RPM Range in which max. Battery charging can be obtained

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ganderpe

Member
Joined
Jan 27, 2014
Messages
16
I'm wondering in what range RPM wise the generator is able to charge the battery "optimally"?

Will it be able to delivre enough power with a speed of 65km/h (ICE @ 1680 rpm) in paralell mode?

Peter
 
ganderpe said:
I'm wondering in what range RPM wise the generator is able to charge the battery "optimally"?

Will it be able to delivre enough power with a speed of 65km/h (ICE @ 1680 rpm) in paralell mode?

Peter

The short answer is no.

At that speed most of the engine power goes to moving the car.
 
The short answer is no.

At that speed most of the engine power goes to moving the car.

Yes, thats true, the maximum power @ about 1/3 of max. RPM will be even less then 1/3 of max. power (89kW).
But my thinking was going more electrical wise: ICE @ 1680 rpm (imagine no mechanical power direct to the front axle) will the generator be able to draw so much power to bring the ICE near the sweet spot?
 
ganderpe said:
The short answer is no.

At that speed most of the engine power goes to moving the car.

Yes, thats true, the maximum power @ about 1/3 of max. RPM will be even less then 1/3 of max. power (89kW).
True, but the power needed to overcome air resistance will be reduced even more, when speed is reduced to 1/3rd o max speed.
 
You know that the answer is in this diagram anko ;)


PHEV10_zpsulhffcvo.jpg


... but we have to convert the torque in power (for both green and red curves). Visually I think that the maximum available power for the generator is at 120 kph rather than at 65 kph. And it is only theoretical of course.
 
Hmm, forgot about the graph. Nice addition to this topic. I agree with you, about the visual interpretation. But ...

Torque at 65 km/h seems to be about 155 Nm.
Torque at 120 km/h seems to be about 175 Nm.

So, available power at 120 km/h is about 120/65 * 175/155 as much as it is at 65 km/h, so overall 2,08 times as much.
Air resistance at 120 km would be (120/65)^2 times as much as at 65 km/h, 3,40 times as much.

Now, I do know that there are other types of resistance to overcome that scale more linear with speed. But all combined, I would think the the amount of power needed to maintain 120 km is much more than 2.08 times the amount of power needed to drive 65 .....

I wonder, where does my line of thinking fail?

On a site note: the battery will only be recharged when driving itself requires less than 75% of engine power available at that particular speed.
 
Thanks a lot for your infos, the plot say a lot! Includes even the sweet spot! :)

The Outlander PHEV uses an extremly advanced drivesystem, and though simple from the parts point of view!
I like it! :)
 
I think that irrespective of the speed of the vehicle the computer will optimise the use of the engine.
I have noticed this whilst driving.
At 65KPH there is no guarantee the wheels will be directly driven by the engine (which would limit the engine rpm).
If you press the Charge button the computer may decide it is more efficient to disconnect the engine from the wheels and run it at high rpm to both charge the battery and provide power to the wheels through the electric motors. I notice it does this particularly if you are driving on the flat or downhill.
It mostly seems to connect the engine to the wheels when going uphill and as the slope gets steeper.
Once over 80KPH it seems to couple the wheels to the engine most of the time but occasionally stops this if driving downhill.
I drive with ACC on most of the time but I don't know if that makes a difference.
 
You do realise the graphics on the dashboard are not always truthful, when it comes to operating modes? I am asking because it sounds as if your car is quite different from mine.

If mine does not have enough punch to charge in parallel mode, it will simply not charge. Not even in Charge mode. As a matter of fact, it will stop charging when more than approx. 75% of the available engine power is needed for charging (because the engine aims at operating in the sweetspot). It will even start to discharge the battery when more than 100% of available engine power at limited RPMs is needed for propulsion. When 100% of available engine power + 60 kW from the battery together are not enough for propulsion, only then mine will disengage parallel mode. Typically, this requires either very strong acceleration, heavy towing or steep climbing. One other reason for disengaging parallel mode would be if SOC had dropped to approx 20 - 22% and the limited RPMs ice driving conditions would prevent recharging the battery. But this is irrespective of the Charge button and 20 - 22% SOC is very hard to 'accomplish'.

Yours seems to be to maintain parallel mode driving up-hill and to drop out of it when driving down hill. Rather the opposite of what I experience.

Also be aware, at 65 km/h you are right at the border line between serial and parallel hybrid mode. Maybe you are crossing the boundary without noticing?
 
anko said:
. When 100% of available engine power + 60 kW from the battery together are not enough for propulsion, only then mine will disengage parallel mode. Typically, this requires either very strong acceleration, heavy towing or steep climbing. One other reason for disengaging parallel mode would be if SOC had dropped to approx 20 - 22% and the limited RPMs ice driving conditions would prevent recharging the battery. But this is irrespective of the Charge button and 20 - 22% SOC is very hard to 'accomplish'.

?

So, since you have experience towing a van right at the limit, how much power comes out of the battery when moving at say 100km/h on the level, assuming no headwind? Can the engine do it by itself most of the time?
 
On the flat? Most of the time, yes. All the time, no. You will not be able to maintain SOC.

Most of the time, you will need between 75 and 100% of available power. Very seldom, you will need less and be able ti put a little charge back into the battery. More often, you will need more than 100% and you will loose some SOC.

But what the dash won't tell you is that, when the engine is operating with a load between 75 and 100%, the electric power needed to eliminate E-drag in the motors and even in the generator are coming from the battery and not the generator. So, you loose a little bit of charge almost constantly.

How fast to drain? I have been able to do 250 - 300 km before the battery was depleted. And this is with a 2.30 m wide van of 1500 km. cruising at apporx. 94 km/h GPS. Two tips:

- Stay away from 4WD Lock as it will drain your battery so fast, the gauge almost can't keep up :mrgreen: (see other topic that came by today, about se of 4WD Lock).
- Stay away from the B levels (and thus from CC). When you lift the throttle while your B level is not 0, instead of charging the battery big time, the engine will go idle. Same happens when you cancel out of CC (by means of cancel button or touching the brake pedal). When this happens you can switch to B0 and shortly hit the throttle, to wake up the engine again.
 
And this is with a 2.30 m wide van of 1500 km. cruising at apporx. 94 km/h GPS.

@anko: do you have an idea how much the air drag was?
When driving with an average speed of 94km/h: was the outlander decide to go paralell or serial mode?

With our caravan we're allowed to drive 80km/h only, this could probably help to charge the battery before a steep passage?


Peter
 
ganderpe said:
And this is with a 2.30 m wide van of 1500 km. cruising at apporx. 94 km/h GPS.

@anko: do you have an idea how much the air drag was?
When driving with an average speed of 94km/h: was the outlander decide to go paralell or serial mode?

With our caravan we're allowed to drive 80km/h only, this could probably help to charge the battery before a steep passage?


Peter
At 94 km/h it will choose parallel mode. It will only drop out during acceleration or climbing. Or when SOC drops to 22%. Don't forget: 60 kW from the battery plus 30 from the engine is normally more than enough. Even at speeds higher than that.

But once the battery has dropped all the way to 22%, you will get in a 'secondary' hysteresis cycle: serial mode to get it back to 25% and then parallel mode during which it will drop back to 22%.

At 80 km/h you should have a better chance of mantaining SOC. I have read awhile ago ipon the Dutch Mitsubishi web site the the car is week equipped to tow a trailer at speeds up to 80 km/h.

At serial mode speeds the car has no issue maintaining SOC. I have done several hours of testing in a hilly environment with several hill starts and SOC kept going up during the day.
 
anko said:
Hmm, forgot about the graph. Nice addition to this topic. I agree with you, about the visual interpretation. But ...

Torque at 65 km/h seems to be about 155 Nm.
Torque at 120 km/h seems to be about 175 Nm.

So, available power at 120 km/h is about 120/65 * 175/155 as much as it is at 65 km/h, so overall 2,08 times as much.
Air resistance at 120 km would be (120/65)^2 times as much as at 65 km/h, 3,40 times as much.

Now, I do know that there are other types of resistance to overcome that scale more linear with speed. But all combined, I would think the the amount of power needed to maintain 120 km is much more than 2.08 times the amount of power needed to drive 65 .....

I wonder, where does my line of thinking fail?

On a site note: the battery will only be recharged when driving itself requires less than 75% of engine power available at that particular speed.

My line of thinking is simpler :

- torque available (for charging) @65 km/h : 155 - 45 = 110 N.m
- """"""""""""""""""""""""""""""""" @120 km/h : 175 - 100 = 75 N.m

---> Torque ratio 75/110

Power ratio = Torque ratio x speed ratio = 75/110*120/65 = 1.26

So the graph shows that the system has ~ 25% more power available for charging at 120 km/h than at 65 km/h.

Theoricaly... ;)
 
Undoubtedly (and not maybe) you're right.

So your remark generates a new question about the title of this subject : what is "max. battery charging" signifying ?

Max. battery charging per minute or max. battery charging per km ? For me it's obviously "per km", since our trips have (generally) a fixed/mandatory distance, not a fixed time ;)
 
My line of thinking is simpler :

- torque available (for charging) @65 km/h : 155 - 45 = 110 N.m
- """"""""""""""""""""""""""""""""" @120 km/h : 175 - 100 = 75 N.m

---> Torque ratio 75/110
Power ratio = Torque ratio x speed ratio = 75/110*120/65 = 1.26
So the graph shows that the system has ~ 25% more power available for charging at 120 km/h than at 65 km/h.
Theoricaly... ;)

The values
@65 km/h : 155 - 45 = 110 N.m
@120 km/h : 175 - 100 = 75 N.m
are from the pic on page two...

But theese values represent a car probabely on a flat road.
With a caravan and/or while climbing the green graph representing the air and roll drag, hav quite higher levels and the difference between red and green are melting away...

peter
 
Not "probably" Peter ! I would say undoubtedly it's on a flat road.
The only reference for a lot of things is of course a flat road (acceleration measurement, max speed etc ...) ;)
 
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