We know the resistive forces a vehicle encounters while driving down the road. We discussed the details in our chapter on dynamics with variable forces. Given this knowledge and our understanding of work and energy, we can very easily predict the fuel economy of the vehicle in steady state driving conditions. Steady state means conditions are not changing. In the case of a vehicle it would imply a level road (no hills) and constant speed travel. This is representative of highway cruising. [br][br]When we charge a battery and subsequently draw current from it, we ultimately draw energy from it. When we fill a tank of gasoline, we are buying the gasoline and burning it in the engine for its energy content. It turns out that one gallon of gasoline (the non SI unit by which it is sold in the US) contains 1.2x10[sup]8[/sup]J (joules) of energy. That's 120 million joules per gallon of gasoline. That energy allows your car to pull you along the road and up hills, etc. In other words that energy does work against the resistive forces that the vehicle encounters. One gallon of diesel fuel contains more energy per gallon than gasoline. It contains 1.36x10[sup]8[/sup]J/gal. This fact alone hints to one reason why diesel vehicles tend to have better fuel economy.[br][br][b]Thermodynamic efficiency[/b] is the fraction of the heat of combustion (energy from the gasoline burning) that is actually converted to motion going down the road, where the rest of the energy becomes engine heat and exhaust gas heat. The symbol used for thermodynamic efficiency is the Greek lower case eta, [math]\eta.[/math] For gasoline engines, [math]\eta_{gasoline}=0.25-0.30.[/math] For diesel fuel, due to several factors related to burn characteristics of the fuel and the differing engine designs, [math]\eta_{diesel}=0.40-0.50.[/math] (Please note that those are ranges, and not subtraction.) [br][br]So diesel fuel contains more energy per gallon, and more of it is converted to usable energy as you drive down the road. Taking these two factors into consideration, it should not be surprising that diesel power tends to increase efficiency (MPG) by as much as 50% over a gasoline counterpart. [br][br]There are two important things to note here: That the majority of the energy from fuel (that which is not used to propel you down the street) is lost as heat (70-75% for gasoline and 50-60% for diesel), and that small variations in thermodynamic efficiency have large impacts on fuel economy. For instance, going from 30% to 45% percent efficiency leads to a 50% increase in usable energy and a corresponding 50% increase in fuel economy. That's because 45%/30%=1.5.[br][br]The fact that so much of the energy of combustion is lost to heat has led engine makers and automotive companies to try to capture and reuse some of that heat. One such idea which has not made it to production vehicles is to run a little steam turbine off the heat. Another more ordinary use of heat is where Toyota uses the exhaust heat in its Prius line to heat the cabin in cold weather - something which other manufacturers accomplish through engine coolant heat. While both are accomplishing the same thing once the engine is up to operating temperatures, Toyota's solution is much better during the initial minutes of running the engine during which exhaust gases are already hot but engine coolant is still cold.

There is a simple relationship between fuel economy - which we tend in this country to measure in miles per gallon of fuel - and the resistive forces. It follows directly from our work definition, [math]W=\oint\vec{F}\cdot\vec{ds}.[/math] [br][br]Recall that in cases where the two vectors are parallel, and when the force is constant in magnitude along the path, that the work reduces to [math]W=Fs[/math] where 's' is the path length or distance over which the force acted. In the present context, the force is the tractive force with which the vehicle propels itself and the distance 's' is the distance it travels. In steady state conditions, the vehicle is in dynamic equilibrium (net force=0), so the tractive force equals the sum of environmental resistive forces. The work term is the work done by the engine while combusting fuel. Therefore the value of W is just the amount of usable energy provided by the fuel.[br][br]In equation form, this all becomes:[br][center][math]W=F_{tractive}s \\[br]W=F_{resistive}s \\[br]\text{Usable energy in fuel}=F_{resistive}s. \\[br][/math][/center][br]Our ultimate goal in calculating fuel economy is to calculate how far a vehicle can travel per unit energy. In the equation above it is s/W. In standard SI units fuel economy would be measured in meters per joule, or m/J. In this country we instead ask about miles per energy provided by a gallon of gasoline. In that sense we are really faced with a unit conversion problem. [br][br]Given that we want fuel economy in the English unit of miles traveled per one gallon of gasoline, here is how we substitute the terms:[br][br][center][math]\text{energy per gallon of gasoline (J/gal)}\text{1 gal}\eta=F_{resistive}s\\[br]\frac{s}{\text{1 gal}}=\frac{\text{energy per gallon of gasoline (J/gal)}\eta}{F_{resistive}} \\[br]\text{Here 's' is in meters. To convert to miles we use 1609m = 1mi.} \\[br]\text{This is accomplished by multiplying both sides by $\frac{\text{1 mi}}{1609m}$.} \\[br]\text{The left side is then in units of miles per gallon, which I'll call MPG as is common practice.} \\[br]MPG = \frac{\text{energy per gallon of gasoline (J/gal)}\;\eta}{F_{resistive}}\frac{\text{1 mi}}{1609m}. \\[br]\text{Given the energy per gallon given above in the text, this gives:} \\[br]MPG = \frac{1.2 \times 10^8 J/gal \;\eta}{F_{resistive}}\frac{\text{1 mi}}{1609m}. \\[br][/math][/center][br][br]This equation, along with the resistive forces described in previous sections, allows us to predict fuel economy for vehicles traveling at steady speed like highway cruising. It does not work well in stop and go scenarios like city driving because much of the energy is lost to heat when brakes are applied, and because the resistive and tractive forces are not equal while accelerating. [br][br]In hybrid electric vehicles the loss of energy as heat while braking, is largely avoided. Rather than using friction to stop such vehicles (as non-hybrid vehicles do), they use the electric motors as electric generators to recharge the batteries while braking. Such systems are called regenerative braking systems. Similar systems are even implemented in formula one racing these days. They give them the fancy name of KERS, or Kinetic Energy Recovery System. See this [url=https://en.wikipedia.org/wiki/Kinetic_energy_recovery_system]link[/url].[br][br]To calculate steady speed economy we add up all the resitive forces and solve the above equation for MPG. If we use appropriate values for the coefficients in the equations, we should expect a very accurate result.

The economy of purely electric vehicles is calculated as a mile per gallon gasoline equivalent, or MPG[sub]e[/sub], sometimes also written MPG[sub]ge[/sub].  The idea is to take into account how far you can go in an electric car based on what you pay for electricity to charge it, and then converting that cost to equivalent amount of gasoline.  [br][br]The unit of energy that the electric companies charge you for is the kilowatt-hour, or kW-hr.  This is energy enough to run an appliance that consumes 1000 watts of power for the duration of one hour.  This is not a standard SI unit of energy, but 1kW-hr=3.6x10[sup]6[/sup]J.  [br][br]While you may think that you should pay for electric energy like you do gasoline, which is linearly - meaning twice as much costs twice as much - you do not usually.  Power companies tend to have tiered pricing schedules.  For instance, if you have a look at this [url=https://www.sce.com/wps/portal/home/regulatory/tariff-books/rates-pricing-choices/residential-rates/!ut/p/b1/tVJNU8IwEP01PYYsbemHtw44WBxUBEbaC5OGtI22SUmDqL_ewHDQEUQO5pRs3r7d93Zxihc4FeSVF0RzKUi1e6fechQPou7QteOhPxpA9DDwB_NHr-vedg0gMQA4cSI4l_-EU5xSoRtd4qSlbEml0EzoJRMWHO4WKFZsKqKlerdAE8XzHGVSvrTmh2jWokZxykWBaCk5Zbswa_nKpHJSoT1kV6ahfIUTBhklK7-HoBe6yA1tGwU2eCgM7LzrkRX4vo1Hf9Btq3F_XBhaokvERS7x4mtveHG0NxP-0Zsh48_rdRoZK3aS3zRe_KcXe9ONvnAI1zeje4iHs4kDsTOBu2kUOQDeAfDLXI1DRSWz_Y4kkcicwFihWM4UU52NMuFS66a9ssCC7XbbKaQsKtahsrbgWEopW6P6OxInZrv8k2Pou3h64VzPEHoXEzb1vA6cXlUEOoxRmr07H7O8rqNPXdvXlQ!!/dl4/d5/L2dBISEvZ0FBIS9nQSEh/]Edison link[/url] you can look up the various rate plans that are available to customers in southern California.  If they use less energy than what Edison calls the baseline (the definition of which I can't find in the documents), they pay 8.8 cents per kW-hr any time of day or year.  Over the baseline, rates go to 16 cents per kW-hr, and over 200% of baseline they pay 22.4 cents per kW-hr.  In other locations the pricing goes the other way where the higher tiers are actually cheaper, but the California pricing scheme is more common.[br][br]Power companies sometimes offer electric vehicle charging rate plans. Such plans in California have rates that depend on time of day without any tiers.  The problem is they start at 17 cents per kW-hr during daytime and pay 9.1 cents at night.  These tiered or timed plans often cancel much of the expected savings from driving an electric vehicle.  I hope this issue is fixed in upcoming years. Edison electric and other power companies can do this because they have no free market competition. They are monopolies. Such monopolies are government granted exceptions to anti-trust laws. Feel free to research more about government-granted monopolies elsewhere. They are also called de jure monopolies.[br][br]Calculating MPG[sub]e [/sub]is the same as the usual calculation of MPG shown above, but then accounting for the difference in the cost of energy purchased as electricity versus as gasoline.  To do this, we need to know how much energy costs in the form of electricity versus in the form of gasoline.

[u]The Cost of Electric Energy[/u][br]Let's assume you pay 13 cents per kW-hr, or the average of night and day costs on the electric vehicle plan according to SoCal Edison. This is 13 cents per 3.6 MJ. Converted to dollars per joule we get [math]\frac{\$ 0.13}{3.6\times10^6J}=3.6\times10^{-8}\frac{\$ }{J}.[/math][br][br][u]The Cost of Gasoline Energy[/u][br]Recall from above that the energy content of gasoline is 1.2x10[sup]8[/sup]J/gal. As of today, gasoline here is around $3/gal. This means the cost of energy in the form of gasoline is [math]\frac{\$ 3}{1.2\times 10^8(30 \%)}= 8.3\times 10^{-8} \frac{ \$ }{J}.[/math][br][br]

Once those two energy costs are established, to calculate MPG[sub]E[/sub] you just take a standard MPG calculation and multiply by the ratio of gasoline energy cost to electric energy cost. The idea is that if energy purchased as gasoline is, for instance, twice as expensive as energy purchased as electricity, then buying energy as electricity will get you effectively twice as far for your money. [br][br]Given the energy costs calculated above, that ratio is 2.31. So in our local market a gasoline-powered car with 40MPG based on its resistive forces would be considered to make 40MPG(2.31)=92.2 MPG[sub]E[/sub]. [br][br]Note that in the calculation, an assumed value of [math]\eta[/math] must be used. I don't know what the industry standard is for the assumed value in such calculations. The other thing to note is that the price of gasoline fluctuates much more quickly than the cost of electricity. Therefore the ratio (2.31 above) is not constant. Thus the value of MPGe is technically not a constant in time nor by region of the country.