Horsepower of the engines.

[pgfpro]

Thanks Wuzak and I agree with your answer.

General Note.
Everyone here seems to use "ICE" to refer to power from the crankshaft and I am happy to go along with that. In reality the term "Internal Combustion Engine" is any engine which uses air as both the oxygen source for combustion and the working fluid. ICE therefore refers to piston engines, Wankels, gas turbines, etc and compunded combinations of these. By that definition the ICE power would include crankshaft power AND any surplus power available from the turbine.

It is confusing trying to understand what each person is referring to when it comes to power output?
So for now on I will say as an example "the ICE is 800HP total with 50HP turbine surplus.

[ringo]

I have done two things:

take the ICE efficiency as the crank output and the turbine's shaft work to the MGUH against the heat input across the ICE and the Turbine. Which is debatable, but i found it does make some sense.

An alternative approach i took was to say efficiency is really the useful energy over the energy put in in a general sense; which should include the electrical power indiscriminately.
Since it's electrical and thermal they need to be in an equivalent form to include both...

So we can turn the electric power output to an equivalent specific power:

120kW from MGUK divided by the actual air and fuel mass flow in kg/s to get an equivalent kJ/kg. Now that it's in the same form as the enthalpy from the engine add it to the specific power (enthalpy) produced by the ICE; also in kJ/kg. This covers all power produced.

All left for the efficiency now is we have the fuel's LHV (kJ/kg) and the air to fuel ratio and mechanical efficiency to factor in:

useful energy produced / energy put in...

kJ/kg from PU / KJ/kg available from mass of fuel in the air and fuel mix X mechanical efficiency and we get the total PU efficiency.

Now these are two methods, i'm not sure which is more correct.

[ringo]

Thanks Wuzak and I agree with your answer.

General Note.
Everyone here seems to use "ICE" to refer to power from the crankshaft and I am happy to go along with that. In reality the term "Internal Combustion Engine" is any engine which uses air as both the oxygen source for combustion and the working fluid. ICE therefore refers to piston engines, Wankels, gas turbines, etc and compunded combinations of these. By that definition the ICE power would include crankshaft power AND any surplus power available from the turbine.

It is confusing trying to understand what each person is referring to when it comes to power output?
So for now on I will say as an example "the ICE is 800HP total with 50HP turbine surplus.

So you are saying 800hp at the crank separate and apart from 50hp self sustaining?

[gruntguru]

Now these are two methods, i'm not sure which is more correct.

Efficiency = (useful stuff out)/(input cost)
For a heat engine this becomes W/Q
where W = useful work out (mechanical plus electrical) and Q = heat in (from fuel).
For a heat engine you would consider a system (imaginary box) which includes the entire engine and turbomachinery and measure the flow of fuel into and work out-of that system.
The total energy in the system must not change (no storage or retrieval) so you would not include the ES or the MGUK in the system.
It doesnt matter where the work is going - if the electrical output happens to be more than the MGUK can handle (120 kW), it is still useful work being produced by the heat engine.

[pgfpro]

Thanks Wuzak and I agree with your answer.

General Note.
Everyone here seems to use "ICE" to refer to power from the crankshaft and I am happy to go along with that. In reality the term "Internal Combustion Engine" is any engine which uses air as both the oxygen source for combustion and the working fluid. ICE therefore refers to piston engines, Wankels, gas turbines, etc and compunded combinations of these. By that definition the ICE power would include crankshaft power AND any surplus power available from the turbine.

It is confusing trying to understand what each person is referring to when it comes to power output?
So for now on I will say as an example "the ICE is 800HP total with 50HP turbine surplus.

So you are saying 800hp at the crank separate and apart from 50hp self sustaining?

Correct.

[ringo]

It is confusing trying to understand what each person is referring to when it comes to power output?
So for now on I will say as an example "the ICE is 800HP total with 50HP turbine surplus.

So you are saying 800hp at the crank separate and apart from 50hp self sustaining?

Correct.[/quote]

Ok i follow you. I agree. However i think based on the MGUH technology more than 50hp have be harvested.

[ringo]

Now these are two methods, i'm not sure which is more correct.

Efficiency = (useful stuff out)/(input cost)
For a heat engine this becomes W/Q
where W = useful work out (mechanical plus electrical) and Q = heat in (from fuel).
For a heat engine you would consider a system (imaginary box) which includes the entire engine and turbomachinery and measure the flow of fuel into and work out-of that system.
The total energy in the system must not change (no storage or retrieval) so you would not include the ES or the MGUK in the system.
It doesnt matter where the work is going - if the electrical output happens to be more than the MGUK can handle (120 kW), it is still useful work being produced by the heat engine.

What i have done is include the MGUK power that is fed directly from the MGUH. Ideal case this is 120kW over the generating efficiency. That would be what is taken off the turbine shaft. So yes it can be looked at in that way.

The electrical output would be more than the MGUK, the output will be what the system is design for and the excess energy will simply further accelerate the turbine and go out the tailpipe.

[OO7]

How much of this power is estimated to only be available for 1 or 2 qualification laps?

[wuzak]

Now these are two methods, i'm not sure which is more correct.

Efficiency = (useful stuff out)/(input cost)
For a heat engine this becomes W/Q
where W = useful work out (mechanical plus electrical) and Q = heat in (from fuel).
For a heat engine you would consider a system (imaginary box) which includes the entire engine and turbomachinery and measure the flow of fuel into and work out-of that system.
The total energy in the system must not change (no storage or retrieval) so you would not include the ES or the MGUK in the system.
It doesnt matter where the work is going - if the electrical output happens to be more than the MGUK can handle (120 kW), it is still useful work being produced by the heat engine.

What i have done is include the MGUK power that is fed directly from the MGUH. Ideal case this is 120kW over the generating efficiency. That would be what is taken off the turbine shaft. So yes it can be looked at in that way.

The electrical output would be more than the MGUK, the output will be what the system is design for and the excess energy will simply further accelerate the turbine and go out the tailpipe.

If the MGUH could develop more than 120kW the excess would be used to charge the battery (allowed to deploy 4MJ from battery, but can only recover 2MJ through MGUK and most circuits struggle to see that without lift and coast). All the teams have their warning light flashing towards the ends of straights, which suggests they switch the MGUH supply from the MGUK to the ES at that point.

[mrluke]

How much of this power is estimated to only be available for 1 or 2 qualification laps?

I think we are discussing the "self sustaining" mode i.e. power that is available for the whole race. Max power for qualy will be the full MGUK output plus maybe a little bit more from the ICE by taking a bit of life out of it.

[Juzh]

Max power for qualy will be the full MGUK output plus maybe a little bit more from the ICE by taking a bit of life out of it.

Plus reduced back pressure from open wastegates and running mgu-h off the energy store. Also only available in qually, and maybe sometimes during the race (rarely). I remember Bottas being instructed to use "mode 1" in russia for final 2 laps before the accident. But that does count as an ICE output in the end.

[mrluke]

http://www.jamesallenonf1.com/2016/02/h ... gine-base/

Interesting quote:

The headline, Cowell says, is that the V6 hybrid turbo is now the most powerful F1 engine ever – even greater than the 2005 V10s that revved to 20,000rpm.
Meanwhile the technology has advanced the efficiency of engines to 50 per cent, meaning that 50 per cent of the potential power than can be derived from a unit of petrol is being converted.

I think that exceeds most of the forums expectations. I know many members struggle to believe more than 40% is even possible.

[wuzak]

50% thermal efficiency would mean 830hp....

And if that is without the MGUK, the total would be 990hp!!!!

[OO7]

50% thermal efficiency would mean 830hp....

And if that is without the MGUK, the total would be 990hp!!!!

Now just imagine if the F.I.A allowed the development of rotary valves!

[Blackout]

So if the FIA listens to the fans who want 1000hp engines, thay can simply increase fuel flow to 110kg/h so the PU can make around 820hp and increase the ERS power (that's what Magneti Marelli wants for example) so the Kers can deploy 180hp instead of 160
Or 'just' boost the ERS to 250hp