Subtopic: Sources of power on the farm – human, animal, mechanical, electrical, wind, solar and biomass
Question Statement:
A multi-blade agricultural water-pumping windmill has a rotor diameter of $D = 4.0\text{ m}$. The prevailing wind speed is $v = 6.0\text{ m/s}$ and air density is $\rho = 1.20\text{ kg/m}^3$. The aerodynamic power coefficient of the rotor is $C_p = 0.32$ and the mechanical transmission-pump efficiency is $\eta_m = 70\%$. The actual water pumping power delivered by the windmill in $\text{W}$ is ________ (round off to integer).
Answer Key:
365
Step-by-Step Solution & Explanation
```
Windmill Power Aerodynamics:
Rotor Diameter D = 4.0 m
Swept Area A = pi * D^2 / 4
Wind Speed v = 6.0 m/s
C_p = 0.32, eta_m = 0.70
```
**Step-by-step Calculation:**
1. Swept rotor area ($A$):
$A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (4.0)^2 = 4\pi = 12.5664\text{ m}^2$
2. Total kinetic power available in the wind stream ($P_{\text{wind}}$):
$P_{\text{wind}} = \frac{1}{2} \rho A v^3 = \frac{1}{2} \times 1.20 \times 12.5664 \times (6.0)^3$
$P_{\text{wind}} = 0.60 \times 12.5664 \times 216 = 1628.6\text{ W}$
3. Aerodynamic power captured by rotor ($P_{\text{rotor}}$):
$P_{\text{rotor}} = P_{\text{wind}} \times C_p = 1628.6 \times 0.32 = 521.15\text{ W}$
4. Useful water pumping power output ($P_{\text{pump}}$):
$P_{\text{pump}} = P_{\text{rotor}} \times \eta_m = 521.15 \times 0.70 = 364.8\text{ W} \approx 365\text{ W}$