Post on 21-Jan-2016
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Wind turbine design according to Betz and Schmitz
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Energy and power from the wind
• Power output from wind turbines:
• Energy production from wind turbines:
3
p
vPower A c
2
Energy Power Time
v
A
3
Stream Tube
V
4
Extracted Energy and Power
2321ex
23
21ex
vvm2
1E
vvm2
1E
Where:Eex = Extracted Energy [J]Eex = Extracted Power [W]m = Mass [kg]m = Mass flow rate [kg/s]v = Velocity [m/s]
•
•
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Extracted Energy and Power
• If the wind was not retarded, no power would be extracted
• If the retardation stops the mass flow rate, no power would be extracted
• There must be a value of v3 for a maximum power extraction
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Extracted Energy and Power
2 2p p p
• The retardation of the wind cause a pressure difference over the wind turbine
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We assume the following:• There is a higher pressure right upstream the turbine (p-2) than the
surrounding atmospheric pressure
• There is a lower pressure right downstream the turbine (p+2) than the surrounding atmospheric pressure
• Since the velocity is theoretically the same both upstream and downstream the turbine, the energy potential lies in the differential pressure.
• The cross sections 1 and 3 are so far away from the turbine that the pressures are the same
A1
A2
A3
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Continuity (We assume incompressible flow)
332211 AvAvAv
A1
A2
A3
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1 1 1 2 2 2 3 3 3F v A v (p p ) A v A v
Impulse force Impulse forcePressure force
Balance of forces: (Newton's 2. law)
Because of the differential pressure over the turbine, it is now a force F = (p-2 – p+2)∙A2 acting on the swept area of the turbine.
A1
A2
A3
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Energy flux over the wind turbine:(We assume incompressible flow)
2231
1 1 1 1 1 3 3 3 3 3
2231
1 1 3 3
vvE v A p v A v A p v A
2 2
vvE v A v A
2 2
A1
A2
A3
1 1 2 2 3 3
1 3
v A v A v A
p p
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Energy flux over the wind turbine:(We assume incompressible flow)
A1
A2
A3
2 22 2
2 2 2 2 2 2 2 2 2 2
2 2 2 2
v vE v A p v A v A p v A
2 2
E v A p p
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Energy flux over the wind turbine:(We assume incompressible flow)
A1
A2
A3
2231
2 2 2 2 1 1 3 3
vvE v A p p v A v A
2 2
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Continuity:
Balance of forces:
Energy flux:
332211 AvAvAv
1 1 1 2 2 2 3 3 3v A v (p p ) A v A v
2231
2 2 2 2 1 1 3 3
vvv A p p v A v A
2 2
If we substitute the pressure term; (p-2-p+2) from the equation for the balance of forces in to the equation for the energy flux, and at the same time use the continuity equation to change the area terms; A1 and A3 with A2 i we can find an equation for the velocity v2:
222 231
1 1 3 3 2 1 1 3 3
22231
2 2 2 2 2 2 1 3
2 2 1 31 3 2 1 3 1 3 1 3 2
vvv A v A v v A v A
2 2
vvv A v A v A v v
2 2
v v1 1v v v v v v v v v v
2 2 2
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Power CoefficientRankine-Froude theorem
We define the Power Coefficient:
2231
1 1 3 3
p 21
1
vvv A v A
2 2cvv A
2
3 1v x v
In the following, we assume that the velocity v3 can be expressed as v3=x·v1, where x is a constant.
We substitute:
From continuity:
1 3 1 32 2 2 21 2 1
1 1 1
1 3 1 32 2 2 2 23 2 3
3 3 3
v v v vA A A AA v A 1 x
v 2 v 2 v 2
v v v vA A A A A1 1 xA v A 1
v 2 v 2 v 2 x 2 x
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We insert the expressions for A1 and A3 in to the equation for the power coefficient.We will end up with the following equation:
Power CoefficientRankine-Froude theorem
2231
1 1 3 3
p 21
1 2
2 22 21 31 1 1 3 3 3 1 1
p 2 21 1 2 1 1 2
22231 1 2
p 2 21 1 2 2
2 3p
vvv A v A
2 2cvv A
2
v vv v A v v A v Ac
v v A v v A
vv A A1c 1 x 1 x
v v A A 2
1c 1 x x x
2
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Maximum power coefficient:
2p
3 x 2 x 1c0
x 2
1x 1 x 3
3 2
p
x x x 1c
2
Maximum Power CoefficientRankine-Froude theorem
3 2
pmax
1 1 1 13 3 3 32c 0,59
2 54
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Power Coefficient
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The Betz Power
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Thrust
22 2 23 1 2 1 3 2 1 1
2 2 22 1 2 1 T
1 1T m v v A v v A v x v
2 2
1 1T A v 1 x A v c
2 2
v2 T
At maximum power coefficient we have the relation: x =1/3
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8
9
11x1c 2
T
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Example
22 2
2 1 T 1 T
22
1 DT A v c v c
2 2 4
1,2 100 8T 20 1676,5 kN
2 4 9
Find the thrust on a wind turbine with the following specifications:
v1 = 20 m/sD = 100 mcT = 8/9