FLUID DYNAMICS OCEAN 511
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This 6 page Class Notes was uploaded by Aurore MacGyver Sr. on Wednesday September 9, 2015. The Class Notes belongs to OCEAN 511 at University of Washington taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/192149/ocean-511-university-of-washington in Oceanography at University of Washington.
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Date Created: 09/09/15
AMATHATMOS 505 OCEAN 511 Waves TRY NOT TO SPLASH TOO MUCH If any instructions for wave generation are unclear7 please ask for a demo Propagating waves In the long tank7 dip the wooden paddle to about half the depth of the water Move it back and forth a couple of times then stop7 to create a nite packet or group of waves 1 Observe Explain what you observe in at least three sentences 2 Repeat the paddle movement This time have one person follow7 with their nger7 the leading edge ofthe disturbance7 and have another person follow a wave crest in the wave packet Does the crest near the leading edge of the wave packet propagate all the way to the other side of the tank What happens to it group velocity vs phase velocity 9 Now create a bore Pile up77 the water on one end of the tank then release7 sending a step function of water out from the end of the tank Observe What happens to the bore after it re ects from the ends of the tank a few times You may think of this in terms of the shape it starts with versus the shape you notice after a few bounces 4 Why do you think this what you observe in 3 happens Standing waves In the approximately 4 foot tank7 use the paddle to make a standing wave7 ie one whose pattern does not change over time Crests and troughs will alternate in any given location It may be helpful to aim to create 2 wavelengths 2 crests and 2 troughs in the tank 5 Note how the resistance changes depending on whether you are forcing at the same frequency as the standing wave or not Be sure to let others in the group try this Standing waveparticle trajectory In the narrow taller tank7 rst stir up the particles from the bottom of the tank7 then use the paddle to make a standing wave of exactly 1 wavelength in the tank 6 Sketch the particle trajectories Compare this to Figure 714 in Kundu 7 How do they compare to the trajectories you expect in a traveling wave Kundu Fig 76 Propagating waves against a mean ow In the recirculating ume with the large mountain 00 Use the paddle to make propagating waves from upstream of the mountain7 towards the mountain How far do these waves travel 3 Now make propagating waves from the downstream side of the mountain7 back towards the mountain How far do these waves go What factor is limiting the wave propagation 10 If you have time7 note the change in wavenumber as the waves approach the mountain Amath Atm Sci SOSOcean 511 Bretherton Lecture 22 Swirling ow in a narrowing tube HWSh The Problem Consider an inviscid uid of constant density p owing steadily though a cylindrical pipe of radius R aligned along the x axis as in the le half of the gure The ow includes a component U along the pipe and a swirlng component in solid body rotation with angular velocity Q gure shows the case Q lt 0 In cylindrical coordinates X y r cos 9 z r sin 9 the ow velocity is uUiQre5 Ui QZj i ka Use Bernoulli and conservation of mass momentum and circulation around circular material rings to determine the velocity components along the pipe and swirlng around the pipe after the pipe narrows In particular show that the ow along the centerline will move faster in the x direction than the ow at the walls of the pipe Sketch what the vortex lines look like within the narrower part of the pipe Solution Strategy In thinking about any problem we rst should consider xymmelries The ow in this problem should remain radially symmetric as it swirls down the tube as it starts this way upstream of the constriction and the circular crosssection of the tube should preserve this symmetry It is also o en useful to nondimensionalize a problem especially if we will need to solve it numerically and determine the nondimensional parameters governing the solution This is best done a er we stare a little at the governing equations Where there is ambiguity we will use an underline to indicate dimensional elds We let subscript l denote upstream variables and 2 denote downstream variables In an inviscid constantdensity steady ow we can apply the Lagrangian concepts of conservation of circulation and of Bernoulli function also making use of the radial symmetry In particular a uid element and a material circle starting at some upstream radius 1 alongtube velocity g1 U and tangential veloCity X1 251 will end up at a unique downstream radius 52 with alongtube velocity M52 and tangential velocity 223 Governing equations There are two associated boundary conditions First by radial symmetry uid on the centerline of the pipe remains on the center ine 1 2 0 BCl Second the uid on the outside of the tube upstream of the narrong ows tangent to the outside of the tube and hence must remain on the outside of the tube downstream of the narrowing 51 gt 52R2 BC2 We apply our uids knowledge to derive equations relating the downstream quantities to the upstream quantities 1 Mass conservation The upstream mass uX alM across the annulus Q lt r lt 1 1 must equal the downstream mass uX across the annulus r lt r lt r g 61M Pu12 rl Q1 woman 1 2 Circulation The circulation F around the material ring starting upstream with radius 11 is conserved as it moves through the narrowing and takes on its downstream radius 12 r new vz2nrz 2 3 Bernoulli Conservation of the Bernoulli function along a streamline starting upstream at 1 B Blp 12 2152 Bzp a 1252 1 4 Radial force balance Radial force balance in the swirling upstream and downstream ow implies p e 39DEZDt PszVz d22 dzz and similarly with 2 replaced by l in the upstream ow In the upstream ow we can substitute the known swirl velocity 1 911 into this equation to obtain 39P QZEL 396121 61D 321 20 P QzllzZ Q Here 20 is the upstream centerline pressure at 1 0 an arbitrary constant that does not in uence any of the ow dynamics Nondimensionalization We nondimensionalize these equations using the upstream ow Lengths are nondimensionalized by dividing by the upstream pipe radius R x cR r1 rlR r2 rzR From now on we ll specify such nondimensionalizations by writing the dimensional variable in terms of the corresponding dimensionless variable multiplied by the dimensional scaling factor K Rx 1RV1 2 er because this makes it easy to substitute into the dimensional equations 15 to derive nondimensional analogues If we view the upstream radius of a material circle as an unknown function of its downstream radius w imply the nondimensional boundary conditions r1rz 0 0 BCl r1rz Rz 1 R2 05 BC2 Here R2 05 is the radius ratio of the downstream tube radius to the upstream tube radius We nondimensionalize uid velocities using the in ow velocity U 2Uu222UV2 1Uu121UV1 The nondimensional upstream velocity components are obtained from their dimensional counterparts 1 U 3 H1 1 219Z13 UV1 QRV13 V1 Srl Here S is the nondimensional swirl ratio of the upstream tangential velocity at the pipe wall to the upstream pipe radius We nondimensionalize pressures using upstream kinetic energy 212 20 PU2 P1 The nondimensional upstream pressure can be found from its dimensional form Q to be 21 20 pU2 p1 20 p 92532 10 p Qszr122 2 p1 S2r122 UP Now we nondimensionalize the equations Straightforward substitution of the nondimensionalizations of the individual variables into the dimensional equations Q Q yields Mass uzrz drz um drl r1 drl 3 drldrz uzrzrl 1 Circulation vzrz v1r1 Sn2 3 V2 SrlzVz 2 Bernoulli p uzz v222 p1 ulz v122 Szrl2 2 uzz 2S2r12 p2 vi 3 Radial force dpz drz vzz r2 4 This can be regarded as a system of four equations for four unknown functions of r2 namely n m vz and p2 The two equations 2 and 3 are algebraic and can be used to express u and v in terms of the other two unknowns V1 and p2 The two equations 1 and 4 can then be regarded as a system of 2 coupled firstorder ODEs for V1 and p2 as functions of r2 This ODE system needs two boundary conditions these are BCl2 Hence we appear to have a mathematically well posed problem The solution depends on two nondimensional parameters R2 the pipe radius ratio which is 05 for our geometry and S the swirl ratio which we can vary Solution of equations One complication is that both ODE boundary conditions are on the variable r1 a boundary value problem at different values of r2 rather than having both variables V1 and p2 specified at one value of r an initial value problem Since it is easy to numerically solve ODE initial value problems we use an iterative method called a shooting method We start at r 0 where BCl tells us that r10 0 guess an initial value pm for pz0 numerically integrate shoot this ODE BVP out to r 05 and calculate the BC2 error 8 p20 lr105ll Using a couple of initial guesses pm we can use an numerical iteration method called the secant or modified Newton method to converge to the pm for which 8 p20 0 and BC2 is satisfied My ODE solver uses a secondorder accurate midpoint RungeKutta method and a grid spacing Arz 001 This shooting method is implemented in the function hw5hsolnm on the class web page which is called with S 2 as the first part of the script swirldemom The input parameters are R2 S and a pair of initial guesses p20 to get the iteration started Swirl S 2 Pressure 14 c 12 M olt 1 3 I r 5 08 I r x I o z o 6 I I 10 o 4 C 15 upstream 02 downstream p20 187388 0 material radius A an 0 01 02 03 04 05 0 02 04 06 08 1 r2 r Along pipe velocity Tangential swirl velocity 7 upstream 5 dovmstream 4 5 0 material radius 4 3 1 gt 3 2 2 4 1 u m 0 02 04 06 08 1 02 04 06 08 1 Fig 1 Numerical solution of swirling ow problem The output solution is shown in Fig l The upper left panel shows the convergence of the shooting solutions red dashed lines to the correct solution r1rz which satis es r105 l and the p20 corresponding to this solution The upper right panel shows the nondimensional upstream blue and downstream red radial pressure distribution The downstream radial pressure gradients are large to support the stronger downstream swirling velocities The overall pressure is lower downstream corresponding via Bemoulli s theorem to higher downstream uid speeds The circles correspond to a set of material radii showing the strong reduction in radius downstream This corresponds to a strong downstream jet especially near the centerline as required by conservation of mass lower left panel and large increases in the swirling velocity as required by conservation of circulation lower right panel The increased swirl induces the jet because it causes the downstream pressure drop to be larger at the center than at the edges and because this pressure drop mainly goes into alongpipe velocity at the center but mainly accelerates swirl at the pipe edge Three typical streamlines through the narrowing found with the function plotstreamm are shown in Fig 2 They are all helical Upstream from the narrowing all uid elements are swirling at the same rate and moving downstream at the same rate Downstream the streamlines re ect the angular swirl velocity and alongpipe jet being strongest in the center Fig 3 created using the function plotvortexm shows a segment of a vortex line advected in from upstream of the narrowing where the vorticity is 2Qi and the vortex lines are in the x direction Since the vortex line segment is a material line it starts to bend when the uid elements at its leading edge reach the narrowing and begin to swirl faster via conservation of circulation This twists the vortex line into a helical spiral downstream of the narrowing The tangential component of the vorticity is associated with the radial shear around the centerline jet Fig 2 Fluid trajectories starting upstream of the narrowing at x 4 and three different radii r1 l 071 and 025 Each trajectory is integrated forward 45 time units In the first four units all move along the pipe at unit speed but in the last 05 time units the trajectory closest to the center is swept into the central jet core and moves rapidly downstream while the outer trajectory sees little downstream acceleration Fig 3 Segment ofa vortex line corresponding to an upstream radius r1 071 The segment starts parallel to the x axis including 85 lt x lt 4 A material point is placed every 05 units in x along the upstream vortex line After advecting for 45 time units the vortex line is advected as a material line into the narrowing where its leading edge begins to swirl faster and accelerate causing vortex tilting and stretching to produce a helical vortex line of radius r 03 with a larger vorticity vector downstream of the narrowing
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