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Car Forum / Driving, Maintenance, Tuning / Maintenance and Repair / February 2008regenerative braking
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? -- Rich | What is the efficiency of regenerative brakes? | [quoted text clipped - 7 lines] | 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. >| What is the efficiency of regenerative brakes? >| [quoted text clipped - 14 lines] >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. 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. An Analysis of Hybrid Electric Propulsion Systems for Transit Buses Milestone Completion Report by OKeefe 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. | >| What is the efficiency of regenerative brakes? | >| [quoted text clipped - 19 lines] | 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. b) radiated energy. 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 One improves the efficiency of radiation by adding the correct length 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. The question asked was: " 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. > | 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. 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 a.s. Moron. And Guys! Please leave your liberal educations behind. In Science and 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! >>| 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 [quoted text clipped - 23 lines] > high currents even assuming the regenerate system is capable of those > currents. Physicists trying to do engineering! This is a hoot! Let's return to body-on-frame construction, with a ladder type frame. All frame connections will be bolted and electrically isolated. The driver's side frame rail will be the positive bus bar and the passenger side will be the negative bus bar. The capacitors will bolt directly to the frame rails, as will the motor/generators at each end of the car. Problem solved! In fact, the capacitors can act as the crossmembers, so no actual crossmembers will be required. Make sure to throw the main disconnect before placing your jack stands. nate (guess I better get started on this patent application doomaflatchey...)  Signaturereplace "roosters" with "cox" to reply. http://members.cox.net/njnagel >> | What is the efficiency of regenerative brakes? >> | [quoted text clipped - 19 lines] > EE that "proves" no more than 1/2 the energy stored in a capacitor can > be recovered. 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). But even using the latest version of Panasonic NiMH batteries with 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 > proof, but that would imply something like 40% of the KE absorbed by > the regenerative brakes could be recovered. [quoted text clipped - 4 lines] > http://tinyurl.com/292xyt gives the efficiency of regenerative braking > as 49-50% at best and 39% as current practice. > 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. Maybe at some EE correspondence school in outer Elbonia, but not anywhere creditable. Most capacitors are nearly 100% efficient at low frequency and moderate charge/discharge rates, I have no idea where you came up with some "proof" that no more than half can be recovered. Even in very stressing conditions, such as advanced dielectric capacitors under high discharge rates (pulsed power systems, for example) you can get well over 90% of the energy back out of a capacitor. If you're losing energy in a capacitor, then the capacitor was poorly selected for the application- such as using an electrolytic cap at radio frequencies or higher. Batteries are indeed a different story, though. I agree that current hybrids are probably achieving well below 50%. >> 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 [quoted text clipped - 5 lines] > frequency and moderate charge/discharge rates, I have no idea where you > came up with some "proof" that no more than half can be recovered. 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 > in very stressing conditions, such as advanced dielectric capacitors > under high discharge rates (pulsed power systems, for example) you can [quoted text clipped - 5 lines] > Batteries are indeed a different story, though. I agree that current > hybrids are probably achieving well below 50%. >>> 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 [quoted text clipped - 10 lines] > discharge and known to every electric engineer: > http://www.smpstech.com/charge.htm But we're not talking about the first-year method of charging a capacitor from an ideal voltage source through a resistor are we? Nor are we talking about discharging through a resistor to a purely resistive load, are we? > 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. EXACTLY my point when I made the first rebuttal!! There is no "theorem" that states with any sort of generality that "no more than 1/2 the energy stored in a capacitor can be recovered." In fact you can get nearly 100% out as heat, should that be your goal (as it often is in pulsed welding systems, for example). Period. End of story. >>>> 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 [quoted text clipped - 15 lines] > are we talking about discharging through a resistor to a purely > resistive load, are we? 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 >> As other people have stated earlier, this applies only for straight >> linear capacitor / resistor systems without inductance. [quoted text clipped - 7 lines] > nearly 100% out as heat, should that be your goal (as it often is in > pulsed welding systems, for example). Period. End of story. | >>> 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 [quoted text clipped - 27 lines] | nearly 100% out as heat, should that be your goal (as it often is in | pulsed welding systems, for example). Period. End of story. At the EE correspondence school in outer Elbonia, if you drain a tank of water you can only put half the water back, easily provable based on the differential equation describing alcohol ingestion and known to every hydraulic engineer that is currently unemployed and wondering why he can't get a job even as a plumber's mate in inner Elbonia, let alone as a car mechanic's tea boy and general gofer in London. Might I suggest that Evgenij Barsukov is ineducable and suitable plonk material? > | >>> 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 [quoted text clipped - 38 lines] > Might I suggest that Evgenij Barsukov is ineducable and suitable > plonk material? You want to measure credentials or salary? | > | >>> 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 [quoted text clipped - 40 lines] | | You want to measure credentials or salary? Sorry man, it might not be obvious, but this effect is false, easily provable based on conservation of energy as known to every mechanical engineer. http://www.kettering.edu/~drussell/Demos/SHO/damp.html > | > | >>> Is that a fact and reason based answer or just a guess? A battery > is [quoted text clipped - 51 lines] > mechanical engineer. > http://www.kettering.edu/~drussell/Demos/SHO/damp.html I knew you will not accept the challenge. You remind me of that blue cube/spring oscillator in your link - pops up without any obvious reason, scrabbles a few meaningless one-liners and than disappears again. Why don't you come to the table and play, instead of just bringing the ping-pongs to the real players? Regards, Evgenij | > | > | >>> Is that a fact and reason based answer or just a guess? A battery | > is [quoted text clipped - 57 lines] | Why don't you come to the table and play, instead of just bringing | the ping-pongs to the real players? You remind me of those cretins that change names constantly - pops up without any obvious reason, scrabbles a few meaningless one-liners such as "easily provable based on the differential equation describing capacitor discharge" and thEn disappears again. Let's see your proof that half the charge on a capacitor will be lost in an inductive load such as the motor on a Prism. There's the challenge; salary is irrelevant, you haven't got the job yet. [snip] >> Sorry man, it might not be obvious, but this effect is false, easily >> provable based on conservation of energy as known to every [quoted text clipped - 6 lines] > Why don't you come to the table and play, instead of just bringing > the ping-pongs to the real players? When you play with people who don't play by rules, you lose. > Regards, Don't. For instance: http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Androrgasm.html > Evgenij Enjoy some more of http://users.telenet.be/vdmoortel/dirk/Physics/ImmortalFumbles.html :-) Cheers, Dirk Vdm > At the EE correspondence school in outer Elbonia, if you drain > a tank of water you can only put half the water back, Well actually in outer Elbonia you only have a hole in the ground to store the water when it's out of the tank so you usually can't even get half back in. -jim >> 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 [quoted text clipped - 3 lines] >Maybe at some EE correspondence school in outer Elbonia, but not >anywhere creditable. http://www.iop.org/EJ/abstract/0031-9120/40/4/008 Two theorems on dissipative energy losses in capacitor systems Ronald Newburgh 2005 Phys. Educ. 40 370-372 doi:10.1088/0031-9120/40/4/008 "Abstract. This article examines energy losses in charge motion in two capacitor systems. In the first charge is transferred from a charged capacitor to an uncharged one through a resistor. In the second a battery charges an originally uncharged capacitor through a resistance. Analysis leads to two surprising general theorems. In the first case the fraction of energy dissipated in the resistor depends solely on the ratio of the two capacitances. The values of the original charge and the resistance play no role. In the second case half of the energy supplied by the battery is dissipated and half is stored in the capacitor. The values of the battery emf and the resistance play no role." If you don't have an account, I will try to sketch the proof. In simple RC charging, the power/ current relationship is invariant up to the limit 0/0 of no resistance and down to the 0/0 instance of infinite resistance/conductance. The proof is a matter of generalizing this result along the lines of the Norton and Thevenin theorems. >Most capacitors are nearly 100% efficient at low >frequency and moderate charge/discharge rates, I have no idea where you [quoted text clipped - 5 lines] >the application- such as using an electrolytic cap at radio frequencies >or higher. By the way, the problem is getting the power IN to the capacitor as well as getting it out! Of course you can get nearly all power out of the capacitor, its how to avoid putting half of it into a resistor--sooner or later. >Batteries are indeed a different story, though. I agree that current >hybrids are probably achieving well below 50%. http://www.sandia.gov/pv/docs/PDF/batpapsteve.pdf A Study of Lead-Acid Battery Efficiency Near Top-of-Charge and the Impact on PV System Design by Stevens and Corey "Notice also that the overall efficiency shows high values, with full charge represented by approximately 85% efficiency, a commonly used value for battery charge efficiency. More importantly, notice the dramatically lower efficiencies for the increments above about 80% state of charge, where most values are below 60% efficiency, and full charge is represented by less than 50% efficiency." >Maybe at some EE correspondence school in outer Elbonia, but not >anywhere creditable. If I didn't eschew personal attacks, I would invite you to compare degrees. Only one of mine is an Ivy League school. >> Is that a fact and reason based answer or just a guess? Bluster, puffery, and obfuscation prevail. Dear John Bailey: ... > If you don't have an account, I will try to sketch the proof. > In [quoted text clipped - 5 lines] > Thevenin > theorems. 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 Dear David A. Smith, On Feb 5, "N:dlzc D:aol T:com \(dlzc\)" <dl...@cox.net> wrote: > Dear John Bailey: > > If you don't have an account, I will try to sketch the proof. [quoted text clipped - 9 lines] > capacitor has losses, and the conductors do too. But this can be > brought to less than 20% losses without too much difficulty. Inductors do not source or dissipate real power. -- Rich Dear RichD: Dear David A. Smith, On Feb 5, "N:dlzc D:aol T:com \(dlzc\)" <dl...@cox.net> wrote: >> Dear John Bailey: >> > If you don't have an account, I will try to sketch [quoted text clipped - 10 lines] >> conductors do too. But this can be brought to less >> than 20% losses without too much difficulty. > Inductors do not source or dissipate real power. Really? http://en.wikipedia.org/wiki/Superconducting_magnet#Magnet_quench ... resulting in explosion. Don't get too hung up names. http://en.wikipedia.org/wiki/Energy_density David A. Smith "RichD" <r_delaney2001@yahoo.com> wrote in message news:6713da05-f9fa-4495- > 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. Inductors do not source or dissipate real power. -- Rich An ideal inductor doesnt. And you can get close enough to ideal for most practical purposes. > http://www.iop.org/EJ/abstract/0031-9120/40/4/008 > Two theorems on dissipative energy losses in capacitor systems [quoted text clipped - 12 lines] > stored in the capacitor. The values of the battery emf and the > resistance play no role." > >Most capacitors are nearly 100% efficient at low > >frequency and moderate charge/discharge rates, I have [quoted text clipped - 11 lines] > the capacitor, its how to avoid putting half of it into a > resistor--sooner or later. 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? -- Rich >> By the way, the problem is getting the power IN to the capacitor as >> well as getting it out! Of course you can get nearly all power out of [quoted text clipped - 12 lines] > model charging a non-ideal battery? Does it > look capacitive? Dynamic braking makes use of both the generated EMF and the motor inductance. The R is just one of the unavoidable losses. Assuming the motion generated EMF is less than the storage voltage, the braking controller shorts the motor, briefly, allowing the motion generated EMF to drive a current ramp up to some upper limit. The rate of that ramp is dependent on the motion generated EMF and the motor inductance. Then the short opens and allows an inductively generated EMF to add to the motion generated EMF, to drive that peak current back into the storage battery (or capacitor). The motor current then ramps down at a rate dependent on the inductance and the difference between the storage device voltage and the motion generated EMF. Once the current falls to a lower limit, the short is reapplied and another current peak is produced. The motor braking torque is proportional to the average current during this cycle, which happens many times per second, and the upper and lower current limits are chosen by the controller to produce the commanded braking torque. This works till the motion generated EMF falls too low to achieve the the desired peak current when the motor is shorted. Then mechanical brakes have to take over to finish the stop and hold the vehicle still. Note that there is no intentional loss in this process. The switch across the motor is ideally either a short (zero ohms) or and open (infinity ohms), and real switches come satisfyingly close to these two states. The largest loss is usually the motor winding resistance, which can be quite low. Capacitors take and release energy more efficiently than batteries (which exhibit a little hysteresis effect when the energy flow changes directions), but with more voltage change. I think vehicles now in production, charge the batteries, directly with dynamic braking, but I would be happy to learn that I am wrong on this. SignatureRegards, John Popelish > What is the efficiency of regenerative brakes? > [quoted text clipped - 10 lines] > -- > Rich 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 > What is the efficiency of regenerative brakes? > [quoted text clipped - 7 lines] > I'm not looking for a theoretical discussion, just a > number. Anybody know the number? 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 > -- > Rich > > Let's say we have a 3000 lb. vehicle, traveling 30 mph. > > It hits a red lght. That's 1350000 lb-(mi/hr)^2 [quoted text clipped - 6 lines] > part is dissipated as heat. > http://www.wired.com/cars/energy/news/2004/05/63541 *************************************** 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? -- Rich >>> Let's say we have a 3000 lb. vehicle, traveling 30 mph. >>> It hits a red lght. That's 1350000 lb-(mi/hr)^2 [quoted text clipped - 17 lines] > -- > Rich Hopefully that engineer was joking! > >> Depends on the breaking rate... to fast and a large > >> part is dissipated as heat. [quoted text clipped - 11 lines] > > Hopefully that engineer was joking! "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?" -- Rich >What is the efficiency of regenerative brakes? > [quoted text clipped - 7 lines] >I'm not looking for a theoretical discussion, just a >number. Anybody know the number? 20% "Regenerative braking can be extremely powerful. According to Craig Van Batenburg, who teaches Honda and Toyota hybrid service at Automotive Career Development Center in Worcester, MA, no more than 17 percent of its capability is used in these cars to avoid putting people into the windshield. Even at that low level of use, in a typical mixture of highway and around-town driving, regenerative braking can recover about 20 percent of the energy normally wasted as brake heat. This reduces the drawdown of the battery charge, extends the overall life of the battery pack and reduces fuel consumption." quoted from: Regenerative Braking Charges Ahead by Jacques Gordon in Motor Age http://www.search-autoparts.com/searchautoparts/article/articleDetail.jsp?id=68244 Don't believe anything from the web without citations, data, and reasonable logic. Checking further, it is somewhat suprising that regenerative braking can even work much of the time. Batteries are not happy with large surges of current, preferring a metered trickle, preferably controlled by a microprocessor. The high tech solution to that is to add an ultracapacitor. from the same article: "An ultracapacitor has a surface area that is several orders of magnitude greater than conventional types, and the separation is less than 10 angstroms (one angstrom is one ten-billionth of a meter). It can hold a pretty big charge, and its voltage output and discharge rate can be controlled with external circuitry. For instance, large ultracapacitors are often used to provide hours of backup power for computer systems after a general power failure. That is a relatively long and slow discharge when compared to capacitors used to start a motor. However, ultracapactors also can be used to deliver very high voltage for a shorter period of time." That then runs into Newburgh's theorems: http://www.iop.org/EJ/abstract/0031-9120/40/4/008 Two theorems on dissipative energy losses in capacitor systems Ronald Newburgh 2005 Phys. Educ. 40 370-372 doi:10.1088/0031-9120/40/4/008 "Abstract. This article examines energy losses in charge motion in two capacitor systems. In the first charge is transferred from a charged capacitor to an uncharged one through a resistor. In the second a battery charges an originally uncharged capacitor through a resistance. Analysis leads to two surprising general theorems. In the first case the fraction of energy dissipated in the resistor depends solely on the ratio of the two capacitances. The values of the original charge and the resistance play no role. In the second case half of the energy supplied by the battery is dissipated and half is stored in the capacitor. The values of the battery emf and the resistance play no role." I seem to recall that Newburgh's theorems can be bypassed by using a large inductance to limit the charging current. Perhaps a clever engineer can figure out how to use the inherent inductance of one of the motors to provide that ballast. John >>What is the efficiency of regenerative brakes? >> [quoted text clipped - 9 lines] > >20% And another calculation gives an answer of 80%? Makes me very confident of the accuracy. On Feb 4, 8:20 am, david.bostw...@chemistry.gatech.edu (David Bostwick) wrote: > In article <nm3eq3dvj0o9u5f05nb7m82b8fn87i4...@4ax.com>, John Bailey <john_bai...@rochester.rr.com> wrote: > [quoted text clipped - 14 lines] > And another calculation gives an answer of 80%? Makes me very confident of > the accuracy. 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? > What is the efficiency of regenerative brakes? > [quoted text clipped - 10 lines] > -- > Rich 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 >What is the efficiency of regenerative brakes? [...] >I'm not looking for a theoretical discussion, just a >number. Anybody know the number? I don't know *the* number. But I do know *a* number. The distance that an electric vehicle can travel in city traffic on a given set of batteries is extended by at best 10% if it uses regen braking. Hardly worth it for the extra complexity and stress on the electrics. Economies in other areas can had to achieve this figure or better. I remember this figure from some years ago; I was disappointed to hear it to be so low. SignatureJohn Savage (my news address is not valid for email) |
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