Diagenesis: compaction, cementation and dissolution

Quick refresher

Effective stress, temperature, time and the composition of the pore waters control diagenesis. Near the surface the pore waters are renewed, and material is brought in and carried out. On burial the exchange slows, and the system is almost closed for silica and aluminium (almost nothing enters or leaves).
Mechanical compaction of rigid grains quickly reduces the intergranular volume (the volume between the grains) to about 28% by 1.5 km, and almost stops at about 26%. Ductile fragments are crushed further. The greatest effective stress the rock has experienced sets the compaction, not the present depth: on uplift the pores do not return.
Quartz cement grows on the clean surface of quartz grains. Its rate rises by about a factor of 1.7 for every 10 °C. The volume of cement depends on the time spent in the hot zone, the surface area of the grains and the share of the surface covered by coats. Pressure solution (grains dissolving where they press on each other) at grain contacts and at stylolites supplies the silica.
Dissolution of feldspars and carbonates creates pores that are visible in thin section. But in a closed system the products, kaolinite and quartz, precipitate nearby, and the gain in porosity is several times smaller than the volume dissolved. Real secondary porosity exists where solutions carry the material away.
Carbonates change earlier and more strongly than sandstones. They show dissolution of aragonite, early cement, pressure solution at shallow depths, and dolomitization (calcite turning into dolomite). When dolomite replaces calcite “mole for mole”, the volume of the solid phase falls by almost 13%. But the change needs magnesium to flow in: hundreds or thousands of pore volumes of seawater.
The order of events decides the result. Early cement can save the framework from compaction. An early arrival of oil can slow cementation, according to several studies. We build the prediction with compaction and cementation models calibrated at wells, not by extending a trend.

Key relationships

\mathrm{IGV} = \mathrm{IGV}_f + (P_i - \mathrm{IGV}_f)\,e^{-\beta\sigma'}

intergranular volume after mechanical compaction; $\sigma'$ is the greatest effective stress

\sigma' = (G_{ob} - G_p)\,z

effective vertical stress from the gradients of overburden and pore pressure

r = a\cdot 10^{bT}

rate of quartz precipitation per unit area of grain surface; $T$ in °C

V_q = \phi_0\left(1 - e^{-E}\right)

volume of quartz cement; $E$ grows with the surface area of quartz and the time in the hot zone

\mathrm{2\,KAlSi_3O_8 + 2\,H^+ + H_2O \rightarrow Al_2Si_2O_5(OH)_4 + 4\,SiO_2 + 2\,K^+}

dissolution of K-feldspar with precipitation of kaolinite and quartz

Common pitfalls

Treating porosity as a function of the present depth: compaction is set by the greatest effective stress, and quartz cement by the time spent in the warmth.
Expecting abnormal pressure to preserve the porosity of a quartz sandstone at any depth: it holds back compaction, but for quartz cement what matters is temperature and time.
Counting all the dissolution porosity visible in thin section as a gain: in a closed system the pores are mostly redistributed.
Transferring a cementation rate from one basin to another without allowing for the heating history, the grain size and the coats on the grains.
Confusing the “diagenesis” of English-language reports with the Russian-language one: most of the changes on burial are called catagenesis in Russian-language practice.

After this lesson you can

tell early diagenesis, changes on burial and changes on uplift apart, and match them with the Russian-language terms — diagenesis, catagenesis, hypergenesis
calculate the intergranular volume from the mechanical compaction curve and estimate how ductile grains and abnormal pore pressure change it
explain how pressure solution and stylolites supply silica, and calculate the volume of quartz cement from a kinetic model for a given heating history
explain why grain coats, coarse grains and rapid burial preserve porosity, and why abnormal pressure protects against compaction but not against quartz cement
make a volume balance for the dissolution of feldspar and for dolomitization, and distinguish new porosity from a redistribution of pores
describe how the paragenetic sequence (the order of diagenetic events) is reconstructed, matched with the time of oil arrival and used to predict reservoir quality away from wells

What diagenesis is: stages, terms and driving forces

Fresh sand on the bed of a river or sea is a loose packing of grains with a porosity of about 40–45%, filled with water. After tens of millions of years at a depth of 4 km it becomes a sandstone with a porosity of 5–25%, whose grains are pressed together, partly fused and surrounded by new minerals. Everything that happened between these two states, apart from metamorphism, is diagenesis. The lesson on reservoir quality gives an overview of how diagenesis changes reservoir quality. This lesson covers the processes themselves.

It is convenient to distinguish three settings, by whether the pore waters communicate with the surface. Early diagenesis (eodiagenesis) takes place near the surface. This is usually in the first hundreds of metres, or, by a broader definition, down to 1–2 km and about 70 °C. Here the pore water is connected with the water of the basin or with groundwater on land, so material is easily brought in and carried out. Burial diagenesis (mesodiagenesis) takes place in rock that is cut off from the surface. Temperature and effective stress rise, and the exchange of material with the surroundings is slow. Late changes on uplift (telodiagenesis) happen when rock is brought up to the surface, or just beneath an unconformity (an old erosion surface with a gap in the rock record), and meets fresh waters again.

Terms: English-language and Russian-language

English-language term: eodiagenesis
Where and when: near the surface, pore waters are renewed
Russian-language term: diagenesis

English-language term: mesodiagenesis
Where and when: on burial: heating, growing stress
Russian-language term: catagenesis (early and late)

English-language term: telodiagenesis
Where and when: on uplift and erosion, inflow of fresh water
Russian-language term: close to hypergenesis

English-language term	Where and when	Russian-language term
eodiagenesis	near the surface, pore waters are renewed	diagenesis
mesodiagenesis	on burial: heating, growing stress	catagenesis (early and late)
telodiagenesis	on uplift and erosion, inflow of fresh water	close to hypergenesis

The correspondence is approximate. In Russian-language lithology, the transformation of a sediment into a rock is diagenesis. The changes of the rock on burial are catagenesis. Still deeper, at the transition to metamorphism, they are metagenesis. Changes near the surface under the action of the atmosphere and fresh waters are hypergenesis. English-language literature speaks of the late stage of diagenesis instead of catagenesis. There, the word catagenesis usually means the maturation of organic matter (its change into oil and gas).

There are four driving forces. Effective stress is the part of the weight of the rocks that the framework of grains carries (effective stress). It controls mechanical compaction. Temperature controls the rate of chemical reactions, such as the growth of quartz cement and the transformations of clay minerals. Time decides how far these reactions have gone: at the same temperature, an ancient rock is changed more than a young one. The composition of the pore waters (marine, fresh or brine, with dissolved gases and acids) decides which minerals dissolve and which precipitate (come out of solution). And the original composition and texture of the sediment set what these forces act upon: sorting, grain size, and the share of ductile fragments, feldspars and aragonite.

SchematicTypical temperature ranges of the main processes of sandstone diagenesis (schematic). The boundaries are approximate and shift with the composition of rocks and waters and with the burial history; depths are for a geotherm of 15 °C at the surface and 30 °C per kilometre.

The order in which the events took place is called the paragenetic sequence of diagenetic events. We reconstruct it from the relationships between minerals: what grew on what, what replaced what, which cement filled the pores before compaction and which after it. For this we use several tools. Thin sections (slices of rock seen under a microscope) and point counting show the minerals and pores (intergranular volume and cement in thin section). Cathodoluminescence (the glow of minerals under an electron beam) tells a quartz overgrowth from the grain it grew on. The scanning electron microscope shows clay minerals. For temperatures and ages, we use fluid inclusions in cement (tiny drops of fluid trapped in the crystals), isotopic composition and the radiometric age of authigenic illite (illite that grew in the rock).

Mechanical compaction: intergranular volume and compaction curves

While the sand is not cemented, the grain contacts carry the weight of the overlying rocks. As the effective stress grows, grains slip and rotate to pack more tightly, and brittle ones crack. Soft ones (shale fragments, micas, altered volcanic fragments) are crushed and squeezed into neighbouring pores. They turn into pseudomatrix. We measure compaction by the intergranular volume (IGV): the fraction of the rock that lies between the framework grains, whether pores, cement or matrix fill it.

Observations of weakly cemented sandstones show the general picture. In well-sorted sands of rigid grains, the intergranular volume falls from 40–42% at the surface to about 28% at a depth of 1.5 km. After that it hardly changes and stays at about 26%. The framework has become stable, and pores are lost further only through chemical processes. An exponential in effective stress describes such a curve conveniently.

The mechanical compaction curve

\mathrm{IGV} = \mathrm{IGV}_f + (P_i - \mathrm{IGV}_f)\,e^{-\beta\sigma'}, \qquad \sigma' = (G_{ob} - G_p)\,z

where:

$\mathrm{IGV}$ intergranular volume after compaction, % of the rock volume
$P_i$ initial (depositional) porosity — the intergranular volume at deposition, %
$\mathrm{IGV}_f$ the intergranular volume to which compaction tends at very high stresses, %
$\beta$ compaction coefficient, MPa⁻¹
$\sigma'$ the greatest effective vertical stress that the rock has experienced, MPa
$G_{ob},\ G_p$ gradients of overburden and pore pressure, kPa/m = MPa/km
$z$ depth of burial, km

Units: percent of the rock volume; stresses in MPa; depth in km

Holds when

The form is empirical. It is used in published models of sandstone compaction. We fit $P_i$ , $\mathrm{IGV}_f$ and $\beta$ to thin sections from wells for each type of sand, by sorting and the share of ductile grains.
The sand is not cemented: cement that grew early props up the framework, and compaction goes no further.
Compaction is irreversible: on uplift and erosion the intergranular volume is not restored, which is why the formula contains the greatest stress, not the present one.
There is no matrix; clay matrix compacts by its own laws.

What counts is stress, not depth. At the same burial depth, abnormally high pore pressure reduces $\sigma'$ , so the sand is only as compacted as it would be at a shallower depth (undercompaction). The exponential porosity trend with depth (the compaction trend) has the same nature, but it mixes compaction with cementation.

Worked exampleIntergranular volume at depth: rigid grains, ductile fragments and abnormal pressure

Given (assumed data)

Sand at deposition: intergranular volume $P_i = 41$ %
Curve of rigid quartz sand: $\mathrm{IGV}_f = 26$ %, $\beta = 0.11$ MPa⁻¹ — chosen to pass about 28% at 1.5 km and to tend to 26%
Curve of sand rich in ductile fragments: $\mathrm{IGV}_f = 12$ %, $\beta = 0.07$ MPa⁻¹ (assumed)
Pressures: overburden 22.6 kPa/m; normal pore pressure 10.4 kPa/m; in the abnormal-pressure zone 16.0 kPa/m

Solution

Effective stress at normal pressure grows by $22.6 - 10.4 = 12.2$ MPa per kilometre: at 1, 2 and 3 km it is 12.2, 24.4 and 36.6 MPa.
Rigid sand: $\mathrm{IGV} = 26 + 15\,e^{-0.11\sigma'}$ . At 1 km: $26 + 15\cdot e^{-1.342} = 26 + 15\cdot 0.2613 = 29.9$ %; at 2 km: $26 + 15\cdot 0.0683 = 27.0$ %; at 3 km: $26 + 15\cdot 0.0178 = 26.3$ %.
Sand with ductile fragments at 3 km: $12 + 29\cdot e^{-0.07\cdot 36.6} = 12 + 29\cdot 0.0772 = 14.2$ %.
Abnormal pressure at 3 km: $\sigma' = (22.6 - 16.0)\cdot 3 = 19.8$ MPa — as at a depth of $19.8/12.2 = 1.62$ km at normal pressure. Rigid sand: $26 + 15\cdot 0.1133 = 27.7$ %; with ductile fragments: $12 + 29\cdot 0.2501 = 19.3$ %.
Porosity loss by compaction, in percent of the original rock volume, by the formula of the lesson on reservoir quality: rigid sand at 3 km $100\cdot(41 - 26.3)/(100 - 26.3) = 19.9$ %; with ductile fragments $100\cdot(41 - 14.2)/(100 - 14.2) = 31.2$ %, and with abnormal pressure 26.9%.

Answer: Rigid sand has almost exhausted mechanical compaction by 2 km (27.0%), and from 2 to 3 km it loses only another 0.7% of the rock volume. Sand with ductile fragments keeps only 14.2% at 3 km. Abnormal pressure adds 1.4% of the rock volume to the intergranular volume of rigid sand, and 5.1% to that of sand with ductile fragments.

What it means. For quartz sand below 2 km, compaction no longer decides how many pores are left. Cement does (next sections). The framework is stable, and the difference in porosity between beds is a difference in cement. For sands with shale fragments and micas, the main enemy is compaction. By 3 km they lose about one and a half times as many pores, and abnormal pressure saves a noticeable part of them by lowering the effective stress. So the practical rule is this: look for pores preserved by abnormal pressure in sands that are still compacting, that is, loose ones and those rich in ductile grains. The parameters of the curves are assumed. In a real basin, we fit them to the intergranular volume in thin sections, plotted against the greatest effective stress reconstructed from the burial history. The data in this example are assumed for illustration.

CalculatedCalculated compaction curves of the example. The lines show the intergranular volume of rigid quartz sand (solid) and of sand rich in ductile fragments (dashed) at normal pore pressure. The dots show the same sands at 3 km in the abnormal-pressure zone.

Chemical compaction: pressure solution and stylolites

The weight of the rocks passes through the contacts between grains. These are small areas where the stress is many times greater than the average over the rock. A mineral under high stress dissolves more readily. So at the contacts quartz slowly goes into solution, and the dissolved silica diffuses (spreads) to the free surface of grains in neighbouring pores and is precipitated there. This is pressure solution, or chemical compaction. The grains grow into each other: point contacts become linear, then concavo-convex and sutured (toothed). The intergranular volume shrinks beyond the limit of mechanical compaction.

Dissolution is uneven. Clay particles and mica at the contact of quartz grains speed it up markedly. So quartz dissolves first of all along clay laminae and mica laminae, and there stylolites form. These are sutured (toothed) seams along which from fractions of a centimetre to several centimetres of rock have dissolved. The insoluble residue (clay, iron oxides, organic matter) has gathered in a thin film. The dissolved silica is precipitated nearby, in neighbouring beds of sandstone, as quartz cement. In carbonates, pressure solution begins at smaller depths and lower temperatures than in quartz sandstones, and stylolites are found everywhere there.

Source of cement. A sandstone in which stylolites have “eaten” part of the rock cements itself: the silica travels centimetres or decimetres, not kilometres. So we calculate the volume of quartz cement from the surface of quartz on which it precipitates and from time, not from an inflow from outside (next section).
Baffles to flow (barriers that slow it). A stylolite with a clay film is poorly permeable across it, and the cement that has grown nearby makes the adjacent beds denser. In carbonates such seams divide a reservoir into packages that communicate poorly (Reservoir architecture and performance).
Signs in core. Dark, toothed (sutured) seams, usually parallel to the bedding. The teeth are from fractions of a millimetre to centimetres high. Next to the seams the rock is more compacted and less porous.

Quartz cement: temperature and time

Quartz precipitates from pore water onto the surface of quartz grains and continues their crystal lattice. The grain gets a quartz overgrowth (a regenerated rim) and tends towards the shape of a crystal with faces. The boundary of the old grain is often visible as a “dusty” rim of inclusions, and under cathodoluminescence the grain and the cement glow differently. Cement grows where there is a clean quartz surface. On grains of feldspar and rock fragments, and on grains covered with coats of clay minerals, it hardly starts to grow at all.

There is little silica in pore water, so cement grows slowly, over millions of years. A widely used kinetic model (a model based on reaction rates) assumes that the slowest step is the growth of the crystals themselves. There is enough silica from contacts and stylolites, and the rate of precipitation on the surface grows with temperature by an exponential law. For the sandstones on which it has been tested, the model describes the volume of cement well.

Rate of quartz cement precipitation (Walderhaug's model)

r = a\cdot 10^{bT}

where:

$r$ rate of quartz precipitation per unit area of grain surface, mol/(cm²·s)
$T$ temperature, °C
$a,\ b$ constants: in the original calibration of the model on Jurassic sandstones of the Norwegian shelf $a = 1.98\cdot 10^{-22}$ mol/(cm²·s), $b = 0.022$ °C⁻¹, mol/(cm²·s); °C⁻¹

Units: mol/(cm²·s), °C

Holds when

The rate is set by precipitation on the surface, not by the delivery of silica: the source is pressure solution nearby.
Quartz grains with a clean surface; a surface covered by coats takes no part in precipitation.
The constants were found for particular sandstones. For another basin, check them against the volume of cement in thin sections and the temperatures of fluid inclusions.

Every 10 °C speeds up precipitation by a factor of $10^{0.22} \approx 1.66$ . At 80 °C the rate is about $1.1\cdot 10^{-20}$ mol/(cm²·s), at 130 °C it is $10^{1.1} \approx 12.6$ times greater. So noticeable cement grows only above about 70–80 °C, and what matters most is the time the sandstone has spent in the hot zone.

Cement precipitates onto the surface of the quartz grains. The area of this surface per unit volume of rock is what matters. For spheres of diameter $D$ occupying a fraction $f$ of the rock volume, it equals $6f/D$ . Coats covering a fraction $C$ of the surface leave only a fraction $1 - C$ free. As the cement grows, the free surface shrinks along with the remaining pores. We add up the precipitation over the whole time of heating and get the volume of cement (the derivation is in the advanced section):

Volume of quartz cement at a constant heating rate

V_q = \phi_0\left(1 - e^{-E}\right), \qquad E = \frac{M\,a\,A_0}{\rho\,\phi_0\,b\,\ln 10\,H}\left(10^{bT_2} - 10^{bT_1}\right), \qquad A_0 = \frac{6f}{D}(1 - C)

where:

$V_q$ volume of quartz cement in 1 cm³ of rock, fractions of the rock volume
$\phi_0$ porosity at the start of cementation — the intergranular volume left by compaction, fraction
$M,\ \rho$ molar mass and density of quartz: 60.08 g/mol and 2.65 g/cm³, g/mol; g/cm³
$A_0$ area of the free surface of quartz grains in 1 cm³ of rock, cm²
$f$ fraction of the rock volume occupied by detrital quartz grains, —
$D$ grain size, cm
$C$ fraction of the quartz surface covered by coats on the grains, —
$H$ heating rate of the rock, °C/s
$T_1,\ T_2$ temperature at the start of cementation and the present (or final) one, °C

Units: cm, cm², g, mol, s, °C; $V_q$ and $\phi_0$ as fractions of the rock volume

Holds when

The heating rate is constant. For a complex history, take the integral in time steps.
The free surface decreases in proportion to the remaining pores: $A = A_0(\phi_0 - V_q)/\phi_0$ .
Mechanical compaction after the start of cementation is not taken into account — the cement props up the framework.
The sandstone is quartzose; feldspars and rock fragments are not included in $f$ .

The more quartz surface (fine grains, a lot of quartz, no coats) and the longer the rock was heated, the more cement: the factor $1/H$ is the time spent at each degree. At $E \ll 1$ the cement grows almost in proportion to $E$ , at large $E$ the pores almost disappear.

Worked exampleHow much quartz cement will grow by 129 °C

Given (assumed data)

Sand: the same rigid quartz sand; grains 0.20 mm (0.020 cm); detrital quartz is 60% of the rock volume; no coats on the grains
History: now at a depth of 3.8 km; geotherm 15 °C at the surface and 30 °C/km, as in the lesson on reservoir quality; uniform heating, 2 °C per million years (burial of about 67 m per million years)
Start of cementation: 80 °C — an assumption of the calculation
Model constants: $a = 1.98\cdot 10^{-22}$ mol/(cm²·s), $b = 0.022$ °C⁻¹; quartz 60.08 g/mol, 2.65 g/cm³; 1 million years is $3.156\cdot 10^{13}$ s

Solution

The sand passed 80 °C at a depth of $(80 - 15)/30 = 2.167$ km, where $\sigma' = 12.2\cdot 2.167 = 26.4$ MPa and $\mathrm{IGV} = 26 + 15\cdot e^{-0.11\cdot 26.4} = 26 + 15\cdot 0.0548 = 26.8$ %: $\phi_0 = 0.268$ . The present temperature is $15 + 30\cdot 3.8 = 129$ °C.
Quartz surface in 1 cm³ of rock: $A_0 = 6\cdot 0.60/0.020 = 180$ cm².
Heating rate $H = 2/(3.156\cdot 10^{13}) = 6.34\cdot 10^{-14}$ °C/s. The numerator is $M a A_0/\rho = 60.08\cdot 1.98\cdot 10^{-22}\cdot 180/2.65 = 8.08\cdot 10^{-19}$ cm³/s; the denominator is $\phi_0 b\ln 10\,H = 0.268\cdot 0.022\cdot 2.303\cdot 6.34\cdot 10^{-14} = 8.61\cdot 10^{-16}$ ; their ratio is $K = 9.38\cdot 10^{-4}$ .
$E = K\,(10^{0.022\cdot 129} - 10^{0.022\cdot 80}) = 9.38\cdot 10^{-4}\cdot(688.7 - 57.5) = 0.592$ .
Cement: $V_q = 0.268\cdot(1 - e^{-0.592}) = 0.268\cdot 0.4468 = 0.120$ , that is, 12.0% of the rock volume; porosity $26.8 - 12.0 = 14.8$ %.
Chlorite coats on 85% of the surface: $E$ is multiplied by 0.15 and equals 0.0888; $V_q = 0.268\cdot 0.0850 = 0.0228$ — 2.3%, porosity 24.5%.
Heating twice as slowly (1 °C per million years) doubles $E$ : 1.184, $V_q = 0.268\cdot 0.6939 = 0.186$ , porosity 8.2%. Twice as fast (4 °C): $E = 0.296$ , $V_q = 0.268\cdot 0.2562 = 0.0687$ , porosity 19.9%.
What took more — compaction or cement — in percent of the original rock volume, by the formulas of the lesson on reservoir quality: compaction $100\cdot(41 - 26.8)/(100 - 26.8) = 19.4$ %, cement $12.0\cdot(100 - 41)/(100 - 26.8) = 9.7$ %; with slow heating the cement is $18.6\cdot 59/73.2 = 15.0$ %.

Answer: By 129 °C about 12.0% of quartz cement will grow, and porosity will fall from 26.8 to 14.8%. With chlorite coats, porosity will be preserved at 24.5%. With heating twice as slow, it will fall to 8.2%, and with heating twice as fast, only to 19.9%. Compaction took 19.4% of the original volume, and cement 9.7% (15.0% with slow heating).

What it means. Compaction took more pores from this sand than cement did. But compaction is the same in all rigid sands, and cement is not. The same depth and the same temperature give a porosity anywhere from 8.2 to 19.9%, depending on how long the sandstone spent in the hot zone. An old, slowly subsiding bed is cemented more strongly than a young, rapidly buried one. The same factor applies to grain size: a fine-grained sand with 0.10 mm grains is cemented in the same way as this one under twice as slow heating, because it has twice the surface. Continuous coats decide most of all: they leave quartz only 15% of the surface. The calculation is optimistic in two places. It leaves out cement that precipitates below 80 °C (starting the calculation at 60 °C, one would get 0.5% more cement by 129 °C). And it leaves out mechanical compaction after 80 °C (by 3.8 km the intergranular volume of rigid sand would shrink by another 0.7%). The model constants were found for particular sandstones. In a new basin, we check them against the volume of cement and the temperatures of inclusions in wells. The data in this example are assumed for illustration.

CalculatedCalculated porosity of the rigid quartz sand of the example as it is heated above 80 °C. The curves are for heating of 1, 2 and 4 °C per million years without coats on the grains (the numbers beside the curves), and for 2 °C per million years with chlorite coats on 85% of the surface (dashed). The vertical mark is 129 °C, the present temperature of the sand in the example.

Coats on grains. Continuous coats of chlorite, less often of microcrystalline quartz, cover the surface, and cement grows only in their gaps (clay minerals in the pores). The share of covered surface directly reduces $E$ : at 50% by half, at 85% by almost a factor of seven. So what matters is not whether coats are present but whether they are continuous. We assess continuity over many grains.
Grain size and composition. Coarse grains and a lot of feldspar and rock fragments mean less quartz surface and less cement. Fine-grained quartz sandstones are cemented the most.
Rapid burial. Less time in the hot zone means less cement. That is why young deep sandstones are often more porous than old ones at the same depth.
Abnormal pressure. It holds back mechanical compaction, but in the model the quartz cement does not depend on stress. The intergranular volume that is preserved can be filled with quartz if there are no coats.
Oil. Oil that arrived in the trap early, according to several studies, slows cementation, but by how much is a matter of debate (the section on the timing of events).

Carbonate cement and authigenic clay minerals

Carbonate cement in sandstones is often early. Calcite, dolomite or siderite precipitates right at the surface, often around the remains of shells and organic matter. It cements separate areas completely, and this is how concretions (hard lumps) and dense layers arise. Large calcite crystals that enclose many grains at once (poikilotopic cement) are sometimes visible on a broken surface by the shine of cleavage planes (flat faces along which a crystal splits). Early cement props up the framework, so a large intergranular volume is preserved under it, but it is already filled with cement. On burial, late ferroan (iron-bearing) carbonates, such as ferroan calcite and ankerite, can precipitate, while part of the early carbonates dissolves.

Clay minerals in the pores grow at different stages. The lesson on reservoir quality covers how each of them reduces permeability. For the processes, what matters is when they grow and what they grow from:

Authigenic minerals of sandstones: when they grow and what they do

Mineral: Calcite, dolomite, siderite (early)
When it grows: near the surface, the first hundreds of metres
Effect on the pores: concretions and dense layers; prop up the framework

Mineral: Chlorite coats
When it grows: more often early: iron-rich clay coats of the sediment turn into chlorite on burial
Effect on the pores: get in the way of quartz cement

Mineral: Kaolinite
When it grows: on flushing by fresh water, replacing feldspars; near the surface and on uplift
Effect on the pores: booklets of plates in the pores, micropores

Mineral: Illite from smectite
When it grows: on burial, at about 60–120 °C, in clays and clay matrix
Effect on the pores: releases water and silica

Mineral: Fibrous illite
When it grows: above about 120 °C, from kaolinite and K-feldspar
Effect on the pores: bridges in the pores; permeability falls by an order of magnitude

Mineral: Ankerite, ferroan calcite
When it grows: on burial
Effect on the pores: fill the remaining pores

Mineral	When it grows	Effect on the pores
Calcite, dolomite, siderite (early)	near the surface, the first hundreds of metres	concretions and dense layers; prop up the framework
Chlorite coats	more often early: iron-rich clay coats of the sediment turn into chlorite on burial	get in the way of quartz cement
Kaolinite	on flushing by fresh water, replacing feldspars; near the surface and on uplift	booklets of plates in the pores, micropores
Illite from smectite	on burial, at about 60–120 °C, in clays and clay matrix	releases water and silica
Fibrous illite	above about 120 °C, from kaolinite and K-feldspar	bridges in the pores; permeability falls by an order of magnitude
Ankerite, ferroan calcite	on burial	fill the remaining pores

The temperatures are approximate and depend on the composition of the waters and rocks. For example, kaolinite turns into illite only if potassium is available. If it is lacking, kaolinite persists deeper.

Dissolution and secondary porosity

Pore water dissolves whatever is unstable in the new conditions: feldspars, volcanic rock fragments, carbonate cement, shells, salt and gypsum. Pores are left where the dissolved material was. This is secondary porosity. We recognise it by its form: pores larger than the neighbouring grains, voids in the shape of grains, partly dissolved grains with a “honeycomb” residue, and grains that “float” in pores without touching their neighbours.

The main question is where the dissolved material goes. Dissolving feldspar needs acid. Carbonic acid supplies it, including carbonic acid in fresh waters, and so do organic acids released when source rocks mature. The dissolution also yields products that must go somewhere: potassium, aluminium and silica. Aluminium is almost insoluble in pore water and usually precipitates at once as kaolinite. Silica precipitates as quartz cement. For K-feldspar the reaction is:

Dissolution of K-feldspar with precipitation of kaolinite and quartz

\mathrm{2\,KAlSi_3O_8 + 2\,H^+ + H_2O \;\rightarrow\; Al_2Si_2O_5(OH)_4 + 4\,SiO_2 + 2\,K^+}

where:

$\mathrm{KAlSi_3O_8}$ K-feldspar (microcline, orthoclase), mol
$\mathrm{Al_2Si_2O_5(OH)_4}$ kaolinite, mol
$\mathrm{SiO_2}$ quartz (if the silica has precipitated nearby), mol
$\mathrm{H^+},\ \mathrm{K^+}$ hydrogen (acid) and potassium ions in the pore water, mol

Units: numbers of moles of reactants and products

Holds when

Silica precipitates within the same volume of rock as quartz, and aluminium as kaolinite: the system is closed for silicon and aluminium.
Potassium leaves with the water. On burial deeper than about 120 °C, it may itself go into the growth of illite.

Two moles of dissolved feldspar yield a mole of kaolinite and four moles of quartz. Kaolinite is porous (there are micropores between its plates), but these pores are tiny and give hardly any permeability.

Worked exampleFeldspar has dissolved — how many pores have actually been added

Given (assumed data)

Sandstone: 10% of the rock volume, occupied by K-feldspar, has dissolved
Minerals: K-feldspar 278.33 g/mol and 2.56 g/cm³; kaolinite 258.16 g/mol and 2.60 g/cm³; quartz 60.08 g/mol and 2.65 g/cm³

Solution

Molar volumes $V = M/\rho$ : feldspar $278.33/2.56 = 108.72$ , kaolinite $258.16/2.60 = 99.29$ , quartz $60.08/2.65 = 22.67$ cm³/mol.
For 2 mol of feldspar — $2\cdot 108.72 = 217.44$ cm³ — a mole of kaolinite (99.29) and 4 mol of quartz ( $4\cdot 22.67 = 90.68$ ) precipitate: together 189.97 cm³. The volume freed is $217.44 - 189.97 = 27.47$ cm³, that is, 12.6% of the volume of the dissolved feldspar.
Closed system: 10% of feldspar gives $10\cdot 99.29/217.44 = 4.57$ % of kaolinite and $10\cdot 90.68/217.44 = 4.17$ % of quartz cement, and porosity gains only $10\cdot 0.126 = 1.26$ %.
Open system — the water carries away both aluminium and silica: porosity gains the whole 10%.

Answer: In a closed system the dissolution of 10% of feldspar adds only about 1.26% of porosity. The rest of the space is taken by 4.57% of kaolinite and 4.17% of quartz. An open system would give the whole 10%.

What it means. In thin section, such a rock looks “porous”: voids are visible where feldspar grains were. But beside them, booklets of kaolinite and quartz overgrowths have grown in the pores. The total porosity has hardly changed, and permeability has more likely fallen, because large connected pores have been replaced by the micropores of kaolinite. This is a redistribution of pores, not new porosity. A system is open where waters flow through a bed for a long time and in large amounts: near the surface, beneath unconformities, along faults. At great burial the exchange is slow, and the system is close to closed for silicon and aluminium. How close is a matter of long debate, so we estimate the gain in porosity from dissolution at depth with caution. The data in this example are assumed for illustration; the molar masses and densities are rounded reference values.

Carbonate diagenesis

Carbonates change earlier and more strongly than sandstones. Aragonite and magnesian calcite (forms of calcium carbonate) are unstable, and carbonate dissolves and is redeposited easily (carbonate minerals). Diagenesis creates or destroys a considerable part of the porosity of a carbonate reservoir, and the traces of its stages are visible in the same rock.

On the sea floor. Fibrous aragonite and magnesian calcite cement grows on grains even before burial. When sedimentation pauses, the sea floor hardens. Such cement makes the rock rigid early, but it can close part of the pores.
In the fresh-water zone. When sea level falls, shoals and reefs are exposed. Aragonite dissolves, leaving moulds, and the dissolved carbonate precipitates nearby as calcite cement. This is an example of the redistribution of pores, covered in the lesson on carbonates. Karst (a landscape and voids shaped by dissolution) develops beneath unconformity surfaces: vugs, channels, caves and collapse breccias (rubble from collapsed cave roofs).
On burial. Pressure solution and stylolites appear already at hundreds of metres, and the dissolved carbonate cements the neighbouring pores. Late solutions, which are acidic and rich in carbon dioxide or hydrogen sulfide, can both dissolve and cement.
Fractures. Rigid cemented carbonates break in a brittle way, and fractures link isolated pores into a network (Fractures and naturally fractured reservoirs).

Dolomitization is the replacement of calcite by dolomite. Magnesium is brought by seawater; by brines (very salty waters concentrated by evaporation in lagoons and sabkhas, or salt flats) that drain downwards through carbonates; or by hot solutions rising along faults. Dolomite forms where brine or solution flows through the rock for a long time. So dolomitized bodies follow the paths of flow: the margins of platforms, faults, permeable packages. How many pores result depends on how the replacement proceeds.

Dolomitization “mole for mole”

\mathrm{2\,CaCO_3 + Mg^{2+} \;\rightarrow\; CaMg(CO_3)_2 + Ca^{2+}}

where:

$\mathrm{CaCO_3}$ calcite, mol
$\mathrm{CaMg(CO_3)_2}$ dolomite, mol
$\mathrm{Mg^{2+}},\ \mathrm{Ca^{2+}}$ magnesium and calcium ions in the pore solution, mol

Units: numbers of moles

Holds when

All the carbonate of the dolomite is taken from the replaced calcite: only magnesium is brought in, only calcium is carried out.
This is the limiting case, with the greatest gain of pores. If the solution brings carbonate as well, dolomite may replace limestone “volume for volume”, and no pores are added.

Two moles of calcite occupy more space than a mole of dolomite, so the volume of the solid phase decreases by almost 13% (molar volume), and the space freed may become pores.

Worked exampleDolomitization of limestone: how much space and how much magnesium

Given (assumed data)

Limestone: porosity 20%, calcite framework
Replacement: “mole for mole”: two moles of calcite give one mole of dolomite; the rock volume does not change
Molar volumes: as in the lesson on carbonates: calcite $100.09/2.71$ , dolomite $184.40/2.86$ cm³/mol
Seawater: magnesium 52.8 mmol per kg, density 1025 kg/m³

Solution

Molar volumes: calcite 36.93, dolomite 64.48 cm³/mol. The volume of the solid phase changes by $64.48/(2\cdot 36.93) - 1 = -12.7$ %.
1 m³ of limestone contains 0.80 m³ of calcite; after replacement it is $0.80\cdot 0.873 = 0.698$ m³ of dolomite, and the porosity is $1 - 0.698 = 0.302$ , that is, 30.2%.
1 m³ of rock contains $0.80\cdot 10^6/36.93$ — about 21,660 mol of calcite; half as much magnesium is needed — 10,830 mol.
1 m³ of seawater contains $0.0528\cdot 1025 = 54.1$ mol of magnesium. The volume of water needed is $10{,}830/54.1 = 200$ m³ — 1000 pore volumes, if the water gives up all its magnesium.

Answer: The limiting “mole for mole” replacement would raise the porosity from 20 to 30.2%. But for this, at least 200 m³ of seawater must pass through every cubic metre of limestone. That is a thousand pore volumes, and only if the water gives up all its magnesium.

What it means. Two conclusions follow. First, dolomitization is a process of flow, not of stagnation. It proceeds where a solution is pumped through the rock for a long time, and we build the map of dolomites from the paths of this flow. Second, 13% is an upper estimate. Often dolomite grows “volume for volume”, at the expense of brought-in carbonate, and the porosity of dolomite is largely inherited from the limestone. Early dolomite that grew before compaction can even reduce porosity, and excess dolomite that goes on growing in the pores can destroy it. Dolomite resists compaction and pressure solution better than limestone, so at great depths dolomites are often more porous than limestones. The composition of seawater is the modern one. The data in this example are assumed for illustration.

Timing of events, oil arrival and prediction of quality

We tie the sequence of diagenetic events to time through the burial history (burial history curve and trap timing). Fluid inclusions in a quartz overgrowth were trapped while it grew. The homogenisation temperature of an inclusion (the laboratory temperature at which its contents become one phase) shows at what temperature the cement grew, and the burial curve shows when. The radiometric age of fibrous illite gives the time of its growth. We compare these data with the time of oil generation and migration, and find out which came first: cement or oil.

The question is not an idle one. Suppose oil came into the trap early. In the oil-saturated part, the pore water then remains only as thin films, and according to several studies the delivery of silica to the surface of grains slows. Other studies show that in water-wet sandstones (where water coats the grains) cement goes on growing under oil as well. The example below shows what this dispute costs.

Worked exampleOil has arrived — will the cement stop?

Given (assumed data)

Sandstone: from the quartz cement example, without coats; heating 2 °C per million years; now 129 °C
Oil: arrived when the bed was at 100 °C
Two limits: (a) oil completely stopped the growth of quartz in the oil-saturated part; (b) oil had no effect on the cement

Solution

Limit (a): cement grew only from 80 to 100 °C. $E = 9.38\cdot 10^{-4}\cdot(10^{0.022\cdot 100} - 57.5) = 9.38\cdot 10^{-4}\cdot(158.5 - 57.5) = 0.0947$ ; $V_q = 0.268\cdot(1 - e^{-0.0947}) = 0.268\cdot 0.0904 = 0.0242$ — 2.4%, porosity $26.8 - 2.4 = 24.4$ %.
Limit (b) is the calculation of the previous example: 12.0% of cement, porosity 14.8%.
The difference between the limits is 9.6% of the rock volume. In the water-bearing part of the bed, below the contact, limit (b) is correct.

Answer: In the oil-saturated part, the porosity lies between 14.8 and 24.4%, depending on how far the oil holds back the cement. In the water-bearing part below the contact, it is about 14.8%.

What it means. Here the dispute about oil and cement is worth 9.6% of the rock volume in porosity. That is the difference between a good and a mediocre reservoir. We test it on the accumulation itself. We compare the porosity and the volume of cement of the same sandstone above and below the oil–water contact, and we compare the temperatures of inclusions in the cement with the time when oil arrived. If there is noticeably less cement above the contact, oil held it back. If there is the same amount, it did not. Something else follows for development too: the water-bearing part of the bed beneath the accumulation may turn out to be noticeably denser than the oil-saturated part. This is important for aquifer influx (water moving in from the aquifer) and water injection. The data in this example are assumed for illustration.

How reservoir quality is predicted from processes

1
Facies and composition
From core and the facies model we learn where the different sands are, and their grain size, sorting, share of quartz, feldspars and ductile fragments, and the likelihood of coats on the grains (quality prediction).
2
Burial and temperature history
A burial curve and geotherm checked against vitrinite reflectance (a measure of how strongly the rock has been heated) and inclusion temperatures (Source rocks, maturation and migration). From them we get the greatest effective stress and the time spent at each temperature.
3
Compaction
A curve of intergranular volume against stress for each type of sand, fitted to thin sections from wells.
4
Quartz cement
A kinetic model. The quartz surface comes from the grain size and composition, and the share of coats from the facies. We check the constants against cement in wells.
5
Other cements and dissolution
Early carbonates, illite, dissolution — by facies and the history of the waters, usually as probabilities rather than as an exact calculation.
6
Permeability and uncertainty
We take permeability from the relationship of its own rock type (rock types). The result is a range, not a single number, together with an answer to the question: what is different on the new structure from the calibration wells?

Terms and symbols

Diagenesis and catagenesis: All the changes of a sediment and rock after deposition and before metamorphism; in Russian-language lithology the changes of the rock on burial are distinguished as catagenesis.
Intergranular volume (IGV): The fraction of the rock between the framework grains — pores, cement and matrix; a measure of mechanical compaction.
Pseudomatrix: Crushed ductile fragments squeezed into the pores during compaction; it looks like clay matrix but arose after deposition.
Pressure solution: Dissolution of grains at loaded contacts and along stylolites with redeposition nearby; chemical compaction.
Quartz overgrowth: Quartz cement growing on quartz grains as a continuation of their lattice; the rate rises with temperature.
Grain coats: Thin crusts of chlorite, less often of microcrystalline quartz or other minerals, covering the surface of the grains and hindering quartz cement.
Secondary porosity: Pores that arose after deposition — by dissolution, dolomitization, fractures.
Redistribution of pores: Dissolution with precipitation of the products nearby: visible pores appear, while the total porosity hardly changes.
Dolomitization: Replacement of calcite by dolomite; “mole for mole” it reduces the volume of the solid phase by almost 13% and needs a large inflow of magnesium.
Paragenetic sequence: The order in which compaction, cementation, dissolution and replacement took place, reconstructed from the relationships of minerals.

Common mistakes and misconceptions

Wrong

“Porosity is set by depth: at one depth one sandstone has one porosity.”

In fact

Compaction is set by the greatest effective stress, and quartz cement by the time in the hot zone. In the lesson's example the same sand at 129 °C has a porosity of from 8 to 20% depending on the heating rate.

Wrong

“10% of feldspar has dissolved, so porosity has grown by 10%.”

In fact

Only in an open system. If aluminium and silica have precipitated nearby as kaolinite and quartz, the gain is about 1.3%, and permeability may even fall.

Wrong

“Under abnormal pressure a quartz sandstone will keep its porosity at any depth.”

In fact

Abnormal pressure lowers the effective stress and holds back mechanical compaction, but for quartz cement what matters is temperature and time. The preserved intergranular volume will be overgrown with quartz if there are no coats on the grains.

Wrong

“Dolomitization always improves a reservoir.”

In fact

A gain of up to 13% of the volume is an upper estimate for “mole for mole” replacement. Often the replacement is “volume for volume”, early dolomite may reduce porosity, and excess dolomite may fill the pores.

Wrong

“The report says diagenesis, so it means early diagenesis in the Russian-language sense.”

In fact

In English-language literature diagenesis includes the changes on burial that Russian-language lithology calls catagenesis. Check which stage and which temperatures are meant.

In depth: advanced detail

Where the formula for the volume of quartz cement comes from

Let the free quartz surface in 1 cm³ of rock be $A$ and the precipitation rate $r$ mol/(cm²·s). In a time $\mathrm{d}t$ , $r A\,\mathrm{d}t$ mol of quartz precipitates, with a volume of $(M/\rho)\,r A\,\mathrm{d}t$ cm³. The surface shrinks along with the remaining pores: $A = A_0(\phi_0 - V_q)/\phi_0$ . We get $\dfrac{\mathrm{d}V_q}{\mathrm{d}t} = \dfrac{M A_0}{\rho\,\phi_0}\,r\,(\phi_0 - V_q)$ .

At constant heating $T = T_1 + H t$ , and $\mathrm{d}t = \mathrm{d}T/H$ . Separating the variables: $\dfrac{\mathrm{d}V_q}{\phi_0 - V_q} = \dfrac{M A_0 a}{\rho\,\phi_0 H}\,10^{bT}\,\mathrm{d}T$ . The integral is $\int 10^{bT}\mathrm{d}T = 10^{bT}/(b\ln 10)$ , so $-\ln\dfrac{\phi_0 - V_q}{\phi_0} = E$ with $E$ from the main section, and therefore $V_q = \phi_0(1 - e^{-E})$ .

The formula shows at once what depends on what. $E$ is proportional to $f(1 - C)/(D H)$ . So doubling the grain size has the same effect as doubling the heating rate, and coats on 85% of the surface have the same effect as an almost sevenfold increase in the heating rate. Also, $E$ is inversely proportional to $\phi_0$ : for the same quartz surface, a smaller pore volume is overgrown faster in relative terms. For a complex history, with breaks and uplifts, we take the integral in steps, each with its own heating rate. On uplift and cooling the growth almost stops.

Connections to other topics

Reservoir quality: geological controls on porosity and permeability — an overview of the factors of reservoir quality, intergranular volume in thin section, porosity losses by compaction and cementation (COPL and CEPL), rock types and prediction.
Porosity — primary and secondary porosity and the exponential compaction trend.
Pore pressure and overburden pressure — overburden pressure and effective stress — the force of mechanical compaction.
Abnormal formation pressures: origin, prediction and recognition — undercompaction and abnormal pore pressure.
Carbonate sediments, environments and carbonate porosity — carbonate minerals, pore types and the subsalt buildups of the Pre-Caspian Basin.
Minerals and rocks — minerals of sandstones and carbonates, and molar volumes.
Traps and timing — the burial history curve and the timing of trap formation.
Source rocks, maturation and migration — temperature and time in the maturation of organic matter, and the heating history.
Sources of formation damage — how authigenic clays and cements behave in drilling, injection and acid treatments.

Check yourself

1
Two identical quartz sandstones are now at the same depth at 125 °C. The first was buried for 20 million years, the second for 80 million years, both uniformly. Which has more quartz cement?
1. A.It cannot be said without knowing the pressure
2. B.The second: it spent longer in the hot zone
3. C.The same: the temperature is the same
4. D.The first: it is younger
Show the answer
Correct answer: B. The second: it spent longer in the hot zone
The volume of cement is set by the time spent at each temperature: in the formula $E \propto 1/H$ , and the second sandstone was heated four times more slowly. At the same present temperature it is more strongly cemented. Pressure does not affect cement in the kinetic model.
2
For the rigid sand of the lesson ( $P_i = 41$ %, $\mathrm{IGV}_f = 26$ %, $\beta = 0.11$ MPa⁻¹) find the intergranular volume at a depth of 2.5 km at normal pressure and after this sand has been uplifted to 1.5 km.
Show the answer
$\sigma' = 12.2\cdot 2.5 = 30.5$ MPa; $\mathrm{IGV} = 26 + 15\cdot e^{-3.355} = 26 + 15\cdot 0.0349 \approx 26.5$ %. After uplift to 1.5 km, the intergranular volume does not return to 28%, the value for a sand that has never been buried deeper than 1.5 km. Compaction is irreversible and set by the greatest stress, so the same 26.5% remains.
3
A sandstone contains 9% of quartz cement, detrital quartz occupies 60% of the rock volume, and all the silica came from stylolites in the same bed. How much rock dissolved at the stylolites per metre of section, and how much is that per seam, if the seams are 0.3 m apart?
Show the answer
The quartz needed is 0.09 m³ per 1 m³ of rock. With a quartz fraction of 0.6, the dissolved rock is $0.09/0.6 = 0.15$ m per metre of section. There are three or four seams per metre (one every 0.3 m), so each has “eaten” about 4–5 cm of rock, as much as well-developed stylolites do. The insoluble residue (clay, mica) remains as a film on the seam.
4
In a sandstone 6% of the rock volume of K-feldspar has dissolved, and aluminium and silica have precipitated nearby. By how much has porosity grown?
1. A.By about 6%
2. B.It has not changed
3. C.By about 0.8%
4. D.By about 3%
Show the answer
Correct answer: C. By about 0.8%
In a closed system about 12.6% of the volume of the dissolved feldspar remains free: $6\cdot 0.126 \approx 0.76$ %. The rest is taken by kaolinite (about 2.7%) and quartz (about 2.5%). The voids visible in thin section where the feldspar was are misleading: the pores are mostly redistributed.
5
Which sign says most strongly that a deep quartz sandstone has preserved its porosity despite a temperature above 130 °C?
1. A.A lot of kaolinite in the pores
2. B.Abnormally high pore pressure
3. C.Continuous chlorite coats on the grains
4. D.A high share of feldspars in thin section
Show the answer
Correct answer: C. Continuous chlorite coats on the grains
Quartz cement grows on a clean quartz surface, and continuous coats leave it only their gaps. In the lesson's example, the porosity at 129 °C is 24.5%, against 14.8% without coats. Abnormal pressure holds back compaction but not cement. Feldspars reduce the quartz surface, but they dissolve themselves and give kaolinite. Kaolinite fills the pores.
6
A limestone with a porosity of 30% is completely dolomitized “mole for mole” without a change in the rock volume. What will the porosity be, and why is it usually lower in real dolomites?
Show the answer
The framework of 0.70 m³ shrinks by a factor of $0.873$ to 0.611 m³, so the porosity is about 38.9%. In reality, the replacement often proceeds “volume for volume”, at the expense of brought-in carbonate. Dolomite may continue to grow in the pores, and the rock itself compacts. Early dolomite also cements pores before the replacement. So 13% is a limit, not a rule.

Updated: 26 October 2026