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AskPablo: Well to Wheel Efficiency Part I

| Monday April 16th, 2007 | 4 Comments

hummer.jpgWith 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.
aerodynamic%20drag%20equation.jpgFirst, 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…
Pablo Päster
Sustainability Engineer


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  • http://www.liquididea.com William Hertling

    Hi Pablo,
    First off, let me commend you on an excellent job with these calculations. I think they are very educational in that the various elements of the calculations help you see how different aspects of driving style and vehicle choice affect vehicle efficiency.
    However, the point of driving anywhere is generally to get you from Point A to Point B (as you also point out in next week’s column.
    But note that the purpose is to get YOU the driver (or the driver and passengers) from Point A to Point B, and not the vehicle. We rarely drive just to move our car from one place to another.
    So to get a true efficiency calculation, we need to add in the factor of the driver’s weight compared to the vehicle’s weight. I couldn’t find gross vehicle weight for the Matrix XR online, but let’s assume that it is 2500 pounds. Then let’s assume that the average driver is 175 pounds and that, for the sake of this calculation the overwhelming majority of trips are single occupancy trips. With those three assumptions, we see that the driver makes up only 7% of the total mass being moved from one place to another.
    So I think the true efficiency of your vehicle is closer to 19.9% * 7% = 1.3% total system efficiency.

  • http://www.et3.com/ Daryl Oster

    Actually any transportation system that uses energy has zero transportation energy efficiency!
    The reason is that for true efficiency measure one must divide into the energy used by the best theoretical alternative — and the best alternative uses ZERO ENERGY for transportation.
    The earth is one such approximation — all 6+ billion humans have spent their entire lives moving at 67,000mph in orbit around the sun. And our solar system orbits the milky way galaxy at 200,000mph, and our galaxy orbits the apparent center of the known universe at about 400,000mph.
    Therefore the average human travels more than 250 billion miles in their life time yet uses no energy to do so! — by comparison, any earth bound vehicle has almost zero TRANSPORTATION efficiency!
    So we can only compare energy USE (not efficiency) to travel a given distance. Of course thermal efficiency effects energy use, as does aerodynamic drag, and rolling resistance. And it is proven that both of these terms can be practically reduced to almost zero in the case of Evacuated Tube Transport (ETT) tm.

  • Jonathan Goldman

    Sorry Pablo…energy is in kWh, not kW.
    Care to recalculate?

  • Jonathan Goldman

    Sorry Pablo…energy is in kWh, not kW.
    Care to recalculate?