What can throwing a balsa glider (or paper airplane) teach us about the fundamental energy cost of air transport?
When you throw a well-balanced paper airplane or model glider into still air at an appropriate speed, it flies in a straight line at a constant speed at an angle to the horizontal known as the “glide slope” and eventually hits the ground. Needless to say, the smaller this angle is, the better the glider. If you measure the mass of the glider, and the speed and the angle at which it flies, you can determine the rate at which it is losing potential energy. This is the power one would have to supply to the plane in order to keep it flying level.
The most important measure of energy use in transportation is energy consumed per unit mass per unit distance travelled; this is known as the transport cost[fn]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [11 Sept 2009]. [/fn]. It can be determined for the glider, and compared with full-sized aircraft. We call it Etrans (to emphasize that it is a measure of energy and not monetary cost) and the formal units are J/kg/m, kWh/tonne/km etc. but as we will see, physically it is just a constant times g, the acceleration due to gravity.
Experiment: Determining the glide slope
Find a balsa glider or make a paper airplane. Take it in a large space (an empty lecture theatre is ideal) and throw it a few times in such a way that it flies in a straight line at a constant speed. You will rapidly discover that it has a natural flying speed and that it flies in the straightest line if you throw it at that speed. Using a tape measure and stop watch, or digital camera, measure the flying speed v and the glide slope(vertical displacement (d)/horizontal displacement (h)). Determine Etrans, in kWh/tonne/km and as a multiple of g.
Sample calculation
Say the glide slope is one in five, i.e. the glider loses h = 1 m in height for every d = 5 m (travelled horizontally). The potential energy loss is mgh, so
$E_{trans} = \dfrac {mgh}{md} = \dfrac{gh}{d} = 0.2 g$
So, for our glider, the transport cost is about 0.2 g ~ 2 m/s2. Let’s convert this to more conprehensible units of kWh/tonne/km. A kWh is 3.6 MJ, a tonne is 1000 kg, and a km 1000 m, so a kWh/tonne/km is 3.6 J/kg/m = 3.6 m/s2.
In other words, the transport cost for the glider is 2/3.6 ~ 0.6 kWh/tonne/km.
The transport cost of a typical long-range jetliner like the 747-400 flying is about 0.5 kWh/tonne/km[fn] Wikipedia. Boeing 747-400 (online). http://en.wikipedia.org/wiki/Boeing_747-400.[28 August 2009][/fn].
Remarkable eh? The comparison of these two numbers provokes differing reactions from different people:
\begin{array}{|lll|}
\hline
\textnormal{Observation} & \textnormal{Physicist} & \textnormal{Non-physicist}\\ \hline \hline
\textnormal{0.5 vs 0.6} & \textnormal{Almost the same!} & \textnormal{Not the same} \\ \hline
\textnormal{The two transport costs } & \textnormal{Why should they be} & \textnormal{Why shouldn’t they be?} \\
\textnormal{are about the same.} & \textnormal{almost the same?} & \\ \hline
\end{array}
How do your reactions compare with this table?
Comparison between balsa glider and Boeing 747
The approximate equality of these two transport costs is in fact something of a fluke. If the patterns of airflow around a balsa glider and a 747 were the same, these two numbers would be the same. However, air flows much more cleanly around a 747, and its glide slope is about 1/15, or about 3 times better than the glider. On the other hand, the mechanical efficiency of its engines is only about 30%, so the advantage due to the airflow is almost precisely cancelled out. The table below illustrates the differences between the balsa glide and the Boeing 747[fn value=2][/fn],[fn]Elert, Glenn. Energy Density of Aviation Fuel – Kerosene energy content (online). http://hypertextbook.com/facts/2003/EvelynGofman.shtml [11 Sept 2009].[/fn]. The transport cost is used to calculate the total energy needed to fly a 747 across the Atlantic Ocean, which is also converted into the amount of fuel burnt and amount of CO2 per passenger.
\begin{tabular}{|l|l|}
\hline
\multicolumn{2}{|c|}{Transport cost of a paper airplane or balsa glider} \\
\hline
Distance travelled & 5 m \\
Height drop & 1 m \\
Glide slope & 0.2 \\
\mathnormal{g} & 10 m/s^2 \\
Transport cost & 0.56 kWh/tonne/km \\
Transport cost & 2 MJ/tonne/km \\
\hline
\multicolumn{2}{|c|}{747 flying transocean} \\
\hline
Mass} & 400 tonnes \\
Number of Passengers & 400 \\
Distance travelled & 14,000 km \\
Transport cost & 0.5 kWh/tonne/km \\
Transport cost & 1.8 MJ/tonne/km \\
Total energy & 10,080,000 MJ \\
Kerosene energy content & 45,000 MJ/tonne \\
Total fuel burnt & 224 tonnes \\
CO_{2} \textnormal{produced} & 704 tonnes \\
CO_{2} \textnormal{per passenger} & 1.8 tonnes (one-way) \\
\hline
\end{tabular}
The sharp-eyed amongst you will note that 224 tonnes is more than half the starting mass of the aircraft, so the mass mid-way through the flight will be much less than 400 tonnes. To do this right requires some algebra, or a simple interative calculation as shown in Balsa_Gliders_and_747s_Spreadsheet.xls. If you take 1.8 MJ/tonne/km as a given transport cost, you end up with 175 tonnes of fuel burnt (or 1.4 tonnes carbon dioxide per passenger). An improvement, but still none-too-green.
But what about kinetic energy here? Haven’t we forgotten that the aircraft has to be accelerated up to speed? A Boeing 747 cruises at about 250 m/s, i.e. the kinetic energy is ½ mv 2 = 12.5 GJ, or less than 0.2% of the energy required for the journey. The energy used to get to cruising height (which we get back during descent at the end of the trip anyway), which is about 10 km, is also a small percentage of the total energy required. Potential energy = mgh = 39 GJ, or 0.5% of the energy required for the journey. All the energy used in a journey by air goes into overcoming drag – friction drag, and induced drag (caused by deflecting air downwards in order to keep the aircraft up).