2014-2020 Formula One 1.6l V6 turbo engine formula

[WhiteBlue]

Where did you get your peak fuel flow numbers from?

2013 V8 uses 150-160 kg of fuel for the race. Peak flow estimate 150 kg/h for the same race profile
2014 V6 uses 100 kg of fuel for the race. Peak flow must be 100 kg/h

Performance is supposed to be roughly similar so the proportion of average flow and peak flow is roughly the same for the same average race profile.

But lets work with what you have:
485 kJ/s / .0417kg/s = 11630 kJ/kg of fuel , that's 11630kj/kg/46000kJ/kg = 25.2% BTE

So I have used the 150 kg number but made a miscalculation somewhere along the way. For the peak flow rate of 150 kg/h we get indeed 25.2% BTE for the old V8. ==> 150 kg/h peak flow must be too high for some mistake in the assumption. WE have a pritty good estimate that the V8 had 29-30%.

But the V8 efficiency wasn't at the core of our disagreement. The difference in efficiency was. With my mistake the difference is even higher so the original point stands that the two engine types must be different ball park.

I think I found the discrepancy. We assumed that the hp figures given by Marmorini and quoted below are break horse power. This appears logical because he also quoted mechanical horse power of the MGU-K which is definitely at the brake.

If I take Marmorini's power figures I get 485 kW and 448kW respectively. Fuel flow of 27.78 g/s @ 46 kJ/g gives you 1278 kW thermal power from the fuel flow. By doing the ratio I get 35% and 38% BTE.

If you compare break power and thermal power of the peak fuel rate as we did you must get break thermal efficiency. This tells me that you still have a bug in your calculation unless of course you can demonstrate a second time that I have made a miscalculation.

[ringo]

Performance is supposed to be roughly similar so the proportion of average flow and peak flow is roughly the same for the same average race profile.

You can't say that and somehow the laws of physics bend around that command.

But lets work with what you have:
485 kJ/s / .0417kg/s = 11630 kJ/kg of fuel , that's 11630kj/kg/46000kJ/kg = 25.2% BTE

So I have used the 150 kg number but made a miscalculation somewhere along the way. For the peak flow rate of 150 kg/h we get indeed 25.2% BTE for the old V8. ==> 150 kg/h peak flow must be too high for some mistake in the assumption. WE have a pritty good estimate that the V8 had 29-30%.

Nope you need to account for that estimate, I can't believe it without some support.

But the V8 efficiency wasn't at the core of our disagreement. The difference in efficiency was. With my mistake the difference is even higher so the original point stands that the two engine types must be different ball park.

I think I found the discrepancy. We assumed that the hp figures given are break horse power and I still believe in this. If you compare break power and thermal power of the peak fuel rate as we did you must get break thermal efficiency. This tells me that you still have a bug in your calculation unless of course you can demonstrate a second time that xpansive and I have made a miscalculation.

No i have no bug in my calculation. I have two efficiencies. One where i look only at the indicated power and one where i look at the brake power at the flywheel.
the cycle efficiency is 34 to 36% and my brake thermal efficiency is in the range of 27% to 31% depending on how it's manipulated.

I don't buy 800hp from the engine, I'd accept this however:



Compounded power. Constant supply of MGUH as long as traction permits.
I'll read the marmori thing later, however i'd say he quoted a compounded power, which is probably a white lie, since the power is technically still flywheel power.
edit: having read the quote, marmori said 600 to 640hp. this is likely. no mention of 800hp like some other sources.

[WhiteBlue]

Sorry to correct you.

Next year, with an engine having somewhere between 600 and 650 horsepower and an additional 160 horsepower coming from the ERS, if you add the two it’s very similar to what you have today.

He said 600-650 ICE power with another 160 hp [for a compounded figure of 760-810 hp]. And those figures were obviously at the brake or he would not have quoted the mechanical power of the ERS.

[pgfpro]

How are you guys calculating break thermal efficiency??? It seems you are calculating it to different ways???

[wuzak]

edit: having read the quote, marmori said 600 to 640hp. this is likely. no mention of 800hp like some other sources.

That is for the ICE. That is, without the 160hp from the MGUK.

[WhiteBlue]

How are you guys calculating break thermal efficiency??? It seems you are calculating it to different ways???

Not really! It is a bit different in routing but the result is the same, at least in the computation where ringo used my figures for the V8. He is still reluctant to validate my and Marmorini's figures for the V6 though.

[ringo]

Just to be clear.

I believe in 600 to 650 Brake Horse power, aka power at flywheel with no MGUK power involved.

What i don't believe in is 800 Brake Horse power aka power at flywheel with no MGUK power involved.

I do agree with marmorini that when the electrical power is applied it will be an additional 160hp, i have agreed with this notion throughout.

Now my graph has around 560hp with 160hp added. I'm not a ferrari engineer, i can't squeeze out 650hp, i've come up 580hp with 30 degree intake air, but it is what it is. lol. close enough to 600!
But at the end of the day i'm in the ball park. And it's clear to all that 800hp from the engine alone is not possible.

even when i use 800hp,
for thermal efficiency
586kw/0.0417 = 14052.59 14052.59/46000 = 30.5%

[wuzak]

Just to be clear.

I believe in 600 to 650 Brake Horse power, aka power at flywheel with no MGUK power involved.

What i don't believe in is 800 Brake Horse power aka power at flywheel with no MGUK power involved.

I don't think anyone was talking 800hp without the 160hp from the MGUK.

Pirelli mentioned as much as 900hp - that would be 740hp without the MGUK. But I don't believe that is possible.

[pgfpro]

OK guys I need your help. I have a couple spreadsheets that I have built over the years and now after looking at all this
"break thermal efficiency" posts its got me concern.

In one of my spreadsheets when BTE is needed I use this formula that i got the info from the book
"Engineering Fundamentals of the Internal Combustion Engine" by Willard W. Pulkrabek on page 73 the formula to figure out BTE goes like this.

Using his example:
(77.3 kW) / (0.0060 kg/sec) (44000 kJ/kg) (0.97)
(77.3) / (256.08)
= 0.302 = 30.2%

77.3 = power in kW
0.0060 = mf mass flow rate of fuel
44000 = Qhv heating value of fuel
0.97 = Nc fuel conversion efficiency

so then when I use some of the numbers that you guys have brought up as a example;

even when i use 800hp,
for thermal efficiency
586kw/0.0417 = 14052.59 14052.59/46000 = 30.5%

His formula;
(586 kw) / (0.417 kg/sec) (44000 kJ/kg) (0.97)
(586 kw) / (17797.56)
= 0.0329 = 33 %

Seems like this formula is about 3 points higher.

I'm confused. Anyone lol

[ringo]

It may be related to fuel atomization?

[CBeck113]

Just to be clear.

I believe in 600 to 650 Brake Horse power, aka power at flywheel with no MGUK power involved.

What i don't believe in is 800 Brake Horse power aka power at flywheel with no MGUK power involved.

I don't think anyone was talking 800hp without the 160hp from the MGUK.

Pirelli mentioned as much as 900hp - that would be 740hp without the MGUK. But I don't believe that is possible.

Maybe they want to be on the safe side for once...

[Chuckjr]

Ringo and WB, fascinating conversation. Really enjoying the interplay.

Deleted the rest. Answered my own question. Sorry.

[xpensive]

I don't know, what ever happened to power output over (flow times specific energy) makes efficiency?

Works for me anyway, 450 kW (610 Hp) output calls for 35% efficiency at 46 MJ/kg.

[pgfpro]

OK I think I figured it out. Referring to same book he says...

"A combustion efficiency (Nc) is defined to account for the fractions of fuel that burns. Typically (Nc) has values in the range of 0.95 to 0.98 when an engine is operating properly. I noticed when I lower the Nc number it raises the efficiency rate. This didn't make sense to me. its seems if the combustion efficiency is a higher number the BTE should also be higher and not lower.

So I think the formula should be...

Using his example:
(77.3 kW) / ((0.0060 kg/sec) (44000 kJ/kg) / (0.97))
(77.3) / (272.16)
= 0.284 = 28.4%

[WhiteBlue]

I think there are two different deviations pgfpro.

1. He is using a different lower caloric value for the fuel.
2. He introduces an artificial Nc figure which further reduces the theoretical caloric value

Ad 1:
I think our figure is more appropriate for F1 because we have to consider that the fuel suppliers tweak the caloric value to the upper end. F1 fuel is reportedly different from other fuels in that regard.

Ad 2:
Ringo, xpensive and I have all compared the raw caloric energy of the fuel without an assumption for incomplete combustion with the mechanical energy or power at the brake. In my view any incomplete combustion should be included in the generated power figure and not in the caloric value of the fuel. It is an excuse for the engine maker to make him look better. There is no written law in nature that you cannot burn the fuel to 100%. There are known processes that do exactly this. So in my view your book is wrong.