Sucker-rod pumping

Quick refresher

The installation is a drive at the wellhead (the pumping unit turns the rotation of the motor into a rocking motion), a rod string, and a plunger pump with two ball valves: the travelling valve in the plunger and the standing valve at the bottom of the barrel (the cylinder of the pump).
On the upstroke the travelling valve is closed, and the rods carry the weight of the liquid column above the plunger. On the downstroke it is open, and the column rests on the standing valve, that is, on the tubing. This transfer of load stretches and releases the rods and the tubing twice in every cycle.
The theoretical displacement is the plunger area multiplied by the stroke length and the number of strokes per day. The actual delivery is lower. The volumetric efficiency takes four losses into account: stroke lost to the stretch of the rods and tubing, incomplete filling of the barrel, leaks, and shrinkage of the oil (oil that has lost its dissolved gas takes up less volume).
The greatest load on the polished rod is the weight of the rods in liquid, plus the fluid load (the pressure difference across the plunger multiplied by its area), plus inertia (the extra force that speeds the moving rods up and slows them down). The smallest load is the weight of the rods in liquid minus inertia.
A dynamometer card is a record of the load on the polished rod along its stroke. In the ideal case it is a parallelogram. Its distortions show gas in the barrel, incomplete filling, leaking valves, friction and parted rods.
The capacity of the pump is usually larger than the inflow. If the liquid level is pumped down to the intake, the barrel does not fill completely, and the plunger hits the liquid. We choose the operating mode from the dynamic fluid level and the dynamometer card. We vent the gas into the annulus (the space between the tubing and the casing, the steel pipe that lines the well), and we make the clearance volume (the dead space between the valves) of the pump as small as possible.

Key relationships

q_{\text{th}} = 1440\,A_{\text{pl}}\,S\,n

theoretical displacement, m³/d: the plunger area in m², the stroke in m, and the number of strokes per minute

\alpha_{\text{p}} = \dfrac{q}{q_{\text{th}}}

pump volumetric efficiency

P_{\text{liq}} = (p_{\text{out}} - p_{\text{in}})\,A_{\text{pl}}

fluid load

P_{\uparrow} = P'_{\text{rod}} + P_{\text{liq}}, \qquad P_{\downarrow} = P'_{\text{rod}}

static loads on the polished rod on the upstroke and the downstroke

S_{\text{pl}} = S - \dfrac{P_{\text{liq}} L}{E}\left(\dfrac{1}{A_{\text{rod}}} + \dfrac{1}{A_{\text{tbg}}}\right)

plunger stroke: the stroke of the polished rod minus the stretch of the rods and of the unanchored tubing

Common pitfalls

Calculating the fluid load from the pump setting depth instead of the dynamic fluid level. The column of liquid in the annulus pushes the plunger up from below.
Forgetting that the plunger stroke is shorter than the polished rod stroke, especially with a large plunger, a deep pump and unanchored tubing.
Comparing pressures as gauge values (measured above the atmosphere) instead of absolute values (measured from zero) when calculating the compression of gas in the barrel.
Raising the number of strokes when the barrel is already not filling completely. The delivery will not grow, but impacts and wear will.
Judging a pump from one dynamometer card without knowing the liquid level, the annulus pressure and the rate.

After this lesson you can

name the parts of the installation from the electric motor to the standing valve, and explain what each one does
follow the pumping cycle: which valve is open, and what carries the weight of the liquid column, on the upstroke and on the downstroke
calculate the theoretical displacement and the pump volumetric efficiency (the ratio of the actual delivery to the theoretical one), and split the loss of delivery into its parts
estimate the loads on the polished rod (the top rod of the string, which passes through the wellhead seal), the stretch of the rods and tubing, and the stroke length of the plunger
read a simple dynamometer card, and tell normal operation from the effect of gas, an incompletely filled barrel, leaking valves and parted rods
explain how pumping is controlled by the dynamic fluid level (how deep the liquid stands in the well while the pump works), and what is used against gas, sand, deposits and wear in a deviated wellbore

What the installation consists of

The lesson on choosing an artificial lift method describes the sucker rod pump in a few words: a drive at the wellhead moves the rod string, and the string moves the plunger of the pump at depth. Here we look at the installation part by part, from top to bottom, along the path that the energy takes from the power grid to the liquid.

SchematicA sucker rod pumping installation (a diagram, not to scale; the well is shortened). The pumping unit turns the rotation of the motor into the rocking of the walking beam. The rod string inside the tubing passes the motion to the plunger. The liquid rises up the tubing into the flowline, and the gas that separates at the pump intake collects in the annulus above the dynamic fluid level (the gas outlet from the annulus is not shown in the diagram).

The pumping unit (pumpjack) is a mechanism that turns rotation into a rocking motion. An electric motor (less often a gas engine) turns at a few hundred to fifteen hundred revolutions per minute. A V-belt drive and a gear reducer slow this down to a few revolutions per minute, usually 3 to 12 and rarely more than 15–20. Cranks with counterweights sit on the output shaft of the reducer. Pitmans (connecting rods) join them to the rear arm of the walking beam, a beam that rocks on a post. The horsehead is fixed to the front arm. Its arc is drawn from the centre of the rocking, so the bridle (a wire-rope hanger) that wraps around the horsehead moves strictly vertically, along the axis of the well.

The polished rod is fixed to the bridle. It is the top member of the rod string. It is made smooth so that it can be sealed: at the wellhead the packing of the stuffing box hugs the rod tightly. The stuffing box holds the well pressure while the rod slides through it up and down. Below the stuffing box there is an outlet to the flowline (the pipe that takes the production away from the well), and a separate outlet leads from the annulus.

Counterweights are needed because without them the motor would work in jerks. On the upstroke it would lift the rods and the liquid, and on the downstroke the rods would pull the mechanism by themselves. A counterweight stores energy when the rods go down and gives it back when they go up, like the counterweight of an elevator. A correctly balanced unit loads the motor about equally on both strokes (the calculation is in the advanced section on counterbalance).

Parts of the installation along the path of the energy, from the grid to the liquid

Part: Electric motor, belt drive, gear reducer
What it does: give a slow rotation with a large torque
Industry term: prime mover, V-belt, gear reducer

Part: Crank, connecting rod, rocking beam
What it does: turn the rotation into a rocking motion
Industry term: crank, pitman, walking beam

Part: Counterweights
What it does: even out the load on the motor between the upstroke and the downstroke
Industry term: counterweights

Part: Beam head, wire-rope hanger
What it does: guide the rod strictly vertically
Industry term: horsehead, bridle

Part: Smooth top rod, seal at the wellhead
What it does: pass the motion into the well and seal the wellhead
Industry term: polished rod, stuffing box

Part: String of rods
What it does: passes the motion and the force to the plunger
Industry term: sucker rod string

Part: Cylinder, piston, two valves
What it does: lift the liquid
Industry term: barrel, plunger, travelling and standing valves

Part	What it does	Industry term
Electric motor, belt drive, gear reducer	give a slow rotation with a large torque	prime mover, V-belt, gear reducer
Crank, connecting rod, rocking beam	turn the rotation into a rocking motion	crank, pitman, walking beam
Counterweights	even out the load on the motor between the upstroke and the downstroke	counterweights
Beam head, wire-rope hanger	guide the rod strictly vertically	horsehead, bridle
Smooth top rod, seal at the wellhead	pass the motion into the well and seal the wellhead	polished rod, stuffing box
String of rods	passes the motion and the force to the plunger	sucker rod string
Cylinder, piston, two valves	lift the liquid	barrel, plunger, travelling and standing valves

The equipment is standardised. The sizes of rods, pumps and pumping units are set by the standards of the American Petroleum Institute (API) and by the standards of the CIS countries (the former Soviet republics). So the parts of different manufacturers can be exchanged, which is a rare convenience in artificial lift. Besides beam pumping units, there are drives without a walking beam: chain and hydraulic drives, including long-stroke ones with a stroke of 6 m or more. They use the same pump and rods.

The rod string

Sucker rods are round steel bars about 8 m long (in American practice 25 or 30 ft, that is 7.6 or 9.1 m) with threaded ends. They are screwed together with couplings. The usual diameters are 16, 19, 22 and 25 mm (5/8, 3/4, 7/8 and 1 inch), and thinner and thicker ones also exist. In a well one and a half kilometres deep, the rod string is a steel “string” of almost two hundred pieces. It pulls, acts as a spring, and rubs against the tubing, all at the same time.

Rods work under fatigue (repeated loading): in every cycle the load changes from the smallest to the largest and back. At six strokes per minute this is more than three million cycles a year. So the strength of the string depends on two things: the largest stress (the force per unit of steel area), and the size of its swing over the cycle. The bigger the swing, the lower the allowed largest stress (the Goodman diagram describes this relationship). In a corrosive environment, such as salty water and especially in the presence of hydrogen sulfide, the allowed stress is lowered further. Parted rods and rods that unscrew are among the most frequent reasons for workovers (repair jobs that need a rig) on wells with sucker rod pumps.

The top rods carry the weight of all the rods below them and the weight of the liquid. The bottom rods carry almost only the load from the liquid. So the string is often made tapered: rods of a larger diameter at the top and a smaller diameter at the bottom. The string is lighter for the same strength, and the pumping unit and the reducer are loaded less.

Besides steel rods, fibreglass rods are used. They are light and resistant to corrosion, but they tolerate compression badly. There are also continuous rods with no couplings, which wear the tubing less. At the very bottom of the string, a few heavy rods of a large diameter (sinker bars) are often installed. Their weight helps the string to move down and keeps the lower rods from being compressed and bent.

The rod string moves down only under its own weight. We cannot push it from above, because rods that come under compression bend and quickly break from fatigue. This is where the limit on the number of strokes comes from. At the top of the stroke the polished rod starts moving down with an acceleration, and the rods must be able to keep up with it.

The limiting number of strokes set by the free fall of the rods

n_{\text{lim}} = \frac{30}{\pi}\sqrt{\frac{2g}{S}} \approx \frac{42.3}{\sqrt{S}}

where:

$n_{\text{lim}}$ the number of strokes per minute at which the acceleration of the polished rod at the top is equal to the acceleration of free fall, min⁻¹
$S$ length of the stroke of the polished rod, m
$g$ acceleration of free fall, 9.81 m/s², m/s²

Units: $S$ in metres, $g$ in m/s², $n$ in strokes per minute

Holds when

the polished rod moves in a harmonic (sine-wave) motion: the greatest acceleration is $\omega^2 S/2$ , where $\omega = \pi n/30$ is the angular velocity of the crank in radians per second
the rods fall with the acceleration $g$ , with no liquid and no friction. In liquid they fall more slowly, because buoyancy reduces their weight and friction resists the motion. So the formula gives an upper limit, not a working value.

At a higher number of strokes the polished rod goes down faster than the rods can fall. The bridle goes slack, and then the load returns with a blow. The longer the stroke, the lower the allowed number of strokes. In practice, for an ordinary pumping unit, we take no more than about 0.7 of this limit, and even less in viscous oil, where the rods “hang”.

In other unit systems (1)

Oilfield units

n_{\text{lim}} = \sqrt{\frac{70{,}500}{S\,[\text{in}]}}

$S$ in inches, $n$ in strokes per minute.

$70{,}500 \approx (30/\pi)^2\cdot 2g$ with $g = 386$ in/s². The exact value is about 70,400; reference books use the rounded number.

The pump and its cycle

The pump itself is simple. A hollow plunger hangs on the rods and moves in the barrel (the cylinder). There are two valves, and both are ball valves. The ball lies on a seat. A flow from below lifts the ball and passes, and a reverse flow presses the ball onto the seat. The travelling valve is in the plunger and moves with it. The standing valve does not move and closes the bottom of the barrel. The plunger is sealed in the barrel by a very small clearance, from a hundredth to a tenth of a millimetre, or by rings or flexible cup seals.

By the way they are run into the well, pumps are of two types. In a tubing pump the barrel is run on the tubing as part of the tubing string, and the plunger is run later on the rods. The plunger can be almost as wide as the inside diameter of the tubing, so the delivery is larger. But to change the barrel, the tubing must be pulled out of the well. An insert pump is assembled at the surface and run as a whole on the rods inside the tubing, where it sits in a seating nipple (a short fitting in the tubing that locks the pump in place). To change it, it is enough to pull the rods, but the plunger diameter is smaller.

SchematicHow the valves of a sucker rod pump work (a diagram). On the upstroke the travelling valve is closed, the plunger lifts the column of liquid, and the barrel fills through the open standing valve. On the downstroke the standing valve is closed and holds the column of liquid, and the plunger moves down through the liquid, letting it pass through the open travelling valve.

The pumping cycle

1
Upstroke: the rods take the load
The plunger starts to move up. The pressure of the liquid column from above presses the ball of the travelling valve onto its seat. The valve is closed, and the plunger lifts the whole column of liquid in the tubing like a piston. The weight of the column hangs on the rods, less the force with which the liquid under the plunger pushes it up from below.
2
Upstroke: the barrel fills
The pressure under the plunger falls. When it becomes lower than the pressure at the pump intake, liquid from the well lifts the ball of the standing valve and follows the plunger into the barrel. The column of liquid in the tubing rises, and its top part goes into the flowline.
3
Downstroke: the tubing takes the load
The plunger starts to move down. The pressure under it rises. The standing valve closes and now holds the column of liquid by itself, and through the barrel its weight passes to the tubing.
4
Downstroke: the liquid passes to the space above the plunger
When the pressure under the plunger becomes equal to the pressure above it, the travelling valve opens, and the plunger moves down through the liquid, leaving it above itself. The rods carry only their own weight in liquid. At the bottom point the cycle starts again.

Pump delivery and the volumetric efficiency

A sucker rod pump is a positive-displacement pump, which pushes out a fixed volume on every stroke cycle (a double stroke, up and down). This volume is equal to the plunger area multiplied by the stroke length. We get the theoretical displacement by multiplying this volume by the number of strokes per day.

Theoretical displacement of a sucker rod pump

q_{\text{th}} = 1440\,A_{\text{pl}}\,S\,n, \qquad A_{\text{pl}} = \frac{\pi D^2}{4}

where:

$q_{\text{th}}$ theoretical displacement: the volume that the plunger would displace in a day if its stroke were equal to the stroke of the polished rod, m³/d
$A_{\text{pl}}$ cross-sectional area of the plunger, m²
$D$ plunger diameter, m
$S$ length of the stroke of the polished rod, m
$n$ number of strokes (double strokes) per minute, min⁻¹

Units: metres, square metres, strokes per minute; the result is in m³/d

Holds when

the plunger stroke is equal to the stroke of the polished rod, the barrel fills completely with liquid, and there are no leaks; the volumetric efficiency allows for the departures from this
in the practice of the CIS countries, the stroke of the polished rod is put into the formula. In the API method the volume is calculated from the plunger stroke, and the loss of stroke is not part of the volumetric efficiency. When comparing coefficients from different sources, this must be checked.

The displacement is proportional to the square of the plunger diameter, to the stroke length and to the number of strokes. The same displacement can be given by different combinations of these three quantities. The choice between them is decided by the loads (see the next section).

In other unit systems (2)

Metric oilfield units

q_{\text{th}}\,[\text{m}^3/\text{d}] = 1.131\cdot 10^{-3}\;D^2\,[\text{mm}^2]\;S\,[\text{m}]\;n\,[\text{min}^{-1}]

$D$ in mm, $S$ in m, $n$ in strokes per minute, $q_{\text{th}}$ in m³/d.

$1.131\cdot 10^{-3} = 1440\cdot(\pi/4)\cdot 10^{-6}$ : 1440 minutes in a day, $\pi/4$ converts the diameter to the area, and $10^{-6}$ converts mm² to m².

Oilfield units

q_{\text{th}}\,[\text{bbl/d}] = 0.1166\;D^2\,[\text{in}^2]\;S\,[\text{in}]\;n\,[\text{min}^{-1}]

$D$ and $S$ in inches, $n$ in strokes per minute, $q_{\text{th}}$ in barrels per day.

$0.1166 = (\pi/4)\cdot 1440/9702$ : a barrel holds 42 gallons of 231 cubic inches each, that is 9702 cubic inches.

The actual delivery is usually lower than the theoretical one. The ratio of the actual delivery to the theoretical one is the pump volumetric efficiency (Russian-language practice calls it the delivery factor). English-language literature uses the same name for a close concept that is defined a little differently (see the second assumption of the equation above).

Pump volumetric efficiency and its components

\alpha_{\text{p}} = \frac{q}{q_{\text{th}}} = \alpha_{\text{s}}\,\alpha_{\text{f}}\,\alpha_{\text{leak}}\,\alpha_{\text{sh}}

where:

$\alpha_{\text{p}}$ pump volumetric efficiency, —
$q$ actual delivery of liquid, measured at the surface, m³/d
$\alpha_{\text{s}}$ the share of the stroke of the polished rod that reaches the plunger: $S_{\text{pl}}/S$ ; the rest is the elastic stretch of the rods and tubing, —
$\alpha_{\text{f}}$ fillage: the share of the volume swept by the plunger that is occupied by liquid; the rest is free gas (gas that is not dissolved in the liquid), —
$\alpha_{\text{leak}}$ the share that is left after leaks through the plunger clearance, the valves and the tubing threads, —
$\alpha_{\text{sh}}$ shrinkage, $1/B_{\text{liq}}$ : the liquid at the intake contains dissolved gas and takes up more volume than at the surface; $B_{\text{liq}}$ is the formation volume factor of the liquid at the intake conditions, —

Units: dimensionless fractions

Holds when

the factors are treated as independent; this is a convenient approximation for finding the main cause of the losses, not an exact law
the delivery was measured in steady operation, not just after a start-up

For an installation in good order the volumetric efficiency usually lies between 0.6 and 0.8. What matters more than the number itself is what it is made of. The loss of stroke is calculated from Hooke's law (a steel rod stretches in proportion to the load), the fillage shows on the dynamometer card, the shrinkage comes from the properties of the oil (Black oil model: solution gas–oil ratio and formation volume factors), and the leaks take what is left. This remainder grows as the pump wears.

Worked exampleTheoretical displacement and volumetric efficiency

Given (assumed data)

Plunger diameter $D$: 44 mm
Stroke of the polished rod $S$: 2.5 m
Number of strokes $n$: 6 per minute
Measured delivery of liquid $q$: 22 m³/d

Solution

Plunger area: $A_{\text{pl}} = \pi\cdot 0.044^2/4 = 15.21$ cm².
Volume per stroke: $15.21\cdot 10^{-4}\cdot 2.5$ m³ $= 3.80$ L. Strokes per day: $6\cdot 1440 = 8640$ .
Theoretical displacement: $q_{\text{th}} = 1.131\cdot 10^{-3}\cdot 44^2\cdot 2.5\cdot 6 = 32.84$ m³/d.
The same in oilfield units: $D = 1.732$ in and $S = 98.43$ in. Then $0.1166\cdot 1.732^2\cdot 98.43\cdot 6 = 207$ bbl/d. This is the same value, because a cubic metre holds 6.29 barrels.
Volumetric efficiency: $\alpha_{\text{p}} = 22/32.84 = 0.67$ .
The limit on the number of strokes from the free fall of the rods: $42.3/\sqrt{2.5} = 26.8$ per minute. As a rule of thumb, we work at about 0.7 of it, that is 19. Six strokes is well within the limit.

Answer: The theoretical displacement is 32.84 m³/d and the actual delivery is 22 m³/d, so the volumetric efficiency is 0.67.

What it means. The pump delivers a third less than the calculated volume. This is a usual picture, and by itself it does not point to a fault. Where this third goes will be shown by the calculation of loads and of the plunger stroke in the next section, and by the dynamometer card. The data in this example are assumed for illustration.

Loads on the rods and the plunger stroke

The weight of the rod string, the load from the liquid, and the forces that come from the motion (inertia, friction and vibrations) act on the polished rod. We start with statics, as if the pump were working very slowly.

The weight of the rods in liquid. The rods are immersed in the liquid that fills the tubing, and the buoyant force reduces their weight, as with a casing string in drilling fluid (Casing design). The weight in liquid is the weight in air multiplied by the buoyancy factor.

The fluid load. On the upstroke the column of liquid in the tubing presses on the plunger from above. This is the pressure at the pump discharge. From below, the pressure at the intake presses on it. The difference of the pressures multiplied by the plunger area is the fluid load. The intake pressure is made by the column of liquid in the annulus, from the dynamic fluid level down to the pump, and by the pressure of the gas above it. So the pump lifts the liquid not from its own depth, but in effect from the dynamic fluid level (pressure rise of the pump).

Static loads on the polished rod

\begin{aligned} P_{\uparrow} &= P'_{\text{rod}} + P_{\text{liq}}, \qquad P_{\downarrow} = P'_{\text{rod}},\\ P'_{\text{rod}} &= m_{\text{rod}}\,g\,L\,k_{\text{b}}, \qquad P_{\text{liq}} = (p_{\text{out}} - p_{\text{in}})\,A_{\text{pl}} \end{aligned}

where:

$P_{\uparrow},\ P_{\downarrow}$ load on the polished rod on the upstroke and on the downstroke, N
$P'_{\text{rod}}$ weight of the rod string in liquid, N
$m_{\text{rod}}$ mass of one metre of rods with couplings, kg/m
$L$ length of the rod string (along the wellbore), m
$k_{\text{b}}$ buoyancy factor, $1 - \rho/\rho_{\text{steel}}$ ; $\rho$ is the density of the liquid in the tubing, and $\rho_{\text{steel}} = 7850$ kg/m³ is the density of steel, —
$P_{\text{liq}}$ fluid load, N
$p_{\text{out}},\ p_{\text{in}}$ pressure at the pump discharge and at the pump intake, Pa
$A_{\text{pl}}$ plunger area: the full area, without subtracting the cross-section of the rods, m²

Units: SI; both pressures gauge or both absolute, because only their difference enters the formula

Holds when

very slow motion: no inertia, friction or vibration
the well is vertical; in a deviated well the weight of the rods is calculated from the vertical setting depth, not from the length of the string, and the friction of the rods against the tubing there increases the load on the upstroke and reduces it on the downstroke
the rod string has one diameter; for a tapered string the weights of the tapers are added
the fluid load is calculated on the full plunger area precisely because the weight of the rods is taken in liquid. If the weight of the rods is taken in air, the discharge pressure is multiplied by the plunger area less the cross-section of the rods, and the intake pressure still by the full area. The result is practically the same (the derivation is in the advanced section)
the force with which the wellhead pressure pushes the rod out of the stuffing box is not included; it is usually a fraction of a kilonewton

The difference between the loads on the upstroke and on the downstroke is equal to the fluid load. It grows with the size of the plunger and with the depth of the dynamic fluid level. The weight of the rods grows with the setting depth and with the rod diameter. The pumping unit must withstand the greatest load, and the rods must withstand millions of cycles from the smallest load to the greatest.

In other unit systems (2)

Metric oilfield units

P_{\text{liq}}\,[\text{kN}] = 0.1\,(p_{\text{out}} - p_{\text{in}})\,[\text{MPa}]\;A_{\text{pl}}\,[\text{cm}^2]

pressures in MPa, plunger area in cm², load in kN.

1 MPa on an area of 1 cm² is 100 N, that is 0.1 kN.

Oilfield units

P_{\text{liq}}\,[\text{lbf}] = 0.340\;\gamma\;D^2\,[\text{in}^2]\;H_{\text{dyn}}\,[\text{ft}]

$\gamma$ is the specific gravity of the liquid (water = 1), $D$ is the plunger diameter in inches, and $H_{\text{dyn}}$ is the depth of the dynamic fluid level in feet; the wellhead pressures in the tubing and in the annulus are taken as equal.

$0.340 = 0.433\cdot\pi/4$ : 0.433 psi per foot is the pressure gradient of fresh water (psi is pound-force per square inch), and $\pi/4$ converts the square of the diameter to the area.

Dynamics. A rod string is several tonnes of steel, which stops and speeds up in the opposite direction twice in every cycle. The simplest way to estimate the inertia is to assume that the rods move as one piece and that the polished rod moves harmonically. Then the greatest acceleration, expressed as a fraction of $g$ , gives the acceleration factor.

Acceleration factor and loads with inertia

\begin{aligned} K_{\text{a}} &= \frac{\omega^2 S}{2g} = \frac{S\,n^2}{1790},\\ P_{\max} &\approx P'_{\text{rod}} + P_{\text{liq}} + K_{\text{a}}\,P_{\text{rod}}, \qquad P_{\min} \approx P'_{\text{rod}} - K_{\text{a}}\,P_{\text{rod}} \end{aligned}

where:

$K_{\text{a}}$ acceleration factor: the greatest acceleration of the polished rod as a fraction of $g$ , —
$\omega$ angular velocity of the crank, $\pi n/30$ , rad/s
$S$ length of the stroke of the polished rod, m
$n$ number of strokes per minute, min⁻¹
$P_{\text{rod}}$ weight of the rods in air, $m_{\text{rod}}\,g\,L$ : inertia depends on the mass, and the buoyant force does not affect it, N
$P_{\max},\ P_{\min}$ estimates of the greatest and the smallest load on the polished rod in a cycle, N

Units: $S$ in metres, $n$ in strokes per minute; the loads in any identical units

Holds when

harmonic motion of the polished rod; in an ordinary pumping unit the acceleration at the bottom is larger than the harmonic one, and at the top it is smaller
the rods are rigid and move as one piece; elastic vibrations, the inertia of the liquid and friction are not included
the estimate is good for small depths and small numbers of strokes. For design, engineers use methods based on solving the wave equation for the rod string (for example API RP 11L, now designated TR 11L)

The greatest load comes at the start of the upstroke: the rods already carry the liquid and speed up upwards. The smallest load comes at the start of the downstroke. The inertia grows in proportion to the stroke length and to the square of the number of strokes: if we double the number of strokes, the extra load becomes four times bigger.

In other unit systems (2)

Metric units

K_{\text{a}} = \frac{S\,[\text{m}]\;n^2}{1790}

$S$ in m, $n$ in strokes per minute.

$1790 = 1800\,g/\pi^2$ with $g = 9.81$ m/s² (more exactly 1789). Russian-language reference books often give $S\,n^2/1440$ . This expression includes a correction for the finite length of the pitman: when the ratio of the crank radius to the pitman length is 1 : 4, the acceleration at the start of the upstroke is 1.25 times the harmonic one ( $1800/1.25 = 1440$ ).

Oilfield units

K_{\text{a}} = \frac{S\,[\text{in}]\;n^2}{70{,}500}

$S$ in inches, $n$ in strokes per minute.

the same expression with $g = 386$ in/s²: the exact value is about 70,400, and reference books use 70,500.

The loss of stroke. Under the fluid load the rods stretch, following Hooke's law (the stretch is proportional to the load). On the downstroke the load passes to the tubing: now the tubing stretches and the rods shorten. While the load passes from the rods to the tubing and back, the plunger does not move relative to the barrel. So the plunger stroke is shorter than the polished rod stroke by the sum of these stretches.

Stretch of the rods and tubing, and the plunger stroke

\lambda_{\text{rod}} = \frac{P_{\text{liq}}\,L}{E\,A_{\text{rod}}}, \qquad \lambda_{\text{tbg}} = \frac{P_{\text{liq}}\,L}{E\,A_{\text{tbg}}}, \qquad S_{\text{pl}} = S - \lambda_{\text{rod}} - \lambda_{\text{tbg}}

where:

$\lambda_{\text{rod}},\ \lambda_{\text{tbg}}$ elastic stretch of the rod string and of the tubing string under the fluid load, m
$E$ Young's modulus of steel (a measure of how stiff it is), about 207,000 MPa, Pa
$A_{\text{rod}}$ cross-sectional area of the body of a rod, m²
$A_{\text{tbg}}$ cross-sectional area of the metal of the tubing (the wall), m²
$L$ length of the rod string and of the tubing down to the pump, m
$S_{\text{pl}}$ length of the plunger stroke relative to the barrel, m

Units: SI

Holds when

the rod string has one diameter; for a tapered string the stretches of the tapers are added: $\lambda_{\text{rod}} = (P_{\text{liq}}/E)\sum L_i/A_i$
the tubing is not anchored; if the bottom of the tubing is fixed with an anchor, $\lambda_{\text{tbg}} = 0$
statics: at large depths and numbers of strokes the inertia of the rods adds to the plunger stroke (advanced section)
the rods are steel; for fibreglass rods the elastic modulus is several times smaller, and the stretch is correspondingly larger

The loss of stroke grows with the setting depth and with the plunger diameter (through the fluid load), and it falls as the rod cross-section grows. It does not depend on the stroke length, so a long stroke loses a smaller share of itself.

In other unit systems (1)

Oilfield units

\lambda\,[\text{in}] = \frac{12\;P_{\text{liq}}\,[\text{lbf}]\;L\,[\text{ft}]}{E\,[\text{psi}]\;A\,[\text{in}^2]}

the load in pounds-force, the string length in feet, the area in square inches, and the stretch in inches.

12 is the number of inches in a foot; Young's modulus of steel is about $30\cdot 10^6$ psi.

Worked exampleLoads and plunger stroke for the lesson's well

Given (assumed data)

Pump: 44 mm plunger ( $A_{\text{pl}} = 15.21$ cm²) at a depth $L = 1200$ m; the well is vertical
Operating mode: $S = 2.5$ m, $n = 6$ min⁻¹; measured delivery 22 m³/d
Rods: steel, 22 mm, one taper; 3.1 kg/m with couplings
Tubing: 73 × 5.5 mm, without an anchor
Liquid: density 950 kg/m³; dynamic fluid level 950 m
Wellhead pressures (gauge): in the tubing 1.0 MPa, in the annulus 0.2 MPa
For splitting the volumetric efficiency: formation volume factor of the liquid at the intake 1.04; leaks about 3% of the delivery

Solution

The discharge pressure is the wellhead pressure plus the column of liquid in the tubing: $p_{\text{out}} = 1.0 + 950\cdot 9.81\cdot 1200\cdot 10^{-6} = 12.18$ MPa. At the intake it is the pressure in the annulus plus the column from the liquid level down to the pump (250 m): $p_{\text{in}} = 0.2 + 2.33 = 2.53$ MPa. The pressure difference across the plunger is 9.65 MPa.
Fluid load: $P_{\text{liq}} = 0.1\cdot 9.65\cdot 15.21 = 14.68$ kN.
Weight of the rods in air: $P_{\text{rod}} = 3.1\cdot 9.81\cdot 1200 = 36{,}493$ N, that is 36.49 kN. Buoyancy factor: $1 - 950/7850 = 0.879$ . Weight in liquid: $P'_{\text{rod}} = 36{,}493\cdot 0.879 = 32{,}077$ N, that is 32.08 kN.
Static loads: on the upstroke $32.08 + 14.68 = 46.76$ kN, and on the downstroke 32.08 kN.
Acceleration factor: $K_{\text{a}} = 2.5\cdot 6^2/1790 = 0.0503$ . The inertia addition is $0.0503\cdot 36.49 = 1.8$ kN. The estimate is $P_{\max} \approx 49$ kN and $P_{\min} \approx 30$ kN.
Cross-section of the rods: $\pi\cdot 0.022^2/4 = 3.80$ cm². Cross-section of the tubing wall: 11.66 cm². Stretch of the rods: $\lambda_{\text{rod}} = \dfrac{14{,}678\cdot 1200}{2.07\cdot 10^{11}\cdot 3.80\cdot 10^{-4}} = 0.224$ m. For the tubing, the same calculation with its cross-section gives $\lambda_{\text{tbg}} = 0.073$ m.
Plunger stroke: $S_{\text{pl}} = 2.5 - 0.224 - 0.073 = 2.20$ m, that is $\alpha_{\text{s}} = 0.88$ . With an anchor on the tubing it would be 2.28 m.
The volumetric efficiency of this well is 0.67 (the previous example). The other three factors together account for $0.67/0.88 = 0.76$ . With a formation volume factor of the liquid of 1.04 and leaks of about 3%, the fillage is $\alpha_{\text{f}} = 0.76\cdot 1.04/0.97 = 0.81$ .

Answer: The pumping unit must withstand about 49 kN on the rod. Of the 2.5 m of stroke, 2.20 m reaches the plunger. About 19% of the barrel volume is taken by gas.

What it means. The elasticity of the rods and tubing takes 12% of the stroke. An anchor on the tubing would give back about 25% of this loss. But the main loss here is the fillage: about a fifth of the barrel volume is taken by gas. We check this assumption with the dynamometer card (next section). The greatest load is only 4% above the static one: at six strokes per minute the inertia is small. The data in this example are assumed for illustration.

Choosing the operating mode. The same theoretical displacement can be given by different combinations of the plunger diameter, the stroke length and the number of strokes. They are not equal, and this is clear when we compare the loads.

Worked exampleThree ways to get one delivery

Given (assumed data)

Well, rods and tubing: as in the previous example; stroke $S = 2.5$ m
Plunger options: 32, 44 and 57 mm
Theoretical displacement in all the options: 32.84 m³/d, as for the 44 mm pump at 6 strokes per minute

Solution

For the same displacement the number of strokes is inversely proportional to the plunger area, that is, to the square of the diameter. For 32 mm it is $6\cdot(44/32)^2 = 11.3$ , for 44 mm it is 6, and for 57 mm it is $6\cdot(44/57)^2 = 3.6$ per minute.
At the same pressure difference of 9.65 MPa, the fluid load is proportional to the plunger area: 7.76, 14.68 and 24.63 kN.
The loss of stroke to the stretch of the rods and tubing is proportional to this load: 0.16, 0.30 and 0.50 m, that is 6, 12 and 20% of the stroke.
The static load on the upstroke is 39.8, 46.8 and 56.7 kN. With inertia ( $K_{\text{a}} = S n^2/1790$ ) it is 46.4, 48.6 and 57.4 kN.
Load cycles of the rods per year: 5.96, 3.15 and 1.88 million.

Answer: A small plunger needs 11.3 strokes per minute instead of 6, and a large one only 3.6. But the large plunger loads the rods up to 56.7 kN instead of 46.8, and loses 20% of the stroke to stretch.

What it means. With the small plunger the static load is clearly smaller, but inertia eats up most of the gain, and the rods go through almost twice as many cycles in a year. The large plunger with rare strokes is more economical in energy, but it loads the rods, the pumping unit and the reducer most, and loses a fifth of the stroke to stretch. Usually engineers prefer a long stroke and a moderate number of strokes. They take the smallest plunger diameter that gives the needed delivery without excessive speed. The data in this example are assumed for illustration.

These calculations also show the limit of what a sucker rod pump can do. The deeper the pump, the larger the part of the allowed rod load that goes to the weight of the rods themselves, and the less is left for the liquid. At the same time the loss of stroke grows. So from a depth of about two kilometres, a sucker rod pump can lift no more than a hundred to a hundred and fifty cubic metres of liquid per day. With growing depth, the possible delivery falls quickly (the application ranges of the lift methods).

The dynamometer card

We cannot look inside a pump that is deep in the well, but the load on the polished rod tells us how it works. A dynamometer card is a graph of the load on the rod against its position during one cycle. A dynamometer records it. This is an instrument with sensors for force and position, fitted on the bridle or on the rod.

We can predict the shape of an ideal card. Suppose the rods were perfectly stiff. Then the load would jump at the two ends of the stroke. The rod would go up under the load $P_{\uparrow}$ and down under $P_{\downarrow}$ , and the card would be a rectangle. Elasticity turns the rectangle into a parallelogram. At the start of the upstroke the rods stretch and the tubing shortens while the fluid load passes to the rods. In that time the polished rod travels the distance $\lambda_{\text{rod}} + \lambda_{\text{tbg}}$ . At the start of the downstroke everything happens in the reverse order.

CalculatedCalculated static dynamometer card of the example well: stiff rods (dashed line) and elastic rods with unanchored tubing (solid line). A is the bottom of the stroke, where the travelling valve has closed. A–B: the rods take up the fluid load. B: the standing valve has opened and the plunger moves up. C is the top of the stroke. C–D: the load passes to the tubing. D: the travelling valve has opened and the plunger moves down. Inertia and vibrations are not included.

The bottom line (D–A) is the weight of the rods in liquid. The top line (B–C) is the same weight plus the fluid load. The vertical distance between the two lines is the fluid load, that is, the pressure difference across the plunger. So we can use it to estimate the pump intake pressure.
The sloping lines show the taking up of the load (A–B) and the release of the load (C–D). Their horizontal length is the loss of stroke. The length of the horizontal lines is the plunger stroke.
The area of the figure is the work that the polished rod does in one cycle. For an ideal card it is equal to $P_{\text{liq}}\,S_{\text{pl}}$ , the work of lifting the liquid.

A real card differs from a parallelogram even for a pump in good order. The inertia of the rods and the liquid and the elastic vibrations of the string are added to the static loads. Stretching waves run along the rods at the speed of sound in steel, about 5 km/s. They are reflected from the plunger and come back. The lines become wavy, and the figure is skewed. The deeper the pump and the higher the number of strokes, the stronger the distortion. So for deep wells the surface card is converted into a pump card (also called a downhole card). This is the load and the position at the pump itself, found by solving the wave equation for the rod string. On the pump card the faults of the pump show up without interference.

Each fault changes the shape of the card in its own way, and this makes the card the main diagnostic tool for sucker rod pumps. The most frequent case is a barrel that is not completely full of liquid.

CalculatedCalculated static dynamometer card of the same well when the barrel is 81% full of liquid. The rest of the volume holds gas at the intake pressure (solid line). On the downstroke the plunger first compresses the gas, so the load is not released at once and not along a straight line. The travelling valve opens at point D′, 0.32 m later than with a full barrel (dashed line). The compression of the gas is taken as isothermal (at constant temperature), and the clearance volume as full of liquid.

How faults change the dynamometer card

Fault: Gas in the barrel (gas interference)
What happens in the pump: on the downstroke the plunger first compresses the gas, so the travelling valve opens late
What the card shows: the load is released gradually, along a smooth curve; the useful stroke is shorter

Fault: Incomplete filling with the level pumped down (fluid pound)
What happens in the pump: above the liquid in the barrel there is gas at a very low pressure; the plunger reaches the liquid and hits it
What the card shows: the load holds for part of the downstroke, then drops almost vertically; after the impact there are oscillations

Fault: Leak in the travelling valve, worn plunger
What happens in the pump: on the upstroke liquid slips past to the space under the plunger, and the plunger cannot hold the column at low speed
What the card shows: the load builds up more slowly and is lost before the end of the upstroke: the upper corners are cut off

Fault: Leak in the standing valve
What happens in the pump: on the downstroke liquid leaves the barrel and goes back into the well
What the card shows: the load is released more slowly and comes back before the end of the downstroke: the lower corners are cut off

Fault: Higher friction: paraffin, a viscous emulsion, a deviated wellbore
What happens in the pump: the rods rub against the tubing or stick in deposits and in viscous liquid
What the card shows: on the upstroke the load is higher than calculated, and on the downstroke it is lower: the figure is stretched vertically

Fault: Parted or unscrewed rods
What happens in the pump: the plunger is no longer connected to the polished rod
What the card shows: a narrow band at the level of the weight of the rods in liquid, or lower. The higher the break, the lower the band, because the polished rod carries only part of the string

Fault: Sticking plunger
What happens in the pump: the plunger moves stiffly in the barrel because sand or deposits get in the way
What the card shows: sharp spikes of load at some places along the stroke

Fault	What happens in the pump	What the card shows
Gas in the barrel (gas interference)	on the downstroke the plunger first compresses the gas, so the travelling valve opens late	the load is released gradually, along a smooth curve; the useful stroke is shorter
Incomplete filling with the level pumped down (fluid pound)	above the liquid in the barrel there is gas at a very low pressure; the plunger reaches the liquid and hits it	the load holds for part of the downstroke, then drops almost vertically; after the impact there are oscillations
Leak in the travelling valve, worn plunger	on the upstroke liquid slips past to the space under the plunger, and the plunger cannot hold the column at low speed	the load builds up more slowly and is lost before the end of the upstroke: the upper corners are cut off
Leak in the standing valve	on the downstroke liquid leaves the barrel and goes back into the well	the load is released more slowly and comes back before the end of the downstroke: the lower corners are cut off
Higher friction: paraffin, a viscous emulsion, a deviated wellbore	the rods rub against the tubing or stick in deposits and in viscous liquid	on the upstroke the load is higher than calculated, and on the downstroke it is lower: the figure is stretched vertically
Parted or unscrewed rods	the plunger is no longer connected to the polished rod	a narrow band at the level of the weight of the rods in liquid, or lower. The higher the break, the lower the band, because the polished rod carries only part of the string
Sticking plunger	the plunger moves stiffly in the barrel because sand or deposits get in the way	sharp spikes of load at some places along the stroke

The shapes are described for a static card or a pump card. On a surface card, vibrations distort them. We confirm the diagnosis by measuring the level, the pressures and the rate.

The dynamic fluid level and the pumping mode

A well gives its highest rate at the lowest bottomhole pressure (the pressure at the bottom of the well while it produces; Inflow performance relationship (IPR)). That is when the dynamic fluid level stands a little above the pump intake. But the capacity of the pump is usually chosen with a margin, so it is larger than the inflow. Then the level slowly falls to the intake. The barrel no longer fills completely, and for part of every stroke the pump works for nothing. The plunger hits the liquid (fluid pound), and these blows damage the pump, the rods and the gear reducer.

There are three ways to deal with this. The first is to reduce the delivery, by lowering the number of strokes or the stroke length, until it equals the inflow. The second is to switch the well to intermittent pumping. The pump runs until it has pumped out the liquid that has collected, and then it stands idle while the level recovers. A timer sets the running and idle times, using the results of level surveys. The third way is a pump-off controller. The control panel records the dynamometer card all the time. It stops the drive or lowers the number of strokes as soon as it sees incomplete filling.

The level is measured with an echometer, an acoustic level meter. The survey itself is called an acoustic fluid-level survey. At the wellhead we send a sound pulse into the annulus, by firing a blank cartridge or releasing a puff of compressed gas, and a microphone records the reflections. Weak, frequent echoes come from the tubing collars (the couplings between the tubing joints), and a strong echo comes from the surface of the liquid. The depth of the level is half of the distance travelled by the sound.

The depth of the level from the echometer record

H_{\text{dyn}} = \frac{v\,t}{2}, \qquad v = \frac{2\,N_{\text{c}}\,l_{\text{tbg}}}{t_{\text{c}}}

where:

$H_{\text{dyn}}$ depth of the liquid level in the annulus, measured from the wellhead, m
$t$ time from the pulse to the arrival of the echo from the level, s
$v$ average speed of sound in the gas of the annulus, m/s
$N_{\text{c}}$ number of collars whose echoes have been counted on the record, —
$l_{\text{tbg}}$ average length of one tubing joint, m
$t_{\text{c}}$ arrival time of the echo from the last counted collar, s

Units: metres, seconds, metres per second

Holds when

the speed of sound is the same over the whole height of the gas column. In fact it depends on the composition, the pressure and the temperature of the gas, and it is found from the upper part of the wellbore
if the echoes from the collars cannot be read, the speed is taken from the composition and pressure of the gas, or from the echo of a reference marker (a reflector fixed on the tubing at a known depth)
the depth is measured along the wellbore; in a deviated well it is converted to vertical depth with the directional survey

If we know the level and the gas pressure in the annulus, we can estimate the pump intake pressure and the bottomhole pressure (how to do it, and with what reservations).

Worked exampleThe level from an echometer record, and the intake pressure

Given (assumed data)

Echo from the 50th collar: after 2.20 s
Average length of a tubing joint: 9.6 m
Echo from the level: after 4.35 s
Pump and liquid: setting depth 1200 m; liquid density 950 kg/m³
Wellhead pressures (gauge): in the tubing 1.0 MPa, gas in the annulus 0.2 MPa

Solution

Speed of sound: $v = 2\cdot 50\cdot 9.6/2.20 = 436.4$ m/s.
Depth of the level: $H_{\text{dyn}} = 436.4\cdot 4.35/2 = 949$ m, which is 950 m rounded.
The pump is submerged below the level by $1200 - 950 = 250$ m. The pump intake pressure is $p_{\text{in}} = 0.2 + 950\cdot 9.81\cdot 250\cdot 10^{-6} = 2.53$ MPa (gauge).
If the level fell to the intake, the intake pressure would drop to the gas pressure in the annulus, 0.2 MPa. The fluid load on the 44 mm plunger would then grow from 14.7 to 18.2 kN.

Answer: The level is about 950 m. The pump is submerged 250 m below it, and the intake pressure is about 2.53 MPa.

What it means. A submergence of 250 m is a margin in the inflow: the pump has not yet pumped the well down, and the delivery can be raised until the level comes close to the intake. The estimate of the intake pressure from the density of degassed liquid is too high if the liquid in the annulus is full of gas bubbles. The data in this example are assumed for illustration.

Gas, sand, deposits and a curved wellbore

Gas. Free gas that gets into the barrel takes the place of liquid, and it can also be compressed. On the downstroke the plunger first compresses the gas up to the discharge pressure, and only then does the travelling valve open. On the upstroke the gas left under the plunger first expands, and the standing valve opens late. Both delays take away useful stroke. The leftover gas stays in the volume between the two valves when the plunger is at the bottom of its stroke. This volume is called the clearance volume (the dead space of the pump).

The clearance volume: the delay of the intake and the condition for gas lock

x_{\text{open}} = l_{\text{cl}}\left(\frac{p_{\text{out}}}{p_{\text{in}}} - 1\right), \qquad \text{gas lock when}\quad 1 + \frac{S_{\text{pl}}}{l_{\text{cl}}} \le \frac{p_{\text{out}}}{p_{\text{in}}}

where:

$x_{\text{open}}$ the distance that the plunger travels from the bottom of the stroke before the standing valve opens, m
$l_{\text{cl}}$ height of the clearance volume: its volume divided by the plunger area, m
$p_{\text{out}},\ p_{\text{in}}$ pressure at the pump discharge and at the pump intake, both absolute, Pa
$S_{\text{pl}}$ length of the plunger stroke, m
$1 + S_{\text{pl}}/l_{\text{cl}}$ compression ratio of the pump: the ratio of the largest volume under the plunger to the smallest, —

Units: any consistent units; the pressures are absolute

Holds when

the clearance volume is completely filled with gas. This is the worst case: liquid that stays in it reduces the volume of gas and the delay
an ideal gas, with isothermal (constant-temperature) compression and expansion. When the gas is compressed fast, it heats up and the pressure grows faster, so the estimate is on the safe side
the valves are tight, and the pressure losses in them are not included

The gas in the clearance volume must expand from the discharge pressure to the intake pressure before a new portion of liquid can enter the barrel. If the plunger stroke is not long enough for this, the valves stop opening. The plunger compresses and expands the same portion of gas again and again, and delivery stops. This is gas lock. It is more likely when the intake pressure is lower and the clearance volume is larger.

Worked exampleThe clearance volume and gas lock

Given (assumed data)

Discharge pressure: 12.18 MPa (gauge)
Intake pressure: 2.53 MPa at a level of 950 m; 0.2 MPa if the level is pumped down to the intake (gauge)
Plunger stroke $S_{\text{pl}}$: 2.20 m
Height of the clearance volume $l_{\text{cl}}$: 0.25 m and 0.05 m
Atmospheric pressure: 0.10 MPa

Solution

Absolute pressures: at the discharge $12.18 + 0.10 = 12.28$ MPa, and at the intake $2.53 + 0.10 = 2.63$ MPa. The pressure ratio is 4.67.
Clearance volume of 0.25 m: the compression ratio is $1 + 2.20/0.25 = 9.8$ . This is more than 4.67, so there is no gas lock. But if the clearance volume is filled with gas, the standing valve opens only after $x_{\text{open}} = 0.25\cdot(4.67 - 1) = 0.92$ m. The pump loses 42% of the plunger stroke.
Clearance volume of 0.05 m: $x_{\text{open}} = 0.05\cdot(4.67 - 1) = 0.18$ m, a loss of 8% of the stroke.
The level is pumped down to the intake: the intake pressure is $0.2 + 0.10 = 0.30$ MPa, and the pressure ratio is 40.9. The compression ratio 9.8 is less than this ratio. A pump with a clearance volume of 0.25 m locks up once it has taken in gas. With 0.05 m the compression ratio is $1 + 2.20/0.05 = 45$ . This is more than 40.9, and the pump copes (for simplicity, the plunger stroke is kept the same).

Answer: At a level of 950 m there will be no gas lock, but a clearance volume of 0.25 m filled with gas takes away 42% of the stroke. With the level pumped down, such a pump locks up with gas, but a pump with a clearance volume of 0.05 m does not.

What it means. A clearance volume of a few centimetres hardly matters. A quarter of a metre takes more than a third of the stroke from a pump that handles gas, and with the level pumped down it leads to gas lock. So we set the plunger as close to the standing valve as possible. For wells with gas, we choose pumps with the travelling valve in the lower part of the plunger, and a long stroke. The data in this example are assumed for illustration.

Venting gas into the annulus. The annulus is connected to the flowline, and its pressure is kept as low as possible. The gas that separates at the intake goes up the annulus and bypasses the pump.
Setting the pump below the perforated interval. The liquid flows to the intake from above, going downwards, while the gas bubbles rise. The casing works as a natural separator. This also gives the largest drawdown (the pressure drop between the reservoir and the well). It is not always possible, because the wellbore must be deep enough below the perforations.
A gas anchor is a separator at the pump intake. The flow is made to turn downwards. The liquid goes to the intake, and the bubbles rise into the annulus. In a packer-type anchor, the production is led through a bypass tube into the annulus above the intake. The liquid then runs down to the pump already free of gas.
A small clearance volume and a long stroke, so that gas that has got into the barrel anyway is pushed out of it on every stroke.
Enough submergence below the level. The higher the intake pressure, the smaller the volume of free gas (the share of free gas at the intake). The price is a smaller drawdown on the reservoir.

A curved wellbore. In a deviated or curved wellbore the rods are pressed against the tubing, and they wear the tubing and themselves (Directional wells: applications and profile). Rod guides (they keep the rods away from the tubing wall) made of wear-resistant plastic are fitted on the rods. Rod rotators are used too: on every stroke they turn the string a little, so that the wear is even. These measures do not remove the wear completely, and in strongly curved wellbores other lift methods take the place of the sucker rod pump.

Sand. At small rates the liquid moves slowly in the tubing, and it carries the sand out badly. It settles above the pump, jams the plunger, and wears the valves, the seats and the barrel. Wear-resistant materials and special pump designs help, but the possibilities of the sucker rod pump are limited here.

Paraffin and salts. Deposits of paraffin wax on the rods and in the tubing raise the friction and the loads. Russian-language practice calls them asphaltene–resin–paraffin deposits. Salts (scale) block the valves and the pump intake. Paraffin is removed with scrapers on the rods and by washing with hot oil or a solvent. Chemicals (inhibitors, which stop paraffin, scale and corrosion) are usually dosed from the wellhead into the annulus, and through it they reach the pump intake. A chemical can be delivered below the intake only through a separate small tube, and this makes the well design more complicated. Scale in the reservoir itself and at the bottom of the well (Formation damage and damage skin) is prevented by squeezing an inhibitor into the reservoir. From there it is slowly carried out with the production.

Repair. A strong point of the sucker rod pump is that downhole repair is simple. To change an insert pump or to replace a parted rod, a light hoisting unit (a small workover rig) is enough. Such a repair is fairly cheap and quick, and this is one of the reasons why sucker rod pumps stay economic on low-rate wells.

Terms and symbols

Sucker rod pump (SRP): A plunger pump driven by a rod string. The name is also used for the whole installation: the drive, the rod string and the pump.
Pumping unit: The beam drive of a sucker rod pump at the wellhead. It turns the rotation of the motor into the up-and-down motion of the polished rod.
Polished rod: The top member of the rod string, which moves through the stuffing box. The load and the stroke are measured on it.
Travelling and standing valves: The ball check valves of the pump. The first is in the plunger, and the second is at the bottom of the barrel.
Theoretical displacement $q_{\text{th}}$: The volume swept by the plunger in a day when its stroke is equal to the stroke of the polished rod.
Pump volumetric efficiency $\alpha_{\text{p}}$: The ratio of the actual delivery to the theoretical one. It accounts for the loss of stroke, the filling of the barrel, the leaks and the shrinkage.
Fluid load $P_{\text{liq}}$: The pressure difference across the plunger multiplied by its area. The rods carry it on the upstroke, and the tubing carries it on the downstroke.
Dynamometer card: A record of the load on the polished rod against its position during one cycle.
Clearance volume: The volume between the valves when the plunger is at the bottom of its stroke. The gas that stays in it delays the intake of liquid.
Gas lock: A condition in which the plunger compresses and expands the gas in the barrel, and the valves do not open.
Submergence below the dynamic fluid level: The height of the liquid column in the annulus above the pump intake. Together with the gas pressure in the annulus, it sets the intake pressure.

Common mistakes and misconceptions

Wrong

“The fluid load on the rods is the weight of the liquid column in the tubing from the pump to the wellhead.”

In fact

From below, the plunger is pushed up by the intake pressure, that is, by the column of liquid in the annulus above the pump. The fluid load is set by the depth of the dynamic fluid level, corrected for the wellhead pressures, and not by the setting depth.

Wrong

“To raise the rate, it is enough to increase the number of strokes.”

In fact

If the barrel is already not filling completely, the delivery will not grow. What will grow are the inertia loads (in proportion to the square of the number of strokes), the number of cycles and the impacts. First we find out what limits the rate: the pump or the inflow.

Wrong

“The plunger travels as far as the polished rod.”

In fact

The plunger stroke is shorter by the stretch of the rods and of the unanchored tubing under the fluid load. The loss grows with depth and with the plunger diameter. In the example of this lesson it is 12% of the stroke.

Wrong

“A volumetric efficiency of 0.65 means that the pump is worn.”

In fact

Usually most of the losses come from the loss of stroke, gas and the shrinkage of the oil, and the leaks account for what is left. We suspect wear when the volumetric efficiency falls with time at the same level and the same gas, and we confirm it with a dynamometer card.

Wrong

“Gas lock depends only on the amount of gas.”

In fact

It depends on the ratio of the discharge and intake pressures and on the clearance volume. The pump locks up when its compression ratio is less than this ratio. The pressures must be absolute.

In depth: advanced detail

Where the load formulas come from

Take the rod string together with the plunger and add up the vertical forces. Two forces act downwards: the weight of the rods in air, and the pressure of the liquid on the top face of the plunger, which is the ring around the rod. Two forces act upwards: the load on the polished rod, and the pressure in the barrel, which acts on the whole area of the plunger from below.

The force balance of the rod string with the plunger

P = P_{\text{rod}} + p_{\text{out}}\,(A_{\text{pl}} - A_{\text{rod}}) - p_{\text{bar}}\,A_{\text{pl}}

where:

$P$ load on the polished rod, N
$P_{\text{rod}}$ weight of the rods in air, N
$p_{\text{out}}$ pressure above the plunger, that is, the pump discharge pressure, Pa
$p_{\text{bar}}$ pressure in the barrel under the plunger: on the upstroke it is equal to the intake pressure, and on the downstroke it is equal to the discharge pressure, Pa
$A_{\text{pl}},\ A_{\text{rod}}$ areas of the plunger and of the cross-section of the rod, m²

Units: SI; gauge pressures

Holds when

statics; rods of one cross-section; the pressure of the liquid on the side surface of the rods gives no vertical force
the pressure losses in the valves are not included

On the downstroke $p_{\text{bar}} = p_{\text{out}}$ , and what is left is $P_{\downarrow} = P_{\text{rod}} - p_{\text{out}} A_{\text{rod}}$ . The discharge pressure is the wellhead pressure plus the pressure of the liquid column: $p_{\text{out}} = p_{wh} + \rho g L$ . The term $\rho g L A_{\text{rod}}$ is the buoyant force, so $P_{\text{rod}} - \rho g L A_{\text{rod}}$ is the weight of the rods in liquid. A small extra term $-p_{wh} A_{\text{rod}}$ is left, and we neglect it. On the upstroke $p_{\text{bar}} = p_{\text{in}}$ , and $(p_{\text{out}} - p_{\text{in}})\,A_{\text{pl}}$ is added to the same quantity. This is the fluid load on the full area of the plunger.

Counterbalance and power

On the upstroke the polished rod does work against the load $P_{\uparrow}$ . On the downstroke the rods give work back under the load $P_{\downarrow}$ . The counterweight, converted into a force at the polished rod, acts as a constant force directed against the weight of the rods. We choose it so that the motor does the same work on both strokes.

The counterbalance effect and the power at the polished rod

C = P'_{\text{rod}} + \frac{P_{\text{liq}}}{2}, \qquad N_{\text{rod}} = \frac{P_{\text{liq}}\,S_{\text{pl}}\,n}{60}

where:

$C$ counterbalance effect: the force of the counterweights, converted into a force at the polished rod, N
$N_{\text{rod}}$ power at the polished rod: the work per cycle (the area of the dynamometer card) multiplied by the number of cycles per second, W
$P'_{\text{rod}},\ P_{\text{liq}}$ weight of the rods in liquid and the fluid load, N
$S_{\text{pl}}$ length of the plunger stroke, m
$n$ number of strokes per minute, min⁻¹

Units: SI; $n$ in strokes per minute

Holds when

an ideal static dynamometer card, with no inertia, friction or gas in the barrel. The real power is found from the area of the recorded dynamometer card
in practice the balance is checked by comparing the largest motor current on the upstroke with that on the downstroke

With this counterweight the motor does the work $P_{\text{liq}} S_{\text{pl}}/2$ on each stroke. The counterweight takes the whole weight of the rods and half of the fluid load. The power at the polished rod is the pressure difference across the plunger multiplied by the volume that the plunger sweeps per second. That is the hydraulic power of the pump with a full barrel.

Worked exampleThe counterweight and the power for the lesson's well

Given (assumed data)

Weight of the rods in liquid and fluid load: 32.08 and 14.68 kN
Plunger stroke and number of strokes: 2.20 m; 6 per minute
Measured delivery and pressure difference across the pump: 22 m³/d; 9.65 MPa

Solution

Counterbalance effect: $C = 32.08 + 14.68/2 = 39.4$ kN.
Work per cycle from the ideal dynamometer card: $14.68\cdot 2.20 = 32.3$ kJ. Power at the polished rod: $N_{\text{rod}} = 32.3\cdot 6/60 = 3.23$ kW.
Hydraulic power from the measured delivery: $N = 22\cdot 9.65/86.4 = 2.5$ kW.
The motor is usually chosen to be about twice as powerful as the calculated power at the polished rod: about 6.5 kW.

Answer: The counterweights must give about 39.4 kN at the polished rod. The power at the polished rod is about 3.23 kW, and the motor is of the order of 6.5 kW.

What it means. The hydraulic power from the measured delivery is lower than the power from the ideal dynamometer card. Part of the barrel volume is taken by gas, part of the liquid leaks away, and the volume of degassed liquid at the surface is smaller than it is in the pump. The double margin of the motor covers the friction of the rods and of the liquid, the losses in the reducer and the belts, and the unevenness of the load over the cycle. The data in this example are assumed for illustration.

The inertia of the rods and the added plunger stroke

In statics the plunger stroke is always shorter than the stroke of the polished rod. In motion there is also an opposite effect. At the bottom of the stroke the rods slow down and speed up again upwards. The force of inertia is added to their weight, and the string stretches further, in the same way as a bar stretches under its own weight. At the top of the stroke everything is the other way round, and the string shortens. So the plunger goes lower and higher than it would in statics. This extra stroke is called plunger overtravel.

An estimate of the added plunger stroke from inertia

\lambda_{\text{ov}} \approx \frac{\rho_{\text{steel}}\,g\,L^2}{E}\,K_{\text{a}}, \qquad S_{\text{pl}} = S - \lambda_{\text{rod}} - \lambda_{\text{tbg}} + \lambda_{\text{ov}}

where:

$\lambda_{\text{ov}}$ added plunger stroke due to the inertia of the rods, m
$\rho_{\text{steel}}$ density of steel, 7850 kg/m³, kg/m³
$L$ length of the rod string, m
$E$ Young's modulus of steel, Pa
$K_{\text{a}}$ acceleration factor, $S n^2/1790$ , —

Units: SI

Holds when

a steel string of constant cross-section; harmonic motion of the polished rod
the acceleration is the same along the whole length of the string, so there are no wave effects. This is an estimate of the order of magnitude, not a design formula
a bar of length $L$ under its own weight stretches by $\rho_{\text{steel}} g L^2/(2E)$ . Inertia adds the fraction $K_{\text{a}}$ of this stretch at the bottom of the stroke and takes away the same amount at the top, so the total change is doubled

The added stroke grows with the square of the depth and with the square of the number of strokes. In shallow wells with a small number of strokes it is almost unnoticeable. In deep wells with a large number of strokes it gives back a noticeable part of the lost stroke.

In other unit systems (1)

Oilfield units

\lambda_{\text{ov}}\,[\text{in}] \approx \frac{40.8\;L^2\,[\text{ft}^2]\;K_{\text{a}}}{E\,[\text{psi}]}

$L$ in feet, $E$ in psi, the result in inches.

$40.8 = 0.2836\cdot 144$ : the specific weight of steel is 0.2836 pound-force per cubic inch, and 144 is the square of the number of inches in a foot.

Worked exampleThe added stroke in a shallow and in a deep well

Given (assumed data)

The lesson's well: $L = 1200$ m, $S = 2.5$ m, $n = 6$ min⁻¹
A deep well with fast pumping: $L = 2400$ m, $S = 3.0$ m, $n = 10$ min⁻¹

Solution

The lesson's well: $K_{\text{a}} = 0.0503$ . Then $\lambda_{\text{ov}} = \dfrac{7850\cdot 9.81\cdot 1200^2}{2.07\cdot 10^{11}}\cdot 0.0503 = 0.027$ m, against 0.30 m of static loss of stroke.
The deep well: $K_{\text{a}} = 3.0\cdot 10^2/1790 = 0.1676$ . Then $\lambda_{\text{ov}} = \dfrac{7850\cdot 9.81\cdot 2400^2}{2.07\cdot 10^{11}}\cdot 0.1676 = 0.36$ m.

Answer: In the lesson's well, inertia adds about 3 cm to the plunger stroke. In the deep well with fast pumping it adds about 36 cm.

What it means. Twice the depth, a stroke of 3.0 m instead of 2.5 m, and 10 strokes instead of 6 have raised the added stroke by more than ten times. So in deep wells a static calculation of the plunger stroke is too pessimistic, and the stroke is found by methods that take the vibrations of the string into account. The data in this example are assumed for illustration.

Connections to other topics

Artificial lift: when it is needed and how to choose the method — where a sucker rod pump is the right choice and where another method is better; the pressure rise of the pump and the dynamic fluid level.
Inflow performance relationship (IPR) — how much the reservoir gives as the bottomhole pressure falls: the limit for any pump delivery.
Nodal analysis of the production system — the pump as an element of the producing system: the steady operating mode is where the delivery equals the inflow.
Black oil model: solution gas–oil ratio and formation volume factors — the formation volume factor and the solution gas–oil ratio of the oil: shrinkage and free gas at the intake.
Well completion: opening the reservoir and the tubing string — the tubing, the annulus and the well design into which the pump is run.
Directional wells: applications and profile — the well profile, on which the friction and the wear of rods and tubing depend.
Electric submersible pumps (ESP) — electric submersible pumps, for larger deliveries than sucker rod pumps can give.
Progressing cavity pumps and hybrid systems — progressing cavity pumps with a rod drive, for viscous oil and sand.

Check yourself

1
The plunger moves up. Which valves are open, and what carries the weight of the liquid column in the tubing?
1. A.Both valves are open; the tubing carries the column
2. B.Both valves are closed; the rods carry the column
3. C.The travelling valve is closed and the standing valve is open; the rods carry the column
4. D.The travelling valve is open and the standing valve is closed; the tubing carries the column
Show the answer
Correct answer: C. The travelling valve is closed and the standing valve is open; the rods carry the column
On the upstroke the pressure of the column presses the ball of the travelling valve onto its seat, and the plunger lifts the column of liquid. Its weight hangs on the rods. The pressure under the plunger falls, the standing valve opens, and the barrel fills. On the downstroke it is the other way round: the column rests on the closed standing valve, that is, on the tubing.
2
A plunger of 57 mm diameter, a stroke of 3.0 m, and 5 strokes per minute. What is the theoretical displacement? What is the volumetric efficiency if the measured delivery is 40 m³/d?
Show the answer
$q_{\text{th}} = 1.131\cdot 10^{-3}\cdot 57^2\cdot 3.0\cdot 5 \approx 55.1$ m³/d. The volumetric efficiency is $40/55.1 \approx 0.73$ , which is within the usual limits for an installation in good order.
3
A 44 mm plunger (area 15.2 cm²) at a depth of 1500 m, a dynamic fluid level of 1100 m, a liquid density of 900 kg/m³, and equal wellhead pressures in the tubing and in the annulus. Find the fluid load and the stretch of a string of 19 mm steel rods (cross-section 2.84 cm²).
Show the answer
The pressure difference across the plunger is equal to the pressure of a liquid column whose height runs from the wellhead to the dynamic fluid level: $900\cdot 9.81\cdot 1100 \approx 9.71$ MPa. The fluid load is $0.1\cdot 9.71\cdot 15.2 \approx 14.8$ kN. The stretch of the rods is $\lambda = \dfrac{14{,}800\cdot 1500}{2.07\cdot 10^{11}\cdot 2.84\cdot 10^{-4}} \approx 0.38$ m. The setting depth enters the stretch (as the length of the rods), but it does not enter the fluid load.
4
On the dynamometer card the load stays high on the downstroke almost to the middle of the stroke and then falls vertically. The echometer shows the level at the pump intake. What is happening, and what should be done?
1. A.A leak in the standing valve; the pump must be replaced
2. B.The barrel fills only halfway and the plunger hits the liquid; reduce the delivery or switch to intermittent pumping
3. C.The pump is working normally
4. D.Parted rods; a workover is needed
Show the answer
Correct answer: B. The barrel fills only halfway and the plunger hits the liquid; reduce the delivery or switch to intermittent pumping
The level has been pumped down to the intake. The inflow is less than the delivery of the pump, so the barrel fills only partly. The plunger travels part of the downstroke under the full load until it hits the liquid. We bring the delivery into line with the inflow: we reduce the number of strokes or the stroke length, or we switch the well to intermittent pumping. With parted rods the card would be a narrow band below the weight of the rods.
5
The number of strokes is raised from 6 to 9 per minute with the same stroke length. How do the theoretical displacement and the inertia addition to the load change?
Show the answer
The theoretical displacement is proportional to the number of strokes, so it grows 1.5 times. The acceleration factor $K_{\text{a}} = S n^2/1790$ is proportional to the square of the number of strokes, so it grows $1.5^2 = 2.25$ times. The actual delivery will grow less if the inflow or the filling of the barrel does not allow it.
6
The height of the clearance volume is 0.2 m, the plunger stroke is 1.8 m, and the discharge pressure is 10 MPa (absolute). At what intake pressure does a pump that has filled with gas lock up?
Show the answer
The compression ratio of the pump is $1 + 1.8/0.2 = 10$ . Gas lock sets in when the ratio of the absolute discharge and intake pressures reaches the compression ratio. That is at an intake pressure of $10/10 = 1.0$ MPa (absolute) and lower. If the clearance volume is reduced to 0.05 m, the compression ratio becomes 37, and the limit falls to about 0.27 MPa.

Updated: 3 October 2026