Gas lift design · MunaiHub

Quick refresher

A design starts with the operating mode: at what gas rate and at what depth to inject the gas. The greatest depth at which gas, at a given pressure, can still enter the tubing is the point of balance. The operating valve is placed a little above it, which leaves a margin in pressure.
The unloading valves are spaced from the top down. The first goes where the kill fluid, pushed out by the gas, balances the gas pressure (with a small margin). Each next valve goes where the liquid that is left below the previous valve balances the gas. By then the tubing above the previous valve already holds a mixture of gas and liquid, and its pressure is taken from the design line, which is the pressure in the tubing at the design operating mode.
Each next valve is set to open at a lower gas pressure at the wellhead, lower by one pressure step $\Delta p_{\text{step}}$ . This way the upper valve closes before the lower one, and in operation the gas pressure stays below the closing pressures of all the unloading valves. The price of the steps is pressure: with $N$ unloading valves, the gas in operation loses $N\,\Delta p_{\text{step}}$ , and the operating valve is placed higher.
With depth the valves come closer and closer together. They are not placed closer than the minimum spacing, so the lower valves are moved up, and we check again that the gas will open each next one.
The design is calculated for the whole life of the well. As the reservoir pressure falls and the gas rate grows, the injection can be moved deeper, if spare mandrels have been left below the operating valve. The depth of the move is limited not only by the gas pressure in operation but also by unloading: the unloading valves must bring the gas to the new mandrel.
The energy for compressing gas grows slowly with the system pressure, as a logarithm. The system pressure is limited by the cost of the compressor and by the pressures that the pipelines, the wellhead equipment and the casing are designed for.

Key relationships

p_{\text{g}}(H_{\text{op}}) - \Delta p_{\text{valve}} = p_{\text{t}}(H_{\text{op}})

point of balance: gas can still enter the tubing at the operating mode

p_{\text{t,des}}(H_n) + \rho_{\text{kill}} g\,(H_{n+1} - H_n) = p_{\text{g}}\big(H_{n+1};\ p_{\text{g,wh},n+1}\big) - \Delta p_{\text{margin}}

greatest depth of the next unloading valve

p_{\text{g,wh},n} = p_{\text{ko}} - (n - 1)\,\Delta p_{\text{step}}

gas pressure at the wellhead at which valve $n$ opens

p_{\text{g,wh}}^{\text{op}} < p_{\text{ko}} - N\,\Delta p_{\text{step}}

limit of the gas pressure in operation with $N$ unloading valves

Common pitfalls

Placing the operating valve exactly at the point of balance. With the slightest fall of pressure in the system, the gas moves to the upper valve.
Setting all the unloading valves to one pressure. The upper ones will not close, and gas will enter the tubing in several places.
Rounding the depth of a valve down the wellbore. Gas will not open a valve that is deeper than the calculated depth, and unloading will stop.
Designing only for today's conditions and leaving no mandrels below the operating valve.
Judging the pressure of the gas system by the energy of compression. It grows only as a logarithm of the pressure, and what decides is the cost and strength of the equipment.

After this lesson you can

list the data needed for the design, and explain what goes wrong if there is an error in each of them
find the point of balance (the greatest depth at which gas can be injected for a given gas pressure and gas rate) and choose the depth of the operating valve with a margin
space the unloading valves that are operated by the gas pressure: find the depth of the first one from the kill fluid (the heavy fluid that fills the well before start-up), and the depths of the next ones from the design line (the tubing pressure at the design operating mode), with a pressure step for each valve and a minimum distance between valves
check the design: the order in which the unloading valves close, the limit of the gas pressure in operation, and the margin to the opening pressures
estimate what a higher pressure in the gas system gives, and why the equipment, not the energy, decides
allow in the design for falling reservoir pressure and a rising water cut (spare mandrels, which are extra mandrels left below the operating valve; moving the operating valve; checking unloading for future conditions), and know what changes in the design of intermittent gas lift with a plunger

What the design decides and where it starts

A gas-lift installation is the tubing with side-pocket mandrels (side-pocket mandrels), the valves in these mandrels, and the gas system at the surface. The design answers four questions. What operating mode does the well need (the gas rate and the depth of injection)? How is it to be started (where to place the unloading valves and what to set them to)? What is needed from the surface (the pressure and the volume of gas)? And how will the installation work years from now? The principle of gas lift and the choice of the operating mode are explained in the lesson Gas lift: principles and choice of operating conditions. The construction of valves, unloading and setting on the test rack are in the lesson Unloading a gas lift well: valves and mandrels. Here they come together in the calculation of one well.

Data for a gas-lift design

Data: Well design: true vertical and measured depths, tubing, casing, packer, type of mandrels
What depends on it: pressures in the tubing and annulus, and where mandrels can be placed
If it is wrong: a mandrel in a section of strong curvature that wireline tools cannot pass

Data: Reservoir: pressure, productivity index, bubble point pressure
What depends on it: flow rate and pressure in the tubing at the injection depth
If it is wrong: tubing pressure higher than calculated: the gas will not reach the operating valve

Data: Production: water cut, producing gas–oil ratio, densities, viscosity
What depends on it: weight of the column and friction in the tubing
If it is wrong: the same; the water cut grows with time, so a design for today will become out of date

Data: Temperature: reservoir temperature and the flow temperature at the wellhead at different flow rates
What depends on it: nitrogen pressure in the valve domes
If it is wrong: the valves will open at the wrong pressures, and unloading will stop

Data: Gas: the pressure and volume the system can supply, composition, moisture
What depends on it: injection depth, number of unloading valves
If it is wrong: less gas or a lower pressure: the operating valve turns out to be too deep

Data: Kill fluid and its level
What depends on it: depth and spacing of the unloading valves
If it is wrong: heavier than calculated: the gas will not push it out as far as the next valve

Data: Tubing pressure at the wellhead, separator, flowline
What depends on it: the whole pressure profile in the tubing
If it is wrong: higher back pressure: lower flow rate, and the valves work differently

Data	What depends on it	If it is wrong
Well design: true vertical and measured depths, tubing, casing, packer, type of mandrels	pressures in the tubing and annulus, and where mandrels can be placed	a mandrel in a section of strong curvature that wireline tools cannot pass
Reservoir: pressure, productivity index, bubble point pressure	flow rate and pressure in the tubing at the injection depth	tubing pressure higher than calculated: the gas will not reach the operating valve
Production: water cut, producing gas–oil ratio, densities, viscosity	weight of the column and friction in the tubing	the same; the water cut grows with time, so a design for today will become out of date
Temperature: reservoir temperature and the flow temperature at the wellhead at different flow rates	nitrogen pressure in the valve domes	the valves will open at the wrong pressures, and unloading will stop
Gas: the pressure and volume the system can supply, composition, moisture	injection depth, number of unloading valves	less gas or a lower pressure: the operating valve turns out to be too deep
Kill fluid and its level	depth and spacing of the unloading valves	heavier than calculated: the gas will not push it out as far as the next valve
Tubing pressure at the wellhead, separator, flowline	the whole pressure profile in the tubing	higher back pressure: lower flow rate, and the valves work differently

The model of flow in the tubing used for the calculation is checked against pressure measurements in working wells of the same field.

Data are never exact, so we first choose the design case. A design “for the expected conditions” places the operating valve as deep as possible according to the best estimates. The well then produces more, but if the tubing pressure turns out higher or the kill fluid heavier, the gas will stop at the top valve, and wireline work will be needed. A design “with a margin” builds margins into pressures and depths. The well will almost certainly start, but the operating valve is higher and production is lower. The choice depends on the cost of an intervention. On a land well pad, wireline work is comparatively quick and cheap. On an offshore well, and even more on a subsea well, each tool run is expensive, so there the design is more cautious.

Order of the design

1
Well and model
Collect the data, check the model of inflow and of flow in the tubing against measurements, and make sure that without gas the well does not work or works poorly.
2
Operating mode
From the gas pressure and the gas rate that the system can give, find the point of balance and place the operating valve a little above it (next section).
3
Unloading valves
Find the depths of the unloading valves from the top down, with a pressure step and a minimum spacing. Add spare mandrels below the operating valve.
4
Setting
From the pressures at the valves and the temperature, calculate the nitrogen pressures and the test-rack opening pressures (setting on the test rack), and choose the port of the operating valve.
5
Checks
The order in which the valves close, the gas pressure in operation below the closing pressures, unloading in a colder well, and the operating mode after several years.
6
Surface
The gas pressure and volume for all the wells, the compressor stages, and the pressures that the lines and the wellhead equipment are designed for.

Operating mode: the point of balance and the depth of the operating valve

Gas enters the tubing where its pressure in the annulus exceeds the pressure in the tubing by the pressure drop that the operating valve needs (condition for gas injection). The tubing pressure at the injection depth depends on the operating mode. The deeper the injection, the larger the flow rate and the higher the tubing pressure at the valve, while the gas pressure grows slowly with depth. So for a given gas pressure at the wellhead and a given gas rate there is a greatest injection depth, the point of balance. Deeper than this, gas will not enter the tubing in the operating mode.

It is convenient to find the point of balance on a graph of pressure against depth. For each gas rate we draw a curve: the gas pressure needed at the valve if injection is at this depth (the tubing pressure in the operating mode plus the pressure drop across the valve). Beside it we draw lines of the gas pressure in the annulus for different pressures at the wellhead. The crossing point is the point of balance. More gas means a lighter mixture, lower tubing pressure and a deeper point of balance. A higher gas pressure at the wellhead also gives a deeper point.

CalculatedThe point of balance of the well of the examples (calculated with the teaching model of the lesson). The solid curves are the gas pressure needed at the operating valve for injection at the given depth, for 30, 50 and 70 thousand m³/d of gas. The dashed lines are the gas pressure in the annulus at 8.1, 9.1 and 10.1 MPa at the wellhead. At 9.1 MPa and 50 thousand m³/d, the gas enters the tubing no deeper than 2501 m (the dot).

Worked exampleWhere to place the operating valve

Given (assumed data)

Well: vertical; the middle of the perforated interval at 2650 m, the packer at 2615 m; 89 mm tubing (76 mm inside diameter)
Reservoir: $\bar p_R = 22.0$ MPa, $J = 14$ m³/(d·MPa); the bottomhole pressure is above the bubble point
Production: water cut 50%, producing gas–oil ratio 60 m³/m³ (so the gas–liquid ratio of the reservoir fluid is 30 m³/m³), liquid density 1000 kg/m³; tubing pressure at the wellhead 1.6 MPa
Gas: specific gravity 0.7, average temperature in the annulus 323.15 K; pressure drop across the operating valve 0.5 MPa; the field can give the well up to 50 thousand m³/d; the gas pressure at the annulus wellhead in operation is not higher than 9.1 MPa (0.5 MPa below the system pressure of 9.6 MPa, which leaves room for the steps of the unloading valves; this is checked in the next example)
Model: as in the lesson Gas lift: principles and choice of operating conditions: a drift-flux model for the tubing, and a column of gas with an average $Z$ ; the data are assumed for illustration

Solution

Static fluid level: $H_{\text{st}} = 2650 - \dfrac{(22.0 - 0.1)\cdot 10^6}{1000\cdot 9.81} = 418$ m below the wellhead. Without gas the well does not work.
Points of balance at 9.1 MPa at the wellhead (table below): with 40 thousand m³/d of gas, 2404 m and 122.2 m³/d of liquid; with 50, 2501 m and 134.1 m³/d; with 60, 2570 m and 142.6 m³/d.
Around 50 thousand m³/d, each extra thousand cubic metres of gas gives $\dfrac{142.6 - 122.2}{60 - 40} = 1.02$ m³ of liquid. This is several times more than is needed to pay for the gas: $\dfrac{20}{251.6\cdot 0.5} = 0.159$ m³ per thousand m³, at the assumed oil price and gas cost of the lesson on gas lift. The gas is worth it, and the limit is how much the field can give: 50 thousand m³/d.
The operating valve is placed 50 m above the point of balance, in the mandrel at 2450 m. The mode there is a flow rate of 131.3 m³/d and a tubing pressure at the valve of 10.657 MPa. The gas at the valve must be at a pressure of $10.657 + 0.5 = 11.157$ MPa.
The gas pressure at the wellhead from the formula for a column of gas, with an average $\bar Z = 0.8072$ : the exponent is $\dfrac{0.03418\cdot 0.7\cdot 2450}{0.8072\cdot 323.15} = 0.2247$ , and $p_{\text{g,wh}} = 11.157\,e^{-0.2247} = 8.912$ MPa. This is 0.19 MPa below the limit of 9.1 MPa.

Answer: The operating valve is in the mandrel at 2450 m, and the gas rate is 50 thousand m³/d. The well will give about 131 m³/d of liquid (66 m³/d of oil) at a gas pressure at the wellhead of 8.91 MPa.

What it means. The 50 m above the point of balance cost 2.8 m³/d of liquid (2%). But they give a margin of 0.19 MPa, for swings of the pressure in the system and for errors in the productivity index and in the flow model. If the valve were exactly at the point of balance, the slightest fall of the gas pressure or rise of the tubing pressure would move the injection to the upper valve. The mandrels deeper than 2450 m do not stand empty: blanking plugs (plugs that close the pocket) are left in them for the future (see the section on the whole life of the well). The data in this example are assumed for illustration.

Point of balance at 9.1 MPa at the annulus wellhead: the well of the example

Gas, thousand m³/d: 30
Point of balance, m: 2267
Liquid, m³/d: 105.4
Oil, m³/d: 52.7

Gas, thousand m³/d: 40
Point of balance, m: 2404
Liquid, m³/d: 122.2
Oil, m³/d: 61.1

Gas, thousand m³/d: 50
Point of balance, m: 2501
Liquid, m³/d: 134.1
Oil, m³/d: 67.1

Gas, thousand m³/d: 60
Point of balance, m: 2570
Liquid, m³/d: 142.6
Oil, m³/d: 71.3

Gas, thousand m³/d: 70
Point of balance, m: 2600 (bottom mandrel)
Liquid, m³/d: 147.4
Oil, m³/d: 73.7

Gas, thousand m³/d	Point of balance, m	Liquid, m³/d	Oil, m³/d
30	2267	105.4	52.7
40	2404	122.2	61.1
50	2501	134.1	67.1
60	2570	142.6	71.3
70	2600 (bottom mandrel)	147.4	73.7

With 70 thousand m³/d the gas would reach the bottom mandrel at 2600 m; there are no mandrels deeper. Calculated with the teaching model of the lesson; the pressure drop across the operating valve is 0.5 MPa. The data are assumed for illustration.

Spacing the unloading valves

For the gas to reach the operating valve, the kill fluid is pushed out in parts (why the gas does not reach it at once). Spacing answers the question: how deep can each next valve be placed so that the gas can still open it? Below the working valve (the valve that is passing gas at that moment), the annulus and the tubing still hold kill fluid. Above it, the tubing holds a mixture of gas and liquid. The gas will open the next valve if, at its depth, the gas pressure is higher than the pressure of this liquid.

What tubing pressure should we take at the working unloading valve? It is not known in advance: it depends on how much gas goes through the valve and what the reservoir gives at that moment. The usual choice is the design line: the pressure in the tubing at the design operating mode (the flow rate and the gas rate, with injection through the operating valve). While the gas enters higher up, the flow rate is lower than the design value, and the tubing pressure at the upper valves is usually below the line. So the calculation leaves a margin.

Greatest depth of the next unloading valve

p_{\text{t,des}}(H_n) + \rho_{\text{kill}}\,g\,(H_{n+1} - H_n) = p_{\text{g}}\big(H_{n+1};\ p_{\text{g,wh},n+1}\big) - \Delta p_{\text{margin}},\qquad p_{\text{g,wh},n} = p_{\text{ko}} - (n - 1)\,\Delta p_{\text{step}}

where:

$H_n,\ H_{n+1}$ true vertical depths of the working valve and of the next one, m
$p_{\text{t,des}}(H_n)$ tubing pressure at depth $H_n$ on the design line; for the first valve, the tubing pressure at the wellhead $p_{wh}$ at $H_0 = 0$ , MPa
$\rho_{\text{kill}}$ density of the kill fluid; $\rho_{\text{kill}} g$ is converted to MPa/m with the factor $10^{-6}$ , kg/m³
$p_{\text{g}}(H;\ p_{\text{g,wh}})$ gas pressure in the annulus at depth $H$ for a pressure $p_{\text{g,wh}}$ at the wellhead, from the formula for a column of gas (gas pressure in the annulus), MPa
$p_{\text{g,wh},n}$ gas pressure at the annulus wellhead at which valve $n$ opens, MPa
$p_{\text{ko}}$ kick-off pressure: the gas pressure at the wellhead during unloading, MPa
$\Delta p_{\text{step}}$ pressure step: the amount by which each next valve opens at a lower pressure at the wellhead, MPa
$\Delta p_{\text{margin}}$ margin: the amount by which the gas at the opening valve must exceed the pressure of the liquid, MPa

Units: metric oilfield units; pressures are absolute

Holds when

The well is vertical. (In a deviated well, true vertical depths are converted into measured depths along the hole for the mandrels.)
Below the working valve, the annulus and the tubing hold kill fluid, and the level in the tubing stays at this valve. The friction of the liquid while it is pushed out is not included, and the margin covers it.
The tubing pressure at the working valve is taken from the design line, and enough gas passes through the valve for this mode.

On the left is the pressure of the kill fluid at the depth of the next valve, and on the right is the gas pressure less the margin. The depth at which they are equal is the greatest; the valve is placed at it or higher. The heavier the kill fluid, the higher the tubing pressure at the working valve, and the larger the step and the margin, the closer the next valve is to the working one.

Why a step? When the gas has reached valve $n + 1$ and goes into the tubing through two valves, the gas pressure in the annulus falls. Valve $n$ must close before valve $n + 1$ . For this, the closing pressure of the upper valve, converted to the wellhead, must be higher than that of the lower one. The closing pressure of a valve is below its opening pressure by the difference $k_{\text{t}}(p_{\text{dome}} - p_{\text{t}})$ (the difference between the opening and closing pressures), and for the upper valves this difference is larger, because the tubing pressure there is lower. The step covers this difference and the errors of setting. It is chosen from the characteristics of the valves. The step is larger when the valve is more sensitive to the tubing pressure and when the data are poorer. For ordinary valves operated by the gas pressure, designs use steps of about 0.14–0.34 MPa (20–50 psi). In the example of this lesson a smaller one is taken, 0.1 MPa. The check of the data sheet below shows that it is enough for these valves, but with a small margin. The price of the step is pressure: in operation the gas must be below the closing pressures of all the unloading valves, that is, below $p_{\text{ko}} - N\,\Delta p_{\text{step}}$ with $N$ unloading valves.

The margin $\Delta p_{\text{margin}}$ covers errors in the density of the kill fluid and in the depths, friction while the fluid is pushed out, and the pressure drop needed for the gas to pass through the seat.
Minimum spacing. Valves are not placed closer than a certain distance. The pressure difference between close valves becomes smaller than the errors of the calculation, and the gas starts to go through two valves. The spacing depends on the accuracy of the data: the less well the pressures, the temperatures and the kill fluid are known, the larger the spacing must be. In the example of this lesson, 75 m is taken.
Rounding. The mandrels sit between pipe joints, and their depths are fitted to the lengths of the pipes. A valve can be moved only up the wellbore: gas will not open it deeper than the greatest depth.
From the bottom up. With depth the valves come together: the tubing pressure on the design line grows faster than the gas pressure. When a spacing comes out smaller than the minimum, the lower valves are moved up, starting from the operating valve, and we check again that the gas will open each next one.

Worked exampleUnloading valves for the well of the previous example

Given (assumed data)

Operating valve and mode: mandrel at 2450 m, 50 thousand m³/d of gas, 131.3 m³/d of liquid; the design line is the tubing pressure in this mode
Unloading: the well is filled up to the wellhead with brine of density 1030 kg/m³, and the tubing is open to the flowline (1.6 MPa); the kick-off gas pressure at the annulus wellhead is the system pressure, 9.6 MPa
Design rules (assumed): pressure step 0.1 MPa, margin 0.15 MPa, minimum spacing 75 m, mandrel depths rounded up the wellbore to 5 m

Solution

Gradient of the kill fluid: $1030\cdot 9.81\cdot 10^{-6} = 0.010104$ MPa/m.
Valve 1: $1.6 + 0.010104\,H = 9.6\,e^{H/10{,}910} - 0.15$ , where $10{,}910 = \dfrac{0.8078\cdot 323.15}{0.03418\cdot 0.7}$ m is the height over which the gas pressure grows by a factor of $e$ , with an average $\bar Z = 0.8078$ . By trial, $H = 854.3$ m, and both sides are about 10.23 MPa. The valve is placed at 850 m.
Valve 2: at 850 m the tubing pressure on the design line is 4.188 MPa. $4.188 + 0.010104\,(H - 850) = 9.5\,e^{H/10{,}873} - 0.15$ with $\bar Z = 0.8050$ . By trial, $H = 1500.0$ m, and both sides are about 10.76 MPa. The valve is placed at 1495 m.
Valves 3, 4 and 5 are found in the same way (gas at the wellhead 9.4, 9.3 and 9.2 MPa). The greatest depths are 1948.0, 2232.8 and 2397.6 m, and after rounding 1945, 2230 and 2395 m.
From valve 5, gas at 9.1 MPa at the wellhead would open the operating valve. But the distance between them is only $2450 - 2395 = 55$ m, which is less than 75 m. Valve 5 is moved up to $2450 - 75 = 2375$ m. Check: from 2375 m (tubing pressure 10.303 MPa) the gas at 9.1 MPa will push the liquid out down to 2471 m, which is deeper than the operating valve.

Answer: Five unloading valves at 850, 1495, 1945, 2230 and 2375 m, and the operating valve at 2450 m. The spacings are 850, 645, 450, 285, 145 and 75 m. In operation the gas pressure at the wellhead must be below

9.6 - 5\cdot 0.1 = 9.1

MPa, as was assumed in the first example.

What it means. The first valve is set by the kick-off pressure and the kill fluid, and the next ones by the design line: the higher the tubing pressure at the working valve, the shorter the next step. The spacings shrink with depth, and near the operating valve the rule of minimum spacing has to step in. If six unloading valves were needed, the limit of the gas pressure in operation would be 9.0 MPa, and the operating valve would have to be moved up. So the depth of the operating valve and the number of unloading valves are found together, sometimes in several passes. The data in this example are assumed for illustration.

CalculatedSpacing of the valves on a graph of pressure against depth (calculated with the teaching model of the lesson). The solid curve is the tubing pressure in operation: above the operating valve it is the design line, below it is a column of liquid. The dashed segments are the kill fluid below the wellhead and below each unloading valve. The grey dashed lines are the gas in the annulus during unloading, at 9.6 MPa at the wellhead (opening of valve 1) and at 9.1 MPa (operating valve). The orange line is the gas in operation, 8.91 MPa at the wellhead. On the left is the tubing string: the numbers are the unloading valves, O is the operating valve, and the empty squares are spare mandrels.

Spacing of the valves: greatest depths and the depths chosen

Valve, depth: 1, 850 m
Greatest depth, m: 854.3
Spacing, m: 850
Gas at the wellhead at opening, MPa: 9.6
Tubing pressure on the design line, MPa: 4.188

Valve, depth: 2, 1495 m
Greatest depth, m: 1500.0
Spacing, m: 645
Gas at the wellhead at opening, MPa: 9.5
Tubing pressure on the design line, MPa: 6.521

Valve, depth: 3, 1945 m
Greatest depth, m: 1948.0
Spacing, m: 450
Gas at the wellhead at opening, MPa: 9.4
Tubing pressure on the design line, MPa: 8.368

Valve, depth: 4, 2230 m
Greatest depth, m: 2232.8
Spacing, m: 285
Gas at the wellhead at opening, MPa: 9.3
Tubing pressure on the design line, MPa: 9.632

Valve, depth: 5, 2375 m
Greatest depth, m: 2397.6
Spacing, m: 145
Gas at the wellhead at opening, MPa: 9.2
Tubing pressure on the design line, MPa: 10.303

Valve, depth: operating, 2450 m
Greatest depth, m: 2471.2
Spacing, m: 75
Gas at the wellhead at opening, MPa: 9.1 during unloading
Tubing pressure on the design line, MPa: 10.657

Valve, depth	Greatest depth, m	Spacing, m	Gas at the wellhead at opening, MPa	Tubing pressure on the design line, MPa
1, 850 m	854.3	850	9.6	4.188
2, 1495 m	1500.0	645	9.5	6.521
3, 1945 m	1948.0	450	9.4	8.368
4, 2230 m	2232.8	285	9.3	9.632
5, 2375 m	2397.6	145	9.2	10.303
operating, 2450 m	2471.2	75	9.1 during unloading	10.657

The greatest depth of each valve is found from the previous valve in the position chosen for it. Valve 5 is moved up from 2395 to 2375 m to keep the 75 m spacing to the operating valve. Below the operating valve there are spare mandrels at 2525 and 2600 m with blanking plugs. The data are assumed for illustration.

The data sheet of the installation and design checks

When the depths are known, a seat is chosen for each unloading valve and its opening pressure is set. The seat must pass enough gas that the tubing pressure at the working valve does not exceed the design line. Otherwise the next valve would turn out to be deeper than the gas can open it. We take the smallest seat that does this. For the opening pressure: at valve $n$ the gas at the wellhead is $p_{\text{g,wh},n}$ , and the tubing pressure comes from the design line. The force balance gives the nitrogen pressure in the dome at the temperature of the valve, which is also the closing pressure at the valve. The temperature correction then gives the test-rack opening pressure (force balance, setting on the test rack). The valve temperatures are taken on a straight line from the flow temperature at the wellhead at the design flow rate to the reservoir temperature: in the example, from 35 °C to 88 °C at 2650 m.

Data sheet of the gas-lift installation of the well of the examples

Valve: depth; seat: 1: 850 m; 4.8 mm
Opening at the wellhead, MPa: 9.60
Nitrogen = closing at the valve, MPa: 10.154
Closing at the wellhead, MPa: 9.395
Opening on the rack, MPa gauge: 9.141

Valve: depth; seat: 2: 1495 m; 4.8 mm
Opening at the wellhead, MPa: 9.50
Nitrogen = closing at the valve, MPa: 10.742
Closing at the wellhead, MPa: 9.365
Opening on the rack, MPa gauge: 9.271

Valve: depth; seat: 3: 1945 m; 6.4 mm
Opening at the wellhead, MPa: 9.40
Nitrogen = closing at the valve, MPa: 11.060
Closing at the wellhead, MPa: 9.250
Opening on the rack, MPa gauge: 9.555

Valve: depth; seat: 4: 2230 m; 6.4 mm
Opening at the wellhead, MPa: 9.30
Nitrogen = closing at the valve, MPa: 11.308
Closing at the wellhead, MPa: 9.209
Opening on the rack, MPa gauge: 9.598

Valve: depth; seat: 5: 2375 m; 7.9 mm
Opening at the wellhead, MPa: 9.20
Nitrogen = closing at the valve, MPa: 11.339
Closing at the wellhead, MPa: 9.113
Opening on the rack, MPa gauge: 9.895

Valve: depth; seat: operating: 2450 m; port 8.9 mm
Opening at the wellhead, MPa: 8.91 (in operation)
Nitrogen = closing at the valve, MPa: —
Closing at the wellhead, MPa: —
Opening on the rack, MPa gauge: —

Valve: depth; seat: 2525 and 2600 m
Opening at the wellhead, MPa: —
Nitrogen = closing at the valve, MPa: blanking plugs
Closing at the wellhead, MPa: —
Opening on the rack, MPa gauge: —

Valve: depth; seat	Opening at the wellhead, MPa	Nitrogen = closing at the valve, MPa	Closing at the wellhead, MPa	Opening on the rack, MPa gauge
1: 850 m; 4.8 mm	9.60	10.154	9.395	9.141
2: 1495 m; 4.8 mm	9.50	10.742	9.365	9.271
3: 1945 m; 6.4 mm	9.40	11.060	9.250	9.555
4: 2230 m; 6.4 mm	9.30	11.308	9.209	9.598
5: 2375 m; 7.9 mm	9.20	11.339	9.113	9.895
operating: 2450 m; port 8.9 mm	8.91 (in operation)	—	—	—
2525 and 2600 m	—	blanking plugs	—	—

The unloading valves are bellows valves operated by the gas pressure, with a bellows of 5.0 cm². For seats of 4.8, 6.4 and 7.9 mm, $R$ = 0.0362; 0.0643; 0.0980. The valve temperatures are 52.0; 64.9; 73.9; 79.6; 82.5 °C, and the factors $C_T$ are 0.8764; 0.8397; 0.8158; 0.8014; 0.7942. The test-rack opening pressure is $C_T\,(p_{\text{dome}} - 0.101)/(1 - R)$ , rounded to 0.001 MPa. The port of the operating valve passes 50 thousand m³/d at a pressure drop of 0.5 MPa and 84.0 °C (gas rate through a port). The data are assumed for illustration.

Order of closing. The closing pressures, converted to the wellhead, fall from the top down: 9.395; 9.365; 9.250; 9.209; 9.113 MPa. So when gas goes through two neighbouring valves and the pressure falls, the upper valve closes first. And if the pressure falls below both, the lower one will open again before the upper one, because its opening pressure is lower too.
The first valves are the tightest spot. At valve 1, the difference between the opening and closing pressures is 0.22 MPa at the valve, because the tubing pressure there is low, 4.19 MPa. Converted to the wellhead, the differences for valves 1 and 2 are 0.205 and 0.135 MPa. So the window between their closing pressures is only $0.1 - (0.205 - 0.135) = 0.030$ MPa. A step smaller than 0.07 MPa would break the order of closing.
Pressure in operation. We need 8.91 MPa at the wellhead, which is 0.20 MPa below the closing pressure of the lowest unloading valve. In operation all of them are closed, and the gas enters only through the operating valve.
Seats. For the tubing pressure at the working valve not to exceed the design line, valves 2–5 must pass 27; 37; 44; 48 thousand m³/d of gas, and the deeper the valve, the more. This rule gives no answer for valve 1: with gas injected at 850 m, the well starts to work only at 20 thousand m³/d, and the tubing pressure at the valve is then already below the design line. The smallest seat is enough for it. The 4.8 mm seat at its opening pressure passes 33; 33; 31; 26; 21 thousand m³/d. This is enough only for valves 1 and 2. Valves 3 and 4 need 6.4 mm, and valve 5 needs 7.9 mm. With a smaller seat unloading would stop: there would not be enough gas, the tubing pressure at the valve would rise above the design line, and the gas would not reach the next valve. A larger seat increases $R$ and the difference between the opening and closing pressures, so the order of closing is checked with the seats already chosen.
Unloading in a cold well. During unloading the flow rate is low and the well is colder than in operation. The nitrogen in the domes is colder, and the valves open and close at a lower gas pressure. Unloading is also checked at lower valve temperatures. If the margins are not enough, the opening pressures are corrected (which temperature to use).

Pressure in the gas system and the compressor

The gas pressure at the wellheads is set when the compressor station and the gas pipelines are designed, and afterwards it is hardly ever changed. The higher it is, the deeper the operating valve and the fewer unloading valves are needed: the top valve goes deeper, and the steps eat a smaller share of the pressure. The compressor, however, must deliver the wellhead pressure plus the losses in the gas pipeline and at the control choke, and the equipment must withstand it with a margin.

Worked exampleWhat one extra megapascal in the system gives

Given (assumed data)

Well and design rules: from the previous examples; 50 thousand m³/d of gas; the operating valve is 50 m above the point of balance (rounded up to 25 m); the limit of the pressure in operation is the system pressure minus one step for each unloading valve
System pressure at the wellhead during unloading: 8.6, 9.6 or 10.6 MPa
Compressor: suction (inlet) pressure 0.7 MPa, discharge (outlet) pressure equal to the wellhead pressure plus 0.4 MPa of losses in the gas pipeline; it uses $1/0.65$ times as much as ideal isothermal compression, as in the lesson on gas lift; electricity at an assumed US$0.06 per kWh

Solution

The design is repeated in full for each pressure (table below). At 8.6 MPa: 5 unloading valves, the operating valve at 2175 m, 115.0 m³/d of liquid. At 9.6: 5 valves, 2450 m, 131.3 m³/d. At 10.6: 4 valves, 2550 m, 136.8 m³/d.
Compression power for 50 thousand m³/d: $N = \dfrac{1.173\cdot 50\cdot\ln(p_{\text{disch}}/0.7)}{0.65}$ , which is 230.4, 239.9 and 248.5 kW at a discharge pressure of 9.0, 10.0 and 11.0 MPa.
From 8.6 to 9.6 MPa: there is more oil by $(131.3 - 115.0)\cdot 0.5 = 8.15$ m³/d, which is $8.15\cdot 251.6 = 2051$ US$/d. Electricity costs more by $(239.9 - 230.4)\cdot 24\cdot 0.06 = 13.7$ US$/d.
From 9.6 to 10.6 MPa: there is more oil by $(136.8 - 131.3)\cdot 0.5 = 2.75$ m³/d (692 US$/d), and energy costs more by $(248.5 - 239.9)\cdot 24\cdot 0.06 = 12.4$ US$/d. The gain in oil is 3.0 times smaller, because the gas has already reached almost the bottom of the well.

Answer: Each megapascal of system pressure costs 12.4–13.7 US$ of electricity per day per well, and in this example it gives from 692 to 2051 US$ of extra oil, as long as the operating valve has not reached the bottom of the well.

What it means. The energy of compression grows as a logarithm of the pressure and hardly depends on whether the outlet is at 9 or 11 MPa. The system pressure is limited by other things: the cost of the compressor (compression stages, drive), the pressures that the gas pipelines, the gas distribution manifold, the wellhead equipment and the casing are designed for, and safety. So the pressure is chosen for the whole field: how many wells gain from deeper injection, and how much the equipment for this pressure costs. The oil price is the assumed price of the library's example project, and the data in this example are assumed for illustration.

The design at different pressures of the gas system: the well of the examples, 50 thousand m³/d of gas

System pressure, MPa: 8.6
Unloading valves: 5
Operating valve, m: 2175
Liquid, m³/d: 115.0
Compression, kW: 230.4

System pressure, MPa: 9.6
Unloading valves: 5
Operating valve, m: 2450
Liquid, m³/d: 131.3
Compression, kW: 239.9

System pressure, MPa: 10.6
Unloading valves: 4
Operating valve, m: 2550
Liquid, m³/d: 136.8
Compression, kW: 248.5

System pressure, MPa	Unloading valves	Operating valve, m	Liquid, m³/d	Compression, kW
8.6	5	2175	115.0	230.4
9.6	5	2450	131.3	239.9
10.6	4	2550	136.8	248.5

The limit of the gas pressure in operation is 8.1; 9.1; 10.2 MPa respectively. In operation, 7.91; 8.91; 9.28 MPa is needed at the wellhead. At 10.6 MPa the point of balance is at the bottom mandrel, and the operating valve is placed 50 m above it. The data are assumed for illustration.

The requirements for the compressor are added up over all the wells. The volume is the sum of the design gas rates, plus gas for starting wells after shutdowns, and for losses. The pressure is the highest that the wells need during unloading, plus the losses in the gas pipelines. From the suction pressure to the discharge pressure, the gas is compressed in several stages, with cooling between them. The compression ratio (the outlet pressure divided by the inlet pressure) of one stage of a piston compressor is usually not more than three to four, and the number of stages is chosen so that the gas at the outlet of each stage does not overheat. For 10 MPa with a suction pressure of 0.7 MPa, the total compression ratio is $10/0.7 \approx 14.3$ . Two stages of $\sqrt{14.3} \approx 3.8$ each are at the upper limit, and three of $\sqrt[3]{14.3} \approx 2.4$ each leave a margin. The calculation of the discharge temperature decides (discharge temperature). How compressors are built is explained in the same lesson.

Design for the whole life: falling pressure and a rising water cut

A well is designed for years. The reservoir pressure falls and the water cut grows: the liquid gets heavier, there is less reservoir gas, and the flow rate decreases. To keep up the oil rate, more gas is given, and the point of balance goes deeper. The fall of the reservoir pressure helps too, because the tubing pressure at the valve falls. The injection can be moved deeper with wireline tools, without pulling the tubing (changing valves), but only into a mandrel that is already there. So below the operating valve, all the way down to the packer (the seal at the bottom of the annulus), spare mandrels (also called bracketing mandrels) with blanking plugs are installed.

Two conditions limit how deep the injection can be moved. In operation, the gas must enter through the new mandrel at a wellhead pressure below the closing pressures of the unloading valves. And at each start-up, the unloading valves must bring the gas to this mandrel. With the same steps, each extra unloading valve lowers the allowed pressure in operation by one more step. When the injection is moved, the valves are rearranged and reset for the new conditions, with the same calculation as in the design.

Worked exampleThe well after eight years: what spare mandrels give

Given (assumed data)

Installation: from the previous examples: unloading valves at 850, 1495, 1945, 2230 and 2375 m, the operating valve at 2450 m, spare mandrels at 2525 and 2600 m; kick-off pressure 9.6 MPa, step 0.1 MPa, margin 0.15 MPa, brine 1030 kg/m³
Forecast (assumed): reservoir pressure 22.0 → 21.0 → 20.0 → 19.0 → 18.0 MPa and water cut 50 → 60 → 68 → 75 → 80% after 0, 2, 4, 6 and 8 years; gas 50 thousand m³/d, and 70 from the fourth year, when the gas is reallocated between the wells of the field (Gas lift operation and optimisation)
Liquid: oil 860 and water 1140 kg/m³, so the density of the mixture grows with the water cut; the productivity index is the same

Solution

For each year and each mandrel we check the unloading (the unloading valves go into the deepest mandrels that the gas will reach; the seats follow the rule of the data sheet, and the closing pressures fall from the top down) and the operating mode (the wellhead pressure is below the closing pressure of the lowest unloading valve). See the table below.
For the first two years the operating valve stays at 2450 m. With 50 thousand m³/d, gas into the mandrel at 2525 m would need 9.19 and 9.22 MPa at the wellhead, which is above the limit of 9.1 MPa.
In the fourth year, at 70 thousand m³/d, in operation the gas would enter even the bottom mandrel, which needs 9.03 MPa at the wellhead. But the unloading valves will bring it only to 2525 m: from the mandrel at 2525 m the gas pushes the brine out only down to 2573 m. The operating valve is moved to 2525 m, and a sixth unloading valve is placed in the mandrel at 2450 m. The well gives 114.4 m³/d of liquid, against 109.5 m³/d if the injection had stayed at 2450 m.
By the eighth year the pressure in the reservoir and in the tubing is lower, and during unloading the gas reaches 2600 m. The well gives 93.7 m³/d of liquid instead of 83.5, and 18.74 m³/d of oil instead of 16.70 (+12.2%).

Answer: Spare mandrels let us move the injection deeper from the fourth year, to 2525 and then to 2600 m, and keep 4.5–12.2% more oil than with injection at 2450 m.

What it means. The gain from moving the injection is small in each year, but it adds up over the years and costs only a few wireline runs. The example also shows something that is easy to miss: the injection depth is limited not only by the gas pressure in operation but also by unloading. In the fourth year the gas pressure in operation would be enough to use the mandrel at 2600 m, but unloading with steps of 0.1 MPa could not reach it. Only unloading with nitrogen or a recalculation of the steps would open it, not a simple rearranging of the valves. The data in this example are assumed for illustration.

The operating valve by year: the well of the examples

Year: 0
Reservoir: pressure, MPa; water cut, %: 22.0; 50
Gas, thousand m³/d: 50
Operating valve, m (unloading valves): 2450 (5)
Liquid, m³/d: with spare mandrels / only 2450 m: 131.3 / 131.3

Year: 2
Reservoir: pressure, MPa; water cut, %: 21.0; 60
Gas, thousand m³/d: 50
Operating valve, m (unloading valves): 2450 (5)
Liquid, m³/d: with spare mandrels / only 2450 m: 116.1 / 116.1

Year: 4
Reservoir: pressure, MPa; water cut, %: 20.0; 68
Gas, thousand m³/d: 70
Operating valve, m (unloading valves): 2525 (6)
Liquid, m³/d: with spare mandrels / only 2450 m: 114.4 / 109.5

Year: 6
Reservoir: pressure, MPa; water cut, %: 19.0; 75
Gas, thousand m³/d: 70
Operating valve, m (unloading valves): 2525 (5)
Liquid, m³/d: with spare mandrels / only 2450 m: 101.1 / 96.1

Year: 8
Reservoir: pressure, MPa; water cut, %: 18.0; 80
Gas, thousand m³/d: 70
Operating valve, m (unloading valves): 2600 (6)
Liquid, m³/d: with spare mandrels / only 2450 m: 93.7 / 83.5

Year	Reservoir: pressure, MPa; water cut, %	Gas, thousand m³/d	Operating valve, m (unloading valves)	Liquid, m³/d: with spare mandrels / only 2450 m
0	22.0; 50	50	2450 (5)	131.3 / 131.3
2	21.0; 60	50	2450 (5)	116.1 / 116.1
4	20.0; 68	70	2525 (6)	114.4 / 109.5
6	19.0; 75	70	2525 (5)	101.1 / 96.1
8	18.0; 80	70	2600 (6)	93.7 / 83.5

The unloading valves are reset each time to the conditions of the year, and the seats are chosen by the rule of the data sheet (the upper valves, like valve 1 in the data sheet, need only the smallest seat). Mandrels that unloading does not use get blanking plugs. In the fourth and eighth years the new lowest unloading valve (at 2450 and 2525 m) needs about 65 thousand m³/d of gas, but a 7.9 mm seat passes only 59 and 55. So it gets a seat of 9.5 mm of the same valve (passing 85 and 80), and the order of closing and the margin to the pressure in operation are kept with it. The density of the liquid follows the water cut, and the productivity index is kept constant (a simplification). The data are assumed for illustration.

CalculatedLiquid rate of the well of the examples by year (calculated with the teaching model of the lesson): with the operating valve moved into the spare mandrels, and with injection only through the mandrel at 2450 m. The labels show the mandrel through which the gas enters.

Intermittent gas lift and the plunger: what changes in the design

For low-rate wells, intermittent gas lift is designed (continuous and intermittent gas lift). The task is different: not to lighten the flow, but to push out a slug of liquid quickly. This leads to differences in the design. The operating valve is placed as deep as possible, almost at the packer, with a check valve below it so that the gas does not push the slug back into the reservoir. The valve must have a large flow area so that the gas enters the tubing quickly. These are valves with a large seat, pilot valves (two-stage valves that snap fully open), or valves operated by the tubing pressure, which open when a slug has collected above them (valves operated by the tubing pressure). Gas is supplied at the wellhead either by the timer of an intermitter (a timer-controlled valve on the gas line) or through a choke, and the valve itself sets the moment of ejection.

When the reservoir pressure is very low, an accumulation chamber, a section of larger diameter, is placed at the bottom of the tubing. The same slug then takes less height and presses less on the reservoir. And so that the gas does not break through the slug and part of the liquid does not run back, a plunger, a free piston, is placed in the tubing. It needs a bumper spring (a spring stop) at the bottom, and a catcher (which holds the plunger at the top) and an arrival sensor (which detects the plunger) at the wellhead. The tubing must have the same inside diameter, with no narrowings, along the whole length of the plunger's travel.

Worked exampleWhat a plunger gives a low-rate well

Given (assumed data)

Well and cycle: from the example of the lesson on intermittent gas lift: operating valve at 1750 m, 73 mm tubing; a slug is ejected when there are 300 m of liquid above the valve; 15 min for injection, lifting and bleeding; a quarter of the slug falls back
With a plunger (assumed): only 5% of the slug falls back; the plunger falls to the bottom while the next slug builds up, so the 15 min for lifting do not change; the same gas per cycle is needed, because it fills the same tubing

Solution

Without a plunger (the example of the lesson on gas lift): 14.7 cycles per day of 0.679 m³ each, which is 10.0 m³/d, with 2.70 thousand m³/d of gas and 271 m³ per cubic metre of liquid.
With a plunger, a cycle lifts out $0.95\cdot 300\cdot 0.003019 = 0.8604$ m³. But the next slug grows from 15 m, not from 75 m: $t = 5.4712\cdot 60\cdot\ln\dfrac{1083 - 15}{1083 - 300} = 101.9$ min ( $\tau$ in hours; 1083 m is the height to which the liquid would rise).
A cycle lasts $101.9 + 15 = 116.9$ min, or $1440/116.9 = 12.32$ cycles per day. The liquid is $12.32\cdot 0.8604 = 10.6$ m³/d.
The gas per cycle is the same, 184 m³: $12.32\cdot 184 = 2.27$ thousand m³/d, and the injection gas–liquid ratio is 214 m³/m³.

Answer: A plunger adds about 6% of liquid and cuts the gas used per cubic metre of liquid by 21%.

What it means. The main gain is in gas: with no fallback, each cubic metre of gas lifts more liquid. The flow rate grows less, because there are fewer cycles per day: the slug has to be built up almost from zero. The plunger must have time to fall to the bottom while the slug builds up. Here it has 101.9 min for this. The fall time depends on the design of the plunger, and it is measured with an echo sounder (a device that times the echo of a sound pulse) that follows the falling plunger. The travel of the plunger also scrapes wax off the tubing walls. The share of fallback with and without a plunger is assumed. The data in this example are assumed for illustration.

How the cycle is chosen and tuned on a working well is explained in the lesson Gas lift operation and optimisation. Plunger lift without gas injection, using the energy of the gas of the well itself, is used in gas wells where liquid accumulates (plunger lift).

Terms and symbols

Point of balance: The greatest depth at which gas, at a given wellhead pressure and gas rate, can still enter the tubing in the operating mode.
Design tubing pressure line: The pressure in the tubing at the design operating mode. The depths and settings of the unloading valves are calculated from it.
Gas-lift valve spacing: The calculation of the depths of the unloading valves from the top down, with a pressure step, a margin and a minimum spacing.
Pressure step $\Delta p_{\text{step}}$: The amount by which each next unloading valve opens at a lower gas pressure at the wellhead.
Margin $\Delta p_{\text{margin}}$: The excess of the gas pressure over the pressure of the kill fluid at the valve that is opening.
Kick-off pressure: The gas pressure at the annulus wellhead during unloading. It sets the depth of the top unloading valve.
Spare mandrels: Side-pocket mandrels below the operating valve with blanking plugs (also called bracketing mandrels). They let us move the gas injection deeper without pulling the tubing.
Data sheet of the installation: The depths of the mandrels, the types, seats and setting pressures of the valves, the port of the operating valve, and the design operating mode.

Common mistakes and misconceptions

Wrong

“Put the operating valve at the point of balance: that is the deepest.”

In fact

Without a margin, the slightest fall of the gas pressure or rise of the tubing pressure moves the injection to the upper valve. In the example of this lesson, the 50 m of margin cost 2.8 m³/d of liquid and give 0.19 MPa of margin in pressure.

Wrong

“We will set all the unloading valves to the same opening pressure, so that they all open in the same way.”

In fact

Then, when the gas moves to the lower valve, the upper valve will not close. The gas pressure will not fall below its closing pressure while it is passing gas itself. The pressure step from valve to valve provides the order of closing.

Wrong

“The calculated depth of the valve is 1500 m, so we will put the mandrel at the nearest pipe joint, 1505 m.”

In fact

Gas will not open a valve deeper than the greatest depth, because the pressure of the liquid there is already higher than the gas pressure. Mandrels are fitted to the pipe joints only upward along the wellbore.

Wrong

“We will design for today's data and recalculate later.”

In fact

The setting of the valves can be recalculated, but the depths of the mandrels can be changed only by pulling the tubing. In the example of this lesson, the spare mandrels below the operating valve gave 4.5–12.2% more oil from the fourth to the eighth year.

Wrong

“It does not pay to raise the pressure in the gas system: the compressor will eat the whole gain.”

In fact

The energy of compression grows as a logarithm of the pressure. In the example of this lesson an extra megapascal costs 12–14 US$ of electricity per day per well. What decides is the cost of the equipment and the pressures that it is designed for.

In depth: advanced detail

Spacing by hand: constant gradients

If we treat the gas pressure in the annulus as growing along a straight line with a constant gradient $g_{\text{g}}$ , and take the design line from the graph, the depth of the next valve is found without trial. The gas pressure at depth $H_{n+1}$ is $p_{\text{g,wh},n+1} + g_{\text{g}}H_{n+1}$ . Setting it equal, less the margin, to the pressure of the kill fluid, we get:

Spacing to the next valve with constant gradients

H_{n+1} - H_n = \frac{p_{\text{g,wh},n+1} + g_{\text{g}}H_n - \Delta p_{\text{margin}} - p_{\text{t,des}}(H_n)}{\rho_{\text{kill}}\,g - g_{\text{g}}}

where:

$g_{\text{g}}$ pressure gradient of the gas in the annulus, taken as constant, MPa/m
$\rho_{\text{kill}}\,g$ gradient of the kill fluid, MPa/m
$p_{\text{g,wh},n+1},\ \Delta p_{\text{margin}},\ p_{\text{t,des}}(H_n)$ as in the equation for the greatest depth of the next valve, MPa

Units: metric oilfield units; pressures are absolute

Holds when

The gas gradient is constant. It is taken from the formula for a column of gas at the average pressure over the interval, so the error is small.
For the first valve $H_0 = 0$ and $p_{\text{t,des}}(0) = p_{wh}$ .

The numerator shows how much the gas at the working valve is “stronger” than the tubing pressure. The denominator shows how much faster the liquid gains pressure than the gas. This shows why the spacings shrink: with depth the design line catches up with the gas line, and the numerator gets smaller.

Worked exampleChecking the two upper valves by hand

Given (assumed data)

Data: from the spacing example; the gas gradient is taken at the wellhead pressure: $g_{\text{g}} = p_{\text{g,wh}}\cdot 0.03418\cdot 0.7/(Z\cdot 323.15)$ , where $Z$ is taken at the wellhead pressure

Solution

Valve 1: at 9.6 MPa, $Z = 0.8131$ and $g_{\text{g}} = \dfrac{9.6\cdot 0.03418\cdot 0.7}{0.8131\cdot 323.15} = 0.000874$ MPa/m. $H_1 = \dfrac{9.6 - 0.15 - 1.6}{0.010104 - 0.000874} = 850.5$ m.
Valve 2: at 9.5 MPa, $Z = 0.8146$ and $g_{\text{g}} = 0.000863$ MPa/m. $H_2 - 850 = \dfrac{9.5 + 0.000863\cdot 850 - 0.15 - 4.188}{0.010104 - 0.000863} = 638.0$ m, so $H_2 = 1488.0$ m.

Answer: The hand calculation gives 850.5 and 1488.0 m against 854.3 and 1500.0 m from the formula for a column of gas: 4 and 12 m higher.

What it means. The gas gradient grows with pressure, and so with depth. Taken at the wellhead pressure, it underestimates the gas pressure at the bottom, and the valve comes out a little higher, which is an error on the safe side. But if we put in a gradient chosen for another gas pressure, the error can also be on the unsafe side. For each 10% of error in the gradient, the top valve moves by about 8 m, and the shifts of the lower valves add up. The data in this example are assumed for illustration.

Valves operated by the tubing pressure: spacing by a variable gradient

A valve operated by the tubing pressure opens under the pressure in the tubing, and the gas pressure hardly acts on it (opening pressure). So it does not need steps of gas pressure at the wellhead: the order of opening and closing is set by the tubing pressures at the valves. Such valves are spaced along a line of tubing pressure that is higher than the working one. The line starts at a wellhead pressure that is raised above the working one by a part of the difference between the gas pressure and the working tubing pressure (the designer chooses this part). It ends at the expected tubing pressure at the injection depth. Such a line places the valves closer together, and more of them are needed, but the gas in operation does not lose pressure on steps.

Spacing by a variable gradient is also used as a cautious option for valves operated by the gas pressure, when the data about the well are unreliable: there are more valves, but unloading is more reliable. In a well with two tubing strings sharing one annulus, valves operated by the gas pressure are put in one string, and valves operated by the tubing pressure in the other, so that all the gas does not go into one of them (Gas lift operation and optimisation).

Connections to other topics

Gas lift: principles and choice of operating conditions — the operating point, the gas pressure in the annulus, the performance curve of the well, and intermittent gas lift.
Unloading a gas lift well: valves and mandrels — unloading, how valves are built, the force balance and setting on the test rack: the basis of the data sheet of the installation.
Gas lift operation and optimisation — how the installation works in practice: measurements, problems, and the sharing of gas across the field.
Completion string equipment — side-pocket mandrels, the packer and other equipment of the tubing string.
Wireline operations — changing and moving gas-lift valves on wireline.
Multiphase flow in tubing — the pressure in the tubing with gas, which is the basis of the design line.
Chokes, surface pressure losses and well temperature — the temperature of the flow in the well, with which valves are set.
Gas processing and compression — compressors, compression stages and power.

Check yourself

1
Why is each next unloading valve operated by the gas pressure set to open at a lower gas pressure at the wellhead?
1. A.So that, when the gas moves to the lower valve, the upper valve closes first, and so that in operation the gas pressure is below the closing pressures of all the unloading valves
2. B.Because a deeper valve is hotter and opens at a lower pressure by itself
3. C.So that the operating valve opens first during unloading
4. D.So that less gas is used at the wellhead
Show the answer
Correct answer: A. So that, when the gas moves to the lower valve, the upper valve closes first, and so that in operation the gas pressure is below the closing pressures of all the unloading valves
The closing pressure of a valve is below its opening pressure, and for the upper valves the difference is larger. The step from valve to valve makes the closing pressures, converted to the wellhead, fall from the top down. When the gas has started to go through two valves and the pressure falls, the upper valve closes first. The price is pressure: with $N$ unloading valves the gas in operation must be below $p_{\text{ko}} - N\,\Delta p_{\text{step}}$ . The temperature of the valves is taken into account when setting them on the test rack, and it does not replace the step.
2
Unloading valve 1 is at 905 m, and the tubing pressure there on the design line is 4.6 MPa. The kill fluid is 1200 kg/m³, valve 2 must open at 11.3 MPa at the wellhead, the margin is 0.2 MPa, and the gas gradient in the annulus is 1.0 kPa/m (constant). What is the greatest depth at which valve 2 can be placed?
Show the answer
The gradient of the liquid is $1200\cdot 9.81\cdot 10^{-6} = 0.01177$ MPa/m. The spacing is $\dfrac{11.3 + 0.0010\cdot 905 - 0.2 - 4.6}{0.01177 - 0.0010} = \dfrac{7.405}{0.01077} \approx 688$ m. So valve 2 can be placed no deeper than about 1593 m. It is placed at the nearest pipe joint above this depth.
3
A designer has placed the operating valve exactly at the point of balance for the gas pressure in operation. What happens if the pressure in the gas system falls by 0.1 MPa?
1. A.The flow rate will grow, because there will be less gas in the tubing
2. B.The well will stop flowing at once and will never start again
3. C.The gas will not be able to enter through the operating valve in the same mode. The pressure in the annulus will rise until the upper unloading valve opens, and the injection will move higher up
4. D.Nothing: the operating valve is an orifice, and it passes gas at any pressure
Show the answer
Correct answer: C. The gas will not be able to enter through the operating valve in the same mode. The pressure in the annulus will rise until the upper unloading valve opens, and the injection will move higher up
At the point of balance the gas pressure at the operating valve is exactly equal to the tubing pressure plus the needed pressure drop. With no margin, the gas pressure will not be enough to pass the same rate. The annulus will keep “pumping up” until the nearest unloading valve opens, and the gas will go in higher up. The flow rate will fall, and operation through two valves is possible. So the operating valve is placed above the point of balance.
4
The kick-off pressure at the wellhead is 10.0 MPa, the step is 0.12 MPa, and the spacing gave five unloading valves. At the design gas rate, the operating valve needs 9.55 MPa at the wellhead. Is the design acceptable, and what can be done?
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The limit of the pressure in operation is $10.0 - 5\cdot 0.12 = 9.40$ MPa. At 9.55 MPa the lower unloading valves will not close, and the gas will go through several valves. The options are to raise the operating valve (it needs a lower pressure there), to use a smaller step if the valves allow it, to reduce the number of unloading valves by using a lighter kill fluid or a higher kick-off pressure, or to give more gas, which moves the point of balance deeper. Each option is checked again by spacing.
5
Why are mandrels with blanking plugs installed below the operating valve, all the way down to the packer?
1. A.They are needed only for measuring pressure
2. B.So that gas enters the tubing in several places at once
3. C.To strengthen the tubing
4. D.So that in the future the gas injection can be moved deeper with wireline tools, without pulling the tubing
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Correct answer: D. So that in the future the gas injection can be moved deeper with wireline tools, without pulling the tubing
As the reservoir pressure falls and the gas rate grows, the point of balance goes deeper. The depths of mandrels can be changed only by pulling the tubing, while the valves in existing mandrels can be changed on wireline in a day. In the example of this lesson the injection was moved to 2525 and then to 2600 m. Each time, unloading is also checked, because the unloading valves must bring the gas to the new mandrel.
6
In intermittent gas lift a quarter of the slug of liquid falls back, and with a plunger 5%. How will the gas per cubic metre of liquid lifted change, if the same gas is needed per cycle and the slug size is the same?
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The gas per cubic metre is inversely proportional to the part of the slug that is lifted: $0.75/0.95 \approx 0.79$ , so the gas per cubic metre falls by about 21%. The same came out in the example of this lesson. There the number of cycles per day became smaller, but the gas per cycle and the lifted part of the slug do not depend on the number of cycles.

Updated: 18 October 2026