basti313 wrote:godlameroso wrote:basti313 wrote:
Water is not compressible...not sure how you can get the value. I would rather see it for a compressible fluid like fuel, I think your order of magnitude is ok. You can just roughly calculate it with the volume difference, which is not big in compresses liquids in comparison to gases.
It's still tricky because increasing the pressure also inhibits the ability to vaporize fuel, because as we all know, boiling points tend to rise with pressure.
Well, vaporizing the fuel is not really a problem. The compressed air is always hot enough - the air needs to be cooled by the fuel for more performance... The question is when it exactly the fuel vaporizes and when it starts burning. In modern F1 this is not only a fuel supplier problem, but when we look at the knocking, all comes together:
- fuel
- injection strategy
- compaction ratio
- air temperature
Especially the last point is hard to change...if your intercooler is too week you may be loosing performance, just because you can not make your fuel and injection strategy work.
In response to various points above.
1. Everything is compressible - including solids and liquids. The heat of compression comes from the work required which is the integral of P.dV For liquids and solids, dV (the change in volume) is very small so the work done is also very small (eg Compressors need a lot more energy than pumps. Air gets very hot when compressed in a compressor or a piston engine compression stroke).
2. The value can calculated knowing the compressibility u of the liquid 0.46E-6 (kPaE-1) for water and 1.2E-6 for n-octane
So the work is W=(P2+P1)/2 x (V2-V1) (trapezoidal area under a straight line on a P-V chart)
= (P2+P1)/2 x uV(P2-P1)
= P2/2 x uVP2 (P1<<P2)
= (uVP2^2)/2
= (0.46E-6 x 0.001 x (50E3)^2)/2 = 0.575 KJ/kg (water) and 1.5 KJ/kg (octane)
Specific heat for water is 4.2 kJ/(kg.K) so each kilogram of water will increase temp by 0.575/4.2 =
0.13*C
Specific heat for octane is 2.1 kJ/(kg.K) so each kilogram will increase temp by 1.5/2.1 =
0.71*C
This the temperature rise for a 100% efficient pump, however pumps tend to have high efficiency (above 90%) so the actual heating will be no more than 10% higher.
I will fix the error in my earlier post.
3. Vaporisation occurs after the injector nozzle when the pressure is much lower than 500 bar. Vaporisation before the nozzle is extremely undesirable.
4. Intercooling is not required to increase air density (higher boost will create all the airflow needed for the limited fuel). Intercooling has a negative effect on energy recovery and on fuel vaporisation. (Honda found it desirable to reduce the intercooling to improve efficiency on the RA168e). Intercooling is needed mainly to control knock and limit combustion chamber thermal stress.