Progressing cavity pumps and hybrid systems

Quick refresher

The rotor is a single-lobe screw, and the cavity of the stator has two lobes and twice the pitch (the pitch is the length of one full turn of the helix). The rotor is offset from the axis of the stator by the eccentricity, and it touches the elastomer along sealing lines that separate the cavities.
In one revolution the pump displaces the volume $4\,e\,d\,t_{\text{s}}$ : the free cross-section $4ed$ multiplied by the pitch of the stator. The delivery is proportional to the speed, and it hardly pulsates.
Some of the liquid slips back through the sealing lines. The slippage grows with the pressure drop, gets smaller as the viscosity and the interference grow, and hardly depends on the speed, so at low speeds the volumetric efficiency is lower.
The torque is made of the hydraulic part $V\Delta p/2\pi$ , the friction of the rotor in the stator, and the viscous friction of the rods. The rod string twists by tens of turns and stores energy, and after a stop this energy turns it back.
The elastomer limits the temperature and the fluid composition: aromatic hydrocarbons and gas cause swelling and blistering, and running without liquid burns the stator in minutes.
Combined systems, with a submersible motor instead of rods or with gas lift above the pump, remove the limits of one method at the cost of more complicated equipment. Insert pumps are changed without pulling the tubing (the pipe string that carries the production up the well).

Key relationships

V = 4\,e\,d\,t_{\text{s}}, \qquad q_{\text{th}} = 1440\,V n

the displacement per revolution (m³) and the theoretical delivery (m³/d) with $n$ in rpm

q = q_{\text{th}} - s\,\Delta p_{\text{pump}}, \qquad \eta_{\text{v}} = \dfrac{q}{q_{\text{th}}}

the actual delivery with slippage, and the volumetric efficiency

M_{\text{h}} = \dfrac{V\,\Delta p_{\text{pump}}}{2\pi}, \qquad N = M\,\omega

the hydraulic torque and the power at the polished rod

\theta = \dfrac{M L}{G J_p}, \qquad J_p = \dfrac{\pi d_{\text{rod}}^4}{32}

the twist of the rod string

Common pitfalls

Carrying the volumetric efficiency from the test stand into the well as a constant. It depends on the pressure drop, the viscosity, the interference and the speed.
Taking the pressure drop of the pump from the weight of the liquid column alone, and forgetting the friction of viscous oil in the gap between the rods and the tubing.
Pumping the well dry. Without liquid the elastomer burns in minutes.
Choosing the elastomer and the interference without tests in a sample of the oil at the reservoir temperature.
Walking up to the drive right after a stop. The twisted rods and the draining liquid turn it backwards.

After this lesson you can

explain how the rotor and the stator form moving closed cavities, and why the delivery of a progressing cavity pump is even and proportional to the speed
calculate the displacement per revolution of a pump from its geometry, and its theoretical and actual delivery, and explain what the slippage (liquid that leaks back through the pump) and the volumetric efficiency (the actual delivery divided by the theoretical one) depend on
calculate the torque and the power of the drive, and estimate the twist of the rod string
choose the elastomer and the interference fit for the temperature, the oil composition and the gas, and name the causes of stator failure
estimate how the viscosity of the oil and sand change the pressure drop, the torque and the mode of the pump
describe combined systems (an installation with a submersible motor, and a pump together with gas lift) and insert pumps, and estimate what gas injected above the pump gives

How the installation is built

The lesson on choosing an artificial lift method describes the progressing cavity pump briefly: a metal screw turns in an elastomer sleeve, and the cavities between them lift the liquid. Here we look at the installation in detail. We start with the most common kind, with a rod drive from the surface.

SchematicA progressing cavity pump installation with a rod drive (a diagram, not to scale; the well is shortened). The drive at the wellhead turns the polished rod and the rod string. The rotor at the end of the rods turns in the stator, which is fixed to the tubing. A torque anchor stops the reaction torque from unscrewing the tubing. The liquid rises through the ring-shaped gap between the rods and the tubing.

The drive head stands right on the wellhead equipment. An electric motor turns the hollow shaft of the head through a V-belt drive or a gear reducer. The polished rod passes through this shaft and is held in a clamp. The bearing unit of the head carries the weight of the rod string and the axial force from the pressure drop on the rotor. A stuffing box seals the turning rod. The speed is usually from a few dozen to a few hundred revolutions per minute. It is changed by changing the pulleys or with a variable speed drive. The drive must have a backspin brake (a brake that stops the string from spinning backwards; see the section on torque).

The stator is run on the tubing. It is a steel tube with an elastomer lining bonded (vulcanised) to it, and the helical cavity is moulded in the lining. The rotor is a steel screw with a chromium or other wear-resistant coating. It is run on the rods inside the tubing and put into the stator. The height at which the rotor sits is measured from the tag bar (a stop) below the stator. The rotor must be wholly inside the stator, but in operation it must not rest on the stop. We must allow for the rods stretching under load.

The rotor turns clockwise, seen from above, and it drags the stator with it: the reaction torque tries to turn the stator, and with it the bottom of the tubing, in the same direction. The top of the tubing is fixed at the wellhead, and such a turn of the bottom unscrews the threads. So a torque anchor that presses against the casing is installed under the stator. In deviated sections the rods are fitted with rod guides, so that they do not wear the tubing. If the production contains gas, a gas anchor (a separator that sends the gas into the annulus) is installed under the pump, or the pump is set below the perforated interval (venting the gas).

Parts of the installation, from the grid to the liquid

Part: Electric motor, belt drive or reducer
What it does: set the speed of the rotor
Industry term: prime mover, belts, gear reducer

Part: Drive head
What it does: carries the weight of the rods, seals the rod, brakes backspin
Industry term: drive head, stuffing box, backspin brake

Part: Polished rod and rod string
What it does: pass the torque to the rotor
Industry term: polished rod, rod string

Part: Tubing
What it does: the path of the liquid to the wellhead; the stator hangs on it
Industry term: tubing

Part: Stator with an elastomer lining
What it does: the fixed part of the pump
Industry term: stator

Part: Rotor
What it does: the turning part of the pump
Industry term: rotor

Part: Tag bar and torque anchor
What it does: set the position of the rotor and keep the tubing from unscrewing
Industry term: tag bar, torque anchor

Part	What it does	Industry term
Electric motor, belt drive or reducer	set the speed of the rotor	prime mover, belts, gear reducer
Drive head	carries the weight of the rods, seals the rod, brakes backspin	drive head, stuffing box, backspin brake
Polished rod and rod string	pass the torque to the rotor	polished rod, rod string
Tubing	the path of the liquid to the wellhead; the stator hangs on it	tubing
Stator with an elastomer lining	the fixed part of the pump	stator
Rotor	the turning part of the pump	rotor
Tag bar and torque anchor	set the position of the rotor and keep the tubing from unscrewing	tag bar, torque anchor

Progressing cavity pumps are usually run in wells up to 1500–2000 m deep, with rates from a few to a few hundred cubic metres per day (ranges of use of the methods). Some installations work beyond these limits. As the depth grows, the pressure drop and the torque grow, and with them the load on the rods and on the elastomer.

The rotor, the stator and the cavities

The rotor is a single-lobe screw. In any cross-section it is a circle of diameter $d$ , whose centre is offset from the axis of the rotor by the eccentricity $e$ . From one cross-section to the next the circle turns, and its centre describes a helix with the pitch of the rotor $t_{\text{r}}$ . The cavity of the stator has two lobes. In cross-section it is a slot of width $d$ and length $d + 4e$ with half-round ends, and it makes a full turn over the length of the pitch of the stator, $t_{\text{s}} = 2t_{\text{r}}$ .

SchematicThe geometry of a progressing cavity pump (a diagram). On the left is the cross-section: the slot of the stator of width

d

and length

d + 4e

, and the circle of the rotor of diameter

d

. The free area (hatched) is equal to

4ed

. On the right are the cross-sections of one pump at one moment, taken at every quarter of the rotor pitch

t_{\text{r}} = t_{\text{s}}/2

. The slot turns by half a revolution along the pump, and the circle of the rotor moves from one end of the slot to the other, but the free area is the same in every cross-section. With time the picture in each cross-section changes in the same way: the cavities move along the pump.

The assembled rotor touches the lining along continuous sealing lines. They separate the closed cavities, each one a stator pitch long. A pump a few pitches long contains several cavities, one above the other. When the rotor turns, its axis goes round the axis of the stator on a circle of radius $e$ , in the opposite direction. The cross-section of the rotor slides back and forth along the long axis of the slot. The cavities do not change either their shape or their volume. They move along the pump, like a nut along a screw: in one revolution of the rotor, by one pitch of the stator.

From this comes the main property: in any cross-section the free area between the rotor and the stator is the same. It is the area of the slot minus the area of the circle, that is, $4ed$ (see the advanced section). So the flow does not pulsate, and the delivery is proportional to the speed. This is a positive-displacement pump, like a plunger pump (Sucker-rod pumping). Unlike a centrifugal pump, its delivery hardly depends on the pressure against which it works.

The displacement per revolution and the theoretical delivery of a progressing cavity pump

V = 4\,e\,d\,t_{\text{s}}, \qquad q_{\text{th}} = 1440\,V n

where:

$V$ displacement per revolution: the volume displaced in one turn of the rotor, m³
$e$ eccentricity: the offset of the axis of the rotor from the axis of the stator, m
$d$ diameter of the cross-section of the rotor, equal to the width of the slot of the stator, m
$t_{\text{s}}$ pitch of the stator, twice the pitch of the rotor, m
$q_{\text{th}}$ theoretical delivery, m³/d
$n$ rotational speed of the rotor, rpm

Units: metres and cubic metres; $n$ in rpm, $q_{\text{th}}$ in m³/d (1440 is the number of minutes in a day)

Holds when

a single-lobe rotor in a two-lobe stator is the most common geometry; for pumps with more lobes the displacement is found in a different way
the cavities are completely filled with liquid, and there is no slippage; the volumetric efficiency allows for the departures from this

The displacement is set by the geometry of the cross-section and by the pitch: the larger the eccentricity, the diameter and the pitch, the larger the delivery per revolution. Manufacturers give it in their catalogues, in litres per revolution or as the delivery at 100 rpm. The head is set not by the cross-section but by the length of the pump.

The sealing lines are not perfectly tight. Under a pressure difference the elastomer is pressed back a little, and part of the liquid flows into the next cavity. So the pressure difference is spread over several cavities in a row. A stage is a section of the pump one stator pitch long. Each stage holds a pressure difference of the order of a few tenths of a megapascal. The allowed pressure difference of the pump (its rated head) is proportional to the number of stages, that is, to the length. The manufacturer gives it in metres of water column or in megapascals. We choose the pump with a margin in head.

Worked exampleThe delivery of a pump from its geometry

Given (assumed data)

Rotor: cross-section diameter $d = 40$ mm, eccentricity $e = 5$ mm
Stator: pitch $t_{\text{s}} = 300$ mm
Operating mode: 150 rpm; the volumetric efficiency from tests is 0.80

Solution

Free cross-section: $4ed = 4\cdot 5\cdot 40 = 800$ mm², that is 8.0 cm².
Displacement per revolution: $V = 8.0\cdot 30 = 240$ cm³, or 0.24 L per revolution.
Theoretical delivery for each revolution per minute: $1440\cdot 0.00024 = 0.3456$ m³/d. At 150 rpm, $q_{\text{th}} = 51.84$ m³/d.
Actual delivery: $q = 0.80\cdot 51.84 = 41.5$ m³/d. To get 35 m³/d at the same efficiency, we need $35/(0.80\cdot 0.3456) = 127$ rpm.

Answer: The pump displaces 0.24 L per revolution, which is about 34.6 m³/d for every 100 rpm. At 150 rpm and a volumetric efficiency of 0.80 it will deliver about 41.5 m³/d.

What it means. It is convenient to adjust the delivery of a progressing cavity pump to the inflow: it changes in proportion to the speed. Here we take the volumetric efficiency of 0.80 as given. The next section shows that it is not constant: it depends on the pressure drop, the viscosity and the speed itself. The data in this example are assumed for illustration.

Slippage and the characteristic of the pump

The liquid that flows through the sealing lines from a cavity with a higher pressure into a cavity with a lower pressure is called slippage. It makes the delivery smaller than the theoretical one:

The actual delivery and the volumetric efficiency of a progressing cavity pump

q = q_{\text{th}} - s\,\Delta p_{\text{pump}}, \qquad \eta_{\text{v}} = \frac{q}{q_{\text{th}}} = 1 - \frac{s\,\Delta p_{\text{pump}}}{1440\,V n}

where:

$q$ actual delivery at intake conditions, m³/d
$q_{\text{th}}$ theoretical delivery, m³/d
$s$ slippage coefficient: how much liquid flows back for each megapascal of pressure drop, m³/(d·MPa)
$\Delta p_{\text{pump}}$ pressure drop of the pump: the difference between the discharge and intake pressures, MPa
$\eta_{\text{v}}$ volumetric efficiency, the same concept as the pump volumetric efficiency of a sucker rod pump, —
$V,\ n$ displacement per revolution and rotational speed, m³, rpm

Units: m³/d, MPa, rpm

Holds when

the slippage depends linearly on the pressure drop. This is an approximation within the rated head. Near the limiting pressure drop, and in a worn pump, the slippage grows faster
the coefficient $s$ is found on a test stand for a given liquid, temperature and interference. It falls as the viscosity and the interference grow
there is no free gas in the liquid: gas takes up part of the cavities and reduces the delivery of liquid still more

The slippage is set by the pressure drop and not by the speed. So at low speeds it takes a larger share of the delivery, and the volumetric efficiency is lower. Viscous oil “plugs” the gaps, so the efficiency on it is higher than on water.

CalculatedThe calculated characteristic of the pump of the lesson's examples (0.24 L per revolution): the delivery against the pressure drop at 100, 200 and 300 rpm on water (solid lines, slippage 2.5 m³/d per MPa) and on viscous oil (dashed lines, 0.8 m³/d per MPa). The lines are parallel for each liquid: the slippage depends on the pressure drop and not on the speed. The intercepts on the delivery axis are the theoretical deliveries. The coefficients are assumed.

The manufacturer tests every pump on a stand: it measures the delivery of water at several speeds and pressure drops. The results show whether the pump holds its rated head, and we choose the interference from them. But in the well the pump works on oil of a different viscosity and at a different temperature, and the volumetric efficiency from the stand is carried over with corrections.

Worked exampleThe volumetric efficiency on water and on viscous oil

Given (assumed data)

Pump: from the previous example, 0.24 L per revolution
Pressure drop: 8.0 MPa
Slippage coefficients from the stand: water 2.5, viscous oil 0.8 m³/(d·MPa)

Solution

Slippage at 8.0 MPa: on water $2.5\cdot 8.0 = 20.0$ m³/d, and on oil $0.8\cdot 8.0 = 6.4$ m³/d.
At 100 rpm, $q_{\text{th}} = 34.56$ m³/d. On water $q = 14.56$ m³/d ( $\eta_{\text{v}} = 0.42$ ), and on oil 28.16 m³/d (0.81).
At 300 rpm, $q_{\text{th}} = 103.68$ m³/d. On water it is 83.68 m³/d (0.81), and on oil 97.28 m³/d (0.94).

Answer: On water the volumetric efficiency grows from 0.42 at 100 rpm to 0.81 at 300 rpm. On viscous oil it is high at any speed, from 0.81 to 0.94.

What it means. The volumetric efficiency cannot be carried from the stand into the well as a constant: it depends on the pressure drop, the speed and the viscosity. On a low-viscosity liquid it pays to run the pump at higher speeds. On viscous oil the speed is limited by other causes: the filling of the cavities and the torque (the sections below). The data in this example are assumed for illustration.

The elastomer and the interference fit

The elastomer is the heart of a progressing cavity pump and its most vulnerable part. It must be elastic enough to fit the rotor tightly, strong enough to resist wear by sand, and resistant to oil, water, gas and temperature.

The rotor is made a little larger than the cavity of the stator. The difference in size is the interference fit, and it makes the sealing lines tight. A larger interference means less slippage, but more friction torque and more heating. On every revolution the lining is squeezed and released, part of the energy turns into heat inside the rubber, and the rubber carries the heat away badly. The interference also changes in the well: the elastomer swells in oil and expands when it is heated. So for hot wells and for oils that cause swelling, we take a smaller rotor. We choose the interference itself from tests of the elastomer in a sample of the well oil at the reservoir temperature.

Swelling. Aromatic hydrocarbons (benzene, toluene, xylene) of light oils and condensate, and carbon dioxide, penetrate the elastomer, and it swells. The interference grows, the friction and the heating grow, and the lining tears.
Gas and a fast fall of pressure. Under pressure, gas dissolves in the elastomer. If the pressure is lowered fast, for example at a stop, when the pump is pulled, or when the level is pumped down, the gas expands inside the rubber and blows it up into bubbles (explosive decompression).
Gas in the cavities. Free gas is compressed as the cavities move along, so the pressure drop is not spread evenly along the length. The lower stages create a smaller part of it and the upper stages a larger part, and they are loaded more and heat up more.
Hydrogen sulfide makes butadiene-nitrile rubber stiffer and more brittle, and it cracks.
Temperature speeds up ageing and reduces strength. Each elastomer has its own limit.
Running without liquid. Without the liquid that lubricates and cools the lining, the friction of the rotor on the dry elastomer heats it and burns it in minutes.
Chemical treatments (acids, solvents, inhibitors) can also damage the elastomer. Compatibility is checked beforehand.

Stator materials (guidelines)

Material: Nitrile butadiene rubber (NBR)
Temperature, up to: ≈ 90–100 °C
Strengths: elastic, wear-resistant, cheap
Weaknesses: swells in aromatics, hardens with hydrogen sulfide

Material: NBR with a high acrylonitrile content
Temperature, up to: ≈ 90–100 °C
Strengths: swells less in oil and aromatics
Weaknesses: less elastic

Material: Hydrogenated nitrile (HNBR)
Temperature, up to: ≈ 130–150 °C
Strengths: heat-resistant, resistant to hydrogen sulfide
Weaknesses: more expensive

Material: Fluoroelastomer (FKM)
Temperature, up to: ≈ 150 °C and higher
Strengths: resistant to aromatics and high temperature
Weaknesses: expensive, mechanically weaker

Material: Metal stator
Temperature, up to: 300 °C and higher
Strengths: steam and other hot wells
Weaknesses: more slippage

Material	Temperature, up to	Strengths	Weaknesses
Nitrile butadiene rubber (NBR)	≈ 90–100 °C	elastic, wear-resistant, cheap	swells in aromatics, hardens with hydrogen sulfide
NBR with a high acrylonitrile content	≈ 90–100 °C	swells less in oil and aromatics	less elastic
Hydrogenated nitrile (HNBR)	≈ 130–150 °C	heat-resistant, resistant to hydrogen sulfide	more expensive
Fluoroelastomer (FKM)	≈ 150 °C and higher	resistant to aromatics and high temperature	expensive, mechanically weaker
Metal stator	300 °C and higher	steam and other hot wells	more slippage

The limits are approximate and depend on the formulation of the particular material. The elastomer and the interference are chosen from tests in a sample of the oil at the reservoir temperature and from the manufacturer's data.

Viscous oil and sand

Heavy oil is the main field for progressing cavity pumps. The pump does not mix the liquid hard, as a centrifugal pump does, and it does not make stable emulsions. Viscosity reduces the slippage, and the moving parts are slow. But viscosity changes the calculation in three places.

Friction in the tubing. The viscous liquid rises to the wellhead slowly, in laminar flow (smooth flow in layers), through the ring-shaped gap between the turning rods and the tubing. The friction losses can be a noticeable part of the pressure drop of the pump, while for a low-viscosity oil they can be ignored.
Filling of the cavities. Viscous oil flows slowly into the cavities through the intake. At high speeds they do not have time to fill, and the delivery stops growing with the speed. So on heavy oil the pumps run at low and medium speeds, usually not more than two or three hundred revolutions per minute, and with a margin of submergence below the level.
Torque. Rods turning in viscous liquid need extra torque, and oil that has cooled after a stop makes starting harder.

Laminar flow in the ring-shaped gap between the rods and the tubing

\frac{\Delta p_{\text{fr}}}{L} = \frac{8\,\mu\,Q}{\pi\left[r_2^4 - r_1^4 - \dfrac{(r_2^2 - r_1^2)^2}{\ln(r_2/r_1)}\right]}

where:

$\Delta p_{\text{fr}}$ pressure loss from friction over the length $L$ , Pa
$L$ length of the tubing from the pump to the wellhead, m
$\mu$ dynamic viscosity of the liquid, Pa·s
$Q$ flow rate of the liquid, m³/s
$r_2$ inside radius of the tubing, m
$r_1$ radius of the body of the rod, m
$\ln$ natural logarithm, —

Units: SI

Holds when

laminar flow of a Newtonian liquid (one whose viscosity does not depend on the flow rate): the Reynolds number $Re = \rho v\,(d_2 - d_1)/\mu$ (the number that tells smooth flow from turbulent flow) is less than about 2000
the rods are coaxial with the tubing. For the axial flow of a Newtonian liquid their rotation has no effect. Rods pressed against the tubing wall reduce the losses, and couplings and rod guides increase them, so the formula gives an estimate
the viscosity is constant along the length; in fact the oil cools towards the wellhead and thickens

The losses are proportional to the viscosity and the flow rate, and they depend very strongly on the gap: at the same flow rate in a tubing of a larger diameter they are several times smaller. At a low viscosity and a high flow rate the flow becomes turbulent (the Reynolds number is more than about 2000–4000), and then the losses are found from a friction factor, as in the lesson on nodal analysis.

Worked exampleThe friction of viscous oil in the tubing with rods

Given (assumed data)

Well: pump at 850 m, rate 35 m³/d
Liquid: heavy oil with 20% water: viscosity in the tubing 1000 mPa·s, density 970 kg/m³; for comparison, a low-viscosity oil of 5 mPa·s
String options: 89 mm tubing (inside diameter 76 mm) with 25 mm rods, or 73 mm tubing (62 mm) with 22 mm rods

Solution

Velocity in the 76/25 mm gap: $v = \dfrac{35/86{,}400}{\pi\,(0.076^2 - 0.025^2)/4} = 0.1001$ m/s. The Reynolds number is $970\cdot 0.1001\cdot 0.051/1.0 = 5.0$ , so the flow is laminar.
By the formula for the ring-shaped gap, the pressure gradient is 1812 Pa/m, and over 850 m the friction loss is 1.54 MPa.
In the 73 mm tubing with 22 mm rods the gap is smaller: 3.85 MPa, which is 2.5 times more.
A low-viscosity oil (5 mPa·s) in the 89 mm tubing: the flow is laminar here too ( $Re \approx 991$ ), and the losses are about 0.008 MPa, which is very small.
For comparison, the column of liquid from the pump to the wellhead weighs $970\cdot 9.81\cdot 850\cdot 10^{-6} = 8.09$ MPa.

Answer: The viscous oil adds 1.54 MPa to the pressure drop of the pump in the 89 mm tubing and 3.85 MPa in the 73 mm tubing. This is 19 and 48% of the weight of the liquid column.

What it means. For heavy oil, the diameter of the tubing and of the rods is a design choice just like the pump itself. In a narrow gap, almost half of what is needed to lift the liquid goes into friction. The calculation is made for smooth coaxial rods. Couplings, rod guides and the cooling of the oil towards the wellhead increase the losses. The data in this example are assumed for illustration.

Sand. The elastomer gives way elastically under a grain of sand and lets it pass without scoring. So a progressing cavity pump carries sand better than a plunger pump or a centrifugal pump. Viscous oil also holds the sand in suspension and lubricates the rotor–stator pair. In cold production of heavy oil with sand (production with no heat added to the reservoir), the sand is deliberately carried out with the oil, and progressing cavity pumps are the main way of lift there. There are risks too. After a stop the sand settles in the tubing above the pump, and it may jam the rotor at the start. The abrasive wears the coating of the rotor and the lining, and the slippage grows. What helps is running without frequent stops, washing, a softer elastomer and a wear-resistant coating of the rotor.

Torque, power and the rod string

The work of the pump in one revolution is the volume displaced against the pressure drop: $V\Delta p_{\text{pump}}$ . If we divide it by the angle of a revolution, $2\pi$ , we get the hydraulic torque. To it we add the friction torque of the rotor in the stator, which is given by stand tests and grows with the interference and when the elastomer swells, and the viscous torque of the rods turning in the liquid.

The torque and the power at the polished rod

M = M_{\text{h}} + M_{\text{fr}} + M_{\text{rod}}, \qquad M_{\text{h}} = \frac{V\,\Delta p_{\text{pump}}}{2\pi}, \qquad N = M\,\omega, \qquad \omega = \frac{\pi n}{30}

where:

$M$ torque at the polished rod, N·m
$M_{\text{h}}$ hydraulic torque, N·m
$V$ displacement per revolution of the pump, m³
$\Delta p_{\text{pump}}$ pressure drop of the pump, Pa
$M_{\text{fr}}$ friction torque of the rotor in the stator (from stand tests), N·m
$M_{\text{rod}}$ viscous friction torque of the turning rods, N·m
$N$ power at the polished rod, W
$\omega$ angular velocity, rad/s
$n$ rotational speed, rpm

Units: SI

Holds when

the hydraulic torque is found from the theoretical displacement: the pressure drop acts on the rotor whatever amount of liquid slips back
the viscous torque of a smooth rod, coaxial with the tubing, in a Newtonian liquid: $M_{\text{rod}} = \dfrac{4\pi\mu\omega\,r_1^2 r_2^2}{r_2^2 - r_1^2}\,L$ (the symbols are as in the formula for friction in the ring-shaped gap). Couplings, rod guides and the pressing of the rods against the tubing in a curved wellbore increase it
the friction of the rods against the tubing in deviated sections is not included

The hydraulic torque depends only on the displacement and the pressure drop, and not on the speed. The power grows in proportion to the speed. A pump with twice the displacement gives the same delivery at half the speed, but it needs twice the torque.

The rod string passes this torque from the drive to the rotor, and it twists like a long spring. We estimate the twist with the formula for the torsion of a bar:

The twist of the rod string and the stored energy

\theta = \frac{M L}{G J_p}, \qquad J_p = \frac{\pi d_{\text{rod}}^4}{32}, \qquad W = \frac{M\,\theta}{2}

where:

$\theta$ angle of twist of the string, rad
$M$ the torque that is passed, N·m
$L$ length of the rod string, m
$G$ shear modulus of steel (its resistance to twisting), about $8\cdot 10^{10}$ , Pa
$J_p$ polar moment of inertia of the cross-section of the rod (how the cross-section resists twisting), m⁴
$d_{\text{rod}}$ diameter of the body of the rod, m
$W$ elastic energy stored in the twisted string, J

Units: SI

Holds when

solid rods of one diameter; couplings and upset ends (the thickened ends of the rods) reduce the twist a little
the torsion is elastic: the shear stresses in the rods are below the yield strength (the stress at which steel starts to deform permanently)

The twist is proportional to the torque and the length, and inversely proportional to the fourth power of the diameter: 25 mm rods at the same torque twist 1.7 times less than 22 mm rods. The strength of the string is checked for the combined effect of the torque, the weight of the rods themselves and the axial force from the pressure drop on the rotor.

When the drive stops, the stored energy is released: the string untwists in the opposite direction. In addition, the column of liquid in the tubing runs back through the pump into the well and turns the rotor like a hydraulic motor, also in the opposite direction, until the levels equalise. Without braking, the reverse speed can exceed the working speed, and the pulleys and couplings of the drive are destroyed. So every drive of a progressing cavity pump is equipped with a backspin brake, which limits the speed of untwisting and lets the string and the liquid give up their energy calmly.

Worked exampleThe operating mode, torque and power of a pump for heavy oil

Given (assumed data)

Well: a pump (0.24 L per revolution) at 850 m, a dynamic fluid level of 600 m (the depth of the liquid surface in the annulus, the space between the tubing and the casing); wellhead pressures 1.0 MPa in the tubing and 0.2 MPa in the annulus (gauge)
Production: 35 m³/d, density 970 kg/m³, viscosity in the tubing 1000 mPa·s; 89 mm tubing, 25 mm rods
Pump from the stand: slippage on this oil 0.8 m³/(d·MPa), friction torque of the rotor in the stator 80 N·m
Drive and rods: efficiency of the transmission 0.92, of the electric motor 0.88; shear modulus of steel 80 GPa

Solution

Intake pressure: $0.2 + 970\cdot 9.81\cdot 250\cdot 10^{-6} = 2.58$ MPa. The pressure drop of the pump, with the friction from the previous example, is $\Delta p_{\text{pump}} = 1.0 + 8.09 + 1.54 - 2.58 = 8.05$ MPa.
Slippage: $0.8\cdot 8.05 = 6.4$ m³/d. Speed: $n = (35 + 6.4)/0.3456 = 120$ rpm. The volumetric efficiency is 0.84.
Hydraulic torque: $M_{\text{h}} = 0.00024\cdot 8.05\cdot 10^6/(2\pi) = 307.5$ N·m. The viscous torque of the rods, from the formula given in the assumptions of the torque equation, is 23.5 N·m. In all, with the friction in the stator, $M = 307.5 + 80 + 23.5 = 411$ N·m.
Power at the polished rod: $\omega = \pi\cdot 119.9/30 = 12.56$ rad/s, and $N = 411\cdot 12.56\cdot 10^{-3} = 5.16$ kW. The hydraulic power is $35\cdot 8.05/86.4 = 3.26$ kW (the divisor $86.4 = 86{,}400/1000$ converts the product of m³/d and MPa into kilowatts). So the efficiency of the pump with the rods is 63%. From the grid, $5.16/(0.92\cdot 0.88) = 6.4$ kW.
Twist of the rods: $J_p = \pi\cdot 2.5^4/32 = 3.83$ cm⁴, and $\theta = \dfrac{411\cdot 850}{8\cdot 10^{10}\cdot 3.83\cdot 10^{-8}} = 114$ rad, about 18 turns. The stored energy is $411\cdot 114/2\cdot 10^{-3} = 23.4$ kJ.

Answer: The pump must turn at about 120 rpm. The torque at the polished rod is about 411 N·m, and the power is 5.16 kW (6.4 kW from the grid). The rod string is twisted by about 18 turns.

What it means. Three quarters of the torque is hydraulic, and it is set by the pressure drop, in which the friction of the viscous oil makes up a noticeable share. We choose the motor of the drive with a margin for the start, because oil that has cooled after a stop, and sand that has settled, raise the starting torque. The energy of the twisted string is like that of a load of one tonne lifted by 2.4 m. At a stop, the backspin brake must take it up. The data in this example are assumed for illustration.

A curved wellbore. Rods that turn in deviated and curved sections are pressed against the tubing, and they wear both it and themselves. The wear of the tubing is one of the main worries of installations with a rod drive. The pump can be set in a strongly deviated section too, sometimes with an inclination above 70°. But the dogleg severity (how sharply the wellbore bends) is limited to a few degrees per 30 m (Directional wells: applications and profile), and the rods are fitted with rod guides. We use non-rotating guides and continuous rods without couplings. The stator hangs on the tubing, so to replace the pump we pull both the rods and the tubing, usually with a light hoist (a small hoisting rig). Such wells are shallow and do not flow by themselves.

The operating mode and monitoring

The main rule of operation: the pump must always be submerged in liquid. We match the delivery to the inflow by choosing the speed so that the dynamic fluid level stays above the intake with a margin. We check the level with an echometer or with an intake pressure gauge. Thousands of pumps work with no sensors, but then we must watch with special care for the well being pumped down. The signs can be seen at the surface too. The pump is a positive-displacement one, so when there is not enough liquid, the torque at the polished rod and the motor current fall, and with gas and a pumped-down level they fluctuate.

Signs of faults in a progressing cavity pump installation

What is seen: The torque and the current fall and fluctuate, and the level is at the intake
Probable cause: the pump has pumped the well down, and gas comes to the intake
What is done: lower the speed

What is seen: The torque grows slowly at the same operating mode
Probable cause: swelling of the elastomer, paraffin in the tubing, sand above the pump
What is done: test the elastomer on a sample of the oil, wash, remove the paraffin

What is seen: The delivery falls at the same speed
Probable cause: wear of the rotor and the stator: the slippage grows; a leak in the tubing
What is done: check the tubing; replace the pump

What is seen: The torque has fallen to the torque of turning the rods alone
Probable cause: parted or unscrewed rods
What is done: stop and pull the rods

What is seen: Stators fail often
Probable cause: the elastomer is not suitable, overheating, running without liquid
What is done: test the elastomers, choose the interference, protect against pumping-down

What is seen	Probable cause	What is done
The torque and the current fall and fluctuate, and the level is at the intake	the pump has pumped the well down, and gas comes to the intake	lower the speed
The torque grows slowly at the same operating mode	swelling of the elastomer, paraffin in the tubing, sand above the pump	test the elastomer on a sample of the oil, wash, remove the paraffin
The delivery falls at the same speed	wear of the rotor and the stator: the slippage grows; a leak in the tubing	check the tubing; replace the pump
The torque has fallen to the torque of turning the rods alone	parted or unscrewed rods	stop and pull the rods
Stators fail often	the elastomer is not suitable, overheating, running without liquid	test the elastomers, choose the interference, protect against pumping-down

Combined systems

Combined (hybrid) systems join two lift methods, or parts of two installations, to remove the limits of each. Two combinations are common: a progressing cavity pump with a submersible motor, and a pump with gas lift. We look at them, and also at insert pumps, which are changed without pulling the tubing.

A submersible motor instead of rods: the electric progressing cavity pump installation. A submersible electric motor turns the rotor, as in an electric submersible pump installation (ESP), with the same protector, cable and control station. But a centrifugal pump needs thousands of revolutions per minute, and a progressing cavity pump needs hundreds. So a reducer is put between the motor and the pump, or a permanent-magnet motor is used that runs at low speed from a variable speed drive. The rotor moves in a planetary way (its axis goes round the axis of the stator), and it is joined to the shaft through a flexible shaft or a hinged coupling. With no turning rods, there are no parted rods, no wear of the tubing and no loss of torque to the friction of the rods. The installation can also be run into a deviated or horizontal wellbore. Most often it is the progressing cavity pump itself that is changed in such an installation, because it wears faster than the motor.

Insert pumps. To change a worn pump without pulling the tubing, the stator is run inside the tubing on the rods together with the rotor, and it is seated in a lock seat, as with an insert sucker rod pump. In installations with a submersible motor, the same is given by pumps that are brought out with wireline equipment (equipment run on a cable). An enlarged tubing lets the pump pass, and the connection to the motor shaft at the bottom is opened and closed again. Such an installation costs more, but every change of the pump is cheaper and quicker, and over the whole service life the costs can be lower (costs over the whole service life).

A pump with gas lift. Gas injected into the tubing above the pump makes the column of liquid above it lighter, and the pump needs a smaller pressure drop: fewer stages and a smaller motor are enough. The pump itself can be set deeper, and a larger drawdown can be taken. If the pump fails, for example because sand wears it, gas lift goes on lifting, though at a smaller rate. To keep the injected gas out of the pump intake, the annulus above the pump is closed off by a packer (a seal that closes the annulus) with a cable lead-through, and the gas lift valves are set above the packer (Gas lift: principles and choice of operating conditions). Gas lift is most often combined with an ESP, and less often with progressing cavity pumps, because gas lift lifts viscous oil badly.

CalculatedThe calculated pressure in the tubing and under the pump against depth for the well of the ESP lesson. A: the pump alone, 150 m³/d. B: the pump with gas lift, 15 thousand m³/d of gas through a valve at 1900 m. The rate and the intake pressure are the same, but the discharge pressure is almost 40% lower, and the pressure drop that the pump creates is half as large. C: gas lift without a pump (the losses in the stopped pump are not included), about 79 m³/d. The gas–liquid flow is calculated with the teaching model of the lesson on multiphase flow.

Worked exampleGas lift above an ESP: what it gives and what it costs

Given (assumed data)

Well and pump: from the ESP lesson: 150 m³/d of liquid (152.4 m³/d at the intake), the intake at 1950 m, intake pressure 5.77 MPa; with no gas the pressure rise of the pump is 15.55 MPa, the number of stages is 342, and the shaft power is 46.1 kW
Gas lift: 15 thousand m³/d of gas through a valve at 1900 m above the packer
Flow in the tubing: the teaching model of the lesson Multiphase flow in tubing: 62 mm tubing, gas of specific gravity 0.7, $Z = 0.9$ , average temperature 60 °C; the gas that comes out of the oil itself is not included, because there is little of it compared with the injected gas
Compressor: takes the gas at 0.7 MPa (absolute pressure)

Solution

The injection gas–liquid ratio in the tubing above the valve (cubic metres of injected gas per cubic metre of liquid): $15{,}000/152.4 = 98.4$ m³/m³. By the model, the pressure in the tubing at the valve is 12.54 MPa, instead of 20.82 MPa with no gas.
The discharge pressure of the pump is the pressure at the valve plus 50 m of liquid column: $12.54 + 0.51 = 13.05$ MPa. The pressure rise of the pump is $\Delta p_{\text{pump}} = 13.05 - 5.77 = 7.28$ MPa, which is 53% less.
A pump of the same stages needs only 160 stages instead of 342. The shaft power is 21.6 kW instead of 46.1, a saving of 24.5 kW.
The gas must be delivered to the valve at a pressure above 12.54 MPa. Even an ideal isothermal compressor (one that keeps the gas at a constant temperature) would spend on compressing 15 thousand m³/d from 0.7 to 12.64 MPa (absolute) $0.1013\cdot 10^6\cdot\dfrac{15{,}000}{86{,}400}\cdot\ln\dfrac{12.64}{0.7}\cdot 10^{-3} = 51$ kW. This is more than the pump saves, and a real compressor spends still more. The weight of the gas column in the annulus lowers the pressure needed at the wellhead a little, but it does not change the conclusion.
If the pump stops, gas lift alone, from the crossing of the inflow and outflow curves, will give about 79 m³/d at a bottomhole pressure of 12.4 MPa (the losses in the stopped pump are not included).

Answer: Gas cuts the pressure rise of the pump from 15.55 to 7.28 MPa and the shaft power from 46.1 to 21.6 kW. If the pump fails, it keeps about 79 m³/d out of 150.

What it means. The electric energy of the pump is not saved for nothing: compressing the gas costs more. So this combination is justified where gas at a high pressure is already available, on a field with a gas lift system or with gas wells, or where reliability matters most. When the pump often fails because of sand, gas lift keeps up production until the repair, and a pump with a smaller pressure rise has an easier life. The data in this example are assumed for illustration.

Terms and symbols

Progressing cavity pump (PCP): A positive-displacement pump made of a metal rotor in an elastomer stator. With a rod drive the rods turn it from the surface. With a submersible motor (an electric progressing cavity pump installation) the motor turns it.
Rotor and stator: A single-lobe steel screw, and a steel tube with an elastomer lining in which a two-lobe helical cavity with twice the pitch is moulded.
Eccentricity $e$: The offset of the axis of the rotor from the axis of the stator. The slot of the stator is longer than the cross-section of the rotor by $4e$ .
Displacement per revolution $V$: The volume displaced in one revolution of the rotor: $4\,e\,d\,t_{\text{s}}$ .
Stage: A section of the pump one stator pitch long. The allowed pressure drop is proportional to the number of stages.
Slippage: The flow of liquid back through the sealing lines under the pressure drop.
Volumetric efficiency $\eta_{\text{v}}$: The ratio of the actual delivery to the theoretical one.
Interference fit: The excess of the size of the rotor over the size of the cavity of the stator, which makes the sealing lines tight.
Stator elastomer: The rubber-like material of the lining. We choose it for the temperature and the fluid composition.
Torque anchor: An anchor under the stator that stops the reaction torque from unscrewing the tubing.
Backspin: The untwisting of the rods and the turning of the rotor in the opposite direction after a stop. The brake of the drive stops it.
Hybrid artificial lift: Two lift methods in one well: gas lift above a pump, or a progressing cavity pump with a submersible motor.

Common mistakes and misconceptions

Wrong

“The volumetric efficiency of a pump is a fixed number from its data sheet.”

In fact

It depends on the pressure drop, the viscosity, the interference and the speed. The slippage is set by the pressure drop and not by the revolutions, so at low speeds the efficiency is lower. We recalculate the stand data on water for the oil of the well.

Wrong

“A progressing cavity pump is a positive-displacement pump, so it can work with no liquid.”

In fact

Without liquid the elastomer is neither lubricated nor cooled, and it burns in minutes. We keep the level above the intake, and we match the delivery to the inflow.

Wrong

“The larger the interference, the better: there are fewer leaks.”

In fact

A larger interference raises the friction torque and the heating of the elastomer, and in the well the interference grows further from swelling and temperature. We choose the interference from tests of the elastomer in a sample of the oil.

Wrong

“The pressure drop of the pump is the weight of the liquid column up to the wellhead.”

In fact

The wellhead pressure and the friction are added. For viscous oil, the friction in the gap between the rods and the tubing can be comparable with the weight of the column. The head of liquid in the annulus above the intake reduces the pressure drop.

Wrong

“Gas lift above the pump saves energy.”

In fact

It reduces the power of the pump, but compressing the gas, as a rule, takes more energy than the pump saves: gas lift lifts liquid with a lower efficiency than a pump. The gain is in reliability, in a reserve if the pump fails, and where gas at a high pressure is already available.

In depth: advanced detail

Why the free area of the cross-section is

4ed

The slot of the stator in cross-section is two half-circles of diameter $d$ , whose centres are $4e$ apart, and a rectangle between them. Its area is the area of a circle of diameter $d$ plus the area of the rectangle $4e\times d$ . Taking away the area of the cross-section of the rotor, the same circle, we get the free area:

The free area of the cross-section of a progressing cavity pump

S_{\text{free}} = \left(\frac{\pi d^2}{4} + 4e\,d\right) - \frac{\pi d^2}{4} = 4\,e\,d

where:

$S_{\text{free}}$ area filled by liquid in the cross-section, m²
$d,\ e$ diameter of the cross-section of the rotor, and the eccentricity, m

Units: any consistent units

Holds when

a single-lobe rotor in a two-lobe stator; the interference and the deformation of the elastomer are not included.

The area does not depend on where in the slot the circle of the rotor is, so it is the same in all cross-sections and at all moments. This is where the even delivery comes from. In one revolution the cavities move on by one stator pitch, and the displacement per revolution is equal to $4\,e\,d\,t_{\text{s}}$ .

Why does the cross-section of the rotor move along a straight line? The centre of the cross-section is $e$ away from the axis of the rotor, and it turns with the rotor. The axis of the rotor itself goes round the axis of the stator on a circle of the same radius $e$ in the opposite direction. The sum of two equal rotations in opposite directions is an oscillation along a straight line with a swing from $-2e$ to $+2e$ : the circle of the rotor goes from one end of the slot to the other. This is also where the length of the slot, $d + 4e$ , comes from.

Connections to other topics

Artificial lift: when it is needed and how to choose the method — when a progressing cavity pump is the right choice, and how it compares with the other methods.
Sucker-rod pumping — sucker rod plunger pumps: venting gas, the echometer, the rod string.
Electric submersible pumps (ESP) — electric submersible pumps: the motor, the protector and the cable are the same as in installations with a submersible motor.
Gas lift: principles and choice of operating conditions — gas lift is the basis of combined systems with a pump.
Multiphase flow in tubing — gas–liquid flow in the tubing, from which the gain from gas above the pump is calculated.
Nodal analysis of the production system — the operating point of a well with a pump and without it.
Directional wells: applications and profile — the inclination and the dogleg severity of the wellbore, on which the wear of rods and tubing depends.

Check yourself

1
A rotor of 30 mm diameter with an eccentricity of 4 mm, and a stator pitch of 240 mm. What is the theoretical delivery of the pump at 200 rpm?
Show the answer
The displacement per revolution is $V = 4\cdot 4\cdot 30\cdot 240 = 115{,}200$ mm³ $= 0.1152$ L. The theoretical delivery is $1440\cdot 0.1152\cdot 10^{-3}\cdot 200 \approx 33.2$ m³/d.
2
On the stand, at 300 rpm, a pump gave a volumetric efficiency of 0.80 on water. What will it be at 100 rpm and the same pressure drop, if the slippage does not depend on the speed?
1. A.0.40
2. B.0.27
3. C.0.93
4. D.0.80
Show the answer
Correct answer: A. 0.40
At 300 rpm the slippage is 20% of the theoretical delivery, that is, $0.2\cdot 3 = 0.6$ of the theoretical delivery at 100 rpm. At 100 rpm what is left is $1 - 0.6 = 0.40$ .
3
Why must a progressing cavity pump not be started or left running with the valve at the wellhead closed?
1. A.It will overheat, like a centrifugal pump against a closed valve
2. B.The pressure will grow until the elastomer, the rods or the wellhead equipment are destroyed
3. C.The rotor will start to turn in the opposite direction
4. D.Nothing will happen: the pump will simply stop delivering
Show the answer
Correct answer: B. The pressure will grow until the elastomer, the rods or the wellhead equipment are destroyed
A positive-displacement pump displaces liquid at any discharge pressure, so the pressure grows until the weakest element is destroyed. The protection is a pressure sensor that switches off the drive, and a relief valve.
4
In a month the torque at the polished rod grew from 400 to 550 N·m, with the rate, the speed and the wellhead pressure unchanged. What causes are possible, and what should be checked?
Show the answer
The pressure drop and the speed are the same, unless the friction of the flow in the pipes has grown, so the hydraulic torque has not changed. The friction torque has grown. The causes: swelling of the elastomer (aromatics, gas, temperature) has increased the interference; sand has settled above the pump or in the cavities; paraffin has deposited on the rods and in the tubing, which increases the viscous torque of the rods and, by narrowing the passage, the friction of the flow and with it the pressure drop. We check the elastomer on a sample of the oil, wash the well and remove the paraffin. If the elastomer is swelling, we choose another elastomer or a rotor with a smaller interference.
5
A rod string of 22 mm rods, 1000 m long, passes a torque of 450 N·m. By how many turns is it twisted, and what changes with 25 mm rods?
Show the answer
$J_p = \pi\cdot 0.022^4/32 \approx 2.30\cdot 10^{-8}$ m⁴, and $\theta = 450\cdot 1000/(8\cdot 10^{10}\cdot 2.30\cdot 10^{-8}) \approx 245$ rad, about 39 turns. For 25 mm, $J_p \approx 3.83\cdot 10^{-8}$ m⁴ and $\theta \approx 147$ rad, about 23 turns, which is $(25/22)^4 \approx 1.7$ times less.
6
What justifies gas lift above an ESP if compressing the gas needs more energy than the pump saves?
1. A.The gas cools the motor
2. B.Without gas an ESP does not work
3. C.The availability of gas at a high pressure, a reserve if the pump fails, and a smaller load on the pump
4. D.Nothing: it is a design error
Show the answer
Correct answer: C. The availability of gas at a high pressure, a reserve if the pump fails, and a smaller load on the pump
The combination pays when gas at a high pressure is already available on the field, when reliability matters (gas lift keeps up production if the pump fails), and when a pump with a smaller pressure rise has an easier life. In energy terms alone it does not win.

Updated: 5 October 2026