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Need to calculate power expended by the motor

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data

    i need to calculate the power used up to carry a vehicle with 433mm diameter wheels,mass of 163kg up a slope of 3%. The horizontal distance of the slope is 2.9km. The surface of the slope is asphalt with rolling resistance of 0.22

    From the following motor specs:

    Rated Power = 1800 W
    Nominal Speed = 1060 RPM
    Nominal speed= 111 rad/s
    Nominal torque = 16.2 Nm
    Nominal Torque = 16.2 Nm
    Continuous Torque = 42 Nm (5.0 kW)

    2. Relevant equations

    P = Energy expended(E) / time (T);

    where E = mgh*sin(α)

    and T = height(H)/ average speed(Vavg);

    3. The attempt at a solution

    P = mgHsin(α)/ (H/Vavg)

    -> The H cancels and are left with

    P = mg*Vavg*sin(α)

    → One of the problems i have is i dont know how to find the average velocity but i think:

    Nominal speed * radius = m/s

    ∴Vavg = 111*0.5*433mm = 24m/s

    ∴ P = 163*9.81*sin(3%)*24
    = 20W

    → i am also not sure what to do about the rolling resistance but i think:

    Rolling resistance Force(Frr = mg*rolling resistace(Crr))

    Frr = 163*9.81*0.22
    = 351N

    And i know W = F*d

    ∴W = 351*2.5km = 877.5KJ

    and i know P = w/t

    and that t = d/s = 2.9Km/24m/s = 120.8s

    ∴P = W/t = 877.5KJ/120.8 = 7.26KW

    Dont know if i should add or subtract from the previous power...

    PLEASE HELP ANY1:cry:...

    Thanks in advance:smile:
     
  2. jcsd
  3. Mar 31, 2014 #2

    jack action

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    Gold Member

    It will be simpler to use P = F X V (which is the same as E X d / t).

    According to the problem, it seems the speed is constant, so what do you think the average speed is?

    You do not calculate the angle of slope correctly. Grade (Slope) on Wikipedia

    The force F you are trying to find is actually a summation of 2 forces: the resistance due to the slope and the resistance due to the rolling resistance. Do they add up or are they in opposite directions? Do they oppose the motion of the vehicle, or do they help propel the vehicle? Answering these questions will help you find out if you need to add them or substract one from the other.
     
  4. Apr 1, 2014 #3
    @jack action thanks for your help:

    1)I am assuming since you said all this forces are resistances then they should be "added" .

    2)I didn't have to calculate the grade slope.. it was given as is , the car is going up a racing route with details given here [URL: http://www.mapmyride.com/za/sandton-gauteng/94-7-cycle-challenge-route-56201790] [Broken] at the bottom of the page!

    What is also worrying me is that the "hypotenuse" of this mountain is not really a straight-line like it would be in a conventional triangle... is the another to accommodate for this.. THANKS.
     
    Last edited by a moderator: May 6, 2017
  5. Apr 1, 2014 #4

    jack action

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    Yes

    You've miss the point. Going back to the link I gave you, the angle of the slope is found with:

    c9023169d6099ec637cab0bf42227400.png

    So, your equation is:

    sin α

    Not:

    sin(3%)

    Although, for very small value of α, you can assume that sin α ≈ tan α ≈ α = %slope/100; But that is no what you did. You calculated sin(0.03) as if it was 0.03 deg, not 0.03 rad.

    Another point: I don't know where that problem comes from, but the 0.22 rolling resistance coefficient is abnormally high for asphalt. It is a value more realistic for a wheel on sand. You can find some more realistic values for rolling resistance coefficient here.
     
    Last edited by a moderator: May 6, 2017
  6. Apr 1, 2014 #5
    Thanks, one other thing though: on that wapsite there is both Friction Coefficient: 1.00 & Rolling Resistance Coefficient: 0.014,

    How does the Friction Coefficient affect my vehicles motion, does it bring some kind of force?,...does it affect the power the motor uses up in this case!???
     
  7. Apr 1, 2014 #6

    jack action

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    Science Advisor
    Gold Member

  8. Apr 1, 2014 #7
    Thanks for everything...

    @

    "jack action"
     
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