Topic: Fluid Mechanics
If $U_{\infty}=$ free stream velocity, $u=$ velocity at y, and $\delta=$ boundary layer thickness, then in a boundary layer flow, the momentum thickness $\theta$ is given by-
- A. $\theta=\int_{0}^{\delta}\frac{u}{U_{\infty}}(1-\frac{u}{U_{\infty}})dy$
- B. $\theta=\int_{0}^{\delta}\frac{u}{U_{\infty}}(1-\frac{u^{2}}{U_{\infty}^{2}})dy$
- C. $\theta=\int_{0}^{\delta}\frac{u^{2}}{U_{\infty}^{2}}(1-\frac{u}{U_{\infty}})dy$
- D. $\theta=\int_{0}^{\delta}(1-\frac{u}{U_{\infty}})dy$
Answer: A. $\theta=\int_{0}^{\delta}\frac{u}{U_{\infty}}(1-\frac{u}{U_{\infty}})dy$
Explanation: Momentum thickness ($\theta$) is defined as the distance by which the solid boundary should be displaced to compensate for the reduction in momentum of the flowing fluid due to boundary layer formation. It is mathematically given by $\theta=\int_{0}^{\delta}\frac{u}{U_{\infty}}(1-\frac{u}{U_{\infty}})dy$. Displacement Thickness $\delta^{*}=\int_{0}^{\delta}(1-\frac{u}{U_{\infty}})dy$. Energy Thickness $\delta^{**}=\int_{0}^{\delta}\frac{u}{U_{\infty}}(1-\frac{u^{2}}{U_{\infty}^{2}})dy$.