Does ECO mode in electric cars make sense?

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somega
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I just saw a video about an electric car which has an ECO mode.

In ECO mode the car has less power.

So let's say in normal mode the car has 100 kW and in ECO mode 60 kW.

But why should the range of the car be greater if it has less power?

Because as far as I know to reach a speed of v it takes this energy:

[tex]E=m v^2/2[/tex]

As you can see the acceleration or power is not part of the equation.

It should not matter if you reach the speed with the power of 100kW or 60kW.

So why is the power reduced in ECO mode?
 
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For gasoline vehicles we all know the reasons. For an electric car the question does not seem at all simple to me. There are issues motor efficiency which are far beyond my knowledge but the I believe the penalty for "jackrabbit" starts is far less than for gas engine.
Suppose we wish to traverse a city block (from red light to red light) in a fixed time say equal to distance/posted speed limit.. Considering v2 air drag the optimal route is to accelerate instantly to speed limit and decelerate instantly. Diminishing the power will generate slower acceleration requiring faster max speed and more overall drag. But this is what ECO mode will lead to...so its seems a perverse incentive to me
 
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Battery efficiency decreases at high rates of charge and discharge. Good search term to learn more is lithium ion battery discharge efficiency. A good source is: https://batteryuniversity.com/learn/article/bu_808c_coulombic_and_energy_efficiency_with_the_battery. A quote from that source: The energy retrieved after a charge is always less than what had been put in. Parasitic reaction that occurs within the electrochemistry of the cell prevents the efficiency from reaching 100 percent. Ultra-fast charging and heavy loading also reduces the energy efficiency. This also contributes to battery strain by reducing cycle life.

So Eco mode not only help improve kilowatt mileage, but also improves battery life. Especially for leadfoot drivers.
 
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somega said:
$$E=mv^2/2$$
As you can see the acceleration or power is not part of the equation.
But your equation doesn't take into account the distance, which depends on the acceleration, which depends on the power. Yes it takes a given amount of energy to reach a certain velocity, but the time it takes dictates the power needed.

If you are reaching velocity ##v##, it implies necessarily an acceleration ##a##. The power ##P## is defined by:
$$P = \frac{dE}{dt} = \frac{Fdx}{dt} = ma\frac{dx}{dt} = mav$$
Or:
$$a = \frac{P}{mv}$$
But we also know that:
$$a = \frac{dv}{dt} = \frac{vdv}{dx}$$
Or:
$$dx = \frac{v}{a}dv = \frac{mv^2}{P}dv$$
The distance traveled ##s## is:
$$s= \int dx = \frac{mv^3}{3P}$$
Thus, the lower the power, the smaller the acceleration and the longer the distance traveled for a given final velocity ##v##.
 
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