Hello, I did some research and a little study to understand the evolution of the conso of a thermal vehicle. Well I simplified a little (very little) ...
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http://www.ecolo.org/documents/document ... -elect.htm
We write:
One liter of gasoline can provide 10 thermal kWh. One could believe that a liter of gasoline would travel 100 km 100 km / h. But ... the maximum yields are from 23% (gasoline) to 28% (diesel). These yields are rarely achieved: cold engines, non-optimized diets. For urban driving, it is estimated that the actual yield is around 10%. The weakness of this real return is mainly due to accelerations.
In addition, the consumption stops decreasing below 60 km / h (because it is necessary that the engine turns ..). As a result, few vehicles consume less than 5 l / 100 km at 100 km / h and less than 7.5 l / 100 km in the city. The air conditioning is added to that (1 / 3 consumption in town, 1 / 6 on the road).
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Sure :
http://philippe.boursin.perso.sfr.fr/pdgmoteu.htm
We read: -Consumption specific:
The Cse is the mass of fuel (in grams) that the engine would consume to deliver a power of 1 kW for one hour (ie a job of 3600 kJ).
The specific consumption is calculated by dividing the hourly consumption by the effective power.
Cse (g / kW.h) = mc (g) / (P (kW) * t (hour))
with fuel density ρ (740 kg / m3) and volume of fuel consumed V
Cse (g / kW.h) = V (cm3) * ρ (g / cm3) * 3600 / (P (kW) * t (s))
Cse (g / ch.h) = V (cm3) * ρ (g / cm3) * 3600 / (P (ch) * t (s))
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Sure :
http://insee.fr/fr/themes/document.asp? ... f_id=18067
We read :
The graphical representation of kilometric consumption (L / 100km) as a function of speed is not a straight line, but a U-shaped curve, which has a minimum around 80 km / h for a light vehicle (70 km / h for a heavy weight). This speed of 80 km / h at which the kilometer consumption is minimal is therefore the energy optimum.
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The expertise of "danielj" :
Consumption in liters per hundred kilometers and liters per hour of a thermal engine vehicle.
As soon as a heat engine is running, with the vehicle stationary in neutral, it consumes fuel. This is the minimum consumption for the engine to work and overcome its own friction, maintain its temperature, drive accessories, camshaft, water pumps, oil pump, alternator, etc.
It is not possible to go below this minimum consumption in liters per hour.
This results in consumption in
liters to the hundred kilometers goes through a minimum when the stabilized speed varies from zero to maximum!
I show it to you for a particular case (but extensible in all the cases):
I start from two measures:
1) The consumption of the vehicle
90km / h (stabilized) is 5.5 liters / 100km (C).
2) Measured consumption of engine idling (vehicle stationary in neutral),
2 liters per hour. (measure of 0,5 l in 15mn)
1 ° -
to 90km / hto make a hundred km, we put (60 / 90) x100 = 66,66 mn.
I use for the minimum operation of the out-of-the-money engine to move the car forward: (2 / 60) x66,66 =
2,2 liters. (A)
The rest of consumption is therefore used to advance the vehicle
either: 5,5 - 2,2 =
3,28 liters. (B)
This consumption is mainly used to overcome the resistance of the air (and friction proportional to the speed
who are very weak and that I integrate with the resistance of the air as a first approximation).
The resistance of the air is proportional to the square of the speed.
2 ° -
to 70km / h, To know the consumption due to resistance of the air
at another speed it suffices to make the ratio of the squares of the velocities; for 70 km / h one will have: conso 2 / conso 1 = 70 squared / 90 squared = 4900 / 8100 = 0,605. Let 3,28 x 0,605 =
1,984 liters (B ')
Consumption for the minimum operation of the engine out of expense to move the car (A ') is calculated as before, time to make 100km = (60 / 70) x 100 = 85,71 mn; min conso x time per hundred km, that is (2 / 60) x 85,71 =
2,857 liters (A ').
Hence the consumption at 70 km / h of 1,984 + 2,857 = 4,841 liters per hundred km (C ').
3 ° -
at 50 km / hwe find in the same way:
4 liters (A '') and
1,010 liters (B '') {50 squared divided by 90 squared = 0,308 of coef x 3,28 = 1,010}.
the conso to 50 km / h is 1,010 + 4 = 5,010 liters to the hundred km (C '').
So higher than 70 km / h! ! !
4 ° -
at 30 km / hwe find in the same way:
6,666 liters (A '' ') and
0,364 liters (B '' ') {30 squared divided by 90 squared = 0,111 coef x 3,28 = 0,364}.
the conso to 30 km / h is 0,364 + 6,666 = 7,030 liters to the hundred km (C '' ').
So higher than 50 km / h! ! !
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Allure of the consumption curve in liters per hour (fcV):
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Allure of the consumption curve in liters per hundred kilometers (fcV):
WE HAVE A CONSO MINI FOR SOME SPEED (60 ... 80 km / h + -). CONSOTING IN LITERS PER HOUR INCREASES ABOVE AND INCREASES BELOW!
Other times the autojournal gave consumption tables 100km and I saw that there was this optimal speed!
a+