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

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You are right about optionality of MGU-H usage, I missed that so thanks for setting it straight. Indeed also there is not a word in regulations that forbids the use of a wastegate. I have a feeling in that case, that some duality of boost control my take place, where MGU-H as well as a wastegate are applied. At least in such configuration, I feel the boost control will be possible in logical fashion.

In conditions of diminished boost requirement towards the top RPM, the excess exhaust will have to wastegated OUT to prevent from choking the engine on the exhaust side. Choking of any kind increases pumping losses and lowers BSFC. IMHO :D
wuzak wrote:
321apex wrote:The rules allow the following fuel flow:
9500 - 91 kg/h
10000 - 96
10500 - 100 = <<<-------GOLDEN POINT

The higher the engine speed with limited fuel, the higher are the friction losses (FMEP)... and LESS brake POWER. So ideal power peak is at 10500 RPM, however you must factor in the RPM spread for the gearing, which will force you to rev the engine in a zone of DIMINISHING FUEL efficiency [BSFC].
In theory - not necessarily in practice.

As we have theorised before, in turbo-compound mode the peak power will be moved up the rpm range.

The reason for this that the constant fuel flow from 10,500rpm dictates a constant mass air flow from the turbo and a falling boost requirement.
Which means that with falling boost, the power would also be falling... and BSFC will be going in wrong direction, getting worse.
wuzak wrote:The exhaust energy doesn't fall as much as the boost, so there is more power to use for the MGU-H. And that the increased MGU-H power more than offsets the increased friction.
How do you know that exhaust energy will not be falling as much as boost? Can you prove this without the aid of Cosworth cartoon? :mrgreen:
wuzak wrote:The Cosworth graphs seem to confirm a 1000rpm+ shift for peak power to higher rpm.
This graph is treated as if it was some Gods truth, where it is completely unjustified. This is simply a theoretical study on a very greay area of unknown technology (MGU-H) and it is full of fudge factors.

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

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Holm86 wrote:
321apex wrote:
chip engineer wrote:Also, with the fuel flow limit, there is no air flow increase needed at high rpm.
Since when??? Are you inventing new principles of Otto cycle engine operation? Maybe you can provide some citations from technical literature to support this.

Although some measure of charge stratification is possible with direct fuel injection, however what you are expecting isn't supported by any thermodynamic principles.
The fuel flow is completely flat from 10500 rpm to 15000. So the airflow will also be completely flat. Thats why the boost pressure needs to drop from 10500 rpm to 15000 rpm.
With that I agree, but you fail to notice that by going as you say from 10500 to 15k, you are producing less and less power, which throws your precious BSFC out the window. Not long ago, someone was talking in this thread that this new F1 engine will be achieving Prius level fuel efficiency. What will happen to that, if you want to make BSFC progressively worse by capriciously clinging to the 15k allowable limit ?

RPM is COSTLY!!! The higher you go the more you pay in fuel and in hardware cost.
It is unnecessary to rev high if you don't need to.

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Holm86
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Re: Formula One 1.6l V6 turbo engine formula

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You do NOT use more fuel at 15000 rpm than at 10500 rpm!! You only get a decrease in power becaust of the increased friction. And the question is if you have less power at 15000 rpm than at 10000 rpm.

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

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Abarth wrote:
321apex wrote:[[...]Rest assured, that I am not a pioneer in using analogies drawn from one area of science to explain others. In this case, the NA analogy with Turbo is quite nearly identical if you understand the subject. If my argument makes little sense to you, then you should review in detail the information I provided along until you understand it. What I share here with you is something that was never given to me on a platter. You are getting the gist and are not liking it. [...]
What about continue learning instead of throwing tons of theory together and missing the whole picture?

Why should a configuation which peaks at say 15'000 in N/A not be able to peak at 10'500 when overcharged? Ever heard of overboost mapping, decreasing the boost arbitrarily?

Hint, as Holm86 already pointed out: You need to decrease boost going from 10'500 to 15'000 in these fuel flow restricted engines.
Unless you deliberately are "choking" them earlier, needing to rise the boost again. Not THAT efficient in terms of pumping losses though.... :? .
I thought you didn't want to get into a discussion with me.

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Holm86
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Re: Formula One 1.6l V6 turbo engine formula

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apex you need to remember that from 10500 rpm to 15000 the airflow through the engine will be the same. But from 10500 rpm and upwards the engine itselfs sustains more and more of the airflow which means that the turbo is suppling less and less boost. The flow and energy through the exhaust should stay the same. Though the exhaust gasses would travel faster, slightly decreasing the pressure which is why you dont need to wastegate it.

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

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Holm86 wrote:apex you need to remember that from 10500 rpm to 15000 the airflow through the engine will be the same. But from 10500 rpm and upwards the engine itselfs sustains more and more of the airflow which means that the turbo is suppling less and less boost. The flow and energy through the exhaust should stay the same. Though the exhaust gasses would travel faster, slightly decreasing the pressure which is why you dont need to wastegate it.
I accept what you are saying and will ponder it very carefully :D .

Any interesting news from Bahrain yet?

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

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321apex wrote:You are right about optionality of MGU-H usage, I missed that so thanks for setting it straight. Indeed also there is not a word in regulations that forbids the use of a wastegate. I have a feeling in that case, that some duality of boost control my take place, where MGU-H as well as a wastegate are applied. At least in such configuration, I feel the boost control will be possible in logical fashion.

In conditions of diminished boost requirement towards the top RPM, the excess exhaust will have to wastegated OUT to prevent from choking the engine on the exhaust side. Choking of any kind increases pumping losses and lowers BSFC. IMHO :D
Have you been talking to Ringo?

Letting the exhaust out the wastegate is a waste of energy. And energy is precious in the 2014 F1 cars.

The wastegate will be there as a backup safety device. The turbine will have been matched to the engine, so it shouldn't give more back pressure than intended.

321apex wrote:
wuzak wrote: In theory - not necessarily in practice.

As we have theorised before, in turbo-compound mode the peak power will be moved up the rpm range.

The reason for this that the constant fuel flow from 10,500rpm dictates a constant mass air flow from the turbo and a falling boost requirement.
Which means that with falling boost, the power would also be falling... and BSFC will be going in wrong direction, getting worse.
As Holms has said, constant fuel = constant air flow (mass air flow) requirements = less boost required.

In terms you have already used, IMEP should be the same. Friction will rise, so power will fall, but only in line with increased friction, not with decreased boost.

The exhaust energy should be, more or less, the same. Thus the turbine power will be more or less the same. But compressor power will be lower, so more power for the MGU-H.

321apex wrote:
wuzak wrote:The Cosworth graphs seem to confirm a 1000rpm+ shift for peak power to higher rpm.
This graph is treated as if it was some Gods truth, where it is completely unjustified. This is simply a theoretical study on a very greay area of unknown technology (MGU-H) and it is full of fudge factors.
Not God's truth, just a good guide as to what these engines will do.

What is so unknown about the MGU-H? It is a motor/generator. They have been around for years. As has electricity. And turbines, and compressors. And internal combustion (Otto) engines.

The difficult part of the MGU-H is not extracting energy from the turbine. It is the control needed to keep the turbo at the correct operating condition depending on rpm, boost requirements and throttle position.

You keep talking about "fudge factors". What, precisely, would they be? I'm quite sure that Cosworth have excellent engine simulation software.

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

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wuzak wrote: In terms you have already used, IMEP should be the same. Friction will rise, so power will fall, but only in line with increased friction, not with decreased boost.
How do you get constant IMEP with rising RPM and not rising air mass flow rate?
That's NOT how it works.

For example:
- at 12k, you have 100x6=600 combustion cycles per second in the V6
- at 15k, you have 125x6=750 comb. cycles/sec

If you divide the same mass flow of air/fuel for each case, then you will get 1-600/750=0.2, or 20% theoretical reduction in IMEP if we ignore tuning aspects impacting VE, which are quite real.

Which means that you would produce at least that much LESS power coupled with added frictional losses.

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rscsr
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Re: Formula One 1.6l V6 turbo engine formula

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321apex wrote:
wuzak wrote: In terms you have already used, IMEP should be the same. Friction will rise, so power will fall, but only in line with increased friction, not with decreased boost.
How do you get constant IMEP with rising RPM and not rising air mass flow rate?
That's NOT how it works.

For example:
- at 12k, you have 100x6=600 combustion cycles per second in the V6
- at 15k, you have 125x6=750 comb. cycles/sec

If you divide the same mass flow of air/fuel for each case, then you will get 1-600/750=0.2, or 20% theoretical reduction in IMEP if we ignore tuning aspects impacting VE, which are quite real.

Which means that you would produce at least that much LESS power coupled with added frictional losses.
Power should stay the same as it is effectively IMEP x rpm (before accounting friction)

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Abarth
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Joined: 25 Feb 2011, 19:47

Re: Formula One 1.6l V6 turbo engine formula

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321apex wrote:
Abarth wrote:[....]Why should a configuation which peaks at say 15'000 in N/A not be able to peak at 10'500 when overcharged?[...]
I thought you didn't want to get into a discussion with me.
You are right. Thank you for remembering me :)

Nevertheless, please feel free to answer my question....oh...with peaking I mean power, not torque btw.

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Abarth
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Re: Formula One 1.6l V6 turbo engine formula

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rscsr wrote:[...]Power should stay the same as it is effectively IMEP x rpm (before accounting friction)
Yes. And as has been pointed out, the compounded system may very well be more efficient at 12'000 than 10'500 because of the more or less constant exhaus gas flow (assuming about same Lambda) and lower boost requirement at higher revs.
This leaves more exhaust energy to generate power with MGU-H for the MGU-K in the self sustained mode.

wuzak
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Re: Formula One 1.6l V6 turbo engine formula

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321apex wrote:
wuzak wrote: In terms you have already used, IMEP should be the same. Friction will rise, so power will fall, but only in line with increased friction, not with decreased boost.
How do you get constant IMEP with rising RPM and not rising air mass flow rate?
That's NOT how it works.

For example:
- at 12k, you have 100x6=600 combustion cycles per second in the V6
- at 15k, you have 125x6=750 comb. cycles/sec

If you divide the same mass flow of air/fuel for each case, then you will get 1-600/750=0.2, or 20% theoretical reduction in IMEP if we ignore tuning aspects impacting VE, which are quite real.

Which means that you would produce at least that much LESS power coupled with added frictional losses.
Sorry, meant Indicated Horsepower (ihp)

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

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Abarth wrote:
rscsr wrote:[...]Power should stay the same as it is effectively IMEP x rpm (before accounting friction)
Yes. And as has been pointed out, the compounded system may very well be more efficient at 12'000 than 10'500 because of the more or less constant exhaus gas flow (assuming about same Lambda) and lower boost requirement at higher revs.
This leaves more exhaust energy to generate power with MGU-H for the MGU-K in the self sustained mode.
Can you please explain your point in some scientific way? Graphs, tangible theories?

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rscsr
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Re: Formula One 1.6l V6 turbo engine formula

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321apex wrote:
Abarth wrote:
rscsr wrote:[...]Power should stay the same as it is effectively IMEP x rpm (before accounting friction)
Yes. And as has been pointed out, the compounded system may very well be more efficient at 12'000 than 10'500 because of the more or less constant exhaus gas flow (assuming about same Lambda) and lower boost requirement at higher revs.
This leaves more exhaust energy to generate power with MGU-H for the MGU-K in the self sustained mode.
Can you please explain your point in some scientific way? Graphs, tangible theories?
Do you actually try to understand other posts?

How often should everything repeated. Lower Boost requirement at the same exhaust energy means more energy to recover.

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Abarth
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Re: Formula One 1.6l V6 turbo engine formula

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321apex wrote:
Abarth wrote:
rscsr wrote:[...]Power should stay the same as it is effectively IMEP x rpm (before accounting friction)
Yes. And as has been pointed out, the compounded system may very well be more efficient at 12'000 than 10'500 because of the more or less constant exhaus gas flow (assuming about same Lambda) and lower boost requirement at higher revs.
This leaves more exhaust energy to generate power with MGU-H for the MGU-K in the self sustained mode.
Can you please explain your point in some scientific way? Graphs, tangible theories?
No, I because I think that the vast majority here understand what is meant, and as you are used to learn without being fed from a plate (as per your statement), homework for you.

What is the exhaust mass flow of a gasoline engine, runnig at a fuel - air mass ratio of 1:15 when 100 kg fuel per hour are fully burnt in this very engine.
- at 5'000 rpm
- at 10'000 rpm
- at 15'000 rpm