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 condition...

Ngày tải lên: 08/08/2014, 11:21

... 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...

Ngày tải lên: 08/08/2014, 11:21

... 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 Sin...

Ngày tải lên: 08/08/2014, 11:21

... 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 a...

Ngày tải lên: 08/08/2014, 11:21

... 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...

Ngày tải lên: 08/08/2014, 11:21

... 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 c...

Ngày tải lên: 08/08/2014, 11:21

... 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...

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 - - - ... 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) -...

Ngày tải lên: 08/08/2014, 11:21

... 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+...

Ngày tải lên: 08/08/2014, 11:21

... 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...

Ngày tải lên: 08/08/2014, 11:21