2014-2020 Formula One 1.6l V6 turbo engine formula

All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
mrluke
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Re: Formula One 1.6l V6 turbo engine formula

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Holm86 wrote:
mrluke wrote:
Tommy Cookers wrote: an earlier post mentions 3.5 bar boost ie 3.5 bar absolute (if that can be believed)
many have calculated that as a turbo engine around only 2 bar abs is needed (the 1988 engines were 'only' 2.5 bar abs)
though I have evidenced the efficiency benefits of backpressure, 3.5 bar does not suggest backpressure any time soon
3.5 bar boost typically refers to 3.5 bar above atmospheric. i.e. 4.5 bar abs.

This becomes obvious when you talk about an engine running 0.5 bar boost, it is not being feed with lower than atmos pressure air.

As far as I understand it the mgu-h generates power by braking the turbo and thus controlling boost. This is going to cause higher back pressure pre turbo than in a traditional setup where the excess boost is bled out of the system via a wastegate. Clearly there must be a limit on how much load the turbine can deal with before a wastegate will need to open.

Feel free to correct If i have misunderstood.
I calculated the boost to be 1.25 (2.25 absolute) bars of boost @ 10.500 rpm falling to 0.6 (1.6 absolute) bars of boost @ 15.000 rpm. This was based on a static temperature and a stoichiometric air fuel ratio.
Hmm if that is correct then it explains why the cars sound much tamer than in the 80s. I was expecting a much more "boosted" sound. Without doing any math I would have expected the PUs to need to run higher boost levels in order to make so much power at low rpm but if you have worked it through then I wont disagree with you :)

ppj13
ppj13
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Re: Formula One 1.6l V6 turbo engine formula

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Holm86 wrote:
I calculated the boost to be 1.25 (2.25 absolute) bars of boost @ 10.500 rpm falling to 0.6 (1.6 absolute) bars of boost @ 15.000 rpm. This was based on a static temperature and a stoichiometric air fuel ratio.
I did the math also, those figures seem correct to me.

I mean, I get closer to 2.5 absolute, but it's simply a matter of how much you guess air is cooled down and how bad the VE is ruined by small heads and backpressure. 2.25 is plausible.

Even if exhaust pressure is less than that, which I don't think possible, venting all pressure after the turbine and using the MGUH to power the compressor will be beneficial in short burst in qualy (when electric energy is abundant).

bhall
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Re: Formula One 1.6l V6 turbo engine formula

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Powershift wrote:
munudeges wrote:Keep it civilised? You're either wrong or you're not. Talking about torque when you mean power is not an opinion. It's simply wrong.
It is you that is wrong, the instantaneous force of the wheel to the track is torque and not power, If you are referring to how long it takes to get down the straight then yes power is what needs to be known(as long as the torque doesn't have you wheel spinning the whole time), but it is the torque of the system that determines if there is wheelspin or not.

At the start of a drag strip where dragsters can develop enough torque to flip the cars, the road takes much more of a beating than at the end of the 1/4 mile when the engines are producing far more horsepower.

Listen to Hamilton explain the feeling with the much higher torque produced by these PU's.
https://www.youtube.com/watch?v=ZjaOVaFn2Gk
Not so much.

Wheel torque is engine torque x gear ratio x final drive. Acceleration and top-speed are determined by a car's power-to-weight ratio. These are important.

A vehicle equipped with a lower-torque engine will nevertheless produce the same wheel torque as a vehicle equipped with a higher-torque engine, provided the two vehicles are the same weight, are geared to reach the same speeds, and both have engines with the same horsepower. (And if they have equal cd, but let's not talk aero.)

The only starring role played by engine torque in this scenario is to determine the unique gear ratios needed by each car in order for it to match the ideal performance of the other, which, again, is the same by virtue of their equal power-to-weight ratios.

The need for different gearing between the two is necessary, because the higher-torque engine will invariably have a lower redline than its lower-torque counterpart. The inverse is also true. Otherwise, they wouldn't have the same horsepower.

To that end, the higher-torque/lower-RPM vehicle will be geared longer to sacrifice acceleration in order to match the top-speed of the lower-torque/higher-RPM vehicle, which, in turn, will be geared shorter to sacrifice top-speed in order to match the other vehicle's acceleration. When the two meet in the middle with an ideal combination of both acceleration and top-speed, they will each produce the same wheel torque.

This is the magic of gearboxes.

I apologize if this seems terse. The increasingly frequent attribution to an engine's torque output of what rightly belongs to its power-band has made me a little nutty.

OK, nuttier.

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Blackout
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Re: Formula One 1.6l V6 turbo engine formula

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Wasnt the V8 thermal efficiency 35%?
Because without Kers and 40% we would get 693ps with these V6...

Tommy Cookers
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Re: Formula One 1.6l V6 turbo engine formula

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ppj13 wrote:
Tommy Cookers wrote:@ ppj (mostly)
AFAIK ....... a turbo is normally operating in blowdown turbine condition
blowdown 'pulses' drive the turbine, the mean exhaust pressure is not raised relative to the induction pressure (it can fall)
backpressure means the mean exhaust pressure is higher than the induction pressure, this is called pressure turbine condition (the exhaust valve motions of course isolate the combustion chamber from the backpressure)
1. Pressure inside exhaust manifold will be close to pressure at intake manifold (let's say I:3.0, E:2.5bar) when turbine is powering the compressor. If turbine is turbocompounding (powering the MGUH), probably higher.
2. Pressure inside exhaust manifold will be close to zero if you open a big wastegate. Pumping alone, doing very quick math, will take 20hp less than regular ICE, 40hp less than turbocompounding ICE. That extra power will "go" to the flywheel.
3. Air mass flow can be kept with the MGUH at the same levels. Boost required to keep AMF will be lower, but that's not related to the point.
Tommy Cookers wrote: mgu-k recovery is torque limited by rule so that 161 hp recovery is only allowable over about 5500 crank rpm, below that it falls
surely there will be little or no capacity provided eg in the motor drives to allow higher total power (than this 161 hp) from storage ?
I don't follow. If you mean there is no energy in the batteries to power both MGUK and MGUH consistently during a race in the max power mode, then I fully agree with you. If you mean you can't ever draw 250hp from the batteries, I don't agree. You can for short burst. There is no limit to that, not in the rules, not in the physics, as long as you plan for it.
there is a level of exhaust turbine power recovery that doesn't increase mean exhaust pressure
so there is no loss of crankshaft power ie on the exhaust scavenge upstroke (this level we might call the blowdown level)
according to Wright, who made 10000+ such engines, to NACA who researched it, and to Buchi who invented it ?
they show that around 10% power can be added by the tc without loss of crankshaft power or any increase in fuel consumption
they used axial flow turbines, mechanically driven centrifugal superchargers and rather low CRs

in 2014 we are using some of the turbine power to drive a centrifugal supercharger ie turbocharging
this will give a nice forward pressure when we choose to allow this ie at low mgu-h recovery
increasing turbine power % is possible, but forward pressure is reduced as mean exhaust pressure must rise
increase beyond this gives backpressure, some crankshaft power is lost, but compounded power is maintained and sfc increased

so the crankshaft power temporarily liberated by wasting exhaust and supercharging electrically depends on the operating condition
ie compounding does not necessarily mean any raising of exhaust pressure (this was my original point)
as rpm rises over 10500 rpm less supercharging power will be needed, gain from cutting the compounding may be less than hoped


the motors are not driven directly from the battery
to provide always the desired torque at every PU rpm the motor voltage must be continuously varied
this requires a motor drive, the designer will choose some limit to its useable current and related losses (self-heating)
also inversion will likely be needed, and the drives are bidirectional
your eg 250 hp drives regime will need to be bigger and will need more cooling than eg a 160 hp drive regime
transistors in the drive will have low thermal time constants, there is little scope for temporarily overload (unlike the MGs)

given that the dominant jobs of the electrical system are near-continuous motor action of an MG limited to 160 hp
and near-continuous generator action of an MG surely sized for this at around 100 hp
I just can't see it having particularly a 250 hp capability for some cunning but occasional activity of poor efficiency
Last edited by Tommy Cookers on 18 Mar 2014, 11:05, edited 5 times in total.

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GitanesBlondes
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Re: Formula One 1.6l V6 turbo engine formula

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bhallg2k wrote:
Powershift wrote:
munudeges wrote:Keep it civilised? You're either wrong or you're not. Talking about torque when you mean power is not an opinion. It's simply wrong.
It is you that is wrong, the instantaneous force of the wheel to the track is torque and not power, If you are referring to how long it takes to get down the straight then yes power is what needs to be known(as long as the torque doesn't have you wheel spinning the whole time), but it is the torque of the system that determines if there is wheelspin or not.

At the start of a drag strip where dragsters can develop enough torque to flip the cars, the road takes much more of a beating than at the end of the 1/4 mile when the engines are producing far more horsepower.

Listen to Hamilton explain the feeling with the much higher torque produced by these PU's.
https://www.youtube.com/watch?v=ZjaOVaFn2Gk
Not so much.

Wheel torque is engine torque x gear ratio x final drive. Acceleration and top-speed are determined by a car's power-to-weight ratio. These are important.

A vehicle equipped with a lower-torque engine will nevertheless produce the same wheel torque as a vehicle equipped with a higher-torque engine, provided the two vehicles are the same weight, are geared to reach the same speeds, and both have engines with the same horsepower. (And if they have equal cd, but let's not talk aero.)

The only starring role played by engine torque in this scenario is to determine the unique gear ratios needed by each car in order for it to match the ideal performance of the other, which, again, is the same by virtue of their equal power-to-weight ratios.

The need for different gearing between the two is necessary, because the higher-torque engine will invariably have a lower redline than its lower-torque counterpart. The inverse is also true. Otherwise, they wouldn't have the same horsepower.

To that end, the higher-torque/lower-RPM vehicle will be geared longer to sacrifice acceleration in order to match the top-speed of the lower-torque/higher-RPM vehicle, which, in turn, will be geared shorter to sacrifice top-speed in order to match the other vehicle's acceleration. When the two meet in the middle with an ideal combination of both acceleration and top-speed, they will each produce the same wheel torque.

This is the magic of gearboxes.

I apologize if this seems terse. The increasingly frequent attribution to an engine's torque output of what rightly belongs to its power-band has made me a little nutty.

OK, nuttier.

Nice explanation...unreal people here are still going on about engine torque.
"I don't want to make friends with anybody. I don't give a sh*t for fame. I just want to win." -Nelson Piquet

bhall
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GitanesBlondes wrote:Nice explanation...unreal people here are still going on about engine torque.
Thanks. It is what it is. You live and learn.

Here's something else that just crossed my mind, though. Devil's advocate:

If by some miracle of engineering the 2014 cars did, in fact, have more wheel torque than last year's cars, it would only be the case for first gear, because wheel torque progressively decreases with every upshift.

That means the blessed soul driving the miraculous machine - on water, I might add - would find that his gilded torque chariot supplies less wheel torque in 2nd gear than the heretical 2013 cars did in 1st gear. So, for that ratio, and for every one thereafter, the suddenly fallible jalopy would create wheel torque in amounts no greater than what was regularly experienced last year.

Yeah, boyeeeeeeee!

mrluke
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bhallg2k wrote:Not so much.

Wheel torque is engine torque x gear ratio x final drive. Acceleration and top-speed are determined by a car's power-to-weight ratio. These are important.

A vehicle equipped with a lower-torque engine will nevertheless produce the same wheel torque as a vehicle equipped with a higher-torque engine, provided the two vehicles are the same weight, are geared to reach the same speeds, and both have engines with the same horsepower. (And if they have equal cd, but let's not talk aero.)

The only starring role played by engine torque in this scenario is to determine the unique gear ratios needed by each car in order for it to match the ideal performance of the other, which, again, is the same by virtue of their equal power-to-weight ratios.

The need for different gearing between the two is necessary, because the higher-torque engine will invariably have a lower redline than its lower-torque counterpart. The inverse is also true. Otherwise, they wouldn't have the same horsepower.

To that end, the higher-torque/lower-RPM vehicle will be geared longer to sacrifice acceleration in order to match the top-speed of the lower-torque/higher-RPM vehicle, which, in turn, will be geared shorter to sacrifice top-speed in order to match the other vehicle's acceleration. When the two meet in the middle with an ideal combination of both acceleration and top-speed, they will each produce the same wheel torque.

This is the magic of gearboxes.

I apologize if this seems terse. The increasingly frequent attribution to an engine's torque output of what rightly belongs to its power-band has made me a little nutty.

OK, nuttier.
This is true for the peak power of both 2013 and 2014 power units but not so for the remainder of the rev range.

The v8 engines were VERY peaky. Theres a reason that last year there were lots of mid corner gear changes whereas this year they wont need to, because they have max power on demand all of the time.

munudeges
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Powershift wrote:It is you that is wrong, the instantaneous force of the wheel to the track is torque and not power...
What happens at the wheel is completely irrelevant with regards to torque at that point. Torque without considering RPM is really quite meaningless, and that means power.
If you are referring to how long it takes to get down the straight then yes power is what needs to be known(as long as the torque doesn't have you wheel spinning the whole time), but it is the torque of the system that determines if there is wheelspin or not.
Ahhhh, the old chestnut of straight line speed being about power and acceleration being about torque. Please......

http://www.epi-eng.com/piston_engine_te ... torque.htm
It often seems that people are confused about the relationship between POWER and TORQUE. For example, we have heard engine builders, camshaft consultants, and other technical experts ask customers:

"Do you want your engine to make HORSEPOWER or TORQUE?"

And the question is usually asked in a tone which strongly suggests that these experts believe power and torque are somehow mutually exclusive.

In fact, the opposite is true, and you should be clear on these facts:

POWER (the rate of doing WORK) is dependent on TORQUE and RPM.
TORQUE and RPM are the MEASURED quantities of engine output.
POWER is CALCULATED from torque and RPM...
It seems as though many 'experts' even in Formula 1 are becoming confused. They can't wrap their heads around having more lower end power but having an engine that produces the same amount of power overall - so they believe it must be caused by something else.

In reality you can have apparently completely different but strangely equivalent engines meeting in the middle, but it will be they will produce varying amounts of power at different engine speeds - the power band in other words. What nature gives you with one hand it takes away with the other.

bhall
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Re: Formula One 1.6l V6 turbo engine formula

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mrluke wrote:This is true for the peak power of both 2013 and 2014 power units but not so for the remainder of the rev range.

The v8 engines were VERY peaky. Theres a reason that last year there were lots of mid corner gear changes whereas this year they wont need to, because they have max power on demand all of the time.
Though the power-bands are obviously much wider now, a development which has dramatically enhanced the PU's response to driver input, it is my sincere hope that teams have still geared their cars for peak power. Otherwise, this is gonna be a really weird season, yanno?

R_Redding
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Re: Formula One 1.6l V6 turbo engine formula

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More data about the fuel flow sensors..

Tech Requirment and the procurement details..

http://www.fia.com/sites/default/files/ ... eter_0.pdf
Accuracy..
Gill Sensors themselves quote" that 52 per cent of its meters are with a 0.1 per cent accuracy reading, with 92 per cent within 0.25 per cent".

After being calibrated by Gill Sensors ...they are then re-calibrated by Calibra Technology.

http://www.calibratechnology.com/

Calibra Technology will also help the FIA enforce the new rules by providing random checks of flow meters throughout the season.

The fuel sensor is not calibrated during the race..

All communication to the sensor are encrypted to stop the team having access to the flow meters internal memories..and calibration data.

Any offsets provided by the Fia ...are to the SECU ..to make the max injector squirt duration times and hence consumption comply with the reading they are obtaining from the fuel sensor.


Rob

ppj13
ppj13
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Tommy Cookers wrote:
turbocompounding doesn't increase mean exhaust pressure at all, so there is no loss of crankshaft power from it
according to Wright, who made 10000+ such engines, to NACA who researched it, and to Buchi who invented it
around 10% power can be added by the tc without loss of crankshaft power or any increase in fuel consumption
(when using axial flow turbines on engines with mechanically driven centrifugal superchargers and rather low CRs)
somewhere above this % some crankshaft power is lost, but compounded power is maintained and sfc increases (backpressure)
a non-compounded engine is dumping exhaust pressure from 7-8 bar to 1 at blowdown to no effect
a (blowdown) compounded engine exports some of this pressure energy, the end state is still exhaust gas at 1 bar
the crankshaft is on the upstroke/scavenge stroke exposed to the same 1 bar pressure, so loses no more power
Semantics, again. We both say the truth. Except in one point:

Extracting energy from the exhaust increases mean exhaust pressure. The only way a turbomachine has to quickly extract mechanical energy from a fluid is to expand it (reduce its pressure). The expanded state is the atmosphere, so the unexpanded has to have more pressure. And that pressure is seen by the piston all the time during the scavenge stroke. In a typical turbo engine, exhaust pressure is somewhere close to boost pressure.

Big simplification ahead, but still very reasonable:

In a properly designed exhaust, NA, the exhaust discharge pressure is slightly above ambient pressure (say 1.05 bar). In the ideal otto cycle, at every scavenge stroke, the work you do per turn of the engine (4T) is P*deltaV=P*s*c*1/2 (c being the stroke of the engine, s being the area of the cylinder). The power you need to pump exhaust gases out of the cylinder is just that, times speed of the engine.

So if exhaust pressure goes up to, say, 3.15 bar, the pumping power is 3 times bigger. That does not mean that using exhaust energy create a loss in flywheel power. Only if you use it wrong (let's say, to power a big horn). If you use that power to charge the air or to propel the flywheel (turbocompound), the final power will be higher. Efficiency also, all things equal. Even volumetric efficiency (sometimes), because the pump "losses" is the work you need to pump the exhaust out plus the one you need to suck fresh air in. Turbocharging makes the latter negative.

It is very cheap, but it does not mean it is absolutely "free". If you get the power from somewhere else (batteries), forced induction is even more advantageous in terms of raw power.
Tommy Cookers wrote: the motors are not driven directly from the battery
to provide always the desired torque at every PU rpm the motor voltage must be continuously varied
this requires a motor drive, the designer will choose to limit its useable current and related losses (self-heating)
also inversion will likely be needed, and the drives are bidirectional
your eg 250 hp drives regime will need to be bigger and will need more cooling than eg a 160 hp drive regime
a drive will have much lower thermal time constants then eg the motors, so there is little ability to temporarily overload the drives

given that the dominant jobs of the electrical system are near-continuous motor action of an MG limited to 160 hp
and near-continuous generator action of an MG surely sized for this at around 100 hp
I just can't see it having particularly a 250 hp capability for some cunning but occasional activity of poor efficiency
Again 99% agreed.

The "only" part that needs upgrading to support this mode is the electronic power unit (Batteries have a thermal limit on power, so it may need not beefing up, as in qualy they are not very hot anyway, and both MG will not see different load). The electronic power unit is what, about 3kg?. Would you trade 3 extra kg for a short burst of 40 extra hp, qualy only? I would. You wouldn't. That's ok, that's what forums are about. :D

321apex
321apex
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Re: Formula One 1.6l V6 turbo engine formula

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R_Redding wrote:More data about the fuel flow sensors..

Tech Requirment and the procurement details..

http://www.fia.com/sites/default/files/ ... eter_0.pdf
Accuracy..
Gill Sensors themselves quote" that 52 per cent of its meters are with a 0.1 per cent accuracy reading, with 92 per cent within 0.25 per cent".

After being calibrated by Gill Sensors ...they are then re-calibrated by Calibra Technology.

http://www.calibratechnology.com/

Calibra Technology will also help the FIA enforce the new rules by providing random checks of flow meters throughout the season.

The fuel sensor is not calibrated during the race..

All communication to the sensor are encrypted to stop the team having access to the flow meters internal memories..and calibration data.

Any offsets provided by the Fia ...are to the SECU ..to make the max injector squirt duration times and hence consumption comply with the reading they are obtaining from the fuel sensor.


Rob
Instrument vendors want to sell their product and as such will make all sorts of technical claims.
Liquid flow rate measurement is a complex science riddled with error if you deviate from controlled laboratory conditions. Such measurement does not belong on board of a race car.

Key factors having impact on reading accuracy are:
- aeration of fuel (small air bubbles within it) which is common in racing cars
- temperature
- viscosity which is affected by temp
- density

All these factors are not equal from car installation to car installation and there could be ways to fool the sensor into allowing a larger effective flow under acceptable indicated flow rate.

This is a pandoras box of cheating possibilities, where fuels can be brewed according to fuel flow rate meter needs.

Tommy Cookers
Tommy Cookers
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Re: Formula One 1.6l V6 turbo engine formula

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ppj13 wrote: Extracting energy from the exhaust increases mean exhaust pressure. The only way a turbomachine has to quickly extract mechanical energy from a fluid is to expand it (reduce its pressure). The expanded state is the atmosphere, so the unexpanded has to have more pressure. And that pressure is seen by the piston all the time during the scavenge stroke. In a typical turbo engine, exhaust pressure is somewhere close to boost pressure.
Wright showed the world (in a paper that was linked in this thread) that their compounding did not increase exhaust pressure
which was my point, that 2014 F1 can have turbocharging and some simultaneous compounding without raised exhaust pressure
so your scheme would liberate power only according to the the compounding etc state in use
and this might be different for qually, compared to the race-optimum

the argument you present above was campaigned against and apparently disproved by Wright evidence 60+ years ago
tc uses some of the exhaust pressure in the blowdown event ('pulse') that is otherwise totally lost
the pressures at EO and bdc are the same in their tc and non-tc engines, the tc work is simply free
all related to blowdown being a choked process ?
granted the their tc engine has necessarily 12 of the 18 cylinders using long (tuned length ?) exhaust pipes (not so the non-tc)

granted also that to go above a certain level of recovery the mean exhaust pressure must rise
and 2014 will be doing this for much of the race ?
eventually (or now even ?) this may extend to actual backpressure, when EV closure timing will become critical

FWIW I think the penalty of a temporary 250 hp ES and motor drive capability is a lot more than 3 kg

@ dren or anyone
any more/clearer pics of the MB 'log' exhaust manifold ??
IIRC short systems can still give some tuned-length effect

ppj13
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@ Tommy,

I didn't know the paper. I just finished reading it. Thanks for mentioning, now my education is better.

What the paper says: They took a supercharged engine, recovered some exhaust energy and did not increase the pump losses. They tunned previously untunned exhausts? They didn't mention.

There is a diagram in the paper in which they say the backpressure is 0.21bar at "rated power": harvesting 300hp from a 2400hp engine. To do that, they used huge turbines (290mm diameter) spining at 15000rpm. (each dealing with 800hp).

If I had to do a turbo engine with almost stationary modes, I'll use the biggest turbine possible because they have better efficiency (less backpressure).

That's ok, although this airplane engine setup is so strange to me that I do not know if this is possible. What I suspect is that F1 teams cannot use a 290mm turbine, packaging aside (the housing will be like 500mm), because if MGUH fails they will have to apply full throttle at the end of the time practice in order to have some boost at the start of the race.

Smaller turbines, needed for relatively short transients, do produce backpressure.

But you are right, F1 can have turbocharging and compounding with very little backpressure. They need huge turbines with low expansion and high area. The huge inertia that implies can be dealt with the MGUH. This will bring big hp gains over teams with more "conventional sized" turbines.

Did some team bit the bullet?