petrol consumption

[karl]

IMHO there is no more drag in B5 than in B0 (at the same stable speed of course). Bx only affects the effects of the positions of the accelerator pedal.
At the same speed, the pedal will be more depressed in B5 than in B0 and give the same result.

Exactly, it just moves the "zero" position up or down on the accelerator.

 

[ian4x4]

The aerodynamic drag relation to speed has been quoted as a squared relation. It is actually a cubic relation. .

Actually it is a squared relation (various sites):

"In fluid dynamics, the drag equation is a formula used to calculate the force of drag experienced by an object due to movement through a fully enclosing fluid. The formula is accurate only under certain conditions: the objects must have a blunt form factor and the fluid must have a large enough Reynolds number to produce turbulence behind the object. The equation is

D = Cd * A * .5 * r * V^2
D is the drag force, which is by definition the force component in the direction of the flow velocity,[1]
r is the mass density of the fluid, [2]
V is the flow velocity relative to the object,
A is the reference area, and
Cd is the drag coefficient – a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. "

The power of the wind vs wind speed is a cubic relationship. This is used for calculating wind farm locations. The same air that a car has to use power to push itself through. I suggest anyone interested, performs a dimension analysis of any equations and there will be parts of the equation not just V^2 that equate to another V.

http://www.iowaenergycenter.org/wind-energy-manual/wind-and-wind-power/wind-speed-and-power/

I won't be debating this as it has been done to death on windmill forums elsewhere.

Surely we are talking of drag here which is proportional to V squared.
I understood that 'windmill type generators' were influenced by the lift generated by the airscrew moving through the air giving the extra component making it V cubed. I read vertical 'drag type generators' are less efficient and only in proportion to V squared wind speeds.

And on the other point of all electric motors produce back emf (Flemings left and right hand rules work against each other) and inductive losses (like in transformers and coils).

What interesting reading the Technical Highlights Manual makes.

 

[gwatpe]

I was hoping a technical moderator would step in and do this.

Engineering students are a good substitute and many publish on the internet and make info available, even specific for the aerodynamics of vehicles, and allow their referenced work to be distributed.

Here is an extract from a paper on turbocharger design, with a section on car aerodynamic drag for a HONDA.

references

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The energy of the wind vs speed is a cubic relation, and yes a car has the same air to push through at different speeds, so yes, the power required to move through the air will increase with the same relation. There are many other factors affecting a car apart from the aerodynamic drag, but these were not important in my discussion. They are not a part of any equations supplied in prior discussion relating to aerodynamic drag either.

 

[biosci]

The energy of the wind vs speed is a cubic relation,


Yes the power of wind is cubic - but the drag on an automobile is a squared function.

I do not know why your reference has a cubic term but any physics book, web sites, and even wikipedia show the correct relationship for moving through a fluid (air) is v^2.

 

[biosci]

The energy of the wind vs speed is a cubic relation,


Yes the power of wind is cubic - but the drag on an automobile is a squared function.

I do not know why your reference has a cubic term but any physics book, web sites, and even wikipedia show the correct relationship for moving through a fluid (air) is v^2.

To clarify, the drag force or the force required to move through air increases as squared factor. However the amount of power consumed or required adds another velocity term since the power required to move against wind drag Is consumed faster with increased speed.
Essentially: Power(to overcome wind drag) = force(drag as v^2) x velocity

 

[Grigou]

Good point biosci, I was preparing a similar answer before reading yours

In such a discussion it's better to define all terms, including "drag" !

 

[ian4x4]

Sorry for wading in without reading the question properly.

Of course drag is a force and should be measured in similar units as torque, and not power, which I think is a measurement of the rate of delivering that torque.

Wisely, the graph on page 27 of the PHEV Technical Highlights Manual, charts drag against the torque, and is worth a look.

 

[gwatpe]

Have moved way off topic, but have to agree that I had referred to power to overcome the aerodynamic drag, and the op had just quoted a drag component only. We are really only interested in the amount of power and how this changes with speed for a particular car. We drivers, have few analytical tools available in the car as supplied and so there will continue to be conjecture.

My original post was about how my PHEV performed when driven with petrol only consumption compared to a similar weight MMC vehicle, a petrol Challenger.

 

[PeteInOz]

Thanks jaapv for your reply.

There is no doubt, as you say, that there is a great deal of complexity and seamless interaction between the various drive trains, and that Mitsubishi must have worked tirelessly to fine tune all the systems at hand to produce the final product. I have only raised some of the points I have so that members can consider more fully some of the possible technical nuances that may underpin and affect their vehicles.

Thanks also to Grigou for his input on his feelings on the affects (or not) of the various regenerative break settings.

Cheers PeteInOz

 

[lalacurf121]

Wow, that great information for me , Thank you