1 GEG 124: Energy Resources Name

Home , National Wind, Twin Groves Wind Farm

GEG 124: Energy Resources Name: ______Lab #10: Wind Day: ______Recommended Textbook Reading Prior to Lab:  Chapter 8: Wind. Energy Resources by Theodore Erski o Wind’s Capacity Growth o Air Pressure, Wind & Power o Wind Farms o Wind’s Virtues and Vices

Goals: after completing this lab, you will be able to:  Create a sketch of how various components are wired together on the dedicated Rutland 503 Windcharger turbine cart.  Measure and record wind speed using a Kestrel 3500 Weather Meter.  Evaluate the charge entering a 12 volt battery from the Rutland 503 Windcharger.  Create a wind rose that illustrates wind direction and frequency using selected and compiled data of a theoretical site in the American Midwest.  Calculate wind power density using the standard formula used across the wind industry.  Graph wind power density with increasing wind speed.  Differentiate between linear and exponential growth.  Evaluate sites for a wind farm, and judge them based on wind power density calculations.  Calculate and compare the average wind power within various wind power classes.  Compose a descriptive narrative of the average wind speeds across the United States after examining a national wind map illustrating wind speeds 80 meters above the ground.  Draw isotachs across Illinois using wind speed data provided by the National Renewable Energy Laboratory, and classify wind speed areas across the state using colored pencils.  Analyze a wind speed map of the state of Texas, and describe the likely locations for wind farms.  Calculate wind turbine capacity using the standard formula used across the wind industry.  Calculate the annual power production from the Twin Groves wind farm, and compare this production to that produced by the Duck Creek coal-fired power plant.  Reflect upon why, even with lower power production, the people of Illinois (and other states) encourage the construction and expansion of wind farms.

Key Terms and Concepts:  Betz Limit  Wind maps  capacity  Wind power class  Capacity factor  Wind power density  Isotachs  Wind rose

Required Materials:  Calculator  Colored pencils  Color printing for this lab  High-speed internet connection (for Module #3)  Kestrel 3500 Weather Meter (for Module #1 only)  Rutland 503 Windcharger turbine (for Module #1 only)

Problem Solving Module #1: Examining, Diagraming and Using a Wind Turbine

The Rutland 503 Windcharger is assembled, fully functional, and is built into its own dedicated rolling cart. It demonstrates how an off-grid site can generate and store electricity. The major components are:

 Windcharger generator: Six blades capture the power of the wind and spin ceramic magnets in close proximity to wound copper coils (the magnets and copper coils are inside the nacelle). Alternating Current (AC) electricity is generated when the turbine’s rotor spins. The nacelle contains an AC-to-DC converter (rectifier) which converts the electricity into Direct Current (DC).  Regulator: This regulates the current entering the battery and thus prevents the battery from becoming overcharged. There are two LEDs on the regulator. The LED on the top is the “charging” LED, and it indicates if current is flowing into the battery. When this LED is green, the regulator is allowing current to enter the battery. When it is red, the regulator is shunting excess current, thereby slowing the generator, and preventing current from entering the battery. The LED on the bottom indicates the battery’s charge.  A 12-volt, 40 amp hour battery: This stores electricity for use-on-demand.  Power inverter: Converts direct current (DC) electricity from the battery into alternating current (AC) electricity. Most appliances in a home run on AC.  Lamp: An appliance running on AC.

1. Closely examine the wind charger and all its components, paying especially close attention to the wiring. In Figure 1, sketch how all the components are electrically connected. You do not need to include the cart in your sketch. Instead, concentrate on the various components on the cart and how the wiring flows to and from these components.

Figure 1

2. Roll the windcharger cart outside and have the turbine capture the wind. Measure the wind speed in knots using the Kestrel 3500 Weather Meter. How fast is the wind blowing in knots?

Answer: will vary with each class.

Figure 2 illustrates the power curve for the Rutland 503 Windcharger in ideal, non-turbulent conditions. Notice that as wind speed increases, the current entering the 12 volt battery also increases.

Figure 2

3. According to your measured wind speed and Figure 2, how much charge is flowing into the battery?

Answer: Will vary with each class depending on wind speed.

4. Examine the regulator. What does the “charging” LED tell you is happening to the battery, and why is this happening?

Answer: Will vary with each class. Depending on the battery’s charge, it could be charging or regulating. If the regulator’s “charging” LED is green, it is allowing current to flow into the battery. If the “charging” LED is red, it is shunting current, thereby slowing the generator, and preventing current from entering the battery.

5. Speculate about some locations where having this sort of setup might be useful.

Answer: Will vary with each student. Look for comments about off-grid sites, such as cabins, utility sheds, boats, RV campers, etc.

Problem Solving Module #2: Wind Frequency, Speed and Power Density

Wind roses are graphic renditions of the frequency of wind blowing from particular directions at a particular site. They are constructed after a meteorological tower (a “met” tower) records wind speed and direction every ten minutes, for an entire year, at 10, 32, 40, and 58 meters above the ground (some met towers can be hundreds of meters tall, but for practical purposes most are between 60-80 meters tall). Wind roses are constructed on a circular layout, where the circle’s center illustrates “zero frequency” and subsequent concentric circles illustrate increasing frequency. Spokes are drawn along the sixteen cardinal directions (N, NNE, NE…etc.) to designate direction from which the wind blows.

Table 1 is shows wind direction and frequency from a hypothetical met tower placed somewhere in the American Midwest. The information in Table 1 is a compilation of thousands of lines of data collected over a year.

Wind Direction Frequency (%) at 58 Meters High 0° (N) 7 22.5° (NNE) 3 45° (NE) 3 67.5° (ENE) 1 90° (E) 0 112.5° (ESE) 0 135° (SE) 1 157.5° (SSE) 1 180° (S) 12 202.5° (SSW) 11 225° (SW) 10 247.5° (WSW) 14 270° (W) 26 292.5° (WNW) 4 315° (NW) 5 337.5° (NNW) 2

Table 1

6. Use the data in Table 1 to complete the wind frequency rose illustrated in Figure 3. One spoke is already done for you.

Figure 3

7. Write a descriptive narration of Figure 3. Be sure to include comments about cardinal directions, as well as about the frequency of wind from specific cardinal directions.

Suggested Answer: Figure 3 is a wind rose that illustrates the direction and frequency of wind at a particular site. The site has 270° (western) winds blowing 26% of the time. Higher-frequency winds also blow from 180°, 202.5°, 225°, and 247.5°. The overall impression is of south, southwest, and western prevailing winds. This impression is verified by the graphic in Figure 1, as well as the tabular data in Table 1 where we see that a full 73% of the wind comes from between the cardinal directions of 180° (south) and 270° (west).

8. In the United States we typically measure speed in miles per hour (mph). In the wind industry, however, speed is measured in meters per second (m/s). Complete Table 2 using the provided relationship between mph and m/s.

Miles per Hour (mph) Meters per Second (m/s) 1 0.45 5 2.25 10 4.50 15 6.75 20 9.00 25 11.25

Table 2

Wind power density (the power contained in the wind) is calculated in Watts/m2. The formula is:

W/m2 = 1.91 x 0.5 x ρ x V3

Where:  1.91 = A multiplier used in this lab to account for the fact that wind never blows constantly, at an average speed, at any site. Half of the time, winds are faster than the site’s average, and faster winds contain much more power than slower winds because wind power varies with the cube of wind speed.*  0.5 = A constant value derived from the equation used to calculate kinetic energy (K= ½ mv2).  ρ = 1.225 (air’s density (kg/m3) at 15 °C, kept constant for this lab).  V = Wind’s velocity (m/s).

*For more information about wind’s power density, see:  The Iowa Energy Center: http://www.iowaenergycenter.org/wind-energy-manual/wind-and- wind-power/wind-speed-and-power/  The Danish Wind Industry Association: http://www.motiva.fi/myllarin_tuulivoima/windpower%20web/en/tour/wres/powdensi.htm

9. Use the above formula to complete Table 3.

V (m/s) Wind Power Density (W/m2) 1 1 2 9 3 32 4 75 5 146 6 253 7 401 8 599 9 853 10 1,170 11 1,557

Table 3

10. Graph the data in Table 3 onto Figure 4. To begin, place the wind velocity values (V) on the “X” (horizontal) axis. Then, place the wind power density values (W/m2) on the “Y” (vertical) axis. Finally, place the data from Table 3 into your graph and connect the data points with a smooth solid line.

Figure 4

11. Complete the following paragraph by filling in the blanks: A 1 m/s increase in wind velocity, from 2 m/s to 3 m/s, increases wind-power density by 23 W/m2. A 1 m/s increase in wind velocity from 3 m/s to 4 m/s increases wind power density by 43 W/m2. A 1 m/s increase in wind velocity from 4 m/s to 5 m/s increases wind power density by 71 W/m2. A 1 m/s increase in wind velocity from 5 m/s to 6 m/s increases wind power density by 107 W/m2.

12. A linear growth rate increases by the same value with each step. When graphed, this creates a straight line at a constant slope. An exponential growth rate is proportional to a function’s current value. When graphed, this creates a curved, steeply increasing trend line. What kind of growth rate do you see in Table 3 and Figure 4?

Exponential growth rate

13. Assume a developer had three potential sites on which to build a wind farm. Site #1 has an average wind speed of 7.3 m/s. Site #2 has an average wind speed of 7.5 m/s. Site #3 has an average wind speed of 7.7 m/s. Assuming all other variables are equal, calculate the power density for all three sites, and then explain why the developer would insist on locating the wind farm at site #3, even though all three sites have very similar wind speeds.

Site #1 wind power density: W/m2 = 1.91 x 0.5 x 1.225 x 7.33 = 455 W/m2 Site #2 wind power density: W/m2 = 1.91 x 0.5 x 1.225 x 7.53 = 494 W/m2 Site #3 wind power density: W/m2 = 1.91 x 0.5 x 1.225 x 7.73 = 534 W/m2

The developer would insist on Site #3 over Site #1 because Site #3 has 79 W/m2 more power density than Site #1. The developer would insist on Site #3 over Site #2 because Site #3 has 40 W/m2 more power density than Site #2.

Wind speed is often classified into “Wind Power Classes” based on the power within the wind 50 meters above the ground. Thus, based on the information in Table 4, the following narrative can be stated: “A wind power class of 3 is categorized as “Fair”, because 50 meters above the ground wind consistently blows between 6.4 to 7 meters per second and contains 300-400 Watts per square meter.”

Wind Power Name Wind Speed (m/s) at Wind Power Density at Class 50 m 50 m (W/m2) 1 Poor 0.0-5.6 0-200 2 Marginal 5.6-6.4 200-300 3 Fair 6.4-7.0 300-400 4 Good 7.0-7.5 400-500 5 Excellent 7.5-8.0 500-600 6 Outstanding 8.0-8.8 600-800 7 Superb >8.8 >800

Table 4

14. If the wind 50 meters above the ground consistently blows at 10 miles per hour, what is its Wind Power Class?

Poor

15. If the wind 50 meters above the ground consistently blows at 16 miles per hour, what is its Wind Power Class?

Good

16. Based on the information in Table 4, what happens to the power within the wind as wind speed increases?

The power within the wind increases as wind speed increases.

17. Complete the following sentences by filling in the blanks: The average power in the “Fair” wind power class is 350 Watts/m2. The average power in the “Excellent” wind power class is 550 Watts/m2. Therefore, every square meter of wind in an “Excellent” wind power class contains 200 Watt/m2 more power compared to every square meter of wind in a “Fair” wind power class.

18. Construct a narrative of the “Outstanding” wind power class following the above provided example and using the information in Table 4.

“A wind power class of 6 is categorized as “Outstanding”, because 50 meters above the ground wind consistently blows between 8 to 8.8 meters per second and contains 600-800 Watts per square meter.”

Problem Solving Module #3: Wind Maps

Figure 5 illustrates annual average wind speeds across the United States 80 meters above the ground.

Credit/courtesy of The National Renewable Energy Laboratory (NREL)

Figure 5

19. Write a very general, descriptive narration of the map’s content in Figure 5 using the terms “the West” “the Midwest” and “the East” to broadly describe locations in the United States. Be sure to include specific values from the map’s key.

Suggested answer: The West, including the states of California, Oregon, Nevada and Arizona, has generally slow annual average winds 80 meters above the ground. Winds in the West generally blow 5.5 m/s or slower. The East, including the states of Pennsylvania, Virginia, Alabama, Georgia and Florida, also has generally slow annual average winds 80 meters above the ground. Winds in the East generally blow 5.5 m/s or slower. The Midwest, including the states of North and South Dakota, Nebraska, Kansas, Oklahoma and Texas, has generally fast annual average wind speeds 80 meters above the ground. Winds in the Midwest blow 7.5 to 8.5 m/s.

Isotachs are lines connecting places of equal wind speed. Isotach creation follows these guidelines:

 Isotachs never cross each other.  Isotachs are often drawn in 0.5 m/s intervals.  Drawing isotachs requires estimating their placement on a map.  Widely-spaced isotachs indicate little wind speed difference across an area.  Narrowly-spaced isobars indicate a lot of wind speed difference across an area.

Figure 6 is a before and after example of how to draw isotachs and label wind speed ranges.

Iowa base map credit/courtesy of The National Renewable Energy Laboratory (NREL)

Figure 6

20. Explain how a wind farm developer can benefit from a map showing isotachs.

Suggested answer: Isotachs allow map users to quickly assess wind speed patterns across large areas. A wind farm developer will want to place a wind farm in the best possible location. Seeing a map with isotachs allows a wind farm developer to quickly assess what are generally the best and worst locations in a state. The developer can then focus on developing wind farms only in those areas of a state that have the best speeds.

21. In Figure 7, draw isotachs across Illinois using the provided point data and the map’s key to demarcate isotachs intervals. After drawing the isotachs, shade the wind speed areas with colored pencils according to the map’s key. Notice that the 7-7.5 m/s isotachs, as well as its area shading, is already done for you.

Illinois base map credit/courtesy of The National Renewable Energy Laboratory (NREL)

Figure 7

22. Based on your isotachs and shading work for Illinois, why is it unlikely that wind farms would be built in Southern Illinois? Be sure to specify wind speeds in your commentary.

Wind speeds in southern Illinois are too slow compared to other locations within the state. Wind speeds in southern Illinois vary between 4.5 to 5.5 m/s, which is much slower than the 6.5 to 7.5 m/s wind speeds seen in the central and northern parts of the state.

23. Figure 8 is a wind map of Texas. Based on its information, in what region of Texas would you expect to find the most wind farms, and what cities would likely see the most growth due to this energy resource? Be sure to explain your reasoning.

I would expect to see the most wind farms in the northwest region of Texas because the winds are highest here, blowing on average between 8-9 m/s. The cities of Amarillo, Abilene and San Angelo would likely see the most growth due to this energy resource because they are located in the northwest part of the state.

Texas base map credit/courtesy of The National Renewable Energy Laboratory (NREL)

Figure 8

Go to Wikipedia and enter “Wind power in Texas” into its search engine (or type the following url into your browser’s window: http://en.wikipedia.org/wiki/Wind_power_in_Texas). Follow the link and examine the “Location Map”. Use this map to answer questions 24-26.

24. Is the region you selected in the above question correct?

Suggested answer: The student will likely have selected the northwest region of Texas. Thus this answer will likely be a simple “yes”.

25. What are the names of some of the operating wind farms in the area you selected?

Suggested answers: Wildorado, Brazos, Hackberry, Sweetwater, Roscoe, Elbow Creek, Trent, Lone Star, Horse Hollow, Capricorn Ridge.

26. Compare the operating wind farms in the Texas Panhandle, where Amarillo is located, to those further south, near Abilene. Given the average wind speeds near each city, what is a question we might ask?

Suggested answer: Why are there more operating wind farms near Abilene instead of near Amarillo, when the area around Amarillo does not have as consistently high winds compared to the Texas Panhandle?

Problem Solving Module #4: Calculating Wind Turbine Capacity and Power Production

Wind turbine capacity is calculated in Watts. The formula is:

3 CapacityWatts = 0.5 x ρ x A x V x Cp

Where:  0.5 = A constant value derived from the equation used to calculate kinetic energy (K= ½ mv2).  ρ = 1.225 (air’s density (kg/m3) at 15 °C, kept constant for this lab).  A = Swept area of turbine’s blades (m2).*  V = Velocity of wind (m/s).  Cp = 0.4 (power coefficient (efficiency), kept constant for this lab).**

*To calculate A we must have the radius of the turbine. We obtain the radius by measuring the length of a single turbine blade. For example, if a single turbine blade is 38 meters long, then:

 A = πr2  = 3.14 x 382  =3.14 x 1,444  =Area swept by turbine’s blades is 4,534 m2

**The theoretical maximum power coefficient (efficiency) is called the Betz Limit, and is 59.3%. This literally means that no more than 59.3% of the wind’s kinetic energy can be converted to mechanical energy by a rotor. Another way to say this is that the theoretical maximum efficiency of any wind turbine is 59.3%. In the real world, this value usually fall between 30-45% (0.30 – 0.45).

27. Calculate the capacity of a turbine with 42 meter long blades designed for 9 m/s wind. Be sure to show your work

3 Capacity = 0.5 x ρ x A x V x Cp = 0.5 x 1.225 x (π422) x 93 x 0.4 = 0.5 x 1.225 x (3.14 x 1,764) x 729 x 0.4 = 0.5 x 1.225 x 5,542 x 729 x 0.4 = 989,829 Watts = 989,829 / 1,000,000 = 1 MW

28. Calculate the capacity of a turbine with 51.2 meter long blades designed for 9 m/s wind. Be sure to show your work

3 Capacity = 0.5 x ρ x A x V x Cp = 0.5 x 1.225 x (π51.22) x 93 x 0.4 = 0.5 x 1.225 x (3.14 x 2,621) x 729 x 0.4 = 0.5 x 1.225 x 8,234 x 729 x 0.4 = 1,470,634 Watts = 1,470,634 Watts / 1,000,000 = 1.5 MW

29. What happens to the capacity of a turbine if the length of the blades increases and all other variables are held constant?

The capacity increases.

30. Calculate the capacity of a turbine with 50 meter long blades designed for 11 m/s wind. Be sure to show your work

3 Capacity = 0.5 x ρ x A x V x Cp = 0.5 x 1.225 x (π502) x 113 x 0.4 = 0.5 x 1.225 x (3.14 x 2,500) x 1,331 x 0.4 = 0.5 x 1.225 x 7,854 x 1,331 x 0.4 = 2,561,150 Watts = 2,561,150 / 1,000,000 = 2.6 MW

31. Calculate the capacity of a turbine with 50 meter long blades designed for 12.5 m/s wind. Be sure to show your work

3 Capacity = 0.5 x ρ x A x V x Cp = 0.5 x 1.225 x (π502) x 12.53 x 0.4 = 0.5 x 1.225 x (3.14 x 2,500) x 1,953 x 0.4 = 0.5 x 1.225 x 7,854 x 1,953 x 0.4 = 3,758,021 Watts = 3,758,021 Watts / 1,000,000 3.8 MW

32. What happens to the capacity of a turbine if velocity of the wind increases and all other variables are held constant?

The capacity increases.

Recall from Lab #2 that “capacity factor” is the percent of time that a power plant operates at full capacity. No power plant operates at 100% capacity because of variables such as maintenance, refueling, weather and safety inspections. In the case of wind turbines, weather is a critical variable because wind is an intermittent energy resource. Sometimes a site has consistent, high winds, but those conditions will change and slower, inconsistent wind will inevitably arrive. The recent, average capacity factor of utility scale wind turbines in the United States is 31%.

33. The Twin Groves Wind Farm in McLean County, Illinois contains 240 turbines. Each turbine has a capacity of 1.65 MW. What is the total installed capacity of this wind farm?

396 MW Because 240 turbines x 1.65 MW = 396 MW installed capacity

34. What is the annual expected, real-world generation from the Twin Groves Wind Farm? Be sure to show your work.

1,075,378 MWh Because: 396 MW x 8,760 hours = 3,468,960 MWh 3,468,960 MWh x 0.31 = 1,075,378 MWh

35. The Duck Creek coal-fired power plant in Canton, Illinois has a total installed capacity of 425 MW. The recent, average capacity factor of coal-fired power plants in the United States is 63.8%. What is the annual expected, real-world generation from the Duck Creek power plant? Be sure to show your work.

2,375,274 MWh Because: 425 x 8,760 hours = 3,723,000 MWh 3,723,000 MWh x 0.638 = 2,375,274 MWh

36. How much more electricity will the Duck Creek coal-fired power plant generate in one year compared to the Twin Groves wind farm? Be sure to show your work.

1,299,896 MWh Because: 2,375,274 MWh - 1,075,378 MWh = 1,299,896 MWh

37. Assume that typical American homes use about 10.8 MWh of electricity in a year. How many more homes will the Duck Creek coal-fired power plant be able to supply with electricity compared to the Twin Groves wind farm? Be sure to show your work.

120,361 more homes

Because:

Twin Groves can supply electricity to 99,572 homes in one year Because: 1,075,378 MWh / 10.8 MWh = 99,572 homes

Duck Creek can supply electricity to 219,933 homes in one year Because: 2,375,274 MWh / 10.8 MWh = 219,933 homes

So… 219,933 homes - 99,572 homes = 120,361 homes

Wind farms stretch out across large land areas, but most often that land is dual-use, meaning it is used to generate electricity and grow crops. The physical infrastructure of a wind farm (access roads, turbines and substations) is much smaller than the total land area at a wind farm site. In row cropped areas, the actual physical infrastructure of a wind farm claims about 0.6 acres/MW.

38. The total land area of the Twin Grove Wind Farm site is 22,000 acres. Given the total installed capacity of this wind farm (you calculated this in question #33), how many acres does the physical infrastructure of this wind farm actually occupy? Be sure to show your work.

238 acres Because: 396 MW x 0.6 acres/MW = 238 acres

39. The Duck Creek coal-fired power plant occupies about 640 acres. How does this compare to the acres that the actual physical infrastructure of the Twin Grove Wind Farm occupies? Be sure to show your work.

The Duck Creek coal-fired power plant occupies 402 acres more land than the actual physical infrastructure of the Twin Grove Wind Farm because 640 acres – 238 acres = 402 acres.

40. Why would the people of Illinois (and other states) encourage the construction and expansion of wind farms?

Suggested answer: The people of Illinois (and other states) would encourage the construction and expansion of wind farms because of the environmental benefits of wind generation over coal-fired generation. Coal generation produces carbon dioxide emissions, ash, sulfur and nitrogen emissions, and particulate pollution. It is environmentally expensive to produce electricity with coal. Wind, by contrast, produces no pollution and is thus environmentally inexpensive, even benign. People are increasingly concerned that “cheap” energy production via coal is in fact not actually cheap when considering all the environmental externalities.

Summary of Key Terms and Concepts:  The Betz Limit is the theoretical maximum power coefficient (efficiency) of a wind turbine, and is 59.3%.  Capacity (also called “nameplate capacity”) is the maximum amount of electricity a power plant can produce if operating 100% of the time.  Capacity factor is the percent of time that a power plant operates at full capacity.  Isotachs are lines connecting places of equal wind speed.  Wind maps are graphical renditions of average wind speeds across an area such as a state or a region.  Wind speed is often categorized into “Wind Power Classes” on wind maps in order to best communicate a message. This classification scheme is often based on the power within the wind 50 meters above the ground. Wind power classes often range from “Poor” to “Superb”, and have specific wind speeds associated with each classification.  Wind power density is the power contained in the wind is calculated in Watts/m2.  Wind roses are graphic renditions of the frequency of wind blowing from particular directions at a particular site.