Air Transport

The central question when considering any form of transport is how much energy $E_C$  is required to move a unit mass a unit distance. This will determine the resources and environmental consequences of moving goods and people around the globe. The units of $E_C$ are J/kg/m (or MJ/tonne/km), which reduces to m/s², so clearly a lot depends on g.

A little aerodynamics[note] J.D.Anderson, Introduction to Flight, McGraw-Hill, 6th Edition (2008) [/note]

The total drag on an aircraft is equal to sum of the “parasite drag” and “induced drag”. The “parasite drag” $D_p$ is caused by airflow being slowed as it passes the body. It has the form (Eqn.1):

$\begin{equation} D_p = \dfrac{1}{2} C_{D,0} \rho S v^2 \tag{1} \end{equation}$

Here $C_{D,0}$ is a dimensionless coefficient of drag that is determined by the overall shape of the aircraft (the “zero” means drag at zero lift – see below), $\rho$ is the density of air, $S$ the wing plan area (normally this would be the frontal area, but in aerodynamics the convention is different; the change makes $C_D$ much smaller than you might expect), and $v$ the air speed.

The lift $L$ has the same form, with a coefficient of lift $C_L$ replacing that of drag. You can imagine that the total aerodynamic force on the wing is mostly vertical but angled backwards a little due to air being deflected downwards (to generate the lift). Thus the lift contributes to the total drag force, and this is called induced drag, $D_i$, which is a dependent on the effective aspect ratio of the wing (length / mean chord), $A$. Expressions for this and total drag $D$ are given in Eqns.2-5:

$\begin{equation} L = \dfrac{1}{2} C_L \rho S v^2 \tag{2} \end{equation}$

$\begin{equation} D_i = \dfrac{1}{2} C_{D,i} \rho S v^2 \tag{3} \end{equation}$

$\begin{equation} C_{D,i} = \dfrac{C^2_L}{\pi A} \tag{4} \end{equation}$

$\begin{equation} D = D_p + D_i = \dfrac{1}{2} (C_{D,0} + \dfrac{C^2_L}{\pi A})\rho S v^2\tag{5} \end{equation}$

In level flight (i.e. most of most journeys), $L=W$, the weight of the aircraft. This constraint means that there is preferred flying speed where the drag is at a minimum; at higher speeds, the parasite drag becomes large; at lowers speeds, the induced drag becomes large.

At the minimum total drag speed, $v_{MD}$, the specific energy cost of transport is at a minimum. A little algebra gives Eqn.6:

$\begin{equation} v^2_{MD} = \dfrac{W}{\rho S}\sqrt{\dfrac{1}{\pi A C_{D,0} }}\tag{6} \end{equation}$

Evaluating the energy cost at this speed gives the simple result (Eqn.7):

$\begin{equation} E_C = \dfrac{D}{m} = 2g\sqrt{\dfrac{C_{D,0}}{\pi A}} \tag{7} \end{equation}$

One major factor has been forgotten here: the mechanical efficiency $\eta$ of the aircraft’s engines. This is important because so far we have only considered the mechanical energy, not the total energy expended. The ratio of the two, $\eta$ , is generally about 1/3; i.e. 2/3 of the chemical energy produced by burning fuel is lost as useless heat. Hence, Eqn.7 is really just the mechanical energy cost of transport; the real thermal energy cost of transport is about three times higher and is given by Eqn.8:

$\begin{equation} E_C = \dfrac{D}{m} = \dfrac{2g}{\eta} \sqrt{ \dfrac{C_{D,0}}{\pi A}} \tag{8} \end{equation}$

The above result contains all the engineering complexity of the airframe in just two numbers; the drag coefficient $C_{D,0}$ depends on the shape, and is 0.0155[note] Sometimes expressed as the minimum drag coefficient, which is double the zero-lift coefficient.[/note] for a 747[note] Boeing 747, http://en.wikipedia.org/wiki/Boeing_747-400 [2019-09-20].[/note] (slightly lower for more modern airliners); the effective aspect ratio $A$ is 7.4 (slightly higher for more modern airliners). All the thermodynamic complexity of the engines is rolled into the efficiency $\eta$.

Putting in the numbers: E_c = 1.5 J/kg/m = 1.5 MJ/tonne/km

Three other factors we have ignored:

• Airliners are flown at about Mach 0.85, and at this speed, air passing the airframe approaches the speed of sound in places, so supersonic wave drag is starting to rear its head, raising the total drag. The economics of the airline industry is different from that of cargo ships; staying well clear of the speed of sound would mean unacceptably long flights. At 0.85M , the 747 has a $L/D$ ratio of 17.

• Acceleration to cruise speed – energy added is ½ mv²; how important is this?

Put in some numbers for a 400 tonne Boeing 747 cruising at 250 m/s (900 km/h):

Kinetic energy = ½ mv² = 12.5 GJ

This energy has to be dissipated with air-brakes on approach to landing. But to set it in context, how far would you have to fly to expend 12.5 GJ overcoming the total minimum drag force (286 kN)?

Distance = 12.5 GJ/286 kN  = 44 km

i.e. the fuel expended to accelerate to cruising speed is tiny compared to a long haul flight of many 1000s of km.

• Potential energy added to gain cruising altitude – mgh.

The potential energy of a 747 at 10,000 m is (400,000 kg)(9.8 m/s²)(10,000 m) = 40 GJ. This is larger than the kinetic energy, and the aircraft is not operating optimally during the climb, but you get it all back during final descent (gravity is a “conservative field”), so it counts little in the overall energy cost of the flight.

Comparison with real life data

A Boeing 747-400 has a range maximum range of 13,450 km burning approximately 200 kL of fuel (with a heat of combustion of 36 MJ/L).  At the start of a long flight the aircraft has a mass of 400 tonnes, and at the end, 250 tonnes. Taking a mean mass of 325 tonnes, these data yield:

E_c = 1.8 MJ/tonne/km

Environmental Cost

A 747’s tanks carry about 150 tonnes of fuel. How much CO2? Multiply this mass by x 44/14[note] 14.17 would have been a slightly more accurate number – see https://c21-wp.phas.ubc.ca/article/fossil-fuels-for-transport/ [/note] and you get 470 tonnes CO2; multiply by 18/14 and you get 190 tonnes H2O (10 km up where it shouldn’t be). See Fig.3.

A note on “Fuel Distance”

David MacKay[note] D.MacKay, Without Hot Air: withouthotair.com/cC/page_276.shtml [2019-09-20].[/note] introduced the idea of an intrinsic fuel distance:

d_Fuel = C/g

Here C is the enthalpy of the fuel, 45 MJ/kg. Hence d_Fuel = 4,600 km. MacKay argues that if you multiply this by the maximum L/D, and by the mass fuel fraction of the aircraft (or bird) and efficiency of the engines, you get the range. In the 747 case, (4,600 km x 17 x 175/400 x 1/3) = 11,400 km. The record non-stop flight for a bird is about the same, for precisely the same reasons.

 

Conclusion

Our simple-minded result for the energy cost of air transport of 1.5 MJ/tonne/km is remarkably close to operating values, and only slightly lower for airliners more modern than the 50-year-old 747. The operating value for an Airbus 380 (almost 50% more passengers than the 747) is 1.6 MJ/tonne/km. The energy cost per seat-km has, however dropped by a factor 3 in the last 50 years; the major factors being engine efficiency and airframe mass per seat. Changes due to improved aerodynamics have been smaller, and of course, the energy content of the fuel in MJ/kg has remained constant.

 

Solar Powered Aircraft?

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)[fn]Boeing.  747-8 Technical Characteristics (online).  http://www.boeing.com/commercial/747family/747-8_fact_sheet.html [27 May 2010].[/fn].  Its typical cruising speed is Mach 0.8, or about 800 km/h, which is 0.22 km/s[fn]Wikipedia.  Speed of Sound (online).  http://en.wikipedia.org/wiki/Speed_of_sound [27 May 2010].[/fn]. The specific energy of jet fuel is 36 MJ/L[fn]/article/useful-numbers[/fn]. This means:

Total energy used in flight:

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

Total time over which the energy is used:

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

Average Power:

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

Solar Panels on the Wings?

On a bright sunny day at noon, the solar intensity will be about 1 kW/$^2$. Typical solar panels are about 10% efficient. That means if you manage to arrange the solar panels to face the Sun perpendicularly, you will need 1,000,000 m$^2$ of solar panels to generate 100 MW, and then only around the noon hour. The wing area of a Boeing 747-8 is 550 m$^2$, i.e. several thousand times too small even to consider the possibility of commercial solar powered flight.

 

Updated 2019-09-20