Cor van de Water wrote: > you have a total efficiency of: > 90%(batt) x 95%(controller) x 85%(motor) x 90%(drivetrain) = 65%
Excluding the battery efficiency for simplicity and considering the situation from the point of view of energy provided by or received by the battery: Consider a 1000kg (approx. 2200lb) vehicle travelling 50km/h (approx. 30mph); it has a kinetic energy of 96450.6J. Using your numbers above, 96450.6J/(0.95 x 0.85 x 0.90) = 132715J were provided by the battery to achieve this. Regenning to a complete stop provides 96450.6J * (0.95 x 0.85 x 0.90) = 70095.5J to the battery. So, (0.95 x 0.85 x0.90) = 72.7% of the available kinetic energy is provided to the battery, however only 70095.5J returned / 132715J consumed = 0.528, or 52.8% of the battery energy consumed in achieving that kinetic energy level is recovered. Joules aren't a unit that means a lot to some of us (though SI units are easy to work this example in ;^), so let's convert it to something more familiar: 1J = 1W-second, so 70095.5J = 19.47Wh. If we assume that you accelerate to 50km/h and then regen to a stop every 500m (so you start from a stop at 0m and are again stopped at 500m), then out of every 500m travelled you recover 19.47Wh. So, how much does this extend your range? If your vehicle consumes 200Wh/mi (=200Wh/1609m) at 50km/h, then every 500m you recover enough energy to travel a further 156.6m: 156.6m/500m = 31% increase in range. Another way to look at it is that accelerating to speed after the stop consumes 132715J from the battery; this is 36.86Wh, or enough energy to travel 296.6m at constant speed. So, regenning to a stop allows you to recover enough energy to travel another 156.6m, but avoiding the stop in the first place saves you enough energy to travel 296.6m further. Obviously, the shorter the distance between each start and stop, the greater the % range increase appears. How high can we push it? Let's assume it takes us 5s to accelerate to 30mph (50km/h) and 5s to regen back to a stop. Let's assume that we accelerate to speed and then immediately regen to a stop, and continuously repeat this pattern. During each acceleration we travel 34.7m and during each deceleration we travel another 34.7m. So, for every 69.4m travelled, we consume 132715J to accelerate, plus about 31055J (69.4m @ 200Wh/mi) = 163770J. We recover 70095.5J, which is enough to travel (70095.5/163770) = 42.8% further in this inefficient driving pattern. Finally, remember that for simplicity I've ignored battery efficiency. The battery is less than 100% efficient at discharging or charging, and since the energy recovered during regen must be stored in the battery until you are ready to use it, less is actually available than my simplified example suggests. Hope this helps, Roger. _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA)
