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#### mpoulton

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18 amps X 230 volts = 4140 watts, or 4.14kW. 11 cents/kWh X 4.14kW = 45.54c/hour. 6 hours X 45.14c/hour = 273.24c, or \$2.73.

Now, it's actually going to cost less than that. The 5HP / 18A rating for the pump is the absolute maximum amount of power it can draw in operation. Depending on the flow rate and pressure, the pump may use considerably less power than that. You'd have to measure the actual current in operation to know for sure. Also, the calculation above doesn't consider "power factor". The watts = volts X amps calculation is always off by a bit in AC circuits with certain types of loads, motors included. It will actually use about 20% less power than the volts X amps calculation would indicate.

So to get a better idea of how much it will really cost to run, measure the current and the voltage in operation. Multiply them, and multiply by 0.8. That will give you a much more accurate value for power consumption.

#### mpoulton

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That pump seems very small for irrigation purposes. When I was in Nebraska, the typical electric well pumps were at least 60HP, and often 100HP. They each had their own dedicated 480V 3-phase service. A 5HP well pump will not move much water.

#### mpoulton

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if i remember its going to cost about \$22 per 6hrs of use. & im thinking 3 pulls per grid, so 20x3=\$60. am i figuring right ? ( ex. 16amp x 230v = 3680 / 1000 = 3.68 x 6 = \$22 ?
? My calculation I already posted is correct (based on 18A load, not 16A). Your calculation above is missing some steps. Where are you getting dollars from? You calculated the load as 3.68kW, then multiplied by 6 hours. That gives you 22kWh, not \$22. Multiply by the cost per kWh (11 cents) to get the cost of operation per pump, per 6 hour period: \$2.42.

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