Well control: primary barrier and kick indicators

Quick refresher

While the pressure of the mud column is higher than the pore pressure of every permeable formation that has been opened, no fluid enters the well. This is primary control. If an influx has started, we shut in the well with the blowout preventer equipment (secondary control) and kill it.
The bottomhole pressure at rest depends only on the mud density and the height of its column. A kick starts if the formation is “stronger” than expected, or if the mud density or the height of the column has fallen.
The main causes are: mud that is too light; swabbing (the string pulling fluid into the hole like a piston) when the string is pulled out; failure to fill up the hole as the string comes out; lost circulation (mud escaping into the rock); gas entering the mud; and stopping the circulation when only the equivalent circulating density (ECD, the density that the mud effectively has while it circulates) held the formation back.
Direct signs are: a rise of the flow rate at the outlet with an unchanged pump rate; a gain in the volume of mud in the pits (the mud tanks at the rig); flow with the pumps stopped; and a difference between the actual fill-up (or displaced) volume and the calculated one during trips (pulling the string out and running it back in).
Gas expands as it rises, almost in inverse proportion to the absolute pressure (the pressure counted from zero, not from the atmosphere), and most of the expansion happens in the upper few hundred metres. In a shut-in well the rising gas cannot expand, and it raises the pressure in the whole well.
The sooner a kick is noticed, the smaller the volume of the influx, the lower the pressures when the well is shut in and killed, and the lower the risk of rupturing the formation under the casing shoe.

Key relationships

p_{bh} = 0.00981\,\rho_m\,h > p_p

condition of primary control at rest (MPa, g/cm³, m)

\Delta p = 0.00981\,\rho_m\,\Delta h

loss of pressure at the bottom when the mud level falls by $\Delta h$

p_1 V_1 = p_2 V_2

expansion of gas: the pressures are absolute, the temperature and $Z$ are constant

\Delta p_{bh} = \left(\frac{\rho_m}{\rho_s} - 1\right) p_a \ln\frac{p_{bh}}{p_a}

fall of the bottomhole pressure because the mud is gas-cut (the pressures are absolute)

Common pitfalls

Estimating the fall of the bottomhole pressure from the density of gas-cut mud measured at the outlet.
Not comparing the actual fill-up volume with the calculated one on every trip out.
Explaining a flow after pumping a heavy slug (a small batch of heavier mud), or while the mud is warming up, by “safe” causes without a check.
Putting gauge pressure (measured above the atmosphere) instead of absolute pressure into Boyle's law.

After this lesson you can

explain what primary and secondary well control are, and which barriers provide them
list the causes of the loss of primary control and estimate their effect on the bottomhole pressure (the pressure at the bottom of the hole)
tell apart direct and indirect signs of a kick, and name the events that can imitate them
explain how gas behaves in a well: why gas-cut mud (mud with gas mixed into it) lowers the bottomhole pressure only a little, why gas expands near the wellhead, and why its rise in a shut-in well raises the pressures
explain why early detection of a kick decides how difficult and dangerous the killing will be

Primary and secondary control

Fluid flows from a formation into the well if the pressure in the hole opposite a permeable formation (one that lets fluid through) is lower than the pore pressure. Such an influx is called a kick. If it is not stopped, the formation fluid pushes mud out of the well, the pressure in the hole falls even more, and the influx speeds up, until the fluid comes out at the surface out of control. The uncontrolled flow of formation fluid out through the wellhead (the top of the well at the surface) is called a blowout. If fluid from one formation goes along the hole into another formation without coming to the surface, it is an underground blowout, a crossflow between formations.

Two levels of control protect the well from this.

Primary control means that the pressure of the column of drilling mud is higher than the pore pressure of the opened formations (see the lesson on pore and overburden pressure). While this holds, no fluid enters the well, and drilling goes on as usual.
Secondary control: if primary control is lost and an influx has begun, we seal the well at the wellhead with the blowout preventer (BOP) equipment. This stops the influx, but it does not remove its cause. Then we circulate the formation fluid out of the well, keeping the bottomhole pressure above the pore pressure, and we replace the mud with a heavier one. This is killing the well (Well kill methods). After that, primary control works again.

If these fail as well, we turn to emergency measures. For example, we drill an inclined relief well (a second well that reaches the wild well, the one out of control) to kill the wild well through the formation. Such measures are sometimes called tertiary control.

Modern practice describes the same thing through well barriers: between the formation and the surface there must be at least two independent barriers. While we drill, the first barrier is the mud column. The second is an envelope made of the casing (the steel pipe that lines the hole) with its cement, the wellhead and the blowout preventer equipment, which can be closed. The loss of one barrier is a signal to restore it at once, and not a reason to go on working on the one that is left.

How severe a kick is depends on several factors. The permeability and thickness of the formation: a sandstone gives up fluid much faster than a shale. The drawdown (the pressure difference that pushes fluid into the hole): the more the pressure in the hole falls below the pore pressure, the stronger the influx. The kind of fluid: gas, because of its very low viscosity (it flows very easily), enters much faster than oil and water, and as it rises it also expands. And time: the longer an influx goes unnoticed, the more fluid will be in the hole.

Why primary control is lost

The pressure at the bottom of a column of mud at rest is $p_{bh} = 0.00981\,\rho_m\,h$ (MPa, g/cm³, m). So primary control is lost only if the formation turns out to be “stronger” than expected, if the mud density has fallen, or if the height of its column has fallen. To these three ways we add short pressure changes when the pipe moves and when the pumps are stopped.

Causes of kicks and what exactly they change

Cause: The pore pressure is higher than expected
What gets smaller: the margin of the mud pressure over the pore pressure
How it happens: an unforeseen overpressure zone, or the top of a tall gas accumulation, has been opened (Abnormal formation pressures: origin, prediction and recognition)

Cause: The mud density falls
What gets smaller: the density $\rho_m$
How it happens: the solids control equipment (shale shakers and other cleaning devices) removes the weighting material (the heavy powder that makes the mud dense) together with the cuttings (rock chips); the mud is diluted with too much water; gas enters the mud; in deep hot wells the mud expands as it heats

Cause: The hole is not filled up while pipe is pulled
What gets smaller: the height of the column $h$
How it happens: pipe is pulled out of the well, and its volume is not replaced with mud, so the level goes down

Cause: Lost circulation
What gets smaller: the height of the column $h$
How it happens: mud goes into a fractured or cavernous (full of cavities) formation, or into fractures that were opened by too high a pressure in the hole

Cause: Swabbing
What gets smaller: the pressure under the bit during the pull
How it happens: the string that is being pulled acts as a piston, and the mud has no time to flow under the bit

Cause: Stopping the circulation
What gets smaller: the addition to the bottomhole pressure from friction in the annulus (the space between the string and the wall of the hole)
How it happens: if only the ECD held the formation back, the bottomhole pressure falls below the pore pressure when the pumps stop

Cause	What gets smaller	How it happens
The pore pressure is higher than expected	the margin of the mud pressure over the pore pressure	an unforeseen overpressure zone, or the top of a tall gas accumulation, has been opened (Abnormal formation pressures: origin, prediction and recognition)
The mud density falls	the density $\rho_m$	the solids control equipment (shale shakers and other cleaning devices) removes the weighting material (the heavy powder that makes the mud dense) together with the cuttings (rock chips); the mud is diluted with too much water; gas enters the mud; in deep hot wells the mud expands as it heats
The hole is not filled up while pipe is pulled	the height of the column $h$	pipe is pulled out of the well, and its volume is not replaced with mud, so the level goes down
Lost circulation	the height of the column $h$	mud goes into a fractured or cavernous (full of cavities) formation, or into fractures that were opened by too high a pressure in the hole
Swabbing	the pressure under the bit during the pull	the string that is being pulled acts as a piston, and the mud has no time to flow under the bit
Stopping the circulation	the addition to the bottomhole pressure from friction in the annulus (the space between the string and the wall of the hole)	if only the ECD held the formation back, the bottomhole pressure falls below the pore pressure when the pumps stop

The first two causes weaken the barrier along the whole height of the hole. The others act through the level of the mud or through short pressure changes.

A lower density. We watch the density of the mud entering and leaving the well all the time, because it may fall during normal work. Badly chosen solids control equipment removes the weighting material (barite) together with the cuttings. When the mud is treated, it may be diluted with more water than needed. Gas may enter the mud. A separate section below is devoted to gas-cut mud: its effect on the bottom is much smaller than the measurement at the wellhead suggests.

Not filling up the hole. When pipe is pulled, the metal of the pipes leaves the well, and its volume must be replaced with mud at once. This is filling up the hole. If we do not do it, the level goes down, and the pressure at the bottom goes down with it. We calculate the volume to be added in advance from the length of the pipe pulled and its displacement, the volume of metal per metre. When the pulled pipes are full of mud, the volume is that of the metal and of the inner space.

Fill-up volume and fall of the level when pipe is pulled

V = L\, d_s, \qquad \Delta h = \frac{V}{A_c - d_s}, \qquad \Delta p = \rho_m\, g\, \Delta h

where:

$V$ volume of mud that must be added, m³
$L$ length of the pipe pulled, m
$d_s$ displacement of the pipe: the volume of metal per 1 m of length, with the tool joints counted. Numerically it equals the cross-section area of the metal, m³/m
$A_c$ cross-section area of the hole (the inner section of the casing) where the level goes down, m²
$\Delta h$ how far the level goes down if we do not fill up, m
$\Delta p$ fall of the bottomhole pressure, Pa

Units: SI; it is convenient to take $d_s$ in L/m (1 L/m = 0.001 m²) and to calculate $\Delta p\,[\text{MPa}] = 0.00981\,\rho_m\,[\text{g/cm}^3]\;\Delta h\,[\text{m}]$

Holds when

the pipe is pulled “dry”: the mud drains from it back into the well, and the level goes down equally in the pipe and in the annulus;
the denominator is the section of the hole less the section of the metal of the pipes that stay in the well at the level of the mud;
mud does not go into the formation and does not come from it. The deviation of the actual fill-up volume from the calculated one is exactly the sign of trouble.

The fill-up is a “bill” for every metre of metal pulled out. If the well takes less mud than it should, the free volume has been taken by formation fluid. If it takes more, mud is going into the formation.

Worked exampleWhat happens if we do not fill up the well while pulling

Given (assumed data)

Mud density $\rho_m$: 1.22 g/cm³
Static overbalance at the bottom: 1.4 MPa
Displacement of the drill pipe with tool joints $d_s$: 3.2 L/m (pulling “dry” pipe)
Inner diameter of the casing: 220 mm
Pulled without a fill-up: 280 m of pipe

Solution

The volume that should have been added: $V = 280\cdot 3.2 = 896$ L, which is 0.896 m³.
The area in which the level goes down: $A = \pi\cdot 0.220^2/4 - 0.0032 = 0.0348$ m².
The fall of the level: $\Delta h = 0.896/0.0348 = 25.7$ m. The fall of the bottomhole pressure is $\Delta p = 0.00981\cdot 1.22\cdot 25.7 = 0.31$ MPa.
The overbalance that is left: $1.4 - 0.31 = 1.09$ MPa. Pulling about 1270 m of pipe without filling up would “eat” the whole overbalance.

Answer: Without a fill-up the level would go down by 25.7 m, and the bottomhole pressure by 0.31 MPa, which is 22% of the margin.

What it means. Every stand (three pipes joined together) pulled without a fill-up takes away part of the margin, and during the pull swabbing lowers the pressure under the bit for a while as well. So we fill up the well as we pull, and we always compare the actual fill-up volume with the calculated one. If the well “takes” less than it should, the free volume has been taken by formation fluid. This is one of the direct signs of a kick. The data in this example are assumed for illustration.

Swabbing. When the string is pulled, a volume is freed under the bit, and the mud must have time to flow into it from above, through the narrow gap between the string and the wall of the hole. This flow uses up a pressure difference, so during the pull the pressure under the bit is lower than the hydrostatic pressure. The string works like the piston of a syringe. The effect is stronger in these cases: a fast pull; a mud with high viscosity and gel strength (how stiffly the mud sets when it stands still); a small gap between the string and the hole; a thick mudcake; and balling (rock stuck to the bit and the assembly). If the fall of the pressure exceeds the overbalance, the formation “sucks” fluid into the well. When the string is run in, the opposite happens: the pressure under the bit rises (surge). When we run in fast, we may rupture the formation and cause lost circulation.

Lost circulation. Mud may go into a formation with natural fractures and caverns, or into a very permeable formation. It may also go into fractures that the mud itself has opened, when the pressure in the hole exceeds the fracture pressure (the pressure at which the rock cracks). The causes are too high a density, a high ECD, or a sudden run-in of pipe. The level in the well falls, and the pressure at all depths decreases. Another formation with a higher pressure in the same open hole may begin to kick. So lost circulation and a kick often come as a pair. The measures against lost circulation (the smallest density that is enough, a moderate ECD, smooth running in, enough clearance) protect against kicks at the same time.

Stopping the circulation. During circulation the pressure at the bottom is greater than the hydrostatic pressure by the friction pressure losses in the annulus (Circulating pressure losses). If the mud density is chosen only just high enough, a formation may be held by this addition alone, and it may start to kick every time the pumps are stopped, for example when a stand is added to the string. A characteristic sign is gas that comes to the surface in portions after every connection (adding a pipe to the string).

Gas-cut mud: the wellhead frightens more than the bottom

Gas gets into the mud in different ways: from the drilled rock of a gas-bearing formation (even at the right mud density), from swabbing during pulls and connections, and from a real influx. At the surface gas-cut mud looks frightening: its density at the outlet may fall by a third and more. If we multiply this drop in density by the depth of the well, we get an enormous fall of the bottomhole pressure. In reality it is tens of times smaller.

The reason is the compressibility of gas (gas shrinks easily when the pressure rises). At depth, under a pressure of tens of megapascals, the bubbles take up a tiny volume, and the density of the mud is almost the same as the original. They expand mostly in the last tens of metres before the wellhead, where the pressure comes close to atmospheric. The mud is light only in the upper part of the annulus, and almost the whole column stays practically gas-free.

CalculatedThe calculated density of gas-cut mud against depth for the data of the example below: 0.90 g/cm³ at the outlet, with an original density of 1.30 g/cm³. Already deeper than 85 m the density differs from the original by less than 4%.

Fall of the bottomhole pressure because the mud is gas-cut

\Delta p_{bh} = \left(\frac{\rho_m}{\rho_s} - 1\right) p_a \ln\frac{p_{bh}}{p_a}

where:

$\Delta p_{bh}$ by how much the bottomhole pressure is lower than with gas-free mud, MPa
$\rho_m$ density of the mud without gas, g/cm³
$\rho_s$ density of the gas-cut mud measured at the outlet at atmospheric pressure, g/cm³
$p_a$ absolute pressure at the outlet: atmospheric, about 0.101 MPa, MPa
$p_{bh}$ absolute pressure at the bottom, MPa

Units: any identical units of pressure; the pressures are absolute, the logarithm is natural

Holds when

the gas is ideal (it follows the simple gas law), and the temperature is constant along the hole;
the gas moves with the mud and does not dissolve in it, and its mass is negligible;
the gas-cutting is steady: the same amount of gas for each unit volume of mud along the whole annulus;
$p_{bh}$ stands under the logarithm, so we may put in its place the pressure of the gas-free column. The error is negligible.

The fall is proportional to the atmospheric pressure (tenths of a megapascal), and it grows with depth only logarithmically. Even strong gas-cutting lowers the bottomhole pressure by tenths of a megapascal, not by tens. The derivation of the formula is in the advanced section.

Worked exampleGas-cut mud: the naive estimate and the right calculation

Given (assumed data)

Density of the gas-free mud $\rho_m$: 1.30 g/cm³
Density of the mud at the outlet $\rho_s$: 0.90 g/cm³
True vertical depth of the well: 3200 m
Atmospheric pressure $p_a$: 0.101 MPa

Solution

Hydrostatic pressure of the gas-free mud: $0.00981\cdot 1.30\cdot 3200 = 40.81$ MPa.
The naive estimate from the density at the outlet: $0.00981\cdot 0.90\cdot 3200 = 28.25$ MPa, as if the pressure had fallen by 12.6 MPa.
By the formula for gas-cut mud: $\rho_m/\rho_s - 1 = 1.30/0.90 - 1 = 0.444$ . The absolute pressure at the bottom is about $40.81 + 0.101 = 40.91$ MPa. $\Delta p_{bh} = 0.444\cdot 0.101\cdot\ln(40.91/0.101) = 0.27$ MPa.
The density profile (see the plot) shows that already deeper than 85 m the density of the mud differs from the original by less than 4%.

Answer: The pressure at the bottom fell by only 0.27 MPa, and not by 12.6 MPa, as the density at the outlet suggests.

What it means. Gas-cut mud by itself rarely causes a kick, but it is a signal that must be explained. If the gas comes from the drilled rock, it is useless to make the mud heavier: the gas will keep entering while the gas-bearing formation is drilled. If the gas comes in portions after connections and trips, or there is more and more of it, the pressure in the hole may already be close to the pore pressure. Besides, gas at the surface is dangerous in itself (fire, and hydrogen sulfide, a poisonous gas), so a degasser (a device that separates gas from the mud) takes it out. The data in this example are assumed for illustration.

Signs of a kick

An influx begins at the bottom, hundreds and thousands of metres from the drilling crew. So we judge it only by signs at the surface. They are divided into direct signs (primary indicators), which directly mean that an extra volume is entering the well, and indirect signs (secondary indicators), which may point to a kick but may have other causes.

Direct signs of a kick

Sign: A rise of the flow rate at the outlet with an unchanged pump rate
Why it appears: the fluid that comes from below pushes mud out of the annulus in addition to what is pumped in
What can imitate it: a change of the pump rate, a faulty sensor

Sign: A gain in the volume of mud in the receiving pits
Why it appears: the volume of the influx is added to the volume of the circulating mud; the gain equals the volume of the influx, and it is written down for the killing calculations
What can imitate it: transfers between pits, addition of water and chemicals, work of the solids control equipment

Sign: Flow with the pumps stopped
Why it appears: without circulation, only the formation can keep the flow from the well going
What can imitate it: expansion of the mud as it warms up; displacement of mud by a heavy slug by the U-tube effect (the heavy slug sinks and pushes mud out of the annulus, like water in a U-shaped tube); return of mud by a “breathing” formation (one that takes in mud during circulation and gives it back afterwards)

Sign: The fill-up volume is less than the calculated one (when running in, the displaced volume is greater than the calculated one)
Why it appears: part of the freed volume has been taken by formation fluid
What can imitate it: an error in the calculated displacement of the pipe, pulling pipe with mud inside

Sign	Why it appears	What can imitate it
A rise of the flow rate at the outlet with an unchanged pump rate	the fluid that comes from below pushes mud out of the annulus in addition to what is pumped in	a change of the pump rate, a faulty sensor
A gain in the volume of mud in the receiving pits	the volume of the influx is added to the volume of the circulating mud; the gain equals the volume of the influx, and it is written down for the killing calculations	transfers between pits, addition of water and chemicals, work of the solids control equipment
Flow with the pumps stopped	without circulation, only the formation can keep the flow from the well going	expansion of the mud as it warms up; displacement of mud by a heavy slug by the U-tube effect (the heavy slug sinks and pushes mud out of the annulus, like water in a U-shaped tube); return of mud by a “breathing” formation (one that takes in mud during circulation and gives it back afterwards)
The fill-up volume is less than the calculated one (when running in, the displaced volume is greater than the calculated one)	part of the freed volume has been taken by formation fluid	an error in the calculated displacement of the pipe, pulling pipe with mud inside

We treat any direct sign as a kick until a check proves the opposite.

The simplest check of a direct sign is to watch the well with the pumps stopped (a flow check). If the mud goes on coming out, either the formation is at work, or one of the “imitators” listed above is. The procedure for such a check, and for actions when an influx is confirmed, is set by the approved procedures of the rig. Shutting in and killing the well are covered in the lesson on killing the well.

Indirect signs of a kick

Sign: A sudden rise of the rate of penetration (a drilling break)
How it is linked with a kick: the bit has entered a porous undercompacted formation, or a formation with a pressure higher than expected: at a smaller overbalance the drilled rock comes off the bottom more easily
Other possible causes: a change of lithology, for example from shale into sandstone

Sign: A rise of the gas content in the mud
How it is linked with a kick: gas from a formation, especially in portions after connections and trips, shows a small margin of pressure
Other possible causes: a gas-bearing formation that was drilled out, with the mud density correct

Sign: A fall of the pump pressure with a rise of the pump strokes
How it is linked with a kick: a light, low-viscosity fluid in the annulus lowers the pressure that is needed for circulation, and the mud in the pipe begins to “fall through”
Other possible causes: a washout in the drill string (a hole worn in the pipe), or the string breaking

Sign: A change of the cuttings, the torque, the overpull, the properties of the mud
How it is linked with a kick: large splintery shale cuttings and a rise of the torque are signs of an unstable hole at a small overbalance; a change of the salinity of the mud shows an influx of formation water
Other possible causes: many technological causes

Sign	How it is linked with a kick	Other possible causes
A sudden rise of the rate of penetration (a drilling break)	the bit has entered a porous undercompacted formation, or a formation with a pressure higher than expected: at a smaller overbalance the drilled rock comes off the bottom more easily	a change of lithology, for example from shale into sandstone
A rise of the gas content in the mud	gas from a formation, especially in portions after connections and trips, shows a small margin of pressure	a gas-bearing formation that was drilled out, with the mud density correct
A fall of the pump pressure with a rise of the pump strokes	a light, low-viscosity fluid in the annulus lowers the pressure that is needed for circulation, and the mud in the pipe begins to “fall through”	a washout in the drill string (a hole worn in the pipe), or the string breaking
A change of the cuttings, the torque, the overpull, the properties of the mud	large splintery shale cuttings and a rise of the torque are signs of an unstable hole at a small overbalance; a change of the salinity of the mud shows an influx of formation water	many technological causes

Indirect signs are a reason to be on guard and to check the well, not a proof of an influx.

SchematicWhere the signs of a kick can be seen on a rig: 1 — the flow rate of the mud at the outlet; 2 — the volume of mud in the receiving pit; 3 — the trip tank for accounting for volumes during trips; 4 — the standpipe pressure (the pump pressure read on the pipe that feeds the string) and the pump stroke rate; 5 — gas in the mud. At the bottom: an influx from a permeable formation into the annulus. The diagram is not to scale.

A special case is oil-based muds. Gas dissolves well in their oil base. So a gas influx gives a much smaller gain of volume in the pits than in a water-based mud, and it may stay unnoticed for a long time. The dissolved gas comes out only in the upper part of the hole, when the pressure falls below the bubble point pressure (the pressure at which gas begins to come out of the liquid), but it does so quickly and in a large volume. With such muds the control of volumes must be especially careful.

Trips: the most vulnerable time

During trips (pulling the string out and running it in) the risk of a kick is especially high. The pumps are stopped, and the flow rate at the outlet cannot be measured. The level in the well keeps changing. Pulling the string causes swabbing. And if an influx does start, the bit may be far from the bottom, and it will be impossible to circulate the fluid out from the bottom. By industry experience, a considerable share of kicks, including severe ones, is linked with pulling the tools.

So we control volumes during trips with the trip tank, a narrow, tall tank connected to the well. In it even a small change of volume gives a clear change of level. When we pull, we fill up the well from it. When we run in, the displaced mud goes into it. After every few stands we compare the actual volume with the calculated one.

The usual precautions when pulling follow directly from the physics explained above:

before the pull we make sure that the well is quiet: the mud is uniform, there is no gas in it, and the well does not flow;
we pump a small slug of heavier mud into the pipe, so that we can pull the pipe “dry”. Otherwise the mud pours out of the pipe onto the rig floor and makes it hard to keep track of the volumes. We remember that by the U-tube effect the heavy slug will push some mud out of the annulus (see the advanced section);
we limit the speed of the pull, so that swabbing does not “eat” the overbalance;
if there are doubts, we make a short trip: we pull the tools up part of the way, for example to the casing shoe, and run them back to the bottom. Gas or a gain of volume after such a trip shows swabbing.

Gas in the well: expansion and rise

A gas influx behaves differently from an influx of oil or water, because gas is compressible. At a constant temperature its volume is inversely proportional to the absolute pressure (Boyle's law). For a real gas we add corrections for the gas deviation factor $Z$ (a number that shows how far a real gas departs from the simple gas law) and for temperature (Ideal and real gas: the gas deviation factor (Z)).

Expansion of gas as the pressure falls

V_2 = V_1\,\frac{p_1\, Z_2\, T_2}{p_2\, Z_1\, T_1} \;\approx\; V_1\,\frac{p_1}{p_2}

where:

$V_1,\ V_2$ volume of the gas in states 1 and 2, m³
$p_1,\ p_2$ absolute pressures, MPa
$Z_1,\ Z_2$ gas deviation factors, —
$T_1,\ T_2$ absolute temperatures, K

Units: any identical units; the pressure and temperature are absolute

Holds when

the mass of the gas does not change: the gas does not dissolve in the mud and does not come out of it;
the approximate form is for a constant temperature and $Z$ . At the bottom the gas is hotter, so as it rises to the wellhead its volume grows a little less than by the approximate formula, and the change of $Z$ may shift the result in either direction.

Gas lifted from the bottom, where the pressure is 35 MPa, to a depth where the pressure is 3.5 MPa, will grow in volume about 10 times, and at the wellhead hundreds of times.

In an open well (during circulation, or if the well has not been shut in) the gas expands as it rises and pushes mud out of the annulus, so the gain of volume in the pits speeds up. Most of the expansion happens in the upper few hundred metres. The mud that is pushed out lowers the pressure of the column at the bottom, the formation gives up still more gas, and the process speeds up by itself. So a gas influx that is noticed late develops especially fast.

CalculatedThe calculated growth of the volume of a gas pocket as it rises in an open well 3000 m deep with mud of 1.20 g/cm³ (the pressures are absolute; the temperature and

Z

are constant; the lightening of the column by the gas itself is not counted). By the middle of the way the volume only doubles. At a depth of about 290 m it has grown 10 times, and at the wellhead about 350 times.

In a shut-in well the picture is the opposite. Gas is lighter than the mud, and it rises through it even without circulation. If the well is shut in, and the mud and the walls are almost incompressible, the gas has nowhere to expand. Its volume stays almost the same, and so does the pressure in it: the pressure that it had at the bottom. As the pocket rises, it carries this pressure with it. The column of mud above it, up to the wellhead, gets shorter, so the pressure at the wellhead rises. Below it, the column of mud that the pocket has passed through adds its weight, so the pressure at the bottom and in the whole hole rises, including the weakest place, the rock under the shoe of the last casing string.

Worked exampleRise of a gas pocket in a shut-in well

Given (assumed data)

True vertical depth of the well and of the casing shoe: 3000 m and 1800 m
Mud density: 1.20 g/cm³
Pressure in the annulus at the wellhead right after shutting in: 1.5 MPa
Fracture pressure under the shoe: an equivalent mud weight of 1.63 g/cm³ (the density of the mud that would give the same pressure)
Rise of the gas pocket without expansion: by 600 m; the pocket stays below the shoe

Solution

The pressure in the pocket stays the same, while the column of mud above it has become 600 m shorter. So the pressures at the wellhead, under the shoe and at the bottom have grown by the weight of this column: $0.00981\cdot 1.20\cdot 600 = 7.06$ MPa.
The pressure under the shoe before the rise: $1.5 + 0.00981\cdot 1.20\cdot 1800 = 22.69$ MPa (an equivalent mud weight of 1.28 g/cm³). After the rise: $22.69 + 7.06 = 29.75$ MPa (1.68 g/cm³).
The fracture pressure under the shoe: $0.00981\cdot 1.63\cdot 1800 = 28.78$ MPa. It is exceeded by 0.97 MPa. The limit was already reached after a rise of about 520 m.
The pressure at the wellhead has grown to $1.5 + 7.06 = 8.56$ MPa.

Answer: A rise of 600 m raised the pressure in the whole well by 7.06 MPa. Under the shoe it (29.75 MPa) exceeded the fracture pressure (28.78 MPa).

What it means. The influx has already been stopped, but the danger grows. The rock under the shoe may rupture, and then lost circulation begins, with a risk of an underground blowout, or of fluid coming to the surface behind the casing. So we cannot simply leave a shut-in well with gas in it. We follow the pressure and manage it by special methods that let the gas expand in a controlled way (Well kill methods). In reality the small compressibility of the mud and the elasticity of the casing strings let the gas expand a little, and the rise of the pressure is a little less than calculated. The data in this example are assumed for illustration.

Why it is important to notice a kick early

From what has been said it is clear why time decides everything. A small influx that is noticed at once takes a short interval of the annulus. After the well is shut in, the pressures at the wellhead are low, and when the influx is circulated out, the pocket expands moderately. A large influx takes hundreds of metres of the annulus. It lightens the column more, it gives higher pressures when the well is shut in, and, when it is circulated out, the gas expands so much that the pressure under the shoe may exceed the fracture pressure.

A simple estimate shows how this depends on the volume of the influx. A column of light fluid of height $h_k$ in the annulus “falls short” of the hydrostatic pressure by $(\rho_m - \rho_k)\,g\,h_k$ compared with mud. In the simplest case, after the well is shut in, the pressure in the annulus at the wellhead is greater than the pressure in the drill pipe by exactly that amount. The details are in the lesson on killing the well.

Worked exampleA small and a large gas influx

Given (assumed data)

Volume of the influx: 1.5 m³ (noticed at once) and 6.0 m³ (noticed late)
Volume of the annulus at the bottom: 20 L per metre
Density of the mud and of the gas at bottomhole conditions: 1.20 and 0.25 g/cm³

Solution

The height of the pocket in the annulus is $h_k = V/C$ : $1.5/0.020 = 75$ m and $6.0/0.020 = 300$ m.
The shortfall of the hydrostatic pressure: $0.00981\cdot(1.20 - 0.25)\cdot 75 = 0.70$ MPa and $0.00981\cdot(1.20 - 0.25)\cdot 300 = 2.80$ MPa.

Answer: An influx that is noticed late gives a four times larger addition to the pressure in the annulus after shutting in: about 2.80 MPa against 0.70 MPa.

What it means. This is only the initial difference. When the pocket is circulated out, it rises and expands, and the pressure in the annulus grows further, much more for a large pocket than for a small one. If the expected pressure under the shoe exceeds the fracture pressure, the well can no longer be safely shut in and circulated clean in the usual way. So when each section is designed, we estimate the kick tolerance: the largest volume of influx that we can still shut in and circulate out without rupturing the formation under the shoe (Fracture gradient and leak-off testing). The data in this example are assumed for illustration.

Terms and symbols

Kick: An influx of formation fluid (gas, oil or water) into the well, when the pressure in the hole opposite a permeable formation is lower than the pore pressure.
Blowout: The uncontrolled flow of formation fluid out through the wellhead of the well.
Underground blowout: The uncontrolled flow of fluid along the hole from one formation into another.
Primary and secondary control: Holding the formation by the pressure of the mud column; sealing the wellhead with the blowout preventer equipment, if primary control has been lost.
Fill-up, trip tank: Replacing the volume of the pulled pipe with mud; a narrow tank for accurate accounting of these volumes.
Swabbing: A fall of the pressure under the bit when the string is pulled, because of which a formation may “suck” fluid into the well. When the string is run in, the opposite effect is a rise of the pressure (surge).
Gas-cut mud: Mud with gas bubbles. Its density at the outlet is much lower than its density deep in the hole.
Kick tolerance: The largest volume of influx that can be shut in and circulated out without rupturing the formation under the casing shoe.

Common mistakes and misconceptions

Wrong

“The mud at the outlet is a third lighter, so the pressure at the bottom has fallen by a third.”

In fact

Gas expands mostly near the wellhead, and the bottomhole pressure falls by tenths of a MPa. But gas in the mud is a signal: we must find out where it comes from.

Wrong

“The well is shut in, so the danger is over.”

In fact

Rising gas in a shut-in well raises the pressures in the whole well, including the rock under the shoe. We must manage the pressures by the established methods.

Wrong

“A flow after the pumps are stopped is just the mud warming up or a heavy slug.”

In fact

Such explanations are possible, but we check them: a flow from the U-tube effect or from heating dies away and matches the calculation. Until the cause is proved, we treat the flow as a kick.

Wrong

“A couple of stands without a fill-up will change nothing.”

In fact

The shortfall builds up and adds to swabbing, and without accounting for volumes we lose the main way to notice an influx during a pull.

Wrong

“Let us make the mud much heavier, and the problem of kicks is solved.”

In fact

A mud that is too heavy causes lost circulation, and lost circulation lowers the level and itself leads to a kick. The density must lie in the window between the pore pressure and the fracture pressure.

In depth: advanced detail

Where the formula for gas-cut mud comes from

Suppose there are $a$ cubic metres of gas at atmospheric pressure $p_a$ for each cubic metre of mud at the outlet. We neglect the mass of the gas, so the density of the mixture at the outlet is $\rho_s = \rho_m/(1 + a)$ , which gives $a = \rho_m/\rho_s - 1$ . At a depth where the absolute pressure equals $p$ , the same gas takes the volume $a\,p_a/p$ , and the density of the mixture is $\rho_m/(1 + a\,p_a/p)$ . We put this into the equation of hydrostatics $\dd p = \rho\, g \dd z$ and integrate from the wellhead ( $z = 0$ , $p = p_a$ ) to the depth $z$ . We get the pressure profile:

Pressure and density in a column of gas-cut mud

\rho_m\, g\, z = (p - p_a) + a\, p_a \ln\frac{p}{p_a}, \qquad \rho(p) = \frac{\rho_m}{1 + a\, p_a/p}

where:

$z$ true vertical depth from the wellhead, m
$p$ absolute pressure at the depth $z$ , Pa
$a$ volume of gas at atmospheric pressure per unit volume of mud, $\rho_m/\rho_s - 1$ , —
$\rho(p)$ density of gas-cut mud at pressure $p$ , kg/m³
$\rho_m,\ p_a$ density of the mud without gas and atmospheric pressure, kg/m³, Pa

Units: SI; the pressures are absolute

Holds when

the same as for the formula of the fall of the bottomhole pressure: an ideal gas, a constant temperature, and gas that moves with the mud and does not dissolve in it.

The first equation gives the depth at which the pressure reaches $p$ . The plot of the density profile is built from it. For a gas-free mud we would have $\rho_m g z = p - p_a$ , so at the bottom the difference is $a\,p_a\ln(p_{bh}/p_a)$ , and this is the fall of the bottomhole pressure.

The U-tube: the pipe and the annulus

The drill string and the annulus are connected at the bit, like the arms of a U-shaped tube. When the fluids in them are the same, the levels stand at the same height. If there is a heavier fluid in the pipes, it “outweighs” the fluid in the annulus. Mud comes out of the annulus at the surface, and the level in the pipes goes down until the pressures at the bit balance. This is the basis of pumping a heavy slug before a pull, and the same effect explains a flow that is easily taken for a kick. In a shut-in well the same principle links the pressures in the pipe and in the annulus, and the calculations of killing are built on it.

Worked exampleHow much mud a heavy slug pushes out

Given (assumed data)

Mud density: 1.22 g/cm³
Slug in the pipe: 1.5 m³ of mud of density 1.50 g/cm³
Inner volume of the drill pipe: 7.4 L/m

Solution

The height of the slug in the pipe: $1.5/0.0074 = 203$ m.
The pressures at the bit balance when the level in the pipe has gone down by $y$ , at which $1.22\,y = (1.50 - 1.22)\cdot 203$ : $y = 46.5$ m.
In doing so, $46.5\cdot 7.4 = 344$ L of mud comes out of the annulus.

Answer: After the pumps are stopped, the well will “overflow” about 344 L, and the flow will stop.

What it means. Such a flow is expected and calculated in advance. If the flow does not die away, or its volume is clearly larger than calculated, the slug is not the cause, and we treat the flow as a sign of a kick. The data in this example are assumed for illustration.

Connections to other topics

Pore pressure and overburden pressure — hydrostatics, pore pressure and the choice of the mud density.
Abnormal formation pressures: origin, prediction and recognition — how overpressure zones are predicted and recognised: the main source of unexpected kicks.
Well kill methods — what we do after the well is shut in: the pressures in the pipe and the annulus, killing.
Blowout prevention equipment — the design of the blowout preventer equipment, the secondary barrier.
Fracture gradient and leak-off testing — the fracture pressure under the shoe and the kick tolerance.
Circulating pressure losses — the ECD, and why the bottomhole pressure changes when the pumps stop.

Check yourself

1
Which of the signs is a direct sign of a kick?
1. A.A rise of the rate of penetration
2. B.A gain in the volume of mud in the receiving pits with an unchanged pump rate
3. C.A fall of the pump pressure
4. D.A rise of the gas content in the mud
Show the answer
Correct answer: B. A gain in the volume of mud in the receiving pits with an unchanged pump rate
A gain in the volume in the pits directly means that an extra volume, formation fluid, has come into the system. The other signs are indirect: they have other explanations too, such as a change of rock, a washout in the string, or a gas-bearing formation that was drilled out.
2
The well was not filled up while the string was pulled, and the level of the mud went down by 60 m. The mud density is 1.25 g/cm³, and the static overbalance was 1.0 MPa. Has primary control been kept?
Show the answer
The loss of pressure at the bottom is $0.00981\cdot 1.25\cdot 60 \approx 0.74$ MPa. The overbalance that is left is about 0.26 MPa. Formally control has been kept, but the margin is almost used up, and swabbing during the pull lowers the pressure under the bit still more. We must fill the well up, and find out why the actual fill-up volume differs from the calculated one.
3
Mud of 1.30 g/cm³ comes out of a well 3000 m deep gas-cut, with a density of 1.00 g/cm³. By about how much has the bottomhole pressure changed?
1. A.It fell by about 2 MPa
2. B.It fell by about 8.8 MPa
3. C.It fell by about 0.2 MPa
4. D.It has not changed at all
Show the answer
Correct answer: C. It fell by about 0.2 MPa
By the formula for gas-cut mud: $(1.30/1.00 - 1)\cdot 0.101\cdot\ln(38.4/0.101) \approx 0.3\cdot 0.101\cdot 5.94 \approx 0.18$ MPa. The naive calculation from the density at the outlet would give $0.00981\cdot 0.30\cdot 3000 \approx 8.8$ MPa, almost 50 times more. Gas expands mostly near the wellhead, and deep down the mud is hardly lightened.
4
The well was shut in after a gas kick, and the influx stopped. Why does the pressure at the wellhead begin to rise after a while, although nothing is being done?
Show the answer
Gas is lighter than mud, and it rises. In a shut-in well it has nowhere to expand, so the pressure in the pocket stays almost equal to the original pressure at the bottom. As the pocket rises, the column of mud above it gets shorter, so the pressure at the wellhead rises, and the weight of the column it has passed is added below it, so the pressure at the bottom and under the shoe rises. If we do not manage the pressure, the rock under the shoe may be ruptured.
5
A slug of heavier mud was pumped into the pipe, the pumps were stopped, and mud began to flow out of the annulus. What is true?
1. A.This is certainly a kick, there is no other explanation
2. B.This is the mud warming up, and we can pay no attention to it
3. C.This may be the displacement of mud by a heavy slug by the U-tube effect: such a flow dies away and matches the calculated volume. If it does not die away, or the volume is larger, it is a kick
4. D.A flow without the pumps working is impossible
Show the answer
Correct answer: C. This may be the displacement of mud by a heavy slug by the U-tube effect: such a flow dies away and matches the calculated volume. If it does not die away, or the volume is larger, it is a kick
A heavy slug in the pipe outweighs the mud in the annulus and pushes it out, until the pressures at the bit balance. This volume is calculated in advance. If the flow does not stop, or its volume is larger than expected, the cause is the formation, and we treat the flow as a kick.
6
Why is early detection of a kick so important, if we can shut the well in with the blowout preventer equipment in any case?
Show the answer
The longer an influx goes unnoticed, the more fluid is in the hole. A large influx lightens the column more, gives higher pressures at the wellhead when the well is shut in, and, when it is circulated out, the gas expands so that the pressure under the shoe may exceed the fracture pressure. A small influx that is noticed at once fits within the kick tolerance, and we can shut it in and circulate it out safely.

Updated: 22 September 2026