Buckling of Tubing in Progressing Cavity Pumping Wells S.Vagapov and Sh. Vagapov, (Scientific and Production Enterprise Limited Liability Company

Burintekh, Ufa, Republic of Bashkortostan, Russian Federation)

Abstract

It is theoretically explained why the bottom portion of freely suspended tubing buckles in a progressing

cavity pumping well. In contrast to the sucker rod pumping system, there are two energy sources for tubing

buckling in progressing cavity pumping system:  potential energy of fluid in the tubing and the elastic strain

energy of twist stored in the rod and tubing strings. This means that the tubing buckles, and much more severely,

than in a sucker rod pumping system for the same pumping condition. Possibly the rod string is bent by the

buckled tubing.

The pump pressure load is proportional to the cross-sectional area of the stator multiplied by the

differential pressure across the pump. Generalized force make it easy to calculate the buckling effect of the

internal pressure and the elastic strain energy of twist. Tubing buckles as if subjected to an upward generalized

force.

Field evidence of the tubing buckling in progressing cavity pumping wells is given.

Tubing anchor must resist the torque and the upward axial load. Tubing buckling and its detrimental

effects may be entirely prevented by anyone of the following means: (a) the tubing anchor; (b) tubing anchor and

the sucker rod guides spaced between the pump and neutral point. The use of rod guides decreases the necessary

tension in the tubing. Detrimental effect of tubing buckling may be minimized by the sucker rod guides spaced

between the pump and neutral point.

Introduction

It is well known that the severe rod-tubing wear occurs near the pump even in vertical wells. At first

glance, the eccentric motion of the rotor seems to be the main reason for severe rod-tubing wear above the pump.

However, we assume that the buckling of tubing and rod string is also responsible for severe rod-tubing wear

above the pump (Vagapov 2004).

It is clear, that there is a similarity between a sucker rod pumping system and a progressing cavity

pumping system. Firstly, by their nature the sucker rod pumps and the progressing cavity pumps are both

positive displacement pumps. The major components, such as the tubing and rod strings are the same. Secondly,

the surface drive systems (pump jack and drive heard) transfer the energy to the positive displacement pumps

through the sucker rod string. The rotor is driven by a sucker rod string inside the tubing, which is rotated by a

surface drive system. The pump plunger is driven by reciprocating motion of a rod string inside the tubing. In

contrast to the cyclic rod string loading that occurs in rod pumping, the rod string is exposed a relatively constant

loading in progressing cavity system. Generally, the progressing cavity pump can be considered as a sucker rod

pump having endless pump upstroke.

__________________________________________________________________________________________ This paper has not been published elsewhere and is not under consideration by another journal. This paper has not been peer reviewed.

Possibility of tubing buckling in progressing cavity pumping wells was first theoretically considered by Vagapov in paper «Buckling of Tubing and

Rod String in Progressing Cavity Pumping System» (in Russian). April 2004, Neftyanoe Khozyaistvo - Oil Industry.

It is well known, that the lower part of freely suspended tubing buckles in sucker rod pumping well

(Lubinski and Blenkarn 1957). The buckling occurs during pump upstroke, when the weight of the column of

fluid in the tubing is transferred to the pump plunger and the rod string. When consideration is given to a

progressing cavity pumping system, the question may be asked, "Whether the tubing be buckled in progressing

cavity pumping wells or not?"

Obviously, the previous theoretical findings apply to the tubing in progressing cavity pumping system.

The lower portion of freely suspended tubing buckles since the weight of the column of fluid in the tubing rests

on the rotor and the rod string. During buckling, the stator moves upward, therefore, the rotor undergoes

displacement with respect to the stator. However, the motion of the rotor, unlike the plunger, has only one degree

of freedom. For this reason, the rotor rotates with respect to the stator during this motion. At the same time, in

contrast to the sucker rod pumping system, the tubing is buckled permanently, when a progressing cavity

pumping system is in operation.

The published works do not take into account buckling of the tubing in progressing cavity pump system.

For the progressing cavity pumping system, the influence of internal pressure and the elastic strain energy on a

tubing string instability has not been investigated yet. In this paper, the phenomenon of buckling of tubing string

in progressing cavity pump system is explained.

Solution was obtained by use the principle of virtual work, Timoshenko strain energy method, thought

experiment method, principle of solidification and principle of superposition.

Let us recall the strain energy method. The most well-known mathematical methods of the buckling

analysis is

• Solving differential equation of the deflection curve

• Strain energy method.

Solving differential equation of the deflection curve. The curve AB in Fig.1 represents the shape of the

axis of the bar after bending. This curve is called the deflection curve. The critical buckling load is derived by

solving the differential equation of the deflection curve

EI𝑑!yd!x

= P(δ − y(x))

However, there are cases in which this equation is complicated and exact solution becomes difficult.

The energy method can help to solve more complex problem.

Strain energy method. The energy method is an approximate method. However, Timoshenko (1955,

1956) wrote that the error arising from approximation is only about 1 percent. Take a bar loaded at the end as

shown on the Fig.2. Under the action of this load a buckling of the bar will take place. Let δ be a deflection at

the top of the bar during buckling. The point of application of the load move in a downward direction and the

work will be done by the load during this motion. The work done by the load during the motion is the decrease

in the potential energy of the load. Let U be potential energy of bending and U1 the decrease in the potential

energy of the load. When U1 > U the straight form of the equilibrium is unstable. If the U1 < U, the straight form

of the equilibrium is stable. The critical equilibrium state is obtained when U = U1.  

The investigation was based upon the following assumptions.

1. Rod and tubing strings behave as linear elastic bodies.

2. Wellbore is assumed to be straight and vertical.

3. Tubing string is freely suspended in the well.

4. Progressing cavity pumping system operates in the clockwise direction, therefore, the rotor and the

stator has a right hand helix.

5. It is assumed that the progressing cavity pump is the single lobe pump.

6. Friction forces between the rotor and stator may be neglected.

7. Dynamic load may be neglected.

8. Elastic behavior induced by stretch may be neglected.

9. It is assumed that the tubing volume is kept constant during buckling.

10. It is assumed that area corresponding to tubing OD and rod OD  is kept constant during buckling.

11. Fluid level is near the pump intake.

12. No tail pipe below the pump.

13. Non-tapered tubing and rod strings.

Energy Source of Tubing Buckling in Progressing Cavity Pumping Wells

In order to identify the root cause of buckling of tubing, the following question shall be answered first:

• What is the source of energy for buckling of tubing in progressing cavity pumping wells?

Let us consider a progressing cavity pumping well as shown in Fig.3. Consider now equilibrium

condition of the system between the cross sections m1 n1 and m2 n2 . We assume that the lower part of freely

suspended tubing buckles when a progressing cavity pumping system is in operation. We assume that the rod

and tubing strings are subjected to the twisting moment in position Fig.3a. Let the tubing will buckle and moves

from the position Fig.3a to the position Fig.3b. Fig.3b shows the situation when the sucker rod string stops the

tubing deflection.

Let the following designations be made.

δ1- displacement the lower end of tubing due to buckling

δ2- displacement the center of gravity of buckled tubing

h1 - height of the liquid column inside tubing above the pump

h2 - height of the liquid column outside tubing above the pump

L - length of the of buckled tubing (measured along its axis)

ρi - density of liquid in the tubing

ρo - density of liquid in the annulus

ρ - density of steel

A1 - area corresponding to tubing outside diameter (OD)

A2 - area corresponding to tubing inside diameter (ID)

A3 - plunger cross-section area

A4- area corresponding to rod OD (rod area)

Pi - pressure above fluid surface inside the tubing

Po- pressure above fluid surface outside the tubing

P1- pump intake pressure

∆P - pump differential pressure

d – rotor diameter

e - eccentricity

A!" - cross section area of the stator

A  ! - area corresponding to rotor OD

I p - polar moment of inertia

G – modulus of elasticity in shear

I p G – torsional rigidity

dφ  - total angle of twist

dφ! - angle of twist of the stator during displacement δ!

dφ! - angle of twist of the rotor during displacement δ!

Pst - stator pitch length

U - the strain energy of bending

Ut - the strain energy of twist

Tr - pump torque

Q - pump flow rate

W - virtual work

W! - pump hydraulic power

n - pump rotational speed

F – force (a fictitious force)

Qi - generalized force

Tubing Weight. Determine the work done by the tubing weight during buckling. Between the conditions

of Figs.3a and 3b, the tubing shortens due to buckling. We see that the center of gravity of the tubing is raised.

Therefore, the point of application of tubing weight undergoes displacement δ2. The work done by the tubing

weight during displacement δ2 is

−δ!     A! − A! L  ρ ………………………………………………………………………………..……(1)

The point of application of the tubing weight moves in upward direction and does the negative work

during this motion.

Weight of Fluid Inside Tubing. Determine the work done by the weight of a column of fluid inside

tubing during buckling. The rotor can be considered as a bolt or a screw and a stator can be considered as a nut

into which the rotor is inserted. The stator is rising but the rotor is stationary with respect to ground coordinate

system. Note that the motion of the helical steel rotor, unlike the plunger, has only one degree of freedom.

Consequently, the helical steel rotor rotates with respect to the stator during this motion. Since the stator moves

upward, the internal volume of the stator above the rotor is increased.

In order to give an illustration, let us carry out a thought experiment (Fig.4). It is well known that if a

deformable body is in a state of static equilibrium, it would also be in static equilibrium if the body were rigid

(this axiom is known as the principle of solidification). Imagine that the water inside the pump is frozen in the

position Fig.4a. It is evident that the rotor and frozen water cavities conform to the shape of the stator. Let the

tubing will buckle and moves from position Fig.4a to position Fig.4b. We can see that the stator is rising but the

rotor and the ice cavities are stationary with respect to ground coordinate system. For this reason, the helical

steel rotor with ice cavities screws onto the stator during this motion and the distance between the end of the

rotor and the intake or suction end of a stator is increased (δ!, Fig.4b). Determine the increase in internal volume

of the stator above the rotor during buckling.

Case 1. The length of the rotor is equal to the length of the stator (Fig.4a). Determine the amount of

water V! in a tube of length L in the position, as shown in (Fig.4a).

We have

V  ! =  V!  −  V!  ………………………………………………………………………………………….(2)

Where Vt is the internal volume in the tubing,

Vr is the amount of water displaced by the rod string (the length of the rod string is L.

It is assumed that the tubing volume is kept constant during buckling (see assumptions). Because of

this, the amount of water  V! in a buckled tubing above the rotor in the position, as shown in (Fig.4b) is

V  ! =  V!  + V!" −  V!  ….…………………………………………….………………………..…….…...(3)

Where Vst is the internal volume in the portion δ! of the stator above the rotor.

The difference between V  !   and V  ! is

V  ! −  V  ! =  V!  + V!" −  V!  −  V!  +  V! =  V!"  ……………………………………………………..….(4)

Case 2. The length of the stator is smaller than the length of the rotor (Fig.4c). Determine the amount of

water  V! in a tube in the position, as shown in (Fig.4c).

We have

V  ! =  V!  −  V!  −  V!"#    ………………………………………………….………………………..…….(5)

Where Vrot is the amount of water displaced by the portion Lrot of the rotor.

The amount of water  V! in a buckled tubing in the position, as shown in (Fig.4d) is

V  ! =  V!  + V!" −  V!  − V!"#….……………………………..…….…………………………………….(6)

Where Vst is the internal volume in the portion δ!of the stator above the ice (Fig.4d).

The difference between V  !   and V  ! is

V  ! −  V  ! =  V!  + V!" −  V!  −  V!"#  −  V!  +  V! + V!"# = V!"  …………………….…………….………(7)

On the other hand

V!"   =  A!"  δ!  …………………………………………………………………………………………....(8)

Where A!" - cross section area of the stator.

Comparing Eq.8 with the Eq.7 and Eq.4 we can conclude, that the increase of the internal volume

(V  ! −  V  !) in the stator does not depend upon the rotor sizes but only upon the cross section area of the stator

A!"  and the displacement δ!of the lower end of tubing.

As a result, the fluid is flowing down from tubing into the stator during buckling of tubing. This means

that the center of gravity of a column of fluid inside tubing is lowered. Hence, the point of application of weight

of a column of fluid inside tubing undergoes displacement δ1. The work done by the weight of a column of fluid

inside tubing during displacement δ1 is

A!"δ!P! + A!"  δ!h!ρ! ....…………………………………………………………………………………(9)

The point of application of weight of a column of fluid inside tubing moves in downward direction and

positive work will be done during this motion.

To determine the work done by the weight of a column of fluid inside tubing during displacement δ2 ,

we can break buckling process down into two steps that are more easily solved. It is does not matter that the

virtual displacements (steps) look unrealistic, because the work of force acting on a body that moves from a start

to an end position is determined only by these two positions, and does not depend on the trajectory of the body.

Consider the following hypothetical scenarios of buckling.

Step 1. It is assumed that both strings are buckled (Fig.3c). Let the rod string axis coincides with the

tube axis. It is evident that the point of application of weight of a column of fluid inside tubing undergoes

displacement δ2. Hence, the work done by the weight of a column of fluid inside tubing in this case is

−δ! A! − A!  L  ρ! ………………………………………………………………………..……………(10)

The point of application of the weight of a column of fluid inside tubing moves in upward direction and

negative work will be done during this motion.

Step 2. The rod string becomes straight and moves to an end position (Fig.3b). This means that the rod

string displaces an amount of fluid in upward direction. Because of this, the center of gravity of a column of

fluid inside tubing is raised to reflect the new volume of displaced fluid. Hence, the point of application of the

weight of a column of fluid inside tubing undergoes displacement δ2. The work done by the weight of a column

of fluid inside the tubing in this case is

−δ!A!Lρ! ………………………………………………………………………………..……………(11)

The point of application of the weight of a column of fluid inside tubing moves in upward direction and

negative work will be done during this motion.

Generally, the work done by the weight of a column of fluid inside tubing during displacement δ2 will

be found by using Eqs.10 and 11.

−δ! A! − A! Lρ! − δ!A!Lρ! = −δ!A!Lρ! ………………………………………………………….(12)

Weight of Fluid Outside Tubing. Determine the work done by the weight of a column of fluid outside

tubing during buckling. To determine the work done by weight of a column of fluid outside tubing during

displacement δ2, we can break buckling process down into two hypothetical scenarios of buckling.

Step 1. It is assumed that the stator is rising but stationary with respect to the rotor (Fig.3d). This means

that the tubing string displaces an amount of fluid outside tubing in downward direction. Therefore, the point of

application of the weight of a column of fluid outside tubing undergoes displacement δ2. The work done by the

weight of a column of fluid outside tubing in this case is

 δ!A!Lρ! ………………………………………………………………………………………………(13)

The point of application of the weight of a column of fluid outside tubing moves in downward direction

and positive work will be done during this motion.

Step 2. Let us imagine, that the water inside the pump is frozen in the position Fig.3d. Let us imagine,

that the rotor moves to an end position (Fig.3b) in downward direction. Therefore, the rotor with ice cavities

screws onto the stator during this motion. Thus, the distance between the lower end of the rotor and the intake or

suction end of the stator is increased (δ!,  Fig.4b). It is evident that the rotor and frozen water cavities conform to

the shape of the stator. This means that the rotor with ice cavities displace an amount of fluid outside tubing in

upward direction. Hence, the point of application of the weight of a column of fluid outside tubing undergoes

displacement δ1. The work done by the weight of a column of fluid outside tubing in this case is

−A!"δ!P! ………………………………………………………………………………………………(14)

where P1 = (h2 ρo + Po) - pump intake pressure .

The point of application of the weight of a column of fluid outside tubing moves in upward direction

and negative work will be done during this motion.

Generally, the work done by the weight of a column of fluid outside tubing during displacement δ2 will

be found by using Eqs.13 and 14.

 δ!A!Lρ! −A!"δ!P!……………………………………………………………………………………(15)

Elastic Strain Energy of Bending in Tubing. Determine the work done by the internal elastic forces

(elastic strain energy in bending) in the tubing during buckling. When a progressing cavity pumping system is in

operation, a tubing string is twisted by a torque reaction applied to the lower end. Therefore, there is case in

which a tubing buckles by combination of bending and twisting. Compare a torsional rigidity of the tubing string

and a torsional rigidity of the rod string (see Appendix). The data shows that the torsional rigidity of the rod

string is much less than the torsional rigidity of the tubing string (less than 1 percent). This means that the

shearing stress in twist of the tubing string is much less than the shearing stress in twist of the rod string.

Therefore, the influence of a torque reaction on the tubing buckling may be neglected. However, it should be

noted that a torque applied onto the tubing can promote buckling (Mitchell 2004).

Following the energy method (Timoshenko 1956), the elastic strain energy of bending stored in the

tubing during deflection, may be obtained by assuming that the tubing is only compressed by a centrally applied

load F. Consequently, the work done by the load F during displacement δ1 will be transformed into potential

energy of strain. Hence, the work done by the elastic force in the tubing or the strain energy of bending stored in

the tubing during displacement δ1 is

−Fδ! …………………………………………………………………………………………..............(16)

The point of application of the load moves in upward direction and negative work will be done during

displacement δ1.

Elastic Strain Energy of Twist in Tubing and Rod Strings. When a progressing cavity pumping system is

in operation, a large amount of the elastic strain energy of twist is stored in the rod and tubing strings (no torque

anchor below the pump, see assumption). Determine now the work done by the internal elastic forces in the

twisted tubing and twisted rod strings (elastic strain energy in twist) during buckling of tubing.

The rod string is subjected to the twisted moment T!  (Fig.5a). At the same time, the tubing string is

subjected to the twisting moment T!  (Fig.5b). It has already been said in this paper, the rotor can be considered as

a bolt or a screw and a stator can be considered as a nut into which the rotor is inserted. For this reason, the

helical steel rotor screws onto the stator during displacement δ!. Because of this, the helical steel rotor rotates

with respect to the stator and the stator also rotates with respect to the rotor during this motion. The total angle of

twist is

dφ = dφ! + dφ! =!!!!"  2π  .………………………..…………………………………….……………(17)

Where P!" is stator pitch length.

It is interesting to note, that  the total amount of elastic strain energy of twist does not depend upon the

torque anchor at the pump. In presence of the torque anchor (dφ! = 0), the amount of elastic strain energy of

twist can be stored only in the rod string (dφ = dφ!). Let us assume, for simplicity, that a rotating motion of the

lower end of tubing string is prevented, however, an axial motion is permitted. In this case, the amount of elastic

strain energy of twist may be calculated from diagram of twist only for the rod string (Fig.6). The torque is

represented by the ordinates and the angle of twist by the abscissas. The total elastic strain energy of twist stored

in the rod string is given by the area of the triangle OAB

!!  T!  φ! ……………………………………….….…...……….………………………………………(18)

Where T!   is the pump torque.

It is clear, that the work done by the internal elastic forces in the twisted rod string during a decrease

dφ! is given by the small area of the trapezoid ABCD. Let's make the following assumption: the value  φ! is

large compared dφ!, therefore, the work done by the pump torque T!  may be calculated

T!dφ! …………………………………………………………………………………………..….…..(19)

Substituting Eq.17, and observing that dφ! = 0  we obtain the work done by the internal elastic forces in

the twisted rod string during a decrease dφ!

T!!!!!"  2π  …………………………………………………………...……………….………....….……(20)

It is well known that the pump torque is

T!   =!!

!"! …………………………………………..…………………..……………………………...(21)

Where  W! - pump hydraulic power

n - pump rotational speed.

On the other hand, the pump hydraulic power is

W! = Q∆P ……………………………………………………….....…..………….……………...….(22)

Where Q - pump flow rate,

∆P - pump differential pressure.

For the single lobe pump (see assumptions) the flow rate is (Krylov 1962)

Q = n(A!" − A!)P!" ………………………………………….…………………………………..……(23)

Where A  ! - cross section area of the rotor.

It has already been said in this paper that the total amount of elastic strain energy of twist does not

depend upon the torque anchor at the pump. Therefore, substituting Eqs.21, 22 and 23 into Eq.20 we obtain the

work done by the internal elastic forces in the twisted tubing and twisted rod strings (elastic strain energy in

twist) during buckling

δ!(A!" − A!)  ∆P ...…………………………………………………………………………………….(24)

Determine whether this work is positive or not? Progressing cavity pumping system operates in the

clockwise direction, therefore, the rotor and the stator has a right hand helix (see assumptions). When a

progressing cavity pumping system is in operation, the rod string is subjected to the twisting moment T!  (Fig.5a).

The clockwise torque T!  is applied to the rod string at the surface from the wellhead drive unit and the rotating

movement is transmitted from the surface to the pump through a sucker rod string. The counter clockwise

resistive torque T!  is applied at the lower end of the rod string. Therefore, there will be a rotation of the lower

(upper) end of the rod string with reference to the upper (lower) end through an angle of twist  φ!(Fig.6). In this

case, the area OAB represents the total elastic strain energy of twist stored in the rod string during the angle of

twist  φ!.

It is well known that the theory of bending is based on the assumption that the beam consists of many

fibers aligned longitudinally. A disc, isolated as in Fig.5a, will be in the following state of strain. We see that the

strain energy of the twisted rod string (see the marked fiber) causes the lower end of the rod string to try

spinning in the clockwise direction (torque  T!", Fig.5a). At the same time, the rotor rotates with respect to the

stator during buckling. The rotor has a right hand helix and rotates also in the clockwise direction during this

motion. This means, that the rod string tends to decrease the elastic strain energy of twist during a decrease dφ!.

The area ODC (Fig.6) represents the total elastic strain energy of twist stored in the rod string during the angle of

twist φ!! . During this reversing deformation the amount of elastic strain energy of twist stored in the rod string is

lowered. For this reason, the potential energy of system is lowered.

The tubing string may be considered as a circular shaft built in at the upper end and twisted by a counter

clockwise torque reaction T! applied to the lower end (Fig.5b). Due to the twist, the lower end of the tubing

string rotates with respect to the upper end through an angle of twist  φ!.  The elastic strain energy of twist stored

in the tubing string causes the lower end of the tubing string to try spinning in the counter clockwise direction

(torque  T!", Fig.5b). It has been explained previously, the rotor rotates in the clockwise direction during

buckling, however, the stator rotates in the counter clockwise direction. As a result the tubing string tends to

decrease the elastic strain energy of twist during a decrease dφ!. During this reversing deformation the amount

of energy of twist stored in the tubing string is lowered. For this reason, the potential energy of system is

lowered. In summary, we can conclude that the work done by the internal elastic forces in the twisted tubing and

twisted rod strings during buckling are positive.

Finally, the total work done by the forces within on a system for the progressing cavity pumping well is

W = W! =   − δ!     A! − A! L  ρ + A!"δ!P! + A!"  δ!h!ρ! −δ!A!Lρ! +  δ!A!Lρ! − A!"δ!P! − Fδ!

+  δ!(A!" − A!)  ∆P……………………..……………………………………….……………………..………..(25)

The principle of virtual displacement states that the virtual work done by the actual forces is zero if and

only if the body is in equilibrium,  W = W! =  0.

Using the notation ∆P = P! +h!P!   − P!  and q= A! − A! ρ + A!ρ! − A!ρ! this equation may be

transformed into the significantly simpler form

δ!A!"∆P + δ!(A!" − A!)  ∆P − δ!Lq − Fδ! = 0….…….…………..….………….…………...……...(26)

Eq.26 may be rewritten

δ!(2A!" − A!)  ∆P − δ!Lq − Fδ! = 0  …………………………...…….……………………………...(27)

It should be noted, that ΔP is the pump differential pressure, q is the tubing weight with fluid per unit

distance in fluid.

To better understand the basic reasons for buckling of tubing, let us consider Eq.26 in more detail.

The term δ!A!"∆P is the positive work done by the force A!"∆P during displacement δ1. The term

δ!(A!" − A!)  ∆P is the positive work done by the force (A!" − A!)  ∆P during displacement δ1. The plus sign

indicates that the positive works tend to decrease the stability of the tubing string. Therefore, the positive works

done by the forces A!"∆P and  (A!" − A!)  ∆P exert a buckling effect on the tubing string.

The sign of the two terms δ2Lq and Fδ1 are negative. The minus sign indicates that the negative works

tend to increase the stability of the tubing string in the straight position. Therefore, the negative works done by

the forces Lq and F  try to straighten up the tubing string.

It is well known that following the energy method, buckling occurs when the positive works done by

the forces A!"∆P and  (A!" − A!)  ∆P exceed to the sum of potential energy of bending (F δ1) and work done by

the tubing weight with fluid (Lq) during displacement δ1.

We have

δ!A!"∆P + δ!(A!" − A!)  ∆P   > δ!Lq + Fδ!……………………………………….………………….(28)

Divide both side of the equation by δ!

A!"∆P + (A!" − A!)  ∆P   >!!!!Lq + F  …………………………………………………………………(29)

Let us estimate a magnitude of the term (!!!!Lq + F). The axial compressive force F must exceed some

critical value F!"#  in order to buckle the tubing. For tubing size 2 7/8 (ID = 2.441 in, ID = 2.441 in, q = 5.545

lb/ft) the critical force F!"# = 423  lb (Vagapov 2016). Noting, that !!!!≈ 1, we obtain (!!

!!Lq + F) ≈ 2100 lb (take

for instance, L = 300 ft). It is obviously, that the magnitude of the term A!"∆P + (A!" − A!)  ∆P is much greater

than 2100 lb. in the great majority of pumping wells. For this reason we can conclude, that the lower part of

freely suspended tubing buckles in progressing cavity pumping wells.

The term δ!A!"∆P represents the buckling effect produced by the internal pressure. The term δ!(A!" −

A!)  ∆P represents the buckling effect produced by the strain energy of twist. In the other words, the tubing string

buckles under the influence of internal pressure (the term δ!A!"∆P) and the strain energy of twist (the term

δ!(A!" − A!)  ∆P). Due to the superposition principle (the material follows Hooke’s law, see assumptions), each

of these terms can be analyzed separately:

1. Consider the effect of internal pressure on the buckling (the term δ!A!"∆P). Let us assume that the

tubing will buckle and moves from position Fig.3a to position Fig.3b. The tubing buckles while the rod string is

straight during this motion. For this reason, the stator is rising but the rotor is stationary with respect to the

ground coordinate system. As the stator continues to move upward, the rotor undergo displacement δ!with

respect to the stator, therefore, the internal volume of the stator above the rotor is increased. Because of this, the

column of fluid inside tubing moves in downward direction and does the positive work δ!A!"∆P to induce

buckling. As a result, the fluid level in tubing drops down, therefore, the center of gravity and potential energy of

system is lowered.

2. Now let add to the effect of internal pressure, the effect of strain energy of twist (the term δ!(A!" −

A!)  ∆P). The helical steel rotor rotates with respect to the stator during displacement δ! (the helical steel rotor

has only one degree of freedom). It has been explained previously, the rotor rotates in the clockwise direction

during this motion. In contrast, the stator rotates in the counter clockwise direction during this motion. Therefore

the rod and tubing strings tend to decrease the strain energy of twist during buckling. Because of this, the elastic

force in the twisted rod and the tubing strings does the positive work δ!(A!" − A!)  ∆P during buckling.

Therefore, the potential energy of system is lowered.

Based upon the foregoing, one may conclude that there are two energy sources for tubing buckling in

progressing cavity pumping system:  potential energy of fluid in the tubing and the elastic strain energy of twist

stored in the rod and tubing strings.

Vagapov (2016) proposed a similar approach to calculate the total work done by the forces within on a

sucker rod pumping system during buckling. For the sucker rod pumping system, the total work done by the

forces within on a system is

δ!A!∆P − δ!Lq − Fδ! = 0…………………………………………………………..………………..(30)

Where A! - plunger cross-section area.

For the progressing cavity pumping system, the total work done by the forces within on a system is (see

Eq.26)

δ!A!"∆P + δ!(A!" − A!)  ∆P − δ!Lq − Fδ! = 0

We see, that there is a similarity between Eq.30 and Eq.26. Comparing Eq.30 with Eq.26, the following

conclusions are reached:

1. The primary sources of energy for buckling of tubing in sucker rod and progressive cavity pumping

systems are the potential energy of fluid in the tubing (the term  δ!A!∆P in Eq.30 and term δ!A!"∆P in Eq.26).

There is another source of energy (the term δ!(A!" − A!)  ∆P) in progressing cavity pumping system: elastic

strain energy of twist stored in the rod and the tubing strings. So one can conclude that the tubing will buckle,

and much more severely in a progressive cavity pumping system, than in a sucker rod pumping system for the

same pumping condition (see the numerical example below).

2. Possibly, the rod string is bent by the buckled tubing in progressing cavity pumping system. In the

case of the sucker rod pumping system, the bottom portion of a rod string is bend by the buckled tubing in a

sucker rod pumping system. This phenomenon of simultaneous buckling was explained by the Vagapov (2016).

On the other hand, the tubing will buckle and much more severely in a progressive cavity pumping system, than

in a sucker rod pumping system for the same pumping condition. Therefore, it can be expected that the buckled

tubing in progressing cavity pumping system bend the rod string. So one conclude, that the buckled sucker rod

string rotates inside a buckled tubing string.1

It is well known that a torque applied onto the tubing or the rod strings can promote buckling (Mitchell

2004). The rod string is subjected to the twisting moment when a progressing cavity pumping system is in

operation. Based upon the foregoing, one may add another argument supporting the simultaneous buckling of

tubing and rod strings in progressing cavity pumping system.

3. The term δ!A!∆P is the positive work done the column of fluid inside the tubing during displacement

δ1 for the sucker rod pumping system. The term δ!A!"∆P is the positive work done the column of fluid inside the

tubing during displacement δ1 for the progressing cavity pumping system. Noting that A!is the plunger cross-

section area and ∆P is the pump differential pressure, we assume that the pump pressure load (pump axial load)

in progressing cavity pumping system is proportional to the cross-sectional area of the stator A!" multiplied by

the differential pressure across the pump ∆P.

Lubinski A. and Blenkarn K.A. (1957) proposed so-called "fictitious force" in the analysis of tubing

buckling in sucker rod pumping wells. During pump upstroke the tubing will buckle as if subjected to an upward

fictitious force and equal to

F = A!∆P …………………………………………………………………………………………….(31)

On the other hand, for the sucker rod pumping system (see Eq.30)

δ!A!∆P − δ!Lq − Fδ! = 0

For the progressing cavity pumping system (see Eq.27)

δ!(2A!" − A!)  ∆P − δ!Lq − Fδ! = 0

We see, that there is a similarity between Eq.30 and Eq.27. We propose a similar approach in the

analysis of tubing buckling in progressing cavity pumping system. It has already been said in this paper, the

motion of the helical steel rotor has only one degree of freedom. Therefore, based on the terminology of

Lagrangian mechanics, the force (2A!" − A!)  ∆P can be considered as a generalized force and the displacement

δ1 can be considered as a generalized displacement.

The generalized force makes it easy to calculate the buckling effect of the internal pressure and the

elastic strain energy of twist.

__________________________________________________________________________________________ 1 The phenomenon of simultaneous buckling in progressing cavity pumping system is not considered in detail in this paper, because this analysis can seriously

complicate the present paper.

When a progressing cavity pumping system is in operation, the tubing will buckle as if subjected to an

upward generalized force

Qi = (2A!" − A!)  ∆P………………………………………………………….…….….………………(32)

It is well known, that the tubing is buckled below the neutral point. For the progressing cavity pump

system, the location of the neutral point is obtained from the fact that the weight in fluid of the portion of the

tubing below the neutral point is equal to the generalized force. The distance to the neutral point is

L = (!"!"!!!)  ∆!!

……………………………………………..………………………………………….(33)

Consider the following example.

Determine the fictitious force F (for the sucker rod pumping system) and the generalized force Qi (for

the progressing cavity pumping system) if the pump displacements (the flow rate) and the differential pressure

across the pumps are about the same. Tubing size, 2 7/8 (OD = 2.875, ID = 2.441) in., q t = 6.4 lb/ft. Differential

pressure across the pump ∆P = 2,000 psi (working fluid level 4,000 ft., gradient 0.5 psi/ft, i.e., fluid specific

gravity 1.154). Pump depth 4,000 ft. (fluid level is near the pump intake).

Solution.

1. Sucker rod pumping system:

Plunger cross-section area, A3 = 2.4 in2. (1 ¾ in. plunger).

Plunger stroke, 100 in.

Strokes per minute, 12.

Pump displacement is

PD = 0.1166 x 100 x 12 x 1 ¾ 2 = 429 B/D

Fictitious force is

F = 2.4 x 2,000 = 4,800 lb.

2. Progressing cavity pumping system:

Rotor minor diameter, 1.378 in (35 mm).

Eccentricity, 0.315 in (8 mm).

Stator pitch length, 11.8 in (300 mm).

Rotational speed, 140 rpm.

Flow rate is (Appendix, Eq.36)

Q = 4 x 8 x 35 x 300 x 140 x 1440 = 67.7 x 109 mm3 per day or 426 B/D

Area corresponding to the rotor minor diameter is

A! = 𝜋 !.!"#!

! = 1.491 in2.

Cross-sectional area of the stator is (Appendix, Eq.37)

A!" = 4 x 0.315 x 1.378 + 𝜋 !.!"#!

! = 3.227 in2.

Generalized force (see Eq.32) is

Qi = (2 x 3.227 – 1.491) 2,000 = 9,926 lb.

Obviously, the generalized force is much greater than the critical force for the tubing.

The first source of energy for buckling of tubing

A!"∆P = 3.227 x 2,000 = 6,454 lb (65% of the generalized force).

The second source of energy for buckling of tubing

(A!" − A!)  ∆P = (3.227 – 1.491) 2,000 = 3,472 lb (35% of the generalized force).

The distance to the neutral point is

L = !,!"#!.!

= 1551 ft.

Ratio of the generalized force to the fictitious force is !,!"#!,!""

= 2.07

So one can conclude that the tubing will buckle, and much more severely in a progressive cavity

pumping system, than in a sucker rod pumping system for the same pumping condition.

Field Evidence of Tubing Buckling in Progressing Cavity Pumping Wells

For progressing cavity system, an analysis of tubing wear failures in Suriname’s Tambaredjo Field has

been published by Daniel, N., Henk Chin A Lien and Sanjai, K. (2014) and Carpenter, C. (2015). The single lobe

pumps are installed at an average depth of 1,000 ft (shallow vertical wells). In the period 2008-2012 an average

580 downhole failures occurred annually, 54% was caused by tubing leaks, with a repetitive frequency of up to 6

failures per year on well. The majority of the tubing leakages occurred immediately above the pump.  Most of the

tubing wear occurs at the tubing body in direct contact with the rod couplings.

From the 1,359 wells with leaking tubing, almost half (49%) of the leakage occurred at the lower

section (between the 23th and the 36th tubing) of the tubing string. The majority of tubing failures are

mechanical wear of tubing caused by the friction between the sucker rod string and the tubing.

The tubing was inspected visually in 12 pilot wells with the highest leaking tubing rate. Inspection on

the inner wall of the tubing was made possible by cutting open the tubing to identify any internal damage. It was

found out, that the abrasive wear is seen only on one side of the inner wall of the tubing, and on the other side is

a residual deposit. The photographs in the articles clear illustrate a typical example of mechanical wear of tubing.

It is thought that the eccentric motion of the rotor seems to be the main reason for severe rod-tubing

wear above the pump. However, we assume that the buckling of tubing is responsible for severe rod-tubing wear

above the pump. The fact that the wear occurred only on one side of the inner wall of the tubing suggests that the

buckling of tubing is caused the trouble. There is also some field evidence in support of this speculation, namely,

that the deposits occur on the other side of the inner wall of the tubing. This may add another argument in favor

of the buckling of tubing above the pump. Based upon the foregoing, we assume that the buckling of tubing

above the pump is the primary source of tubing wear failures above the pump of pilot wells.

Let us estimate the distance to the neutral point of pilot wells. The following data concerning wells are

taken from the papers.

Tubing size, 2 7/8 (OD = 2.875, ID = 2.441) in., q t = 6.4 lb/ft.

Assuming, that the length of pipe is 32 ft, the pump setting depth is 36 x 32 = 1,152 ft.

Fluid specific gravity is 0.88 (estimated), the gradient is 0.382 psi/ft.

Assuming, that the working fluid level is at the pump, the pump differential pressure is

ΔP = 1,152 x 0.382 = 440 psi.

Assuming, that the rotor OD is 1.378 in. (actual rotor size is not known), the cross section area of the

rotor A! is 1.491  in!.

Assuming, that the eccentricity is 0.315 in. (actual eccentricity is not known), the cross section area of

the stator is

A!"= 4 x 0.315 x 1.378 + !  !.!"#!

! = 3.227  in2.

Substitution of these values into Eq.33 gives the distance to the neutral point

L   =   !!"(!  !  !.!!"  !  !.!"#)!.!

 = 341 ft.  

On the other hand, the portion of tubing string reported as subjected to a fast wear (between the 23th

and the 36th tubing)

L = 36 x 32 – 23 x 32 = 416 ft.

We see, that the correlation between the distance from pump to neutral point and the zone of fast wear

of tubing is satisfactory. A more perfect correlation would hardly be expected, because several factors were not

known and were only estimated. Therefore, the satisfactory correlation indicates that the phenomenon of

buckling is responsible for the wear of tubing string above the pump.

Obviously, the friction due to buckling increases a torque applied to the polished rod at the surface.

However, the buckling occurs above the pump, therefore, the friction forces tend to be concentrated at the lower

portion of the tubing. Calculations show that the portion of the buckled tubing is about 20…25 % of the total

tubing length in the great majority of pumping wells. This indicates, that it can be difficult to separate the  hydraulic pump torque (and the friction pump torque) from the friction torque due to buckling. Thus, one can

easily reach incorrect conclusions to diagnose certain pump characteristics. It should be noted, that because of

the fact that the portion of the buckled tubing is 20…25 % of the total tubing length, the friction due to buckling

can increase elastic strain energy of twist stored in the rod string.

Means For Prevention of Buckling and Minimize of Buckling Detrimental Effects

Tubing buckling and its detrimental effects may be prevented by the tension and torque anchors; the

sucker rod guides and the tubing anchors; a sufficient weight of tail pipe below the pump.

Tension and Torque Anchors. "A tubing anchor which permits the tubing to elongate, but prevents

shortening, is called a tension anchor " (Lubinski 1957, 1962). To avoid the tubing back-off problems a tubing

anchor must resist torque for progressing cavity pumping system. One might think that a torque anchor can

prevent the buckling effect caused by the elastic strain energy of twist (the term δ!(A!" − A!)  ∆P). However, this

is not so; as already explained, the total amount of elastic strain energy of twist does not depend upon the torque

anchor at the pump. In presence of the torque anchor, the total amount of elastic strain energy of twist can be

stored only in the rod string.

At the same time the tubing anchor must resist the upward axial load. If the tubing is caught in its most

extended position and the tension is sufficient, the tubing anchor will entirely prevent the buckling. Therefore,

buckling will not occur if the buckling effect of generalized force Qi is counterbalanced by the straightening

effect obtained by increasing the tension in the tubing. In this case, the amount of necessary tubing pickup may

be obtained by the equations, which are not considered in this paper. However, it is interesting to note, that the

use of rod guides may decrease the amount of necessary tension in the tubing, because the first energy source of

buckling (the term δ!A!"∆P) disappears after the rod guides installation.

So one conclude, that the tubing anchor must resist the torque and must resist the upward axial load for

a progressing cavity pumping system.

Sucker Rod Guides. It has already been said in this paper that the tubing buckles while the rod string is

straight. For this reason, the stator is rising but the rotor is stationary with respect to ground coordinate system.

Because of this, the column of fluid inside tubing moves in downward direction and does the positive work

δ!A!"∆P to induce buckling. Let us imagine the hypothetical scenarios of buckling in which the rod string axis

coincides with the tubing axis during buckling. In this case, the stator is rising but the rotor is stationary with

respect to barrel, therefore the fluid level in tubing does not drop down and the fluid column does not work.

Therefore, when setting rod guides, the energy source of buckling δ!A!"∆P disappears. For this reason, we

assume that the rod guides help to minimize buckling, because the one of the energy sources of buckling

disappears after the rod guides installation. In this case, only strain energy of twist (the term δ!(A!" − A!)  ∆P)

will exert a buckling effect on the tubing string. However, the buckling effect of strain energy of twist may be

counterbalanced by the straightening effect obtained by increasing the tension in the tubing. Therefore, in this

case the use of rod guides decreases the necessary tension in the tubing.1

So one conclude that the detrimental effect of tubing buckling may be minimized by the sucker rod

guides spaced between the pump and neutral point. The detrimental effect of tubing buckling may be entirely

prevented by the tubing anchor with the rod guides.

Tubing Centralizers and Tail Pipe.2 As already pointed out in this paper, the progressing cavity

pump can be considered as a sucker rod pump having endless pump upstroke. Therefore, there is no "tubing

breathing" effect in progressing cavity pumping system. The detrimental effect of tubing buckling may be

entirely prevented by the sufficient weight of tail pipe below the pump. Possibly the detrimental effect of tubing

buckling may be prevented by the tubing centralizer.

Improvements in Progressing Cavity Pump Design.3 The detrimental effect of tubing buckling may

be minimized by improvement in progressing cavity pump design. Possibly the buckling of tubing hastens pump

wear. We believe that the bottom and top tag systems cannot prevent the buckling of tubing in progressing cavity

pumping system.

Conclusions

1. The lower portion of freely suspended tubing buckles when a progressing cavity pumping system is

in operation.

2. In contrast to the sucker rod pumping system, there are two energy sources for tubing buckling in

progressing cavity pumping system:  potential energy of fluid in the tubing and the elastic strain energy of twist

stored in the rod and tubing strings. This means that the tubing buckles, and much more severely, than in a

sucker rod pumping system for the same pumping condition.

3. Possibly the rod string is bent by the buckled tubing. This means that the buckled sucker rod string

rotates inside buckled tubing.

4. Tubing buckling is accompanied by a lowering of the point of application of the weight of a column

of fluid inside tubing. The fluid column inside tubing moves in downward direction and does the positive work

necessary to induce buckling. This means that the center of gravity of system is lowered.

__________________________________________________________________________________________ 1 The amount of necessary tubing pickup may be obtained by the equations which are not considered in this paper.

2 Buckling issues concerning an influence the  tubing centralizers and a tail pipe are not considered in this paper.

3 We do not discuss these issues in this paper.

5. Tubing buckling is accompanied by a lowering of elastic strain energy of twist stored in the rod and

tubing strings. The internal elastic forces in the twisted rod and tubing strings do the positive work to induce

buckling. This means that the potential energy of system is lowered.

6. Pump pressure load is proportional to the cross-sectional area of the stator multiplied by the

differential pressure across the pump.

7. Generalized force makes it easy to calculate the buckling effect of the internal pressure and the elastic

strain energy of twist. Tubing buckles as if subjected to an upward generalized force.

8. Tubing anchor must resist the torque and the upward axial load.

9. Tubing buckling and its detrimental effects may be entirely prevented by anyone of the following

means: (a) the tubing anchor; (b) tubing anchor and the sucker rod guides spaced between the pump and neutral

point. The use of rod guides decreases the necessary tension in the tubing.

10. Detrimental effect of tubing buckling may be minimized by the sucker rod guides spaced between

the pump and neutral point.

Acknowledgment The authors would like to express their appreciation to the colleagues from Burintekh company.

References

1. Baldenko, D.F., Baldenko, F.D. and Gnoevykh , A.N. 2005. Single-screw Hydraulic Mashine:

Progressing Cavity Pumps (in Russian). Vol.1. IRTs Gasprom.

2. Carpenter, C. 2015. Reducing Tubing Failures in the Tambaredjo Field, Suriname.  Journal of

Petroleum Technology, SPE-0615-0107-JPT.

  3. Daniel, N., Henk Chin A Lien and Sanjai, K. 2014. Case Study for Reducing Tubing Failures in

Suriname''s Tambaredjo Field.  SPE Energy Resources Conference, 9-11 June, Port of Spain, Trinidad and

Tobago SPE-169978-MS.   4.  He, L., Pei, X., Peng, B. et al. 2008. The Mechanism of PCP Wells' Tubing and Rod Wear Issue in

Polymer Flooding in Daqing Oil Field. SPE Progressing Cavity Pumps Conference, 27-29 April, Houston,

Texas, USA, SPE-113114-MS.

5. Krylov, A.V. 1962. Single Lobe Progressing Cavity Pumps (in Russian). Gostoptehizdat.

6. Lubinski, A. and Blenkarn, K. A. 1957. Buckling of Tubing in Pumping Wells, Its Effect and Means

for Controlling It. Trans. AIME (1957) 210, 73. SPE-672-G.  

7. Lubinski, A., Althouse, W.S. and Logan, J.L. 1962. Helical Buckling of Tubing Sealed in Packers. J.

Pet. Tech. 14 (6): 655-670. SPE-178-PA.

8. Mitchell, R.F. 2008. Tubing Buckling-The State of the Art. SPE Drill & Compl 23 (4): 361-370.

SPE-104267-PA.

9. Mitchell, R.F. 2004. The Twist and Shear of Helically Buckled Pipe. SPE Drill & Compl

19(01). SPE-87894-PA.

10. Timoshenko, S. 1st Ed. 1930, 2nd Ed. 1940, 3rd Ed. 1955. Strength of Materials, Part I, Elementary

Theory and Problems, D. Van Nostrand Company.

11. Timoshenko, S. 1st Ed. 1930, 2nd Ed. 1941, 3rd Ed. 1956. Strength of Materials, Part II, Advanced

Theory and Problems, D. Van Nostrand Company.

12.Vagapov, S.Y. 2004. Buckling of Tubing and Rod String in Progressing Cavity Pumping System (in

Russian). April. Neftyanoe Khozyaistvo - Oil Industry.

13.Vagapov, S.Y. and Vagapov, Sh.S. 2016. Simultaneous Buckling of Tubing and Rod Strings in

Pumping Wells.  https://dx.doi.org/10.6084/m9.figshare.3498734.v1.

 

Appendix

The polar moment of inertia of rod

I!" =!  !!

!

!".………………………...…………….……………………………….……………….……(34)

Where d! - rod outside diameter (for example, the rod size, ¾ (0.75) in.).

I!" =!  !.!"!

!" = 0.03 in4.

The polar moment of inertia of tubing

I!" =!  !!!

!"1 − !!

!

!!!.…………………………………………………………….……………….……(35)

Where D! – tubing outside diameter (for example, the tubing size, 2 7/8 (OD = 2.875, ID = 2.441)0

d! – tubing inside diameter.

I!" =π  2.875!

321 −

2.441!

2.875!= 3.22  in!

Compare a torsional rigidity of the rod string and a torsional rigidity of the tubing string

I!"GI!"G

=0.03G3.22G

= 0.009

Volumetric displacement of a single lobe pump is

Q  =  4 e d Pst n…………………………………………………………………………………………(36)

Where e – eccentricity,

d – rotor diameter,

Pst - stator pitch length,

n - pump rotational speed.

Cross-sectional area of the stator is

A!" =  4  e  d   + !  !!

! …………………………..…………………………..……………………………(37)

Fig. 1

Fig. 2

(a) (b) (c) (d)

Fig. 3 – Real (b) and hypothetical (c, d) buckling scenarios

(a) (b) (c) (d)

Fig. 4 – Scheme of the thought experiments

(a) (b)

Fig.5 – Rotation the rotor (a) and stator (b) during buckling

Fig.6 – Diagram of twist