Let's break down the problem step by step: ### **Given Data** - **Cylinder diameter, \( d \) = 1 m** - **Cylinder length, \( l \) = 10 m** - **Rotational speed, \( N \) = 100 rpm** - **Freestream velocity, \( U = 100 \) km/hr = \( \frac{100 \times 1000}{3600} = 27.78 \) m/s** - **Altitude = 2 km (standard atmosphere)** - **Ignore end effects** --- ## **Step 1: Find Air Density at 2 km Altitude** From the standard atmosphere table: - At 2 km altitude, \( \rho \approx 1.007 \) kg/m³ --- ## **Step 2: Rotational Speed in Radians per Second** \[ \text{Angular velocity},\ \omega = 2\pi \times N / 60 \] \[ \omega = 2\pi \times 100 / 60 = 10.47\ \text{rad/s} \] --- ## **Step 3: Surface Velocity of Cylinder** \[ V_{\text{cylinder}} = \omega \times R \] Where \( R = d/2 = 0.5 \) m: \[ V_{\text{cylinder}} = 10.47 \times 0.5 = 5.24\ \text{m/s} \] --- ## **Step 4: Circulation (\( \Gamma \))** For a rotating cylinder, \[ \Gamma = 2\pi R \cdot V_{\text{cylinder}} \] \[ \Gamma = 2\pi \times 0.5 \times 5.24 = 2\pi \times 2.62 = 16.45\ \text{m}^2/\text{s} \] --- ## **Step 5: Lift Per Unit Length (Kutta-Joukowski Theorem)** \[ L' = \rho U \Gamma \] \[ L' = 1.007 \times 27.78 \times 16.45 = 460.78\ \text{N/m} \] --- ## **Step 6: Total Lift (\( L \))** \[ L = L' \times l = 460.78 \times 10 = 4607.8\ \text{N} \] --- ## **Step 7: Locate the Stagnation Points** For a rotating cylinder, the stagnation points are given by: \[ \sin \theta_s = \frac{V_{\text{cylinder}}}{U} \] \[ \theta_s = \sin^{-1}\left(\frac{5.24}{27.78}\right) = \sin^{-1}(0.1886) = 10.87^\circ \] There are two stagnation points, one on each side: - **\( \theta = 10.87^\circ \)** - **\( \theta = 180^\circ + 10.87^\circ = 190.87^\circ \)** --- ## **Final Answers** ### **1. Maximum lift force:** \[ \boxed{4608\ \text{N}} \] ### **2. Location of stagnation points:** \[ \boxed{10.87^\circ \text{ and } 190.87^\circ \text{ from the front stagnation point (relative to the freestream direction)}} \] --- **Let me know if you'd like a diagram or further clarifications!**