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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. = vnﬂ. = {—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; {BJszxyi idiv=y2e 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} 6143}: ﬁx? JIM6y ﬂy x+y or v xy+ 2 +ﬁx} or [c] 2 [d] g+gzﬂ=%+%UZiﬁ; 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 Vimf] 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+ {fKinZﬂll = —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" ﬂihﬁEXHUZlellyll=%+#{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, noslip: 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‘dit2 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(x25} Arts. {a} NOTE: w is negative {down} 2."
The wall Slope is E Hall: —i§, or marrungs: 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 Emoment 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 hvodimensional 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] priceanoteﬁst
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 putV LITJ?— +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 ginVie +peH= 12m+ EtaVin were will:
2a Introduce Vrﬂ = [l and ‘91:“ sz to obtain 2.1 = H Arts. 4.91 Consider 2D 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 341
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
“bizIi): 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}. ElfFE 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 NavierStekes: 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)
lullPi“: h ...
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 Spring '08
 Seshadri,K

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