Calculating Power Usage of PC Pump for Oil Well Setting

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Discussion Overview

The discussion focuses on calculating the power usage of a progressive cavity pump in an oil well setting. Participants analyze the relationship between torque, RPM, and power consumption, while addressing potential errors in the conversion of power units and energy calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a formula for calculating power based on torque and RPM, expressing uncertainty about the large result obtained.
  • Another participant clarifies that a kilowatt-hour (kWh) is defined as 1 kW for 1 hour, suggesting that the multiplication by 3600 is unnecessary.
  • A participant confirms their original calculation was for horsepower and discusses the conversion to kilowatts, questioning the need for the multiplication by 3600.
  • Further clarification is provided that to convert kW to kWh, one simply uses the time in hours, emphasizing the importance of unit consistency.
  • A participant proposes an example calculation, suggesting that if their power calculation yields 10 kW, running continuously for 24 hours would result in 240 kWh/day.

Areas of Agreement / Disagreement

Participants generally agree on the definition of kWh and the error in multiplying by 3600, but there is no consensus on the initial power calculation or the implications of the large result.

Contextual Notes

Limitations include potential misunderstandings of unit conversions and the initial assumptions regarding the power calculation formula. The discussion does not resolve the accuracy of the original power calculation.

simulation135
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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.
 
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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.
 
simulation135 said:
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
 
russ_watters said:
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.
 

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