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Wind Energy

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Submitted By lkjds
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PART A - Understanding the wind resource
Q1 - Weather and climate [8 marks] a In less than 100 words, explain the difference between how climate and weather influence thewind regime of a given location.[4 marks]
Climate is the long-term description of variations of temperature and wind while weather is the short-term description. Climate is classified as the average and typical range of different variables, while weather is the state of atmosphere for day to day. Climate is what you expect, weather is what your get. A designer use climate data to design the capacity of wind farm, actually weather determines the actual power output in a given period. b Consider Fig. 1. This figure is produced by taking a long timeseries of wind speed data fora given location and analysing it to determine what timescales (or frequencies) areassociatedwith the most prominent variations in wind speed. The x-axis measures period (or frequency)and the y-axis measures \Spectral Density", which in this context just means how prominenta given timescale/frequency is in producing variations in wind speed.
In 100 words or less explain the shape of the graph. In 50 words or less, explain theimplicationsfor a wind farm operator.[4 marks]
The spectrum – frequency diagram measures the variations of wind speed. The peak of the wave indicates a significant change in wind speed over the corresponding time period. The four peaks mean: seasonal wind speed change in a year, wind speed change in few days (4), day and night wind speed change in a day, and randomly wind speed change in a minute. The one hour gap means wind speed does not vary too much in a short period.
For wind farm operator, it is noticed that wind speed change seasonally and daily. The wind speed also varies between day and night, but in a typical day of one day, wind speed is constant and reliable.

Q2 - Reading a weather map [27 marks] a In 50 words or less, describe how the Coriolis force effects the movement of the winds. Whatdirection does it act in: the Southern Hemisphere, the Northern Hemisphere?[3 marks]
The coriolis force is an apparent force due to the rotating frame of reference, which deflect the motion of fluid to the right in the northern hemisphere and to the left in the southern hemisphere. b In 50 words or less, with reference to the Navier Stokes equations, explain what“geostrophicbalance" means. What are the two major assumptions required toassume geostrophic balance?Present a mathematical expression for the geostrophic balance.[4 marks]

Assume that the velocity gradients are small as air is not accelerating, and the viscous effects are negligible, then the equation is simplified to:

Left side is pressure gradient force and right side is coriolis force which is constant depending on latitude value. c In 50 words or less, with reference to the geostrophic balance, explain the relationship betweenthe spacing of the isobars and wind speed.[2 marks]

Wind speeds are proportional to the pressure gradients. Closely spaced isobars means fast winds when widely spaced isobars means slow winds. d See Figures 2, 3, 4, and 5 which contains a series of synoptic charts (weather maps) andhas 3 locations marked on each chart. (Location 1 is near Port Augusta, SA; Location 2 isin northern Tasmania; Location 3 on the Central Tablelands, NSW) On the map, draw thevectors of pressure gradient force (PGF), Coriolis force (CF) and wind direction (U) for eachlocation for each chart.[12 marks]

e See Figures 2, 3, 4, and 5. Consider that you are a wind farm operator with find farms locatedat the marked locations. In 100 words or less each, describe theperformance of two groupingsof wind farms over this period.

i Location 1 and Location 2[2 marks]
Location 1 has moderate wind speed and the direction of wind speed does not change much.
Location 2 has higher wind speed but the direction changes ii Location 2 and Location 3[2 marks]
Location 2 has high gradient wind forces and higher wind speed.
Location 3 is near the high pressure centre and does not have high wind speed. iii Briefly compare the two groupings[2 marks]
Location 2 is the best site for wind farms.

Q3 - Basics of atmospheric physics [24 marks] a Calculate the density of the air at hub height.[6 marks]
Pz=P0exp-zH
P=ρRT
Tz=T0-∀(z-z0)
For location 1: P = 990 * exp (-1050 / 8000) = 868.2hPa T = (273 + 8) – 0.01 *50 = 280.5K ρ = 868.2 / 280.5 / 286.9 * 100 = 1.08kg/m3
For location 2: P = 1022 * exp (-60 / 9000) = 1015hPa T = (273 +25) – 0.005 * 50 = 297.8K Ρ = 1015 / 297.8 / 286.9 * 100 = 1.19kg/m3 b What is the approximate wind direction at each location? (do not mark on map)[1 marks]
Location 1: WNW
Location 2: NE c Estimate the gradient wind speed using the isobar spacing and the equation for geostrophicbalance and the data provided for the gradient height in Fig. 7. Clearly mark on the map thex selected to evaluate P.[6 marks]

By measure the lines in location 1, location2 and relative ruler, get the distance.
Location 1: x is estimated as 280km, the latitude is 35°S, actual gradient height is 300m f=2ωsin⁡(φ) v=1ρf*∆P∆x f = 1.458e-4 * sin(35) = 8.363e-5s-1
P = 990 * exp (-1300 / 8000) = 841.5hPa
T = (273 + 8) – 0.01 * 300 = 278K ρ = 841.5 / 278 / 286.9 * 100 = 1.055kg/m3 v = 1 / 1.055 / 8.363e-5 *400 / 280000 = 16.2m/s
Location 2: x is estimated as 163km, the latitude is 41°S, actual gradient height is 400m f = 1.458e-4 * sin (41) = 9.565e-5 s-1
P = 1022 *exp (-410 / 9000) = 976.5hPa
T = (273 + 25) -0.005 * 400 = 296K ρ = 976 / 296 / 286.9 * 100 = 1.150kg/m3 v = 1 / 1.150 / 9.565e-5 * 400 / 163000 = 22.3m/s d Based on your answer to part c) and using the data provided for the roughness length scalesin Fig. 7, estimate the wind speed at hub height. When selecting a roughness scale, pick amid-point value.[4 marks]
Location 1: roughness length is selected as 50mm uz=uref*lnzz0lnzrefz0 VH = 16.2 * ln (50 / 0.05) / ln (300 / 0.05) = 12.9m/s
Location 2: roughness length is selected as 600mm
VH = 22.3 * ln (50 / 0.6) / ln (400 / 0.6) = 15.2m/s e From your answer to part d) what strength is this wind speed classed as in relation to theBeaufort scale[1 marks]
Location 1: Strong breeze
Location 2: Near gale f Determine the wind power potential at the hub height (kW/m2).[1 marks]
P=1/2*ρ*V3
Location 1: P = ½ * 1.08 * 12.93 = 1159W/m2 = 1.16kW/m2
Location 2: P = ½ * 1.19 * 15.23 = 1896W/m2 = 1.90kW/m2 g In 20 words or less each, list and briefly describe three independent factors that influence thewind power potential.[3 marks]
Air density, it affects the mass of air and is constant in a given location.
Elevation, high elevation reduces the pressure and temperature of the air.
Wind speed, wind speed determines the energy transferred from the air. h In 50 words or less, how useful is this analysis in determining the relative wind power potentialof the two locations for a prospective wind farm? [2 marks]
The analysis gives the hub wind speed and wind power potential of these locations, which can use the data directly to determine the size of the wind farm and choose the turbine which best matches the data.

Q4 - Characterising the wind resource [18 marks] a In 50 words or less, explain why we need to characterise the wind resource using a long termrecord of wind data? What do you think is a sufficiently long record of wind data?[1 marks]
The wind data may vary seasonally, and there is a period for the change of global climate like the active of sun or Elnino-Lanino. A long term record can reduce the randomness of the data sample. Five year record is necessary. b In 50 words or less, explain what the parameters 'c' and 'k' represent in the Weibull distribution.[1 marks]
K determines the shape, c is the scaling factor. Large k value represents a peak in PDF diagram while the wave of low k value is flat. Large c value represents peak in large wind speed, but the peak is flat. c In 50 words or less, explain what is the value in obtaining the Weibull distribution for a givenlocation when planning a wind farm there.[3 marks]
The Weibull distribution gives the average wind speed, the range of wind speed, the most common wind speed (peak), the power output, and the variation of data. d Consider the data in excel spread sheet which represents recordings of wind speedat a fixed location. Determine the factors 'c' and 'k'. Show all relevant working.[3 marks]
By using =average (A2:A7702), get average wind speed is 4.079m/s
By using =STDEV (A2:A7702), get standard deviation is 2.107m/s k = (2.107 / 4.079) ^ -1.086 = 2.05 c = 4.079 * (0.568 + (0.433 / 2.05)) ^ (-1 / 2.05) = 4.61m/s e In 50 words or less, explain why wind data available from a weather station should not bedirectly used when predicting the wind resource for a nearby wind farm.[2 marks]
The wind data from a weather station is recorded at 10m elevation. The elevation is different from the turbine of wind farm, and the roughness factor may also different at weather station and the wind farm. The translation of data should go as a reference of free stream wind. fThe \c" and \k" parameters that you calculated in Q4-d were obtained from a bureau ofmeteorology (BoM) weather station. The data was recorded at 10 metres elevation and thenearby terrain is suburban with many buildings nearby. You are considering installing windturbines at a nearby location which is predominantly open grassland. The turbines you wishto install will have a hub-height of 100 metres. i Determine the \c" and \k" parameters that are relevant for your prospective wind farm.Show all working.[5 marks]
Use the log law and the reference of gradient height
CFS= 4.61* ln (400 / 0.6) / ln (10 / 0.6) = 10.7m/s
CZH = 10.7 * ln (100 / 0.05) / ln (300 / 0.05) = 9.35m/s
K will not change, k = 2.05 ii Estimate the ratio of wind power potential that would be predicted from Q4-a to Q4-f-i.In 50 words or less, comment.[3 marks]
As P = ½ρv3, the ratio of wind power potential should be proportional to the cubic of the ratio of wind speed, R = (CZH / C) 3 = (9.35/4.61)^3 = 8.34

Q5 - Prediction of wind farm performance [34 marks]
You are given the Weibull parameters calculated in questions PartA-Q4-d and PartA-Q4-f-i, andthe wind turbine power curve in Table 2. a Present the following graphs for both sets of Weibull parameters: iWeibull PDF (present results of both sets of Weibull parameters on one set of axes).[2 marks]

iiWeibull CDF (present results of both sets of Weibull parameters on one set of axes).[2 marks]

b For only the Weibull parameters calculated in question PartA-Q4-f-i,determine (show all relevant working): i The number of hours per year that the wind is blowing below the cut in speed.[2 marks]
Assume cut in speed is 3.5m/s, use formula =weibull.dist(3.5, 2.05, 9.35, true)*365*24, get the answer is 1094hr. (2.05 is k, 9.35 is c, true means CDF)
Enter the formula in Excel to get the answer. ii The number of hours per year that the wind is blowing between the cut in and the cutout speed.[2 marks]
Assume cut out speed is 25.5m/s, use formula = (weibull.dist (25.5, 2.05, 9.35, true)-weibull (3.5, 2.05, 9.35, true))*365*24, get the answer is 7662hr. iii The number of hours per year that the wind is blowing between the rated and the cut outspeed.[2 marks]
Assume rated speed is 13.5m/s, use formula = (weibull.dist (25.5, 2.05, 9.35, true) -weibull (13.5, 2.05, 9.35, true))*365*24, get the answer is 1044hr. iv The number of hours per year that the wind is blowing above the cut out speed.[2 marks]
Use formula = (1-weibull.dist (25.5, 2.05, 9.35, true))*365*24, get the answer is 3.52hr. c First: bin wind speeds in 1 m/s increments and calculate the probability of each binned valueof wind speed using the Weibull CDF for both sets of Weibull parameters. Use this informationto determine the number of hours per year that the wind is blowing within each bin range. Then, for both sets of Weibull parameters: i Present the velocity-duration curve (present results of both sets of Weibull parameters onone set of axes).[4 marks]

ii Calculate the average power, annual energy production, and capacity factor for a singleturbine.[4 marks]
Annual energy production: the sum of (duration*power).
Series 1 (c=4.61): 1.81MWh
Series 2 (c=9.35): 8.27MWh
Average power: annual energy production divides by time (h).
Series 1: 207W
Series 2: 945W
Capacity factor: the ratio of average power and rated power.
Series 1: 10.3%
Series 2: 47.2% iii In 50 words or less, comment on these results.[2 marks]
When c is higher, the average power and capacity factor increase significantly, capacity factor is 4.6 times larger when the ratio of c is 2. The reason for lower ratio than calculated value is the cut out speed and the limited rated power. d You are considering installing a wind farm consisting of 3 rows of 4 turbines. You mayassumethat the wind consistently blows from the same direction normal to the orientation of the rows.For only the Weibull parameters from PartA-Q4-f-i, calculate: the total annual energy, totalaverage power, and total capacity factor of the wind farm. Show all relevant working. Performyour calculations under two assumptions:
X/D = 5 (X/D is the ratio of the row spacing to the turbine diameter)
X/D = 10[10 marks]
1-UxU0=1-1-CT1+2*k*XD2
When X/D =5, right side = 0.1667
Value of c in row 2 is scaled to 9.35*(1-0.1667) = 7.79m/s
Value of c in row 3 is scaled to 7.79*(1-0.167) = 6.49m/s
Use similar method above, get
Average power at row 2: 711W
Average power at row 3: 495W
Average power of farm: (945+711+495)*4 = 8604W
Capacity factor: 8604/2000/12 = 35.9%
Annual energy of farm: 8604*365*24 = 7.537e7Wh = 75.4MWh
When X/D =10, right side = 0.07407
Value of c in row 2 is scaled to 9.35*(1-0.07407) = 8.66m/s
Value of c in row 3 is scaled to 8.66*(1-0.07407) = 8.02m/s
Use similar method above, get
Average power at row 2: 846W
Average power at row 3: 748W
Average power of the farm: (945+846+748)*4 = 10156W
Capacity factor: 10156/2000/12 = 42.3%
Annual energy from farm: 10156*365*24 = 8.897e7Wh = 8.90MWh
If you are planning a wind farm project, there are many ways to evaluate and select a windturbine from a large range of possibilities.
In 50 words or less, explain why comparing annual energy production and capacity factor mayproduce different results. Which do you think is a more useful metric?[2 marks]
It is different due to different selections of wind turbines which have different rated power. Capacity strongly depends on rated power of turbine/farms when annual energy production may be similar due to the power-wind speed curve of the selected turbines.

Part B - Basic fluid mechanic principles of wind turbines without wakerotation
Q1 - Momentum analysis [18 marks] a What is the axial induction factor? Give an expression and explain in one sentence.[2 marks] a=v1-v2v1 Axial induction factor is the fractional decrease in wind velocity between the free stream and the rotor plane. b Derive an expression for the power generated by a horizontal axis wind turbine only in termsof: the far upstream wind speed U, the axial induction factor a, the rotor cross sectional areaA, and the air density . Clearly state all assumptions.[6 marks]

Assume that:
The static pressure in and out of the slipstream far ahead of and behind the rotor is equal to the free stream pressure p1 = p4
Thrust loading is uniform over the rotor disk.
Fluid velocity remains constant throughout actuator disk, U2 = U3
No rotation is imparted to the flow by disk the flow behind the rotor was assumed to be non-rotating.
Flow is homogenous, incompressible and steady state.
The power is equal to thrust force T times the velocity at disk U2
T=A*(p2-p3)
By applying Bernoulli equation p1 + ½ρ U12 = p2 + ½ρ U22 p3 + ½ρ U32 = p4 + ½ρ U42
And recall that a=(U1-U2)/U1=(U3-U4)/U3, U4= (1-2a)*U1 p2-p3 = ½ρ U1 (4a-4a2)
T = ½ρ A U2 4a (1-a)
P = T * U2 = ½ρ A U3 4a (1-a)2 c Based on your previous answer, what is the expression for the differential force (or thrust, dT)generated by an annular section of the rotor?[2 marks]
T = ½ρ A U2 4a (1-a) d Derive an expression for the efficiency, , of the horizontal axis wind turbine only in terms of:a (the axial induction factor).[2 marks]
HAWT power divides by wind power ½ρ A U3, get the answer is 4a*(1-a)2 e Determine the value of axial induction factor which maximises the efficiency of the horizontalaxis wind turbine.[3 marks]
Differentiate 4a*(1-a)2, get 12a2 - 16a + 4, let it equals to 0 and get a= 1/3 f What is the maximum possible efficiency (i.e. the Betz limit)? List three assumptions requiredfor this analysis to be true.[3 marks]
When a= 1/3, efficiency η=59.26%.
No friction loss. Pressures at far upstream and downstream are the same. Homogenous, incompressible and steady state wind flow.

Q2 - Airfoils [41 marks] a Draw a labelled diagram of a blade element, containing the following information:

b In 20 words or less each, define the meaning of the following angles: , p;0, T, p, and .[5 marks] φ: the angle of relative velocity and the plane of rotation θp0: the blade pitch angle at tip θT: the blade twist angle θp: The section pitch angle which is the angle between the chord line and the pane ofrotation α: The angle of attack which is the angle between the chord line and relative velocity c Express a relationship between: iUrel, U, a, and [1 marks]
Urel=U(1-a)sinφ
ii, a, and r (the ratio of the blade element speed to the wind speed, related to the bladetip speed by r = rR)[1 marks] tanφ=U(1-a)Ωr(1+a')=1-aλr, as a’=0 due to no wake rotation d What angle is relevant for determining the lift and drag coefficients? In 20 words or less explainwhy.[2 marks]
Angle α, the angle of attack, is between lift force/normal force and drag force/tangential force. e What angle is relevant for determining the resulting thrust force dFn and dFt? In 20 words orless explain why.[2 marks]
Angle φ, relative wind angle, which is the angle measured from rotation plane (force exerted on) and wind velocity. f Now imagine that you have B identical blades on a wind turbine. Derive expressions fordFN and dFT in terms of only: ,Urel, B, Cl, Cd, , c, r, B, and dr. [6 marks]
As φ is the angle between thrust force and lift/drag force, then: dFN=dFLcosφ+dFDsinφ dFT=dFLsinφ-dFDcosφ
Where dF_L and dF_D are lift force and drag force: dFL=12ρCLUrel2cdr dFD=12ρCDUrel2cdr
The turbine has B blades, and integrates two equations: dFN=12BρUrel2CLcosφ+CDsinφcdr dFT=12BρUrel2CLsinφ-CDcosφcdr g In 100 words or less, explain why turbine blades are twisted (diagrams may be useful here).Isthe maximum twist applied at the hub or at the tip of the blade?[4 marks]
Rotor blades for large wind turbines are always twisted. Seen from the rotor blade, the windwill be coming from a much steeper angle (more from the general wind direction in the landscape), as you move towards the root of the blade, and the centre of the rotor.
A rotor blade will stop giving lift, if the blade is hit at an angle of attack which is too steep.Therefore, the rotor blade has to be twisted, so as to achieve an optimal angle of attackthroughout the length of the blade.
Maximum twist is applied at the tip of the blade. h Consider a single-blade wind turbine (B = 1) which is a type NACA-23012 airfoil, with liftand drag coefficients provided in Fig. 9. iUrel,[4 marks]
Tip speed ratio is 2*Pi*r/t/U = 2*Pi*5/3/7 = 1.496 tanφ=U(1-a)Ωr(1+a')=1-aλr φ = arctan((1-1/3)/1.496)=24°
Urel=U(1-a)sinφ
Urel = 7*(1-1/3)/sin(24) = 11.5 m/s ii α[2 marks] α = φ–φT–φp0 = 24-4-2 = 18° iiidFD and dFL,[2 marks]
From the graph, CL is 1.75 and CD is 0.021 dFL=12ρCLUrel2cdr dFL = ½ *1.25*1.75*11.52*1*0.2 = 28.9N dFD=12ρCDUrel2cdr dFD = ½ *1.25*0.021*11.52*1*0.2 = 0.347N ivdFN, dFT, dQ, and dP (power).[5 marks] dFN=12BρUrel2CLcosφ+CDsinφcdr = 26.6N dFT=12BρUrel2CLsinφ-CDcosφcdr = 11.4N dQ=12BρUrel2CLsinφ-CDcosφcrdr = 57.2Nm dP = Ω*dQ = 57.2*2*Pi/3 = 120W

Q3 - Blade element theory [25 marks]

λr=rR*λ tanφ=U(1-a)Ωr(1+a')=1-aλr Urel=U(1-a)sinφ c=8*Pi*r*sinφ3*B*CL*λr dFN=12BρUrel2CLcosφ+CDsinφcdr dFT=12BρUrel2CLsinφ-CDcosφcdr dQ=12BρUrel2CLsinφ-CDcosφcrdr dP = Ω*dQ
Assume α(angle of attack) is ideal at 8° which has highest C_L/C_D ratio. b From the results in part a), what is the total power generated by the turbine? Does thisnumber seem realistic? Justify your answer in 50 words or less. [5 marks]
8.1kW
It is not real as assumptions like: no wake rotation, ideal attack of angle, no twist on the tip of the blade and operate at Betz limit. These assume higher ideal power.

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...would happen to our world if our energy unexpectedly disappeared? Fossil fuels are very important to our world because they provide an overwhelming majority of our world energy. So, what happens when they run out? As the world enters a new century, new source of energy such as wind power need to be used more to help the world keep running on all cylinders. According to the American Wind Energy Association, “Wind power captures the natural wind in our atmosphere and converts it into mechanical energy then electricity. People started using wind power centuries ago with windmills, which pumped water, ground grain, and did other work. Today's wind turbine is a highly evolved version of a windmill. Modern wind turbines harness wind's kinetic energy and convert it into electricity.” Although wind turbines are always reliable, we should invest is wind energy because it creates safer jobs, it does not pollute, and will get more than your investment back. If you invest properly in wind power today and take advantage of any grants and tax incentives, even if your electric rates do not go up as they are forecast too, you will get your money back over time, well within the lifetime of the equipment, and sooner if there is a rate increase in the future. There are also environmental benefits. At one time there was an argument that a wind turbine will never produce as much power as was used to manufacture it. First of all, this is not correct. The, "Embodied Energy," in a turbine is earned back...

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Wind Power as a Source of Energy

...other sources of energy in the near future, like wind power or solar power. These changes would also help reduce the amounts of greenhouse (A gas in the atmosphere that absorbs and emits radiation within the thermal infrared range) gasses being emitted into the air, which would help slow down the effects of global warming. Wind is air in motion (Casper), and wind power is a large percent of current energy produced. It can help reduced emitted greenhouse gasses and slow down the effects of global warming. The global wind energy potential is roughly five times current global electricity use (Flavin). Also the price of wind generated energy fell more steeply than any other energy source over the past 30 years (Burnett). The cost of wind power fell from approximately 25 cents per kilowatt per hour (kwh) in the early 1980s to between 5 cents and 7 cents per kwh (Burnett). Advocates argue that a new generation of turbines will bring down the cost to below 5 cents per kwh (Burnett). This would make it much easier for people to be able to afford it. Wind power could provide as much as 6 percent of the nation’s energy (Burnett). It is currently providing less than 1% of the U.S. power supply but could double its shares within 10 years (Burnett). Furthermore, wind power plants receive accelerated depreciation (Burnett). As of 2006, wind power was the cheapest form of large scale electric generation to install (Lerner). This led to the growth of wind power plants. Wind power is the earliest...

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How Green Is Wind Energy

...able to gather for energy purposes. Some are more eco-friendlier than others. Man has learned that its next major energy source needs to be safe for the environment causing minimal harm to the environment. Everyone is becoming environment-friendly and is looking for options that will help in the upkeep of the environment. Many are interested in the source of renewable energy so that natural resources are not diminished. There are many alternatives in the market to choose from. Wind turbines are one of alternatives (Linzmeier, 2009). Now the question that comes to mind is, is wind energy green? Wind turbines produce cleaner energy & bring no hazardous effects to the atmosphere by giving any pollutant emissions; however, it does come with casualties. As the U.S. energy industry inches away from dependence on fossil fuels, experts have announced various greener technologies, such as ethanol, solar and geothermal power, as choice alternatives. And while each of those options holds recognizable environmental gains over oil and coal, none compare to wind power’s squeaky clean performance (Conger, 2010). Mark Jacobson an engineer at Stanford University states that after compiling evidence of major energy technologies and environmental impact, wind comes out on top. Jacobson and associates from University of California drafted a blueprint for switching 100 percent of the world’s energy to renewable sources. Wind energy was in the forefront...

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...Research Topic: Wind Energy and Sense of Place Feeling comfortable in different cities and communities may mean different things to different people. There are many reasons someone may feel uncomfortable in a certain location and through designing a research topic that helps to look at how windmills can affect someone’s sense of place. Through wind energy and sense of place I want to look at how wind energy can affect how people feel within a community. Sense of place can be affected by how someone feels that their health may be at risk, noise pollution, or other people in their life’s opinions on certain topics. With this research topic I also want to investigate if the way windmills appear to a person will change how they feel near them....

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...Wind Power as an Alternative Energy Source Although fossil fuels drive our economy, it is rather expensive and significantly contributes to pollution (Lad 2). A few alternatives include hydroelectricity, solar energy, and wind. Solar and hydroelectric energy are good choices, but can be relatively costly (“Top 5 Misconceptions of Solar Energy” 5), (“Hydro Electricity”). Wind, on the other hand, is free and available in most areas. However, wind power has not been fully utilized in the United States. The purpose of this paper is to give an overview of wind power in order to bring to light its usefulness as an alternative energy solution. Uses Wind power can be used to generate electricity, charge batteries, grind grain, etc. (Deshmukh 3). Turbines can be used alone or on a wind farm and can also be connected to a utility power grid to sell electricity back to the power company. According to the literature, wind technologies are most useful in the Midwest and along the coast. Pros and Cons Although wind turbines can kill birds and be damaged in severe storms (Deshmukh 11), wind turbines can provide many advantages. For example, wind turbines can harness the power of the wind, producing electricity, charging batteries, etc. Unlike fossil fuels, wind power consumes no fuel, and emits no air pollution. In addition, wind can be accessible in nearly all areas. Types There are several types of wind turbines, each having its own set of advantages and disadvantages...

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Case Study on the Danish Wind Energy System

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