Honda Power Unit Hardware & Software

All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
anthonyfa18
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henry
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dsdevries wrote:
13 Mar 2019, 16:23
restless wrote:
13 Mar 2019, 14:00
offtopic, but - is there a description how MGU-H works, what is used for energy harvesting?
I wonder if these ways are set in stone or someone can find additional routes for harvest.
The MGU-H is basically a dynamo positioned between the turbine and the compressor of the Turbocharger.

In a normal turbo the hot gasses from the exaust spin a turbine which is directly connected a compressor. That compressor compresses the intake air before it is injected into the cylinder. This compressed air will produce a more powerfull bang inside the cylinder and thus the engine has more power.
So far so good but:

But, when the turbo spins while no power is needed by the engine, then all that energy is wasted. So the MGU-H will disengage the compressor and engages a dynamo which will charge the battery. That battery can power multiple things, but it is mainly used to spin the turbo at lower refs when the exaust gases alone are not powerfull enough. It is also used by the bobine to produce a more powerfull spark.
Is not quite right.

When the engine is running at high power the turbine generates more power than is needed to drive the compressor, so it can also drive the MGU-H making power which can either drive the MGU-K or be stored in the Energy Store, battery in your example.

As you say the ES is used to drive the compressor when there is insufficient exhaust gases but it’s main job is to drive the MGU-K.

I have not heard of it’s use to increase ignition intensity, your bobine, but if it is I doubt it is significant.
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Truth is confirmed by inspection and delay; falsehood by haste and uncertainty : Tacitus

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godlameroso
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restless wrote:
13 Mar 2019, 14:00
offtopic, but - is there a description how MGU-H works, what is used for energy harvesting?
I wonder if these ways are set in stone or someone can find additional routes for harvest.
The MGU-H as the name implies harvests heat energy. The heat created by the combustion engine precisely. The turbine on a turbo "blocks" the expanding heated air that just left the combustion chamber, the energy left over after the engine has combusted its fuel drives the turbine. The MGU-H is connected to this turbine. The MGU-H is also connected to the compressor, the compressor feeds the engine with oxygen to be mixed with fuel so that combustion can take place. It's a very elegant loop, the MGU-H can not only drive the compressor, it can also slow it down, thus the MGU-H is an electric motor that regulates the speed of the turbo.

It is an incredibly powerful device, because as stated, it can maintain boost pressures within a very tight range at all engine RPMs, that alone brings incredible efficiency benefits. The next way it benefits efficiency is that it can charge the battery, and it can do this any number of ways. First, when the car is at full power and the exhaust gases flow in excess of what is needed by the turbo to run the engine, the MGU-H can slow the turbo down and convert that load into electrical energy to be deployed later.

The MGU-H can transfer energy to and from the MGU-K, so the MGU-K can power the MGU-H similar to the way a supercharger works(ie with a pulley spinning the compressor). Therefore you can use the engine, to send power to the MGU-K, which then sends power to the MGU-H, which can either be used to run the turbo, or to harvest energy to the battery. This means that you can get around the 2MJ per lap harvest limit put on the MGU-K, by sending the energy to the MGU-H instead and then to the battery. The downside of this is that it costs fuel, and engine power to charge the battery this way, and it's incredibly inefficient.


However, having a lot of energy harvested in the battery(within the 4MJ pool) is a good thing because it allows you to run both the MGU-K and the MGU-H without wasting any engine power. That means the turbo is no longer choking the engine, and the crank is being helped to the tune of 160hp by the MGU-K, the result means the engine is ~50% efficient.
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henry
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@godlamerosa

That’s a nice fairly comprehensive summary.

However, the extra harvest route, MGU-K > MGU-H > ES is not incredibly inefficient, it’s probaby only a few percentage points worse than direct K to H. It depends on the chosen charge rate. The higher the rate the less efficient.
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Truth is confirmed by inspection and delay; falsehood by haste and uncertainty : Tacitus

Tommy Cookers
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the rules don't allow the designer to spend fuel energy in the most race-efficient way but ....
they allow the designer to spend energy in the most race-efficient way after conversion of fuel energy to electrical energy

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henry
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Tommy Cookers wrote:
13 Mar 2019, 20:58
the rules don't allow the designer to spend fuel energy in the most race-efficient way but ....
they allow the designer to spend energy in the most race-efficient way after conversion of fuel energy to electrical energy
I think the efficiency @godlamerosa raised was conversion to electricity.

You and I have previously debated the point you are making now and I seem to remember agreed.
Fortune favours the prepared; she has no favourites and takes no sides.
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godlameroso
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Makes me wonder if when they talk about power gains what they mean is an increase in average output over a race distance instead or a fixed peak number.
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Bandit1216
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gruntguru wrote:
12 Mar 2019, 00:47
Bandit1216 wrote:
11 Mar 2019, 09:40
100 kW to drive the compressor can't be right. Oké, it's quite big and boost is high, but this just can't be true. That means you would need an ordinary 1.6 roadcar engine to drive the compressor wheel. I think someone added a zero.
For a compressor with 80% efficiency:
Mass flow (kg/s)  MAP (bar abs)  Compressor Power (kW)
0.5                3.0                69
0.67               4.0               122
0.83               5.0               181
Who. I didn't realize it was that much. Should have done my calc. first :shock:
But just suppose it weren't hypothetical.

Gibbs
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Since we're on this topic about thermal efficiency and the MGU's, a friend and I were having a discussion about how the waste heat of the thermal efficiency would be affected by the electric motors. He argued that a standard thermal efficiency equation (using BSFC and a heating value) still held true for a V6 hybrid, as the work done by the motors was still the result of the fuel. My knowledge on the MGUH/K and ES and the sources of energy is very very primitive but as stated above the H and K are capable of extracting heat from the engine and utilising it as power, - and from my knowledge - also kinetic energy from the brakes and drivetrain(?).

My question is, does a standard thermal efficiency equation still hold true for the V6H, if some of the power being output by the electric motors is being generated by a source which is not fuel? And if so, how would the thermal efficiency be calculated or affected?

p.s sorry if everything I've said doesn't make sense, I'm still learning :D

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godlameroso
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Technically its all from burning fuel. The burned air fuel mixture gets converted to kinetic energy. You had to burn fuel in order to give the car kinetic energy to harvest.
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henry
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Gibbs wrote:
15 Mar 2019, 13:51
Since we're on this topic about thermal efficiency and the MGU's, a friend and I were having a discussion about how the waste heat of the thermal efficiency would be affected by the electric motors. He argued that a standard thermal efficiency equation (using BSFC and a heating value) still held true for a V6 hybrid, as the work done by the motors was still the result of the fuel. My knowledge on the MGUH/K and ES and the sources of energy is very very primitive but as stated above the H and K are capable of extracting heat from the engine and utilising it as power, - and from my knowledge - also kinetic energy from the brakes and drivetrain(?).

My question is, does a standard thermal efficiency equation still hold true for the V6H, if some of the power being output by the electric motors is being generated by a source which is not fuel? And if so, how would the thermal efficiency be calculated or affected?

p.s sorry if everything I've said doesn't make sense, I'm still learning :D
Personally when thermal efficiencies are given for these power units I view it as the output energy rate in self sustaining mode measured at the crank divided by the input energy rate from the fuel.

The crank power is composed of the work done on the pistons directly plus the work done by the MGU-K which is driven solely by the MGU-H which is in turn driven by the exhaust gases acting on the turbine. This is usually referred to as self-sustaining mode.

I usually assume the max energy input rate at 1250 kW. So when they quote 50% efficiency I take it to be 625 kW, (850hp). The share of this from the two sources is debatable, maybe 565 from the pistons and 60 from the exhaust gases via the electrical components.

There are a number of other discrete crank outputs depending on various combinations of ICE, H, K and Energy Store outputs and inputs but I don’t think any of them can be considered when calculating thermal efficiency as its normally interpreted.
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Truth is confirmed by inspection and delay; falsehood by haste and uncertainty : Tacitus

Jaisonas
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No vibrations =D>

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HPD
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Image
Image
Aus FP2

maguetox
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Interesting how the cooling piping routes and their location in the radiators itself are so different in both teams. Different cooling packaging philosophies.

anthonyfa18
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Mercedes: Red Bull "very different" after "huge" Honda step

https://ca.motorsport.com/f1/news/merce ... t/4354032/