2014-2020 Formula One 1.6l V6 turbo engine formula

[chip engineer]

You cannot really control back pressure. Remember it's a result of a restriction in the exhaust system. Assuming we're not using waste gates, you cannot control back pressure.

It seems to me that you might be able to tune the exhaust to minimize instantaneous back pressure at say 14,000 rpm (or whatever the max useful rpm is) where you don't need as much boost and can extract the most power from the MGU-H.
Then at 10,500 rpm you let the turbo rpm increase to produce the needed boost and to reduce back pressure (not counting being off optimum tuning).

[ringo]

True, but that seems more of an effect than a cause.

[wuzak]

You cannot really control back pressure.

I think the point is that with the MGUH you really can.

[wuzak]

There have been some excited utterances in this last couple of pages, so lets try for a reality check.

If you increase back pressure you reduce engine power but, potentially, increase MGUH power. But that will be less power than the normal engine with normal back pressure whose MGUH power delivered to the MGUK is supplemented with energy from the ES.

If you overboost the compressor, you require more power to drive the compressor. This is more likely more than the increased turbine power will give you, so you will end up with less power for the MGUH. It will also mean the engine will have to run lean, unless you use the excess air with some funky valve timing to helps with scavenging the exhaust. This is what the Rolls-Royce Crecy did, but that was a 2 stroke and the induction system was entirely dependent on the supercharger.

I don't believe that there will be 100kW + power recovered from the MGUH. More likely about half that. This will enable the ES energy to be spread out for a longer period over a lap. ie giving the MGUK's maximum 120kW for a much higher proportion of the lap.

Also, Ringo, I don't believe wastegates will be used in normal operation of these engines. Wasting any energy would be competitive death in the 2014 engine formula.

[WhiteBlue]

I don't believe that there will be 100kW + power recovered from the MGUH. More likely about half that. This will enable the ES energy to be spread out for a longer period over a lap. ie giving the MGUK's maximum 120kW for a much higher proportion of the lap.

That is a very sensible idea. If I look at a qualifying lap I have 4 MJ from the ES to spread out. But I may only need 6 MJ to cover a 65% WOT operation of the lap. So effectively my MGU-H need only to supply 40 kW brake power to fill the budget. That means at 90% efficiency of the conversion I need only 44.5 kW mechanical energy from the MGU-H. Compared to my nominal ICE power of 485 kW that is less than 10%. If we think that 7% compounding is possible with 1950 technology why not 9.4% in 2014. That would mean I need a BTE of 41.6% for the over all compounded process. Or in other words the the compounding needs to increase my BTE by 3.6%. It sounds much more doable.

The caveat is that I get full performance with 605 kW at he crank only in qualifying over one lap, on relatively medium power circuits with medium lap length. For a Monza type lap the ES stored energy would not cover my budget.

[wuzak]

Some basic calcs.

Energy required to run MGUK at 100% for a percentage of a lap:
Varies from 3.6MJ for a 60s lap and 50% full power to 10.8MJ for a 120s lap and 75% of lap at full MGUH power.
4.62MJ for 70s lap and 55% of lap at full MGUH power (eg Monaco)
6.24MJ for 80s lap and 65% of lap at full MGUH power
7.8MJ for 100s lap and 65% of lap at full MGUH power (eg Spa)

This equates to MGUH power (assuming 2MJ taken from ES per lap)
37.4kW for 70s lap and 55% of lap at full MGUH power (eg Monaco)
53.0kW for 80s lap and 65% of lap at full MGUH power
58.0kW for 100s lap and 65% of lap at full MGUH power (eg Spa)

The variation, as above, is from 26.7kW to 73.3kW.

If you have 4MJ stored, you can't use MGUH power for a 60s lap below 55% of lap at maximum power of MGUK. A 120s lap with 75% would required 56.7kW generated from the MGUH.

[dren]

If using the MGUH load for the backpressure increase past 10.5, you could get an increase in recovery to the redline. This would put the MGUH load at some percentage at 10.5k rpm, increasing to 100% at around redline. The powerband would increase even without added fuel, and you would gain with improvements in bsfc. The energy storage in the battery could be used in the lower revs.

So maybe 7% recovery at 10.5k rpm, then around as high as 20% at 15k rpm. 695hp to 780hp

I say this because I don't know how else to vary the backpressure other than loading the MGUH.

Not happening.

You cannot really control back pressure. Remember it's a result of a restriction in the exhaust system. Assuming we're not using waste gates, you cannot control back pressure. To intention increase it is also bad for the engine, it will reduce the crankshaft power, as the pumping losses will increase on the exhaust stroke, this increase friction power hence reducing brake power. It shouldn't be the goal to increase back pressure in the manifold. It should be the goal to increase boost pressure and reduce back pressure at turbine exit.
Remember you are doing this to increase pressure ratio of the turbine. That's what this whole back pressure thing is about. Pressure ratio is simply input pressure divided by exhaust pressure. In the Wright TC they were ultimately looking at that aspect.

Now that thing they call a waste gate.. nah! useless piece of equipment.

Maybe I was confused, but I thought tha'ts what TC was getting at above, basically robbing power from the crank and transfering it to the MGUH at a roughly 1:1 proportion since it aided in increased efficiency in overall total power output, which would show as a power gain over a steady fuel rate. The backpressure in the manifold decreases your blowdown losses over the exhaust valves. Maybe I'm confused?

[WhiteBlue]

Some basic calcs.

Energy required to run MGUK at 100% for a percentage of a lap:
Varies from 3.6MJ for a 60s lap and 50% full power to 10.8MJ for a 120s lap and 75% of lap at full MGUH power.
4.62MJ for 70s lap and 55% of lap at full MGUH power (eg Monaco)
6.24MJ for 80s lap and 65% of lap at full MGUH power
7.8MJ for 100s lap and 65% of lap at full MGUH power (eg Spa)

This equates to MGUH power (assuming 2MJ taken from ES per lap)
37.4kW for 70s lap and 55% of lap at full MGUH power (eg Monaco)
53.0kW for 80s lap and 65% of lap at full MGUH power
58.0kW for 100s lap and 65% of lap at full MGUH power (eg Spa)

The variation, as above, is from 26.7kW to 73.3kW.

If you have 4MJ stored, you can't use MGUH power for a 60s lap below 55% of lap at maximum power of MGUK. A 120s lap with 75% would required 56.7kW generated from the MGUH.

The middle case basically confirms for race trim what I have already calculated for qualifying. Approximately 6MJ MGU-K energy budget for full 120 kW and 65% WOT. Depending on race or qualy trim you can take either 2 or 4 MJ out of the ES. It is anybodies guess how much power the MGU-H will have @ 650 bhp for the ICE. I wanted to look at a possible worst case scenario with just 60 bhp or 44.5 kW MGU-H power which is an additional 9.2% of the 650 bhp of the ICE. So the total compounded power would be 530 kW or 710 bhp. But with the augementation of the ES in quali mode we would be getting 605 kW or 811 bhp. Is there anybody out there who thinks we will get 650 bhp ICE and 60 bhp from the MGU-H?

[dren]

I just did some quick calculations for brake recovery energy at the Nurburgring based on a total car weight of 785 kg. I looked at only the 6 major braking zones since this was easy info to gather speeds. I came up with a total of 11.4 MJ.

I don't know what kind of brake balance front to rear cars typically run on average, so I'll throw a guess at 60/40. So that lowers the recovery to about 4.6 MJ. I expect there to always be some percentage of split between brake and MGUK load from 100% brakes to maybe a guess of 50/50. This would allow around 2.3 MJ of harvesting max, account for losses, and we get close to the 2 MJ limit. Of course there are a lot of assumptions here, but it shows that just from braking, it will be possible to aquire the full ES limit.

So we should expect very little to no MGUH energy transfer to ES. It will all go straight to the MGUK or take from the ES unless when being used as a load in place of a waste gate.

[dren]

WB, after reading several pages from several posts, that seems to be one reasonable outcome. The sizing of the MGUH, electronics, cooling and batteries will likely play a major role in how much the MGUH will recover. More from the MGUH means smaller ES.

[xpensive]

I just did some quick calculations for brake recovery energy at the Nurburgring based on a total car weight of 785 kg. I looked at only the 6 major braking zones since this was easy info to gather speeds. I came up with a total of 11.4 MJ.

I don't know what kind of brake balance front to rear cars typically run on average, so I'll throw a guess at 60/40. So that lowers the recovery to about 4.6 MJ. I expect there to always be some percentage of split between brake and MGUK load from 100% brakes to maybe a guess of 50/50. This would allow around 2.3 MJ of harvesting max, account for losses, and we get close to the 2 MJ limit. Of course there are a lot of assumptions here, but it shows that just from braking, it will be possible to aquire the full ES limit.

So we should expect very little to no MGUH energy transfer to ES. It will all go straight to the MGUK or take from the ES unless when being used as a load in place of a waste gate.

I believe there's an easier way to calculate available KERS-energy, as MGU-K power (120 kW) times total braking time?

120 kW times what, 20 seconds, that's 2400 kWs, or 2.4 MJ?

[dren]

Yup, far easier, or just take the max allowable 120kw and 2MJ per lap and back figure how much 100% MGUK load time is needed, 16.6 sec. I was curious about the brake bias percentages and how much max when charging the MGUK will drag.

The Mercedes FRIC system may allow them a little quicker recovery at the back if the car doesn't pitch as much as others.


I just looked at James Allen's website, says about 18% of the lap at the Nurburgring is spent braking. this leads to around 16 to 17 seconds total braking if lap times are in the 1:34 range. The calculations I was getting using an average of 20 m/s^2 were around 13 seconds total braking. I left off two small turns, so 16 seconds sounds about right. So we are almost right at total braking time needed, but that's assuming all braking energy is going directly into the ES, which it isn't. May need at least 2 laps to fully supply the ES from the MGUK, or supplement with the MGUH, which we were figuring would be almost solely supplying the MGUK under acceleration. I would think the accumulation through the waste gate action of the MGUH would be negligible here.

[WhiteBlue]

I'll make an estimate then for what I expect in terms of power and efficiency in 2014 to be reached. This estimate is based on the upper Marmorini ICE (650 bhp) figure and what I think would be a reasonable minimum compounded power in order to meet the claim that total power will be very similar in 2014 to 2013.

P0 = ICE base brake power at best rpm = 650 bhp = 485 kW
fuel flow f=m/t= 100 kg/h = 27.78 g/s
e = specific fuel energy = 46 MJ/kg
E = Thermal power of fuel flow = f*e = 1278 kW
Turbine power to MUG-H = 9.2% P0 = 60 bhp = 44.5 kW
P1 = compounded brake power = 650 (ICE)+ 54 bhp (MGU-K) = 704 bhp = 485 + 40 kW = 525 kW
note: MGU-H -> MGU-K -> crank reduces 44.5 kW @ turbine to 40 kW brake power

====>
Brake specific fuel consumption
BSFC0 = f/P0 = 57.3 g/kJ
BSFC1 = f/P1 = 52.9 g/kJ

======>
Brake thermal efficiency
BTE0 = P0/E = 38 %
BTE1 = P1/E = 41%

========>
Electric compounding:
MGU-K energy budget without ES (80s lap @ 62.5% of lap = 50s WOT) = 2 MJ
Net power at MGU-K from MGU-H = 40 kW

Race:
MGU-K energy budget Race (80s lap @ 62.5% of lap = 50s WOT) = 4 MJ
Race power Pr= 650 + 107 bhp = 757 bhp = 485 + 80 kW = 565 kW

Qualifying:
MGU-K energy budget Quali (80s lap @ 62.5% of lap = 50s WOT) = 6 MJ
Qualifying power Pq = 650 + 161 bhP = 811 bhp = 485 + 120 kW = 605 kW

For comparison a V8 with 750 bhp and 0.4 MJ ES /lap
KERS energy budget Race/Quali (80s lap @ 62.5% of lap = 50s WOT) = 0.4 MJ
Power in 2014 dual torque mode: 750 + 11 bhp = 761 bhp = 559 +8 kW = 567 kW

If the V8 is run in dual torque mode and not in push to pass mode it has very nearly the same power as the 2014 power train in race trim. In qualifying the 2014 power train has 7% more power according to this estimation. Some assumptions were made:

• 2 MJ ES loading is accomplished on this track by MGU-K harvesting
• No spool up needed as ICE is always running above critical rpm

And for comparison the Honda RA168E engine in race trim:
Pr = 611 bhp = 455.6 kW
Fuel was 84% Toluene + 16% n-Heptane
e(TH-fuel) = 17.669 BTU/lb = 41.1 MJ/kg
BTE = 30.6%
f (RA168E) = 36.22 g/s
BSFC (RA168E) = f (RA168E)/Pr = 79.5 g/kJ
Thermal power 555.6 kW/0.306 = 1489 kW
Hypothetical petrol fuel flow = 1489 kW/ 46 MJ/kg = 32.37 g/s (2% rich)
BSFC (RA168E@petrol) = f (RA168E@petrol)/ Pr = 71.0 g/kJ

====> If the RA168E was run on F1 petrol at the same efficiency it was run with Heptane/Toluene it would consume 16.5% more fuel than a 2014 F1 V6

[Dragonfly]

You guys rock in this thread. I try to follow with my limited knowledge in ICE theory and English terminology and abbreviations. I think it would be fun to see who was closer to reality when the new power units hit the track.

[trinidefender]

I'm hearing 2 numbers being thrown around that should probably be revised. The numbers for the turbo compounding. 7% power recovery is given for the low end of the scale and 22% is given for the high. However the blowdown turbines they were using were much less efficient than the ones they would use now. So my thinking goes that those numbers should be revised slightly.

Unless I have missed something crucial here of course