1
$\begingroup$

An aircraft's wing generates force of lift by directing the oncoming airflow down. How is this momentum balanced when the aircraft is not climbing, but maintains level flight?

$\endgroup$
1
0
$\begingroup$

Lift and gravity are "balancing" forces, just as thrust and drag.

The engine provides energy to the system. When there is a "surplus" of energy, it is converted to lift and speed. There is a statement in the aviation world, "The throttle provides, the elevator divides." As power is increased, if one desires to maintain level flight, the elevator is deflected appropriately, and the speed will increase.

Depending on one's perspective, some of the lift is generated by a reduction of pressure above the wing surface. This also applies to the above statements. It is not necessarily all deflection downward.

$\endgroup$
0
$\begingroup$

An airplane in level flight, cruising, needs lift to fight the gravity. It lacks buoyancy as opposed to blimps.

It sometimes needs even more power when cruising in not the optimal configuration of payload and misplaced CG of the load.

So even when a plane is not climbing it needs lift. That lift is provided by changing the airflow momentum.

$\endgroup$
0
$\begingroup$

The angle of a typical wing in flight is slightly greater than zero. So the wing/airfoil deflects the air downwards thus creating the necessary Lift for keeping the airplane up.

enter image description here

Regarding the momentum, in order for the air to change direction because a force is applied on it. Due to the Newton's second law of motion, an equal and opposite force is applied by the air on the foil. That force is the aerodynamic force, which is decomposed to lift and drag.

enter image description here

In order for the plane to maintain height and speed, the following forces are at equilibrium:

  • On the horizontal axis: Thrust and Drag
  • On the vertical axis: Lift and Weight

enter image description here

$\endgroup$
3
  • $\begingroup$ again, if you're in level flight the wing is not pushed up, but merely stays at the altitude. But the air is still pushed down. Something else has to gain upward momentum $\endgroup$ – Francis L. Dec 8 '20 at 23:58
  • $\begingroup$ If I start with 3 units of still air, and I accelerate 1 of them to a velocity of 2 and the other 2 to a velocity of -1, I have conserved momentum, and the forces balance, but notice that I have not conserved kinetic energy. So when you conserve kinetic energy and momentum, forces are unbalanced. We do not need to unbalance momentum to create a force. We can redistribute energy while conserving momentum and create forces. There is downwash behind the wing between the wingtips, but there is upwash in front of the wing and outboard of the wingtips. $\endgroup$ – Phil Sweet Dec 12 '20 at 3:02
  • $\begingroup$ Both linear and angular momentum are conserved in all axes in this problem. The lift force comes from the creation of a pair of counterrotating wake flows in the transverse plane. These flows perfectly cancel each other. There is no change in momentum in the flow domain. I'll add that last bit to my answer. $\endgroup$ – Phil Sweet Dec 12 '20 at 3:06
0
$\begingroup$

TLDR: If I stand in front of two wheels and push equally so that they counterrotate, the linear momentum starts and ends at zero, and the rotational momentum starts and ends at zero (counterrotation cancels momentum of individual wheels), but there is a force exerted on me.

In the 2D flow model
When you consider a large control boundary around the wing, all the vertical moment vanishes in the flow. Behind the foil a sufficient distance, the flow is once again perfectly horizontal - it's identical to the upstream flow. This is because the flow field is modeled as the sum of two items. The first is the farfield flow - straight, uniform flow from front to back. The second is a circulation flow. From a distance, this looks like a circular vortex centered on the wing, and the velocity trends to zero as you get further away. All the vertical momentum terms in the flow are contained in this circulation flow, and there is as much going up in front of the wing as there is going down behind the wing.

The combined fields yield a force on the wing, due to the different velocities and pressures on the top and bottom surfaces of the wing. But there is no work being done because there is no translation in the direction of the force. And there is no momentum change in the fluid mass taken as a whole. The flow far downstream is identical to the flow far upstream. This is true not just in steady state, but in startup and slowdown as well.

In the 3D flow model
Things get a bit more complicated. Here, there is a change between the upstream far field velocity and the downstream far field velocity . But just as in the 2D case, there is no energy exchange between the wing and the flow. So the transverse momentum in the wake has to have come from the fluid itself. What happens is that the wake is travelling a bit slower, on average, and this momentum deficit is manifested in the transverse flow that persists downstream of the wing. But the momentum in the transverse plane of the wake still balances. There is as much going up as going down, and there is as much going right as going left.

So for real-world objects in a breeze, if there is a lift force

  1. There will be a net momentum change in the downstream flow that corresponds to that force. There will be persistent downstream transverse flows.

  2. The energy needed to produce the transverse flow in the wake comes from the only available energy source - the initial velocity of the upstream flow. So there is an inescapable stream-wise energy and momentum defect in the wake, and this produces drag on the object.

Momentum comes in two flavors - linear and angular momentum. Both are conserved quantities when no work is done on the system, as is the case here.

In the 3D case, where the wake flow does not return to the upstream condition, both types of momentum are none the less conserved.

Viewing the transverse velocities in a transverse plane far behind the wing, there is equal linear momentum up and down, and equal linear momentum right and left. Summed over the domain, the transverse linear momentum is unchanged. Between the wingtips, there is downwash. Outboard of the wingtips, there is upwash.

The rotational momentum starts at zero in front and must remain at zero behind the wing. The wake consists of a pair of counterrotating vortexes that perfectly cancel each other's angular momentum regardless of any asymmetry of the wing. Creating these wake features requires a force. That force is mostly in the transverse plane, and we call the in-plane component lift, but that force can't ever be entirely in the transverse plane. It always has a drag component.

$\endgroup$
1
  • $\begingroup$ thank you, you seem to be the only person that has understood the question $\endgroup$ – Francis L. Dec 9 '20 at 12:33

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.