Wind turbine power
Work, Energy & Power - OCR A-Level Physics
Key Idea
Power available in a wind stream
The maximum power a wind turbine can extract is found by assuming all the kinetic energy of the air sweeping through the rotor is captured. Real turbines reach only a fraction of this ideal.
The maximum power a wind turbine can extract is found by assuming all the kinetic energy of the air sweeping through the rotor is captured. Real turbines reach only a fraction of this ideal.
$$P = \tfrac{1}{2}\rho A v^{3}$$
Derivation
In time $t$, a cylinder of air of length $vt$ and cross-sectional area $A$ sweeps through the blades. Its volume is $Avt$, so its mass is $m = \rho A v t$, where $\rho$ is the air density. The kinetic energy of that air is $E_k = \tfrac{1}{2}mv^{2} = \tfrac{1}{2}\rho A v t \cdot v^{2}$. Dividing by $t$ gives the power: $P = E_k/t = \tfrac{1}{2}\rho A v^{3}$.
Variables
$P$ is the ideal power output in W. $\rho$ is the density of air in $\text{kg m}^{-3}$ (typically $1.2 \text{ kg m}^{-3}$). $A$ is the area swept by the blades in $\text{m}^{2}$, equal to $\pi r^{2}$ for blades of length $r$. $v$ is the wind speed in $\text{m s}^{-1}$.
- Power scales with the cube of wind speed. Doubling $v$ multiplies $P$ by 8. This is why wind farms are sited where mean wind speeds are highest.
- Power scales with the square of blade length (since $A = \pi r^{2}$). A turbine with blades twice as long captures four times the power for the same wind.
- The Betz limit shows that no turbine can extract more than $\sim 59\%$ of the available wind power, because the air behind the blades must keep moving to make way for the next mass of air. Real turbines reach $\sim 35\%$ to $45\%$ efficiency.
- A turbine with $r = 30 \text{ m}$ in a $12 \text{ m s}^{-1}$ wind: $A = \pi \times 30^{2} \approx 2830 \text{ m}^{2}$, so $P_{\text{ideal}} = 0.5 \times 1.2 \times 2830 \times 12^{3} \approx 2.9 \text{ MW}$.
Common Mistake
HIGH
Wrong: Using $v^{2}$ in the formula because it appears in the kinetic energy equation $E_k = \tfrac{1}{2}mv^{2}$, giving $P = \tfrac{1}{2}\rho A v^{2}$.
Right: One factor of $v$ comes from the kinetic energy, and a second factor of $v$ comes from the mass flow rate ($m/t = \rho A v$). The two combine to give $v^{3}$. Always check: doubling wind speed multiplies power by 8, not 4.
Right: One factor of $v$ comes from the kinetic energy, and a second factor of $v$ comes from the mass flow rate ($m/t = \rho A v$). The two combine to give $v^{3}$. Always check: doubling wind speed multiplies power by 8, not 4.
Worked Example [3 marks]
A wind turbine has blades of length $16 \text{ m}$. The wind speed is $17 \text{ m s}^{-1}$ and the density of air is $1.2 \text{ kg m}^{-3}$. Calculate the ideal power available from the wind.
Show Solution
1
Swept area
$A = \pi r^{2} = \pi \times 16^{2} = 804 \text{ m}^{2}$.
[1]2
$P = \tfrac{1}{2}\rho A v^{3} = 0.5 \times 1.2 \times 804 \times 17^{3}$.
[1]3
$P = 0.5 \times 1.2 \times 804 \times 4913 \approx 2.4 \times 10^{6} \text{ W} = 2.4 \text{ MW}$.
[1]Answer
$2.4 \text{ MW}$ (ideal)
Examiner Tips and Tricks
- "Show that" wind turbine questions usually want the full derivation. Start from $m/t = \rho A v$, then multiply by $\tfrac{1}{2}v^{2}$ to get $P = \tfrac{1}{2}\rho A v^{3}$.
- If the question quotes a stated power output that is much less than $\tfrac{1}{2}\rho A v^{3}$, the difference is the turbine's efficiency. Use $\eta = P_{\text{stated}} / P_{\text{ideal}}$.