Energy consumption of cars

The energy consumption of cars, either with internal combustion engine or with electric motor, is mainly due to the following three processes:

1. Air drag
2. Breaking
3. Accelerating

Less important, but not negligible, is the loss due to rolling resistance. The first process depends only on the size of the car, or more precisely on its effective cross section (surface area of its front) and is independent of its mass. The second and third processes depend on the mass of the car and through its mass on its engine. An electric motor weighs less in general than a combustion engine, but this is offset to some extent by the mass of the batteries, especially when these are old-fashioned lead-acid batteries. Application of modern lightweight Li-ion batteries gives the electric car an advantage in breaking and accelerating over the combustion-engine car. Another energy advantage of the electric car (and of hybrid cars) is the easy application of regenerative breaking, some (say about 50%) of the kinetic energy that would be lost in heating up the breaks can be re-funneled back into the batteries.

An important difference between electric- and combustion-engine cars is the thermodynamic efficiency in the generation of their propelling power. Electric power is usually generated in big (500 to 1000 MW) power stations fed by fossil fuels and operating at an efficiency of about 38%, which means that about 38% of the heat of combustion of the fuel (coal, natural gas, oil, etc.) is converted into electric energy. The relatively small combustion engines of cars, on the other hand, operate at an efficiency of around 25%.

Other energy losses—that are difficult to quantify, however—are in the production of gasoline (or other fuels used in combustion engines such as diesel), the transport of electricity from power station to electric outlet, and losses in the charging of the batteries of electric cars.

Numerical example

It is known that an average car runs 12 km (7.5 mile) on one 1 liter (0.26 gallon) of gasoline (28 mile/gallon). The energy content of gasoline is 10 kWhour (= 3600 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:

1. 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; breaking throws that kinetic energy away.
2. 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.
3. The rolling resistance force is proportional to the weight m_cg, where g ≈ 10 ms⁻² is the gravitational acceleration and m_c is the mass of the car.

If ρ 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 consumption P (energy consumed per unit of time) by the air drag is

$P_{\mathrm {drag} }={\frac {1}{2}}v^{3}A\rho .$

The power required for starting and stopping every d meter is

$P_{\mathrm {acc} }={\frac {1}{2}}v^{3}{\frac {m_{\mathrm {c} }}{d}}.$

And the power consumed by the tires rolling over the road is

$P_{\mathrm {rol} }=rvm_{\mathrm {c} }g,\;$

where r is the rolling resistance coefficient.

For a typical car: m_c = 1000 kg and A = 1 m²; the density of air ρ = 1.3 kg/m³. A typical value for the dimensionless variable r is 0.01. For freeway driving: d = 0 and v = 70 miles per hour (110km/h). This gives (with the factor 4 accounting for the thermodynamic efficiency):

$P_{\mathrm {drag} }+P_{\mathrm {rol} }=4\left({\frac {1}{2}}\,v^{3}\,A\,\rho +rvm_{\mathrm {c} }g\right)=86.4\;\mathrm {kW} .$

With an energy content of gasoline of 10 kWh/l this means that the car needs 8.6 liter to drive 110 km, which amounts to 12.8 km/l.

For city driving we take v = 50 km/h and d = 175 m, again the efficiency is 25%,

$P_{\mathrm {tot} }=4\left[{\frac {1}{2}}v^{3}{\Big (}{\frac {m_{\mathrm {c} }}{d}}+A\rho {\Big )}+rvm_{\mathrm {c} }g\right]=43\;\mathrm {kW} .$

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 87 kWh for a combustion car (this includes the 38% energy loss at the power station). The difference of 30kWh reflects the higher efficiency of the power station over the car engine.

If we assume that an electric car has regenerative breaking that channels back to the batteries 50% of the kinetic (breaking) 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, then

$P_{\mathrm {tot} }=1/(0.38)\left[{\frac {1}{2}}v^{3}{\big (}{\frac {1}{2}}{\frac {m_{\mathrm {c} }}{d}}+A\rho {\big )}+rvm_{\mathrm {c} }g\right]=18.3\;\mathrm {kW} ,$

which is almost a factor 2.5 more economical than a gasoline car. Clearly the regenerative breaking is important for this factor. Without it, the factor would be 0.38/0.25 = 1.5.

To drive 100 km with an electric car one needs beforehand to charge the battery with an amount of energy of about 50*0.38 = 19 kWh. If one draws from an electric outlet, say, 4 kW power—which at 220 volt is about 18 ampere—then it will take 4.75 hours to charge the battery with the amount of energy sufficient for a drive of 100 km. (It is assumed that the battery can stand an 18 ampere charging current). This long charging time exhibits one great bottleneck hampering the introduction of electric driving. Gasoline worth of 19 kWh is 2 liter, about half a gallon, which requires a loading time of a few seconds, not 4.75 hours.

Finally, it must be reiterated that it is assumed in this example that the electricity for the car is generated by a power station fed by fossil fuels. When in the future sufficient electricity is generated "greenly", a different comparison is called for.

(To be continued),

see David MacKay