View source for Energy consumption of cars

==Numerical example==
It is known that an average car runs 12 km (7.5 [[mile]]) on one liter (0.26 [[gallon]]) of gasoline (28 mile/gallon). The energy content of gasoline is 10 [[Kilowatt hour|kWh]] (= 36000 kJ) per liter. The question is: do the effects mentioned in the previous section account for this gasoline/energy consumption?  

In the next section equations will be derived on basis of the following assumptions:
# The driver accelerates rapidly up to a cruising speed ''v'', and maintains that speed for a distance ''d'', which is the distance between traffic lights (or other events requiring a full stop and acceleration back to ''v'').  When the driver stops he slams on the brakes turning all his kinetic energy into heating  the brakes. Then he accelerates back up to his cruising speed, ''v''. This acceleration gives the  car kinetic energy; braking throws that kinetic energy away. 
# While the car is moving, it pushes a volume of air to a speed ''v''.  This costs the energy that earlier is referred to as air drag. The power required is proportional to the mass of the air that is moved, i.e., it is proportional to  the volume times the density.
# The rolling resistance force is proportional to the weight ''m''<sub>c</sub>''g'', where ''g'' &asymp; 10 ms<sup>&minus;2</sup> is the [[Acceleration due to gravity|gravitational acceleration]] and ''m''<sub>c</sub> is the mass of the car.  

If &rho; is the density of air and ''A'' is the effective cross section of the car (the base of the volume of air that is pushed by the car), one can derive that the power  ''P'' (energy consumed per unit of time) of the air drag is
:<math>
P_\mathrm{drag} = \frac{1}{2} v^3\; A \rho.
</math>
The power required for starting and stopping every ''d'' meter is
:<math>
P_\mathrm{acc} = \frac{1}{2} v^3\; \frac{m_\mathrm{c}}{d}.
</math>
And the power consumed by the tires rolling over the road is
:<math>
P_\mathrm{rol} =  r v  m_\mathrm{c} g, \;
</math>
where ''r'' is the rolling resistance coefficient.

When filling out numbers it is most convenient to use [[SI]] units and to express all relevant quantities in meter, second, and kilogram. The answer then appears automatically in watt (W).

For a typical car: ''m''<sub>c</sub> = 1000 kg and ''A'' = 1 m<sup>2</sup>; the density of air  &rho; = 1.3 kg/m<sup>3</sup>. A typical value for the dimensionless variable ''r'' is 0.01.  For freeway driving: ''d'' &asymp; ∞ and ''v'' = 110km/h (70 miles per hour), which gives ''v'' = 30.56 m/s. In total,  (with  the factor 4 accounting for the thermodynamic efficiency): 
:<math>
P_\mathrm{drag}+P_\mathrm{rol} = 4\left( \frac{1}{2}\,v^3\,A\,\rho + r v  m_\mathrm{c} g\right)
=  86,395\;\mathrm{W} \approx  86.4\; \mathrm{kW},
</math>
of which 12.2 kW is due to the rolling resistance.  With an energy content of gasoline of 10 kWh/l, this implies that the car needs 8.6 liter to drive 110 km, which amounts to 12.8 km/l. This ''a priori'' computation agrees nicely with the empirical fact that a car drives about 12 km on a liter of gasoline.

For city driving we take ''v'' = 50 km/h = 13.89 m/s  and ''d'' = 175 m, again the efficiency is 25%,
:<math>
P_\mathrm{tot} = 4 \left[ \frac{1}{2} v^3 \Big(\frac{m_\mathrm{c}}{d}+ A\rho\Big) + r v  m_\mathrm{c} g\right] = 43,141 \;\mathrm{W} \approx  43\; \mathrm{kW} .
</math>
The car needs 4.3 l to drive 50 km, which amounts to 11.6 km/l.

When you drive an electric car  of the same mass (including batteries) for an hour on the freeway with 110 km/h you spend 57 kWh as compared to 86.4 kWh for a combustion car (this includes the 38% energy loss at the power station). The difference of 29.4 kWh reflects the fact that the efficiency of the power station is higher than of the car's combustion engine.   

If we assume that an electric car has regenerative braking that channels back to the batteries 50% of the kinetic (braking) energy, then in city driving the electric car is considerably more economical than the gasoline car. Take a car (including batteries) of 1000 kg, drive 50 km/h and take again ''d'' = 175 m, ''A'' = 1 m<sup>2</sup>, then
:<math>
P_\mathrm{tot} = 1/(0.38) \left[ \frac{1}{2} v^3 \big(\frac{1}{2}\frac{m_\mathrm{c}}{d}+ A\rho\big) + r v  m_\mathrm{c} g\right] 
 = 18.3 \;\mathrm{kW} ,
</math>
which is almost a factor 2.5 more economical than a gasoline car. Clearly the regenerative braking is important for this factor. Without it, the factor would be 0.38/0.25 = 1.5. 

Driving  an electric car without recharging for 100 km on a freeway, one spends 57*(100/110) = 52 kWh. Since this number includes the efficiency of the power station, we must correct: 52&times;0.38 = 20 kWh and it follows that the battery must contain at least 20 kWh before the 100 km journey is undertaken. If the charging of the battery draws, say, 4 kW power—which at 220 [[volt]] is about 18 [[ampere]]—then it will take ''five hours'' to charge the battery with the amount of energy sufficient for a freeway drive of 100 km. (It is assumed that the battery can stand an 18 ampere charging current). This long charging time exhibits one of the greatest bottlenecks in the introduction of electric driving. Gasoline good for 100 km is  8 liter (2 gallons), which requires a loading time of less than a minute, which is far less than five hours.