a frictional resistance of 5 Nm

    I need to find:
    1) Input Power at Shaft 1.
    2) Efficiency of the gear train as a percentage.
    3) Determine an equation for the efficiency of the gear train in terms of the load (torque) on shaft 2 (all other factors remaining constant).

    2. The attempt at a solution

    1) I've started by converting all RPM to rad/s -

    Gear A = 100 rpm = 10.47 rad/s
    Gears B & C = 160 rpm = 16.76 rad/s
    Gear D = 400 rpm = 41.89 rad/s

    Then calculated power lost in each shaft:

    Gear A = 100 rpm = 10.47 rads/s then 10.47 * 5Nm = 52.4W
    Gear B = 160 rpm = 16.76 rads/s then 16.76 * 5NM = 83.8W
    Gear C = 160 rpm = 16.76 rads/s then 16.76 * 5Nm = 83.8W
    Gear D = 400 rpm = 41.89 rads/s then 41.89 * 5Nm = 209.5W
    Total Power Loss in Gearbox = 430W

    Calculated Power Output:

    P = W * T
    P = 41.89*200
    P = 8378W

    Calculated Power Input by adding on the total losses in the gearbox:

    8378 + 430= 8808W

    2) For the efficiency it's simply:

    Efficiency = Power Out/Power In * 100
    Efficiency = 8378/8808 * 100
    Efficiency = 95%

    3) Now this is the question I am not getting at all. The way i'm starting to interpret it is I need to calculate the efficiency using only the 200Nm load on shaft 2. Surely I've already done this with the above efficiency calculation?

    Any help will be majorly appreciated, no matter how small. Just need that lightbulb moment.
     
  2. jcsd
  3. Jan 6, 2015 #2

    Bystander

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    Not going to tell you you've screwed up, and not going to say you haven't. Just went through from the output end looking at torque: step 1) is 200 + 5; step 2) torque on shaft c is 2.5 x 205 + 5; .... ; and got 833.0 required on the input shaft. This would be compared to 800 just for the 4:1 step up. Comes up 96%. I guarantee nothing, but it's offered as an alternate solution.
     
  4. Jan 6, 2015 #3

    CWatters

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    No.

    The losses in the gearbox are constant so the efficiency won't be constant, rather it will vary with output load. They want an equation that would allow you to plot a graph of efficiency vs load torque.

    For example what happens if the output load torque is reduced until the output power is much less than the losses in the gearbox (which are still 430W). The efficiency would be very poor right.
     
  5. Jan 6, 2015 #4

    CWatters

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    PS I agree with your sums for parts 1) and 2).
     
  6. Jan 6, 2015 #5

    CWatters

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    Doing it that way I get..

    Shaft D torque = 200 + 5 = 205Nm
    Shaft C torque = (205*50/20) + 5 = 517.5Nm
    Shaft B torque = (517.5 * 50/50) + 5 = 522.5Nm
    Shaft A torque = (522.5 * 80/50) + 5 = 841Nm

    Input power = 841 * 10.47 = 8805W

    Close enough.
     
  7. Jan 10, 2015 #6
    Excellent, thanks for your response!

    I've come up with this equation, which is just a broken down version of the efficiency formula:

    Efficiency = W * T / W * T + PL

    E = 41.89 * 200 / 41.89 * 200 + 430

    Then multiplied by 100 to get percentage efficiency...95%

    Halving the torque load on shaft 2:

    E = 41.89 * 100 / 41.89 * 200 + 430

    Then multiplied by 100 to get percentage efficiency...90.7%

    Would this be correct? I think I was overthinking the question if so...

    Cheers!
     
  8. Jan 10, 2015 #7
    Correcting Typo -

    E = 41.89 * 100 / 41.89 * 100 + 430

    Then multiplied by 100 to get percentage efficiency...90.7%
     
  9. Jan 12, 2015 #8

    CWatters

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    Yes that's the right method.
     
  10. Sep 4, 2015 #9

    P G

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    Can any body help me to analyse wagon r gear train?
     
  11. Sep 4, 2015 #10

    CWatters

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    Welcome to the forum.

    Please start a new thread for your question. Use the homework template that appears when you click the "post new thread" button.
     
  12. Mar 10, 2017 #11
    41.89 * 100 / 41.89 * 100 +430 = 431? how is this correct CWatters?
    I get 97.5% from Output Power/Input Power * 100.

    Just to Clarify, I am on the same question of the same assignment and I am trying to verify my method before I submit the assignment. Also Correct me If I am wrong but question D remained unanswered in this thread?
     
  13. Mar 10, 2017 #12
    Looks like you've miscalculated:

    4189/4620 = 0.9067

    0.9067 * 100 = 90.7%

    Question D was indeed unanswered, it's not exactly explained very clearly. The only advice I was given from my lecturer:

    "Here we need an equation in which the only variable is the load torque, the other factors (including losses) are constants."

    Hope it's all going well for you.
     
  14. Mar 10, 2017 #13
    So your answer for B and C were correct but the answer d was never submitted?

    I presume you 41.89 was reference to the angular velocity of Gear D as I have 41.88 rad Sec for that, where did you get your 4620 from??

    Thanks
     
  15. Jul 14, 2017 #14
    Hi. Could you explain why the output torque equals the load? If gear D in your case turns in the opposite direction than the load, does the output torque still equal the load? Thanks!
     
  16. Jul 14, 2017 #15

    CWatters

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    Read the original post. It says that gear D is the output that drives the load...

    Quote..

    "Gear D = 400 rpm 20 Teeth (Output against a load of 200 Nm)"
     
  17. Jul 14, 2017 #16
    Hi CWatters. Thanks for your reply. Yes, I understand that. But, does that necessarily mean that the output torque equals the load? Isn't the output torque the one that drives gear D, not the one that gear D is against on? Back to my last question, if gear D turns in the opposite direction than the load, then the output torque doesn't equal the load, does it? Sorry, I'm a bit confused with the definition of the load. If you could give any textbook references, I would really appreciate it! Thanks again!
     
  18. Jul 15, 2017 #17

    CWatters

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    Sometimes people are a bit casual with the use of the word "same" or "equals" - when they mean the magnitude is the same but the direction or sign is different.

    In this case we are talking about a system that has constant rpm (no acceleration).

    You should be familiar with..

    T = Iα

    where
    T is the net torque
    I is the moment of inertia
    α is the angular acceleration

    If the speed (rpm) of the output shaft is constant then the angular acceleration is zero so the net Torque must be zero. That means the output torque turning the shaft must be of equal magnitude but opposite sign (direction) to the load torque. eg

    Toutput + Tload = 0
    or
    Toutput = - Tload

    So same magnitude but opposite direction
     
    Last edited: Jul 15, 2017
  19. Jul 15, 2017 #18

    CWatters

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    Gear D turns in the same direction as the output shaft and the load.
    The torque required to turn Gear D has the same magnitude but opposite direction to the load.

    Think about what the load might be.. It could be a disk welded to the output shaft with brake pads either side. It could be an electric generator delivering power to a town. In both cases these loads resist attempts to turn them. The torque they produce is in the opposite direction to the direction they are being turned.
     
  20. Jul 15, 2017 #19
    Thanks a lot for your reply. I really appreciate it. But, I'm sorry, I still have the same question. I hope you could help me figure this out.
    We both agree that the load opposes the motion. So, how can gear D (or any other gears) turn in the same direction as the load?
    Sorry, it may be a stupid question, but I just can't wrap my head around it. Thanks a lot CWatters!
     
  21. Jul 15, 2017 #20

    CWatters

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    Gear D and the load are fixed to the same shaft (the output shaft). So they must turn in the same direction.

    I can make a diagram if it helps?
     
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