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9.19 Coursework Example: Subsonic Military Aircraft

305

Table 9.19. Vigilante wing CFw correction (3-D and other

 

shape effects)

 

 

 

 

 

 

 

 

 

 

 

Item

CFw

% of CFwbasic

 

Supervelocity

0.0003850

7.00

 

Pressure

0.0000136

0.04

 

Interference (wing–body)

0.0000328

3.00

 

Flap/Slat Gap

 

2.00

 

Total CFw

 

12.04

 

 

 

 

 

Empennage

Because it is the same procedure as for the wing, it is not repeated. The same percentage increment as for the wing is used for the coursework exercise. In the industry, engineers must compute systematically as shown for the wing.

V-tail:

wetted area, AwVT = 235.33 ft2

basic CF Vtail = 0.00277

fVT = 1.12 × 0.00277 × 235.33 = 0.73 ft2

H-tail:

wetted area, AwHT = 388.72 ft2

basic CF Htail = 0.002705

fHT = 1.12 × 0.002705 × 388.72 = 1.18 ft2

9.19.4 Summary of Parasite Drag

The wing reference area Sw = 700 ft2, CDpmin = f/Sw . Table 9.20 summarizes the Vigilante parasite drag. As indicated in Section 9.16, [3] provides a correlated factor of 1.284 to include all the so-called other effects. Therefore, the final flat-plate equivalent drag is faircraft = 1.284 × 8.87 = 11.39 ft2, 28.4% = 11.62 ft2, to include military aircraft excrescence. This gives CDpmin at Mach 0.6 = 11.39/700 = 0.01627 ([3] uses 0.1645). This is the CDpmin at the flight Mach number before compressibility effects begin to appear; that is, it is seen as the CDpmin at incompressible flow. At higher speeds, there is a CF shift to a lower value. The CDpmin estimation must be repeated with a lower CF at Mach 0.9 and Mach 2.0. To avoid repetition in accounting for compressibility, a factor of 0.97 is used (i.e., a ratio of values at Mach 0.9 and Mach 0 in Figure 9.20b – a reduction of 3%) is taken at 0.9 Mach. A factor of 0.8 (i.e., a reduction of 20%) is taken at Mach 2.0, as shown in Table 9.16. At

Table 9.20. Vigilante parasite drag summary

Fuselage

3.66 ft2

Wing

3.30 ft2

V-Tail

0.73 ft2

H-Tail

1.18 ft2

Total

8.87 ft2

 

 

306

 

 

 

 

 

 

 

Aircraft Drag

 

Table 9.21. Vigilante CDp estimation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

CL

0

0.10

0.16

0.20

0.30

0.40

0.50

0.60

 

 

CDp

0.00080

0.00015

0

0.00010

0.00080

0.00195

0.00360

0.00600

 

 

 

 

 

 

 

 

 

 

 

 

the compressible flow, the wave drag is added. At supersonic speed, shock waves contribute to it.

To correlate with the methodology presented herein, the following values ofCDp were extracted from [3].

9.19.5 CDp Estimation

The data for CDp given in Table 9.21 were extracted from [3] and are approximate.

9.19.6 Induced Drag

The formula for induced drag used is:

CDi = CL2/(3.14 × 3.73) = CL2 /11.71 (Table 9.22)

9.19.7 Supersonic Drag Estimation

Supersonic flight would have a bow shock wave that is a form of compressibility drag, which is evaluated at zero CL. Drag increases with a change of the angle of attack. The difficulty arises in understanding the physics involved with an increase in the CL. Clearly, the increase – although lift-dependent – has little to do with viscosity unless the shock interacts with the boundary layer to increase pressure drag. Because the very purpose of design is to avoid such interaction up to a certain CL, this book addresses the compressibility drag at a supersonic speed composed of compressibility drag at a zero CL (i.e., CD shock) plus compressibility drag at a higher CL (i.e., CDw ).

To compute compressibility drag at a zero CL, the following empirical procedure is adopted from [3]. The compressibility drag of an object depends on its thickness parameter; for the fuselage, it is the fineness ratio and for the wing it is the t/c ratio. The fuselage (including the empennage) and wing compressibility drags are computed separately and then added in with the interference effects. Graphs are used extensively for the empirical methodology (Figures 9.19 through 9.26). Compressibility drag at both Mach 0.9 and Mach 2.0 is estimated.

Drag estimation at Mach 0.9 follows the same method as worked out in the civil aircraft example and is tabulated in Section 9.19.8. For the fuselage compressibility drag (including the empennage contribution) at Mach 2.0, the thickness parameter is the fuselage fineness ratio.

Table 9.22. Vigilante induced drag

CL

0.2

0.3

0.4

0.5

0.6

0.7

CDi

0.00342

0.00768

0.01370

0.02140

0.03070

0.04180

9.19 Coursework Example: Subsonic Military Aircraft

307

Table 9.23. Vigilante supersonic drag summary

 

 

 

 

 

 

 

 

 

CDw at Mach 2.0

 

 

 

 

Fuselage and empennage contribution

0.01271

 

Wing contribution

0.00458

 

Wing–fuselage interference (supersonic only)

0.00075

 

Total

0.01804

 

 

 

 

Step 1: Plot the fuselage cross-section along the fuselage length as shown in Figure 9.17 and obtain the maximum cross-section Sπ = 45.25 ft2 and the fuselage base Sb = 12 ft2. Find the ratios (1 + Sb/Sπ ) = 1 + 12/45.25 = 1.27 and Sπ /Sw = 45.25/700 = 0.065.

Step 2: Obtain the fuselage fineness ratio l/d = 73.3/7.788 = 9.66 (d is minus the intake width). Obtain (l/d)2 = (9.66)2 = 93.3.

Step 3: Use Figure 9.21 to obtain CDπ (l/d)2 = 18.25 at M= 2.0 for (1 + SD/Sπ ) = 1.27. This gives CDπ = 18.25/93.3 = 0.1956. Convert it to the fuselage contribution of compressibility drag expressed in terms of the wing reference area: CDw f = CDπ × (Sπ /Sw ) = 0.1956 × 0.065 = 0.01271.

For the wing compressibility drag at Mach 2.0, use the following steps:

Step 1: Obtain the design CL DES from Figure 9.22 for the supersonic aerofoil for the AR × (t/c)1/3 = 3.73 × (0.05)1/3 = 1.374. This gives CL DES = 0.352. Test data of CL DES from [3] gives 0.365, which is close enough and used here.

Step 2: Obtain from Figure 9.23 the two-dimensional design Mach number, MDES 2D = 0.784. Using Figure 9.24, obtain MAR = 0.038 for 1/AR =

0.268. Using Figure 9.24, obtain MD 1/ = 0.067 for 1/ = 37.5 deg.

4 4

Figure 9.17. Vigilante RA-C5 fuselage cross-section area distribution

308

 

 

 

 

 

 

Aircraft Drag

Table 9.24. Vigilante total aircraft drag coefficient, CD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

At Mach 0.6

 

 

 

 

CL

0.0

0.1

0.2

0.3

0.4

0.5

0.6

CDpmin

 

 

 

0.01627

 

 

 

 

CDp

0.00080

0.000150

0.00010

0.0008

0.00195

0.00360

0.0060

CDi

0

0.000854

0.00342

0.00768

0.01370

0.02140

0.0307

 

Aircraft CD

0.01707

0.017270

0.01980

0.02475

0.03192

0.04127

0.0523

 

@ Mach 0.6

 

 

 

 

 

 

 

 

 

 

 

At Mach 0.9

 

 

 

 

CL

0.0

0.1

0.2

0.3

0.4

0.5

0.6

CDpmin

 

0.97 × 0.01627 = 0.01582 (Ref. [3] gives 0.01575)

 

 

CDp

0.00080

0.000150

0.00010

0.00080

0.00195

0.00360

0.0060

CDi

0

0.000854

0.00342

0.00768

0.01370

0.02140

0.0307

 

Wave drag,

0.000300

0.00100

0.00200

0.00320

0.00550

0.01000

0.0200

 

CDw

 

 

 

 

 

 

 

Aircraft CD

0.01737

0.01827

0.02180

0.02795

0.03742

0.05127

0.0723

 

@ Mach 0.9

 

 

 

 

 

 

 

 

CDw versus CL is from CFD/test data. Here, it is reduced from Ref. [3].

 

 

 

 

 

 

At Mach 2.0

 

 

 

CL

0

0.1

0.2

0.3

0.4

0.5

0.6

 

CDpmin

 

0.8 × 0.01627 = 0.01300 (Ref. [3] gives 0.01302)

 

 

CDp

0.0008

0.000150

0.00010

0.00080

0.00195

0.0036

0.0060

CDi

0

0.000854

0.00342

0.00768

0.01370

0.0214

0.0307

 

Shock drag

 

 

0.01804 at zero CL (CD shock)

 

 

 

Wave drag,

0

0.003000

0.01100

0.02300

0.04100

 

 

 

CDw

 

 

 

 

 

 

 

 

Aircraft CD

0.0318

0.035000

0.04550

0.06250

0.08700

 

 

 

@ Mach 2.0

 

 

 

 

 

 

 

CDw versus CL is to be taken from CFD/test data.

Figure 9.18. Vigilante RA-C5 drag polar

9.19 Coursework Example: Subsonic Military Aircraft

309

(a) Local skin friction

 

(b) Mean skin friction

Figure 9.19. Flat-plate skin friction coefficient CF variation

Step 3:

Make the correction to obtain in the design Mach as MDES =

 

MDES 2D + MAR + MD 1/4 or MDES = 0.784 + 0.0.038 + 0.067 =

 

0.889. Then, M = MMDES = 2.0 0.889 = 1.111.

Step 4:

Compute (t/c)5/3 × [1 + (h/c)/10)] = (0.05)5/3 × [1 + (0)/10)] =

 

0.00679.

Step 5:

Compute AR tan LE = 3.73 × tan 43 = 3.73 × 0.9325 = 3.48.

Step 6:

Use Figure 9.25 and the values in Step 5 to obtain [ CDC WING/

 

{(t/c)5/3 × [1 + (h/c)/10]}] = 0.675. Compute CDC WING = 0.675 ×

 

0.00679 = 0.00458.

Finally, the interference drag at supersonic speed must be added to the fuselage and wing compressibility drag. Following is the procedure for estimating the wing– fuselage interference drag:

Step 1: Compute (fuselage diameter at maximum area/wing span) = 7.785/53.14 = 0.1465.

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