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Discussion Starter #1
In a recent trip, my friend Slyastro (who owns a Ford Focus Electric and a 2012 Volt) did a climb of Mt. Washington autoroad with his FFE.
He started at the bottom of the hill, freshly fully recharged, climbed to the top , took a picture of his energy screen :

We can see that he had driven 12.7km at a rate of 732Wh per km, and a total energy expense of 9.3kWh.

He then regened back to the bottom of the hill, where he took another picture:

We can also see the trip total driven was now at 24.8km, with an average of 167 Wh/km and a total energy use of 4.1kWh, since the reset of the trip counter.

On level ground, the FFE consumes about 150Wh per km. So, to displace the car of 24.8km took about 3.72kWh.
The difference between the 4.1kWh minus the 3.72kWh (0,38kWh) means that the regen efficiency is so high that it lost only
0,38kWh during the descent! if we take the potential energy (7.2kWh) that was regened from the top of the mountain to the bottom, I compute an 0,38kWh / 7,2kWh x 100 = 5,3% loss or an 94.7 % regen efficiency, which I find extremely high.

(Mass of the FFE is 1674kg, plus 90kg driver + 80kg photograph + 40kg of luggage)
Height difference between the base and summit: 1402m
gravitationnal constant: 9.8 m/s2

Did the Ford's engineers did such a very good job? (Or I mislooked something?)

Francois
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Perhaps they didn't take battery round trip losses into account. No existing battery has those numbers even if there are zero losses (wind, rolling, generator I^2 R) everywhere else.

GM evidently does take the round trip losses into account in their measurements, and the net is rarely as good as 70% all things taken into account..
 

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Volt can do 4-6 miles per kWh
 

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Agree with DCFusor. The other way to look at this is that for this to be true the drive train would have to be 95% efficient. That's not a reasonable number.
 

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Just eyeballing it, it looks like he put back into the battery about half of what he consumed during the climb (assuming that the battery was 100% at the start of the climb). At 90%+ regen efficiency, wouldn't it be much higher?
 

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Too many assumptions to make any kind of valid calculation about actual regen efficiency -- particularly the "level ground" assumed efficiency. And 95% seems unrealistic anyway.

BTW, what exactly is a "80kg photograph"???
 

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Discussion Starter #8
Perhaps they didn't take battery round trip losses into account. No existing battery has those numbers even if there are zero losses (wind, rolling, generator I^2 R) everywhere else.

GM evidently does take the round trip losses into account in their measurements, and the net is rarely as good as 70% all things taken into account..
I took off the rolling losses, by removing the energy used to ride the 24.8 km on a flat level road.

Our datapoints are the two provided by the photos of the energy display of the dash of the car.

What I would like, is another scientific point of view to calculate if I am off or in the bull's eye about this regen.
If you calculate the potential energy, (= m . g. h) at the top of the mountain, I arrive at 25954104 Joules which converts to 7.2kWh.
So, the regen can only put back that much energy in the battery during the 12.4 km down trip. And it effectively seems
to do so.

Such a high efficiency number (95%) disturbs me, but nonetheless, the numbers are there.

Francois
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Discussion Starter #9
Just eyeballing it, it looks like he put back into the battery about half of what he consumed during the climb (assuming that the battery was 100% at the start of the climb). At 90%+ regen efficiency, wouldn't it be much higher?
Let's not forget that a certain amount of energy is used to simply move forward the car, even on flat, level ground. On empirical measurements of the FFE, the average is of 0,150kWh per km. So, when you remove the 12.4km x 0,150kWh/km = 1,86kWh of energy is required to move the car.
What is put back in the battery is what is regened from the potential energy (from altitude).

Francois
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Discussion Starter #10
Too many assumptions to make any kind of valid calculation about actual regen efficiency -- particularly the "level ground" assumed efficiency. And 95% seems unrealistic anyway.

BTW, what exactly is a "80kg photograph"???
She is the lady that took the pictures of the car at the top of Mt Washington and posted them on their facebook page.
http://www.facebook.com/photo.php?fbid=10151026138000718&set=a.10151026137420718.423183.76723350717&type=1&relevant_count=1

Francois
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That's about 55-60% max regen right there. Just about the same as the engineers at GM says the Volt has.
 

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Discussion Starter #12
Volt can do 4-6 miles per kWh
Maybe, but I got only 2.85 last month. (Measured at the wall plug.)
those are good datapoints:
2.85 miles per kWh translates to 4.56 km per kWh, or 0.219 kWh per km.
4.0 miles per kWh translates to 6.48 km per kWh, or 0.154 kWh per km
6.0 miles per kWh translates to 9.72 km per kWh, or 0.103 kWh per km driving very very conservatively.

So, for the 24.8km distance that was made, that means that 5.43 kWh was used, for the first case,
3.82 kWh for the second case, and 2.55kWh for the third case.
The round trip took 4.1kWh (as displayed on the picture of the bottom of the hill).
So, for the first case, he CREATED energy (very unlikely! ;-) )
for the second case (which was our estimate used) he lost 4.1-3.82 = 0,28kWh with the regen!!! over 7.2kWh of potential energy to recover, that
means a 3.8% loss or a 96.2% regen efficiency...
for the third case, he would have lost 4.1-2.55 = 1.55kWh that means 21.5% losses or 78.5% efficiency, which is more a number that
we could "accept" but that is the worst case... the true number probably lies somewhere in between the 78.5% and 96.2% estimates.

Francois
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Perhaps they didn't take battery round trip losses into account. No existing battery has those numbers even if there are zero losses (wind, rolling, generator I^2 R) everywhere else.

GM evidently does take the round trip losses into account in their measurements, and the net is rarely as good as 70% all things taken into account..
I would disagree about the battery when you talk about DC-DC energy conversion. Lithium ion's typically have very low losses at > 95% with 98-99% being typical. Compared to the horrid lead acids at around 70-80%. When you start adding all the converters and wiring losses, that's where the issue lies, not the actual battery.
 

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Discussion Starter #14
That's about 55-60% max regen right there. Just about the same as the engineers at GM says the Volt has.
Bonaire, how do you arrive at those numbers?

I would tend to think that the regen efficiency is calculated from the potential energy (from the high point of altitude) that effectively gets regened back to the battery, removing the rolling energy to ride the 24.8kms

Francois
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So, Jedi, if Li batteries are that good (sorry...) then why does it take 13.3 kwh to charge back 10kwh? Surely you're not saying the built in charger is only 70% efficient! As an EE, I know better than that one, and as a lifelong fooler-arounder-with-batteries-all-kinds, I haveta call bull on the 95 percent round trip efficiency on ANY battery. It just ain't so, Joe. Sure, Li's better than lead acid, no argument, but not that close to perfect by a long shot.
 

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I would tend to think that the regen efficiency is calculated from the potential energy (from the high point of altitude) that effectively gets regened back to the battery, removing the rolling energy to ride the 24.8kms
Theoretically that's true but as DCFusor points out, you're suggesting that regen, which involves changing kenetic energy to chemical energy, is vastly more efficient than changing electrical energy to chemical energy. This is very hard to believe given that for regen you first have to change kinetic energy to electrical energy. Only after this conversion can you begin changing electrical energy to chemical. Basically you're suggesting that doing Conversion A involves greater losses than first doing Conversion B and then doing Conversion A, which flies in the face of knowing that every conversion leads to greater losses.

The assumption has to be that the kWh number is bogus, which isn't that hard to believe.
 

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Discussion Starter #17 (Edited)
If we assume the numbers given by the Focus's dash are wrong, then this would have to be verified by other means.
But, in the meantime, we have no indication that the datapoints are wrong or skewed.

About the energy conversions:
In the Focus, there is only one permanent magnet electrical motor, that turns to a generator when the car goes into regen mode.
As it's the same motor to put the car in movement, would it's efficiency to convert electrical energy to movement would be the same
as to convert movement to electrical energy? And we all know that those motors could be very very efficient.

We don't have the exact number of Wh per km to figure out the energy consumption taken away from rolling and that does not get
to be regened. This uncertain number leads to a regen efficiency range between 78 to 96% as posted on post #12 above.

But, to give another point about the FFE's regen efficiency, it is to be noted that the same driver, Sylvain Juteau (slyastro on this site)
did contest the Green energy Rallye of Montreal this week end. http://www.rallyevert.com/
The Focus Electric he was competing with was a Ford Canada's owned Focus.
He was competing against 3 iMievs. Well, he did finish FIRST for the energy consumption used.
And as the rallye did use a road with a LOT of stops, the regen efficiency of the Focus probably made it outperform
the iMiev, thus winning first place for energy consumption.


Francois
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You'd need much more data in order to determine the actual regenerative efficiency, but if it were truly 95%, the total battery charge would be much higher at the end of the trip.
 

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Discussion Starter #19
What would be a good experimental setup to collect such datas and what would be the pertinent datas to collect,
so to be able to evaluate the regen efficiency?

Descending a hill on a long distance seemed to me the best way to minimize roundoff errors. A 12.4 km descent is not a short descent!

Francois
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What would be a good experimental setup to collect such datas and what would be the pertinent datas to collect,
so to be able to evaluate the regen efficiency?

Descending a hill on a long distance seemed to me the best way to minimize roundoff errors. A 12.4 km descent is not a short descent!

Francois
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I'm far from an expert on this, but the first thing you'd need to identify is exactly what efficiencies you are trying to measure (much like the difference between wall-to-wheel versus battery-to-wheel efficiencies). Then you'd need to do is identify all of the energy sinks (e.g., rolling resistance, aerodynamic drag, conversion losses). At that point, you'd know what you're actually measuring.
 
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