# regenerative braking

Discussion in 'Misc Electronics' started by RichD, Feb 3, 2008.

1. ### RichDGuest

What is the efficiency of regenerative brakes?

Let's say we have a 3000 lb. vehicle, traveling 30 mph.
It hits a red lght. That's 1350000 lb-(mi/hr)^2
kinetic energy, dissipated through the disc brakes.

Now assume it's a Prius - how much is recovered
into the batteries?

I'm not looking for a theoretical discussion, just a
number. Anybody know the number?

2. ### AndroclesGuest

| What is the efficiency of regenerative brakes?
|
|
| Let's say we have a 3000 lb. vehicle, traveling 30 mph.
| It hits a red lght. That's 1350000 lb-(mi/hr)^2
| kinetic energy, dissipated through the disc brakes.
|
| Now assume it's a Prius - how much is recovered
| into the batteries?
|
| I'm not looking for a theoretical discussion, just a
| number. Anybody know the number?
|

80%. Friction with air still causes some braking but most
of the KE is recovered and stored as useful energy to get
going again. Remember that no energy is required to continue
in a vacuum, so all the energy a car uses is wasted as heat
directly through the exhaust pipe and radiator and the rest
warms the atmosphere through friction.

3. ### AndroclesGuest

| On Sun, 03 Feb 2008 11:52:51 GMT, "Androcles"
|
| >
| >| >| What is the efficiency of regenerative brakes?
| >|
| >|
| >| Let's say we have a 3000 lb. vehicle, traveling 30 mph.
| >| It hits a red lght. That's 1350000 lb-(mi/hr)^2
| >| kinetic energy, dissipated through the disc brakes.
| >|
| >| Now assume it's a Prius - how much is recovered
| >| into the batteries?
| >|
| >| I'm not looking for a theoretical discussion, just a
| >| number. Anybody know the number?
| >|
| >
| >80%. Friction with air still causes some braking but most
| >of the KE is recovered and stored as useful energy to get
| >going again. Remember that no energy is required to continue
| >in a vacuum, so all the energy a car uses is wasted as heat
| >directly through the exhaust pipe and radiator and the rest
| >warms the atmosphere through friction.
| >
| Is that a fact and reason based answer or just a guess? A battery is
| not as efficient as a capacitor and there is a theorem from sophomore
| EE that "proves" no more than 1/2 the energy stored in a capacitor can
| be recovered.

Sounds like a sophomore's proof.

What you will not be able to recover is
a) the heat lost to resistance.

I mention the second because it is less obvious than the first.
http://www.androcles01.pwp.blueyonder.co.uk/AC/oscillator.JPG
http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.gif

of antenna to the circuit above and then you call it a transmitter.

More personal research would be needed to recover the
| proof, but that would imply something like 40% of the KE absorbed by
| the regenerative brakes could be recovered.

If the objective is to burn fuel then cars are 100% efficient.
If the objective is to convert chemical energy to mechanical
energy then cars are 18% efficient, measured as lifting its
own weight (and that of its passengers) against gravity.
By driving the car off a cliff most of that energy can be
recovered. If the car is used as a pile driver that would
be useful work.
A brake does not assist the car in doing useful work.
" What is the efficiency of regenerative brakes? "

Firstly we have to decide if stopping the car is efficient,
because clearly if that is the only purpose then a friction
brake is 100% efficient for succeeding or 0% efficient
for wasting useful kinetic energy. Indeed, locking the brake
will result in no heat loss at all, that will be transferred to the
tyres skidding against the road surface, but few would
call that efficient.
A regenerative brake returns some of the energy to the
battery without the corresponding heat loss and is therefore
82% efficient, the other 18% being lost to air resistance.
This can be improved upon by streamlining all cars to
look like aircraft. Of the 82%, 1-2% will be lost heating
the cables between the brake and the battery.
The figure of 80% is necessarily approximate since
an identical brake fitted to a different vehicle will change
the overall efficiency.

|
| An Analysis of Hybrid Electric Propulsion Systems for Transit Buses
| Milestone Completion Report by O'Keefe and Vertin of the
| National Renewable Energy Laboratory
| http://tinyurl.com/292xyt gives the efficiency of regenerative braking
| as 49-50% at best and 39% as current practice.

If you look at figure 46 of that document you'll see many variable
parameters including vehicle dimensions, location, time of year etc.
and we can argue the efficiency of brakes on roller coasters in the
Swiss Alps or kids burning rubber and doing wheelies on motor
cycles or the space shuttle deploying a parachute on landing.
http://static.howstuffworks.com/gif/space-shuttle-landing3.jpg
Regenerative braking on Airbus 308 would not be efficient, lifting
batteries to 30,000 feet is a waste of fuel and it doesn't have
electric turbo fans to recover the energy.

A Prius is not a transit bus and a transit bus is only efficient when
fully laden, it serves no purpose when driven around empty.

Hard data is ever harder to find. Back-of-the-envelope estimates seem
to be running at up to 20%
http://geekbuffet.wordpress.com/2007/06/10/getting-back-what-you-put-in/

Tom Davidson
Richmond, VA

5. ### Sam WormleyGuest

Depends on the breaking rate... to fast and a large
part is dissipated as heat.

http://www.wired.com/cars/energy/news/2004/05/63541

6. ### David BostwickGuest

And another calculation gives an answer of 80%? Makes me very confident of
the accuracy.

7. ### Don Stauffer in MinnesotaGuest

And another calculation gives an answer of 80%? Makes me very confident of
the accuracy.[/QUOTE]

I had earlier heard of estimates in the 80% range (from theory), so
this sounds good to me. I would also assume that it depends somewhat
on driving style. Locking brakes does not provide good efficiency. I
believe all the hybrids have hydraulic backup, don't they?

8. ### Evgenij BarsukovGuest

Battery is storing energy chemically at almost constant voltage,
therefore proof you are talking about does not apply. In fact
a traditional 18650 Li-ion battery at low current rate (say C/10 rate)
can be charged/discharged at 98% energy efficiency.
Main energy loss is due to IR drop both during charge and discharge,
which declines with I^2. E_loss = I^2*R

Unfortunately during regenerative breaking currents are very high
so batteries have to be designed with very low R (that is why
only Panasonic and Sanyo can make suitable for HEV NiMH batteries).
further reduced R, 80% efficiency of both energy recovery + subsequent
reuse (which both have losses) appears very optimistic. I would bet on
60%, but I would also appreciate somebody posting the real number.

Regards,
Evgenij

More personal research would be needed to recover the

9. ### Evgenij BarsukovGuest

Sorry man, it might not be obvious, but this effect is true, easily
provable based on the differential equation describing capacitor
discharge and known to every electric engineer:
http://www.smpstech.com/charge.htm

As other people have stated earlier, this applies only for straight
linear capacitor / resistor systems without inductance.

Switching systems with inductance is a different story and they can
be 90-95% efficient.

It does not apply to batteries because they are governed by different
differential equations. Their recharge can be approximated as as highly
non-linear capacitors, and because of this non-linearity can be shown
that they can be much more efficient than straight capacitors (mostly
because they maintain almost constant voltage).

Regards,
Evgenij

Even

10. ### Guest

It depends on how fast you decelerate.

Energy from an emergency stop is not worth it.

But if you only have to decelerate a bit to take a turn and you start
breaking in time then you can get most of the energy back.

My sail car design uses a 250 watt motor as electrical assist with
cycling. It looks to me like my 4 wheeled bicycle is going to do a lot
more breaking then a conventional bike. Hitting the regen-breaks would
probably take some meters to stop the thing with such small motor.
Most electric motors can take big surges over a short duration, should
be the same when working as a generator. Then the batteries and the
capacitors determine how much current one can store.

http://wind-car.go-here.nl

11. ### N:dlzc D:aol T:com \(dlzc\)Guest

Dear John Bailey:

....
But we don't have a resistor, we have an inductor adding or
removing kinetic energy from the system. The dielectric of the
capacitor has losses, and the conductors do too. But this can be
brought to less than 20% losses without too much difficulty.

http://www.ecass-forum.org/eng/faq/faq001.html
https://www.electrochem.org/dl/ma/201/pdfs/0233.pdf

The deal breakers will be charge / discharge rate (aka. current).

David A. Smith

12. ### RichDGuest

***************************************
Toyota engineer Dave Hermance said drivers who
slowly roll through intersections using "California
stops" are decreasing their mileage. "If you don't
stop, you don't get the free energy of regenerative braking."
***************************************

How did Toyota get to #1 with such numbskull
engineers?

13. ### RichDGuest

Dear David A. Smith,

Inductors do not source or dissipate real power.

14. ### RichDGuest

Interesting. It is a schoolboy level proof.

But the automotive circuit does not conform
to the models discussed. The car battery is not
charging/discharging a capacitor, but an
electric motor/generator, which resembles a
R-L circuit with a back emf (i.e. another battery).

However, it might still partially apply... how to
model charging a non-ideal battery? Does it
look capacitive?

15. ### Sam WormleyGuest

Hopefully that engineer was joking!

17. ### HLSGuest

Inductors do not source or dissipate real power.

18. ### RichDGuest

"So how does my exam look, Doc?"
"You're fine, so I decided to inject you with AIDS."
"You did what?!?!"
"Well, how do you expect to enjoy the benefits of
modern medicine if you're healthy?"

19. ### Evgenij BarsukovGuest

I would not downplay the generality of this relationship. It has
nothing to do with complexity of the system.
It generally applies to arbitrary systems that consists of (as many
as you want, and arbitary connected) resistors and capacitors, and
arbitrary path of discharge (e.g. it does not have to be constant current).
As long as both capacitors and resistors are linear, you can not beat
this rule.

The only way to beat it is to add inductors to the mix, or to make
capacitors non-linear (e.g. batteries).
It is important to understand this conceptually to avoid some
attempts to beat the rule by making some complex systems (but without
inductors), which would be a bit like trying to beat the Karnot's
efficiency law by using alcohol instead of water in a boiler.

Regards,
Evgenij

20. ### BenjGuest

Great 'theorem"! Tell you what. I've got a real nice large HV
capacitor. To prove your "theorem" I'll charge it up to a nice voltage
and then use a resistor to remove half the energy stored in it. At
that point we'll apply the terminals to your ass.

Moron.

Engineering Quantity Matters! No, Virginia, the world CANNOT be saved
by windmills!

One key parameter in regenerative braking is the rate at which energy
produced and returned to the storage system. If you are going 60 and
just slow down, you can put large percentage of the energy back into
the system (except for the friction and other losses you are
overcoming, natch) but if you come up to a read light and lock the
wheels, you get next to nothing back. Even with rapid stops there is
the question how rapidly and efficiently batteries can be charged at
high currents even assuming the regenerate system is capable of those
currents. Physicists trying to do engineering! This is a hoot!