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PC Pump Power Usage

  1. Jul 29, 2010 #1
    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. jcsd
  3. Jul 29, 2010 #2

    russ_watters

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    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.
     
  4. Jul 29, 2010 #3
    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.
     
  5. Jul 29, 2010 #4

    russ_watters

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    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
     
  6. Jul 30, 2010 #5
    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.
     
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