Let's work through each step using the given data and the Blasius solution for laminar boundary layers: --- ## **Given Data** - **Kinematic viscosity,** \( \nu = 8. \times 10^{-5} \) m²/s - **Plate length,** \( L = 1.5 \) m - **Free stream velocity,** \( U_\infty = 2 \) m/s --- ## **1. Reynolds Number at the End of the Plate (\( x = L \))** The Reynolds number is calculated as: \[ Re_L = \frac{U_\infty L}{\nu} \] **Plug in the values:** \[ Re_L = \frac{2 \times 1.5}{8. \times 10^{-5}} = \frac{3.}{8. \times 10^{-5}} = 3.75 \times 10^4 \] --- ## **2. Boundary Layer Thickness (\( \delta \)) at the Trailing Edge** For a laminar flow over a flat plate (Blasius solution): \[ \delta = \frac{5.x}{\sqrt{Re_x}} \] At \( x = L \): \[ \delta_L = \frac{5.L}{\sqrt{Re_L}} \] **Plug in the values:** \[ \delta_L = \frac{5. \times 1.5}{\sqrt{3.75 \times 10^4}} = \frac{7.5}{193.6} \approx .0388~\text{m} = 38.8~\text{mm} \] --- ## **3. Local Wall Shear Stress (\( \tau_w \)) at \( x = L \)** Blasius solution for wall shear stress: \[ \tau_w = .332 \rho U_\infty^2 \sqrt{\frac{\nu}{U_\infty L}} \] **Note:** Density (\( \rho \)) of engine oil is not given, so keep it as \( \rho \) or assume a typical value if needed. \[ \tau_w = .332 \rho (2)^2 \sqrt{\frac{8. \times 10^{-5}}{2 \times 1.5}} \] Calculate the square root: \[ \frac{8. \times 10^{-5}}{3.} = 2.667 \times 10^{-5} \] \[ \sqrt{2.667 \times 10^{-5}} = 5.16 \times 10^{-3} \] Plug values into the equation: \[ \tau_w = .332 \rho \times 4 \times 5.16 \times 10^{-3} = .332 \times 4 \times 5.16 \times 10^{-3} \rho = 1.328 \times 5.16 \times 10^{-3} \rho = 6.85 \times 10^{-3} \rho~(\text{N/m}^2) \] --- ## **4. Total Friction Drag (\( F_D \)) on One Side per Unit Width** Total drag per unit width (Blasius): \[ F_D = C_f \times \frac{1}{2} \rho U_\infty^2 \times L \] Where the average skin friction coefficient for a laminar plate: \[ C_f = \frac{1.328}{\sqrt{Re_L}} \] Calculate \( C_f \): \[ C_f = \frac{1.328}{\sqrt{3.75 \times 10^4}} = \frac{1.328}{193.6} = .00686 \] Now, plug values into drag equation: \[ F_D = .00686 \times \frac{1}{2} \rho (2)^2 \times 1.5 \] \[ = .00686 \times .5 \times 4 \times 1.5 \rho = .00686 \times 3 \rho = .0206 \rho~\text{N/m} \] --- ## **Summary Table** | Step | Formula | Result (in terms of \( \rho \)) | |------|---------|-------------------------------| | 1. Reynolds number | \( Re_L = \frac{U_\infty L}{\nu} \) | \( 3.75 \times 10^4 \) | | 2. Boundary layer thickness | \( \delta_L = \frac{5L}{\sqrt{Re_L}} \) | \( .0388 \) m (38.8 mm) | | 3. Wall shear stress | \( \tau_w = .332 \rho U_\infty^2 \sqrt{\frac{\nu}{U_\infty L}} \) | \( 6.85 \times 10^{-3} \rho \) N/m² | | 4. Total friction drag | \( F_D = C_f \frac{1}{2} \rho U_\infty^2 L \) | \( .0206 \rho \) N/m | --- ### **Let me know if you need further clarification or if you want to plug in a typical value for oil density!**

# Fluid Mechanics Analysis of Engine Oil Flow Let's analyze the flow of engine oil over a flat plate using the Blasius solution for laminar boundary layers. ## Given Data - **Kinematic viscosity,** \( \nu = 8.0 \times 10^{-5} \) m²/s - **Plate length,** \( L = 1.5 \) m - **Free stream velocity,** \( U_\infty = 2 \) m/s --- ## 1. Reynolds Number at the End of the Plate (\( x = L \)) The Reynolds number is calculated using the formula: \[ Re_L = \frac{U_\infty L}{\nu} \] Substituting the given values: \[ Re_L = \frac{2 \times 1.5}{8.0 \times 10^{-5}} = \frac{3}{8.0 \times 10^{-5}} = 3.75 \times 10^4 \] --- ## 2. Boundary Layer Thickness (\( \delta \)) at the Trailing Edge For laminar flow, the boundary layer thickness at the end of the plate is given by: \[ \delta = \frac{5L}{\sqrt{Re_L}} \] Calculating \( \delta_L \): \[ \delta_L = \frac{5 \times 1.5}{\sqrt{3.75 \times 10^4}} = \frac{7.5}{193.6} \approx 0.0388 \text{ m} \text{ (or 38.8 mm)} \] --- ## 3. Local Wall Shear Stress (\( \tau_w \)) at \( x = L \) The wall shear stress is computed as: \[ \tau_w = 0.332 \rho U_\infty^2 \sqrt{\frac{\nu}{U_\infty L}} \] Calculating: \[ \tau_w = 0.332 \rho (2)^2 \sqrt{\frac{8.0 \times 10^{-5}}{2 \times 1.5}} \] Calculate the square root: \[ \frac{8.0 \times 10^{-5}}{3} = 2.667 \times 10^{-5} \quad \Rightarrow \quad \sqrt{2.667 \times 10^{-5}} = 5.16 \times 10^{-3} \] Substituting back: \[ \tau_w = 0.332 \rho \times 4 \times 5.16 \times 10^{-3} = 6.85 \times 10^{-3} \rho \text{ (N/m²)} \] --- ## 4. Total Friction Drag (\( F_D \)) on One Side per Unit Width The total drag per unit width can be estimated using: \[ F_D = C_f \times \frac{1}{2} \rho U_\infty^2 \times L \] Where the average skin friction coefficient for laminar flow is: \[ C_f = \frac{1.328}{\sqrt{Re_L}} \] Calculating \( C_f \): \[ C_f = \frac{1.328}{\sqrt{3.75 \times 10^4}} = \frac{1.328}{193.6} \approx 0.00686 \] Substituting back into the drag equation: \[ F_D = 0.00686 \times \frac{1}{2} \rho (2)^2 \times 1.5 \] \[ = 0.00686 \times 0.5 \times 4 \times 1.5 \rho = 0.0206 \rho \text{ (N/m)} \] --- ## Summary of Results 1. **Reynolds Number:** \( 3.75 \times 10^4 \) 2. **Boundary Layer Thickness:** \( 0.0388 \) m (or 38.8 mm) 3. **Local Wall Shear Stress:** \( 6.85 \times 10^{-3} \rho \) N/m² 4. **Total Friction Drag:** \( 0.0206 \rho \) N/m --- Feel free to ask if you have questions or need further assistance!