Electric submersible pumps (ESP)

Quick refresher

The installation consists of the surface equipment (a control station, usually with a variable speed drive, and a step-up transformer), a cable along the tubing, and the submersible unit. The unit has a pump of hundreds of stages, an intake module or a gas separator, a protector, a submersible motor and sensors.
A centrifugal pump is a dynamic pump: its rate depends on the head against which it works. The performance curve shows the head, in metres of liquid column, that a stage gives at each rate. The pump should work in the operating range near the maximum efficiency.
For a low-viscosity liquid the head in metres does not depend on the density, but the pressure and the power are proportional to it: $\Delta p = \rho g h$ .
The required head is the height to which the liquid must be lifted from the dynamic fluid level (the depth of the liquid surface in the annulus, the space between the tubing and the casing), plus the difference between the wellhead pressure and the annulus pressure, plus the friction. The last two are expressed in metres of liquid column. The number of stages is the required head divided by the head of one stage at the design rate.
By the affinity laws the rate is proportional to the speed, the head to its square and the power to its cube. A variable speed drive changes the operating mode without lifting the pump out of the well. But the operating point moves along the well curve, and the motor power and the gas at the intake limit the frequency.
Free gas is the main enemy of an ESP. At only 10–15% by volume the head already falls noticeably, and at a larger share the pump may lose its delivery (gas lock). The ways to fight gas are the intake pressure, separation and gas-handling stages.

Key relationships

H_{\text{req}} = H_{\text{dyn}} + \dfrac{p_{wh} - p_{\text{ann}} + \Delta p_{\text{fr}}}{\rho g}

the required head of the pump, m

z_{\text{stg}} = \dfrac{H_{\text{req}}}{h(q_{\text{in}})}

the number of stages (rounded up)

N_{\text{shaft}} = \dfrac{\rho g\,q_{\text{in}}\,H_{\text{pump}}}{\eta}

the power at the pump shaft

\dfrac{q_2}{q_1} = \dfrac{n_2}{n_1}, \quad \dfrac{h_2}{h_1} = \left(\dfrac{n_2}{n_1}\right)^{2}, \quad \dfrac{N_2}{N_1} = \left(\dfrac{n_2}{n_1}\right)^{3}

the affinity laws for the corresponding points of the performance curves

Common pitfalls

Choosing a pump from the rate at the surface and forgetting that the pump handles the volume at intake conditions: the surface volume multiplied by the oil formation volume factor (how much the volume of oil changes between the surface and the intake), plus the free gas that has not been separated.
Setting the design rate at the edge of the operating range. When the inflow falls, the pump leaves the range.
Expecting the rate to change in proportion to the frequency. The operating point also depends on the well curve.
Raising the frequency without checking the motor power and the intake pressure.
Starting the motor while the liquid draining back through the pump is turning it backwards.

After this lesson you can

name the parts of an ESP, from the control station to the sensors under the motor, and explain what each one does
explain how a centrifugal stage works, and what its performance curve shows: head (the height to which the pump can lift the liquid), efficiency, power and operating range
size a pump for a well: calculate the required head, the number of stages and the power, choose the motor and check its cooling
use the affinity laws (the laws of similarity) to estimate how the supply frequency changes the rate, head and power, and find the operating point of the pump in the well at another frequency
calculate the share of free gas (gas that is not dissolved in the liquid) at the intake, and choose measures against gas
estimate the motor current and the losses in the cable, and recognise from the current and the pressures that the well is pumped down, that gas is present, and other faults

What the installation consists of

The lesson on choosing an artificial lift method describes the ESP in a few words: a multistage centrifugal pump with a submersible motor that is powered through a cable. Here we look at the installation part by part, along the path that the electric power takes from the grid to the liquid.

SchematicA well with an electric submersible pump installation (a diagram, not to scale; the well is shortened). The pump, the intake module or gas separator, the protector and the motor are assembled into one submersible unit at the end of the tubing. The unit is set above the perforated interval, so the liquid flows to the intake along the motor and cools it. The gas that separates at the intake rises in the annulus (the space between the tubing and the casing) above the dynamic fluid level.

Surface equipment. The control station switches the installation on and off and protects it. It guards against motor overload and underload, against a fall in the resistance of the insulation, and it reacts to signals from the sensors. Most often the station contains a variable speed drive (VSD). This changes the frequency of the supply current, and with it the speed of the motor and the pump. The station usually works at the grid voltage, and between it and the well stands a step-up transformer. It gives such a voltage that, after the losses in the cable, the motor receives its rated voltage. This voltage ranges from hundreds of volts to several kilovolts. Between the transformer and the wellhead there is a junction box, also called a vent box. Gas from the well can creep along the wires of the cable. The box stops it before it reaches the electrical equipment, where a spark at a switch could set it on fire.

The cable runs from the wellhead along the tubing and is fastened to the pipes with metal bands. Protectors are fitted where it is easy to damage it while the tubing is run in. It passes through the wellhead equipment in a sealed lead-through. If the well has a packer (a seal that closes the annulus), the cable passes through the packer too. Along the submersible unit, where the gap between the unit and the casing is small, the round cable is replaced by a flat motor lead extension. It ends with the cable connector of the motor, the pothead.

The pump is built of sections. Each section holds dozens of stages, and the whole pump may have from several dozen to several hundred. A stage is an impeller on the common shaft and a fixed diffuser, which is fastened in the housing. The impeller speeds up the liquid. The diffuser slows it down, turns the speed into pressure, and leads the liquid to the next impeller. The stages are connected in series: the outlet of one is the inlet of the next. So the head of the pump (the pressure it gives, expressed as a height of liquid column) is proportional to the number of stages, and the sizes of the impellers set the rate. Pumps are made in size groups (called series in the industry). A size group is set by the largest outside diameter that passes into a production casing of a given size. The bigger the size group, the larger the rate and the head of one stage, and the higher the efficiency.

Under the pump there is an intake module with a screen, through which the liquid enters the pump. Instead of it there may be a gas separator, which separates part of the free gas and releases it into the annulus (see the section on gas).

The protector (also called the seal section) joins the shaft of the motor to the shaft of the pump. It does four jobs. It separates the clean motor oil from the liquid of the well. It takes up the thermal expansion of the oil when the motor heats up. It equalises the pressure inside the motor with the pressure in the well, so that the seals do not work under a large pressure difference. And its thrust bearing takes the axial force (the force along the shaft) that the impellers of the pump create.

The submersible motor is long and fairly thin, and it is filled with oil that insulates the winding and lubricates the bearings. It is usually a three-phase induction motor (the usual kind of AC motor). At 50 Hz its shaft turns at about 2900 revolutions per minute. Its power ranges from ten kilowatts to several hundred. If one motor is not enough, two or three are connected in series. Permanent-magnet motors (brushless motors, which have magnets on the rotor) are used more and more. They have a higher efficiency, and they work over a wider range of speeds. The motor can give its heat only to the liquid that flows past it on the way to the intake. This is the reason for the requirement on the velocity of the flow along the motor.

Under the motor a downhole sensor unit is fixed. This is a combined pressure and temperature gauge system (downhole telemetry). It measures the pressure and the temperature at the intake, the temperature of the motor and the vibration, and sometimes the pressure at the discharge. It sends the data through the power cable. Above the pump, a check valve is fitted in the tubing. It stops the liquid from running back from the tubing through the pump when the pump stops. Above the check valve there is a drain valve. It is opened before the pump is pulled out of the well, so that the tubing does not have to be lifted full of liquid.

Parts of an ESP, from the grid to the liquid

Part: Control station, variable speed drive
What it does: start, protection, control of the speed
Industry term: switchboard, variable speed drive

Part: Step-up transformer
What it does: gives the voltage that the motor needs, allowing for the losses in the cable
Industry term: step-up transformer

Part: Junction box
What it does: keeps gas that has come along the cable away from the electrical equipment
Industry term: vent box

Part: Cable, extension, cable connector
What it does: bring the current to the motor
Industry term: power cable, motor lead extension, pothead

Part: Check and drain valves
What it does: keep the liquid in the tubing, and drain it before the pump is pulled
Industry term: check valve, drain valve

Part: Pump made of sections of stages
What it does: raises the pressure of the liquid
Industry term: multistage centrifugal pump

Part: Intake module or gas separator
What it does: lets the liquid in, and separates gas
Industry term: intake, gas separator

Part: Protector
What it does: protects the motor, takes the axial force
Industry term: protector, seal section

Part: Submersible electric motor
What it does: turns the shaft
Industry term: submersible motor

Part: Sensor unit
What it does: pressure, temperature, vibration
Industry term: downhole sensor

Part	What it does	Industry term
Control station, variable speed drive	start, protection, control of the speed	switchboard, variable speed drive
Step-up transformer	gives the voltage that the motor needs, allowing for the losses in the cable	step-up transformer
Junction box	keeps gas that has come along the cable away from the electrical equipment	vent box
Cable, extension, cable connector	bring the current to the motor	power cable, motor lead extension, pothead
Check and drain valves	keep the liquid in the tubing, and drain it before the pump is pulled	check valve, drain valve
Pump made of sections of stages	raises the pressure of the liquid	multistage centrifugal pump
Intake module or gas separator	lets the liquid in, and separates gas	intake, gas separator
Protector	protects the motor, takes the axial force	protector, seal section
Submersible electric motor	turns the shaft	submersible motor
Sensor unit	pressure, temperature, vibration	downhole sensor

In Russian-language practice, the code of an installation usually gives the size group, the rated rate in cubic metres per day and the head in metres. For example, the code “5-125-1500” means a pump of size group 5, for 125 m³/d, with a head of 1500 m. The rated rate is the rate at the maximum efficiency (the best operating mode of the pump), on water at 50 Hz.

The stage and the performance curve

A centrifugal pump does not displace a fixed volume per turn, as a plunger pump or a progressing cavity pump does. It gives energy to the liquid through speed. So its rate depends on the head against which it works: the more the pressure has to be raised, the smaller the rate, and the other way round. If we close the valve at the discharge, the pump keeps turning and gives its largest head at zero rate. It cannot run like this for long, because without flow the pump and the motor overheat. But the pressure does not grow without limit, as it would with a positive-displacement pump (one that displaces a fixed volume on every turn).

Head is the energy that the pump gives to each newton of weight of the liquid, expressed as the height of a column of this liquid in metres. Head is convenient because, for low-viscosity liquids, it does not depend on the density. A stage lifts water and oil to the same height. But the pressure that it creates is proportional to the density.

The pressure rise, the power and the efficiency of a stage

\Delta p_{\text{stg}} = \rho g h, \qquad N_{\text{stg}} = \rho g\,q_{\text{in}}\,h, \qquad \eta = \frac{N_{\text{stg}}}{N_{\text{shaft,stg}}}

where:

$\Delta p_{\text{stg}}$ pressure rise in one stage, Pa
$\rho$ density of the pumped liquid at the conditions in the pump, kg/m³
$g$ acceleration of free fall, 9.81 m/s², m/s²
$h$ head of the stage at the rate $q_{\text{in}}$ , from the performance curve, m
$q_{\text{in}}$ rate: the volume of liquid that passes through the pump, at the intake conditions, m³/s
$N_{\text{stg}}$ hydraulic (useful) power of the stage, W
$N_{\text{shaft,stg}}$ shaft power spent on the stage, W
$\eta$ efficiency of the stage, which is also the efficiency of a pump made of identical stages, —

Units: SI

Holds when

a low-viscosity liquid (its viscosity is close to that of water) with no free gas. The manufacturer measures the curves on water, and viscosity and gas reduce the head and the efficiency
the head and the efficiency are taken from the performance curve at the rate and the speed at which the stage works

For a pump made of identical stages, the head and the power are as many times larger as the number of stages, and the efficiency is the same. The difference between the shaft power and the useful power goes into heat: friction of the liquid in the channels, leaks between the impeller and the diffuser, and friction in the bearings.

In other unit systems (2)

Metric oilfield units

\Delta p_{\text{stg}}\,[\text{MPa}] = \frac{\gamma\,h\,[\text{m}]}{102}

$h$ in metres, $\gamma$ is the specific gravity of the liquid (water = 1), $\Delta p$ in MPa.

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

Oilfield units

\Delta p_{\text{stg}}\,[\text{psi}] = 0.433\,\gamma\,h\,[\text{ft}]

$h$ in feet, $\gamma$ is the specific gravity of the liquid, $\Delta p$ in psi.

0.433 psi per foot is the pressure gradient of fresh water.

The manufacturer measures the performance curve on a test stand, on water, at the rated frequency: 50 Hz (about 2900 rpm) or 60 Hz (about 3500 rpm). The catalogue gives it for one stage or for a pump with a set number of stages. There are three curves, all against the rate:

The head is largest at zero rate. It falls as the rate grows, down to zero at the largest rate.
The efficiency is zero at zero rate (there is no useful work) and at the largest rate (there is no head). Between them it goes through a maximum. This is the best efficiency point (BEP), and the rate at this point is called the rated rate.
The shaft power of ordinary stages grows with the rate. So the motor is overloaded if the pump works at a rate above the design rate. This happens, for example, when the tubing leaks and the liquid returns to the well, and the pump has to create a smaller head.

CalculatedAn assumed stage curve on water at 50 Hz (the curves are invented for the lesson, but they look like catalogue curves): the head is

h = 6.4\,[1 - (q/280)^2]

m, and the efficiency is

\eta = 0.60\,x(2 - x)

, where

x = q/140

. The best efficiency point (marked BEP in the figure) is 140 m³/d at a head of 4.8 m and an efficiency of 0.60. The operating range, where the efficiency is at least 90% of the maximum, is from about 96 to 184 m³/d.

The pump must work in its operating range (the recommended operating range). This is an interval of rates around the best efficiency point, and the manufacturer gives it. It is usually from about 0.7 to 1.3 of the rated rate. Inside it the efficiency is close to the maximum. To the left, at small rates, the flow presses the impellers down, so the lower thrust washers wear, and the liquid circulates inside the stage and heats up. To the right, at large rates, the impellers lift up, and the efficiency falls fast. In the assumed curve of this lesson, the operating range is the set of rates where the efficiency is at least 90% of the maximum: from 0.68 to 1.32 of the rated rate.

Sizing a pump for a well

To size an ESP means to find a pump that, at the design rate, gives exactly the head that the well requires, and that works in its operating range. Then we choose the motor, the protector and the cable for it. The input data are the inflow performance relationship (IPR, the curve that shows how much the reservoir gives at each bottomhole pressure) and the wanted rate, the setting depth and the design of the well, the properties of the fluids, and the pressures at the wellhead and in the annulus.

The required head is made of the same parts as the pressure rise of the pump: lifting the liquid from the dynamic fluid level to the wellhead, the difference between the wellhead pressure and the annulus pressure, and the friction in the tubing. Expressed in metres of the pumped liquid, it is the head that the pump must give.

The required head of the pump

H_{\text{req}} = \frac{p_{\text{out}} - p_{\text{in}}}{\rho g} = H_{\text{dyn}} + \frac{p_{wh} - p_{\text{ann}} + \Delta p_{\text{fr}}}{\rho g}

where:

$H_{\text{req}}$ the head that the pump must create at the design rate, m
$p_{\text{out}},\ p_{\text{in}}$ pressures at the pump discharge and at the pump intake, Pa
$H_{\text{dyn}}$ vertical depth of the dynamic fluid level in the annulus, m
$p_{wh}$ pressure at the wellhead in the tubing, Pa
$p_{\text{ann}}$ gas pressure in the annulus at the wellhead, Pa
$\Delta p_{\text{fr}}$ pressure loss from friction in the tubing, from the pump to the wellhead, Pa
$\rho$ density of the liquid in the tubing and in the annulus below the level, kg/m³
$g$ acceleration of free fall, 9.81 m/s², m/s²

Units: SI; the pressures are all gauge or all absolute

Holds when

the liquid has no free gas and the same density in the tubing and in the annulus below the level. Gas in the tubing makes the column lighter. Then the discharge pressure is found from multiphase flow, and the required head is smaller
the depths are vertical, and the friction is found from the length of the tubing: $\Delta p_{\text{fr}} = \lambda\,\dfrac{L}{d}\,\dfrac{\rho v^2}{2}$ , as in the lesson on nodal analysis
the weight of the gas column in the annulus above the level is small

The required head grows when the dynamic fluid level goes down, that is, when more liquid is produced, and it grows with the wellhead pressure. The performance curve tells us how many metres of head one stage gives at the design rate, and this gives the number of stages.

The number of stages and the shaft power of the pump

z_{\text{stg}} = \frac{H_{\text{req}}}{h(q_{\text{in}})}, \qquad N_{\text{shaft}} = \frac{\rho g\,q_{\text{in}}\,H_{\text{pump}}}{\eta(q_{\text{in}})}

where:

$z_{\text{stg}}$ number of stages; the result is rounded up, —
$h(q_{\text{in}})$ head of one stage at the rate $q_{\text{in}}$ and at the frequency at which the pump will work, m
$q_{\text{in}}$ rate at the intake conditions: the oil and water rates multiplied by their formation volume factors, plus the free gas that has not been separated, m³/s
$H_{\text{pump}}$ head of the pump, $z_{\text{stg}}\,h$ ; for an exact choice it is equal to the required head, m
$N_{\text{shaft}}$ power at the pump shaft, W
$\eta(q_{\text{in}})$ efficiency of the pump at this rate, —
$\rho,\ g$ density of the pumped liquid at the conditions in the pump, and the acceleration of free fall, kg/m³, m/s²

Units: SI

Holds when

all stages are identical, the liquid has a low viscosity, and the share of gas is small. We allow for viscosity and gas with corrections or with extra stages
to the power of the pump we add what the gas separator and the protector use, usually a few per cent

Stages add head, not rate. The shaft power is proportional to the rate and the head. We choose the motor with a margin of power, and we check it for cooling, for the temperature in the well and for starting conditions.

In other unit systems (1)

Metric oilfield units

N_{\text{shaft}}\,[\text{kW}] = \frac{q_{\text{in}}\,[\text{m}^3/\text{d}]\;\Delta p_{\text{pump}}\,[\text{MPa}]}{86.4\;\eta}

$q_{\text{in}}$ in m³/d at intake conditions, $\Delta p_{\text{pump}} = \rho g H_{\text{pump}}$ in MPa, $N_{\text{shaft}}$ in kW.

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

The sizing procedure

1
The operating mode
from the IPR we find the bottomhole pressure at the wanted rate, and from it and the setting depth the intake pressure. We compare it with the bubble point pressure (the pressure below which gas comes out of the oil). Below this pressure, free gas comes to the intake.
2
The volume at the intake
the oil and water rates are converted to intake conditions with the formation volume factors (Black oil model: solution gas–oil ratio and formation volume factors), and the free gas that the separator will not remove is added.
3
The pump
a size group that passes into the casing, and a type of stage whose design rate falls in the operating range, better in its right half. When the inflow falls, the operating point will move to the left.
4
Head and stages
the required head, the number of stages from the performance curve, and the shaft power.
5
Motor, protector, cable
a motor with a margin of power; cooling of the motor; a cable chosen for the current, the temperature and the allowed losses (see the section on the motor and the cable).
6
A check for the future
with nodal analysis: the operating modes at another reservoir pressure, productivity and water cut, and at different frequencies. We also decide whether a variable speed drive is needed, and for what range of frequencies. Detailed recommendations are given, for example, by the recommended practice of the American Petroleum Institute for sizing ESP installations, API RP 11S4.

Worked exampleSizing an ESP for a well with a high water cut

Given (assumed data)

Well: vertical; the middle of the perforated interval is at 2100 m, and the pump intake is at 1950 m; casing 146 × 7.7 mm (inside diameter 130.6 mm), tubing 73 mm (inside diameter 62 mm)
Inflow: $\bar p_R = 18.0$ MPa, $J = 14$ m³/(d·MPa), bubble point pressure 5.0 MPa
Wanted operating mode: 150 m³/d of liquid at a water cut of 80%; at intake conditions the formation volume factors are $B_o = 1.08$ (oil) and $B_w \approx 1$ (water)
Liquid in the tubing and the casing: $\rho = 1030$ kg/m³; friction factor in the tubing $\lambda = 0.022$
Wellhead pressures (gauge): in the tubing 1.5 MPa, gas in the annulus 0.6 MPa
The assumed stage (figure above): on water at 50 Hz $h = 6.4\,[1 - (q/280)^2]$ m, $\eta = 0.60\,x\,(2 - x)$ , $x = q/140$ , with the rate $q$ in m³/d
Motor: outside diameter 117 mm

Solution

Bottomhole pressure: $p_{wf} = 18.0 - 150/14 = 7.29$ MPa. This is above the bubble point pressure, so the straight-line inflow performance relationship applies. The intake pressure is smaller by the weight of a liquid column 150 m high: $p_{\text{in}} = 7.29 - 1.52 = 5.77$ MPa. This is also above 5.0 MPa, so there is no free gas at the intake.
Rate at intake conditions: $q_{\text{in}} = 30\cdot 1.08 + 120 = 152.4$ m³/d. Dynamic fluid level: $H_{\text{dyn}} = 1950 - \dfrac{(5.77 - 0.6)\cdot 10^6}{1030\cdot 9.81} = 1438$ m.
Friction in the tubing: the average velocity is $v = 0.584$ m/s, and $\Delta p_{\text{fr}} = 0.022\cdot\dfrac{1950}{0.062}\cdot\dfrac{1030\cdot 0.584^2}{2}\cdot 10^{-6} = 0.122$ MPa.
Discharge pressure: $p_{\text{out}} = 1.5 + 19.703 + 0.122 = 21.325$ MPa, where 19.703 MPa is the liquid column from the pump to the wellhead. The pressure rise is $\Delta p_{\text{pump}} = 21.325 - 5.77 = 15.555$ MPa.
Required head: $H_{\text{req}} = \dfrac{15.555\cdot 10^6}{1030\cdot 9.81} = 1539$ m. A check through the level: $1438 + \dfrac{(1.5 - 0.6 + 0.122)\cdot 10^6}{1030\cdot 9.81} = 1539$ m.
From the performance curve at $q_{\text{in}} = 152.4$ m³/d: $h = 6.4\,[1 - (152.4/280)^2] = 4.504$ m, $x = 1.089$ , and $\eta = 0.595$ . The rate is inside the operating range (96–184 m³/d), to the right of the best efficiency point.
Number of stages: $z_{\text{stg}} = 1539.4/4.504 = 341.8$ , and we take 342.
Power: the hydraulic power is $N = 152.4\cdot 15.555/86.4 = 27.4$ kW, and the shaft power is $N_{\text{shaft}} = 27.4/0.595 = 46.1$ kW. With a margin of 15% the motor must give at least 53 kW. We take 56 kW, which we assume to be the nearest larger size in the manufacturer's range.
Cooling of the motor: the velocity of the liquid in the gap between the casing and the motor is $v_{\text{cool}} = \dfrac{152.4/86{,}400}{\pi\,(0.1306^2 - 0.117^2)/4} = 0.67$ m/s. This is more than the 0.3 m/s that is usually required.

Answer: A pump of 342 assumed stages (a head of about 1539 m at 152.4 m³/d at the intake) and a motor of 56 kW. The shaft power of the pump is 46.1 kW, and the dynamic fluid level is about 1438 m.

What it means. Almost all of the head goes into lifting the liquid from the dynamic fluid level. At this rate the friction in the tubing is less than one per cent. We put the design rate to the right of the best efficiency point on purpose. When the reservoir pressure falls, the required head will grow and the operating point will move to the left, through the best efficiency point and not past the edge of the operating range. Rounding the number of stages up gives a small excess of head, and the pump will work slightly to the right of the design point. The crossing of the curves in the next section shows where exactly. The data in this example are assumed for illustration.

Rotational speed: the affinity laws and frequency control

If we change the speed of the impeller, all the velocities of the liquid in the stage change in proportion, and the picture of the flow stays similar to the old one. This gives the affinity laws (laws of similarity) for one and the same stage:

The affinity laws of a centrifugal pump when the speed changes

\frac{q_2}{q_1} = \frac{n_2}{n_1}, \qquad \frac{h_2}{h_1} = \left(\frac{n_2}{n_1}\right)^{2}, \qquad \frac{N_2}{N_1} = \left(\frac{n_2}{n_1}\right)^{3}, \qquad \eta_2 = \eta_1

where:

$q_1,\ q_2$ rates at similar (corresponding) points of the curves, m³/d
$n_1,\ n_2$ rotational speeds. For a motor fed by a variable speed drive they are almost proportional to the frequency $f$ of the current, rpm
$h_1,\ h_2$ heads of the stage at these points, m
$N_1,\ N_2$ shaft powers, kW
$\eta_1,\ \eta_2$ efficiencies, —

Units: the ratios are dimensionless; the quantities with the subscripts 1 and 2 are in the same units

Holds when

one and the same stage, a low-viscosity liquid with no gas. When the frequency is lowered a lot (below about 30–35 Hz, when the rated frequency is 50 Hz), the efficiency falls and the laws hold less well
the laws link the corresponding points of two curves, not the operating points in the well. We find the operating point at a new frequency as the crossing of the new pump curve with the well curve

A rise of the frequency by 10% gives, at a corresponding point, 10% more rate, 21% more head and 33% more power. The whole curve moves up and to the right, and the best efficiency point moves along the parabola $h \propto q^2$ .

A variable speed drive gives three benefits. A smooth start. An induction motor connected straight to the grid takes, in the first seconds, a current several times the rated one. The drive speeds the motor up gradually, by raising the frequency step by step. Adjustment to the inflow. The reservoir pressure and the productivity change with time, and with the frequency we can bring the pump back into its operating range without lifting the installation out of the well. A margin for uncertainty. For a new well whose inflow is not known exactly, one pump covers several possible operating modes. The frequency is usually changed within a range from about 30–35 to 60–70 Hz. The price is the losses in the drive itself (a few per cent) and the distortions of the shape of the current. These are filtered out, so that they do not harm the motor.

The motor limits the frequency too. If the voltage is raised in proportion to the frequency, the motor keeps its rated torque, and the power that it can give grows in proportion to the frequency. But the power that the pump needs grows about as the cube of the frequency. So when the frequency is raised, the margin of motor power runs out fast. The second limit from above is the intake pressure: producing more liquid lowers it, and gas may appear. Which of the two limits comes first depends on the well (in the example below it is the gas). From below, the frequency is limited by the operating range of the pump and by the cooling of the motor, because the flow along the motor becomes slower.

CalculatedCalculated head curves of the pump from the sizing example (342 assumed stages) at 40, 45, 50 and 55 Hz, and the well curve, which is the head that the well requires at each rate (dashed line). The crossing points are the operating modes. The horizontal axis is the liquid rate at standard conditions. The rate of the pump at intake conditions is 1.6% larger.

Worked exampleThe same pump at 45 and 55 Hz

Given (assumed data)

Pump: from the sizing example: 342 assumed stages, and a motor of 56 kW at 50 Hz
Well: the same; its curve is the required head at each rate (figure above)
Motor at another frequency: the voltage is changed in proportion to the frequency

Solution

The affinity laws for the best efficiency point of the stage: at 45 Hz ( $n_2/n_1 = 0.9$ ) it moves from the point 140 m³/d, 4.80 m to the point 126 m³/d, 3.89 m. At 55 Hz it moves to 154 m³/d, 5.81 m. The power at corresponding points is multiplied by $0.9^3 = 0.729$ and by $1.1^3 = 1.331$ .
The operating points are the crossings of the pump curve and the well curve. At 50 Hz it is 150.1 m³/d (the rounding of the number of stages has added 0.06 m³/d to the calculated 150), at 45 Hz it is 122, and at 55 Hz it is 178 m³/d.
The intake pressure is 7.74, 5.77 and 3.80 MPa. At 55 Hz it is below the bubble point pressure of 5.0 MPa. Free gas will come to the intake, and this calculation does not allow for it, so the operating mode at 55 Hz is only an approximate estimate. The intake pressure falls to the bubble point pressure already at 51.9 Hz.
The shaft power is 32.5, 46.1 and 63.0 kW. At 55 Hz the motor of 56 kW can give $56\cdot 1.1 = 61.6$ kW, which is less than the 63.0 kW that is needed. The power limit is reached at about 54.4 Hz.

Answer: At 45 Hz the rate falls to 122 m³/d (by 18%). At 55 Hz it grows to 178 m³/d (by 18%), but the motor is overloaded and gas comes to the intake. Without changing the equipment, it is sensible to keep the frequency in this well no higher than about 52 Hz.

What it means. By the affinity laws we might expect $150\cdot 0.9 = 135$ m³/d at 45 Hz, but the well gave less. At the corresponding point, at 135 m³/d, the pump gives only $0.9^2 = 81$ % of the old head. But at this rate the well needs 93% of the old head. This is because part of the required head does not depend on the rate: the lift from the static fluid level (the level in a shut-in well) to the wellhead, and the difference between the tubing and annulus pressures at the wellhead. The head is not enough, and the operating point moves further to the left. The frequency is a convenient control, but it is limited from both sides: from below by the operating range of the pump and the cooling of the motor, and from above by the motor power and the gas at the intake. The data in this example are assumed for illustration.

Gas at the intake

Free gas that has come to the intake is the main cause of unstable work of an ESP. In the impeller the heavy liquid is thrown to the outer edge, and the light gas collects near the inlet of the impeller, where the pressure is lowest. The bubbles clog the channels, and the head of the stage falls. The more gas there is, the greater the fall. At some share of gas the stages stop creating head altogether. The gas locks the pump, and it stops delivering liquid. This is gas lock, or loss of delivery. The motor current falls and fluctuates, no liquid arrives, and the motor, without the cooling flow, overheats.

How much gas the pump can carry depends on the design of the stage, the intake pressure and the rate. As a guide, the head falls noticeably already at 10–15% of free gas by volume. Ordinary radial-flow stages lose their delivery at about 20–25%. Mixed-flow (diagonal) stages carry about twice as much, and special gas-handling (multiphase) stages carry even more. The exact limit is given by the manufacturer.

We find the share of gas at the intake in the same way as in the lesson on choosing a method, from the solution gas–oil ratio of the oil (the gas dissolved in each cubic metre of oil) and the formation volume factors at the pressure and the temperature of the intake. Here we add the part of the gas that goes into the annulus and bypasses the pump.

The share of free gas at the pump inlet, allowing for separation

\varphi_{\text{g}} = \frac{(1 - E_{\text{sep}})\,q_o\,(R - R_s)\,B_g}{(1 - E_{\text{sep}})\,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 that enters the pump, —
$E_{\text{sep}}$ the share of free gas that goes into the annulus, by natural separation at the intake and through the gas separator, —
$q_o,\ q_w$ 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 the pressure and the temperature of the intake, m³/m³
$B_g$ formation volume factor of the gas at the intake conditions: $B_g = \dfrac{p_{\text{sc}}}{p}\cdot\dfrac{T}{T_{\text{sc}}}\,Z$ (the pressures are absolute, the temperatures are in kelvins, and Z is the gas compressibility factor, a correction for a gas that is not ideal), m³/m³
$B_o,\ B_w$ formation volume factors of oil and water at the intake conditions, m³/m³

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

Holds when

the gas that is not dissolved in the oil at the intake pressure comes to the intake together with the liquid; the solubility of gas in water is not included
the share $E_{\text{sep}}$ is an estimate: it depends on the rate, the size of the bubbles, the viscosity and the design of the separator. At $E_{\text{sep}} = 0$ the formula is the same as the formula of the lesson on choosing a method

The volume of liquid $q_o B_o + q_w B_w$ and the remaining gas pass through the pump, and we choose the pump for this volume. The separated gas goes up the annulus, and it needs an outlet to the flowline.

Intake pressure. The deeper the pump is submerged below the dynamic fluid level, the higher the intake pressure and the more gas stays dissolved. The price is a smaller drawdown on the reservoir, so this is a compromise between rate and gas.
Natural separation. Gas goes into the annulus if the bubbles rise faster than the liquid goes down to the intake. This works better at small rates, with a wide gap between the unit and the casing, and with large bubbles. It works best if the intake is below the perforated interval. Then the unit is put into a shroud (a sleeve around the unit) with an inlet at the bottom, so that the liquid goes to the intake along the motor and cools it.
The gas separator. A rotary gas separator spins the flow. The heavy liquid is thrown to the wall and goes into the pump, and the gas collects near the shaft and is thrown out into the annulus. In favourable conditions it removes most of the free gas, about 80–90%, and two separators in series remove even more. But its rotating parts are worn by sand, and they get out of balance when salts deposit on them. It is one more part that can fail.
Gas-handling stages and dispersers. Mixed-flow stages, multiphase stages and dispersers (devices that break the gas into fine bubbles) let the pump carry more gas. The lower part of the pump is sometimes built of gas-handling stages, and the upper part of ordinary ones. As the pressure grows along the pump, the gas is compressed and dissolves, and the upper stages have less of it.
An outlet for the gas from the annulus. The separated gas must go freely from the annulus into the flowline. If it collects there, the annulus pressure grows and pushes the level down to the intake, and the gas goes into the pump again.

Worked exampleHow much gas will come to the intake

Given (assumed data)

Well: oil rate 40 and water rate 60 m³/d, producing gas–oil ratio 120 m³/m³
Downhole sensor at the intake: 4.0 MPa (gauge), 75 °C
Properties at intake conditions: solution gas–oil ratio 35 m³/m³, $B_o = 1.15$ , $B_w \approx 1$ , $Z = 0.88$
Gas separator: sends 85% of the free gas into the annulus (an estimate)

Solution

Free gas at standard conditions: $40\cdot(120 - 35) = 3400$ m³/d.
The formation volume factor of the gas, from the absolute pressure $4.0 + 0.1013 = 4.101$ MPa and the temperature 348 K: $B_g = \dfrac{0.1013}{4.101}\cdot\dfrac{348.15}{293.15}\cdot 0.88 = 0.02582$ . At intake conditions this is 87.8 m³/d of gas.
The liquid at the intake is $40\cdot 1.15 + 60 = 106$ m³/d. The share of gas without separation is $\varphi_{\text{g}} = \dfrac{87.8}{87.8 + 106} = 0.453$ , that is 45%.
With the gas separator, $0.15\cdot 87.8 = 13.2$ m³/d of gas goes into the pump: $\varphi_{\text{g}} = \dfrac{13.2}{13.2 + 106} = 0.11$ , that is 11%. The volume through the pump is 119 m³/d, which is 19% more than the liquid at the surface.

Answer: Without separation almost half of the volume at the intake is gas (45%), and an ordinary pump will lose its delivery. With the gas separator the share falls to 11%, which is at the edge of what radial-flow stages can carry.

What it means. This well needs a gas separator and a pump with mixed-flow stages, or with a gas-handling module in the lower part. We choose the pump for 119 m³/d, and not for 100 m³/d of liquid. The separated gas, about 2890 m³/d at standard conditions, must go freely from the annulus into the flowline. Another way is to set the pump deeper below the level: the intake pressure grows, and there is less gas. The data in this example are assumed for illustration.

The motor, the protector and the cable

A submersible motor gives its heat only to the liquid that flows past it. If the flow along the motor is too slow, the winding overheats, the insulation ages, and the life of the motor is cut several times over. This happens when the intake is below the perforations and there is no shroud, when the rate is small, or when the casing is too wide. The usual requirement is that the velocity of the flow along the motor is not less than about 0.3 m/s. In hot wells and for a viscous liquid it must be higher.

The velocity of the liquid along the submersible motor

v_{\text{cool}} = \frac{4\,q_{\text{in}}}{\pi\left(D_{\text{cas}}^2 - D_{\text{mot}}^2\right)}

where:

$v_{\text{cool}}$ average velocity of the liquid in the annular gap along the motor, m/s
$q_{\text{in}}$ rate at the intake conditions, m³/s
$D_{\text{cas}}$ inside diameter of the casing (or of the shroud), m
$D_{\text{mot}}$ outside diameter of the motor, m

Units: SI

Holds when

all the liquid that goes to the intake passes the motor: the motor is below the intake, and the inflow comes from below
free gas cools badly, so the velocity is estimated for the liquid

The wider the gap, the slower the flow. In a wide casing at a small rate, a shroud of a smaller diameter raises the velocity.

The protector is the part on which the life of the motor largely depends. Most often a motor fails when the liquid of the well gets into the oil through worn seals. The chambers of the protector are of two kinds. In a labyrinth chamber the oil and the well liquid are kept apart by the difference in their densities, and this works reliably only in a vertical position. A chamber with an elastic diaphragm also works in a deviated wellbore. The two kinds are often connected in series. The second frequent cause of motor failures is the quality of the electric power: voltage surges, unbalanced phases and higher harmonics (distortions of the current). This happens especially where the power is generated on site and not taken from the grid.

The cable is chosen for the motor current, for the temperature in the well (the materials of the insulation and of the protective sheaths depend on it), and for the allowed losses. The copper conductors, a couple of kilometres long, have a resistance of a few ohms. They lose power in proportion to the square of the current, and the voltage at the motor is lower than at the output of the transformer. This is why submersible motors are made high-voltage. At twice the voltage, the current for the same power is half as large, and the losses in the cable are four times smaller.

The motor current and the losses in the cable

I = \frac{N_{\text{shaft}}}{\sqrt{3}\,U\,\eta_{\text{mot}}\cos\varphi}, \qquad \Delta N_{\text{cab}} = 3 I^2 R_{\text{cond}}, \qquad R_{\text{cond}} = \rho_{\text{el}}\,\frac{L_{\text{cab}}}{S}\,\bigl[1 + \alpha\,(t - 20)\bigr]

where:

$I$ current in each conductor of the cable, A
$N_{\text{shaft}}$ power at the motor shaft, W
$U$ line voltage at the motor, V
$\eta_{\text{mot}},\ \cos\varphi$ efficiency and power factor of the motor (the power factor shows how much of the current does useful work), —
$\Delta N_{\text{cab}}$ power loss in the three conductors of the cable, W
$R_{\text{cond}}$ resistance of one conductor at its average temperature, Ω
$\rho_{\text{el}}$ resistivity of copper at 20 °C, 0.0175, Ω·mm²/m
$L_{\text{cab}},\ S$ length of the cable and cross-section of the conductor, m, mm²
$\alpha$ temperature coefficient of the resistance of copper, 0.0039, 1/°C
$t$ average temperature of the conductor along the well, °C

Units: SI; the conductor cross-section in mm², the resistivity in Ω·mm²/m

Holds when

a balanced three-phase load; the motor works near its rated mode, and its efficiency and power factor are the nameplate values
only the resistance of the conductors is included. The drop of the line voltage is estimated as $\Delta U \approx \sqrt{3}\,I R_{\text{cond}}\cos\varphi$ , and the inductive reactance of the cable increases it a little

The losses grow as the square of the current. We choose the voltage at the output of the transformer so that, after the drop in the cable, the motor receives its rated voltage.

Worked exampleThe current, the losses in the cable and the energy per cubic metre

Given (assumed data)

Motor: 56 kW, 1600 V, efficiency 0.84, $\cos\varphi = 0.83$ ; the load is 46.1 kW from the sizing example
Cable: copper, three conductors of 16 mm² each, length 2000 m, average conductor temperature 60 °C
Surface equipment: the transformer and the control station with the drive together have an efficiency of 0.96

Solution

Current: $I = \dfrac{46.1\cdot 10^3}{\sqrt 3\cdot 1600\cdot 0.84\cdot 0.83} = 23.9$ A.
Resistance of a conductor: $0.0175\cdot 2000/16 = 2.188$ Ω at 20 °C, and at 60 °C it is $2.188\cdot(1 + 0.0039\cdot 40) = 2.53$ Ω.
Losses in the cable: $3\cdot 23.9^2\cdot 2.53\cdot 10^{-3} = 4.3$ kW. Voltage drop: $\sqrt 3\cdot 23.9\cdot 2.53\cdot 0.83 \approx 87$ V, so about 1690 V is needed at the output of the transformer.
The motor takes $46.1/0.84 = 54.9$ kW. Together with the cable it is 59.2 kW, and from the grid, allowing for the surface equipment, it is 61.7 kW.
The efficiency of the system is $27.44/61.7 = 0.445$ . Per cubic metre of liquid, $61.7\cdot 24/150 = 9.9$ kWh is used.

Answer: The current is about 23.9 A. The cable loses 4.3 kW (7.3% of the power supplied to the cable). The installation takes 61.7 kW from the grid, and the efficiency of the system is 45%.

What it means. Less than half of the energy taken from the grid goes into lifting the liquid. The rest is lost in the pump (the largest loss), the motor, the cable and the surface equipment. A motor of 800 V with the same power would need twice the current and would lose four times more in the cable, about 17 kW. So for deep wells we choose high-voltage motors, and a cable with conductors of a large enough cross-section. The data in this example are assumed for illustration.

Starting, monitoring and faults

After a stop, the column of liquid in the tubing, if no check valve holds it, runs back through the pump into the well. Like a turbine, it turns the pump and the motor in the opposite direction. This is backspin, also called turbine rotation. Starting the motor at this moment almost certainly burns it out or breaks the shaft. So the control station forbids a restart for a set time after a stop, or on a signal from a backspin sensor. A check valve is also fitted in the tubing above the pump. At the first start we check the direction of rotation too. If the phases are connected in the wrong order, the pump turns the wrong way. It also delivers something, but much less, and we notice the error from the rate and the current.

In the past, the work of an ESP was judged almost only from the record of the motor current over a day and from rare measurements of the rate. The current is still the most important sign: in normal work it is steady. The data of the downhole sensors and of remote monitoring have been added to it: the pressure and temperature at the intake, the motor temperature, the vibration, the insulation resistance, the frequency and the voltages, and the rate and the water cut from the metering unit. The sensors at the intake make it possible to measure the bottomhole pressure without running downhole gauges, and even to do pressure build-up tests.

Signs of ESP faults

What is seen: The intake pressure falls slowly, then the current falls and fluctuates, and the station trips the installation on underload
Probable cause: the pump has pumped the well down: the inflow is less than the delivery, and gas comes to the intake
What is done: lower the frequency, install a pump of smaller delivery, treat the reservoir; for the time being use the periodic mode

What is seen: The current and the discharge pressure fluctuate at a normal level
Probable cause: gas at the intake
What is done: a gas separator, gas-handling stages, a higher intake pressure, an outlet for the gas from the annulus

What is seen: The current is higher than usual, and the wellhead pressure and the rate are lower
Probable cause: a leak in the tubing: the pump works at a larger rate and a smaller head
What is done: check the tightness of the tubing and of the drain valve

What is seen: The rate falls gradually, and the intake pressure grows
Probable cause: wear of the stages by sand, salts in the channels, a blocked intake
What is done: washing, a chemical treatment, replacing the pump

What is seen: The motor temperature grows
Probable cause: too little flow along the motor, deposits on the housing
What is done: check the rate and the level; a shroud, washing

What is seen: The insulation resistance falls
Probable cause: water in the motor oil, damage to the cable
What is done: prepare a repair before the insulation breaks down

What is seen	Probable cause	What is done
The intake pressure falls slowly, then the current falls and fluctuates, and the station trips the installation on underload	the pump has pumped the well down: the inflow is less than the delivery, and gas comes to the intake	lower the frequency, install a pump of smaller delivery, treat the reservoir; for the time being use the periodic mode
The current and the discharge pressure fluctuate at a normal level	gas at the intake	a gas separator, gas-handling stages, a higher intake pressure, an outlet for the gas from the annulus
The current is higher than usual, and the wellhead pressure and the rate are lower	a leak in the tubing: the pump works at a larger rate and a smaller head	check the tightness of the tubing and of the drain valve
The rate falls gradually, and the intake pressure grows	wear of the stages by sand, salts in the channels, a blocked intake	washing, a chemical treatment, replacing the pump
The motor temperature grows	too little flow along the motor, deposits on the housing	check the rate and the level; a shroud, washing
The insulation resistance falls	water in the motor oil, damage to the cable	prepare a repair before the insulation breaks down

If the inflow is too small even for the smallest suitable pump, the well is switched to short-cycle intermittent operation. The pump pumps out the liquid that has collected and stops, and when the level has recovered, it starts again. The stop and the start are set by the intake pressure (stop when it falls to one set point, start when it rises to another) or by a timer. But every start loads the motor and the cable, so a correctly sized pump is better than frequent starts.

The run life of an ESP (the time between failures) ranges from several months to several years. It depends strongly on the conditions: the temperature, sand and salts, corrosion (hydrogen sulfide, carbon dioxide), the quality of the electric power, and how carefully the installation was assembled and is operated. A noticeable share of failures comes in the first weeks after the unit is run into the well: damage to the cable and the unit during the run, and errors in sizing. So where ESPs are only beginning to be used, the run life is usually low at first, and it grows as experience builds up. Average figures for different well stocks should be compared with care, because in hard conditions and in easy ones they differ several times over.

Strengths and limits of an ESP

Strengths: large liquid rates, high head
Limits: works badly at small rates

Strengths: works in deviated and horizontal wells
Limits: tolerates free gas and sand badly

Strengths: the delivery is controlled by the frequency
Limits: viscous oil reduces the head and the efficiency, and emulsions form

Strengths: system efficiency of about 35–60%
Limits: the tubing has to be pulled to replace it

Strengths: little space at the wellhead, no moving parts at the surface
Limits: needs a reliable power supply; high temperature shortens the life of the motor and the cable

Strengths: sensors at the intake, real-time data
Limits: the cable has to be passed through the wellhead and the packer, and it is easy to damage while the unit is run into the well

Strengths	Limits
large liquid rates, high head	works badly at small rates
works in deviated and horizontal wells	tolerates free gas and sand badly
the delivery is controlled by the frequency	viscous oil reduces the head and the efficiency, and emulsions form
system efficiency of about 35–60%	the tubing has to be pulled to replace it
little space at the wellhead, no moving parts at the surface	needs a reliable power supply; high temperature shortens the life of the motor and the cable
sensors at the intake, real-time data	the cable has to be passed through the wellhead and the packer, and it is easy to damage while the unit is run into the well

Special arrangements and uses

Deviated and horizontal wells. The unit can be run into a wellbore with a large inclination, up to almost horizontal, but not through sharp bends. A long stiff unit that is dragged through a section with a high dogleg severity (how sharply the wellbore bends; Directional wells: applications and profile) bends itself, and its bearings and seals wear faster. The pump itself is set in a straight section. In a horizontal well, a pump set closer to the reservoir gives a larger drawdown. The gas in the gently sloping part collects at the upper wall and is easier to separate.
Access below the pump. The unit is hung to the side on a Y-shaped crossover (a Y-tool), and a bypass pipe is run beside it down to the bottom. Through this pipe we can log and survey the well, perforate, set bridge plugs (plugs that seal off part of the well) to shut off water, and treat the reservoir, all without pulling the pump. The price is a smaller allowed diameter of the pump and the motor in the same casing.
Two pumps in one well. Two units are installed in series when one motor does not have enough power. Two pumps for two reservoirs make a multiple-zone completion (separate production from two reservoirs at the same time). Each reservoir works with its own pump and has its own measurement of the rate. In wells with a lot of sand, after a workover a lower “sacrificial” pump is sometimes installed. It takes the first, dirtiest period, and the upper pump is switched on later.
Running on coiled tubing. The unit is run on a string of coiled tubing (a continuous flexible pipe wound on a reel). The cable is fixed outside the pipe, and then the production goes through the pipe. Or the cable is laid inside the pipe. Then the production goes up the annulus, and the unit can be run into a live well (a well under pressure) without killing it (stopping the flow with a heavy kill fluid). This is quicker and cheaper than a workover with a hoist, especially on small offshore platforms.
Pumping and injecting water. An ESP lifts water from a water-bearing reservoir and delivers it to injection wells or straight into another reservoir of the same well. In this case the arrangement is turned upside down, and the pump stands under the motor. Horizontal pump sets on the surface work as pressure maintenance pumps and as booster pumps for pipelines.
Water separation in the well. At a water cut above about 80%, the water can be separated right in the well, by hydrocyclones (devices that spin the flow so that heavy and light liquids part) fed by the same pump, and injected into a reservoir below. Oil with a smaller share of water is lifted to the surface. The technology is complicated: it needs a suitable reservoir for injection and water free of sand.
A pump together with gas lift is one of the combined systems. They are described in the lesson on progressing cavity pumps and hybrid systems.

Terms and symbols

ESP: An electric submersible pump installation. The name is also used for the multistage pump itself.
Submersible motor: An oil-filled motor, cooled by the flow of liquid to the intake.
Protector: The part between the motor and the pump. It protects the motor oil, equalises the pressure and takes the axial force.
Gas separator: A module at the intake that separates free gas and releases it into the annulus.
Stage: An impeller and a diffuser. The stages are connected in series, and their heads add up.
Performance curve: The head, power and efficiency against the rate, measured on water at the rated frequency.
Head $h$ , $H$: The energy given to a unit weight of liquid, in metres of its column. The pressure is $\rho g H$ .
Best efficiency point, operating range: The rate at the maximum efficiency (the rated rate), and the interval of rates around it in which the pump is meant to work.
Affinity laws: The rate is proportional to the speed, the head to its square, and the power to its cube.
Variable speed drive: It changes the supply frequency and with it the rotational speed. It gives a smooth start and control of the delivery.
Gas lock (loss of delivery): The delivery stops when gas locks the stages of the pump.
Backspin (turbine rotation): The rotation of the pump and the motor in the opposite direction, driven by the liquid that runs back from the tubing after a stop.
Downhole sensor unit: The pressure, temperature and vibration sensors under the motor.

Common mistakes and misconceptions

Wrong

“The more stages, the larger the rate.”

In fact

Stages add head, not rate. Extra stages move the operating point to a larger rate only as far as the inflow allows. A pump with an excess of head pumps the well down until it loses its delivery, or it works outside the operating range.

Wrong

“The catalogue curve is what the pump will give in the well.”

In fact

The curves are measured on water with no gas at the rated frequency. In the well they are recalculated for the frequency (the affinity laws), for viscosity and for gas, and the pressure is found from the density of the pumped liquid.

Wrong

“Let us raise the frequency by 10% and get 10% more oil.”

In fact

At a corresponding point the rate grows in proportion to the frequency. But the operating point in the well is set by the crossing with the well curve. The power grows about as the cube of the frequency, the intake pressure falls, and gas may appear.

Wrong

“The current has fallen, so the pump is unloaded and everything is fine.”

In fact

A fall or fluctuations of the current usually mean that the pump delivers less: the level is pumped down, there is gas at the intake, or the shaft is broken. This is a reason to check the intake pressure and the rate.

Wrong

“We calculate the delivery of the pump from the rate at the surface.”

In fact

The volume that passes through the pump is the volume at intake conditions: oil with dissolved gas (its formation volume factor is greater than one) and free gas that has not been separated. In a well with gas it can be tens of per cent larger than the rate at the surface.

In depth: advanced detail

Where the affinity laws come from

The head that an impeller gives to the liquid is set by the change of the angular momentum of the liquid in the impeller (how strongly the liquid swirls around the shaft). For an ideal impeller, with an infinite number of thin vanes and no losses, this is the Euler equation for pumps:

The theoretical head of an impeller (the Euler equation)

H_{\text{th}} = \frac{u_2\,c_{u2} - u_1\,c_{u1}}{g}

where:

$H_{\text{th}}$ theoretical head of the impeller, m
$u_1,\ u_2$ circumferential speeds of the impeller at the inlet and at the outlet, $u = \pi D n/60$ , m/s
$c_{u1},\ c_{u2}$ circumferential components of the absolute velocity of the liquid at the inlet and at the outlet, m/s
$g$ acceleration of free fall, m/s²

Units: SI

Holds when

an ideal impeller: the flow follows the vanes exactly, and there are no losses. The real head is lower because of the finite number of vanes, friction and leaks
with an inlet without swirl ( $c_{u1} = 0$ ), what is left is $H_{\text{th}} = u_2 c_{u2}/g$

All the velocities in the impeller are proportional to the rotational speed $n$ (for similar velocity triangles), so the head is proportional to $n^2$ . The rate is a velocity multiplied by the flow area, so it is proportional to $n$ . The power $\rho g q H$ is proportional to $n^3$ .

The same reasoning for geometrically similar impellers of different diameter $D$ gives $q \propto nD^3$ , $H \propto n^2D^2$ and $N \propto n^3D^5$ . This is why stages of a larger size group give more rate and head at the same frequency. The shaft of an ESP is turned at a high speed, several thousand revolutions per minute, to get the head. An impeller of a small diameter, one that fits into the well, would otherwise give too little head per stage.

Viscous oil and emulsions

As the viscosity grows, the friction of the liquid against the walls of the channels and the discs of the impellers grows. The head and the rate fall, and the power grows, so the efficiency falls especially strongly. The small stages that are run into wells are more sensitive than large ones. The corrections are noticeable already at a viscosity of about a hundred mPa·s, and at several hundred the efficiency falls several times over. The curve for a viscous liquid is found by recalculating the catalogue curve with standardised methods, or from the manufacturer's data.

Emulsions can be worse than viscous oil. In fast-turning stages, oil and water are broken into fine drops and form stable emulsions (mixtures in which one liquid is spread as droplets in the other). Near the phase inversion point (where oil-in-water turns into water-in-oil), the viscosity of such an emulsion can be many times greater than the viscosity of both the oil and the water (Multiphase flow in tubing). It helps to feed a demulsifier (a chemical that breaks the emulsion) to the intake. For heavy oil, progressing cavity pumps are more often chosen: they move the liquid slowly and hardly mix it. At a high water cut, when the continuous phase is water, the viscosity of the mixture is again close to that of water, and the ESP works well.

Connections to other topics

Artificial lift: when it is needed and how to choose the method — when an ESP is the right choice, and how it compares with the other methods.
Nodal analysis of the production system — the operating point of the pump in the well is the same crossing of the inflow and outflow curves.
Inflow performance relationship (IPR) — how much the reservoir gives as the bottomhole pressure falls, including below the bubble point pressure.
Multiphase flow in tubing — gas in the tubing above the pump makes the column lighter; friction in the pipes, and emulsions.
Black oil model: solution gas–oil ratio and formation volume factors — the solution gas–oil ratio and the formation volume factors at intake conditions.
Sucker-rod pumping — sucker rod pumps, for small rates and shallow wells.
Progressing cavity pumps and hybrid systems — progressing cavity pumps for viscous oil and sand, and combined systems, including gas lift above an ESP.
Gas lift: principles and choice of operating conditions — gas lift, a method of lift that tolerates gas well.
Directional wells: applications and profile — the dogleg severity of the wellbore through which the submersible unit passes.

Check yourself

1
A pump creates a head of 1500 m on water. What pressure rise will it create on a liquid of density 1100 kg/m³ with the same viscosity, at the same rate?
1. A.16.2 MPa
2. B.It cannot be said without tests on this liquid
3. C.13.4 MPa
4. D.14.7 MPa
Show the answer
Correct answer: A. 16.2 MPa
For a low-viscosity liquid the head in metres does not depend on the density, and the pressure is proportional to it: $1100\cdot 9.81\cdot 1500 \approx 16.2$ MPa. On water it would be about 14.7 MPa. The shaft power would also grow, 1.1 times.
2
At one point of its curve at 50 Hz, a pump gives 120 m³/d and 1300 m of head, and takes 40 kW at the shaft. What will it give at the corresponding point at 60 Hz, and will a motor of 45 kW (at 50 Hz) cope?
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The rate is $120\cdot 1.2 = 144$ m³/d, the head is $1300\cdot 1.2^2 = 1872$ m, and the power is $40\cdot 1.2^3 \approx 69$ kW. With the voltage raised in proportion, a motor of 45 kW can give about $45\cdot 1.2 = 54$ kW at 60 Hz, so it will be overloaded. Besides, the operating point in the well will be different: it is found as the crossing of the new pump curve with the well curve.
3
Why is the design rate of an ESP often chosen to the right of the best efficiency point of the stage?
1. A.The motor heats up less this way
2. B.To the right of the best efficiency point the power is smaller
3. C.To the right of the best efficiency point the pump is not sensitive to gas
4. D.When the inflow falls, the operating point moves to the left and stays in the operating range
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Correct answer: D. When the inflow falls, the operating point moves to the left and stays in the operating range
The reservoir pressure falls with time, the required head grows, and the operating point moves to the left along the curve. If we start to the right of the best efficiency point, the pump passes through it and stays longer in the operating range.
4
The pump intake was set below the perforated interval, with no shroud. What is good about this, and what is dangerous?
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It is good for gas: the liquid goes to the intake from above, going down, and the bubbles rise into the annulus, so the casing works as a separator. It is dangerous for the motor: it stands below the intake, and the liquid does not flow past it, so the motor overheats without cooling. The unit is put into a shroud with an inlet at the bottom, so that the flow goes along the motor.
5
The intake pressure is 3 MPa, and the share of free gas without separation is 30%. Which measure will NOT reduce the share of gas that enters the pump?
1. A.A gas separator
2. B.Setting the pump deeper below the level
3. C.Raising the rotational speed
4. D.An intake below the perforations, in a shroud
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Correct answer: C. Raising the rotational speed
Raising the frequency increases the withdrawal. The dynamic fluid level goes down, the intake pressure falls, and there is more free gas. The other measures either separate the gas or raise the intake pressure.
6
The current of an ESP motor is steady for several hours, then starts to fluctuate, and the station trips the installation on underload. After a restart the picture repeats. What is happening, and what should be done?
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The pump delivers more than the reservoir gives. The level goes down to the intake (the well is pumped off), the intake pressure falls below the bubble point pressure, gas comes to the intake, and the pump loses its delivery (gas lock). We must bring the delivery into line with the inflow: lower the frequency, install a pump of smaller delivery, or increase the inflow by treating the reservoir. Frequent starts shorten the life of the motor and the cable.

Updated: 5 October 2026