Energy consumption of cars: Difference between revisions
imported>Paul Wormer No edit summary |
imported>Paul Wormer No edit summary |
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see [http://www.withouthotair.com/download.html David MacKay] | see [http://www.withouthotair.com/download.html David MacKay] | ||
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==Numerical example== | |||
It is known that an average car goes 12 km (7.5 mile) on one 1 liter (0.26 gallon) of gasoline (28 mile/gallon). The energy content of gasoline is 10 [[Watt|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 a formula 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 start). 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 friction. If ρ is the density of air and ''m''<sub>c</sub> is the total mass of the car, one can derive that the power consumption ''P'' (energy consumed per unit of time) is | |||
:<math> | |||
P = \frac{1}{2} v^3\left(\frac{m_\mathrm{c}}{d} + A \rho\right), | |||
</math> | |||
where ''A'' is the effective cross section of the car (the base of the volume of air that is pushed by the car). | |||
For a typical car: ''m''<sub>c</sub> = 1000 kg and ''A'' = 1 m<sup>2</sup>; the density of air ρ = 1.3 kg/m<sup>3</sup>. For freeway driving: ''d'' = 0 and ''v'' = 70 miles per hour = 110km/h = 31 m/s. This gives (with a factor 4 accounting for the thermodynamic efficiency): | |||
:<math> | |||
P = 2v^3A\rho = 2\cdot (31)^3 \cdot 1 \cdot 1.3\; \mathrm{W} = 77456\; \mathrm{W} \approx 80\; \mathrm{kW}. | |||
</math> | |||
For an electric car (40% efficiency) of the same weight (including batteries) the corresponding number is 50 kW. If you drive your car for an hour on the freeway, you cover 110 km and spend 80 kWh which costs you 8 l, or 14 km/l. | |||
For city driving we take ''d'' = 750 m, ''v'' = 50 km/hour = 14 m/s and obtain | |||
:<math> | |||
P = 2 (14)^3 \left( \frac{1000}{750} + 1\dot 1.3\right) = 14624 \mathrm{W} \approx 15 \mathrm{kW}. | |||
<math> | |||
This amounts to 15 kWh for 50 km | |||
From MacKay: | From MacKay: | ||
Revision as of 07:23, 20 December 2009
The energy consumption of cars, either with internal combustion engine or with electric motor, is mainly due to the following three processes:
- Friction with air
- Breaking
- Accelerating
Less important is the loss due to rolling resistance. The first process depends only on the size of the car and is independent of its weight (mass). The second and third process depend on the weight of the car and through its weight on its engine. An electric motor is in general lighter than a combustion engine, but this is offset to some extent by the weight 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 gasoline car. Another energy advantage of the electric car is the easy application of regenerative breaking, some of the kinetic energy that would be lost in heating up the breaks can be re-funneled into the batteries.
An important difference between electric- and combustion-engine cars is the thermodynamic efficiency of 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 40%, which means that about 40% of the heat of combustion of the fuel (coal, natural gas, oil, etc.) is converted into electric energy. The relative small combustion engines of cars, on the other hand, operate at an efficiency of around 25%.
Other energy losses—but that are difficult to quantify—are in the production of gasoline (or other fuels such as diesel used in combustion engines), the transport of electricity from power station to electric outlet, and losses in the charging of the batteries of electric cars.
(To be continued),
see David MacKay