Energy Use in Cars 4: Regenerative Braking Systems

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What single system could be added to a gasoline car to improve its city driving fuel economy by 30-40%?

Big Ideas: 
  • The fuel economy benefits of regenerative braking in stop-and-go driving.

What single system could be added to a gasoline 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.

First, a quick review: When you are driving a car, the energy from the fuel goes into four main places:

  1. Accelerating the car up to its cruising speed. A moving car has kinetic energy, and it needs to get this energy from the engine. Once we are at a constant speed, we do not need to add any more kinetic energy, but we still need our foot on the gas because of the next two issues.
  2. Air resistance. Driving a car makes the air around it swirl around, and this takes energy. Driving faster makes the air swirl much more.
  3. Rolling resistance. This accounts for all of the small bits of friction within the car, as well as resistance due to the tires on the road.
  4. Heat. Burning fuel doesn't make the car move directly. It creates a lot of heat, and then the engine has to convert that heat into motion. However there is still a lot of heat that gets carried away by the radiator and the exhaust, so not ALL of it gets converted into motion.

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).

However, a Regenerative Braking 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.

  1. Electric energy storage. When you hit the brakes, this system engages an integrated motor-generator which is connected to the spinning wheels. The motor slows down the car and converts the motion into electrical energy, which is stored in batteries. These systems typically need a large payload of batteries or supercapacitors to store this energy, so they are usually only used on vehicles that are already hybrid or electric. These systems can capture and return around 50% of the energy lost in braking1.
  2. Compressed Gas energy storage. When you hit the brakes, this system engages a pump which forces compressed air into a tank. This converts the mechanical energy of motion into elastic energy in the gas. When you want to re-accelerate, the gas is let back out through the pump, which works in reverse to accelerate the car. These systems can capture and return around 70% of the energy lost in braking2.
  3. Flywheel energy storage. When you hit the brakes, this system engages a clutch which transfers the mechanical energy of motion into a single spinning disc called a flywheel. The disc is held on very smooth bearings so it can be accelerated to spin at a very high rate. When you brake the car the flywheel gets spun up. When you want to accelerate the flywheel is connected to the wheels and the energy from the flywheel can get you started. These systems can capture and return around 70% of the energy lost in braking1.

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/h3.

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, but how can we figure out how much energy this stop-and-go costs us on an overall average basis?

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.

2001 Toyota Camry Specifications

city mileage4: 0.103 Litre/km

empty mass5: 1420 kg

CD6: 0.29

Frontal Area6: 2.42 m2

Coefficient of Rolling Resistance7: 0.015

Fuel Consumption due to Rolling Resistance

We can calculate the fuel consumption per kilometre by following the procedure we developed in Energy Use in Cars 3. Let's assume the car is carrying one passenger (70 kg) and a full tank of gas (40 kg).

<br />
\begin{eqnarray}<br />
   \textnormal{Force of rolling resistance} & =& (\textnormal{Coefficient of rolling resistance})(\textnormal{mass})(g) \nonumber \\<br />
   & = &  (0.015)(1420 \textnormal{ kg} + 70 \textnormal{ kg} + 30 \textnormal{ kg}) (9.8 \textnormal{ m}/\textnormal{s}^2) \nonumber \\<br />
   & =& 223 \textnormal{ Newtons} \nonumber<br />
   \end{eqnarray}<br />

<br />
\begin{eqnarray}<br />
   \textnormal{Work done against rolling resistance } & =& (\textnormal{Force of rolling resistance})(\textnormal{distance})\nonumber \\<br />
   & = &  (223 \textnormal{ N}) (1000 \textnormal{ m}) \nonumber \\<br />
   & =& 223 \textnormal{ kJ} \nonumber<br />
   \end{eqnarray}<br />

Considering the efficiency of a typical fuel engine gives us the necessary fuel energy input:

<br />
\begin{eqnarray}<br />
\textnormal{Fuel Energy Input} & =& \dfrac{\textnormal{Work Output}}{\textnormal{Efficiency}} \nonumber \\<br />
& =& \dfrac{223 \textnormal{ kJ}}{25 \%} \nonumber \\<br />
& =& 892 \textnormal{ kJ} \nonumber</span></p>
<p>   \end{eqnarray}<br />

And to provide this amount of energy we need to use

<br />
\begin{eqnarray}<br />
\textnormal{Energy per litre} &=& \dfrac{\textnormal{\# of Joules}}{\textnormal{\# of litres}} \nonumber \\<br />
\textnormal{\# of litres} &=& \dfrac{\textnormal{\# of Joules}}{\textnormal{Energy per litre}} \nonumber \\<br />
&=& \dfrac{892 \textnormal{ kJ}}{32 \textnormal{ MJ/L}} \nonumber \\<br />
&=& 0.028 \textnormal{ L} \nonumber</span></p>
<p>   \end{eqnarray}<br />

So, 0.028 L of gasoline is required to overcome rolling resistance for each kilometre the car travels.

Fuel Consumption due to Air Drag

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 ground8.

Following our procedure from Energy Use in Cars 2, we see the work done against air resistance is:

<br />
\begin{eqnarray}<br />
\textnormal{Work done against air resistance} &=& \dfrac{1}{2} \rho A_{car} C_D d v^2 \nonumber \\<br />
& =& \dfrac{1}{2} (1.3 \textnormal{ kg/m}^3)(2.42 \textnormal{ m}^2)(0.29)(1000 \textnormal{ m})(14 \textnormal{ m/s})^2 \nonumber \\<br />
& =& 89.4 \textnormal{ kJ} \nonumber<br />
   \end{eqnarray}<br />

We can figure out how much fuel is required for this using the efficiency formula

<br />
\begin{eqnarray}<br />
\textnormal{Efficiency} & =& \dfrac{\textnormal{Work Output}}{\textnormal{Fuel Energy Input}} \nonumber \\<br />
\textnormal{Fuel Energy Input} & =& \dfrac{\textnormal{Work Output}}{\textnormal{Efficiency}} \nonumber \\<br />
& =& \dfrac{89.4 \textnormal{ kJ}}{25 \%} \nonumber \\<br />
& =& 357.6 \textnormal{ kJ} \nonumber<br />
   \end{eqnarray}<br />

And to provide this amount of energy we need to use

<br />
\begin{eqnarray}<br />
\textnormal{Energy per litre} & =& \dfrac{\textnormal{\# of Joules}}{\textnormal{\# of litres}} \nonumber \\<br />
\textnormal{\# of litres} & =& \dfrac{\textnormal{\# of Joules}}{\textnormal{Energy per litre}} \nonumber \\<br />
& =& \dfrac{357.6 \textnormal{ kJ}}{32 \textnormal{ MJ/L}} \nonumber \\<br />
& =& 0.011 \textnormal{ L} \nonumber<br />
   \end{eqnarray}<br />

So, 0.011 L is required to overcome air drag for each kilometre that the car travels.

Fuel Consumption due to repeated Acceleration

Subtracting the above figures from the total fuel consumption for the Camry allows us to figure out the fuel consumption just for the repeated acceleration due to stop-and-go driving. This gives

<br />
\begin{eqnarray}<br />
\textnormal{Consumption due to repeated acceleration} & =& \textnormal{Total Camry fuel consumption}  \nonumber \\<br />
&& - \textnormal{Fuel consumed by rolling resistance} \nonumber \\<br />
&& - \textnormal{Fuel consumed by air drag} \nonumber \\<br />
& =& 0.103 \textnormal{ L/km} - 0.028 \textnormal{ L/km} - 0.011 \textnormal{ L/km} \nonumber \\<br />
& =& 0.064 \textnormal{ L/km} \nonumber<br />
  \end{eqnarray}<br />

This tells us that repeated acceleration is responsible for about 0.064 / 0.103 = 62% of the Camry's city fuel economy.

Potential Savings due to Regenerative Braking

If we added a flywheel regenerative braking system, we could recover 70% of this energy, thereby saving (70%) (62%) = 43% of the total city fuel economy! This number is higher than the 30% savings quoted by David MacKay1, which suggests there may be some other complexities we haven't considered. However, we are reasonably close to his number and these savings are definitely something to get excited about. While these technologies have not yet been integrated into production gasoline cars, they have the potential to significantly improve their mileage. However regenerative braking is a common feature of electric and hybrid cars, and is one of the reasons for those vehicles' low fuel consumption.

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