HOW DO YOU CALCULATE POWER NEEDS FOR PUMPING WATER?

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To start calculating your power needs, remember it takes 1 horsepower of energy to lift 2 cubic feet of water a vertical distance of 1 foot per second if the pumping efficiency is 100 percent. However, pumping efficiency is never 100 percent due to friction, temperature, engine wear, etc; 70 percent is considered very good and 20 percent is considered very poor.

You may think the energy needed to pump 900 gallons per minute from a well with an estimated pumping efficiency of 50 percent would be 2 horsepower since 900 gallons per minute is equivalent to 2 cubic feet per second.

This would be true if you were simply lifting 2 cubic feet from some well depth as in a bucket. However, since you are pumping a water column, the power needs increase with depth. For example, if you were pumping through a pipe that holds one cubic foot of water per linear foot and from a well where the lift was 10 feet, this well column would contain a weight equivalent to 10 cubic feet of water or about 75 gallons.

Actual horsepower required to lift 900 gallons per minute to a height of 10 feet at an efficiency of 50 percent is about 4.6. The energy to pump 900 gallons per minute from a depth of 50 feet is 22.7 horsepower. This is not a linear relationship and hydraulic engineers have developed charts for energy needs for lifting water various distances under a variety of different delivery efficiencies.

There are only three (four if using diesels) additional numbers needed to calculate the cost to pump it. For electric-powered pumping plants these are the amount of pressure you need, the pump efficiency, and the unit cost of electric power. The equation to use for electric power is:

  • UC = 2.36 * H * UCE / PE
    Where:
    • UC = unit pumping cost, $/acre-foot H = pumping head, pounds per square inch
    • UCE = unit electric power costs,
    • $/KWH (this is an average cost based on power and demand charges)
    • PE = pumping plant efficiency, a decimal normally in the .5 to .7 range

For example:

10 psi unit-cost, assume that the average electric power cost is $.07/KWH. The pumping plant efficiency is estimated at .6.

Thus,UC = 2.36 * H * UCE / PE UC = 2.36 * 10 * .07 / .6 = $2.75 10 PSI = $2.75/AF

A handy unit-cost is the money required to pressure water to 10 pounds-per-square-inch (PSI). (This is only the direct power costs. Capital costs for the engine/motor and maintenance are not considered.) Since we are talking about unit costs, we already know how much water we have to move, one acre-foot. If you utilize the above formulas, you should be able to accurately predict your power needs for pumping water.

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