# PC Pump Power Usage

Hi I am trying to determine the power usage of a progressive cavity pump in an oil well setting. I have some data to analyze, where I know the amount of viscous rod torque (ft lbs) per 1000 ft of rod length, I can use various rod lengths, and I know the RPM. Calculating the total Rod Torque I need to determine the power consumption of the motor turning the PC pump. I am using the formula
Power = (Torque X RPM)/3300
where power is in Horse Power

I then converted the HP into KW using a conversion of 1 HP per 0.746 KW, then multiplying by 3600 to get KW hours. Finally I multiplied by \$0.07 per KW hour as is the going rate for electricity. However the answer I got was extremely large, I believe the problem lies with the formula I am using. Any help would be greatly appreciated.

russ_watters
Mentor
Welcome to PF.

A kWh is 1 kW for 1 hour. So you don't need to multiply by 3600. You calculated kiloJoules.

My original Power calculation was for Horse Power, I converted the Horse Power into KW. Then I multiplied by 3600 to get kwh. Unless you're saying that i converted from Horse Power to kwh with a factor of 0.746.

russ_watters
Mentor
Then I multiplied by 3600 to get kwh.
That's the error. To convert kW to kWh you multiply by *1*. Just look at the units: kW*hr=kWh.

There's a wiki on it too:
Energy in watt hours is the multiplication of power in watts and time in hours.
http://en.wikipedia.org/wiki/Kilowatt_hour

That's the error. To convert kW to kWh you multiply by *1*. Just look at the units: kW*hr=kWh.

There's a wiki on it too: http://en.wikipedia.org/wiki/Kilowatt_hour

Ok so then say my power calculation came out to 10 KW using the formula in my original post, and my machine is running 24 hours a day. Then you're saying to determine how much power in terms of KWh I multiply the 10KW by 1 hour, and then by 24 hours in a day, meaning my machine would use 240 KWh/day?

Thank You for the clarification as well.