Hydraulic pumps: piston and jet

Quick refresher

The energy reaches the pump at depth in the liquid. The power pump at the surface creates the pressure, the column of power fluid in the tubing (the pipe run inside the well) adds to it, and at the pump the power fluid gives its energy to the production.
In the open system, the spent power fluid mixes with the production and rises with it. In the closed system it returns through a separate channel and stays clean. A jet pump works only in the open system.
The hydraulic piston pump is a positive-displacement pump (it pushes out a fixed volume with every stroke): the pistons of the engine and of the pump sit on one rod. The balance of work gives $p_{\text{pf}} - p_{\text{sp}} = (p_{\text{out}} - p_{\text{in}})\,(P/E) + F$ , where $F$ is the friction in the unit, written as a pressure. The larger the P/E, the higher the pressure of the power fluid and the smaller its flow rate.
The jet pump has no moving parts: a jet from the nozzle drags the production into the mixing chamber, and in the diffuser the speed of the mixture turns into pressure. Its efficiency does not exceed about a third.
A large ratio of the nozzle area to the area of the mixing chamber gives a large head (pressure rise) at a small production rate, and a small ratio does the opposite. The ratio is chosen from the required head and checked for cavitation: the pressure at the inlet of the chamber must not fall to the vapour pressure (the pressure at which the liquid boils).
A hydraulic piston pump needs a flow of power fluid of the order of the rate of the well, more or less depending on P/E, while a jet pump needs several times more. The power fluid is cleaned, and the separators and the metering are designed for the total flow.

Key relationships

p_{\text{pf}} - p_{\text{sp}} = (p_{\text{out}} - p_{\text{in}})\,\dfrac{P}{E} + F

the pressure balance of the hydraulic piston pump

q_{\text{pf}} = \dfrac{q}{\alpha_{\text{p}}\,\alpha_{\text{e}}\,(P/E)}

the flow rate of the power fluid of a hydraulic piston pump

R = \dfrac{A_{\text{n}}}{A_{\text{t}}},\quad M = \dfrac{q}{q_{\text{pf}}},\quad H = \dfrac{p_{\text{out}} - p_{\text{in}}}{p_{\text{pf}} - p_{\text{out}}}

the dimensionless quantities of a jet pump

\eta = M H

the efficiency of a jet pump

q_{\text{cav}} = A_{\text{s}}\sqrt{\dfrac{2\,(p_{\text{in}} - p_{\text{v}})}{\rho\,(1 + K_{\text{s}})}}

the largest production rate without cavitation

Common pitfalls

Forgetting the column of power fluid in the tubing. The pressure at the pump is tens of megapascals higher than at the wellhead, and the power pump gives only what is missing.
Choosing P/E by pressure alone. A smaller P/E lowers the wellhead pressure, but it raises the flow rate of the power fluid and the load on its treatment and on the separators.
Choosing a jet pump by its efficiency alone, without checking cavitation. The most economical pair of nozzle and chamber often cavitates.
Expecting a jet pump to pump the well down to a low pressure. Cavitation limits the intake pressure from below.
Feeding the unit with badly cleaned power fluid. Solid particles quickly wear the pistons and valves of a hydraulic piston pump.

After this lesson you can

describe an installation with a power fluid, which has a power pump, treatment of the fluid and channels in the well, and explain the difference between the open and the closed system
derive the pressure balance of a hydraulic piston pump through the ratio of the areas of the pump end (the pumping half of the unit) and the engine (P/E), and use it to calculate the flow rate and pressure of the power fluid, the number of strokes and the power
explain how a jet pump passes the energy of the jet to the production, and read its dimensionless characteristics: the ratios of areas, flow rates and pressures, and the efficiency
choose the nozzle and the mixing chamber of a jet pump for a given rate, and check them for cavitation
compare hydraulic piston pumps and jet pumps with each other and with other artificial lift methods in terms of energy, reliability and conditions of use

Energy through pipes: how the installation is arranged

In a sucker rod pump the energy goes to the pump along the rod string, and in an electric submersible pump it goes along the cable (artificial lift methods). Hydraulic pumps get it through pipes. At the surface, a power pump (most often a multi-plunger pump, sometimes a multistage centrifugal pump) injects the power fluid into the well. The pressure is usually from 10 to 25–30 MPa. At depth the power fluid gives its energy to the downhole unit, and the unit gives it to the production of the well.

SchematicAn installation of a hydraulic pump with an open system (a diagram, not to scale). The power fluid goes down the tubing, and a mixture of the spent power fluid and the production rises up the annulus (the space between the tubing and the casing). At the surface, part of the liquid from the separator is cleaned and injected into the well again by the power pump.

The pressure at the pump is made of the pressure of the power pump and the weight of the column of power fluid in the tubing: at a depth of 2000 m a column of liquid with a density of 950 kg/m³ adds 18.6 MPa. But the mixture that rises to the wellhead presses on the pump discharge with the same weight. So when the densities are equal, the weights of the two columns cancel each other. The power pump pays only for the work that the unit does on the production, the friction losses and the pressure in the flowline (the surface pipe that takes the production away from the well).

Treatment of the power fluid. It is taken from the production after the separator (a vessel that separates the gas, oil and water). Large particles settle in a settling tank, a hydrocyclone (a device that spins the liquid to throw out the sand) removes the sand, and a filter holds back the rest. The cleaner the liquid, the longer the pistons and valves of the hydraulic piston unit last. A jet pump is less demanding.
Distribution. One power pump feeds one well or several wells through a distribution unit with regulators and flow meters. The flow rate of the power fluid to each well sets the mode of its pump.
Channels in the well. The power fluid is fed down the tubing. It comes back up a second string of tubing run alongside, or up the annulus. The gas that separates from the production is, where possible, led up the annulus separately, so that it does not enter the pump.

A very useful feature is the free pump. The downhole unit is not fixed to the tubing. It is dropped into the tubing, and the flow of power fluid carries it down to a seat. To bring the unit up, we reverse the direction of the flow. We feed the fluid through the return channel, and it carries the unit up to the wellhead, where a catcher takes it. A check valve under the seat keeps the liquid from going into the reservoir. Changing a pump takes a few hours and needs no workover crew. This is valuable on offshore platforms and in remote areas.

The power fluid: open and closed systems

Where does the power fluid go after it has given up its energy? There are two ways.

Open power-fluid system: The spent power fluid mixes with the production and rises with it in the same channel. Only two channels are needed in the well. The separator receives the sum of the power fluid and the production, and the power fluid has to be separated and cleaned again each time. A jet pump works only in this way, because in its mixing chamber the liquids mix.
Closed power-fluid system: The power fluid returns to the surface in a separate, third channel, and does not mix with the production. It stays clean, it needs less treatment, and the production is easier to measure. But one more channel is needed, which means more pipe and a more expensive well. This system suits only hydraulic piston and hydraulic turbine units.

Oil or water as the power fluid

: Lubrication of pistons and valves
Oil: good
Water: additives are needed

: Leaks in the unit
Oil: less, because oil is more viscous
Water: more

: Fire danger at the wellhead
Oil: higher
Water: lower

: Metering of production
Oil: harder, because a large volume of oil circulates in the system
Water: easier

: Corrosion and deposits
Oil: less
Water: inhibitors are needed

	Oil	Water
Lubrication of pistons and valves	good	additives are needed
Leaks in the unit	less, because oil is more viscous	more
Fire danger at the wellhead	higher	lower
Metering of production	harder, because a large volume of oil circulates in the system	easier
Corrosion and deposits	less	inhibitors are needed

A qualitative comparison. On fields with a high water cut (a large share of water in the liquid), the power fluid is more often treated produced water (the water that comes up with the oil). It is convenient to add inhibitors of corrosion and deposits (chemicals that stop them) to the power fluid, because they go straight into the well.

A hydraulic piston pump needs about as much power fluid as the production it lifts, more or less depending on P/E. A jet pump needs several times more (see the examples below). All this flow passes through the separators, the treatment plant and the power pumps, and this must be allowed for in the design of the surface facilities. If the power fluid is oil, a large volume of oil circulates in the system, and the oil production of the well is measured with a correction for it.

The hydraulic piston pump

A hydraulic piston pump is two piston devices on one rod. At the top is the hydraulic engine. Its piston moves under the pressure of the power fluid. At the bottom is the pump end. Its piston draws in the production and pushes it out to the discharge. At the end of each stroke a spool valve (the engine valve, a sliding valve that switches the flow) switches the supply of power fluid to the other side of the engine piston, and the pistons go back. The unit is double-acting: the pump delivers production on the upstroke and on the downstroke. The valves of the pump end, suction and discharge valves on both sides of the piston, direct the flow.

SchematicA double-acting hydraulic piston unit on the upstroke (a diagram, not to scale). The power fluid pushes on the engine piston from below, and the piston of the pump end draws the production in below itself and pushes it out above itself. On the downstroke the spool valve changes the direction of the supply, and the roles of the chambers change places.

The main characteristic of the unit is the P/E ratio (pump-to-engine ratio). This is the ratio of the working volume of the pump end to the working volume of the engine, that is, of their effective areas at the same stroke. If P/E is less than one, the area of the engine is larger than the area of the pump. Then the unit creates at the pump a pressure rise that is larger than the pressure drop across the engine. It is like a lever that gains in force and loses in distance. The price is the flow rate: the engine passes more power fluid than the pump delivers production.

The pressure balance of a hydraulic piston pump

p_{\text{pf}} - p_{\text{sp}} = (p_{\text{out}} - p_{\text{in}})\,\frac{P}{E} + F

where:

$p_{\text{pf}}$ pressure of the power fluid at the inlet of the engine, MPa
$p_{\text{sp}}$ pressure of the spent power fluid at the outlet of the engine, MPa
$p_{\text{out}}$ pressure at the discharge of the pump end, MPa
$p_{\text{in}}$ pressure at the pump intake, MPa
$P/E$ ratio of the working volumes (effective areas) of the pump end and of the engine, —
$F$ friction losses in the unit, expressed as an extra pressure drop of the power fluid, MPa

Units: any consistent units of pressure

Holds when

work per cycle: the engine receives $(p_{\text{pf}} - p_{\text{sp}})V_{\text{e}}$ , the pump end gives $(p_{\text{out}} - p_{\text{in}})V_{\text{p}}$ , and the difference is friction. $P/E = V_{\text{p}}/V_{\text{e}}$ (the derivation is in the advanced section)
leaks are included separately, in the flow rates; the inertia of the masses and the pressure oscillations during the stroke are neglected
$F$ grows with the number of strokes and with the viscosity of the liquid; it is taken from the manufacturer's data

In the open system the spent fluid goes into the same channel as the production, and $p_{\text{sp}} = p_{\text{out}}$ . In the closed system $p_{\text{sp}}$ is the pressure in the return channel at the pump: the pressure at its wellhead plus the weight of the column and the friction.

The delivery, the number of strokes and the flow rate of the power fluid

q = \alpha_{\text{p}}\,1440\,V_{\text{p}}\,n, \qquad q_{\text{pf}} = \frac{1440\,V_{\text{e}}\,n}{\alpha_{\text{e}}} = \frac{q}{\alpha_{\text{p}}\,\alpha_{\text{e}}\,(P/E)}

where:

$q$ delivery of production at intake conditions, m³/d
$q_{\text{pf}}$ flow rate of the power fluid, m³/d
$V_{\text{p}},\ V_{\text{e}}$ volume displaced by the pump end and by the engine in one double stroke, m³
$n$ number of double strokes per minute, min⁻¹
$\alpha_{\text{p}}$ pump volumetric efficiency of the pump end: leaks and incomplete filling, —
$\alpha_{\text{e}}$ the share of the power fluid that does work in the engine; the rest is leaks past the piston and the spool valve, —

Units: metric oilfield units; 1440 is the number of minutes in a day

Holds when

a double-acting unit: $V_{\text{p}}$ and $V_{\text{e}}$ are the volumes over a full cycle, up and down
there is no free gas (gas that is not dissolved in the liquid) at the intake; with gas the volumetric efficiency falls, as with a sucker rod pump (Sucker-rod pumping)

The delivery is set by the number of strokes, and the number of strokes by the flow rate of the power fluid that the power pump feeds. By adjusting this flow at the surface, we change the operating mode of the unit. We choose the unit so that the calculated number of strokes is well below the rated one, because then it lasts longer.

The wellhead pressure follows from the pressure at the engine: $p_{\text{surf}} = p_{\text{pf}} - \rho g H_{\text{pump}} + \Delta p_{\text{fr}}$ , where $\rho g H_{\text{pump}}$ is the weight of the column of power fluid in the tubing down to the pump setting depth $H_{\text{pump}}$ , and $\Delta p_{\text{fr}}$ is the friction in it. If the densities of the power fluid and of the mixture in the annulus are equal, and the friction in the annulus can be neglected, the weights of the columns cancel. For the open system we get a simple rule: the pressure of the power pump equals the pressure in the flowline plus $(p_{\text{out}} - p_{\text{in}})\,P/E$ plus $F$ and the friction in the tubing.

Worked exampleA hydraulic piston pump for a well: the number of strokes, the power fluid, the pressure and the power

Given (assumed data)

Well: vertical, with the pump at 2000 m; the power fluid goes down 73 mm tubing (inside diameter 62 mm), and the mixture comes up the annulus of 168 mm casing (open system); the pressure in the flowline at the wellhead is 1.0 MPa
Operating mode: 60 m³/d of production at an intake pressure of 4.0 MPa, with no free gas; the production and the power fluid are both 950 kg/m³
Unit (assumed): $P/E = 0.75$ ; the pump end displaces 1.5 L per double stroke; the rated number of strokes is 50 per minute; $\alpha_{\text{p}} = 0.90$ , $\alpha_{\text{e}} = 0.95$ ; friction in the unit $F = 1.0$ MPa
Power pump with its drive: efficiency 0.8; friction in the tubing from the Darcy–Weisbach formula (the standard pipe friction formula), $\lambda = 0.022$ ; the pressures are absolute

Solution

The discharge pressure is the wellhead pressure plus the column in the annulus (the friction in the wide annulus can be ignored): $p_{\text{out}} = 1.0 + 950\cdot 9.81\cdot 2000\cdot 10^{-6} = 1.0 + 18.639 = 19.639$ MPa. The pump must raise the pressure of the production by $19.639 - 4.0 = 15.639$ MPa.
Delivery and number of strokes: $q_{\text{th}} = 60/0.90 = 66.7$ m³/d, and the number of double strokes is $n = \dfrac{66.7}{1440\cdot 0.0015} = 30.9$ per minute, which is 62% of the rated number.
Power fluid: the engine displaces $66.7/0.75 = 88.9$ m³/d. With leaks, $q_{\text{pf}} = 88.9/0.95 = 93.6$ m³/d, which is 1.56 times the well rate.
Pressure at the engine (the spent fluid goes into the annulus at $p_{\text{out}}$ ): $p_{\text{pf}} = 19.639 + 15.639\cdot 0.75 + 1.0 = 32.37$ MPa.
At the wellhead: the column of power fluid in the tubing is 18.639 MPa, and the friction at a velocity of 0.36 m/s is 0.043 MPa. So $p_{\text{surf}} = 32.37 - 18.639 + 0.043 = 13.77$ MPa. By the rule, the same is $1.0 + 11.729 + 1.0 + 0.043 = 13.77$ MPa.
Power: the hydraulic power at the wellhead is $(13.77 - 0.1)\cdot 93.6/86.4 = 14.81$ kW, and from the grid it is $14.81/0.8 = 18.51$ kW. The useful power is $60\cdot 15.639/86.4 = 10.86$ kW, so the efficiency of the system is $10.86/18.51 = 59$ %.

Answer: The unit will work at about 31 double strokes per minute and will need 94 m³/d of power fluid at a wellhead pressure of 13.77 MPa. It takes about 19 kW from the grid, and the efficiency of the system is 59%.

What it means. The weight of the column of power fluid, almost 19 MPa, comes to the power pump “for free”, because the column of mixture in the annulus balances it. The efficiency of the system in this example is like that of a good sucker rod pump or ESP (efficiency guidelines): a positive-displacement unit loses comparatively little. But the example is favourable, because there is no free gas at the intake. In the field, gas and the wear of pistons and valves usually lower the efficiency noticeably. The friction in the unit $F$ and the volumetric efficiencies are assumed here. In practice we take them from the data sheet of the unit and correct them with measurements. The data in this example are assumed for illustration.

Choosing P/E for the well of the example: the same pump end, different engines

P/E: 0.50
Power fluid, m³/d: 140
Wellhead pressure, MPa: 9.9
Power at the wellhead, kW: 15.9

P/E: 0.75
Power fluid, m³/d: 94
Wellhead pressure, MPa: 13.8
Power at the wellhead, kW: 14.8

P/E: 1.00
Power fluid, m³/d: 70
Wellhead pressure, MPa: 17.7
Power at the wellhead, kW: 14.3

P/E: 1.25
Power fluid, m³/d: 56
Wellhead pressure, MPa: 21.6
Power at the wellhead, kW: 13.9

P/E	Power fluid, m³/d	Wellhead pressure, MPa	Power at the wellhead, kW
0.50	140	9.9	15.9
0.75	94	13.8	14.8
1.00	70	17.7	14.3
1.25	56	21.6	13.9

The pressures are absolute; the other data are as in the example. The power changes little: P/E redistributes it between pressure and flow rate. The choice is set by the pressure that the surface equipment can take and by the volume of power fluid that the treatment system is ready to accept. The data are assumed.

The jet pump

A jet pump has not a single moving part. The power fluid at high pressure passes through a nozzle and leaves it as a jet at a speed of hundreds of metres per second. Its pressure then falls below the intake pressure. In the suction chamber around the jet, the production is drawn into the flow, and it enters the mixing chamber (the throat) together with the jet. The throat is a cylindrical channel. In it the fast jet gives part of its momentum (the “push” of a moving mass) to the slow production, and the mixture evens out its speed. In the diffuser, a widening channel, the speed of the mixture falls and the pressure rises to the discharge pressure.

SchematicA jet pump (a diagram, not to scale; in the well it stands vertically) and the change of pressure along it. The power fluid speeds up in the nozzle, and the pressure is lowest at the inlet of the mixing chamber. In the chamber and the diffuser it rises to the discharge pressure.

The pump passes on energy in a different way from a piston or a wheel. The jet and the production exchange momentum by turbulent mixing, and some energy is inevitably lost in this, much as in an inelastic collision. So the efficiency of a jet pump is low: even the best combinations of nozzle and mixing chamber give about a third. But the pump is simple and small, it tolerates sand, gas and high temperature, and it can be changed without pulling the tubing.

Area ratio $R = A_{\text{n}}/A_{\text{t}}$: The exit area of the nozzle divided by the area of the mixing chamber. This is the main geometric characteristic: pumps with the same $R$ are similar, and one dimensionless curve describes them.
Flow ratio $M = q/q_{\text{pf}}$: The production rate per unit rate of power fluid. The volumes are at pump conditions.
Pressure ratio $H$: $(p_{\text{out}} - p_{\text{in}})/(p_{\text{pf}} - p_{\text{out}})$ : how much the pump raises the pressure of the production, compared with the pressure that the power fluid loses.
Efficiency $\eta = M H$: The useful power passed to the production, divided by the power given up by the power fluid.

The dimensionless characteristic of a jet pump

\begin{aligned} H &= \frac{B - S}{1 + K_{\text{n}} - B},\\ B &= 2R + \frac{2M^2R^2}{1 - R} - (1 + K_{\text{td}})(1 + M)^2R^2,\\ S &= (1 + K_{\text{s}})\,\frac{M^2R^2}{(1 - R)^2} \end{aligned}

where:

$H$ pressure ratio $(p_{\text{out}} - p_{\text{in}})/(p_{\text{pf}} - p_{\text{out}})$ , —
$R$ ratio of the areas of the nozzle and of the mixing chamber, —
$M$ ratio of the flow rates of the production and of the power fluid, —
$B$ pressure rise from the inlet of the mixing chamber to the discharge, in fractions of the velocity head of the jet $\rho v_{\text{n}}^2/2$ , —
$S$ fall of the pressure of the production from the intake to the inlet of the mixing chamber, in the same fractions, —
$K_{\text{n}},\ K_{\text{s}},\ K_{\text{td}}$ loss coefficients in the nozzle, at the suction inlet of the production, and in the mixing chamber with the diffuser (found by experiment; in the lesson 0.03, 0 and 0.30), —

Units: dimensionless quantities

Holds when

one-dimensional flow of an incompressible liquid; the densities of the power fluid and of the production are the same
the mixing chamber is cylindrical, and the mixing in it is complete; the velocities at the intake and at the exit of the diffuser are small
there is no free gas at the intake; if there is gas, its volume is added to the production rate, and the curve is only approximate
the loss coefficients are found by tests; here the jet leaves the nozzle at the pressure at the inlet of the mixing chamber. In common field methods, the pressure drop in the nozzle is counted from the intake pressure. Then the denominator contains $B - S$ , and at the same $M$ the pressure ratio comes out lower (see the advanced section).

The curve $H(M)$ falls: the more production the jet drags along, the less pressure it can add to it. At $M = 0$ the head is largest (the pump works “against a closed valve”), and at $H = 0$ the production rate is largest. Large $R$ gives a high head at a small production rate, and small $R$ gives a large production rate at a small head. The derivation is in the advanced section.

CalculatedDimensionless characteristics of a jet pump for three area ratios, with the loss coefficients of the lesson: the pressure ratio

H

and the efficiency against the flow ratio

M

. The dot is the operating mode of the pump in the example (

R =0.40

), and the dashed line is the cavitation limit for its pressures. The highest efficiency is 34% at

R = 0.25

, 32% at

R = 0.40

, and 29% at

R = 0.50

The flow rate of the power fluid through the nozzle sets the speed of the jet. We find it from the energy balance in the nozzle: the pressure drop from the inlet of the nozzle to the inlet of the mixing chamber goes into speeding up the jet and into losses.

The flow rate of the power fluid through the nozzle

q_{\text{pf}} = A_{\text{n}}\sqrt{\frac{2\,(p_{\text{pf}} - p_{\text{in}})}{\rho\,(1 + K_{\text{n}} - S)}}

where:

$q_{\text{pf}}$ flow rate of the power fluid, m³/s
$A_{\text{n}}$ exit area of the nozzle, m²
$p_{\text{pf}},\ p_{\text{in}}$ pressure of the power fluid before the nozzle, and the intake pressure, Pa
$\rho$ density of the power fluid, kg/m³
$K_{\text{n}},\ S$ loss coefficient in the nozzle, and the fall of the pressure of the production at the inlet of the chamber (from the characteristic), —

Units: SI

Holds when

the velocity before the nozzle is small compared with the velocity of the jet
$S$ is usually a few hundredths, and in estimates it is dropped. The flow rate then comes out slightly smaller

At given pressures the flow rate grows in proportion to the nozzle area. So the nozzle is chosen for the needed flow rate of power fluid, and the mixing chamber for the area ratio.

In other unit systems (1)

Metric oilfield units

q_{\text{pf}}\,[\text{m}^3/\text{d}] = 122.2\,A_{\text{n}}\,[\text{mm}^2]\sqrt{\frac{\Delta p\,[\text{MPa}]}{\rho\,[\text{kg/m}^3]\,(1 + K_{\text{n}} - S)}}

$A_{\text{n}}$ in mm², $\Delta p = p_{\text{pf}} - p_{\text{in}}$ in MPa, $\rho$ in kg/m³.

$122.2 = 86{,}400\cdot 10^{-6}\cdot\sqrt{2\cdot 10^{6}}$ : the seconds in a day, mm² to m², and MPa to Pa under the root.

The lowest pressure in the pump is at the inlet of the mixing chamber: there the production has speeded up, but the jet has not yet given up its energy. If this pressure falls to the vapour pressure of the liquid (the pressure at which it boils at that temperature), vapour bubbles form in it. They collapse further downstream, in the zone of high pressure. This is cavitation. The production rate stops growing, and the blows of the collapsing bubbles destroy the walls of the mixing chamber. The production speeds up in the ring-shaped gap between the jet and the wall of the chamber, with an area $A_{\text{s}} = A_{\text{t}} - A_{\text{n}}$ , and this gap limits the production rate.

The largest production rate of a jet pump without cavitation

q_{\text{cav}} = A_{\text{s}}\sqrt{\frac{2\,(p_{\text{in}} - p_{\text{v}})}{\rho\,(1 + K_{\text{s}})}}, \qquad A_{\text{s}} = A_{\text{t}} - A_{\text{n}}

where:

$q_{\text{cav}}$ production rate at which the pressure at the inlet of the mixing chamber falls to the vapour pressure, m³/s
$A_{\text{s}}$ area of the ring-shaped suction inlet of the production into the mixing chamber, m²
$p_{\text{in}}$ intake pressure, Pa
$p_{\text{v}}$ vapour pressure of the liquid at the temperature at the intake (absolute), Pa
$\rho,\ K_{\text{s}}$ density of the production and the loss coefficient at its suction inlet, kg/m³, —

Units: SI; the pressures are absolute

Holds when

a one-dimensional estimate. In fact cavitation begins earlier, because in the boundary layer of the jet the pressure is lower than the average. So we leave a margin in the production rate
there is no free gas. If the production contains gas, it expands at the inlet of the chamber and limits the production rate in a similar way

The limit depends only on the area of the production inlet and on the intake pressure, and not on the nozzle. To take more production, or to work at a lower intake pressure, a larger mixing chamber is needed with the same nozzle. This means a smaller $R$ , and so a smaller head. This is why a jet pump cannot pump a well down to a low pressure.

In other unit systems (1)

In dimensionless form

M_{\text{cav}} = \frac{1 - R}{R}\sqrt{\frac{(1 + K_{\text{n}})\,(p_{\text{in}} - p_{\text{v}})}{(1 + K_{\text{s}})\,(p_{\text{pf}} - p_{\text{v}})}}

all the pressures are absolute, in the same units.

It follows from the equations of the nozzle and of the inlet of the production if we set the pressure at the inlet of the mixing chamber equal to the vapour pressure.

Worked exampleChoosing the nozzle and the mixing chamber

Given (assumed data)

Well and mode: from the previous example: the pump at 2000 m, 60 m³/d at an intake pressure of 4.0 MPa, an open system, and both liquids at 950 kg/m³
Power pump: wellhead pressure 20 MPa (absolute); efficiency with its drive 0.8
Pump (assumed): loss coefficients $K_{\text{n}} = 0.03$ , $K_{\text{s}} = 0$ , $K_{\text{td}} = 0.30$ ; vapour pressure 0.1 MPa

Solution

The pump must raise the pressure of the production by $19.639 - 4.0 = 15.639$ MPa. Before the nozzle, $p_{\text{pf}} = 20 + 18.639 - \Delta p_{\text{fr}}$ . The friction depends on the flow rate of the power fluid, so the calculation is repeated. In the end, $\Delta p_{\text{fr}} = 0.174$ MPa, $p_{\text{pf}} = 38.46$ MPa, and $H = \dfrac{15.639}{38.46 - 19.639} = 0.831$ .
At $R = 0.30$ the pump gives only $H(0) = 0.883$ even at zero production, which is a little more than we need. To take 60 m³/d it would need a huge amount of power fluid. With that, the friction in the tubing would grow and the pressure before the nozzle would fall, so there is no solution. At $R = 0.35$ there is a solution, but the efficiency is low (the table below).
$R = 0.40$ : from the characteristic $M = 0.3199$ , so $q_{\text{pf}} = 60/0.3199 = 187.6$ m³/d, which is 3.1 times the well rate. The efficiency of the pump is $0.3199\cdot 0.831 = 0.266$ .
Nozzle: $S = 0.0455$ , and the jet velocity is $v_{\text{n}} = \sqrt{2\,(38.46 - 4.0)\cdot 10^6/(950\cdot(1.03 - 0.0455))} = 271.5$ m/s. The area is $A_{\text{n}} = 187.6/86{,}400/271.5\cdot 10^{6} = 8.00$ mm² (diameter 3.19 mm). Mixing chamber: $A_{\text{t}} = 8.00/0.40 = 20.0$ mm² (5.05 mm). The suction inlet of the production is $A_{\text{s}} = 12.0$ mm².
Cavitation: $q_{\text{cav}} = 12.0\cdot 10^{-6}\sqrt{2\cdot(4.0 - 0.1)\cdot 10^6/950}\cdot 86{,}400 = 94$ m³/d, which is 1.6 times the well rate. The pressure at the inlet of the chamber is 2.41 MPa. At $R = 0.45$ the efficiency is higher, 0.288, but the cavitation margin is only 15%, and at $R = 0.50$ the pump cavitates.
Power: at the wellhead $(20 - 0.1)\cdot 187.6/86.4 = 43.2$ kW, and from the grid 54.0 kW. The efficiency of the system is $10.86/54.0 = 20$ %.

Answer: We choose

R = 0.40

: a nozzle of about 8.00 mm² (3.19 mm) and a mixing chamber of about 20.0 mm² (5.05 mm). The power fluid is 188 m³/d, and about 54 kW is taken from the grid, three times more than for the hydraulic piston pump (19 kW).

What it means. The jet pump needs twice as much power fluid as the hydraulic piston pump, and three times as much energy, but there is not a single moving part in the well. The one-dimensional estimate of cavitation is optimistic, so a pair with a small margin is not taken, even if its efficiency is a little higher. Nozzles and chambers are made in series in which $R$ takes several standard values. We choose the nearest pair and recalculate the operating mode, and we check the calculation from the dimensionless curve against the manufacturer's characteristic. The data in this example are assumed for illustration.

Choosing the area ratio: wellhead pressure 20 MPa, 60 m³/d at 4.0 MPa intake pressure

$R$: 0.30
Power fluid, m³/d: —
Pump efficiency, %: —
Cavitation limit, m³/d: —
Conclusion: not enough head

$R$: 0.35
Power fluid, m³/d: 256
Pump efficiency, %: 20
Cavitation limit, m³/d: 162
Conclusion: suitable, margin 169%

$R$: 0.40
Power fluid, m³/d: 188
Pump efficiency, %: 27
Cavitation limit, m³/d: 94
Conclusion: suitable, margin 57%

$R$: 0.45
Power fluid, m³/d: 173
Pump efficiency, %: 29
Cavitation limit, m³/d: 69
Conclusion: cavitation margin 15% — small

$R$: 0.50
Power fluid, m³/d: 175
Pump efficiency, %: 28
Cavitation limit, m³/d: 56
Conclusion: cavitation

$R$	Power fluid, m³/d	Pump efficiency, %	Cavitation limit, m³/d	Conclusion
0.30	—	—	—	not enough head
0.35	256	20	162	suitable, margin 169%
0.40	188	27	94	suitable, margin 57%
0.45	173	29	69	cavitation margin 15% — small
0.50	175	28	56	cavitation

Calculated from the dimensionless characteristic of the lesson. The larger the $R$ , the smaller the inlet of the production into the mixing chamber and the earlier cavitation begins. At small $R$ the pump does not create the needed head. The data are assumed.

Worked exampleHow far the intake pressure can be lowered

Given (assumed data)

Pump: from the previous example: the inlet of the production into the mixing chamber is 12.0 mm²
Task: keep 60 m³/d while lowering the intake pressure; vapour pressure 0.1 MPa, liquid 950 kg/m³

Solution

The velocity of the production in the inlet gap at 60 m³/d: $v_{\text{s}} = 60/86{,}400/(12.0\cdot 10^{-6}) = 57.9$ m/s.
Cavitation begins when the intake pressure becomes equal to the vapour pressure plus the velocity head: $p_{\text{in,min}} = 0.1 + 950\cdot 57.9^2/2\cdot 10^{-6} = 1.69$ MPa.
With a margin of 20% in the production rate, in case cavitation begins earlier than the one-dimensional estimate says, the intake pressure should not be lowered below $0.1 + (1.69 - 0.1)\cdot 1.44 = 2.4$ MPa.

Answer: Cavitation will not stop this pump from giving 60 m³/d as long as the intake pressure is not lower than about 1.69 MPa, or 2.4 MPa with a reasonable margin. The larger head needed for this is provided by raising the pressure of the power fluid.

What it means. To lower the intake pressure further, a larger mixing chamber is needed, which means a smaller $R$ , a smaller head and more power fluid. Even then a limit will remain. A hydraulic piston pump, like a sucker rod pump, can work at an intake pressure of a few tenths of a megapascal if the gas is vented. So wells with a low reservoir pressure are not pumped down deeply with a jet pump. The data in this example are assumed for illustration.

Piston or jet: which to choose

The hydraulic piston pump and the jet pump

Feature: System efficiency
Hydraulic piston pump: higher; with no gas, like positive-displacement pumps
Jet pump: low: about a third in the pump, less in the system

Feature: Intake pressure
Hydraulic piston pump: can be lowered a long way if gas is vented
Jet pump: limited by cavitation

Feature: Power fluid
Hydraulic piston pump: clean; its flow rate is comparable with the well rate
Jet pump: tolerates solid particles; its flow rate is several times the well rate

Feature: Gas and sand
Hydraulic piston pump: gas reduces the filling, sand wears the unit
Jet pump: tolerates them well: there are no moving parts

Feature: System
Hydraulic piston pump: open or closed
Jet pump: open only

Feature: Repair
Hydraulic piston pump: a complicated unit, brought out by the flow
Jet pump: the nozzle and the chamber are changed in a few hours

Feature	Hydraulic piston pump	Jet pump
System efficiency	higher; with no gas, like positive-displacement pumps	low: about a third in the pump, less in the system
Intake pressure	can be lowered a long way if gas is vented	limited by cavitation
Power fluid	clean; its flow rate is comparable with the well rate	tolerates solid particles; its flow rate is several times the well rate
Gas and sand	gas reduces the filling, sand wears the unit	tolerates them well: there are no moving parts
System	open or closed	open only
Repair	a complicated unit, brought out by the flow	the nozzle and the chamber are changed in a few hours

A qualitative comparison. The ranges of use by rate and depth are in the lesson Artificial lift: when it is needed and how to choose the method.

Compared with other methods, hydraulic pumps have their own advantages. They suit deviated and deep wells, where rods wear the tubing. They suit remote and offshore wells, where it is important to change the pump without a workover crew. One surface plant can serve several wells. The power fluid dilutes viscous oil. And corrosion is fought with an inhibitor in the power fluid. Jet pumps are also widely used when wells are brought into production and tested, because they are easy to run in, start and change. Their weak points are surface equipment for high pressure, the treatment of large volumes of power fluid, and the load on the separators.

There is also a third kind of downhole unit, the hydraulic turbine pump. The power fluid turns a turbine, and the turbine turns a multistage centrifugal pump, like that of an ESP (an electric submersible pump installation, Electric submersible pumps (ESP)). The turbine turns fast, so few stages are needed, and the unit is short and light and can be brought out by wireline equipment (a cable run from the surface). The power fluid itself cools and lubricates it, and there is no electric motor or cable in the well.

Terms and symbols

Power fluid: Oil or water that is injected into the well at high pressure to drive the downhole unit.
Power pump: A surface pump that injects the power fluid into the wells.
Hydraulic piston pump: A piston hydraulic engine and a piston pump on one rod. The delivery is proportional to the number of double strokes.
P/E ratio: The ratio of the working volumes of the pump end and of the engine of a hydraulic piston unit.
Jet pump: A pump with no moving parts: a nozzle, a suction chamber, a mixing chamber and a diffuser.
Area ratio $R$: The area of the nozzle divided by the area of the mixing chamber. It sets the shape of the characteristic of a jet pump.
Cavitation: The formation and collapse of vapour bubbles where the pressure falls to the vapour pressure. It limits the production rate and destroys metal.
Open and closed system: The spent power fluid either mixes with the production (open system) or returns through a separate channel (closed system).
Free pump: A downhole unit that is carried to its seat and brought out by the flow of power fluid.

Common mistakes and misconceptions

Wrong

“The power pump has to create a pressure equal to the pressure at the downhole unit.”

In fact

Most of the pressure is given by the column of power fluid in the tubing: in the example of this lesson it is almost 19 MPa out of 32. At the wellhead we need only what the unit passes to the production, plus the friction and the pressure in the flowline.

Wrong

“The larger the P/E, the more economical the unit.”

In fact

P/E redistributes the power between the pressure and the flow rate of the power fluid, and it hardly changes the power. The choice is set by the pressure that the surface equipment can take and by the volume of fluid that the treatment system is ready to accept.

Wrong

“Let us take the combination of nozzle and chamber with the highest efficiency.”

In fact

The most economical pairs often cavitate, because their inlet of production into the mixing chamber is small. First we check cavitation, with a margin, and then we choose by efficiency.

Wrong

“A jet pump will pump a well down as well as a piston pump, because it tolerates gas and sand.”

In fact

The intake pressure of a jet pump is limited by cavitation. In the example of this lesson, at 60 m³/d it must not be lower than about 1.7 MPa, or about 2.4 MPa with a margin. A hydraulic piston pump, if the gas is vented, works at a few tenths of a megapascal.

Wrong

“The power fluid is just water going round in a circle, and we need not think about it.”

In fact

It must be cleaned. Its volume (several times the well rate for a jet pump) passes through the separators and the power pumps and changes the whole balance of the surface facilities.

In depth: advanced detail

Where the pressure balance of the hydraulic piston pump comes from

In one double stroke the engine passes a volume $V_{\text{e}}$ of power fluid. It takes it in at the pressure $p_{\text{pf}}$ and gives it back at $p_{\text{sp}}$ , so it receives the work $(p_{\text{pf}} - p_{\text{sp}})\,V_{\text{e}}$ . In the same time the pump end displaces a volume $V_{\text{p}}$ , raising its pressure from $p_{\text{in}}$ to $p_{\text{out}}$ , so it does the work $(p_{\text{out}} - p_{\text{in}})\,V_{\text{p}}$ . The difference goes into the friction of the seals, the spool valve and the valves. We divide the work balance by $V_{\text{e}}$ and write the friction losses as an extra pressure drop $F$ . This gives $p_{\text{pf}} - p_{\text{sp}} = (p_{\text{out}} - p_{\text{in}})\,V_{\text{p}}/V_{\text{e}} + F$ , which is the equation of the lesson with $P/E = V_{\text{p}}/V_{\text{e}}$ .

The force balance on the pistons gives the same result. The pressure drop across the engine piston, multiplied by its effective area, balances the pressure drop across the pump piston, multiplied by its area, plus the friction force. The effective areas allow for the cross-section of the rod, which enters and leaves the chambers. This is why manufacturers give P/E and the working volumes per stroke, and not the diameters of the pistons. Leaks do not take part in the pressure balance. They are included in the flow rates through $\alpha_{\text{p}}$ and $\alpha_{\text{e}}$ .

In the closed system the spent fluid rises in its own channel, and $p_{\text{sp}} = p_{\text{sp,wh}} + \rho g H_{\text{pump}} + \Delta p_{\text{fr,sp}}$ , where $p_{\text{sp,wh}}$ is the pressure in the return line at the wellhead. The production then rises alone, without the power fluid, and $p_{\text{out}}$ is found from its density and flow rate.

Where the jet pump characteristic comes from

Consider the flow section by section, and write the velocity head of the jet as $Z = \rho v_{\text{n}}^2/2$ . The velocity of the jet is $v_{\text{n}} = q_{\text{pf}}/A_{\text{n}}$ , that of the production in the suction gap is $v_{\text{s}} = q/A_{\text{s}}$ , and that of the mixture in the chamber is $v_{\text{t}} = (q + q_{\text{pf}})/A_{\text{t}}$ . In terms of $M$ and $R$ , $v_{\text{s}}/v_{\text{n}} = MR/(1 - R)$ and $v_{\text{t}}/v_{\text{n}} = (1 + M)R$ .

The nozzle. The pressure drop from the inlet of the nozzle to the inlet of the mixing chamber speeds up the jet: $p_{\text{pf}} - p_{\text{1}} = (1 + K_{\text{n}})\,Z$ , where $p_{\text{1}}$ is the pressure at the inlet of the chamber.
The inlet of the production. The production speeds up from an almost zero speed at the intake: $p_{\text{in}} - p_{\text{1}} = (1 + K_{\text{s}})\,\rho v_{\text{s}}^2/2 = S\,Z$ .
The mixing chamber. The pressure difference between its ends, multiplied by the area, is equal to the difference between the flows of momentum at the exit and at the inlet, plus the friction against the wall. The flows at the inlet are $\rho q_{\text{pf}} v_{\text{n}}$ and $\rho q\,v_{\text{s}}$ , and at the exit it is $\rho\,(q + q_{\text{pf}})\,v_{\text{t}}$ . The jet is faster than the mixture, so the pressure in the chamber rises.
The diffuser. The speed of the mixture falls almost to zero, and the pressure rises by $(1 - K_{\text{d}})\,\rho v_{\text{t}}^2/2$ .

Adding the rise in the chamber and in the diffuser and writing everything through $Z$ , we get $p_{\text{out}} - p_{\text{1}} = B\,Z$ , with $B$ from the equation of the lesson ( $K_{\text{td}}$ is the sum of the losses in the chamber and the diffuser). Then $p_{\text{out}} - p_{\text{in}} = (B - S)\,Z$ and $p_{\text{pf}} - p_{\text{out}} = (1 + K_{\text{n}} - B)\,Z$ , and their ratio is the characteristic $H(M)$ . If the pressure drop in the nozzle is instead counted from the intake pressure, as is usual in common field sizing methods, the denominator contains $B - S$ in place of $B$ . At $M = 0$ the two forms agree, and as $M$ grows they diverge. In the operating mode of the example ( $M =0.32$ ) the pressure ratio by the field form is 8% lower, and the highest efficiency is lower by a few percentage points. The field form is more cautious: sizing by it needs more power fluid. A calculation by either form is checked against the manufacturer's characteristic.

The same system gives the cavitation limit. If $p_{\text{1}}$ is equal to the vapour pressure $p_{\text{v}}$ , then $Z = (p_{\text{pf}} - p_{\text{v}})/(1 + K_{\text{n}})$ and $S\,Z = p_{\text{in}} - p_{\text{v}}$ , and this is where $M_{\text{cav}}$ comes from.

Connections to other topics

Artificial lift: when it is needed and how to choose the method — the place of hydraulic pumps among the artificial lift methods, and the choice of method.
Sucker-rod pumping — a positive-displacement pump with a rod drive: delivery, volumetric efficiency, gas at the intake.
Electric submersible pumps (ESP) — a centrifugal pump with an electric drive; the hydraulic turbine pump is its hydraulic relative.
Progressing cavity pumps and hybrid systems — progressing cavity pumps for viscous oil and sand are one more alternative.
Gas lift: principles and choice of operating conditions — another method that tolerates gas and sand, and that also leaves a high bottomhole pressure.
Multiphase flow in tubing — friction and the weight of the mixture in the tubing.

Check yourself

1
How does the open power fluid system differ from the closed one, and which pump can work only in an open system?
Show the answer
In the open system the spent power fluid mixes with the production and rises with it, so it has to be separated and cleaned again each time, but only two channels are needed. In the closed system it returns through a separate channel and stays clean. A jet pump mixes the liquids in its mixing chamber, so it works only in the open system.
2
A hydraulic piston unit with $P/E = 0.6$ raises the pressure of the production by 12 MPa. The friction in the unit is 0.8 MPa, and the system is open. What pressure drop of the power fluid is needed across the engine?
1. A.12.8 MPa
2. B.7.2 MPa
3. C.20.8 MPa
4. D.8.0 MPa
Show the answer
Correct answer: D. 8.0 MPa
$p_{\text{pf}} - p_{\text{sp}} = 12\cdot 0.6 + 0.8 = 8.0$ MPa. The area of the engine is larger than that of the pump, so the pressure drop across the engine is smaller than across the pump. The price is the flow rate of the power fluid: about $1/0.6 \approx 1.7$ times the delivery, not counting leaks.
3
A unit delivers 45 m³/d of production, $P/E = 1.2$ , the pump volumetric efficiency is 0.9, and the share of power fluid that works in the engine is 0.95. How much power fluid is needed?
Show the answer
$q_{\text{pf}} = 45/(0.9\cdot 0.95\cdot 1.2) \approx 43.9$ m³/d. When P/E is greater than one, even less power fluid is needed than production, but its pressure is higher.
4
A jet pump has $M = 0.5$ and $H = 0.6$ . What is its efficiency, and what happens if we try to take twice as much production without changing the pressures?
Show the answer
The efficiency is $\eta = MH = 0.30$ . The characteristic $H(M)$ falls: at twice the production rate the pump can develop a smaller pressure ratio, and at the same pressures such a production rate cannot be reached. We need another pair of nozzle and chamber, or another pressure of the power fluid. Besides, twice the production rate doubles the velocity of the production in the inlet gap, and cavitation may be near.
5
Why can the pair of nozzle and mixing chamber with the largest area ratio $R$ turn out to be unsuitable, although it creates the largest head?
1. A.At a large $R$ the nozzle passes too little power fluid
2. B.The area ratio does not affect the work of the pump
3. C.At a large $R$ the ring-shaped inlet of the production into the mixing chamber is small, and at the needed production rate the pressure at the inlet falls to the vapour pressure: cavitation
4. D.At a large $R$ the efficiency is always above 50%
Show the answer
Correct answer: C. At a large $R$ the ring-shaped inlet of the production into the mixing chamber is small, and at the needed production rate the pressure at the inlet falls to the vapour pressure: cavitation
The suction inlet of the production $A_{\text{s}} = A_{\text{t}} - A_{\text{n}}$ is small at a large $R$ , the velocity of the production in it is high, and the pressure at the inlet of the chamber falls further. The largest production rate without cavitation is proportional to $A_{\text{s}}$ , so we check a pair with a large $R$ for cavitation first.
6
The inlet of the production into the mixing chamber is 15 mm², the intake pressure is 3.0 MPa (absolute), the vapour pressure is 0.1 MPa, the density of the liquid is 900 kg/m³, and $K_{\text{s}} = 0$ . Estimate the largest production rate without cavitation.
Show the answer
The velocity at the cavitation limit is $\sqrt{2\cdot 2.9\cdot 10^6/900} \approx 80.3$ m/s, and the production rate is $15\cdot 10^{-6}\cdot 80.3\cdot 86{,}400 \approx 104$ m³/d. The one-dimensional estimate is optimistic, so in practice we design for a production rate that is noticeably smaller.

Updated: 7 October 2026