{"id":754,"date":"2010-05-25T16:24:21","date_gmt":"2010-05-25T23:24:21","guid":{"rendered":"https:\/\/c21-wp.phas.ubc.ca\/index.php\/energy-use-in-cars-4-regenerative-braking-systems"},"modified":"2023-03-31T14:19:42","modified_gmt":"2023-03-31T21:19:42","slug":"old-energy-use-in-cars-4-regenerative-braking-systems","status":"publish","type":"article","link":"https:\/\/c21.phas.ubc.ca\/article\/old-energy-use-in-cars-4-regenerative-braking-systems\/","title":{"rendered":"OLD Energy Use in Cars 4: Regenerative Braking Systems"},"content":{"rendered":"
What single $<\/u>100 system could be added to a gasoline $<\/u>700,000 car to improve the city fuel economy by 30-40%? A calculation of the fuel consumption due to stop-and-go driving, for the purposes of estimating potential savings from a regenerative braking system.<\/p>\n
First, a quick review: <\/b>When you are driving a $5 car, the energy from the fuel goes into four main places:<\/p>\n Under normal circumstances, all of the kinetic energy that we build up when accelerating to cruising speed gets “lost” when we brake back down to a stop. (It’s not really lost, but it gets converted to heat in the braking system, and isn’t any good for moving the car anymore).<\/p>\n However, a Regenerative Braking<\/b> system can actually slow us down by transforming our kinetic energy into potential energy of some kind, and then we can use that energy later on to re-accelerate up to speed. There are several types of regenerative braking systems, each using a different type of energy storage mechanism.<\/p>\n Of these three, the Flywheel technology is most suited to being added to a gasoline car. One particular manufacturer makes a flywheel unit that weighs only 25 kg but can store enough energy to accelerate a car up to 90 km\/h[note]Flywheel Regenerative Braking: Flybrid Systems. Flybrid Kinetic Energy Recovery System <\/i> (online) http:\/\/www.flybridsystems.com\/F1System.html<\/a> [28 August 2009]. Broken Link<\/strong>[\/note].<\/p>\n So, if we had one of these regenerative braking systems, how good would our fuel economy be? The primary improvement would be in city driving where there is a lot of stop-and-go traffic. We’ve already seen how to compute the energy cost of a single stop<\/a>, but how can we figure out how much energy this stop-and-go costs us on an overall average basis?<\/p>\n For the purposes of this calculation let’s use a particular car: a 2001 Toyota Camry. This car was chosen because it’s relatively common and because the detailed specifications we need for our calculation were available. The relevant specs of this car are listed below.<\/p>\n 2001 Toyota Camry Specifications<\/b><\/p>\n City Mileage[note]Toyota Camry city mileage: Microsoft. 2001 Toyota Camry Engines and Fuel Economy<\/i> (online). http:\/\/autos.msn.com\/research\/vip\/spec_engines.aspx?year=2001&make=Toyota&model=Camry&trimid=-1 <\/a> [28 August 2009]. Broken Link<\/strong>[\/note]: 0.103 Litre\/km<\/p>\n Empty Mass[note]Toyota Camry curb weight: CARS-DIRECTORY.NET. 2001 Toyota Camry<\/i> (online). http:\/\/www.cars-directory.net\/specs\/toyota\/camry\/2001_9\/1456\/<\/a> [28 August 2009].[\/note]: 1420 kg<\/p>\n $C_D$[note]Toyota Camry aerodynamic information: Wikipedia. Automobile drag coefficient<\/i> (online). http:\/\/en.wikipedia.org\/wiki\/Automobile_drag_coefficient<\/a>.[28 August 2009][\/note]: 0.29<\/p>\n Frontal Area[note]Toyota Camry aerodynamic information: Wikipedia. Automobile drag coefficient<\/i> (online). http:\/\/en.wikipedia.org\/wiki\/Automobile_drag_coefficient<\/a>.[28 August 2009][\/note]: 2.42 m2<\/sup><\/p>\n Coefficient of Rolling Resistance[note]Rolling Friction Coefficients: http:\/\/auto.howstuffworks.com\/tire4.htm<\/a> [28 August 2009].[\/note]: 0.015<\/p>\n Fuel Consumption due to Rolling Resistance<\/b><\/p>\n We can calculate the fuel consumption per kilometre by following the procedure we developed in Energy Use in Cars 3<\/a>. Let’s assume the car is carrying one passenger (70 kg) and a full tank of gas (40 kg).<\/p>\n \\begin{eqnarray} \\begin{eqnarray} Considering the efficiency of a typical fuel engine gives us the necessary fuel energy input:<\/p>\n \\begin{eqnarray} And to provide this amount of energy we need to use<\/p>\n \\begin{eqnarray} So, 0.028 L of gasoline is required to overcome rolling resistance for each kilometre the car travels.<\/p>\n Fuel Consumption due to Air Drag<\/b><\/p>\n In order to calculate the effect of air drag we need to choose a typical speed the car will be traveling at. Let’s choose 50 km\/h. Sometimes the car will be traveling faster or slower than this, but seeing as the city mileage guidelines are generated using a range of speeds up to 90 km\/h, this seems like a reasonable middle ground[note]The actual choice of which “average speed” to use is a little more complicated because the fuel consumption due to air drag varies as the velocity cubed: Wikipedia. Fuel economy in automobiles<\/i> (online). http:\/\/en.wikipedia.org\/wiki\/Mileage#United_States_EPA_fuel_economy_ratings<\/a> [28 August 2009]. Broken Link<\/strong>[\/note].<\/p>\n\n
\n
\n\\text{Force of rolling resistance} & =& (\\text{Coefficient of rolling resistance})(\\text{mass})(g) \\nonumber \\\\
\n& = & (0.015)(1420 \\text{ kg} + 70 \\text{ kg} + 30 \\text{ kg}) (9.8 \\text{ m}\/\\text{s}^2) \\nonumber \\\\
\n& =& 223 \\text{ Newtons} \\nonumber
\n\\end{eqnarray}<\/p>\n
\n\\text{Work done against rolling resistance } & =& (\\text{Force of rolling resistance})(\\text{distance})\\nonumber \\\\
\n& = & (223 \\text{ N}) (1000 \\text{ m}) \\nonumber \\\\
\n& =& 223 \\text{ kJ} \\nonumber
\n\\end{eqnarray}<\/p>\n
\n\\text{Fuel Energy Input} & =& \\dfrac{\\text{Work Output}}{\\text{Efficiency}} \\nonumber \\\\
\n& =& \\dfrac{223 \\text{ kJ}}{25 \\%} \\nonumber \\\\
\n& =& 892 \\text{ kJ} \\nonumber\\end{eqnarray}<\/p>\n
\n\\text{Energy per litre} &=& \\dfrac{\\text{# of Joules}}{\\text{# of litres}} \\nonumber \\\\
\n\\text{# of litres} &=& \\dfrac{\\text{# of Joules}}{\\text{Energy per litre}} \\nonumber \\\\
\n&=& \\dfrac{892 \\text{ kJ}}{32 \\text{ MJ\/L}} \\nonumber \\\\
\n&=& 0.028 \\text{ L} \\nonumber\\end{eqnarray}<\/p>\n