Sand production: types, prediction and cost of the problem

Quick refresher

Sand is produced when two conditions are met: the rock at the cavity has failed, and the flow is able to carry the broken grains away. So sand often appears not at once, but when the rate rises, when water breaks through, or after a shut-in and restart.
At the wall of a cavity there is no lateral confinement: the effective radial stress there is zero, and the hoop stress (the stress along the circumference of the hole) is at its maximum. A rock that stands tens of megapascals deep in the reservoir loses this support right at the wall. This is why sand starts at cavities.
Drawdown and depletion raise the effective hoop stress at the wall. For an open vertical wellbore with equal horizontal stresses it is $2\sigma'_h + (2 - A)(\bar p - p_{wf})$ , where $\sigma'_h$ grows as the reservoir is depleted.
A cavity stands better than a sample in uniaxial compression, because its wall is supported by the neighbouring rock. We estimate this strength by the thick-wall cylinder test, which reproduces the geometry of the cavity. The strength from a log is an estimate with a scatter of a factor of one and a half or more in each direction. We calibrate it against core and the experience of neighbouring wells.
Sand production can be transient (after perforating and after a change of regime), continuous, or catastrophic. What decides is the allowable sand concentration and where the sand goes.
Managing sand means a smooth ramp-up to the regime, a limit on the drawdown, not perforating the weakest layers, monitoring with sensors, and durable equipment. If this is not enough, we use mechanical or chemical sand control.

Key relationships

\sigma'_1 = C_0 + q\,\sigma'_3,\quad q = \tan^2\!\left(45^\circ + \varphi/2\right)

Mohr–Coulomb criterion in principal effective stresses

\sigma'_\theta = 2\sigma'_h + (2 - A)(\bar p - p_{wf})

effective hoop stress at the wall of an open wellbore during production

\Delta p_{\text{cr}} = \dfrac{C_w - 2\sigma'_{h,i} - 2(1-A)(p_i - \bar p)}{2 - A}

critical drawdown with depletion taken into account

C_w \approx 2.25\,\sigma_{\text{TWC}}

a first estimate of the strength of the cavity wall from the thick-wall cylinder test

v_s = \dfrac{g\,d^2(\rho_s - \rho_l)}{18\,\mu}

settling velocity of a grain in a fluid (Stokes' law)

Common pitfalls

Judging the strength by one correlation with a log, without calibration on core from one's own field.
Assuming that a well that did not produce sand at first will not do so later. Depletion and water cut lower the critical drawdown.
Taking the strength of the “average” rock of the interval. The sand comes from the weakest layer, and it is the one that is least often brought up with the core.
Changing the regime sharply. Every sharp start or rise of production destroys the arches of grains that have formed, and gives a new portion of sand.
Installing sand control “just in case”. It reduces the productivity for ever and makes workovers harder.

After this lesson you can

explain why sand production needs both the failure of the rock and the transport of the broken grains, and which operating conditions bring the failure closer
describe the tests of rock strength (uniaxial and triaxial compression, the thick-wall cylinder, a test that copies the geometry of a hole in the rock), find the parameters of the Mohr–Coulomb criterion (a rule that links the stress at which rock fails to the confining stress), and explain why the wall of a cavity is stronger than a sample in uniaxial compression
estimate the strength from a log, and understand how large the uncertainty of such an estimate is
calculate the effective stress at the wall of a cavity in a producing well, the critical drawdown (the largest pressure difference that the wall can stand) and the critical depletion
tell apart transient, continuous and catastrophic sand production, and estimate its consequences and cost
list the measures for managing sand production, and understand when sand control cannot be avoided

Why the reservoir gives sand

A sandstone is grains held together by cement and friction. The weight of the overlying rocks is shared between the skeleton of grains and the fluid in the pores (Rock compressibility and the effect of stress on core properties), and the skeleton gets the effective stress. Over time, under load and at high temperature, the grains partly dissolve at their contacts, and the dissolved material is deposited in the pores as quartz overgrowths (thin rims of quartz on the grains), calcite and clay minerals. This is how a loose sediment becomes a strong rock. Young, shallow sands have not had time to go this way. Sands in zones of abnormally high pressure do not have time either: the fluid there carries most of the load, the skeleton is lightly loaded, and compaction and cementation lag behind (Abnormal formation pressures: origin, prediction and recognition).

For sand to get into the well, two events are needed. First comes the failure: at the wall of the wellbore or of a perforation tunnel the stresses concentrate, and a rock that cannot stand them loses its cohesion. Then comes the transport: the flow must tear off the broken grains and carry them into the well. Without the second event the broken rock stays in place, and may even become stronger by settling into an arch at the entrance of the tunnel. So weak reservoirs often work for years without sand, and then start to produce it after what looks like a small change of regime.

Weak rock. Little cement, high porosity, well-rounded uniform grains. Poorly sorted angular grains interlock better than rounded ones.
Drawdown. The lower the bottomhole pressure compared with the reservoir pressure, the more strongly the wall is loaded (section on the stresses at the cavity), and the more strongly the flow pulls on the grains.
Depletion. As the reservoir pressure falls, the effective stress grows throughout the reservoir, and so does the load on the wall. Sand that was not there in the first years often appears in the second half of development.
Water cut. In a water-wet sand, thin films and menisci (curved surfaces) of bound water pull the grains together by capillary forces. When water flows in the reservoir, the menisci disappear. Clay cement softens. To keep the former oil rate, we have to raise the liquid rate and the drawdown.
Cyclic loads. Starts and stops, a change of choke, and a pump working in jerks loosen the grains and destroy the arches.
High velocities. At the entrance of a perforation tunnel the flow converges, especially in gas wells, and carries away even slightly broken material.

SchematicA perforation tunnel in a weak sandstone (schematic, not to scale). Around the tunnel is a zone of failed rock, and at the entrance of the tunnel the grains have settled into an arch, which holds at a constant regime. If the failed zones of neighbouring tunnels merge, a cavity forms behind the casing.

Rock strength: uniaxial and triaxial compression, the thick-wall cylinder

The simplest test is uniaxial compression. A cylindrical sample, about twice as long as its diameter, is squeezed along its axis without lateral confinement (nothing presses on its sides), at a constant rate, until it fails. The greatest stress is the unconfined compressive strength (UCS), $C_0$ . For weakly cemented sandstones it is a few megapascals, and for strong ones tens of megapascals and more than a hundred.

In the reservoir the rock is confined from all sides, and this makes it much stronger. For the grains to slide past each other, we have to overcome not only the cement but also the friction, and friction grows with the pressing force. This is shown by the triaxial test: a sample in a rubber sleeve is confined with a lateral pressure $\sigma_3$ and squeezed along the axis until it fails. The greater the confinement, the greater the axial stress at failure. For most rocks, in the working range, the relation between the two stresses is close to a straight line. This is the Mohr–Coulomb criterion.

Mohr–Coulomb strength criterion in principal effective stresses

\sigma'_1 = C_0 + q\,\sigma'_3,\qquad q = \tan^2\!\left(45^\circ + \frac{\varphi}{2}\right) = \frac{1 + \sin\varphi}{1 - \sin\varphi},\qquad C_0 = 2S_0\sqrt{q}

where:

$\sigma'_1,\ \sigma'_3$ largest and smallest effective principal stresses at failure, MPa
$C_0$ unconfined compressive strength (UCS), MPa
$q$ slope of the line: how much confinement strengthens the rock, —
$\varphi$ angle of internal friction, °
$S_0$ cohesion: the shear strength with no normal load, MPa

Units: any identical units of stress. The stresses are effective, that is, less the pore pressure

Holds when

The intermediate principal stress $\sigma'_2$ does not affect the strength (in reality it affects it a little, usually strengthening the rock).
The straight line is an approximation. At large confinements the strength curve flattens, and in loose sands the cohesion is close to zero.
The strength is the peak one, the greatest stress of the test. After the peak a weak rock goes on carrying part of the load.

This is the same line that touches the Mohr circles on the plane of normal against shear stress: $\tau = S_0 + \sigma'_n\tan\varphi$ . For sandstones the friction angle is usually 25–40°, so $q$ is about 2.5–4.6: every megapascal of confinement adds 2.5–4.6 MPa to the strength. So confinement decides everything. At every point of the wall where the radial effective stress is zero, the rock holds only $C_0$ . The cavity itself stands longer: the failure of separate points of the wall moves the load deeper into the rock, and the test on a thick-wall cylinder shows by how much.

CalculatedTests of four samples of a weak sandstone from the example below: the axial effective stress at failure against the lateral confinement, and the Mohr–Coulomb line. Marked are the strength without confinement, as at the wall of a cavity, and at a confinement equal to the effective horizontal stress in the reservoir.

Worked exampleMohr–Coulomb parameters from triaxial tests

Given (assumed data)

Lateral confinement $\sigma'_3$: 0; 2; 4; 8 MPa
Axial stress at failure $\sigma'_1$: 3.6; 10.1; 16.4; 28.5 MPa
Effective horizontal stress in the reservoir: 7.26 MPa (the example on the critical drawdown)

Solution

A line through the four points (least squares): $\sigma'_1 = 3.78 + 3.106\,\sigma'_3$ , that is, $C_0 = 3.78$ MPa and $q = 3.106$ .
The angle of internal friction: $\sin\varphi = (q - 1)/(q + 1) = 2.106/4.106 = 0.513$ , $\varphi = 30.9°$ . The coefficient of friction is $\tan\varphi = 0.60$ .
Cohesion: $S_0 = C_0/(2\sqrt q) = 3.78/(2\cdot1.762) = 1.07$ MPa.
The strength under the confinement that exists deep in the reservoir: $3.78 + 3.106\cdot7.26 = 26.3$ MPa, which is 7 times more than at the wall without confinement.

Answer:

C_0 \approx 3.78

MPa,

\varphi \approx 30.9°

and

S_0 \approx 1.07

MPa. The rock confined by the reservoir stands about 26.3 MPa.

What it means. This is a weak sandstone: without confinement it holds less than 4 MPa, and its friction angle is the usual one for sands. Deep in the reservoir nothing threatens it. Only the wall of the wellbore and of the perforation tunnels is dangerous, because there is no confinement there. Four samples are the minimum. The scatter of the points around the line shows how reliable the line itself is, and samples from weaker layers that could not be taken would have given a lower line. The data in this example are assumed for illustration.

Closest to the conditions at a cavity is the thick-wall cylinder test (thick-wall cylinder, TWC). An axial hole is drilled in the sample, usually one third of the diameter of the sample. The sample is confined from outside and from the ends with the same pressure, and the pressure is raised until the wall of the hole fails and the sample starts to “collapse”. This pressure is called the strength of the thick-wall cylinder, $\sigma_{\text{TWC}}$ . The wall of the hole in the test is loaded in the same way as the wall of a perforation tunnel or a wellbore: the radial stress at it is zero, and the hoop stress is the greatest.

Hoop stress at the wall of the hole of a thick-wall cylinder

\sigma_\theta = \frac{2\,P\,\rho^2}{\rho^2 - 1},\qquad \rho = \frac{D}{d}

where:

$\sigma_\theta$ hoop stress at the wall of the hole, MPa
$P$ external pressure on the sample, MPa
$D,\ d$ outer diameter of the sample and diameter of the hole, mm

Units: any identical units of pressure; the diameters in the same units

Holds when

The rock is elastic up to failure. The pressure in the hole is atmospheric (the pore pressure is not maintained in the test).
This is the Lamé solution for a thick-wall pipe under external pressure. The axial stress does not enter the formula.

At $\rho = 3$ the factor is 2.25: at the moment of failure the hoop stress at the wall of the hole, in an elastic calculation, is $2.25\,\sigma_{\text{TWC}}$ . This is the first estimate of the effective strength of the cavity wall $C_w$ , which we compare with the hoop stress at the wall in a well. It is much larger than $C_0$ : the wall of a small hole is supported by the surrounding rock, and the stress falls quickly with depth into the rock. The factor 2.25 is an assumption and not a constant. Different methods convert the strength of the cylinder into the strength of the wall with a factor of roughly 2 to 3, and refine it from the history of sand production in neighbouring wells. Small holes are stronger than large ones, so for an open wellbore, much wider than the hole in a sample, such an estimate is more likely overstated.

Hardness by pressing a ball into a sawn core, and scratching the surface of the core with a cutter while measuring the force, give the strength almost continuously along the core without breaking it. They are calibrated against compression tests.
Core description is the fastest check. If the grains fall off when rubbed with a finger, and the sample crumbles in the hand, sand production is quite likely. If the core cannot be brought up at all without a rubber sleeve, the rock is loose.

Strength from logs, and its uncertainty

Core is not available in every well or from every layer, and we need the strength over the whole section. We estimate it from logs. What makes a rock strong (cement, compaction, low porosity) also makes it stiff, and sound travels faster in it. So the strength correlates with the interval transit time of the compressional wave $\Delta t$ (the time that sound takes to cross a unit length of rock), with the porosity and with the dynamic Young's modulus (Sonic log). We build the relation from the core of our own field. The published relations for sandstones, obtained in different basins, differ by several times at the same $\Delta t$ .

CalculatedThick-wall cylinder strength of core samples of the field against the interval transit time (logarithmic scale of strength), the relation built from them by least squares, and the band between the 10th and the 90th percentile of the scatter around it. The estimate for a new well at

\Delta t

= 390 µs/m is marked.

Worked exampleStrength from the interval transit time, and its scatter

Given (assumed data)

Core of the field: 14 samples: $\Delta t$ from 305 to 458 µs/m, and thick-wall cylinder strength from 2.6 to 39.3 MPa (graph)
Sandstone in the new well: $\Delta t$ = 390 µs/m

Solution

A relation of the form $\sigma_{\text{TWC}} = a\,e^{-b\,\Delta t}$ is built as a straight line through the logarithms of the strength: $a = 2204$ MPa, $b = 0.01413$ m/µs.
The estimate for the new well: $2204\cdot e^{-0.01413\cdot390} = 8.9$ MPa. In microseconds per foot this is 119 µs/ft.
The scatter: the standard deviation of the logarithm of strength about the line is $s = 0.34$ . Between the 10th and the 90th percentile the factor is $e^{\pm1.2816\,s} = 1.546^{\pm1}$ : from 5.8 to 13.8 MPa.

Answer: The expected thick-wall cylinder strength is about 8.9 MPa, but with a probability of about 80% it lies between 5.8 and 13.8 MPa.

What it means. A scatter of a factor of one and a half in each direction is normal for such correlations. The strength depends on the kind and amount of cement, the size and roundness of the grains, and the shale content, and the interval transit time reflects them only partly. In the section on the critical drawdown we see what this scatter turns into: from “the wall will fail without any drawdown” to “it will stand more than 10 MPa”. Moreover, core is biased towards strong rock: the weakest layers break down when the core is taken and do not get into the sample. So the correlation is more likely to overstate the strength of weak intervals. The data in this example are assumed for illustration.

Dynamic and static properties. The moduli from sonic logs refer to small, fast strains and are higher than the static ones. Going from the moduli to the strength needs its own calibration.
Depth matching. A sample a few centimetres in size is compared with a log that averages tens of centimetres. In a thinly layered section the log does not resolve a thin weak layer.
Heterogeneity. The sand comes from the weakest layer of the interval, not from the average one. So we build the estimates over the whole section and look at the minimum values.

Stresses at a cavity and the critical drawdown

The wellbore and the perforation tunnel are cavities in loaded rock. Around them the stresses concentrate as the Kirsch solution describes (stresses at the wall of a well): at the wall the hoop stress is the greatest, and the radial stress equals the pressure in the cavity. In a producing well the fluid enters the wall, and the pore pressure at it equals the bottomhole pressure $p_{wf}$ , so the effective radial stress at the wall is zero. The wall is the place where the rock is loaded most strongly and, at the same time, has no confinement.

How does the load change during production? Lowering the bottomhole pressure raises the hoop stress at the wall. But the pore pressure around the well also falls. The rock there tends to contract, the surrounding reservoir prevents it, and this partly unloads the wall, by $A(\bar p - p_{wf})$ , where $A$ is the poroelastic coefficient. So every megapascal of drawdown adds not two but $2 - A$ megapascals to the effective hoop stress.

Effective hoop stress at the wall of an open wellbore during production

\sigma'_\theta = 2\sigma'_h + (2 - A)\,(\bar p - p_{wf}),\qquad A = \frac{\alpha(1 - 2\nu)}{1 - \nu}

where:

$\sigma'_\theta$ effective hoop stress at the wall (less the pore pressure at the wall, which equals $p_{wf}$ ), MPa
$\sigma'_h$ effective horizontal stress in the reservoir at the current reservoir pressure, $\sigma_h - \bar p$ , MPa
$\bar p$ average (current) reservoir pressure, MPa
$p_{wf}$ bottomhole pressure during production, MPa
$A$ poroelastic stress coefficient; $\alpha$ is the Biot coefficient and $\nu$ is Poisson's ratio, —

Units: any identical units of pressure

Holds when

A vertical open hole, with both horizontal stresses equal. The rock is elastic, and the Biot coefficient in the strength criterion is 1.
Steady radial inflow, and the pore pressure at the wall equals the bottomhole pressure.
The stress along the axis of the well at the moment of failure is intermediate and does not affect the failure. After noticeable depletion this may stop being true: the effective axial stress at the wall becomes the greatest (example below), and the formula is optimistic.
With unequal horizontal stresses, in place of $2\sigma'_h$ we have $3\sigma'_H - \sigma'_h$ , and for a perforation tunnel $3\sigma'_1 - \sigma'_3$ , where $\sigma'_1$ and $\sigma'_3$ are the two effective stresses in the plane of its cross-section.

The first term is the load that the cavity itself creates in the loaded reservoir. The second is the addition from the drawdown. The wall fails when $\sigma'_\theta$ reaches the effective strength of the wall $C_w$ . The radial stress at the wall is zero, so its strength is the strength without confinement, at the scale of the cavity.

During depletion the effective horizontal stress grows. The overburden pressure is constant, and the total horizontal stress in uniaxial strain of the reservoir falls by only the share $A$ of the fall of the reservoir pressure (poroelastic model). So $\sigma'_h = \sigma'_{h,i} + (1 - A)(p_i - \bar p)$ , where the index $i$ is the initial state. If we substitute this into the formula for the wall and set $\sigma'_\theta = C_w$ , we get the drawdown at which failure begins.

Critical drawdown and critical depletion (open hole)

\Delta p_{\text{cr}} = \frac{C_w - 2\sigma'_{h,i} - 2(1 - A)(p_i - \bar p)}{2 - A},\qquad (p_i - \bar p)_{\text{cr}} = \frac{C_w - 2\sigma'_{h,i}}{2(1 - A)}

where:

$\Delta p_{\text{cr}}$ critical drawdown $\bar p - p_{wf}$ , at which the wall starts to fail, MPa
$C_w$ effective strength of the cavity wall; a first estimate is $2.25\,\sigma_{\text{TWC}}$ , MPa
$\sigma'_{h,i}$ effective horizontal stress at the initial reservoir pressure, MPa
$p_i,\ \bar p$ initial and current reservoir pressure, MPa
$(p_i - \bar p)_{\text{cr}}$ critical depletion: the fall of the reservoir pressure at which the wall fails even without drawdown, MPa

Units: any identical units of pressure

Holds when

The same as for the formula for the wall. The depletion is uniform over the reservoir, and the reservoir deforms only vertically.
$C_w$ does not change with time. In reality water cut and load cycles lower it.
A negative $\Delta p_{\text{cr}}$ means that the wall has already failed without drawdown.

The critical drawdown falls linearly with depletion: for every megapascal of fall of the reservoir pressure it falls by $2(1-A)/(2-A)$ megapascals. The bottomhole pressure below which we must not go, $p_{wf} = \bar p - \Delta p_{\text{cr}}$ , also falls, but more slowly than the reservoir pressure. The window of the safe regime narrows until, at the critical depletion, it disappears altogether.

CalculatedThe region of work without failure of the wall (by the hoop stress) for the wellbore of the example below: the lowest allowed bottomhole pressure against the reservoir pressure as the reservoir is depleted (solid line), and the same with a wall strength 20% lower, for example after water breakthrough (dashed). The thin line is the bottomhole pressure equal to the reservoir pressure (no drawdown). Regimes above the solid line and below the thin one are safe.

Worked exampleCritical drawdown, its fall with depletion, and after water breakthrough

Given (assumed data)

Reservoir: depth 1500 m; average overburden gradient 21.5 kPa/m; initial reservoir pressure 15.3 MPa
Rock: $\nu = 0.30$ ; Biot coefficient 1; the horizontal stresses are equal, no tectonics
Thick-wall cylinder strength from core: 9.0 MPa
Wellbore: vertical, open (uncased) across the reservoir

Solution

Overburden pressure $21.5\cdot1500/1000 = 32.25$ MPa. Effective vertical stress $32.25 - 15.3 = 16.95$ MPa. Effective horizontal stress $0.3/0.7\cdot16.95 = 7.264$ MPa, so $2\sigma'_{h,i} = 14.53$ MPa.
$A = (1 - 0.6)/0.7 = 0.571$ , $2 - A = 1.429$ , $1 - A = 0.429$ . The wall strength is $C_w = 2.25\cdot9.0 = 20.25$ MPa.
The critical drawdown at the initial pressure: $(20.25 - 14.53)/1.429 = 4.0$ MPa. The bottomhole pressure must not be lower than 11.3 MPa.
After depletion to 12 MPa: $(20.25 - 14.53 - 2\cdot0.429\cdot3.3)/1.429 = 2.0$ MPa. The bottomhole pressure must not be lower than 10.0 MPa.
The critical depletion: $(20.25 - 14.53)/(2\cdot0.429) = 6.7$ MPa. At a reservoir pressure below 8.6 MPa the wall fails even without drawdown.
If water breakthrough lowers the strength of the wall by 20%: $C_w = 16.2$ MPa, and at the initial pressure $\Delta p_{\text{cr}} = 1.2$ MPa.
If, instead of the measurement, we take the strength from the log (previous example): at 5.8 MPa (the low end of its range) $C_w = 13.05$ MPa, which is less than $2\sigma'_{h,i}$ , and the wall fails without drawdown. At 13.8 MPa (the high end) $C_w = 31.05$ MPa and $\Delta p_{\text{cr}} = 11.6$ MPa.

Answer: At the initial pressure the wall stands a drawdown of about 4.0 MPa. After depletion to 12 MPa it stands only 2.0 MPa, and with depletion of more than 6.7 MPa the failure will start at any drawdown.

What it means. Of the three “enemies” of the wall (drawdown, depletion and water cut), the fastest is water breakthrough: the loss of a fifth of the strength took away 70% of the allowed drawdown. Depletion acts more slowly, but it is unavoidable. The scatter of the strength from the log turns the answer into a range “from zero to 11.6 MPa”. This is why we calibrate the forecast against wells that already work: against sand sensors, regimes and the history of depletion. The calculation itself is cautious: it predicts the start of failure of the wall, and noticeable sand production begins later, because the failed zone redistributes the load and settles into arches. But it does not take into account weak layers, cycles of starts, unequal horizontal stresses, or the fact that the wellbore is much wider than the hole in a sample, and all this moves the answer in the dangerous direction. The axial stress, which the formula does not check, acts in the same direction: the effective vertical stress grows by the whole fall of the reservoir pressure, and at 12 MPa it is already 20.25 MPa. This is as much as $C_w$ , and more than the hoop stress at the wall. If we compare it too with the strength of the wall, the limit comes even without drawdown. The factor 2.25 between the strength of the cylinder and the strength of the wall is also an assumption, and it affects the answer as strongly as the strength itself. The data in this example are assumed for illustration.

A perforation tunnel is loaded more than the wellbore

A horizontal tunnel in a vertical well sees the vertical stress in its cross-section. For the reservoir of the example, the largest effective hoop stress at its wall, even without drawdown, is $3\sigma'_v - \sigma'_h \approx 43.6$ MPa, against $14.5$ MPa at the wellbore. By an elastic calculation, the wall of a tunnel in such a rock fails right after perforating. And indeed, tunnels in weak sandstones widen, and in the first hours and days of work the well produces the broken material. Then the failed zone stabilises. So for perforated wells we choose the strength $C_w$ in the formula from the history of neighbouring wells, remembering that a small cavity is stronger than a large one.

Transient, continuous and catastrophic sand production

Three kinds of sand production

Kind: Transient
What it looks like: a burst after a start or a step up in production, falling away over hours to weeks
Where it comes from: rock broken by perforating; rearrangement of arches
What to do: ramp up to the regime in steps

Kind: Continuous
What it looks like: a steady background with no decline
Where it comes from: erosion of the failed zone, weak layers
What to do: keep below the allowable level, catch the sand at the surface

Kind: Catastrophic
What it looks like: an avalanche-like growth, the wellbore fills
Where it comes from: collapse of a cavity, water breakthrough, large drawdown
What to do: cut production or stop, clean out, decide about sand control

Kind	What it looks like	Where it comes from	What to do
Transient	a burst after a start or a step up in production, falling away over hours to weeks	rock broken by perforating; rearrangement of arches	ramp up to the regime in steps
Continuous	a steady background with no decline	erosion of the failed zone, weak layers	keep below the allowable level, catch the sand at the surface
Catastrophic	an avalanche-like growth, the wellbore fills	collapse of a cavity, water breakthrough, large drawdown	cut production or stop, clean out, decide about sand control

Transient sand production in gas wells dies out fastest, and in wells with viscous oil more slowly: a viscous fluid pulls the grains harder.

SchematicHow the sand concentration changes with time for the three kinds of sand production (a schematic, with no scale). The steps under the upper graph are increases in production.

The maximum sand-free rate is found by testing: we raise production in steps and watch the sand with sensors. While each step gives only a transient burst, the regime is acceptable. When the background after a step does not return to its former level, the threshold has been passed. This rate is not constant: with depletion and water cut it falls, and we repeat the measurements.

The allowable sand concentration is set not by the reservoir but by what the equipment will stand and what the field can process: the durability of chokes and bends, the capacity of separators, and the options for disposal. For gas wells the limit is much stricter than for oil wells: gas carries the grains at a high velocity, and erosion is faster. The concentration in the fluid is expressed in grams per cubic metre. In English-language practice it is in pounds per thousand barrels: 1 pound per thousand barrels is 2.85 g/m³.

Consequences and cost

The reservoir and the casing. The failed tunnels merge into a cavity, and the casing loses its support and may buckle or bend. The well has to be repaired or is lost. A cavity increases inflow, but it no longer lets us isolate the layers from each other.
The wellbore and the downhole equipment. Sand settles on the bottom and covers the perforated interval. It forms plugs in the tubing (the pipes that carry the oil up) and jams safety valves and downhole tools. Pumps wear out: the stages of an electric submersible pump (a pump driven by an electric motor at the bottom of the well) are abraded, and the plunger and barrel of a sucker-rod pump (a pump driven from the surface by a string of rods) jam (Electric submersible pumps (ESP), Sucker-rod pumping).
The surface. An abrasive jet erodes chokes, bends, tees and valves. Sand accumulates in separators and tanks and reduces their working volume, and they have to be stopped for cleaning.
Disposal. The produced sand is coated with oil. It is washed and disposed of as waste.
The cost of protection. Sand control costs money too: the equipment and the work at completion, the loss of productivity, and less freedom for workovers (repairs of a producing well). A gravel pack (gravel packed around a screen to hold the sand back) in a cased hole adds a skin (extra flow resistance) that grows as the pack gets plugged (losses in a gravel pack). If the productivity is lower, more wells are needed for the same production, and artificial lift (pumps or gas lift) has to be introduced earlier.

Worked exampleHow much sand a “small” concentration produces

Given (assumed data)

Liquid production: 160 m³/d
Sand concentration: 100 g/m³
Sand: grain density 2650 kg/m³; bulk density 1.6 t/m³
Production casing: inside diameter 150 mm

Solution

Per day: $160\cdot100 = 16{,}000$ g, that is, 16 kg. Per year: $16\cdot365/1000 = 5.84$ t.
By volume in the fluid the sand takes up $0.1/2650 = 0.0038$ %. We cannot see it in a sample by eye.
The bulk volume per year: $5.84/1.6 = 3.65$ m³. If all of it settled in the casing, it would fill $3.65/0.01767 = 207$ m of the wellbore.
In units of pounds per thousand barrels this is $100/2.85 = 35$ .

Answer: One well with a concentration of 100 g/m³ produces about 5.84 t of sand a year, which is 3.65 m³.

What it means. A concentration that cannot be told apart in a sample gives cubic metres of sand in a year: a plug hundreds of metres long if it stays in the wellbore, and tonnes of waste for every well. So what matters is not only the amount of sand produced but also where the sand goes: whether it reaches the surface, and where it settles there. The data in this example are assumed for illustration.

Whether the sand reaches the surface depends on the ratio of the flow velocity to the settling velocity of the grain. In a low-viscosity fluid a small grain settles by Stokes' law.

Settling velocity of a grain in a fluid at rest (Stokes' law)

v_s = \frac{g\,d^2\,(\rho_s - \rho_l)}{18\,\mu}

where:

$v_s$ settling velocity of the grain, m/s
$g$ acceleration of gravity, 9.81, m/s²
$d$ diameter of the grain, m
$\rho_s,\ \rho_l$ density of the grain and of the fluid, kg/m³
$\mu$ viscosity of the fluid, Pa·s

Units: SI

Holds when

A single spherical grain. The Reynolds number of the grain, $\rho_l v_s d/\mu$ , is less than one. Otherwise the resistance is greater, and the formula overstates the velocity.
The fluid is at rest and the wellbore is vertical. At a high concentration the grains hinder each other and settle more slowly, and in a deviated wellbore they quickly slide to the lower wall.

The settling velocity grows as the square of the grain size and is inversely proportional to the viscosity: in viscous oil the sand stays in suspension, and in water and gas it settles fast. Angular grains settle more slowly than spheres of the same size, so for them the formula gives an upper estimate.

Worked exampleWill the flow lift the sand

Given (assumed data)

Grain: $d$ = 0.25 mm; $\rho_s$ = 2650 kg/m³
Oil in the wellbore: $\rho_l$ = 860 kg/m³; $\mu$ = 4 mPa·s
Liquid rate: 60 m³/d at wellbore conditions
Pipes: tubing with an inside diameter of 62 mm; below its shoe (its lower end), a casing of 150 mm

Solution

The settling velocity: $v_s = 9.81\cdot(0.25\cdot10^{-3})^2\cdot(2650 - 860)/(18\cdot0.004) = 0.0152$ m/s, that is, 15.2 mm/s. The Reynolds number of the grain is $860\cdot0.0152\cdot0.00025/0.004 = 0.82$ , so Stokes' law applies.
The velocity of the liquid in the tubing: $60/86{,}400/(\pi\cdot0.062^2/4) = 0.23$ m/s, which is 15 times the settling velocity.
In the casing below the tubing: 0.039 m/s, which is only 2.6 times larger.
If the rate is lowered to 20 m³/d, the velocity in the casing falls to 0.013 m/s, which is less than the settling velocity.

Answer: In the tubing the flow carries the sand up with confidence. In the wide casing below it the margin is small, and at a rate of 20 m³/d the sand starts to settle on the bottom.

What it means. Sand produced from the reservoir does not always reach the surface. It accumulates where the flow is slower: in the casing below the tubing, in the sump (the part of the well below the perforations), on the lower wall of a deviated wellbore, and at the surface in the separators. So the shoe of the tubing is set closer to the perforated interval, and when the rate is lowered we watch for a sand plug. In water the same grain settles several times faster, and the Reynolds number goes outside the limits of Stokes' law. The data in this example are assumed for illustration.

Managing sand production

The alternative to continuous protection is sand management: we let the well work without mechanical protection, but in a regime in which sand is not produced, or is produced in allowable amounts, and we are ready to catch it. It is cheaper and it keeps the productivity, but it needs discipline in operation and constant monitoring.

What sand management includes

1
Forecast
The strength from core and logs, the stresses, the critical drawdown and how it changes with depletion and water cut. A comparison with the behaviour of neighbouring wells.
2
Opening the reservoir
Do not perforate the weakest layers if the reservoir is heterogeneous. Choose the direction of perforation so that the tunnels are loaded less. Do not open an interval from which water will come.
3
Ramp-up
Raise production in steps over hours or days, giving the arches time to form. Avoid frequent stops and starts.
4
Regime
Keep the drawdown below the critical one, and revise the limit as the reservoir is depleted and waters out. Maintain the reservoir pressure by injection: this keeps both the energy of the reservoir and the margin of strength of the wall.
5
Monitoring
Acoustic sensors on the flowlines (the pipes from the well to the separator) that register the impacts of grains. Erosion probes that signal when their wall is worn through. Sampling. Measuring the sand level in the separators. Checking the wall thickness at the places where the flow turns.
6
Equipment
Chokes and valves made of durable materials, smooth bends instead of right angles, sand traps, and washing out the separators while they work. Among the methods of artificial lift, gas lift (lifting the oil with injected gas) carries sand better than the others (there are no moving parts in the flow), and so do progressing cavity pumps (screw-type pumps) (Progressing cavity pumps and hybrid systems, Gas lift: principles and choice of operating conditions).
7
Protection
If the allowable drawdown is too small for a profitable rate, or the sand production cannot be held within allowable limits, protection is needed: screens (filters in the well), gravel pack, frac-pack (hydraulic fracturing combined with a gravel pack), and chemical consolidation of the rock (Sand control methods).

The decision “install protection at once, or live with sand” is an economic one. On one pan of the scales are the cost of protection and the lost productivity of all the wells. On the other are the probability that protection will be needed later, its cost, and the losses before it. Late protection in a well that has already produced sand and has a cavity behind the casing works less well than in a new one, and this is built into the calculation. We compare the options by net present value (Net present value (NPV)), and we account for the uncertainty of the strength and of the critical drawdown with scenarios.

Terms and symbols

Sand production: The arrival in the well of the load-carrying grains of the reservoir, after the rock at a cavity has failed and the flow has carried the grains away.
Unconfined compressive strength $C_0$: The greatest stress that a sample stands without lateral confinement (UCS).
Mohr–Coulomb criterion: A linear relation of the stresses at failure: $\sigma'_1 = C_0 + q\sigma'_3$ . The parameters are the cohesion $S_0$ and the angle of internal friction $\varphi$ .
Thick-wall cylinder strength $\sigma_{\text{TWC}}$: The external pressure at which the wall of an axial hole in a sample fails. It is the closest to the conditions at a cavity.
Effective strength of the cavity wall $C_w$: The effective hoop stress at the wall, in an elastic calculation, at which the wall starts to fail. A first estimate is $2.25\,\sigma_{\text{TWC}}$ .
Critical drawdown: The drawdown at which the wall starts to fail. It falls with depletion and water cut.
Critical depletion: The fall of the reservoir pressure after which the wall fails even without drawdown.
Maximum sand-free rate: The highest rate at which, after ramp-up to the regime, the sand production returns to the background. We find it by testing.
Sand arch: An arch of broken grains at the entrance of a perforation tunnel that holds at a constant regime.

Common mistakes and misconceptions

Wrong

“The sandstone in the reservoir stands 26 MPa, so sand production does not threaten it.”

In fact

26 MPa is the strength under confinement, as deep in the reservoir. At the wall there is no confinement: in uniaxial compression the same rock holds less than 4 MPa, and the wall of a cavity holds about 20 MPa of hoop stress (an estimate from the thick-wall cylinder). The stresses in the reservoir already create 14.5 MPa at the wall, and every megapascal of drawdown adds another 1.4.

Wrong

“The well has worked for five years without sand, so it will not need protection.”

In fact

The critical drawdown falls with depletion, and water cut can take away most of it at once. In the lesson's example the loss of a fifth of the strength reduced the allowable drawdown from 4.0 to 1.2 MPa.

Wrong

“The strength from a correlation with the sonic log is a reliable basis for a decision.”

In fact

The scatter of such correlations is a factor of one and a half or more, and the core is biased towards strong rock. We calibrate the estimate against the core of our own field and the behaviour of working wells.

Wrong

“Sand is not seen in a sample, so there is none.”

In fact

100 g/m³ is less than four thousandths of a per cent by volume, but it is about 6 t of sand a year per well at a production of 160 m³/d. We monitor sand production with sensors and measurements in the separators, not by eye.

Wrong

“If sand has started to come, raise production at once to wash out the wellbore.”

In fact

A jump in production destroys the arches and can turn transient sand production into catastrophic sand production. We change the regime in steps, watching the sensors. The plug is washed out separately.

In depth: advanced detail

Where the factor (2 − A) comes from

We add two solutions. The first is the Kirsch solution for a well with the pressure $p_{wf}$ in a reservoir with a uniform pore pressure $\bar p$ : the hoop stress at the wall is $2\sigma_h - p_{wf}$ (the horizontal stresses are equal). The second is a correction for the fact that the pore pressure around the well has fallen from $\bar p$ to $p_{wf}$ at the wall. For such an axisymmetric change of pressure (the same in all directions around the well) in the surrounding reservoir, the solution of the elasticity problem gives at the wall, where the radial stress does not change, a change of the hoop stress of $A\,\Delta p = -A(\bar p - p_{wf})$ (derivation in the lesson on rock mechanics).

In all, the total hoop stress at the wall is $2\sigma_h - p_{wf} - A(\bar p - p_{wf})$ . The effective one, less the pore pressure at the wall, which equals $p_{wf}$ , is $2\sigma_h - 2p_{wf} - A(\bar p - p_{wf})$ . If we substitute $\sigma_h = \sigma'_h + \bar p$ , we get $2\sigma'_h + (2 - A)(\bar p - p_{wf})$ . With unequal stresses the first solution gives, at the most loaded point, $3\sigma_H - \sigma_h - p_{wf}$ , and in place of $2\sigma'_h$ we have $3\sigma'_H - \sigma'_h$ . If the wall were impermeable and the pore pressure at it stayed equal to the reservoir pressure (this happens during drilling, when a cake separates the wall), there would be no poroelastic correction, and every megapascal of fall of the pressure in the well would add exactly a megapascal to the effective hoop stress. During production the pore pressure at the wall falls together with the bottomhole pressure, and the addition is $2 - A$ .

Other criteria and models of sand production

The Mohr–Coulomb criterion ignores the intermediate principal stress. Criteria that take it into account (Drucker–Prager, Mogi–Coulomb and others) usually give a higher strength for the wall of a well: the axial stress along the wellbore also confines the rock. For strongly fractured and weathered rocks we use the nonlinear Hoek–Brown criterion. The choice of the criterion can shift the forecast of the critical drawdown by megapascals, so we calibrate it on the same wells as the strength.

An elastic calculation answers only the question of when the wall will start to fail. How much sand comes afterwards depends on how the zone of failed rock grows and how the flow carries grains out of it. This is calculated with numerical models in which the rock gradually softens after the peak strength, and the production of grains is tied to the flow velocity. Such models are calibrated against tests of samples with a hole through which fluid is pumped, and against measurements of sand in wells. Besides failure by shear, at the wall the grains can be torn off by tension at a very large pressure gradient, and a loose sand that has no cohesion at all can erode.

Connections to other topics

Rock compressibility and the effect of stress on core properties — effective stress, its growth with depletion, and uniaxial strain of the reservoir.
Rock mechanics for hydraulic fracturing — the Kirsch solution, the poroelastic coefficient and the model of horizontal stresses.
Sonic log — the interval transit time and the dynamic moduli from which we estimate the strength.
Core analysis — how core is taken and tested, and why the weak layers are lost.
Formation damage and damage skin — fines migration, another problem with similar signs.
Well completion: opening the reservoir and the tubing string — designs of the bottom of the well: open hole, perforated casing, screen.
Effect of completion on inflow — how much productivity a gravel pack takes away.
Sand control methods — screens, gravel pack, frac-pack and chemical consolidation.

Check yourself

1
What distinguishes sand production from the production of fines?
1. A.The size: sand is larger than 1 mm
2. B.Only gas wells produce sand
3. C.The grains that carry the load in the skeleton of the rock are produced, after the rock has failed
4. D.Nothing: they are the same
Show the answer
Correct answer: C. The grains that carry the load in the skeleton of the rock are produced, after the rock has failed
Fines lie loose in the pores and carry no load. They migrate with the flow and plug the pores. Sand is the framework of the rock itself. For it to move, the rock at the cavity must fail, and the flow must carry the grains away. So the remedies also differ: sand control must hold back the load-carrying grains and let the fines pass.
2
From triaxial tests, $C_0$ = 5 MPa and the angle of internal friction is 30°. What strength does the rock have at an effective confinement of 6 MPa, and what at the wall of a cavity?
Show the answer
$q = (1 + \sin 30°)/(1 - \sin 30°) = 1.5/0.5 = 3$ . At a confinement of 6 MPa: $5 + 3\cdot6 = 23$ MPa. At the wall the radial effective stress is zero, and the strength without confinement is 5 MPa, which is 4.6 times less. So sand starts at the wall and not deep in the reservoir. (The cavity as a whole stands longer than a point of its wall, because the neighbouring rock supports it. The thick-wall cylinder strength takes this into account.)
3
The effective strength of the wall of a wellbore is 24 MPa, the initial effective horizontal stress is 8 MPa, and $A$ = 0.5. Find the critical drawdown at the start of development and after depletion by 4 MPa.
Show the answer
At the start: $(24 - 2\cdot8)/(2 - 0.5) = 8/1.5 \approx 5.3$ MPa. After depletion the effective horizontal stress grows by $(1 - 0.5)\cdot4 = 2$ MPa, to 10 MPa: $(24 - 20)/1.5 \approx 2.7$ MPa, which is half as much. The critical depletion is $(24 - 16)/(2\cdot0.5) = 8$ MPa.
4
Sand started to come after production was increased. What is the most sensible action?
1. A.Install a gravel pack at once
2. B.Increase production still more, to wash out the wellbore
3. C.Go back to the former regime and ramp up to the new one in steps, watching the sand sensors
4. D.Alternate stops and starts to “shake out” the sand
Show the answer
Correct answer: C. Go back to the former regime and ramp up to the new one in steps, watching the sand sensors
The jump in production destroyed the arches of grains. Going back to the former regime and ramping up smoothly to the new one lets the arches form again. If the sand production falls to the background after every step, the regime is acceptable. Sharp changes and cycles of starts only loosen the rock. Protection is needed when the allowable drawdown is too small for a profitable rate.
5
Why does water cut often start sand production in a well that worked without it for years?
1. A.Water is heavier than oil and presses on the rock
2. B.Water cut does not affect sand production
3. C.The capillary bonds between the grains disappear, the clay cement softens, and for the former oil rate the liquid production and the drawdown grow
4. D.Water dissolves quartz
Show the answer
Correct answer: C. The capillary bonds between the grains disappear, the clay cement softens, and for the former oil rate the liquid production and the drawdown grow
In a water-wet sand, bound water pulls the grains together with menisci. When water flows in the pores, these forces disappear. The clay cement softens and weakens. For the former oil rate we have to produce more liquid at a larger drawdown. All this lowers the margin to the critical drawdown, sometimes by most of it at once, as in the example of the lesson.
6
A grain of sand in oil settles at 12 mm/s. The inside diameter of the casing below the tubing is 150 mm. At what liquid rate does the lifting velocity in the casing become equal to the settling velocity?
Show the answer
The cross-section area is $\pi\cdot0.15^2/4 = 0.0177$ m². The rate is $0.012\cdot0.0177 = 2.1\cdot10^{-4}$ m³/s, which is about 18 m³/d. At a lower rate the sand will accumulate in the casing below the tubing, and a reliable lift needs a velocity several times the settling velocity.

Updated: 9 October 2026