## Lift more complex than Bernoulli?

 As discussed above, it's an approximation. If fluid is well approximated as inviscid and incompressible in given problem, it's a good approximation. But the actual, physical Bernoulli Effect precisely at the surface is always zero.
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.

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 Quote by rcgldr Absent gravity, the frame of an aircraft would be accelerating and non-inertial, which creates issues. An external force, normally gravity, is required to prevent the aircraft from accelerating when diverting the air flow. The source of that force needs to be included to end up with a closed system, one where momentum and energy are conserved (in order to use the aircraft as an inertial frame of reference).
 Quote by K^2 Gravity also does zero work, so including an external force does not change the argument.
My point was that gravity would be doing work on the earth (not the aircraft) in the situation where the aircraft was not accelerating.

 Quote by K^2 There is no work being done by aircraft on the medium in frame of reference attached to the aircraft. Total energy in the medium is definitely conserved.
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?

 Quote by K^2 The only time you get supersonic stream over a wing is in transonic region, and that results in separation, turbulence, wave drag, loss of control, and many other unpleasant things. During supersonic flight, the flow over the wing is subsonic. That's what allows supercruise in the first place.
Sorry, but this is incorrect. The flow over a wing in supersonic flight is supersonic, aside from the fluid inside the (thin) boundary layer.

 Quote by K^2 The difference between oblique and normal shock waves is that a normal shock wave guarantees subsonic flow behind it at any air speed. The oblique shock wave slows the flow down as well, but whether the flow behind it is subsonic will depend on the speed.
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.

 Quote by K^2 Simple check. Take an object at Mach 1+. It generates a 45° oblique shock. Can the flow behind that be supersonic? No. Because if it's just supersonic before the shock and just supersonic behind, there cannot be a shock in between. You need a sharp change in velocity which you can't get. So for low Mach numbers, somewhere within Mach 1-2 region, you do get a good range of air speeds where the flow is entirely subsonic within the cone.
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.

 Quote by K^2 So what happens if you go even faster? Naturally, at high enough speed the flow within the oblique shock becomes supersonic. But the flow over the wing is still going to be subsonic. Why? Because the wing is going to generate its own bow shock with a subsonic flow behind it.
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.
 Quote by K^2 This might break down for hypersonic flights. I know nothing about hypersonic aerodynamics. But a typical modern supersonic jet does not deal with supersonic flows over the surfaces except for transition regions. Oh, and the wings are made thin primarily to reduce drag in transonic regions, providing for better control and making it easier to punch through to supersonic.
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.

 Quote by rcgldr Absent gravity, the frame of an aircraft would be accelerating and non-inertial, which creates issues. An external force, normally gravity, is required to prevent the aircraft from accelerating when diverting the air flow. The source of that force needs to be included to end up with a closed system, one where momentum and energy are conserved (in order to use the aircraft as an inertial frame of reference).
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.

 Quote by rcgldr From what I recall, in addition to friction drag, induced drag also contributes to slowing down the air's speed when the flow is diverted (using the aircraft as a frame of reference). Friction drag converts the energy into heat, but it's not clear to me where the induced drag energy loss goes. My guess is that the energy loss related to induced drag corresponds to an increase in energy of the source of the force that keeps the aircraft from accelerating. In the normal case where the earth is the source of the external force via gravity, the energy could be going into the earth, since the earth would accelerate towards the aircraft, ignoring interactions between air and earth.
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.

 Quote by rcgldr As an alternative example, imagine an ice boat on a huge flat block of ice, that rests on a frictionless surface. Say the ice boat is moving perpendicular to the wind, accelerating the huge block of ice in the direction of the wind, in order to keep the ice boat moving at constant velocity with respect to some inertial frame (the "angle of attack" of the skates on the ice boat would be increasing over time as the huge block of ice accelerates). From the ice boats frame of reference, the energy of the block of ice is increasing, and the source of that energy is the energy being extracted from the air.
I'm having a hard time envisioning this example (or its relevance) here...

 Quote by 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.
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.

 Quote by K^2 In most supersonic aircraft, the wing operates within the subsonic stream behind the shockwave generated by the nose of the aircraft. If you look at geometry of supersonic fighters, for example, you can tell the design cruise speed by the angle of the cone within which the airplane fits.
 Quote by K^2 The only time you get supersonic stream over a wing is in transonic region, and that results in separation, turbulence, wave drag, loss of control, and many other unpleasant things. During supersonic flight, the flow over the wing is subsonic. That's what allows supercruise in the first place.
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.

 Quote by K^2 The difference between oblique and normal shock waves is that a normal shock wave guarantees subsonic flow behind it at any air speed. The oblique shock wave slows the flow down as well, but whether the flow behind it is subsonic will depend on the speed.
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.

 Quote by K^2 Simple check. Take an object at Mach 1+. It generates a 45° oblique shock. Can the flow behind that be supersonic? No. Because if it's just supersonic before the shock and just supersonic behind, there cannot be a shock in between. You need a sharp change in velocity which you can't get. So for low Mach numbers, somewhere within Mach 1-2 region, you do get a good range of air speeds where the flow is entirely subsonic within the cone.
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.

 Quote by K^2 So what happens if you go even faster? Naturally, at high enough speed the flow within the oblique shock becomes supersonic. But the flow over the wing is still going to be subsonic. Why? Because the wing is going to generate its own bow shock with a subsonic flow behind it.
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.

 Quote by K^2 This might break down for hypersonic flights. I know nothing about hypersonic aerodynamics. But a typical modern supersonic jet does not deal with supersonic flows over the surfaces except for transition regions.
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.

 Quote by K^2 Oh, and the wings are made thin primarily to reduce drag in transonic regions, providing for better control and making it easier to punch through to supersonic.
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.

 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.