Aerodynamics for engineering students - part 5 pot
... (2)m=a =5. 5[1 +- 5 55 ~~’8 (31 = 5. 5[1 - 0. 054 55 cos B] ao =5. 5 [ 1 5* 55T: 5 (31 = 5. 5[1 + 0.363 64 cos e] 238 Aerodynamics for Engineering Students I 44 J. J ti444 tJ4J ... - - At any section z from the centre-line [B from the wing-tip] [ 3.048 - 1 .52 4 (;)I chord c = 3.048 1 - = 3.048[1 + OSCOSB] 3.048 (2)...
Ngày tải lên: 08/08/2014, 11:21
... Equating (3 .53 ) and (3 .54 ): Potential flow 137 (3 .55 ) The rate of change of total pressure His and substituting for Eqn (3 .55 ): aff - 4: dqt - P- r + P4t - dr = pqt Now for this ... - In - = - In - * Here TO is the radius of the equipotential q5 = 0 for the isolated source and the isolated sink, but not for the...
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... 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; ... v2 u2 -+ - +- 2 7-1 2 y-1 Pmiq I Fig. 6.37 (a) 'Incompressible' flow. (b) Compressible subscritical flow. (c) Critical flow 324 Aerodynamic...
Ngày tải lên: 08/08/2014, 11:21
Aerodynamics for engineering students - part 1 pdf
... 50 0 50 4 50 5 50 5 50 7 51 2 51 4 51 5 51 7 52 0 52 2 52 2 52 5 52 7 52 7 52 7 53 3 53 3 53 4 53 5 53 5 53 5 53 8 53 9 53 9 53 9 54 1 54 1 54 2 54 9 54 9 55 0 55 0 55 1 55 2 55 4 55 8 56 ... 0. 85, a = 20. 05( 273 - 56 .5) '12 = 297ms-' V = 0. 85 x 297 = 252 m s-' 7160 = 0.1 151 kgm-3 P p =-= RT 287.3 x 21...
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Aerodynamics for engineering students - part 2 pptx
... 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 ay az (2.47a) ... 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...
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Aerodynamics for engineering students - part 4 doc
... corresponding to x = pc. Two-dimensional wing theory 207 -Accurate computation 0 Experimental data Thin-aerofoil theory a 64panels -1 .00 -2 0 -1 5 -1 0 -5 0 5 10 15 20 Angle of attack,degrees ... 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 f...
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Aerodynamics for engineering students - part 7 docx
... 0.2 0.12 0.3 0.08 0 .5 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 " 0.0 -4 .57 " -6 .84" -9 .09" -1 1.31" ne 0" ... 1.983 2. 153 2.240 2.330 2.421 h P 4.193 4.6 95 5. 2 65 5. 930 7.626 9.938 11.3 85 13.104 15. 102 0.233 0.208 0.186 0.1 65 0.128 0.09...
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Aerodynamics for engineering students - part 8 pps
... 65 = [ 35. 5p5f8 x 16480 + 55 .8 x lo5 - 55 .8 x I05pj4f5 55 .8 x 1 05 i.e. 55 .8 4 65 5f4 - 55 .8 x lo5 = ( 35. 6 - 55 .8)1 05 ( 0.1488 ) 5. 84 - 55 .8 x 1oSp = or 55 .8~ ... 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....
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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...
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Aerodynamics for engineering students - part 10 docx
... ,ii = 5 = u/* 650 0 6 750 7000 7 250 750 0 7 750 8000 8 250 850 0 8 750 9000 9 250 950 0 9 750 10 000 10 250 10 50 0 10 750 11 000 11 50 0 12 000 12 50 0 13 000 13 50 0 14 ... 14 50 0 150 00 15 500 16 000 16 50 0 17 000 17 50 0 18 000 18 50 0 19000 19 50 0 20 000 20 50 0 21 000 21 50 0 22 000 22 50 0 23 000 23 50 0...
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