Let's solve the problem step by step. Given velocity profile for laminar boundary layer: \[ \frac{u}{U_\infty} = \sin\left(\frac{\pi}{2} \frac{y}{\delta}\right) \] Where: - \( u \) = local velocity at distance \( y \) from the plate - \( U_\infty \) = free stream velocity - \( \delta \) = boundary layer thickness Let's find the required expressions: --- ### (i) **Boundary Layer Thickness (\(\delta\)):** The boundary layer thickness, \(\delta\), is typically found by integrating the momentum integral equation (von Kármán equation): \[ \frac{d\theta}{dx} = \frac{\tau_w}{\rho U_\infty^2} \] where, \(\theta\) is the momentum thickness: \[ \theta = \int_0^\delta \frac{u}{U_\infty}\left(1-\frac{u}{U_\infty}\right) dy \] Let's calculate \(\theta\): \[ \theta = \int_0^\delta \sin\left(\frac{\pi}{2} \frac{y}{\delta}\right) \left[1 - \sin\left(\frac{\pi}{2} \frac{y}{\delta}\right)\right] dy \] Let \( \eta = \frac{y}{\delta} \implies dy = \delta d\eta \), limits: \( y=0 \Rightarrow \eta=0 \), \( y=\delta \Rightarrow \eta=1 \): \[ \theta = \delta \int_0^1 \sin\left(\frac{\pi}{2} \eta\right) \left[1 - \sin\left(\frac{\pi}{2} \eta\right)\right] d\eta \] Expand: \[ = \delta \left[ \int_0^1 \sin\left(\frac{\pi}{2} \eta\right) d\eta - \int_0^1 \sin^2\left(\frac{\pi}{2} \eta\right) d\eta \right] \] Now, solve each integral: 1. \(\int_0^1 \sin\left(\frac{\pi}{2} \eta\right) d\eta\): Let \( u = \frac{\pi}{2} \eta \implies d\eta = \frac{2}{\pi} du\), when \( \eta = 0 \), \( u = 0 \), when \( \eta = 1 \), \( u = \frac{\pi}{2} \): \[ = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \sin u du = \frac{2}{\pi} [-\cos u]_0^{\frac{\pi}{2}} = \frac{2}{\pi} [-(0-1)] = \frac{2}{\pi} (1) = \frac{2}{\pi} \] 2. \( \int_0^1 \sin^2\left(\frac{\pi}{2} \eta\right) d\eta \): \[ \sin^2 x = \frac{1}{2} (1 - \cos 2x) \] So, \[ \int_0^1 \sin^2\left(\frac{\pi}{2} \eta\right) d\eta = \int_0^1 \frac{1}{2} \left[1 - \cos(\pi \eta)\right] d\eta = \frac{1}{2} \int_0^1 d\eta - \frac{1}{2} \int_0^1 \cos(\pi \eta) d\eta \] \[ = \frac{1}{2} [1 - 0] - \frac{1}{2} \left[ \frac{1}{\pi} \sin(\pi \eta) \bigg|_0^1 \right] = \frac{1}{2} - \frac{1}{2} \left[ \frac{1}{\pi} (0-0) \right]= \frac{1}{2} \] So, \[ \theta = \delta \left( \frac{2}{\pi} - \frac{1}{2} \right) \] --- ### (ii) **Shear Stress at the Wall (\(\tau_w\)):** Shear stress at the wall is: \[ \tau_w = \mu \left( \frac{\partial u}{\partial y} \right)_{y=0} \] From the velocity profile: \[ u = U_\infty \sin\left( \frac{\pi}{2} \frac{y}{\delta} \right) \] \[ \frac{\partial u}{\partial y} = U_\infty \cos\left( \frac{\pi}{2} \frac{y}{\delta} \right) \cdot \frac{\pi}{2} \cdot \frac{1}{\delta} \] At \( y = 0 \): \[ \left. \frac{\partial u}{\partial y} \right|_{y=0} = U_\infty \cos(0) \cdot \frac{\pi}{2} \cdot \frac{1}{\delta} = U_\infty \cdot 1 \cdot \frac{\pi}{2} \cdot \frac{1}{\delta} = \frac{\pi U_\infty}{2\delta} \] So, \[ \tau_w = \mu \cdot \frac{\pi U_\infty}{2\delta} \] --- ### (iii) **Average Coefficient of Drag (\(C_f\)):** The local skin friction coefficient is: \[ C_f = \frac{\tau_w}{\frac{1}{2} \rho U_\infty^2} \] Insert the value of \(\tau_w\): \[ C_f = \frac{\mu \frac{\pi U_\infty}{2\delta}}{\frac{1}{2} \rho U_\infty^2} = \frac{\pi \mu U_\infty}{2\delta} \cdot \frac{2}{\rho U_\infty^2} = \frac{\pi \mu}{\rho U_\infty \delta} \] The average coefficient of drag over a length \( L \): \[ \bar{C_f} = \frac{1}{L} \int_0^L C_f dx \] But for a laminar boundary layer, \(\delta\) grows as: \[ \delta \sim \sqrt{\frac{x \nu}{U_\infty}} \] But for this specific profile, let's relate \(\delta\) to \(x\) using the momentum integral equation. Recall: \[ \frac{d\theta}{dx} = \frac{\tau_w}{\rho U_\infty^2} \] Using values from above: \[ \theta = \delta \left( \frac{2}{\pi} - \frac{1}{2} \right) \] \[ \tau_w = \mu \frac{\pi U_\infty}{2 \delta} \] So, \[ \frac{d}{dx} \left[ \delta \left(\frac{2}{\pi} - \frac{1}{2}\right) \right] = \frac{\mu \frac{\pi U_\infty}{2 \delta}}{\rho U_\infty^2} \] \[ \left(\frac{2}{\pi} - \frac{1}{2}\right) \frac{d\delta}{dx} = \frac{\mu \frac{\pi}{2}}{\rho U_\infty \delta} \] \[ \frac{d\delta}{dx} = \frac{\mu \frac{\pi}{2}}{\rho U_\infty \delta \left( \frac{2}{\pi} - \frac{1}{2} \right)} \] \[ \delta \frac{d\delta}{dx} = \frac{\mu \frac{\pi}{2}}{\rho U_\infty \left( \frac{2}{\pi} - \frac{1}{2} \right)} \] \[ \frac{1}{2} \frac{d (\delta^2)}{dx} = \frac{\mu \frac{\pi}{2}}{\rho U_\infty \left( \frac{2}{\pi} - \frac{1}{2} \right)} \] \[ \frac{d (\delta^2)}{dx} = \frac{\mu \pi}{\rho U_\infty \left( \frac{2}{\pi} - \frac{1}{2} \right)} \] Integrate from \( x=0, \delta=0 \): \[ \delta^2 = \frac{\mu \pi}{\rho U_\infty \left( \frac{2}{\pi} - \frac{1}{2} \right)} x \] \[ \delta = \sqrt{ \frac{\mu \pi x}{\rho U_\infty \left( \frac{2}{\pi} - \frac{1}{2} \right) } } \] --- ## **Summary of Final Answers:** ### (i) **Boundary Layer Thickness:** \[ \boxed{ \delta = \sqrt{ \frac{ \mu \pi x }{ \rho U_\infty \left( \frac{2}{\pi} - \frac{1}{2} \right) } } } \] --- ### (ii) **Shear Stress at the Wall:** \[ \boxed{ \tau_w = \frac{ \mu \pi U_\infty }{ 2 \delta } } \] --- ### (iii) **Average Coefficient of Drag:** \[ \boxed{ C_f = \frac{ \pi \mu }{ \rho U_\infty \delta } } \] Where \(\delta\) is as given above. --- Let me know if you need the numerical values or further simplification!