How to calculate fuel to accelerate a vehicle

  • Thread starter IMK
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  • #1
IMK
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Homework Statement


Hello,
This is not a home work question but as I am not a math or physics major I need some help moving this problem forward.

I am trying to calculate how much fuel (petrol/diesel/LPG) it takes to accelerate a vehicle and well to cut a long story sort I seem to be getting my units mixed up also whether I should us small or big calories. Lastly is the incline and issue that I also need to consider and how do I apply it please?

So was wondering if someone could get me off in the right direction please?

Many thanks in advance



Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rock.freak667
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What measurements do you have to work with?

Also if you are confused with the units, convert all your units to SI units. So if you convert your mass from grams to kilograms (your mass is now in kg), you use big calories (Cal).
 
  • #3
IMK
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Hello rock.freak667 and many thanks for your reply.

I am sorry for my late response but has been busy day here and I just got home.

As far as units are concerned as follows are the variables I would be working with:
The Acceleration in g. Say 0.1g
The Acceleration duration in seconds. For 3 seconds
(Although I guess I could use a single value for the above two as n G for a second.)
The Inclination in degrees. (I guess positive for up and negative for down)
The Fuel type (Petrol, Diesel or LPG) I guess this could be in calories as I think this would be a good commons unit.

What I am looking for a result is the amount of the fuel that would be used (in mL) assuming the system is 100% efficient.

I guess later I could get some real system efficiency data and apply it to the result for completeness.

I did have a go at this a while ago but as I am not a math or physics major I have found myself rapidly bogged down in trying to figure how to approach the problem.

Any help would really be appreciated.

Many thanks in advance.
 
  • #4
rock.freak667
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About the incline, do you know the length or height? The coefficient of friction?

I am asking since well the fuel will have a caloric value, so you will need to find some way to relate everything back to energy.

EDIT: I also may not be able to give you a definitive answer to arrive at the correct method. But I will try as best as I can.

EDIT2: Do you know the initial position of the car?
 
Last edited:
  • #5
IMK
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I guess to keep it simple for the time being and say the the system is frictionless and wind speed or air resistance also plays no part. As for the length or height I think we could just use the acceleration duration on a constant incline.

???
 
  • #6
rock.freak667
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I guess to keep it simple for the time being and say the the system is frictionless and wind speed or air resistance also plays no part. As for the length or height I think we could just use the acceleration duration on a constant incline.

???

Ah in that case you could try to find the change in height and equate that with the mass of fuel burnt (I think they give you the caloric value of fuel in J/g or kJ/kg something so?)
 
  • #7
IMK
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Ah in that case you could try to find the change in height and equate that with the mass of fuel burnt (I think they give you the caloric value of fuel in J/g or kJ/kg something so?)

Again many thanks for the input,
It is late here and I must go see the sandman, if you have more on the subject do please post.
 
  • #8
rock.freak667
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Again many thanks for the input,
It is late here and I must go see the sandman, if you have more on the subject do please post.

Well what I posted, in my head, should give you the energy used in moving down a certain distance.
 
  • #9
IMK
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Hello rock.freak667

So I have had another look at this and so it looks like it comes down to joules to find the work done per second? But how do I get the angle of inclination in? as work (J) done going up must be greater than work (J) done going down

J=N·m=(m·2)·(kg·(s·2)) but event this does not look right as there is no acceleration?

Many thanks
 
  • #10
rock.freak667
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Hello rock.freak667

So I have had another look at this and so it looks like it comes down to joules to find the work done per second? But how do I get the angle of inclination in? as work (J) done going up must be greater than work (J) done going down

J=N·m=(m·2)·(kg·(s·2)) but event this does not look right as there is no acceleration?

Many thanks


You don't need to directly relate the units. You have the acceleration so you can get the force and if you have the force and velocity, you can get the power. But what I was getting at was if you get the change in vertical height, you can get the energy needed to move that distance.
 
  • #11
IMK
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You don't need to directly relate the units. You have the acceleration so you can get the force and if you have the force and velocity, you can get the power. But what I was getting at was if you get the change in vertical height, you can get the energy needed to move that distance.


But what if there is no change in vertical height?
 
  • #12
rock.freak667
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But what if there is no change in vertical height?

Remember, using trigs, you can relate the length of the incline, to a change in height. The length of the incline would be the hypotenuse of a right angled triangle.
 
  • #13
IMK
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Remember, using trigs, you can relate the length of the incline, to a change in height. The length of the incline would be the hypotenuse of a right angled triangle.

I mist something as.
force = mass x acceleration
work = force x distance (distance is a function of acceleration and time)

So we have work. I guess somehow I can relate work to Energy Required or Calories to do the work these Calories I guess I can relate to the fuel type.

But I still don’t see how to apply inclination to the above?
Although I think I am getting closer to a solution than I have ever been.

Again many thanks
 
  • #14
rock.freak667
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I mist something as.
force = mass x acceleration
work = force x distance (distance is a function of acceleration and time)

So we have work. I guess somehow I can relate work to Energy Required or Calories to do the work these Calories I guess I can relate to the fuel type.

But I still don’t see how to apply inclination to the above?
Although I think I am getting closer to a solution than I have ever been.

Again many thanks

Oh well if you want to get the work done like that, then it would work fine. My method did not really include forces and such. But that is a better way to go.

I assumed your caloric value of the fuel was measured in kJ/kg. So if you wanted to get the mass, you'd need to do something like

Mass of fuel*Caloric value of fuel = Work done by car in moving distance.
 

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