MEN 321 Problems For Chapter 9 with Solution, Fall 2009

“9.3 The pressure distribution on the 1-m-diameter circular disk in Fig. P9.3 is given in the table. the disk. Determine the drag on D reo Pp = —5 kN/m2 FIGURE P9.3 D=(pdh-(pdh=( pletrdr) —p, £0? since df) =20rdr ! 2 r=o Thus, 05m D=20( pr dr — (~544)2 (in?) =20{ pr dr + 3.93 kN ° (7) where pp~ i rem Evaluate the integral numerically ysing the following integrand: rym £0 ._kN/m r(m) p (kN/m?) 0 4.34 0 0 0.05 4.28 0.05 0.2/4 ; ; 0.10 4.06 0.10 0.406 3.72 0.15 . 0.15 0.$58 0.20 3.10 0.20 0.620 0.25 2.78 0.25 0.695 0.30 2.37 0,30 0.7/1 0.35 1.89 0.35 0.662. 0.40 1.41 0.40 0. S64 0.45 0.74 O48 0-333 0.50 0.0 0.50 0.000 = Using a standard numerical integration technique with the abwe integrand gives oF = 5.43 kV 9-3 9.4 A 0.10 m-diameter circular cylinder moves through air with a speed U. The pressure distribution on the cylinder’s 1 surface is approximated by the three straight line segments shown in Fig, P9.4. Determine the drag coefficient on the cylinder. Neglect shear forces. 20\ 40 60 80 100 120 140 160 180 HFIGURE P9.4 b= (pbr cose de = br (pcose do TT or ef Break up the integration into the following three segments: D) 0<0¢70=/,222rad where p= -7390+3 where o~rad. ée. pl =2 and pl =-6 6=0 @=/.222 2) 70°< p<l00’ or 1,222<6<1.7%5 rad where f) = 8.590 -16.5 M where O~rad te. pl =-6 and pI =-/5 @, deg O=/.222 O=|.745 and 3) 100°<p </80° or 1.745 <0=3./%4 rad where P =-1s Mw, Thus, - joo? 190° Lh = 2br| S pcvs0 de +p cas6 do +{ pcs do) =2br[ I, +L+L] ty) 70 100° where (con't) 9.4 M(con't) Via 12.22 = §(- 730 +3) caso dé =[-7.39(cas6 +8 sind)+3 sind| 0.791 174s © ious 1,={ (8.596 - /6.5) cos@ do =[8.54(coso +0 sin8 ~165 sin9| =-0.260 a 4,222 and 3.1% 3.14 I,= (15) c0s@ dO) = -/,5 sin9| = /,477 1745S L7ES Hence, oly = 2brfo.7a1 ~0.260 +1477] = 0.852br or with C= Lp - 0.852br _ 0,426 > Zev" ~ Feuvr(zbr) ~ For But the pressure at @=0, the stagnation point, is 3 me. Thus, ¢pU'<3 H. so that _ 0426 G, = 3 = OS42Z 9.5 | 9.5 Repeat Problem 9.1 if the object is a cone made by rotating the triangle A~B-C about the horizontal axis A—C rather than the wedge-shaped bar. 3 2 = 40U7AC, where G,=0.5 and Az (0.3c) = 0.097 c* «) Also b= “Lpyq A + { Pac sin6 dA where dA=21rdx =2m (Xx sin®) dx ence, X= 0.36/5/90 od = ~(-0.2) £0U" (0.097 c2) +{ Pac sino |2™(x sind)dx | @ x20 or with fy,= average pressure = constant Ey (2) becomes Lf = (0.2) PU (0.097 c2) t Por sin @ can) { x dx Xoo = (0.2) 2 pV (0.092c*) + fp, sin?O (217) (0.36/sin6) /2 or i =[(o2z pu'+ sc ](0.09 7c?) (3) By combining Eqs. ()) and (2) we obtain £ eV" (0.097?) (0.5) =[(o24 eU* + fg ] (0.09 Tc) which gives Pac = (0.3) zeU* 9.1 9.1 The wedge-shaped bar shown in Fig. P9.1 moves through a U fluid with its top (surface A-C) parallel to the free stream. The drag coefficient is 0.5 and the pressure on the base (surface A~B) is less than the free stream pressure by an amount equal to 20% of the dynamic pressure. Viscous effects are negligible. Determine the average pressure on surface B—C. BFIGURE P9.1 Also, = “fae Abe t Pac Ag, sin@, where frre = (-0.2)% U~ and Abe = Ang /sin9 Hence, 2 -(-0.2)4 0U "Aya + Poe fag sind = zeU Age (0.5) or (0.2)40U*+p,, =(0.5)£0U* or fe, =(0.3) 400" 9.2. Pave = —1.2 kN/m? . Tave = 5.8 x 10-2 kN/m2 9.2 The average pressure and shear stress acting on the surface of the 1-m-square flat plate are as indicated in Fig. P9.2. u a=7° Determine the lift and drag generated, Determine the lift and drag_§ ———> Oo) if the shear stress is neglected. Compare these two sets of results, (2 Paye = 2.3 KN/m2 Tave = 7.6 x 10-2 kNém2 BFIGURE P9.2 Since Sp dA= Pye and (%.dA=%yg A if follows that = -PA, sink +p, A, sinh +T A, cosk +& Ay cosd or with 4,=Az =/m* and « = 7° L= A, sind (p,-p) 14, cosale+%) = (Im) sin?” (2.3 ~(-1.2)) 44 +(Im*) cos7°(s.9x10* + 26x10 7) AM = 0.427kN + 0./33 kN = 0.560 kN Note, if shear stress is neglected 7 = 0.427 kN (ce, 7 =% =) Also, £L= -p,4, Cosx +p, Az cos — TA, sing - Te Ay sind or L= A, cosx (p,- p,)-A, sinx(T +%) = (Im?) cos7°"(2.3 --1.2)) £4 —(Im®) sin 7°(5.8x102+ 7.6 xj?) 4M = 3.47 kN - 0.013 kN = 3.45 kN Note , 'f shear stress is neglected X= 3.47kN Mote: If the general expressions d) = if f 6080 dh +(% sin@ dA and X=~-\p sinddA +f % cos@ dA are use, be careful abovt the signs involved. On the yooer surtace 6,=97° and and % are positive as indicated ee ce in the figure. On the lower surface 0,= 277° and p and %, are positive as indicated in the lower figure. lo ip For example, with this notation 2% <0 on the lower surtece. (li - - X= —(-1.2 1h) 51 97°(1m*) -(2.3 4) sin 277° (1m?) O,=277° + (5.0x107 4K ) 295 97°(1m?) +(-7.6x10 24M ) £95277?) = 3Y¥S5 kV, as obtained above. 9.6 | 9.6 A 5-ft-long porpoise swims with a speed of 30 ft/s. Would a boundary layer type flow be developed along the sides of the animal? Explain. Re= or with L= SH, U= 308, and v= L2xloSE (i.e, 60°F water) Re= isi gen = /,24x/0" This Reynolds number is Jarge enough _to ensure boundary layer flow. (Re */000 is often assumed to be the lower limit.) [9.7 X7 Typical values of the Reynolds number for various an- imals moving through air or water are listed below. For which cases is inertia of the fluid important? For which cases do vis- cous effects dominate? For which cases would the flow be lam- inar; turbulent? Explain. - (e) rccisiosioAd cin (a) (b) (c) (d) Wi dN CN Animal Re large whale 300,000,000 flying duck 300,000 large dragonfly 30,000 invertebrate larva 0.3 0.00003 bacterium (imams Inertia important if Rex/ (c.e whale, duck, dragonfly) Viscous effects dominate if Res! (¢.¢ larva, bacterivm) Boundary layer flow becomes turbulent for Re on the order of 10° to 105 (¢.e, whale and perhaps the duck) The flow would be laminar for the dragontly, larva, and bacterivm and perhaps the duck. 7-3 L.9 9.9 Approximately how fast can the wind blow past a 0.25-in.-diameter twig if viscous ef- fects are to be of importance throughout the en- tire flow field (i.e., Re < 1)? Explain. Repeat for a 0.004-in.-diameter hair and a 6-ft-diameter smokestack. Re=2 <) or U<& if viscms effects are to be important throughout the flow. For standard air V=4LS7x16 *£° Thos, ~¥ 57X10 V <Be , where D is the diameter in feet. object |D, ft | U, ft : twig 2,08XI0"| 7 S4¥x/07 hair 3.33x/0"| 0.47/ smokestack | 6 2.62K/0> 1-9 9.10 A viscous fluid flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm. Determine the boundary layer thickness at distances of 0.20, 2.0, and 20 m from the leading edge. Assume laminar flow. For laminar flow &=CYX , where C is a constant, Thus, 3 C= oe = PEM _ 0.0105 on $= 0.0105 1X where X~m §~m Vx" yl.3m ? x,m | 6,m | b,mm 0.2. 0.90¥70| “470 2.0 0.0/48 148 20.0 0.0470 | 420 — LAL 9.11 If the upstream velocity of the flow in Problem 9.10 is U = 1.5 m/s, determine the ki- nematic viscosity of the fluid. _ U5 2) (12x10) —6 m* =6.65X/0 = 25 (1.3m) 6.65X10