gruntguru wrote:If MAP = BP the turbo machinery is a simple Brayton cycle - the combustor being replaced by the piston engine as its heat source.
In the F1 case, the heat input to this Brayton cycle (Gas Turbine) is relatively fixed wrt PR. The surplus work (Wt - Wc) however, is dependent on PR. The theory behind this is well established.
There are useful calculators here
https://www.engineering-4e.com/calc4.htm for compressor and turbine power (you can use the isentropic compression calculator for expansion as well or use the expansion calculator further down the page). The calculators are for 100% isentropic efficiency so the actual turbine work will be about 0.8 times the output from the calculator and the compressor work will be equal to the calculator output divided by 0.8.
Have fun playing. Don't forget to adjust the massflow and turbine inlet temp when you change the PR (boost).
SIMPLE EXAMPLE.
Compressor Massflow = 0.55 kg/s
Turbine Massflow = 0.578 (AFR = 19.6:1)
PR = 3.3
T comp inlet = 298 K
T turbine inlet = 1000 K
Calculator gives Wcomp = 66.8 kW and Wturb = 161.9 kW (Need to use 57.8 kg/s to get 3 significant figures into calculator)
At 80% eff for comp and turb Pcomp = 66.8 x 1/0.8 = 83.5 kW and Pturb = 161.9 x 0.8 = 129.5 kW
Pnet = Pturb - Pcomp = 129.5 - 83.5 =
46 kW surplus.
Now try it at PR = 2.5 and compressor massflow 0.5 x 2.5/3.3 = 0.378 (assuming intercooling to ambient)
Turbine massflow is 0.406 and AFR = 14.5:1
T turbine inlet will be higher - say 1220 K
Calculator gives Wcomp = 33.8 kW and Wturb = 114.5 kW
At 80% eff for comp and turb Pcomp = 33.8 x 1/0.8 = 42.3 kW and Pturb = 114.5 x 0.8 = 91.6 kW
Pnet = Pturb - Pcomp = 91.6 - 42.3 =
45.4 kW surplus.
So interestingly, at 80% isentropic efficiency for the turbine and compressor, changing the PR has little effect on surplus energy to the MGUH. At lower efficiencies lower PR will be favoured and at higher efficiencies, higher PR will produce more surplus energy.
@gg
these calculations do not seem to me supportive of the view that there is any recovery benefit in lean mixtures (driving higher PR)
(and your PR = 2.5 calc has a mistake in the massflow, stoichiometry is actually reached around PR = 2.25)
they suggest that recovery is not higher with a lean mixture ie mixture strength (AFR) is unimportant
this at least is useful eg allowing us to hold PR constant over 10500
if the engines don't at 10500 use much leaning, the downsides (bigger machinery and charge cooler) may be less
the charge cooling need not be great, higher charge temperature may help lean combustion as such (aviation often avoided cooling)
but (with leaning) the bigger charge will still need more cooling (unless you have convinced people that no cooling is needed)
also the machinery will be bigger and less responsive
anyway, my interest is mainly whether or not using extra air materially increases recovery
a PR of 2.5 corresponds to 10% lean at 10500 assuming perfect breathing
Gilles Simon wrote that valve sizes may be compromised to allow the highest CR (smallest chamber) ....
this would depress breathing/VE and need a MEP raised above the usual
having at 10500 a tuned-length exhaust -P occuring at tdc (and not at 12500) helps to make the high -dP workable
so keeping MAP constant to 12500 with maybe little further leaning (as Abarth says, the VE will fall)
F1 could be running 3 bar for some leanness essentially to allow greater CR and so help ICE efficiency
(for a given charge temperature, combustion will produce a lower temperature, this also helps the turbine)
ie not because they think it increases recovered power
anyway at these exhaust pressures the recovery will be quite large
but to the extent that VE is depressed leanness cannot be judged from MAP
is 3 bar 'boost' does not mean the AFR is 30% lean, it could be far less lean than that
also, valve seats can have generous width, so over-scavenge will be unneccessary and some underscavenge will be acceptable
