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Unformatted text preview: 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
 Sakar

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