Honda Power Unit Hardware & Software

[godlameroso]

a 25kg battery pack at ~350w hrs/kg(for current top tier battery tech) so the battery packs are around 6-8kW hr, improving the energy density of the cells will mitigate the heat generated from using higher voltage. The latest cutting edge battery tech coming out this year should see energy densities reaching 400-450w hrs/kg. Remember you're only allowed 4MJ per lap wich comes out to being around 1.25kW hr so teams could probably make due with a much smaller battery pack but the increased capacity improves efficiency and allows higher voltages and charge rates.

[GoranF1]

After Hasaegawa confirmed no token upgrade in next two races, we now have a confirmation of new fuel in Canada.


http://www.motorsport.com/f1/news/techn ... ss-735059/

[Andres125sx]

After Hasaegawa confirmed no token upgrade in next two races, we now have a confirmation of new fuel in Canada.


http://www.motorsport.com/f1/news/techn ... ss-735059/

This is interesting

"Theoretically there are still some big gains to be had: both in terms of fuel, on the combustion side, and also on the friction side, on the engine as well," he said.

"In the development process we are a little behind because the programme started later than our rivals, so you might expect us to be making more progress. But certainly we are working on it right now.

"We are looking from a friction perspective, improving the efficiency of the engine by taking friction out. We've already identified that there are some gains to be had there: so it is just getting that in to the system."

[gruntguru]

though current per se is not bad ie a 500 V system is not inherently less efficient than a 1000 V system though maybe a bit heavier eg loosely, a 1000 V machine will need 4x the resistance of an equivalent 500 V machine
so the resistive power loss will be the same in both

Current is bad. Higher current = higher resistive energy losses.
The system doesn't "need" any resistance. Resistance is bad and must be minimized throughout the circuit including transmission lines and motor/generator windings.
For similar circuit resistance the lower voltage system will have significantly higher losses.

[gruntguru]

the design will involve matching ie what crankshaft-equivalent rpm is chosen to match the battery voltage chosen. Remember that charging (likely) involves the generator voltage exceeding the battery voltage (that somewhat varying with charge state) and vice versa when motoring imo the match rpm will be held down to around 12500 not eg 15000

Good point. How this pans out depends on inverter topology but I guess you are saying the system can only regenerate at 120 kW or more above 12,500 and motor at 120 kW or more below 12,500? Of course some rpm overlap would be possible if the MGU itself is rated above 120 kW (but only operated at 120 or less).

To increase the overlap further would require a transformer - probably in the regen' direction.

[Andres125sx]

though current per se is not bad ie a 500 V system is not inherently less efficient than a 1000 V system though maybe a bit heavier eg loosely, a 1000 V machine will need 4x the resistance of an equivalent 500 V machine
so the resistive power loss will be the same in both

Current is bad. Higher current = higher resistive energy losses.
The system doesn't "need" any resistance. Resistance is bad and must be minimized throughout the circuit including transmission lines and motor/generator windings.
For similar circuit resistance the lower voltage system will have significantly higher losses.

Sorry for the OT, but I have to make a question related to efficiency being improved with higher voltage. This is something known on the drone world I work in, but would like to know if someone could give me an idea about the real improvement I would see if switching to higher voltage with my drone.

It is a 6,5kg octocopter currently working with 4s batteries (14,8V nominal, 16,8V fully charged), wich draws 120A at whole open throttle, and around 65A while hovering. Flying time is only 6-7 minutes and power to weight ratio is quite poor for a multirotor, more than enough to fly safely, but I know I could make a better use of my design (frame designed by myself to improve some common problems from comercial drones) than I do with a better power/weight ratio.

So I´m considering if switching to 5s batteries wich would not require big modifications (18.5V nominal, 22,2V fully charged) and I´ve found some props wich would keep current on same level wich currently is on the limit for the distribution board so I cannot increase current no matter what´s the voltage

Another option is switching to 6s batteries wich is a very common mod, but that would require some mods on the drone. Not big modifications, but some modifications, the biggest problem is with 6s I would need new motors (with 5s I can still use the same). Considering I invested more than 700€ on good motors I´d prefer if I don´t need to replace them obviously, but if the improvement is worth, I´d consider it


Does someone have an idea about the difference/improvement I should expect on each case? Biggest problem with this is I will need a battery replacement, and I have around 16 batteries, so it will be a very expensive change even if I keep same motors, so I need to be sure it will be worth, and I´m not

Any help will be welcome

[Tommy Cookers]

though current per se is not bad ie a 500 V system is not inherently less efficient than a 1000 V system though maybe a bit heavier eg loosely, a 1000 V machine will need 4x the resistance of an equivalent 500 V machine
so the resistive power loss will be the same in both

Current is bad. Higher current = higher resistive energy losses.
The system doesn't "need" any resistance. Resistance is bad and must be minimized throughout the circuit including transmission lines and motor/generator windings.
For similar circuit resistance the lower voltage system will have significantly higher losses.

with 2 otherwise similar real machines you cannot have a lower and a higher voltage machine with similar resistances
it's differences in armature resistance (or equivalent) that makes the machines behave similarly despite the voltages being different
the 1000V machine will fail catastrophically on 1000 V if it has the same resistance as the 500 V machine

btw
resistance is what governs current motion in phase with voltage and this is the source and sink of mechanical power ie motor and MG action
current motion out-of-phase with voltage has no work transfer ie gives only a false ('wattless') power
so don't we need voltage drop (ie resistance) across the conductor (armature or equivalent) within the field to give MG action ??
if so superconductivity at this point (ie current limited externally) wouldn't give MG action
and even a mechanically frictionless machine could never be 100% efficient
(superconductivity in the field winding is course attractive)

not btw
of course the cabling of the lower voltage system would need to be heavier to have similar cabling losses to the higher voltage system
the batteries may be heavier in the high voltage system - but there's a minimum battery weight rule which negates this issue

[R_Redding]

the design will involve matching ie what crankshaft-equivalent rpm is chosen to match the battery voltage chosen. Remember that charging (likely) involves the generator voltage exceeding the battery voltage

Good point.

To increase the overlap further would require a transformer

Neither points are correct.

TC is correct when he notes that the charging voltage has to exceed the battery voltage to allow charging , and when the generated motor emf is higher than the battery voltage , standard regeneration will occur.

But...When the generator voltage is lower than the battery voltage , regeneration can still be made to occur by using DC-DC boost switching techniques, using the motor as an inductor to create voltages higher than VBattery to facilitate charging.

The technique uses the bottom elements in a 3PH H bridge ,anchoring one side of the coil,and pulsing the other to induce a higher voltage across the winding, This is done rotationally and in synch with the field... and a higher voltage DC pulse pattern is produced.

Rob.

[gruntguru]

Current is bad. Higher current = higher resistive energy losses.
The system doesn't "need" any resistance. Resistance is bad and must be minimized throughout the circuit including transmission lines and motor/generator windings.
For similar circuit resistance the lower voltage system will have significantly higher losses.

with 2 otherwise similar real machines you cannot have a lower and a higher voltage machine with similar resistances
it's differences in armature resistance (or equivalent) that makes the machines behave similarly despite the voltages being different
the 1000V machine will fail catastrophically on 1000 V if it has the same resistance as the 500 V machine

btw
resistance is what governs current motion in phase with voltage and this is the source and sink of mechanical power ie motor and MG action
current motion out-of-phase with voltage has no work transfer ie gives only a false ('wattless') power
so don't we need voltage drop (ie resistance) across the conductor (armature or equivalent) within the field to give MG action ??
if so superconductivity at this point (ie current limited externally) wouldn't give MG action
and even a mechanically frictionless machine could never be 100% efficient
(superconductivity in the field winding is course attractive)

Voltage drop due to circuit resistance results entirely in heat loss - bad. Current in electric motors is controlled primarily by inductance not resistance. This energy goes into building a magnetic field in the windings and ultimately into turning the motor shaft.

Electric motors or generators with superconducting windings come closest to eliminating resistive losses.

[Tommy Cookers]

.....Current in electric motors is controlled primarily by inductance not resistance........

this is true (of permanent-field motors) 'only' for the types that use some type of pulsed excitation
the inductance is small of course, but sufficient due to the continuously high (pulse) rate-of-change of voltage

ok here even I assume they are using such types of machine
as highly responsive (for shifting) as older types but plausibly more efficient in the general low response running mode

though they still have resistance of course
and (for a given power) current must rise as rpm falls below some chosen 'match' rpm - whatever system voltage we have designed for


anyway, the main electrical inefficiency is in the ES and there's maybe an efficacy issue in its functional matching with the MGs

[haza]

https://twitter.com/Simonagius/status/7 ... 46496?s=09 Honda RA616H

[PlatinumZealot]

If you charge a lithium battery at a higher voltage you'll have more current going into it, not less = more heat, it's the energy store they're talking about, not the motor/generators.

I thought I=P/V

160000W/1000v=160A
160000W/500v=320A

Same power, higher voltage=less current.

The Voltage is usually fixed for the motor speed you want. But anyway it is not a matter of playing with the equation. It is making sure that the voltage and current that you use at the motor translates to the power at the motor shaft. If you have the same exact motor, more voltage equals more current equals more power and more heat.

[Mclarettino]

http://www.motorsportmagazine.com/opini ... ine-update

Mercedes, Ferrari and Renault all seem to be introducing/upgrading some form of TJI/HCCI system in Canada.

Honda's update planned for Canada too. But nowhere can find any mention of Honda working on any TJI/HCCI system of their own. Are they just keeping quiet about it?

[NL_Fer]

If they really switch nu an external turbo next year, i think this years updates will be allot of ICE updates.

[GoranF1]

Can anyone say wHat is the difference between SCCI and HCCI?