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: Write the shaft power available from the wind.
Shaft power after drivetrain losses is \(P = \eta_d \, C_p \, \tfrac{1}{2}\rho A V^3\), where \(\eta_d\) is drivetrain efficiency, \(C_p\) is the power coefficient, \(\rho\) is air density, \(A\) is the swept rotor area, and \(V\) is wind speed.

Step 2: Solve for the swept area at \(V = 7\ \text{m/s}\).
\(2000 = 0.90 \times 0.36 \times 0.5 \times 1.225 \times A \times 7^3\).
The constant multiplier is \(0.90 \times 0.36 \times 0.5 \times 1.225 \times 343 = 68.07\).
\(A = \dfrac{2000}{68.07} = 29.38\ \text{m}^2\).

Step 3: Convert area to rotor diameter.
\(A = \dfrac{\pi D^2}{4}\), so \(D = \sqrt{\dfrac{4A}{\pi}} = \sqrt{\dfrac{4 \times 29.38}{3.14}} = \sqrt{37.43} = 6.12\ \text{m}\).
So \(X = 6.12\).

Step 4: Find the shaft power when wind speed drops to \(6\ \text{m/s}\).
Keeping the same rotor, \(C_p\), and \(\eta_d\), power now scales only with \(V^3\).
\(P_2 = 0.90 \times 0.36 \times 0.5 \times 1.225 \times 29.38 \times 6^3 = 0.19845 \times 29.38 \times 216\).
\(P_2 = 5.830 \times 216 = 1259\ \text{W} \approx 1.26\ \text{kW}\), so \(Y \approx 1.3\).

Step 5: Match to the closest option.
Options B and C keep \(X\) or \(Y\) badly off; option D overstates \(Y\) at 1.50. The pair \(X = 6.12\), \(Y \approx 1.28\) fits best.

Final Answer:
The nearest correct combination is \(X = 6.12\) m and \(Y = 1.28\) kW. \[ \boxed{X \approx 6.12\ \text{m},\ Y \approx 1.28\ \text{kW}} \]