AskPablo: Wind Farming

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Last year I was asked to help get water to a village. The only problem was that this village was thousands of miles away in the Panjshir Region of Afghanistan and I had no travel budget…
To make matters worse I couldn’t just call up my Rain-For-Rent franchise for a mobile pump-truck. I had to design a system to pump water up to a village and I had to do it with locally available materials. I also couldn’t just specify an electric pump because the village had no power. I couldn’t specify an electric generator to power that electric pump since fuel is in short supply as well. If we had a running creek in the village we could use hydro-power, but then we wouldn’t need to pump any drinking water, now would we? Since Afghanistan is known for its abundant wind it seems like a natural choice. Since bicycle parts are widely available I decided to design a windmill made from bicycle parts.


There are two important factors in wind power; the efficiency of the windmill and the wind speed. You see, the ability to derive energy from wind is not directly proportional to wind speed. As wind speed doubles, its force quadruples. This is why hurricanes get exponentially more destructive as their wind speed increases. The equation for power (P, in kW/m^2) is the efficiency of the turbine, times the theoretical maximum efficiency of wind generation ( 0.59), times the mechanical efficient (0.90), times 6.1 x 10^-4 (1/2 the density of air), times the air speed cubed (v^3, in m/s). Deep breath…
The windmill that I designed used a bicycle wheel with sheet metal or wooden blades in a similar style to the “American Multivane” type (like you typically see on farms), with a blade diameter of 1 meter. This type of windmill has a maximum efficiency of 31%, or 0.31. The average wind speed at the village is 7 m/s (15 mph). So, using the formula above: (0.31) x (0.59) x (0.9) x 6.1 x 10^-4 x (7)^3 = 0.344 kW/m^2. The area of the windmill is 0.7854 m^2, so 0.344 kW/m^2 x 0.7854 m^2 = 0.2705 kW, or 270.5 W (roughly equivalent to 5 incandescent light bulbs).
Now lets see how much water that will pump. Power is defined as W/t (W=work, t=time). Work is equal to mass, time gravity (9.8 m/s^2), times height (mgH). If the pump is 60% efficient, we have 162.3 W to work with (270.5 W x 0.60). We can rely on 8 hours per day of wind, or 28,800 seconds (8h x 3600 s/h) and the village is 10 meters above the water source. So, using our available power (162.3 W), we can pump “?” kg x 9.8 m/s^2 x 10 m over 28,800 seconds. Solving for “?” kg of water we get: (162.3 x 28,800)/(9.8 x 10) = 47,696 kg/day, or 47,696 liters. This should be enough for our village. If the village were any higher we could use multiple windmills and a more powerful pump or we could have a series of windmills and pumps. Since wind is not constant we would want to have a tank or cistern in the village to store the water for when it is needed.
That was fun, wasn’t it? Well, please send me any comments and/or questions for next week to pablo(dot)paster(at)gmail(dot)com. Thanks for reading www.AskPablo.org!
Pablo Päster, MBA
Sustainability Engineer