Question

Consider a boundary-layer velocity profile: 
\[ \frac{u}{U} = \begin{cases} \left( \frac{y}{\delta} \right)^2 & y \le \delta \\ 1 & y > \delta \end{cases} \] The shape factor (ratio of displacement thickness to momentum thickness) is \(\underline{\hspace{2cm}}\) (round off to 2 decimal places).

Hint:

Polynomial velocity profiles make $\delta^\*$ and $\theta$ integrals easy using nondimensional substitution $\eta = y/\delta$.

Answer 1

Correct Answer: 4.8

Displacement thickness: \[ \delta^\* = \int_0^\delta \left( 1 - \left(\frac{y}{\delta}\right)^2 \right) dy = \delta \int_0^1 (1 - \eta^2)\, d\eta \] \[ = \delta \left[ 1 - \frac{1}{3} \right] = \frac{2\delta}{3} \] Momentum thickness: \[ \theta = \int_0^\delta \frac{u}{U} \left(1 - \frac{u}{U}\right) dy = \delta \int_0^1 \eta^2 (1 - \eta^2)\, d\eta \] \[ = \delta \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{2\delta}{15} \] Shape factor: \[ H = \frac{\delta^\*}{\theta} = \frac{ \frac{2\delta}{3} }{ \frac{2\delta}{15} } = 5 \]