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Thread: Demand factor for multiple car charging station

  1. #41
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    Quote Originally Posted by Ingenieur View Post
    work (or energy depending on context) = F d = m a d
    d = v t and 1/2 v = a t or a = 1/2 v/t
    substituting
    work = m 1/2 v/t v t = 1/2 m v^2
    2 work / m = v^2, not work/d, that equals Force
    Your derivation shows that absent outside forces, the work done on an object starting at rest will equal the resulting kinetic energy of the object. Which agrees with conservation of energy.

    Air resistance is a dissipative force like friction, energy conservation arguments aren't going to work. The energy lost to drag F over a distance d is just F * d. And your initial observation that F varies as v2 means that the energy you need to overcome drag for a given distance also varies as v2. In other words, at speeds where the energy required to keep the car moving is dominated by air resistance, the range you'll have for a fixed size energy source varies as v2.

    Cheers, Wayne

  2. #42
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    Quote Originally Posted by wwhitney View Post
    Your derivation shows that absent outside forces, the work done on an object starting at rest will equal the resulting kinetic energy of the object. Which agrees with conservation of energy.

    Air resistance is a dissipative force like friction, energy conservation arguments aren't going to work. The energy lost to drag F over a distance d is just F * d. And your initial observation that F varies as v2 means that the energy you need to overcome drag for a given distance also varies as v2. In other words, at speeds where the energy required to keep the car moving is dominated by air resistance, the range you'll have for a fixed size energy source varies as v2.

    Cheers, Wayne
    energy conservation better work
    you heat the air/body surface and friction tires, bearings, etc

    the point is going down the road at a constant speed
    engine T x tran ratio x diff ratio / tire rad
    = road thrust = drag + friction
    apply more T and you accelerate

    if you increase speed/rpm by 10% fuel rate increases ~ the same
    engine T ~ (V x P x eff) / (4 Pi)
    V = displacement, P = mean eff press ~ comp ratio x atm press, eff ~ vol eff
    mean eff press ~ charge volume ~ throttle
    consider boost on an


    engine speed is linear and a/f constant at ~13:1 (varies with comb mode)
    as does T to offset friction + drag
    rpm linear with air, and a/f is constant, and T ~ charge
    consider turbo/supercharger boost, a psi is ~ 7% more T/P
    basically increasing vol eff and hence mean eff press
    so even though drag F ~ v^2 the additional fuel required is linear with vel

    all this varies with speed, typ fric > drag if v< 70 mph
    gearing, etc
    The difference between genius and stupidity is that genius has its limits.

  3. #43
    Join Date
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    my car gets 27 mpg at 70 mph, 1 hr, 70 miles, 2.6 gal, 0.037 gpm
    if sq with v at 140 mph it would get 6.75 mpg, 1 hr, 140 mile, 20.7 gal, 0.148 gpm
    car would run dry in 50-55 min or so
    from videos it actually gets 15 based on videos

    According to studies backed by the department of energy, the average car will be at its advertised MPG at 55 mph. But as the speed increases:

    - 3% less efficient at 60 mph
    - 8% less efficient at 65 mph
    - 17% less efficient at 70 mph
    - 23% less efficient at 75 mph
    - 28% less efficient at 80 mph

    80/55^2 ~ 2.1, it does not use twice as much fuel
    only ~ 30% more
    The difference between genius and stupidity is that genius has its limits.

  4. #44
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    This was supposed to be about battery chargers.
    If you go and decide to dance with a gorilla the dance ain't over till the gorilla decides it's over.

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