Aerodynamics for engineering students - part 7 docx

... 0.0 0 .7 -0 .08 0.8 -0 .12 0.9 -0 .16 1.0 -0 .20 e 11.31" 9.09" 6.84" 4. 57& quot; 0.0 -4 . 57& quot; -6 .84" -9 .09" -1 1.31" ne 0" 2.22" 4. 47& quot; ... 0.104 0.0008 -0 .0831 -0 .1166 -0 .1 474 -0 . 175 4 WP)li7l 0.228 0.183 0.138 0.092 0 -0 .098 -0 .138 -0 .183 -0 .228 Wings of finite span When the component of the free-stream velocity ... 0. 070 2 bar. Thus = -0 .161 (0. 070 2/0.1 277 ) - 1 cp3= 0 .7 x 22 (Using the linear theory, Eqn (6.145) gives 2E -2 x (lO?T/180) c- = -0 .202) p 3-4 T5T= dK-i There is an oblique...

Aerodynamics for engineering students - part 10 docx

... 0.8082 0.8025 0 .79 69 0 .79 12 0 .78 56 0 .77 99 0 .77 43 0 .76 87 0 .76 30 0 .75 74 0 .75 17 Constant in stratosphere Constant in stratosphere 0.4344 0.4195 0.4050 0.3910 0. 377 3 0.3640 0.351 ... 0.8 471 0.8423 0.8 374 0.8325 0.8 276 0.82 27 0.8 177 0.8128 0.8 078 0.8028 0 .79 78 0 .79 28 Constant in stratosphere Constant in stratosphere 1 .73 03 1 .77 02 1.8113 1.8536 1.8 972 1.9421 ... 2.2420 2.2 976 2.3549 2.4141 2. 475 2 2.5383 2.6034 2. 670 7 2.88 97 3.1268 3.3833 3.6608 3.981 1 4.2860 4.6 376 5.0180 5.42 97 5. 875 1 6.3 570 6. 878 5 7. 44 27 8.0532 8 .71 38 9.4286...

Aerodynamics for engineering students - part 1 pdf

... = 273 °C and po = 1 .71 x kgm-ls-l (i) Full-scale aircraft M = 0.85, a = 20.05( 273 - 56.5)'12 = 297ms-' V = 0.85 x 2 97 = 252m s-' 71 60 = 0.1151 kgm-3 ... This equation is F - A = P(;) Hence [ML-'T-2] = [p][LT-'L-'] = /p][T-'] Thus [p] = [ML-lT-'] and the units of p are therefore kgm-ls-l; in the SI system ... 3.3 Two-dimensional flow from a source (or towards a sink) 52 52 52 53 54 56 56 62 62 62 64 64 66 68 68 71 72 73 73 75 76 78 83 83 83 84 85 86 86 87 87 89 89...

Aerodynamics for engineering students - part 2 pptx

... direction. For three-dimensional flows Eqns (2.45) and (2.46) are written in the forms: 1 at ax ay az (ax ay az ap ap ap ap au av aw -+ u-+v-+w-+p -+ - +- =o au av aw -+ - +-= o ax ... second. 68 Aerodynamics for Engineering Students Therefore, true air speed = Ma = 0 .72 8 x 340.3 248 m s-' = 89 1 km h-' In this example, ~7 = 1 and therefore there ... becomes -6 xSy aP x 1 at Equating (2.42) and (2.43) gives the general equation of continuity, thus: aP a(Pu) a(P.1 - 0 -+ - + at ax ay This can be expanded to: -+ u-+v-+p -+ - =o...

Aerodynamics for engineering students - part 3 pot

... Therefore p-po=-pU 1- 2sin0f- 2 2[ ( 2rua >’ (3.49) Potential flow 129 Now and Therefore r; = x2 + y2 - 2xc + c2 m x2+y 2-2 xc+c2 47r x2 + y2 + 2xc + c2 $I=-ln ... - In - = - In - * Here TO is the radius of the equipotential q5 = 0 for the isolated source and the isolated sink, but not for the combination. 120 Aerodynamics for Engineering ... Putting the line along the x-axis as $ = 0 128 Aerodynamics for Engineering Students Doublet axis Fig. 3.20 Streamlines due to a doublet - 2c - -x- Fig. 3.21 Consider again...

Aerodynamics for engineering students - part 4 doc

... angle Q to Ox 178 Aerodynamics for Engineering Students The lift per unit span = apU2clT(l +cosO)dO = 7i-apU2c It therefore follows that for unit span I CL = ($q) =27ra The moment ... dx H = -Jhge where p = pUk and x’ = x - (1 - F)c. Putting C C c x/ =-( i -cose) (i-cos~) =-( cos~-cose) 2 2 2 and k from Eqn (4.51): +,( (1 -; ) COS4Il - (1 where ... some general form given by, say, (4.61) V - = B~ + CB, cosne V Fig. 4.19 170 Aerodynamics for Engineering Students iy z plane 0 U Fig. 4.8 Zhukovsky transformation, of the...

Aerodynamics for engineering students - part 5 pot

... Table 5.1 7~ /8 0.382 68 0.923 88 0.923 88 0.382 68 0.923 88 ~14 0 .70 7 11 0 .70 7 11 -0 .70 7 11 -0 .70 7 11 0 .70 7 11 3~18 0.923 88 -0 .382 68 -0 .38268 0.923 88 0.38268 75 12 1 .ooo 00 - 1 .ooo ... 0.00 473 9 = 0.22 079 A1 + 0.89202 A3 + 1.251 00 A5 + 0.66688 A7 0.0116 37 = 0.663 19 A1 f0.989 57 A3 - 1.315 95A5 - 1.64234 A7 0.0216 65 = 1.1 15 73 A1 - 0. 679 35 A3 - ... 2.688 78 A7 0.032998 = 1.343 75 AI - 2.031 25 A3 - 2 .71 8 75 A5 - 3.40625 A7 These equations when solved give A1 = 0.020 329, A3 = -0 .000 955; A5 = 0.001 029; A7 = -0 .000...

Aerodynamics for engineering students - part 6 potx

... v2 u2 -+ - +- 2 7- 1 2 y-1 Pmiq I Fig. 6. 37 (a) 'Incompressible' flow. (b) Compressible subscritical flow. (c) Critical flow 324 Aerodynamics for Engineering Students ... x0)x + [xo(C + XO) - A] = 0 (6.86) having the formal solution x=J [-( C+XO) f .\/(C+XO)( ~-~ XO) +4A] (6. 87) 306 Aerodynamics for Engineering Students Now for values of MI near ... Rankine- Hugoniot relations Eqn (6.42): -= or rearranged For air -/ = 1.4 and p2 6M: P1 5+M? -= - Reversed to give the ratio in terms of the exit Mach number - P1 - - (Y+ w;...

Aerodynamics for engineering students - part 8 pps

... Aerodynamics for Engineering Students Pi 0.9 0.8 0 .7 - 0.6 0.5 - 0.4 - 0.3 - - - U - 0 0.1 0.2 0.3 0.4 0.5 0.6 0 .7 0.8 1.0 0.9 0.8 0 .7 - 0.6 0.5 - 0.4 - 0.3 - - - ... method for laminar boundary layers is to use the approximate expressions (7. 64a’,b’,c’) with Eqn (7. 59) (or 7. 98) rewritten as (H + 2) db’ - Cf V, e dU, - - - - - - - - dx 2 ... (7. 1 27) are multiplied by 2UeS/v, whereby with use of Eqn (7. 128) it becomes Ue dS2 2VSS F1 (A) - - = F2 (A) - - v dx V 444 Aerodynamics for Engineering Students Fig. 7. 36...

Aerodynamics for engineering students - part 9 ppt

... a = 4 - 2e - 2d - f b = 2 - d - 2e - f c=l-e Substituting these in Eqn (9.15) gives T= c~ 4-2 -2 d-f 2-d-2e-f I-e d e f n p VKV = CpnZD4f [(z)d(L)e (-3 f] D2n pD2n2 ... [(L)"(T)-b(ML-3)c(L2T-1)d(ML-'T-2)e(LT-1~] Separating this into the three fundamental equations gives (M) I=c+e (L) 1 =a-3c+2d-e+f (T) 2=b+d+2e+f (9.15) 508 Aerodynamics for Engineering Students ... Dominy (1992) &apos ;Aerodynamics of Grand Prix Cars', Proc. 1. Mech. E., Part D: J. of Automobile Engineering, 206, 26 7- 2 74 51 8 Aerodynamics for Engineering Students compliant...

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