Aerodynamics for engineering students - part 1 pdf

... 95 95 96 97 10 1 10 4 10 4 10 4 10 5 10 6 10 7 10 9 11 0 11 0 11 2 Contents Preface xlll 1 Basic concepts and defdtions Preamble 1. 1 Units and dimensions 1. 1 .1 Fundamental dimensions ... 716 0 = 0 .11 51 kgm-3 P p =-= RT 287.3 x 216 .5 e=($&) 314 = 1. 19 P 1. 44 x kgm-ls-l 17 1 1. 19 = - 1 0-5 = Consider a dimension that, on the aircraft, has a length of 10 ... Exercises 1 1 1 2 2 3 4 4 4 5 5 8 8 8 10 11 15 15 17 19 19 22 26 26 28 29 30 35 38 41 44 44 50 8 Aerodynamics for Engineering Students 1. 2.4 Temperature In any form...

Aerodynamics for engineering students - part 2 pptx

... 325Nm-'. Therefore + 1 = 1. 4 21 po - 42670 p 10 1325 Therefore 1 5 1 - M2 = 0 .10 6 5 M' = 0.530 M = 0.728 1 + - M2 = (1. 4 21) 2'7 = 1. 106 The speed of sound ... equation of state for a perfect gas is P/(m = R Substituting forplp in Eqn (1. 11) yields Eqn (1. 13) and (1. 14), namely ~p - cv = R, cP = - R cy =- ‘R Y -1 Y -1 * It should be ... 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...

Aerodynamics for engineering students - part 3 pot

... 8 = 1. Then Eqn (3.49) becomes 1 2 PT -PO =-pU2 (1 - [2+B]’) (3.50) 1 2 = pU2(3+4B+BZ) At the bottom p = p~ when 8 = -n/2 and sin O = - 1 : (3. 51) 1 2 PB -PO = ... the line along the x-axis as $ = 0 12 8 Aerodynamics for Engineering Students Doublet axis Fig. 3.20 Streamlines due to a doublet - 2c - -x- Fig. 3. 21 Consider again a ... Therefore p-po=-pU 1- 2sin0f- 2 2[ ( 2rua >’ (3.49) Potential flow 12 9 Now and Therefore r; = x2 + y2 - 2xc + c2 m x2+y 2-2 xc+c2 47r x2 + y2 + 2xc + c2 $I=-ln...

Aerodynamics for engineering students - part 4 doc

... C C c x/ =-( i -cose) (i-cos~) =-( cos~-cose) 2 2 2 and k from Eqn (4. 51) : +,( (1 -; ) COS4Il - (1 where ’t (4.54) Fig. 4 .16 17 6 Aerodynamics for Engineering Students and ... 64panels -1 .00 -2 0 -1 5 -1 0 -5 0 5 10 15 20 Angle of attack,degrees Fig. 4.26 Variation of lift and moment coefficients with angle of attack for NACA 4 412 aerofoil Exercises 1 A ... J (4 .11 la) (4 .1 1 lb) Making a comparison between Eqns (4 .10 9) and (4 .1 11) for the vortices and the [VxQIVOflkXS [V.QISOLUL%S and [VYQIVOI'tiL%S = -[ vxQISOllrCeS (4 .11 2) With...

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 0.707 11 0.707 11 -0 .707 11 -0 .707 11 0.707 11 3 ~18 0.923 88 -0 .382 68 -0 .38268 0.923 88 0.38268 7 512 1 .ooo 00 - 1 .ooo ... 0.22079 A1 + 0.89202 A3 + 1. 2 51 00 A5 + 0.66688 A7 0. 011 637 = 0.663 19 A1 f0.98957 A3 - 1. 315 95A5 - 1. 64234 A7 0.0 216 65 = 1. 1 15 73 A1 - 0.679 35 A3 - 0.896 54 ... - PlTl P2T2 PI4 41 -P2&2+P1A1 -p2A2+4(p1+p2)(A2-A1) =o (6 .13 ) (6 .14 ) (6 .15 ) (6 .16 ) a n s M= 0 Fig. 6 .1 One-dimensional isentropic expansive flow 240 Aerodynamics for Engineering...

Aerodynamics for engineering students - part 6 potx

... rule - the application of linearized theories of subsonic flow Consider the equation (6 .1 18) in the subsonic two-dimensional form: (1 -Mk )-+ -= O @4 a24 (Eqn(6 .11 8)) ax2 ay2 For a ... 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; ... by the factor 1/ C. With D = 1, i.e. 17 = y, Eqn (6 .13 1) gives (6 .13 1) Thus the solution to Eqn (6 .1 18) or (6 .1 19) is found by applying the transfonnation X and, for this and the...

Aerodynamics for engineering students - part 7 docx

... 0 .1 0 .16 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& quot; 9.09" 6.84" 4.57" 0.0 -4 .57" -6 .84" -9 .09" -1 1. 31& quot; ... 0.225 0 .16 3 0 .10 4 0.0008 -0 .08 31 -0 .11 66 -0 .14 74 -0 .17 54 WP)li7l 0.228 0 .18 3 0 .13 8 0.092 0 -0 .098 -0 .13 8 -0 .18 3 -0 .228 Wings of finite span When the component of the free-stream ... 37 .16 " M 1. 59 1. 666 1. 742 1. 820 1. 983 2 .15 3 2.240 2.330 2.4 21 h P 4 .19 3 4.695 5.265 5.930 7.626 9.938 11 .385 13 .10 4 15 .10 2 0.233 0.208 0 .18 6 0 .16 5 0 .12 8 0.098 0.086...

Aerodynamics for engineering students - part 8 pps

... A/Ae = 1 + 1. 178~ where suffix i denotes the value at the entry section. Also A, U, = A U, A 48 u Iu - e - A - 1 + 1. 178~ Then -= due -4 8 x 1. 178 (1 + 1. 178~ )-& apos; ... 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 - - - U - 0 0 .1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1. 0 Fig. ... (19 81) 'H-R, method for predicting transition', AZAA J., 19 , 810 412 . 420 Aerodynamics for Engineering Students Aerodynamics for Engineering Students Pi 0.9 0.8 0.7 - 0.6...

Aerodynamics for engineering students - part 9 ppt

... - 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, 206, 26 7-2 74 51 8 Aerodynamics for Engineering Students...

Aerodynamics for engineering students - part 10 docx

... 9750 10 000 10 250 10 500 10 750 11 000 11 500 12 000 12 500 13 000 13 500 14 000 14 500 15 000 15 500 16 000 16 500 17 000 17 500 18 000 18 500 19 000 19 500 ... 0.50 kT 0 .11 8 0 .11 5 0 .11 2 0 .10 9 0 .10 6 0 .10 3 kQ 0. 015 7 0. 015 4 0. 015 0 0. 014 5 0. 013 9 0. 013 2 548 Aerodynamics for Engineering Students Suitable values for initial guesses for a and ... 5. 018 0 5.4297 5.87 51 6.3570 6.8785 7.4427 8.0532 8. 713 8 9.4286 10 .202 11 .039 11 .945 12 .924 13 .985 15 .13 2 16 .373 17 . 716 19 .16 9 20.742 22.443 24.284 554 Aerodynamics for Engineering...

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