Question

A horizontal axis wind turbine with 24 m diameter blades, running with an average wind velocity of 6.0 m.s\(^{-1}\), is used for pumping irrigation water. The average air density is 1.23 kg.m\(^{-3}\). Considering coefficient of power as 0.3, transmission efficiency as 90%, pump efficiency as 60%, acceleration due to gravity as 9.81 m.s\(^{-2}\) and density of water as 1000 kg.m\(^{-3}\), the discharge of the pump for a total head of 20 m is _________ L.s\(^{-1}\). (Rounded off to 2 decimal places)

Hint:

To calculate discharge in wind-powered pumps, first calculate the mechanical power delivered to the pump, adjust for efficiencies, and then use the hydraulic power formula to find the discharge.

Answer 1

Given: Diameter of blade, $D = 24$ m $\Rightarrow$ Radius, $r = 12$ m
Wind velocity, $V = 6$ m/s
Air density, $\rho_a = 1.23$ kg/m³
$C_p = 0.3$, $\eta_t = 0.90$, $\eta_p = 0.60$
$g = 9.81$ m/s², $\rho_w = 1000$ kg/m³, $H = 20$ m
Step 1: Power available in wind

\[ P_{wind} = \frac{1}{2} \rho_a A V^3, \quad A = \pi r^2 = \pi (12)^2 = 452.39 \, {m}^2 \] \[ P_{wind} = 0.5 \cdot 1.23 \cdot 452.39 \cdot (6)^3 = 60076.73 \, {W} \]

Step 2: Useful power at pump

\[ P_{useful} = P_{wind} \cdot C_p \cdot \eta_t \cdot \eta_p = 60076.73 \cdot 0.3 \cdot 0.9 \cdot 0.6 = 9730.37 \, {W} \]

Step 3: Use hydraulic power equation to find $Q$

\[ P = \rho_w g Q H \Rightarrow Q = \frac{P}{\rho_w g H} = \frac{9730.37}{1000 \cdot 9.81 \cdot 20} = \frac{9730.37}{196200} \approx 0.0496 \, {m}^3/{s} \] \[ Q = 0.0496 \cdot 1000 = \boxed{49.60 \, {L/s}} \]