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PCP system design

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Progressing cavity pump systems are, in general, highly flexible in terms of their ability to function effectively in a diverse range of applications. As with other artificial-lift systems, the basic objective in the design of a PCP system is to select system components and operating parameters (e.g., pump speed) that can achieve the desired fluid production rates while not exceeding the mechanical performance capabilities of the equipment components to facilitate optimal service life and system value. When a PCP system is designed for a particular application, both the system components and operating environment must be considered to ensure that a suitable system design is achieved.

Overview of the design process

Fig. 1 presents a “design process” flow chart that outlines the many factors and considerations that should be addressed in the selection of an effective overall system configuration and operating strategy. At each step, the designer selects certain operating parameters or specific equipment components and must then assess the impacts of these decisions on system performance. For example, selection of a particular tubing size is based on such design considerations as flow losses and casing size. Some considerations apply to more than one decision, as is the case with flow losses that affect pump, tubing, and rod-string selection. Other design considerations may produce conflicting results, which complicates the decision-making process. For example, the use of rod-string centralizers may minimize wear but may also increase flow losses. As with other artificial-lift systems, the design process is generally iterative, and individual parameters are often adjusted to achieve an optimal design for a particular application. As Fig. 1 shows, the primary design considerations for a PCP system are:

The first step in the design process is to gather information for the application of interest. Past experience, fluid properties, production, well records, and reservoir data are all useful sources of relevant information. Next, it is necessary to determine the anticipated fluid rates. These can be estimated from historical data or by setting a dynamic fluid level and calculating production rates based on reservoir data and an inflow performance relationship.[1] Initial values must then be set for the:

  • Wellbore geometry
  • Pump-seating location
  • Dynamic fluid level
  • Tubing size
  • Rod-string configuration

If the design is for an existing well, some of these parameters may already be constrained.

Once these equipment and operating parameters are established, flow losses can be calculated. If the estimated flow losses are unacceptably high, they can be reduced by increasing the tubing size, reducing the rod-string induced flow restrictions, or decreasing the fluid rate. Next, initial values for pump intake and discharge pressures, net lift, pump speed, and pump displacement can be set. This allows the designer to select a range of pump models capable of satisfying the desired pump displacement and lift requirements. However, if there are no pumps available that meet both requirements, then the prescribed pump displacement and lift specifications must be relaxed by decreasing the fluid rate expectations, increasing pump speed, reducing discharge pressure requirements, increasing pump intake pressure, or by implementing some combination of these changes. The individual pumps that satisfy the requirements are then evaluated on the basis of geometric design and fluid considerations to select the most appropriate pump model.

Once a specific pump model has been selected, rod loading, rod-string/tubing wear, and surface equipment requirements can be evaluated. If the calculated rod stresses exceed the allowable value, then either the rod-string strength must be increased through the use of a larger rod or higher-strength material, or the loading must be decreased through a reduction in the net lift requirements or the use of a smaller-displacement pump. Similarly, if the predicted rod-string/tubing wear rates are unacceptable, then steps must be taken to reduce axial loads (e.g., use of a smaller pump), or the rod string must be reconfigured so that it is less prone to wear. After the rod loading and wear considerations are satisfied, the final step in the design process is the selection of surface equipment. If the available surface equipment cannot meet the polished-rod power requirements, then the design process must be repeated to configure a downhole system or operating parameters that result in reduced system loads. For example, reduced power requirements can be achieved by lowering the pump speed (which will also likely lead to a lower differential pump pressure) or by selecting another pump with a smaller displacement. Once a final system design has been established, any areas of potential concern should be re-evaluated to confirm that the design satisfies the functional requirements of the application within acceptable operating guidelines.

It is quite apparent that the interdependency between the numerous equipment selection and well completion options, variations in operating conditions, and complex fluid flow and mechanical interactions that affect system loading and performance can make the assessment and design of PCP systems difficult and time consuming. In new applications, numerous iterations may be required just to establish a workable system. Because design optimization based on manual calculations is usually impractical, computer programs have been developed to help designers work faster and more effectively. The following sections provide further details on specific design parameters.

Design example

Problem statement

A vertical well is expected to produce 100 m3/d [629 B/D] of 12°API oil and no water, gas, or sand. The well is cased with 177.8 mm [7 in.] OD casing perforated at 1000 m [3,281 ft] from surface. At the desired flow rate, the fluid level is expected to be 600 m [1,968 ft] from surface. The casing is vented to atmosphere, while the flowline pressure is 1500 kPa [218 psi]. The oil viscosity is 1,000 cp [1000 mPa•s].


Design a PCP system to produce this well with the following constraints. The pump should be set below the perforations at 1010 m [3,312 ft]; its speed should not exceed 350 rpm; and the pump should not be loaded above its rated pressure. The rod stress should be < 80% of yield (assume API Grade D rods).


The following pumps are available. Assume that any of these pumps will operate at 85% volumetric efficiency under downhole conditions and that the friction torque will be 20% of the hydraulic torque at the pump’s rated pressure.


Pump A Pump B Pump C Pump D Pump E
Displacement, m3/d/rpm 0.15 0.30 0.45 0.70 1.00
Pressure rating, kPa 12,000 12,000 18,000 15,000 12,000
Major diameter, mm 50 54 52 58 74
Minor diameter, mm 38 41 35 44 51
OD, mm 88.9 95 108 114.3 114.3
Length, m 4.0 4.5 8.0 12.0 10.0


Solution

Using Eq. 1,

Vol4 page 0787 eq 001.png....................(1)

we can determine the minimum displacement required to achieve the desired flow rate without exceeding the specified maximum pump speed:

Vol4 page 0830 eq 001.png

The pump displacement must be > 0.336 m3/d/rpm. This eliminates Pumps A and B from further consideration.


The next step is to determine the differential pressure on the pump using Eqs. 2 through 4.

Vol4 page 0788 eq 001.png....................(2)

Vol4 page 0789 eq 001.png....................(3)

Vol4 page 0789 eq 002.png....................(4)

The pump intake pressure is:

Vol4 page 0830 eq 002.png

Casing-head pressure was defined in the problem statement to be atmospheric pressure, or 0 kPa (gauge pressure). The gas and liquid hydrostatic pressures can be calculated from the gas and liquid densities and the column heights. The pump intake is 1010 m from surface, and the fluid level is 600 m from surface. This means that there is 600 m of gas column and 410 m of liquid column. An average gas density can be estimated from the pressure at surface: 0.8 kg/m3.

Vol4 page 0830 eq 003.png

This gives a gas column hydrostatic pressure of 5 kPa. (Note that this method is an approximation; the actual gas density will change as the pressure increases, but because the value is so small relative to the other pressures in the system, the error introduced by this approximation is small.) The density of 12°API oil is 984 kg/m3, so the liquid hydrostatic pressure is 3958 kPa:

Vol4 page 0831 eq 001.png

In this case, the produced oil must flow from the perforations past the pump to reach the intake. Any flow losses here must also be considered in calculating the pump intake pressure. However, because the distance is small and there is a large clearance between the pump and casing, these losses are small and can be neglected. Note that if 139.7 mm OD casing had been used instead, there would be a very small annulus between the casing and the pump, and the flow losses between the perforations and the intake could be quite significant.


The pump discharge pressure is calculated from:

Vol4 page 0831 eq 002.png

The tubing-head pressure is given as 1500 kPa. The liquid hydrostatic head will depend on the location of the top of the pump. The pump is seated at 1010 m (intake depth), but the three pump alternatives have different lengths, so the top will be at a different location in each case. Also, the flow loss will depend on the selection of tubing and rods. The solution process will be iterative; it is necessary to calculate these values for one set of equipment and then redo the calculation if it appears that the selected equipment may not be the best choice. If the pump length is 8 m, the top of the pump will be at 1002 m, and the hydrostatic head of the liquid in the tubing is 9673 kPa:

Vol4 page 0831 eq 003.png

The calculation of flow losses was not described in detail in this chapter, but many different formulations are available in the literature, including this Handbook. For now, we will consider the use of 88.9 mm × 13.8 kg/m tubing, with 25.4 mm rods, 7.62 m in length, with standard couplings (55.6 mm diameter, 101.6 mm length). For the specified well depth, 131 couplings are needed, for a total length of 13.3 m; the remaining 988.7 m (assuming that the top of the pump is at 1002 m) is covered by rod segments. We can calculate the flow losses past 988.7 m of rod and 13.3 m of coupling separately and then add the two results together to obtain the total flow loss. This approximation neglects the flow effects at the ends of the couplings, but it should still provide adequate results. The ID of 88.9 mm × 13.8 kg/m tubing is 76.0 mm, and the drift diameter is 72.82 mm. For flow calculations, it is recommended that the ID instead of drift diameter be used. Assuming that the rods and couplings are concentric, the flow losses can be calculated (using one method) to be 5223 kPa past the rod body and 841 kPa past the couplings for a total of 6064 kPa. If, as normally expected, the rods and couplings are not concentric within the tubing, the flow losses would be somewhat reduced, but such a reduction will not be considered here, so the results are conservative.


We can now calculate the pump discharge pressure,

Vol4 page 0831 eq 004.png

and the pump differential pressure,

Vol4 page 0832 eq 001.png

The pump is required to work against a differential pressure of 13 274 kPa. Only Pumps C and D have a pressure rating exceeding this value. Also, note that Pump E cannot be used with this tubing because the major rotor diameter is larger than the drift diameter of the tubing. However, if a larger tubing size that would accommodate the large rotor diameter were used, the flow losses would be reduced, possibly to the point that the pressure rating of Pump E would not be exceeded. Therefore, Pump E will continue to be considered a potential candidate. All of the pumps have an OD that is less than the drift diameter of even the heaviest-wall 177.8 mm casing. Therefore, none of these pumps must be eliminated due to casing size.


The next task is to estimate the torque in the rods. The torque on the pump is given by:

Vol4 page 0832 eq 002.png

The friction torque was estimated in the problem statement to be 20% of the hydraulic torque at the pump’s rated pressure. Hydraulic torque is calculated from Eq 5:

Vol4 page 0789 eq 003.png....................(5)

Vol4 page 0832 eq 003.png

From this, we can estimate the friction torque for each pump. For example, for Pump C,

Vol4 page 0832 eq 004.png

Accordingly, the values are: Pump C - 180 N•m; Pump D - 233 N•m; and Pump E - 266 N•m.


Next, the hydraulic torque for a differential pressure of 13 274 kPa is calculated for each of these pumps as

Vol4 page 0832 eq 005.png

Thus, the hydraulic torque values are as follows: Pump C - 663 N•m; Pump D - 1026 N•m; and Pump E - 1470 N•m.


The torque on the rod string includes the pump torque plus torsional loading of the rod string resulting from mechanical interaction (friction) with the tubing and the resistance to rotation caused by the fluid viscosity. In a vertical well, the tubing friction loads can usually be considered negligible. The resistive torques for each of these pumps can be calculated at the speed at which they would run to produce the required amount of oil[2]: Pump C - 69.4 N•m; Pump D - 44.4 N•m; and Pump E - 31.2 N•m. The total rod torque is then the sum of the respective pump friction, hydraulic torque and rod resistive torque values: Pump C - 912 N•m; Pump D - 1304 N•m; and Pump E - 1768 N•m.


When considering rod loading, we must calculate the axial load in the rods and the torque. Axial load can be found from Eq. 6:

Vol4 page 0832 eq 006.png....................(6)

Calculation of the uplift forces will be neglected for this example, providing a slightly conservative result. The rod weight is easily calculated from the specific weight of steel and the rod volume, neglecting the additional weight from the couplings and upsets. For a 25.4 mm rod that is 1002 m long (Pump C) with a steel specific weight of 77 kN/m3, the rod weight is

Vol4 page 0833 eq 001.png

For pumps D and E, with their respective rod lengths, the rod weights are 38.9 kN and 39.0 kN.


Pump load is given by Eq. 7 as

Vol4 page 0833 eq 002.png....................(7)

where C = 7.9 × 10–4 when p is in Newtons, d and e are in millimeters, and pressures are in kPa.


To get the eccentricity values for the pumps for use in this equation, we must recognize that the major diameter is equal to the minor diameter plus twice the eccentricity. Therefore, the eccentricities for Pumps C, D, and E are 8.5, 7.0, and 11.5 mm, respectively. At a discharge pressure of 17 238 kPa and intake pressure of 3963 kPa, the axial load at the pump is as follows: Pump C - 38.2 kN; Pump D - 45.5 kN; and Pump E - 85.0 kN. So, neglecting the uplift forces, the total axial rod loads corresponding to the three pumps are: Pump C - 77.3 kN; Pump D - 84.4 kN; and Pump E - 124.1 kN.


The total stress of the rods can now be determined. For Pump C, this gives:

Vol4 page 0833 eq 003.png

The maximum stress is 514 MPa, which is 88% of the minimum yield for Grade D rods [586 MPa]. Note that the rod stresses exceed the yield capacity if the other two pumps are used. This condition would be in violation of the 80% loading criterion included in the problem statement.


To redesign the system to produce the well within the specified parameters, it appears that two viable options would be to decrease the differential pressure on the pump, or to increase the strength of the rods. There is nothing that can be done to reduce the hydrostatic head on the system while maintaining the same flow rate. However, a decrease in flow rate would reduce flow losses and would cause the fluid level in the casing to rise, thus increasing the pump intake pressure and decreasing the pump differential pressure. The tubing-head pressure can typically be reduced significantly only through changes in the gathering system to reduce the flowline pressure, through the addition of a transfer pump, or through the use of viscosity-reducing chemicals at surface.


Another way to decrease the pressure on the pump is to reduce the flow losses, which accounted for almost half the differential pump pressure. This can be achieved either through diluent injection or by increasing the flow area in the production tubing by using a larger-diameter pipe or a smaller-diameter rod string. Although using smaller rods would reduce flow losses, the load capacity would also be reduced (assuming the same material), so this does not appear to be a viable option, although the use of higher-strength rods may be an option in some cases. However, the use of a larger tubing string seems quite practical in this case.


The flow losses with 114.3 mm tubing with 25.4 mm rods would be ≈ 1338 kPa. This reduces the differential pressure to 8548 kPa, producing a corresponding reduction in the pump hydraulic torque values. The resistive torque is also reduced slightly. The total torque on the rod string for the three pump candidates can be recalculated to give the following values: Pump C - 671 N•m; Pump D - 935 N•m; and Pump E - 1241 N•m. The total axial load on the rod string can be recalculated as follows: Pump C - 63.0 kN; Pump D - 67.5 kN; and Pump E - 93.0 kN. With the lower torque and axial loads, the peak rod stress in the three cases is as follows: Pump C - 382 MPa (65%); Pump D - 520 MPa (89%), and Pump E - 693 MPa (118%). Pump C now gives a rod stress that is below 80% of yield, the criterion in the problem statement; the other two pumps will still cause the rod stress to exceed the specified criterion. Note that Pump C will operate at 261 rpm to produce 100 m 3 /d/rpm at a volumetric efficiency of 85%.


At this point in a typical system design, the pump, tubing, and rods have all been selected. The surface-drive system must now be established. The rod-string axial load at the surface is 63 kN, the torque is 671 N•m, and the operating speed of the polished rod is 261 rpm. A suitable drive can be selected from any manufacturer’s catalog by comparing these values to the published load and speed limits. The type of drive head (right angle or vertical, solid or hollow shaft, direct electric or hydraulic, etc.) normally is based on user preferences and field characteristics. For example, if electricity is not available, then an internal combustion engine must be used, which normally leads to the selection of a hydraulic system because otherwise the belts would typically have to be very long. If electricity is available but electronic speed control systems are not available or used in the area, hydraulic systems are often still preferred if regular speed adjustments are anticipated; otherwise, direct electric drives with a fixed selection of belts/sheaves or gears are typically used.


This example problem did not address wear and fatigue considerations because a vertical well was specified. In directional wells, however, wear- and fatigue-related problems can be significant. Estimating fatigue life and wear rates is quite difficult and is beyond the scope of this chapter. The example problem also did not consider the many issues that can arise when wells produce gas. The presence of gas affects both the frictional pressure losses and the hydrostatic gradient, and the corresponding calculations are much more complex. Pump efficiency is also significantly affected by any free gas that enters the pump intake. Most pump vendors have access to software tools that can be used to complete a system design evaluation for these more complex applications.


References

  1. Vogel, J.V.: Inflow Performance Relationships for Solution-Gas Drive Wells,” JPT (1968) 83.
  2. Matthews, C.M., Skoczylas, P., and Zahacy, T.A.: Progressing Cavity Pumping Systems: Design, Operation and Performance Optimization: Short Course Notes, CFER Technologies (2001).

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See also

Downhole PC pump selection and sizing

Rod and tubing design for PCP systems

Alternate PCP system configurations

Progressing cavity pump (PCP) systems

PEH:Progressing Cavity Pumping Systems