stevesingo wrote: ↑04 Jan 2018, 15:02
Question
1, Will the increase in MGU-H rpm also increase the amount of work done by the compressor?
2, Will energy inputted from MGU-K be equal to increased compressor work + inertia?
3, Will the added compressor load assist the reduction of rpm of the MGU-H during recovery of stored inertial energy.
4, Electrical losses aside what will the ratio of % energy in to % energy out?
My opinion, each question in turn:
1. We're talking about varying the compressor speed by <1% either side of what you'd otherwise have for a fraction of a second each time, so the additional effect of this is negligible on the compressor or turbine.
2. Yes, although as stated in (A1), the effect is negligible.
3. The compressor load, averaged out over even a short amount of time like 0.1s, is the same as without the flywheel system. We don't vary the compressor speed by more than 1% either side of optimal, and the compressor spends most of its time in the middle half of that 1% range, either side of the optimal amount. The effect in terms of "additional pumping losses" will therefore be quite small. I suppose it is possible that the compressor loses some efficiency through "chop", if poorly designed, but if that were the case, the compressor would also be inefficient even if we weren't using it as a flywheel.
4. Good question.
For context, I think K direct to ES and back to K is probably in the region of 90-95% roundtrip efficiency.
I further think that the flywheel method won't be as efficient, but perhaps in the region of 80-85% at a guess?
Edited to add: the first 4MJ of energy deployed via the K will be via the direct route of ES->K which is likely >95% efficient.
How you calculate the roundtrip efficiency of energy which nominally went the K->H->ES->K route is quite a complex topic, even if you ignore that it is pulsed in one direction, constant in another, and there's compounding in there too.