How to calculate power required by vehicles

[sophiecentaur]

I agree with most of that; at least, I did, initially.
But you can't just write them off:
At present, all motors and generators are designed with constant supply volts in mind and that is clearly the most 'convenient' but the fact that capacitors can dump and supply considerable dollops of energy, rapidly seems to be what appeals about them. The energy per kg figures are around that of many rechargeable cells.
The pedal pressure thing wouldn't be a factor if the back pressure were supplied by a servo; whatever you were dumping the energy into, you'd still have a similar problem and you'd still need some Forodo somewhere in the system - just to be safe.
If all the control of charge and discharge were carried out using (a suitable) high frequency AC, the values of inductors could be a lot lower. Aero AC systems use 400Hz, I believe; why not 40kHz, for instance? High power semiconductors are pretty impressive these days. The fact that the available Super Capacitors have only a few volts maximum working conditions needs some ingenuity but they could well get better. Conventional Electrolytics work at much higher voltages than when they were first available.

I was referring to exponential discharging / charging through a resistor. But, of course,
E = C Vsquared, if you're talking about stored energy. It's just not a friendly device to use, like a car battery.

 

[Naty1]

it takes more energy to charge the batteries then they acquire

I haven't read much about the latest in battery development but older style lead acid batteries lose about 15% to 20% of charging power to heat...the battery gets noticeably warmer when charged. AGM batteries with extremekly low internal resistance lose only a few percent, maybe 2% to 4% to heating when charged.

In addition AGM batteries can accept a charge up to their full rating in amp hours...so,for example, a 200 amp hour AGM battery can be charged at up to 200 amps...a tradional wet cell lead acid battery, like the ones in most cars, will only accept about 25% of their rating in amps...about 50 amps for a 200 AH rated battery.

In addition an alternator, likely loses another 15% or more of engine power in belt friction, heating, and internal frictional and air movement losses. Likewise battery chargers have their own losses...as do inverters if boosting battery voltages to higher ac levels...these also are only about 80 or 85% efficient.

I opted for a lot of battery sourced power aboard my boat when cruising both for quiet and to avoid even greater losses and noise in running a gas or diesel generator. I got my quiet for reding physics, but the losses were substantial no matter what when away from commercial dockside power.

 

[Phrak]

I agree with most of that; at least, I did, initially.
But you can't just write them off:

I find capacitors as applied to braking systems inferior to batteries for the numerous reasons I've already give. In nearly parallel argument they are inferior to batteries in supplying power. Without evidence to the contrary I don't find them worthy of consideration.

At present, all motors and generators are designed with constant supply volts in mind and that is clearly the most 'convenient' but the fact that capacitors can dump and supply considerable dollops of energy, rapidly seems to be what appeals about them. The energy per kg figures are around that of many rechargeable cells.

I understand that. This is the initial attraction to capacitors. But you have to carry the analysis beyond the favorably qualities to the disadvantageous qualities if you are interested in a fair comparison

The pedal pressure thing wouldn't be a factor if the back pressure were supplied by a servo;

Pedal pressure is roughly equivalent to braking power. This manifests as a potential across the drive motor now used as a generator. This power, in a dynamic braking system, in whole or in part, is to be sent to a storage device. Whatever voltage is developed in this process needs to be matched to the voltage of the device that will be doing the storing. The unmatched voltage times the current generated is simply more lost heat.

With capacitors, the variation in voltage is compounded as the product by both the generator's swing and the swing in capacitor voltage. I've already explained this above in shorter words.

So is the lossive inductive element that will withstand these variations bigger or smaller than a bread basket?

I have a way-to-long background in power conversion electronics.
Though I am not sufficiently motivated actually solve this particular problem numerically, given some theoretical values, I'm quite confident that all current power conversion electronics are dismally insufficient to make this a practically competitive alternative to batteries.

 

[sophiecentaur]

You're probably right in the end. Dissipating hundreds of kW is hard enough (with brakes) - storing energy at that rate sounds even harder. Bigger than a breadbasket by quite a bit, I should imagine.
I still don't get your point about brake pedal force. Of course it applies in basic brakes but, in all other systems - including your everyday servo - it is made artificially so because we like it that way. I have driven a truck with some air brakes which were more or less on/off and there was virtually no feel back. A nightmare when you are not carrying a load.

 

[likephysics]

Just a question about regenerative braking - are the caps charged directly from the braking mechanism, Brakes--->dynamo-->charging ckt-->caps
or is it like
Brakes-->some mechanical winding system(like a spring or a flywheel)--> the spring unwinds slowly -->dynamo-->charging ckt-->caps.

 

[Phrak]

sophiecentaur. Sorry, by variable brake peddle pressure I was referring to, in rough terms to the variable power presented to the conversion circuit. The actual pressure has nothing to do with it. In effect, your foot pressure is just a mean to tell the braking circuitry how much power needs to be absorbed to reduce the vehicle's kinetic energy at any given time.

likephysics. Good thinking! I don't know if it's a viable idea, but I've only been thinking in the electronics box. Maybe you found a way around it.

Say the flywheel can quickly absorb a vehicle's kinetic energy during a braking episode. Now you may not have to dump it so quickly into electrical storage. The longer the time you have to do this, the smaller the currents that are involved. The smaller the currents, the more efficient, and the smaller, simpler and lighter the electronics that is required to do this.

 

[sophiecentaur]

It's amazing to think that, in a very few years, we are all likely to be driving vehicles which use something like we're discussing. Something else to go wrong, I suppose!

I am reassured by my (sailing) boat engine (diesel) which is 35ys old, has a flywheel that you can hardly lift and one bit of electronics - the rev counter. It seems to be refusing to die. (Refuses to start, too, on occasions)

 

[mheslep]

And in the end, the bottom line is not BTU comparison but cost.

In which case the energy cost for an electric mile is well established to be 3-4X less expensive than a gasoline powered mile, given the average efficiency of the US gasoline vehicle and the current price of gasoline.

 

[mheslep]

Someone here at the forum pointed out to me that there is development on fast charging LiPo batteries:..

Almost, that was Lithium Ion Iron Phosphate, LiFePO, that has demonstrated high power, not the polymer version of Li ion which is inexpensive but lower power as mentioned above.

 

[mheslep]

Adding a bit more to your fine response...

At some point air resistance dominates rolling friction.

Yes, break even (drag vs rolling resistance ) is about 40 km/h (25mi/h) in a sedan, and interestingly only 10 km/h for a bicyclist.

That's because it is quadratic with velocity. A quick measure of air resistance is given by the Bernoulli pressure 1/2*(rho)*(v^2). For 40 m/sec this gives approx 1000 N/m^2. Take this over the cross section of the vehicle (approx 1 sq meter) at the velocity in question and you get a power of 40 kW or about 50 horsepower.

Yes, just so for 1 M^2 at 40m/s = 144 km/h = ~90 mi/h, moving right along. The average car effective drag area is actually a bit lower, Cd*A = ~0.8 M^2, and at the highway speed of 60 mi/h the tractive load for air drag alone is about 18HP. The rolling resistance coeffient for a sedan is about Crr=0.01. So for a one ton vehicle the rolling resistance force is MgCrr = 1000kg x 10 x 0.01 = 100N, a constant over any speed. The power load for rolling resistance is also F x V, so at 60 mi/h we have 100N * 26.6M/s = 2.66kW = 3.6HP. Total tractive load, drag and RR, at 60 mi/h level road, constant speed, 1 ton sedan is ~22HP. Another metric is energy cost per distance, in this case that is 0.27 kWh per mile ( about 3 cents per mile at my electric rates).

Streamlining helps only up to a point. That's because the friction of sliding sideways through the air eventually equals the amount of friction of butting headlong into the air. This point occurs for a cylinder approx 40 times its diameter in length. In the above example, then for a cylindrical car 40 meters long and 1 meter in diameter, you would have 100 horsepower consisting of 50 horsepower butt-end friction and 50 horsepower irreducible sliding friction.

True, though there's a 10x gap between the Cd of the average sedan (0.3) and the average airplane (0.03)

From D. McCay's excellent reference on this subject.
The energy cost per 100 km here includes a engine efficiency of 0.25. To get actual tractive load divide by 4.

http://www.inference.phy.cam.ac.uk/withouthotair/cA/page_254.shtml