energy from wind. power power: rate at which energy is delivered power = energy time measured in...
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![Page 1: Energy from Wind. Power Power: Rate at which energy is delivered Power = Energy Time Measured in Watts (W), kilowatts (kW), or horsepower Power is an](https://reader036.vdocument.in/reader036/viewer/2022062716/56649e0c5503460f94af5227/html5/thumbnails/1.jpg)
Energy from Wind
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Power• Power: Rate at which energy is delivered
Power = Energy Time
• Measured in Watts (W), kilowatts (kW), or horsepower
• Power is an instantaneous quantity• Power does not accumulate• Think gallons per minute
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Energy• Energy: Ability to do something
• Measured in kilowatt Hours (kWhrs)• Why?
– Since Power = Energy/Time,
then Power Time = Energy
• Energy does accumulates over time• Think gallons• Gallons = (gallons/min) minutes
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PowerkW
(kilowatts)
EnergykWh
(kilowatt hours)
Think gpm
Think gallons
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Wind Resource• At any instant, the only question that makes
sense is “What’s the power of the wind?”• Answer depends on 2 quantities
– Instantaneous wind speed, v– Air density, , which depends on
• Elevation• Temperature• Weather• At sea level and 77F (standard conditions), air density =
1.225 kg/m3
• At 5,000 ft elevation, is ~16% less than at sea level
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Power Density of the Wind• Power Density: P/A
P/A = ½ v3 (in W/m2)
• Example: Suppose the wind speed is 8.0 m/s, and air density is 1.0 kg/m3, then
P/A = ½ (1.0 kg/m3)(8.0 m/s)3 = 256 W/m2
– For each square meter of area, there are 256 W of power– Use Metric Units!– If wind speed doubles, power density increases by 8
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Swept Area• The single most important parameter of a wind
turbine is its rotor’s swept area
A
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Power of a Wind Turbine• The power of a wind turbine is
P = ½ v3 A CP
A: swept area of rotorCP: rotor efficiency
• Example: A 2.5 m diameter turbine with a 25% efficient rotor in our 8.0 m/s wind will have
P = ½ (1.0 kg/m3)(8.0 m/s)3 [ (2.5 m/2)2](0.25)
= 314 W
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How NOT to estimate energy in the wind
• How much energy can this turbine produce? • Need a constant wind speed and time• Example: If the wind speed is a constant 8.0
m/s, then in 1 month our turbine will produce– (314 W)(30 days)(24 hrs/day) = 226 kWhrs/month– The average home in NC uses around 850
kWhrs/month• The wind speed is not constant
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10 Minute Wind Data
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0 10 20 30 40 500
2
4
6
8
Fre
qu
ency
(%
)
Probability Distribution Function
50WS HI (mph)
Actual data Best-f it Weibull distribution (k=2.04, c=15.96 mph)
Wind Speed Distributions
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Using the Annual Average Wind Speed to Calculate Energy Production is Problematic
• Using the average Annual wind speed will under estimate energy production because of the cubic relationship between wind speed and power.
• Need to cube each 10 minute wind speed• The average of the cubes is greater than the
cube of the average
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Cube of Average vs Average of Cubes for site with 6.5 m/s average annual wind speed
• Cube of the Average– Class 3 site @ 30 meters
= 6.5 m/s– P/A = .6125 x 6.53
– P/A = 168 watts/m2
Too Low
• Average of CubesP/A of 10.0 m/s = 612P/A of 5.0 m/s = 76.56P/A of 4.6 m/s = 59.6
19.6/3748.16/3
6.5 m/s 249 w/m2
Energy Pattern Factor (EPF) = Average of cubes / cube of average = 249 / 168 = 1.48
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10 minute datamph20 std dir F
10/1/2006 0:00 1.00 0.6 0 5010/1/2006 0:10 1.00 0.5 202 5010/1/2006 0:20 3.10 1 270 5010/1/2006 0:30 3.60 0.9 248 5010/1/2006 0:40 4.00 1.6 225 5110/1/2006 0:50 6.70 2.4 225 5310/1/2006 1:00 5.50 2.1 202 5410/1/2006 1:10 8.90 2.5 202 5410/1/2006 1:20 8.50 2.2 202 5510/1/2006 1:30 7.50 2.8 225 5510/1/2006 1:40 5.40 1.9 225 5510/1/2006 1:50 4.50 1.9 225 5510/1/2006 2:00 4.00 2 270 55
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Average of Cubes is Greater than Cube of Average
Time Stamp Speed (mph) m/s P/A1/1/2006 15:50 7.6 3.39 23.921/1/2006 16:00 8.2 3.66 30.051/1/2006 16:10 9.2 4.11 42.441/1/2006 16:20 10.5 4.69 63.091/1/2006 16:30 10.6 4.73 64.911/1/2006 16:40 9.8 4.38 51.291/1/2006 16:50 10.3 4.60 59.551/1/2006 17:00 10.6 4.73 64.911/1/2006 17:10 12.4 5.54 103.901/1/2006 17:20 10.9 4.87 70.571/1/2006 17:30 11.4 5.09 80.741/1/2006 17:40 12.2 5.45 98.96
average speed 4.60
P/A of Average 59.69 watts/m2Average of Cubes 62.86 watts/m2
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Energy Pattern Factor
• Average of Cubes divided by Cube of Average• 62.86 / 59.69 = 1.05• EPF = 1.05• Typical EPF = 1.9• Multiply power density calculated from
average annual wind speed by 1.9 to get more accurate average annual power density
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Estimating Average Annual Power Density from Annual Average Wind Speed
• What would be a reasonable estimate of an annual average power density when the average annual wind speed was 12 mph (5.35 m/s) and elevation was 4,000’
• Annual Average P/A = ½ Density x V3 (in meters/sec) x 1.9• AA P/A of 12 mph = ½ (1.225 x .88) x 5.353 x 1.9• AA P/A of a 12 mph wind at 4,000’ = 156 watts/m2
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Air Density Changes with Elevation
Density Change with Elevation
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
9,000
10,000
70 75 80 85 90 95 100
Density Change Compared to Sea Level, %
Ele
vati
on
, ft
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Swept Area Method of Estimating Energy Production (AEO)
• AEO = (Average annual power density x 1.9) x area of rotor (m2) x efficiency x hours/year
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Swept Area• Power is directly related to the area intercepting the
wind• Doubling the swept area will double power available
to it• Nothing tells you more about a wind turbines
potential than area swept by rotor• Area = πr2 or πd2/4• Relatively small increases in blade length produce
large increase in swept area• Doubling diameter will quadruple swept area
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Credit: Paul Gipe
Swept Area
A = Pi D2 / 4
1 m = 3.3 ft
Area = πr2
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Swept Area of Bergey XL.1
• Bergey XL.1 has three blades each 4’ long and a rotor diameter of 8.2’
• 8.2’ / 3.28 (ft/m) = 2.5 meter diameter
• Radius = 1.25 meter• Area = πr2
• Area = πr2 = π 1.252 = 4.9 m2
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Power Intercepted by Bergey XL1 with 4.9 m2 of Wind Power at 4,000’, 00, in 7 m/s wind
• Power = ½ density x area x velocity3
• Power = ½ (1.218 kg/m2) x 4.9 m2 x 73
• Power = .609 x 4.9 m2 x 73
• Power = .609 x 4.9 x 343• Power = 1,023 watts
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Estimating Annual Energy Output of XL.1 with Swept Area Method @ class 3 site; 6.5 m/s @ 5,000’
• AEO in watts = Annual Average P/A x Swept Area x efficiency x hours per year
• AEO = (1/2 air density) x (v3) x (1.9) x 4.9 x .20 x 8760
• AEO = ½ (1.225 x .860) x (6.53) x 1.9 x 4.9 x .20 x 8760
• AEO = 2,359 Kwh
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Power Curve Method or Method of Bins
2 Things Needed
Need to know (or approximate) your wind distribution
Power Curve of turbine
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Wind Distribution• Wind is known to follow a Weibull distribution
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240
500
1000
1500
2000
2500
3000
Distribution of Wind Speeds
Frequency
Wind Speed (m/s)
# o
f O
ccu
rren
ces
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Wind Distribution• Wind is known to follow a Weibull distribution • =WEIBULL(c, k, vavg)• Rayleigh Distribution if k=2
Credit: Paul Gipe
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Wind Speed Distributions
k = 2
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Pro
b.
den
sity
k = 3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Pro
b.
den
sity
k = 1.5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Pro
b.
den
sity
• Wind is empirically known to follow a Weibull probability distribution
• Weibull curve: has shape parameters: c & k• Average k in US: k = 2 (Raleigh distribution)
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Method of BinsWind Distribution: From your logger!
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Power Curve• The turbine’s manufacturer will provide you
with its power curve
Bergey XL.1
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Whisper Power Curves
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Utility Scale Power Curve (GE)
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Method of BinsPower Curve (kW)
Wind Distribution (hrs)
AEO (kWhrs)H
ours
Ener
gy (k
Whr
)
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Method of Bins• Calculate Energy = Power Time for each wind
speed bin• Sum ‘um up!
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Charts from Manufacturer