“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

## Wednesday, July 17, 2013

### Regenerative braking in the Lexus CT 200h

I've been driving my Lexus CT 200h for about two years and about 39,000 miles. In that time, I've learned a lot about driving techniques to minimize specific fuel consumption (g.p.m., gallons per mile). I've also given some thought to what it is about a hybrid that makes it more fuel efficient. One of those items is regenerative braking, where some of the kinetic energy in the moving vehicle is used to charge the battery rather than to heat the brake rotors. This is done by having the energy of the moving vehicle turn the electric motor backwards, thus making it a generator and thereby charging the battery. Of course, friction brakes are also used.

I've wondered just how much of the braking energy goes into the battery and have been hard pressed to find data for this. However, the CT 200h has a display option for "Consumption" (see photo at left). It's difficult to see (click to enlarge) but there are small boxes in the vertical bars that indicate mileage by the minute. Each complete box, according to the legend, represents 50 watt hours of energy (180,000 joules). At the top of a hill, I applied sufficient braking to keep my speed at approximately 35 m.p.h. At the bottom of the hill, a stoplight brought me to a stop.

I can calculate the energy difference from 35 m.p.h. at the top of the hill to 35 m.p.h. at the bottom of the hill by using Google Earth to find the elevation change. I determined it to be 103 meters. Because my speed didn't change, neither did my kinetic energy, therefore the reduction in my potential energy went to some combination of heating my brake rotors and charging my battery.

My best estimate of the mass of the vehicle with the 1/4 tank of gasoline and myself and my baggage is 1,600 kg. Therefore, the potential energy lost in the descent is $E=mgh=1600kg*9.8\frac m{s^2}*103 m=1.62*10^6joules$. The display shows "E" boxes and fractions of "E" boxes and my best estimate, assuming that 3.5 "E" boxes are shown is that $3.5*50Wh=175Wh$ or $630,000joules$ were sent to the battery. I don't think it could be lower than $585,000joules$ or higher than $675,000 joules$.

Assuming that I'm interpreting the cryptic display correctly (and that Ed Davies doesn't haul me up short!), about $630000/1620000=38.9\%$ of the potential energy went to charge the battery. The rest was dissipated as thermal energy in the disc brake rotors and, ultimately to the atmosphere. The battery pack in the CT 200h is a 1.3 kWh Ni metal hydride battery. The 630,000 joules equal 0.175 kWh or 13.5% of a full charge for the battery. Per the owners' manual, I'm able to drive in "EV mode" (battery only) for two miles, but I'd best accelerate slowly even by my standards, and not exceed about 20 m.p.h.

As an aside, this is the energy in about 18 cm^3 of gasoline. Figure I'd have to burn about four times that, or 72 cm^3 to charge the battery with the engine at 25% efficiency. And, of course, the regenerative braking isn't effective when the battery is fully charged, isn't used (much) in hard stops, etc. Still, it does increase overall fuel efficiency.

Finally all of these figures have large "error bars," the regenerated energy on the display, the elevations from Google Earth, the mass of the vehicle, and the ability to stay at precisely 35 m.p.h. (although really, all I need is to be going at the same speed when I stop logging as when I start so that the kinetic energy is unchanged). Still, it's enough for me to have a good idea of what the regenerative braking can give me.

Update: Based on a comment by Gabriel Grosskopf, I measured the distance over which I descended. It was 1,530 meters, thus the slope is 3.86 degrees (0.0673 rad). The typical instrument landing system glideslope is 3 degrees though a few, such as VNY - Van Nuys - at 3.9 degrees, are steeper. Assuming that I drove the 1530 meters at 35 m.p.h. or 15.6 m/s, it took me 98 seconds to put 630,000 joules into the battery. This is a charging rate of 6,440 watts or 6.4 kW. As mentioned in a previous post, when I put fuel in my gasoline tank, I'm adding energy at a minimum rate of 11 mW, about 1,700 times as fast. To be fair, we'll divide that by four since IC engines are much less efficient than electric motors. So we're adding useful energy at 2.75 Mw or 430 times as fast. Coincidentally, the Nissan Leaf touts a 6.6 kW charger to charge its 24 kWh battery.

Gabriel Grosskopf said...

Your regen % is lower than I expected. 103m is quite a serious elevation change. You do not say what the slope is, but I am curious of either
1) the battery reached full charge
2) the regen peaked at the battery maximum charge rate.

Both these are areas where the new plug-in hybrids have an advantage over the regular hybrids

Rob Ryan said...

I can definitely say that the battery didn't reach full charge. The slope is fairly steep, I'll run calculate it tomorrow. I couldn't say what the battery's maximum charge rate. I'd have considered a plug in, but I got the vehicle in 2011 and the Prius PHEV wasn't out yet. There's still not a CT200h PHEV. The Volt was overpriced IMHO. Anyway, look for a small supplement tomorrow.

Gabriel Grosskopf said...

That is an interesting update. I have flown 3% approaches, so I have a feel for how steep it is. And VNY sounds like minimum power and extra flaps.

6.4kW is a quite a feat for a battery of that size, and not sustainable for long.

The 6.6kW limit of the Nissan Leaf is not a battery limit, but the limit of the charger electronics. There are a number of public charging stations that supply DC and then the battery can be charged at 40kW until it is 80% full, after which the rate drops off quite a lot.

Gabriel Grosskopf said...

One of the reasons why I follow well researched blogs like yours is that they make me think, and this post is a good example of that ...

I kept on pondering it and this morning I suddenly realized that you did not take into consideration the rolling resistance of the car for the mile (almost) that it rolled down the hill.

A convenient guess is that a small EV would have consumed approx 175Wh to do this at 35 mph, which means that the regen and rolling resistance consumed about the same fraction of the potential energy.

Now since you already calculated 175Wh is 630kJ, I can take that off the total, and then the regen % becomes 630kJ / 990kJ, or just over 63%.

Rob Ryan said...

Hmmm.... well, assuming a Crr 0f 0.015 and a mass of 1,600 kg, a speed of 15.6 m/s, normal component of mg using cos(3.86 degrees) I calculate 3,661 watts. Doing this for 98 seconds, I calculate 99.7 watt hours(call it 100 because all of these estimates are subject to large errors).

But I don't consider it relevant. We know that the potential energy is converted to overcoming rolling resistance, overcoming aero drag, overcoming friction in the vehicle components, heating the rotors, and charging the battery. We know how much potential energy is lost in the descent, and we know (sort of) how much went to the battery.

So I guess that your point is that you consider only the portion that could either go to the brakes or the battery to be pertinent and I considered the portion that goes to the battery vs. the total available potential.

To address your figure, I estimate that 30 watt hours go to aero drag (I can, as they say in class, "show my figures" if desired). I don't know the resistance to motion of the drive train, but let's say I lose 5% or so there.

Doing all of this, my estimate is that the 450 watt hours of potential energy are converted as follows:
Aero drag: 30 watt hours
Rolling resistance: 100 watt hours
Drive line: 25 watt hours
Brakes: 120 watt hours
Battery: 175 watt hours

So, IF I want to know that portion of energy not used to overcome the various resistances, my estimate is 175/(120+175)=.593 or 59%.

I'd love it if someone with better figures from more accurate measurements would dive in.

Rob Ryan said...

Hopefully, it was obvious that I meant "...that portion of energy not used to overcome the various resistances that goes to charge the battery..."