Honda Power Unit Hardware & Software

[bill shoe]

@ringo
imo
the elasticity of the load path gives isolation from 20-40 Hz ripple in PU torque

Do you mean elasticity between the MGU-K output and the ICE (crankshaft) input? If so I think there would not naturally be this kind of elasticity there, so there would have to be a vibration absorber like the springs in a clutch.

[PlatinumZealot]

@ringo
imo
the elasticity of the load path gives isolation from 20-40 Hz ripple in PU torque

Do you mean elasticity between the MGU-K output and the ICE (crankshaft) input? If so I think there would not naturally be this kind of elasticity there, so there would have to be a vibration absorber like the springs in a clutch.

This has been discussed before too bad the thread is so long. I would dig it up for you guys.

The translation of the japanese article pretty much explains it. When in extra harvest mode the MGUH is alternately charging and discharging in 20Hz to 40Hz intervals.

For example: The mguk motors the mguh on the "uptick" the cycle (this may or may not add energy to help turn the compressor).

Then the remaining extra inertia of the mugh goes into charging the energy store on the "down tick" of the cycle.

This happens twenty times to forty times a second.


It's like a light bulb.It is flashing on and off sixty times a second but your eyes are not fast enough to see it. In the same wasy the MGUH can be charged and disharged so rapidly that the inertia of the turbo does not feel it.. But the electrical system does!

Again charging the enegy store thru the mugk is limited. However charging thru the mugh is unlimited. And this is how you get around that barrier.

[Tommy Cookers]

the K motors the H by K generation of electrical energy mechanically loading the ICE
at 20 or 40 Hz

if we pulsed the ICE from 0 to 100% torque demand at 3 Hz the car would surge violently via the trans etc elasticity to Earth
3 Hz being around the natural frequency of the system between PU and Earth
so if we pulse the K from 0 to 100% generation at 3 Hz the car would surge at some significant fraction of this violence

pulsing at 20 Hz it wouldn't as the PU torque variation is elastically isolated from Earth at such a high frequency

iirc whether 20 or 40 Hz depends on whether we are cycling K motoring/K generating or cycling K idle/K generating

[bill shoe]

@ringo
imo
the elasticity of the load path gives isolation from 20-40 Hz ripple in PU torque

Do you mean elasticity between the MGU-K output and the ICE (crankshaft) input? If so I think there would not naturally be this kind of elasticity there, so there would have to be a vibration absorber like the springs in a clutch.

This has been discussed before too bad the thread is so long. I would dig it up for you guys.

The translation of the japanese article pretty much explains it. When in extra harvest mode the MGUH is alternately charging and discharging in 20Hz to 40Hz intervals.

I understand the high-frequency (and clever) ripple technique to transfer energy while complying with FIA rules that seem to prohibit those energy flows.

What is TC saying was elastic? The H rotational speed? Perhaps I misunderstood the components.

[PlatinumZealot]

I understand the high-frequency (and clever) ripple technique to transfer energy while complying with FIA rules that seem to prohibit those energy flows.

What is TC saying was elastic? The H rotational speed? Perhaps I misunderstood the components.

If i understand him, he implied that the engine won't display jerky rpm during the high oscillating charging from the mguk because of the elasticity and momentum of the drivetrain components. You got pistons firing, two big ole tyres at the back, the crankshaft.

Well.. I thought about it and i would agree if the charging from the MGUK was only taking a small amount of energy during each pulse. But i think it is not and the pulses are quite slow too only 1200 to 2400 pulses per minute.

So i would theorise that it is not the drivetrain that absorbs these ripples but the turbocharger via the MGUH. The turbocharger is the perfect "energy sink" and it was said in the article that the extra energy goes to the turbo. So no jerky motions at the MGUK and ICE during extra harvest.

[ringo]

the K motors the H by K generation of electrical energy mechanically loading the ICE
at 20 or 40 Hz

Isn't mechanically driving the ICE which in turn is depending on the exhaust gases to power the H with that little energy from the K inefficient?
Why not send the power from the K back through the cables that go from the MGUH directly to the MGUK?
If that makes sense.

[Thunder]

I know it's been a long 2 Days since i last commented here but yes, thisTopic is still a technical one about the PU itself. For everything else there is a General Honda F1 Topic here: viewtopic.php?f=1&t=26921

[johnny comelately]

Regarding "muramasa"'s brilliant post:

"Fuel pump is positioned on cylinder head cover, same as RA616H. But the fuel line, which was placed along the ignition coils for RA616H, cannot be seen at the same location with RA617H. Therefore RA617H is not top injection.
　　RA615H: side injection on inlet port side.
　　RA616H: top injection
　　RA617H: The author is speculating that it's side injection on exhaust port side (can be seen in pics anyways)

THROUGH TO

"We've trialed dozens of different configurations for the sub chamber, and much more than that for the injector. If you consider increasing combustion efficiency and thermal efficiency under the current reg, pre-chamber is the path you must take. Current regulation will continue until 2020. We are going to continue development of increasing combustion speed, which includes increasing compression ratio, beyond 2018."

EXHAUST SIDE injection is very effective at solving " increasing air homogeneity" in a non-traditional approach we have had great success with.
And not being of the Mahle variety with separation from the pre-combustion chamber giving far more latitude with ignition point mixture. With previous photos from "Wazari" showing a separate ingress (valve controlled?) to the multi egresses for the flames. Whilst this is very good, it is seriously constrained by the FIA rule saying only one.

[hurril]

the K motors the H by K generation of electrical energy mechanically loading the ICE
at 20 or 40 Hz

Isn't mechanically driving the ICE which in turn is depending on the exhaust gases to power the H with that little energy from the K inefficient?
Why not send the power from the K back through the cables that go from the MGUH directly to the MGUK?
If that makes sense.

Because it is an accounting exercise more than anything. The MGU-h is employed as a switch, routing energy from the MGU-k on to the ES. At least this is my understanding of what goes on.

[henry]

the K motors the H by K generation of electrical energy mechanically loading the ICE
at 20 or 40 Hz

Isn't mechanically driving the ICE which in turn is depending on the exhaust gases to power the H with that little energy from the K inefficient?
Why not send the power from the K back through the cables that go from the MGUH directly to the MGUK?
If that makes sense.

The direct route from K to ES is limited by regulation to 2 mj per lap. The route from K to ES via the H is unlimited.

This second , unlimited, route is only used once the 2 mj limit has benn used up. It’s less efficient than simply driving the K to charge the ES which happens under acceleration before the driver reaches WOT.

More subtly the unlimited route may be used before the 2 mj limit has been reached in order that the full 2 mj is achieved using the more efficient, direct driving, route.

[henry]

Imagine this block diagram

CAR < — ICE —> K —> ( T -> C -> H ) -> ES. T C H is the turbine, MGU-H, compressor assembly

To charge the ES.

At high speed the K is switched off and the CAR is driven by the ICE alone (say 500 kw) the turbine drives the compressor and the excess goes to the H, the ES gets charged at , say, 50 kw.

1. The K then switches on and the ICE drives the CAR at 380 Kw and the TCH at 120 kw. The H switches to being driven by the K, motoring. The TCH assembly increases speed because it is driven by the excess turbine power plus the 120 kw from the K ( maybe 170 kw) it gains kinetic energy.

2. The K switches off, the H switches from motoring to generating driven by the excess kinetic energy and the the excess turbine power and sends this to the ES. It absorbs the kinetic energy of the TCH which slows.

Repeat 1 and 2 at 20 to 40 hz.

The transmission, tyres and vehicle inertia see power fluctuating between 380 and 500 kw. The K drive train sees a power fluctuation of 0 to 120. The crank sees both, good luck to it.

Earlier in this topic I posted some thoughts on the numbers involved for the THC assembly. I could refine these but the principles hold, and the TCH speed variation will be in the right ballpark. I include that here. The jump will return to the right section for other posts on this topic.

Craigy’s posts are very thought provoking.

They reminded me that with the advent of this formula I tried to understand the power requirements for the MGU-H to defeat lag. I think the numbers are also relevant to the “flywheel” discussion

Rotrex advertise an electrically driven centrifugal supercharger, C38-61, which has similar sort of flow/pressure profiles to the F1 numbers. It has a rotational inertia of 9.0E-3 kgm2. I doubled that to allow for the turbine and MGU-H itself giving 1.8E-2 to be accelerated.

In the driving part of the cycle this inertia will be driven by the MGU-K, 120 kw, plus the excess power from the turbine, that is the power in excess of that to drive the compressor. Obviously the latter is very variable, I’ll use 30kw for calcs. So the driving phase is 150 kw. Assuming 20 hz and symmetrical drive and recovery, and a starting speed of 110,000 rpm the speed will go up only 180 rpm. So the fluctuations will be relatively small.

Assuming symmetry the recovery will be 150 x .025 = 3.75 kJ. This gives 75 kJ/sec.

This would mean the MGU-H outputting 150kw in this mode vs the much lower, maybe 60 kw, in sustain mode.

Craigy made the good point that the cycle does not need to be symmetrical. If the MGU-H could generate at 300kw the output per cycle would double.

[johnny comelately]

Imagine this block diagram

CAR < — ICE —> K —> ( T -> C -> H ) -> ES. T C H is the turbine, MGU-H, compressor assembly

To charge the ES.

At high speed the K is switched off and the CAR is driven by the ICE alone (say 500 kw) the turbine drives the compressor and the excess goes to the H, the ES gets charged at , say, 50 kw.

1. The K then switches on and the ICE drives the CAR at 380 Kw and the TCH at 120 kw. The H switches to being driven by the K, motoring. The TCH assembly increases speed because it is driven by the excess turbine power plus the 120 kw from the K ( maybe 170 kw) it gains kinetic energy.

2. The K switches off, the H switches from motoring to generating driven by the excess kinetic energy and the the excess turbine power and sends this to the ES. It absorbs the kinetic energy of the TCH which slows.

Repeat 1 and 2 at 20 to 40 hz.

The transmission, tyres and vehicle inertia see power fluctuating between 380 and 500 kw. The K drive train sees a power fluctuation of 0 to 120. The crank sees both, good luck to it.

Earlier in this topic I posted some thoughts on the numbers involved for the THC assembly. I could refine these but the principles hold, and the TCH speed variation will be in the right ballpark. I include that here. The jump will return to the right section for other posts on this topic.

Craigy’s posts are very thought provoking.

They reminded me that with the advent of this formula I tried to understand the power requirements for the MGU-H to defeat lag. I think the numbers are also relevant to the “flywheel” discussion

Rotrex advertise an electrically driven centrifugal supercharger, C38-61, which has similar sort of flow/pressure profiles to the F1 numbers. It has a rotational inertia of 9.0E-3 kgm2. I doubled that to allow for the turbine and MGU-H itself giving 1.8E-2 to be accelerated.

In the driving part of the cycle this inertia will be driven by the MGU-K, 120 kw, plus the excess power from the turbine, that is the power in excess of that to drive the compressor. Obviously the latter is very variable, I’ll use 30kw for calcs. So the driving phase is 150 kw. Assuming 20 hz and symmetrical drive and recovery, and a starting speed of 110,000 rpm the speed will go up only 180 rpm. So the fluctuations will be relatively small.

Assuming symmetry the recovery will be 150 x .025 = 3.75 kJ. This gives 75 kJ/sec.

This would mean the MGU-H outputting 150kw in this mode vs the much lower, maybe 60 kw, in sustain mode.

Craigy made the good point that the cycle does not need to be symmetrical. If the MGU-H could generate at 300kw the output per cycle would double.

Is this electrically induced (not induction ) power fluctuation effect at the tyre similar
to the flat plane, cross plane, big bang, smooth bang engine configuration effect? or is it similar at a different frequency (save me the calcs)

[Tommy Cookers]

roughly it's similar to 1 or 2 cylinders cutting in out then in 20 or 40 times a second

1 if at eg 12000 ICE rpm
2 if at eg 6000 ICE rpm

200 Nm K torque (referred to crankshaft) at any ICE rpm from 5710 roughly to 15000

but 1 or 2 cylinders cutting in and out at 20-40 times a second is ok - but eg 2 - 4 times a second is mechanically embarassing
like the difference between driving your car at 30 mph in top gear and driving it at 3 mph in top gear

[johnny comelately]

it's similar to 1 or 2 cylinders cutting in out then in 20 or 40 times a second

1 if at eg 12000 ICE rpm
2 if at eg 6000 ICE rpm

OK.
Then it is a fairly significant force from a traction point of view, as crankshaft torque peak pulses are significant.
And not of a resonant nature?

[Tommy Cookers]

resonance is any excursion beyond the deflection associated with an equal change in steady load

at our 20 and 40 Hz the deflection doesn't exceed that corresponding to 120 kW change in steady load ie 0 Hz
at some Hz between 0 Hz and 20 Hz(I suggest) it does

excess or unplanned torsional deflection may impede the gear shifting
there's lots of torsion in the load path through the gearbox shafts