With gasoline prices as high as they are many people are concerned about vehicle efficiency. Other people who are concerned about their impact on the future of our climate care about vehicle efficiency as well. Where does the energy that we put into our cars actually go? And what is the overall efficiency of a car? Since I have pretty extensive data on my own vehicle, a 2005 Toyota Matrix XR, I will use it as an example in this week’s AskPablo.
We are all familiar with the standard measure of vehicle efficiency, mpg (or liters per 100 km in Europe). Miles per gallon, or the CO2 emissions derived from it does not show the whole picture. The drilling, pumping, transporting, and refining of petroleum products such as gasoline and diesel requires additional energy that we often overlook. By some estimates this “well to tank” phase adds 15-20% to the emissions/energy use. With that in mind let’s move on and explore where the energy goes once it gets in your tank.
First, my car must overcome aerodynamic drag in order to maintain a given velocity. For this example we will assume that I am trying to maintain a velocity of 65 mph since I am a good, law-abiding driver. Aside from my velocity, I will also need to know my vehicle’s coefficient of drag (0.32, ), my vehicle’s cross sectional area (roughly 2.5827 m^2), and the density of air (1.293 kg/m^3). By plugging all of this into the aerodynamic drag equation we get 451 newtons (101 pound-force). According to the Green Car Company you can decrease your fuel use by 20% by slowing down from 70 mph to 60 mph. Multiplying the coefficient of drag by the cross sectional area results in a good indicator of how aerodynamic a car is. My car scores 8.896 while the Honda Insight scores 5.10 and the Hummer H2 scores a pathetic 26.3 (values in ft^2).
The next important factor is your vehicle’s rolling resistance. To use the rolling resistance equation we need the mass of the vehicle (1,250 kg), gravity ( 9.807 m/s^2), the tire deformation (0.01 m), and half of the tire radius (0.2032 m). This gives us 123 newtons (27.6 pounds-force). The rolling resistance for a Hummer H2 is around 550 newtons.
So, to maintain a velocity of 65 mph the vehicle uses 574 newtons of force (129 pounds-force). To find out how much “work” is done (in Joules) over an hour of driving at this speed we multiply the force by the distance (in meters). So 574 newtons x 104,607 m = 60,044,418 Joules. Dividing this by the time (in seconds) gives us 16,679 W (60,044,418 J / 3,600 s), 16.68 kW. The specs for the car show a maximum power of 97 kW, so the car is using only 17% of its full potential to maintain a constant highway velocity. This is why car companies like Chrysler are now using cylinder deactivation so that you aren’t running all 8 cylinders while maintaining a constant speed.
But how much energy is contained in the fuel? Averaging about 28 mpg over one hour requires 2.32 gallons (65 miles / 28 mpg). Gasoline has an energy density of 130 MJ/gallon, so the energy contained in one hour’s worth of fuel is 301.6 MJ (130 MJ/gallon x 2.32 gallons), which is equal to 83.78 kW (301.6 MJ / 3,600 s). To find the tank to wheel efficiency we simply need to compare the power used (16.68 kW) with the potential power of the fuel used (83.78 kW) to get 19.9% efficiency (16.68 kW /83.78 kW).
To be continued next week…