Artificial lift: when it is needed and how to choose the method

Quick refresher

A well flows naturally as long as the reservoir pressure, minus the losses in the reservoir, is higher than the pressure needed to lift the fluid column, overcome friction and push against the wellhead pressure. Depletion, a rising water cut and a falling gas–oil ratio break this balance.
A pump adds pressure in a single step at its setting depth. Gas lift lightens the column above the point where gas is injected. Both lower the outflow curve, so an operating point appears where natural flow is impossible.
A pump lifts liquid from the dynamic fluid level (the level of the liquid in the annulus while the well is producing). Its pressure rise is roughly the weight of the liquid column from that level to the wellhead, plus the difference between the wellhead pressure and the annulus pressure, plus friction. (The annulus is the space between the tubing and the casing.) Hydraulic power is the flow rate multiplied by this pressure rise.
How much you gain from lowering the bottomhole pressure depends on the shape of the IPR. For a straight line the rate grows in proportion to the drawdown. Below the bubble point (the pressure at which gas starts to come out of the oil) it grows by less.
Gas is the main enemy of pumps and the main helper of gas lift. Sand wears out centrifugal and plunger pumps, while progressing cavity pumps and gas lift tolerate it better. Viscous oil is lifted best by progressing cavity and sucker rod pumps.
Over the service life, operating costs (energy, repairs, downtime) are usually bigger than capital costs. Choose the method with the lowest life-cycle cost within the given limits, not the one that is cheapest to install.

Key relationships

H_{\text{st}} = H - \dfrac{\bar p_R - p_{\text{ann}}}{\rho g}

depth of the static fluid level

\Delta p_{\text{pump}} \approx p_{wh} - p_{\text{ann}} + \rho g H_{\text{dyn}} + \Delta p_{\text{fr}}

pump pressure rise in terms of the dynamic fluid level

N = q\,\Delta p_{\text{pump}},\qquad N_{\text{el}} = \dfrac{N}{\eta}

hydraulic power and the power drawn from the grid

\varphi_{\text{g}} = \dfrac{q_o (R - R_s) B_g}{q_o (R - R_s) B_g + q_o B_o + q_w B_w}

free gas fraction at the pump intake

Common pitfalls

Choosing a pump for today's operating mode without looking at the forecast of rate, water cut and reservoir pressure.
Extending a straight-line IPR below the bubble point. The gain from the pump will then be overstated.
Forgetting about free gas (gas that is not dissolved in the liquid) at the pump intake.
Comparing methods by the cost of the equipment rather than by life-cycle cost.
Planning gas lift without a source of high-pressure gas, or electric submersible pumps without a reliable power supply.

After this lesson you can

tell from the pressures whether a well can flow naturally, and explain why artificial lift is needed as the reservoir is depleted, the water cut (the share of water in the produced liquid) rises and the gas–oil ratio (the volume of gas produced with each volume of oil) falls
explain how a pump and gas lift change the pressure distribution in the well, and calculate the pressure rise a pump must produce and its hydraulic power (the useful power the pump gives to the liquid)
estimate how much the rate rises when the bottomhole pressure is lowered, for a straight-line and for a Vogel inflow performance relationship (IPR, the curve of rate against bottomhole pressure)
describe the principle, strengths and limitations of gas lift, sucker rod pumps, electric submersible pumps, progressing cavity pumps, jet pumps and plunger lift
rule out unsuitable methods by rate, depth, gas, sand, wellbore deviation, power and infrastructure, and compare the remaining ones by life-cycle cost (the total cost over the whole service life)

When a well needs artificial lift

A well flows naturally if the fluid that comes from the reservoir to the bottom of the hole has enough pressure to rise up the tubing to the wellhead (the top of the well) and on to the gathering point. In the language of nodal analysis, this means that the inflow curve crosses the outflow curve (well performance diagram). On a new field this is usually the case. But as development goes on, the balance breaks down. Without pressure maintenance (injecting water or gas to hold up the pressure), the reservoir pressure falls. The water cut rises, and the column in the tubing gets heavier. The gas–oil ratio falls, and there is less gas to lighten the column (Multiphase flow in tubing). Eventually the curves stop crossing, and the well dies (stops flowing).

The well has stopped flowing, or flows at a rate too low to pay.
More is needed than natural flow gives. A pump can lower the bottomhole pressure far below what natural flow achieves and take a higher rate from the reservoir.
The reservoir has little energy from the start. In small fields without pressure maintenance, artificial lift is sometimes needed from the first day.
The gathering point is far away. A well whose production goes through a long flowline — a remote or subsea well, for example — needs extra pressure at the wellhead.

A simple test is the static fluid level: the level at which the liquid stands in a shut-in well. The column of liquid from this level down to the reservoir balances the reservoir pressure. If the level is below the wellhead, the reservoir cannot hold a column of liquid up to the surface, and without gas the well will not flow. Sometimes the well can still be started, if the produced fluid contains a lot of gas and the productivity is high. First you induce inflow (by swabbing, which means pulling fluid out of the well, or by injecting nitrogen). Then the gas lightens the column. But you cannot rely on this for long.

Depth of the static fluid level

H_{\text{st}} = H - \frac{\bar p_R - p_{\text{ann}}}{\rho g}

where:

$H_{\text{st}}$ true vertical depth of the static fluid level below the wellhead, m
$H$ true vertical depth of the middle of the perforated interval, m
$\bar p_R$ average reservoir pressure, Pa
$p_{\text{ann}}$ gas pressure in the annulus at the wellhead (gauge; zero if the annulus is open), Pa
$\rho$ density of the liquid in the well, kg/m³
$g$ acceleration due to gravity, 9.81 m/s², m/s²

Units: SI; pressures are gauge (above atmospheric)

Holds when

The well is shut in and the bottomhole pressure has equalised with the reservoir pressure.
Below the level there is liquid of constant density with no gas. The weight of the gas column above the level is small.
Depths are true vertical depths. In a deviated well, a level measured along the hole is converted using the directional survey (the record of the well's path).

A level below the wellhead means that the reservoir pressure is not enough to support a column of liquid up to the surface. The higher the water cut (the heavier the liquid) and the lower the reservoir pressure, the deeper the level.

In other unit systems (1)

Metric oilfield units

H_{\text{st}} = H - \frac{102\,(\bar p_R - p_{\text{ann}})}{\gamma}

$H$ in m, pressures in MPa, $\gamma$ is the specific gravity of the liquid (water = 1).

$102 = 10^6/(1000\cdot 9.81)$ : a column of fresh water 102 m high produces a pressure of 1 MPa.

How a pump and gas lift put a well back to work

On the nodal analysis diagram, artificial lift lowers the outflow curve. To produce the same rate, the well now needs a lower bottomhole pressure (what shifts the curves). A pump and gas lift do this in different ways.

A pump adds pressure in a single step. Above the pump, the pressure in the tubing increases with depth, starting from the wellhead pressure, as in an ordinary well filled with liquid. At the pump it drops sharply to the pump intake pressure. Below the pump it increases again, down to the bottomhole pressure. In effect, the pump takes the weight of the column above it off the reservoir. So the bottomhole pressure can be brought very low, as low as the gas and the inflow allow. The deeper the pump is set, the lower the bottomhole pressure can go. The best position is at the perforated interval or below it.

Gas lift lightens the column. High-pressure gas is fed into the annulus and enters the tubing through a valve at depth. Above the injection point, a gas–liquid mixture rises that is much lighter than the liquid. But the column can be lightened only up to a limit: as the gas rate increases, friction increases (the outflow curve and its minimum). That is why gas lift does not lower the bottomhole pressure as far as a pump does. On the other hand, gas and sand do not trouble it (Gas lift: principles and choice of operating conditions).

CalculatedCalculated pressure against depth for the well in the example below. A: the shut-in well. The liquid in the annulus stands up to the static fluid level at about 580 m, with gas at a pressure of 0.3 MPa above it. At the bottom of the hole the pressure is the reservoir pressure, 12 MPa. B: operation with a pump at a depth of 1750 m. The pressure in the tubing increases from the wellhead to the pump, drops in a single step at the pump to the intake pressure, and the bottomhole pressure is 6.0 MPa. C: gas lift with gas injected at 1600 m and an assumed average mixture density of 400 kg/m³. The bottomhole pressure is about 9.2 MPa. Friction is not included.

Pressure rise produced by the pump

\begin{aligned} \Delta p_{\text{pump}} &= p_{\text{out}} - p_{\text{in}},\\ p_{\text{out}} &= p_{wh} + \rho g H_{\text{pump}} + \Delta p_{\text{fr}},\\ p_{\text{in}} &= p_{wf} - \rho g\,(H - H_{\text{pump}}) \end{aligned}

where:

$\Delta p_{\text{pump}}$ pressure rise the pump must produce, Pa
$p_{\text{out}}$ pump discharge pressure — what is needed to lift the liquid to the wellhead, Pa
$p_{\text{in}}$ pump intake pressure, Pa
$p_{wh},\ p_{wf}$ tubing pressure at the wellhead and bottomhole pressure at the middle of the perforated interval, Pa
$H_{\text{pump}},\ H$ true vertical depths of the pump setting and of the middle of the perforated interval, m
$\rho$ density of the liquid in the tubing and below the pump, kg/m³
$\Delta p_{\text{fr}}$ friction losses in the tubing, Pa

Units: SI (any consistent system of units)

Holds when

The liquid contains no free gas and has the same density in the tubing and below the pump. If there is gas in the tubing, the column is lighter, and its weight is calculated with multiphase flow correlations.
Depths are true vertical depths. Friction is calculated along the length of the tubing.
The formula also holds for a pump set below the perforated interval. Then $H - H_{\text{pump}} < 0$ , and the intake pressure is higher than the bottomhole pressure.

The pump raises the pressure from the intake pressure to the pressure needed to bring the liquid to the wellhead. The lower the bottomhole pressure you want, the lower the intake pressure, and the greater the pressure rise and the power.

In a well with a pump, the annulus above the intake is usually open to gas, and the liquid in it stands up to the dynamic fluid level. The intake pressure equals the gas pressure in the annulus plus the weight of the liquid column from the level to the intake. This gives a clear rule of thumb. The pump lifts the liquid from the dynamic level, and its pressure rise is roughly the weight of the liquid column from this level to the wellhead, plus the difference between the wellhead pressure and the annulus pressure, plus friction. An echometer measures the dynamic level: it sends an acoustic pulse down the annulus and times the reflection off the liquid surface. Engineers use the level to monitor the operating mode. A level that has dropped to the intake means that the pump has pumped the well “dry”.

Hydraulic power and power consumption

N = q\,\Delta p_{\text{pump}},\qquad N_{\text{el}} = \frac{N}{\eta}

where:

$N$ hydraulic power — the useful work of lifting the liquid per unit of time, W
$q$ volumetric flow rate through the pump at intake conditions, m³/s
$\Delta p_{\text{pump}}$ pressure rise of the pump, Pa
$N_{\text{el}}$ power drawn from the grid (or the equivalent of the energy expended), W
$\eta$ system efficiency: the share of the energy supplied that goes into lifting the liquid, —

Units: SI

Holds when

The flow is steady. The flow rate is the volume of liquid at intake conditions. Free gas, if it passes through the pump, is compressed, and for it the formula gives only an upper bound (a value that is too high).
The system efficiency includes the losses in the motor, the transmission (rods, cable), the pump and the tubing.

The useful power depends only on how much liquid has to be lifted and by what pressure rise. It is the same for any method. The energy consumption, however, depends on the system efficiency, which differs several times over between methods (see the section on costs).

In other unit systems (2)

Metric oilfield units

N\,[\text{kW}] = \frac{q\,[\text{m}^3/\text{d}]\;\Delta p_{\text{pump}}\,[\text{MPa}]}{86.4}

$q$ in m³/d at intake conditions, $\Delta p_{\text{pump}}$ in MPa, $N$ in kW.

$86.4 = 86{,}400/1000$ : the number of seconds in a day, and MPa·m³/s = MW = 1000 kW.

Oilfield units

N\,[\text{hp}] = 1.70\cdot 10^{-5}\,q\,[\text{bbl/d}]\;\Delta p\,[\text{psi}]

$q$ in barrels per day, $\Delta p$ in psi, $N$ in mechanical horsepower (1 hp = 745.7 W).

$1.70\cdot 10^{-5} = \dfrac{0.15899}{86{,}400}\cdot\dfrac{6894.8}{745.7}$ — converting bbl/d to m³/s, psi to Pa and W to hp.

Worked exampleA well that cannot flow naturally, and a pump for it

Given (assumed data)

Well: vertical, the middle of the perforated interval at a depth of $H = 1800$ m
Reservoir: $\bar p_R = 12.0$ MPa (gauge); single-phase inflow, $J = 8$ m³/(d·MPa), bubble point 5.0 MPa
Liquid (oil with water, water cut 60%): $\rho = 980$ kg/m³
Pump: set at $H_{\text{pump}} = 1750$ m; target bottomhole pressure 6.0 MPa
Wellhead pressures (gauge): 1.0 MPa in the tubing, 0.3 MPa for the gas in the annulus

Solution

Static fluid level: $H_{\text{st}} = 1800 - \dfrac{(12.0 - 0.3)\cdot 10^6}{980\cdot 9.81} = 583$ m. This is below the wellhead: the reservoir pressure is enough only for a column of liquid 1217 m high, and without gas the well cannot flow naturally.
Rate at a bottomhole pressure of 6.0 MPa: $q = 8\cdot(12.0 - 6.0) = 48$ m³/d.
The column of liquid between the intake and the middle of the interval (50 m) exerts a pressure of 0.48 MPa, so $p_{\text{in}} = 6.0 - 0.48 = 5.52$ MPa. This is above the bubble point, so there is no free gas at the intake.
Dynamic fluid level: $H_{\text{dyn}} = 1750 - \dfrac{(5.52 - 0.3)\cdot 10^6}{980\cdot 9.81} = 1207$ m.
The discharge pressure is the wellhead pressure plus the pressure of the column of liquid from the pump to the wellhead: $p_{\text{out}} = 1.0 + 16.82 = 17.82$ MPa. Friction at this rate in 62 mm tubing is about 0.01 MPa, and we ignore it. The pressure rise of the pump is $\Delta p_{\text{pump}} = 17.82 - 5.52 = 12.30$ MPa. The dynamic-level rule gives the same: $1.0 - 0.3 + 11.60 = 12.30$ MPa.
Hydraulic power: $N = 48\cdot 12.30/86.4 = 6.8$ kW. At a system efficiency of 0.5, the well draws 13.7 kW from the grid. That is about 6.8 kWh per cubic metre of liquid.
For comparison, gas lift: the gas is injected at a depth of 1600 m, and the average density of the mixture above the injection point is assumed to be 400 kg/m³. The pressure at the injection point is $1.0 + 6.28 = 7.28$ MPa, the bottomhole pressure is 9.20 MPa, and the rate is only $8\cdot(12.0 - 9.20) = 22$ m³/d.

Answer: A well with its fluid level at a depth of 583 m will not flow on its own. A pump at 1750 m will deliver 48 m³/d of liquid at a pressure rise of 12.30 MPa, drawing about 13.7 kW. Gas lift under the same conditions gives only about 22 m³/d.

What it means. A column of liquid from the wellhead to the reservoir exerts a pressure of 17.3 MPa, while the reservoir pressure is only 12 MPa. The pump takes over the weight of the column above it, so the bottomhole pressure can be held at 6 MPa. Gas lift lightens only the part of the column above the gas injection point, so the bottomhole pressure stays high. For a well with a low reservoir pressure and not much gas, a pump is the natural choice. Here it is not worth lowering the bottomhole pressure below the bubble point, because free gas would go into the intake. The data in this example are assumed for illustration.

What lowering the bottomhole pressure will give

How much a pump adds depends not only on the pump but also on the reservoir, through the shape of the IPR curve. If the oil is undersaturated (its pressure is above the bubble point) and the bottomhole pressure stays above the bubble point, the curve is a straight line, and the rate grows in proportion to the drawdown. If the bottomhole pressure is below the bubble point, gas comes out of solution near the well, and the mobility of the oil (how easily it flows) falls. Each further drop in pressure then yields less and less (Vogel's equation). On top of that, the gas that has come out of solution arrives at the pump intake.

Worked exampleStraight-line and Vogel curves: what a pump will give

Given (assumed data)

Both wells now: $\bar p_R = 18$ MPa, bottomhole pressure 13 MPa, oil rate 50 m³/d
Well A: bubble point 6 MPa: the oil is undersaturated
Well B: bubble point equal to the reservoir pressure, 18 MPa
The pump will allow the bottomhole pressure to be lowered to: 7 MPa

Solution

Well A: the bottomhole pressure stays above the bubble point, so the curve is a straight line. $J = 50/(18 - 13) = 10$ m³/(d·MPa), and at 7 MPa the rate is $10\cdot(18 - 7) = 110$ m³/d (+120%).
Well B: now $x = p_{wf}/\bar p_R = 13/18 = 0.722$ . The fraction of the absolute open flow (the rate at zero bottomhole pressure) is $1 - 0.2x - 0.8x^2 = 0.438$ . So $q_{max} = 50/0.438 = 114.1$ m³/d.
At 7 MPa: $x = 0.389$ , the fraction is 0.801, and the rate is $114.1\cdot 0.801 = 91.4$ m³/d (+83%).
A straight line extended for well B would promise 110 m³/d, which is 20% more than the well will give.

Answer: Lowering the bottomhole pressure from 13 to 7 MPa will increase the rate of well A by 120% and that of well B by only 83%.

What it means. In well B, below the bubble point, a pump will give less, and there will be more gas at the intake. So calculate the expected gain from the curved IPR, and choose the pump with the gas in mind, or choose a method that tolerates gas. The data in this example are assumed for illustration.

Artificial lift methods

All the methods solve the same problem: delivering energy to the liquid at depth. They differ in how they transmit the energy: by rods from a drive at the surface, by an electric cable to a motor in the well, by high-pressure gas, or by high-pressure liquid. Pumps are of two kinds. Positive-displacement pumps (plunger and progressing cavity pumps) push out a portion of liquid with every stroke or turn. Dynamic pumps speed the liquid up: centrifugal pumps do this with impellers (rotating wheels with blades), and jet pumps are also dynamic.

How energy reaches the liquid

Method: Gas lift
Drive and energy transmission: compressor or gas from gas wells; gas through the annulus
What operates in the well: gas lift valves in side-pocket mandrels (pockets in the tubing string that hold the valves)

Method: Sucker rod pump (SRP)
Drive and energy transmission: pumping unit (a rocking-beam machine) or another drive at the wellhead; string of rods
What operates in the well: plunger pump with discharge and suction valves

Method: Electric submersible pump (ESP)
Drive and energy transmission: power grid and control panel; cable run along the tubing
What operates in the well: multistage centrifugal pump with a submersible motor

Method: Progressing cavity pump (PCP)
Drive and energy transmission: wellhead drive and rotating rods, or a submersible motor
What operates in the well: metal rotor in a stator lined with elastomer (a rubber-like material)

Method: Hydraulic pump
Drive and energy transmission: high-pressure pumps at the surface; power fluid through the tubing
What operates in the well: jet pump or hydraulic piston unit

Method: Plunger lift
Drive and energy transmission: energy of the well's own gas
What operates in the well: free plunger in the tubing

Method	Drive and energy transmission	What operates in the well
Gas lift	compressor or gas from gas wells; gas through the annulus	gas lift valves in side-pocket mandrels (pockets in the tubing string that hold the valves)
Sucker rod pump (SRP)	pumping unit (a rocking-beam machine) or another drive at the wellhead; string of rods	plunger pump with discharge and suction valves
Electric submersible pump (ESP)	power grid and control panel; cable run along the tubing	multistage centrifugal pump with a submersible motor
Progressing cavity pump (PCP)	wellhead drive and rotating rods, or a submersible motor	metal rotor in a stator lined with elastomer (a rubber-like material)
Hydraulic pump	high-pressure pumps at the surface; power fluid through the tubing	jet pump or hydraulic piston unit
Plunger lift	energy of the well's own gas	free plunger in the tubing

Gas lift: Gas is fed into the annulus and enters the tubing through valves. Continuous gas lift lightens the column all the time. Intermittent gas lift pushes the liquid out in slugs. Gas lift tolerates sand and gas, and it works in deviated wells and over a wide range of rates. The valves are changed on a wireline (a steel cable run into the well), so no workover crew (a team that repairs the well) is needed, and the tubing stays open for surveys. Limitations: it needs a gas source and compressors. The bottomhole pressure stays higher than with a pump. It lifts viscous oil and emulsions poorly. Hydrates are a threat at low temperature. The casing must withstand the gas pressure. More in Gas lift: principles and choice of operating conditions.
Sucker rod pump (SRP): A drive at the wellhead, most often a pumping unit, moves the string of rods up and down. The rods move the pump plunger at depth. On the upstroke, the plunger lifts the liquid above it, and a fresh portion enters beneath it through the suction valve. On the downstroke, this portion passes through the discharge valve in the plunger into the space above it. The pump is simple, cheap and easy to repair. It draws the well down to a low bottomhole pressure effectively, and it lifts viscous and hot oil. Engineers diagnose the pump from dynamometer cards, which are records of the rod load over one stroke. Limitations: rod loads limit the combination of depth and rate. Rods and tubing wear in deviated wellbores. Free gas reduces how fully the pump fills, up to “gas lock” (the pump holds only gas and stops lifting liquid). A heavy surface drive is unsuitable offshore and inconvenient in a town. More in Sucker-rod pumping.
Electric submersible pump (ESP): A multistage centrifugal pump (a stack of spinning impeller stages) with a submersible electric motor and a protector (seal section) sits at the lower end of the tubing. A cable clamped to the tubing supplies the power. A variable speed drive lets you change the delivery rate, and sensors at the intake report pressure and temperature. The ESP gives the highest liquid rates, works at great depths and in deviated wells, and takes up almost no space at the wellhead. Limitations: free gas reduces the head (the pressure the pump can add) and can make the pump stop delivering, so gas separators are needed. Sand wears the impellers. Viscous oil sharply reduces head and efficiency. High temperature shortens the life of the motor and cable. The pump performs poorly at low rates. Replacing it needs a workover crew. A reliable power supply is needed. More in Electric submersible pumps (ESP).
Progressing cavity pump (PCP): A metal screw, the rotor, turns inside an elastomer sleeve, the stator. The cavities between them move the liquid upwards. The drive is either rotating rods from the surface (rod-driven PCP) or a submersible motor (an electric submersible PCP, ESPCP). A PCP lifts viscous oil and liquid with sand better than the other methods, handles emulsions gently and has a high efficiency. Limitations: high temperature, aromatic hydrocarbons (solvent-like components of oil) and free gas destroy the elastomer. The head and the depth are lower than for an ESP. Rotating rods wear the tubing in deviated wellbores. More in Progressing cavity pumps and hybrid systems.
Jet pump: A power fluid (oil or water) is delivered at high pressure from the surface to a nozzle. The fast jet drags the produced fluid along, and in the diffuser (a widening passage) the velocity of the mixture is converted into pressure. There are no moving parts in the well. The pump is lifted out by the flow of fluid, without pulling the tubing. It tolerates gas and solids, and it works in deviated wellbores. Limitations: low efficiency; high-pressure pumps and power-fluid treatment at the surface; large volumes of liquid in the gathering system. At a low intake pressure, cavitation occurs (vapour bubbles form and damage the pump), so the well cannot be drawn down far. Hydraulic piston pumps are more economical but are sensitive to solids. More in Hydraulic pumps: piston and jet.
Plunger lift: A free plunger travels in the tubing between a stop at the bottom and a shock absorber at the wellhead. While the well is shut in, gas builds up in the annulus. When the well is opened, the gas pushes the plunger up together with the slug of liquid above it. Plunger lift uses the energy of the well's own gas. It is used in gas wells where liquid builds up, and in oil wells with a high gas–oil ratio and a low rate.

By number of wells, sucker rod pumps predominate in old onshore fields, including in Kazakhstan. Most such wells have low rates, and an SRP is simple, cheap and familiar to maintain. Electric submersible pumps go into wells with high liquid rates, including those with a very high water cut. Progressing cavity pumps are used for viscous oil and in wells that produce sand. In offshore fields, and where there is a lot of gas, gas lift and ESPs are used more often.

Selection criteria

First, pick out the methods that can technically work in the given well. Then choose among them on cost. The criteria fall conveniently into groups.

The well and the reservoir. Liquid rate now and in the future; setting depth; inclination and dogleg severity (how sharply the wellbore bends); the diameter of the production casing, which limits the size of the pump and tubing; the shape of the IPR curve and the bottomhole pressure you want; temperature.
Fluids. The gas–liquid ratio and free gas at the intake; sand and other solids; viscosity and emulsions; wax, salts, hydrates; corrosiveness (carbon dioxide, hydrogen sulfide, and water salinity, that is, how salty the water is). Temperature, corrosion and erosion also decide the choice of materials.
Location. Onshore or offshore (size, weight, footprint); closeness to housing (noise and appearance); remoteness and climate; the source of power and its quality (voltage surges shorten the life of submersible motors); availability of high-pressure gas; distance to the gathering point.
Operation. Reliability and mean time between failures; what is used for repairs (a workover crew or wireline equipment); monitoring and automation; staff skills and the service base; the same type of equipment across the field.
Economics. Capital and operating costs; energy efficiency; the cost of failures and of the production lost during them; the effect on surface facilities.

Approximate application ranges

Method: SRP
Liquid rate, m³/d: 1–150; at shallow depths up to 500
Setting depth, m: up to 2,500–3,000
System efficiency, %: 45–60

Method: ESP
Liquid rate, m³/d: 30–1,500; some up to 5,000 or more
Setting depth, m: up to 3,500–4,500
System efficiency, %: 35–60

Method: PCP
Liquid rate, m³/d: 1–300
Setting depth, m: up to 1,500–2,000
System efficiency, %: 50–70

Method: Continuous gas lift
Liquid rate, m³/d: 30–3,000 or more
Setting depth, m: up to 3,500–4,500
System efficiency, %: 10–30

Method: Jet pump
Liquid rate, m³/d: 50–2,000
Setting depth, m: up to 4,000
System efficiency, %: 15–30

Method: Plunger lift
Liquid rate, m³/d: a few, up to 30
Setting depth, m: up to 3,000–4,000
System efficiency, %: —

Method	Liquid rate, m³/d	Setting depth, m	System efficiency, %
SRP	1–150; at shallow depths up to 500	up to 2,500–3,000	45–60
ESP	30–1,500; some up to 5,000 or more	up to 3,500–4,500	35–60
PCP	1–300	up to 1,500–2,000	50–70
Continuous gas lift	30–3,000 or more	up to 3,500–4,500	10–30
Jet pump	50–2,000	up to 4,000	15–30
Plunger lift	a few, up to 30	up to 3,000–4,000	—

The figures summarise industry practice. The limits are approximate and depend on the manufacturer, the casing size and the fluid properties, and individual installations work well beyond them. System efficiency is the share of the energy supplied from the grid, or spent on compressing gas, that goes into lifting the liquid.

How the methods tolerate complications

Method: SRP
Free gas: fair
Sand: fair
Viscous oil: good
Deviated wellbore: fair

Method: ESP
Free gas: poor
Sand: poor
Viscous oil: poor
Deviated wellbore: good

Method: PCP
Free gas: fair
Sand: good
Viscous oil: good
Deviated wellbore: fair

Method: Gas lift
Free gas: good
Sand: good
Viscous oil: poor
Deviated wellbore: good

Method: Jet pump
Free gas: good
Sand: good
Viscous oil: good
Deviated wellbore: good

Method: Plunger lift
Free gas: needed
Sand: poor
Viscous oil: poor
Deviated wellbore: fair

Method	Free gas	Sand	Viscous oil	Deviated wellbore
SRP	fair	fair	good	fair
ESP	poor	poor	poor	good
PCP	fair	good	good	fair
Gas lift	good	good	poor	good
Jet pump	good	good	good	good
Plunger lift	needed	poor	poor	fair

The ratings are qualitative. Protective measures can improve the “fair” ratings and some of the “poor” ones: a gas anchor or gas separator (devices that keep gas out of the pump), setting the pump deeper below the dynamic fluid level or below the perforations, wear-resistant stages, centralisers (guides that keep the equipment in the middle of the hole), and rod protection.

Gas calls for a calculation. A pump suffers not from gas in general but from free gas, which arrives at the intake together with the liquid. To find the free gas fraction, use the producing gas–oil ratio, the solution gas–oil ratio of the oil at the intake pressure (how much gas stays dissolved there), and the gas formation volume factor (Black oil model: solution gas–oil ratio and formation volume factors).

Free gas fraction at the pump intake

\varphi_{\text{g}} = \frac{q_o\,(R - R_s)\,B_g}{q_o\,(R - R_s)\,B_g + q_o B_o + q_w B_w}

where:

$\varphi_{\text{g}}$ volume fraction of free gas in the flow at the intake, —
$q_o,\ q_w$ oil and water rates at standard conditions, m³/d
$R$ producing gas–oil ratio, m³/m³
$R_s$ solution gas–oil ratio of the oil at intake pressure and temperature, m³/m³
$B_g$ gas formation volume factor at intake conditions: $B_g = \dfrac{p_{\text{sc}}}{p}\cdot\dfrac{T}{T_{\text{sc}}}\,Z$ , m³/m³
$B_o,\ B_w$ oil and water formation volume factors at intake conditions, m³/m³

Units: any consistent units; pressures in $B_g$ are absolute, temperatures are in kelvins

Holds when

All the gas that is not dissolved in the oil at the intake pressure goes to the intake together with the liquid. If part of it is vented into the annulus, put only the remaining part into the formula.
We ignore the solubility of gas in water.

When the free gas fraction exceeds about 10–15%, a centrifugal pump loses head and may stop delivering, and a sucker rod pump fills less. There are three ways to reduce the fraction. Set the pump below the perforated interval (the gas rises in the annulus). Use a gas separator. Or keep a higher intake pressure, at the price of a smaller drawdown.

Some circumstances decide the choice at once. An offshore platform has little space and weight allowance, so a pumping unit is unsuitable. A subsea well tens of kilometres from the platform needs a method that puts no working agent into the flowline. The gas of gas lift or the power fluid of a hydraulic pump would add friction in the long pipeline. So engineers choose an ESP or a multiphase pump on the seabed. In residential areas, noise and appearance matter. In remote areas, simplicity and reliability matter. In a cold climate, protection from hydrates and freezing matters.

How a method is chosen

Selection procedure

1
Forecast
Forecast the liquid rate, water cut, gas–oil ratio and reservoir pressure for the whole life of the well, and the IPR now and in the future (forecasting the IPR).
2
Hard constraints
What rules methods out at once: the site, the power supply, a gas source, the well design and safety requirements.
3
Technical screening
By the application ranges and by calculation: whether the method can deliver the required rate at the required bottomhole pressure at this depth, how much free gas there will be at the intake, and how to deal with sand and wellbore deviation.
4
Sizing and checking
For the methods that remain, size the equipment, and check the operating points now and several years ahead by nodal analysis.
5
Cost comparison
Capital cost, energy, repairs, downtime and lost production, and the effect on surface facilities — over the whole service life.
6
Decision and experience
Trial installations, staff training, monitoring of operation, and a review of the choice when conditions change.

There are two extreme approaches. The first is to install, from the start, equipment for the whole life of the well, sized for the large future withdrawal of watered-out liquid. Then in the first years it runs underloaded and inefficiently. The second is to size the equipment for today's operating mode. Then you will have to replace it, and every replacement means a workover (a repair job on the well) and downtime. In practice, engineers look for flexibility: variable speed control of ESPs, gas lift valves at several depths, and a planned change from one method to another. You must think about this when the well is designed. A production casing chosen to save on drilling can limit the size of the pump for good.

Worked examplePreliminary choice of a lift method for a well

Given (assumed data)

Well: middle of the perforated interval at a true vertical depth of 2300 m; the pump can be set at a true vertical depth of 2200 m, where the wellbore is inclined at 30°
Target operating mode: 150 m³/d of liquid, water cut 60%; intake pressure 5.0 MPa, wellhead pressure 1.5 MPa (gauge)
Fluids: producing gas–oil ratio 90 m³/m³; at intake conditions the solution gas–oil ratio is 35 m³/m³, $B_o = 1.15$ , $B_w \approx 1$ ; temperature 80 °C, $Z = 0.9$
Other: slight sand production; a power grid is available, there is no high-pressure gas
For the calculation: liquid in the tubing 1000 kg/m³, friction in the tubing 0.3 MPa; system efficiency: ESP 0.45, jet pump 0.25; electricity — assumed at US$0.06 per kWh

Solution

Rates: oil 60, water 90 m³/d. Free gas at standard conditions: $60\cdot(90 - 35) = 3300$ m³/d.
The gas formation volume factor at the intake uses the absolute pressure $5.0 + 0.1 = 5.1$ MPa and a temperature of 353 K: $B_g = \dfrac{0.1013}{5.1}\cdot\dfrac{353}{293}\cdot 0.9 = 0.02154$ . So the free gas at the intake is 71.1 m³/d.
Liquid at the intake: $60\cdot 1.15 + 90 = 159$ m³/d. Free gas fraction: $\varphi_{\text{g}} = \dfrac{71.1}{71.1 + 159} = 31$ %. With a gas separator that diverts 80% of the gas into the annulus, it is 8%.
Pressure rise of the pump: $p_{\text{out}} = 1.5 + 21.58 + 0.3 = 23.38$ MPa, and $\Delta p_{\text{pump}} = 23.38 - 5.0 = 18.38$ MPa.
Hydraulic power with 159 m³/d through the pump: $N = 159\cdot 18.38/86.4 = 33.8$ kW. Power drawn from the grid: ESP 75 kW (659 MWh a year), jet pump 135 kW (1185 MWh a year). The difference is about US$32 thousand a year.
Ruling out: SRP: 150 m³/d from a depth of 2200 m, in a deviated wellbore at that, is outside its usual range. PCP: a pressure rise of about 18 MPa, together with gas making up a third of the intake volume, is too hard on the elastomer. Gas lift: there is no gas source.

Answer: A reasonable choice is an ESP with a gas separator, wear-resistant stages and variable speed control. The fallback is a jet pump, which tolerates gas and sand more easily but uses 1.8 times as much energy.

What it means. Without separation, almost a third of the volume at the intake is gas, and a centrifugal pump would lose head and stop delivering. The pressure rise has a margin built in: the gas that stays in the oil comes out of solution as the fluid rises and lightens the column in the tubing. The choice does not end here. For the chosen method, select the pump, motor and cable, and check the operating points by nodal analysis, now and at the future water cut. The data in this example are assumed for illustration.

Life-cycle costs

The price of the equipment catches the eye. But over the years of a well's life, more money usually goes into running it: electricity or gas, repairs and replacement of equipment, monitoring and staff. Production is also lost while the well is down. That is why it often pays to spend more on reliable or energy-efficient equipment (Project cash flow).

You can estimate energy consumption in advance. Calculate the hydraulic power from the flow rate and the pressure rise. The system efficiency differs several times over between methods. Positive-displacement pumps and ESPs put roughly a third to two thirds of the energy supplied into the liquid. Jet pumps and gas lift usually put in no more than a third. Where energy is expensive, this changes the choice. You cannot calculate reliability in this way. Take it from failure statistics for similar wells. The mean time between failures depends not only on the equipment but also on how it was chosen, installed and run. Staff training and monitoring of the operating mode raise it as much as new equipment does.

The lift method also affects the surface facilities. With pumps, only the produced fluids of the wells pass through the processing plant. Gas lift needs compressors, high-pressure gas pipelines and a gas distribution system. The injected gas comes back with the produced fluids and loads the separators and compressors. Hydraulic pumps circulate power fluid through the system in volumes comparable with production. This fluid has to be treated, and if it is oil, it must also be separated out when production is measured.

Worked exampleThe cost of failures: a standard pump or a more reliable one

Given (assumed data)

Well: with an ESP, oil production 60 m³/d
Option 1: unit US$120 thousand, mean time between failures 14 months, repair of a failed unit US$60 thousand
Option 2: wear-resistant unit US$160 thousand, mean time between failures 28 months, repair US$80 thousand
Each failure: workover of the well US$30 thousand and 7 days of downtime
Net revenue: the assumed price of $50/bbl in the example project that runs through these lessons, less a payment of 10% of revenue and variable costs of $5/bbl: $252 per cubic metre of oil

Solution

Production lost per failure: $7\cdot 60 = 420$ m³ of oil, or US$105.7 thousand of net revenue.
Cost of one failure (US$ thousand): option 1 is $30 + 60 + 105.7 = 195.7$ , and option 2 is $30 + 80 + 105.7 = 215.7$ .
Failures per year: $12/14 = 0.86$ and $12/28 = 0.43$ . The cost of failures per year is US$168 thousand and US$92 thousand.
Option 2 saves US$75 thousand a year. The extra payment of US$40 thousand is paid back in about 6 months.

Answer: The more expensive, more reliable unit saves about US$75 thousand a year and pays back its extra cost in about 6 months.

What it means. About half of the cost of a failure is lost production. That is why reliability is especially valuable in wells with high rates, and where repair is slow and expensive, for example offshore or in remote areas. In a well that produces only a few cubic metres a day, the conclusion may be the opposite. The data in this example are assumed for illustration.

Artificial lift is increasingly built into the field development plan from the very beginning. Mature fields water out. Small accumulations are developed without pressure maintenance. Remote and subsea wells need boosting. And a high early rate improves the economics of the project. Sensors at pump intakes and telemetry (remote data transmission) let engineers watch every well in real time, notice failures sooner and avoid wasting energy. Models that link the reservoir, wells, gathering system and processing plant let them compare lift methods by life-cycle cost (Field development concepts: surface facilities).

Terms and symbols

Artificial lift: Operating a well with a pump or gas lift, which add energy to the energy of the reservoir.
Static fluid level: The liquid level in a shut-in well. The column from it down to the reservoir balances the reservoir pressure.
Dynamic fluid level: The liquid level in the annulus of a producing well with a pump. It is used to estimate the intake pressure and the bottomhole pressure.
Pump intake pressure: The pressure of the liquid at the pump intake. The free gas at the intake and the pressure rise the pump must produce depend on it.
SRP, ESP, ESPCP, rod-driven PCP: Sucker rod pump, electric submersible pump unit, and progressing cavity pump units with a submersible electric motor and with a rod drive.
System efficiency: The share of the energy supplied from the grid, or spent on compressing gas, that goes into lifting the liquid.
Mean time between failures (MTBF): The average running time of a unit between failures. It is the main measure of the reliability of artificial lift.

Common mistakes and misconceptions

Wrong

“Let's install a more powerful pump — the rate will rise as much as we like.”

In fact

The reservoir limits the rate. Even at zero bottomhole pressure the rate will not exceed the absolute open flow (the rate at zero bottomhole pressure). Below the bubble point, each drop in pressure gives less and less oil and sends more and more gas to the intake. A pump that has pumped the well “dry” runs erratically.

Wrong

“If there is a lot of gas in the produced fluid, the choice is obvious — gas lift.”

In fact

Gas lift copes well with gas, but it leaves the bottomhole pressure high, uses a lot of energy, and needs a gas source and compressors. For a well with a low reservoir pressure, a pump with a gas separator, or a pump set below the perforations, may give more.

Wrong

“We choose whatever is cheaper to buy and install.”

In fact

Over the service life, energy, repairs and the production lost in failures usually cost more than the equipment. Compare the methods on life-cycle costs.

Wrong

“Gas in the produced fluid does not bother a pump as long as there is enough liquid.”

In fact

Free gas at the intake reduces the head of a centrifugal pump and the fill of a sucker rod pump, up to a complete loss of delivery. Calculate the gas fraction and reduce it by separation, by setting the pump below the perforations, or by keeping a higher intake pressure.

Wrong

“A pump is selected once for the whole life.”

In fact

Rate, water cut and reservoir pressure change, so equipment chosen for today may be unsuitable in a year or two. Build in flexibility, and review the choice as development goes on.

In depth: advanced detail

Bottomhole pressure from the dynamic fluid level

Not every well with a pump has a downhole gauge. So engineers often estimate the bottomhole pressure from echometer data, that is, from the dynamic fluid level and the gas pressure in the annulus:

Bottomhole pressure from the dynamic fluid level

p_{wf} \approx p_{\text{ann}} + \rho_{\text{liq}}\,g\,(H - H_{\text{dyn}})

where:

$p_{wf}$ bottomhole pressure at the middle of the perforated interval, Pa
$p_{\text{ann}}$ gas pressure in the annulus at the wellhead, Pa
$\rho_{\text{liq}}$ average density of the liquid in the annulus below the level, kg/m³
$H,\ H_{\text{dyn}}$ true vertical depths of the middle of the perforated interval and of the dynamic fluid level, m

Units: SI

Holds when

The weight of the gas column above the level is small. At a high annulus pressure it is already a few per cent, and then it is added.
The liquid in the annulus contains no gas and has a known density.

The formula is simple, but its assumptions are often broken. Above the pump intake, the liquid in the annulus is full of gas bubbles rising upwards. It is lighter than degassed liquid, so the simple calculation gives a bottomhole pressure that is too high. For a “foamy” column, engineers apply corrections for the gas content, which they estimate from the pressure build-up in the closed annulus. Also, the pump draws in water, and above the intake the annulus often holds almost pure oil. The echometer gives a reliable picture when these corrections are applied and the results are compared with downhole gauge measurements, at least in some of the wells.

Connections to other topics

Nodal analysis of the production system — how a pump and gas lift shift the outflow curve and where the operating point ends up.
Well inflow: the well performance diagram — the well performance diagram and why the rate of a naturally flowing well falls faster than the reservoir pressure.
Inflow performance relationship (IPR) — how much the reservoir will give as the bottomhole pressure is lowered, including below the bubble point.
Multiphase flow in tubing — gas–liquid flow in the tubing and the minimum of the outflow curve, on which gas lift is based.
Gas lift: principles and choice of operating conditions — gas lift: the principle, and the continuous and intermittent modes.
Sucker-rod pumping — sucker rod pumps: design, dynamometer cards, calculation of delivery and loads.
Electric submersible pumps (ESP) — electric submersible pumps: performance curve, choosing the stages, motor and cable.
Progressing cavity pumps and hybrid systems — progressing cavity pumps for viscous oil and sand.
Hydraulic pumps: piston and jet — jet and hydraulic piston pumps.

Check yourself

1
A well 2000 m deep (true vertical) is shut in; the reservoir pressure is 14 MPa, the liquid density is 1000 kg/m³, and the annulus is open. Where is the static fluid level, and can the well flow naturally without gas?
1. A.Above the wellhead — it can
2. B.It cannot be said without knowing the rate
3. C.About 573 m below the wellhead — it cannot
4. D.About 1427 m below the wellhead — it cannot
Show the answer
Correct answer: C. About 573 m below the wellhead — it cannot
The reservoir pressure supports a column of liquid $14\cdot 10^6/(1000\cdot 9.81) \approx 1427$ m high, counting from the bottom of the hole. So the level is at a depth of $2000 - 1427 \approx 573$ m. Since the level is below the wellhead, the liquid will not reach the surface without gas to lighten the column.
2
The dynamic fluid level is at 1200 m, the liquid density is 900 kg/m³, the wellhead pressure is 1.2 MPa, the annulus pressure is 0.3 MPa, and friction can be neglected. What pressure rise must the pump produce, and what is the hydraulic power at a delivery rate of 60 m³/d?
Show the answer
The column from the level to the wellhead exerts $900\cdot 9.81\cdot 1200 \approx 10.6$ MPa. The pressure rise is $\approx 1.2 - 0.3 + 10.6 = 11.5$ MPa. The power is $60\cdot 11.5/86.4 \approx 8.0$ kW. At a system efficiency of 0.5, about 16 kW will be drawn from the grid.
3
Viscous oil (about 1000 mPa·s) with sand production, setting depth 800 m, rate 30 m³/d. Which method suits best?
1. A.ESP
2. B.Gas lift
3. C.Plunger lift
4. D.Progressing cavity pump
Show the answer
Correct answer: D. Progressing cavity pump
A progressing cavity pump lifts viscous oil and liquid with sand better than the others, and the depth and rate are within its usual range. A centrifugal pump on viscous oil loses head and efficiency and is worn by sand, gas lift lifts viscous oil poorly, and plunger lift needs gas.
4
A subsea well 25 km from the platform; there is no high-pressure gas, but electric power is available. Which method is preferable?
1. A.Plunger lift
2. B.Sucker rod pump
3. C.ESP or a multiphase pump on the seabed
4. D.Gas lift
Show the answer
Correct answer: C. ESP or a multiphase pump on the seabed
An ESP produces a large head to overcome friction in the long pipeline, runs on electricity and is controlled remotely, and adds neither gas nor power fluid to the flowline. Gas lift would add gas and friction to the line and needs a gas source; a pumping unit is impossible on a subsea well.
5
The reservoir pressure equals the bubble point, 16 MPa. The well produces 40 m³/d at a bottomhole pressure of 12 MPa. How much will it produce at 4 MPa according to a straight-line IPR and according to Vogel's equation?
Show the answer
On the straight line: $J = 40/4 = 10$ m³/(d·MPa), so the rate is $10\cdot 12 = 120$ m³/d. By Vogel: at $x = 12/16 = 0.75$ the fraction is $1 - 0.15 - 0.45 = 0.4$ , so $q_{max} = 100$ m³/d. At $x = 0.25$ the fraction is $1 - 0.05 - 0.05 = 0.9$ , so the rate is 90 m³/d. The straight line overstates the gain by a factor of 1.6: 80 m³/d instead of 50.
6
Why are lift methods compared by life-cycle cost and not by the price of the equipment?
1. A.Because energy, repairs and production lost in failures usually cost more than the equipment over the years
2. B.Because equipment gets cheaper with time
3. C.Because capital costs do not affect the economics
4. D.Because tax rules require it
Show the answer
Correct answer: A. Because energy, repairs and production lost in failures usually cost more than the equipment over the years
Capital costs are paid once, whereas energy, repairs and downtime are paid for all the years the well operates. A method that is more expensive to install but economical and reliable often turns out cheaper over its service life.

Updated: 19 October 2026