mass yield and minimo consumption

[mir]

Hi.
question that abstains from above:

with my car I go to the highway and once I reach the maximum performance point of the engine I begin to measure the consumption that will be x.

now I go back and do the same path in the first flush of gas (as if I were in column). I will have a consumption y

x is greater or less than y?

 

[Er Presidente]

Hi.
question that abstains from above:

with my car I go to the highway and once I reach the maximum performance point of the engine I begin to measure the consumption that will be x.

now I go back and do the same path in the first flush of gas (as if I were in column). I will have a consumption y

x is greater or less than y?

depends.
maximum performance means less consumption in proportion to work rendered.
the minimum consumption is obtained with the motor at the minimum (the full lasts three days), at maximum yield the full lasts four hours to the maximum.
But with the engine at the minimum and the machine stops the performance will be zero, at the maximum performance will be close to 30%.

 

[Fulvio Romano]

what he says is correct, but incomprehensible.
I think he meant:

y will be smaller than x. but how, x wasn't at maximum performance? Yes, but with x you had a higher speed, and the resistance of the car with the air goes with the square of the speed, so on the return you will have a lower yield, but you will have spent even less energy, and therefore consumed less.

This is a pretty classic mistake, made so many people I talked to. When we talk about efficiency we talk about work produced at fuel costs. when talking about cars instead it tends to focus on the kilometers paths. but given a path, the energy to travel it is not uniquely determined. depends on several factors, first of which is the speed with which you travel.

 

[mir]

I also found the same problem when I was looking for the answer... so, from your reasoning, x would be less than y if there were no friction. correct?

the limit in my reasoning was precisely in the fact that the performance is not tied to time so if I put myself more or less time to do something does not affect the performance I do... but then the frictions arrive.

 

[cacciatorino]

I think he meant:

It can't be him. He must be a hacker who stole his account.

 

[Er Presidente]

what he says is correct, but incomprehensible.
I think he meant:

y will be smaller than x. but how, x wasn't at maximum performance? Yes, but with x you had a higher speed, and the resistance of the car with the air goes with the square of the speed, so on the return you will have a lower yield, but you will have spent even less energy, and therefore consumed less.

This is a pretty classic mistake, made so many people I talked to. When we talk about efficiency we talk about work produced at fuel costs. when talking about cars instead it tends to focus on the kilometers paths. but given a path, the energy to travel it is not uniquely determined. depends on several factors, first of which is the speed with which you travel.

actually now is all clearer.

 

[PierArg]

Hi, mir,
First of all, the President and also fulvio.
In particular, when it says that the km routes should not be taken into account because they are misleading.


in all cases your question has the answer in the listed plans.
are curves that indicate consumption zones based on gear and speed.
http://www.skuola.net/universita/dispense/motore-diesel-piano-quotato#p1

 

[cacciatorino]

In addition to what is said, it must be decided whether what interests us is the fuel consumption to make a certain journey, or to keep the machine running for a certain time, i.e. the liters/km or the liters/hour.

 

[MBT]

the question is asked incorrectly and generates strange answers. . .
In fact, as it exposes it mir, the variables are too many.. .
the condition is of motion with a high march, to an intermediate regime of turns, with a considerable demand of power and, consequently, a consumption x
the condition b is of motion with a low march, to a minimum regime of turns, with a minimum demand of power and a consumption y.

In addition, it should be explicit if we talk about hourly consumption or consumption per kilometer. .

too many variables in play to give a theoretical answer
by making some reasoning, however, we can say that:
the "odyerni" engines and in particular the turbodiesel, have a particularly flat torque and performance curve. the specific consumption, therefore, varies little in the usual operating arc of the motor.
Therefore, with all the beautiful approximations of the case, we can also say that consumption is proportional to the rotation regime of the motor and the applied load, with good peace of theories and yields.
then, by extension, we can also say that consumption Timetable case a must be greater than that of the case b as the engine turns at higher speeds and with load applied greater than the case b
Well, what about consumption in the kilometer? ? ?
Well... The story becomes a little more complicated... should be analyzed the reports of the change, if nothing else
if I have a very first court change compared to having a first long
I think I can risk that, on an average car, consumption in both higher and in b... but it is really a risk

 

[Fulvio Romano]

Well, what about consumption in the kilometer? ? ?
Well... history becomes a little more complicated. . .

I said there were so many other variables. for example the number of surprises made!

 

[PierArg]

I knew that who commands is the quoted plan that is the function of performance.
consumption is solo a consequence...it is true that they are all experimental stuff (and not hypothesized data a priori) that also takes into account the influence of the moon or if the driver went "body" before driving.

this page is taken from the jacose dante
==as you can see it is about specific consumption, i.e. grams of fuel per power unit.

 

[Er Presidente]

go from roma to milan with the motor (ideally) in condition of maximum performance port necessarily to do that job with the least possible consumption.

if in a conversion of energy the thermodynamic efficiency, or thermodynamic efficiency, is the relationship between mechanical work accomplished and the energy baked in the system, then if I go from roma to milan "with a wire of gas" I could also consume less (use less fuel in absolute) but the work done will be lower, I went, yes, to thousand but in a greater time. then the "consumption", at equal work, will be increased.

That's why I brought the absurd example of being stuck with the engine to the minimum, I have the minimum possible consumption (time) but the work is zero and therefore the performance is null.


Sorry I'm so confused, but I can't do better.

 

[Antosacca]

Hi.
question that abstains from above:

with my car I go to the highway and once I reach the maximum performance point of the engine I begin to measure the consumption that will be x.

now I go back and do the same path in the first flush of gas (as if I were in column). I will have a consumption y

x is greater or less than y?

the answer, fully disputed:
depends on cx which is the only unknown of the problem.
if the cx of your car is very welcome x>y,
if the cx of your car is small x<y.
the scissors between the two consumptions, i.e. at what speed and at what engine regime the inequalities are reversed, it is a little difficult to know.
My preference to consume and pollute less is to travel near the maximum torque, of course with the most suitable gear. make modest accelerations and choose with a very broad view of what is happening around and do not exceed where the 95-105 km/h is allowed.
beyond this speed, due to the resistance to progress that does not change only to the square, consumption increases a lot.

 

[MBT]

if in a conversion of energy the thermodynamic efficiency, or thermodynamic efficiency, is the relationship between mechanical work accomplished and the energy baked in the system, then if I go from roma to milan "with a gas wire" I could also consume less (use less fuel in absolute) but the work done will be lower, I went, yes, to Milan but in a greater time. then the "consumption", at equal work, will be increased.

Gotcha!
the work done is exactly the same. I go from Rome to Milan.
what changes between doing it in step of snail or at supersonic speed is the power necessary to do that job... .
In fact, I require power to the engine, and the consumption that follows it is according to the specific consumption (motor terminal and how it is working) and the power supplied

the answer, fully disputed:
depends on cx which is the only unknown of the problem.
if the cx of your car is very welcome x>y,
if the cx of your car is small x<y.
the scissors between the two consumptions, i.e. at what speed and at what engine regime the inequalities are reversed, it is a little difficult to know.
My preference to consume and pollute less is to travel near the maximum torque, of course with the most suitable gear. make modest accelerations and choose with a very broad view of what is happening around and do not exceed where the 95-105 km/h is allowed.
beyond this speed, due to the resistance to progress that does not change only to the square, consumption increases a lot.

say that, with a certain approximation, the power required for vehicle advancement varies "approximately" with speed cube. at equal vehicle and boundary conditions, of course. . .

 

[Fulvio Romano]

if in a conversion of energy the thermodynamic efficiency, or thermodynamic efficiency, is the relationship between mechanical work accomplished and the energy baked in the system, then if I go from roma to milan "with a wire of gas" I could also consume less (use less fuel in absolute) but the work done will be lower, I went, yes, to thousand but in a greater time.

Gotcha!
the work done is exactly the same. I go from Rome to Milan.
what changes between doing it in step of snail or at supersonic speed is the power necessary to do that job... .
In fact, I require power to the engine, and the consumption that follows it is according to the specific consumption (motor terminal and how it is working) and the power supplied

the work done is less. the work is done against the field* (non-preservative) of the forces of friction. such field is function of speed. If I make the way in less time, regardless of the necessary power that will be greater, but I will find myself spending more energy to win greater forces, and then I do more work.
the work in the conservative field (the gravitational one) we consider it null, considering points of departure and arrival at the same quota.

then there is also to say that from roma to milan the road is uphill, on the return is downhill... (Smile)

say that, with a certain approximation, the power required for vehicle advancement varies "approximately" with speed cube. at equal vehicle and boundary conditions, of course. . .

Why the cube?
forces should be:
- turning friction (depending on speed)
- Pneumatic deformation energy (pocus dependent on speed, at least for low speed)
- viscous friction (which goes with the square of speed)

Or not? Where does this cube go?

(*)
If you challenge me that the forces of friction do not represent a field, I leave with a skewer to make you rizzare the hair. Don't challenge me! :biggrin:

 

[Er Presidente]

Gotcha!
the work done is exactly the same. I go from Rome to Milan.
what changes between doing it in step of snail or at supersonic speed is the power necessary to do that job. . .
In fact, I require power to the engine, and the consumption that follows it is according to the specific consumption (motor terminal and how it is working) and the power supplied

True, I was wrong with the term that was to be mechanical energy and not just work, which is independent from time. I ask for venia.

say that, with a certain approximation, the power required for vehicle advancement varies "approximately" with speed cube. at equal vehicle and boundary conditions, of course. . .

at speeds below 100 km/h varies roughly with the speed square, as well as the cube (not for nothing all performance performances are declared to 90 km/h).

 

[Er Presidente]

the work done is less. the work is done against the field* (non-preservative) of the forces of friction. such field is function of speed.
the work in the conservative field (the gravitational one) we consider it null, considering points of departure and arrival at the same quota.

then there is also to say that from roma to milan the road is uphill, on the return is downhill... (Smile)


Why the cube?
forces should be:
- turning friction (depending on speed)
- Pneumatic deformation energy (pocus dependent on speed, at least for low speed)
- viscous friction (which goes with the square of speed)

Or not? Where does this cube go?

(*)
If you challenge me that the forces of friction do not represent a field, I leave with a skewer to make you rizzare the hair. Don't challenge me! :biggrin:

the cube is justified when aerodynamic resistance becomes preponderant.

 

[PierArg]

the work done is less. ...

(*)
If you challenge me that the forces of friction do not represent a field, I leave with a skewer to make you rizzare the hair. Don't challenge me! :biggrin:

fulvio launches the challenge glove :biggrin::tongue:


Maybe I missed something at work.
work is defined
♪
(I can't put vector signs...:smile

where the force is to advance the vehicle.
of course the force changes... but also the aerodynamics.
We must see what contribution is important.

 

[Fulvio Romano]

Maybe I missed something at work.
work is defineddl=f(v[o3])*ds= f(v[o3])*vdt(I can't put vector signs...:smile

where the force is to advance the vehicle.
of course the force changes... but also the aerodynamics.
We must see what contribution is important.

Right, so?
If you do the free body diagram on the vehicle, you will have that instant force on the vehicle itself is nothing. We also assume constant speed, so from the free body diagram the apparent forces disappear.

Now, you have in one verse the force of the engine that performs work, and on the other the resistant forces, whose resulting is equal and contrary to that of the engine.

from how you wrote the equation would seem that total work is equal to strength (constant) for speed (constant) for the integral of time. Obviously if you double the speed, and then half the length of integration, the result does not change.
if however you see in bold, f is function (increasing) of v... and the accounts return

fulvio launches the challenge glove :biggrin::tongue:

It's an old diatribe with physics theorists. they say that non-conservative forces do not constitute a field, because the force field is a vector field dependent solely on the position in space.

I argue that everything depends on what you mean by "space". if you mean the euclideous (etheadimensional) space, ok. but, we suppose to freeze a certain phenomenon. the set of motion acts of all points represent a vector space of an eulerian type that has the same mathematical characteristics of the euclideous one. so I can consider with "space" a 12-dimensional vector space where at each point I associate six euclide coordinates and six eulerians. in this space I can define a "field" of non-conservative forces.

we have supposed to "freeze the phenomenon", but this is not strictly necessary. As the question is purely cinematic and we do not talk about energy transformations, the concept of entropy is not applicable. if it is not defined the entropy does not even define the arrow of time, then "freeze" a future phenomenon is not anti-usal action, and therefore it is lawful in the context in which we are operating.

and if some theoretical physicist has something to say, we open a special thread! :finger:

 

[MBT]

the work done is less. the work is done against the field* (non-preservative) of the forces of friction. such field is function of speed. If I make the way in less time, regardless of the necessary power that will be greater, but I will find myself spending more energy to win greater forces, and then I do more work.
the work in the conservative field (the gravitational one) we consider it null, considering points of departure and arrival at the same quota.

Um, I have to digest this...
perhaps there is a misunderstanding in the terms used
I have understood "work" as what I need to go from to b and should be independent from the fate that is done at a speed or another. I mean by work the energy spent to move the body between two points having different potential.

Why the cube?
forces should be:
- turning friction (depending on speed)
- Pneumatic deformation energy (pocus dependent on speed, at least for low speed)
- viscous friction (which goes with the square of speed)

Or not? Where does this cube go?

at speeds below 100 km/h varies roughly with the speed square, as well as the cube (not for nothing all performance performances are declared to 90 km/h).

"the thermodynamic god is always worth, everywhere and anyway! also in a pisci@t@ in the middle of the desert losses are proportional to the square of speeds!" [cit.]
se le losses are proportional to the square of speeds, the power power spending must be proportional to the cube.... Am I wrong?? ?

to do fleas, in fact this is not true.
aerodynamic resistors are minimal at low speeds and are heard later, the frictions of the tires have their trend, the coefficients of friction volving "would" decrease with the increase of speeds etc...
I, who do not dare engineer and like practical calculations made on the tip of the fingers, arrived saying that the power needed to advance the vehicle varies "approximately" with the cube of speed... knowing that if I'm wrong, I'm mistaken for excess (and therefore I'm cautionary)

True, I was wrong with the term that was to be mechanical energy and not just work, which is independent from time. I ask for venia.

I was doubtful whether to report the error or not...