Wind energy

Assumptions

Ideal rotor as an actuator disc, steady, incompressible, frictionless flow, uniform velocity over the disc and no rotation of the wake. Let $v_1$ be the upstream (free) wind speed, $v$ the speed through the disc of area $A$, and $v_2$ the speed far downstream.

  v1 ->  \   |  /  -> v2 (slow)
 --------\--|--/--------
 ->      | disc |  v
 --------/--|--\--------

Power from momentum

Mass flow through the disc: $\dot m = \rho A v$.

Force on the disc (change of momentum): $F = \dot m (v_1 - v_2)$.

Power extracted: $P = Fv = \rho A v^2 (v_1 - v_2)$.

Power also equals the loss of kinetic energy of the stream:

$$P = \tfrac{1}{2}\dot m (v_1^2 - v_2^2) = \tfrac{1}{2}\rho A v (v_1 - v_2)(v_1 + v_2)$$

Equating the two expressions for $P$:

$$v = \frac{v_1 + v_2}{2}$$

So the wind speed at the disc is the mean of the upstream and downstream speeds.

Power coefficient

Substituting $v$:

$$P = \tfrac{1}{4}\rho A (v_1 + v_2)(v_1^2 - v_2^2)$$

Let $a = v_2/v_1$. Then

$$P = \tfrac{1}{2}\rho A v_1^3 \cdot \tfrac{1}{2}(1+a)(1-a^2)$$

The available power in the wind is $P_0 = \tfrac{1}{2}\rho A v_1^3$, so

$$C_p = \frac{P}{P_0} = \tfrac{1}{2}(1+a)(1-a^2)$$

Maximum

$$\frac{dC_p}{da} = \tfrac{1}{2}\left(1 - 2a - 3a^2\right) = 0 \Rightarrow 3a^2 + 2a - 1 = 0$$

giving $a = \tfrac{1}{3}$ (the other root is $-1$, not physical). Then

$$C_{p,max} = \tfrac{1}{2}\left(\tfrac{4}{3}\right)\left(\tfrac{8}{9}\right) = \frac{16}{27} = 0.593$$

Meaning

No wind turbine can capture more than 59.3 % of the kinetic energy of the wind passing through its swept area, because the air must keep moving behind the rotor. Practical large turbines reach $C_p$ of 0.40–0.50 due to blade drag, tip losses and wake rotation; the maximum for a Savonius rotor is about 0.15–0.20.

Tip speed ratio is $\lambda = \dfrac{\omega R}{v}$, where $\omega R$ is the blade tip speed. The curve of $C_p$ against $\lambda$ shows how a rotor type behaves.

HAWT versus VAWT

Types of VAWT

• Savonius: two or three half-cylinder cups; a drag device; high starting torque, low speed, $\lambda \approx 1$, $C_p \approx 0.15$–0.2; used for water pumping and ventilation.
• Darrieus: curved or straight blades (H-rotor) with aerofoil section; lift device; $\lambda \approx 4$–6, $C_p \approx 0.35$–0.40; needs starting help.

$C_p$-$\lambda$ characteristics

 Cp
 0.5|          ___ 3-blade HAWT
    |        /     \
 0.4|      /  _.-Darrieus
    |     / /      \
 0.2|  Savonius       \
    | _/_.-'''-.       
    +---------------------- lambda
     1   2   4   6   8

• Multi-bladed American windmill: peak at $\lambda \approx 1$, high torque, low $C_p$ (about 0.3) and good for pumping.
• Three-blade HAWT: peak $C_p$ at $\lambda \approx 6$–8, high speed, suitable for electricity.
• Each rotor has one optimum $\lambda$; away from it $C_p$ drops, so variable-speed operation keeps $\lambda$ near the optimum.

A wind farm is a group of grid-connected wind turbines installed at one site to produce bulk electric power, sharing roads, substation and transmission line.

Site selection factors

1. Wind resource: annual mean speed above about 6 m/s at hub height from at least one year of measurement; wind rose, Weibull distribution and low turbulence.
2. Terrain: open land, ridges, hill tops or coast; avoid trees and buildings that cause turbulence and speed-down.
3. Grid access: nearness to a substation or line of enough capacity.
4. Access and ground: roads for heavy transport, firm soil for foundations.
5. Environment and society: noise (distance from houses), bird and bat routes, visual effect, shadow flicker, land cost and acceptance.
6. Extreme conditions: lightning, icing, high winds, earthquakes.

Layout

Turbines are placed in rows at right angles to the prevailing wind. Typical spacing is 3–5 rotor diameters (D) between turbines crosswise and 5–10 D along the wind.

 wind -->   o     o     o     o   (row 1)
 
              o     o     o     o (row 2, staggered)
         <--- 5-10 D --->

Wake effect

Air behind a rotor has lower speed and higher turbulence. A turbine inside this wake produces less power and suffers higher fatigue loads. Power loss from wake is commonly 5–15 % of the farm output.

Reduction: adequate spacing, staggering of rows, laying rows across the dominant wind direction, use of wake models (e.g. Jensen) to optimise the layout, taller towers and control strategies that yaw turbines slightly off the wind.

(a) Powers

Swept area $A = \dfrac{\pi D^2}{4} = \dfrac{\pi (40)^2}{4} = 1256.6\ \text{m}^2$.

$$\begin{aligned} P_w &= \tfrac{1}{2}\rho A v^3 = 0.5(1.225)(1256.6)(12)^3 \\ &= 1.330\times10^{6}\ \text{W} = 1330\ \text{kW} \\ P_m &= C_p P_w = 0.40 \times 1330 = 532\ \text{kW} \\ P_e &= 0.90 \times 532 = 479\ \text{kW} \end{aligned}$$

(b) Tip speed ratio

$$\omega = \frac{2\pi N}{60} = \frac{2\pi(20)}{60} = 2.094\ \text{rad/s}$$ $$\lambda = \frac{\omega R}{v} = \frac{2.094 \times 20}{12} = 3.49$$

(c) Wind shear

$$\frac{v_h}{v_{10}} = \left(\frac{h}{10}\right)^{\alpha} \Rightarrow v_{80} = 5\,(8)^{0.14} = 5(1.338) = 6.69\ \text{m/s}$$

Power density (power per unit swept area):

$$\frac{P}{A} = \tfrac{1}{2}\rho v^3 = 0.5(1.225)(6.69)^3 = 183\ \text{W/m}^2$$

(d) Annual energy

$$E = P_{rated}\times CF \times 8760 = 479 \times 0.30 \times 8760 = 1.26\times10^{6}\ \text{kWh}$$

Answer: $P_w = 1330$ kW, $P_m = 532$ kW, $P_e = 479$ kW; $\lambda = 3.49$; $v_{80} = 6.69$ m/s with 183 W/m²; annual energy $\approx 1.26$ GWh (1 258 000 kWh).