2014-2020 Formula One 1.6l V6 turbo engine formula

[WhiteBlue]

One other thing that i was puzzling. The batteries should work better at hotter temperatures. Cooling them is counter intuitive, could cooling be an indication that the C rating is pushed beyond normal limits?

Battery chemistry is an exothermal reaction. It means that you increase heat rejection the more intense you drive the battery. The increase in temperature is the result of higher charging currents not the condition for it.

The better you cool the batteries the more heat gets extracted and lower temps will result at the same power level. So cooling the batteries by liquid cooling is a condition to high power rates.

[ringo]

But also, heating the batteries increase the rate of chemical reaction. Internal resistance also drops with increased temperature.
I guess it's a give and take.

Now about the KERS batteries, what are the general specs?

[747heavy]

Internal resistance also drops with increased temperature.

Sorry Ringo, are you are sure about this?
Supra conductivity is normaly achieved at 0K.
It is correct, that a battery looses performance when it is cold, because you need a minimum temperatue/energy for the chemical reaction to take place, but the resistance of most materials increases with temperature.
As WB pointed out the internal resistance of the battery means it will heat up during charging/discharging. As lower the internal resistance of a battery as lower the temperature involved and as lower the voltage drop.

R= U/I and P=U*I
which means for a know restitance and current you can calculate the voltage drop and then the power (loss) which get´s turned into heat.

To achieve the charging rates WB is looking for we need:
4000KJ (4MJ) / 12s = 333KJ (kW/s)
at an voltage of 305V that makes 333000W/305V = 1092A /3 (because we have 3 cells in parallel)= 364A

Now, we need to know the internal resitance of an A123 cell (I will do some calcs later), but for now I will use the 140mOhm claimed in one of the KERS papers I have posted earlier.
This would mean 364A*0.14Ohm= 50.96V (this is the voltage drop at the battery)

50.96V*364A= ~18550W or 18.5kW

As Xpensive would say, that´s a lot of toasters, and would require some massive cooling.

This energy is a "loss" so you would need to account for it, when calculating your havesting&release rate.

For fun you could calculate the cable diameter required for 1092A (from the MGU to the accu).
An acceptable voltage droop for such an application is normally 0.25V.
(this means 1092A*0.25V=273W your cable gets pretty warm as well, but its managable)

Let´s assume we have 1m distance and two cables (+/-) that makes 2 meter and we use silver cable (which is common for these applications).

Silver cable has a specific resistance of ~0,0161Ohm*m/mm^2.
With this you can calculate the area (mm^2) needed for your cable, and then, assuming it´s a round cable, the diameter needed.

We talk about some pretty serious installations here.
Just think about the electronics to control 333kW!!! and the connectors involved.

[ringo]

Sorry about the internal resistance thing, that is for a lead acid battery, with flowing liquid. Not solid state batteries.
But generally performance drops with temperature. A little heat isn't bad.
The Power losses may actually do a little good heating the battery.
Though most of it may be to much heat, which must be rejected with cooling.

747, are you using kW/s for any reason? do you mean kJ/s?
I now see what the calculation is doing.

333kW that's 446hp, i don't see any reason the KERS should charge and discharge at a rate where, only 120kW is required.

Anyway, just as a note, Power = V x Charge/ time.

charge = 333,000 x 12s / 305V = 13101.6 Couloumbs or 3.64 Ah.

[747heavy]

this is a point of discussion, where WB never came back with an answer.
If (what was disscused before)the max KERS power is limited to 120kW (like it is now to 60kW) you will never be able to harvest 4MJ.
Because 120kW x 12s (our average braking time per circuit)= 1440kWs = 1440KJ (1.44MJ)

Yes sorry I wrote the unit wrong - appologies
its kWs=KJ (not kW/s)

I think this is a realistic target for 2013.
You could perhaps increase the storage capacity to 2MJ and let people "harvest energy" over 2-3 laps and then they are able to use it for longer.
2000kWs(KJ)/120kw = 16.66 sec

But with an max power of 120kW you can only harvest 1.44MJ/lap on a average circuit.

You would need ~5-6 laps (and 7.5MJ storage) to be able to use the full 120kW for the rest of an lap, when you are not braking (63sec).
As you will not need the extra power all the time, a sensible compromise would be (IMHO) ~3MJ storage and a 120kW power limit.
Collect for ~2.5-3 laps and make strategic use of it for a fast lap or an overtake.

[747heavy]

O.K. here we have two examples of the Brembo braking data

Canada:
Laptime: ~75s
Braking points: 7
Time under brakes: 12.94s
average braking time: 1.85s
brakings shorter then average: 4



Barcelona:
Laptime: ~80s
Braking points: 8
Time under brakes: 10.37s
average braking time: 1.29s
brakings shorter then average: 3



If somebody is interested, he can look up the rest of the races and do some analysis, but I think laptime 75-80s and braking time of 10-12s per lap is a pretty good estimate.

[747heavy]

Anyway, just as a note, Power = V x Charge/ time.
charge = 333,000 x 12s / 305V = 13101.6 Couloumbs or 3.64 Ah.

Yes, that looks all right to me.
If we had 1h time we could achieve the same charge with 3.64A charging current
Because we have 3 cells in parallel this makes 1,21A per cell.
The cells are ~11Ah(C), so this would be a relaxed charge with ~C/9
(C/10 would be the "classical" charging rate 1h with 1.1A = 11Ah)
But because we only have 12s to get the job done, we need to giddy up a bit.
3.64Ah = 13104As
13104As/12s = 1092A

1092A/3 = 364A which means charging with ~33C

[pipex]

747, the braking power figures from Brembo are average or maximum? Either way, they look very big... To recover energy at that amount of power the system should work at very high voltages to make it work I think...

[747heavy]

As I see it, it´s the max power needed to stop (slow) the car at this point.

The figure is high, because the time is so short, and the energy (K.E.) of the car given.

If we take the 12s for braking in an 75s lap example (Barcelona), we have 12s vs. 63s it´s a ratio of ~5.25.
So our engine has ~5 times as much time to build up the K.E. using the power it has.
Now our brakes have much less time, so they need to be 5 times as powerful.

This is a bit simplyfid, because it does not accout for aero drag etc., which also slows the car down, but the principle is correct-IMHO.
If we include aero drag and some other factors I think we can say, that our brakes need to be 3-4 times as powerful as our engine.

This is the "problem" with KERS we have only 1/5th of the time, trying to harvest as much energy back as we can.
That´s why it has to be so powerful.

[Blackout]

Has someone some photos showing the Williams JPH01 Roll hoop-airbox/turbo layout, or another car with similar characteristics ? thanks

[godlameroso]

Remember it's not going to be just KERS in 2013 but rather a combination of different systems all charging the battery to different capacities. If braking only provides 1.44MJ per lap, what about the other systems, they would only need to contribute 25% more to get close to 2MJ per lap. This will definitely be an interesting engineering challenge. Maybe the batteries can ionize the air in some weird way and create more downforce, who knows?

[747heavy]

Has someone some photos showing the Williams JPH01 Roll hoop-airbox/turbo layout, or another car with similar characteristics ? thanks

Don´t know if this is what you are looking for Blackout
WB had posted a photo in the other thread.
I will see, if I can find a better photo from the engine in the car

the old Formula Palmer with Audi turbo engine

[Blackout]

viewtopic.php?p=215149#p215149

Yes I'm looking for this kind of photo with a full mounted but naked car. Thanks.
I heard the R18's Turbo is placed 'inside' the V of the V6. But how it will be placed in a modern l4 F1 car ? No one tried before to place it behind the engine (like in Ringo's drawing*) or above an inclined engine ? :

* http://i69.servimg.com/u/f69/14/79/55/26/turbop10.jpg

[WhiteBlue]

this is a point of discussion, where WB never came back with an answer. If (what was disscused before)the max KERS power is limited to 120kW (like it is now to 60kW) you will never be able to harvest 4MJ.

The rules limit only the power you apply for acceleration, not the breaking power.

O.K. here we have two examples of the Brembo braking data Canada:

Canada is really heavy breaking. There are 16.7 MJ to harvest if you just figure the seven corners. 4MJ should not be a problem at all on that track.

[xpensive]

There you go WB, 4 MJ in 12.94 seconds, makes for just 309 kW on average, piece of cake really.