What will be the momentum thickness for the velocity distribution in the boundary layer of:

$\genfrac{}{}{0ex}{}{u}{U/}=\; y/\delta$, where u = Velocity at a distance y from the plate and $u=U$ at $y=\delta$ (Boundary layer thickness)

Concept:

The momentum thickness, θ is given by the following expression:

\(\theta = \mathop \smallint \nolimits_0^\delta \frac{u}{U}\left( {1 - \frac{u}{U}} \right)dy\)

Calculation:

Given:

\(\frac{u}{U} = \frac{y}{\delta }\)

Momentum thickness is given by:

\(\theta = \mathop \smallint \nolimits_0^\delta \frac{u}{U}\left( {1 - \frac{u}{U}} \right)dy\)

\(\theta = \mathop \smallint \nolimits_0^\delta \frac{y}{\delta }{\rm{\;}}\left( {1 - \frac{y}{\delta }{\rm{\;}}} \right)dy\)

\(\theta = \mathop \smallint \nolimits_0^\delta \left\{ {\frac{y}{\delta } - {{\left( {\frac{{\rm{y}}}{{\rm{\delta }}}} \right)}^2}} \right\}dy\)

\(\theta = \left\{ {\frac{{{y^2}}}{{2\delta }} - \frac{{{y^3}}}{{3{\delta ^2}}}} \right\}|_0^\delta \)

\(\theta = {\rm{\;}}\left\{ {\frac{{{{\rm{\delta }}^2}}}{{2{\rm{\delta }}}} - \frac{{{{\rm{\delta }}^3}}}{{3{{\rm{\delta }}^2}}}} \right\}\) = δ/6

∴ Momentum thickness, θ = δ/6

Important Points 

1. Displacement thickness:

\({\delta ^*} = \;\mathop \smallint \nolimits_0^{\rm{\delta }} \left( {1 - \frac{{\rm{u}}}{{\rm{U}}}} \right){\rm{dy}}\)

2. Energy thickness:

\({\delta ^{**}} = \;\mathop \smallint \nolimits_0^\delta \frac{u}{U}\left( {1 - {{\left( {\frac{{\rm{u}}}{{\rm{U}}}} \right)}^2}} \right)dy\)