BOUNDARY LAYER THEORY - Anasayfa

1 HIGH RENOLDS NUMBER FLOW BOUNDARY LAYERS(Re ) BOUNDARY LAYERThin region adjacent to surface of a body where viscous forces dominate over inertia forcesRe =Re>> 1 inertiaforcesviscousforces BoundarylayerseparationWake: viscouseffectsnot importantvorticitynot zeroFlowfieldaroundan arbitraryshapeInnerflowStrongviscouseffe ctsOuterflowViscouseffectsnegligibleVort icityzero(Inviscidpotentialflow) BOUNDARY LAYER THEORYS teady ,incompressible 2-D flow with no body forces. Valid for laminar flow for To solve eq.

2 We first assume an approximate velocity profile inside the the wall shear stress to the velocity fieldTypically the velocity profile is taken to be a polynomial in y,and the degree of fluid this polynominaldetermines the number of BOUNDARY conditions which may be satisfied EXAMPLE:LAMINAR FLOW OVER A FLAT PLATE: *021(2)ddUdxdxU ++=0()nudy ()x 2()uabcfU =++=U U 0,99U ReUL =High Reynolds Number Flow Laminar BOUNDARY LAYER predictable Turbulent BOUNDARY layerpoor predictability Controlling parameter To get two BOUNDARY LAYER flows identical match Re(dynamic similarity) Although BOUNDARY LAYER s and prediction are complicated,simplifythe N-S equations to make job easier2-D , planar flow u* = , x*, y*= *,uvvU =,xyLDimensionless gov.

3 ;Y; Na ve way of solving problem forIf you drop the viscous term Euler s eqs. (inviscidfluid)22221()RePutxyyxy ++= ++ .0V =**2*2**22viscous terms1()ReuuuPuuutxyxxy ++= ++ *2 PPU =1 0Re Re We can not satisfy all the BOUNDARY order of eqs. Reduces by 1 Inside B-L can not get rid of viscous termsDerivation of B-L eqs. From the N-S eqs Physically based argument :determine the order of terms in N-S Limiting procedure as Re eqs. and throw out small termsU U (x,y)yLU *1100L = Assumption 1 TermOrder*v*1L = **ux **y **x 2**2uy **dudt(1)1(1)=**1 =**1 =2*1 **1uux = * (1) (1)Neglect since of order>>>122**2*2**1()RePutxyyxy ++= ++ **2*2**221()ReuuuPuuutxyxxy ++= ++ **11 =(1)1(1)=2* 2(1)(1)*2(1)() *2(1)() Also for y direction**()(1)(1)() **()()()() *2**2*2*(){}(1)()() + *() small relative toTo good approximationpressure at the edge of B-L.

4 Is equal to pressure onboundary LAYER . Time dependant known from the other flow Pressure at all points is the same Only need to consider x-direction B-L. eqs.**Py *() **Px (1) ()PPx (,)PPxt Prandtl(1904)0uvxy += Outerflow(inviscid)yx2-D planar1)2)221uuuPuuvtxyxy ++= + non-linearbut parabolic typeunknowsu,v(x,y,t)known from the potential flow(,)PPxt Need (timedependant) 2-D, steadyBCs u= =0 at y=0 u=u(y) at x=0 u= (x)y (y ) marching condition B-L.

5 Eqs. can be solved exactly for several cases Can approximate solution for other casesLimitation of egs.:where they fail?(1) Abrupt chances U (2) Eqs. are not applicable near the leading edge*1L = L is small invalid(3)Where the flow separates not valid beyond the separation pointSeparationpointBernouillieqs. =constant 11202dPdVUdxdx +=2constant2PV +=Valid along the streamlinessubstitute the eqsu,vcan be foundknown0dpdx=1dPdUUdxdx =SIMILARITY SOLUTION TO EQSE xample 1 Flow over a semi-infinite flat plateZero pressure gradient p = constantSteady ,laminar & U=constant( )0dpdx= Bernouillieqs.

6 , Governing ( eqs.) become22uuuuvxyy += +=0dpdx=0uxy += (1)(2) y=0 u= v =0 (no-slip) & y , u U x=0 u=UBlasuis(1908) the stream function (x,y) Recall ; uy = x = note that satisfies cont. eqs. substitute mom. Eqs = (2 ) Now, assume that we have a similarity stretching variable, which has all velocity profiles on plate scaling (,,)gUx = ()uyfU =y xdimensional analysis()(Re)Uxggx ==21()Re xU [] both() Viscous dif.

7 Depthy =ReUx =5xU Let[-] similarity variableUyx =()ufU =Use similarity profile assumption to turn 2 1 ()()yyxudyUfdyUfdU === xfixeduy = = 0()()UxfdUxF == ()F ()UxF =()UxF =Uyx =00yudy = ddydxyx =+ Now, substitute into for (x,y) to get for F( ) '12 UFUxFxxx =+ 'dFFd =2''2dFFd =2''2 UUFyx = 11122 Uyxxxx = = '1()2 UFFxx = ''UUxFUFyx == 2''2 UFxyx = 23'''3 UFyx = Substituting into eq. (2 )211221'(''')()(')()'''''22 UUUUUFFFFUFF xxxx = or2''2 UFx 21'2 UFx 21''2 UFFx +2'''UFFx ='''F1'''''02 FFF+=blasiuseq.

8 3rd order , non linear ODE000yyuy == == Note: for BVP'''''0 FFF+=2 Uyx =BC s areAt y=0 u=v=00 =BC 1)0'0UF ==F (0)=0BC 2)00y ==1(')02 UFFx =F(0)=0BC 3)(x,y )U yUy 'UFU ='()F 1'()1F =OrAt x=0 F ( )=1 same with BC 3) Matching Solution to blasiusega)powerseries b)runge-kutta results tabulated form for F,F ,F , ( ) dimensionless function uU =0'xUFU = = ##00 0 = From the =F'uFU =''F## Velocity profile5[]121(')21Re'2xUFFxxFFU = = = 1 ( ) = =RexU Uyx =5F =uU Shear stress distribution along the flat plate**() (,) uxyyxuuyxy =+ 461 Re10 Re10 = = = = At the wall(y=0)00()yuxy = = ()wx 20200()''yUxUFyx == == 30()''(0)

9 UxFx =Distribution along the dimensionalize:022''(0) Re1 ReRe2fxxxFUxCU ==== =Friction : 0 x valid near the leading edgexUp tothe point we are consideringDrag force acting on the flat plateWe have to integrate shear stress00 unit width ()xperdDF = ()DFbUx =x xDdimensionless drag coef.( C)we have 2 wetted sides2562 A= valid for laminar flow for to 10 DDDxxFCUACRe ==<Width normal to the blackboard6xfor Re >10 turbulent drag becomes considerably greater BOUNDARY LAYER Thickness : at 5 (Table)55 ReRexxUuyyxUUUxxx == = = =.

10 Defined as the distance from the wall for which u= BOUNDARY LAYER Parameter (thicknesses)Most widely used is but is rather arbitrary y= when u= U hard to establish more physical parameters are neededDisplacement thickness: U U * * an imaginary displacement of fluid from the surface to account for lost mass flow in BOUNDARY LAYER **.000**00 or() (1)totyUmudyUdyUdyUdyuUUudydyU = === = = *if . always by definition cons =>Momentum thickness: U an imaginary displacement of fluid of velocity to account for lost momentum due to the formation of a BOUNDARY LAYER velocity profileU 200 Mass flow in Possible momentum actual momentum ()() UudyUudyu = "lost" momentum 0(1) will occur in = ** Various thinknesses defined above are,to some extend,an indication of the distancenoover which viscous effects extend.