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Gear train torque / power and load?

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data

    This is a final hurdle of a long question designing a simple gear train. That is done but now am not sure of my calculations for power in / out and torque in and out, so hopefully someone can help?

    OK - I have 4 spur gears inline, Gear A = 100 rpm 80 Teeth (input motor - unknown power?)
    Gear B = 160 rpm 50 Teeth (idler)
    Gear C = 160 rpm 50 Teeth (idler)
    Gear D = 400 rpm 20 Teeth (output against a load of 200 Nm)
    And ALL shafts carry a frictional resistance of 5 Nm

    1/ What is the input power at shaft 1 (gear A)?
    2/ What is the efficiency percentage?

    2. Relevant equations

    P = TW
    T = P/W
    T1 x W1 = P input

    3. The attempt at a solution

    I have done the following:

    (TA x WA) - (TB x WB) - (TC - WC) - (TD - WD) = ZERO
    And TA x WA = Power input

    Convert the rpm to rad s:
    Gear A = 100 rpm = 10.47197551 rad s
    Gears B and C = 160 rpm = 16.75516082 rad s
    Gear D = 400 rpm = 41.88790205 rad s

    Input Power (Pi) = (TB x WB) + (TC x WC) + (TD + WD)
    = (5 Nm x 16.75516082 rad s) + (5 Nm x 16.75516082 rad s) + (5 Nm + 200
    Nm x 41.88790205 rad s)
    Input Power = 8.755 Kw

    Torque in = Input power / WA = 836 Nm
    Torque out = 205 Nm
    Power out = TW = 205 Nm x 41.88790205 rad s) = 8.587 Kw

    Efficiency = Output power / Input power = 98 %


    Does this look correctly tackled? Any advice very welcome!!

    I like the 205 Nm out as the gear ratio is 1 : 4 and the output is 1/4 input, which makes sense?
     
    Last edited: Oct 26, 2013
  2. jcsd
  3. Oct 26, 2013 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    As I see it the power lost in each gear shaft will be it's angular velocity * frictional torque. So for each shaft the losses are..

    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

    Add it up and Total lost in the gearbox = 430W approx.

    Power in must = Power Out + losses in gearbox

    Power out = 41.89 * 200 = 8378W
    So
    Power in = 8378 + 430 = 8808W

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

    Not really my field but I cant see anything I've missed.
     
  4. Oct 26, 2013 #3
    Thank you so much!

    A concise and helpful reply - I've 'thanked' you.

    I think 95% sounds much more realistic, so I'll take a look and do it your way.

    Thanks again!
     
  5. Jan 14, 2014 #4
    LDC1972 following this answer and was it correct? I found the efficiency to be different to yours....
     
  6. Jan 14, 2014 #5
    Any help going folks???
     
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