Question

A small horizontal axis wind turbine (HAWT) is designed to deliver 2 kW shaft power at a steady wind speed of \(7\ \text{m.s}^{-1}\). The HAWT operates with an aerodynamic power coefficient of 0.36, a drivetrain efficiency of 0.90, and an air density of \(1.225\ \text{kg.m}^{-3}\). Neglecting other losses, the required rotor diameter is \(X\) m. When the wind speed drops to \(6\ \text{m.s}^{-1}\), the shaft power is \(Y\) kW. Identify the nearest correct option for the \(X\) and \(Y\) combination (take \(\pi = 3.14\))

• A. \(X = 6.12\) m
  \(Y = 1.28\) kW
• B. \(X = 3.06\) m
  \(Y = 0.63\) kW
• C. \(X = 6.12\) m
  \(Y = 2.00\) kW
• D. \(X = 6.75\) m
  \(Y = 1.50\) kW

Hint:

Use the wind power formula with the power coefficient and drivetrain efficiency to size the rotor area.

Answer 1

The Correct Option is A

Step 1: Get the rotor area and diameter as before.
From $P_1 = \eta_d C_p \tfrac{1}{2}\rho A V_1^3$ at $V_1 = 7$ m/s: $A = \dfrac{2000}{0.90 \times 0.36 \times 0.5 \times 1.225 \times 343} = \dfrac{2000}{68.07} = 29.38\ \text{m}^2$.
$D = \sqrt{\dfrac{4A}{\pi}} = \sqrt{\dfrac{4 \times 29.38}{3.14}} = 6.12$ m, so $X = 6.12$.

Step 2: Use a direct power ratio instead of recomputing from scratch.
Since $A$, $C_p$, and $\eta_d$ stay fixed, $\dfrac{P_2}{P_1} = \left(\dfrac{V_2}{V_1}\right)^3$.
$\dfrac{P_2}{P_1} = \left(\dfrac{6}{7}\right)^3 = \dfrac{216}{343} = 0.6297$.

Step 3: Scale the known rated power.
$P_2 = 2\ \text{kW} \times 0.6297 = 1.26$ kW, so $Y \approx 1.3$.

Final Answer:
This ratio shortcut gives the same pair, $X = 6.12$ m and $Y \approx 1.28$ kW, so option A is correct. \[ \boxed{X \approx 6.12\ \text{m},\ Y \approx 1.28\ \text{kW}} \]