Quote:
Originally Posted by gwyllion
Note that the more powerful hybrids appear to get slightly more energy per lap. However, if we assume that only 95% of the hybrid energy can converted into mechanical energy, this is no longer the case: - no ERS
- petrol: 148.0 * 0.41375 = 61.2 MJ
- diesel: 140.3 * 0.43634 = 61.2 MJ
- 2 MJ ERS
- petrol: 143.5 * 0.41375 + 0.95 * 2 = 61.3 MJ
- diesel: 138.4 * 0.43634 + 0.95 * 2 = 62.3 MJ
- 4 MJ ERS
- petrol: 139.0 * 0.41375 + 0.95 * 4 = 61.3 MJ
- diesel: 134.0 * 0.43634 + 0.95 * 4 = 62.3 MJ
- 6 MJ ERS
- petrol: 134.5 * 0.41375 + 0.95 * 6 = 61.3 MJ
- diesel: 129.7 * 0.43634 + 0.95 * 6 = 62.3 MJ
- 8 MJ ERS:
- petrol: 132.2 * 0.41375 + 0.95 * 8 = 62.3 MJ
- diesel: 125.4 * 0.43634 + 0.95 * 8 = 62.3 MJ
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One more thing: interestingly if you not calculate with 8 MJ, but only 7 MJ, you will end up 61,3 MJ/lap too. I tried it, because now we have 7 braking points with 500 kJ releasable energy between to braking point which is only 3,5 MJ. Toyota said, they could do twice as much as now, which is 7 MJ, but i'm not sure they could go up really to 8 MJ and use all energy given by the rules. And i'm guess 7 braking points will last for 2014 too.
And still interesting why we have k factor for 2-4-6 MJ hybrid diesels, and why we don't have it for 0 and 8 MJ case.
And also i'm a bit confused by the efficiency numbers. Efficiencies for the engines must be indicated eff. numbers. In this case, we cannot have 95% mechanical efficiency for ERS, that should be indicated eff too. But what can we call indicated efficiency for the ERS? (it's only a theoretical question, the calculations are correct of course.
And me too thanks for them, they helped me a lot!)
And the last one: are the ERSs can have 95% eff in real life? We have flybrid, supercapacitors, etc... They are not the same. Can someone tell me, which is their efficiency? 95% is correct for all of them?