Energy consumption of cars

The energy consumption of cars, irrespective of whether they are driven by an internal combustion engine or by an electric motor, is mainly due to the following three physical processes: The third effect is less important than the other two; it depends on the weight (mass times gravitational pull) of the car. 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 power (energy per time unit) due to the first and second process  depends on the speed cubed of the car, while the consumption due the third process is proportional to speed. These dependences of power usage on speed are physical laws that hold for any vehicle, independent of its means of propulsion. The power consumption due to the three processes holds alike for electrically and gasoline driven automobiles. 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. Hence application of modern lightweight Li-ion batteries decreases the energy lost by braking and accelerating. (Apart from the additional advantage that Li-ion batteries have a larger energy content per battery mass than lead-acid batteries).
 * 1) Air drag
 * 2) Braking and accelerating
 * 3) Rolling resistance

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

Another important—but frequently overlooked—difference between electric and combustion-engine cars 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%. Without regenerative breaking, this difference in thermodynamic efficiency is the only energetic advantage of an electric car over a combustion-engine car; it makes the electric car about 50% more efficient. When one is interested in a comparison of the "carbon footprint"  (overall emission of CO2) of the electric versus the combustion-engine car, the efficiency of fossil-fuel fed power stations must obviously  enter the equation. (That is, as long as the present situation persists that the bulk of the electric power is generated by combustion of 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 US gallon) of gasoline (28 mile/gallon). The energy content of gasoline is roughly 10 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:
 * 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 mcg, where g &asymp; 10 ms&minus;2 is the gravitational acceleration and mc 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

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} = r v  m_\mathrm{c} g, \; $$ 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: mc = 1000 kg and A = 1 m2; the density of air &rho; = 1.3 kg/m3. A typical value for the dimensionless variable r is 0.01. For freeway driving: d &rarr; ∞ and v = 110 km/h (70 miles per hour), which gives v = 30.56 m/s. In total, (with  the factor 4 accounting for the thermodynamic efficiency):

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}, $$ 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%,

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}. $$ The car needs 4.3 l to drive 50 km, which amounts to 11.6 km/l.

Electric cars
When one drives an electric car of the same mass (1000 kg including batteries) for an hour on the freeway with 110 km/h one spends 57 kWh as compared to 86.4 kWh for a combustion car. The difference of 29.4 kWh is due to the 38% efficiency of the power station versus the 25% efficiency of a small 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 m2, 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) + r v m_\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.

Driving an electric car without recharging for 100 km on a freeway, one spends 57*(100/110) = 52 kWh. Since this number is not corrected for the energy loss of the power station, we compute 52&times;0.38 = 20 kWh. In other words, an electric car battery must contain at least 20 kWh before a non-stop journey of 100 km is undertaken. If the charging of the battery is performed at home with 4 kW—which at 230 volt (common in European homes) is about 17 ampere—it will take five hours to charge the battery with the 20 kWh needed for a freeway drive of 100 km. (It is assumed that the battery can stand an 17 ampere charging current). This long charging time forms 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 a few seconds. The energy content of gasoline being so large, the introduction of electric driving will be an uphill battle (provided gasoline remains available at an affordable price).

The difference between loading times of gasoline and electric cars can also be looked at from a slightly different angle. The energy content of one US gallon (3.8 liter) of gasoline is about 36 kWh. This means that when an electric car is to be charged in one hour time with the same amount of energy as contained in one gallon of gasoline, the charging station has to deliver 36 kW of power. Using W = I&sdot;V (power W is equal to strength I of electric current times voltage V), one finds that at 360 volt (a common voltage for power hungry consumer appliances) the current I to be maintained during the hour of charging must be 36,000/360 = 100 ampere. Most non-industrial (home) electric installations do not allow more than, say, 20 to 25 ampere, so that either charging times must be increased by a factor four to five (four to five hours to load at home the energy-equivalent of one gallon of gasoline) or the voltage and current strength must be increased, which requires a heavier electric installation than is now common in most homes. Note that this argument depends only on laws of nature: technological improvements in batteries and electricity generation do not influence the numbers. It is a simple fact that decrease in charging times requires increase in electric power: high voltages (with batteries that can stand it) and high electric currents (with electric installations that can stand it) are needed. For example, loading the energy-equivalent of one gallon of gasoline at a 100 kW outlet (e.g., 100 ampere at 1 kilovolt) takes 22 minutes (and gives the energy for about 100 mile of non-stop freeway driving at 65 mile/h).

Clearly, if/when electric driving becomes common, an infrastructure of many electric power stations with large-capacity power lines must be constructed and&mdash;unless electric transport by superconductive lines becomes possible&mdash;loss of energy due to Ohm's heating of the lines (proportional to the square of current strengths) will be non-negligible.

Equations
Energy is force times distance, E = F &Delta;x. Power is energy per time interval, P = E/&Delta;t. Speed is distance covered per time interval, v = &Delta;x/&Delta;t. 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 F v dt = m \int_0^t \frac{dv}{dt} v dt =  m \int_{v(0)}^{v(t)} v dv = \frac{1}{2}m v(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 &frac12;mcv2 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 &Delta;t = 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 &Delta;t seconds the car pushes the air in front of the car over a distance &Delta;x. The volume of air pushed is &Delta;V = A &times; &Delta;x and the mass Mair of air pushed in a time interval &Delta;t is &Delta;V &times; &rho;, where &rho; is the density of air. This volume of air gets the speed v of the car and hence the kinetic energy &frac12; v2Mair 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 = &Delta;x/&Delta;t.

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} = r v m_\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.