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Power needed to accelerate up incline

  1. Dec 17, 2007 #1
    1. The problem statement, all variables and given/known data
    a vehicle of mass 200kg accelerates from rest up to 40km/h in 15seconds up an incline of 1 in 8. The resistance to motion can be considered to be in two parts, the first being due to friction and constant at 50 N/tonne with the second part being due to air resistance and being equal to 0.02v2 + v Newtons, where v is in km/hr.
    a) Calculate the resistance to motion at 40km/h
    b) Calculate the effort required to achieve 40km/h
    c) Calculate the power required

    d) we are later asked as part of coursework to choose a electric motor size for vehicle


    2. Relevant equations



    3. The attempt at a solution

    a)whilst travelling at constant velocity

    Force of air resistance (N) = 0.02*40^2+40 = 72 N
    Rolling resistance (N) = 50 * 0.2t = 10 N
    Force of incline (N) = 200*9.81*1/8 = 245.25

    total resistance to motion at 40km/h (N) = 72+10+245.25 (N) = 327.25 N

    b) Effort required = total resistance to motion + ma

    to find a

    v = 40km/h = 11.1m/s
    u = 0 m/s
    t = 15 s
    a = ? m/s^2

    v = u + at
    11.1 (m/s)= a (m/s^2)*15(s)
    a = 0.74 m/s^2

    effort required = E = 327.25 (N)+ (200(kg)*0.74(m/s^2))
    E = 475.25 N

    c)
    to find work done we need distance travelled

    s = ut + 0.5at^2
    s = 0.5*0.74(m/s^2)*15^2(t)
    s = 83.25 m

    workdone = 475.25(N)*83.25(m) = 39564.6 joules

    Power = workdone/time = 39865.6(j)/15(s) = 2637.6 j/s (watts)

    Power (hp) = 2598.2 / 745.7 = 3.49 hp

    the answer I obtained doesn't take into account the varing air resistance would i need to find an average and recalculate?

    in terms of selecting a suitable size motor if i work out the power needed at 39.9km/h with a acceleration of 0.74m/s^2 the power needed is

    power = (total resistance + ma)*v

    power = (475.25(N)+0.74(m/s^2)*200(kg))*11.1(m/s)
    power = 6918 j/s

    power (hp) = 6918/754.7 = 9.28 hp

    I'm a little confused as why the power required to reach 40km/h is so less than power needed at 39.9km/h with a acceleration 0.74m/s^s. have i done calc wrong? what size motor do i go for? please help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 17, 2007 #2

    Hootenanny

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    Your solution to (a) is correct. However, for (b) is the resistance to motion going to be constant throughout the acceleration?
     
  4. Dec 17, 2007 #3
    hi, thanks, do I need to try and find an average for air resistance?
     
  5. Dec 17, 2007 #4

    Hootenanny

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    The effort required is going to be a function of velocity. Tell me have you solved any first order differential equations before? If not, I may be sending you off on a wild goose chase.
     
  6. Dec 17, 2007 #5
    no haven't done any 'first order differential equations before' will this be hard? the main of the cousework was to select a suitable motor and battery supply.
     
  7. Dec 17, 2007 #6

    Hootenanny

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    What's your level? Have you done any calculus previously?
     
  8. Dec 17, 2007 #7
    it is BEng degree level (UK), I have gone back into education after many years so i'm quite rusty. but i know we haven't covered any differentiating in this course yet.
     
  9. Dec 17, 2007 #8

    Hootenanny

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    I think I've over-complicating things a little here anyway. What you've done for (b) looks okay to me. However, for (c) I believe that you're going to have to take into account the varying air resistance as you say. Note that for motion in one dimension we define the work done by a variable force thus;

    [tex]W = \int^{x_1}_{x_0}F\cdot dx[/tex]
     
  10. Dec 17, 2007 #9
    thanks, i'll some research into this, see if i can sovle it now
     
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