One of the more confusing things about electricity is how it’s measured and how we pay for it. We would like to take a moment to clear up any misunderstandings about electrical power, especially in terms of calculating the electrical cost of operating a submersible pump. We all pay for the Kilo-watt hours on our electrical bill.

kW hours are calculated by multiplying the watts time hours of operation, all of this divided by 1000 to equate to kW hours.

kW-Hr = (Motor Watts * Hours of operation)/1000.

With the exception of the 1/2HP 115V 3 W motor Franklin watts (what we pay for) are lower than the competitive motor.  Many people assume that amp draw is the key factor, when in fact, the real issue is how well you convert the amps to output, otherwise known as efficiency of operation.

Assuming the average household consumes 500 Gal of water per day, using a 20 GPM 2W 1.5HP unit pump, this would require 25 minutes of operation per day or 750 minutes or 12.5 hours of operation per month. Assuming a kW cost of $0.10/kW-Hr, it would cost 2180*12.5*0.1/1000=$2.73, the physical cost, to operate an Franklin Electric equipped unit or $3.15 for the competitive unit.

Kilowatts are a measure of power, amps are a component of power and the other key is the materials and efficiencies associated with converting the power to rotational energy.  So, next time your customer wants to know how much it costs to operate their private water system, you can show them the figures and assure them that a submersible pump from Franklin Electric is one of the best deals around.