Energy Use in Cars 2: Constant Speed Cruising

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If a body in motion tends to stay in motion, why do we need to burn gas to travel at highway speeds?

Big Ideas: 
  • Air resistance affects the motion of a car.
     

Why do we need to burn gas to keep travelling at the same speed? The basic answer is "because if we didn't, eventually the car would stop." In everyday life, there is always friction and air resistance that opposes any motion, and if you leave a moving object alone, this friction and drag will eventually cause it to stop. This lecture looks at how this drag impacts a car.

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, this doesn't take any more energy, but we still need our foot on the gas because of the next two issues.
  2. Overcoming 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 in the exhaust gases that gets pumped out the back of the car, so not ALL of it gets converted into motion.

So if we are travelling at a constant speed, we don't need to worry about #1 on the list. We have already accelerated up to speed, so that part is taken care of. However we need to figure out how to understand the other ways that energy is used.

Item number 4 is taken care of by the notion of the efficiency of the car's engine. For a typical gasoline engine, only around 25% of the heat energy from the fuel gets converted into mechanical energy which gets used for the first three items on this list1.

The energy content of gasoline is about 32 x 106 J / litre, but because of the engine efficiency only 25% of this chemical energy gets converted to mechanical energy2.

So what about Air Resistance?

When we drive a car we leave behind us a big tube of air that is swirling around (See Figure 1). The passage of the car is what makes the air swirl around, so our car engine needs to provide all the energy for all of that swirling. Figuring out all the details of exactly which air is swirling where is not important; we just want to make a reasonably accurate estimate of how much energy this will cost us, so we'll develop the following model.

The swirling air is confined to some region near the path of the car. Let's imagine this region is a long tube, with a cross sectional area Atube, and that the passage of the car makes it swirl with velocity v, which is the same velocity as the car. The area Atube is similar to the frontal area of the car, but not exactly the same. A more streamlined car will have Atube slightly smaller than the frontal area of the car. The ratio of Atube / Acar is called the Drag Coefficient (CD).  For a typical family sedan, CD= 0.33 and for a cyclist, CD= 0.93

Figure 1. Tube of air swirling around a moving car.

We want use this idea to figure out how much energy it costs the car per kilometre travelled. We can figure out how much energy the car loses to the air by figuring out the kinetic energy of this tube of moving air. To figure out kinetic energy we just need the mass and the volume of the tube of air. A car travelling at speed v will also make the air travel at speed v, so all we need to do is get the mass.

Say the car travels for some distance d. The length of the tube of air that the car encounters in that distance will be the same d:

$ \textnormal{Length} = d $

So the total volume of this tube will be:

$ \textnormal{Volume} = \textnormal{(Area)(Length)} = A_{tube}d $

And the mass of the tube will be:

$ \textnormal{Mass} = \textnormal{(Density)(Volume)} = \rho A_{tube}d $

So now the kinetic energy of the tube will be:

<br />
\begin{eqnarray}<br />
   KE &=& \dfrac{1}{2} m v^2 \nonumber \\<br />
   &=& \dfrac{1}{2} A_{tube} d v^2 \nonumber \\<br />
   &=& \dfrac{1}{2} \rho A_{car} C_D d v^2 \nonumber<br />
  \end{eqnarray}<br />

Given that the area of a typical family sedan is

$ A = (2 \textnormal{ m})(1.5 \textnormal{ m}) = 3 \textnormal{ m}^2 $

let's see how much work is done against air resistance for each kilometre a typical car driving at 50 km/h (14 m/s) travels.

<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)(3 \textnormal{ m}^2)(0.33)(1000 \textnormal{ m})(14 \textnormal{m/s})^2 \nonumber \\<br />
   &=& 126,126 \textnormal{kg•m}^2 / \textnormal{s}^2 \nonumber \\<br />
	&=& 126 \textnormal{ kJ} \nonumber<br />
  \end{eqnarray}<br />

So, for each kilometre travelled, 126 kJ of work is done against air resistance.

We can figure out how much fuel is required for each kilometre travelled using the efficiency formula:

<br />
\begin{eqnarray}<br />
   \textnormal{Efficiency} &=& \dfrac{\textnormal{Work Output}}{\textnormal{Work Input}}  \nonumber \\<br />
   &=& \dfrac{\textnormal{Work Output}}{\textnormal{Fuel Energy Input}} \nonumber \\<br />
    \textnormal{Fuel Energy Input} &=& \dfrac{\textnormal{Work Output}}{\textnormal{Efficiency}} \nonumber \\<br />
&=& \dfrac{126 \textnormal{ kJ}}{25 \%} \nonumber \\<br />
	&=& 505 \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{505 \textnormal{ kJ}}{32 \textnormal{ MJ\L}} \nonumber \\<br />
	&=& 0.016 \textnormal{ L} \nonumber<br />
  \end{eqnarray}<br />

So, 0.016 L of fuel is required to drive 1 km.

If we compare this with our earlier rule of thumb that the typical fuel consumption of a car is 0.076 L/km4.  We see that air resistance is only accounting for 21% of the energy cost. This is because we did the calculation at 50 km/h. At this speed, air friction is really a very small part of the fuel requirements of a car, which is why sometimes we choose to neglect it in our calculations. However, because the fuel consumption depends on the velocity squared, air resistance becomes much more important at higher speeds.

At 100 km/h, the fuel consumption will be FOUR times higher, or 0.064 L/km. This is much closer to 0.076 L/km. To get an even better understanding of energy consumption in cars, we can also take into account the rolling resistance of the car. We'll get into that in the next mini-lecture [Energy Use in Cars 3: Rolling Resistance].

Remember that we are considering only the energy needed to keep the vehicle moving at a constant speed. Most automobiles have many other systems  that consume fuel as well (e.g. drive-train losses, standby, accessories such as air conditioning). Further, engine efficiency will not be a constant in reality, but rather optimized for certain speeds. Nevertheless, our calculations provide a very useful lower bound for fuel economy.

 

Summary:

    Can we use these ideas to figure out how to get more mileage out of our cars? Well it looks like when you're travelling at high speed MOST of the gasoline goes into making the air swirl around.

 

Resources
Take Home Experiment: 
Energy Use in Cars Take-Home Experiment

Comments

"At 100 km/h, the fuel

"At 100 km/h, the fuel consumption will be FOUR times higher, or 0.064 L/km". Indeed, because Ekin = 0,5mv^2, so the energy requirement is not doubled but done *4 (v^2).

However, this is theory. Can anyone explain why reality is rather different? See e.g. http://en.wikipedia.org/wiki/Fuel_economy_in_automobiles#Speed_and_fuel_...

As from the optimal speed (around 55 miles/hour or 90 km/h), the line (fuel consumption) goes down almost in a linear fashion. And then the line under 55 m/h is even more different.
All in all, this is really different from a quadratic relationship between energy use and speed. What explains this? Factors such as heat and other friction loss? But since the air friction is so important at higher speeds, I thought the graph would not be (almost) lineair. Or isn't it?

We are considering only the

We are considering only the energy needed to keep the vehicle moving at a constant speed. Most automobiles have many other systems that consume fuel as well (e.g. drive-train losses, standby, accessories such as air conditioning). Further, engine efficiency will not be a constant in reality, but rather optimized for certain speeds. Nevertheless, our calculations provide a very useful lower bound for fuel economy.

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