turbine hw1

8/7/2019 turbine hw1

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Solution to problem 1

a. Let us first find the available biggest wind turbine. By searching internet, it was found thatcurrently the biggest wind turbine under operation is Emercon E- 126 model[1]. This model has

a capacity of 7 MWand its rotor diameter is 126 m.

For calculations of total land area required we will use a 4 by 10 spacing [2], which means the

spacing between two rows is 10 times rotor diameter and the spacing between two turbines is a

row is three times the rotor diameter.

No of turbines required = 830 MW/ 7 MW ~= 120

So in a single row there will be 60 turbines.

Row spacing = 10* rotor dia = 10* 126 m ~= 1.25 km

Total space required for each row ~= 60*(4*rotor dia) = 240*126 m ~= 30.25 km

Land required = 1.25*30.25 sq km = 37.8 sq km ~= 9,500 acres

b. In this case the power output from each turbine will be = 0.6*7 MW = 4.2 MWHence number of turbine required will be = 830 MW/4.2 MW = 198

So in a single row there will be 99 turbines

Row spacing = 10* rotor dia = 10* 126 m ~= 1.25 km

Total space required for each row ~= 99*(4*rotor dia) = 396*126 m ~= 49.9 km

Land required = 1.25*49.9 sq km 62.375 sq km ~= 9,500 acres ~= 15,500 acres

c. From available data, it is found that the maximum efficiency of commercial solar panels isaround 15% (module: HIP- 200 BA 19/20, manufactured by Saniyo Electric) [3]. Also the

intensity of solar radiation in Orlando, FL is 0.22 kW/m2

[4].

So using this data the land area required for 830 MW of power would be

= 830 x 10^6/(0.15 x0.22 x 10^3)

~= 6250 acres


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References

[1]. http://www.metaefficient.com/news/new-record-worlds-largest-wind-turbine-7-

megawatts.html

[2]. http://www.powernaturally.org/programs/wind/toolkit/13_windpowerproject.pdf

[3]. http://sroeco.com/solar/most-efficient-solar-panels

[4]. http://www.solarpanelsplus.com/solar-insolation-levels/


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2006 energy consumption in United States based on different sources

2030 projection of energy consumption in United States based on different sources

Reference:

http://www.worldenergyoutlook.org/docs/weo2008/WEO2008.pdf

coal

gas

oil

nuclear

hydro

bio

solar

wind

coal

gas

oil

nuclear

hydro

bio

solar

wind


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Here we do the problem based on the data in problem 2. Hence we need to make some

assumptions, while some of the conclusions are quite obvious from problem 2. For example, as

we know the fuel coal cannot be used in any kind of reciprocating engines. Hence all the

portions of this fuel must go to the turbomachinery part, for example steam turbine where theboiler produces superheated steam by burning coal. For oil and gas part we assume that they

are equally used both in turbomachinery and in other engines. Hence the obtained pie-charts

are presented below.

2006 energy consumption in United States based on different conversion mechanism

turbine

genset

photovoltaic

electromechanical


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2030 projection of energy consumption in US based on different conversion mechanism

turbine

genset

photovoltaic

electromechanical


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From references [1-7] the average wind speed along I-10 is taken as 10.0 mph. Given the

average wind speed and the space of the median in the high ways, a good model as found by

searching on internet is Southeast Air companys Air X Model [8]. This models rotor diameter is

1.15 m. The performance curve for the model as found in [9] is shown below.

So for 10 mph average wind speed the monthly output from one turbine is nearly 20 kWh.

Hence the annual output would be around 240 kWh.

Now the total no of turbines along I-10 with a spacing of 8 rotor dia would be

= total length of I- 10/ spacing

= 3959 x 10^3/(8 x 1.15)

= 430326.087


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So the total annual output from all the turbines = 430326 x 240 kW-hr

~= 103.25 GW-hr

References

[1]. http://www.windpoweringamerica.gov/wind_maps.asp

[2]. http://www.windpoweringamerica.gov/pdfs/wind_maps/us_windmap.pdf



[5]. http://www.weatherimages.org/data/imag123.html

[6]. http://en.wikipedia.org/wiki/Interstate_10

[7]. http://www.sercc.com/climateinfo/historical/avgwind.html

[8]. http://www.ecobusinesslinks.com/small_wind_turbines_small_wind_generators_gallery.htm

[9]. http://www.windenergy.com/documents/spec_sheets/0083_rev_c_air-x_single.pdf


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As stated in the solution of problem-1, from available data, it is found that the maximum

efficiency of commercial solar panels is around 15% (module: HIP- 200 BA 19/20, manufactured

by Saniyo Electric) [1]. Also the intensity of solar radiation in Orlando, FL is 0.22 kW/m2[2].

Now,

The energy from 1 m2 area per year = ( 0.15 x 0.22 ) KW x (365 x 24) hrs

= 289 KWh

Now,

50,000 sq ft = 50,000 x (12/39.37) x (12/39.37) sq meter = 4645.171 sq m

Therefore the total energy output from an area of 5000 sq ft per year would be

= 289 x 4645.171 KW hr

= 1342454.419 KW hr

References




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The curve is plotted for different values of the shape parameter while the two other parameters are

fixed.

Now, the curve is plotted for different values of the scale parameter while the two other parameters are

fixed.

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0 100 200 300 400

beta = 0.75

beta = 1.5

beta = 2

beta = 0.5

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

0 100 200 300 400

eta = 100

eta = 90

eta = 75

t

(t)

t

(t)


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Finally, the curve is plotted for different values of the time parameter while the two other parameters

are fixed.

-0.002

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0 50 100 150 200 250 300 350

t0 = 10

t0 = 20

t0 = 40

t

(t)

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turbine hw1