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# Computational Fluid Dynamics EOC 6189

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This 12 page Class Notes was uploaded by Gino Zemlak on Monday October 12, 2015. The Class Notes belongs to EOC 6189 at Florida Atlantic University taught by Ananthakrishnan in Fall. Since its upload, it has received 28 views. For similar materials see /class/221661/eoc-6189-florida-atlantic-university in Engineering Ocean at Florida Atlantic University.

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Lecture Notes 7 1 P Ananthakrishnan 82307 E00618 Computational Fluid Dynamics Fall 2007 Chapter 1 Exact Solutions to Model Partial Differential Equations Let us begin with exact analysis of certain familiar partial differential equations that represent physical processes of uid dynamics such as advection diffusion and dispersion The equations governing uid dynamics problems are not so simple nevertheless understanding of the analysis and solution of model partial differential equation will enable one to e iciently develop or better understand algorithms for approximate solutions of uid ow problems i Advection Equation A simple partial differential equation governing advection is given by where c which represents the speed of advection is a constant and u E uwt Let the known initial value of u be As you may have studied in your earlier mathematics courses this equation can be solved by a range of methods For example let us consider the one based on the Fourier transform method Let u u 672 dw x27r if 6 dk 1 u 7 V 27r Substituting the Fourier integral representation of u in the given partial differential equation one can obtain the following ordinary differential equation for the transform 2K du dt ciku 0 solution of which is ux A eiikct where A is the integration constant At t 0 u fw or M f where fk is the Fourier transform of f Therefore the integration constant A f Thus ux f4lt eiikct Substituting the above in the Fourier inverse transform we get 71 M 6 dk V2W 1 73919 t 39k f4lt e 2 0 62 1 x27r 1 4lt ikzict 7 dk m f 6 fw0t The solution of the partial differentail equation is thus was t m 7 ct The solution can be interpreted as follows If the argument cc 7 ct is constant then u will be also constant In other words7 on lines on which dw 7 ct 07 u will be consant Lines on which solution it remains constant are called characteristic lines The slope of the characteristic lines on the characteristic t X plane is c To an observer moving at ET c7 u will appear to be stationary Stated yet again differently7 u is advected with velocity cl The solution for a typical initial value is illustrated in the following gure ft X0 One could have also obtained the solution by assuming existence of the characteristics a priori Since on the characteristics it is constant an an duct dw 5 dt 0 Or slope of the characteristics is given by div 7 git3t dt T git8w which by using the given partial differential equation can be written as div 7 dt 7 C In other words7 with respect to frame moving with dwdt c7 u will appear to be stationary That is7 u is advected with speed c Lecture Notes 7 2 P Ananthakrishnan 82507 EOC6189 Computational Fluid Dynamics Fall 2007 ii Diffusion Equation Next7 let us consider the one dimensional diffusion equation given by an 7 8271 at Vex where 1 is a positive constant known as the diffustion coe icient In view of simulations to be later considered in the course7 let the boundary conditions be uy0t 07 fortlt0 UL77 fortZO uy gt oot 0 The above system of equations would represent the ow caused by an in nitely long plate y0 impulsively set into motion from time t 07 as illustrated on the left of the gure In this case it will correspond to horizontal ow velicity and UL7 the velocity of the plate The equations could also represent diffusion of a substance brought into a contact with a medium for example7 diffusion of Cl ions into concrete exposed to seawater from time t 07 as illustrated on the right of the gure In this case7 u will correspond to concentration massvolume of the Cl and UL7 the Cl concentration in sea water touching the concrete Velocity Profile att Concentration Profile att gt Plate sliding With velocity U0 Sea water with Cl content U0 The above equation can be solved by similitude transformation The similitude variable can be determined using the Buckingham s Pi theorem that you must have learnt in undergraduate Fluid Mechanics course Let us then assume that u depends on y t 1 and U0 ie H fytvy U0 which is a functional relation involving N 5 variables Of the ve variables letting brie y u to be anything velocity concentration temperature cat whatever there are at the most 3 variables ie M3 which are dimensionally independent For example 3 t and UL7 is one such subset of dimensionally independent variables Then by Pi theorem the above functional relation can be replaced by a relation involving N 7 M 2 non dimensional variables H1 FH2 where u u H1 a and You may want to review the subject of Dimensional Analysis in particular the Pi theorem and the exponent method in case the above steps are not clear The second non dimensional variable can be also taken as Thus by Pi theorem With taken as the similitude variable 7 the above relation can be written as In other words the non dimensional uUL7 depends only on one variable namely the similitude variable 7 even though the dimensional u is function of y and t This suggests the possibility of obtaining an ordinary differential equation for the non dimensional u As shown below this is just a straightforward exercise involving chain rule of differentiation As L TFM 3 dan Similarly7 a d an 53 571 1 d W617 a 7 a 1d 332 ayx dn d 16187 a i a lol2 2 With the above transformation relations for derivatives the diffusion equation 8 8 182 718532 becomes 1 ldu i 1 d271 7m V Or d271 1 du i 0 Wi Above equation can be solved as follows Let Then the above differential equation for u becomes7 dvl 70 dry 27 solution of which is 11 A 6 7724 Where A is constant of integration Since 1 dudn n u A 67024 da uy gt oot 0 and because 7 77 u Aeiu 24 da The integral corresponds to error function in statistics The integral can be split as 77 D u A 67024 daiA 67024 da D co lln 67024 daiO 67024 dd D 00 The second integral is equal to Therefore uAn 67024 daixg 0 From the wall condition u UL7 on 3 ie 7 0 Therefore and thus the solution is 1 7 u 2 i177 eia4da U0 l 0 Graphical representation of the solution is shown below ln uid mechanics the above problem also sheds light on viscous stress in impulsively started ows Shear stress my is given for this ow by an my pzaiy where p is uid density Using the solution obtained above du 37 739 Vi i W dn 82 pyii 64724 x W V One can thus observe that viscous shear stress on the plate ie 7 0 is lxt singular in time Later in the course we will be discussing simulation of impulsively started motions in viscous uid when the present result will help to understand small time numerical solutions better Lecture 7 4 P Ananthakrishnan 82907 E00618 Computational Fluid Dynamics Fall 2007 Boundary Conditions for Equations Governing Incompressible Fluid Flow Next let us examine the boundary conditions for the equations of uid motion discussed in the previous class The boundary conditions for an inviscid uid ow are as follows On a rigid surface the velocity must satisfy the no ux condition ie 12 7 Vn where Vn denotes the normal velocity of the surface On a free surface which say is denoted by z n the no ux condition implies 132 i n 0 Dt In other words a free surface is a material surface The stress vector must be continuous across a free surface this means that in the absence of surface tension and viscous stress 13 pawn onz 7 where plum denotes the atmospheric pressure In terms of gage pressure the above dynamic con dition can be written as p00nzn In the particular case of a potential ow the above inviscid uid boundary conditions reduce to the following The no ux condition on a rigid surface becomes Vn 871 where Vn denotes the normal velocity of the surface The freesurface kinematic condition remains essentially the same 132 i n 7 0 Dt Substitution of the dynamic condition of p O on the free surface in the Euler s integral gives 8 1 V 2gn0 011272 Boundary Conditions VISCOUS Fluid Flow In the case of viscous ow uid velocity must satisfy both the no slip and no ux conditions on a a rigid surface Therefore on a body that is moving with a velocity V 12 l7 on the body surface S The form of free surface kinematic condition remains the same at least in the modeling of one layer Howl ie D 2 7 0 The dynamic condition corresponds to the continuity of the stress vector 7 across the free surface In the absence of atmospheric forcing the atmospheric pressure on the free surface is zero gage and atmospheric viscous stress is also zero Therefore the stress vector 7 must be equal to zero on the free surface In a two dimensional ow this means 7 f1 0 tdiu f O on the free surface Where 73f denote unit normal and tangential vectors on the surface respectively With the following relations already discussed 7j 7 a h and 87 au aij 7p6jlulta the above conditions can be expanded as ea lama 7 5 am auj 7 132 Hltaj 71271 7w emf l lp i ltaj 82 My 0 Where the repeated indices imply summation 1In a twolayer ow modeling the velocity vector must be continuous across the interface see Wehausen Surface Waves Open Boundary Conditions lnviscid Potential Flow To numerically solve ows in in nite domain one has to truncate the physical domain using arti cial boundaries known as open boundaries As these are not actual physical boundaries one has to develop boundary conditions that will make them appear as open In other words the open boundary conditions if properly constructed will not cause any re ection and will transmit the ow without any modi cation across the open boundary Let us consider a body motion in a free surface causing radiation of waves as illustrated in the gure below The lateral open boundaries are denoted by EL and ER and the bottom boundary by BZZ In the far eld per Sommerfeld condition waves must be propagating outward One can therefore use the advection equation refer to Lecture Notes 1 to model EL and ER as 845 all EC O OHER 845 845 E CQ O ODZL Here C denotes the wave speed and phi the velocity potential Above equations imply that waves are propagating to the right outward at ER and to the left outward at EL In the case of three dimensional wave motion problem the correspondng open boundary condition can be written as 345 345 EC 70 onE where R denotes the radial direction from the body towards the far eld open boundary The above method of modeling the open boundary for transient wave motion problems is referred to as the Orlanski condition in the literature on CFD In the case of linear timeharmonic wave radiation problem in which the unknown potential can be decomposed as wyzt ltIgtwyz 6721 where 2 E xjl and a wave or body motion frequency the advection equation can be reduced to the followin equations In the case of 2D problem on ER 345 345 7 Ci 0 at 8R 8ltIgt gt 72w of am 0 3w gt 87 7 ikltIgt 0 since C ak where kdenotes the wave number Similarly one can obtain for EL open bondary 8ltIgt 4 E2klt1gt70 In the case of threedimensional timeharmonic wave problem the corresponding open boundary condition is given by 8ltIgt 4 izk io where R denotes the radial outward direction By taking the bottom open boundary B to be at a depth that is larger than onehalf the wave length so that the domain can be a deep water domain one can set either 45 or to be zero on B Lecture 7 16 P Ananthakrishnan 110106 E00618 Computational Fluid Dynamics Fall 2006 BoundaryFitted Coordinates Transformation of Flow Equations Let us now consider ow equations transformed to the computational space For illustration let us consider the classical lid driven cavity ow problem as shown in the gure Y Physical Space Computational Space As can be seen the grid spacing is not uniform it is coarse in the middle and ne at the corners presumably so to resolve vortices of all scales well Let the grids to be time dependent even though the ow domain remains constant imagine that grids are generated adpatively as the ow structures evolve Governing Equations in the Physical Space Xyt The incompressible Navier Stokes equations governing the above ow are an 311 7 0 330 33 7 an an an 7 1 30 2 Eua 11 7 igauvzyu 311 311 311 i 1 30 2 Eu 11 7 payVsz7J and the unknowns are velocity components u and 1 and pressure p In the above equations 0 denotes density and 1 E up the coe icient of kinematic viscosity with 1 being the coe ient of dynamic viscosity If acceleration of gravity were acting along the negative y direction 13 will denote the dynamic pressure Transformation Governing Equations to the Computational Space n739 Using the trans formation relations for derivatives obtained earlier cf Lecture Notes 15 one can transform the above equations to an uniform computational space For compactness let us use subscript notation to denote derivatives for example an E 37 uT E 277 etc Equation of Continuity 977 7 316 7 W175 535 0 X component of the equation u v 7 117 36 1 M f Mug 7 315 J 7 5 005 707J Mag 7 315 1 ng ug ngn m Augg 7 23719 Cum gt 3u7w wv73 00501731 315u7w 1 u7 7 J T 7 7 J 7 5 J 77 J T un7p7Jnps7yspngt 11 ng ug ngn m Augg 7 23719 Cum y component of the equation u 7 w 11 7 1 1 7 7J7 3177175 7 315 ifT 7 5 755 707 770nm 7551 1 ng 115 ngn 11 A1155 7 23115 0117quot gt Wu 7 007 7 M71 7 17 7amp7 J J 00501731 315u7w 1 5 J T 7 J T np7J7wnPEwEPn 11 ng 115 ngn 11 A1155 7 23115 011 In the above AE B E wEwni yiyn C E J 2 7 J 2 7 J 2 7 The equations get lengthier upon transformation a small price worth paying to be able to tackle arbi trary boundary and to be able to resolve ow gradients e iciently and yet be able to solve the problem on an uniform and rectangular mesh in the computational space In the following chapters we shall discuss various algorithms for the solution of the incompressible Navier Stokes equations

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