Transportation

Energy Use in CarsSolar Powered AirplanesEnergy Cost of TransportFlyingPitot Tubes
When does cycling cost more fuel than driving? People recommend commuting by bicycle rather than by car to save energy. However sometimes when a cyclist pushes the button to make a traffic light change, it means that a bus full of people has to stop. When a cyclist makes the bus stop like this, is the extra fuel it has to burn to get back up to speed more than the fuel saved by choosing to commute by bicycle rather than car?

To tell whether stopping the bus uses more fuel than one person driving to work we need to figure out how much fuel it costs to get the bus back up to speed and how much fuel is burnt in a typical daily commute. Let’s look at the bus first.

We know that driving uses fuel, because we keep having to buy it at the gas station! We burn the fuel to get heat energy, and then the bus’s engine converts heat energy into motion of the bus and some other things. The energy from the fuel goes into four main places:

  1. Accelerating the bus 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 don’t need to spend any more energy accelerating but we still need our foot on the gas because of the next two issues.
  2. Air resistance. Driving a bus 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 bus, as well as resistance due to the tires on the road.
  4. Heat. Burning fuel doesn’t make the bus 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 bus, so not ALL of it gets converted into motion.
Part 1: Stop-and-Go Driving
Question: Where do you think most of the energy from the fuel ends up? (choose 1-4 above)

Answer: 4. Heat. 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 list. One way of thinking about an engine is that it transforms the chemical energy from the fuel into mechanical energy and heat, like this:

We would say that this engine has an efficiency of 25%.

$$\text{Engine Efficiency} = \dfrac{ \text{Mechanical Energy Out}}{\text{Fuel Energy In}}$$

To answer our question about the bus we are mostly concerned with how much energy it takes to get back up to speed so let’s just look at the energy cost for that. In city driving, this is a very important part of your fuel consumption, especially if you are moving at typical traffic speeds (below 50 km/h). Then, once we know how much mechanical energy is required, we can use the efficiency of the engine to determine how much fuel needs to be burned.

Imagine the bus is initially moving at 50 km/h. When the cyclist makes the light change it decelerates using the brakes. This converts ALL of the kinetic energy of the bus into heat energy in the brakes. Then after it stops it must re-accelerate.

The mass of an empty 99 bus is 19,820 kg[1]New Flyer. Hybrid Transit Solutions (online). http://www.newflyer.com/pix/Brochures/hybridbrochure.pdf [6 May 2010].

The passengers weigh an average of 70 kg each, and at rush hour the bus is likely full to capacity with 120 passengers[2]New Flyer. Hybrid Transit Solutions (online). http://www.newflyer.com/pix/Brochures/hybridbrochure.pdf [6 May 2010]..

So the total mass of the bus is

$$19,820 \text{ kg} + (120 \text{ passengers})\times\left(\dfrac{70 \text{ kg}} {\text{passenger}}\right) = 28,220 \text{ kg}$$

Now we need to know how much energy it takes to get this mass moving again at 50 km/h (14 m/s). The kinetic energy of a moving object is given by $KE = \frac12mv^2$, so the total kinetic energy required is

$$\begin{eqnarray}
KE &=& \dfrac{1}{2} m v^2 \nonumber \\
&=& \dfrac{1}{2} (28,220 \text{ kg})\times(50 \text{ km/h})^2 \nonumber \\
&=& \dfrac{1}{2} (28,220 \text{ kg})\times(14 \text{ m/s})^2 \nonumber \\
&=& 2.77 \times 10^6 \text{ kg} \text{ m}^2 \text{/s}^2 \nonumber \\
&=& 2.77 \times 10^6 \text{ J} \nonumber
\end{eqnarray}$$

In order to create this much mechanical energy, we need to burn additional fuel. We can see from the efficiency diagram that

$\text{Mechanical Energy} = \text{(Engine Efficiency)\times(Fuel Energy)}$

The energy content of diesel is 38.7 x 106 J / litre[3]Energy Information Administration. Converting Energy Units 101(online). http://www.eia.doe.gov/basics/conversion_basics.html [21 August 2009]. and since Diesel engines are significantly more efficient than gasoline engines, we use an efficiency estimate of 37% for the bus[4]Soimar M. The Challenge Of CVTs In Current Heavy-Duty Powertrains (online). Diesel Progress North American Edition. http://findarticles.com/p/articles/mi_m0FZX/is_4_66/ai_62371160/ [21 August 2009].. Now we can calculate how much Fuel Energy is required.

$$\begin{eqnarray}
\text{Fuel Energy} &=& \dfrac{\text{Mechanical Energy}}{37 \%} \nonumber \\
&=& \dfrac{2.77 \times 10^6 \text{ Joules}}{0.37} \nonumber \\
&=& 7.49 \times 10^6 \text{ Joules}\nonumber
\end{eqnarray}$$

To determine how many litres of fuel this is, we calculate

$$\begin{eqnarray}
\text{Energy Content of diesel} &=& \dfrac{\text{# of Joules}}{\text{# of litres}} \nonumber \\
\text{# of litres} &=& \dfrac{\text{# of Joules}}{\text{Energy Content of diesel}} \nonumber \\
&=& \dfrac{7.49 \times 10^6 \text{ J}}{38.7 \times 10^6 \text{ J/litre}} \nonumber \\
&=& 0.19 \text{ litres of diesel} \nonumber
\end{eqnarray}$$

Now we want to compare this to the amount of fuel that the cyclist is SAVING. To do that we’ll just use a simple estimate of typical car efficiency. The average gasoline car consumes 0.076 litres of fuel per kilometer traveled[5]MacKay DJC. Sustainable Energy – Without the Hot Air (online). UIT Cambridge. p.31. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/1.112.pdf [21 August 2009]. . So if your commute is 10 km long, you would be saving 0.76 litres of gasoline by riding your bike. If you made the bus stop once on your 10 km trip, you would be costing 0.19 litres in order to save 0.76 litres, so it is still a good idea to ride your bike.

Question: Given that stopping a bus costs 0.19 litres of fuel, and cars consume 0.076 litres/ km, how long would your commute need to be to save an amount of gasoline equivalent to stopping the bus once?

  1. 0.014 km
  2. 0.26 km
  3. 0.42 km
  4. 1.0 km
  5. 2.5 km

Answer:  E. $\dfrac{0.19\text{ L}}{0.076 \text{ L/km}} = 2.5\text{ km}$

So how can we interpret this result?

The first thing we might think is that if your commute is shorter than 2.5 km, then the amount of fuel that you save by riding your bike is smaller than 0.19 litres, so if you cause the bus to stop once during your bike commute then you will have caused a larger consumption of fuel than you save. However there are a couple of other factors to consider.

Firstly, when you hit the light for the crosswalk often you stop MORE than just a bus. For example, there might be a few cars behind the bus that also have to stop. This would make us think that the amount of gas wasted is more than 0.19 litres.

However, sometimes when you push the light and you stop the bus the timing of lights is such that it would have stopped in a block or two anyways. So in this case, pushing the light doesn’t really make any difference to the fuel consumption of the bus.

These two effects are very difficult to model, and they act in opposite directions, so we might be tempted to conclude that we can’t know what the effect of pushing the crosswalk button is. However, we can see from this calculation that some of the time it will cost a significant amount of fuel, and therefore it is a good thing to avoid if you can avoid it.

It also suggests that in order to design efficient systems for transit and bicycles, it is a good idea to design crossing lights so the cyclists don’t cause the bus to stop. One way of doing this is to have transponders in the buses that make sure that lights up ahead stay green for them. Another way is to have completely separate cycling routes that don’t interfere with the transit, similar to systems used in many European countries.

Part 2: Constant Speed Cruising
If we are traveling 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 list[6]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.262. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [25 August 2009]..

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 energy[7]Wikimedia Foundation Inc. Gasoline (Online). http://en.wikipedia.org/wiki/Gasoline [25 August 2009]..

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 $A_\text{tube}$, and that the passage of the car makes it swirl with velocity $v$, which is the same velocity as the car. The area $A_\text{tube}$ is similar to the frontal area of the car, but not exactly the same. A more streamlined car will have $A_\text{tube}$ slightly smaller than the frontal area of the car. The ratio of $A_\text{tube}$ / $A_\text{car}$ is called the Drag Coefficient ($C_D$).  For a typical family sedan, $C_D = 0.33$ and for a cyclist, $C_D = 0.9$[8]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.257. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [25 August 2009].

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 traveled. 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 traveling 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 $d$:

$\text{Length} = d$

So the total volume of this tube will be:

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

And the mass of the tube will be:

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

So now the kinetic energy of the tube will be:

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

Given that the area of a typical family sedan is

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

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.

\begin{eqnarray}
\text{Work done against air resistance} &=& \dfrac{1}{2} \rho A_\text{car} C_D d v^2 \nonumber \\
&=& \dfrac{1}{2} (1.3 \text{ kg/m}^3)(3 \text{ m}^2)(0.33)(1000 \text{ m})(14 \text{m/s})^2 \nonumber \\
&=& 126,126 \text{kg\cdot m}^2 / \text{s}^2 \nonumber \\
&=& 126 \text{ kJ} \nonumber
\end{eqnarray}

So, for each kilometre traveled, 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:

\begin{eqnarray}
\text{Efficiency} &=& \dfrac{\text{Work Output}}{\text{Work Input}} \nonumber \\
&=& \dfrac{\text{Work Output}}{\text{Fuel Energy Input}} \nonumber \\
\text{Fuel Energy Input} &=& \dfrac{\text{Work Output}}{\text{Efficiency}} \nonumber \\
&=& \dfrac{126 \text{ kJ}}{25 \%} \nonumber \\
&=& 505 \text{ kJ} \nonumber
\end{eqnarray}

And to provide this amount of energy we need to use

\begin{eqnarray}
\text{Energy per litre} &=& \dfrac{\text{# of joules}}{\text{# of litres}} \nonumber \\
\text{# of litres}&=& \dfrac{\text{# of joules}}{\text{Energy per litre}} \nonumber \\
&=& \dfrac{505 \text{ kJ}}{32 \text{ MJ/L}} \nonumber \\
&=& 0.016 \text{ L} \nonumber
\end{eqnarray}

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/km[9]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.31. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/1.112.pdf [25 August 2009]..  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.

Part 3: Rolling Resistance
Natural Resources Canada recommends keeping your tires inflated to their maximum pressure to conserve gasoline. Why does this matter? A rough calculation of rolling resistance in cars explores the impact of having underinflated tires.

If we are traveling 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 list[10]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.262. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [25 August 2009]. . The energy content of gasoline is about 32 x 106 J/litre, but because of the engine efficiency only 25% of that gets converted to mechanical energy[11]Wikimedia Foundation Inc. Gasoline (Online). http://en.wikipedia.org/wiki/Gasoline [25 August 2009]. .

So what about Rolling Resistance?

In cars rolling resistance comes from the fact that the tires are soft, and get deformed as we drive forward, costing the car some energy. The effect of this depends on the inflation of the tire, what kind of tire you have, and how fast you are going. A common approximation which is reasonably accurate is just that the rolling resistance is a constant frictional force that depends on the weight of the car (similar to any other kind of friction).

Force due to Rolling Resistance $F_{RR}$
= Coefficient of Rolling Resistance $\mu_{RR}$
$\times$ Mass of vehicle
$\times$ Acceleration of gravity $g$

The Coefficient of Rolling Resistance is usually written as $\mu_{RR}$, and it has different values for different types of vehicles. Some example values of rolling resistance are given in the table below[12]A Discovery Company. How Tires Work (online). http://auto.howstuffworks.com/tire4.htm [25 August 2009]..

\begin{array}{|c c|}
\hline
\textbf{Tire Type} & \textbf{Coefficient of Rolling Friction} \\ \hline
\text{Low rolling resistance car tire} & 0.006\, – 0.01 \\
\text{Ordinary car tire} & 0.015 \\
\text{Truck tire} & 0.006\, – 0.01 \\
\text{Train wheel} & 0.001 \\
\hline
\end{array}

So what does this tell us? In order to figure out how this force impacts our fuel economy we need to figure out how much energy is required to overcome it. For this we use the Work-Energy principle, which tells us how much energy a force will add to a system.

$\text{Work = (Force)}\times\text{(Distance)}$

Because the rolling friction opposes the motion of the car, it actually subtracts energy from the car. This energy needs to be made up by burning more fuel.

A typical sedan has a mass of around 1200 kg. For this car, plus a single driver (70 kg) the force of rolling resistance will be:

\begin{eqnarray}
F_{RR} & =& \mu_{RR} m g \nonumber \\
& =& (0.015)(1270 \text{ kg})(9.8 \text{ m/s}^2) \nonumber \\
& =& 187 \text{ Newtons} \nonumber
\end{eqnarray}

Over the course of driving one kilometre, this will require extra energy given by:

\begin{eqnarray}
W & =& (F_{RR})\times(\text{Distance}) \nonumber \\
& =& (187 \text {N}) \times (1000 \text{ m}) \nonumber \\
& =& 187,000 \text{ N m} \nonumber \\
& =& 187 \text{ kJ for each kilometre driven} \nonumber
\end{eqnarray}

We can figure out how much fuel is required to drive one kilometre by using the efficiency formula:

\begin{eqnarray}
\text{Efficiency} & =& \dfrac{\text{Work Output}}{\text{Work Input}} \nonumber \\
& =& \dfrac{\text{Work Output}}{\text{Fuel Energy Input}} \nonumber \end{eqnarray}

\begin{eqnarray}
\text{Fuel Energy Input} &=& \dfrac{\text{Work Output}}{\text{Efficiency}} \nonumber \\
& =& \dfrac{187 \text{ kJ}} {25 \%} \nonumber \\
& =& 748 \text{ kJ} \nonumber
\end{eqnarray}

And to provide this amount of energy we need to use

\begin{eqnarray}
\text{Energy per litre} & =& \dfrac{\text{# of Joules}}{\text{# of litres}} \nonumber \\
\text{# of litres} & =& \dfrac{\text{# of Joules}}{\text{Energy per litre}} \nonumber \\
& =& \dfrac{748 \text{ kJ}}{32 \text{ MJ/L}} \nonumber \\
& =& 0.023 \text{ L} \nonumber
\end{eqnarray}

So, 0.023 L of fuel is required to drive 1 km.

Remember, this result is just to overcome the rolling friction. If we add this to the 0.064 L/km highway mileage we calculated in Constant Speed Cruising (taking air drag into account) this comes out to a total of 0.087 L/km. This is a little bit higher than the reported average of 0.076 L/km[13]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. p.31. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/1.112.pdf [25 August 2009]., which seems reasonable as the highway mileage was calculated at a speed of 100 km/h which is perhaps a bit fast.

So, now what would be the impact of having low air pressure in our tires? Let’s imagine that having your air pressure reduced by 5% would result in a 5% increase in the coefficient of rolling resistance. A typical car’s tires are inflated to around 40 psi, so this would correspond to being 2 psi lower than average. A 5% increase in the Coefficient of Rolling Resistance would bring it up to 0.01575, and the associated fuel consumption would increase to 0.024 L/km. This is an extra 0.01 L/km, or approximately an extra 1% of fuel mileage. This corresponds closely with the guidelines published by Natural Resources Canada[note value=1][/note].

This extra drag starts to add up when your tires are really low on air. If they are 10 psi low, that would correspond to an extra 5% fuel mileage!

Part 4: Regenerative Braking Systems
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.
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 braking[14]MacKay DJC. Sustainable Energy – Without the Hot Air (online). UIT Cambridge. p. 125-126. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/113.252.pdf [21 August 2009]..
  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 braking[15]Hydraulic / Compressed Gas Regenerative Braking: Eaton Corporation. Hydraulic Launch Assist (online). http://www.eaton.com/EatonCom/ProductsServices/Hybrid/SystemsOverview/HydraulicHLA/index.htm [28 August 2009]..
  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 braking[note value=1][/note].

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[16]Flywheel Regenerative Braking: Flybrid Systems. Flybrid Kinetic Energy Recovery System (online) http://www.flybridsystems.com/F1System.html [28 August 2009]..

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 Mileage[17]Toyota Camry city mileage: Microsoft. 2001 Toyota Camry Engines and Fuel Economy (online). http://autos.msn.com/research/vip/spec_engines.aspx?year=2001&make=Toyota&model=Camry&trimid=-1 [28 August 2009].: 0.103 Litre/km

Empty Mass[18]Toyota Camry curb weight: CARS-DIRECTORY.NET. 2001 Toyota Camry (online). http://www.cars-directory.net/specs/toyota/camry/2001_9/1456/ [28 August 2009].: 1420 kg

$C_D$[19]Toyota Camry aerodynamic information: Wikipedia. Automobile drag coefficient (online). http://en.wikipedia.org/wiki/Automobile_drag_coefficient.[28 August 2009]: 0.29

Frontal Area[note value=6][/note]: 2.42 m2

Coefficient of Rolling Resistance[20]Rolling Friction Coefficients: http://auto.howstuffworks.com/tire4.htm [28 August 2009].: 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).

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

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

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

\begin{eqnarray}
\text{Fuel Energy Input} & =& \dfrac{\text{Work Output}}{\text{Efficiency}} \nonumber \\
& =& \dfrac{223 \text{ kJ}}{25 \%} \nonumber \\
& =& 892 \text{ kJ} \nonumber\end{eqnarray}

And to provide this amount of energy we need to use

\begin{eqnarray}
\text{Energy per litre} &=& \dfrac{\text{# of Joules}}{\text{# of litres}} \nonumber \\
\text{# of litres} &=& \dfrac{\text{# of Joules}}{\text{Energy per litre}} \nonumber \\
&=& \dfrac{892 \text{ kJ}}{32 \text{ MJ/L}} \nonumber \\
&=& 0.028 \text{ L} \nonumber\end{eqnarray}

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 ground[21]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 (online). http://en.wikipedia.org/wiki/Mileage#United_States_EPA_fuel_economy_ratings [28 August 2009]..

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

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

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

\begin{eqnarray}
\text{Efficiency} & =& \dfrac{\text{Work Output}}{\text{Fuel Energy Input}} \nonumber \\
\text{Fuel Energy Input} & =& \dfrac{\text{Work Output}}{\text{Efficiency}} \nonumber \\
& =& \dfrac{89.4 \text{ kJ}}{25 \%} \nonumber \\
& =& 357.6 \text{ kJ} \nonumber
\end{eqnarray}

And to provide this amount of energy we need to use

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

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

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

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 MacKay[note value=1][/note], 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.

Part 5: Gasoline Cars Vs. Electric Cars
Did you know that electric cars driven in BC produce 40 times less greenhouse gas than gasoline cars?

Our prior analysis has looked at the fuel consumption of gasoline cars. But what about cars that don’t use gasoline at all? Electric cars are becoming more and more common, and with good reason: they offer significant energy and pollution savings over traditional gasoline cars.

In order to compare the two types of cars, we need to get away from looking strictly at the fuel consumption, as we can’t make a direct comparison along these lines. Rather, we’ll look at two other factors: the total amount of energy consumed and the greenhouse gas produced.

Energy Consumption

The average fuel consumption of gasoline cars is about 0.076 L/km traveled[22]MacKay DJC. Sustainable Energy – Without the Hot Air (online). UIT Cambridge. p. 29-31. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/1.112.pdf [21 August 2009].. To convert this figure to energy we multiply by the energy content of gasoline: 32 MJ/L[23]Energy Content of Gasoline: http://en.wikipedia.org/wiki/Gasoline [4 November 2009]. Therefore:

\begin{eqnarray}
\text{Energy consumption of a gasoline car} & =& (0.076 \text{ L/km})(32 \text{ MJ/L}) \nonumber \\
& =& 2.4 \text{ MJ for each km traveled} \nonumber
\end{eqnarray}

If we take a typical mechanical efficiency of 0.25, then the total mechanical energy requirement is $2.4 \text{ MJ/L} \times 0.25 = 0.6 MJ$ per km traveled. Let us assume that the mechanical energy is the same for an electric car and consider the efficiency of the electric car’s systems to calculate the total energy consumption of an electric car. In this case, we need to consider the efficiency of charging and discharging the car’s batteries and the mechanical efficiency of the electric motor[24]Electric battery and motor efficiencies: http://www.stanford.edu/group/greendorm/participate/cee124/TeslaReading.pdf [4 November 2009].

\begin{eqnarray}
\text{Charging/discharging} & =& \dfrac{\text{Useful Electrical Power}}{\text{Power Required to Charge}} \nonumber \\
& =& 86 \% \nonumber
\end{eqnarray}

\begin{eqnarray}
\text{Electric Motor} & =& \dfrac{\text{Mechanical Power}}{\text{Electrical Power Input}} \nonumber \\
& =& 85 \% \nonumber
\end{eqnarray}

\begin{eqnarray}
\text{Energy Consumption of Electric Car}& =& \dfrac{0.6 \text{ MJ}}{(0.86)(0.85)} \nonumber \\
& = & 0.82 \text{ MJ for each km traveled} \nonumber
\end{eqnarray}

This is 1/3 of the energy consumption of an equivalent gasoline-powered car, and it is due to the overall efficiency of an electrical system compared to a heat engine. In addition, electric cars also typically have regenerative braking systems which recovers half of the kinetic energy of the car when it brakes. We can calculate the impact of regenerative braking on the car’s energy consumption. From our earlier calculations we know that stopping and starting is responsible for 60% of the energy consumption of a car in city driving. Regenerative braking recovers 50% of that, for a savings of 30%. This gives a total of 0.7 x 0.82 MJ = 0.57 MJ for each km travelled.

This result is pretty close to the average consumption for electric cars which is 0.54 MJ for each km travelled[note value=1][/note]. This suggests that we are correct in assuming that the effect of the battery efficiency, motor efficiency, and regenerative braking are largely responsible for the savings of electric cars over gasoline cars..

Greenhouse Gas Production

Electric cars certainly consume less energy than gasoline cars, but where does that energy come from? In order to compare the pollution from these two types of transportation we need to understand how much greenhouse gas is produced when the electricity is generated.

In British Columbia, much of our electricity is produced in hydro-electric power stations that have extremely low greenhouse gas emissions. The average greenhouse gas production of electricity generated in BC is 7.8 g CO2/MJ and in Canada is 64 g CO2/MJ[25]Canada Greenhouse Gas Emissions: http://www.ec.gc.ca/ges-ghg/default.asp?lang=En&n=EAF0E96A-1#section11 [2012.09.27], so the equivalent emissions for an electric car are:

\begin{eqnarray}
\text{Greenhouse gas emissions in BC for each km traveled} & =& (0.57 \text{ MJ})(7.8 \text{ g CO}_2 \text{/MJ}) \nonumber \\
& =& 4.4 \text{ g CO}_2 \text{ for each km traveled} \nonumber
\end{eqnarray}

$\text{Canadian average} = (0.57 \text{ MJ})(64 \text{ g CO}_2 \text{ /MJ)}$

Given that the GHG production due to gasoline is 2.32 kg CO2\L[26]Greenhouse Gas Equivalence of Gasoline: http://www.epa.gov/cleanenergy/energy-resources/refs.html [4 November 2009]., the corresponding emissions for a gasoline car are:

\begin{eqnarray}
\text{Greenhouse gas emissions for each km traveled} & =& (0.076 \text{ L})(2.32 \text{ kg CO}_2 \text{/L}) \nonumber \\
& =& 176 \text{ g CO}_2 \text{ for each km traveled} \nonumber
\end{eqnarray}

Summary

Therefore, greenhouse gas emissions for a gasoline car are over 40 times higher than for an electric car (in BC). Even if we use the Canadian average of greenhouse gas emissions for electricity generation, gasoline cars still emit over 5 times as much greenhouse gas as electric cars.

Part 6: Gasoline Cars vs. Bicycles
When does a car have the same fuel efficiency as a bicycle?

In comparing bicycles and cars, it is tempting to assume that the energy consumption of a cyclist is nearly zero. However, that’s not the case. A bicycle is powered by a person, and much like an internal combustion engine that person needs to convert chemical energy into mechanical energy in order to move. Let’s consider the case of movement at a steady speed and see how a bicycle stacks up against a standard car. Bicycles and cars will be compared in terms of efficiency, air resistance and rolling resistance.

Efficiency

As we’ve discussed previously (See Energy Use in Cars 1) much of the chemical energy that is consumed in a car gets converted to heat. On average, only 25% of a car’s chemical energy gets turned into useful mechanical energy, the rest being used to heat the surrounding air and exhaust gases. Surprisingly, it turns out that the efficiency of a cyclist is about the same[27]P.E. di Prampero, G. Cortili, P. Mognoni and F. Saibene, J Appl Physiol 47: 201-206, 1979.! A detailed analysis of energy use in cycling shows that people are about 25% efficient in converting food energy into mechanical energy, the rest being used to maintain the function of the body and generate heat. So there is no clear advantage to either form of transportation on the efficiency front.

Air Resistance

Our previous look into air resistance (See Energy Use in Cars 2) shows that the work done against air resistance when traveling a distance $d$ is given by:

$\text{Work done against air resistance} = \dfrac{1}{2} \rho A C_D d v^2 $

So for a given distance, the only values that differ between bikes and cars are the Area ($A$), Coefficient of Drag ($C_D$), and speed ($v$). Let’s look at the term $A \times C_D$. This product is also called the Drag Area, and it’s the area of the tube of air that gets dragged along with a moving object.

For a typical family sedan,

$A = 2 \text{ m} \times 1.5 \text{ m} = 3 \text{ m}^2;\ C_D = 0.33$

so, the drag area is

$A \times C_D = (3 \text{ m}^2)\times(0.33) = 1 \text{ m}^2$

For a cyclist[28]MacKay DJC. Drag Coefficients: Sustainable Energy Without the Hot Air (online). UIT Cambridge. p. 257 http://www.inference.phy.cam.ac.uk/withouthotair/cA/page_257.shtml [4 November 2009].,

$A \approx 1 \text{ m}^2; \ C_D = 0.9$

 so the drag area is

$A \times C_D = (1 \text{ m}^2)\times(0.9) = 0.9 \text{ m}^2 $

This means that if they travel at the same speed, cars and bicycles have nearly the same air resistance. Because cars are so much more streamlined (smaller $C_D$), they have the same air resistance as bicycles despite being so much larger.

However, cars and bicycles rarely travel the same speed. The typical cruising speed for a bicycle is around 20 km/h, and that for a car is around 50 km/h. Going 2.5 times faster means that typically the air resistance for a car will be (2.5)2 = 6.25 times higher.

Rolling Resistance

So, is the higher energy cost of a car solely due to going faster than bicycles? Let’s look at the last factor in constant-speed energy consumption: rolling resistance. In a previous article (See Energy Use in Cars 3) we saw that the work done against rolling resistance was given by:

\begin{eqnarray}
\text{Work done against rolling resistance} &=& \text{Coefficient of Rolling Resistance } \times \text{ mass} \nonumber \\
& & \times \text{ acceleration of gravity}(g) \times \text{ distance} \nonumber
\end{eqnarray}

This is where the differences between cars and bicycles really stand out. A car with a mass of 1200 kg will need 187 kJ for each km traveled, whereas a bicycle with a mass of 10 kg will only need 4 kJ for each km traveled. This means that even if they travel the same speed, a car will have a much higher energy consumption than a bicycle.

Adding it all up:

The work done against both air resistance and rolling resistance for each km traveled is:

$\text{Work} = \text{(Coefficient of Rolling Resistance)(mass)(gravity)} + \dfrac{1}{2} \rho A C_D v^2$

The coefficient of rolling resistance for a bicycle is 0.005[29]Wikimedia Foundation Inc. Rolling Resistance (Online). http://en.wikipedia.org/wiki/Rolling_resistance [12 May 2010]., so the total work done at 20 km/h is:

\begin{eqnarray}
\text{Work} & =& (0.005)(80 \text{ kg})(9.8 \text{ m/s}^2) + \dfrac{1}{2} (1.3 \text{ kg/m}^3)(1 \text{ m}^2)(0.9)(5.55 \text{m/s})^2 \nonumber \\
& =& 4 \text{ N} + 18 \text{ N} \nonumber \\
& =& 22 \text{ N} \nonumber\end{eqnarray}

 

For a car, which has a coefficient of rolling resistance of 0.015[30]Rolling Friction Coefficients: http://auto.howstuffworks.com/tire4.htm [4 November 2009]., at 20 km/h this works out to:

\begin{eqnarray}
\dfrac{\text{Energy}}{\text{distance}} & =& (0.015)(1270 \text{ kg})(9.8 \text{ m/s}^2) + \dfrac{1}{2} (1.3 \text{ kg/m}^3)(3 \text{ m}^2)(0.33)(5.55 \text{m/s})^2 \nonumber \\
& =& 187 \text{ N} + 20 \text{ N} \nonumber \\
& =& 207 \text{ N} \nonumber
\end{eqnarray}

Summary

So even if the car goes the same speed as a bicycle (or slower), the bicycle will always have better energy economy because of rolling resistance.

Introduction
This module introduces, somewhat incidentally, the concept of estimation, both in the sense of rounding extremely precise numbers off to remove excessive significant figures, and simplifying a complex process (the flight of a 747) to a simpler one.
Roughly how much power is necessary to keep a 747 aloft?

Flight is a complex process, and the amount of power needed for a 747 during flight will vary depending on what it’s doing. Whether or not the plane is rising, falling, or cruising at a constant altitude, the density of air at the altitude it’s flying, the mass of its payload, and prevailing winds will all affect the energy necessary to keep the plane going.  We don’t want to deal with any of these complications, so we’ll make a very rough estimate of the amount of power needed to keep a 747 flying.

A modern passenger 747 has a maximum range of around 15,000 km, during which it uses 200,000 L of jet fuel (this assumes that travelling the maximum range requires the entire tank of fuel). Its typical cruising speed is Mach 0.8, or about 800 km/h, which is 0.22 km/s. The specific energy of jet fuel is 36 MJ/L. This means:

$$ \text{E} = 2\times 10^{5} \text{ L } \times 36 \times 10^{6} \text{ J/L } = 7.2 \times 10^{12} \text{ J } $$

Total time over which the energy is used:

$$ \text{t} = \frac{15000 \text{ km}}{0.22 \text{ km/s}} = 6.8 \times 10^{4} \text{ s} $$

Average Power:

$$ P_{ave} = \frac{E}{t} = \frac{7.2 \times 10^{12} \text{ J}}{6.8 \times 10^{4} \text{ s}} = 100 \text{ MW} $$

Solar Panels: How Do They Work?

The sun radiates power toward us, in the form of sunlight: this power is what drives everything from the water cycle to the growth of plants (and the creation of fossil fuels). The power from the sun given to an area of one square meter, assuming the sun is directly above us, is 1365 W/m^2 (this is known as a “flux”). The goal of solar power generation is to turn this incident sunlight into power we can use, such as mechanical or electrical power. A number of methods exist for extracting power from the sun, the two must straightforward being solar thermal power (using sunlight for cooking, or for heating water to run a turbine) and the growing of food and biofuel.

Solar panels are slabs of photovoltaic cells, and use the photoelectric effect to generate electricity from sunlight. Here’s how they work: light from the sun hits the photovoltaic cell. The result is a transfer of energy into the electrons of the cell, raising them into an “excited state”. Due to the material properties of photovoltaic cells (they’re semiconductors) excited electrons are free to travel through the cell, but only along one direction. This travel is a (DC) electric current, which can be attached to a load, such as a light bulb or motor.

Of course, we have to worry about efficiency, define to be the ratio between the power output and the power input. The efficiency limit for solar panels is around 30%; use of concentrators may increase this to around 60%. Common photovoltaics have efficiencies of around 10%, but we can assume 20%, which can be reached by the most expensive photovoltaics available today.

Powering a 747 With Solar Panels

Figure 1.  The schematics of at Boeing 747-8 7.

We will cover the top surface of the wings of a 747 with solar panels. The wing area is 525 m2 8

This gives a power of:

$$\text{P} = 1365 \text{ W/$m^{2}$} \times 525 \text{ $m^{2}$} = 7.1 \times 10^{5} \text{ W}$$

This is less than 1% of the needed power for the plane to take flight. To power an airplane, we would need solar panels covering an area:

$$\text{A} = \frac{1.0 \times 10^ 8 \text{ W}}{1.365 \times 10^2 \text{ W/$m^2$}} = 7.3 \times 10^4 \text{ $m^2$} = (270 \text{ m})^2$$

This is about the footprint of BC Place in Vancouver, which doesn’t sound too bad until we consider that at any given moment tens of thousands of planes are aloft in North America alone.

Figure 2. A to-scale comparison of a Boeing 747-87 to the area (in blue) solar panels would need to cover in order to power a 747’s flight (270 m)2.

Complications

There are lots of complications to this problem, of course.  Several were mentioned throughout the document, and a few more are listed below:

  • At high latitudes, the sun isn’t directly overhead, even at midday. In fact, the angle of incoming solar radiation depends not only on latitude but also on the time of day and the season. An airplane flying level to the ground at high latitudes will get much less than 1365 W/m2 (simple trigonometry shows that the flux from the Sun at latitude θ should be 1365 x cos θ where θ is the angle between the normal to the plane of the solar panel and the direction of the solar flux).
  • Flying any solar-powered airplane at night is impossible, since the flux from the Moon is orders of magnitude smaller than the flux from the sun.  Flying below clouds is also problematic, since clouds reduce the power from the Sun by a factor of 106.
  • If we decided instead to use ground-based solar power to fuel our airplanes, there are other forms of solar power generation, such as the Stirling engines mentioned earlier, that potentially have higher efficiencies than photovoltaics10.
Introduction

How much energy does it take to move a person or a tonne of freight from A to B?

Big Ideas:

  • When comparing the energy cost of different modes of transport, it is important to use the appropriate units for a more accurate description
  • When comparing the energy cost of different modes of transport, it is important to consider the size of the vehicle and the number of people it can carry
Work and Energy

The amount of energy required to get from A to B is equal to whatever force is required to push the object in the direction it is going in, multiplied by the distance traveled, i.e. the mechanical work done. How much force do you need? If you are moving at a constant speed, then the only reason a force is needed is to overcome friction, or drag. For travelling on land or water, physics puts no constraints on how small this force can be. For flight, physics imposes fairly strict limitations on how low the transport cost can be (because one has to move fast to stay aloft). However, when the transport cost is expressed in appropriate units, remarkable similarities can be seen in very different modes of locomotion.

Units

When dealing with generalities, the most useful units for the expression of transport costs are something like:

$$\frac{\text{energy}}{\text{distance $\times$ mass}} $$

Energy can be given in J, kJ, MJ, GJ or kWh

Distance is usually given in km

Mass, in kg or tonnes, can refer to that of the entire vehicle or only that of the useful load. Or, it can be a single passenger or 100 passengers.

Dimensional Analysis

If you want to make an educated guess about what the energy cost of transport should be, consider a dimensional analysis approach:

$$\frac{J}{kg \cdot m} = \frac{kg \cdot m^2 / s^2}{kg \cdot m} = m/s^2 $$

So transport cost has the same dimensions as acceleration. What is the most relevant acceleration when considering moving stuff from A to B? The acceleration due to gravity, g, is the most likely culprit. If g were equal to 0, you could push ten-tonne loads around with your little finger, albeit rather slowly.

The acceleration due to gravity in energy units is 9.8 J/(kg·m) or 9.8 MJ/(tonne·km). This would be the energy cost of lifting an object vertically, or of sliding it along the floor if the coefficient of kinetic friction, μ k = 1. Of course we can do better than this, as we have wheels and wings. However, the efficiency of our engines,  the ratio of mechanical work done to chemical energy in the fuel burnt to do it, is never much bigger than about 30 or 40%, and this works against us. Hence the energy cost of most forms of transport is somewhere between 3 and 6 times better than our dimensional estimate, as can be seen from the following table.

Energy cost of transport per total mass of laden vehicle (direct fuel consumption only)
Mode of Transport Energy cost (MJ/tonne/km)
Walking (5 km/h)3 3
Boeing 747-3004 1.8
Cycling (human powered, 20 km/h)3 1.5
2005 Honda Civic (2 persons)5 1.4
Electric bicycle6 0.4

Some questions arise instantly:

Why is walking at 5 km/h almost twice as costly as a Boeing 747 flying at 900 km/h, tonne for tonne and km for km?

(a) Walking is a rather inefficient means of getting from A to B, as you cannot “free-wheel” or “glide”. Note how much more costly walking is compared to cycling. (b) When walking slowly, about half of your power output is simply going into keeping your body functioning and warm. (c) Comparing by the tonne is misleading: a 400-tonne 747 carries about 400 people, so it takes a tonne of 747 to lift and move one person; when walking, the “vehicle” and the “passenger” are, of course, one and the same.

Why is human-powered cycling three times as costly, energy-wise, as an electric bike?

Here we are comparing food energy input to electrical energy input. The former has a mechanical efficiency of 20-30%. The mechanical efficiency of an electric motor is much higher. If the electricity used to charge the batteries comes from a thermal power station, and we use the total chemical energy input instead of the electrical, the electric bike would look worse than the human-powered one (due to transmission and charging losses). On the other hand, if we considered the energy used to produce the food the human ingested … it gets very complicated (but I don’t eat any less when I don’t cycle, I just put on weight).

Cost per Passenger

While MJ/tonne/km makes sense physically, MJ/km/(100 passengers) is a more meaningful unit economically (and environmentally). Redoing the table in these terms:

Energy cost of transport per passenger (direct fuel consumption only)

Mode of Transport Energy cost (MJ/(100 passengers)/km)
Walking 20
Boeing 747-300 (400 pass – 100% capacity) 140
Boeing 747-300 (240 pass – 60% capcity) 230
Cycling (human powered) 16
2005 Honda Civic (1 person – 20% capacity) 200
2005 Honda Civic (2 persons – 40% capacity) 100
2005 Honda Civic (4 persons – 80% capacity) 50
Electric bicycle 6 (Ref.6)
BC Transit Bus (Trolley, 55 persons – 100% capacity) 187
BC Transit Bus (Diesel, 60 persons – 100% capacity) 407
BC Transit Skytrain (Mark 1, 80 persons – 100% capacity) 117
Intercity rail 20 (Shinkansen data2)-170 (Amtrak8)
Tram (total system cost) 30 (Croydon tramlink2)
BC Transit Sea Bus (400 persons – 100% capacity) 757
Ship 200 (freight2)-1000 (cruise ship9)

This makes much more sense intuitively. Now the most interesting comparison is between a fully-laden 747 and a Honda Civic; the energy cost is the same for a plane as for the Civic with one and a half occupants, although the plane is moving at 10 times the speed. How come? Simply put, the 747 has no rolling resistance, is a much more aerodynamic shape, of necessity, and doesn’t stop at traffic lights. In both cases it takes about a tonne of vehicle to move one person. A 60%-full 747 (200 passengers) needs almost the same amount of energy to get it from A to B as a full one, because the passengers account for such a small fraction of its total mass, so the energy required per 100 passengers per km is about 1/0.6 ≈ 1.7 times as much as for the full aircraft. The average load factor (ratio of filled seats to available seats) for US scheduled airlines is about 60%10.

Notice how surprisingly bad sea travel is; this is mostly because it takes a very large mass of boat to float one passenger (5 tonnes for a BC Ferry11, 40 tonnes for the QE2 cruiseliner12) or one tonne of freight.

The bicycle remains unbeatable for short journeys. Trams are best for journeys of intermediate length, high-speed rail for longer ones; for low rolling resistance you cannot do better than steel on steel13. In the absence of public transport, pile as many people as you can into a Honda Civic.

      

Order of preference

  

Floating global warming

Introduction

It takes energy to fly. How much?

Big Ideas:

  • To fly, you have to move in a fluid, and move that fluid around, which takes energy.
Physics of Flight

The basic physics of flying has been well understood since about 19171. We do not intend to go into these details here; we are going to concentrate on the energy and environmental cost of air transport. The energy cost of transport is how much energy it takes to move a given mass a given distance2.

For an unconventional approach to the pedagogy of flight, see Ref.3

Why do we need energy to fly?

In order to fly, aircraft or birds need to continually push air downwards so that the deflected air pushes back and provides a lift force. This continual addition of momentum to the air requires a continual input of mechanical energy. For straight and level flight, this energy comes from the burning of fossil fuels in the engines of an aircraft, or the conversion of food energy in the muscles of a bird. For gliding flight, the energy comes from the loss of potential energy as the glider descends.

Measuring the energy cost of flying

Consider the forces acting on an aircraft in level flight at a constant velocity. All forces have to sum to zero, so we can be sure that the lift L equals the weight W, and that the thrust from the engines T equals the total drag D. The best measure of the quality of an aircraft, from the point of view of minimizing the energy cost, is the ratio of lift to drag, L/D. Plainly, the less drag you have for a given weight, the less thrust, and therefore energy, you need to get the aircraft from one place to another. The newest airliners like the Boeing 787 have lift to drag ratios of about 20. High performance sailplanes (i.e. gliders) have much higher ratios, but they are not configured to carry useful loads.

 

Now consider what happens if we turn off the engines and let the aircraft glide. For the sum of the forces to be zero, the aircraft pitches down and assumes a glide slope of angle θ. You can see from the diagram below that tan θ = D/L.

 

So an aircraft with L/D = 20 will glide at an angle tan θ = 1/20, i.e. θ ≈ 3°. It is no coincidence that all airports require airliners to approach the runway for landing at an angle 3°, for at this angle airliners are almost gliding, and the engines are therefore fairly quiet.

Given L/D it is a simple step to calculate the energy cost of transport. If an airliner has a glide slope of 1 in 20, that means it loses potential energy mgh for every distance of d = 20h travelled. Hence the energy cost of transport, energy divided by (mass times distance) = mgh/(md) = gh/(20h) = g/20 ≈ 0.5 m/s2 = 0.5 MJ/(tonne.km).

Now look up the fuel and range statistics for a Boeing 747-300 (the long-haul version, which we choose because short-haul versions spend a larger proportion of their time and energy taxiing around airports, accelerating and braking):4

Maximum mass5: 378 tonnes

Fuel capacity: 199,000 L

Range: 12,400 km

If we reckon on the aircraft carrying, on average, half its fuel capacity, and fuel has a density of 0.7 kg/L, then at the mid-point of a long flight, the mass of aircraft is about 300 tonnes. Jet fuel has a heat of combustion of around 34 MJ/L.

Energy cost = (127,000 L)(34 MJ/L)/((200 tonnes)(15,700 km)) = 1.8 MJ/(tonne.km)

Hmm. This nothing like the 0.5 MJ/(tonne.km) we estimated from the glide slope. We need to take account of the fact that the engines are not 100% efficient, in fact they are only about 35% efficient6, so for every 100 J of fuel burnt, only 35 J goes into pushing the aircraft forward; the rest just heats the environment (directly).

Hence a better calculation from the glide slope (which is 1 in 18 for the 7477) would be:

Energy cost = (g/18)/(0.35) ≈ 1.6 MJ/(tonne.km)

This is about 10% less than the in-service data – not bad for a simple calculation. No wading through tables of data. Just two numbers: the glide slope and the engine efficiency. This instantly tells us the two things that have to be improved to reduce the energy cost in terms of MJ/(tonne.km): the glide slope and the engine efficiency. However, each is a long slog. The glide slope is determined by the aerodynamic cleanliness of the aircraft, particularly the slenderness of the wings. However, the overall shape of a jet airliner has changed little since the Boeing 707 prototype first flew in 1954, and so there have not been enormous improvements in L/D.

Lift-to-drag ratios for aircraft and birds (years shown are first-flight dates)
Type (L/D)max
$5 balsa glider8 4
Sparrow9 4
Gull8 11
Albatross8 20
Piper Warrior (shown in movie) 10
Boeing 707 (1954)7 18.5
Boeing 747 (1969, shown in movie)7 18
Boeing 787 (2009)10 21.5

Note how small things like sparrows and balsa gliders tend not to fly very well. Big birds like the Albatross fly much better than small ones. Ditto for aircraft. How to set up a simple experiment to measure the glide slope and L/D ratio for a balsa or paper glider can be found in our “Balsa Gliders and 747s” article11.

Engine efficiency is determined in part by how hot one can run the combustion chamber, and this is determined by the quality of materials used. A factor of two has been gained in the last 50 years; efficiencies have risen from about 17% to 35%6.

Energy per passenger-km or per tonne of freight

So far we have only considered the energy cost per tonne of aircraft, as that is the number basic physics tells us. Plainly a more telling quantity is the energy per passenger-km. Given that the energy per tonne-km is closely constrained by physics, the next most consideration in airliner design is the mass of the aircraft divided by the number of passengers. The more passengers you can get into a lighter aircraft the better. Here, materials science is having an effect. The Boeing 787 is largely built of composite materials while the 747 and earlier models were entirely aluminum. Comparing the long-haul 787-9 with the long-haul 747-300 we see that the former has a mass of 0.7 tonnes per seat, and the latter 0.8 tonnes per seat – a 14% improvement12.

The Jevons Paradox

Will these improvements in efficiency help the environment? Unlikely. As W. Stanley Jevons noticed when studying the British coal industry in the 1860s13, the more efficiently a resource is used, the more that resource gets used. In other words, improvements in the energy efficiency of flying are unlikely to reduce the carbon footprint of the global aviation business. The more efficiently planes fly, the cheaper flying will become, and the more people will fly, and the greater will be the GHG emissions. Too bad.

150 tonnes fuel x 44/14 = 470 tonnes CO2

They still do it better than we do

Introduction

How can wasps endanger a commercial airline flight?

Big Ideas:

  • Pitot tubes are used to measure speed of fluid.
  • Pitot tubes are applied to determine the air speed of aircraft.
  • Bernoulli’s equation forms the basic explanation of how a Pitot tube measures fluid speed.

On November 21, 2013, an A330 flight had to make an emergency landing, shortly after takeoff at Brisbane Airport 1 . The problem was with a Pitot tube used to measure air speed of the aircraft, which is the speed of the aircraft relative to the air.   Fig. 1 shows a picture of the underside of the airplane where there are multiple Pitot tubes installed.  The Captain’s Pitot tube was giving false readings of air speed causing the emergency landing.  The malfunction was due to blockage of the Pitot tube by a nest of a mud-dauber wasp 2.   Why would such a blockage cause a malfunction?  In this article we discuss what a Pitot tube is and how it works.

Fig. 1.  The underside of the A330 involved in an emergency landing.  Multiple Pitot tubes are installed underneath.  The Captain’s Pitot tube was giving false readings of air speed.  The cause of malfunction was a blockage by a nest of a mud-dauber wasp.   Photo credit: Australian Transport Safety Bureau 3

What is a Pitot tube?

Pitot tubes are named after Henri Pitot, a French engineer in the 18th century.  The basic design of a Pitot tube is shown in Fig. 2.  A Pitot tube has a tube into which fluid enters and is quickly decelerated to zero velocity at the entrance of the tube because the  tube is not open on both ends.  Pitot tubes are used to determine the speed at which fluid flows relative to the Pitot tube.  Notice that a Pitot tube can generally be used to measure the speed of liquids and gases, but in the following we will discuss the example of air.  The pink and dashed stream line in Fig. 2 shows air travelling outside the tube with speed $v_{1}$ relative to the Pitot tube. The pressure outside the Pitot tube is $P_{1}$, which is the atmospheric pressure surrounding the Pitot tube. The air along the stream line shown in Fig. 2 is decelerated to zero velocity, relative to the Pitot tube, at the entrance of the Pitot tube and the pressure in the tube is $P_{2}$. Often a Pitot tube will have another tube around it that has slits in it, as shown in grey in Fig. 2. Since the streaming air is not entering this outer tube, the pressure inside the outer tube is just the atmospheric pressure, $P_{1}$, at the airplane’s current elevation. As we will see it is the difference in pressure between $P_{1}$ and $ P_{2} $ that will be useful to determine the air speed, $v_{1}$.

Fig. 2.  A cartoon drawing of a Pitot tube.  Fluid along a stream line, shown in pink and dashed, enters the inside tube at speed $v_{1}$. The fluid is quickly decelerated to zero speed, $ v_{2} = 0 $, at the entrance of the tube. The pressure of the fluid outside the Pitot tube is $ P_{1} $. In the case of our airplane example, $ P_{1} $ is the atmospheric pressure at the airplane’s current elevation. The pressure inside the inner tube into which the fluid decelerates is $ P_{2} $. The pressure in the outer tube is the atmospheric pressure, $ P_{1} $, because no fluid flows directly into it. Bernoulli’s equation relates the speeds $ v_{1} $ and $ v_{2} $ and the pressures $ P_{1} $ and $ P_{2} $. From this relation we can find the speed $ v_{1} $ of the fluid relative to the Pitot tube.

 

Fig. 3.  A Pitot tube installed on a Kamov Ka-26 helicopter.  The small holes are to measure the atmospheric pressure outside of the Pitot tube.  Photo credit:  Zátonyi Sándor 4 .

How does a Pitot tube work?

The basic principle of how a Pitot tube can be used to measure the relative speed of a fluid entering the tube involves Bernoulli’s equation.  Bernoulli’s equation relates the pressure and velocity of fluid along a stream line.  A stream line indicates the path of travel of fluid.  For the pink and dashed stream line in Fig. 2, Bernoulli’s equation states that

$$ P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho g y_{1} = P_{2} + \frac{1}{2} \rho v_{2}^{2} + \rho g y_{2} $$ (Eq. 1)

where $ y_{1} $ and $ y_{2} $ are vertical height positions along the stream line and $ \rho $ is the density of the fluid. Here $ y_{1} = y_{2} $ because there is no height differences involved along the stream line. The speed of the fluid inside the Pitot tube, relative to the Pitot tube, is $ v_{2} = 0 $ so that Eq. 1 becomes

$$ P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} $$. (Eq. 2)

We can now solve for the speed $ v_{1} $ of the fluid outside of the Pitot tube from Eq. 2 which gives

$$ v_{1} = \sqrt{\frac{2}{\rho}(P_{2} – P_{1})} $$. (Eq. 3)

The speed of the fluid relative to the Pitot tube, $ v_{1} $, can be determined from the difference in pressure outside and inside the Pitot tube. This difference in pressure can be measured with a pressure measurement device.

A Pitot tube on an aircraft can become blocked by ice, dirt, or in the case of the incident mentioned in the introduction, a nest from an insect. In this case, the pressure reading $ P_{2} $ inside the Pitot tube is affected, resulting in incorrect air speed measurements (measurements of $ v_{1} $). Correct air speed readings are critical for maintaining lift of the airplane and for flight safety, which makes properly functioning Pitot tubes an essential part of the aircraft.

Pitot tubes are not only used for air speed, they are also used on boats for water speed, and various industrial applications where the speed of fluid needs to be measured. Bernoulli’s equation as used in this article still applies even though the fluid is different.

Footnotes   [ + ]

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