# Wind Turbine Problem: Step-by-Step Solution Given Data: - Rated electrical power, \( P_{el} = 750 \) kW \( = 7,500,000 \) W - Wind speed at rating, \( v = 11.5 \) m/s - Air density, \( \rho = 1.25 \) kg/m³ - Tip speed ratio (design), \( \lambda_a = 7 \) - Slip ratio, \( a = .5 \) - Number of blades, \( z = 3 \) - Generator efficiency, \( \eta_{gen} = .96 \) - Gearbox efficiency, \( \eta_{gb} = .98 \) - Power coefficient at design (\( \lambda_a=7 \)), \( C_{p,Schmitz} = .55 \) - Generator synchronous speed, \( n_{gen} = 300 \) rpm --- ## a) Actual Power Coefficient of the Rotor (\( C_{p,actual} \)) **Step 1:** Find total mechanical power at rotor (\( P_{rotor} \)): \[ P_{el} = \eta_{gen} \cdot \eta_{gb} \cdot P_{rotor} \] \[ P_{rotor} = \frac{P_{el}}{\eta_{gen} \cdot \eta_{gb}} = \frac{7,500,000}{.96 \times .98} \approx 7,969,072 \text{ W} \] **Step 2:** Power available in wind intercepted by rotor: \[ P_{wind} = \frac{1}{2} \rho A v^3 \] But \( P_{rotor} = C_{p,actual} \cdot P_{wind} \), so \[ C_{p,actual} = \frac{P_{rotor}}{P_{wind}} \] Since area \( A \) is not given, \( C_{p,actual} \) cannot yet be numerically calculated until part (b) is solved. **Answer:** \[ C_{p,actual} = \frac{P_{rotor}}{.5 \rho A v^3} \] The value will be calculated after determining rotor diameter in part (b). --- ## b) Required Diameter of the Rotor (\( D \)) **Step 1:** Express \( P_{rotor} \): \[ P_{rotor} = C_{p,Schmitz} \times \frac{1}{2} \rho A v^3 \] \[ A = \frac{P_{rotor}}{.5 \rho v^3 C_{p,Schmitz}} \] **Step 2:** Substitute values: \[ A = \frac{7,969,072}{.5 \times 1.25 \times (11.5)^3 \times .55} \] Calculate \( v^3 \): \[ (11.5)^3 = 11.5 \times 11.5 \times 11.5 = 1,520.875 \] Now, plug in: \[ A = \frac{7,969,072}{.5 \times 1.25 \times 1,520.875 \times .55} \] \[ A = \frac{7,969,072}{.625 \times 1,520.875 \times .55} \] \[ .625 \times 1,520.875 = 950.547 \] \[ 950.547 \times .55 = 522.801 \] \[ A = \frac{7,969,072}{522.801} \approx 15,247.6 \text{ m}^2 \] **Step 3:** Area of a circle \( A = \pi (D/2)^2 \): \[ \frac{\pi D^2}{4} = 15,247.6 \] \[ D^2 = \frac{4 \times 15,247.6}{\pi} = \frac{60,990.4}{3.1416} \approx 19,409.7 \] \[ D = \sqrt{19,409.7} \approx 139.3 \text{ m} \] **Answer:** **Required rotor diameter \( D \approx 139.3 \) m** --- ## a) **(continued) Actual Power Coefficient** Now that \( A \) is known: \[ C_{p,actual} = \frac{7,969,072}{.5 \times 1.25 \times 15,247.6 \times 1,520.875} \] \[ .5 \times 1.25 = .625 \] \[ 15,247.6 \times 1,520.875 = 23,203,335 \] \[ .625 \times 23,203,335 = 14,502,084 \] \[ C_{p,actual} = \frac{7,969,072}{14,502,084} \approx .55 \] **Answer:** **The actual power coefficient of the rotor: \( C_{p,actual} \approx .55 \)** --- ## c) Rotational Speed at Design Point (\( n_{rotor} \) in rpm) **Tip speed ratio:** \[ \lambda_a = \frac{\omega R}{v} \] - \( R = D/2 = 139.3/2 = 69.65 \) m \[ \omega = \frac{\lambda_a v}{R} = \frac{7 \times 11.5}{69.65} = \frac{80.5}{69.65} \approx 1.156 \text{ rad/s} \] Convert to rpm: \[ n_{rotor} = \frac{\omega \cdot 60}{2\pi} = \frac{1.156 \times 60}{6.2832} \approx \frac{69.36}{6.2832} \approx 11.04 \text{ rpm} \] **Answer:** **Rotor rotational speed at design point: \( n_{rotor} \approx 11. \) rpm** --- ## d) Torque on Rotor Shaft (\( T_{rotor} \)) \[ P_{rotor} = T_{rotor} \cdot \omega \] \[ T_{rotor} = \frac{P_{rotor}}{\omega} = \frac{7,969,072}{1.156} \approx 6,893,000 \text{ Nm} \] **Answer:** **Torque on the rotor shaft: \( T_{rotor} \approx 6.89 \) MNm** --- ## e) Thrust Force on Tower (\( F_{thrust} \)) **Thrust force formula:** \[ F_{thrust} = 2 a \rho A v^2 \] Given \( a = .5 \): \[ F_{thrust} = 2 \times .5 \times 1.25 \times 15,247.6 \times (11.5)^2 \] \[ 2 \times .5 = 1 \] \[ (11.5)^2 = 132.25 \] \[ F_{thrust} = 1 \times 1.25 \times 15,247.6 \times 132.25 \] \[ 1.25 \times 15,247.6 = 19,059.5 \] \[ 19,059.5 \times 132.25 \approx 2,521,186 \text{ N} \] **Answer:** **Thrust force on tower: \( F_{thrust} \approx 2.52 \) MN** --- ## f) Required Gear Ratio (\( i \)) \[ i = \frac{n_{gen}}{n_{rotor}} = \frac{300}{11.04} \approx 272 \] **Answer:** **Gearbox ratio required: \( i \approx 272 \)** --- ## g) Shutdown Wind Speed for 3x Thrust Let maximum thrust be \( 3 \times 2,521,186 = 7,563,558 \) N. Thrust is proportional to \( v^2 \): \[ F_{thrust} = K v^2 \] \[ K = 1.25 \times 15,247.6 = 19,059.5 \] From above, \[ 2,521,186 = 19,059.5 \times 132.25 \implies K = 19,059.5 \] Set \( F_{thrust} = 7,563,558 \): \[ 7,563,558 = 19,059.5 \times v^2 \] \[ v^2 = \frac{7,563,558}{19,059.5} \approx 397 \] \[ v = \sqrt{397} \approx 19.94 \text{ m/s} \] **Answer:** **System must be shut down at wind speed \( v \geq 19.9 \) m/s** --- # **Summary Table** | Part | Quantity | Value | Unit | |------|------------------------------|----------------|---------------| | a | Actual power coefficient | .55 | - | | b | Rotor diameter | 139.3 | m | | c | Rotor speed at design | 11. | rpm | | d | Rotor torque | 6,893,000 | Nm | | e | Thrust force | 2,521,000 | N | | f | Gear ratio | 272 | - | | g | Shutdown wind speed (3x thrust)| 19.9 | m/s |