Energy consumption of cars

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

1. Air drag
2. Braking and accelerating
3. Rolling resistance

The third effect is less important than the other two; it depends on the weight (mass times gravitational pull) of the car and is proportional to speed. 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 process depends on the mass of the car and through its mass on its kind of engine. The consumption of energy (per time unit) of the first and second process depends on the speed cubed of the car.

An electric motor weighs less in general than a combustion engine, but this is offset to large extent by the mass of the batteries, especially when these are old-fashioned lead-acid batteries. Application of modern lightweight Li-ion batteries is advantageous for braking and accelerating, apart from the additional advantage that Li-ion batteries have a larger energy content (run a longer distances between recharges) per battery mass than lead-acid batteries.

A crucial energy advantage of electric (and hybrid) cars over combustion-engine cars is the easy application of regenerative braking: kinetic energy that would be lost in heating up the brakes can easily be re-channeled into the batteries.

An important difference between electric and combustion-engine cars—one that is frequently overlooked—is the thermodynamic efficiency of the generation of their motive power. Electric energy 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%. When one is interested in a comparison of the "carbon footprint" (emission of CO₂) of the electric versus the combustion-engine car, the efficiency of fossil-fuel fed power stations must obviously be included in the equation. That is, as long as the present situation persists that the bulk of the electricity originates from fossil fuels.

Other energy losses—that are difficult to quantify—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 outlets, 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 liter (0.26 gallon) of gasoline (28 mile/gallon). The energy content of gasoline is 10 kWhour (= 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:

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; braking 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 P (energy consumed per unit of time) of 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):

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

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.

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 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, 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 braking is important for this factor. Without it, the factor would be 0.38/0.25 = 1.5.

To drive, without recharging, 100 km with an electric car, the battery needs to contain at least 50×0.38 = 19 kWh. If the charging draws, say, 4 kW power—which at 220 volt is about 18 ampere—then it will take close to five 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 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, not five 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.

Equations

Energy is force times distance, E = F Δx. Power is energy per time interval, P = E/Δt. Speed is distance covered per time interval, v = Δx/Δt. Hence power is force times speed, P = F v. Integration of power over a time interval gives the kinetic energy increase,

$E_{\mathrm {kin} }=\int _{0}^{t}Fvdt=m\int _{0}^{t}{\frac {dv}{dt}}vdt=m\int _{v(0)}^{v(t)}vdv={\frac {1}{2}}mv(t)^{2},$

where Newton's second law F = m dv/dt is used and it is assumed that speed at time zero is zero, v(0) = 0.

The amount ½m_cv² of kinetic energy must be transmitted to a car every time it has stopped and accelerates again to cruising speed v. How often does this happen per unit of time? A car driving at speed v (m/s) covers a distance d (m) in Δt = d / v seconds. Supposing that the car stops every d meter, the power (energy per time) spent by the car is

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

With regard to air resistance: Assume that in time Δt seconds the car pushes the air in front of the car over a distance Δx. The volume of air pushed is ΔV = A × Δx and the mass M_air of air pushed per time interval Δt is ΔV × ρ, where ρ is the density of air. This volume of air gets the speed v of the car and hence the kinetic energy ½ v²M is conveyed to the air and the power is

$P_{\mathrm {drag} }={\frac {1}{2}}v^{2}{\frac {M_{\mathrm {air} }}{\Delta t}}={\frac {1}{2}}v^{2}A\rho {\frac {\Delta x}{\Delta t}}={\frac {1}{2}}v^{3}A\rho$

where it is used that v = Δx/Δt.

Finally, the rolling resistance is simply proportional to the graviational force pulling the car on the road. The power is this force times the speed of the car. The dimensionless proportionality constant is denoted by t. Hence

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

External link

This article rests to a large extent on a chapter from the book Sustainable Energy – without the hot air by David J.C. MacKay. A copy can be bought as hardcover, paperback, or pdf file. It can be downloaded free of charge from David MacKay's site.