rburt07 — Your new data correspond (according to my calculations) to "heretical" mode operation with MG1 acting as a motor at a speed of -1492 rpm. Your fuel consumption of 0.7 galUS/h = 2.65 L/h corresponds to FE of 3.3 L/100 km = 71 mpgUS. This is in much better agreement with your SG measured result of 67 mpgUS. Perhaps the non-use of air conditioning (whose electricity usage has to be made up by consuming fuel) made the difference. Or else, last time you may have been burning some fuel to top up the NiMH battery or 12-V battery. My calculations assumed that all electricity generated was being shuttled as electrical power between the MGs — it's close to the steady-state situation, and the only case that I can analyze. I'm also assuming 100% efficiency in the MGs — reality seems to be around the 95% mark for each, from both my experience and from what I've read.
The attached excerpts might help explain how heretical mode can improve engine efficiency by (in most cases) forcing down the ICE's speed while increasing its load, to produce the required power at the greatest possible efficiency. They are taken from Toyota's US patent #6 131 680, posted earlier by me in the thread "
Heretical Mode." Fig.12 shows as dotted lines C1-C3 the constant-power curves of an ICE. They are hyperbolas since (ICE torque
Te) x (ICE rpm
Ne) = (ICE power
Pe). At any required engine power, one can in principle run the engine at any point along the dotted curve corresponding to the needed ICE power (e.g., for cruising at the given speed). The closed curves labelled with engine efficiency numbers
a% represent combinations of engine torque and rpm at which it has the same thermodynamic efficiency. The ICE's highest possible efficiency occurs at point A1 on power curve C1. If that's the power you need, then the ICE should be operated at the torque/speed combination represented by point A1. Usually, you need less power than this, say a power represented by curve C3. Then point A3 is the point on C3 corresponding to the greatest efficiency possible at that power. Curve A is the engine's "maximum efficiency operating curve," and is stored in a lookup table in the ECU. For each power requirement, the ECU determines where best to operate the engine. Curve B is the boundary of the operable range of the engine. Fig.13 plots how the efficiency varies with engine rpm as one moves along each of the curves C1, C2, and C3. The peak efficiencies occur at points A1, A2, and A3 respectively. This is explained in more detail in column 25, lines 26-58 in the patent.
The point is that, at power requirements less than C1, it's most efficient to force the engine speed
Ne down and its torque
Te up. This is what is happening in heretical mode. End of lesson!
alan_in_tempe — What you're saying is not really in conflict with what I'm saying, although I do disagree with some of your specific statements. I am relying on basic mechanics to draw physical conclusions about energy flow. They are independent of the precise electrical mechanisms used to drive the MGs in order to achieve this end. You are right about how the MGs must be driven, but this is immaterial to my argument. What I'm saying
must be true if physics is to remain valid here. Let me elaborate.
First, there must be a balance of torques at each shaft 'S', 'C', and 'R' in the nomograms; for example, the torque exerted by the engine on MG1 must (by Newton's third law) be equal and opposite to the torque exerted by MG1 on the engine. Since the planetary-gear's "torque-split" is always 72% to the ring-gear (MG2) and 28% to the sun-gear (MG1), the ICE's torque
Te always splits this way. This split is shown by vectors
Tes and
Ter in my diagrams. The torque
Tm1 always opposes
Tes to provide balance at shaft 'S'. The torque
Tm2 can either reinforce or oppose
Ter depending on the operating mode. And the wheel torque
Tr combines with
Ter and
Tm2 to provide balance at shaft 'R'. It is not necessary to consider shaft 'C' any further in this analysis, but if you like, you can add the needed vectors to see that there's balance there too.
Let me address a couple of your specific claims:
- No, both MG1's and the ICE's torques cannot be negative at the same time, for then balance wouldn't hold at shaft 'S'.
- If, for example, the product (torque Tm1) x (rpm Ns) is positive, then power is flowing into MG1, and so it is acting as a motor. If Tm1 x Ns is negative, then power is flowing out of MG1, and so it is acting as a generator. This is true irrespective of any considerations about MG1's drive voltage. Yes, the voltage must be chosen correctly in order to achieve the desired torque Tm1, and this is the function of the ECU, but the voltage per se is irrelevant to my physical argument.
- The MGs are rated at up to 650 Vac, and MG2 is rated at 105 kW, although the combined (engine + battery) power sent to it can't reach this figure.
Stan
Re: Which Mode was I in
Re: Which Mode was I in
Re: Which Mode was I in
