PCP system designProgressing 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. Contents [hide] Overview of the design process Design example o 2.1 Problem statement o 2.2 Solution References Noteworthy papers in OnePetro External links See also Category 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 and individual parameters are often adjusted to achieve an optimal design for a particular application. and reservoir data are all useful sources of relevant information. well records. production. fluid properties. As with other artificial-lift systems. it is necessary to determine the anticipated fluid rates. For example. some of these parameters may already be constrained. Other design considerations may produce conflicting results. 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.size is based on such design considerations as flow losses and casing size. As Fig. the primary design considerations for a PCP system are: Fig. Past experience. as is the case with flow losses that affect pump. and rod-string selection. Pump selection and sizing Fluid flow effects and considerations Rod loading Rod fatigue Power transmission selection The first step in the design process is to gather information for the application of interest. 1 shows.[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. tubing. 1—Flowchart illustrating the design process for PCP systems. which complicates the decision-making process. Next. . the design process is generally iterative. Some considerations apply to more than one decision. the use of rod-string centralizers may minimize wear but may also increase flow losses. . initial values for pump intake and discharge pressures. then the design process must be repeated to configure a downhole system or operating parameters that result in reduced system loads. The pump should be set below the perforations at 1010 m [3. reducing the rod-string induced flow restrictions. The well is cased with 177. its speed should not exceed 350 rpm. The oil viscosity is 1. variations in operating conditions. Once a final system design has been established. or sand. However. In new applications. increasing pump speed. Because design optimization based on manual calculations is usually impractical. and pump displacement can be set.000 cp [1000 mPa•s]. For example. pump speed. . After the rod loading and wear considerations are satisfied.8 mm [7 in. while the flowline pressure is 1500 kPa [218 psi]. If the calculated rod stresses exceed the allowable value. or the loading must be decreased through a reduction in the net lift requirements or the use of a smallerdisplacement pump. 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.281 ft] from surface. if there are no pumps available that meet both requirements. Design example Problem statement A vertical well is expected to produce 100 m3/d [629 B/D] of 12°API oil and no water. or decreasing the fluid rate. This allows the designer to select a range of pump models capable of satisfying the desired pump displacement and lift requirements.] OD casing perforated at 1000 m [3.968 ft] from surface. It is quite apparent that the interdependency between the numerous equipment selection and well completion options. and the pump should not be loaded above its rated pressure. net lift. then either the rod-string strength must be increased through the use of a larger rod or higher-strength material. flow losses can be calculated. rod loading. numerous iterations may be required just to establish a workable system.g. 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. reducing discharge pressure requirements. and surface equipment requirements can be evaluated. rod-string/tubing wear. The individual pumps that satisfy the requirements are then evaluated onTHE BASIS OF geometric design and fluid considerations to select the most appropriate pump model. computer programs have been developed to help designers work faster and more effectively. the fluid level is expected to be 600 m [1. then the prescribed pump displacement and lift specifications must be relaxed by decreasing the fluid rate expectations. they can be reduced by increasing the tubing size. The rod stress should be < 80% of yield (assume API Grade D rods).312 ft]. If the available surface equipment cannot meet the polished-rod power requirements. Next. if the predicted rod-string/tubing wear rates are unacceptable. Similarly. gas. 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. The casing is vented to atmosphere. At the desired flow rate. or by implementing some combination of these changes. The following sections provide further details on specific design parameters. then steps must be taken to reduce axial loads (e. use of a smaller pump). the final step in the design process is the selection of surface equipment. If the estimated flow losses are unacceptably high. or the rod string must be reconfigured so that it is less prone to wear. Design a PCP system to produce this well with the following constraints. increasing pump intake pressure. Once a specific pump model has been selected.Once these equipment and operating parameters are established. .9 95 108 114... mm 88...... 1..0 4.5 8..... kPa 12...(2) .The following pumps are available... .. mm 50 54 52 58 74 Minor diameter..70 1.3 114..000 12. 2 through 4... m 4.....(3) .. . mm 38 41 35 44 51 OD..000 18..0 12....000 15....336 m3/d/rpm..... 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..15 0.........0 10..00 Pressure rating...45 0....30 0..000 12.. m3/d/rpm 0...000 Major diameter...3 Length............(4) .. The next step is to determine the differential pressure on the pump using Eqs.0 Solution Using Eq.(1) we can determine the minimum displacement required to achieve the desired flow rate without exceeding the specified maximum pump speed: The pump displacement must be > 0..... Pump A Pump B Pump C Pump D Pump E Displacement. This eliminates Pumps A and B from further consideration. but many different formulations are available in the literature. 101. the remaining 988. with standard couplings (55.6 mm diameter.8 kg/m 3. The liquid hydrostatic head will depend on the location of the top of the pump. and the fluid level is 600 m from surface. there would be a very small annulus between the casing and the pump. the actual gas density will change as the pressure increases. so the top will be at a different location in each case. the top of the pump will be at 1002 m. For now. Note that if 139.7 mm OD casing had been used instead. (Note that this method is an approximation.3 m of coupling separately and then add the two results together to . The gas and liquid hydrostatic pressures can be calculated from the gas and liquid densities and the column heights.7 m of rod and 13. 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.8 kg/m tubing. However. For the specified well depth.7 m (assuming that the top of the pump is at 1002 m) is covered by rod segments. This means that there is 600 m of gas column and 410 m of liquid column. but because the value is so small relative to the other pressures in the system. This gives a gas column hydrostatic pressure of 5 kPa. we will consider the use of 88. the error introduced by this approximation is small.6 mm length).4 mm rods. The pump is seated at 1010 m (intake depth). Also. the flow loss will depend on the selection of tubing and rods. If the pump length is 8 m.9 mm × 13. 7.The pump intake pressure is: Casing-head pressure was defined in the problem statement to be atmospheric pressure.) The density of 12°API oil is 984 kg/m3. with 25. Any flow losses here must also be considered in calculating the pump intake pressure. The pump intake is 1010 m from surface. these losses are small and can be neglected. An average gas density can be estimated from the pressure at surface: 0. because the distance is small and there is a large clearance between the pump and casing. The solution process will be iterative.62 m in length. and the flow losses between the perforations and the intake could be quite significant. for a total length of 13. We can calculate the flow losses past 988. and the hydrostatic head of the liquid in the tubing is 9673 kPa: The calculation of flow losses was not described in detail in this chapter. but the three pump alternatives have different lengths. 131 couplings are needed. including this Handbook. the produced oil must flow from the perforations past the pump to reach the intake.3 m. or 0 kPa (gauge pressure). The pump discharge pressure is calculated from: The tubing-head pressure is given as 1500 kPa. so the liquid hydrostatic pressure is 3958 kPa: In this case. The next task is to estimate the torque in the rods. Also. However.266 N•m. Therefore. and the pump differential pressure... 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. Assuming that the rods and couplings are concentric. as normally expected. and Pump E . Accordingly.. if a larger tubing size that would accommodate the large rotor diameter were used.. All of the pumps have an OD that is less than the drift diameter of even the heaviest-wall 177. the values are: Pump C . the hydraulic torque values are as follows: Pump C . note that Pump E cannot be used with this tubing because the major rotor diameter is larger than the drift diameter of the tubing. Pump D . The ID of 88. and the drift diameter is 72. none of these pumps must be eliminated due to casing size.663 N•m. If. The torque on the pump is given by: The friction torque was estimated in the problem statement to be 20% of the hydraulic torque at the pump’s rated pressure. Only Pumps C and D have a pressure rating exceeding this value. The pump is required to work against a differential pressure of 13 274 kPa.. For example. This approximation neglects the flow effects at the ends of the couplings. Hydraulic torque is calculated from Eq 5: . it is recommended that the ID instead of drift diameter be used.. the flow losses would be somewhat reduced. the hydraulic torque for a differential pressure of 13 274 kPa is calculated for each of these pumps as Thus. For flow calculations. Next. Therefore. but it should still provide adequate results.1026 N•m.obtain the total flow loss.9 mm × 13. Pump D . and Pump E 1470 N•m... we can estimate the friction torque for each pump.. the rods and couplings are not concentric within the tubing. We can now calculate the pump discharge pressure...180 N•m..8 kg/m tubing is 76.. possibly to the point that the pressure rating of Pump E would not be exceeded. so the results are conservative.233 N•m.. .8 mm casing...82 mm.0 mm..(5) From this. Pump E will continue to be considered a potential candidate. the flow losses would be reduced. but such a reduction will not be considered here.. for Pump C.. .(6) Calculation of the uplift forces will be neglected for this example.. 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 . and E are 8. 6: .4 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. d and e are in millimeters...9 × 10–4 when p is in Newtons.3 kN.. the rod weights are 38.45.84.0.124. The total stress of the rods can now be determined. 7.1304 N•m.. For a 25.31. we must calculate the axial load in the rods and the torque. with their respective rod lengths.5 kN...4 kN.. neglecting the uplift forces..85.. Pump D .77.(7) where C = 7. this gives: The maximum stress is 514 MPa. Note that the rod stresses exceed the yield capacity if the other two pumps are used. Pump D ..1768 N•m. D. the rod weight is For pumps D and E. This condition would be in violation of the 80% loading criterion included in the problem statement. hydraulic torque and rod resistive torque values: Pump C ..5 mm. When considering rod loading...0 kN.1 kN. Pump D .. we must recognize that the major diameter is equal to the minor diameter plus twice the eccentricity.. the total axial rod loads corresponding to the three pumps are: Pump C ..2 N•m.0 kN. providing a slightly conservative result... and Pump E ..44..9 kN and 39. Therefore. The rod weight is easily calculated from the specific weight of steel and the rod volume...4 mm rod that is 1002 m long (Pump C) with a steel specific weight of 77 kN/m3. For Pump C. and pressures are in kPa.. So. To get the eccentricity values for the pumps for use in this equation..2 kN.. and Pump E . At a discharge pressure of 17 238 kPa and intake pressure of 3963 kPa.5. and 11.... neglecting the additional weight from the couplings and upsets...912 N•m.. In a vertical well.38. Axial load can be found from Eq.. Pump D . The total rod torque is then the sum of the respective pump friction. and Pump E . Pump load is given by Eq. . 7 as . and Pump E .. the tubing friction loads can usually be considered negligible. respectively. the axial load at the pump is as follows: Pump C .69..4 N•m. which is 88% of the minimum yield for Grade D rods [586 MPa].. the eccentricities for Pumps C. The total axial load on the rod string can be recalculated as follows: Pump C . through the addition of a transfer pump.4 mm rods would be ≈ 1338 kPa. and the corresponding calculations are much more complex. although the use of higher-strength rods may be an option in some cases. The flow losses with 114. Another way to decrease the pressure on the pump is to reduce the flow losses. the torque is 671 N•m. For example. tubing.671 N•m. Estimating fatigue life and wear rates is quite difficult and is beyond the scope of this chapter.0 kN. 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. which accounted for almost half the differential pump pressure.0 kN. and Pump E . The resistive torque is also reduced slightly. the other two pumps will still cause the rod stress to exceed the specified criterion. The type of drive head (right angle or vertical. .To redesign the system to produce the well within the specified parameters. otherwise.67. Pump efficiency is also significantly affected by any free gas that enters the pump intake.382 MPa (65%). Pump D . The rod-string axial load at the surface is 63 kN. There is nothing that can be done to reduce the hydrostatic head on the system while maintaining the same flow rate. wear. and Pump E . If electricity is available but electronic speed control systems are not available or used in the area. Pump C now gives a rod stress that is below 80% of yield. The presence of gas affects both the frictional pressure losses and the hydrostatic gradient. In directional wells. and the operating speed of the polished rod is 261 rpm. it appears that two viable options would be to decrease the differential pressure on the pump. or to increase the strength of the rods. and Pump E . Although using smaller rods would reduce flow losses. then an internal combustion engine must be used. a decrease in flow rate would reduce flow losses and would cause the fluid level in the casing to rise. The tubing-head pressure can typically be reduced significantly only through changes in the gathering system to reduce the flowline pressure.1241 N•m. Most pump vendors have access to software tools that can be used to complete a system design evaluation for these more complex applications. The surfacedrive system must now be established. which normally leads to the selection of a hydraulic system because otherwise the belts would typically have to be very long. the use of a larger tubing string seems quite practical in this case. Pump D . solid or hollow shaft. the pump.63. if electricity is not available. and rods have all been selected. A suitable drive can be selected from any manufacturer’s catalog by comparing these values to the published load and speed limits. direct electric or hydraulic. or through the use of viscosityreducing chemicals at surface. the load capacity would also be reduced (assuming the same material). The total torque on the rod string for the three pump candidates can be recalculated to give the following values: Pump C . hydraulic systems are often still preferred if regular speed adjustments are anticipated. however. With the lower torque and axial loads. the criterion in the problem statement.693 MPa (118%). so this does not appear to be a viable option. 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 example problem also did not consider the many issues that can arise when wells produce gas. However. Pump D .520 MPa (89%). thus increasing the pump intake pressure and decreasing the pump differential pressure. producing a corresponding reduction in the pump hydraulic torque values. the peak rod stress in the three cases is as follows: Pump C . etc.5 kN.93.) normally is based on user preferences and field characteristics.3 mm tubing with 25.and fatigue-related problems can be significant. This example problem did not address wear and fatigue considerations because a vertical well was specified.935 N•m. However. This reduces the differential pressure to 8548 kPa. direct electric drives with a fixed selection of belts/sheaves or gears are typically used. V. ↑ Vogel.1. 2.A.. C. CFER Technologies (2001).: Inflow Performance Relationships for Solution-Gas Drive Wells.M.7 Progressing cavity pumps YR q . Operation and Performance Optimization: Short Course Notes. P. Skoczylas. ↑ Matthews. T. and Zahacy. Noteworthy papers in OnePetro Use this section to list papers in OnePetro that a reader who wants to learn more should definitely read External links Use this section to provide links to relevant material on websites other than PetroWiki and OnePetro 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 Category Categories: 3. J..References 1.: Progressing Cavity Pumping Systems: Design.” JPT (1968) 83.