[Q] Use the control volume shown to derive the definition of the momentum thickness, $\theta$, for flow over a flat plate $$ \rho U^{2}\theta =\rho U^{2}\int_{0}^{h}\frac{u}{U}\!\left(1-\frac{u}{U}\right)\,dy =\frac{\text{Drag force on the plate from zero to }x}{\text{unit depth into the page}} =\int_{0}^{x}\tau_{w}\,dx $$

The words in the figure describe the upper and lower control volume boundaries

[Sol] \(\because \text{The integral form of the continuity equation} \underbrace{\rho U h_0}_{\text{inlet mass flow}} =\underbrace{\rho\displaystyle\int_{0}^{h} u(y)\,dy}_{\text{outlet mass flow}} \)
\(\because \text{The pressure is the same everywhere}\)
\(\therefore x-\text{momentum balance} \overbrace{-\,\rho U^{2}h_0}^{\substack{\text{inlet momentum flux}\\ \text{(enters, minus)}}} +\overbrace{\rho\displaystyle\int_{0}^{h} u^{2}(y)\,dy}^{\substack{\text{outlet momentum flux}\\ \text{(exits, plus)}}} =\underbrace{-\displaystyle\int_{0}^{x}\tau_w\,dx}_{\substack{\text{wall shear force on fluid}}} \)
\(\therefore \text{Eliminate } h_0 \text{u sing continuity } \big(Uh_0=\int_{0}^{h}u\,dy\big): -\rho U\,\overbrace{\int_{0}^{h}u(y)\,dy}^{Uh_0} +\rho\int_{0}^{h}u^{2}(y)\,dy =-\int_{0}^{x}\tau_w\,dx \)
\(\therefore \text{The momentum thickness } \theta: \underbrace{\rho\int_{0}^{h} u(y)\big(U-u(y)\big)\,dy}_{\text{momentum deficit}} =\underbrace{\int_{0}^{x}\tau_w\,dx}_{\substack{\text{drag force on the plate}\\ \text{from }0\text{ to }x\ \text{(per unit depth)}}}, \)
\(\qquad \rho U^{2}\, \underbrace{\int_{0}^{\infty}\frac{u(y)}{U}\!\left(1-\frac{u(y)}{U}\right)dy}_{\displaystyle \theta} =\int_{0}^{x}\tau_w\,dx \)
\(\therefore \text{A control volume calculation leads to the following definition } \theta=\int_{y=0}^{\infty}\frac{u}{U_e}\left(1-\frac{u}{U_e}\right)\,dy\)