Laminar Boundary Layer over a Flat Plate

A laminar boundary layer is a thin layer of fluid which is adjacent to the surface of a solid body. In the case of a flat plate, the boundary layer is formed due to the friction between the fluid and the surface of the plate. This friction causes the fluid velocity to decrease from the free stream velocity to zero at the surface of the plate.

Shear Stress

Shear stress is the force per unit area that acts parallel to the surface of the plate. In the case of a flat plate, the shear stress is maximum at the surface of the plate and decreases with distance from the surface.

Ratio of Shear Stresses

In the given problem, we are given that the ratio of shear stresses at two sections 1 and 2 is r1/r2. The distance between these two sections is given by X2=9x1.

We can use the equation for shear stress in a laminar boundary layer to relate the shear stresses at sections 1 and 2.

τ = μ(dU/dy)

where τ is the shear stress, μ is the viscosity of the fluid, U is the velocity of the fluid, and y is the distance from the surface of the plate.

Using this equation, we can write the ratio of shear stresses at sections 1 and 2 as:

r1/r2 = (dU/dy)1/(dU/dy)2

We can use the fact that the velocity profile in a laminar boundary layer is given by:

U/Ue = 2(y/δ) - (y/δ)^2

where Ue is the free stream velocity, δ is the thickness of the boundary layer, and y is the distance from the surface of the plate.

Using this equation, we can relate the velocity gradients at sections 1 and 2 as:

(dU/dy)1/(dU/dy)2 = (y1/y2)(δ2/δ1)

Substituting X2=9x1, we get:

(dU/dy)1/(dU/dy)2 = (1/9)(δ2/δ1)

Since the flow is laminar, we can use the expression for the thickness of the boundary layer:

δ = 5.0*x/Re_x^1/2

where x is the distance from the leading edge, and Re_x is the Reynolds number at that distance.

Substituting this expression for δ in the above equation, we get:

(dU/dy)1/(dU/dy)2 = (1/9)((x2/Re_x2^1/2)/(x1/Re_x1^1/2))

Since Re_x is proportional to x^1/2, we can write:

(dU/dy)1/(dU/dy)2 = (1/9)((x2/x1)^1/2)

Substituting X2=9x1, we get:

(dU/dy)1/(dU/dy)2 = 1/3

Therefore, the ratio of shear stresses at sections 1 and 2 is r1/r2 = 3.