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 ... therefore be calculated by 1 Po - P = AP = 5 PO4 and therefore 1 2 po -p = - x 1 .22 6 (26 4)' = 426 70NmP2 Now In standard conditionsp = 101 325 Nm-'. Therefore ... 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 1 pdf

... 21 3 21 3 21 4 21 5 21 6 22 0 22 2 22 3 22 4 22 6 22 7 22 9 23 4 23 4 23 7 24 0 24 3 24 5 24 9 24 9 25 1 25 5 25 7 25 7 25 9 26 1 26 9 27 0 27 3 27 3 27 4 27 5 27 8 28 1 28 3 28 4 29 4 29 ... group formed. The Eqns (1.40) may then be solved for a, b and c in terms of d and e giving a = 2 -d - 2e b = 2- d c=l-e Substituting these in Eqn (1.39) gives F = v2-d-2eg2-d ... = 27 3°C and po = 1.71 x kgm-ls-l (i) Full-scale aircraft M = 0.85, a = 20 .05 (27 3 - 56.5)' 12 = 29 7ms-' V = 0.85 x 29 7 = 25 2m s-' 7160 = 0.1151 kgm-3...

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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 ... out 4s x2 + y2 + c2 + 2xc On expanding, Therefore: $I =- - 4xc - 1 6x2 c2 - I 47r -[ x2 + y2 + c2 + 2xc 2( x2 + y2 + c2 + 2x 42 Since c is very small 2 can be ... criteria for irrotational continuous flow are that a=+ a=+ +- a24 a24 -+ -= o =- 8x2 ay2 8x2 ay= also written as V2q5 = V2$ = 0, where the operator nabla squared (3.4) a2 a2 v =-+ -...

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Aerodynamics for engineering students - part 4 doc

... 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 ... distribution p does not agree closely with Two-dimensional wing theory 175 and subtracting p2-p1=-pu 2[ 2 ("1 " ;2) + (3 2- ( ? )2] - 2 uu and with the aerofoil thin and at small ... maximum camber- line ordinate. The equations used to define the camber line are: mc Yc = p2 (2P5 - P> ESP mc Yc = - 2 [(I - 2P) + 2PE - c21 5 1 P (1 -P> (4.86)...

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Aerodynamics for engineering students - part 5 pot

... cOsel = cose2 = - &x - Xd2 + (2 - x - x1 J(x - + (z + sz - z1 )2 z + sz - 21 COS e2 + - = - sin 02 = (2 J(x - + (2 + sz - The binomial expansion, i.e. ... theory 25 3 Table 5.1 7~/8 0.3 82 68 0. 923 88 0. 923 88 0.3 82 68 0. 923 88 ~14 0.707 11 0.707 11 -0 .707 11 -0 .707 11 0.707 11 3~18 0. 923 88 -0 .3 82 68 -0 .3 826 8 0. 923 88 0.3 826 8 75 12 1 .ooo ... 0. 924 0.707 0.383 0 Firo 0 0.343 0.383 0. 82 1 .o rm2s-I 0 16.85 28 .7 40 .2 49 .2 Finite wing theoly 23 5 t= 2- 1 1 I A /B x-xj-/4q p1 C Fig. 5 .25 Geometric notation for L-shaped...

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Aerodynamics for engineering students - part 6 potx

... u', 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 ... the form vp-a=q 5-+ (6.68 b) The local velocity may also be expressed in terms of the Mach angle p by rearranging the energy equation as follows: q2 a2 c2 2 7-1 2 -+ - =- but a2 = ... 3 02 Aemdynamics for Engineering Students For air M:+5 M2 - 2- 7 Mf-1 (6.49a) Inspection of these last equations shows that M2 has upper and lower limiting values: For M1 403 M2...

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Aerodynamics for engineering students - part 7 docx

... 0.9853/14.03 = 0.07 02 bar. Thus = -0 .161 (0.07 02/ 0. 127 7) - 1 cp3= 0.7 x 22 (Using the linear theory, Eqn (6.145) gives 2E -2 x (lO?T/180) c- = -0 .20 2) p 3-4 T5T= dK-i There is an ... 0 .22 5 0.163 0.104 0.0008 -0 .0831 -0 .1166 -0 .1474 -0 .1754 WP)li7l 0 .22 8 0.183 0.138 0.0 92 0 -0 .098 -0 .138 -0 .183 -0 .22 8 Wings of finite span When the component of the free-stream ... Aerodynamics for Engineering Students L$t coefficient 4a CL =- drn For M = 1. 72 CL = 2. 86a Drag (wave) coefficient /' [ (0 .28 (1 - $) - a )2+ (0. 12( 1- 2q + a) 2...

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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 - - - ... of Eqn (7. 127 ) 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. ... - - =I-+S-=I-+S dx dx dx dx dx dA dS dx It follows from Eqns (7.62d) and (7.64b’) respectively that dA 2S2 dUe dS v dx 6- = - ~ = 2A so dS db’ dx dx - = 8’1 (A) - (7. 127 )...

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Aerodynamics for engineering students - part 9 ppt

... 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 (9.16) ... [(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 (19 92) &apos ;Aerodynamics of Grand Prix Cars', Proc. 1. Mech. E., Part D: J. of Automobile Engineering, 20 6, 26 7 -2 74 51 8 Aerodynamics for Engineering Students compliant...

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Aerodynamics for engineering students - part 10 docx

... 1.77 02 1.8113 1.8536 1.89 72 1.9 421 1.9884 2. 0361 2. 08 52 2. 1359 2. 1881 2. 2 420 2. 2976 2. 3549 2. 4141 2. 47 52 2. 5383 2. 6034 2. 6707 2. 8897 3. 126 8 3.3833 3.6608 3.981 1 4 .28 60 ... 0.3640 0.351 1 0.3386 0. 326 4 0.3146 0.30 32 0 .29 21 0 .28 13 0 .27 08 0 .26 07 0 .25 09 0 .24 13 0 .23 21 0 .22 32 0 .20 62 0.1906 0.1761 0.1 628 0.1505 0.1390 0. 128 5 0.1 188 0.1098 0.1014 ... 20 000 20 500 21 000 21 500 22 000 22 500 23 000 23 500 24 000 24 500 25 000 0.8533 0.8477 0.8 420 0.8364 0.8307 0. 825 1 0.8194 0.8138 0.80 82 0.8 025 0.7969 0.7912...

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