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![Page 1: Wind Power Energy Sources Fall 2014. Wind Potential Wind energy is the most abundant renewable energy source after solar 120 GW of peak world capacity.](https://reader035.fdocuments.net/reader035/viewer/2022070410/56649eb65503460f94bbf5c9/html5/thumbnails/1.jpg)
Wind Power
Energy Sources
Fall 2014
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Wind Potential
Wind energy is the most abundant renewable energy source after solar
120 GW of peak world capacity installed by 2008
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Drag Wind Power
Earliest type of wind power Velocity limited to wind speed
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Drag Power Mechanics
Force on surface equal to drag D = CD[½ρ(u – v)2cℓ] Power = Drag · Velocity P = Dv
= CD[½ρ(u – v)2cℓ]v= ½ρu3CD(1 – v/u)2cℓ(v/u)= ½ρu3APCD(1 – v/u)2(v/u)
where AP = cℓ is the cross-sectional area
u
v
Dc
ℓ
CD = 1.4 for this shape
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Drag Power Limitations
½ρu3APCD(1 – v/u)2(v/u)At v = 0, power = 0 (from v/u)At v = u, power = 0 (from 1 – v/u)v cannot exceed u
Maximum power at v/u = 1/3Greatest coefficient of power CP = 4/27CD
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Lift-type Wind Turbines
Wind is parallel to axis of rotation and perpendicular to motion of blades
v
crosssection
u
c
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Velocity Triangle
v
u
vr = u - v
β θ
α
v/u = tanβ
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v
β
Force Triangle
L = c·ℓ·½ρvr2·CL
D = c·ℓ·½ρvr2·CD
ℓ is a length in the radial direction
F = L·sinβ – D·cosβ net force in direction of
blade motion
lift, L
drag, D
net force
β
useful force, F
u
r
vr n = 2
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v
β
Power
P = F·v = (L·sinβ – D·cosβ)v = c·ℓ·½ρ · vr
2 · v · (CL·sinβ – CD·cosβ)
= AP·½ρ(vr2/u2)(v/u)·u3·sinβ(CL – CD/tanβ)
= ½ρAP·(1/sin2β)(v/u)u3sinβ(CL – CD/tanβ) Since sinβ = u/vr and tanβ=u/v
lift, L
drag, D
net force
β
useful force, F
u
multiply by u3·u-3
vr = u - v
DL
P
Cu
vC
u
v
u
v
Au
P2
31
21
only a function of v and u
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At any radius, with area dAp
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If (v/u)2 » 1, the “1” in the square root is insignificant and the equation becomes
CP, the dimensionless Coefficient of Power, is the LHS of the equation
Coefficient of Power
DL
P
Cu
vC
u
v
u
v
Au
P2
31
21
DLP C
u
vC
u
vC
2
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Max CP occurs at
The maximum value is
v
uCC
C
C
u
vLD
D
L
3
2
3
2
DLP C
u
vC
u
vC
2
v
u
u
vC
C
CC L
D
LP 3
21
3
22
2
2
2
27
4
3
1
9
4
D
LLP
L
D
LP
C
CCC
CC
CC
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Calculating Power (approx.)
AP, the “plan form” area, is the total solid blade area, n·R·Cave where Cave is computed to ensure that actual blade area seen from the axial direction is equal to R·Cave
A is the swept area
R
Cave
r
dℓ
n = 3
rmin
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Torque
Recall: F = L·sinβ – D·cosβ, u/v = tanβ As radius ↑, v ↑
Torque = Force × radius r Torque = Force × (r/R) × R Both Force and Radius would increase with r
Two reasons why T ↑ as r ↑
We want torque to stay the same over entire blade to minimize internal bending moment
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Keeping Torque Uniform
We can change θ, blade tilt (but it cannot be negative) c, chord (blade width)
only a function of r (and u but wind speed is uniform over blade)
We are assuming that blade shape drag and lift behavior is same.
Rpm is given
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Example
A 20-m radius blade has a coefficient of lift of 1.2 when α = 12º and decreases linearly to CL = 0.4 when the angle of attack is 0. The coefficient of drag is 0.1 over the whole blade, and the stall angle is 12º. The blades begin at radius = 4 m. The wind speed is 7 m/s. What is the rotor speed if the “speed ratio” (blade speed
over wind speed) remains under the max CP value? What is the optimum pitch angle at various radii for the
greatest extraction of power from the wind? Determine the chord at the tip for uniform torque on the
blade (chord at base is 1 m)
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Solution (speed)
Blade velocity is greatest at tip Max CP occurs at
The blade tip moves (relative to ground) at 56 m/s This is 56/20 = 2.8 rad/sec or 26.7 RPM
s
m
C
Cuv
C
C
u
v
D
L
D
L 561.
2.1
3
27
3
2
3
2
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Solution (table)
radius v/u β α θ CL CL/CD X KT/cm º º º N·m/m
20 8 7 7 0.0 .88 0.11 0.6 0.60
10 4 14 12 2.0 1.2 0.08 3.3 1.65
5 2 27 12 14.6 1.2 0.08 2.2 0.6
2 0.8 51 12 39.3 1.2 0.08 1.4 0.143
First, pick a bunch of radii from the base to the tipThen, calculate u/v for each; u is constant and v is angular velocity × radius
u/v = tanβ
βv
u
Now find β, the angle between blade velocity relative and blade velocityThe angle of attack is the smallest of β or the stall angleThe pitch, the angle from the blade profile to blade velocity, is α – β
The lift coefficient scales from 0.4 to 1.2 as α goes from 0º to 12º; CD remains constant
L
DL C
C
u
vC
u
vX 11
2
The specific torque is proportional to (r/R)·X, so torque (per chord length) times an arbitrary constant K, will equal that expression
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Solution (chord length)
The specific torque at the blade tip is 1.43/.60 = 4.2 times that at the base
Therefore, the chord at the base must be 4.2 times that at the tip
Since the base chord is 1 m, the tip chord is 1 ÷ 4.2 = .24 m
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Useful approx for Cp max
Function of B=no of blades, D/L, and tip-speed ratio λ at design wind speed
24
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25
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Wind Characteristics
The relative wind speed (u/c) at a given time follows the Rayleigh distribution
c is scale parameter Shape factor (k)
describes variance k = 2 common for wind
kk
c
u
c
u
c
ku exp)(
1
Φ(u)
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Rayleigh Averages
Mean wind speed ū = cΓ(1 + 1/k) Γ is “gamma function” Γ(1.5) = 0.89
But power is proportional to the cube of wind speed
Average of the cubes ≠ the cube of the averages
k
ncu nn 1
33.12
31 333 ccu
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Power Averages
Power Density [W/m²], ū = c·0.89 → c = ū/0.89
So if the average wind speed ↑ by 15%, the annual energy ↑ by 50% 1.153 ≈ 1.5
33.12
1
2
1 33 cuE
886.133.189.0
33
u
u 21
21
E
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Meteorological Data
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Performance Data39 m Diameter, 500 kW Nameplate Capacity Turbine
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Electrical Output