Patent Publication Number: US-9897083-B2

Title: Calculating downhole cards in deviated wells

Description:
CROSS-REFERENCE TO RELATED APPLICATIONS 
     This application claims the benefit of U.S. Patent Application Ser. No. 61/552,812 entitled “Modified Everitt-Jennings With Dual Iteration on the Damping Factors and Adaptation to Deviated Wells by Including Coulombs Friction” and filed 28 Oct. 2011; Ser. No. 61/598,438 entitled “Modified Everitt-Jennings With Dual Iteration on the Damping Factors” and filed 14 Feb. 2012; Ser. No. 61/605,325 entitled “Implementing Coulombs Friction for the Calculation of Downhole Cards in Deviated Wells” and filed 1 Mar. 2012; and Ser. No. 61/706,489 entitled “Iterating on Damping when Solving the Wave Equation and Computation of Fluid Load Lines and Concavity Testing” and filed 27 Sep. 2012, each of which is incorporated herein by reference in its entirety. This application is also filed concurrently with co-pending application Ser. No. 13/663,161 entitled “Fluid Load Line Calculation and Concavity Test for Downhole Pump Card,” Ser. No. 13/663,167 entitled “Calculating Downhole Pump Card With Iterations on Single Damping Factor,” and Ser. No. 13/663,174 entitled “Calculating Downhole Pump Card With Iterations on Dual Damping Factors,” each of which is incorporated herein by reference. 
    
    
     BACKGROUND OF THE DISCLOSURE 
     A. Sucker Rod Pump System 
     Reciprocating pump systems, such as sucker rod pump systems, extract fluids from a well and employ a downhole pump connected to a driving source at the surface. A rod string connects the surface driving force to the downhole pump in the wall. When operated, the driving source cyclically raises and lowers the downhole pump, and with each stroke, the downhole pump lifts well fluids toward the surface. 
     For example,  FIG. 1  shows a sucker rod pump system  10  used to produce fluid from a well. A downhole pump  14  has a barrel  16  with a standing valve  24  located at the bottom. The standing valve  24  allows fluid to enter from the wellbore, but does not allow the fluid to leave. Inside the pump barrel  16 , a plunger  20  has a traveling valve  22  located at the top. The traveling valve  22  allows fluid to move from below the plunger  20  to the production tubing  18  above, but does not allow fluid to return from the tubing  18  to the pump barrel  16  below the plunger  20 . A driving source (e.g., a pump jack  11 ) at the surface connects by a rod string  12  to the plunger  20  and moves the plunger  20  up and down cyclically in upstrokes and downstrokes. 
     During the upstroke, the traveling valve  22  is closed, and any fluid above the plunger  20  in the production tubing  18  is lifted towards the surface. Meanwhile, the standing valve  24  opens and allows fluid to enter the pump barrel  16  from the wellbore. The highest point of the plunger&#39;s motion is typically referred to as the “top of stroke” (TOS), while the lowest point of the pump plunger&#39;s motion is typically referred to as the “bottom of stroke” (BOS). 
     At the TOS, the standing valve  24  closes and holds in the fluid that has entered the pump barrel  16 . Additionally, at the TOS, the weight of the fluid in the production tubing  18  is supported by the traveling valve  22  in the plunger  20  and, therefore, also by the rod string  12 , which causes the rod string  12  to stretch. 
     During the downstroke, the traveling valve  22  initially remains closed until the plunger  20  reaches the surface of the fluid in the barrel  16 . Sufficient pressure builds up in the fluid below the traveling valve  22  to balance the pressure. The build-up of pressure in the pump barrel  16  reduces the load on the rod string  12  so that the rod string  12  relaxes. 
     This process takes place during a finite amount of time when the plunger  20  rests on the fluid, and the pump jack  11  at the surface allows the top of the rod string  12  to move downward. The position of the pump plunger  20  at this time is known as the “transfer point” because the load of the fluid column in the production tubing  18  is transferred from the traveling valve  22  to the standing valve  24 . This results in a rapid decrease in load on the rod string  12  during the transfer. 
     After the pressure balances, the traveling valve  22  opens and the plunger  20  continues to move downward to its lowest position (i.e., the BOS). The movement of the plunger  20  from the transfer point to the BOS is known as the “fluid stroke” and is a measure of the amount of fluid lifted by the pump  14  on each stroke. In other words, the portion of the pump stroke below the transfer point may be interpreted as the percentage of the pump stroke containing fluid, and this percentage corresponds to the pump&#39;s fillage. Thus, the transfer point can be computed using a pump fillage calculation. 
     If there is sufficient fluid in the wellbore, the pump barrel  16  may be completely filled during an upstroke. Yet, under some conditions, the pump  14  may not be completely filled with fluid on the upstroke so there may be a void left between the fluid and the plunger  20  as it continues to rise. Operating the pump system  10  with only a partially filled pump barrel  16  is inefficient and, therefore, undesirable. In this instance, the well is said to be “pumped off,” and the condition is known as “pounding,” which can damage various components of the pump system. For a pumped off well, the transfer point most likely occurs after the TOS of the plunger  20 . 
     Typically, there are no sensors to measure conditions at the downhole pump  14 , which may be located thousands of feet underground. Instead, numerical methods are used calculate the position of the pump plunger  20  and the load acting on the plunger  20  from measurements of the position and load for the rod string  12  at the pump jack  11  located at the surface. These measurements are typically made at the top of the polished rod  28 , which is a portion of the rod string  12  passing through a stuffing box  13  at the wellhead. A pump controller  26  is used for monitoring and controlling the pump system  10 . 
     To efficiently control the reciprocating pump system  10  and avoid costly maintenance, a rod pump controller  26  can gather system data and adjust operating parameters of the system  10  accordingly. Typically, the rod pump controller  26  gathers system data such as load and rod string displacement by measuring these properties at the surface. While these surface-measured data provide useful diagnostic information, they may not provide an accurate representation of the same properties observed downhole at the pump. Because these downhole properties cannot be easily measured directly, they are typically calculated from the surface-measured properties. 
     Methods for determining the operational characteristics of the downhole pump  20  have used the shape of the graphical representation of the downhole data to compute various details. For example, U.S. Pat. No. 5,252,031 to Gibbs, entitled “Monitoring and Pump-Off Control with Downhole Pump Cards,” teaches a method for monitoring a rod pumped well to detect various pump problems by utilizing measurements made at the surface to generate a downhole pump card. The graphically represented downhole pump card may then be used to detect the various pump problems and control the pumping unit. Other techniques for determining operational characteristics are disclosed in U.S. Patent Publication Nos. 2011/0091332 and 2011/0091335, which are both incorporated herein by reference in their entireties. 
     B. Everitt-Jennings Method 
     In techniques to determine operational characteristics of a sucker rod pump system  10  as noted above, software analysis computes downhole data (i.e., a pump card) using position and load data measured at the surface. The most accurate and popular of these methods is to compute the downhole card from the surface data by solving a one-dimensional damped wave equation, which uses surface position and load as recorded at the surface. 
     Various algorithms exist for solving the wave equation. Snyder solved the wave equation using a method of characteristics. See Snyder, W. E., “A Method for Computing Down-Hole Forces and Displacements in Oil Wells Pumped With Sucker Rods,” Paper 851-37-K, 1963. Gibbs employed separation of variables and Fourier series in what can be termed the “Gibb&#39;s method.” See Gibbs, S. G. et al., “Computer Diagnosis of Down-Hole Conditions in Sucker Rod Pumping Wells,” JPT (January 1996) 91-98;  Trans ., AIME, 237; Gibbs, S. G., “A Review of Methods for Design and Analysis of Rod Pumping Installations,” SPE 9980, 1982; and U.S. Pat. No. 3,343,409. 
     In 1969, Knapp introduced finite differences to solve the wave equation. See Knapp, R. M., “A Dynamic Investigation of Sucker-Rod Pumping,” MS thesis, U. of Kansas, Topeka (January 1969). This is also the method used by Everitt and Jennings. See Everitt, T. A. and Jennings, J. W., “An Improved Finite-Difference Calculation of Downhole Dynamometer Cards for Sucker-Rod Pumps,” SPE 18189, 1992; and Pons-Ehimeakhe, V., “Modified Everitt-Jennings Algorithm With Dual Iteration on the Damping Factors,” 2012 South Western Petroleum Short Course. The Everitt-Jennings method has also been implemented and modified by Weatherford International. See Ehimeakhe, V., “Comparative Study of Downhole Cards Using Modified Everitt-Jennings Method and Gibbs Method,” Southwestern Petroleum Short Course 2010. 
     To solve the one-dimensional wave equation, the Everitt-Jennings method uses finite differences. The rod string is divided into M finite difference nodes of length L i  (ft), density ρ i  (lbm/ft 3 ) and area A i  (in 2 ). If we let u=u(x, t) be the displacement of position x at time t in a sucker rod pump system, the condensed one-dimensional wave equation reads: 
     
       
         
           
             
               
                 
                   
                     
                       v 
                       2 
                     
                     ⁢ 
                     
                       
                         
                           ∂ 
                           2 
                         
                         ⁢ 
                         u 
                       
                       
                         ∂ 
                         
                           x 
                           2 
                         
                       
                     
                   
                   = 
                   
                     
                       
                         
                           ∂ 
                           2 
                         
                         ⁢ 
                         u 
                       
                       
                         ∂ 
                         
                           t 
                           2 
                         
                       
                     
                     + 
                     
                       D 
                       ⁢ 
                       
                         
                           ∂ 
                           u 
                         
                         
                           ∂ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
     where the acoustic velocity is given by: 
             v   =         144   ⁢   Eg     ρ             
and D represents a damping factor.
 
     The first and second derivatives with respect to time are replaced by the first-order-correct forward differences and second-order-correct central differences. The second derivative with respect to position is replaced by a slightly rearranged second-order-correct central difference. 
     In the method, the damping factor D is automatically selected by using an iteration on the system&#39;s net stroke (NS) and the damping factor D. The damping factor D can be computed by the equation: 
     
       
         
           
             
               
                 
                   D 
                   = 
                   
                     
                       
                         
                           ( 
                           550 
                           ) 
                         
                         ⁢ 
                         
                           ( 
                           
                             144 
                             ⁢ 
                             g 
                           
                           ) 
                         
                       
                       
                         
                           2 
                         
                         ⁢ 
                         π 
                       
                     
                     ⁢ 
                     
                       
                         
                           ( 
                           
                             
                               H 
                               PR 
                             
                             - 
                             
                               H 
                               H 
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           τ 
                           2 
                         
                       
                       
                         
                           ( 
                           
                             ∑ 
                             
                               
                                 ρ 
                                 i 
                               
                               ⁢ 
                               
                                 A 
                                 i 
                               
                               ⁢ 
                               
                                 L 
                                 i 
                               
                             
                           
                           ) 
                         
                         ⁢ 
                         
                           S 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
           
         
       
     
     Where H PR  is the polished rod horsepower (hp), S is the net stroke (in), τ is the period of one stroke (sec.), and H HYD  is the hydraulic horsepower (hp) obtained as follows:
 
 H   HYD =(7.36·10 −6 ) QγF   t   (3)
 
where Q is the pump production rate (B/D), γ is the fluid specific gravity, and F t  is the fluid level (ft). The pump production rate is given by:
 
 Q =(0.1166)( SPM ) Sd   2   (4)
 
where SPM is the speed of the pumping unit in strokes/minute, and d is the diameter of the plunger.
 
     Additional details on the derivation of the damping factor D in equation (2) and the original iteration on the net stroke and damping factor algorithm are provided in Everitt, T. A. and Jennings, J. W., “An Improved Finite-Difference Calculation of Downhole Dynamometer Cards for Sucker-Rod Pumps,” SPE 18189, 1992. 
     A modified Everitt-Jennings method also uses finite differences to solve the wave equation. As before, the rod string is discretized into M finite difference elements, and position and load (including stress) are computed at each increment down the wellbore. Then, as shown in  FIG. 2 , an iteration is performed on the net stroke and damping factor, which automatically selects a damping factor for each stroke. 
     The wave equation is initially solved to calculate the downhole card using surface measurements and an initial damping factor D set to 0.5 (Block  42 ). The initial net stroke S 0  is determined from the computed card, and the fluid level in the well is calculated (Block  44 ). At this point, a new damping factor D is calculated from equation (2) (Block  46 ) and so forth, and the downhole card is again computed with the new damping factor D (Block  48 ). Based on the recalculated downhole card, a new net stroke S is determined (Block  50 ). 
     At this point, a check is then made to determine whether the newly determined net stroke S is close within some tolerance ε of the initial or previous net stroke (Decision  52 ). If not, then another iteration is needed, and the process  40  returns to calculating the damping factor D (Block  46 ). If the newly determined net stroke is close to the previously determined net stroke (yes at Decision  52 ), then the iteration for determining the net stroke can stop, and the process  40  continues on to iterate on the damping factor D using the converged net stroke S (Block  54 ). The downhole data is then calculated using the newly calculated damping factor D (Block  56 ), and the pump horsepower H pump  is then calculated (Block  58 ). 
     At this point, a check is made to see if the pump horsepower H pump  is close within some tolerance to the hydraulic horsepower H hyd  (Decision  60 ). If so, then the process  40  ends as successfully calculating the downhole pump card with converged net stroke and damping factor D (Block  62 ). If the pump horsepower H pump  and the hydraulic horsepower H hyd  are not close enough (no at Decision  60 ), then the process  40  adjusts the current damping factor D by a ratio of the pump horsepower H Pump  and the hydraulic horsepower H Hyd  (Block  64 ). The process  40  of calculating the pump card with this adjusted damping factor D is repeated until the values for the pump and hydraulic horsepower HPump and HHyd are close within the specified tolerance (Blocks  56  through  64 ). 
     The advantage of the automatic iteration on the net stroke and the damping factor D as set forth above is that the damping factor D is adjusted automatically without human intervention. Thus, users managing a medium group to a large group of wells do not have to spend time manually adjusting the damping factor D as may be required by other methods. 
     C. Deviated Well Model 
     As noted above, most of the methods presently used to compute downhole data using surface position and load as recorded by a dynamometer system at the surface rely on a vertical-hole model that does not take into consideration deviation of the well. For example,  FIG. 3A  schematically shows a vertical model  30  of a vertical well  18  having a rod string  28  disposed therein. With the well model  30  being vertical, the only relevant friction forces are of viscous in nature. The viscous friction F v  is the result of viscous forces arising in the annular space during a pumping cycle, which are proportional to the velocity of the axial displacement u. 
     However, when dealing with a deviated well such as shown in a deviated model  32  shown somewhat exaggerated in  FIG. 3B , mechanical friction F m  arises from the contact between the tubing  18 , the rod string  28 , and the couplings  29 . Even though those forces F m  can be ignored when the well is mostly vertical, they have to be accounted for when the well is deviated. If the algorithm used to compute the downhole data does not take into consideration the mechanical friction F m  for a deviated well, the resulting downhole card can appear distorted. This condition cannot be helped by changing the viscous damping factor D in the wave equation. 
     Thus, the vertical model is not well-suited for calculating downhole data when the sucker rod pump system  10  is used in a deviated well. Primarily, the dynamic behavior of the rod string  28  is different for deviated wells than for vertical wells. Indeed, in vertical wells, the rod string  28  is assumed to not move laterally. In deviated wells, however, mechanical friction F m  becomes non-negligible because there is extensive contact between the rod string  28 , the couplings  29 , and the tubing  18 . Also, since the well is deviated, some sections of the rod string  28  can be bent between two couplings  29  in the middle of a dog leg turn, which introduces the concept of curvature of the rod string  28  as well. 
     The above equations discussed for the wave equation only consider friction forces of a viscous nature in the vertical model. Yet, the friction forces particular to deviated wells are of viscous and mechanical nature, as detailed above. Although mechanical friction F m  has generally been ignored, it has since been addressed. For example, to deal with the Coulombs friction that results from the mechanical friction in a deviated well, the most well-known technique have been disclosed by Gibbs and Lukasiewicz. See Gibbs, S. G., “Design and Diagnosis of Deviated Rod-Pumped Wells,” SPE Annual Technical Conference and Exhibition, Oct. 6-9, 1991; and Lukasiewicz, S. A., “Dynamic Behavior of the Sucker Rod String in the Inclined Well,” Production Operations Symposium, Apr. 7-9, 1991, both of which are incorporated herein by reference. 
     To deal with mechanical friction in deviated wells, Gibbs modified the wave equation by adding a Coulombs friction term to it. For example, US Pat. Publication 2010/0111716 to Gibbs et al. includes a term C(x) in the wave equation that represents the rod and tubing drag force. By contrast, Lukasiewicz derived equations for axial and transverse displacement of the rod element, creating a system of coupled differential equations. 
     D. Equations for Axial and Transverse Displacement of Rod Element 
     As recognized in Lukasiewicz, a rod string in a deviated well moves longitudinally up and down (i.e., axially) and also moves laterally (i.e., transversely). Thus, the behavior of the axial stress waves as well as the transverse stress waves of a rod element can be analyzed to better characterize the behavior of the rod string  28  in the deviated well. 
     To that end,  FIG. 4  diagrams dynamic behavior of a rod element  34  of a sucker rod pump system for a deviated well model  32 . This diagram shows the various forces acting on the rod element  34  in the axial and transverse directions. As represented here, u(s, t) is the axial displacement of the rod element  34  of length ds, and v(s, t) is the transverse displacement of the rod element  34 . The radius of curvature R φ  can be calculated along with the Cartesian coordinates of the wellbore path using a deviation survey. Several methods are available for such calculations, such as a minimum curvature method or a radius-of-curvature method, as disclosed in Gibbs, S. G., “Design and Diagnosis of Deviated Rod-Pumped Wells”, SPE Annual Technical Conference and Exhibition, Oct. 6-9, 1991, which is incorporated herein by reference. 
     In the diagram of the forces acting on the rod element  34 , the radius of curvature R φ  is displayed as an arrow going from the center of the curvature to the rod element  34  of length ds. The axial force denoted F acts upwards and downwards on the rod element  34 . The axial force, therefore, has an axial component as well as a transverse component. The Coulombs friction force F t  opposes the movement of the rod element  34  at the point of contact between the rod element  34  and the tubing  18 . The weight W is shown as the gravitational force pulling downward on the rod element  34 . A normal force N acts perpendicularly on the rod element  34  facing the center of curvature. Both the weight W and the normal force N have axial and transverse components as well. 
     Thus, the axial direction (i.e., the direction tangential to the rod) can be characterized with the following axial equation of motion: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           ∂ 
                           F 
                         
                         
                           ∂ 
                           s 
                         
                       
                       - 
                       
                         A 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         γ 
                         ⁢ 
                         
                           
                             
                               ∂ 
                               2 
                             
                             ⁢ 
                             u 
                           
                           
                             ∂ 
                             
                               t 
                               2 
                             
                           
                         
                       
                       + 
                       
                         γ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         gA 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         cos 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         θ 
                       
                       - 
                       
                         D 
                         ⁢ 
                         
                           
                             ∂ 
                             u 
                           
                           
                             ∂ 
                             t 
                           
                         
                       
                       - 
                       
                         F 
                         t 
                       
                     
                     = 
                     0 
                   
                   , 
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
     Here, F is the axial force in the rod, u(t) is the axial displacement, A is the rod cross-sectional area, γ is the density, g is the acceleration of gravity, θ is the angle of inclination, D is the viscous damping coefficient, F t  is the friction force from the tubing  18 , s is the length measured along the curved rod, and t is time. 
     As noted above, the force F t  is the Coulombs friction force, which is a nonlinear force that tends to oppose the motion of bodies within a mechanical system. Coulombs friction is representative of dry friction, which resists relative lateral motion of two solid surfaces in contact. The relative motion of the rod string  28 , tubing  18 , and couplings  18  as seen in  FIG. 1  pressing against each other is a source of energy dissipation when the well is pumping. 
     In the transverse direction, the transverse equation of motion can be characterized as: 
     
       
         
           
             
               
                 
                   
                     
                       
                         EI 
                         ⁢ 
                         
                           
                             
                               ∂ 
                               2 
                             
                             
                               ∂ 
                               
                                 s 
                                 2 
                               
                             
                           
                           ⁡ 
                           
                             [ 
                             
                               
                                 
                                   
                                     ∂ 
                                     2 
                                   
                                   ⁢ 
                                   v 
                                 
                                 
                                   ∂ 
                                   
                                     s 
                                     2 
                                   
                                 
                               
                               + 
                               
                                 1 
                                 
                                   R 
                                   φ 
                                 
                               
                             
                             ] 
                           
                         
                       
                       + 
                       
                         γ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         A 
                         ⁢ 
                         
                           
                             
                               ∂ 
                               2 
                             
                             ⁢ 
                             v 
                           
                           
                             ∂ 
                             
                               t 
                               2 
                             
                           
                         
                       
                       + 
                       
                         n 
                         t 
                       
                       + 
                       
                         n 
                         p 
                       
                       + 
                       
                         
                           D 
                           t 
                         
                         ⁢ 
                         
                           
                             ∂ 
                             v 
                           
                           
                             ∂ 
                             t 
                           
                         
                       
                       + 
                       
                         F 
                         R 
                       
                       - 
                       
                         γ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         gA 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         sin 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         θ 
                       
                     
                     = 
                     0 
                   
                   , 
                   
                     
 
                   
                   ⁢ 
                   
                     
                       
                         EI 
                         ⁢ 
                         
                           
                             
                               ∂ 
                               4 
                             
                             ⁢ 
                             v 
                           
                           
                             ∂ 
                             
                               s 
                               4 
                             
                           
                         
                       
                       + 
                       
                         EI 
                         ⁢ 
                         
                           
                             ∂ 
                             2 
                           
                           
                             ∂ 
                             
                               s 
                               2 
                             
                           
                         
                         ⁢ 
                         
                           1 
                           
                             R 
                             φ 
                           
                         
                       
                       + 
                       
                         γ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         A 
                         ⁢ 
                         
                           
                             
                               ∂ 
                               2 
                             
                             ⁢ 
                             v 
                           
                           
                             ∂ 
                             
                               t 
                               2 
                             
                           
                         
                       
                       + 
                       
                         n 
                         t 
                       
                       + 
                       
                         n 
                         p 
                       
                       + 
                       
                         
                           D 
                           t 
                         
                         ⁢ 
                         
                           
                             ∂ 
                             v 
                           
                           
                             ∂ 
                             t 
                           
                         
                       
                       + 
                       
                         F 
                         R 
                       
                       - 
                       
                         γ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         gA 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         sin 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         θ 
                       
                     
                     = 
                     0. 
                   
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
           
         
       
     
     Here, EI is the bending stiffness, E is Young&#39;s modulus of elasticity, I is the bending moment, D t  is the viscous damping factor in the transverse direction, n t  is the transverse normal force from the tubing  18 , an n p  is the transverse normal force from the liquid under pressure p, and 
             1   R         
is an actual radius or curvature given by
 
     
       
         
           
             
               1 
               R 
             
             = 
             
               
                 1 
                 
                   R 
                   φ 
                 
               
               + 
               
                 
                   
                     
                       ∂ 
                       2 
                     
                     ⁢ 
                     v 
                   
                   
                     ∂ 
                     
                       s 
                       2 
                     
                   
                 
                 . 
               
             
           
         
       
     
     As demonstrated by Lukasiewicz, the axial force can be introduced into the axial equation of motion (1) to give: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             
                               ∂ 
                               2 
                             
                             ⁢ 
                             u 
                           
                           ⁢ 
                           
                               
                           
                         
                         
                           ∂ 
                           
                             s 
                             2 
                           
                         
                       
                       + 
                       
                         
                           
                             ∂ 
                             v 
                           
                           
                             ∂ 
                             s 
                           
                         
                         · 
                         
                           
                             
                               
                                 ∂ 
                                 2 
                               
                               ⁢ 
                               v 
                             
                             ⁢ 
                             
                                 
                             
                           
                           
                             ∂ 
                             
                               s 
                               2 
                             
                           
                         
                       
                       - 
                       
                         
                           1 
                           
                             a 
                             2 
                           
                         
                         ⁢ 
                         
                           
                             
                               
                                 ∂ 
                                 2 
                               
                               ⁢ 
                               u 
                             
                             ⁢ 
                             
                                 
                             
                           
                           
                             ∂ 
                             
                               t 
                               2 
                             
                           
                         
                       
                       - 
                       
                         
                           D 
                           AE 
                         
                         ⁢ 
                         
                           
                             ∂ 
                             u 
                           
                           
                             ∂ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           
                             γ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             g 
                           
                           E 
                         
                         ⁢ 
                         cos 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         θ 
                       
                       - 
                       
                         
                           F 
                           t 
                         
                         AE 
                       
                     
                     = 
                     0 
                   
                   , 
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
     Here, α is the acoustic velocity of the rod element  34 . Furthermore, by assuming that the rod element  34  lies on the tubing  18  in between couplings  29 , the axial equation of motion (1) can be written as: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             
                               ∂ 
                               2 
                             
                             ⁢ 
                             u 
                           
                           ⁢ 
                           
                               
                           
                         
                         
                           ∂ 
                           
                             s 
                             2 
                           
                         
                       
                       - 
                       
                         
                           1 
                           
                             a 
                             2 
                           
                         
                         ⁢ 
                         
                           
                             
                               
                                 ∂ 
                                 2 
                               
                               ⁢ 
                               u 
                             
                             ⁢ 
                             
                                 
                             
                           
                           
                             ∂ 
                             
                               t 
                               2 
                             
                           
                         
                       
                       - 
                       
                         
                           D 
                           AE 
                         
                         ⁢ 
                         
                           
                             ∂ 
                             u 
                           
                           
                             ∂ 
                             t 
                           
                         
                       
                       + 
                       
                         
                           μ 
                           R 
                         
                         ⁢ 
                         
                           
                             ∂ 
                             u 
                           
                           
                             ∂ 
                             s 
                           
                         
                       
                       + 
                       
                         
                           
                             γ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             g 
                           
                           E 
                         
                         ⁢ 
                         cos 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         θ 
                       
                       - 
                       
                         
                           μ 
                           E 
                         
                         ⁢ 
                         
                           ( 
                           
                             γ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             g 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             sin 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             θ 
                           
                           ) 
                         
                       
                     
                     = 
                     0 
                   
                   , 
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
           
         
       
     
     Additional details on these equations and the axial force are disclosed in Lukasiewicz, S. A., “Dynamic Behavior of the Sucker Rod String in the Inclined Well,” Production Operations Symposium, Apr. 7-9, 1991, which has been incorporated herein by reference. 
     As can be seen, axial equation of motion (3) uses the surface position of the rod string to calculate the downhole position at each finite difference node down the wellbore until the node right above the downhole pump. The axial and transverse equations of motion (3) and (2) are combined to form a system of two coupled non-linear differential equations of fourth order. 
     It is important to note that Coulombs friction (i.e., the mechanical friction that arises from the contact between the rods  28 , tubing  18 , and couplings  29 ) can be consequential in a deviated well and cannot be simulated using viscous damping. In particular, the Coulombs friction forces are not proportional to the velocity of the rod element as the viscous friction forces are. In some cases, the viscous damping factor can be increased to remove extra friction, but the downhole friction due to mechanical cannot be removed. If the viscous damping is pushed too far, the effects of the mechanical friction can look like they have been removed, but in reality the downhole data no longer represent what is happening at the downhole pump. 
     In equation (2), the second term is nonlinear and represents the effect of the vertical deflection on the axial displacement. It is noted that the equations given above are the same equations presented by Lukasiewicz, and that the model developed by Gibbs ignores the transverse movement of the rod string  28 . 
     Being able to treat the mechanical friction when dealing with deviated wells has been a growing concern in the industry. Often, users try to remedy the downhole friction on a downhole card by modifying the viscous damping factor or by adding a drag force term (as done by Gibbs). Yet, this can essentially falsify the downhole results and can hide downhole conditions. 
     Although the prior art (and especially Lukasiewicz) has characterized the equations for motion of a rod string in a deviated well, practical techniques for performing the calculations are needed. This is especially true when the calculations are performed by a pump controller or other processing device, which may have limited processing capabilities. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1  illustrates a sucker rod pump system with a controller for controlling the system&#39;s pump. 
         FIG. 2  illustrates iteration on netstroke and damping factor for the modified Everitt-Jennings algorithm to compute a pump card according to the prior art. 
         FIG. 3A  diagrams a vertical well model. 
         FIG. 3B  diagrams a deviated well model. 
         FIG. 4  diagrams dynamic behavior of a rod element of a sucker rod pump system for a deviated well. 
         FIG. 5  illustrates a flowchart of a process for calculating downhole data for a sucker rod pump system in a deviated well. 
         FIG. 6A  illustrates a pump controller according the present disclosure for a sucker-rod pump system. 
         FIG. 6B  illustrates a schematic of the pump controller for controlling/diagnosing the sucker-rod pump system according to the present disclosure. 
     
    
    
     DETAILED DESCRIPTION OF THE DISCLOSURE 
     According to the present disclosure, the modified Everitt-Jennings algorithm is used to compute downhole data from surface data by solving the one dimensional damped wave equation with finite differences. The one-dimensional damped wave equation, however, only takes into consideration friction of a viscous nature and ignores any type of mechanical friction. If the well is substantially vertical, mechanical friction is negligible, and the obtained downhole data may be accurate. However, in deviated or horizontal wells, mechanical friction between the rods, couplings, and tubing needs to be considered. According to this disclosure, the modified Everitt-Jennings method is adapted to utilize finite differences to incorporate mechanical friction factors in the calculation of downhole data in deviated or horizontal wells. 
     To do this, the teachings of the present disclosure use a finite difference approach to treat a system of two coupled non-linear differential equations, which encompass the forces acting on a rod element in a deviated well. The axial displacement and the transverse displacement of the rod element are considered, providing a complete model for analyzing the downhole conditions. As such, the teachings of the present disclosure utilize the equations as derived by Lukasiewicz, which has been described previously and incorporated herein by reference. The axial and transverse equations of motion for the rod element have been noted in the background section of the present disclosure. 
     Referring now to  FIG. 5 , a process  100  is illustrated in flowchart form for solving of the system of coupled differential equations for axial and transverse displacement of a rod element in a deviated well. For the derivatives that appear in the previous equations (2), (3), and (4), Taylor series approximations are used to generate finite difference analogs (Block  102 ). For the first and second derivatives, a first-order-correct central difference and a second-order-correct central difference are used, respectively. For more details on the derivation of the second derivative analog with respect to displacement. See Everitt, T. A. and Jennings, J. W.: “An Improved Finite-Difference Calculation of Downhole Dynamometer Cards for Sucker-Rod Pumps,” SPE 18189, 1988, which is incorporated herein by reference. 
     In particular, the finite difference analogs are as follows (the subscript i represents the node at an axial distance of the rod string and the subscript j represents the timestep). For the space discretization, the finite difference analogs are: 
     
       
         
           
             
               
                 
                   ∂ 
                   u 
                 
                 
                   ∂ 
                   
                     s 
                     
                       i 
                       , 
                       j 
                     
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         u 
                         
                           
                             i 
                             + 
                             1 
                           
                           , 
                           j 
                         
                       
                       - 
                       
                         u 
                         
                           i 
                           , 
                           j 
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     s 
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       s 
                     
                     2 
                   
                   ⁢ 
                   
                     
                       
                         ∂ 
                         2 
                       
                       ⁢ 
                       u 
                     
                     
                       ∂ 
                       
                         s 
                         2 
                       
                     
                   
                 
               
             
             , 
             
               
 
             
             ⁢ 
             
               
                 
                   ∂ 
                   v 
                 
                 
                   ∂ 
                   
                     s 
                     
                       i 
                       , 
                       j 
                     
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         v 
                         
                           
                             i 
                             + 
                             1 
                           
                           , 
                           j 
                         
                       
                       - 
                       
                         v 
                         
                           i 
                           , 
                           j 
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     s 
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       s 
                     
                     2 
                   
                   ⁢ 
                   
                     
                       
                         ∂ 
                         2 
                       
                       ⁢ 
                       v 
                     
                     
                       ∂ 
                       
                         s 
                         2 
                       
                     
                   
                 
               
             
             , 
             
               
 
             
             ⁢ 
             
               
                 
                   
                     ∂ 
                     2 
                   
                   ⁢ 
                   u 
                 
                 
                   ∂ 
                   
                     s 
                     
                       i 
                       , 
                       j 
                     
                     2 
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         u 
                         
                           
                             i 
                             + 
                             1 
                           
                           , 
                           j 
                         
                       
                       - 
                       
                         2 
                         ⁢ 
                         
                           u 
                           
                             i 
                             , 
                             j 
                           
                         
                       
                       + 
                       
                         u 
                         
                           
                             i 
                             - 
                             1 
                           
                           , 
                           j 
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       s 
                       2 
                     
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         s 
                         2 
                       
                     
                     12 
                   
                   ⁢ 
                   
                     
                       
                         ∂ 
                         4 
                       
                       ⁢ 
                       u 
                     
                     
                       ∂ 
                       
                         s 
                         4 
                       
                     
                   
                 
               
             
             , 
             
               
 
             
             ⁢ 
             
               
                 
                   
                     ∂ 
                     2 
                   
                   ⁢ 
                   v 
                 
                 
                   ∂ 
                   
                     s 
                     
                       i 
                       , 
                       j 
                     
                     2 
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         v 
                         
                           
                             i 
                             + 
                             1 
                           
                           , 
                           j 
                         
                       
                       - 
                       
                         2 
                         ⁢ 
                         
                           v 
                           
                             i 
                             , 
                             j 
                           
                         
                       
                       + 
                       
                         u 
                         
                           
                             i 
                             - 
                             1 
                           
                           , 
                           j 
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       s 
                       2 
                     
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         s 
                         2 
                       
                     
                     12 
                   
                   ⁢ 
                   
                     
                       
                         
                           ∂ 
                           4 
                         
                         ⁢ 
                         t 
                       
                       
                         ∂ 
                         
                           s 
                           4 
                         
                       
                     
                     . 
                   
                 
               
             
           
         
       
     
     For the discretization in time, the finite difference analogs are: 
     
       
         
           
             
               
                 
                   ∂ 
                   u 
                 
                 
                   ∂ 
                   
                     t 
                     
                       i 
                       , 
                       j 
                     
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         u 
                         
                           i 
                           , 
                           
                             j 
                             + 
                             1 
                           
                         
                       
                       - 
                       
                         u 
                         
                           i 
                           , 
                           j 
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     t 
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       t 
                     
                     2 
                   
                   ⁢ 
                   
                     
                       
                         ∂ 
                         2 
                       
                       ⁢ 
                       u 
                     
                     
                       ∂ 
                       
                         t 
                         2 
                       
                     
                   
                 
               
             
             , 
             
               
 
             
             ⁢ 
             
               
                 
                   ∂ 
                   v 
                 
                 
                   ∂ 
                   
                     t 
                     
                       i 
                       , 
                       j 
                     
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         v 
                         
                           i 
                           , 
                           
                             j 
                             + 
                             1 
                           
                         
                       
                       - 
                       
                         v 
                         
                           i 
                           , 
                           j 
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     t 
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       t 
                     
                     2 
                   
                   ⁢ 
                   
                     
                       
                         ∂ 
                         2 
                       
                       ⁢ 
                       v 
                     
                     
                       ∂ 
                       
                         t 
                         2 
                       
                     
                   
                 
               
             
             , 
             
               
 
             
             ⁢ 
             
               
                 
                   
                     ∂ 
                     2 
                   
                   ⁢ 
                   u 
                 
                 
                   ∂ 
                   
                     t 
                     
                       i 
                       , 
                       j 
                     
                     2 
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         u 
                         
                           i 
                           , 
                           
                             j 
                             + 
                             1 
                           
                         
                       
                       - 
                       
                         2 
                         ⁢ 
                         
                           u 
                           
                             i 
                             , 
                             j 
                           
                         
                       
                       + 
                       
                         u 
                         
                           i 
                           , 
                           
                             j 
                             - 
                             1 
                           
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       t 
                       2 
                     
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         t 
                         2 
                       
                     
                     12 
                   
                   ⁢ 
                   
                     
                       
                         ∂ 
                         4 
                       
                       ⁢ 
                       u 
                     
                     
                       ∂ 
                       
                         t 
                         4 
                       
                     
                   
                 
               
             
             , 
             
               
 
             
             ⁢ 
             
               
                 
                   
                     ∂ 
                     2 
                   
                   ⁢ 
                   v 
                 
                 
                   ∂ 
                   
                     t 
                     
                       i 
                       , 
                       j 
                     
                     2 
                   
                 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         v 
                         
                           i 
                           , 
                           
                             j 
                             + 
                             1 
                           
                         
                       
                       - 
                       
                         2 
                         ⁢ 
                         
                           v 
                           
                             i 
                             , 
                             j 
                           
                         
                       
                       + 
                       
                         v 
                         
                           i 
                           , 
                           
                             j 
                             - 
                             1 
                           
                         
                       
                     
                     ) 
                   
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       t 
                       2 
                     
                   
                 
                 - 
                 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         t 
                         2 
                       
                     
                     12 
                   
                   ⁢ 
                   
                     
                       
                         
                           ∂ 
                           4 
                         
                         ⁢ 
                         v 
                       
                       
                         ∂ 
                         
                           t 
                           4 
                         
                       
                     
                     . 
                   
                 
               
             
           
         
       
     
     The analogs for the derivatives with respect to time are straight forward. However, the derivatives with respect to space of a degree greater than one preferably have the finite difference analogs split into several equations to accommodate different taper properties of the rod string  28 . Splitting the finite difference analogs into several equations primarily allows one to pick a change in length Δs of the curved rod so that values for the position, load, and stress can be calculated at chosen steps down the wellbore as opposed to having to interpolate between fixed points. This option allows a user more freedom to refine the discretization to optimize stress analysis. 
     To handle the fourth order derivative with respect to displacement, a central finite difference scheme of second order is used: 
     
       
         
           
             
               
                 
                   ∂ 
                   4 
                 
                 ⁢ 
                 v 
               
               
                 ∂ 
                 
                   s 
                   
                     i 
                     , 
                     j 
                   
                   4 
                 
               
             
             = 
             
               
                 
                   
                     v 
                     
                       
                         i 
                         + 
                         2 
                       
                       , 
                       j 
                     
                   
                   - 
                   
                     4 
                     ⁢ 
                     
                       v 
                       
                         
                           i 
                           + 
                           1 
                         
                         , 
                         j 
                       
                     
                   
                   + 
                   
                     6 
                     ⁢ 
                     
                       v 
                       
                         i 
                         , 
                         j 
                       
                     
                   
                   - 
                   
                     4 
                     ⁢ 
                     
                       v 
                       
                         
                           i 
                           - 
                           1 
                         
                         , 
                         j 
                       
                     
                   
                   + 
                   
                     v 
                     
                       
                         i 
                         - 
                         2 
                       
                       , 
                       j 
                     
                   
                 
                 
                   Δ 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     s 
                     4 
                   
                 
               
               - 
               
                 
                   
                     Δ 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       s 
                       2 
                     
                   
                   6 
                 
                 ⁢ 
                 
                   
                     
                       
                         ∂ 
                         6 
                       
                       ⁢ 
                       v 
                     
                     
                       ∂ 
                       
                         s 
                         6 
                       
                     
                   
                   . 
                 
               
             
           
         
       
     
     To run a diagnostic model of a deviated well based on surface measurements and calculate a downhole pump card, the transverse and axial equations of motion (2) and (3) must be solved simultaneously. The teachings of the present disclosure provide a solution to the model for a deviated well, as discussed in detail below. 
     Without loss of generality, as an initialization step (Block  104 ), the rod string  28  can be assumed to lie on the tubing  18  to solve for an initial value for the axial displacement u. In other words, it is assumed that there is not transverse displacement, i.e., v=0. In this case, a value of the coefficient of friction (μ) (for the friction force acting on the rod string  28  from the tubing  18 ) is selected as 0.05, which can be based on empirical evidence or other information (Block  106 ), and the simplified version of the axial equation of motion (4) assuming no transverse motion is solved first (Block  108 ). 
     In particular, introducing the finite difference analogs into the simplified version of the axial equation of motion (4) yields: 
     
       
         
           
             
               u 
               
                 
                   i 
                   + 
                   1 
                 
                 , 
                 j 
               
             
             = 
             
               
                 
                   u 
                   
                     i 
                     , 
                     j 
                   
                 
                 ⁡ 
                 
                   ( 
                   
                     
                       
                         
                           μΔ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             t 
                             2 
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           s 
                         
                         R 
                       
                       + 
                       
                         2 
                         ⁢ 
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           2 
                         
                       
                       + 
                       
                         2 
                         ⁢ 
                         
                           
                             Δs 
                             2 
                           
                           
                             a 
                             2 
                           
                         
                       
                       - 
                       
                         
                           D 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             2 
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           t 
                         
                         AE 
                       
                     
                     
                       
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           2 
                         
                       
                       + 
                       
                         
                           μΔ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             t 
                             2 
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             2 
                           
                         
                         R 
                       
                     
                   
                   ) 
                 
               
               + 
               
                 
                   u 
                   
                     i 
                     , 
                     
                       j 
                       + 
                       1 
                     
                   
                 
                 ⁡ 
                 
                   ( 
                   
                     
                       
                         
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             2 
                           
                         
                         
                           a 
                           2 
                         
                       
                       - 
                       
                         
                           D 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             2 
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           t 
                         
                         AE 
                       
                     
                     
                       
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           2 
                         
                       
                       + 
                       
                         
                           μΔ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             t 
                             2 
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             2 
                           
                         
                         R 
                       
                     
                   
                   ) 
                 
               
               + 
               
                 
                   u 
                   
                     
                       i 
                       - 
                       1 
                     
                     , 
                     j 
                   
                 
                 ( 
                 
                   
                     
                       - 
                       Δ 
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       t 
                       2 
                     
                   
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         t 
                         2 
                       
                     
                     + 
                     
                       
                         μΔ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           2 
                         
                         ⁢ 
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           s 
                           2 
                         
                       
                       R 
                     
                   
                 
                 ) 
               
               + 
               
                 
                   u 
                   
                     i 
                     , 
                     
                       j 
                       - 
                       1 
                     
                   
                 
                 ⁡ 
                 
                   ( 
                   
                     
                       
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           s 
                           2 
                         
                       
                       
                         a 
                         2 
                       
                     
                     
                       
                         Δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           t 
                           2 
                         
                       
                       + 
                       
                         
                           μΔ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             t 
                             2 
                           
                           ⁢ 
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             2 
                           
                         
                         R 
                       
                     
                   
                   ) 
                 
               
               - 
               
                 cos 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   θ 
                   ⁡ 
                   
                     ( 
                     
                       
                         - 
                         
                           
                             μ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             g 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             Δ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               s 
                               2 
                             
                             ⁢ 
                             Δ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               t 
                               2 
                             
                           
                           E 
                         
                       
                       
                         
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             t 
                             2 
                           
                         
                         + 
                         
                           
                             μΔ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               t 
                               2 
                             
                             ⁢ 
                             Δ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               s 
                               2 
                             
                           
                           R 
                         
                       
                     
                     ) 
                   
                 
               
               + 
               
                 γ 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 g 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 sin 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   
                     θ 
                     ⁡ 
                     
                       ( 
                       
                         
                           
                             μ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             Δ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               s 
                               2 
                             
                             ⁢ 
                             Δ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               t 
                               2 
                             
                           
                           E 
                         
                         
                           
                             Δ 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               t 
                               2 
                             
                           
                           + 
                           
                             
                               μΔ 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 t 
                                 2 
                               
                               ⁢ 
                               Δ 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 s 
                                 2 
                               
                             
                             R 
                           
                         
                       
                       ) 
                     
                   
                   . 
                 
               
             
           
         
       
     
     Next, still assuming that there is not transverse displacement, the axial force F is calculated (Block  110 ), and the transverse equation of motion (2) is solved accordingly (Block  112 ). In particular, introducing the finite difference analogs into the axial force and the transverse equation of motion (2) yields: 
     
       
         
           
             
                 
             
             ⁢ 
             
               
                 F 
                 = 
                 
                   
                     AE 
                     ⁡ 
                     
                       [ 
                       
                         
                           ∂ 
                           u 
                         
                         
                           ∂ 
                           s 
                         
                       
                       ] 
                     
                   
                   = 
                   
                     AE 
                     ⁡ 
                     
                       ( 
                       
                         
                           ( 
                           
                             
                               u 
                               
                                 
                                   i 
                                   + 
                                   1 
                                 
                                 , 
                                 j 
                               
                             
                             - 
                             
                               u 
                               
                                 i 
                                 , 
                                 j 
                               
                             
                           
                           ) 
                         
                         
                           Δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           s 
                         
                       
                       ) 
                     
                   
                 
               
               , 
               
                 
 
               
               ⁢ 
               
                 
                   v 
                   
                     
                       i 
                       + 
                       2 
                     
                     , 
                     j 
                   
                 
                 = 
                 
                   
                     4 
                     ⁢ 
                     
                       v 
                       
                         
                           i 
                           + 
                           1 
                         
                         , 
                         j 
                       
                     
                   
                   - 
                   
                     4 
                     ⁢ 
                     
                       v 
                       
                         
                           i 
                           - 
                           1 
                         
                         , 
                         j 
                       
                     
                   
                   - 
                   
                     v 
                     
                       
                         i 
                         - 
                         2 
                       
                       , 
                       j 
                     
                   
                   - 
                   
                     
                       v 
                       
                         i 
                         , 
                         j 
                       
                     
                     ⁡ 
                     
                       ( 
                       
                         6 
                         - 
                         
                           
                             
                               
                                 2 
                                 ⁢ 
                                 γ 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 A 
                               
                               
                                 Δ 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   t 
                                   2 
                                 
                               
                             
                             - 
                             
                               
                                 D 
                                 t 
                               
                               
                                 Δ 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 t 
                               
                             
                           
                           
                             EI 
                             
                               Δ 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 s 
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                         ) 
                       
                       . 
                     
                   
                 
               
             
           
         
       
     
     At this point, initial values for the transverse displacement v and the axial displacement u are available. The axial force F and the friction force F t  are solved (Block  114 ), and the axial equation of motion (3) is solved (Block  116 ). In particular, introducing the finite difference analogs into the axial force F, the friction force F t , and the axial equation of motion (3) yields: 
     
       
         
           
             
                 
             
             ⁢ 
             
               
                 F 
                 = 
                 
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                     . 
                   
                 
               
             
           
         
       
     
     Finally, solving for the displacement u i+1,j  in the above system yields the downhole position at the downhole pump used to calculate the downhole pump card (Block  116 ). Load at the downhole pump is then computed using Hooke&#39;s law (i.e., 
               Load   =     EA   ⁡     (       ∂   u       ∂   s       )         )         
(Block  118 ). Thus, at this point, the solution can follow the form used in the Everitt-Jennings method.
 
     In particular, solving for the displacement u i+1,j  in the above system requires knowing displacement two nodes behind in space, u i,j  and u i−1,j  relative to the node being calculated u i+1,j . To start the solution, the displacements u 0,j  and u 1,j  need to be known for all of the timesteps j. The initial displacement u 0,j  is know from the surface measurements of the sucker rod pump system, but the next node&#39;s displacement u 1,j  is calculated with Hooke&#39;s law when the polished rod load, Load PR  (the surface load minus the buoyed weight of the rods), is substituted for the Load and a first-order-correct forward-difference analog is substituted for 
                 ∂   u       ∂   s       ,         
which yields:
 
     
       
         
           
             
               u 
               
                 i 
                 , 
                 j 
               
             
             = 
             
               
                 ( 
                 
                   
                     Load 
                     
                       PR 
                       , 
                       j 
                     
                   
                   × 
                   
                     
                       Δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       s 
                     
                     EA 
                   
                 
                 ) 
               
               + 
               
                 u 
                 
                   0 
                   , 
                   j 
                 
               
             
           
         
       
     
     Since the numerical methodology for solving the system of coupled nonlinear differential equations is similar to the numerical implementation of the modified Everitt-Jennings method, a similar iterative method can be used to calculate the net stroke and damping factor for the deviated well model disclosed herein. See, e.g., Pons-Ehimeakhe, V., “Modified Everitt-Jennings Algorithm With Dual Iteration on the Damping Factors,” 2012 South Western Petroleum Short Course, Lubbock, Tex., April 18-19. Moreover, to optimize the resolution of the viscous damping in the deviated model disclosed herein, the current algorithm can further include an iteration on single or dual damping factors as disclosed in co-pending application Ser. No. 13/633,167 entitled “Calculating Downhole Pump Card With Iterations on Single Damping Factor” and Ser. No. 13/633,174 entitled “Calculating Downhole Pump Card With Iterations on Dual Damping Factors”, which are incorporated herein by reference. Thus, a single damping factor D that covers viscous damping or dual damping factors D up  and D down  in the above equations for the upstroke and downstroke can be iterated on in conjunction with fluid load line calculations and concavity testing to better converge on the appropriate damping for the downhole pump card generated. 
     Using finite differences to solve the system of coupled differential equations is a useful method for analyzing stress in the sucker rod pump system. Splitting the finite difference analogs for the space discretization allows the model to be valid for a tapered rod string, including steel rods and fiberglass rods with sinker bars. Finally, including Coulombs friction in the analysis of the deviated well model gives a better approximation of the downhole conditions than using a vertical-hole model. 
     The process  100  disclosed herein, when applied as a diagnostic tool, generates a downhole card without the excess downhole friction caused by deviation and with optimal viscous damping. This process  100  is particularly useful for controlling wells based on the downhole data. As will be appreciated, teachings of the present disclosure can be implemented in digital electronic circuitry, computer hardware, computer firmware, computer software, or any combination thereof. Teachings of the present disclosure can be implemented in a computer program product tangibly embodied in a machine-readable storage device for execution by a programmable processor so that the programmable processor executing program instructions can perform functions of the present disclosure. 
     To that end, the teachings of the present disclosure can be implemented in a remote processing device or a pump controller. For example,  FIG. 6A  shows an embodiment of a pump controller  200  installed on a sucker-rod pump system  10 , such as a pump jack commonly used to produce fluid from a well. The pump system  10  includes a walking beam  11  connected to a frame  15 . The walking beam  11  operatively connects to a polished rod  12  connected via a rod string (not shown) to a downhole pump (not shown), which can be any downhole reciprocating pump as discussed herein. A motor control panel  19  controls a motor  17  to move the walking beam  11  and reciprocate the polished rod  12 , which in turn operates the downhole pump. Although a pump jack is shown, other sucker-rod pump systems can be used, such as a strap jack, or any other system that reciprocates a rod string using cables, belts, chains, and hydraulic and pneumatic power systems. 
     In general, sensors  202  and  204  measure load and position data of the pump system  10  at the surface, and the measured data from the sensors  202  and  204  is relayed to the controller  200 . After processing the information, the controller  200  sends signals to the motor control panel  19  to operate the pump system  10 . A particular arrangement of controller  200  and sensors  202  and  204  is disclosed in U.S. Pat. No. 7,032,659, which is incorporated herein by reference. 
     As shown, the controller  200  uses a load sensor  202  to detect the weight of the fluid in the production tubing during operation of the pump system  10  and uses a position sensor  204  to measure the position of the pump system  10  over each cycle of stroke. The position sensor  204  can be any position measurement device used for measuring position relative to the top or bottom of the stroke. For example, the position sensor  204  can be a dual position sensor that produces a continuous position measurement and a discrete switch output that closes and opens at preset positions of the polished rod  12 . 
     Alternatively, the degree of rotation of the pump system&#39;s crank arm can provide displacement data. For example, a sensor can determine when the system&#39;s crank arm passes a specific location, and a pattern of simulated polished rod displacement versus time can be adjusted to provide an estimate of polished rod positions at times between these crank arm indications. In another alternative, a degree of inclination of the walking beam  11  can provide displacement data. For example, a device can be attached to the walking beam  11  to measure the degree of inclination of the pumping unit. 
     Load data of the system  10  can be directly measured using a load cell inserted between a polished rod clamp and carrier bar. Alternatively, the strain on the walking beam  11  can provide the load data. Using a load sensor  202 , for example, the controller  200  can measure the strain on the polished rod  12  and can then control the pump system  10  based on the strain measured. The load sensor  202  may use any of a variety of strain-measuring devices known to a person of ordinary skill in the art. For example, the load sensor  202  can be a load measurement device used on the pump system  10  that includes a load cell installed on the pumping rod  12  or mounted on the walking beam  11 . The load sensor  202  can measure strain in the polished rod  12  and can use a strain-gage transducer welded to the top flange of the walking beam  11 . 
     Alternatively, the load sensor  202  can be a strain measuring device that clamps on to a load-bearing surface of the walking beam  11  or any convenient location as disclosed in U.S. Pat. No. 5,423,224. In another example, the load sensor  202  can use an assembly similar to what is disclosed in U.S. Pat. No. 7,032,659, which is incorporated herein by reference in its entirety. 
     Finally, the amplitude and frequency of the electrical power signal applied to the motor  17  can be used to determine motor rotation (i.e. displacement data) and motor torque (i.e. load data). In this way, the motor speed and the displacement of the polished rod can provide a series of motor speed and displacement data pairs at a plurality of displacements along the polished rod. That displacement data which represents a complete stroke of the pump system  10  can then be converted to load on the rod string and displacement of the rod string at a plurality of displacements along the polished rod, as described in U.S. Pat. No. 4,490,094. 
     Details of the pump controller  200  are schematically shown in  FIG. 6B . In general, the controller  200  includes one or more sensor interfaces  212  receiving measurements from the load and position sensors  202  and  204 . Additional inputs of the controller  200  can connect to other devices, such as an infrared water-cut meter, an acoustic sounding device (ASD) provide real-time data which can be logged for pressure buildup analysis and real-time calibration for fluid-level control. The controller  200  also include a power system (not shown), as conventionally provided. 
     The controller  200  can have software  222  and data  224  stored in memory  220 . The memory  220  can be a battery-backed volatile memory or a non-volatile memory, such as a one-time programmable memory or a flash memory. Further, the memory  220  may be any combination of suitable external and internal memories. 
     The software  222  can include motor control software and pump diagnostic software, and the data  224  stored can be the measurements logged from the various load and position sensors  202  and  204  and calculation results. The data  224  in the memory  220  stores characteristics of the well, including the depth, azimuth, and inclination of points along the well, which can be derived from drilling and survey data. Because the rod string may be tapered as is sometimes the case, the data  224  in the memory  220  can also store characteristics of the sucker rods taper, such as depth, diameter, weight, and length of various sections of the rod. 
     A processing unit  210  having one or more processors then processes the measurements by storing the measurement as data  224  in the memory  220  and by running the software  222  to make various calculations as detailed herein. For example, the processing unit  210  obtains outputs from the surface sensors, such as the load and position measurements from then sensors  202  and  204 . In turn, the processing unit  210  correlates the output from the load sensor  202  to the position of the polished rod  12  and determines the load experienced by the polished rod  12  during the stroke cycles. Using the software  212 , the processing unit  210  then calculates the downhole card indicative of the load and position of the downhole pump. 
     To control the pump system  10 , the pump controller  200  preferably uses an unabbreviated Everitt-Jennings algorithm with finite differences to solve the wave equation. The controller  200  calculates pump fillage and optimizes production on each stroke. This information is used to minimize fluid pounding by stopping or slowing down the pump system  10  at the assigned pump fillage setting. The pump controller  200  can also analyze the downhole pump card and determine potential problems associated with the pump and its operation. This is so because the shape, pattern, and other features associated with the downhole pump card represents various conditions of the pump and its operation. 
     After processing the measurements, the controller  200  sends signals to the motor control panel  19  to operate the pump system  10 . For example, one or more communication interfaces  214  communicate with the motor control panel  19  to control operation of the pump system  10 , such as shutting off the motor  17  to prevent pump-off, etc. The communication interfaces  214  can be capable of suitable forms of communications, and they may also communicate data and calculation results to a remote site using any appropriate communication method. 
     The foregoing description of preferred and other embodiments is not intended to limit or restrict the scope or applicability of the inventive concepts conceived of by the Applicants. It will be appreciated with the benefit of the present disclosure that features described above in accordance with any embodiment or aspect of the disclosed subject matter can be utilized, either alone or in combination, with any other described feature, in any other embodiment or aspect of the disclosed subject matter. 
     In exchange for disclosing the inventive concepts contained herein, the Applicants desire all patent rights afforded by the appended claims. Therefore, it is intended that the appended claims include all modifications and alterations to the full extent that they come within the scope of the following claims or the equivalents thereof.