PC Pump Power Usage

1. Jul 29, 2010

simulation135

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.

2. Jul 29, 2010

Staff: Mentor

Welcome to PF.

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

3. Jul 29, 2010

simulation135

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.

4. Jul 29, 2010

Staff: Mentor

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

5. Jul 30, 2010

simulation135

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.