Test: Boundary Layer Level - 3 - Mechanical Engineering MCQ

Momentum thickness, ?

x_c is the section with critical Re.

x_c = 1 m ∴ x_c < 1="" />

Hence initial 1 m of the surface is exposed to laminar boundary layer and trailing 2 m is exposed to turbulent boundary layer.

∴ By definition of Drag force (FD)

F_D = 0.768 N

= 2 /15 × 1000 × (3)² × 0.25 × 0.05

= 15 N

Hence drag force on both sides of the plate is 30 N.

Likewise the drag force per unit width for the front half portion of the plate is,

Drag force for the rear half portion of the plate is

F₂ = F - F₁

Hence,

= 1.347

μ = 6.62 × 10⁻² kg ⁄ hr-m

The flow Reynolds number is,

R_ex = 0.3 × 1.169 × 1.25 / 0.184 × 10⁻⁴

= 23824

For the given parabolic velocity distribution, the boundary layer thickness is prescribed by the relation,

At any position, the mass flow in the boundary layer is given by the integral

Where the velocity is given by;

Evaluating the integral with this velocity distribution,

Thus the mass exiting the boundary layer between the two sections is:

= 2.393 × 10⁻³ kg ⁄ s

= 8.61 kg ⁄ hr

∴ Mass flow rate entering through a-d

= ρ × area × velocity

= 1000 (2 × 0.125) × 2.5

= 625 kg ⁄ s

Momentum flux entering a-d

= mass flow rate × velocity = 625 × 2.5 = 1526.5 N

At the training section b-c, velocity varies linearly from 0 to 2.5 m⁄s. For an elementary strip of thickness dy at transverse distance y from the plate surface, the velocity

Can be considered to be uniform ∴ Mass flow rate leaving through b-c

Momentum flux leaving b-c

No mass can enter or leave the control volume through the surface a-b. The continuity requirement then stipulates that Mass rate of outward flow through c-d = (mass rate of inflow through a-d) − (mass rate of out flow through b-c) = 625 − 312.5 = 312.5 kg ⁄ s This outward flow takes place with free stream velocity of 2.5 m ⁄ s

∴ Momentum flux leaving section c-d = 312.5 × 2.5 = 781.25 N

In the absence of any pressure and gravity forces, the force exerted by the plate must be balanced by the net momentum change.

∴ Force exerted by the plate

= 520.83 + 781.85 − 1562.5

= −260.42 N

The negative sign signifies that it acts in a direction opposite to that of water flow. The drag force exerted by water on the plate is equal and opposite to it.

Where θ is the momentum thickness for section x

Making the substitution

At the solid surface, Newton’s law of viscosity gives:

The integration constant is obtained from the boundary condition: δ = 0 at x = 0, and that gives C = 0. Therefore,

This can be expressed in the non-dimensional form as

An estimate of the wall shear stress can be made by substituting the value of boundary layer thickness in the expression for wall shear stress.

In the non-dimensional form,

Reynolds number,

Obviously the flow is turbulent in the boundary layer. Assuming that abrupt transition from laminar to turbulent flow occurs at a Reynolds number of 5 × 105, the drag coefficient appropriate to this Reynolds number is given by

= 2.81 × 10⁻³

Frictional drag on the surface of model submarine is,

Total measured drag = skin friction drag + wave drag ∴ Wave drag = 60 − 35.125 = 24.875 N Wave drag coefficient

Skin friction coefficient or local drag coefficient

Substitution values in the equation we get

The boundary conditions are

Applying momentum integral equation

It is assumed that at x = 0, δ = 0 which yields

C = 0. Thus,

In the proclaim, Rex =

= 0.259 cm

or δ = 2.59 mm