Unloading a gas lift well: valves and mandrels

Quick refresher

The kill fluid in the tubing balances the gas pressure. In the example of this lesson, a kick-off pressure of 10.5 MPa pushes the brine (salt water used as the kill fluid) out of the annulus only down to 972 m, while the operating valve is at 2400 m. This is why unloading valves are placed above the operating valve.
Unloading goes from the top down. The gas enters the tubing through the top valve and lightens the column above it, while the liquid from the annulus keeps passing into the tubing through the lower valves. When the gas reaches the next valve, the pressure in the annulus falls and the top valve closes.
A bellows valve opens when the gas pressure on the bellows (the corrugated metal element that works like a spring) and the tubing pressure on the seat (the ring that the valve ball closes against) overcome the nitrogen pressure in the dome (the chamber charged with nitrogen). It closes when the gas pressure falls to the nitrogen pressure. The difference between the opening and closing pressures grows with the ratio $R$ of the seat area to the bellows area.
The unloading valves close one after another if their closing pressures, converted to the wellhead, fall from the top down. During normal operation, the gas pressure is below the closing pressures of all the unloading valves.
The nitrogen pressure rises with temperature. So a valve is charged on the test rack at 15.6 °C (60 °F), with a correction for its temperature in the well. Without the correction, the valve opens in the well at a pressure several megapascals above the design value.
Gas flows through the seat or the port (the opening) of a valve as through a choke (a narrow opening that restricts the flow). The port of the operating valve is chosen to pass the design gas rate with a small pressure drop. A port that is too small raises the gas pressure and opens the unloading valves.

Key relationships

p_{\text{ko}}\,e^{\,0.03418\,\gamma_{\text{g}} H/(\bar Z \bar T)} = p_{wh} + \rho_{\text{kill}} g H

depth to which the kick-off pressure pushes out the kill fluid

p_{\text{g,open}} = \frac{p_{\text{dome}} - R\,p_{\text{t}}}{1 - R},\qquad R = \frac{A_{\text{p}}}{A_{\text{b}}}

gas pressure that opens the valve

p_{\text{g,close}} = p_{\text{dome}},\qquad \Delta p_{\text{spread}} = k_{\text{t}}\,(p_{\text{dome}} - p_{\text{t}}),\quad k_{\text{t}} = \frac{R}{1 - R}

closing pressure, and the difference between the opening and closing pressures

p_{\text{TRO}} = \frac{C_T\,p_{\text{dome}}}{1 - R},\qquad C_T = \frac{1}{1 + 0.00215\,(T_{^{\circ}\text{F}} - 60)}

test-rack opening pressure at 60 °F; the pressures are gauge

Common pitfalls

Placing a single operating valve at the design depth, with no unloading valves, and relying on the working gas pressure.
Setting valves without a temperature correction, or with the temperature of the rock instead of the temperature of the flow.
Taking the closing pressure of a valve to be equal to its opening pressure.
Fitting an operating valve with a port that is “smaller, to save gas”. The gas pressure rises and opens the unloading valves.
Mixing up gauge pressures (measured above atmospheric pressure) and absolute pressures (measured above zero). The force balance is the same in both, but the gas laws and the temperature correction each need one particular kind.

After this lesson you can

explain why gas cannot reach the operating valve at once, and estimate how deep the kick-off pressure (the highest gas pressure available at start-up) can push the kill fluid
describe the unloading of a gas-lift well stage by stage, and how it looks in the wellhead pressures
name the parts of a bellows gas-lift valve (a valve with a spring-like corrugated metal tube, the bellows), and tell valves operated by the gas pressure from valves operated by the tubing pressure
calculate from a force balance the opening and closing pressures of a valve, the difference between them and the effect of the tubing pressure, and check that the unloading valves close one after another
calculate the opening pressure of a valve on the test rack, with a correction for the temperature of the nitrogen, and estimate what happens without it
estimate the gas rate through the seat or the port of a valve, and choose the port size of the operating valve

Why gas cannot reach the operating valve at once

Before a workover, the well is killed: the annulus and the tubing are filled with brine (salt water) or another kill fluid, so that the pressure of its column is higher than the reservoir pressure (killing the well before a workover). After the workover, the gas-lift well stands full of this fluid. Gas is supplied to the annulus, and the liquid in the annulus has nowhere to go except through the gas-lift valves into the tubing and on up to the flowline. The annulus and the tubing work like connected vessels.

While the gas pushes the liquid out, the tubing stays full of heavy liquid. The level in the annulus goes down until the gas pressure at the level equals the pressure of the liquid in the tubing at the same depth. Gas is light, and its pressure hardly grows with depth, while the pressure of a column of brine grows by about 10.6 kPa for every metre. So the level stops at a shallow depth.

Depth to which the kick-off pressure pushes out the kill fluid

p_{\text{ko}}\exp\!\left(\frac{0.03418\,\gamma_{\text{g}}\,H}{\bar Z\,\bar T}\right) = p_{wh} + \rho_{\text{kill}}\,g\,H

where:

$p_{\text{ko}}$ kick-off pressure: the highest gas pressure at the annulus wellhead that the compressor or another source can supply, MPa (abs.)
$H$ true vertical depth of the liquid level in the annulus, m
$\gamma_{\text{g}},\ \bar Z,\ \bar T$ specific gravity of the gas, and the average gas deviation factor and the average absolute temperature of the gas in the annulus, —, —, K
$p_{wh}$ pressure in the tubing at the wellhead (the back pressure of the flowline), MPa (abs.)
$\rho_{\text{kill}}$ density of the kill fluid, kg/m³
$g$ acceleration due to gravity, 9.81 m/s². The product $\rho_{\text{kill}} g H$ is converted from pascals to megapascals, m/s²
$0.03418$ $0.02897\cdot 9.81/8.314$ : the molar mass of air (kg/mol) multiplied by $g$ and divided by the universal gas constant, K/m

Units: metric oilfield units; pressures are absolute

Holds when

The left side is the pressure of the column of gas in the annulus, as in the lesson on gas lift. The right side is the pressure of the liquid in the tubing, which is full up to the wellhead.
The balance is static. During displacement, flow friction adds to the pressure in the tubing, and the level stops a little higher still.
There is a check valve (a one-way valve) at the bottom of the tubing, so the liquid does not go into the reservoir. Otherwise part of it would go into the reservoir, and the reservoir would first return that same liquid.

The solution of the equation is the greatest depth at which the top unloading valve can be placed. The higher the kick-off pressure and the lighter the kill fluid, the deeper the top valve can be.

Worked exampleWhere the gas will push the kill fluid

Given (assumed data)

Well: from the lesson Gas lift: principles and choice of operating conditions: the middle of the perforations at 2500 m, the operating valve at 2400 m; after the workover it is filled with brine of density 1080 kg/m³; the tubing is open to the flowline, and the wellhead pressure is 1.2 MPa
Gas: the compressor gives 10.5 MPa at the annulus wellhead, which is 1.3 MPa more than is needed in operation (9.19 MPa, injection pressure); gas of specific gravity 0.7, average temperature in the annulus 323.15 K

Solution

The pressure in the tubing full of brine at depth $H$ : $p_{\text{t}} = 1.2 + 1080\cdot 9.81\cdot 10^{-6}H = 1.2 + 0.010595\,H$ MPa.
The gas pressure at the same depth: $p_{\text{g}} = 10.5\exp\dfrac{0.03418\cdot 0.7\,H}{\bar Z\cdot 323.15}$ . With an average $\bar Z = 0.7953$ , the exponent is $H/10{,}742$ . The gas pressure grows by a factor of $e$ only every 10,742 m.
The level stops where the pressures are equal: $10.5\,e^{H/10{,}742} = 1.2 + 0.010595\,H$ . By trial, $H = 971.6$ m, and both pressures there are 11.49 MPa.
To push the brine out down to 2400 m at once, the gas pressure there would have to be $1.2 + 0.010595\cdot 2400 = 26.628$ MPa. At the wellhead, with $\bar Z = 0.816$ , this is $26.628\,e^{-0.2178} = 21.42$ MPa.

Answer: A kick-off pressure of 10.5 MPa pushes the brine out of the annulus only down to a depth of about 972 m. For the gas to reach 2400 m at once, we would need 21.42 MPa at the wellhead, twice what the compressor gives.

What it means. The top unloading valve is placed a little above the depth found, at 950 m in the example of this lesson. This leaves a margin for friction, for an error in the density of the liquid, and for swings in the gas pressure. Further unloading valves are placed below it, and their spacing is calculated separately (Gas lift design). The high pressure is needed only during unloading, and building a 21 MPa compressor just for that would cost too much. The data in this example are assumed for illustration.

We can do without unloading valves only if gas at a much higher pressure is available for the start-up: a separate start-up compressor, nitrogen from a mobile unit, or high-pressure gas from a neighbouring field. This is done, for example, in subsea wells, where every extra valve is an extra risk of leakage and a single valve is fitted. In an ordinary gas-lift well, though, the tubing above the operating valve carries several side-pocket mandrels with unloading valves, and in deep wells more than ten (side-pocket mandrels).

Unloading step by step

SchematicStages of unloading a gas-lift well (diagram, not to scale). Blue shows liquid, bubbles show the mixture of gas and liquid, white valves are open and black valves are closed, and the arrows at the valves show where liquid or gas goes. In the example of this lesson, the gas pressure at the wellhead when the unloading valves open is 10.40; 10.25; 10.10; 9.95 MPa, and in operation through the operating valve it is 9.19 MPa.

How unloading goes

1
Before unloading
The annulus and the tubing are full of kill fluid, and the tubing at the wellhead is open to the flowline. Valves with a high column of liquid above them are open, because the pressure of the liquid at them is higher than the nitrogen pressure in the dome. The upper valves open as soon as gas is fed into the annulus and the pressure in it rises.
2
Pushing the liquid out through all the valves
Gas is fed into the annulus through a control choke (an adjustable restriction), and the pressure is raised slowly. The liquid leaves the annulus through all the valves into the tubing and out at the wellhead. If the pressure is raised suddenly, liquid carrying sand and rust flakes (mill scale) passes through the small seats at high speed and erodes them. The valves will then leak gas even when closed.
3
The gas has reached valve 1
The gas enters the tubing through the top valve and the mixture above it gets lighter. The pressure in the tubing at the lower valves falls below the pressure of the liquid in the annulus, so the liquid keeps being pushed out through them. The bottomhole pressure drops below the reservoir pressure, and the reservoir starts to give fluid: first the kill fluid that it took in, then its own production.
4
Handover to valve 2
The level in the annulus reaches the second valve, and the gas goes into the tubing through both valves. The choke at the wellhead limits the gas supply, so the pressure in the annulus falls. When it falls below the closing pressure of valve 1, valve 1 closes and all the gas goes through valve 2.
5
Down to the operating valve
In the same way the gas is handed over to valves 3 and 4 and, finally, to the operating valve. With each handover the pressure in the annulus falls by a step. When the gas goes only through the operating valve, the gas rate is set to the design value, and the well reaches steady operation.

At the wellhead, unloading shows in two pressures. The gas pressure in the annulus first rises to the kick-off pressure. Then at each handover to the next valve it falls by a step, and finally it settles at the working level. The pressure in the tubing at the wellhead jumps about during unloading, because plugs of liquid and gas pass through the tubing. It calms down when the well works through the operating valve. From these records we judge whether the gas has reached the operating valve.

How a gas-lift valve is built

The most common gas-lift valve is the bellows valve with a nitrogen dome, operated by the gas pressure (IPO, injection-pressure-operated). A bellows is a thin corrugated metal tube that works like a spring. The valve goes into the pocket of a side-pocket mandrel. It is changed on wireline (with tools lowered on a steel wire), without pulling the tubing (changing valves in side-pocket mandrels).

SchematicA gas-lift valve operated by the gas pressure (diagram, not to scale). The nitrogen pressure

p_{\text{dome}}

presses the ball against the seat. The gas pressure from the annulus

p_{\text{g}}

presses on the bellows from outside, and the tubing pressure

p_{\text{t}}

presses on the ball through the seat.

Nitrogen dome is charged at the surface with nitrogen at a pressure $p_{\text{dome}}$ . The nitrogen presses on the bellows and, through the stem, presses the ball against the seat. Part of the dome is sometimes filled with a viscous silicone fluid that damps the vibration of the stem.
Bellows is a corrugated metal accordion with an effective area $A_{\text{b}}$ . The gas from the annulus presses on it from outside, and it compresses and lifts the stem. A spring or a stop keeps the bellows from compressing too far and losing its elasticity.
Stem with ball, and seat are the shut-off pair. The area of the seat along the line where the ball touches it is $A_{\text{p}}$ (the port area). Gas passes through the seat into the tubing, and while the valve is closed, the tubing pressure presses on the ball from below.
Windows and seals. Gas from the annulus enters the valve through the windows of the mandrel pocket. The seals above and below the windows separate it from the tubing.
Check valve stops the production from the tubing from passing into the annulus when the pressure in the tubing is higher than the gas pressure.

Types of gas-lift valves

Valve: Bellows valve with a nitrogen dome, operated by the gas pressure
What opens and closes it: gas pressure on the bellows against the nitrogen pressure; the tubing pressure has little effect
Where it is used: unloading valves of continuous gas lift: the main type

Valve: With a spring instead of nitrogen, or together with it
What opens and closes it: gas pressure against the force of the spring; almost independent of temperature
Where it is used: where temperature is hard to account for; at high pressures the spring is added to the nitrogen

Valve: Operated by the tubing pressure
What opens and closes it: tubing pressure on the bellows; the gas pressure acts only on the seat
Where it is used: intermittent gas lift; wells where gas from one annulus feeds two tubing strings

Valve: Orifice (no shut-off part)
What opens and closes it: always open; the flow is set by the orifice and the pressure drop
Where it is used: operating valve of continuous gas lift

Valve: Blanking plug
What opens and closes it: closes the pocket of the mandrel
Where it is used: mandrels that are not needed yet or were installed for the future

Valve	What opens and closes it	Where it is used
Bellows valve with a nitrogen dome, operated by the gas pressure	gas pressure on the bellows against the nitrogen pressure; the tubing pressure has little effect	unloading valves of continuous gas lift: the main type
With a spring instead of nitrogen, or together with it	gas pressure against the force of the spring; almost independent of temperature	where temperature is hard to account for; at high pressures the spring is added to the nitrogen
Operated by the tubing pressure	tubing pressure on the bellows; the gas pressure acts only on the seat	intermittent gas lift; wells where gas from one annulus feeds two tubing strings
Orifice (no shut-off part)	always open; the flow is set by the orifice and the pressure drop	operating valve of continuous gas lift
Blanking plug	closes the pocket of the mandrel	mandrels that are not needed yet or were installed for the future

Valves are made with an outside diameter of about 25 mm (1 in) and 38 mm (1½ in), more rarely smaller. The larger ones pass more gas and allow a seat of larger diameter.

Force balance: opening and closing pressures

While the valve is closed, three forces act on the moving system of bellows, stem and ball. The nitrogen presses on the whole area of the bellows and holds the ball against the seat: the closing force is $p_{\text{dome}}A_{\text{b}}$ . The gas from the annulus presses on the bellows from outside, but only on the ring-shaped area $A_{\text{b}} - A_{\text{p}}$ . Inside the seat, the ball is pushed up from below by the tubing pressure, which acts on the area $A_{\text{p}}$ . The valve opens when the sum of the last two forces becomes equal to the first.

Gas pressure that opens the valve

p_{\text{dome}}A_{\text{b}} = p_{\text{g}}\,(A_{\text{b}} - A_{\text{p}}) + p_{\text{t}}A_{\text{p}}\quad\Rightarrow\quad p_{\text{g,open}} = \frac{p_{\text{dome}} - R\,p_{\text{t}}}{1 - R},\qquad R = \frac{A_{\text{p}}}{A_{\text{b}}}

where:

$p_{\text{dome}}$ nitrogen pressure in the dome at the temperature of the valve in the well, MPa
$p_{\text{g}}$ gas pressure in the annulus at the depth of the valve, MPa
$p_{\text{t}}$ pressure in the tubing at the depth of the valve, MPa
$A_{\text{b}},\ A_{\text{p}}$ effective area of the bellows, and area of the seat along the line where the ball touches it, cm²
$R$ ratio of the seat area to the bellows area; for unloading valves it is usually a few hundredths, —

Units: any consistent units; the pressures are all absolute or all gauge, and the equation does not change, because the areas on each side add up to $A_{\text{b}}$

Holds when

The valve has no spring. The elasticity of the bellows itself is ignored.
The pressures are uniform over the areas on which they act.

The gas pressure that opens the valve is lower the higher the tubing pressure is. The heavier the column in the tubing at the valve, the easier it is for the valve to open. How strong this effect is depends on the area ratio $R$ . With a small seat the tubing pressure has almost no effect, and the valve reacts only to the gas pressure.

Closing pressure, and the difference between the opening and closing pressures

p_{\text{g,close}} = p_{\text{dome}},\qquad \Delta p_{\text{spread}} = p_{\text{g,open}} - p_{\text{g,close}} = k_{\text{t}}\,(p_{\text{dome}} - p_{\text{t}}),\qquad k_{\text{t}} = \frac{R}{1 - R}

where:

$p_{\text{g,close}}$ gas pressure at which the open valve closes, MPa
$\Delta p_{\text{spread}}$ valve spread, the difference between the opening and closing pressures, MPa
$k_{\text{t}}$ production pressure effect factor (TEF): by how much the opening pressure falls when the tubing pressure rises by one unit, —

Units: any consistent units

Holds when

When the ball has moved away from the seat, the gas pressure acts on the whole area of the bellows, and the valve stays open as long as $p_{\text{g}} > p_{\text{dome}}$ .
The valve has no spring. In reality the elasticity of the bellows shifts the closing pressure, as explained below.

An open valve closes at a lower gas pressure than the one at which it opened. The difference grows with $R$ and with the gap between the nitrogen pressure and the tubing pressure. For unloading valves this is the deciding property: for the upper valve to close, the gas pressure must fall by this difference after the handover to the lower valve.

CalculatedThe gas pressure that opens valve No. 2 of the example (nitrogen pressure 11.710 MPa), against the tubing pressure at the valve, for seats of 4.8, 6.4 and 7.9 mm (

R

= 0.036; 0.064; 0.098). The dashed line is the closing pressure, equal to the nitrogen pressure. All the lines meet where the tubing pressure equals the nitrogen pressure.

Worked exampleOpening and closing pressures of an unloading valve

Given (assumed data)

Valve: unloading valve No. 2 at a depth of 1570 m: effective bellows area 5.0 cm², seat diameter 4.8 mm, no spring
The unloading calculation gave: the valve must open when the gas pressure at the annulus wellhead is 10.25 MPa and the tubing pressure at the valve is 7.588 MPa (the design data of this well, assumed; how valves are spaced is explained in the lesson Gas lift design); pressures are absolute
Gas in the annulus: as in the previous example: specific gravity 0.7, average temperature 323.15 K

Solution

Seat area $A_{\text{p}} = \pi\cdot 0.48^2/4 = 0.1810$ cm²; $R = 0.1810/5.0 = 0.0362$ ; $k_{\text{t}} = R/(1 - R) = 0.0376$ .
Gas pressure at the valve with 10.25 MPa at the wellhead: $10.25\exp\dfrac{0.03418\cdot 0.7\cdot 1570}{0.79465\cdot 323.15} = 11.865$ MPa.
The nitrogen pressure in the well at which the valve opens exactly in this way: $p_{\text{dome}} = p_{\text{g}}(1 - R) + R\,p_{\text{t}} = 11.865\cdot 0.9638 + 0.0362\cdot 7.588 = 11.435 + 0.275 = 11.710$ MPa.
The valve closes when the gas pressure at it falls to 11.710 MPa. The difference between the opening and closing pressures is $11.865 - 11.710 = 0.155$ MPa. The same is given by $k_{\text{t}}(p_{\text{dome}} - p_{\text{t}}) = 0.0376\cdot(11.710 - 7.588) = 0.155$ MPa.
The closing pressure converted to the wellhead with the formula for a column of gas is 10.12 MPa. At this gas pressure at the wellhead, valve No. 2 closes.
With a 7.9 mm seat ( $R = 0.0980$ ) and the same nitrogen pressure, the valve would open at $(11.710 - 0.0980\cdot 7.588)/(1 - 0.0980) = 12.158$ MPa, and the difference between the opening and closing pressures would grow to 0.448 MPa.

Answer: The dome of valve No. 2 is charged so that the nitrogen pressure in the well is 11.710 MPa. The valve opens at 11.865 MPa of gas at the valve and closes at 11.710 MPa, which is 0.155 MPa lower.

What it means. The small seat of unloading valves is a deliberate choice. The difference between the opening and closing pressures is small, so the upper valve closes as soon as the gas pressure falls a little after the handover to the next valve, and the tubing pressure hardly shifts its setting. With a large seat the valve would pass more gas, but it would close only at a noticeably lower pressure. And if the tubing pressure rose, it could open again when the gas is already going deeper. The data in this example are assumed for illustration.

Now we can see when unloading goes as planned. When the gas has reached valve 2 and goes through both valves, the gas pressure falls. Valve 1 closes before valve 2 if its closing pressure is higher. So the closing pressures, converted to the wellhead, must fall from the top down, and in operation the gas pressure at the wellhead must be lower than the closing pressures of all the unloading valves.

Unloading valves of the example: opening and closing pressures (absolute)

Valve, depth: 1, 950 m
Opening at the wellhead, MPa: 10.40
Opening at the valve, MPa: 11.360
Nitrogen in the dome = closing at the valve, MPa: 11.132
Closing at the wellhead, MPa: 10.19

Valve, depth: 2, 1570 m
Opening at the wellhead, MPa: 10.25
Opening at the valve, MPa: 11.865
Nitrogen in the dome = closing at the valve, MPa: 11.710
Closing at the wellhead, MPa: 10.12

Valve, depth: 3, 1980 m
Opening at the wellhead, MPa: 10.10
Opening at the valve, MPa: 12.148
Nitrogen in the dome = closing at the valve, MPa: 12.044
Closing at the wellhead, MPa: 10.02

Valve, depth: 4, 2250 m
Opening at the wellhead, MPa: 9.95
Opening at the valve, MPa: 12.273
Nitrogen in the dome = closing at the valve, MPa: 12.203
Closing at the wellhead, MPa: 9.90

Valve, depth: operating, 2400 m
Opening at the wellhead, MPa: 9.19 (in operation)
Opening at the valve, MPa: 11.465
Nitrogen in the dome = closing at the valve, MPa: —
Closing at the wellhead, MPa: —

Valve, depth	Opening at the wellhead, MPa	Opening at the valve, MPa	Nitrogen in the dome = closing at the valve, MPa	Closing at the wellhead, MPa
1, 950 m	10.40	11.360	11.132	10.19
2, 1570 m	10.25	11.865	11.710	10.12
3, 1980 m	10.10	12.148	12.044	10.02
4, 2250 m	9.95	12.273	12.203	9.90
operating, 2400 m	9.19 (in operation)	11.465	—	—

Seat 4.8 mm, $R$ = 0.0362; the tubing pressure at the opening of each valve comes from the unloading calculation. The closing pressures at the wellhead fall from the top down (10.19; 10.12; 10.02; 9.90), and at each handover the upper valve closes before the lower one. In operation the gas pressure at the wellhead, 9.19 MPa, is 0.7–1.0 MPa below the closing pressures of the unloading valves. In operation they could open only at 10.42; 10.27; 10.11; 9.95 MPa at the wellhead. The data are assumed for illustration.

Valves operated by the tubing pressure, and others

In a valve operated by the tubing pressure (also called a fluid-operated valve; PPO, production-pressure-operated), the roles of the pressures are swapped. The pocket of the mandrel is built so that the tubing pressure presses on the bellows from outside, and the gas from the annulus reaches the seat. The force balance comes out the same, except that $p_{\text{g}}$ and $p_{\text{t}}$ change places.

Tubing pressure that opens a valve operated by the tubing pressure

p_{\text{t,open}} = \frac{p_{\text{dome}} - R\,p_{\text{g}}}{1 - R}

where:

$p_{\text{t,open}}$ tubing pressure at the valve at which it opens, MPa
$p_{\text{dome}},\ p_{\text{g}},\ R$ nitrogen pressure at the temperature of the valve, gas pressure at the valve, ratio of the seat area to the bellows area, MPa, MPa, —

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

Holds when

The valve has no spring. The tubing pressure acts on the bellows from outside, and the gas pressure acts on the seat.

The valve opens when the tubing pressure at the valve rises, that is, when liquid has collected above it. So these valves are convenient in intermittent gas lift, and where the gas pressure in the annulus is no good as a signal, for example when gas from one annulus goes to several tubing strings. The valves for such wells are spaced by different rules (Gas lift design).

In valves with a spring, the closing force comes from the spring, and their setting hardly depends on temperature. In high-pressure valves a spring is sometimes added to the nitrogen, because a spring alone does not give the needed force. Proportional valves open not with a jump but gradually: the flow area grows together with the pressure, and the gas rate follows it. They adjust the gas supply more precisely, but they are more complex and need accurate setting.

Setting on the test rack and the temperature correction

The dome of the valve is charged with nitrogen in a workshop. Then the valve is put on a test rack, a bench with a thermostat that keeps the temperature constant. The tubing side of the valve is left open to the air. Gas pressure is raised smoothly on the annulus side until the valve opens. This pressure, the test-rack opening pressure, shows that the dome is charged correctly. Most manufacturers set valves at 60 °F (15.6 °C), and some at 80 °F (26.7 °C). We check with the manufacturer which temperature is used.

Test-rack opening pressure

p_{\text{TRO}} = \frac{p_{\text{dome}}^{\text{rack}}}{1 - R}

where:

$p_{\text{TRO}}$ gas pressure that opens the valve on the test rack, MPa (gauge)
$p_{\text{dome}}^{\text{rack}}$ nitrogen pressure in the dome at the temperature of the test rack, MPa (gauge)
$R$ ratio of the seat area to the bellows area, —

Units: gauge pressures, as shown by the gauges of the rack

Holds when

On the tubing side there is atmospheric pressure, that is, $p_{\text{t}} = 0$ on the gauge. The equation is a special case of the force balance.

The rack pressure is larger than the nitrogen pressure by a factor of $1/(1 - R)$ : with the valve closed, the gas presses only on the ring-shaped area of the bellows.

The nitrogen in the dome is a gas in a closed volume. In a hot well its pressure is higher than on the test rack. The valve must open in the well at the design pressure, so on the rack it is charged to a lower pressure, recalculated to the temperature of the rack. The dome is charged with a gas, not a liquid. A liquid is almost incompressible, and when heated, the pressure in the dome would grow so much that it would burst the bellows. The gas used is nitrogen. It is inert (it does not burn and does not cause corrosion), it is cheap and sold in cylinders, and the rise of its pressure on heating has long been measured and put in tables.

Temperature correction for the nitrogen in the dome

C_T = \frac{p_{\text{dome}}^{\text{rack}}}{p_{\text{dome}}} = \frac{1}{1 + 0.00215\,(T_{^{\circ}\text{F}} - 60)}

where:

$C_T$ correction factor from the temperature of the valve to the rack temperature of 60 °F, —
$p_{\text{dome}}$ nitrogen pressure at the temperature of the valve in the well, MPa (gauge)
$T_{^{\circ}\text{F}}$ temperature of the valve in the well, °F

Units: temperature in degrees Fahrenheit; pressures are gauge, as in the accepted field method

Holds when

It is a straight-line fit to measurements on nitrogen at the constant volume of the dome. The tables of factors for setting valves are made from it.
It is accurate enough at a nitrogen pressure of about 4 MPa. At higher pressures real nitrogen raises its pressure more when heated, and the correction underestimates it (advanced section).
If part of the dome is filled with silicone fluid, it also expands on heating and adds pressure.

The nitrogen pressure on the rack is lower than in the well: at 80 °C, by about a fifth. For an ideal gas at constant volume, the coefficient of $(T - 60)$ would be $1/520 = 0.00192$ per degree Fahrenheit (for absolute pressure). For nitrogen at the pressures in valve domes it is larger.

In other unit systems (1)

Temperature in degrees Celsius

C_T = \frac{1}{1 + 0.00387\,(T_{^{\circ}\text{C}} - 15.6)}

$T$ in °C; pressures are gauge.

$0.00387 = 0.00215\cdot 1.8$ : one degree Celsius is 1.8 degrees Fahrenheit, and 60 °F = 15.6 °C.

Worked exampleSetting an unloading valve on the test rack

Given (assumed data)

Valve: No. 2 from the previous example: nitrogen pressure in the well 11.710 MPa (absolute), $R = 0.0362$
Temperature: the flow in the tubing is 40 °C at the wellhead and 85 °C at the perforations at 2500 m; the valve temperature is taken on the straight line between them (assumed)
Test rack: 60 °F (15.6 °C), atmospheric pressure of 0.101 MPa on the tubing side; the pressures on the rack are gauge

Solution

Valve temperature: $40 + 45\cdot 1570/2500 = 68.26$ °C, or $1.8\cdot 68.26 + 32 = 154.87$ °F.
Correction factor: $C_T = 1/[1 + 0.00215\cdot(154.87 - 60)] = 0.8306$ .
Gauge nitrogen pressure in the well: $11.710 - 0.101 = 11.609$ MPa. At 15.6 °C it will be $0.8306\cdot 11.609 = 9.642$ MPa, and this is what the dome is charged to.
Test-rack opening pressure: $p_{\text{TRO}} = 9.642/(1 - 0.0362) = 10.004$ MPa (gauge).
If we forget the correction and charge the dome to 11.609 MPa, the nitrogen pressure in the well will grow to $11.609/0.8306 = 13.977$ MPa (gauge), and the valve will open at a gas pressure about $(13.977 - 11.609)/(1 - 0.0362) = 2.46$ MPa above the design value.
An error of 10 °C in the valve temperature changes the nitrogen pressure in the well by about $0.00387\cdot 10\cdot 9.642 = 0.37$ MPa.

Answer: Valve No. 2 is set to a test-rack opening pressure of 10.004 MPa (gauge) at 15.6 °C. Without the temperature correction it would open in the well 2.46 MPa above the design value, and the gas from the compressor simply would not open it.

What it means. The temperature correction is not a refinement but the basis of the setting. It shifts the pressures by megapascals, while the opening pressures of neighbouring valves differ by tenths of a megapascal. Even an error of 10 °C in the temperature estimate shifts the nitrogen pressure by more than the step between valves. If the temperature in the well turns out to be higher than assumed, the valves open and close at a higher gas pressure than calculated. The top valve may then fail to open, and unloading will stop. If it is lower, the valves close later, and gas may enter the tubing in several places. The data in this example are assumed for illustration.

Setting the unloading valves of the example on the test rack at 15.6 °C

Valve, depth: 1, 950 m
Temperature, °C: 57.10
Nitrogen in the well, MPa gauge: 11.031
$C_T$: 0.8615
Opening on the rack, MPa gauge: 9.860

Valve, depth: 2, 1570 m
Temperature, °C: 68.26
Nitrogen in the well, MPa gauge: 11.609
$C_T$: 0.8306
Opening on the rack, MPa gauge: 10.004

Valve, depth: 3, 1980 m
Temperature, °C: 75.64
Nitrogen in the well, MPa gauge: 11.943
$C_T$: 0.8113
Opening on the rack, MPa gauge: 10.053

Valve, depth: 4, 2250 m
Temperature, °C: 80.50
Nitrogen in the well, MPa gauge: 12.102
$C_T$: 0.7991
Opening on the rack, MPa gauge: 10.034

Valve, depth	Temperature, °C	Nitrogen in the well, MPa gauge	$C_T$	Opening on the rack, MPa gauge
1, 950 m	57.10	11.031	0.8615	9.860
2, 1570 m	68.26	11.609	0.8306	10.004
3, 1980 m	75.64	11.943	0.8113	10.053
4, 2250 m	80.50	12.102	0.7991	10.034

The nitrogen in the well is the pressure from the previous table minus 0.101 MPa. The nitrogen pressure at 15.6 °C is the product of the third and fourth columns, rounded to 0.001 MPa. The test-rack opening pressure is the same value divided by $1 - R$ = 0.9638. In the well the lower valves must open at a lower gas pressure at the wellhead, but on the rack their opening pressures are almost equal: the deeper the valve, the hotter it is, and the temperature correction eats up the difference. The data are assumed for illustration.

CalculatedHow much higher the pressure of real nitrogen in the dome is on heating than the 0.00215 correction predicts, at three charging pressures on the rack (absolute). Calculated with the Peng–Robinson equation of state (a formula for how real gases behave) from the lesson Cubic equations of state. At a charging pressure of about 4 MPa the correction is almost exact. At 8–12 MPa and 80 °C it underestimates the pressure by 1.7–3%.

How much gas a valve passes

Gas flows through the seat or the port of a valve in the same way as through a choke (gas rate through a choke). While the tubing pressure is above the critical value (about 0.55 of the gas pressure for natural gas), the flow is subcritical and the gas rate grows with the pressure drop. Below that value the flow is critical: the gas reaches the speed of sound in the narrowest part of the port, and the gas rate no longer depends on the tubing pressure (critical and subcritical flow). In gas lift practice, this formula for the port of a valve is named after its authors, Thornhill and Craver (the Thornhill–Craver equation).

CalculatedGas rate through the port of the operating valve at 2400 m against the tubing pressure, at a gas pressure at the valve of 11.465 MPa, for ports of 8.0, 9.7 and 11.1 mm (discharge coefficient 0.85, which shows how much of the ideal flow the port really passes; adiabatic exponent 1.27, which is the ratio of the specific heats of the gas). To the left of the dashed line the flow is critical. The dot is the working mode of the lesson: 60 thousand m³/d at 10.965 MPa in the tubing.

Worked exampleWhat port the operating valve needs

Given (assumed data)

Operating valve: a sharp-edged orifice at 2400 m, through which 60 thousand m³/d of gas must pass at a gas pressure at the valve of 11.465 MPa and a tubing pressure of 10.965 MPa, which is a pressure drop of 0.5 MPa, as assumed in the lesson Gas lift: principles and choice of operating conditions
Gas: specific gravity 0.7, adiabatic exponent 1.27, temperature at the valve 83.2 °C; discharge coefficient of the port 0.85 (assumed)

Solution

The pressure ratio $10.965/11.465 = 0.956$ is greater than the critical ratio $(2/2.27)^{1.27/0.27} = 0.551$ . The flow is subcritical, and the gas rate depends on the tubing pressure.
By the formula for the gas rate through a choke, 60 thousand m³/d passes at these pressures through a port of diameter 9.70 mm.
If the tubing pressure falls below $0.551\cdot 11.465 = 6.32$ MPa, the flow becomes critical, and the port passes no more than 138 thousand m³/d.
A 7.0 mm port would pass 60 thousand m³/d only at a gas pressure at the valve of 12.83 MPa, which is 10.25 MPa at the wellhead. This is higher than the pressures at which valves No. 3 and No. 4 open in operation: 10.11 and 9.95 MPa at the wellhead. (In operation the tubing pressure at them is lower than during unloading, and their opening pressures are a little higher than the unloading ones.)

Answer: The operating valve needs a port of about 9.7 mm. A 7.0 mm port would need a gas pressure of 10.25 MPa at the wellhead. Valves No. 3 and No. 4 would open, and the gas would enter the tubing in three places at once.

What it means. The port of the operating valve is the main “regulator” of the operating mode. It must pass the design gas rate at a small pressure drop, otherwise the gas pressure rises into the range where the unloading valves are set. In subcritical flow the gas rate through the port depends on the tubing pressure, so surges of flow in the tubing also swing the gas supply. This is one of the causes of unstable operation (Gas lift operation and optimisation). Ports with a specially shaped channel (a nozzle with a diffuser) reach critical flow at a pressure drop of only about 10%, according to manufacturers, and supply the gas more evenly. The port is also chosen with the future in mind: as the water cut grows, more gas will be needed, and the valve can be changed on wireline. The data in this example are assumed for illustration.

A bellows valve is more complicated. The ball does not move off the seat to its full travel at once, but gradually as the gas pressure grows, and at the start of the travel the valve passes less than an open seat would. How much less is shown by flow tests of the valve on gas (the method is set by the recommended practice API RP 11V2). From them we find the stiffness of the valve (the pressure change needed to move the stem by one unit of travel) and the flow capacity at different travels. A simple calculation with the full area of the seat is acceptable for small seats. For large ones we allow a margin: a smaller fraction of the calculated gas rate, or a seat one size larger.

Reliability of unloading and control at the wellhead

Margins in the calculation. The opening pressures of neighbouring valves are set apart with a margin, to cover errors of setting on the rack, the inaccuracy of the temperature, swings of the gas pressure in the network, and the pressure drop needed for the stem to travel through its stroke. The more accurate the data, the smaller the margin that is needed, and the deeper the gas can be injected (Gas lift design).
Valve interference. If the tubing pressure at an upper valve has risen (for example, because the flow rate grew when the gas started to go deeper), its opening pressure falls and it may open again. Valves with a small seat are insensitive to this: in the example of this lesson, for valve No. 4 to open again in operation, the tubing pressure at it would have to grow to tens of megapascals.
Signal valve at the bottom (also called a flag valve). The bottom valve is sometimes set to a noticeably lower gas pressure than the others. Then the wellhead pressure shows at once that the gas is going through it, provided there is no leak in the tubing higher up.
Check against measurements. Where the gas really enters is shown by records of the wellhead pressures, by downhole pressure and temperature surveys in the tubing (at the injection point the gas cools the flow), and by calculating the operating mode with a well model. Problems and their signs are explained in the lesson Gas lift operation and optimisation.

Terms and symbols

Gas-lift unloading: Pushing the kill fluid out with gas through the unloading valves, from the top down to the operating valve.
Kick-off pressure: The highest gas pressure at the annulus wellhead that is available during unloading. It sets the depth of the top unloading valve.
Unloading and operating valves: Unloading valves work only during unloading and close when the gas has gone deeper. Gas goes through the operating valve all the time.
Injection-pressure-operated valve: A bellows valve that is opened by the gas pressure in the annulus. The tubing pressure has little effect on it.
Production-pressure-operated valve: The bellows of the valve is loaded by the tubing pressure. The valve opens when the tubing pressure rises.
$p_{\text{dome}}$ , $A_{\text{b}}$ , $A_{\text{p}}$ , $R$: The nitrogen pressure in the dome, the effective area of the bellows, the area of the seat, and the ratio of the seat area to the bellows area.
Valve spread: The difference between the opening and closing pressures, $k_{\text{t}}(p_{\text{dome}} - p_{\text{t}})$ . It is small in valves with a small seat.
Production pressure effect factor (TEF): $k_{\text{t}} = R/(1 - R)$ : by how much the opening pressure falls when the tubing pressure rises by one unit.
Test-rack opening pressure: The gauge gas pressure that opens the valve on the test rack at 15.6 °C and atmospheric pressure on the tubing side.
$C_T$: The correction factor that converts the nitrogen pressure from the temperature of the valve in the well to the temperature of the test rack.
Multipointing: Gas enters the tubing through several valves at the same time. It is a sign of wrong setting or of a problem.

Common mistakes and misconceptions

Wrong

“The compressor gives enough pressure for operation, so it will start the well through the operating valve.”

In fact

In operation the gas works against a light mixture, but during unloading it works against a column of kill fluid. In the example of this lesson, the pressure that is enough for operation through a valve at 2400 m (9.19 MPa) pushes the brine out only down to 821 m during unloading, and the kick-off 10.5 MPa pushes it down to 972 m. Without unloading valves we would need 21.4 MPa.

Wrong

“A valve closes at the same pressure at which it opened.”

In fact

An open valve closes at a gas pressure equal to the nitrogen pressure, which is below the opening pressure by $k_{\text{t}}(p_{\text{dome}} - p_{\text{t}})$ . In the example of this lesson it is 0.155 MPa for a 4.8 mm seat and 0.448 MPa for a 7.9 mm seat.

Wrong

“The temperature correction is a detail that can be ignored.”

In fact

Without it, valve No. 2 of the lesson would open in the well almost 2.5 MPa above the design value, at pressures that the compressor cannot give. Even an error of 10 °C in the temperature shifts the nitrogen pressure by more than the step between the pressures of neighbouring valves.

Wrong

“It is better to take a smaller port for the operating valve, so that less gas goes through.”

In fact

The gas rate is set at the surface. A small port only raises the gas pressure that is needed to pass this rate. When the pressure reaches the opening pressures of the unloading valves, they open, and gas goes into the tubing above the operating valve.

In depth: advanced detail

Where the difference between the opening and closing pressures comes from

Let us subtract the closing pressure $p_{\text{dome}}$ from the opening pressure $p_{\text{g,open}} = (p_{\text{dome}} - Rp_{\text{t}})/(1 - R)$ : $\Delta p_{\text{spread}} = \dfrac{p_{\text{dome}} - Rp_{\text{t}} - p_{\text{dome}} + Rp_{\text{dome}}}{1 - R} = \dfrac{R}{1 - R}(p_{\text{dome}} - p_{\text{t}})$ . On the test rack $p_{\text{t}} = 0$ , and the difference equals $k_{\text{t}}\,p_{\text{dome}}^{\text{rack}}$ . This is the difference that is measured when the closing pressure of the valve is also checked on the rack.

The meaning is simple. While the valve is closed, the area inside the seat is taken by the tubing pressure, which is usually lower than the gas pressure, so the gas has to “pay extra” for this area. As soon as the ball moves off the seat, the gas pressure acts on the whole area of the bellows, and less of it is needed for the same balance. The larger the seat relative to the bellows and the lower the tubing pressure, the larger the difference. So unloading valves are given small seats, and for flow capacity a valve of larger diameter with a larger bellows is used.

When the 0.00215 correction is not enough

The correction $C_T$ is a linear approximation fitted to the properties of nitrogen. For an ideal gas at constant volume, the pressure is proportional to the absolute temperature. Real nitrogen at the pressures in valve domes raises its pressure more on heating, and the higher the charging pressure, the more it raises it. A calculation with the Peng–Robinson equation of state (Cubic equations of state) gives, at 80 °C, a pressure ratio of 1.247 for a charge of 4 MPa and 1.277 for 10 MPa, while the 0.00215 correction gives 1.243 and 1.247. An ideal gas gives 1.223. For nitrogen in this range, this equation of state describes the rise of pressure on heating at constant volume with an accuracy of about 0.2%. It gives the densities themselves with an error of 1–2%, but in the ratio of pressures this error almost cancels.

For the valves of the example this means that domes charged with the 0.00215 correction will have, in the well, a nitrogen pressure above the design value: by 0.18 MPa at valve No. 1 and by 0.28 MPa at valve No. 4. All the valves will open and close at gas pressures higher than designed, by about the same amount. This is more than the pressure step between valves (0.15 MPa at the wellhead). The order of closing in the example is kept, but the margin between the closing pressures of valves No. 1 and No. 2 shrinks from 0.075 to 0.041 MPa.

So for deep wells with high charging pressures we use more accurate relations that take the charging pressure into account (built from the properties of nitrogen, for example by Winkler and Eads), or we calculate the nitrogen pressure directly from an equation of state. Even more accurate is to test the valve at a temperature close to that in the well.

Connections to other topics

Gas lift: principles and choice of operating conditions — gas lift, the operating point, the injection pressure and the depth of gas injection, which are the input data for unloading.
Completion string equipment — the side-pocket mandrels in which gas-lift valves are placed.
Wireline operations — changing gas-lift valves on wireline.
Chokes, surface pressure losses and well temperature — gas rate through a choke and critical flow; the temperature of the flow in the well.
Cubic equations of state — the Peng–Robinson equation, with which the nitrogen temperature correction was checked.
Gas lift design — spacing of the unloading valves and the margins in the calculation.
Gas lift operation and optimisation — problems of gas-lift wells and their diagnosis.

Check yourself

1
Why can a gas-lift well not be started by feeding gas straight to the operating valve, if the compressor pressure is enough for operation through that valve?
1. A.During unloading the gas works against a column of kill fluid in the tubing, and in operation against a light mixture of gas and liquid
2. B.Unloading valves are needed only to save gas
3. C.The gas is colder and heavier during unloading
4. D.The operating valve is always closed during unloading
Show the answer
Correct answer: A. During unloading the gas works against a column of kill fluid in the tubing, and in operation against a light mixture of gas and liquid
While the tubing is full of kill fluid, the pressure in it at the depth of the operating valve equals the weight of the whole column of liquid: in the example of this lesson, more than 26 MPa at 2400 m. In operation the tubing above the operating valve holds a light mixture, the pressure in it at the valve is about 11 MPa, and 11.5 MPa of gas is enough. Unloading valves let us push the liquid out in parts, making the column lighter from the top.
2
The kick-off pressure at the wellhead is 7.0 MPa, the kill fluid is brine of density 1150 kg/m³, and the tubing pressure at the wellhead is 0.8 MPa (pressures are absolute). Estimate the depth of the top unloading valve, first ignoring the weight of the gas, then taking into account that the gas pressure grows by a factor of $e$ every 12,000 m.
Show the answer
Ignoring the weight of the gas: $7.0 = 0.8 + 1150\cdot 9.81\cdot 10^{-6}H = 0.8 + 0.01128H$ , so $H = 6.2/0.01128 \approx 550$ m. With the weight of the gas, the gas pressure at 550 m is $7.0\,e^{550/12{,}000} \approx 7.33$ MPa, which is more than the pressure of the brine at the same depth, so the level goes deeper. The equation $7.0\,e^{H/12{,}000} = 0.8 + 0.01128H$ gives $H \approx 580$ m, so the weight of the gas column adds about 30 m. The valve is placed a little higher, leaving a margin.
3
For a valve operated by the gas pressure, the nitrogen pressure in the well is 9.0 MPa, $R = 0.06$ and the tubing pressure at the valve is 5.0 MPa. At what gas pressure does it open, and at what pressure does it close?
Show the answer
Opening: $p_{\text{g}} = (9.0 - 0.06\cdot 5.0)/(1 - 0.06) = 8.7/0.94 \approx 9.26$ MPa. Closing: at a gas pressure equal to the nitrogen pressure, 9.0 MPa. The difference is 0.26 MPa. The same is given by $k_{\text{t}}(p_{\text{dome}} - p_{\text{t}}) = 0.0638\cdot 4.0 \approx 0.26$ MPa.
4
A valve will work at 90 °C, and the nitrogen pressure at this temperature must be 10.0 MPa (gauge); $R = 0.06$ . What test-rack opening pressure is it set to at 60 °F?
Show the answer
90 °C = 194 °F; $C_T = 1/[1 + 0.00215\cdot(194 - 60)] = 1/1.2881 = 0.7763$ . The nitrogen pressure at 60 °F is $0.7763\cdot 10.0 = 7.763$ MPa. The test-rack opening pressure is $7.763/(1 - 0.06) \approx 8.26$ MPa (gauge). Without the correction the valve would open in the well about 3 MPa above the design value: at 90 °C the dome would have $10.0/0.7763 \approx 12.88$ MPa.
5
While the well is working, the gas pressure at the wellhead has risen and become higher than the opening pressures of the two lowest unloading valves. What will happen?
1. A.The unloading valves will open, and the gas will enter the tubing in several places, above the operating valve
2. B.Nothing: the unloading valves are locked after unloading
3. C.The flow rate will certainly grow, because there will be more gas in the tubing
4. D.The operating valve will close
Show the answer
Correct answer: A. The unloading valves will open, and the gas will enter the tubing in several places, above the operating valve
A closed unloading valve opens when the gas pressure at it exceeds the opening pressure, and then stays open as long as the gas pressure is above the nitrogen pressure. With too much gas, or with a port of the operating valve that is too small, the gas pressure rises, the unloading valves open and part of the gas enters above the operating valve. This is multipointing. The gas is used less effectively to lighten the column above the operating valve, and the flow rate usually falls.
6
Why are unloading valves given small seats, although a large seat passes more gas?
Show the answer
With a small seat the area ratio $R$ is small. The difference between the opening and closing pressures is small, so the upper valve closes after a small fall of the gas pressure following the handover to the lower valve. The effect of the tubing pressure is small as well, so the valve does not open again when the tubing pressure rises. Little gas is needed for unloading, and a small seat is enough. The operating valve, through which the gas goes all the time, is given a larger port.

Updated: 19 October 2026