7
$\begingroup$

Consider two possible flight paths for a jet passenger plane; the first is a slow climb, held low over 20 or so miles to avoid other flight paths, the second a more rapid climb over 10 miles to the capacity of the plane rather than restricted.

It has been said to me that the more rapid climb will reduce CO2. But logic to me is that the rapid height gain will burn more fuel and so increase CO2 production (or perhaps are the 2 not linked I wonder). I would appreciate being put out of my mystery on this?

$\endgroup$
  • 3
    $\begingroup$ Short answer: the higher you go (up to a comfortable minimum distance between critical mach and stall) the less fuel you will burn. The sooner you can get there, the better, but there are often other considerations. $\endgroup$ – Terry Jun 28 '14 at 18:33
  • $\begingroup$ Whilst fuel burn at higher altitudes is lower is it too simplistic to compare it to putting your foot down in a car to reach 60 in 8 seconds as appose to taking 20 seconds with a lighter touch to the accelerator? Also are noise levels not higher by pushing to a higher altitude quicker. $\endgroup$ – Andy Jun 28 '14 at 19:23
  • 2
    $\begingroup$ staying low lets the noise pollute a larger area (leading to more complaints) $\endgroup$ – ratchet freak Jun 28 '14 at 21:01
  • $\begingroup$ @Andy The overriding factor, both for fuel burn and noise, is altitude, although I lack the physics background to understand the details of why much less than being able to explain it to others. The fuel burn lessens dramatically with altitude. Using a 747-200 as an example, the fuel burn for each engine on short final for landing will be about 5,000 lbs per hour. That's approximately the same fuel burn as at 39,000 in cruise at mach 0.86. Further, at least as I remember (it's been 15 years), the fuel burn on takeoff after liftoff is around 4 times that. $\endgroup$ – Terry Jun 28 '14 at 22:17
  • 2
    $\begingroup$ @Terry: The fuel burn at takeoff is so much higher because you create more thrust than needed for sustained flight. The excess is used for acceleration and climbing. Also, if you fly fast and low, you are far from the speed of minimum drag, so your plane flies less efficiently, consuming more fuel. This is possible at all because the engines have four times as much thrust available at sea level than in 40.000 ft. $\endgroup$ – Peter Kämpf Jun 29 '14 at 5:51
8
$\begingroup$

You are right, climbing adds to the fuel consumed, and increases the energy which is added to the plane, but this energy is conserved and can be reclaimed later, on descent. There are two kinds of energy to consider, potential and kinetic energy. Potential energy is weight times height, and kinetic energy is mass times speed squared, and both can be interchanged. $$E = \frac{m}{2}\cdot v^2 + m \cdot g \cdot h$$ Nomenclature:
$E \:\:\:$ energy
$m \:\:\:$ mass
$v \:\:\:\:$ speed
$g \:\:\:\:$ gravitational constant
$h \:\:\:\:$ height

The thrust of the engines over what is needed to overcome drag, accumulated over time, builds up energy. Lift and Drag, in turn, are proportional to airspeed squared times density of air, so if the airplane climbs, it needs to fly faster to compensate for the thinner air, and drag will stay roughly constant. Thrust is also proportional to the density of the air around the aircraft, and the same goes for fuel consumption. This means that the aircraft will approximately consume the same amount of fuel over time when it flies low and slow, and when it flies high and fast. Since the task is to transport people or goods to a specific destination, flying high and fast will result in less running time at the same fuel flow. This is the reason why flying high results in less CO2 produced!

When the aircraft comes close to its destination, it can now put the potential energy which has been stored as altitude to good use: In a descent, the thrust generation can be reduced by the amount of energy which is now reclaimed, just as you can ease off the gas pedal in your car when coasting downhill. So for the last half hour of the flight, the aircraft will now consume less fuel than before, and by the time it has reached its destination, the difference is just as big as the excess in consumption which was needed during climb.

For the nitpickers: The aircraft will weigh less at the end of the trip (fuel has been consumed), so the descent will not completely compensate for the climb. On the other hand, the lower temperature at altitude will reduce fuel consumption by letting the engines run more efficiently, so in the big picture a definite gain remains.

$\endgroup$
  • $\begingroup$ Thanks for all these comments. I understand what is being said about the journey as a whole but am really focussing on that take off process. From above, comparing a slow 20 mile climb to a short 10 mile climb to say 7000 feet. Are we saying that the slow climb produces more CO2 and consumes more fuel in this specific period. Also the noise generated over the rapid 10 mile height gain must be higher than over the same 10 mile distance as a slow climb? Thanks for the expert opinion being offered, really useful in helping me understand the issues $\endgroup$ – Andy Jun 29 '14 at 7:12
  • 3
    $\begingroup$ CO2 is mainly proportional to engine running time. Now if you compare a 10 mile climb at full thrust to a 20 mile climb at 80% thrust, still the runnig time will dominate the result. If we use Terry's 747 as an example, it will just stay aloft with three of the four engines running. This means you need to produce somewhat above 75% maximum thrust to climb slowly. Climbing more quickly will always be more efficient because it needs less time, even though the CO2 output per minute is higher than in the slow climb. Noise is also lower, simply because the plane is farther from the ground. $\endgroup$ – Peter Kämpf Jun 29 '14 at 7:23
  • 6
    $\begingroup$ @Andy: The point is that no matter whether you have a long shallow climb or a short steep one, the energy you put into elevating the plane's mass to, say, 10 km is the same. Your spending that energy is compressed into a shorter time for the steep climb, but the total energy it costs to get to a higher altitude is the same. The difference is only in the amount of aerodymanic drag you additionally have to overcome, which is significantly smaller at greater altitudes. $\endgroup$ – hmakholm left over Monica Jun 29 '14 at 16:02
  • $\begingroup$ From looking at emissions data, most engines indeed have very similar CO indices for takeoff vs. climb out, but on some engines one mode is much larger than the other. $\endgroup$ – fooot Apr 15 '15 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.