2014-2020 Formula One 1.6l V6 turbo engine formula

[xpensive]

...
We should agree to 12 s for the typical breaking time per lap. The target should be 4 MJ/lap harvested. We need to consider heavy overload factors for the MGUs and inverters depending of load factors. This is what I can extract from the existing AWKERS thread. So we would have 333 kW average brake power. The load factor would be 12s/80s=15% for breaking. If we consider 75% of the lap timer for feeding the power back we get 67 kW electric power augmentation. That would not make a significant difference to the thermal load of the MGU's and inverters. We can simply neglect it in the figures.

What is certainly more interesting is the maximum power the system needs to absorb. I would assume that the maximum power does not exceed the average power significantly. When we did the 2.3MJ figures it was plausible that the MGUs could always harvest with the average collection rate. So for a first shot lets assume that max power is 333 kW and nominal power of the system would be 170 kW with an overload factor of 1.95.

Yii-ha, we have had our moments in the past haven't we WB? Right, lets take it from there, but when being a strictly nuts and bolts man, what whould the electric such say about the peak in and out of 333 kW in and out within say, 10 sec?

[autogyro]

Sorry WB.
I meant variable vanes for the turbine and compressor, although it is possible to have variable blades at least on an axial flow compressor.
I should not have brought up my ideas here, the system you are looking into is restricted by the regulations and I do not think the aero 'blow' aspect of my idea would be allowed.
I will wait until the regulations are finalised.

[machin]

You increase torque you increase acceleration

Here you are quite correct. You increase the torque at the road wheels by decreasing the speed of rotation of those wheels for a given engine rpm (and hence the road speed also decreases for a given engine rpm). Since power = Torque x speed, The power (and hence the energy) stays the same. (Remember the power and torque characteristics of the engine/gearbox system are inextricably linked -they're not two different aspects, they're two different ways of describing the useful work done by the engine).

You are right about inertia too... in order to increase the speed of rotation of the rotating parts on the car they require some of the energy produced by the engine (since they accelerate rotationally as well as linearly in the direction of car motion). For a lightweight race car this inertia accounts for about 10% of the total energy required to accelerate the car from one speed to another. The actual amount depends on the the relative speed of rotation of the component, the weight of the rotating components and the distribution of that weight about each component's axis of rotation.

In addition to the energy required to increase the rotational speed of rotating components we must also use some energy to overcome aero resistance and rolling resistance, plus there will be some loss of energy in the form of heat in the transmission system. At low speeds by far the biggest amount of energy is required to accelerate the car in accordance with KE=1/2mv^2, hence why a given power will accelerate a car faster at lower speeds.

Let that be an end to it.

I hope people have learnt from our discussion.

[xpensive]

Yadayada, I think we all know that Power equals Torque times Rpm by now, but gimme something to w**k on for the night,
with regards to how to stash 300 kW over two seconds in one place, keep it there, then to let it all out again within two sec?

[747heavy]

If you go back to the Brembo data and look at the braking power and braking time figures for this year, we see that we have app. 700-1100 kW/s which we could try to harvest under braking and transfer into a storage medium.

BTW: what´s the typical voltage of such a KERS system? ~400V?

And what do we know about "peak Lithium", as WB is keen on peak oil, it is maybe worthwhile to consider if we have enough Lithium in/on our planet before we make this the basis for EV´s and the future of mobility.

I guess Bolivia could be the next Saudi Arabia with an estimated 50% of global reserves.

[xpensive]

This is what I'm talking about, there's a 1000 kW out there, but how do we grab it and store it over two measly seconds?

[autogyro]

This is why I suggest ways to directly use the brake energy rather than trying to store it all x.

The Lithium is not used up even when batteries are depleted.
Re-cycling is the answer here, then it will never run out.
When you burn fossil fuels they are gone and you increase CO2 and pollution.
There realy is no comparison. Electric is the future.

[747heavy]

This is why I suggest ways to directly use the brake energy rather than trying to store it all x.

The Lithium is not used up even when batteries are depleted.
Re-cycling is the answer here, then it will never run out.
When you burn fossil fuels they are gone and you increase CO2 and pollution.
There realy is no comparison. Electric is the future.

Not 100% sure, this is as easy and straight forward as you make it sound Autogyro.

Das Recycling von Traktionsbatterien stellt derzeit eine Herausforderung dar: Es existieren heute noch keine wirtschaftlich und ökologisch tragfähigen Lösungen im industriellen Maßstab zur Rückgewinnung von Lithium und anderen Aktivmaterialien, die eine Zurückführung als Sekundärrohstoff in die Batterieherstellung ermöglichten.

In a roundabout way it says:

The recycling of traction batteries presents currently one of the greatest challenges.
At the moment, there are no economical and ecological sensible solutions for the industrial recycling of Lithium and other active materials and their return into the production process of batteries.

As far as I know, you don´t recylce the Lithium, you mainly gain Cobalt,Cooper and Nickel from the recylcling of Lithium-Ionen batteries.
But feel free to come to Wolfsburg on 15.02.2011 to present your solution, I´m sure people will listen with great interest to any sensible and meaningful sugesstions.

some infos on the subject:
http://www.meridian-int-res.com/Projects/EVRsrch.htm
https://www.autouni.de/autouni_publish/ ... 011-1.html

This is not an argument against electric or EV´s, it´s just to make sure people see both sides of the coin.

EV is a solution but it is not the only solution.

Everything has it´s up and it´s downsides, we should not forget to keep both in mind, and try to find a sensible solution forward for the future.
EV´s and electric propulsion is no exception here.
If done the wrong way, we just replace one monopol with another one, and one way to kill the planet/nature with another.

Don´t pretend you have all the answers, because you haven´t even ask all the questions yet.

Anyway Merry Christmas autogyro

[machin]

I wonder if I might include your post when next time I am confronted with the ongoing argument from the EV brigade, who insist on saying a multi stepped gearbox will not improve the performance and efficiency of an EV.

Auto, I thought you might also like this: a tyical power and torque curve for an electric motor. As you can see neither the torque nor the power are constant. The power curve shows it peaks at 50% of the "no load speed". Since the electric motor only produces peak power at one output speed the performance of a car motivated by an electric motor is indeed improved by adding additional selectable gears so that it can transmit this power at various different road speeds; to increase acceleration (assuming the car isn't already grip limited) or top speed (assuming the top speed isn't already limited by the aero resistance).



Just thought you'd find it interesting.

[autogyro]

Unfortunately the current generation of EVs do not have multi ratio gearboxes.
Not because stepped ratio gbs would not improve range and performance generaly and by a large amount but because non of the current transmission systems available in the market are anywhere near good enough.

They are not efficient enough and nobody has yet managed to get them to last with electric traction for more than around 2000 miles.

The develoment of high energy KERS in F1 is essential to overcome this problem and also to explode some of the myths that are kept in place by the fossil fuel supporters.

The new formula for 2013 is a golden opertunity for F1 to lead the transport and energy revolution, that so far few have recognised.

[747heavy]

If some of you would like to do some calcs for the MPU, maybe this is helpful.
If understood, it will explain why a EV can/will benefit from the use of a multi ratio gearbox

Measuring Motor Parameters
With just a few motor parameters, the steady state performance can accurately be calculated. These parameters are the motor’s torque constant (oz-in/A), terminal resistance, and no-load current.

The torque constant and terminal resistance is usually supplied by the motor manufacture, but should be measured to accurately predict motor performance.

Any DC, permanent magnet motor has a linear relationship to motor torque and current. This ratio is called the motor torque constant and is usually in units of oz-in/Amp or NM/Amp. The torque constant is directly proportional to the voltage constant which describes the voltage generated per RPM or per rad/sec. This is also called the back EMF constant. Since the torque constant is difficult to measure directly without sophisticated equipment, it is best to measure the voltage constant and calculate the torque constant.

The best way to measure the voltage constant is to drive the motor at a known constant speed and measure the voltage at the terminals. If you lack the means to back-drive the motor you can use the amplifier and measure the no-load RPM of the motor at a fixed voltage.

Most digital volt meters cannot accurately measure low resistance as is usually the case in the motor’s terminal resistance. Connect a good current source (1A or less) while measuring the voltage drop across the motor terminals. The voltage divided by the current is the terminal resistance.

The no-load current is a combination of a motor’s friction (bearing and/or brush), hysteresis iron loss, eddy current loss and viscous fluid loss. The no-load current should really be thought of as a no-load torque. Although the no-load current varies slightly with RPM, it is more or less a constant torque. Making this assumption greatly simplifies the mathematical model of the motor, but may be inaccurate in some instances. The no-load current should be measured at the RPM at which the motor is intended to run.

Calculating Motor Performance
Use these handy equations to calculate steady state motor performance. A spread sheet will help in visually graphing motor parameters. If the Torque constant is not supplied by the motor manufacturer, you can measure the motors no-load RPM/Volt and use the following equations to calculate the torque constant.
Torque constant: Kt=Kb x 1.345
Current draw of motor: I = [V-(Kb x kRPM)]/Rm
Torque output of motor: J = (Kt x I) - (Kt x Inl)
RPM of motor: kRPM = (V - RmI) / Kb
Power output of motor: Po = (J x RPM)/1345
Power input: Pi = V x I
Motor efficiency: Eff = (Po/Pi) x 100
Current at peak motor efficiency: Ie max = Sqrt [(V x Inl)/Rm]

Symbol Definitions:
Eff = Efficiency
I = Current
Iemax=Most efficient current
Inl = No load current
J = Torque (oz-in/A)
Kb = Voltage constant (Volt/1000 RPM)
Kt = Torque constant (oz-In/A)
Pi = Power input (Watts)
Po = Mechanical power output (Watts)
Rm = Terminal resistance
RPM = Revolutions/minute
V = Voltage

some other KERS related infos

[ringo]

What is far more intriguing however, is how to harvest 4000 kJ per, lap, say 600 per corner, when kinetic energy being availabe in abundance, but how do you store and release the same quickly and efficient enough?

That is indeed the question that battery development has to find. If we go by the 120 kW BMW dual torque KERS system you need only 11 kg batteries to reach 4 MJ storage capacity. Nevertheless they install 85 kg batteries with far greater capacity to their hybrid super car. Is this done to get a decent electric only operating range or due to constraints with load/unload speed?

Renault has an EV on sale next year called the fluence.

Fluence Z.E. is powered by a synchronous electric motor with rotor coil. Peak power is 70kW at 11,000rpm, while maximum torque is 226Nm. The weight of the motor – excluding peripherals – is 160kg. Acceleration performance is crisp and linear, with maximum torque available very early on.


The capacity of Renault Fluence Z.E.'s lithium-ion battery is 22kWh. The battery itself tips the scales at 250kg and is located behind the rear seats in order to free up a boot volume of 300dm3 (VDA/ISO).
An energy recovery system enables the battery to be charged when the car decelerates.

O.3168MJ/kg from the batteries. 4MJ sounds like a 12.62 kg.
I guess the BMW batteries are slightly better, but that's a concept car. The Renault is ready for the road.
11 or 12 sounds good. either way.

[747heavy]

That is indeed the question that battery development has to find.
If we go by the 120 kW BMW dual torque KERS system you need only 11 kg batteries to reach 4 MJ storage capacity.
Nevertheless they install 85 kg batteries with far greater capacity to their hybrid super car.
Is this done to get a decent electric only operating range or due to constraints with load/unload speed?

This may goes a way to answer your question:

* Specific energy density: 150 to 250 W·h/kg (540 to 900 kJ/kg)[1]
* Volumetric energy density: 250 to 530 W·h/l (900 to 1900 J/cm³)
* Specific power density: 300 to 1500 W/kg (@ 20 seconds and 285 W·h/l)[1]

Because lithium-ion batteries can have a variety of cathode and anode materials, the energy density and voltage vary accordingly.

Lithium-ion batteries with a lithium iron phosphate cathode and graphite anode have a nominal open-circuit voltage of 3.2 V and a typical charging voltage of 3.6 V. Lithium nickel manganese cobalt (NMC) oxide cathode with graphite anodes have a 3.7 V nominal voltage with a 4.2 V max charge. The charging procedure is performed at constant voltage with current-limiting circuitry (i.e., charging with constant current until a voltage of 4.2 V is reached in the cell and continuing with a constant voltage applied until the current drops close to zero). Typically, the charge is terminated at 3% of the initial charge current. In the past, lithium-ion batteries could not be fast-charged and needed at least two hours to fully charge. Current-generation cells can be fully charged in 45 minutes or less. Some lithium-ion varieties can reach 90% in as little as 10 minutes

Fast charging/boosting increases the internal resistance of the cells --> limiting the max. possible discharge current --> limiting the max possible performance (shortens the lifespan of the cell).
This is perhaps the main reason that the KERS batteries in F1 have such a short lifecycle (1-2 races).
It would be sensible (for any road car relevant/meaningful development) to combine the minimum KERS lifespan with the engine lifespan ( 4-5 races).
Otherwise this "green technology" is very wasteful with resources.

I don´t think, that the OEM´s want their customes to change batteries every month, therefore higher capacity is needed.

All the figures mentioned BMW & Renault seem to be at the lower end of the possible specific energy density of Lithium-Ion batteries.

* Limit the time at which the battery stays at 4.20/cell. Prolonged high voltage promotes corrosion, especially at elevated temperatures. (Spinel is less sensitive to high voltage than cobalt-based systems).

* 3.92V/cell is the best upper voltage threshold for cobalt-based lithium-ion. Charging batteries to this voltage level has been shown to double cycle life. Lithium-ion systems for defense applications make use of the lower voltage threshold. The negative is reduced capacity.

* The charge current of Li-ion should be moderate (0.5C for cobalt-based lithium-ion).
The lower charge current reduces the time in which the cell resides at 4.20V. It should be noted that a 0.5C charge only adds marginally to the charge time over 1C because the topping charge will be shorter. A high current charge tends to push the voltage up and forces it into the voltage limit prematurely.



A lithium-ion battery provides 300-500 discharge/charge cycles.
The battery prefers a partial rather than a full discharge.
Frequent full discharges should be avoided when possible.
Instead, charge the battery more often or use a larger battery.

more infos here

[WhiteBlue]

If you go back to the Brembo data and look at the braking power and braking time figures for this year, we see that we have app. 700-1100 kW/s which we could try to harvest under braking and transfer into a storage medium.

I don't want to be pedantic but I have to point out that kW/s would not be a power dimension but a rate of power change. kJ/s =kW would be power.

This is what I'm talking about, there's a 1000 kW out there, but how do we grab it and store it over two measly seconds?

I was already calculating the nominal power of the motors that we need. If we assume a break balance of 60/40% front/rear we need 2 x51kW fore the front and 68 kW for the rear wheels. The overload factor would provide us with the full 333 kW or kJ/s which the battery must be capable to absorb. The rate of discharge can be significantly lower. In the BMW design the front synchonous electric MGU is capable of an overload factor of 1.74 for a duration of 10 s. Similar technology should be capable of 1.95 s for multiple peaks of 12 s out of an 80 second cycle. So the motor question is resolved.

PDF of an A123 presentation from 2007

On page 15 of the presentation we find that A123's prismatic cells have a nominal voltage of 3.3V. They can be discharged (and charged) at 109A continuous current which is 10 times the nominal current of 10.9A. This means that the chemical reaction in the battery raises the internal temperature to a tolerable level where the chemical reaction is still stable. Instead of continuous current one can also use intermittent current with a medium current level. So we do not need to dimension for maximum current but for average current in the applicable period. We plan to send 2x4MJ of energy through the cells in a period of 80s which is equivalent to 4MJ in 40s. Lets keep that in mind.

The prismatic cells have a specific energy/weight ratio of 91 Wh/kg=328 kJ/kg. We plan to use 4 MJ. This means we need to install 12.21 kg of cells. At 0.4 kg/cell we need 31 cells. If we arrange the 31 prismatic cells in series we create 102.3V electric voltage. 31 cells have 12.4 kg battery weight. If we are allowed to charge/discharge at any rate we would be done now. A complete cycle of the charge/discharge takes 300s but we have only 40s. It means we have to split our electric current to eight parallel cells to run within thermal specification. We are forced to install an array of 31x8 cells which brings the total up to 248 cells. Our total weight of the battery array would be 99.2 kg.

This is the weight for 2007 technology based on nanophosphate anodes and graphite cathodes. I have not found a source about possible improvements of technology from 2007 to 2012 which is five years. Five years is a pretty long time for the guys to further tweak the systems. If they manage to improve the discharge/charge rate by one third we would obviously get down to a much better weight level of 66 kg. I will regard that as a an upside potential and stick to 99.2 kg for now.

[747heavy]

@WB
Where did I say that kW/s is a power figure?

Anyway, to futher answer your question why Renault & BMW install more capacity then theoretical needed.

Rate of Discharge
The capacity of a battery is not a constant. When we talk about the kiloWatthour capacity of a battery,
this is a nominal figure usually defined at a relatively low discharge rate of C/20, which means it takes
20 hours to discharge the battery.
The more slowly a battery is discharged, the more energy in total it will supply – but it is supplying a
relatively small amount of energy per unit time, i.e. its power delivery is low.
When a battery is discharged quickly at a high rate of power, its total nominal energy capacity falls – in
other words it can deliver high power but for a short period of time and delivers less total energy than if
it was discharged slowly.
The standard discharge rate generally used to analyse the performance of pure battery EVs is C/3
which means that at the “average” discharge rate expected for a BEV, the battery will last 3 hours.
Therefore for a 32 kWh battery expected to deliver on average 3 miles range per kWh or 100 miles in
total, over 3 hours, that equates to an average speed of 33 mph with the battery delivering about 10kW
over that time period. Therefore if the car drives faster, the battery capacity will fall as power delivery
increases and range will fall below 100 miles; conversely, if the car drives more slowly than 33 mph, it
can go further than 100 miles but will obviously take longer to do so.
The problem is exacerbated as the size of the battery becomes smaller. The power needed to drive the
vehicle at any speed remains substantially the same if the battery is smaller but the relative rate at
which the battery is being discharged increases. Therefore its effective capacity falls even further due
to the increased discharge rate.
Thus for a standard hybrid vehicle (HEV0) with a nominal 1.5 kWh battery, not even 5 miles range can
be achieved on battery power alone because the battery capacity is so small compared to the power
demand needed to drive the vehicle: a 10 kW draw at 30 mph is a discharge rate of 10 / 1.5 or about
C7 which means the battery will discharge in 1/7 of an hour or say 9 minutes. So at 30 mph one would
expect a range of 4 - 5 miles. The problem is that the nominal capacity of 1.5 kWh applies at C/20, not
C7. C7 is a discharge rate 21 times as fast as the C/3 discharge rate on a full 32 kWh BEV battery.
Thus the nominal capacity of 1.5 kWh falls even further under this higher discharge rate and the vehicle
only provides 1 mile of range, not 4 or 5.