How Much Charge Does Regen Braking Put Back in the Battery?

[Topher]

Wow. Much ado about something simple. The question the OP wants to ask is: how efficient is regenerative braking? Forget about hills and up and down and all that. The answer is in general, normal driving conditions, including hills (not the most extreme ones) and regular braking (not race track level) regen is in my experience somewhere around 50% efficient.

Given the confusion in this thread, would you care to share how you calculated/measured that?

Thank you kindly.

 

[Cebe]

The number quoted by Tesla is 60%. Don't ask me to find the source of the quote, it's in something I read in the past two years

 

[Johan]

Given the confusion in this thread, would you care to share how you calculated/measured that?

Thank you kindly.

I didn't calculate or measure. Like I said "in my experience". But it harmonizes with the number given by Tesla.

 

[Johan]

Tesla official blog:

The Magic of Tesla Roadster Regenerative Braking

How much energy does it recover?

Unfortunately, the adage “your mileage may vary” applies to regen as well. The amount of energy you can recover depends on how and where you drive. From the powertrain point of view it looks pretty good. The energy conversion efficiencies from chemical to electrical (battery), DC current to AC current (inverter), electrical to mechanical (motor), and torque to force (transmission and wheels) are all quite high and work just as efficiently returning energy into the battery. The bigger problem is aerodynamic losses and higher speeds and rolling friction of the tires. These both act to slow the car, but the energy dissipated cannot be recovered. We must also remember that, even though the battery-to-wheel conversion efficiency is pretty good (up to 80% or so), the energy makes a full circle back into the battery and it gets converted twice for a net efficiency of at most 80% * 80% = 64%.

 

[Uncle Paul]

That is one of the reasons that Hybrid vehicles get better mileage than pure ICE.

When going uphill the battery is partially depleted to assist in the climb. Going back down the hill, instead of riding the brakes, the regeneration will partially refill the assist battery.

Biggest thing is that going downhill the energy reclaimed is not only free, but saves on wear and tear on the brakes. Same thing with city driving. Everytime you slow down with regen instead of using your brakes, the battery gets boosted back up.

No way to have a hard number. Even an ICE engine will burn more fuel on the way up, and coasting downhill can reclaim some of that with hyper mileing. The times on the brakes is when the electric battery gets the edge.

 

[jerry33]

That is one of the reasons that Hybrid vehicles get better mileage than pure ICE.

When going uphill the battery is partially depleted to assist in the climb.

You'll get better mpg to use the gas engine for ascents and acceleration (assumes not a plug-in hybrid). The reason is that every time you use the battery, there is an energy conversion loss, which must be recovered by using the gas engine. Using the battery is only efficient when the power demand is so low that the conversion losses are less than the gas engine's inefficiency at low power.

 

[Topher]

I didn't calculate or measure. Like I said "in my experience". But it harmonizes with the number given by Tesla.

If your experience doesn't provide you with a number, how did you create a number? 50% means nothing if we don't know what you mean by that. For example, I am talking about energy back into the battery pack, not round trip. first because that can actually be observed, and second because all energy in a Telsa comes from the battery, so any flat road mileage number already include those losses.

You'll get better mpg to use the gas engine for ascents and acceleration (assumes not a plug-in hybrid). The reason is that every time you use the battery, there is an energy conversion loss, which must be recovered by using the gas engine. Using the battery is only efficient when the power demand is so low that the conversion losses are less than the gas engine's inefficiency at low power.

Depends. If the engine is tuned to perform best at highway speed on a flat, then using battery power (In addition to the gas engine) for the added load of a hill is perfect, it drains the battery right before there is a influx of regenerative braking energy that needs to be stored.

Thank you kindly.

 

[roblab]

With some 100,000 miles of driving, many hills and flats, many road trips, I have decided that the most important number is starting and ending elevation. I have always calculated an extra ten miles of range needed for every thousand feet up, as a rule of thumb. Boise to Winnemucca is 260 miles, but has a 2000 foot upgrade. It will take 280 miles of range. Going the other way, I calculate 245 or so due to the downhill grade. You don't need exact numbers. You mainly need to know that it takes more to go uphill, as in any car. The extra you get coming down is simply frosting on the cake.

I have quit worrying about how many miles I might regain after going up and down, knowing there is a slight loss for every hill. I watch my buffer and adjust my speed accordingly and arrive at the next SC with range to spare. Hills don't change that a whole lot. But total elevation changes might.

In actuality the regen gives me back maybe 6 - 7%. The problem is that people tend to drive faster downhill without noticing it. Especially as all the gas cars start cranking it up to 80 mph or more, because they could only do 65 on the uphill side, and they tend to push and pass and carry on down hill. This is where cruise comes in handy to keep your speed constant.

 

[mspohr]

Here's my real world experience. Yesterday I drove from home (Tahoe City, CA elevation 6400) over Donner Summit (elevation 7200) to Rocklin, CA (elevation 200).
EVTripPlanner gives me distance 95 miles, 49 rated miles, 154 wh/m, 14.8 kwh, net elevation -6757 (there's a bit of up and down on this route).
Return trip 131 rated miles, 411 wh/m, 39 kwh
Average is 282 wh/m
My actual experience was about the predicted (just confirmed with my VisibleTesla log).
So, it seems to me that you get all of the energy back. (I know this doesn't seem likely since you probably lose about 10% in the round trip through the battery but most of this downhill both ways is coasting rather than regen I was getting most of the energy back as coasting down hill rather than recharging the battery.)
Here's the trip down to Roseville Galleria:

And the return trip... actually not since the forum software is balking at inserting the second image.

 

[CoastalCruiser]

Thank you roblab & mspohr for the real world insights.

mspohr, I have made the run you refer to dozens of times. I recall a number of long downhill stretches west of the Donner Pass that were the perfect grade to coast in neutral and hit the speed limit. In fact I used to turn off the motor of my Volkswagen Beetle and later my RX7 (not recommended) on the longer runs to save gas.

Although your run is a bit off from the nature of my original question it is a helpful real world anecdote. It would be interesting to see the return trip graph if you 'd care to make another post.

No more killing-the-motor-to-save-gas bit once I get my first electric car though. It's funny, I was 100% sold on owning an electric vehicle before I knew anything at all about regenerative braking. But now that I'm in line for an M≡ there's plenty of time to more fully understand this new animal. The overworked expression "paradigm shift" is actually well played here. The entire concept of electric cars, taken as a whole, is truly a new way of thinking about personal transportation

ps - Regarding that speed graph, I sure hope the cops don't find a way to get hold of the telemetry from these cars.

 

[David99]

How much goes back to the battery when you go downhill depends very much on the grade and the speed. You can go in neutral and race down the hill. You won't get anything back. Same on a mild grade. It might not be even enough to bring the car to normal speed. Of course going downhill always uses less energy than going uphill. But that's not the question. The question was how much goes back into the battery. That is not a simple answer as many other factors apply and cannot be isolated. The motor/inverter/battery combination is about 80% efficient when using regen. That means 80% of the energy that is present at the rotor will be converted in usable energy at the battery level. But driving the car uses energy, even when going downhill. So only if the grade is speed enough and the steep and you're going slow enough to have excess energy, that ends up going into regen.

 

[David99]

That last sentence should be "... only if the grade is steep enough and you are going slow enough to have excess energy..."

 

[bxr140]

To sum up all the above, it's physically impossible to have the same consumption on a hilly trip as a flat trip when all else is equal (speed, weather, starting/finishing elevation, etc).

Also, It has very little to do with the % grade or the length of the grade--it's all about elevation gained vs elevation lost. Back before we had the trip planner, we used to budget ~7 miles for every 1000ft gained. There wasn't really a rule of thumb for elevation lost, but if you used 5-6 miles per 1000ft you were usually in the clear.

Once you start using the trip planner (and get over the giddyness of staying in the green on the power meter), it all becomes moot. It's all factored in and you don't even think about it.

 

[mspohr]

Here's the return trip. (VisibleTesla for some reason didn't start collecting data until Colfax but you can see from the blue line a steady uphill to Donner summit then the steep downhill to Truckee. Only at the steepest part from the top of Donner, there is some regen (yellow line) but mostly it's coasting with very low power input.
You'll see you get all of the energy back that was required to lift the car up 6000 feet when you come back down. The regen is somewhat lossy (10-20%) but most of the energy is regained in coasting which is lossless.
(See earlier map and graph for the first part of the trip.)
So, the answer is that you get back almost all of the energy used to go uphill when you come back down. In the real world, most of this is coasting (lossless) rather than regen (10-20% loss).

 

[SageBrush]

I'm sure this question has been posed many times. In googling for an answer I found a thread on Reddit that provides some good info, but not exactly what I'm looking for. I am interested in getting an idea of the recovery rate in say the following scenario:

You are driving up a 1 mile long hill with a 5% grade at 45mph. These are just example numbers. When you reach the crest you drive down the 5% grade hill for 1 mile at the same speed.

So, if using enough just enough regen to not slow the card down to under 45mph (assume no actual use of brakes), how much of the power used to get up the hill is put back into the battery?

The example is deceiving because you ignored the the energy costs of driving on flat terrain. Consider this 3.2 mile drive example:

1. 1 mile flat terrain
2. ~ 1.1 miles: 1 mile flat and 0.1 miles straight up
3. ~ 1.1 miles: 1 mile flat and 0.1 miles straight down

If you avoid regen (which as has been mentioned is somewhere in the 30 - 50% energy recouped range and just lose some speed towards the crest that is then recovered while going down, the total energy used is pretty close to being equal to 3.2 miles of flat driving.

Put another (in a somewhat politically incorrect manner): hills do not have much if any penalty unless you are going down too fast and have to slow down with braking.

Anecdote: I descend ~ 1000 ft net to work and then go up the same amount on the return home. After I account for the elevation change potential energy, I always do ~ 5% better going uphill than down. In the case of my Prius it means that the engine is operating more efficiently on the uphill climb. This difference is much more when I drive the Honda Fit (sorry, I cannot remember numbers) and is from the same effect. Electric motors also have a ~ bell shaped efficiency curve but much wider and bowed compared to an ICE

 

[Chopr147]

OK So this is the first i'm hearing about regenerative braking. I'm thrilled! And interested. So, does the energy go back into the battery as a charge? Or to a capacitor to store the energy and used for A/c, heat , radio etc... This forum suggests the battery is charged. Just want to clear it up.

 

[mspohr]

OK So this is the first i'm hearing about regenerative braking. I'm thrilled! And interested. So, does the energy go back into the battery as a charge? Or to a capacitor to store the energy and used for A/c, heat , radio etc... This forum suggests the battery is charged. Just want to clear it up.

The regen energy goes into the battery (the motor turns into a generator). The round trip efficiency of this is 80-90%.

 

[Chopr147]

I had mentioned mountain roads and there are a few spots going back home with long downhill stretches. As it is today i do the old foot off the pedal and let the vehicle coast for a good mile or 2. I am interested in the science of it and have done research etc... and this is the first i'm hearing of the regeneration. But, it makes sense. Thanks for the info

 

[jerry33]

I had mentioned mountain roads and there are a few spots going back home with long downhill stretches. As it is today i do the old foot off the pedal and let the vehicle coast for a good mile or 2. I am interested in the science of it and have done research etc... and this is the first i'm hearing of the regeneration. But, it makes sense. Thanks for the info

The thing about letting the car speed up is that you're pushing more air so the aerodynamic losses increase rapidly. At some point the aerodynamic losses are greater than the heat losses from regen. I suspect this point isn't very high since aerodynamic losses increase with the square of the speed, but I haven't done the math to determine what the intersect speed would be.

 

[SageBrush]

The thing about letting the car speed up is that you're pushing more air so the aerodynamic losses increase rapidly. At some point the aerodynamic losses are greater than the heat losses from regen. I suspect this point isn't very high since aerodynamic losses increase with the square of the speed, but I haven't done the math to determine what the intersect speed would be.

If we start with a speed where half of total losses is road friction and half from aero, and


1)mgV1 + rho*CdA*V1*V1

.. increase speed 10% ..

2)mg*1.1*V1 + rho*CdA*1.1*V1*1.1*V1

Then losses will increase ~ 15%

.....
Of course if you just bleed off some speed before the crest this arithmetic can be ignored.