Dr D
Lv 7
Dr D asked in Science & MathematicsEngineering · 1 decade ago

Power of an Electric Vehicle (Pt II)?

Follow up to a previous question.


An electric vehicle of mass 1000 kg has a battery (engine) which produces maximum power P. The resistance to motion is proportional to the velocity, and the top speed of the vehicle is 40 m/s.

To prevent very high axle torques at low speeds (as P/v --> ∞), a control mechanism is installed to regulate the power of the battery. This device prevents the tractive force from exceeding the weight of the vehicle at low speeds. You may take the tractive force as being constant (= W) at low speeds.

See http://www.geckodrive.com/photos/Step_motor_basics...

It is desired to have the vehicle accelerate from rest to 25 m/s in 10 seconds on a horizontal road. Assume that road-tyre friction cooperates with us at all times. Calculate the required value of P.


Scarlet: I asked the first question in physics, because as you said, it was purely theoretical (with the infinite acceleration at v=0). This question had an engineering adjustment. But not too many of the engineering folk answered, which incidentally reminds me why I hardly ever visit the engineering section in the first place.

Update 2:

Edward: your drag formula may be more physically realistic, but the reason I specified a linear drag-velocity relation is because I wanted to make the problem solvable. In any event, it is permissible for engineering applications to use linear approximations to non-linear problems.

Update 3:

But just commenting on your answer of 68.6 kW. One would expect that a quadratic drag profile should require less engine power than a linear drag profile.

Quadratic Drag Profile.

m*dv/dt = P/v - Av^2 ...(1)

where P/U = AU^2

where U = max speed

So A = P/U^3

Linear Drag Profile.

m*dv/dt = P/v - Bv ...(2)

where P/U = BU

So B = P/U^2 > A since U > 1

Since B > A, the linear drag results in a lower net acceleration for a given engine power. In other words, your answer should have been less than Scarlets 40.5 kW.

3 Answers

  • 1 decade ago
    Best Answer

    Any reason why the first part was in Physics and the second part is in Engineering? Just wondering.

    Interestingly, the issue that the previous question assumed constant power even at low speeds and hence unrealistic acceleration to start was one I was going to mention in my answer, but didn't because I had to go and catch a train (which actually I missed, oh well.)

    I'm not sure if you intend the tractive force to be the net force or just the component delivered by the motor, with the resistance force to be subtracted to give the net force. I'll assume the latter for now.

    The net force during the regulated period will be W - kv, where as in the previous question k = P/(v_max)^2 = P/1600, and will last until v reaches P/W. So we get

    mv' = gm - kv

    and so v' e^(kt/m) + k/m v e^(kt/m) = g e^(kt/m)

    i.e. v e^(kt/m) = gm/k e^(kt/m) + c

    whence v = gm/k + ce^(-kt/m).

    Since v(0) = 0 we get v(t) = gm/k (1 - e^(-kt/m))

    until time t such that gm/k (1 - e^(-kt/m)) = P/mg

    i.e. 1 - e^(-kt/m) = Pk/(mg)^2

    i.e. t = (-m/k) ln (1 - Pk/(mg)^2).

    In the maximum power phase, we accelerate from P/mg m/s to 25 m/s and the equation is the same as last time, namely P = mv' v + kv^2 and hence v^2 = P/k + ce^(-2kt/m). In this phase we start with v = P/mg, so we get c = (P/mg)^2 - P/k. So v^2 = P/k + ((P/mg)^2 - P/k) e^(-2kt/m).

    As before we know P/k = (v_max)^2 = 1600, so substituting this and other values gives us this equation for the time required to reach 25 m/s, assuming g = 9.81 m/s^2:

    625 = 1600 + ((P/9810)^2 - 1600) e^(-Pt/800000)

    so t = (800000/P) ln ((1600 - (P/9810)^2) / 975).

    The time for the first phase is t = (-m/k) ln (1 - Pk/(mg)^2).

    = (-1600000/P) ln (1 - P^2/(1600.(9810)^2))

    So the total time is the sum of these, which we can simplify to

    -800000/P [ln (975/1600 - 975(P/15696000)^2)]

    and must be 10 seconds, which I won't attempt to solve analytically. ;-) Numerically I get P = 40.5 kW.

    Now, if you meant the tractive force to be the net force in the first part, things are simplified. In the first part we have a constant net force of W and hence a constant acceleration of g, up to the point where v = P/mg. So the time taken is given by gt = P/mg, i.e. t = P/mg^2. The second phase is the same as before, so we get the total time as

    t = P/mg^2 + (800000/P) ln ((1600 - (P/9810)^2) / 975)

    = P/96236 + (800000/P) ln ((1600 - (P/9810)^2) / 975)

    As it happens, this case also yields a power requirement of 40.5 kW (the two cases are 40.48 and 40.47 kW). Presumably this is because the first phase only lasts about half a second, building up to a speed of around 4.1 m/s, so the resistive force doesn't make much difference (less than 5% of the total even at the end of the first phase).

  • 4 years ago

    The coal powered plants do produce a huge amount of pollution, but the question is, is the pollution added to the air from charging you car exceed the polution a conventionally-powered vehicles ever would? You are assuming they would. That may or may not be the case. But it is clear those who push for the electric cars don't want you to think about that. Also, if your power plant is NOT powered by coal, then the electric car most certainly would improve things. I think electric cars are the future, but they are still well in the future. There are too many glitches they haven't worked out yet. Ethanol, on the otherhand, is worse for the earth in many ways! It is bad from an ecological standpoint AND many are starving today thanks to all the corn taken off the market that would have gone to empoverished lands.

  • Edward
    Lv 7
    1 decade ago

    Interesting. I wonder...

    Basics first.

    The total force required is a sum of force F required for the car to accelerate in vacuum and to overcome air resistance Fd. We have

    Ft= F + Fd


    Fd= 0.5 p A Cd V^2

    Fd - drag force ,

    p is the density of the air (at 1 atm and 0C it is 1.3 kg/m^3),

    A is the effective area,

    Cd is the drag coefficient (usually 0.25 to 0.45 for a car)

    V is the car's speed of the car relative to air

    Power required is

    P=Ft V

    P=[m (V2- V1 )/ (t2-t1) + 0.5 p A Cd V^2 ]V

    P=[m V / t + 0.5 p A Cd V^2 ]V

    P=m V ^2/ t + 0.5 p A Cd V^3

    let A= ?

    Cd= ?

    However we know that

    P(max) is at 40m/s and at this point F=0

    P=Fd V=0.5 p A Cd V^3

    pCdA= 2Pmax/V^3

    we have

    P=m V ^2/ t + 0.5 2P(max ) [V / V(max)]^3

    So much for the theoretical here comes anganeering

    let Cp = .3 and A= 2.0 m^2

    P=m V ^2/ t + 0.5 p A Cd V^3

    P=1000 (25)^2/10 + 0.5 x 1.3 x 2.0 x 0.3 x (25)^3

    P=62,500 + 6,094 = 68,600 W

    P(hp) 68,600/ 746 = 92 hp

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