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Solution Manual For Fundamentals of Aerodynamics, 6th Edition

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    CHAPTER 1 " 11 @ p= Ro 12X 0 Ge RT (287)(203) op 1058 rar HRI ® PR (123 x 10°)(1716) SOIR TE . 1.2 N=- [I (pu cos O + 1, sin 0) ds, + [" 0-7,sin0) d 17 I. (p, cos@—1,sinb) ds, (1.7) ds cos 6 =dx ds sin 6 = -dy Hence, N=-[F a+ [7 rr) i =-f,, Guepoder [1 @rr)dy TE TE N= [1 [Pu-p) = pl dt [| (tT, dy Divide by ge S = gs ¢(1) N_ Lge (Bee) (zee) I (vz) ar CP c CIE qe 9. cE \q, dq, Cp = EN (c5, -¢,) dxdt [I (es, +e) dy c c This is Eq. (1.15). 1 I~ StudyXY

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    TE A= [I (-pu Sin® + 1, cosO) ds, ™ . + [I (p, sin® +1, cosd) ds, (1.8) TE TE A= Gepodyt [1 Grn, TE < A= (pup) = -pldy + [mor e,) dx A gm - - . | A _Lpm (2 Po). (2 be) yd J (Ev 2)ox que cE L® q. ced. gq. 1 em 1 rec a= 3 [I (ep, ep) dy + = I. (cq, —cq ) dx This is Eq. (1.16). Mg = [I [(pu cosB + 1, SINB)X — (py Sind - 7, cosO)y] dsy TE . . : + | [-p, cosO +1, sinB)x + (p, sind +1, cosB)y] ds, TE TE Mies [| [pu-pdxdc- [| (tr) xdy TE TE +f esp dydy+ | Gute)yax ” TE TE Mie= [© [(pu-p) =p, Po) Ixdx- [© (ntt)xdy TE TE +f Pep) =, op] yay + [mt ydx Divide by guc’: My 1 m= (Rezee) E53 Lope (= =| — = - xdx- — + Ixd qc’ [ie | 9. q. c? Jie a) 2 — StudyXY

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    1g |[py=Pu) [Pr=Pa 1pm (7, 7% vals ( =e) [2 |r 1 (frye 1 3 TE en, = [fC =Coxax- [ (C +C,) xdy TE © +f (C=C ydy + [7 (Cy 4C,) vax) This is Eq. (1.17). 13 Moss Te —_— Soo Ry ~ ~ ; & / % / / TL VE 4 ~~ ~~ ~ / Ed y C ~ FT Mie=- [* 60-p) @) (x -G,-pa) |] xdx c? Mie == Po) > N= I; ®- pu) dX =~ pu) © 3

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    c? ve [6-05] X= 0 TT Tm 2a) e N (P,P) © 13 For a flat plate, 6 = 0 in Egs. (1.7) — (1.11). Hence, N= [py -padx= [1 [2x10° e1)? + 119% 10°) di oe v3 x 2 1 1 N'=-2x10 [5x +x]! +[1.19x 10%]! = [.12x 10 A= [waxes [17317024288 x0) dx . . AT =[1274 x") =[1274N L'=N' cosa - A’ sina. = 1.12 x 10° cos 10° — 1274 sin 10° =[.105x 10 D' =N' sino + A’ cose = 1.12 x 10° sin 10° + 1274 cosa =2.07x 10°N] Mie= [° fpu-pdxax= [| 2x10° Ge? - 119% 10% x dx . . Lox xP f . +2x 10 C55 ho [0595 x 1038; =[578x 10° N M'ys=M'g +L’ (c/4)=-5.78 x 10° + 1.105 x 10° (0.25) =[3.02x 10"N/ gq _ 4 xp Me C378 x10) _ ge N 112 x 10 4 Study

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    1.5 C= Cp COSUL - Cy SINOL = (1.2) cos 12° ~ (0.3) sina. = C4 = Cp Sina + ¢, COST = (1.2) sin 12° + (0.3) cosa = 1.6 cy =c, cosa + cq sino | Also, using the more accurate N' rather than L’ in Eq. (1.22), we have . So Mao fon) ® 4 ON 4 Ug, Hence: o®) Cn Xep/C -2.0 0.0498 1.09 0 0.25 0.41 2.0 0.44 0.336 4.0 0.639 0.306 6.0 0.846 0.293 8.0 1.07 0.284 10.0 1.243 0.277 12.0 1.402 0.271 14.0 1.52 0.266 5 Study

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    Z2 J Xeo SAE = . LACH R12. order. | . E .0 F- : ROCTRDNRTES REDE A Ee a Salle - . ed i o.4+ ~ -\ i en Er man Ed mee ston ts MEA Pole eden BRI i ! i ! | Zo Zo: &oO. Bo. Ao oll : : . | Note that X., moves forward as a is increased, and that it closely approaches the quarter- chord point in the range of o of 10° to 14°. At higher angles-of-attack, beyond the stall (au > 16°), op will reverse its movement and move rearward as a continues to increase. Compare the above variation with the center-of-pressure measurements of the Wright Brothers on one of their airfoils, shown in Fig. 1.28. 1.7 K=3 (mass, length, and time) fi (D, Pus Veo, €, 2) =0 Hence N=35 We can write this expression in terms of N — K = 5 — 3 = 2 dimensionless Pi products: £2 (IT, Ty) where TT; = £5 (Pes Vos ©, D) Ih = {4 (pw, Veo, €, 8) Let IT =ps V'c!D : 6 Study

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    I=(m £3 tY 5m 2 H=0 mass: a+1=0 a=-1 length: -3a+b+c+1=0 b=-2 time: -b-2=0 c=-2 Hence: PE. Ep pV. c pV 2 D hee Let IL = pw’ Vac? go! 1=(m £30 th 2° t)'=0 mass: a=0 a=0 length: -3a+1+b+d=0 d=-1/2 time: -1-2d=0 b=-1/2 Hence: \ Ih = —= Jeg Thus: D V, £2 (TT, IT)= f; (2%) =0 a.¢* eg or: 1.8 Dw =1fi (Pes Vo, C, aur, Cp, ©) K = 4 (mass, length, time, degrees) 7 Study

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    £5 (Dw, Pas Voor €, 8, Cy €4) = 0 Hence, N =7. This can be written as a function of N — K = 7 — 4 = 3 pi products: f3=(I1;, IT, TT3) = 0 where: Th = £4 (Pees Veo, €, Cp, D) IT = £5 (Pees Veo, ©, Cp, 8) ; 113 = £5 (Peo; Veos €, Cp, Cv) | The dimensions of ¢, and c, are fey] = energy _ (force)(distance) _ (met?) mass(°) mass(°®) m(°) [ep] = £242 (°)" where (°) degrees. For IT: oo pal Vo cp" D=1I1, (m £3 CY (OF (LE) Om £ H=1 mass: i+1=0 i=-1 length: -3i+j+k+2n+1=0 n=0" time: -j-2n-2=0 ji=-2 degrees: -n=0 k=-2 Hence: m= a or fly = = For Ip: Th pe Vio © cas" 8

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    =m £2) (2 eh (ey (2 OF (eel mass: i=0 i=0 length: -3i+1+j+2k+n=0 k=0 time: -1-2k-n=0 n=-1 degrees: -k = 0 j=0 Hence: mee a, For Is: Co I; = Por Vo cp Cy T=(m £2) (LY £5 Or" (LH Oy mass: i=0 i=0 length: -3i+j+k+2n+2=0 n=-1 time: -j-2n-2=0 j=0 degrees: -n—1=0 k=0 Hence: I; = = We can take the reciprocal, and still have a dimensionless product. Hence, ’ . Ih = = =Y » Thus, or, 9

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    ro Mi Via, VT _100 800, M, V,a, V,yT, 200V200 Hence, the Mach numbers of the two flows are the same. Rei _ pve, (2) - ode [b_(12)100) 1) [B00 _ 4354 Re, p Vie, gy, p Ve, VT 1.739/\200/\2/V200 The Reynold’s numbers are different. Hence, the two flows are not dynamically similar. 1.10 Denote free flight by subscript 1, and the wind tunnel by subscript 2. For the lift and drag coefficients to be the same in both cases, the flows must be dynamically similar. Hence M,=M, and Re; =Re; For Mach number: V, _ V, a, a, Since a ow VT , we have Vv, \4 250 JRE SR B.A 15 [©] No V, V,c For Reynolds number: oY PTE Hi Hy Assume, as before, that po JT. Hence £2Ye, _ piVie JnoJT 10 Study

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    or, pV: pV, (& J _ (0414)250) © JIT Le, 223 1 or, LY: C3465 0) Nes Finally, from the equation of state: p, 101 x 10° | T= 2 =—"—"++— =3519 3 p= 357 3) | Egs. (1) — (3) represent three equations for the three unknowns, p,, Va, and T>. They are summarized below: v, —==167 1 3 [©] £22Ys 3465 @ VT pT2=351.9 3) From Eq. 3): 02 =351.9/T, “) Subst. (4) into (2): 9) V. wa) =34.65 (5) TT Subst. (1) into (5): 22 (16.7) = 34.65 2 Hence, 11 Study

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    _ (3519)(167) 5 1,- COND) _ from) PT TT (3465) From Eq. (1): V2=16.7 JT, = 167 1696 =[217.5 = sec 3519 3519 kg From Eq. (3): = = P07 —= Or p= mp 20 pr] 111 py=pa-pgAh =1.01 x 10° - (1.36 x 10*)(9.8)(0.2) po =[7.43 x 10° N/m 1.12 Weight = Buoyancy force + lift w= B +L B=(15000) (1.1117) (98) = 1.634x10°N ———~ — — volume air density acceleration (m®) at 1000m of gravity (kg/m®) (m/sec) Go = : Pu Vil = 5 (1.1117) (30)* = 500 N/m? S = nd¥4 = n(14)%/4 = 153.9 m® L = qa S CL =(500)(153.9)(0.05) = 3487 N Hence: W = 1.634 x 10° + 3847 = [1.67 x 10 1.13 Let us use the formalism surrounding Eq. (1.16) in the text. In this case, cq = c,, and from Eq. (1.16), neglecting skin friction 12 Study

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    1 1 w= I. (c,, -c, )ay 1 From Eq. (1.13) in the text, Eq. (1) above can be written as 1pm . w= [, (Cc, -C,) Csinods) ©) Draw a picture: i -6 Following our sign ? i convention, note 1 that 0 is drawn Vio counterclockwise -_— [47 in this sketch, hence AN it is a negative angle, -6. From the geometry: - B=m-¢ Hence, sin (-8) = - sin 0 = sin (1-8) = cos ¢ Substitute this into Eq. (2), noting also that ds = rd¢ and the chord c is twice the radius, ¢ = 2r. From Eq. (2), 1 TE c= > I. (c,, -C,) cosdprdg 1 TE cq = 5 [I (c,, -c,) cos pd _ 1 TE 1 TE a= I C, cosdd > I. C,, cosdpdo 3 Consider the limits of integration for the above integrals. The first integral is evaluated from the leading edge to the trailing edge along the upper surface. Hence, = 0 at LE and 7 at TE. 13 Study

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    The second integral is evaluated from the leading edge to the trailing edge along the bottom surface. Hence, ¢ = 27 at LE and = at the TE. Thus, Eq. (3) becomes c= 4 |” c, cosodo Le C, cos¢d @) d 2 Jo Pu PRET Pe In Eq. (4), C, =2cos’§ for 0< ¢ <m/2 T C,, =0 for > <g=n 2 3 C,, =2cos"¢ for = <¢=<2m 3z Cc, =0 forns¢s— Thus, Eq. (4) becomes cq = [ cos’ do- [rr cos’ od ¢ Since cos’ ¢pd § = a sing)(cos” ¢ + 2), Eq. (5) becomes 3 a=[(5 sing)eos” 6 + 2127 (5 sind)(cos'p + 21% 1 1 C= 5H) - (3)D@ . 1.14 14 Study

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    19/6 _FLuib 1 — TL fp — _ oA, — A, — 7 [IE - - - — ( Boy aE — 4] - — \ SuBNERGED 1B - - — —| ma thy = — = x Fru | I ho - - — — i I EE oo a4, —h _ _ od _— — — — _ ra \Ff Consider the arbitrary body sketched above. Consider also the vertical cylinder element inside the body which intercepts the surface area dA, near the top of the body, and dA, near the bottom of the body. The pressures on dA; and dA, are p; and p; respectively, and makes angles 0; and 0, respectively with respect to the vertical line through the middle of dA; and dA,. The net pressure force in the y-direction on this cylinder is: dF, =-p) cos 6; dA; +p; cos 0; dA; [€))] Let dA, be the projection of dA; and dA; on a plane perpendicular to the y axis. dA, = cos 0; dA; = cos 0; dA, Thus, Eq. (1) becomes dF, = (py — pi) dA, 2) From the hydrostatic equation h, p-pi= [7 pedy ® Combining Egs. (2) and (3), hy dF,= | Lo pedyda, “ However, dy dA, = dV = element of volume of the body. Thus, the total force in the y direction, Fy, is given by Eq. (4) integrated over the volume of the body 15 Study

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