This is what a colleague at QMUL said: >Remi, >in compressed in air engines regenerative braking is usually considered to >be best achieved by compressing air into a seperate (small) tank. Obviously >this also causes some heating - but the main advantage is that it allows >the extra energy to be used when required - ie when accelerating (much >harder with heat energy). Also - I think it would be difficult to utilise >heat from braking to keep the engine at a stable temperature, as this would >mean that when the driver is not braking sufficiently to meet the design >point, the engine would freeze up. >Thanks though - always on the lookout for new ideas. >H.
I think it is so simple that he is slightly missing the point too. He is talking about a continuous run where the engine and tank have reached a stable temperature (covered in ice and below ambient). N2(l) engines (what he does) use a 'fuel' injector approach into the cylinder and mix it with air. I'm not sure they dehumidify or just spew out fine ice in the exhaust. I'm not sure what toll fine ice grit takes on the engine. I'll ask him. Anyway we are talking compressed air engines. I don't think he is really seeing the suggestion about application in stop-start situations and that really most of the k.e. (>99% ?!! really?!!!) could go back to the tank. I think the suggestion is so simple it is defeating me too. It was conceived late in the day when I was muzzy. I've been a little sleepless thinking through the night. It's now morning and I'm even more muzzy and it just doesn't seem right but then again it does. Once again: Forget the losses from the expansion and the venting and running the motor. Say 80% of the PE in the tank gets to the wheel. Say because we are going so slowly that wind/rolling resistance is not a problem. In stop-start we could get back most of that 80% even it is seen as a "useless dissipative process" and "just waste heat". The temptation would have to been put a heat engine on the waste heat and drive a compressor. No need, a 'useless' dissipative process actually in the limit lets you recover 'all' the KE. The tank looses internal energy U - mc(T2-T1) part of this coupled back so U - mc(T2-T1)+ k* mc(T2-T1). It discharges slower. It can't be this simple. Something has got to be wrong. It seems paradoxical and one has to argue both sides to find the 'why'. -----Original Message----- From: Remi Cornwall [mailto:[EMAIL PROTECTED] Sent: 29 October 2008 02:09 To: vortex-l@eskimo.com Subject: RE: [Vo]:In a nutshell the problem is this. Tata Motors - full of compressed air! In a nutshell the problem is this: 1) We want to preserve the charge on the tank in stop-start situations 2) (Obviously the available work eventually venting to the atmosphere will be less than the potential energy stored in the tank. This is not the question for those not sharp enough to understand what is going on. You can't turn 100% of the pressurized energy into work) 3) When driving a certain amount will be lost to wind, rolling resistance etc. Since this is low speed it ain't much. 4) It is fair to say that in short stop-start conditions most of the kinetic energy at the wheels ends up in the brakes as heat. 5) There is no law preventing us from conducting most of that heat back to the tank but the difference in temperatures and ratios of the heat capacities of the brakes to the tank. On engineering terms 'all' the heat could go back to the tank if the brakes had low heat capacity and were very hot. We could elect to send the heat back with a heat engine and compressor but heat flow does this more gracefully. Not all dissipative heat flow is a dead end for engineers. 6) Imagine a continuous stop-start cycle where we can compute the average energy out from the tank and the average energy back into it. It would seem that if the outflow from the tank is O and the inflow is I, then the new outflow is O - I. 7) Naively 6) should preserve the life of the tank charge: The tank temperature is linear in I, E = mcT And the work from the tank is f(T) and nRTln P1/P2 for an isothermal process at least. So linear in T too. If the whole plant function was very sensitive to the tank temperature (higher order terms in T) then the process would be worthwhile. 8) In the steady state the assumption at 6) is probably correct because of (7) 9) Realistically stop-start cycles won't be 'regular'. The temperature will not reach equilibrium and there will be very little change in the tank temperature at next power demand despite our feedback. In short the thing would be sluggish. It probably would be better to have a SMALL high pressure reserve tank to capture the braking energy that then rapidly give it out on the next power demand. 2pm and that probably is the answer without detailed work. May look at it again in a few days.
