# Step-by-Step Solution for Wind Turbine Power Problem Let's break down the problem and solve it step by step. --- ## **Given Data** - **Blade length (\(r\))**: 90 ft - **Wind velocities (\(v\)):** 16 ft/s and 24 ft/s - **Rotor efficiency:** 30% at 16 ft/s, 32% at 24 ft/s - **Operating hours:** 3000 hours/year at 16 ft/s, 4000 hours/year at 24 ft/s - **Combined gearbox & generator efficiency:** 93% - **Air density (\(\rho\)):** 0.075 lbm/ft³ - **1 kW = 23730 lbm·ft²/s³** --- ## **Step 1: Find the Swept Area** Blade length = radius \( r \) Swept area \( A = \pi r^2 \): \[ A = \pi \times (90~\text{ft})^2 = \pi \times 8100~\text{ft}^2 \approx 25447~\text{ft}^2 \] --- ## **Step 2: Power in Wind** The power available in wind: \[ P_{wind} = \frac{1}{2} \rho A v^3 \] --- ## **Step 3: Power Extracted by Turbine (Mechanical Power)** \[ P_{mech} = P_{wind} \times (\text{rotor efficiency}) \] --- ## **Step 4: Power Output (Electrical Power)** \[ P_{elec} = P_{mech} \times (\text{gearbox \& generator efficiency}) \] --- ## **Step 5: Calculate for Each Velocity** ### **A. At 16 ft/s (for 3000 hours)** 1. **Wind Power:** \[ P_{wind,16} = \frac{1}{2} \times 0.075~\frac{lbm}{ft^3} \times 25447~ft^2 \times (16~ft/s)^3 \] Calculate \( (16)^3 = 4096 \): \[ P_{wind,16} = 0.5 \times 0.075 \times 25447 \times 4096 \] \[ = 0.0375 \times 25447 \times 4096 \] \[ = 954.2625 \times 4096 \] \[ \approx 3,908,668~lbm \cdot ft^2/s^3 \] 2. **Mechanical Power:** Efficiency = 30% = 0.30 \[ P_{mech,16} = 3,908,668 \times 0.3 = 1,172,600~lbm \cdot ft^2/s^3 \] 3. **Electrical Power:** Efficiency = 93% = 0.93 \[ P_{elec,16} = 1,172,600 \times 0.93 = 1,090,518~lbm \cdot ft^2/s^3 \] 4. **Convert to kW:** \[ P_{elec,16} = \frac{1,090,518}{23,730} = 45.96~kW \] 5. **Annual Energy:** \[ E_{16} = 45.96~kW \times 3000~h = 137,880~kWh \] --- ### **B. At 24 ft/s (for 4000 hours)** 1. **Wind Power:** \( (24)^3 = 13,824 \) \[ P_{wind,24} = 0.5 \times 0.075 \times 25447 \times 13824 \] \[ = 0.0375 \times 25447 \times 13824 \] \[ = 954.2625 \times 13824 \] \[ \approx 13,195,298~lbm \cdot ft^2/s^3 \] 2. **Mechanical Power:** Efficiency = 32% = 0.32 \[ P_{mech,24} = 13,195,298 \times 0.32 = 4,222,495~lbm \cdot ft^2/s^3 \] 3. **Electrical Power:** \[ P_{elec,24} = 4,222,495 \times 0.93 = 3,927,920~lbm \cdot ft^2/s^3 \] 4. **Convert to kW:** \[ P_{elec,24} = \frac{3,927,920}{23,730} = 165.54~kW \] 5. **Annual Energy:** \[ E_{24} = 165.54~kW \times 4000~h = 662,160~kWh \] --- ## **Step 6: Total Annual Electricity** \[ E_{total} = E_{16} + E_{24} = 137,880~kWh + 662,160~kWh = \boxed{800,040~kWh} \] --- ## **Final Answer** ### **The wind turbine can create approximately \( \boxed{800,000~kWh} \) of electricity each year.** --- ### **Summary Table** | Wind Speed (ft/s) | Hours/Year | Electrical Power (kW) | Annual Energy (kWh) | |-------------------|------------|-----------------------|---------------------| | 16 | 3000 | 46 | 137,880 | | 24 | 4000 | 166 | 662,160 | | **Total** | | | **800,040** |