## Annual Energy Production Mwh Year Per Turbine

1. Rule of thumb, 5-10 m above the building. So tower should be around 25 m tall, which is 80 ft. Place it 10 m away from the building.

2. Rule of thumb, tower height should be 5-10 m above the trees or move tower farther away from trees. So tower should be 40 m tall. Need to look in a catalog for cost of towers; guyed lattice tower would be the cheapest. Need to check manufacturer's brochure for recommended tower size (strength of structure) for that size wind turbine (example, 10 kW from Bergey Windpower). Stand-alone towers, Rohn SSV, cost more but do not need space for guy wires. Bergey website has tower costs.

3. Building is 15 m high, so the comparison is between additional 10 m of tower height at the building and distance of ten building heights, which gives a power reduction of 17%. Because you are farther away, will need larger size wire (see handbook on wiring or Table 7.1). Need a detailed analysis, as the costs are around the same for taller tower or farther away.

5. Class 3 wind is 150-200 W/m2 at 10 m height. Lower value is 150 W/m2, mid-value is 175 W/m2. Terrain exposure, E is 80 m. For grassland, roughness length = 0.01 m. If you use 0.03 m, that is OK. Hh = 50 m. Use Equation 9.1 and calculate for Pavg for 150 and 175 W/m2.

P = {7.173/6.683} 150 W/m2 = 1.2 * 150 W/m2 = 184 W/m2 P = {7.173/6.683} 175 W/m2 = 1.2 * 175 W/m2 = 210 W/m2

If you used the bottom of the class, you would still be in the same class for an exposure of 80 m. If you used the middle of the class, then you would increase the wind class from 3 to 4.

6. Selected 1 MW, 60 m diameter wind turbine, with 5D by 10D spacing. Fifty wind turbines are arranged in a grid of three rows with 17, 17, 16 turbines in each row. Rotor area = 2,800 m2. Assume capacity factor of 35%. Estimate annual energy production for one turbine.

AM WH = CF * area * WM * 0.00876 = 0.35 * 2,800 * 500 * 0.00876 = 4300 MWh/year.

Use availability of 0.95, 4,000 MWh/year per turbine.

Estimated annual energy production = 50 * 4,000 =200,000 MWh/year.

7. Estimate array losses for 3D by 6D spacing. Array loss for row 2 = 10%, array loss for row 3 = 15%. Turbine in row 2, energy = 0.9*4,000 = 3,600 MWh/year. Turbine in row 3 = 0.85*4,000 = 3,400 MWh/year.

Estimated annual energy production = 17 * 4,000 + 17 * 3,600 + 16 * 3,400 = 183,000 MWh/year. Total estimated annual energy production = 50*4.4* 106 kWh = 220* 106 kWh.

8. Choose 1 MW wind plants, 60 m diameter, spacing 5D by 10D. Each turbine requires a space of 300 by 600 m, or an area of 180,000 m2, which is 18 ha. Total area for fifty turbines is 900 ha.

This does not allow any space for a buffer zone around the wind farm. So you would need over 1,000 ha, or 10 km2. That results in 5 MW/km2.

9. Turbine = 3 MW, 4D by 8D spacing, D = 90 m. Space per turbine = 32 * 90 * 90 = 26 ha. 1 km2 = 100 ha, so could place four turbines per km2, which is 12 MW/km2. Notice this is over double the result for problem 8.

10. Row of 3 MW turbines, 2D spacing, D = 90 m. Space per turbine = 180 m, so 1,000 m/180 m = 5 turbines. That is, 15 MW/km.

11. With complex terrain, the spacing will generally be larger than for plains or rolling hills. In general, the overall spacing could be larger than 10D by 10D. From previous problems, answers will probably vary from 3 to 5 MW/km2.

12. Raster based, takes more data space since every pixel has a value. Vector based, means only need endpoints. In final analysis it primarily depends on cost and ease of use.

14. Estimated spacing 4D by 8D, D = 56 m. Area allocated to a turbine is 100,400 m2 = 10 ha, or 10 MW/km2. 1 sq. mile = 2.6 km2. However, note in Figure 9.11, with proper arrangement you could place eighteen turbines in a square mile (7 MW/km2), so the estimated spacing is a little large.

15. There were two rows, so two roads around 1 mile long.

Road area = 2*7 m * 1,600 m = 22,400 m2 = 2.2 ha. Base area of each turbine = 10 * 10 = 100 m2. So 16 turbines = 16,000 = 1.7 ha. Land area taken out of production = 4 ha.

Because there were county roads in place, the land taken out of production is around 0.25 ha per turbine. However, they had to upgrade the county roads with grading and caliche.

16. Area of wind farm is around 12 square miles. For 80, 1 MW wind turbines, which result in 7 MW/sq. mile.

17. Hard to tell, but guess is around 2D in a row, and 3 to 5D from row to row.

18. This is a difficult problem because of the many parameters. Students will have to do quite a bit of searching.

a. How much land does a 50 MW wind farm occupy? 10-20 km2

b. Type of terrain: plains, hills, passes, ridges, and complex terrain?

c. Size turbines? Hub height?

Again, it is a question of cost and time. What kind of risk are the developer and investment bankers willing to accept? Answer depends on the terrain and the availability of long-term database in similar nearby terrain.

My opinion: However, wind farm developers and meteorologists who consult for wind farm developers make these decisions:

a. Period of data collection: Long-term reference database is available nearby in similar terrain, 1 year. No reference database, 2-3 years.

b. Number of met towers:

Plains terrain: Tall towers, 50 m; hub height, 1-2. Short towers, 20-25 m height; 4-6.

Complex terrain: Tall towers, 50 m; hub height, 2-4. Short towers, 20-25 m height, 6-10.

50 MW wind farm would be fifty 1 MW wind turbines. Would you place a met tower at every proposed location? Would you then move the shorter towers to try to find a better site? For plains, definitely not. For complex terrain, one met tower per five turbines? In California, in complex terrain, some project operators actually moved some small wind turbines after installation to improve energy production. It would have been cheaper to install more met towers and move them. Also, this is impractical for large wind turbines.

Costs will have wide variationm from \$1 to 4 million.

Wade Weichmann's response (from actual wind farm in rolling hills area): I have encountered very different numbers from different sources on the number of met stations. I know that at Lake Benton I, Minnesota, 22 met stations were erected for the 143-turbine site for 3 years (15 square miles). On the other hand, Spera stated in Wind Turbine Technology that it could be cost justified to install one met tower per turbine site for large-scale turbines. I believe the number of stations would depend on the topography of the site. A flat topography would need less met towers to get accurate data that could be correlated to long-term regional data mentioned earlier. One-year duration was also mentioned in the Wind Resource Assessment Handbook to predict power density to within 10%. I believe 2 years would strengthen the predictability of power density. Also, the tallest sensors on the met towers should be placed at hub height.

19. Wind speed is around 5 m/s from the northeast.

20. Data should be collected:

21. Elevation of mesa is around 1,600 m. Trails.com purchased Topozone. Have a free 14-day trial. Go to www.awstruewind.com or to www.newmexico.org/map and use terrain.

22. General rule; can install 5-9 MW/km2 for plains and rolling hills, and 8-12 MW/km for ridgeline.

24. Using general rule, use 8 MW/km2, then could install 8*110,788 = 880,000 MW. Capturable power to be in the same ratio as in Table 9.3; capturable power = 17/54 * 880,000 = 280,000 MW.

25. Mesa Redonda, wind speed at 100 m height, 9.1 m/s.

## Renewable Energy Eco Friendly

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable.

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### Responses

• malcolm
How much air speed at 25m height of windmil tower?
8 years ago
• Philip
How much power is 3600 mwh?
7 years ago
• arabella
How to calculate produced energey mwh of turbine?
7 years ago
• Cataldo
What is the general rule for calculating megawatts per square kilometer in plains and rolling hills?
2 months ago