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Wind Turbine worksheet

Grades 11–12

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Rotor size given as: Swept area
- Swept area: 80 m²
- Wind speed at hub height: 7 m/s
- Air density (typical sea-level value: 1.225): 1.45 kg/m³
- Power coefficient Cp (typical: 0.35): 0.26
- Capacity factor (example: 25%): 22%
Find: Turbine power output
Answer
2.
Problem 2.
- Rotor size given as: Rotor diameter
- Rotor diameter: 9 m
- Wind speed at hub height: 6 m/s
- Air density (typical sea-level value: 1.225): 1.3 kg/m³
- Power coefficient Cp (typical: 0.35): 0.4
- Capacity factor (example: 25%): 20%
Find: Turbine power output
Answer
3.
Problem 3.
- Rotor size given as: Rotor diameter
- Rotor diameter: 10 m
- Wind speed at hub height: 7 m/s
- Air density (typical sea-level value: 1.225): 1.5 kg/m³
- Power coefficient Cp (typical: 0.35): 0.5
- Capacity factor (example: 25%): 37%
Find: Turbine power output
Answer
4.
Problem 4.
- Rotor size given as: Rotor diameter
- Rotor diameter: 12 m
- Wind speed at hub height: 8 m/s
- Air density (typical sea-level value: 1.225): 1.5 kg/m³
- Power coefficient Cp (typical: 0.35): 0.28
- Capacity factor (example: 25%): 27%
Find: Turbine power output
Answer
5.
Problem 5.
- Rotor size given as: Swept area
- Swept area: 72.5 m²
- Wind speed at hub height: 7 m/s
- Air density (typical sea-level value: 1.225): 1.025 kg/m³
- Power coefficient Cp (typical: 0.35): 0.3
- Capacity factor (example: 25%): 22%
Find: Turbine power output
Answer
6.
Problem 6.
- Rotor size given as: Swept area
- Swept area: 100 m²
- Wind speed at hub height: 4 m/s
- Air density (typical sea-level value: 1.225): 1.425 kg/m³
- Power coefficient Cp (typical: 0.35): 0.4
- Capacity factor (example: 25%): 16%
Find: Turbine power output
Answer
7.
Problem 7.
- Rotor size given as: Swept area
- Swept area: 80 m²
- Wind speed at hub height: 9 m/s
- Air density (typical sea-level value: 1.225): 1 kg/m³
- Power coefficient Cp (typical: 0.35): 0.42
- Capacity factor (example: 25%): 20%
Find: Turbine power output
Answer
8.
Problem 8.
- Rotor size given as: Rotor diameter
- Rotor diameter: 9 m
- Wind speed at hub height: 7 m/s
- Air density (typical sea-level value: 1.225): 1.2 kg/m³
- Power coefficient Cp (typical: 0.35): 0.28
- Capacity factor (example: 25%): 20%
Find: Turbine power output
Answer
9.
Problem 9.
- Rotor size given as: Swept area
- Swept area: 74 m²
- Wind speed at hub height: 8 m/s
- Air density (typical sea-level value: 1.225): 1.025 kg/m³
- Power coefficient Cp (typical: 0.35): 0.36
- Capacity factor (example: 25%): 25%
Find: Turbine power output
Answer
10.
Problem 10.
- Rotor size given as: Rotor diameter
- Rotor diameter: 9 m
- Wind speed at hub height: 6 m/s
- Air density (typical sea-level value: 1.225): 1.4 kg/m³
- Power coefficient Cp (typical: 0.35): 0.36
- Capacity factor (example: 25%): 22%
Find: Turbine power output
Answer

Answer key: Wind Turbine worksheet

1. Turbine power output: 5.172 kW
1. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.45 \times 80 \times 7^{3}$ $= 19,894 W$
2. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 19,894$ $= 11,789.04 W$
3. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.26 \times 19,894$ $= 5,172.44 W$
4. Change watts to kilowatts, which are easier to read
  $\frac{5,172.44 W}{1,000}$ $= 5.1724 kW$
5. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $5.1724 kW \times 8,760 h \times 22%$ $= 9,968.33 kWh$
2. Turbine power output: 3.573 kW
1. Find the area of the circle the blades sweep, since that is the wind the rotor can catch
  $π \times {(\frac{9}{2})}^{2}$ $= 63.6173 m²$
2. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.3 \times 63.6173 \times 6^{3}$ $= 8,931.86 W$
3. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 8,931.86$ $= 5,292.96 W$
4. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.4 \times 8,931.86$ $= 3,572.74 W$
5. Change watts to kilowatts, which are easier to read
  $\frac{3,572.74 W}{1,000}$ $= 3.5727 kW$
6. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $3.5727 kW \times 8,760 h \times 20%$ $= 6,259.45 kWh$
3. Turbine power output: 10.102 kW
1. Find the area of the circle the blades sweep, since that is the wind the rotor can catch
  $π \times {(\frac{10}{2})}^{2}$ $= 78.5398 m²$
2. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.5 \times 78.5398 \times 7^{3}$ $= 20,204.37 W$
3. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 20,204.37$ $= 11,972.96 W$
4. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.5 \times 20,204.37$ $= 10,102.18 W$
5. Change watts to kilowatts, which are easier to read
  $\frac{10,102.18 W}{1,000}$ $= 10.1022 kW$
6. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $10.1022 kW \times 8,760 h \times 37%$ $= 32,743.2 kWh$
4. Turbine power output: 12.16 kW
1. Find the area of the circle the blades sweep, since that is the wind the rotor can catch
  $π \times {(\frac{12}{2})}^{2}$ $= 113.0973 m²$
2. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.5 \times 113.0973 \times 8^{3}$ $= 43,429.38 W$
3. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 43,429.38$ $= 25,735.93 W$
4. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.28 \times 43,429.38$ $= 12,160.23 W$
5. Change watts to kilowatts, which are easier to read
  $\frac{12,160.23 W}{1,000}$ $= 12.1602 kW$
6. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $12.1602 kW \times 8,760 h \times 27%$ $= 28,761.37 kWh$
5. Turbine power output: 3.823 kW
1. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.025 \times 72.5 \times 7^{3}$ $= 12,744.59 W$
2. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 12,744.59$ $= 7,552.35 W$
3. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.3 \times 12,744.59$ $= 3,823.38 W$
4. Change watts to kilowatts, which are easier to read
  $\frac{3,823.38 W}{1,000}$ $= 3.8234 kW$
5. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $3.8234 kW \times 8,760 h \times 22%$ $= 7,368.41 kWh$
6. Turbine power output: 1.824 kW
1. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.425 \times 100 \times 4^{3}$ $= 4,560 W$
2. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 4,560$ $= 2,702.22 W$
3. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.4 \times 4,560$ $= 1,824 W$
4. Change watts to kilowatts, which are easier to read
  $\frac{1,824 W}{1,000}$ $= 1.824 kW$
5. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $1.824 kW \times 8,760 h \times 16%$ $= 2,556.52 kWh$
7. Turbine power output: 12.247 kW
1. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1 \times 80 \times 9^{3}$ $= 29,160 W$
2. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 29,160$ $= 17,280 W$
3. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.42 \times 29,160$ $= 12,247.2 W$
4. Change watts to kilowatts, which are easier to read
  $\frac{12,247.2 W}{1,000}$ $= 12.2472 kW$
5. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $12.2472 kW \times 8,760 h \times 20%$ $= 21,457.09 kWh$
8. Turbine power output: 3.666 kW
1. Find the area of the circle the blades sweep, since that is the wind the rotor can catch
  $π \times {(\frac{9}{2})}^{2}$ $= 63.6173 m²$
2. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.2 \times 63.6173 \times 7^{3}$ $= 13,092.43 W$
3. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 13,092.43$ $= 7,758.48 W$
4. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.28 \times 13,092.43$ $= 3,665.88 W$
5. Change watts to kilowatts, which are easier to read
  $\frac{3,665.88 W}{1,000}$ $= 3.6659 kW$
6. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $3.6659 kW \times 8,760 h \times 20%$ $= 6,422.62 kWh$
9. Turbine power output: 6.99 kW
1. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.025 \times 74 \times 8^{3}$ $= 19,417.6 W$
2. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 19,417.6$ $= 11,506.73 W$
3. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.36 \times 19,417.6$ $= 6,990.34 W$
4. Change watts to kilowatts, which are easier to read
  $\frac{6,990.34 W}{1,000}$ $= 6.9903 kW$
5. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $6.9903 kW \times 8,760 h \times 25%$ $= 15,308.84 kWh$
10. Turbine power output: 3.463 kW
1. Find the area of the circle the blades sweep, since that is the wind the rotor can catch
  $π \times {(\frac{9}{2})}^{2}$ $= 63.6173 m²$
2. Find the power in the wind: half the air density, times the area, times the wind speed cubed
  $0.5 \times 1.4 \times 63.6173 \times 6^{3}$ $= 9,618.93 W$
3. Find the most any rotor could catch: 16/27 of that power (the Betz limit)
  $\frac{16}{27} \times 9,618.93$ $= 5,700.11 W$
4. Multiply the wind's power by Cp, the share your turbine turns into electricity
  $0.36 \times 9,618.93$ $= 3,462.81 W$
5. Change watts to kilowatts, which are easier to read
  $\frac{3,462.81 W}{1,000}$ $= 3.4628 kW$
6. For a yearly total, multiply by the 8,760 hours in a year and by the capacity factor
  $3.4628 kW \times 8,760 h \times 22%$ $= 6,673.54 kWh$