Consider flow of fluid, having freestream velocity, over a flat plate or smooth thin plate and placed parallel to the direction for free stream as shown in the figure above.

The velocity of fluid on the surface of plate should be equal to the velocity of the plate, but as the plate is stationary the velocity of fluid on the surface of the plate is zero. But at a distance away from the plate the fluid have certain velocity. Thus, velocity of gradient is step up in the fluid near the surface of the plate.

The fluid retards due to the shear resistance developed by the velocity gradient.

Thus, the boundary layer region begins at the sharp leading edge.

At certain points downstream the leading edge, the boundary layer region increases because the retarded fluid, retards further again.

Near this leading edge, the flow in the boundary layer is laminar though the main flow is turbulent.

This layer of fluid is called as the laminar boundary layer.

Boundary layer thickness

It is defined as the distance between the boundary of solid body which is measured in the Y direction to the point, where the fluid velocity is equal to the 0.99 times of the freestream velocity (V) of fluid.

It is denoted by '$\delta$'

For different flow it is denoted as

1. $\delta_{lam}$ = Thickness of laminar boundary layer.

2. $\delta_{tur}$ = Thickness of turbulent boundary layer.

3. '$\delta$' = Thickness of laminar sublayer.

Momentum thickness

It can be defined as distance which is measured perpendicular to the boundary of solid body, which helps in compensating for the reduction in momentum of flowing fluid on account of formation of boundary layer, only if the boundary is displaced.

It is denoted by $\theta$

Consider the flow over a plate, consider section 1-1 at a distance 'x' from the leading edge. Take elemental strip at a distance 'y' from the place having thickness(dy).

Therefore Momentum of fluid $=\text{mass} \times \text{velocity} = (\rho ubdy)\times u$

The momentum of fluid in absence of boundary layer = $(\rho ubdy)\times V$

Therefore loss of momentum through the elemental strip

$=(\rho ubdy)\times V-(\rho ubdy)\times u=(\rho ub)\times (V-u)dy$

Therefore total loss of momentum /sec through 'BC'

$=\int_0^\delta \rho u b (V-u)dy.................(1)$

Let $\theta$ = distance by which place is displaced when fluid is flowing at constant velocity (V).

Therefore loss of momentum/sec of fluid flowing through ($\theta$) with a velocity (V)

= ($\rho\times$ area $\times$ Velocity) $\times$ velocity

$=[\rho\times\theta\times b\times V]\times V$

$=\rho\theta b V^2............(2)$ [$\because$ Area = $\theta\times b$]

Equating equation (1) and (2), we get

$\rho\theta b V^2=\int_0^\delta \rho u b (V-u)dy$

$\rho\theta b V^2=\rho b\int_0^\delta u (V-u)dy$

Cancelling $\rho b$ on both sides, we get

$\theta V^2=\int_0^\delta u (V-u)dy$

$\theta =\dfrac 1{ V^2}\int_0^\delta u (V-u)dy$

$\theta =\int_0^\delta \dfrac u{ V} \left[1-\dfrac uV\right]dy$