Question

A horizontal axis lift type wind rotor of diameter 4 m is used to pump water at a wind velocity of 15 km/h at standard atmospheric pressure and temperature (density of air is 1.23 kg/m\(^3\)). If velocity of wind leaving the rotor blade is reduced to one-third of the approaching wind velocity, the thrust acting on the blade of the wind rotor in N is \(\underline{\hspace{1cm}}\) (round off to two decimal places).

Hint:

Thrust is computed from the difference between the kinetic energy before and after wind passing through the rotor.

Answer 1

Correct Answer: 118.9

Wind velocity: \(V_w = 15\ \text{km/h} = 4.17\ \text{m/s}\)
Diameter of rotor: \(D = 4\ \text{m}\)
Area of rotor: \(\text{Area} = \frac{\pi D^2}{4} = \frac{\pi (4)^2}{4} = 12.57\ \text{m}^2\)
Thrust on the rotor blade is given by:
\[ T = \frac{1}{2} \rho A (V_w^2 - V_{w1}^2) \] Where \(V_{w1} = \frac{V_w}{3} = \frac{4.17}{3} = 1.39\ \text{m/s}\)
\[ T = \frac{1}{2} (1.23)(12.57)((4.17)^2 - (1.39)^2) \] \[ T = 0.5 \times 1.23 \times 12.57 \times (17.39 - 1.93) \] \[ T = 0.5 \times 1.23 \times 12.57 \times 15.46 = 118.92\ \text{N} \] Thus, thrust is \( \boxed{118.92}\ \text{N} \).