Aerodynamics for engineering students - part 3 pot

... (Fig. 3. 6). 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 ... Fig. 3. 12 Image systems 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): af...

Ngày tải lên: 08/08/2014, 11:21

... 2 53 Table 5.1 7~/8 0 .38 2 68 0.9 23 88 0.9 23 88 0 .38 2 68 0.9 23 88 ~14 0.707 11 0.707 11 -0 .707 11 -0 .707 11 0.707 11 3~ 18 0.9 23 88 -0 .38 2 68 -0 .38 268 0.9 23 88 0 .38 268 7512 1 .ooo 00 - ... 0.004 739 = 0.22079 A1 + 0.89202 A3 + 1.251 00 A5 + 0.66688 A7 0.011 637 = 0.6 63 19 A1 f0.98957 A3 - 1 .31 5 95A5 - 1.64 234 A...

Ngày tải lên: 08/08/2014, 11:21

... v2 u2 -+ - +- 2 7-1 2 y-1 Pmiq I Fig. 6 .37 (a) 'Incompressible' flow. (b) Compressible subscritical flow. (c) Critical flow 32 4 Aerodynamics for Engineering Students ... + XO) - A] = 0 (6.86) having the formal solution x=J [-( C+XO) f .\/(C+XO)( ~-~ XO) +4A] (6.87) 30 6 Aerodynamics for Engineering Students Now for...

Ngày tải lên: 08/08/2014, 11:21

... solved for a, b and e in terms off. The solution is, therefore, b=l-f a=-l-f e=O ( 1 .3 5d) (1 .35 e) (1 .35 f) Substituting these values in Eqn (1 .33 ), (1 .36 ) 22 Aerodynamics for Engineering ... 520 522 522 525 527 527 527 533 533 534 535 535 535 538 539 539 539 54 1 54 1 542 549 549 550 550 551 552 554 558 56 1 5 63 567...

Ngày tải lên: 08/08/2014, 11:21

... 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 ... au dudx audy + + - dt at dxdt aydt =-+ u-+v- au au au at ax ay and in the Oy direction d(v+ Sv) au dv dv -_ - +u-+v- dt at ax ay (...

Ngày tải lên: 08/08/2014, 11:21

... unit span MLE = -lCp dx Changing the sign Therefore for unit span 7i- - - Comparing Eqns (4 .31 ) and (4 .32 ) shows that CL CMLB = 4 (4 .31 ) (4 .32 ) (4 .33 ) The centre of pressure ... 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):...

Ngày tải lên: 08/08/2014, 11:21

... 0.104 0.0008 -0 .0 831 -0 .1166 -0 .1474 -0 .1754 WP)li7l 0.228 0.1 83 0. 138 0.092 0 -0 .098 -0 . 138 -0 .1 83 -0 .228 Wings of finite span When the component of the free-stream velocity ... 1.9 83 2.1 53 2.240 2 .33 0 2.421 h P 4.1 93 4.695 5.265 5. 930 7.626 9. 938 11 .38 5 13. 104 15.102 0. 233 0.208 0.186 0.165 0.128 0.098...

Ngày tải lên: 08/08/2014, 11:21

... 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 - - - ... aa - = 4.959 - 8.607 + 3. 65 8j (E) = 3. 65 W Equation (7.16) gives (1 - 1.65j3 + 4 .30 j2 - 3. 65J)dY...

Ngày tải lên: 08/08/2014, 11:21

... 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...

Ngày tải lên: 08/08/2014, 11:21

... 0.57 13 0.5624 0.5 536 0.5448 0.5 238 0.5 035 0.4841 0.4654 0.4474 0. 430 1 0.4 135 0 .39 75 0 .38 21 0 .36 73 0 .35 31 0 .33 95 0 .32 64 0 .31 38 0 .30 16 0.2900 0.2788 0.2680 0.2576 0.2477 0. 238 1 ... 0.4161 0.4 039 0 .39 21 0 .38 05 0 .36 91 0 .35 81 0 .34 72 0 .33 67 0 .32 64 0 .31 63 0 .30 64 0.2968 0.27 43 0.2 535 0. 234 3 0.2166 0.2001 0....

Ngày tải lên: 08/08/2014, 11:21