I can tell you that in "L", you need more than a 6% grade. But I don't know how much more.
However, after setting the cruise control at 63, and going down a 6% grade (in "L" but not off the accel) for 5 miles, I was able to drive another 2 miles before the ICE came back on. Of course I might have hit a sweet spot in the engine cycling and battery SOC at the time.

 

I live at the top of a 1,000 foot hill with an approx 10% grade and in "L" I max out at about 45 mph. I have calculated that the regen is about 1 kWH when I travel at 45MPH (speed limit) but I use about 0.4 kWH of that to travel the two miles down the hill, leaving me with about 0.6kWH at the bottom of the hill. So, to fully charge my battery I would need a 17,000 foot mountain but will have traveled 34 miles free plus have a 10KWH charge in my battery. I hope that helps.

 

Didn't someone post that a drive down Mt. Washington gave them 10 miles of charge? Something like 3-3.5 kWh?

 

A 5000 ft drop of Volt has a potential energy of almost 10 kWh, which at 50% efficiency should give about half of a full battery charge. PE = mgh = 2000kg x 9.8m/s/s x 1667 m = 33 MJ/(3.6MJ/kWh) = 9.2 kWh. So going down 10,000 feet should charge an empty battery. The grade would influence the rate of charge.

Here is a related thread: http://gm-volt.com/forum/showthread.php?15197-regenerative-braking-when-battery-is-full

Also there is increased air friction with increased speed and increased road friction with increased distance and turning. Hard to say what fraction is lost in propelling the car.

 

Im not sure I understand this.. If you leave your home and go down hill, how can the regen add more charge to the battery? Shouldn't the battery be full after a night of charge? And if you can get another 5 miles range, why isn't the charger do that? hummmm..

 

The Volt battery never charges to 100% capacity just like it never discharges to 0%. The Volt battery only charges to about 85% capacity. So when going down hill after a full charge there is still room to capture more energy.

I'm not sure but I recall if you charge the battery to an even higher state at some point MGB comes on to work against MGA to consume the extra energy so as not to overcharge the battery.

 

#3 answer:

Assumptions:
4000 lb fully loaded volt
12.9 Kwh to charge the battery
100% efficient conversion of potential to energy to Electricity (probably over 90% for the volt)
No rolling resistance, AeroDyanamic drag or Auxillary loads

12.9 Kwh = 34,252,390 ft-lb energy/ 4000 lb = 8563 ft required elevation change

 

So you want a hill that L feet on floor will sustain 65 mph?

From my recollection, with rough numbers... L feet on floor regen at freeway speed is ~40kW electric. Assuming a .92 efficiency for regen, that's 44kW of potential energy. 65 MPH drag&road losses are ~15kW? So you need a road where the potential energy loss of the car going down it at 65mph is ~60kW.

Assume a 2000 kg car (nice round number, typical of the car with ~3 people on board or people and stuff.) Potential energy equals mass times gravitation times distance... We want power, which is energy over time, so both sides are per second... 60kW = 2000 kg * 9.81 m/s^2 * height of drop per second (or height = 60 kW/(2000 kg * 9.81 m/s^2 : since a Watt is a kg m^2/s^3, the answer will be in meters of drop per second.) That gives ~3.06 m/s decent rate.

65 mph = 29.06 m/s along the hypotenuse... This gives about 28.9 m/s horizontally. Road grade is rise/run - so if I didn't make a major mistake, your first answer is you need a ~10% down grade (10.5%, but the assumptions aren't that precise.)

This gives a ~40 kW charge rate. The car has a 10 kWh usable capacity - meaning you need 15 minutes at 40 kW to get a full charge. 15 minutes = 900 seconds at 3.06 m/s gives ~2750 meters of descent - about 9000 feet.

You'd probably need a little bit more for the battery losses, but it gives you a rough idea of the magnitude.

 

65 MPH drag&road losses are ~15kW?

FYI the "official" number -- meaning that it's the number GM has given out -- has been 30 kW at 65 MPH on a slight uphill.

 

Question #1 - you don't want to recover or use any electrons: Thus with the LRR tires, the low drag, and already rolling at 65 mph, then I would say on a 10-30% grade you would just need to put it into neutral and pump the brakes to keep it from going over 65 mph. Since it is neutral you will only use the friction brakes and thus not recover any energy, but would waste energy as heat. If no energy is to be wasted or used, then have to come back to it.

Question #2 - don't know, but coasting downhill in "D" is far more efficient than in "L".

Question #3 - Don't know, but nothing personal Paul*, but I think you may be way off on the assumption for conversion of potential energy to electricity on the Volt - I think more like 5%. Regenerative braking is not very efficient - about a 3-5% boost in range compared to without. The way I see it, plugging the Volt into a 120V outlet, which charges the battery much faster than regenerative braking can, takes 9 hours. Thus, even if we assume the charge rate using regenerative braking was similar to a 120V outlet, you will need to be travel down some very steep hill with the regenerative braking being applied at optimal efficiency for at least 9 hours straight. Not sure what grade and vertical descent would be needed, but would guess several hundred miles of rolling distance.

 

Wrong, irrelevant, and wrong.

The whole point of the question posed was about recovering energy, so I'm not sure why you think he doesn't want to. I have no idea why you're suggesting going to neutral and pumping the brakes, as it wastes energy and increases wear while accomplishing nothing helpful, and your estimate of the slopes involved is grossly off.

Coasting in D may be more efficient than coasting in L, depending on the slope and your goals. Higher speeds mean more energy loss, but converting altitude to momentum is more efficient than converting it to electricity...

Even assuming you were right about regenerative braking only being a 3-5% boost in range (you aren't,) that would only translate into a 5% energy recovery if most or all of the energy consumed by the car was to accelerate and decelerate it (it isn't - most of the energy goes into overcoming wind drag and rolling resistance.) Charging on a 120V outlet is about 1.4 kW - regen on the Volt goes as high as 60 kW - 40 times higher than a 120V outlet...

I laid out the math further up, and since every one of your assumptions was wrong, it isn't surprising that your answer is thoroughly wrong, too.

 

I did the numbers once and lost them so I'm just going the concept route. The energy from the drop has to provide sufficient power to overcome the the drag and rolling resistance forces. Forget rolling resistance because they'll be negligible. The drag equation is 1/2 CdA *P* V^2. The density of air, or p, at 25C is 1.164 kg/m^3. Try the frontal area of the Volt at 1.4m X 1.65m. Walter has given you the mass and speed.

 

Fun stuff.

Other than DonC, I think a lot of the above rough analysis ignores the requirement that the car maintain 65MPH speed. Not all the potential energy is available for regen, only the surplus over what is needed to overcome the baseline friction/drag.

I should add that my requirement that you be in "L" was supposed to be a simplifying assumption. The real constraint is: must maintain 65MPH and must fully charge. I imagine that there is a max/min grade that will meet the requirement-- I'm interested in the the answer that most closely approximates a grade that could actually be built.

I have know idea what the right answer is.

 

Umm... you missed the part where I said I think the road and aero load of 65 MPH on flat ground is around 15 kW in steady state and then added that to the rate of descent requirement? The number could be off by 20% in either direction, but I did include a number in my calculations for it. That's why I needed a hill steep enough to induce 60 kW instead of 44 kW...

 

Cool. So, using your numbers of ~9000 feet of elevation drop = full charge. Can we take from that a rough rule of thumb that a 9000/40 mile = 225 foot elevation change will roughly add or subtract one mile of electric range?

 

As a very rough number, it is likely reasonable. If you're descending at a lower speed, you can use a larger percentage of the potential energy to regenerate, and there could be other factors (a lighter car has less potential energy, and I used a pretty heavy Volt. Also HVAC loads, etc.) But if you think of it as a +/- 50% sort of a number, it should be valid, I'd think.

 

Ask the world record Volt range holder from Colorado (75 ev miles)... He's dropped down around a mile vertical or more....

MrEnergyCzar

 

You already have multiple theoretical answers. Not really one for hypotheticals.
I'm more of an experimentalist.

I have to measurements of regen on real declines.
http://gm-volt.com/forum/showthread.php?15682-Pikes-Peak-Climb-Regen

http://gm-volt.com/forum/showthread.php?8925-75Miles-on-a-single-charge

The I70 route was mostly at 65-75mph (going with the flow) but in D not in L, though at 65mph in Cruise control, I can assure you there is little/no difference.

The pikes peak example above I had OBDII running and measured battery state as well as altitude and speed (with a map showing it).


I don't believe there is any road with sufficiently high grade to maintain 65mph for more than a mile or so, without draining the battery, as it would need to be 10-15% grade which is only on very steep mountain roads that are generally tacking up the hill (so have very very sharp turns on them). About 7500ft provided about 83% of usable battery recharge (39 miles estimated, but that is biased ranged estimate because of the regen downhill). The data clearly shows its a lot more than 5%.. efficient.

 

Nice articles Walter. I’ll start over by correcting my errors as I see it.

Question 1: Maybe cnicholson should restate the question for clarity
.
My Mistake #1: Minor – I was calculating % grade as rise over hypotenuse instead of over run.

My mistake #2: Correct answer but to a different question – my quote of 5% regenerative braking efficiency is incorrect, but instead refers to the avg. % of kwh used during pure EV mode that comes from regenerative braking for the average EV driver.

My mistake #3: I stated charging rate of battery from regen. braking is much less than 120v charging i.e. <<1.6 kw. I was wrong, but I still believe it is >3.3 kw but <6.6 kw and not close to =>50 kw. I will give reasons why I believe this in a later post.

My mistake #4: Correct answer but to a different question – my hill experiment concluding that 22% of the potential energy could be regenerate by braking is incorrect, but instead refers to the % of kwh that can be recovered by regen. braking relative to the amount of kwh used to climb the same hill (all under pure EV mode). While Walter and others provide a number of calculations directly to understand % potential energy converted (I will still need to go over them), in the hill climb expt. I just discussed, it still would require a 20,000 foot vertical drop and 60 miles of driving to charge a full battery. If we bump this speed to 65 mph, then much more potential energy will be lost to drag and thus cnicholson guess of Mt. Everest height seems like it might be in the ball park.

Rich

 

Question 1: Maybe cnicholson should restate the question for clarity

Rich: For #1 my intent was to ask a "baseline" question of what kind of grade would be requires for the car to "coast" at 65MPH (cruise control set at 65MPH, SOC stays constant). Walter's very sensible thought process of saying is takes ~15kW for 65MPH on flat road, so just ask what Potential Energy conversation (loss of altitude) did not occur to me, but seems spot on, intuitively. The 15kW number squares nicely with real-world range.

 

I cannot comment on regen at 65mph in L (I've found any grade that can do that without some accelerator pedal, and I've been down some15-20% grades on Pikes peak. At that pitch the L in regen will not hold CC at 25 but will "coast" over the set speed up to abou at 35-40. THere were a few times it wanted to go faster than 40 but I used breaks as the road has nearly hairpin turns and most are in the very steep sections. Now if only I could drive like the official racers (averaging more than 60mph on the uphill! http://www.greencarreports.com/news...s/1078434_pikes-peak-hill-climb-toyota-mitsubishi-place-electric-cars-in-top-10)


But I'm fairly sure that the decent needed to go from full to empty, with recharge via L at 25-35mph will only be about 10000 ft-12000ft, given that I got an 80% charge on only 7500 ft but few longer are as steep., so someone in HI should be be able to do it on Mauna Kona decent. Maybe on Maui too.

In terms of rate, I gained about 8.6kWh in about an hour. There are larger peaks in there (in the very steep sections), but the max change I measured over any 1 min period was 22.58kWh. I consider anything less than 1 min is an instantaneous, not sustained, regen.

 

I cannot comment on regen at 65mph in L (I've found any grade that can do that without some accelerator pedal, and I've been down some15-20% grades on Pikes peak. At that pitch the L in regen will not hold CC at 25 but will "coast" over the set speed up to abou at 35-40. THere were a few times it wanted to go faster than 40 but I used breaks as the road has nearly hairpin turns and most are in the very steep sections. Now if only I could drive like the official racers (averaging more than 60mph on the uphill! http://www.greencarreports.com/news...s/1078434_pikes-peak-hill-climb-toyota-mitsubishi-place-electric-cars-in-top-10)




But I'm fairly sure that the decent needed to go from full to empty, with recharge via L at 25-35mph will only be about 10000 ft-12000ft, given that I got an 80% charge o only 7500 ft but few longer are as steep., so someone in HI should be be able to do it on Mauna Kona decent. Maybe on Maui too.

In terms of rate, I gained about 8.6kWh in about an hour. There are larger peaks in there (in the very steep sections), but the max change I measured over any 1 min period was 22.58kWh. I consider anything less than 1 min is an instantaneous, not sustained, regen.

I have to concede that regen. braking and/or ICE regen. can charge the battery faster than 6.6 kw given Walter's information, tboult data above and my data from yesterday and today.

My data:
As posted yesterday using "D": I stopped at the base of a hill and drove 3.1 miles to the top of the hill (in "D" with ball in middle, never removed foot from throttle, EV mode whole time, all acc off), avg grade=6%, verticle climb = 976 feet, avg. speed=28 mph, used 2.2 kwh. I then drove down the hill back to the same location (in "D", braking to keep ball in middle, never put foot on throttle, all acc off), avg. speed = 30 mph, regenerated 0.5 kwh.

The trip took about 6 min. thus avg. regen. rate of 5 kw. I would think 200-300% higher charge rate on the steeper slopes of the hill.

Repeated same course in "L" today which required me to apply the throttle lightly to maintain 30 mph down the hill - 2.2 kwh to climb the hill (same as yesterday) but generated 0.8 kwh coming down the hill which is about 60% more energy regen. compared to yesterday coasting in "D" and with gentle and careful application of the brakes. Thus avg. rate of battery regen. was 8 kwh. and again I would think max. charge rate on this hill would be several folder greater than this.

I learned a lot from this question but would like to learn more by finding a steep enough and safe enough road to collect similar data at 65 mph.

Would be nice to try on Mauna Kea or Mauna Kono, I just drove down it in a rental jeep several weeks ago while on vacation. I coasted on one section of the road for about 15 miles in the jeep.

Rich

 

Comment on regen braking eff:

One difficulty I see though out the forum when discussing “efficiency” of any system is that we all (or most of us) don’t mention “efficiency relative to what baseline or subsystem and under what condition”. This results in difference values being posted, many of which are correct.

The problem at hand best describes what I am talking about: I often see an efficiency value of 90% for electricity regen from braking being quoted, but this is only correct when eff is based on the known rotational power being applied at the generator shaft and output electrical power being measured at the wires coming directly out of the generator, under optimal conditions in the lab (full load, optimal cooling). This eff value is the same for any electromagnetic motor or generator and has changed little over the decades. So that brings us to the next common quoted regen brake eff value for the Volt of 70% (posts in this forum that cite www.driveforinnovation.com/regenerative-braking), which is also correct if eff is now based on the known rotation power at the wheel’s axle and I believe output is at the wires of the generator, and measured at under optimal conditions (under full regen load of 0.31g motor induced deceleration, no road resistance, no drag effect). If we now consider the AC power of the generator needs to be inverted to DC (~85-90% eff) and the battery temp regulated to < 122F, you are now in the 60% eff range. This furthers drops when you calculate in the slowing forces of drag (speed dependent) and road/rolling resistance, and take into consideration that significant portion of regen motor induced deceleration is occurring below optimal levels, for the average driver (for the current math question we should assume optimal regen motor deceleration).

Thus, for the question at hand, at low speed (20-30 mph) down a hill, the power need to overcome drag and road resistance is not too great and thus maybe at best you will be able to recover 50-60% of the potential energy. However, the power required to overcome drag becomes substantial at 65 mph since drag power increases with the cube of the speed. Thus if 3 kw power is required to overcome drag at 30 mph, then 10x more power (30 kw) will be required to overcome drag at 65 mph (of course it doesn’t matter if you are on a hill or not). Thus a very large portion of your potential energy will be used up combating drag at 65 mph vs. 30 mph.

In summary and for this problem, if we define the eff of regen braking as the amount of potential + kinetic energy (remember car is already moving at 65 mph) at the top of the hill that is recovered into the battery at the base of the hill, your eff will be much greater at 30 mph than at 65 mph. Our assumptions need to keep this in mind when comparing to real world data. Mt. Everest size mountain still looking like the best guess.

Rich

Edit - my comment regarding kinetic energy is irrelevant since it is assumed that the car will be traveling at 65 mph at the bottom of the hill