# Wind Turbine Blade Length Calculation Let's solve the problem step by step: --- ## **Given Data** - **Required Power Output, \( P_{out} \):** 5 MW = \( 5,000,000 \) W - **Elevation above sea level:** 150 m - **Tower height:** 125 m (for reference, doesn’t affect calculation) - **Average hub height wind speed, \( V \):** 8.5 m/s We'll need to **assume**: - **Power coefficient, \( C_p \):** (Typically max is ~.45 for modern turbines) - **Drivetrain efficiency, \( \eta \):** (Typically 90-95%, let’s use .95) - **Air density, \( \rho \):** We'll calculate this, since it changes with elevation. --- ## **Step 1: Air Density Calculation at 150 m** Standard air density at sea level: \( \rho_ = 1.225 \) kg/m³ The air density decreases with altitude. A common approximation for air density as a function of altitude: \[ \rho = \rho_ \times \left(1 - \frac{.0065 \cdot h}{288.15}\right)^{5.2561} \] Where \( h \) is elevation in meters. Plug in values: - \( h = 150 \) m \[ \rho = 1.225 \times \left(1 - \frac{.0065 \times 150}{288.15}\right)^{5.2561} \] \[ = 1.225 \times \left(1 - \frac{9.75}{288.15}\right)^{5.2561} \] \[ = 1.225 \times (1 - .03385)^{5.2561} \] \[ = 1.225 \times (.96615)^{5.2561} \] \[ = 1.225 \times .838 \] \[ \approx 1.03 \text{ kg/m}^3 \] --- ## **Step 2: Wind Power Equation** The power extracted from wind: \[ P_{out} = \eta \cdot C_p \cdot \frac{1}{2} \rho A V^3 \] Where: - \( A = \pi R^2 \) (R = blade length) Let’s solve for R. --- ## **Step 3: Assume Values** - \( C_p = .45 \) - \( \eta = .95 \) - \( \rho = 1.03 \) kg/m³ - \( V = 8.5 \) m/s - \( P_{out} = 5,000,000 \) W --- ## **Step 4: Rearrange for Area** \[ P_{out} = \eta C_p \frac{1}{2} \rho \pi R^2 V^3 \] \[ R^2 = \frac{2P_{out}}{\eta C_p \rho \pi V^3} \] \[ R = \sqrt{\frac{2P_{out}}{\eta C_p \rho \pi V^3}} \] --- ## **Step 5: Plug in the Numbers** \[ R = \sqrt{ \frac{2 \times 5,000,000}{.95 \times .45 \times 1.03 \times \pi \times (8.5)^3} } \] Calculate denominator: - \( (8.5)^3 = 614.125 \) - \( \pi \approx 3.1416 \) - \( .95 \times .45 \times 1.03 \times 3.1416 \times 614.125 \) First, multiply: - \( .95 \times .45 = .4275 \) - \( .4275 \times 1.03 = .440325 \) - \( .440325 \times 3.1416 = 1.3828 \) - \( 1.3828 \times 614.125 = 849.61 \) Now plug in: \[ R = \sqrt{ \frac{10,000,000}{849.61} } \] \[ = \sqrt{11,770.8} \] \[ \approx 108.5 \text{ meters} \] --- ## **Final Answer** ### **A reasonable blade length (radius) for your 5 MW wind turbine is:** \[ \boxed{109 \text{ meters}} \] (rounded to the nearest meter) --- ### **Summary of Assumptions:** - **Power Coefficient (\( C_p \))**: .45 - **Drivetrain Efficiency (\( \eta \))**: .95 - **Air Density (\( \rho \))**: 1.03 kg/m³ (at 150 m ASL) - **Wind Speed (\( V \))**: 8.5 m/s --- Let me know if you want a blade diameter or want to explore the sensitivity to any of these parameters!