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MAE 101A Summer 2009 Sample Problems Chapter 4

MAE 101A Summer 2009 Sample Problems Chapter 4 - 4.3 A...

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Unformatted text preview: 4.3 A nvo—dirnensional velocity field is given by walrus—{Zoos in arbitrary units. At (1:, y) = [1, 2}, compute {a} the accelerations £11: and a}, {b} the velocity component in the direction ti‘ 2 4i)“, {c} the direction of maximum velocity, and {d} the direction of maximum acceleration. Solution: {a} Do each component of acceleration: du Eu flu dt Ll +v {x y +x){2x+l)+[ ny y){ By) 33: dv Ev Ev E = u—+ v— = {x2 —y2 +x){—2y}+[—2xy—y][—2x— 1): a? At[x,y)=[l,2},weobtain a1=lEi and ay=2fij new} {b} At (x, y] = {1, 2}, V 2 —2i — IS]. A unit vector along a 4i)” line would be n = cos4fi°i + sin4fi°j. Then the velocity component along a 4i)” lirte is vw. = v-nfl. = {—Zi— sji- {cos4fl°i+ sin4i2F’j): 5.39 units Ans. {b} {c} The maximum acceleration is Ema: = [1 32 + 262] “3 = 31.6 units at 555.3“ Ans. (c, a) 4.10 After discarding any constants of integration, determine the appropriate value of the unkmwn velocities u or v which satisfy the equation of tivo—dimensional incompressible continuity for: {cumin (Muzzy; {BJsz-xyi idiv=y2-e Solution: Substitute the given component irtto continuity and solve for the unknown component: y}+—;—=—21}’i or: v=-xy2+[[x} Arts.[a} as sin as e .2 an 2 x3 _ _=fi=_ _ ;_=— l . =-— t‘ A .b {b}§x+fiy §x+fiy[xy} fix I or u 3+[y] m{} on sip a 2 av as» y’ — —=fi=— — —; —=—2 I .. =-2 — A . {c} 614-3}: fix? JIM-6y fly x+y or- v xy+ 2 +fix} or [c] 2 [d] g+gzfl=%+%UZ-ifi; 3:45”; or: u-—2xy+x?+f(y} Ans: {d} 4.11 Derive Eq. {4.12m for cylindrical coordinates by considering the flux of an incompressible fluid in and out of the elemental control volume in Fig. 4.2. Tr Null infinimtmal Iklwnl Solution: For the differential CV shown, gemnzeiem —Zeie.m :13 Fig. 4.2 %[r+%]dti‘drdz+pvrrdzdfi+§[pvr}dr1[r+dr}dzdfi+pvgdzdr ‘3 l ”l '5' l fl — ti'dzdr — dEdr — — dEdr +§E[pv9}d +P": r+ 2 +&{pvz) r+ 2 dr —PVTrdZdE—PngZdI—PVI[I+F] dt‘i'dr: D Cancel [dtfi'drdz] and higher—order (4th—order) differentials such as [drdti'dzdrll and, finally, divide by r to obtain the final result: i ii 13 i =a A §t+ r (3r Q” r wWa)+fisz} m- 4.21 A frictionless, incompressible steady—flow field is given by Vim-f] in arbitrary units. Let the density be pt} =constant and neglect gravity. Find an expression for the pres5ure gradient in the x direction. Solution: For this [gravity—flee} velocity, the momentum equation is p[u%+ %] = —vp. or. p.[i2xv)i2vii+ {-fKin-Zflll = —vp Solve for ?p=—pn{2xy2i+2y3]}: or: fl=-fl23tyz Am. 5!: 4.29 Consider a steady, twe—dirnensienal, ineempressihle flew ei' a newtenian fluid with the treleeity field a = —2xy, v = y2 — x2, and w = e. [a] Does this flew satisfy eenservatien ei‘ mass? {b} Find tlte pressure field fix, y) if tlte pressure at paint [it = t), y = til) is equal te pa. Sulutinn: Evaluate and cheek the ineentpressible eentinuity equatien: fl+fl+fl=e=4y+2y+aae Yes! Ants) ex fly at {b} Firtd the pressure gradients fiptn the Navier—Stekes x— and y—relatiens: p£nfl+vfl+ WE] =—%+F{fi+fi+fiw 0]": as fly at 6x2 eyi eil’ fi—Zm—Zy}+{y2—x2}[—2x}]=—%+p{fi+{l+fi}, an": gz—Zpfiy2+x3] and, similarly fer the y—mernenmrn relatien, [teenage ifl+fl+fll or per a» e eai‘leseéed" fli-hfi-EXHUZ-lellyll=-%+#{-2+2+fiL or. gthtxzywi) The tare gradients Epfé‘x and Epfé‘y may be integrated te find p[x, y}: 2 2 4 P z! gdxly=fwf= —2p[%+x?] +f[y}, fliers diferentt’ate: %=—2p{x2yl+f—é=-Zfltxzy+fla “emf—gym], 0“ fly)???” Wists: p=—§[2x2yz+x4+y4)+C=pfl at [x,y}={{},{l), art CIPE Finally, the pressure iield ferthis flew is given by P= Is. -%p(2szs2 + s‘ + 1"} Am. {b} 4.31 According to potential theory (Chap. 3) for the flow approaching a rounded hvo— dimensional body, as in Fig. P431, the velocity approaching the stagnation point is given by i: = UH — 422,53}, where ii is the nose radius and U is the velocity far upstream. Compute the value and position of the maximum viscous normal stress along this streamline. Is this also the position of maximum fluid deceleration? Evaluate the maximum viscous normal stress if the fluid is SAE 30 oil at 209C, with U: 2 ms and a = 5 cm. Fig. P431 Solution: {a} Along this line of symmetry the convective deceleration is one—dimensional: 2 2 2 4 ax=ufl=U 1—3—2 U 21, =2U2 3—3—3—5 fix it x x x d Thishasamaxi'mumdecelerationat d: 2D, or at x=—'~."[5t3)a=-l.29a Ans. {a} The value of maximum deceleration at this point is an,“ = 41.317: Uzi‘a. {b} The viscous normal stress along this line is given by r’2 2m 4 U in =2y§=2pL a Jwithamaximum rm: 1" at x=-a Aar.{b} x3 n Thus maximum stress does not occur at the same position as maximum deceleration. For SAE 3D oil at 20°C, we obtain the numerical result sale 3Doil, p=att E, ,ii=o.2a 3, rm =wz39Pfl Ans. {Ii} m3 ms [Gilli m} 4.43 For the draining liquid film of Fig. P436, what are the appropriate boundary conditions {a} at the bottom y = [l and {b} at the surface y 2 iii? Fig. Has Solution: The physically realistic conditions at the upper and lower surfaces are: {a} at the bottom, y = D, no-slip: u[l}} = I} Arts. [a]: {b} At the surface, y =h, no shear stress, pg: L'}, or %[h}= [I Ans. {b} 4.80 An oil film drains steadily down the side of a vertical wall, as shown. Afler an initial development at the top of the wall, the film becomes independent of z and of constant thickness. Assume that w = w[x) only that the atmosphere offers no shear resistance to the film. {a} Solve Navier—Stokes for wl[x). [b] Suppose that film thickness and [é‘wfé‘x] at the wall are measured. Find an expression which relates ,u to this slope [fiwffix]. Solution: First, there is no pressure gradient é‘gtfi‘z because of the constant—pressure atmosphere. The Navier—Stokes z—component is ,ud wi-‘di-t2 2 pg, and the solution requires w = D at x = D and {dwr‘dx} = t} [no shear at the film edge} at x = d'. The solution is: w: gx(x-25} Arts. {a} NOTE: w is negative {down} 2." The wall Slope is E Hall: —i§, or mar-rungs: Jul _P—23 Ans. {b} a“ [dw'fihlmn 4.32 A solid circular cylirtder of radius R rotates at angular velocity 1."! in a viscous incompressible fluid which is at rest far from the cylinder, as irt Fig. P432. Make simplifying assumptions and derive the governing differential equation and boundary conditions for the velocity field v; in the fluid. Do not solve unless you are obsessed with this problem. What is the steady—state Fig. p432 flow field for this problem? Solution: We assume purely circulating motion: vz = vr = E} and 6‘63 2 t}. Thus the remaining variables are v9: fcnfr, t) and p = fcn{r, t). Continuity is satisfied identically, and the E-moment um equation reduces to a partial differential equation for v9: ‘3'” .E[li[rfir ]—1£] subjectto v,,{R,t}=nR and v,,{oo,t}=o Am. I' fiprfit‘fir’ I am not obsessed with this problem so will not attempt to find a solution. However, at large times, ort 2 co, the steady state solution is v5: flszr. Ans. 4.65 A hvo-dimensional incompressible flow is defined by Ky Kr ”=2 2 x2+y2 x +;v where K = constant. Is this flow irrotational? If so, find its velocity potential, sketch a few potential lines, and interpret the flow pattern. Fig. v4.55 M:— Snlutinn: Evaluate the angular velocity: av as K 2sz K. ZKyZ 2&l12———= —+ = [I [Irrntatinn al} Aer. Introduce the definition of velocity potential and integrate to get fix, y): “2&2- 2K? ; Vzfiz 2mg, solvefor¢=Ktan'l£l]=Kt9 Am. #3:: x +y2 {93' 1:: +1; 1 The ‘13 lines are plotted above. They represent a counterclockwise line vortex. 4.66 A plane polar—coordinate velocity potential is defined by K E .13: cos K =const r Find the stream function for this flow, sketch some streamlines and potential lines, and interpret the flow pattern. Solution: Evaluate the velocities and thence find the stream function: did Kcoscl _ 1 6y: . V 2: ré‘r r2 ré’fl solve yr: — The streamlines and potential lines are shown above. This pattern is a line doublet. 4.5? A tum—dimensional ineompressihle flow field is defirted by the velocity eomponents ”=21! i—l v=—2P'l L L L where V and L are eonstants. If they exist, firtd the stream function and velocity potential. Solution: First eheek eontinuity and irrotationality: é‘u Ev 2F" 21" , —+—=———=1‘.I|| FMIS‘; fixfiyLL ?xV=k£fl—fl]=k(fi+£] ii] price-anotefist fix By L To find the stream funet ion yr, use the definitions of a and v and integrate: 2 FE: 21"[i—l], IF: 2V[fl—y—]+f{x) 5"? L L L 2L 2? if 2? Evaluate fl=_}’+_f=_v=_y fix L nix L r’ 2‘1 Thus £=D and put-V LIT-J?— +cons£ Aria. xix L L L 4.59 Censider the nee—dimensienal inccmpressihle velccity petential d: .1}: + x2 — yz. {a} Is it true that v2¢= fl, and, if so, what dees this mean? {b} If it exists, find the stream functien MI, y} cf this flew. {c} Find the equatien cf the streamline which passes thrrmgh [I:J’J={2,1II- Sulutinn: {a} First check that lIt'2.;ai-=lill, which means that incumpressihle cuntinuity is satisfied. 2 2 We =¥+¥=e+24 :0 Yes I y {b} New use éte find it and v and then integrate te find at. 2 nzfizyh'llrzfl hence w=%+hy+flx} fix ed 2 v=§=x—2y=-%=—2y—ga hence f[:r)=—%+canrt 1 The final stream function is thus 1? . 30’: —.n") + 2.9- + .nnns: Ans. (h) {c} The streamline which passes threugh (x, y) = {2, l) is feund by setting er: a censtant: Atixarl={2, 1), W=%{12—22)+2{2){1)=—%+42% Thus the preper streamline is yr- it?) —x2)+ 2.1;}:- % Ans. {c} 4.154} Liquid drains flow a small hole in a z tank, as shown in Fig. 114.611, such that the Pm I— yelocity field set up is given by urn: D, we: I), , " Hg 2 aflzfr, where z = H is the depth of the ' water far from the hole. Is this flow pattern rotational or inotational? Find the depth at) u 11 . of the water at the radius r = R. —l: -II<1| Solution: From App. D, the angular Fifi- P430 1trelocity is 1 a" 1 e‘ m: =¥Efiy 91 7E” 9) =11 [IRRDTATIDNAU Incompressihle continuity is valid for this flow, hence Bernoulli’s equation holds at the surface, where p = palm, both at infinity and at r = R: 1 1 11m gin-Vie +peH= 12-m+ Eta-Vin were will: 2a Introduce Vrfl = [l and ‘91:“ sz to obtain 2.1 = H- Arts. 4.91 Consider 2-D incompressible steady V Couette flow hetsveen parallel plates with the upper plate movirtg at speed V, as irt Fig. 4.161 Let fl'te fluid be Honnmtomlm, with stress given by 3-41 rfl=o — r .=.o' — T .=r.x=— —+— , oanrfoo’reoortamflar fir Jr} 01v 1} Jr 2 a” £531" Make all the same assumptions as irt the derivation of Eq. [4.140). {a} Find the veloeity profile “0”)- {b} How does the veloeity profile for this ease oompare to that of a newtonian fluid? Solution: {a} Neglect gravity and pressure gradient. If u = u[y) and v = {III at both walls, then eontinuity speeifies that v = [l everywhere. Start with the x—moment um equation: [smas— Many terms drop out heeause v = [l and m and one: = D {beeause a does not vary with x). Thus we only have gr I I? i=1 Eff—H] zlflI on E=.—:ismstrfl*it, H=Cly+C2 3y dy 2 dy 6?? The boundary eonditions are no—slip at both walls: V V “biz-Ii): fl=C1[-Ii) + CE; a[v=+fi)= V=C1{+Ii) +C2, solve (312%, C2 :3 The final solution for the veloeity profile is: V V ufy}. Elf-FE Arts. {a} This is exactly the same as Eq. [4.14m for the newtonian fluid! Ans. {b} *P435 Twe ittnnissible liquids ef equal thickness I: are being sheared between a fixed and a meving plate, ......................... I :1 Pl: #1 as in Fig. P1195. Gravity is neglected, x — Fixed Fig. 114.95 and there is ne variatien with x. Find an expressien fer [a] the veleeity at the interfaee; and [in] the shear stress in eaeh fluid. Assume steady laminar flew. Solitaire: Treat this as a Ch. dpmfiiem [net Ch. I), use eentinuity and Navier-Stekes: Centinuity: in + E = i) + E 2 ti ; thus v =eenrt = i] ferne—slip at the walls 3135? 5y This tells us that there is ne 1releeitp v, henee we need enly eensider Qty} irt Navier—Stekes: e e e 52 e2 a2 allegwgb—fiwet—bay—S) firifi+fi=fi+m,2{fi+—:l Thus st=n+by The veleeity prefiles are linear in y but have a different slepe in eaeh layer. Let a. he the veleeity at the interfaee. [a] The shear stress is the same irt eaeh layer". 1': Flu—I = ,u; Fifi”! Salve fer u; = ““2 V Ans.{a} h 1’1 P'I +P’2 [£1] In terms ef the upper plate veleeity, V, the shear stress is r = (”'91 )E Annfls) lull-Pi“: h ...
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