Calculating how a car consumes energy (gasoline)

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The discussion focuses on understanding how a car consumes gasoline, particularly during acceleration and braking. The original poster calculated that with 1 liter of gasoline, their car could accelerate from 0 to 40 km/h and stop 60 times, estimating engine efficiency at 20%. However, they were advised that their calculations might overlook additional factors like rolling resistance and drivetrain friction, which also affect fuel consumption. The conversation highlights the complexity of accurately measuring fuel efficiency, emphasizing that engine efficiency varies based on conditions and that more power is needed for acceleration than simple kinetic energy calculations suggest. The thread concludes with suggestions for further resources and discussions on fuel economy.
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Hello everyone!

i am traying to understan how my car spend the gasoline. First of all, i supoused that the main way that my car spend energy is when it have to brake the inerce. for example, when you are in a traffic ligth, you are at 0 km/h and you have to acelérate to be at 40 km/h. that is easy to calculate whit the cinetic energy equation.

But for my surprise, this energy is very low, my results were that whit 1 liter of gasoline, you can brake to 0km/h and acelerate to 40 km/h, by 60 times, its crazy.(i supused that the efficience of my engine to convert the energy of the gasoline is 20%).

After of that, i think in the wind resistence. I have make the calculous and for a velocity of 40 km/h whit a good shape car, the energy is even lower that the before case.

I dont know if i am makeing a mistake, or if i cant recognise a important force.

pd: im from Argentina, im not a good english speaker, i hope that you can understand me :)
good week!
Juli
 
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Can you show your calculations? That will help us understand what factors you are consdiring, and perhaps not considering.

Without even considering energy conversion rates, it seems to me that a small car can go as far as 15-25 kilometres on a litre of gas. That's easily enough to start, get up to 40km/h and stop 60 times in a row, provided you don't waste any distance actually cruising.

If it takes, say, 25m to get up to speed and another 25m to stop again, you could do that in a three kilometre stretch of road.

Why do you think your numbers are off?
 
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DaveC426913 said:
Can you show your calculations? That will help us understand what factors you are consdiring, and perhaps not considering.

Without even considering energy conversion rates, it seems to me that a small car can go as far as 15-25 kilometres on a litre of gas. That's easily enough to start, get up to 40km/h and stop 60 times in a row, provided you don't waste any distance actually cruising.

If it takes, say, 25m to get up to speed and another 25m to stop again, you could do that in a three kilometre stretch of road.

Why do you think your numbers are off?
Hi Dave! thanks for replay!
of course. To the first analysis, i used DEc=1/2 * m * (Vf^2-Vi^2). to the mass, i estimated 2000kg, inicial velocity to 0 and final velocity to 40 km/h (11.1 m/s). Then i setched the specific energy of gasoline (34.78 MJ/Lt) and aplyed the engine efficience (20%). With all, i can found how many liters ofc gasoline i need to move my car from 0 to 40 km/h. the result of it is liters to move mi car= 0.0177L (if you multiplicate it for 60 times, you will got 1 liter aproximaly)
To de air resistence analysis, i used Fd=1/2 * Cd * p * v^2 * A. i supose to my car move at 40 km/h for 100 m. i considered that Cd = 0.31 (my car is similar to toyota corolla), p(air density)=1.2 kg/m3 , A=1.9 m2 (i take the heigth and width and multiply them). whit this i calculated the resistence force, and then whit W=F*x i calculated the work of the force along the 100m. the result of it is W=-4320 J

i think that because to go to my work, i have 10km by city, so i probably do the action of break and acelerate 60 times. i have calculated how many liters of gasoline my car spend, and with one liter i can do 12 km aproximaly
 
You're missing other types of friction: rolling resistance in the wheels and friction in the drivetrain.
 
russ_watters said:
You're missing other types of friction: rolling resistance in the wheels and friction in the drivetrain.
i thinked that thouse things are contemplate in the efficience of the engine. i can decrease that value and see what it produce.
but i would like to have a verificated value of efficience of the engine in the wheel
 
jgassmann said:
i am traying to understan how my car spend the gasoline.
This is a complicated subject. The engine efficiency ranges from zero at idle, to highest at the best efficiency point. This is a thread that discusses engine efficiency and has charts for many engines: https://ecomodder.com/forum/showthread.php/bsfc-chart-thread-post-em-if-you-got-1466.html.

And here is another thread that discusses how to measure rolling friction and air resistance: https://ecomodder.com/forum/showthread.php/coastdown-test-06-gmc-canyon-20405.html. There is also some discussion of calculating gas mileage from the drag measurements.

The fuel used to accelerate from a stop will be more than a simple calculation based on engine efficiency and kinetic energy. When the car is stopped, the engine is burning fuel and not doing any work. At the start of movement, the engine is burning more fuel and the vehicle is not yet moving. The total efficiency is zero when stopped or just starting to move.

If you get even more serious about gas mileage, here is a thread that about the effect of certain changes on gas mileage. Working backward from the changes in gas mileage will allow you to estimate some of the factors that cause fuel to be used: https://ecomodder.com/forum/showthread.php/modding-06-gmc-canyon-17070.html.
 
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With the energy density of fuel ##ED_f##, another way would be to look at your car power ##P##, knowing your fuel consumption ##BSFC##, vehicle speed ##v##, and engine efficiency ##\eta_e##:
$$P = BSFC \times \eta_e \times ED_f \times \frac{v}{3600\ s/h}$$
$$P = \frac{10\ L}{100\ km} \times 0.20 \times 34780\ kJ/L \times \frac{40\ km/h}{3600\ s/h} = 7.7\ kW$$
Which means that a car consuming ##10\ L/100\ km## at ##40\ km/h## produces ##7.7\ kW## (or about ##10\ hp##). A snowblower engine would be sufficient to do the job. Of course, increasing to a constant speed of ##100\ km/h## will require 2.5 times that power.

If you want to accelerate at those speeds, it takes even more power. For example, when at ##40\ km/h##, suddenly accelerating at ##0.3g## for your ##2000\ kg## car will require:
$$P = mav = 2000\ kg \times (0.3 \times 9.81\ m/s^2) \times 40\ km/h \left[\times \frac{1\ m/s}{3.6\ km/h} \times \frac{1\ kW}{1000\ W}\right] = 65.4\ kW\ [\approx 88\ hp]$$
This is in addition to the power calculated above. So it takes a lot more power to accelerate.
 
jrmichler said:
If you get even more serious about gas mileage, here is a thread that about the effect of certain changes on gas mileage.
I once did an experiment to test how much my A/C affected my kilometerage. It dropped it by at least 10%.
 
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jgassmann said:
pd: im from Argentina, im not a good english speaker, i hope that you can understand me :)
good week!
Juli
Welcome, Juli!

Please, see:
https://es.m.wikipedia.org/wiki/Economía_de_combustible_en_automóviles

https://en.m.wikipedia.org/wiki/Fuel_economy_in_automobiles#Energy_considerations

1779px-Energy_flows_in_car.svg.png
 
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