Gas lift operation and optimisation

Quick refresher

The main signs are the gas pressure at the annulus wellhead, the tubing pressure at the wellhead, the gas rate and the flow rate. The gas pressure, converted to the depths of the valves and compared with their opening and closing pressures, shows which valves can be open.
Where the gas really enters is shown by a pressure and temperature survey along the producing well. Above the injection point the pressure grows slowly with depth. Below it, the pressure follows the weight of the liquid. A kink (a sharp bend in the pressure line) that is not at a side-pocket mandrel is a sign of a hole in the tubing.
Casing heading (surging of the pressure in the annulus) is the result of feedback. The annulus stores gas. In subcritical flow through the operating valve (the gas is slower than sound in the port), the lower the tubing pressure, the more gas goes through. The loop is broken by critical flow through the operating valve (the gas moves at the speed of sound in the port) and by a high enough gas rate. Simply throttling the gas at the wellhead moves the well closer to the unstable region.
Limited gas is shared so that the marginal returns (the extra oil from the next thousand cubic metres of gas) of the wells that get gas are equal. For a well that needs a minimum gas rate, we compare the sharing with that well getting gas and without it. The gain against an equal split is a few per cent, and the less gas there is, the larger it is.
The limits of the system change the decision. With a limit on water, it pays more to stop the well with the most water than to cut the gas of all the wells. The separator pressure is a compromise between the back pressure on the wells and the capacity of the compressor.
With the years, more gas is needed and it gives less oil. The sharing is recalculated regularly, and the operating mode, the injection depth and the lift method are reviewed.

Key relationships

p_{\text{g}}(H) = p_{\text{g,wh}}\exp\dfrac{0.03418\,\gamma_{\text{g}}H}{\bar Z\bar T}

gas pressure at a valve from the wellhead pressure, for checking which valves are open

V_{\text{sc}} = V_{\text{ann}}\,\dfrac{\bar p}{p_{\text{sc}}}\,\dfrac{T_{\text{sc}}}{\bar T}\,\dfrac{1}{\bar Z},\qquad \dfrac{\dd\bar p}{\dd t} = \bar p\,\dfrac{q_{\text{in}} - q_{\text{out}}}{V_{\text{sc}}}

stock of gas in the annulus and the rate of change of the pressure in it

\dfrac{\partial q_{o,i}}{\partial q_{\text{g},i}} - \mu\,\dfrac{\partial q_{w,i}}{\partial q_{\text{g},i}} = \beta

sharing of gas with a limit on water: water has its own “price” $\mu$

Common pitfalls

Assuming that a steady gas pressure at the wellhead means injection through the operating valve.
Trying to stop surging by throttling the gas at the wellhead.
Sharing gas equally, or “according to the needs” of the wells.
Keeping within the limit on water by cutting the gas of all the wells a little.
Trusting a sharing that was calculated from well tests that have not been checked for a long time.

After this lesson you can

list what is measured on a gas-lift well, and what each measurement says about the work of the valves
use the gas pressure at the wellhead and the data sheet of the installation to find through which valves the gas can go
find the point of gas injection from a pressure survey along the well, and tell a valve problem from a leak in the tubing
explain why a gas-lift well surges, estimate the stock of gas in the annulus, and choose a way to stop the surging
share limited gas between the wells of a field, allowing for minimum gas rates and a limit on water, and estimate the gain against simple rules
take the gathering network and the compressor into account: how the separator pressure (the pressure in the vessel that splits the production into gas and liquid) changes both the wells and the volume of gas, and what changes over the years

Monitoring: what is measured and what it tells

A gas-lift well hides almost nothing, if you know where to look. Every day, and in modern fields continuously, we record the gas pressure at the annulus wellhead, the tubing pressure at the wellhead (the tubing head pressure), the gas rate to the well (measured at the gas distribution manifold, the unit that divides the gas between the wells) and the temperature of the production at the wellhead. From time to time the well is tested through a test separator (a separator used to measure the output of a single well), and its performance curve is taken at several gas rates. A downhole gauge (a pressure and temperature tool) is run into the producing well. The level of the liquid in the annulus is measured with an echo sounder, which times the echo of a sound pulse (dynamic fluid level). The reference for comparison is a model of the well that has been checked against these measurements: what the well should give at this gas rate and this pressure.

Measurements on a gas-lift well

Measurement: Gas pressure at the annulus wellhead
What it shows: through which valve the gas goes, and whether the mode is steady
Signs of problems: higher than designed: gas also goes through an unloading valve; lower: a hole in the tubing or an eroded valve; swings: surging, or switching of the gas between valves

Measurement: Tubing pressure at the wellhead
What it shows: back pressure, surging of the flow
Signs of problems: rising: separator, flowline, choke, wax (paraffin); swinging: surging

Measurement: Gas rate to the well
What it shows: how much gas is really supplied
Signs of problems: drift away from the set point (the value the controller is told to hold), flow meter errors, hydrates (ice-like solids made of gas and water) in the line

Measurement: Test through the test separator
What it shows: liquid, water cut, total gas
Signs of problems: total gas minus reservoir gas is not equal to the injected gas: a measurement error or a leak

Measurement: Pressure and temperature along the well
What it shows: where the gas enters, the gradients
Signs of problems: a kink that is not at a mandrel: a hole in the tubing; two kinks: gas enters in two places

Measurement: Level in the annulus (echo sounder)
What it shows: to what depth the annulus has been cleared of liquid
Signs of problems: a level above the operating valve: the gas has not reached it

Measurement	What it shows	Signs of problems
Gas pressure at the annulus wellhead	through which valve the gas goes, and whether the mode is steady	higher than designed: gas also goes through an unloading valve; lower: a hole in the tubing or an eroded valve; swings: surging, or switching of the gas between valves
Tubing pressure at the wellhead	back pressure, surging of the flow	rising: separator, flowline, choke, wax (paraffin); swinging: surging
Gas rate to the well	how much gas is really supplied	drift away from the set point (the value the controller is told to hold), flow meter errors, hydrates (ice-like solids made of gas and water) in the line
Test through the test separator	liquid, water cut, total gas	total gas minus reservoir gas is not equal to the injected gas: a measurement error or a leak
Pressure and temperature along the well	where the gas enters, the gradients	a kink that is not at a mandrel: a hole in the tubing; two kinks: gas enters in two places
Level in the annulus (echo sounder)	to what depth the annulus has been cleared of liquid	a level above the operating valve: the gas has not reached it

The signs suggest where to look. The cause is confirmed by a survey along the well or by a calculation with the well model.

Records of pressures and rates are useful only if they can be trusted. The gas flow meters on the manifold are checked against well tests: the gas that the test separator gave, minus the reservoir gas (the producing gas–oil ratio multiplied by the oil rate), must match the injected gas. A difference of tens of per cent almost always means a faulty measurement, not a mysterious well.

Which valve passes the gas: a check by the wellhead pressure

Unloading valves operated by the gas pressure open and close according to the gas pressure at the valve (force balance). The data sheet of the installation (data sheet and design checks) gives, for each valve, the opening pressure at the wellhead (with the tubing pressure taken from the design line) and the nitrogen pressure, which is also the closing pressure at the valve. The gas pressure at the wellhead is converted to the depth of each valve with the formula for a column of gas, and compared. A valve whose opening pressure at the wellhead is below the measured pressure has opened. An open valve stays open as long as the gas pressure at it is above the nitrogen pressure.

The wellhead pressure equals the design value and is steady. The gas most likely enters through the operating valve. This can be confirmed only by a survey along the well: with a hole in the tubing at a suitable depth, the pressure may turn out the same.
Higher than the design value. The operating valve passes less gas than is supplied (it is plugged, its port is too small, or more gas has been supplied), and the pressure has risen to the opening of the lower unloading valves. The gas enters the tubing in two places.
It has reached the system pressure, and almost no gas goes. The gas cannot pass downward: the upper valve has closed and the next one has not opened. This happens during unloading if the valves are set with no margin, or when the well has warmed up and the nitrogen pressure in the domes has risen.
Lower than the design value at the same gas rate. The gas has found an easy way into the tubing: a hole in the tubing, an eroded seat of an unloading valve, or a port of the operating valve that is too large.

Worked exampleThe gas pressure has risen: which valves are open

Given (assumed data)

Well: from the lesson Gas lift design: unloading valves at 850, 1495, 1945, 2230 and 2375 m, the operating valve at 2450 m; the design gas pressure at the wellhead in operation is 8.91 MPa
Data sheet: opening pressures at the wellhead 9.6, 9.5, 9.4, 9.3 and 9.2 MPa; nitrogen pressures (closing at the valve) 10.154, 10.742, 11.060, 11.308 and 11.339 MPa
What we see: the gas pressure at the annulus wellhead has risen to 9.22 MPa and stays there; the gas rate is the same, 50 thousand m³/d; the liquid rate has fallen

Solution

The opening pressure of valve 5 at the wellhead is 9.2 MPa, which is below 9.22, so valve 5 has opened. For valves 1–4 the opening pressures at the wellhead are 9.6–9.3 MPa, and at 9.22 MPa they will not open.
Gas at valve 5 at 2375 m: $9.22\exp\dfrac{0.03418\cdot 0.7\cdot 2375}{0.8031\cdot 323.15} = 9.22\,e^{0.2190} = 11.477$ MPa. This is above the nitrogen pressure of 11.339 MPa, so valve 5 stays open.
Gas at valve 4 at 2230 m: $9.22\,e^{0.2053} = 11.321$ MPa (with $\bar Z = 0.8041$ ), which is a little above the nitrogen pressure of 11.308 MPa. So if the wellhead pressure ever rose to 9.3 MPa, valve 4 opened and is still open. It will close only when the wellhead pressure falls below 9.21 MPa.

Answer: Gas enters the tubing through unloading valve 5 at 2375 m, together with the operating valve or instead of it. And if the pressure has risen to 9.3 MPa, also through valve 4.

What it means. The pressure rose because the operating valve stopped passing the set 50 thousand m³/d at the former pressure: its port is partly blocked by mineral deposits (scale), rust flakes or sand, or more gas has been supplied to the well than was calculated. Valves remember history: an open valve closes at a lower pressure than it opened at. So we look not only at the present pressure but also at the record of the past days. First the gas rate set point is checked. If it is unchanged, the operating valve is changed on wireline. A survey along the well will show through which valves, and in what shares, the gas goes. The data in this example are assumed for illustration.

A survey along the well: where the gas enters

The most direct answer to the question “where does the gas enter” is given by a flowing pressure and temperature survey of the producing well. A tool with pressure and temperature sensors is run on a wire through a lubricator (a pressure lock at the top of the well) (wireline work), with stops at set depths, usually above and below each side-pocket mandrel. Above the injection point the tubing holds a mixture of gas and liquid, and the pressure grows slowly with depth. Below it there is liquid, and the pressure follows its weight, about 10 kPa per metre. The gas that has entered the tubing is also colder than the flow, and it cools further as it expands in the port of the valve. So the temperature record shows a step down at the injection point.

One kink at the operating valve: everything is as designed.
A kink at an unloading valve: gas enters above the operating valve. If there are two kinks, gas enters in two places.
A kink between mandrels: gas does not enter through a valve. There is a hole in the tubing or a leaking connection between pipes.
No kinks, liquid up to the wellhead: no gas enters the tubing at all.

Worked exampleThe gas pressure has fallen and the flow rate has collapsed: looking for the cause

Given (assumed data)

Well: from the lesson Gas lift design; in operation the gas pressure at the wellhead was 8.91 MPa and the liquid rate was 131 m³/d
What we see: at the same gas rate, 50 thousand m³/d, the gas pressure at the wellhead has fallen to 5.62 MPa, and the liquid rate to 25 m³/d
Survey along the well: pressures at the stops are in the table below (the values are calculated with the model of the lesson and rounded)

Solution

Check by pressure: at the operating valve at 2450 m, gas at 5.62 MPa at the wellhead has about 6.9 MPa. This is less than the tubing pressure there even in normal operation (10.7 MPa). Gas cannot go through the operating valve, and at this pressure the unloading valves are closed. So the gas has found another way.
Gradients from the survey: between 600 and 900 m, $\dfrac{3.783 - 3.015}{300} = 0.00256$ MPa/m, a gassy mixture. Between 1200 and 1500 m, $\dfrac{8.955 - 6.012}{300} = 0.00981$ MPa/m, a liquid of density about 1000 kg/m³.
The two straight lines meet where $3.783 + 0.00256\,(H - 900) = 6.012 + 0.00981\,(H - 1200)$ . This gives $H = 998$ m, between the mandrels at 850 and 1495 m. The gas does not enter through a valve.
Size of the hole: the gas at 1000 m is $5.62\,e^{0.0844} = 6.11$ MPa with $\bar Z = 0.8770$ , and the tubing pressure is 4.05 MPa. By the formula for the gas rate through a choke (gas rate through a choke), 50 thousand m³/d at these pressures and a flow temperature of 55 °C passes through a hole about 8 mm in diameter. (The adiabatic exponent 1.27 and the discharge coefficient 0.85 are the same as for the port of the operating valve in the lesson on unloading.)

Answer: There is a hole in the tubing at about 1000 m, with an equivalent diameter of about 8 mm. All the gas enters through it, there is liquid in the tubing below it, and the well gives 25 m³/d instead of 131.

What it means. Below the hole the gas does not lighten the tubing: the 1650 m column of liquid presses on the reservoir in full. A hole of this size gives itself away at once, by the wellhead pressure and the flow rate. A small one is harder to find. A 3 mm hole at the same depth would take 12 thousand m³/d out of the 50, and the flow rate would fall only to 125 m³/d. The gas pressure at the wellhead would even rise, to 9.08 MPa: the operating valve gets less gas, the tubing pressure at it rises, and the gas needs a higher pressure. Such leaks are found by a survey along the well or by a pressure test of the tubing with a plug set in a landing nipple (a profile near the bottom of the tubing that holds plugs and other devices). They are fixed by replacing the pipes or by sealing the place of the leak. The data in this example are assumed for illustration.

Pressure survey along a well with a hole in the tubing

Depth, m: 0
Pressure, MPa: 1.600
Increase over the interval above, kPa/m: —

Depth, m: 300
Pressure, MPa: 2.289
Increase over the interval above, kPa/m: 2.30

Depth, m: 600
Pressure, MPa: 3.015
Increase over the interval above, kPa/m: 2.42

Depth, m: 900
Pressure, MPa: 3.783
Increase over the interval above, kPa/m: 2.56

Depth, m: 1200
Pressure, MPa: 6.012
Increase over the interval above, kPa/m: 7.43

Depth, m: 1500
Pressure, MPa: 8.955
Increase over the interval above, kPa/m: 9.81

Depth, m: 1800
Pressure, MPa: 11.898
Increase over the interval above, kPa/m: 9.81

Depth, m: 2100
Pressure, MPa: 14.841
Increase over the interval above, kPa/m: 9.81

Depth, m: 2400
Pressure, MPa: 17.784
Increase over the interval above, kPa/m: 9.81

Depth, m	Pressure, MPa	Increase over the interval above, kPa/m
0	1.600	—
300	2.289	2.30
600	3.015	2.42
900	3.783	2.56
1200	6.012	7.43
1500	8.955	9.81
1800	11.898	9.81
2100	14.841	9.81
2400	17.784	9.81

The tool stops every 300 m. Pressures are absolute, calculated with the teaching model of the lesson and rounded. Below 1000 m the pressure grows by 9.81 kPa/m, which is the weight of a liquid (water and oil) with a density of 1000 kg/m³. Above 1000 m it grows by 2.3–2.6 kPa/m. The data are assumed for illustration.

CalculatedThe tubing pressure of the well of the design lesson: in normal operation (the gas enters through the operating valve at 2450 m) and with a 8 mm hole at 1000 m (all the gas enters through the hole). The dots are the survey along the well. The thin marks on the left are the depths of the side-pocket mandrels. Calculated with the teaching model of the lesson.

Signs and causes of problems

Below are common signs, their possible causes and checks. Almost any sign allows several explanations, so conclusions are drawn from several measurements at once, going from the simple to the complex.

Common problems of a gas-lift well

What is seen: The gas pressure at the wellhead has reached the system pressure, and little gas goes
Possible cause: gas does not pass downward: the upper valve has closed and the lower one has not opened
How to check: survey along the well, recalculation of the valve settings for the real temperature
What to do: reset and change the valves

What is seen: The gas pressure is higher than designed and steady, the flow rate is lower
Possible cause: gas also enters through an unloading valve: the operating valve is plugged or more gas than calculated has been supplied
How to check: check by pressure (the example above), the rate set point, a survey along the well
What to do: restore the set point; change the operating valve

What is seen: The gas pressure is lower than designed at the same gas rate
Possible cause: a hole in the tubing, an eroded seat, a port of the operating valve that is too large
How to check: survey along the well, pressure test of the tubing
What to do: repair of the tubing, change of the valve

What is seen: The tubing pressure swings, the gas pressure is steady
Possible cause: tubing heading: too little gas, tubing too large
How to check: a record of pressures, a test at several gas rates
What to do: more gas; a choke at the tubing wellhead; tubing of a smaller diameter

What is seen: Both pressures swing in cycles that last from minutes to hours
Possible cause: casing heading: the operating valve is in subcritical flow
How to check: a record of pressures and gas rate
What to do: a smaller port, or a valve with critical flow

What is seen: Hours after unloading the mode “drifts”: the gas pressure rises, and the gas goes to an upper valve
Possible cause: the well has warmed up: the nitrogen in the domes has heated, and the opening and closing pressures have grown
How to check: the temperature along the well in operation against the one assumed in the setting
What to do: reset the valves for the real temperature

What is seen: In winter the gas rate falls and the line pressure jumps
Possible cause: hydrates in the gas pipeline or the choke
How to check: the gas temperature, the water dew point (the temperature at which water starts to condense out of the gas)
What to do: gas drying, methanol, heating of the line

What is seen: The flow rate falls, the pressures are the same
Possible cause: the reservoir has changed, or there are deposits in the tubing: wax, scale, sand
How to check: a well test, a drift run of the tubing (running a gauge to check that the tubing is open), reservoir studies
What to do: recalculate the mode; flushing out the tubing, treatments

What is seen: The well test does not agree with the gas flow meter
Possible cause: a measurement error
How to check: check of the flow meters and the test separator
What to do: calibration, repair of the instruments

What is seen	Possible cause	How to check	What to do
The gas pressure at the wellhead has reached the system pressure, and little gas goes	gas does not pass downward: the upper valve has closed and the lower one has not opened	survey along the well, recalculation of the valve settings for the real temperature	reset and change the valves
The gas pressure is higher than designed and steady, the flow rate is lower	gas also enters through an unloading valve: the operating valve is plugged or more gas than calculated has been supplied	check by pressure (the example above), the rate set point, a survey along the well	restore the set point; change the operating valve
The gas pressure is lower than designed at the same gas rate	a hole in the tubing, an eroded seat, a port of the operating valve that is too large	survey along the well, pressure test of the tubing	repair of the tubing, change of the valve
The tubing pressure swings, the gas pressure is steady	tubing heading: too little gas, tubing too large	a record of pressures, a test at several gas rates	more gas; a choke at the tubing wellhead; tubing of a smaller diameter
Both pressures swing in cycles that last from minutes to hours	casing heading: the operating valve is in subcritical flow	a record of pressures and gas rate	a smaller port, or a valve with critical flow
Hours after unloading the mode “drifts”: the gas pressure rises, and the gas goes to an upper valve	the well has warmed up: the nitrogen in the domes has heated, and the opening and closing pressures have grown	the temperature along the well in operation against the one assumed in the setting	reset the valves for the real temperature
In winter the gas rate falls and the line pressure jumps	hydrates in the gas pipeline or the choke	the gas temperature, the water dew point (the temperature at which water starts to condense out of the gas)	gas drying, methanol, heating of the line
The flow rate falls, the pressures are the same	the reservoir has changed, or there are deposits in the tubing: wax, scale, sand	a well test, a drift run of the tubing (running a gauge to check that the tubing is open), reservoir studies	recalculate the mode; flushing out the tubing, treatments
The well test does not agree with the gas flow meter	a measurement error	check of the flow meters and the test separator	calibration, repair of the instruments

The order of action is from the simple to the complex: first check the measurements and set points at the surface, then the pressures against the data sheet, then a survey along the well, and only then wireline work or a workover.

SchematicTypical records of the gas pressure at the annulus wellhead (orange line) and the tubing pressure at the wellhead (solid line) in different operating modes of a gas-lift well (diagram, not to scale): steady operation, tubing heading, casing heading, and gas that does not pass downward.

A separate topic is gas quality. Wet gas forms hydrates (ice-like solids made of gas and water) in gas pipelines and chokes, especially in winter and downstream of chokes, where the gas expands and cools (temperature margin to hydrates). Carbon dioxide causes corrosion when water is present, and hydrogen sulfide causes sulfide stress cracking of steels (cracks in steel that is under stress). The gas touches not only the tubing but also the inside of the casing. So the gas for gas lift is dried, the content of acid gases is watched, and when necessary inhibitors are injected or corrosion-resistant steels are used. Chemicals can also be sent with the gas, for example a scale inhibitor (a chemical that stops mineral deposits): through the operating valve it goes straight into the flow.

Heading (surging) and how to stop it

A gas-lift well can surge in two ways. In tubing heading, the gas pressure is steady, and the flow in the tubing comes in jerks. At a low gas rate the liquid collects, the column gets heavier, and then the gas throws it out as a slug. This is the same as the instability on the left branch of the outflow curve (stability of the operating point). In casing heading, the whole system swings. The annulus is a large tank of gas, and the gas rate into the tubing swings together with the pressure in it.

The second kind works by feedback. While the flow through the operating valve is subcritical (slower than the speed of sound in the port), the gas rate depends on the tubing pressure at the valve (how much gas a valve passes). Suppose the tubing pressure falls by chance. More gas goes in, the column becomes lighter, and the tubing pressure falls further. The annulus gives out gas faster than it receives it, its pressure slowly falls, and at some moment there is too little gas. The tubing fills with liquid, its pressure rises, almost no gas can enter, the annulus is pumped up again, and everything repeats. The period is from minutes to hours: the more gas is stored in the annulus compared with the gas rate, the more slowly it responds.

The stock of gas in the annulus and the rate of change of the pressure in it

V_{\text{sc}} = V_{\text{ann}}\,\frac{\bar p}{p_{\text{sc}}}\,\frac{T_{\text{sc}}}{\bar T}\,\frac{1}{\bar Z},\qquad \frac{\dd\bar p}{\dd t} \approx \bar p\,\frac{q_{\text{in}} - q_{\text{out}}}{V_{\text{sc}}}

where:

$V_{\text{sc}}$ gas in the annulus, converted to standard conditions, m³
$V_{\text{ann}}$ volume of the annulus from the wellhead to the operating valve, m³
$\bar p,\ \bar T,\ \bar Z$ average absolute pressure of the gas in the annulus, its average absolute temperature and its gas deviation factor, MPa, K, —
$p_{\text{sc}},\ T_{\text{sc}}$ standard conditions: 0.101325 MPa and 293.15 K (20 °C), MPa, K
$q_{\text{in}},\ q_{\text{out}}$ gas rate into the annulus from the surface, and from the annulus into the tubing, at standard conditions, m³/min

Units: any consistent units; pressures are absolute

Holds when

The temperature and $\bar Z$ hardly change during a swing. The pressure in the annulus changes almost equally over the whole height.
The second equality follows from the first: with constant $V_{\text{ann}}$ , $\bar T$ and $\bar Z$ , the pressure is proportional to the amount of gas.

The more gas is stored in the annulus compared with the gas rate, the more slowly the pressure in it changes, and the longer the period of the swings, if they arise. The annulus takes a long time to “notice” that the gas rate into the tubing has moved away from the supply from the surface.

Worked exampleHow much gas there is in the annulus and how fast it responds

Given (assumed data)

Well: from the lesson Gas lift design: 88.9 mm tubing, gas in the annulus down to the operating valve at 2450 m; 168.3 mm casing (150.5 mm inside diameter), assumed in this example
Mode: gas pressure 8.91 MPa at the wellhead and 11.16 MPa at the operating valve; 50 thousand m³/d of gas; temperature 323.15 K; pressure drop across the port of the operating valve 0.5 MPa, subcritical flow

Solution

Cross-section of the annulus: $\dfrac{\pi}{4}(0.1505^2 - 0.0889^2) = 0.011582$ m². The volume down to 2450 m is 28.38 m³.
Average pressure $(8.91 + 11.16)/2 = 10.035$ MPa, $\bar Z = 0.8072$ . Gas in the annulus: $28.38\cdot\dfrac{10.035}{0.101325}\cdot\dfrac{293.15}{323.15}\cdot\dfrac{1}{0.8072} = 3159$ m³. This is enough for $3159/34.72 = 91$ min of injection ( $50{,}000/1440 = 34.72$ m³/min).
If the operating valve takes 10% more gas than is supplied, the pressure in the annulus falls at a rate of $10.035\cdot\dfrac{3.472}{3159} = 0.0110$ MPa/min, that is, by 0.11 MPa in 10 minutes.
But the response of the valve is quick. At an unchanged gas pressure, a fall of the tubing pressure at the valve by 0.1 MPa increases the gas rate through the port from 50 to 54.5 thousand m³/d (+9%), and a rise by 0.1 MPa decreases it to 45.0 (−10%).

Answer: The annulus holds about 3.2 thousand m³ of gas, which is an hour and a half of injection. The pressure in it changes by hundredths of a megapascal per minute, while the gas rate through the operating valve responds to each 0.1 MPa in the tubing by about a tenth.

What it means. This leads to the remedies. While the flow through the operating valve is subcritical, surges of the tubing pressure at once swing the gas supply, and the slow annulus does not have time to damp them. A valve with critical flow breaks this link. It can be a port of a smaller diameter, or a nozzle with a diffuser that reaches critical flow at a small pressure drop. The gas rate through it then depends only on the gas pressure. A larger gas rate helps too: the well moves away from the unstable left part of the performance curve. Throttling the gas with a choke at the wellhead is the simplest thing, but with less gas the well is closer to instability. The data in this example are assumed for illustration.

Tubing heading: more gas; tubing of a smaller diameter at the next workover; sometimes a choke at the tubing wellhead, where the back pressure smooths out the jerks at the price of part of the flow rate.
Casing heading: an operating valve with critical flow, or a port of a smaller diameter, so that the flow is critical at the working pressure drop; then, if needed, more gas.
Switching between valves: recalculate the setting with the real temperature in operation. When the well warms up, the nitrogen pressure in the domes rises, and a valve may open and close in turn.
Unsteady gas supply: swings of the pressure in the gas pipeline, from the compressor or from neighbouring surging wells, are passed on to all the wells. Controlling the gas rate at each well and stopping the surging at the wells that start it calm the whole system.

Gas in a field: how to share it

The rule of equal marginal returns and its derivation are explained in the lesson on gas lift: each next thousand cubic metres of gas goes to the well where it gives the most oil, until the marginal returns of all the wells that get gas are equal. In a field, three more circumstances are added to this.

Minimum gas. A well needs some minimum gas rate to lift liquid at all, or so that the system pressure is enough to inject the gas through the operating valve. The first thousands of cubic metres give such a well almost nothing, so the rule of equal returns does not work for it. We assess the well as a whole. Does the oil from all its gas, including the minimum, pay for that gas at the “price” of gas in oil, $\beta$ ? And we compare two decisions: the well gets gas, or it does not.
The curves must be known. The performance curves of the wells come from models checked by tests at several gas rates. If the tests are old, an exact sharing based on wrong curves gives nothing.
The limits of the system: how much water and gas the treatment plants will accept, the pressure in the gathering lines, and the capacity of the compressors (next section).

Six gas-lift wells of the field of the examples

Well: 1
Depth of gas injection, m: 2350
Reservoir: pressure, MPa; $J$ , m³/(d·MPa): 21.0; 28
Water cut, %: 22
Oil at 40 thousand m³/d of gas, m³/d: 187.7

Well: 2
Depth of gas injection, m: 2500
Reservoir: pressure, MPa; $J$ , m³/(d·MPa): 21.5; 16
Water cut, %: 48
Oil at 40 thousand m³/d of gas, m³/d: 73.7

Well: 3
Depth of gas injection, m: 2250
Reservoir: pressure, MPa; $J$ , m³/(d·MPa): 19.0; 20
Water cut, %: 65
Oil at 40 thousand m³/d of gas, m³/d: 49.0

Well: 4
Depth of gas injection, m: 2500
Reservoir: pressure, MPa; $J$ , m³/(d·MPa): 16.0; 10
Water cut, %: 60
Oil at 40 thousand m³/d of gas, m³/d: 19.7

Well: 5
Depth of gas injection, m: 2400
Reservoir: pressure, MPa; $J$ , m³/(d·MPa): 19.5; 12
Water cut, %: 82
Oil at 40 thousand m³/d of gas, m³/d: 15.6

Well: 6
Depth of gas injection, m: 2600
Reservoir: pressure, MPa; $J$ , m³/(d·MPa): 22.5; 9
Water cut, %: 35
Oil at 40 thousand m³/d of gas, m³/d: 62.8

Well	Depth of gas injection, m	Reservoir: pressure, MPa; $J$ , m³/(d·MPa)	Water cut, %	Oil at 40 thousand m³/d of gas, m³/d
1	2350	21.0; 28	22	187.7
2	2500	21.5; 16	48	73.7
3	2250	19.0; 20	65	49.0
4	2500	16.0; 10	60	19.7
5	2400	19.5; 12	82	15.6
6	2600	22.5; 9	35	62.8

This is a different field, not the one where the well of the design lesson stands (there the gas system gives 9.6 MPa). The tubing is 76 mm inside (62 mm for well 6), the tubing pressure at the wellhead is 1.6 MPa, and the gas is up to 12 MPa at the annulus wellhead. None of the wells works without gas. The smallest gas rate at which each well (1–6) can still lift liquid within these limits is 10; 15; 13; 10; 15; 9 thousand m³/d. Calculated with the teaching model of the lesson Gas lift: principles and choice of operating conditions; the data are assumed for illustration.

Sharing of 220 thousand m³/d of gas between the wells of the example

Well: 1
Equal shares: gas / oil: 36 / 183.7
Best sharing: gas / oil: 58 / 198.9
Marginal return, m³ per thousand m³: 0.40
Water at the best sharing, m³/d: 56.1

Well: 2
Equal shares: gas / oil: 36 / 71.5
Best sharing: gas / oil: 45 / 75.9
Marginal return, m³ per thousand m³: 0.40
Water at the best sharing, m³/d: 70.1

Well: 3
Equal shares: gas / oil: 36 / 47.2
Best sharing: gas / oil: 40 / 49.0
Marginal return, m³ per thousand m³: 0.40
Water at the best sharing, m³/d: 91.0

Well: 4
Equal shares: gas / oil: 36 / 18.7
Best sharing: gas / oil: 29 / 16.4
Marginal return, m³ per thousand m³: 0.39
Water at the best sharing, m³/d: 24.7

Well: 5
Equal shares: gas / oil: 36 / 15.0
Best sharing: gas / oil: 20 / 10.7
Marginal return, m³ per thousand m³: 0.40
Water at the best sharing, m³/d: 48.9

Well: 6
Equal shares: gas / oil: 36 / 62.0
Best sharing: gas / oil: 28 / 59.5
Marginal return, m³ per thousand m³: 0.41
Water at the best sharing, m³/d: 32.0

Well: total
Equal shares: gas / oil: 216 / 398.1
Best sharing: gas / oil: 220 / 410.4
Marginal return, m³ per thousand m³: —
Water at the best sharing, m³/d: 322.8

Well	Equal shares: gas / oil	Best sharing: gas / oil	Marginal return, m³ per thousand m³	Water at the best sharing, m³/d
1	36 / 183.7	58 / 198.9	0.40	56.1
2	36 / 71.5	45 / 75.9	0.40	70.1
3	36 / 47.2	40 / 49.0	0.40	91.0
4	36 / 18.7	29 / 16.4	0.39	24.7
5	36 / 15.0	20 / 10.7	0.40	48.9
6	36 / 62.0	28 / 59.5	0.41	32.0
total	216 / 398.1	220 / 410.4	—	322.8

Gas is in thousand m³/d, and oil in m³/d. With equal shares, each well gets 36 thousand m³/d, and 4 thousand m³/d are left over. The highest gas pressure needed at the wellhead with the best sharing is 11.12 MPa (well 5). The data are assumed for illustration.

CalculatedOil of the six wells against the amount of gas that the compressor gives: with the best sharing and with equal shares (calculated with the teaching model of the lesson). The slope of the upper curve is the common marginal return of gas

\beta

: about 0.77 m³ of oil per thousand m³ at 120 thousand m³/d, and about 0.40 at 220. The less gas there is, the larger the difference: with equal shares there is not enough even for the minimum of the weak wells.

Limits: water, the network and the compressor

Gas-lift wells are linked by more than their shared gas. Their production goes along common gathering lines to a common separator, the water goes to a common treatment plant, and the gas returns to the same compressor. Each of these links can become a limit, and then the “best” sharing of gas changes.

Worked exampleThe treatment plant accepts no more than 300 m³/d of water

Given (assumed data)

Wells and gas: from the previous example: 220 thousand m³/d; with the best sharing the water is 323 m³/d
Limit: the water treatment plant accepts no more than 300 m³/d from these wells (assumed)

Solution

Water per cubic metre of oil at a water cut $\text{WC}$ is $\text{WC}/(1 - \text{WC})$ . For wells 1–6 it is 0.28, 0.92, 1.86, 1.50, 4.56 and 0.54 m³/m³. Well 5 gives the most water for each cubic metre of oil.
The first way is to cut the gas of all the wells. With 186 thousand m³/d of gas (the best sharing), the water is 298 m³/d and the oil is 394.6 m³/d.
The second way is to give water a “price” in oil, $\mu$ , and share the gas by counting for each well its oil minus $\mu$ times its water. Already at $\mu \approx 0.095$ it is best to stop well 5 and give its gas to the others: 64, 50, 42, 32, 0 and 32 thousand m³/d. The water is 280 m³/d and the oil is 406.9 m³/d.

Answer: It pays more to stop the well with the most water and give its gas to the others. The oil is 406.9 m³/d, only 3.5 less than with no limit, and 12.3 m³/d more than if the gas of all the wells is cut.

What it means. Cutting the gas of all the wells means losing oil in the strong wells for the sake of the water of the weak ones. The rule with two “prices”, of gas $\beta$ and of water $\mu$ , shows by itself whom to restrict: the wells where each cubic metre of oil comes with the most water. The limit on gas at the treatment plant is handled in the same way: the reservoir gas differs from well to well, and together with the injected gas it goes to the same separators and compressors. The data in this example are assumed for illustration.

The separator pressure is one more common control. The lower it is, the lower the pressures at the wellheads, and each well gives more at the same gas rate. For gas-lift wells back pressure is especially costly, because the gas in the tubing expands more. But the gas for the compressor comes from the same separator: the lower the suction pressure (the pressure at the compressor inlet), the larger the compression ratio (outlet pressure divided by inlet pressure) and the less gas the compressor can give. The compressor has two limits. One is the volume of gas at suction (a piston compressor at a constant speed pumps almost a constant volume of gas at the inlet, and the lower the suction pressure, the less gas there is in this volume). The other is the power of the drive (for the same gas it grows with the logarithm of the compression ratio).

Worked exampleSeparator pressure: back pressure against the compressor

Given (assumed data)

Wells: the six wells of the previous examples; the tubing pressure at the wellhead is the separator pressure plus 0.4 MPa for the gathering line
Compressor (assumed): suction is the separator pressure minus 0.1 MPa, and discharge is 12.4 MPa. With a separator at 1.2 MPa it gives 220 thousand m³/d and is fully loaded both by volume and by power: about 962 kW goes into compression ( $1/0.65$ times as much as ideal isothermal compression)

Solution

Separator at 0.9 MPa: by volume the compressor will give $220\cdot\dfrac{0.8}{1.1} = 160$ thousand m³/d. By power it could give more, so the limit is the volume. With the wellhead at 1.3 MPa and the best sharing, the wells will give 398.3 m³/d of oil.
Separator at 1.5 MPa: by volume the compressor would pass 280, but the power is enough only for $220\cdot\dfrac{\ln(12.4/1.1)}{\ln(12.4/1.4)} = 244.3$ thousand m³/d. With the wellhead at 1.9 MPa, the wells will give 401.6 m³/d.
Separator at 1.2 MPa: 220 thousand m³/d, and 410.4 m³/d of oil. This is more than in both neighbouring cases (table below).

Answer: The best separator pressure in this example is about 1.2 MPa, where the compressor is loaded both by volume and by power. Below it the volume of gas is not enough. Above it the power is not enough, and the back pressure on the wells does harm.

What it means. The answer depends on which limit the compressor has. If it were limited only by power, it would pay to lower the separator pressure to the lowest values: at 0.6 MPa the oil would be 421.0 m³/d, because the wells gain more from the lower back pressure than they lose from the smaller gas rate. If it were limited only by volume, it would pay to raise the pressure to about 1.5 MPa (412.4 m³/d), and beyond that the back pressure would again outweigh the gain (408.8 m³/d at 1.8 MPa). So the separator pressure, the mode of the compressor and the sharing of gas are chosen together, with a model of the whole system. The data in this example are assumed for illustration.

Separator pressure, compressor gas and the oil of the six wells

Separator, MPa: 0.6
Gas by volume / by power, thousand m³/d: 100 / 166
Compressor gas, thousand m³/d: 100
Oil, m³/d: 372.8

Separator, MPa: 0.9
Gas by volume / by power, thousand m³/d: 160 / 194
Compressor gas, thousand m³/d: 160
Oil, m³/d: 398.3

Separator, MPa: 1.2
Gas by volume / by power, thousand m³/d: 220 / 220
Compressor gas, thousand m³/d: 220
Oil, m³/d: 410.4

Separator, MPa: 1.5
Gas by volume / by power, thousand m³/d: 280 / 244
Compressor gas, thousand m³/d: 244
Oil, m³/d: 401.6

Separator, MPa: 1.8
Gas by volume / by power, thousand m³/d: 340 / 268
Compressor gas, thousand m³/d: 268
Oil, m³/d: 391.9

Separator, MPa	Gas by volume / by power, thousand m³/d	Compressor gas, thousand m³/d	Oil, m³/d
0.6	100 / 166	100	372.8
0.9	160 / 194	160	398.3
1.2	220 / 220	220	410.4
1.5	280 / 244	244	401.6
1.8	340 / 268	268	391.9

The gas is shared between the wells in the best way at each pressure. The highest gas pressure needed at the annulus wellhead is from 9.88 to 11.80 MPa, within the 12 MPa of the system. The data are assumed for illustration.

CalculatedOil of the six wells against the separator pressure for the same compressor (calculated with the teaching model of the lesson): on the left the compressor is limited by the volume of gas at suction, and on the right by the power of the drive.

Disruptions call for a sharing decision too. If a compressor has stopped and there is less gas, the gas is taken first from the wells with the smallest marginal return, and the best wells are kept working. After a shutdown of the whole field, the wells are started one at a time, first those that give more oil per unit of gas. Unloading needs more gas and pressure than operation (unloading step by step), and high-pressure gas in the system is only just building up.

Intermittent gas lift and the plunger in operation

Intermittent gas lift (the principle, what changes in the design) is tuned at the well. The gas is supplied either by an intermitter, a timer-controlled valve on the gas line that opens for a set time, or continuously through a choke. In the second case the moment of ejection is set by the downhole valve itself, which is operated by the tubing pressure and opens when a slug has collected above it.

How the cycle is chosen

1
Injection time
The shortest time at which the slug reaches the wellhead: on the record of the tubing pressure at the wellhead there is a jerk when the slug arrives, and then the gas breaks through. If it is shorter, the slug will not arrive and will fall back; if it is longer, the gas is wasted.
2
Number of cycles
Start with rare cycles and make them more frequent while the daily production grows and the gas used per cubic metre stays acceptable. Cycles that are too frequent lift small slugs with a large fallback, and cycles that are too rare keep a high column over the reservoir.
3
Check
By records of the gas pressure (sawtooth-shaped: a slow rise during injection and a sharp fall at the ejection) and by well tests. The cycle is reviewed when the inflow changes.

A plunger adds one more signal: its arrival at the wellhead, which a sensor in the catcher (the device that stops the plunger at the top) records. The time of the lift shows the speed of the plunger: one that is too fast wears out and hits the equipment, and one that is too slow lets gas pass. The time between cycles must be longer than the time the plunger needs to fall to the bottom. A worn plunger lets gas pass through the gap between it and the tubing wall, and the gain in gas is lost, so plungers are inspected and replaced.

With the years and with new tools

A gas-lift field ages. The water cut grows, the reservoir pressure falls, and each well needs more gas and gives less oil for it. There is less reservoir gas and more lift gas is needed, so the gas balance of the field shifts, and in time gas for gas lift may have to be bought (the gas balance of the field). So the sharing of gas is recalculated regularly, after each round of well tests. Operating valves are moved deeper into the spare mandrels (the extra mandrels left below them) (design for the whole life). Low-rate wells are changed to intermittent gas lift or a plunger, and where the reservoir pressure is no longer enough, to pumps (Artificial lift: when it is needed and how to choose the method). A well is stopped when the oil it gives, at the gas rate that suits it best, no longer pays for the gas and the other costs of keeping it.

Permanent downhole gauges of pressure and temperature at the operating valve show where and how the gas enters, without wireline runs.
Downhole valves controlled from the surface, with an electric drive and a cable. The flow area is changed from a control panel, and sensors at the valve send the pressures and the gas rate. The mode is changed without wireline work.
Operating valves with critical flow: a nozzle with a diffuser that, according to manufacturers, reaches critical flow at a pressure drop of only about a tenth of the gas pressure. The gas rate does not depend on the tubing pressure, the mode is more stable, and the calculation is simpler.
Coiled tubing instead of the tubing, or inside it: a flexible pipe with built-in gas-lift valves lets us install gas lift without pulling the old tubing, for example in a well with a leaking casing.
Two tubing strings from one annulus (dual gas lift): one string for each reservoir layer in a multiple-zone completion (Multiple-zone completions: commingled and segregated production). The gas divides between the strings by itself, and one of them can take almost all the gas. To prevent this, valves of different types are placed in the strings, operated by the gas pressure in one and by the tubing pressure in the other, and the ports of the operating valves are chosen with care.
Real-time optimisation: programs that recalculate the models of wells and of the network from current measurements and suggest set points for the gas rate. Their accuracy is limited by the accuracy of the measurements, above all of the well tests.

Terms and symbols

Multipointing: Gas enters the tubing through several valves at once, usually because the gas pressure is above the opening pressures of the lower unloading valves.
Gas-lift heading: Periodic swings of the pressures and the flow rate. Tubing heading happens at a low gas rate. Casing heading comes from the feedback between the annulus and the operating valve.
Flowing pressure and temperature survey: Running a downhole tool into a producing well. The kinks of the pressure profile and the steps of the temperature show where the gas enters.
Lift gas allocation: Sharing limited gas between wells to get the most oil, allowing for minimum gas rates and the limits of the system.
Minimum gas of a well: The smallest gas rate at which the well lifts liquid within the system pressure. In sharing, the well is assessed together with it.
Intermitter: An actuated valve on the gas line of intermittent gas lift that opens on a timer for a set time.
$\beta$ , $\mu$: The “prices” of gas and water in oil when gas is shared: how much oil one more thousand cubic metres of gas would give, and how much oil a cubic metre of water above the limit costs.

Common mistakes and misconceptions

Wrong

“The gas pressure at the wellhead is steady, so the gas goes through the operating valve.”

In fact

It can be steady with injection through two valves and with a hole in the tubing. It is compared with the data sheet of the installation: in the example of this lesson, 9.22 MPa instead of 8.91 already meant that unloading valve 5 was open. The final answer comes from a survey along the well.

Wrong

“The well is surging, so we will throttle the gas at the wellhead.”

In fact

With less gas the well is closer to the unstable region. The feedback between the annulus and the operating valve is broken by critical flow through the operating valve, and tubing heading is stopped by a larger gas rate.

Wrong

“There is little gas, so we will share it equally, which is fairer.”

In fact

Equal shares means giving the gas to wells where it gives less oil. In the example of this lesson, the best sharing gave 3% more oil with 220 thousand m³/d of gas, and 8% more with 120.

Wrong

“There is more water than the plant accepts, so we will cut the gas of all the wells.”

In fact

It pays more to stop the well that gives the most water per cubic metre of oil and give its gas to the others. In the example of this lesson this gives 12.3 m³/d more oil than cutting the gas of all the wells.

Wrong

“The lower the separator pressure, the better for gas lift.”

In fact

For the wells, yes. But the compressor takes its gas from the same separator: the lower the suction pressure, the less gas. In the example of this lesson the best pressure turned out to be where the compressor is loaded both by volume and by power.

In depth: advanced detail

How gas is shared by search

The rule of equal marginal returns is the condition of a maximum for smooth curves that bulge upward (concave curves). If a well has a minimum gas rate, its curve does not start from zero, and the condition is no longer enough. A reliable way is a search over a grid, which is dynamic programming. The gas is divided into portions, for example of a thousand cubic metres. For the first well, its oil is written down for each possible amount of gas. Then the second well is added: for each total amount of gas, we try how much of it to give to the second well, with the remainder going to the best solution for the first well, and we keep the best option. We go on in this way, one well at a time, to the last. The number of options grows as the number of wells times the square of the number of portions, not exponentially, so six wells and 220 portions are calculated at once.

A limit on water is added without making the search more complicated: water is given a “price” $\mu$ (how much oil a cubic metre of water above the limit costs), and the gas is shared by counting for each well its oil minus $\mu$ times its water. $\mu$ is chosen so that the water fits in the limit. Because of the minimum gas rates, the water changes in jumps. In the example of this lesson, with a suitable $\mu$ a whole well is switched off at once, and the water is left with a margin.

Why the annulus makes the well swing

The feedback loop can be estimated from the numbers of the example of this lesson. The annulus is a tank. The pressure in it changes by $\bar p/V_{\text{sc}}$ for each cubic metre of excess or shortage of gas, which is about 3.18 MPa per thousand cubic metres. The operating valve is an “amplifier”: in subcritical flow, a fall of the tubing pressure by 0.1 MPa gives about 4.5 thousand m³/d of gas above the norm. The tubing is a link with a delay: the mixture passes from the valve to the wellhead in about 14 min and the liquid in about 39 min, so a lighter column does not show in the pressure at the valve at once. If the extra gas that the valve gave lowers the tubing pressure more than the original deviation did, the swings grow. If it lowers it less, they fade away.

For such systems, analytical stability criteria for a gas-lift well have been derived. They compare the response of the inflow, of the column in the tubing and of the annulus to a small deviation of the mode. In practice, dynamic models of multiphase flow that link the annulus, the valve, the tubing and the reservoir are used more often, and they show both the period and the size of the swings. The conclusion of both is the same: stability is helped by critical flow through the operating valve, by a high enough gas rate and by a high productivity of the reservoir. With a high productivity, a short fall of the bottomhole pressure at once adds heavy liquid to the tubing and brings the mode back.

Connections to other topics

Gas lift: principles and choice of operating conditions — the operating point, the performance curve of the well, and the rule of equal marginal returns.
Gas lift design — the data sheet of the installation, valve spacing and spare mandrels, against which the checks are made.
Unloading a gas lift well: valves and mandrels — opening and closing pressures of valves, unloading, and the flow capacity of ports.
Wireline operations — surveys along the well, and changing gas-lift valves on wireline.
Chokes, surface pressure losses and well temperature — gas rate through a choke, hydrates, and temperature in the well.
Nodal analysis of the production system — stability of the operating point, and the instability of the left branch of the outflow curve.
Field development concepts: surface facilities — the gas balance of the field and the capacities of treatment plants.
Gas processing and compression — compressors: stages, power and capacity limits.
Produced water: composition, treatment and disposal — how much water a treatment plant will accept, which is a limit for sharing gas.

Check yourself

1
The gas pressure at the annulus wellhead has risen 0.3 MPa above the design value and stays there, the gas rate is the same, and the flow rate has fallen. What is most likely?
1. A.The reservoir pressure has fallen
2. B.The operating valve passes less gas than is supplied, and the gas has also started to go through a lower unloading valve
3. C.A hole in the tubing above the operating valve
4. D.The gas has become lighter
Show the answer
Correct answer: B. The operating valve passes less gas than is supplied, and the gas has also started to go through a lower unloading valve
With an unchanged supply, the pressure in the annulus rises when gas goes into the tubing more slowly than it arrives. This means the operating valve is plugged, or its port is too small for the set gas rate. The pressure rises to the opening pressure of the lower unloading valves, and gas enters in two places. A large hole in the tubing, on the contrary, lowers the gas pressure, and a fall of the reservoir pressure shows first in the flow rate, not in the gas pressure.
2
In a well test, the test separator showed 49.0 thousand m³/d of gas, with an oil rate of 65 m³/d and a producing gas–oil ratio of 60 m³/m³. The flow meter on the gas distribution manifold shows 50 thousand m³/d. Does the gas balance agree?
Show the answer
Reservoir gas: $65\cdot 60 = 3900$ m³/d, or 3.9 thousand. The injected gas according to the separator is $49.0 - 3.9 = 45.1$ thousand m³/d. This is 4.9 thousand, or almost 10%, less than the flow meter shows. A difference of this size needs a check: calibrate the flow meter on the manifold and the test separator, and check the gas–oil ratio and leaks in the line. Until the measurement is checked, it is too early to share gas on the basis of these data.
3
A survey along the well showed a kink of the pressure profile at 1240 m. The side-pocket mandrels are at 850, 1495, 1945 and 2230 m, and the operating valve is at 2375 m. What does this mean?
1. A.The tool is faulty, because there are no kinks on a profile
2. B.The gas enters the tubing not through a valve: probably a hole in the tubing or a leaking connection near 1240 m
3. C.The gas enters through the operating valve and everything is in order
4. D.The gas enters through the unloading valve at 1495 m
Show the answer
Correct answer: B. The gas enters the tubing not through a valve: probably a hole in the tubing or a leaking connection near 1240 m
A kink of the pressure profile shows where gas enters the tubing: above it the mixture is lighter, and below it heavier. There is no mandrel at 1240 m, so the gas enters through a hole or a leaking connection. This is confirmed by a pressure test of the tubing, and the temperature profile will show a step of cooling at the same place.
4
An annulus of 25 m³ is filled with gas at an average pressure of 9 MPa, an average temperature of 320 K and $\bar Z = 0.82$ . 40 thousand m³/d goes out through the operating valve. For how many minutes of injection is the gas stored in the annulus enough?
Show the answer
Gas at standard conditions: $25\cdot\dfrac{9}{0.101325}\cdot\dfrac{293.15}{320}\cdot\dfrac{1}{0.82} \approx 2480$ m³. The gas rate is $40{,}000/1440 \approx 27.8$ m³/min, so the stock lasts about 89 minutes. Such an annulus responds to changes of the rate through the valve over tens of minutes, which is why the period of casing heading is long.
5
The common “price” of gas in the field is 0.5 m³ of oil per thousand cubic metres. A well needs at least 15 thousand m³/d of gas, and with it gives 6 m³/d of oil, while at 25 thousand m³/d it gives 11 m³/d. Should it be given gas?
Show the answer
We compare the oil with the “price” of all the gas of the well. At 15 thousand it is $6 - 0.5\cdot 15 = -1.5$ m³/d, and at 25 thousand it is $11 - 0.5\cdot 25 = -1.5$ m³/d. In both cases the same gas in other wells would give more oil, so the well is not given gas, although between 15 and 25 thousand its marginal return, $5/10 = 0.5$ , is equal to the common one. While gas is expensive, such a well waits until there is more gas.
6
Why does lowering the separator pressure not always increase the production of gas-lift wells?
Show the answer
A lower separator pressure means lower pressures at the wellheads, and each well gives more at the same gas rate. But the gas for the compressor comes from the same separator. The suction pressure falls, the compression ratio grows, and the compressor gives less gas, especially if its capacity is limited by the volume of gas at suction. When the loss of gas costs more than the gain from the lower back pressure, production falls. The separator pressure, the compressor and the sharing of gas are chosen together.

Updated: 19 October 2026