Will Battery-Powered Cars Ever Be Practical?

I read an article recently about a new development in “battery technology” by a research group in Singapore.  I put that in quotes because what they have developed is not a battery in the conventional sense.  It is a capacitor.  A capacitor with really immense storage capabilities, if you believe their press release.

For electric vehicle (EV) applications, a capacitor is far superior to a battery.  Batteries store energy electrochemically.  Capacitors store it electrostatically.  What’s the difference?  No chemical process is required to charge or discharge a capacitor.  It can be charged and discharged at very high currents, without the heating and efficiency loss problems associated with batteries.

This is ideal for EV’s because starting a vehicle from rest…or braking it with regenerative braking…means that a lot energy must move into and out of the battery in a hurry.

Up to now, nobody has been able to make a capacitor with enough charge capacity to use in EV’s except for very short-term storage in hybrid vehicles.  Making a capacitor with sufficient storage for an EV was out of the question…until now.  Maybe.  If these guys are right.

I tend to be skeptical about these things.  I have seen too many “breakthroughs” in energy technology that have turned out to be hoaxes or wishful thinking.  I hope this one isn’t.

Based on the numbers they supplied, I did a few calculations and determined that an EV powered by their capacitor would need a capacitance value around 7000 farads to achieve a range of 150 miles, which I consider the minimum acceptable EV range for most people.

They quote a cost of $0.72/farad, so that battery would cost around $5K.  Expensive, but the Lithium-ion batteries in current EV’s cost three or four times as much.  Furthermore, batteries can’t handle as large charge/discharge currents, and they have a limited charge cycle lifetime.  Capacitors can be charged and discharged tens or hundreds of thousands of times, and should last the life of the car, or even longer.

So is this the silver bullet that will put an EV in everybody’s garage?

It is far too soon to tell, but if these guys are on the right track, maybe we will finally be on our way to kicking our “oil addiction.”



 The rest of this article deals with the equations that define the requirements for energy storage in an EV.  Skip this if you are not into numbers.

The article claims a cost per farad of 72 cents (1.4 farads/$) and 10-20 watt-hours/$.  Using 10 watt-hours/$, this translates to $100 per KW-h.

Ev’s typically get 4 miles per kilowatt-hour.(KWH)  To get a 150 mile range, a capacity of around 50 KWH is required, so that’s a $5K battery. Expensive, but less than Li-ion batteries.

You might think 50 KWH is excessive, but the voltage drops as the capacitor is discharged, and at some point the voltage will be too low to be usable.  When the voltage reaches somewhere around half of the maximum value it will be too low to give decent performance in the vehicle. But 75% of the stored energy will be used at that point.

So only a little less than 40 KWH is usable in a capacitor holding 50 KWH.

Energy storage in a capacitor is defined by the equation:

E = ½CV2

Where E is energy in joules

C is capacitance in farads

V is voltage in volts

A joule is one watt-second.   1 Watt-hour = 3600  joules.­­­­­­­­­­

Looking at the equation above, the voltage rating is obviously crucial.
Doubling the voltage would quadruple the energy capacity.  The voltage rating in a capacitor is determined by the dielectric…the insulating material between the oppositely charged plates.  The dielectric value is a physical characteristic of the insulating material.

Using their numbers, 1.4 farads per dollar and 10 watt-hours per dollar, we can compute the energy storage per farad = 10/1.4 = 7 w-h/farad

From this we can calculate the voltage on the capacitor when it is fully charged using the above equation.

For a 1 farad capacitor:

7 x 3600 = ½(1)V2

V = SQRT [(2)(7)(3600)] = 224.5 volts

Now, let’s calculate the total capacitance required to store 50 KW-h using the same equation:

50 x (3.6 x 106) = ½(C)(224.52)

C = 7142 farads

If the capacitor is discharged from 224.5 volts to 100 volts, the total energy available is:

E = ½ (7142) (224.52 – 1002) / (3.6 x 106) = 40 KW-h

The article also states that 1 farad requires about 1 cubic inch of volume.  For 7142 farads, the total volume required is 7142 cubic inches, or a little over 4 cubic feet.  That’s probably about the same volume as a conventional engine and transmission.  Of course an electric motor is still required, but the ideal design would use “hub drive” motors at each wheel so that regenerative braking could be used as much as possible.

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