Airfoil Induced Drag Energy Conservation

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  • #1
Supercritical
Consider the production of lift only as the reaction force from displacing a mass of air downwards. No matter the mathematical abstraction (Bernoulli, circulation theory) this Newtonian explanation must hold true.

So now imagine a rectangular wing in a wind tunnel where the wing spans the tunnel (essentially a 2-d wing). Parasitic drag will be disregarded.

Thus, if 1 kg/s of air is influenced by the wing, and it's deflected downwards at an average downward velocity component of 1 m/s, 1 N of lift is created. The same 1 kg/s of air deflected at 1 m/s requires 0.5 Watts of power, manifested as 0.1 N of drag if the wind tunnel freestream velocity is 5 m/s.

I can't imagine how the above scenario would fail to check out, and yet here we have induced drag in the absence of wingtip vortices (impossible according to a good deal of literature).

More importantly, however, something is preventing me from reconciling the above with the induced drag equation. Holding the mean chord, lift coefficient and efficiency factor constant, doubling the wingspan halves the induced drag coefficient (CL^2/pi*e*[highlight]AR[/highlight]), which counters the doubling of drag from the wing area term (0.5*rho*[highlight]S[/highlight]*Cdi*v^2). As a result, it would seem that by holding these items constant, the amount of induced drag a wing creates is independent of its span (the span term also drops out algebraically if you write S as b*c and AR as b/c). From the perspective of a wing in flight, then, past some finite wingspan the power imparted on air accelerated downwards from a standstill would exceed the power required to maintain flight. I can't find a way the transition to a 3-d wing rectifies the issue. There is obviously a deviation from energy conservation somewhere, but I can't find it.
 

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  • #2
rcgldr
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The longer the wing span, the greater the mass of the affected air. If the amount of air (in terms of mass) diverted is doubled, the rate of acceleration, exit velocity, and energy imparted to the air is halved. This is why high end cross country gliders have wing spans of 24.3 meters (80 feet) or more, with overall (the entire glider, not just the wings) lift to drag ratios around 60:1.

I'm not sure why or how useful it is to define induced drag based solely on wingtip vortices. The premise that all downwash is due to wingtip vortices seems to conflict with real world observations.
 
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  • #3
Supercritical
Agreed on all points. Increasing the aspect ratio (higher mass flow) while holding weight constant reduces the exit velocity requirement and its corresponding induced drag.

I was focusing specifically on fixed lift coefficient and chord, and varying the span. There's the baseline Newtonian example I used earlier, where 1 kg/s is deflected downwards at 1 m/s: creating 1 N of lift, consuming 0.5 Watts and exerting 0.1 N of induced drag at 5 m/s freestream velocity.

Holding all else constant, doubling the span doubles the mass flow, power requirement and induced drag.

However, the induced drag equation seems to suggest that the power requirement and induced drag are independent of wingspan.

Induced drag = 0.5*rho*area*Cdi*v^2

Cdi = CL^2/pi*e*AR

Substituting c*b (mean chord times span) for area, and b/c (span divided by mean chord) for aspect ratio yields the following drag equation:

Induced drag = 0.5*rho*c^2*v^2/pi*e

The only possibility I see for this equation to agree with the Newtonian mechanics is for the efficiency number to decrease as the span increases, even though this term is said to remain within a fairly small range (0.7 to 0.9). As a result, this equation violates conservation of energy at arbitrarily high wingspans by imparting more energy to the air than it consumes in the form of induced drag. I feel like I'm making an assumption or missing a critical piece here, but I can't put my finger on it.
 
  • #4
rcgldr
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Isn't induced drag porportional to 1 / v2? If so, then your first equation is wrong.
 
  • #5
AlephZero
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I can't imagine how the above scenario would fail to check out, and yet here we have induced drag in the absence of wingtip vortices (impossible according to a good deal of literature).
I thknk you are making some inconsistent physical assumptions here.

If you assume an inviscid fluid and irrotational flow, you will have no drag, but also no lift, for any shape of airfoil whatever.

If you have a inviscid fluid and a flow with circulation, you can create lift, but in the complete system there most be something that is creating the circulation.

And if you have a viscous fluid, clearly you can't "ignore" the effects of the viscosity like drag and vortices.
 
  • #6
Supercritical
rcgldr said:
Isn't induced drag porportional to 1 / v2? If so, then your first equation is wrong.
It's buried in the equation (for which I mistakenly left out the CL term earlier). Lift coefficient required for level flight is a 1/v2 relationship. In the induced drag equation, the CL quantity is squared (1/v4) and then multiplied by the v2 term from dynamic pressure, restoring a 1/v2 relation. Mind you, this is only for level flight. An airfoil at a fixed CL value will generate exponentially (v2) more induced drag at higher speeds.

Induced drag = 0.5*rho*chord^2*v^2*[highlight]CL^2[/highlight]/pi*e

As far as rotational flow is concerned, circulation is assumed to coincide with the production of lift, by definition. Viscosity is always a factor with parasitic drag, but I was trying to isolate the induced drag term, which would still be generated in an inviscid flow. Viscous drag need not obey any conservation law in connection with the production of lift, while induced drag ought to. The original question, however, arose from the difficulty in reconciling this energy conservation approach with the induced drag equation discussed above.
 
  • #7
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If you double the span but keep CL constant then you also double the lift which means you are accelerating upwards which requires energy.
 
  • #8
Supercritical
If you double the span but keep CL constant then you also double the lift which means you are accelerating upwards which requires energy.
Wings do no useful work in a wind tunnel or in a banked turn. Whatever the answer, the energy needs to be accounted for in a manner that is consistent between the wind tunnel case and free-flight, since the discrepancy applies to both examples.
 
  • #9
Supercritical
Just found the Aerospace Engineering forum. Maybe this would belong over there?
 
  • #10
Supercritical
I think I may have resolved this. It seems I muddled the 3d wing coefficients with the 2d ones, the latter of these used for lift and drag per unit span.

In the process I developed a workaround that I think bridges the classical mechanics with the standard aero equations.

According to thin airfoil theory, the lift coefficient is equal to 2*pi*alpha (radians). Applying the Kutta condition, and assuming attached flow, the downwash velocity, Vd, can be stated as Vd = Vo * tan alpha, where Vo is the freestream velocity. Now, using the small angle approximation (y=alpha is within 3% of y=tan alpha at 15 degrees, where many airfoils stall), this can be rewritten as Vd = Vo*alpha, or alpha = Vd/Vo. Substituting this expression for alpha back into thin airfoil theory yields Cl = 2*pi*Vd/Vo, which is a section lift coefficient.

I then assumed the 2-d flow induced drag coefficient is Cl^2/pi, dropping the aspect ratio and efficiency factors, which I judged are applicable only to 3-d flow.

The equation of drag per unit span for a 2-d airfoil is thus:

Drag/span = 1/2*rho*Vo^2*Cdi*chord

Drag/span = 2*pi*rho*Vd^2*chord

This result yields induced drag linearly proportional to span, as expected from classical mechanics. And, for a fixed wing geometry and airspeed, doubling the lift coefficient quadruples the drag and power requirements, again consistent with the basic physics.
 
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