# Physical Significance

Consider again the basic model underlying Prandtl’s lifting-line theory. Return to Fig­ure 5.13 and study it carefully. An infinite number of infinitesimally weak horseshoe vortices are superimposed in such a fashion as to generate a lifting line which spans the wing, along with a vortex sheet which trails downstream. This trailing-vortex sheet is the instrument that induces downwash at the lifting line. At first thought, you might consider this model to be somewhat abstract—a mathematical convenience that somehow produces surprisingly useful results. However, to the contrary, the model shown in Figure 5.13 has real physical significance. To see this more clearly, return to Figure 5.1. Note that in the three-dimensional flow over a finite wing, the streamlines leaving the trailing edge from the top and bottom surfaces are in different directions; that is, there is a discontinuity in the tangential velocity at the trailing edge. We know from Chapter 4 that a discontinuous change in tangential velocity is theoretically allowed across a vortex sheet. In real life, such discontinuities do not exist; rather, the different velocities at the trailing edge generate a thin region of large velocity gradients—a thin region of shear flow with very large vorticity. Hence, a sheet of vorticity actually trails downstream from the trailing edge of a finite wing. This sheet

 tends to roll up at the edges and helps to form the wing-tip vortices sketched in Fig­ure 5.2. Thus, Prandd’s lifting-line model with its trailing-vortex sheet is physically consistent with the actual flow downstream of a finite wing.

 Consider a finite wing with an aspect ratio of 8 and a taper ratio of 0.8. The airfoil section is thin and symmetric. Calculate the lift and induced drag coefficients for the wing when it is at an angle of attack of 5°. Assume that 5 = r. Solution From Figure 5.18, 5 = 0.055. Hence, from the stated assumption, r also equals 0.055. From Equation (5.70), assuming ao = 2n from thin airfoil theory,

 Example 5.1 _ Uo 1 + a0/+rAR(l + r) = 0.0867 degree-1

 Since the airfoil is symmetric, ctL=o = 0°. Thus,

 CL = act = (0.0867 degree 1 (5°) =

 0.4335

 From Equation (5.61),

 (0.4335)2(1 +0.055)8л 0.00789

 Cn  Consider a rectangular wing with an aspect ratio of 6, an induced drag factor 5 = 0.055, and a zero-lift angle of attack of —2°. At an angle of attack of 3.4°, the induced drag coefficient for this wing is 0.01. Calculate the induced drag coefficient for a similar wing (a rectangular wing with the same airfoil section) at the same angle of attack, but with an aspect ratio of 10. Assume that the induced factors for drag and the lift slope, S and r, respectively, are equal to each other (i. e., 5 = г). Also, for AR = 10, 5 = 0.105.

Solution   We must recall that although the angle of attack is the same for the two cases compared here (AR = 6 and 10), the value of Cl is different because of the aspect-ratio effect on the lift slope. First, let us calculate Cl for the wing with aspect ratio 6. From Equation (5.61),

Hence, CL = 0.423

The lift slope of this wing is therefore dCL 0.423

—– = —————- = 0.078/degree = 4.485/rad

da 3.4° – (-2°) ‘ 6 ‘ The lift slope for the airfoil (the infinite wing) can be obtained from Equation (5.70): dCi ^ «о da 1 + (a0/7rAR)(l + r)

 a0 _ a0 1 + [(1.055)«0/л-(6)] 1 + 0.056a0

 4.485

 Solving for ao, we find that this yields ao = 5.989/rad. Since the second wing (with AR = 10) has the same airfoil section, then a0 is the same. The lift slope of the second wing is given by

 a0 5.989 1 + (a0/TrAR)(l + r) _ 1 + [(5.989)(1.105)/л-(103ї = 0.086/degree

 a

 The lift coefficient for the second wing is therefore

 CL = a (a – aL=0) = 0.086[3.4° – (-2°)] = 0.464

 In turn, the induced drag coefficient is 0.0076 Note: This problem would have been more straightforward if the lift coefficients had been stipulated to be the same between the two wings rather than the angle of attack. Then Equation (5.61) would have yielded the induced drag coefficient directly. A purpose of this example is to reinforce the rationale behind Equation (5.65), which readily allows the scaling of drag coefficients from one aspect ratio to another, as long as the lift coefficient is the same. This allows the scaled drag-coefficient data to be plotted versus CL (not the angle of attack) as in Figure 5.20. However, in the present example where the angle of attack is the same between both cases, the effect of aspect ratio on the lift slope must be explicitly considered, as we have done above. Consider the twin-jet executive transport discussed in Example 1.6. In addition to the infor­mation given in Example 1.6, for this airplane the zero-lift angle of attack is —2°, the lift slope of the airfoil section is 0.1 per degree, the lift efficiency factor r = 0.04, and the wing aspect ratio is 7.96. At the cruising condition treated in Example 1.6, calculate the angle of attack of the airplane.

Solution

The lift slope of the airfoil section in radians is

a0 = 0.1 per degree = 0.1 (57.3) = 5.73 rad From Equation (5.70) repeated below

_ _____ "o____

1 + (a0/7rAR)(l + r) lift distribution reaching farther away from the root. Such wings require heavier internal structure. Hence, as the aspect ratio of a wing increases, so does the structural weight of the wing. As a result of this compromise between aerodynamics and structures, typical aspect ratios for conventional subsonic airplanes are on the order of 6 to 8. However, examine the three-view of the Lockheed U-2 high altitude reconnaissance aircraft shown in Figure 5.24. This airplane has the unusually high aspect ratio of 14.3. Why? The answer is keyed to its mission. The U-2 was essentially a point design; it was to cruise at the exceptionally high altitude of 70,000 ft or higher in order to not be reached by interceptor aircraft or ground-to-air-missiles during overflights of the Soviet Union in the 1950s. To achieve this mission, the need for incorporating a very high aspect ratio wing was paramount, for the following reason. In steady, level flight, where the airplane lift L must equal its weight W, L = W = q. xSCL = p^VlSCL [5.71] As the airplane flies higher, px decreases and hence from Equation (5.71) С/. must be increased in order to keep the lift equal to the weight. As its high-altitude cruise design point, the U-2 flies at a high value of C;, just on the verge of stalling. (This is in stark contrast to the normal cruise conditions of conventional airplanes at conventional altitudes, where the cruise lift coefficient is relatively small.) At the high value of С/, for the U-2 at cruising altitude, its induced drag coefficient [which from Equation (5.62) varies as C} would be unacceptably high if a conventional aspect ratio were used. Hence, the Lockheed design group (at the Lockheed Skunk Works) had to opt for as high an aspect ratio as possible to keep the induced drag coefficient within reasonable bounds. The wing design shown in Figure 5.24 was the result. We made an observation about induced drag Д itself, in contrast to the induced drag coefficient CD, . We have emphasized, based on Equation (5.62), that Cdj can be reduced by increasing the aspect ratio. For an airplane in steady, level flight, however, the induced drag force itself is governed by another design parameter, rather than the aspect ratio per se, as follows. From Equation (5.62), we have

 Three-view of the Lockheed U-2 high-altitude reconnaissance airplane.

 Figure 5.34  