Certainly! Let’s walk through the **step-by-step solution** for this kind of airfoil-wind tunnel streamline problem, where you have a symmetric airfoil and the streamlines show velocity changes at 1/3 and 1/6 of the height of the free-stream tube. **Given/Assumptions:** - Symmetric airfoil in a wind tunnel - Incoming (free-stream) velocity: \( V_0 \) - The width (height) of the streamtube far upstream: \( h_0 \) - As air passes over the airfoil, the streamtube narrows at certain points: - At height \( h_1 = \frac{1}{3} h_0 \), velocity is \( V_1 \) - At height \( h_2 = \frac{1}{6} h_0 \), velocity is \( V_2 \) - Air is **incompressible** and **non-viscous** (use continuity and Bernoulli's principles). Let's solve for: 1. The velocities at the narrowed sections (\( V_1, V_2 \)) 2. The pressure differences at those points compared to the free stream. --- ### **Step 1: Use the Continuity Equation** The **continuity equation** for steady, incompressible flow: \[ A_0 V_0 = A_1 V_1 = A_2 V_2 \] where \( A \) is the cross-sectional area. For a wind tunnel of constant width, the area is proportional to height (\( A = b \cdot h \)), so we can focus on heights. #### At \( h_1 = \frac{1}{3} h_0 \): \[ h_0 V_0 = \frac{1}{3}h_0 V_1 \implies V_1 = 3 V_0 \] #### At \( h_2 = \frac{1}{6} h_0 \): \[ h_0 V_0 = \frac{1}{6}h_0 V_2 \implies V_2 = 6 V_0 \] --- ### **Step 2: Apply Bernoulli’s Equation** Bernoulli’s equation for incompressible, non-viscous flow along a streamline: \[ P + \frac{1}{2}\rho V^2 = \text{constant} \] Let: - \( P_0 \): Pressure in the free stream (upstream, region 0) - \( P_1 \): Pressure at \( h_1 \) - \( P_2 \): Pressure at \( h_2 \) #### At \( h_1 \): \[ P_0 + \frac{1}{2}\rho V_0^2 = P_1 + \frac{1}{2}\rho V_1^2 \] \[ P_1 = P_0 + \frac{1}{2}\rho (V_0^2 - V_1^2) \] \[ V_1 = 3V_0 \implies V_1^2 = 9V_0^2 \] \[ P_1 = P_0 + \frac{1}{2}\rho (V_0^2 - 9V_0^2) = P_0 - 4\rho V_0^2 \] #### At \( h_2 \): \[ P_0 + \frac{1}{2}\rho V_0^2 = P_2 + \frac{1}{2}\rho V_2^2 \] \[ P_2 = P_0 + \frac{1}{2}\rho (V_0^2 - V_2^2) \] \[ V_2 = 6V_0 \implies V_2^2 = 36V_0^2 \] \[ P_2 = P_0 + \frac{1}{2}\rho (V_0^2 - 36V_0^2) = P_0 - 17.5\rho V_0^2 \] --- ### **Step 3: Summarize Results** #### **Velocities:** - At \( h_1 = \frac{1}{3} h_0 \): \( V_1 = 3 V_0 \) - At \( h_2 = \frac{1}{6} h_0 \): \( V_2 = 6 V_0 \) #### **Pressures (relative to \( P_0 \)):** - At \( h_1 \): \( P_1 = P_0 - 4\rho V_0^2 \) - At \( h_2 \): \( P_2 = P_0 - 17.5\rho V_0^2 \) --- ## **Final Answers** ### (1) The velocities at the narrowed sections of streamtube are: - At \( 1/3 \) of the original height: **\( V_1 = 3 V_0 \)** - At \( 1/6 \) of the original height: **\( V_2 = 6 V_0 \)** ### (2) The pressures at those points (relative to free stream) are: - At \( 1/3 \) height: **\( P_1 = P_0 - 4 \rho V_0^2 \)** - At \( 1/6 \) height: **\( P_2 = P_0 - 17.5 \rho V_0^2 \)** --- #### **Explanation** - As the streamtube narrows, velocity increases to conserve mass (continuity). - Bernoulli's equation tells us that as velocity increases, pressure decreases. - The pressure drops are significant at the points where the streamtube is narrowest. **If you have actual values for \( V_0 \), \( h_0 \), or air density \( \rho \), you can plug them in to get numerical results!** --- **Let me know if you want to see a diagram or further explanation of the physical meaning!**