Fluid Physics 8.292J/12.330J % (1)

Transcription

1 Fluid Physics 89J/133J Problem Set 5 Solutios 1 Cosider the flow of a Euler fluid i the x directio give by for y > d U = U y 1 d for y d U + y 1 d for y < This flow does ot vary i x or i z Determie the propagatio ad/or growth/decay of small perturbatios to this flow that are siusoidal i the x directio ad which do ot vary i the z directio Solutio: We begi with the geeral equatio for the structure i y of two-dimesioal disturbaces: % (1) dy U c dv% U yy v k + = Note that, as i the classical Rayleigh problem, the secod term i brackets vaishes withi each of the three regios of the flow As i that problem, we fid solutios to (1) withi each regio ad match across the two boudaries separatig the regios The solutio to (1) i each regio, which satisfy the boudary coditios at y =± have the form ky v% = Ae, y > d, ky ky v% = Be + Ce, y d, ky v% = De, y< () The first matchig coditio is to match the displacemet i y, which i this problem is equivalet to matchig v% itself The secod coditio is to match the fluid pressure, which ca be foud from the origial liearized equatios or from itegratig (1) i y to get cotiuity of

2 U y dv% v% dy U c Applyig these two matchig coditios at y = ad y = d to () yields the dispersio relatio 4ckd U kd = ± + (3) kd 1 9 1kd 4( kd) 8e There is a rage of kd i which the argumet of the square root i (3) is egative; i this rage, c had a imagiary part ad the solutios are expoetially growig or decayig The figure below shows the positive root of the odimesioal growth rate, σ : The upper root of the solutio for the odimesioal phase speed over a larger rage of kd is show below:

3 Note that the phase speeds are always smaller tha the peak velocity of the mea flow Like the classical Rayleigh problem, the two trasitio regios of the mea flow support Rossby waves which, i a arrow regime of wavelegths, may iteract ustably to produce expoetially amplifyig disturbaces

4 Cosider a cylidrical tak of radius r of a icompressible fluid, rotatig at agular velocity Ω ad subject to a uiform gravitatioal acceleratio, g, dowward alog the rotatio axis, as pictured below: a) Derive a expressio for the shape of the free surface Solutio: I equilibrium, the radial pressure gradiet acceleratio balaces the outward cetrifugal acceleratio: which, whe itegrated, yields dh V g = =Ω dr r r, Ω h= h + r, g where h is the height of the surface at the ceter Thus the surface is parabolic

5 b) Suppose that the bottom of the cylider is statioary, so that the fluid is movig with respect to it This relative motio exerts a torque o the fluid, but we ca cosider that this torque is cofied to a thi layer adjacet to the bottom of the tak We further assume that the flow remais circularly symmetric Takig M to be the agular mometum of the fluid per uit mass, M = rv where V is the tagetial compoet of velocity A equatio for the coservatio of agular mometum i the thi boudary layer is M M + u = r τ, t r z (4) where τ is called the "frictioal stress" We assume that this stress vaishes at the top of the boudary layer (i fact, that is the defiitio of the top of the boudary layer), ad that the stress at the surface is give by s D τ = C V (5) It ca be show that if the boudary layer is sufficietly thi, the right side of (4) will be large compared to the first term o the left, ad so, to a good approximatio, M τ u r r z (6) We ca further assume that M does ot vary with altitude withi the thi boudary layer The icompressible mass cotiuity equatio i cylidrical coordiates is 1 w ( ru) + = r r z (7) Takig w to vaish at the bottom of the tak, use (5) - (7) to fid a expressio for w at the top of the boudary layer, i terms of the radial distributio of V (or M ) Evaluate this expressio for the case that V = V r r Solutio: If we itegrate (6) through the depth of the boudary layer ad apply (5), we get rcdv uh =, M r (8)

6 where h is the boudary layer depth Now itegratig (7) over the depth of the boudary layer gives wb = 1 ( ruh), r r (9) where gives w b is the vertical velocity at the top of the boudary layer Combiig (8) ad (9) w b 1 rcv = r r M r D (1) I the case that V = V r, this gives r w b 3 = CDV r (11) r Here it is see that the vertical velocity is, i this case, proportioal by a costat to the swirlig velocity compoet c) Oe problem with this solutio is that it requires radial flow through the outer wall of the tak Cosider istead ad ubouded tak (let r ), but use the followig velocity distributio: V = V r m, r r r m m (1) r V m m, r > r r m where is a umber betwee zero ad oe Fid the depth-averaged radial velocity i the boudary layer ad the vertical velocity at the top of the boudary layer Solutio: For the iterior part of this flow, the solutio is, of course, give by (8) ad (11) Substitutig the secod part of (1) ito (8) ad (1) gives ad CD rm uh = Vmr 1 r rm wb = CDVm 1 r, (13) (14)

7 d) Extra credit: Cosiderig the depth of the tak to be H, which may be take to be large compared to the thickess of the boudary layer, fid a expressio for the istataeous rate of decrease (with time) of agular mometum of the fluid above the boudary layer What value of makes this rate of decrease idepedet of radius? Solutio: Accordig to the Taylor-Proudma Theorem, i a slowly chagig, rotatig, icompressible ad iviscid fluid, the velocity should ot vary alog the directio of the rotatio vector Accordigly, above the boudary layer i the preset flow, the radial ad swirlig velocity compoets should ot vary with height If we use U to deote the radial velocity above the boudary layer, mass coservatio tells us that there ca be o et mass flow across ay cylidrical surface, so that UH = uh, (15) where H is the fluid depth (techically, mius the boudary layer depth, h) Applyig this to the outer solutio, (13), gives U CD rm = Vmr H(1 ) r (16) Now agular mometum is coserved i the iviscid flow above the boudary layer, so the time rate of chage of agular mometum is give by M t M = U r (17) Substitutig (16) for U ad usig (1) to calculate M r i (17) gives CD rm = Vm r M t H r (18) Clearly, whe = 1, this spi dow rate is idepedet of radius