Calculating Power Usage of PC Pump for Oil Well Setting

In summary, the conversation revolves around determining the power usage of a progressive cavity pump in an oil well setting. The formula used is Power = (Torque X RPM)/3300, with power measured in Horse Power. However, the resulting answer is extremely large, and the error is found to be in converting from kW to kWh. The correct conversion is to simply multiply by 1, as shown in the units kW*hr=kWh. Therefore, if the power calculation comes out to 10 kW and the machine is running for 24 hours a day, the total power consumption in terms of kWh would be 240 kWh/day.
  • #1
simulation135
3
0
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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  • #2
Welcome to PF.

A kWh is 1 kW for 1 hour. So you don't need to multiply by 3600. You calculated kiloJoules.
 
  • #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.
 
  • #4
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
 
  • #5
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.
 

What is the purpose of calculating power usage of a PC pump for oil well setting?

The purpose of calculating power usage of a PC pump for oil well setting is to determine the amount of power needed to operate the pump, which is essential for efficient and cost-effective oil extraction. This calculation helps in selecting the appropriate pump size and motor power, as well as estimating the operational costs.

What factors affect the power usage of a PC pump for oil well setting?

There are several factors that can affect the power usage of a PC pump for oil well setting, including the pump size, depth and size of the well, viscosity and density of the fluid being pumped, and the desired flow rate. The efficiency of the pump and motor also plays a significant role in determining power usage.

How is power usage calculated for a PC pump?

The power usage of a PC pump can be calculated using the following formula:
Power (kW) = (Flow rate (m3/hr) x Head (m) x Density (kg/m3)) / (3.6 x Pump efficiency). The pump efficiency can be obtained from the manufacturer or through testing.

What are the units of measurement for power usage of a PC pump?

The power usage of a PC pump is typically measured in kilowatts (kW) or horsepower (hp). However, other units such as watts (W) or British Thermal Units per hour (BTU/hr) may also be used.

How can power usage be optimized for a PC pump in oil well setting?

To optimize power usage for a PC pump in oil well setting, it is important to select a pump and motor with appropriate sizes and efficiency for the specific well conditions. Regular maintenance and proper operation of the pump can also help in reducing power usage. Additionally, using energy-efficient technologies, such as variable frequency drives, can further optimize power usage.

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