Blasius Boundary Layer Solution for Flow Over a Flat Plate: Derivation and Results, Study notes of Computer Aided Design (CAD)

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Blasius Similarity Solution

For Boundary Layer Flow Over a Flat Plate

MEMS1055, Computer Aided Analysis in Transport Phenomena

Dr. Peyman Givi

Seth Strayer

Contents

such that:

v

x U ν = f

y

U

xν

= G(η) (7)

Based on these nondimensional parameters, we assume that the solution is of the form:

u = U F (η), v =

U ν x G(η) (8)

We now use the nondimensionalized quantities to go about solving the Navier-Stokes Equations. From continuity, we have,

∂u ∂x

∂v ∂y

such that:

∂u ∂x

∂x

(U F (η)) = U

∂F

∂η

∂η ∂x

= U

∂F

∂η

y

U

x−^3 /^2

ν

∂u ∂x

U η 2 x

∂F

∂η

Likewise, for the y component:

∂v ∂y

U

x

∂G

∂η

Then from Equations 9, 10, and 11:

U η 2 x

∂F

∂η

U

x

∂G

∂η

Upon factoring, we get:

U x

[

η 2

∂F

∂η

∂G

∂η

]

∂G

∂η

η 2

∂F

∂η

Integrating both sides with respect to η yields: ∫ (^) η

0

∂G

∂η

dη =

∫ (^) η

0

η

∂F

∂η

dη

For the left-hand side, v(0) = 0 due to the no-slip boundary condition implies G(0) = 0 from Equation 8. The right-hand side is integrated by parts. After some simple calculus, the function G(η) is calculated as:

G(η) =

[

ηf ′^ − f

]

Next, we would like to solve the x-component of the momentum equation, presented below.

u

∂u ∂x

• v

∂u ∂y = ν

∂^2 u ∂y^2

From Equation 10, we have:

∂u ∂x

U η 2 x

∂F

∂η

Furthermore:

∂u ∂y

∂y

U F (η)

= U

∂F

∂η

∂η ∂y

= U

∂F

∂η

U

xν

∂^2 u ∂y^2

∂y

∂u ∂y

U 2

xν

∂^2 F

∂η^2

With u and v as prescribed in Equation 8, along with Equations 14, 15, 16, and 17, the momentum equation becomes:

U F (η)

U η 2 x

∂F

∂η

U ν x G(η)

U

∂F

∂η

U

xν

= ν

U 2

xν

∂^2 F

∂η^2

Upon factoring and rearrangement:

U 2 x

[

G(η)

∂F

∂η

ηF (η) 2

∂F

∂η

∂^2 F

∂η^2

]

Since F = F (η) and G = G(η), we may write:

G(η) dF dη

ηF (η) 2

dF dη

d^2 F dη^2

Letting F = (^) dηdf = f ′, and substituing in G(η) = (^12)

ηf ′^ − f

from Equation 13, we have:

1 2

ηf ′^ − f

f ′′^ −

η 2 f ′f ′′^ = f ′′′

ηf ′f ′′^ −

f f ′′^ −

η 2 f ′f ′′^ = f ′′′

f f ′′^ = f ′′′

Finally, we obtain:

f f ′′^ + 2f ′′′^ = 0 (18)

Let us now examine the boundary conditions. At η = 0, we have u, v = 0 due to the no-slip boundary condition. Thus, from Equations 8 and 13:

G(0) = 0 = −

f ⇒ f (0) = 0 (19)

From Equation 8, we have:

f ′(0) = u(0) U

= 0 ⇒ f ′(0) = 0 (20)

Finally, as η → ∞, the velocity u should approach the freestream velocity, u → U such that:

f ′(η → ∞) =

U

U

= 1 ⇒ f ′(∞) = 1 (21)

δ = y = η

xν U

xν U

xν U

3.2 Optimum ∆η

In this section, the optimum step size ∆η is calculated. The term optimized in this report will refer to the last value before which the change in the initial guess h(1) used in regards to the shooting method no longer exceeds the change in step size. Results for varying mesh sizes are given in Figure 2.

Figure 2: Step Size Optimization

From this figure, we can see that the change in the initial guess h(1) never exceeds the change in step size. Thus, decreasing the step size from the initial size of ∆η = 0.01 will only result in more computational expense with nearly the same amount of computational accuracy. Thus, we conclude that the optimum step size for this problem is:

∆η = 0. 01 (24)

3.3 Drag

In this section, the drag on the wall of the plate is calculated. The shear stress on the wall τw is given by:

τw = μ

∂u ∂y

y=

= μU

U

xν

∂F

∂η

η=

= μU

U

xν

f ′′

η=

Then the drag force is given by:

D =

∫ L

0

τwdx =

∫ L

0

μU

U

xν

f ′′

η=

dx = 2μU

U x ν

f ′′

η=

Evaluating for a unit length of L = 1, a freestream velocity equal to unity, dynamic viscosity μ = 1. 01 e − 3, and kinematic viscosity ν = 1. 01 e − 6, we obtain:

D = 2(1. 01 e − 3)(1)

1. 01 e − 6

f ′′

η=

= 2. 01 f ′′

η=0 (27)

From Table 1, f ′′

η=0 = 0.33031. Thus,

D = 2.01(0.33031) = 0. 66391 N (28)

3.4 Shear Stress and Drag Coefficient

In this section, we wish to find relations between the shear stress coefficient and the drag coefficient in terms of Reynolds Number. Consider the shear stress coefficient given by:

Cf =

τw 1 2 ρU^

From Equation 25, we can write:

Cf =

μU

U xν f^

η= 1 2 ρU^

2 =^

ν

U xν f^

η= 1 2 U

Cf = 2

ν U x

f ′′

η=0 =

2 f ′′

√ η= Rex

From Table 1, f ′′

η=0 = 0.33031 such that:

Cf =

Rex

Rex

Thus, the shear stress coefficient can be approximated as:

Cf ≈

Rex

Likewise, the drag coefficient is given by:

Cf m =

D

1 2 ρU^

2 L (32)

From Equation 26, we can write:

Cf m =

2 μU

U x ν f^

η= 1 2 ρU^

2 L =

4 ν

U x ν f^

η= U L

Cf m =

4 f ′′

√ η= ReL

Again, from Table 1, f ′′

η=0 = 0.33031 such that:

Cf m =

ReL

ReL

Thus, the drag coefficient can be approximated as:

Cf m ≈

ReL

5 Appendix

5.1 Additional Tables and Figures

• 1 Introduction
• 2 Formulation
  • 2.1 Analytical Derivation of the Blasius Solution
• 3 Results
  • 3.1 Solution Convergence and Boundary Layer Thickness
  • 3.2 Optimum ∆η
  • 3.3 Drag
  • 3.4 Shear Stress and Drag Coefficient
• 4 Conclusion
• 5 Appendix
  • 5.1 Additional Tables and Figures
  • 5.2 References
• List of Figures - 0 0 0 0. η f ′ f ′′ f ′′ - 0.5 0.04045 0.16503 0. - 1 0.16309 0.32819 0. - 1.5 0.36595 0.48471 0. - 2 0.64403 0.62755 0. - 2.5 0.98864 0.74927 0. - 3 1.38778 0.84452 0. - 3.5 1.8277 0.91205 0. - 4 2.29516 0.95499 0. - 4.5 2.77924 0.97929 0.
  • 4.91 3.1831 0.98991 0.
  • 4.92 3.193 0.99009 0. - 5 3.27225 0.99147 0. - 6 4.26854 0.99898 0. - 7 5.26816 0.99993 0. - 8 6.26814 1 0.
• Table 1: Blasius Solution for ∆η = 0.

Figure 3: MATLAB Script (1)

5.2 References

[1] C¸ engel, Yunus A., and John M. Cimbala. Fluid Mechanics: Fundamentals and Applications. McGraw-Hill Higher Education, 2006.

List of Figures

1 Blasius Solution for ∆η = 0. 01.............................. 4 2 Step Size Optimization................................... 5 3 MATLAB Script (1).................................... 9 4 MATLAB Script (2).................................... 10