MAE 210B. Homework Solution #6 Winter Quarter, U 2 =r U=r 2 << 1; ) r << U : (1) The boundary conditions written in polar coordinates,

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1 MAE B Homewok Solution #6 Winte Quate, 7 Poblem a Expecting a elocity change of oe oe a aial istance, the conition necessay fo the ow to be ominate by iscous foces oe inetial foces is O( y ) O( ) = = << ; ) << : () b The bounay conitions witten in pola cooinates, Dene the steam function > : = ; x; = ; () > : = ; ; = : (3) ; = x ; (4) such that continuity is satise. We know that the steam function satises bihamonic equation fo two-imensional Stokes ow ( ) = : (5) The bounay conitions (taking efeence steam function = at the walls) ae > : = ; = ; > : = ; = ; = x; ; (6) = : (7) c Fom equation an B.Cs, we see that the aiables which can inuence the steam function ae = F (; ; ; ): (8)

2 The numbe of aiables with imensions ( = [L T x ], = [LT x ] & = [L] ) ae n = 3 an the numbe of inepenent imensions ([L] & [T ]) ae k =, whence, the numbe of goup is n x k =. Equation 5, witten in pola cooinates is = f (; ) (9) x x Substituting = f (; ) in the aboe equation, we get = () f i + f + f = : () Assuming the solution of the fom f exp(), we get = ~i; ~i, theefoe the solution fo f is gien by f () = A sin + B cos + C sin + D cos ; () f () = (C x B) sin + (A + D) cos x D sin + C cos ; (3) f () = ; f () = x; f () = ; f () = : (4) It's tiial to see B = an D = x(a + ) using the lowe wall bounay conitions, an the othe bounay conitions leas to Soling fo A an C, we get f () = A sin + C sin x (A + ) cos = ; (5) f () = C sin x (A + ) sin + C cos = cos : (6) A = x Afte a eaangement, we n x sin ; B = ; C = x sin cos x sin ; D = sin x sin (7) f (; ) = x sin [ sin sin( x ) x ( x ) sin ] : (8) Velocities (o not epen on ) ae f (), = xf followe fom thei enitions, x sin = x x sin fsin [sin( x ) x cos( x )] + [sin x ( x ) cos ]g ; (9) f sin sin( x ) x ( x ) sin g : ()

3 Pessue can iectly be obtaine fom steamfunction using the elation! = x only fo two-imensional ow) as shown below, e z (tue = p = ; ) p = z ( e z ); () + + witing the aboe equation in e an e iection gies Integating the st equation, p = ( ) = p = x ( ) = = f + f ; () (f + f ); (3) (f + f ): (4) p(; ) = x (f + f ) + G(); ) p = x (f i + f G ) + an substituting p= into the secon equation gies (5) Thus G = f i + f + f = ; ) G = constant = p : (6) p(; ) x p = x (f + f ): (7) Since f () = x3c sin x(a+3d) cos +D sin xc cos we n f +f = xc sin xd cos, theefoe p(; ) x p = (C sin + D cos ); (8) sin + sin sin( x ) p(; ) x p = x sin : (9) e Pessue on the wipe = is p(; ) x p = sin x sin : (3) To calculate the iscous stesses on the wall, we ene a nomal ecto on the wipe n = xe such that the iscous stess on that iection is y n = x e x e ealuate at =. = + ; = + : (3) 3

4 It is easy to see that is ientically zeo = x f f + = (3) an theefoe only the tangential iscous stess contibutes to the foce, which is = (f + f ) = The iscous stess on the wipe =, p = cos x sin y = n= cos x sin cos( x ) x sin : (33) x sin : (34) We see that both the pessue an the iscous stess acting on the wipe ay as x an hence the foce will ay as ln. Theefoe the foce acting on the wipe fom the oigin to any nite aius will be logaithmically innite. But in eal life, the wipe can nee make complete contact with the plane suface an theefoe a nite stess at the oigin, esulting in a nite foce (which will epen on the gap with an the geomety of intesection). This poblem was st sole by G.I. Taylo as a moel fo scape an he shows that the nomal an tangential stess on the scape euces to that of Lubication theoy when <<. Poblem A moe inole poblem woul be the Stokes ow aoun a ui sphee of abitay iscosity, whee both insie an outsie motion of the sphee has to be sole simultaneously. But hee the iscosity of insie ui is negligible, thus it is enough to consie only outsie ow. We stat with the enition of Stokes steam function, sin ; = x sin The fee steam bounay conition at innity an the no penetation conition at = a still hols fo this poblem as in ow oe a soli sphee, but the non-slip conition nees to be eplace with the stess fee conition in the tangential iection, = at = a i.e., =a = because at = a, 8. + = ; ) =a (35) = (36) Taylo, G.I., 96. On scaping iscous ui fom a plane suface. Miszellangen e Angewanten Mechanik (Festschift Walte Tollmien), pp Taylo, G.I., 97. The Scientic Papes of Si Geoey Ingam Taylo: Volume 4, Mechanics of Fluis: Miscellaneous Papes (Vol. 4). Cambige niesity Pess. 4

5 Stokes steam function o not satisfy bihamonic equation, but satises the following equation, with bounay conitions, E 4 = " sin + = a : = ; sin # = (37) = (38)! : = sin : (39) Assuming the solution of the fom = f ( ) sin, the equation becomes x! f = ; (4) Substituting f ( ) into the aboe Eule equation gies us the oots = x; ; ; 4. The geneal solution is an the bounay conitions ae A f ( ) = = a : f = ; + B + C + D 4 (4) = (4) f! : f = : (43) pon soling, we get A =, B = xa=, C = = an D =. Theefoe Hence the elocity, f ( ) = = ( x a ); (44) ( x a) sin : (45) ( x a) cos ; (46) Equation fo pessue is E = = x x a sin : (47) p = ; (48) f x f a sin = sin (49) 5

6 hence (see Acheson, pp. 4) Integating of st equation, p = sin (E a ) = p = x sin (E a ) = p(; ) = x a cos + G(); ) p a G = sin + 3 cos ; (5) 3 sin : (5) an substituting into the secon equation gies G= = ) G = p. Theefoe the pessue is p(; ) = p x a cos ; p(a; ) x p = x a (5) cos : (53) To calculate iscous foce, we ene a nomal ecto on the sphee n = e, theefoe y n = ( ; ; ). But by enition = at = a, only the nomal iscous stess contibutes to ag. = = cos : (54) =a =a a Dag foce is calculate in the iection of motion so that net stess acting on the sphee in that iection is xp(a; ) + p + j =a cos an with the ieential element on the suface of the sphee = a sin, F D =Z Z = 3aZ xp(a; ) + p + =a Z a sin cos ; (55) sin cos ; (56) F D = 4 a: (57) =3 of F D comes fom iscous stess an =3 fom pessue ag. Poblem 3 The conition neee fo the ow to be ominate by iscosity is O( y ) O( ) R =R R=R << ; ) R << : (58) In this low Re limit, we can use the Stokes equation = xp + in spheical pola cooinates. Due to the symmety of the poblem, we hae = = =, thus we hae (; ) an p(; ). The momentum an momentum euces to p = ; p = ; ) p = p : (59) 6

7 No centifugal foce exists in Stokes ow, hence the constant pessue. The momentum equation then becomes, Bounay conitions ae + sin sin x sin = : (6) = R : = R sin ; (6)! : = : (6) Looking at the bounay conition, we assume the solution of the fom = f ( ) sin. With this equation (6) euces to f + f = ; f (R) = R; f () = : (63) The geneal solution is B f ( ) = A + It's tiial to see A = an B = R 3. Hence the elocity is = R sin R The only non-zeo stess component which can contibute to toque is = : (64) = x3 sin R : (65) 3 ; (66) =R = x3 sin : (67) Thus, the toque on the suface is obtaine fom the pouct of stess an the aial istance R sin fom the axis of otation, integate oe the suface with the ieential suface element = R sin, T =Z Z = x6 R 3 Z R sin R sin ; (68) sin 3 ; (69) T = x8 R 3 : (7) 7

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