MEN 321 Problems For Chapter 6 with Solution, Fall 2009

6.1 a V = 2xti — 2ytj The velocity in a certain two-dimen- sional flow field is given by the equation where the velocity is in ft/s when x, y, and ¢ are in feet and seconds, respectively. Determine expressions for the local and convective compo+ nents of acceleration in the x and y directions. What is the magnitude and direction of the ve- locity and the acceleration at the point x = y = 2 ft at the time t = 0? From expression tor velocity , us 2xt and w= -2gt, Since - gu ou A= Bete Re +e ge then 4, (local) = au = 2x and : au. A, (eons) = u oe +Y Ss = (2ut)(2t) + 6 2y44)(o) = 4xt* Similar ly, 7] a y) a,= Feta ar qnd ww a, (local) = oe 7 24 a, (eonv)= usirv dh = (2xt)o) + (- 2yt)(-2t) = 4yt* At xX=422f¢t @qnd to uz Z(2)l.)=0 vers ~2(2)(o) 20 Se That V =0 anh = 2X axe? = 202)+ ¥@)(o) = 4 ttf? Gi, = 2g rege * —2(2) + ¢@)lo) =-4#tls* Thus, a= YO -4F £e/s* 6-/ Veaybens 2 5.66 Fels with |a| 6.2 6.2 Repeat Problem 6.1 if the flow field is described by the equation = (3x2 + 1) — xj where the velocity is in ft/s when x and y are in feet. ‘ 2 = —bx From expression for velocity , us 5K +1 Ank V bxy, Since Ax (local) = SE = © area 4 (anv)= a ou + + ge = (3x41 ox) —(bxy)(o) = 6x (3x41) = 1e x74 bx Sim) larly, ay Be aR 35 Or" a, (local) = 3 = O Ay (oom) = ust + vee = (3x71) (-b5) + (~bxy)-&) = 1gx*y 6g At xey=! ft and tao az3)*+1 =4 wre—-bli)i)=-6 ” — ue ef f ana I7l=V 4-CbL)* = Tal 44 Gna = 6x (FX741) = 60) [30°41] = 24 Lt/5? lig = 1B APY by = 180'G) — 6G) = 12 $t-/s Fs 24 4129 $47 — a |= VG)? G2) > 24.8 fe/s* 6-2 | | 6.3 The velocity ina certain flow field is given by the equa- tion Vea + x2ef + yek Determine the expressions for the three rectangular components of acceleration. From expression for velocity, uz x v=xXe2 Since - ek gu Ju au ay * ge + * RE tes te Mena. o+ ( x Mi) + &2)C0 ) +G2)(e) Similarly , 0 and ov au ow wegen £ Dx +V ag OS + (XK )bxele OB)o) + 2) O75 = 2xta + x°y2 x - ow ¥” oe7 ay = oO Also, = Our aw aur Jw Gg et Ht 5G tbe So mo o+r (xX )o)+ (x*2) (2) + Ga)(y) 2 = X*2*¢+ 49H iT] 6.4 The three components of velocity in a flow field are given by u=xvrtyt 7 vexytyzt+2 w= -3xz —- 27/2 +4 (a) Determine the volumetric dilatation rate, and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field? (4) Volumetric dilatation vate = Z + are ou (Ee. 6.7) Thas, tov velocity Components given volumetric dilatation rate = 24 + (xr#) + (—3x-2) =_O_ This vresulé indicates That There is nv cChame in the Volume of a Fluid element as /t moves from one location to another. (4) . From £45. 6.12, 6.13, and 6.1% wits The velocity com ponents Given: oo, ¢ (5-55) =2 (9-25) =~ FZ cog (2E- BE) = [o- Coee)] «Er oy 2 4 (38. 2). 4 J az- 632)]= 22 Thas Bo — (Lea)e+ sefoisg — Since W 1s not Aero everywhere the flow field Is not trrotatonal. No. o-¥ 6.5 V= — xi t yj Is the flow irrotational? Determine an expression for the vorticity of the flow field described by S=2a (£4, 6.17) Frem expression tor velocity ) “FR xy, u-= 4” ard w= 0, and with - aur ov Wrz (37-2) CE, 6.13) L au gu ly = al Sss-%) (Ee. 6.4 t [wea w,> 2 (SF - 34 ) (Eg. bi) ‘t follows That > Ay = 0 ) Wy =o ) and ,= f [o- C3) ]=3 xy Thas, -_ S A 3xy k > —_ Is hot Irrotadional, No, 6-5 2 (a0 + w, 3 +a,2) Zz [core + (o)S aye] Since Ff 1s hot Hero everywhere the +tlow pd 6.6 6.6 A one-dimensional flow is described by the velocity field u = ay + by? v=w=0 where a and b are constants. Is the flow irrota- tional? For what combination of constants (if any) will the rate of angular deformation as given by Eq. 6.18 be zero? For trrokatsone/ #lour 20, and for the velocity dis tribution given: Thas , ay 1s not Were everywhere qua the flu is not ivrobational, No. Since (Lrom Eg. 6./8 ) *_ Ow, gu b= ax” by 1b flows for the velocity dst bation given That y= a+zby Thus, there are no values of a and b (exept both C gual fo Hero ) taat will give ¢=o for all yalues of Z. Nene. 6-6 G.7 6.7 For a certain incompressible, two-dimensional flow field the velocity component in the y direction is given by the equation v = 3xy — xy Determine the velocity component in the x direction so that the continuity equation is satisfied. To satisty the continuity eguetion, Since ov = 3x-x" oy Then from &9.C/) Ou _ r C2) mx BX + Xx E guation (2) can be ittegrated with respect to x to obtain f tu = — [sxax + [xdr + Cy) or _~ 3,2, x3 us 72x +s + fy) where Fy) ts an undetermined funetiwn of 4. 6.8 Anincompressible viscous fluid is placed between two large parallel plates as shown in Fig. P6.8. The bottom plate is fixed and the upper plate moves with a constant velocity, U. For these conditions the velocity distribution between the plates is linear, and can be expressed as - yz u=U* Determine: (a) the volumetric dilatation rate, (b) the rotation vector, (c) the vorticity, and (d) the rate of angular deformation. (a) Volumetric dilatation rate = gu + a dur =O (bh) Fr velocity distribution GIvtn s D = Ww, e and .1 for du), . U w,? 4 (2- St) = ze Thus, A —_>__ U dd = ae 2 2 td.) ye yr se Cg, 6.18) Thus, ‘. oO rT 6-3 69 A viscous fluid is contained in the space between con- centric cylinders. The inner wall is fixed, and the outer wall ro- tates with an angular velocity w. (See Fig. P6.9a and Video ¥6.1.) Assume that the velocity distribution in the gap is linear as illustrated in Fig. P6.9b. For the small rectangular element shown in Fig. P6.9b, determine the rate of change of the right angle y due to the fluid motion. Express your answer in terms of ro, Ty, and w. (a) m FIGURE P69 From Eg. 4.1/8 2. ou, ou - + — XY? oxta3 Zor The linear distri bution Me- Ip w 4 h-h% So That ou. — bw Gnk Sinte V=0 pe = aw hh The negative sign indcates that The origina! right angle 1s increasing. (b) 6-9 G./o 6.10 Fora certain incompressible flow field it is suggested that the velocity components are given by the equations u= 2xy v= —-x’y w=0 Is this a physically possible flow field? Explain. Ang physically possible incompressible thw freld must satisty Conservation Of mess es expressed by the velateonsh p Ok a Ow _ ax" 39 * Tz ~O am Fer the velocity chstribution given, Ok 22 or 2 aur ae ee -—- Xx — = a J ay 2 ° Substituhon sate Eg.(1) shows thet 2y-x* +0 #0 Thus, This is not @ physically possitle thu field Wo, 6-lo