From what I understand when the angle of attack of a wing is increased enough the boundary layer air slows down. Eventually the air starts moving backwards and a separation bubble occurs. Why does the increase in angle of attack cause the boundary layer to slow?


Flow separation happens when the pressure gradient of the airflow along the flow path becomes too steep. In subsonic flow, the oncoming air is first decelerated ahead of the wing, then swiftly accelerated when it flows around the strongly curved nose section of a wing and then decelerated over the rest of the airfoil's surface. This acceleration is the consequence of the wing's curvature and its angle of attack. See it this way: If the airflow would be along a straight line, it would move away from the surface, creating a local vacuum. In reality, the air settles at a compromise between the straight path and following the contour, creating decreasing pressure along a surface with increasing curvature and increasing pressure along surfaces with decreasing curvature. More precisely, it is always in an equilibrium between inertial, viscous and pressure forces.

A higher angle of attack requires a larger change to the flow direction and, consequently, causes a lower pressure on the suction side of the wing, for two reasons:

  1. The wing inclination increases,
  2. The starting point for the flow which goes around the upper side of the wing (stagnation point) moves slightly down along the contour of the airfoil nose, so more of the curved nose needs to be negotiated.

This suction not only bends the airflow into following the wing's contour, but also accelerates the air ahead of it. The lower the pressure, the more the air speeds up, such that the total energy of air (the sum of pressure and kinetic energy) stays constant. Therefore, pressure and local speed change in sync.

When the curvature decreases further downstream, the flow path becomes straighter and pressure rises again. However, the air particles close to the wing slow down because of friction. The layer of air where this slowing is noticeable is called boundary layer. In it, the deceleration effects due to pressure rise and due to friction add up. At higher angle of attack the air was accelerated more, so more energy is lost to friction. At some point the kinetic energy is exhausted and the air comes to a standstill relative to the wing. This point moves upstream with increasing angle of attack. When this happens, static air will collect and build up, causing flow separation.

In a low Reynolds number flow, this happens relatively soon, so it is desirable to have a turbulent boundary layer to sustain a stronger pressure rise. The exchange of air across a turbulent boundary layer kicks the slowest air particles downstream, and at moderate angles of attack the air still moves along until it reaches the trailing edge. Only when the suction peak around the nose becomes very high at high angle of attack, the consequent steep pressure rise along the remaining flow path overpowers the possibilities of even the turbulent boundary layer, the air decelerates completely and the flow separates.

If the wing moves at high subsonic speed, the curvature-created suction accelerates the flow such that it reaches supersonic speed. Now something odd happens: Supersonic flow accelerates further when subsonic flow would decelerate. Now we have a supersonic pocket of air on the upper surface of the wing where speed increases and density decreases downstream, and the surrounding subsonic air sees little change in density. This picture should give you some idea how it looks: Lambda shock in high subsonic flow

The whole wing moves at Mach 0.68. Compare the green color at some distance from the wing with the scale on the left side, which gives the Mach number for each color shade. At the airfoil nose, you see a blue area. This is where the air decelerates - it gets pushed together by the approaching wing. Now follow the colors along the upper side - they quickly turn green, yellow and red as the air is accelerated into the low pressure area (remember, low pressure equals high speed, so the reddest area has the highest local flow speed and the lowest pressure). In subsonic flow, the suction peak would be somewhere between 20% and 30% of chord, and the colors would slowly change back to yellow and green if you move further downstream. Now we have local supersonic flow (everything redder than light orange is supersonic here), and instead of slowly decelerating, the air is speeding up to a maximum Mach number of 1.23 at almost 60% of chord length.

This cannot last, and at some point this supersonic pocket collapses. This happens instantly in a shock, and as you know, in a straight shock density increases suddenly and speed decreases such that the Mach number after the shock is the inverse of the Mach number ahead of the shock. In the picture above, boundary layer effects create a lambda shock, which has its name from the Greek letter which looks like the shock pattern here. Aft of the shock, you have subsonic flow again, and a much thicker boundary layer which moves very slowly (blue shade). This is due to the energy conversion through the shock, which converts kinetic energy into heat. But the flow is still attached - even this shock did not cause separation.

If this pressure rise is big enough, the boundary layer will come to an instantaneous standstill, and then the flow separates. This is called shock induced separation.

  • $\begingroup$ Hello? Is anyone going to answer this? $\endgroup$ – Crafterguy Apr 6 '17 at 23:45
  • $\begingroup$ "At higher angle of attack the air was accelerated more, so more energy is lost to friction." If the air is moving faster then would'ent it take a longer time to slow down because it has to bleed off more energy. $\endgroup$ – Crafterguy Apr 9 '17 at 0:42
  • $\begingroup$ I do not see why no one is answering this question or at least commenting. If something is wrong with the question like it don't not show up correctly or something like that please tell me $\endgroup$ – Crafterguy Apr 9 '17 at 0:43
  • $\begingroup$ So you think you are entitled to answers to questions which show that you did not read the existing answer? $\endgroup$ – Peter Kämpf Apr 9 '17 at 5:41
  • $\begingroup$ No I read your answer. I just have a question about it. The last comment made it seem like a wanted another answer but, that was not the case. I wrote that comment at like 12pm so I was kind of tired and not thinking clearly $\endgroup$ – Crafterguy Apr 11 '17 at 22:05

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