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Example: What is the pumping capacity needed for the 100 head of beef cattle in the example above? Pumping capacity = 1,000 gallons/day divided by 5 hours = 200 gallons/hour Estimating the horsepower required Next you need to figure the amount of horse- power the pump will need to have. To estimate the horsepower needed, first convert the pump- ing capacity from gallons per hour to gallons per minute. Example: For the same 100 head of beef cat- tle above, you will need to convert the 200 gal- lons per hour of pumping capacity to gallons per minute: Gallons per minute (GPM) = 200 gallons/hour divided by 60 = 3.33 GPM Next, to calculate the horsepower needed, multiply the pumping capacity by the lift, which is the distance that the water must be lifted; then divide that number by 3,960. HP = Q x H / 3,960 Where: HP = Horse power Q = Pumping capacity, in gallons per minute (GPM) H = lift, in feet The well on that cattle ranch in the example above is 100 feet deep. For the formula above, the factors are: Q = 3.33 GPM Lift = 100 feet The calculation would be: Horsepower = 3.33 GPM x 100 feet/3,960 = 0.084 HP Converting horsepower to watts The horsepower used above describes the mechanical work needed to lift a volume of water per unit of time from the pumping water level up to the storage tank. You can convert this measure of horsepower to watts of electricity by multiplying it by 746: Watts = 0.084 HP x 746 watts = 62.7 watts Next, you need to adjust the wattage to take into consideration the loss of electricity in the cable and controls during transmission and in converting electricity to the mechanical move- ments of the pump. The average efficiency rate of these pumps is about 45 percent. To adjust for this inefficiency, we must recalculate our power input by dividing the number of watts by 0.45: Wattage needed = 62.7 divided by 0.45 = 139.3 watts Last, for a solar system, we must choose the number of solar panels that will produce the number of watts needed by the pump. Solar panels or modules have different capacities. There are modules of 25, 50, 70 or 75 watts. It is less expensive to use a more efficient motor than to add an extra solar panel. For the example above, the rancher could buy six 25- watt panels, but it would be much less expen- sive to buy two 70-watt solar panels to generate the 139.3 watts needed. Estimating the size of the windmill When buying a windmill, you will need to know the lift and daily water requirements. Use the formulas above to determine the horsepower your pump needs to have. The primary components of a windmill are the blades, the tower and engine, pump rod, the drop pipe (usually 2-inch galvanized pipe), suck- er rod (wood pole, steel rod or fiberglass) and the piston pump (see Fig. 1). If there are tall trees in the area, you may need or want a taller tower to raise the fan blades above the trees and into the wind. However, windmills are usually no more than 35 feet tall; otherwise, the towers become very expensive. A windmill’s pumping output is affected by three factors: wind speed, wheel or blade diame- ter, and the diameter of the cylinder (Table 3). Wind speed has an important effect on the pumping output. In fact, the power available from the wind is proportional to the cube of the wind speed. This means that when the wind speed doubles, the power increases eight times. Most windmills do not operate at wind speeds of less than 7 mph or more than 30 mph, as the mill can be damaged by high winds. Example: From Table 3, to pump 470 gallons per hour and lift the water 220 feet, a cylinder diameter of 3 inches would require a blade diameter of 14 feet. 5PDF Image | Using Renewable Energy to Pump Water
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