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 installed by 2008

Drag Wind Power

Earliest type of wind power Velocity limited to wind speed

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

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

Lift-type Wind Turbines

Wind is parallel to axis of rotation and perpendicular to motion of blades

v

crosssection

u

c

Velocity Triangle

v

u

vr = u - v

β θ

α

v/u = tanβ

8

9

10

11

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

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

At any radius, with area dAp

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

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

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

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

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

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)

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

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

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

Useful approx for Cp max

Function of B=no of blades, D/L, and tip-speed ratio λ at design wind speed

24

25

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)

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

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

Meteorological Data

Performance Data39 m Diameter, 500 kW Nameplate Capacity Turbine

Electrical Output