4
$\begingroup$

Given an aircraft type and fixing its wing configuration (flaps, slats) and fixing its speed and altitude and assuming it flies horizontally, what is the relationship between the aerodynamic drag and weight of the aircraft?

Of course this relationship depends on the aircraft itself and on many other things, but what is more or less the power-law?

Say we compare two A320s, the one heavier by a factor of 1.3. Of course the heavier one will fly with a higher angle of attack, because it needs more lift. This will create more drag and the engines will run with higher power to achieve the same speed. How much more drag will it be? If the behaviour is approximated as power-law, what would be the exponent? The cross section increases when we "pitch" the aircraft. Is the aerodynamics in the design process optimised for "half-loaded" aircrafts? Also, how does this translate to fuel consumption, say on 40000 feet with the usual speeds?

$\endgroup$
3
$\begingroup$

If you allow to make a few simplifications, the answer is easy:

  • Friction drag is not affected by the angle of attack change. That means no onset of flow separation on the heavier aircraft.
  • The wing's optimum L/D is reached with the same flap setting in both cases.
  • Lift changes linearly with angle of attack, so the lift coefficient $c_L$ can be expressed by the product of lift curve slope $c_{L\alpha}$ and angle of attack $\alpha$.
  • We neglect the changed lift contribution of engine thrust when the angle of attack is increased.

Now the drag $D$ can be expressed by this equation: $$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\left(c_{D0}+\frac{(c_{L\alpha}\cdot\alpha)^2}{\pi\cdot AR\cdot\epsilon}\right)$$ Per the definition above the term $c_{D0}$ is constant, so the change in drag between the lighter aircraft (index 1) and the heavier aircraft (index 2) will be $$\Delta D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\frac{c_{L\alpha}^2\cdot\left(\alpha_2^2-\alpha_1^2\right)}{\pi\cdot AR\cdot\epsilon}$$ In order to express this drag difference $\Delta D$ in terms of the aircraft's mass, write the lift coefficient $c_L$ as $\frac{2\cdot m\cdot g}{\rho\cdot v^2\cdot S}$: $$\Delta D = \frac{g\cdot\left(m_2^2-m_1^2\right)}{\pi\cdot AR\cdot\epsilon}$$

The other symbols are:
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} S\:\:\:\:\:$ wing surface area
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor
$\kern{5mm} g \:\:\:\:\;$ gravitational acceleration

Now we can answer your questions:

How much more drag will it be?

The drag will increase with the square of the mass increase. The gradient of that increase depends on the span loading of the aircraft.

what would be the exponent?

2

Is the aerodynamics in the design process optimised for "half-loaded" aircraft?

No, always for the fully loaded aircraft, as lighter loads can be tolerated much better than higher loads. However, since fuel burn will cause a change in aircraft mass over time, the aerodynamics must work over a range of altitudes.

how does this translate to fuel consumption, say on 40000 feet with the usual speeds?

Drag is compensated by thrust, so you need more thrust to overcome the higher drag. Fuel consumption goes up linearly, but since cruise at 40.000 ft means that the aircraft is light and the engines are running at close to their maximum sustained thrust, a mass increase by 30% is impossible. For a practical result, the heavier aircraft would fly at the same angle of attack and speed but at a lower altitude where both the increased lift and the increased engine thrust can be provided by the higher air density.

If you need to calculate the trip fuel at different weights: We had a similar question before, so please follow the link for an explanation.

$\endgroup$
  • $\begingroup$ Why do you say, a mass increase by 30% is impossible? What is the mass ratio for commercial aircraft comparing start and end of cruise, when they operate on maximum range? $\endgroup$ – polaris12246 Apr 7 '18 at 5:12
  • $\begingroup$ @AlessandroBrillante: No, all I say is that a mass increase of 30% is impossible for an airliner at its given cruising altitude. $\endgroup$ – Peter Kämpf Apr 7 '18 at 5:15
  • $\begingroup$ +1, but applying it has been bugging me since yesterday, so I just posted a tie-in question: aviation.stackexchange.com/q/50304 $\endgroup$ – ymb1 Apr 7 '18 at 5:30
1
$\begingroup$

It is hard to say what the ratio would be exactly. Total drag consists of parasite drag and induced drag. The influence of angle of attack to parasite drag would depend largely on the aircraft design and is usually designed to be lowest at average flight condition weight and at cruising speed and altitude. However I think the effect of angle of attack to parasite drag will be minor at such low angles.

For the induced drag I can at least give a mathematical prediction. Since everything in the lift equation but the angle of attack stays fixed in our example, angle of attack needs to be proportional to weight in horizontal flight. We know that for small angles of attack the coefficient of lift is directly proportional to the angle of attack. Induced drag is proportional to the square of the lift coefficient. Therefore the induced drag would also be proportional to the square of the angle of attack. So 1.3 times the weight means 1.3 times the AoA means 1.3 times the Cl means 1.3^2 times the induced drag.

However at cruising speed induced drag usually accounts for half of the total drag.

So your formula would be something like: Increase in drag= (50%)+ (50%)*1.3^2

Assuming engine thrust is proportional to its fuel consumption, the same formula could be used for fuel consumption

In conclusion weight plays an even larger role in the climb and during landing and takeoff where induced drag plays a major role. At high speeds weight is not as much of an issue.

$\endgroup$
  • $\begingroup$ "at cruising speed induced drag usually accounts for less than 10% of the total drag" are you really sure about that? We are talking about airliners here, and they fly close to their optimum range polar point where induced drag is half of total drag. $\endgroup$ – Peter Kämpf Apr 6 '18 at 19:54
  • $\begingroup$ Thinking about it now, I think you are right it should be half of total drag. I should have thought about it more deeply instead of just putting in numbers of some article. Somehow didn't occure to me that I wouldn't even have had to look it up in the first place. Thanks for noticing! $\endgroup$ – blenderTyphoon Apr 7 '18 at 22:17
0
$\begingroup$

You kind of nailed it by saying that increased weight will require a higher angle of attack.

A heavier plane will indeed require a higher angle of attack to provide more lift.

You mentioned that the plane is flying horizontally, with a constant speed. In that case, the plane is at an equilibrium, the weight is compensated by the lift and drag is compensated by traction.

Let's say we want to fly the same plane, but with more cargo. We would need to increase lift. What can we act upon ? The lift formula sums it all :

foo+bar

Rz lift , ρ Volumic mass of air, S wing surface, V Speed, Cz Lift coefficient

Therefore a weight increase will either require to increase the volumic mass of air (huh?), the wing surface, the speed of the plane or the lift coefficient.

$\endgroup$
  • $\begingroup$ to increase the density of air, just fly lower. For extreme density, try Siberia in winter. $\endgroup$ – Peter Kämpf Apr 6 '18 at 19:48

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.