Gas lift: principles and choice of operating conditions

Quick refresher

Gas injected into the tubing makes the mixture lighter above the injection point. As the gas rises, it expands and pushes the liquid up in slugs (separate plugs of liquid). Below the injection point the tubing holds liquid, so we inject the gas as deep as possible.
The gas pressure sets the injection depth. At the operating valve (the valve through which the gas enters the tubing), the gas pressure must be higher than the pressure in the tubing by the pressure drop across the valve. In the light column of gas, the pressure rises slowly with depth. In the tubing it rises fast. So each injection pressure has its own greatest injection depth.
The curve of flow rate against gas rate first rises steeply, then flattens and passes a maximum, the technical optimum. Beyond it, extra gas adds more friction than it takes weight off the column. The economic optimum is to the left of it: the last thousand cubic metres of gas must pay for themselves with the extra oil.
When there is not enough gas for all the wells, we share it so that the marginal return (the extra oil from the next thousand cubic metres of gas) is the same in every well that receives gas.
At low flow rates, continuous gas lift wastes gas, because the gas slips through the slow liquid. Intermittent gas lift lets liquid collect above the valve and then pushes it out as one slug.
Gas lift uses a lot of energy: only 10–30% of the compression energy goes into lifting the liquid. But the equipment in the well is simple. It tolerates gas, sand and a curved wellbore, and it works over a wide range of flow rates.

Key relationships

p_{wf} = p_{wh} + \Delta p_{\text{above}} + \Delta p_{\text{below}}

bottomhole pressure: the wellhead pressure plus the pressure losses in the tubing above and below the gas injection point

q = J\,(\bar p_R - p_{wf})

straight-line inflow performance relationship (IPR, the curve of rate against bottomhole pressure); the operating point is where it crosses the outflow curve

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

pressure in the column of gas in the annulus (the space between the tubing and the casing); pressures are absolute

p_{\text{g}}(H_{\text{inj}}) - \Delta p_{\text{valve}} \ge p_{\text{t}}(H_{\text{inj}})

condition for injecting gas at depth $H_{\text{inj}}$

\dfrac{\dd q_o}{\dd q_{\text{g}}} = \dfrac{c_{\text{g}}}{v_o}

economic optimum of the gas rate of a single well

\dfrac{\dd q_{o,i}}{\dd q_{\text{g},i}} = \beta

sharing of limited gas: equal marginal returns

Common pitfalls

Giving a well “a bit more gas just in case”. Beyond the technical optimum the rate falls, and long before it the extra gas no longer pays for itself.
Sharing the gas equally between wells, or giving all of it to the best well, instead of equalising the marginal returns.
Taking the gas pressure at the valve to be equal to the wellhead pressure. The column of gas in the annulus has a noticeable weight: in the example of this lesson it adds almost a quarter to the wellhead pressure.
Putting the gauge pressure from the pressure gauge (pressure above the atmosphere) into gas calculations instead of the absolute pressure (pressure above zero).
Expecting a low bottomhole pressure from gas lift. It lightens only the column above the injection point, and only up to a limit, so the bottomhole pressure stays higher than with a pump.

After this lesson you can

explain how injected gas makes the column of fluid in the tubing lighter, and why gas lift cannot lower the bottomhole pressure as far as a pump at the same depth can
find the operating point of a gas-lift well (the flow rate and bottomhole pressure for a given injection depth and gas rate) and the gas pressure needed at the wellhead
explain how the injection pressure sets the depth at which gas can be injected, and why we try to inject as deep as possible
draw and read the curve of flow rate against gas rate: the minimum gas rate to start the well, and the technical and economic optima
share a limited volume of gas between wells by the rule of equal marginal returns (the extra oil from one more thousand cubic metres of gas)
choose between continuous and intermittent gas lift, and estimate the energy that goes into compressing the gas

How gas lifts liquid

A well flows by itself as long as the reservoir pressure is enough to hold up the column of fluid in the tubing, overcome friction and still leave some pressure at the wellhead (when a well needs artificial lift). As the water cut (the share of water in the produced liquid) grows and the reservoir pressure falls, the column becomes too heavy. In a naturally flowing well, the gas that comes from the reservoir makes the column lighter (Multiphase flow in tubing). Gas lift adds gas from the surface, exactly as much as is needed.

Gas comes from a compressor along a high-pressure pipeline. At the wellhead it passes through a control choke (an adjustable restriction) and a flow meter. It goes into the annulus, the ring-shaped gap between the tubing and the casing (the large pipe that lines the well). At the bottom, a packer (a seal) closes the annulus. So the gas pressure does not reach the reservoir, and the produced fluid does not enter the annulus. The tubing carries gas-lift valves in side-pocket mandrels. A side-pocket mandrel is a thickened section of tubing with a pocket at the side. A wireline crew (a crew that works with tools on a steel wire) can pull the valve out of the pocket. The gas enters the tubing through the lowest open valve, the operating valve. The upper valves, the unloading valves, work only at start-up. At that time the annulus and the tubing are full of kill fluid (a heavy fluid that was put in the well to stop it flowing). Through the unloading valves the gas pushes this fluid out stage by stage, until it reaches the operating valve. The lesson Unloading a gas lift well: valves and mandrels explains how the valves are built and how the well is unloaded at start-up.

SchematicA gas-lift well (diagram, not to scale): gas is fed into the annulus and enters the tubing through the operating valve. Above the valve a mixture of gas and liquid rises; below it the tubing holds liquid.

It helps to look at what the gas does from two sides. Weight. The gas takes up part of the cross-section of the pipe, so there is less liquid in it. The average density of the mixture above the injection point falls by a factor of several. The reservoir has to hold up the weight of exactly this column. Energy. The gas is injected compressed. As it rises, the pressure around it gets lower and it expands, and this speeds up the mixture. The energy of compression turns into the work of lifting. With enough gas, the bubbles merge into large ones that fill the whole width of the pipe. They push the plugs of liquid between them upward, and the flow becomes slug flow (flow regimes).

Below the injection point gas lift changes nothing. From the valve to the bottom, the tubing holds the same liquid, and the whole weight of its column presses on the reservoir. This gives the main rule: inject the gas as deep as possible. There is a second consequence. The mixture can be made lighter only up to a limit, because friction grows as the gas rate grows (the outflow curve and its minimum). A pump takes the weight of the whole column above it off the reservoir. Gas lift only lightens part of the column. So with gas lift the bottomhole pressure is higher than with a pump at the same depth (how a pump and gas lift change the pressure).

Bottomhole pressure of a gas-lift well

p_{wf} = p_{wh} + \Delta p_{\text{above}}\!\left(q,\ \text{GLR}_{\text{res}} + \frac{q_{\text{g}}}{q}\right) + \Delta p_{\text{below}}(q)

where:

$p_{wf}$ bottomhole pressure at the middle of the perforated interval, MPa
$p_{wh}$ pressure in the tubing at the wellhead (tubing head pressure), MPa
$\Delta p_{\text{above}}$ pressure difference in the tubing from the gas injection point up to the wellhead: the weight of the mixture and friction, found from a multiphase flow calculation (gas and liquid flowing together), MPa
$\text{GLR}_{\text{res}}$ gas–liquid ratio of the fluid produced from the reservoir: the producing gas–oil ratio multiplied by the share of oil in the liquid, m³/m³
$q_{\text{g}}$ rate of injected gas at standard conditions (a fixed reference pressure and temperature, so that gas volumes can be compared), m³/d
$q$ liquid flow rate, m³/d
$\Delta p_{\text{below}}$ pressure difference from the bottom to the injection point: the weight of the liquid and friction, MPa

Units: metric oilfield units; pressures are absolute, gas volumes are at standard conditions

Holds when

Steady state; all the injected gas enters the tubing through the operating valve.
Gas from the reservoir is added to the injected gas wherever it is free. Below the injection point the pressure is usually above the bubble point (the pressure at which gas starts to come out of the oil), so the gas is dissolved in the oil.
The pressure differences are calculated with correlations or models of multiphase flow (Multiphase flow in tubing). The examples in this lesson use a simple teaching drift-flux model (a model in which the gas drifts upward through the liquid at its own speed).

Together with the inflow performance relationship (IPR) $q = J(\bar p_R - p_{wf})$ , where $J$ is the productivity index and $\bar p_R$ the average reservoir pressure, this equation gives the operating point: the flow rate at which the reservoir delivers as much as gas lift can lift. Gas injection reduces only $\Delta p_{\text{above}}$ . The column below the injection point stays the same.

Lift gas can come from different sources. In compressor gas lift, the gas is separated from the production in a separator (a vessel that splits the production into gas and liquid), compressed at a compressor station and sent back to the wells. It goes round in a closed loop. In gas lift without compressors, we use gas that is already at high pressure, such as gas from gas wells or from a main gas pipeline. There is also an in-well version: gas from a gas-bearing layer that the same well penetrates flows through a regulator straight into the tubing.

Usually the gas goes down the annulus and the mixture rises up the tubing (in Russian-language terminology this is the “annular system”). At very high rates the opposite is sometimes done: the gas goes down the tubing and the mixture rises up the annulus (the “central system”). The annulus has a larger cross-section, so there is less friction. But the casing wears and corrodes, and when the rate falls, heading (pulsations of the flow) starts earlier.

The operating point of a gas-lift well

The operating point is found in the same way as for a naturally flowing well (Nodal analysis of the production system). The node (the point where the inflow and outflow pressures are matched) is at the bottom of the well. The inflow curve is the inflow performance relationship (IPR). The outflow curve is the pressure needed to lift the flow up to the wellhead. There is one difference: the outflow curve is calculated in two parts. From the wellhead down to the injection point, the fluid carries the reservoir gas and the injected gas. From the injection point down to the bottom, there is no injected gas. Each gas rate has its own outflow curve.

How to find the operating point

1
Input data
The depth of the operating valve, the gas rate, the wellhead pressure, the IPR and the fluid properties. All pressures are absolute.
2
Section above the valve
For a trial flow rate $q$ , calculate the pressure in the tubing from the wellhead down to the valve. Use the gas–liquid ratio $\text{GLR}_{\text{res}} + q_{\text{g}}/q$ .
3
Section below the valve
Continue the calculation down to the bottom with the reservoir gas (or with no gas, if the pressure is above the bubble point).
4
Comparison with the reservoir
Compare the pressure you get with the IPR, and change $q$ until the pressures match. If there are two crossing points, the working one is the right-hand one (stability of the operating point).
5
Check of the gas pressure
From the tubing pressure at the valve, find the gas pressure that is needed in the annulus at this depth and at the wellhead (next section).

All the pressures in this lesson are absolute (measured from zero pressure, not from atmospheric pressure). This is essential in gas calculations. A pressure gauge at the wellhead shows the gauge pressure, which is 0.1 MPa lower.

Worked exampleA well that has stopped flowing, restarted on gas lift: flow rate and gas pressure

Given (assumed data)

Well: vertical; the middle of the perforated interval at a depth of 2500 m, the operating valve at 2400 m, above the packer; 89 mm tubing (76 mm inside diameter)
Reservoir: $\bar p_R = 21.0$ MPa, $J = 30$ m³/(d·MPa); the bottomhole pressure is above the bubble point
Production: water cut 40%, producing gas–oil ratio 40 m³/m³ (so the gas–liquid ratio of the reservoir fluid is 24 m³/m³), liquid density 920 kg/m³; tubing pressure at the wellhead 1.2 MPa
Gas: 60 thousand m³/d, specific gravity 0.7 (air = 1); pressure drop across the operating valve 0.5 MPa; average temperature in the well 50 °C (323 K)
Flow model in the tubing: the drift-flux model from the lesson Multiphase flow in tubing: $C_0 = 1.2$ , $\lambda = 0.022$ , $Z = 0.9$ ( $Z$ is the gas deviation factor, which shows how far the real gas is from an ideal gas)

Solution

Static fluid level (the level at which the liquid stands when the well is shut in, that is, closed and not producing): $H_{\text{st}} = 2500 - \dfrac{(21.0 - 0.1)\cdot 10^6}{920\cdot 9.81} = 184$ m below the wellhead. Without gas the well does not flow. The inflow and outflow curves do not cross, even if we treat all the reservoir gas as free.
We look for the flow rate at which the pressures match. At $q = 273.78$ m³/d the gas–liquid ratio above the valve is $24 + 60{,}000/273.78 = 243$ m³/m³, and a calculation from the wellhead downward gives 10.965 MPa at the valve.
Below the valve there are 100 m of liquid: $920\cdot 9.81\cdot 100\cdot 10^{-6} = 0.903$ MPa, and friction is 0.006 MPa. The bottomhole pressure is $10.965 + 0.903 + 0.006 = 11.874$ MPa.
Check against the IPR: $21.0 - 273.78/30 = 11.874$ MPa. This is the same pressure, so the operating point is found.
The gas pressure at the valve must be at least $10.965 + 0.5 = 11.465$ MPa. The gas in the annulus is much lighter than the liquid, about 97 kg/m³, so less pressure is needed at the wellhead. By the formula for a column of gas (next section), with an average $Z = 0.803$ : $p_{\text{g,wh}} = 11.465\exp\!\left(-\dfrac{0.03418\cdot 0.7\cdot 2400}{0.803\cdot 323}\right) = 9.19$ MPa.
Oil: $273.78\cdot 0.6 = 164$ m³/d. The injection gas–liquid ratio (the volume of injected gas per cubic metre of liquid) is $60{,}000/273.78 = 219$ m³ of gas per cubic metre of liquid.

Answer: With 60 thousand m³/d of gas through the valve at 2400 m, the well produces about 274 m³/d of liquid (164 m³/d of oil) at a bottomhole pressure of 11.9 MPa. The gas must be supplied to the annulus at a pressure of about 9.19 MPa.

What it means. The drawdown (the difference between the reservoir pressure and the bottomhole pressure) is 9.1 MPa, although the reservoir pressure is 21 MPa. Gas lift made the column above the valve lighter. But the 100 m of liquid below it and the weight of the mixture above it still leave almost 12 MPa at the bottom. A pump at the same depth could bring the bottomhole pressure down to a few megapascals. The model is a teaching model. We treated the reservoir gas as free over the whole height above the valve, although it comes out of solution only where the pressure is below the bubble point. This makes the flow rate a little too high. The temperature, $Z$ in the tubing and the friction factor are constant. (With a $Z$ that changes with pressure, the flow rate would come out about 1% lower still.) In the field, the calculation is made with a correlation that has been checked against measurements. The data in this example are assumed for illustration.

CalculatedPressure against depth in the well of the example. A — the shut-in well: the liquid stands up to a static level of about 184 m, and the pressure at the bottom is the reservoir pressure of 21 MPa. B — gas lift, 60 thousand m³/d of gas through the valve at 2400 m: the pressure in the tubing rises gently from the wellhead, and steeply below the valve, following the weight of the liquid. C — the gas in the annulus: from 9.19 MPa at the wellhead to 11.46 MPa at the valve. Pressures are absolute; calculated with the teaching model of the lesson.

Injection pressure and the depth of gas injection

Gas can be injected where the pressure in the annulus is higher than the pressure in the tubing. The gas pressure rises with depth because of the gas's own weight, but gas is light. At 10 MPa and 50 °C, gas of specific gravity 0.7 has a density of about 93 kg/m³. This is about 10 times lower than the density of the liquid. So in the annulus the pressure rises by about 0.9 kPa for every metre. In the tubing it rises by a few kPa/m above the valve and by 9 kPa/m below it. Where these lines meet, the gas cannot go any deeper.

Gas pressure in the annulus at a given depth

p_{\text{g}}(H) = p_{\text{g,wh}}\,\exp\!\left(\frac{M g H}{\bar Z R \bar T}\right)

where:

$p_{\text{g}}(H)$ gas pressure in the annulus at depth $H$ , absolute, Pa
$p_{\text{g,wh}}$ gas pressure in the annulus at the wellhead, absolute, Pa
$M$ molar mass of the gas, $0.02897\,\gamma_{\text{g}}$ , where $\gamma_{\text{g}}$ is the specific gravity of the gas relative to air, kg/mol
$g$ acceleration due to gravity, 9.81 m/s², m/s²
$H$ true vertical depth, m
$\bar Z$ gas deviation factor (Z-factor) at the average pressure and the average temperature of the column, —
$R$ universal gas constant, 8.314 J/(mol·K), J/(mol·K)
$\bar T$ average absolute temperature of the gas in the annulus, K

Units: SI; pressures are absolute

Holds when

The gas in the annulus is almost still. Its velocity is low, so friction is negligible.
The temperature and $Z$ are taken as averages over the column. $Z$ depends on the average pressure, which is itself being found, so the calculation is repeated two or three times.
The same gas fills the whole column. The formula is the same as for any column of gas (column of gas).

The gas pressure rises with depth along an exponential curve, but the exponent is small, so the curve is almost a straight line. In the example of this lesson, at 2400 m the pressure is a quarter higher than at the wellhead. The heavier the gas and the higher the pressure, the bigger this increase. Even in a rough calculation, it must not be dropped, and it must not be found from the density of the liquid.

In other unit systems (2)

Metric oilfield units

p_{\text{g}}(H) = p_{\text{g,wh}}\exp\!\left(\frac{0.03418\,\gamma_{\text{g}}\,H\,[\text{m}]}{\bar Z\,\bar T\,[\text{K}]}\right)

$H$ in m, $\bar T$ in K; pressures in any consistent units, absolute.

$0.03418 = 0.02897\cdot 9.81/8.314$ K/m: the molar mass of air multiplied by $g$ and divided by $R$ .

Oilfield units

p_{\text{g}}(H) = p_{\text{g,wh}}\exp\!\left(\frac{0.01875\,\gamma_{\text{g}}\,H\,[\text{ft}]}{\bar Z\,\bar T\,[^{\circ}\text{R}]}\right)

$H$ in feet, $\bar T$ in degrees Rankine (°F + 460); pressures in psia.

$0.01875 = 28.97/(144\cdot 10.73)$ °R/ft: the molar mass of air, the gas constant 10.73 psia·ft³/(lb-mol·°R) and 144 in² in a square foot.

Condition for injecting gas at a given depth

p_{\text{g}}(H_{\text{inj}}) - \Delta p_{\text{valve}} \ge p_{\text{t}}(H_{\text{inj}})

where:

$H_{\text{inj}}$ depth of gas injection, which is the depth of the operating valve, m
$p_{\text{g}}(H_{\text{inj}})$ gas pressure in the annulus at this depth, MPa
$\Delta p_{\text{valve}}$ pressure drop across the operating valve that is needed to pass the given gas rate, MPa
$p_{\text{t}}(H_{\text{inj}})$ pressure in the tubing at this depth in the working mode, MPa

Units: any consistent units; the pressures are all absolute or all gauge

Holds when

Steady state; gas enters the tubing only through the operating valve.
The pressure drop across the valve depends on the gas rate and on the size of the valve port (its opening). How it is calculated is explained in the lesson Unloading a gas lift well: valves and mandrels.

The equality gives the greatest depth that the gas can reach at a given gas pressure at the wellhead. The pressure in the tubing at the injection depth itself depends on where the gas is injected and how much the well produces. So we find this depth by trial: assume a depth, find the operating point and compare the pressures.

Why not inject the gas right at the bottom? The injection pressure that this needs is chosen when the compressor station is designed, and a higher pressure is expensive: compressors, pipes and wellhead equipment all cost more. Besides, at start-up the gas must push the kill fluid out of the annulus, and this needs a higher pressure than normal operation. So we place unloading valves above the operating valve, and through them the fluid is pushed out in parts. Or we take a compressor with spare pressure, as in subsea wells that have a single valve (Unloading a gas lift well: valves and mandrels).

Worked exampleDeeper means more: what the injection pressure gives

Given (assumed data)

Well and gas: as in the previous example, 60 thousand m³/d of gas
Compressor: gives 7.0 MPa at the annulus wellhead today; after a replacement, as much as is needed
Pressure drop across the operating valve: 0.5 MPa

Solution

For several injection depths we find the operating point, the tubing pressure at the injection depth and the gas pressure needed at the wellhead (table below). The needed gas pressure grows with the injection depth, and so does the flow rate.
With 7.0 MPa at the wellhead, the gas reaches 1850 m. There the tubing pressure is 7.73 MPa, and the gas has $8.23 = 7.73 + 0.5$ MPa. The well produces 221 m³/d.
To inject the gas through a valve at 2400 m, we need 9.19 MPa at the wellhead (previous example). The flow rate is 274 m³/d. That is 53 m³/d (24%) more at the same gas rate.
On average, every 100 m of injection depth adds $53/550\cdot 100 = 9.6$ m³/d.

Answer: At 7.0 MPa the gas enters the tubing only at a depth of about 1850 m, and the well produces 221 m³/d. A compressor giving 9.19 MPa lets us inject gas at 2400 m and gives 274 m³/d, which is 24% more without any extra gas.

What it means. The injection pressure is chosen when the compressor station is designed, and the depths of the side-pocket mandrels are chosen when the tubing is run. It is expensive to change them later. So we plan them for the future: the reservoir pressure will fall, the water cut will rise, and the pressure in the tubing at the same depth will change. Several mandrels around the design depth let us move the operating valve up or down without pulling the tubing (Gas lift design). The data in this example are assumed for illustration.

Injection depth and injection pressure: 60 thousand m³/d of gas, the well of the example

Injection depth, m: 1200
Liquid rate, m³/d: 139
Tubing pressure at the valve, MPa: 4.62
Needed at the annulus wellhead, MPa: 4.64

Injection depth, m: 1600
Liquid rate, m³/d: 192
Tubing pressure at the valve, MPa: 6.44
Needed at the annulus wellhead, MPa: 6.05

Injection depth, m: 2000
Liquid rate, m³/d: 237
Tubing pressure at the valve, MPa: 8.56
Needed at the annulus wellhead, MPa: 7.59

Injection depth, m: 2400
Liquid rate, m³/d: 274
Tubing pressure at the valve, MPa: 10.96
Needed at the annulus wellhead, MPa: 9.19

Injection depth, m	Liquid rate, m³/d	Tubing pressure at the valve, MPa	Needed at the annulus wellhead, MPa
1200	139	4.62	4.64
1600	192	6.44	6.05
2000	237	8.56	7.59
2400	274	10.96	9.19

Pressure drop across the valve 0.5 MPa; pressures are absolute. The pressure needed at the wellhead grows with the injection depth more slowly than the tubing pressure at the valve. Part of the increase is carried by the weight of the gas column. The gas column adds 10% to the wellhead pressure for injection at 1200 m, and 25% for injection at 2400 m. The data are assumed for illustration.

Flow rate and gas rate: the gas lift performance curve

If we calculate the operating point for a series of gas rates, we get the main curve of gas lift: the flow rate against the gas rate. It is called the gas lift performance curve. We use it to choose the operating mode, and from several such curves we share the gas between wells.

CalculatedLiquid rate of the well of the examples against the gas rate, for injection through a valve at 2400 m and at 1800 m (calculated with the teaching model of the lesson). The technical optimum for injection at 2400 m (92.8 thousand m³/d, 281 m³/d) and the economic optimum (80.7 thousand m³/d) are marked. For injection at 1800 m the maximum is 218 m³/d.

Start-up. A well that has stopped flowing needs some minimum gas rate to start working at all. For the wells in this lesson, in the teaching model, it is about 1–3 thousand m³/d. A real start-up needs more, because first the gas has to push out the kill fluid.
Steep rise. The first thousands of cubic metres make the column lighter the most. Each thousand adds tens of cubic metres of liquid.
Flattening. The more gas there is, the smaller the gain from the next thousand: the column is already light, and friction grows. The injection gas–liquid ratio (cubic metres of gas per cubic metre of liquid) increases.
Technical optimum (also called maximum production): the top of the curve. Here the lightening of the column by extra gas is exactly balanced by the growth of friction.
Beyond the top the gas only does harm. Friction grows faster than the weight of the mixture falls, and the flow rate decreases.

Why does the curve have a top? Let us see what the pressure difference in the tubing above the valve is made of, at a constant flow rate, as the gas rate grows.

Weight of the mixture and friction in the tubing above the valve: flow rate 281 m³/d, valve at 2400 m

Gas, thousand m³/d: 20
Weight of the mixture, MPa: 12.29
Friction, MPa: 0.56
Total, MPa: 12.85

Gas, thousand m³/d: 40
Weight of the mixture, MPa: 9.77
Friction, MPa: 0.95
Total, MPa: 10.72

Gas, thousand m³/d: 60
Weight of the mixture, MPa: 8.52
Friction, MPa: 1.35
Total, MPa: 9.87

Gas, thousand m³/d: 93
Weight of the mixture, MPa: 7.51
Friction, MPa: 2.03
Total, MPa: 9.54

Gas, thousand m³/d: 120
Weight of the mixture, MPa: 7.08
Friction, MPa: 2.58
Total, MPa: 9.66

Gas, thousand m³/d: 160
Weight of the mixture, MPa: 6.74
Friction, MPa: 3.39
Total, MPa: 10.13

Gas, thousand m³/d: 200
Weight of the mixture, MPa: 6.56
Friction, MPa: 4.20
Total, MPa: 10.76

Gas, thousand m³/d	Weight of the mixture, MPa	Friction, MPa	Total, MPa
20	12.29	0.56	12.85
40	9.77	0.95	10.72
60	8.52	1.35	9.87
93	7.51	2.03	9.54
120	7.08	2.58	9.66
160	6.74	3.39	10.13
200	6.56	4.20	10.76

The pressure difference from the wellhead to the valve, without the wellhead pressure; acceleration is not included. The weight falls more and more slowly, the friction grows almost linearly, and the sum passes through a minimum at about 90 thousand m³/d. Calculated with the teaching model; the data are assumed for illustration.

Condition for the technical optimum

\left(\frac{\partial \Delta p_{\text{weight}}}{\partial q_{\text{g}}}\right)_{q} + \left(\frac{\partial \Delta p_{\text{fr}}}{\partial q_{\text{g}}}\right)_{q} = 0

where:

$\Delta p_{\text{weight}}$ the part of the pressure difference in the tubing above the valve that comes from the weight of the mixture, MPa
$\Delta p_{\text{fr}}$ the part of the pressure difference that comes from friction, MPa
$q_{\text{g}}$ rate of injected gas, thousand m³/d
$(\cdot)_{q}$ derivative at a constant liquid flow rate, —

Units: any consistent units

Holds when

The acceleration of the mixture is not taken into account. The column below the valve does not depend on the gas rate.
The flow rate is at its maximum when extra gas stops lowering the pressure that is needed to lift this flow rate (the derivation is in the advanced section).

The first term is negative: gas makes the mixture lighter. The second is positive: gas speeds up the flow, so friction grows. While the lightening is stronger, gas lowers the bottomhole pressure and increases the flow rate. At the optimum the two are equal.

It is usually not worth running at the technical optimum. Near the top the curve is flat, so the last thousands of cubic metres of gas give almost nothing, but compressing them costs money. The economic optimum is the gas rate at which the value of the oil produced thanks to the last thousand cubic metres of gas equals the cost of that gas. The injected gas comes back with the production and is compressed again. So its cost is first of all the energy for compression, the maintenance of the compressors and the treatment of the gas. If the gas is bought, or could be sold, its price is included as well.

Economic optimum of the gas rate

\frac{\dd q_o}{\dd q_{\text{g}}} = \frac{c_{\text{g}}}{v_o}

where:

$q_o$ oil rate, $q\,(1 - \text{WC})$ , where $\text{WC}$ is the water cut (the share of water in the liquid), m³/d
$q_{\text{g}}$ rate of injected gas, thousand m³/d
$c_{\text{g}}$ cost of a thousand cubic metres of lift gas: compression, drying and maintenance, and for bought gas its price as well, US$/thousand m³
$v_o$ net value of a cubic metre of oil: the price less the payments taken from revenue and the variable costs, US$/m³

Units: any consistent units: the ratio of the costs is in the same units as the slope of the curve

Holds when

The oil price and the cost of gas do not depend on the volumes.
The gas supply of the system is not limited. If it is limited, the gas is shared between wells (next section).
The limits of the surface facilities on water and gas are not taken into account. They move the optimum to the left.

This is the condition for the maximum of the daily income $v_o q_o - c_{\text{g}} q_{\text{g}}$ . The more expensive the gas and the cheaper the oil, the further to the left the optimum lies. With free gas it would coincide with the technical optimum.

Worked exampleTechnical and economic optimum

Given (assumed data)

Well: from the first example, valve at 2400 m, water cut 40%
Oil: net value 251.6 US$/m³: the assumed price in the library's example project, less the payment taken from revenue and the variable costs
Gas: compression, drying and delivery: an assumed US$20 per thousand m³

Solution

On the performance curve of the well (figure above), the flow rate is highest at 92.8 thousand m³/d of gas: 280.8 m³/d, with an injection gas–liquid ratio of 330 m³/m³. This is the technical optimum.
Extra gas pays for itself as long as each thousand cubic metres gives more than US$20 worth of oil. So the slope of the curve must be at least $\dfrac{20}{251.6\cdot 0.6} = 0.1325$ m³ of liquid per thousand m³ (oil is 60% of the liquid).
The slope falls to this value at 80.7 thousand m³/d: the flow rate is 280.1 m³/d and the injection gas–liquid ratio is 288 m³/m³. This is the economic optimum.
Compared with the technical optimum, there is 12.08 thousand m³/d (13%) less gas and 0.753 m³/d less liquid. The saving on gas is $12.08\cdot 20 = 242$ US$/d, and the oil not produced is worth $0.753\cdot 0.6\cdot 251.6 = 114$ US$/d.

Answer: It pays better to give the well about 80.7 thousand m³/d of gas rather than 92.8. The flow rate is almost the same, and the gain is about US$128 per day.

What it means. The top of the curve is flat. Near the technical optimum the gas does almost nothing useful. This is exactly the gas that is better given to other wells when there is not enough. With expensive gas (for example, bought gas), the economic optimum moves to the left. With expensive oil, it moves to the right. The data in this example are assumed for illustration.

When there is not enough gas: sharing it between wells

In a field, one compressor station supplies the gas for all the gas-lift wells, and its capacity is limited. Usually the sum of the technical optima of all the wells, and often even of the economic optima, is more than the gas available. Then the question is no longer “how much gas should we give a well”, but “how should we share the gas to produce the most oil”.

The curves of the wells give the answer. Each next portion of gas is best given to the well where it yields the most oil. This is the well with the highest marginal return (also called marginal oil per unit of lift gas, or marginal gas utilisation). The marginal return is the increase in oil rate for an extra thousand cubic metres of gas, which is the slope of the curve. As the gas rate grows, the marginal return of each well falls. When the gas is shared in the best possible way, the marginal returns of all the wells that receive gas are equal. If one well has a higher return, moving some gas to it from another well will add oil.

The best sharing of gas between wells

\frac{\dd q_{o,1}}{\dd q_{\text{g},1}} = \frac{\dd q_{o,2}}{\dd q_{\text{g},2}} = \dots = \beta, \qquad \sum_i q_{\text{g},i} = Q_{\text{g}}

where:

$q_{o,i}$ oil rate of well number $i$ , m³/d
$q_{\text{g},i}$ gas rate given to well number $i$ , thousand m³/d
$\beta$ the common marginal return of gas: how much oil one more thousand cubic metres of gas would give, m³ of oil per thousand m³
$Q_{\text{g}}$ all the gas that the system can supply, thousand m³/d

Units: any consistent units

Holds when

The curves of the wells bulge upward (they are concave): each next portion of gas gives less than the previous one.
The wells do not affect each other: the pressure in the gathering system does not change with the way the gas is shared.
A well whose marginal return is below $\beta$ even at a small gas rate gets no gas.
If a well needs a minimum gas rate to start, the simple rule breaks down. We then compare the options with and without that well (below).

$\beta$ is the price of gas expressed in oil. Multiplied by the value of oil, it shows how much one more thousand cubic metres of gas would bring in. If this is more than the cost of compression, extra gas would pay for itself, which is an argument for enlarging the compressor station. If $\beta$ is less than $c_{\text{g}}/v_o$ , there is a surplus of gas, and each well is given its economic optimum.

CalculatedThe marginal return of gas, the increase in oil rate per thousand cubic metres of gas, for the two wells of the example below (calculated with the teaching model of the lesson). The horizontal line at 0.37 m³ per thousand m³ crosses the curves where the gas is shared in the best way: 57 thousand m³/d to well A and 23 to well B, 80 together. The dashed line is the pay-off threshold of the gas,

c_{\text{g}}/v_o

, at which each well would stop if there were enough gas.

Worked exampleTwo wells and one compressor

Given (assumed data)

Well A: from the first example: water cut 40%, technical optimum 93 thousand m³/d of gas
Well B: $\bar p_R = 18.5$ MPa, $J = 12$ m³/(d·MPa), perforations at 2300 m, valve at 2200 m, water cut 75%, liquid 980 kg/m³, the same tubing and gas; technical optimum 87 thousand m³/d
Gas: the compressor supplies 80 thousand m³/d to the two wells together, which is less than half of the sum of their technical optima

Solution

Equal shares, 40 thousand m³/d each: the oil is $153.85 + 26.27 = 180.12$ m³/d. The marginal return is then 0.800 for A and 0.143 for B, in m³ of oil per thousand m³. In well B the gas works 5.6 times worse, so it pays to move part of it to A.
All the gas to well A: the oil is 168.0 m³/d, less than with equal shares. At 80 thousand m³/d the marginal return of A has fallen to 0.09, while the first thousands of cubic metres in well B would give much more.
We move gas from the well with the lower marginal return to the well with the higher one until the returns are equal. A gets 56.5 and B gets 23.5 thousand m³/d, and both have 0.37 m³ of oil per thousand m³. The oil is $163.1 + 22.4 = 185.5$ m³/d.
The price of gas in this sharing is $0.37\cdot 251.6 = 93$ US$ for each extra thousand cubic metres. This is much more than the cost of compression, US$20.

Answer: It is best to give well A about 56.5 thousand m³/d of gas and well B 23.5. This gives 5.4 m³/d more oil than equal shares, and 17.5 m³/d more than giving everything to one well.

What it means. The gain from sharing the gas correctly is a few per cent of production, with no new spending. So on large gas-lift fields the sharing is recalculated all the time, as the water cut and the pressures change, with software that links the models of wells, gathering systems and compressors. And a marginal return of gas that is several times the cost of compressing it is an argument for enlarging the compressor station. The data in this example are assumed for illustration.

In a field the same rule is applied to dozens of wells at once. The curves come from models tuned to measurements, and the sharing is found by software that also takes into account the limits of the gathering and treatment system. For a well that needs a minimum start-up gas rate, the first portions of gas give nothing. We assess it together with this minimum and compare the options with and without it (advanced section). How this is done in practice is explained in the lesson Gas lift operation and optimisation.

Continuous and intermittent gas lift

Everything so far has been about continuous gas lift: gas enters the tubing all the time, and the flow in the pipe is steady. It works well as long as the well gives enough liquid. At low flow rates the mixture in the tubing moves slowly, at a speed comparable to the speed at which bubbles rise, and the gas slips through the liquid without lifting it (slip). The liquid stays behind in the tubing, the column stays heavy, and the gas is wasted.

Intermittent gas lift works in cycles. While no gas is supplied, liquid from the reservoir collects in the tubing above the operating valve. A check valve (a one-way valve) at the bottom stops it from being pushed back into the reservoir. Then gas is let into the tubing quickly and in large amounts. This is done on a signal from an intermitter (a timer-controlled valve on the surface gas line) or through a valve with a large port. The gas throws the collected column up to the wellhead in one slug, like a piston. Part of the liquid runs back down along the walls and through the gas. These are the fallback losses. The gas is bled from the tubing into the line, and the cycle starts again.

The gas used per cycle grows more slowly than the size of the slug. The gas fills the tubing above the valve, so the amount is set by the volume of the pipe and the pressure in it, and this pressure depends only in part on the weight of the slug. This leads to the main choice in tuning. Large slugs save gas. But while the column grows, it presses on the reservoir, and the inflow weakens. Small slugs let the reservoir work at a low pressure, but they need frequent cycles and more gas.

Worked exampleA low-rate well: continuous or intermittent gas lift

Given (assumed data)

Well: the middle of the perforations at 1800 m, the operating valve at 1750 m, a check valve below it; 73 mm tubing (62 mm inside diameter, cross-section 0.003019 m²)
Reservoir and production: $\bar p_R = 11.0$ MPa, $J = 1.5$ m³/(d·MPa); liquid 900 kg/m³, water cut 30%; wellhead pressure 1.0 MPa
Continuous gas lift: by the model of the lesson, with 8 thousand m³/d of gas: 6.0 m³/d
Intermittent gas lift (assumed): a slug is ejected when there are 300 m of liquid above the valve; a quarter of the slug falls back and stays in the tubing; 15 min for injection, lifting and bleeding; the average gas pressure in the tubing at the end of the ejection is 3.5 MPa, $Z = 0.9$ , temperature 323 K

Solution

Continuous gas lift: at the valve the mixture moves at 0.50 m/s, which is similar to the speed at which bubbles rise, 0.26 m/s. The liquid fills 45% of the cross-section, while its share of the flow rate is 5%. The gas shoots through it. The injection gas–liquid ratio is 1331 m³/m³.
Intermittent: while there is no liquid above the valve, the bottomhole pressure is the wellhead pressure plus 50 m of liquid below the valve (we ignore the weight of the gas in the tubing): $1.0 + 900\cdot 9.81\cdot 50\cdot 10^{-6} = 1.44$ MPa. The reservoir could lift the liquid above the valve to a height of $h_\infty = \dfrac{(11.0 - 1.44)\cdot 10^6}{900\cdot 9.81} = 1083$ m.
The column grows more slowly the higher it is: $h(t) = h_\infty - (h_\infty - h_0)\,e^{-t/\tau}$ , where $\tau = \dfrac{A}{J\rho g\cdot 10^{-6}} = 5.47$ h ( $J$ is in m³/(d·MPa); the answer in days is converted to hours). From 75 m (the liquid that fell back in the previous cycle) up to 300 m, the time is $t = \tau\ln\dfrac{1083 - 75}{1083 - 300} = 83$ min.
A cycle lasts $83 + 15 = 98$ min, which is 14.7 cycles per day. Each one lifts out $0.75\cdot 300\cdot 0.003019 = 0.679$ m³, so 10.0 m³/d in total.
In each cycle the gas fills the tubing above the valve, a volume of 5.28 m³: $5.28\cdot\dfrac{3.5}{0.1013}\cdot\dfrac{293}{323}\cdot\dfrac{1}{0.9} = 184$ m³ at standard conditions. Per day this is 2.70 thousand m³, and the injection gas–liquid ratio is 271 m³/m³.

Answer: Intermittent gas lift gives about 10.0 m³/d with 2.70 thousand m³/d of gas. Continuous gas lift gives 6.0 m³/d with 8 thousand m³/d. So intermittent lift gives more liquid with about three times less gas.

What it means. In such a well, continuous gas lift spends gas on slip. Intermittent gas lift lifts the liquid fast, in a slug, and in each cycle it needs just enough gas to fill the tubing. The share of liquid that falls back, the injection time and the gas pressure are assumed here. At the well they are refined from measurements, and the cycle time is tuned. A plunger between the gas and the liquid almost removes the fallback (plunger lift). The data in this example are assumed for illustration.

Intermittent gas lift: slug height and operating mode of the well of the example

Column at ejection, m: 150
Cycles per day: 27.5
Liquid, m³/d: 9.3
Gas, thousand m³/d: 5.06
Injection gas–liquid ratio, m³/m³: 542

Column at ejection, m: 200
Cycles per day: 21.6
Liquid, m³/d: 9.8
Gas, thousand m³/d: 3.98
Injection gas–liquid ratio, m³/m³: 406

Column at ejection, m: 300
Cycles per day: 14.7
Liquid, m³/d: 10.0
Gas, thousand m³/d: 2.70
Injection gas–liquid ratio, m³/m³: 271

Column at ejection, m: 400
Cycles per day: 10.7
Liquid, m³/d: 9.7
Gas, thousand m³/d: 1.97
Injection gas–liquid ratio, m³/m³: 203

Column at ejection, m: 500
Cycles per day: 8.1
Liquid, m³/d: 9.2
Gas, thousand m³/d: 1.49
Injection gas–liquid ratio, m³/m³: 162

Column at ejection, m	Cycles per day	Liquid, m³/d	Gas, thousand m³/d	Injection gas–liquid ratio, m³/m³
150	27.5	9.3	5.06	542
200	21.6	9.8	3.98	406
300	14.7	10.0	2.70	271
400	10.7	9.7	1.97	203
500	8.1	9.2	1.49	162

The other data are as in the example. The gas pressure in the tubing at the end of the ejection is taken as the same for all slugs, 3.5 MPa. In reality it is higher for large slugs, and the saving of gas is smaller than in the table. Production is highest with slugs of about 300 m. Larger slugs save gas, but the column presses on the reservoir for longer. The data are assumed for illustration.

Intermittent gas lift has several variants. With a plunger, a free piston that the gas pushes up together with the slug, the fallback almost disappears, and the regular run of the plunger also cleans wax from the tubing. In wells with a very low pressure, the liquid is collected in an accumulation chamber (chamber lift), a section of larger-diameter pipe at the bottom. For the same slug volume the column is lower and presses less on the reservoir. How the cycles are tuned, and how intermitters and valves for intermittent gas lift are chosen, is explained in the lesson Gas lift operation and optimisation.

Continuous and intermittent gas lift

Feature: Flow rate
Continuous: from tens to thousands of m³/d
Intermittent: from a few to a few tens of m³/d

Feature: Gas supply
Continuous: constant, adjusted with a choke
Intermittent: in portions, by a timer or a valve

Feature: Flow in the tubing
Continuous: steady, a mixture of gas and liquid
Intermittent: a slug of liquid pushed out by gas

Feature: Bottomhole pressure
Continuous: constant
Intermittent: varies during the cycle

Feature: Load on the gathering system
Continuous: even
Intermittent: surges of gas and liquid

Feature	Continuous	Intermittent
Flow rate	from tens to thousands of m³/d	from a few to a few tens of m³/d
Gas supply	constant, adjusted with a choke	in portions, by a timer or a valve
Flow in the tubing	steady, a mixture of gas and liquid	a slug of liquid pushed out by gas
Bottomhole pressure	constant	varies during the cycle
Load on the gathering system	even	surges of gas and liquid

The limits in flow rate are approximate. They depend on the tubing diameter, the depth and the pressure. The change to intermittent mode is checked by calculation and by trial.

The gas lift system at the surface, and energy

Gas lift is not only the well. It is also a system at the surface, usually shared by many wells. The production comes to a separator. The separated gas is compressed at a compressor station, dried and sent along high-pressure pipelines to the gas distribution manifold. This is the unit where the gas is divided between the wells by adjustable chokes and measured at each well. From the annulus the gas enters the tubing, rises with the production and returns to the separator, and so on in a circle.

The injected gas comes back with the production. So the separators, the gas treatment plant and the compressors are sized for the sum of the reservoir gas and the lift gas. In the well of the first example, for 60 thousand m³/d of injected gas there is only about 6.6 thousand m³/d of reservoir gas, so there is 9 times more injected gas. The reservoir gas tops up the system. Part of the gas is burned in the compressor drives and in heaters, and part is lost. With time the balance changes: the water cut grows, there is less oil and less reservoir gas, and in the later years gas for gas lift sometimes has to be brought in from outside. Starting up after a shutdown in which the pressure was bled off is a separate problem. There is then no high-pressure gas in the system, and wells that do not flow on their own cannot supply it. So the first wells need an outside source, a gas pipeline or nitrogen.

In a cold climate the main threat is hydrates. Wet high-pressure gas, when it cools (especially after the chokes, where it expands and gets colder), forms solid compounds with water that can block a gas line or a valve. The gas for gas lift is dried, and when necessary an inhibitor (a chemical that stops hydrates from forming), for example methanol, is injected into the line.

The energy of gas lift is first of all the work of the compressors. The smallest work is needed by ideal isothermal compression, in which the gas is cooled so that its temperature does not rise. (The example with gas lift above a pump in the lesson on hybrid systems uses the same formula.)

Power of ideal isothermal compression of gas

N_{\text{iso}} = p_{\text{sc}}\,q_{\text{g}}\,\ln\frac{p_2}{p_1}

where:

$N_{\text{iso}}$ power of ideal isothermal compression, W
$p_{\text{sc}}$ standard pressure, 101,325 Pa, Pa
$q_{\text{g}}$ gas rate at standard conditions, m³/s
$p_1,\ p_2$ absolute pressures at the compressor inlet and outlet, Pa

Units: SI; pressures are absolute

Holds when

The gas is ideal and the compression is isothermal at 20 °C (an infinite number of stages with cooling), with no losses. This is a lower limit for real compressors. If the gas is compressed at a temperature $T$ , the power is higher by a factor of $T/(293\ \text{K})$ . The compressibility of a real gas ( $Z < 1$ ) reduces it by 5–10%. At 40–50 °C these two corrections almost cancel each other.
A real multistage compressor with its drive uses more. In the example below it is assumed to use $1/0.65$ times as much.

The work grows in proportion to the gas rate and to the logarithm of the compression ratio (the outlet pressure divided by the inlet pressure). To raise the injection pressure from 7 to 9 MPa with a suction pressure (the pressure at the compressor inlet) of 0.7 MPa increases the energy cost by only about 11%, and the gas then enters noticeably deeper.

In other unit systems (1)

Metric oilfield units

N_{\text{iso}}\,[\text{kW}] = 1.173\;q_{\text{g}}\,[\text{thousand m}^3/\text{d}]\;\ln\frac{p_2}{p_1}

$q_{\text{g}}$ in thousand m³/d at standard conditions; pressures in any consistent units, absolute.

$1.173 = 101{,}325\cdot 1000/86{,}400/1000$ : thousand cubic metres per day converted to m³/s, and watts to kilowatts.

Worked exampleHow much energy gas lift uses

Given (assumed data)

Well A at the economic optimum: 280.1 m³/d of liquid with 80.7 thousand m³/d of gas; bottomhole pressure 11.665 MPa; 9.03 MPa is needed at the annulus wellhead
Compressor: takes the gas in at 0.7 MPa; uses $1/0.65$ times as much as ideal isothermal compression (assumed)
Electricity: an assumed US$0.06 per kWh

Solution

The useful work is the work that a pump at the bottom would do, lifting the liquid without gas: $\Delta p = 1.2 + 22.563 - 11.665 = 12.098$ MPa ( $22.563$ MPa is the weight of a 2500 m column of liquid), and $N_{\text{useful}} = 280.1\cdot 12.098/86.4 = 39.2$ kW (hydraulic power).
Compression: $N_{\text{iso}} = 1.173\cdot 80.7\cdot\ln\dfrac{9.03}{0.7} = 242$ kW. For a real unit: $242/0.65 = 372$ kW.
Efficiency of the gas lift system: $39.2/372 = 10.5$ %. A pump with a system efficiency of 0.45 would do the same work with 87 kW.
Energy per thousand cubic metres of gas: $372\cdot 24/80.7 = 111$ kWh, which costs about US$7. This is about a third of the gas cost assumed in the example on the optimum. The rest is the maintenance of the compressors, drying and equipment wear.

Answer: About 10.5% of the energy spent on compressing the gas goes into lifting the liquid. The gas lift of this well uses about 372 kW.

What it means. The low efficiency is the price of simple equipment in the well and of tolerance to gas and sand. Gas lift is justified when high-pressure gas is already available, or is compressed anyway (for treatment, transport or injection into the reservoir), when the production contains a lot of gas, and when the reliability of the downhole equipment matters more than the cost of energy. The data in this example are assumed for illustration.

Where gas lift fits

How gas lift copes with well conditions

Condition: A lot of gas in the production
Gas lift: good
Why: the reservoir gas itself lightens the column, so less gas needs to be injected; free gas hinders pumps

Condition: Sand, solid particles
Gas lift: good
Why: there are no moving parts in the well; worn valves are changed on wireline

Condition: Deviated, curved wellbore
Gas lift: good
Why: there are no rods or rotating parts; changing valves on wireline is limited by the angle of deviation

Condition: Changing flow rate
Gas lift: good
Why: the flow rate is adjusted from the surface by the gas rate, and the equipment in the well stays the same

Condition: Low reservoir pressure
Gas lift: poor
Why: the bottomhole pressure stays high, and at low flow rates only intermittent mode works

Condition: Viscous oil, emulsions
Gas lift: poor
Why: friction is high, and gas does not speed up a viscous mixture well

Condition: Cold climate, wet gas
Gas lift: with complications
Why: hydrates in the gas lines, and wax (paraffin) because the production cools

Condition: No high-pressure gas
Gas lift: expensive
Why: we need compressors, gas pipelines and a source of gas for the whole production life, including start-up

Condition	Gas lift	Why
A lot of gas in the production	good	the reservoir gas itself lightens the column, so less gas needs to be injected; free gas hinders pumps
Sand, solid particles	good	there are no moving parts in the well; worn valves are changed on wireline
Deviated, curved wellbore	good	there are no rods or rotating parts; changing valves on wireline is limited by the angle of deviation
Changing flow rate	good	the flow rate is adjusted from the surface by the gas rate, and the equipment in the well stays the same
Low reservoir pressure	poor	the bottomhole pressure stays high, and at low flow rates only intermittent mode works
Viscous oil, emulsions	poor	friction is high, and gas does not speed up a viscous mixture well
Cold climate, wet gas	with complications	hydrates in the gas lines, and wax (paraffin) because the production cools
No high-pressure gas	expensive	we need compressors, gas pipelines and a source of gas for the whole production life, including start-up

These are qualitative assessments. The choice of method for a particular well is explained in the lesson Artificial lift: when it is needed and how to choose the method.

Gas lift beats pumps where reliability and flexibility matter. The equipment in the well is cheap and hardly ever breaks, valves are changed without pulling the tubing, the flow rate is adjusted from the surface, and surveys can be run through the tubing. Gas lift loses on bottomhole pressure and on energy: a pump set near the perforations takes more from the reservoir and uses less energy to do it. In old fields where the reservoir pressure is falling, gas-lift wells are converted to pumps over time. Sometimes the two methods are combined: gas injected above an electric submersible pump (ESP, a multistage pump driven by an electric motor at the bottom of the well) makes its work easier (hybrid systems).

Terms and symbols

Gas lift: Artificial lift by injecting gas into the tubing: the gas lightens the column above the injection point.
Operating valve: The lowest open gas-lift valve, through which gas enters the tubing all the time. It sets the depth of gas injection.
Unloading valves: Valves above the operating valve. They work during start-up, while the gas pushes the kill fluid out of the annulus.
Point of gas injection: The depth at which gas enters the tubing. The deeper it is, the lower the bottomhole pressure for the same gas rate.
Gas lift performance curve: The flow rate of a well against the gas rate, for a given injection depth.
Technical and economic optimum: The gas rate that gives the highest flow rate, and the gas rate at which the last thousand cubic metres of gas pay for themselves with the oil they add.
Injection gas–liquid ratio: Cubic metres of injected gas per cubic metre (or tonne) of liquid produced.
Marginal return of gas: The increase in oil rate per extra thousand cubic metres of gas, which is the slope of the gas lift curve.
Intermittent gas lift: Lifting liquid in slugs: liquid collects above the valve and is thrown out by the gas.

Common mistakes and misconceptions

Wrong

“More gas means more oil.”

In fact

Only up to the technical optimum. Beyond it the flow rate falls because of friction. And long before the optimum, extra gas stops paying for itself: in the example of this lesson, the last 13% of the gas added less than 0.3% of the liquid.

Wrong

“The gas comes back with the production, so gas lift costs nothing.”

In fact

The gas goes round in a circle, but the energy to compress it is spent every time. In the example of this lesson it is about 110 kWh per thousand cubic metres, and the efficiency of the system is about 10%.

Wrong

“If there is little gas, we will give it all to the best well.”

In fact

The gas is shared so that the marginal returns of the wells become equal. In the example of this lesson, giving all the gas to the best well produced 17.5 m³/d less oil than the right sharing, and even less than an equal split.

Wrong

“The gas pressure at the valve is the same as at the wellhead.”

In fact

A column of gas has weight. Over 2.4 km it adds almost a quarter of the wellhead pressure in the example of this lesson. We calculate it with the formula for a column of gas, using absolute pressures and an average $Z$ .

Wrong

“Gas lift suits any low-rate well.”

In fact

At low velocities the gas slips through the liquid, and continuous gas lift wastes gas. For such wells we use intermittent gas lift, plunger lift or a pump.

In depth: advanced detail

How the flow rate responds to gas

The operating point is the solution of the equation $\bar p_R - q/J = p_{\text{outflow}}(q, q_{\text{g}})$ , where $p_{\text{outflow}}$ is the bottomhole pressure on the outflow curve. Let us take the derivative of both sides with respect to $q_{\text{g}}$ (how each side changes when the gas rate changes), treating the flow rate as a function of the gas rate:

Slope of the performance curve of a gas-lift well

\frac{\dd q}{\dd q_{\text{g}}} = -\,\frac{\partial p_{\text{outflow}}/\partial q_{\text{g}}}{\partial p_{\text{outflow}}/\partial q + 1/J}

where:

$\partial p_{\text{outflow}}/\partial q_{\text{g}}$ how much extra gas lowers the pressure needed to lift the same flow rate (negative up to the technical optimum), MPa per thousand m³/d
$\partial p_{\text{outflow}}/\partial q$ slope of the outflow curve at the given gas rate, MPa per m³/d
$J$ productivity index, m³/(d·MPa)

Units: any consistent units

Holds when

The IPR is a straight line, and the operating point is stable, so the denominator is positive (stability of the operating point).

The flow rate stops growing when the numerator becomes zero, which gives the condition for the technical optimum. The denominator explains why gas lift likes wells with a high productivity. When $J$ is large, the term $1/J$ is small, and every drop of pressure in the tubing turns into a large gain in flow rate. When $J$ is small, the same gas gives little.

Why the marginal returns must be equal

We need to find the gas rates $q_{\text{g},i}$ that make the total oil production $\sum_i q_{o,i}(q_{\text{g},i})$ as large as possible, when $\sum_i q_{\text{g},i} = Q_{\text{g}}$ . A maximum problem with a constraint is solved by the method of Lagrange multipliers. We form the function $\Phi = \sum_i q_{o,i} - \beta\left(\sum_i q_{\text{g},i} - Q_{\text{g}}\right)$ and set its derivatives with respect to each $q_{\text{g},i}$ equal to zero. The result is $\dd q_{o,i}/\dd q_{\text{g},i} = \beta$ for all the wells, which is the equality of the marginal returns.

The multiplier $\beta$ has a meaning: it is the increase in total production for an extra thousand cubic metres of gas in the system. If the curves bulge upward, the point we find is really a maximum. A well whose marginal return is below $\beta$ even at zero gas does not need gas (its gas rate is zero). If a well needs a minimum start-up gas rate, its curve does not bulge upward, and the equality condition is no longer enough. We then compare the ways of sharing the gas with this well and without it.

Connections to other topics

Multiphase flow in tubing — calculating the pressure in the tubing with gas, and the minimum of the outflow curve on which gas lift is based.
Nodal analysis of the production system — the operating point and its stability.
Artificial lift: when it is needed and how to choose the method — how gas lift differs from pumps and how to choose a method.
Inflow performance relationship (IPR) — how much the reservoir will give when the bottomhole pressure is lowered.
Unloading a gas lift well: valves and mandrels — gas-lift valves, side-pocket mandrels and unloading the well.
Gas lift design — valve spacing and the choice of pressures and gas rates in design.
Gas lift operation and optimisation — heading, problems, field optimisation and tuning of intermittent gas lift.
Progressing cavity pumps and hybrid systems — gas lift above a pump in combined systems.
Pore pressure and overburden pressure — pressure in a column of gas.

Check yourself

1
Why can gas lift not lower the bottomhole pressure as much as a pump set at the same depth?
1. A.Gas lightens only the column above the injection point, and only up to a limit beyond which friction grows. A pump takes the weight of the whole column above it off the reservoir.
2. B.Gas dissolves in the oil and makes it heavier.
3. C.The gas-lift valve creates a pressure drop equal to the reservoir pressure.
4. D.A pump always works with a lower wellhead pressure.
Show the answer
Correct answer: A. Gas lightens only the column above the injection point, and only up to a limit beyond which friction grows. A pump takes the weight of the whole column above it off the reservoir.
Below the injection point gas lift does not lighten the column at all, and above it the density of the mixture cannot become as small as we like. As the gas rate grows, friction grows and at some point outweighs the lightening. A pump creates a jump in pressure equal to the weight of the column above it plus the losses, so the pressure at its intake can be very low.
2
The gas pressure at the annulus wellhead is 8.0 MPa (absolute), the specific gravity of the gas is 0.65, the average temperature in the annulus is 320 K and the average $Z = 0.85$ . What is the gas pressure at a depth of 1800 m?
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The exponent is $0.03418\cdot 0.65\cdot 1800/(0.85\cdot 320) = 0.147$ . The pressure is $8.0\cdot e^{0.147} \approx 8.0\cdot 1.158 \approx 9.27$ MPa. The column of gas added almost 1.3 MPa, which is about 16% of the wellhead pressure. If the annulus were full of water, the increase would be about 18 MPa.
3
A well is given gas at the technical optimum. What happens to the liquid rate if the gas rate is increased by another 20%?
1. A.It does not change, because the valve passes no more than a certain rate
2. B.The well stops flowing
3. C.It grows by about 20%
4. D.It decreases a little
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Correct answer: D. It decreases a little
Beyond the top of the performance curve, extra gas increases friction more than it lightens the column. So the bottomhole pressure that is needed grows, and the flow rate decreases. Near the top it decreases slowly, because the curve is flat there. The gas is wasted.
4
Two wells share gas. The marginal return of the first is 0.6 m³ of oil per thousand cubic metres of gas, and that of the second is 0.2. How should the sharing be changed?
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Move some of the gas from the second well to the first. Each thousand cubic metres taken from the second well reduces production by 0.2 m³, and each thousand given to the first increases it by 0.6 m³. Keep moving gas until the marginal returns become equal: the return of the first well will fall, and that of the second will rise. If the return of the second well is below the common level even at a small gas rate, it does not need gas at all.
5
The net value of oil is US$240/m³, the lift gas costs US$30 per thousand cubic metres and the water cut is 50%. At what slope of the curve of liquid rate against gas rate should we stop adding gas?
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A thousand cubic metres of gas pays for itself as long as it gives more than US$30 worth of oil, that is, more than $30/240 = 0.125$ m³ of oil. Oil is half of the liquid, so the slope in terms of liquid must be at least $0.125/0.5 = 0.25$ m³ per thousand cubic metres of gas. Beyond that the gas does not pay for itself, even if the technical optimum is still far away.
6
A well produces 4 m³/d on continuous gas lift with an injection gas–liquid ratio of more than 1000 m³/m³. Why is it so high, and what can we suggest?
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At a low flow rate the mixture moves slowly, at a speed similar to the speed at which bubbles rise, and the gas slips through the liquid. The liquid stays behind in the tubing, the column stays heavy and the gas is wasted. It is better to change to intermittent gas lift (let the liquid collect above the valve and eject it as a slug, with a plunger if needed), or to a pump if the reservoir pressure is low.

Updated: 19 October 2026