Lift more complex than Bernoulli?

Studiot

I've no idea what that means, bu I do know that I can assign precise and exact values to all the variables at the free surface in Bernoulli's Theorem. There are no approximations there.

rcgldr

My point was that gravity would be doing work on the earth (not the aircraft) in the situation where the aircraft was not accelerating.

What about the ice boat analogy (using the acceleration of a second medium to offset the force related to the first medium (as opposed to using gravity))? In the frame of reference of the ice boat, the huge block of ice is gaining energy; what is the source of that energy if not the diverted air flow?

cjl

Sorry, but this is incorrect. The flow over a wing in supersonic flight is supersonic, aside from the fluid inside the (thin) boundary layer.

No, it will depend on the flow turning angle. In almost all cases, an oblique shock has two possible solutions, a strong oblique shock and a weak oblique shock. In the vast majority of cases, the weak oblique shock solution is the one which will be physically realized within a flow, and the weak oblique shock (by definition) has M₂ > 1. This is a good thing too, as this is the weakest possible shock wave for the flow turning angle and incoming mach number, which means the smallest losses within the flow. Smaller flow losses mean less drag.

At M₁ = 1, you don't actually get a shock, you get a mach wave, and it is at a 90° angle to the flow (not 45). To generate a 45 degree oblique shock, you must have the proper combination of incoming mach number and flow deflection angle, which could range from the weakest possible shock (a mach wave) with M₁ = sqrt(2), or you could have an incoming flow at M₁ = 3, with a 25° flow deflection, just to name a couple of examples. You can see more possibilities on the chart here. Interestingly enough, a 45 degree shock will always be a weak shock, and thus in every case, a 45 degree shock will have M₂ > 1 (the flow out the back of the shock will still be supersonic). Strong oblique shocks have shock angles greater than 60°, even for very high M₁ and small turning angles.


Only if it is an extremely poorly designed wing. This would cause an immense amount of drag for no real benefit. The only subsonic flow over a modern supersonic wing should be in the boundary layer.

Note also that there is such a thing as a supersonic leading edge vs a so-called subsonic leading edge on a supersonic aircraft. This does not mean that the flow is subsonic on the wing, even for a subsonic leading edge. A subsonic leading edge means that the leading edge of the wing is swept back sufficiently far that the normal component of the flow velocity is subsonic (or, in other words, the leading edge is swept back more steeply than the mach angle). This allows spanwise disturbances from the root of the wing to propagate in front of the wing itself, which means the flow ahead of the wing is affected by the wing. On a supersonic leading edge, the flow is unaffected until it hits the wing itself. In either case, the flow over the wing is supersonic though - this terminology is referring only to the component of the flow velocity normal to the leading edge of the wing.
The wings are thin primarily to reduce drag in supersonic regions - airfoils made to reduce transonic drag look quite different. This is why, for example, the wings on a modern commercial airliner (made to minimize drag in the transonic region of around M = 0.75 to M = 0.95) look very different from the wings on an F-104 starfighter (made for M = 2 or so). Interestingly, if you look at a modern fighter, the wings are somewhat thicker than older fighters. This is largely for transonic and low supersonic flight - aircraft optimized for purely supersonic operation (SR-71, D-21, XB-70, TU-144, concorde, etc) have extremely thin wings which give substantially poorer performance at low speeds, but for a fully supersonic aircraft, the very thin wing gives much better drag characteristics.

cjl

True, but as far as calculating lift, drag, and other forces on the aircraft, the source of the other force is largely irrelevant. In addition, any acceleration the aircraft itself will undergo is orders of magnitude smaller than the acceleration of the fluid, so the fact that the frame is non-inertial can pretty much always be neglected in calculation. For the purposes of solving the flow around an aircraft, you can pretty much just assume the aircraft is fixed in the air by some force (how doesn't really matter), then solve for the flow around the craft and the resulting forces on the craft.


It's correct that the induced drag contributes to slowing down the air's speed - by a simple momentum balance, all drag must slow down the flow in some way. As for where the energy goes, I believe (though I'm going by memory here, so I could be wrong) that the energy goes into the generation of the wing vortices. The vortices do not help lift, since they contain as much upwards flow as downwards flow, but they definitely contain energy that must have come from the original flow.


I'm having a hard time envisioning this example (or its relevance) here...

cjl

If the aircraft is in steady-state level flight (the usual starting point for this kind of analysis), gravity is doing no work on the earth from the frame of the aircraft, since the force and velocity vectors are normal to each other.

boneh3ad

As cjl mentioned, this is simply not true. It is sometimes true, but not universally. For example, the F-104 Starfighter had a supersonic wing. Shoot, it even had a supersonic fuselage, thus the need for the inlet cones on its engines. In general, the flow behind an oblique/conical shock is still supersonic.

Even on a subsonic airplane you can get supersonic flow locally in the transonic regime. I should point out that this does not necessarily lead to separation (though it certainly can), does not necessarily lead to transition to turbulence and certainly does give rise to wave drag. Of course, on a supersonic aircraft, even if the wing was subsonic, there would still be wave drag because the nose is still creating a shock.

That isn't true. It depends on a number of things. The oblique shock arises from the fact that the flow has to turn, and in general, there are two β solutions (shock angles) to the so-called θ-β-M equations for a given incoming Mach number. The smaller solution is called the ordinary or weak solution, while the larger one is called the extraordinary or strong one. In almost every situation, it is the ordinary solution that forms, and for every ordinary solution, the downstream Mach number is supersonic. You can have an oblique shock that is supersonic if your downstream pressure is high enough, leading to the formation of the shock governed by the extraordinary solution.

If you get slow enough or the deflection angle gets large enough, the two solutions approach one another and eventually no solution exists, signifying that the shock detaches, becoming a bow shock. In that case, you have a much more complicated situation and the flow immediately behind the normal region is subsonic but quickly accelerates back to supersonic speeds most of the time as it curves around the geometry.


Assuming the flow is just above Mach 1 so that a shock actually does form, the angle will not be 45° in general unless the turning angle of the flow is also infinitesimal. Otherwise it is a Mach wave. If we do assume that the turning angle is such that it generates a shock, the angle will not necessarily be 45° and the downstream Mach number will either be supersonic and the shock will be oblique, or else the shock will detach and be locally supersonic before accelerating again. For a small Mach number like M₁=1.1, you would need a turning angle on the order of 1° in order to have an attached shock, and then you will have β≈19° and M₂≈1.03, which is still supersonic.

Of course, as that air downstream of the shock is then expanded around the curved nose cone, it will undergo an isentropic expansion (ignoring the boundary layer of course) and you will see it speed up rather than slow down.

Again, that depends entirely on a number of parameters, including the nose LE radius of the wing, the Mach number approaching the wing and the geometry of the wing behind it. There is a reasonably good chance that in this case, the flow over the wing will still be supersonic. It may not be quite as supersonic as the free stream, but you cannot generally state it is subsonic. Generally speaking, modern supersonic aircraft are designed to avoid this. You can even look at the wing profiles and tell it is supersonic because they are generally long, thin and have very, very little camber since in a supersonic flow, thickness and camber contribute exactly zero to the lift, while having large drag implications.

Hypersonic airfoils follow the same general guidelines as supersonic airfoils. They simply have additional considerations when dealing with thermal management and drag, among other things.

It has nothign to do with "punching" through the sound barrier, which doesn't exist. Transonic airfoils designed to operate in that range are not even generally very thin. Look up supercritical airfoils. They are an entirely different concept. Supersonic airfoils are thin and uncambered because neither thickness nor camber contribute to supersonic lift, but they both contribute to supersonic drag.

EDIT: Just realized that I repeated a lot of what cjl said. Haha.

Stan Butchart

Getting back to the complications of thr Bernoulli principle__
When I decouple the tangetial velocity as it is typically givrnm relative to the wing, I find that most of it comes from the velocity of the wing. The remaining component, which is the actual air velocity, does not describe a relative Bernoulli flow.

Whrn rhe flow bends thereacceleration as it changes its velocity vector. This is the source of the pressure change. It is a centripedal acceleration that creates a normal pressure gradient, defined by the relationship of the tangential velacity and thr curve of the surface.
What saves the day is that when the pressure gradient across the field is integrated it all reduces to the familiar Bernoulli Equation, making it useable.

I accept that you guys may have an application of the Bernoulli Equation that I am not familier with.