Blasius Similarity Solution

1 Blasius Similarity SolutionFor Boundary Layer Flow Over a Flat PlateMEMS1055, Computer Aided Analysis in Transport PhenomenaDr. Peyman GiviSeth Strayer4/9/19 Contents1 Introduction12 Analytical Derivation of the Blasius Solution ..13 Solution Convergence and Boundary Layer Thickness .. Optimum .. Drag .. Shear Stress and Drag Coefficient ..64 Conclusion75 Additional Tables and Figures .. References ..11 List of Figures11i1 IntroductionIn this report we consider the Blasius Boundary Layer Solution for flow over a flat plate of lengthunity. The Similarity variables are given by =y Ux , f ( ) =uU,12[ U x( f f)](1)whereUdenotes the freestream velocity.

2 We would like to analytically show thatff + 2f = 0f(0) =f (0) = 0f ( ) = 1and use these results to calculate a numerical Solution . Using the numerical Solution , we wouldalso like to: Determine a finite such that the Blasius Solution converges and discuss the optimum Calculate the drag on the wall of the plate Determine a relationship for calculating the boundary layer thickness Determine the relationship between the shear stress and drag coefficients and ReynoldsNumber2 Analytical Derivation of the Blasius SolutionIn this section, the Blasius third-order ODE and it s corresponding boundary conditions will bederived. From the definition of Navier-Stokes, we have that:f1(u,x,y, ,U) = 0(2)f2(v,x,y, ,U) = 0(3)Using the Buckingham Pi Theorem, we can find nondimensionless parameters which accuratelydescribe the system presented by Equations 2 and 3.

3 Note that the derivation of these parametersis omitted. With regards tou, 1=uU, 2=y Ux (4)such that:uU=f(y Ux )=F( )(5)With regards tov, 3=v xU , 4=y Ux (6)1such that:v xU =f(y Ux )=G( )(7)Based on these nondimensional parameters, we assume that the Solution is of the form:u=UF( ), v= U xG( )(8)We now use the nondimensionalized quantities to go about solving the Navier-Stokes continuity, we have, u x+ v y= 0(9)such that: u x= x(UF( )) =U F x=U F ( 12y Ux 3/2 ) u x= U 2x F (10)Likewise, for theycomponent: v y=Ux G (11)Then from Equations 9, 10, and 11: U 2x F +Ux G = 0(12)Upon factoring, we get:Ux[ 2 F + G ]= 0 G = 2 F Integrating both sides with respect to yields.

4 0 G d =12 0 F d For the left-hand side,v(0) = 0 due to the no-slip boundary condition impliesG(0) = 0 fromEquation 8. The right-hand side is integrated by parts. After some simple calculus, the functionG( ) is calculated as:G( ) =12[ f f](13)Next, we would like to solve the x-component of the momentum equation, presented u x+v u y= 2u y2(14)2 From Equation 10, we have: u x= U 2x F (15)Furthermore: u y= y(UF( ))=U F y=U F Ux (16) 2u y2= y( u y)=U2x 2F 2(17)Withuandvas prescribed in Equation 8, along with Equations 14, 15, 16, and 17, the momentumequation becomes:UF( )( U 2x F )+ U xG( )(U F Ux )= (U2x 2F 2)Upon factoring and rearrangement:U2x[G( ) F F( )2 F = 2F 2]SinceF=F( ) andG=G( ), we may write:G( )dFd F( )2dFd =d2Fd 2 LettingF=dfd =f , and substituing inG( ) =12( f f)from Equation 13, we have.

5 12( f f)f 2f f =f 12 f f 12ff 2f f =f 12ff =f Finally, we obtain:ff + 2f = 0(18)Let us now examine the boundary conditions. At = 0, we haveu,v= 0 due to the no-slipboundary condition. Thus, from Equations 8 and 13:G(0) = 0 = 12f f(0) = 0(19)From Equation 8, we have:f (0) =u(0)U= 0 f (0) = 0(20)Finally, as , the velocityushould approach the freestream velocity,u Usuch that:f ( ) =UU= 1 f ( ) = 1(21)33 ResultsThe MATLAB Script used to calculate the Solution to Equation 18 is . This scripttakes the third-order ODE and transforms it into three first-order ODE s. It then uses Euler sMethod to solve the first-order ODE s, and also applies the shooting method to determine aninitial value for the unprescribed boundary condition forf (0).

6 Full code listings for this script aregiven in the Appendix, Section , Figures 3 and 4. The exact methodology behind solving athird-order ODE of this nature has been omitted from this Solution Convergence and Boundary Layer ThicknessThe numerical Solution to Equation 18 for a step size of = is shown in Figure 1. We cansee that the normalized velocity profile,f =u/U, asymptotically approaches a value of one,indicating that the velocity approaches the freestream velocity as is increased. Portions of thetabular data are presented in the Appendix, Table 1: Blasius Solution for = the boundary condition presented by Equation 21, we would like to determine a finite valueof where the condition begins to hold.

7 This value determines where convergence of the systemoccurs. Given the asymptotic nature of the Solution , typically is chosen such that the velocity iswithin 99% of the freestream velocity,u= By examining the tabular data presented inTable 1, we observe that at = ,u= This is the first value to come within 99% ofthe freestream velocity. Thus, replacing = with = (finite): = (22)Using this same concept, if we define the boundary layer thickness as the value ofyfor whichu= , along with the definition of in Equation 1, we obtain:4 =y= x U= x U 5 x U(23) Optimum In this section, the optimum step size is calculated. The term optimized in this report willrefer to the last value before which the change in the initial guessh(1) used in regards to theshooting method no longer exceeds the change in step size.

8 Results for varying mesh sizes aregiven in Figure 2: Step Size OptimizationFrom this figure, we can see that the change in the initial guessh(1)neverexceeds the change instep size. Thus, decreasing the step size from the initial size of = will only result in morecomputational expense with nearly the same amount of computational accuracy. Thus, weconclude that the optimum step size for this problem is: = (24) DragIn this section, the drag on the wall of the plate is calculated. The shear stress on the wall wisgiven by: w= u y y=0= U Ux F =0= U Ux f =0(25)Then the drag force is given by:D= L0 wdx= L0 U Ux f =0dx= 2 U Ux f =0(26)Evaluating for a unit length ofL= 1, a freestream velocity equal to unity, dynamic viscosity = 3, and kinematic viscosity = 6, we obtain:D= 2( 3)(1) 1(1) 6f =0= =0(27)5 From Table 1,f =0= Thus,D= ( ) = (28) Shear Stress and Drag CoefficientIn this section, we wish to find relations between the shear stress coefficient and the dragcoefficient in terms of Reynolds Number.

9 Consider the shear stress coefficient given by:Cf= w12 U2(29)From Equation 25, we can write:Cf= U Ux f =012 U2= Ux f =012 UCf= 2 Uxf =0=2f =0 RexFrom Table 1,f =0= such that:Cf=2( ) Rex= Rex(30)Thus, the shear stress coefficient can be approximated as:Cf Rex(31)Likewise, the drag coefficient is given by:Cfm=D12 U2L(32)From Equation 26, we can write:Cfm=2 U Ux f =012 U2L=4 Ux f =0 ULCfm=4f =0 ReLAgain, from Table 1,f =0= such that:Cfm=4( ) ReL= ReL(33)Thus, the drag coefficient can be approximated as:Cfm ReL(34)64 ConclusionIn this report we were able to calculate the Blasius Boundary Layer Solution for flow over a flatplate of length unity.

10 In the formulation we derived the Blasius Equation and its correspondingboundary conditions given byff + 2f = 0f(0) =f (0) = 0f ( ) = 1 The third order differential equation was split up into three first-order differential equations andthe numerical Solution was calculated via the use of Euler s Method and the Shooting Method inthe MATLAB Script . Using the numerical data presented in Table 1, the followingresults were calculated: Finite : = = Boundary Layer Thickness: = 5 x U Optimum step size: = Drag on the plate:D= [N] Shear Stress Coefficient:Cf Rex Drag Coefficient:Cfm ReL75 Additional Tables and Figures f f f 1: Blasius Solution for = 3: MATLAB Script (1)9 Figure 4: MATLAB Script (2) References[1] C engel, Yunus A.