BOUNDARY LAYER THEORY

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. 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.

2 ;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. is equal to pressure onboundary LAYER .

3 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. 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.

4 , 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. 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.

5 (2 )211221'(''')()(')()'''''22 UUUUUFFFFUFF xxxx = or2''2 UFx 21'2 UFx 21''2 UFFx +2'''UFFx ='''F1'''''02 FFF+=blasiuseq. 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)UxFx =Distribution along the dimensionalize.

6 022''(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 == = = =: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 **.

7 000**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.,() only> ()Definition is sametes(remaks) f xalways > or ZPG,APG,FPG,turbulance5 From flat plate analysis Rexx [][]055*'0050 (1) (1)(1) == = = == = = = *0(5) , (1)ReyFxxuuxSimilarlyduux === = * * FALKNER-SKAN SIMILARITY SOLUTIONSS tagnation-point flow (Hiemenzflow) Similarity methodsFlow over a flat plate (Blasiusflow)(,)xy Falkner& Skan(1931) general similarity solution of the B-L of similarity solutions to the 2-D,steady B-L egs.

8 Look for general similarity solutions of the form(,)()'() where (1) () - unspecified function of x which will be determined later()uxyUxfxyx == (2) (,)()()() check : u=()()xyUxxfUxxy == 1'()()fx eqs. (3)or in terms of (,) (3') no-slip, smooth matchinguudUuuvUxydxydUxyUyxyxydxy +=+ =+ ()yx =Substitute eq.(2) into (3 ) 222()()()(,) '," (1'' )dfdfUxxfxyffddUfuydUddfdfUfdUddfUfUfxdx dxdxUxdxdxddxdydddxdxdx ==== == =++ = =+ = []22''''1 ='''""dUfUffUxyxyxdxxdUdfdUdfUdUUdffUfdx dfxydxdxdxxdx = ===+ +=+ []2233222222""''"Substitute above results into (3')''""'1(')"""'(')"'UfUfyydUdUUfyUfyUd dfUfdxdUdUUUfffUffUfdxdxdcdxdUdUddUUUfUf fUffUfdxdxdxdxdUufdxdx == + =+ ==+ = ()()2222()""'To put the eq. into standard form, multiply by "'"1'0 (4)dUddUUUffUfdxdxUdUfUfffdxdx ++ = = + Transformed gov.

9 A similarity solution exists, eq.(4) must be an ODE for the function fin terms of .So, coefficieuts & must be constant for a similarity solution()2 Falker-Skan "'"1'eq. (05)ffff ++ = same as for flat plate (0)'(0)0'()1 : BCs don't depend on ,Exact solutions to the B-L. Eqs. May be obtained by pursuing the following PROremarkStepCEDURE: Select & . (a p ar1fff == 2ticular flow configuration is considered this will not be known a priori but will be exident when step 2 is completed).: Determine U() , (x) , (x) (6a-b)Step 2ddUUdxdx =='''2'': Determine the function f()which is the solution of the following problem'''1()0with BCs (0)(0)0 , ()1 as Calculate the stream funStep ctionin 3 Step h4p :yfffffff ++ = == N()()22sical coord.(,)()()() in step #2 , instead of working with eqs. 6 a-b) (6a)' RemarE 2 (6b)k' Flate Pxample #1stepla te (ZPG)#1yxyUxxfxdUdUdxdx = == ()2' means that flat plate at ZPG1 = , =02 (6a)' 0 (6b)'()0(6) leads to step #2 U=con st0.

10 ThisdUdxdUdxdUxbdx == = 22(6)' 1 : '''''0 (0)'(0)02 ; '1 compare with Blasius sStep #3 Step #olution (,)(),4()dxadxUUfffffyfxUyxxxUyxyUfUfUxU xyUxfy = =+==== == ==same as Blasius solution Ux ()()()22 FLOW OVER WEDGE 1, arbitrary constant (6a' ) Exampl22 e #2 Step (7)#1dUUxdx === = 22(6') Divide eq. (6b') by (7)lnlnln outer flow is that over a wedgeof angle (Fig.)2 112 ()dUUdxxUxdUbdxUcxxc = =+ == 2(1)222221 2 (9) Solve the BVP'''''1(')0(0)'(0)0 '1 Solve numerically to get (),'(),'(2)()Step #3'()dUcxdxffffffasffcfxxf == ++ = == =()()()()()1212: Go back to the physical cStepoordinate (,)()() =2 ()2/ 4yyxyUxxfcxfxxc = STAGNATION-POINT ; FLO=W1 1 =Flow over a wedge Let 1 =2Eq. (8) gives, ()(9)() "'"1(')0 (0)'(0)0 as '()1 (,) x /NotUxcxxffffcfffyxycfc = =++ === = e: See Hiemenz flowExact solution to the full Navier-Stokes equations obtained by Hiemenz for a stagnation IN A CONVERGENT CHANNEL0 , 1 == BOUNDARY LAYER flow on the wall of a convergent : pg.