Reservoir quality: geological controls on porosity and permeability

Quick refresher

Reservoir quality is created twice: at deposition (grain size and sorting, matrix, mineral composition of the framework) and after burial (compaction, cement, dissolution, clay minerals in the pores). The second stage usually makes the result of the first worse, but rarely wipes it out: the best facies (rock types formed in one depositional environment) stay the best.
Porosity is the volume of the pores. Permeability is set by the size of the narrowest places between them. So matrix (the fine mud between the grains), cement in the throats and the fibres of clay minerals reduce permeability much more strongly than porosity.
Mechanical compaction of rigid grains reduces the intergranular volume (the volume between the grains, filled by pores and cement) from 40–45% in fresh sand to about 26% by depths of 1.5–2 km. Beyond that, porosity is taken by pressure solution (grains dissolving where they press on each other) and by cement, above all quartz cement. Quartz cement grows noticeably at temperatures above about 70–80 °C.
A thin section (a slice of rock ground thin enough to see through under a microscope) shows the percentages of pores and cement as they are today. They refer to the compacted volume of the rock. To compare them with the initial porosity, convert them to the initial volume, using the volume of grains, which has been preserved.
Build the porosity–permeability relationship by rock type. A regression of $\lg k$ on $\phi$ gives the median, not the mean. With a large scatter it underestimates the mean permeability several times over.
The flow zone indicator $\mathrm{FZI} = \mathrm{RQI}/\phi_z$ groups samples with the same pore geometry: rocks of one type lie on one $k(\phi)$ curve, while different types at the same porosity differ in permeability by orders of magnitude (factors of ten or more).

Key relationships

\mathrm{COPL} = \dfrac{100\,(P_i - \mathrm{IGV})}{100 - \mathrm{IGV}}

porosity loss by compaction, % of the original rock volume

\mathrm{CEPL} = C\,\dfrac{100 - P_i}{100 - \mathrm{IGV}}

porosity loss due to cement, % of the original volume

\mathrm{RQI} = 0.0314\sqrt{k/\phi},\quad \mathrm{FZI} = \dfrac{\mathrm{RQI}}{\phi/(1 - \phi)}

$k$ in mD, $\phi$ as a fraction, RQI and FZI in µm

k = 1014\,\mathrm{FZI}^2\,\dfrac{\phi^3}{(1 - \phi)^2}

porosity–permeability relationship of one rock type

\bar k = k_G\,e^{\sigma^2/2}

arithmetic mean for a lognormal scatter about the regression line; $\sigma$ in natural logarithms

Common pitfalls

Building a single porosity–permeability relationship for the whole reservoir by mixing rock types: the prediction error is an order of magnitude.
Using the value from the $\lg k$ regression line as the mean permeability of a layer: it is the median.
Subtracting today's percentages of pores and cement from the initial porosity, and forgetting that compaction has reduced the volume of the rock itself.
Judging the influence of clay by shale volume alone, without knowing the mineral and where it lies in the pores.
Extending a porosity trend that was built above the zone of quartz cementation to greater depths and temperatures.

After this lesson you can

explain how the size, sorting, shape and orientation of the grains and the clay matrix set the initial porosity and permeability, and why this mark of the depositional environment often survives burial
tell the main clay minerals of reservoirs apart — kaolinite, illite, chlorite and smectite — and the ways they occur in pores, and explain how each one reduces permeability and raises the irreducible water saturation
describe compaction, cementation and dissolution in broad outline, and share out the porosity loss between compaction and cement, using thin sections and the intergranular volume
build a porosity–permeability relationship, understand what a regression on the logarithm of permeability does and does not tell you, and identify rock types with the flow zone indicator FZI
estimate reservoir quality away from wells from facies, depth and temperature trends, and say in which direction each estimate is wrong

Two histories of one rock: deposition and burial

A sandstone gets its properties in two stages. In the first, currents, waves or wind sort the grains. They decide how much mud is left between the grains and which minerals make up the framework. This produces the initial (depositional) porosity and permeability. In the second stage, the sediment is buried. The weight of the rocks above pushes the grains together, cement (minerals that grow in the pores and glue the grains) forms, clay minerals grow, and some grains dissolve. These changes are called diagenesis (Diagenesis: compaction, cementation and dissolution). Here we cover them only as far as we need to judge reservoir quality.

The second stage usually makes worse what the first created, but rarely wipes it out. A better-sorted and cleaner sand, as a rule, stays the better reservoir at depth as well. That is why, on a porosity–permeability plot, the samples of different facies of one rock unit often fall in separate clouds. Well-sorted aeolian (wind-blown) sands lie higher and to the right of river sands. Poorly sorted debris-flow (mudflow) and alluvial-fan deposits lie lower and to the left. When we have reconstructed the depositional environments (Facies, facies successions and sedimentary logs), we can predict where the reservoir is better. This is the basis of predicting quality away from wells (the last section).

What sets the quality of a sandstone

Factor: Coarser grains at the same sorting
Stage: deposition
Porosity: hardly changes
Permeability: grows roughly as the square of the grain size

Factor: Poorer sorting, more clay matrix
Stage: deposition
Porosity: falls
Permeability: falls more

Factor: Many unstable and ductile grains: feldspars, shale fragments, micas
Stage: deposition (composition)
Porosity: is lost faster on burial
Permeability: falls

Factor: Compaction
Stage: burial
Porosity: falls
Permeability: falls

Factor: Quartz and carbonate cement
Stage: burial
Porosity: falls
Permeability: falls, especially if the cement is in the throats

Factor: Authigenic clay minerals (those that grew in the pores)
Stage: burial
Porosity: falls a little
Permeability: falls several-fold or by orders of magnitude

Factor: Dissolution of grains and cement
Stage: burial
Porosity: grows
Permeability: grows if the new pores are connected

Factor	Stage	Porosity	Permeability
Coarser grains at the same sorting	deposition	hardly changes	grows roughly as the square of the grain size
Poorer sorting, more clay matrix	deposition	falls	falls more
Many unstable and ductile grains: feldspars, shale fragments, micas	deposition (composition)	is lost faster on burial	falls
Compaction	burial	falls	falls
Quartz and carbonate cement	burial	falls	falls, especially if the cement is in the throats
Authigenic clay minerals (those that grew in the pores)	burial	falls a little	falls several-fold or by orders of magnitude
Dissolution of grains and cement	burial	grows	grows if the new pores are connected

Tendencies, not laws: their strength depends on the rock and its burial history. A qualitative summary of the same factors is in the lesson on reservoir and seal; here each is covered in more detail.

Example: the Jurassic sandstones of South Mangyshlak

According to published geological descriptions, the Lower and Middle Jurassic sandstones of South Mangyshlak hold the main reserves of the Uzen and Zhetybai fields. They are fine- to medium-grained and polymictic (made of grains of many different minerals and rocks). They are usually poorly rounded and poorly to moderately sorted, with clay and carbonate cement. The shale volume varies from a few percent to 40–50%. Porosity decreases with depth — from 18–23% at depths of 1050–1300 m (Uzen) to 14–18% at 1950–2650 m (South Zhetybai). Permeability varies from a few mD to 1200 mD and, unlike porosity, depends on the shale volume. The best properties are those of the channel sandstones of braided rivers (wide rivers that split into many shifting channels). These sandstones occur in several beds. All the themes of the lesson can be seen in these facts: the depositional environment, compaction with depth, and the decisive role of clay in permeability.

Sediment texture: what the initial properties depend on

Sediment texture — the size, sorting, shape and packing of the grains — is covered in detail in the lesson Sediment texture: grain size, sorting and shape. Five consequences of it matter for reservoir quality.

Grain size sets the width of the pore channels. For the same packing, permeability grows roughly as the square of the grain size (the Kozeny–Carman model), while porosity does not depend on size. In nature, fresh fine sediment is even more porous than sand: clay flakes and angular silt particles pack loosely, like a house of cards. But on burial such particles compact more strongly. So in the rock the relationship is usually the opposite: fine-grained rocks are less porous and orders of magnitude less permeable.
Sorting is the main factor of initial porosity: about 42% for a very well sorted sand and about 28% for a very poorly sorted one. Permeability falls faster: each worsening of sorting by one φ unit (on the φ scale, one step is a doubling or halving of grain diameter) reduces it roughly 3.7-fold (the Krumbein–Monk formula).
Clay matrix — mud deposited together with the sand — fills the pores of the framework. As long as there is little of it, porosity falls moderately and permeability falls much more, because the flow now goes through the pores of the mud itself (a mixture of large and small grains).
Shape and orientation. Flattened grains pack more loosely at deposition, but they lie flat, with their long axes along the current. The channels between them are elongated too. So fluid finds it easiest to move along the bed, in the direction of the ancient current, and hardest across the beds. Smooth rounded grains slow the flow less than rough angular ones.
Lamination. Laminae (very thin layers) of different texture stack into a package. Its vertical permeability is lower than its horizontal permeability, and the $k_v/k_h$ ratio of the package is set by the densest laminae (anisotropy).

Clay minerals: which mineral, and where it lies

Clay in a sandstone has two origins. Detrital clay was brought in together with the sand: it is the matrix between the grains, thin laminae, lumps and fragments of shale. Authigenic clay grew in the pores after burial, from underground solutions (water with dissolved minerals) and by the breakdown of feldspars and volcanic fragments (authigenic clay minerals). Authigenic clay matters most for quality. It grows on the pore walls and in the throats, that is, precisely where the flow passes.

Permeability is affected both by the mineral itself and by the way it occurs. Authigenic clay grows in the pores in three ways:

Grain coats (pore lining): a thin crust of plates growing perpendicular or parallel to the grain surface; in Russian-language petrography, film cement. Chlorite and smectite usually grow this way.
Pore fill (pore filling): separate clusters of crystals inside the pores. Typical are booklets of kaolinite plates, which look like stacks of coins or worm-like “books”. Micropores (tiny pores) remain between the plates.
Pore bridges (pore bridging): fibres and ribbons stretched across pores and throats from grain to grain — this is how illite often grows. Their volume is small, but they block the flow.

SchematicHow clay occurs in a sandstone — three forms in the pores and a thin layer — and what each does to porosity

\phi

and permeability

k

. Schematic, not to scale.

The main clay minerals of reservoirs

Mineral: Kaolinite
How it grows in the pores: booklets of plates in the pores
Permeability: falls several-fold
What else matters: particles are torn loose by the flow and plug the throats (fines migration)

Mineral: Illite (hydromica)
How it grows in the pores: fibres and plates across the pores
Permeability: falls by an order of magnitude or more with little loss of porosity
What else matters: a lot of bound water; the fibres stick together when core is dried

Mineral: Chlorite
How it grows in the pores: coats on the grains
Permeability: falls moderately
What else matters: continuous coats prevent quartz cement from growing; iron-rich chlorite is sensitive to hydrochloric acid

Mineral: Smectite and mixed-layer clays
How it grows in the pores: coats, pore fill
Permeability: falls; strongly in fresh water
What else matters: swells and breaks up into fine particles in fresh water; high cation exchange capacity

Mineral	How it grows in the pores	Permeability	What else matters
Kaolinite	booklets of plates in the pores	falls several-fold	particles are torn loose by the flow and plug the throats (fines migration)
Illite (hydromica)	fibres and plates across the pores	falls by an order of magnitude or more with little loss of porosity	a lot of bound water; the fibres stick together when core is dried
Chlorite	coats on the grains	falls moderately	continuous coats prevent quartz cement from growing; iron-rich chlorite is sensitive to hydrochloric acid
Smectite and mixed-layer clays	coats, pore fill	falls; strongly in fresh water	swells and breaks up into fine particles in fresh water; high cation exchange capacity

The structure of clay minerals and their swelling are covered in the lesson on water-based muds, the effect of clays on logs and saturation in the lesson Water saturation in shaly sands.

Log analysts describe clay differently — by how it is distributed in the rock volume (shale distribution). It can be laminated (thin laminae between laminae of clean sandstone), dispersed (in the pores) or structural (grains and fragments of shale in place of part of the quartz grains). For flow, the difference is fundamental. Laminated clay reduces the permeability along the layers only in proportion to its share of the thickness, but it reduces the vertical permeability sharply. Dispersed clay reduces permeability in all directions and, at the same shale volume, more strongly than any other form. Structural clay has the least effect, until compaction crushes the fragments into pseudomatrix (soft fragments squeezed between the grains until they look like matrix).

Irreducible water. The huge surface of clay minerals holds water. So in rocks with illite and chlorite, the irreducible water saturation (the water that stays in the pores and cannot be pushed out) can be very high, although the wells produce no water. The resistivity of the formation is low. This is how a low-resistivity pay zone appears (a zone that holds oil or gas but looks like water on resistivity logs).
Core. When core is dried in the ordinary way, illite fibres stick together and lie down on the pore walls. The permeability of the plug (a small cylinder cut from the core) then comes out too high. Such samples are dried by gentle methods (cleaning and drying).
Work in the well. Fresh water and mud filtrate (the liquid from the drilling mud that seeps into the rock) make smectite swell and kaolinite particles break loose. Acids destroy chlorite and may produce secondary precipitation (new solids that form in the pores). So before any fluid is injected, the rock is studied in thin sections and under the electron microscope (Formation damage and damage skin, Acid treatments: principle, economics and candidate selection).

Burial: compaction, cement and dissolution

As the rock is buried, the effective stress grows. This is the part of the weight of the rocks above that the framework of grains carries. First comes mechanical compaction. Grains slide and rotate to pack more tightly, and some break. Ductile (soft) grains, such as shale fragments and micas, are crushed and squeezed into the pores, and form pseudomatrix. Deeper, at higher temperature and stress, chemical compaction is added. At the grain contacts, quartz dissolves under pressure. The contacts become concavo-convex (one grain presses into the next) and sutured (toothed). Stylolites (jagged, seam-like surfaces where rock has dissolved) appear. The dissolved silica comes out of solution nearby and forms cement.

The easy way to measure the degree of compaction is the intergranular volume (IGV). This is the fraction of the rock that lies between the framework grains. Cement only fills the spaces between the grains and does not squeeze them. Compaction squeezes them. So IGV shows how tightly the grains are packed, whether or not cement fills the spaces. Early cement can even keep a large IGV by propping up the framework. In fresh well-sorted sand, IGV is 40–45%. In well-sorted sandstones of rigid grains (quartz and feldspars), mechanical compaction reduces it to about 26% by depths of 1.5–2 km, and after that it changes little. More poorly sorted sands go on losing intergranular volume at greater depths too. Sands rich in ductile fragments compact more strongly and faster.

The main cements of sandstones:

Quartz cement (overgrowths): quartz grows on quartz grains, continuing their crystal lattice, so that the grain is “completed” into a crystal. It forms much faster as the temperature rises and becomes noticeable above about 70–80 °C. So at great depths quartz cement is often the main enemy of porosity.
Carbonate cement — calcite, dolomite, siderite. It is often early and patchy. It cements separate areas and beds completely, creating dense concretions (hard lumps) and thin layers that become baffles (barriers) to flow (Reservoir architecture and performance).
Clay cement — authigenic clay minerals (previous section).

Dissolution works in the opposite direction. Groundwater dissolves (leaches out) feldspars, rock fragments and carbonate cement, and secondary porosity (pores that form after deposition) appears. Permeability grows markedly only when the new voids connect with the earlier pores. Isolated moulds (empty spaces in the shape of dissolved grains) add volume but hardly any flow (primary and secondary porosity).

Porosity is preserved at depth where something slows down compaction or cementation:

Grain coats — continuous chlorite coats and thin coats of microcrystalline (very fine-grained) quartz — make it harder for quartz cement to start growing. The coat must be continuous: quartz grows in its gaps, and a coat that is too thick narrows the throats itself.
Abnormally high pore pressure (fluid pressure in the pores far above normal) reduces the effective stress, so the rock is only as compacted as it would be at a shallower depth (the equivalent depth method).
A little early cement strengthens the framework and slows further compaction. An early arrival of oil in the trap, according to several studies, slows cementation, but how strongly is a matter of debate.

As a result the porosity of sandstones decreases with depth. While mechanical compaction dominates, the decrease is roughly exponential: about the same fraction of the porosity is lost with each extra step of depth (the compaction trend). It becomes faster when quartz cement is added. We build trends separately for rocks of different composition and different facies: polymictic sandstones with ductile fragments lose porosity faster than quartz sandstones.

To tell whether compaction or cement took more of the porosity, we use thin sections (slices of rock seen under a microscope). Point counting (counting what lies under a grid of points on the slice) gives the shares of pores, cement and matrix as they are today. The initial porosity is taken from the texture: usually 40–45% for a well-sorted sand. There is a subtlety. Compaction has reduced the volume of the rock itself, so we cannot subtract today's percentages directly from the initial ones. We convert them to the original volume, using the volume of grains, which has been preserved (the derivation is in the advanced section).

Intergranular volume

\mathrm{IGV} = P_{ig} + C + M

where:

$\mathrm{IGV}$ intergranular volume, % of the present rock volume
$P_{ig}$ intergranular pores, %
$C$ intergranular cement: quartz, carbonates, authigenic clay minerals, %
$M$ detrital clay matrix between the grains, %

Units: percent of the present rock volume, from a point count in thin section

Holds when

Pores inside grains and pores left by dissolved grains are not included in $P_{ig}$ — they are not intergranular.
Cement that has replaced grains is counted with the grains, not in $C$ .

IGV shows how much space is left between the framework grains, that is, how compacted the rock is — whether or not cement fills this space.

Porosity loss by compaction

\mathrm{COPL} = \frac{100\,(P_i - \mathrm{IGV})}{100 - \mathrm{IGV}}

where:

$\mathrm{COPL}$ compactional porosity loss, % of the original rock volume
$P_i$ initial (depositional) porosity, %
$\mathrm{IGV}$ intergranular volume, % of the present volume

Units: percent

Holds when

The volume of the framework grains does not change on burial: grains neither dissolve nor are added.
$P_i$ is taken from the texture of the sediment; an error in it passes directly into COPL (the example below shows by how much).
Cement only fills the intergranular volume without changing it.

Had the grains not moved closer together, the intergranular volume would have remained equal to the initial porosity. Compaction squeezed 100 units of original volume down to $100\,(100 - P_i)/(100 - \mathrm{IGV})$ units. The difference between 100 and this number is the porosity lost.

Porosity loss due to cement

\mathrm{CEPL} = C\,\frac{100 - P_i}{100 - \mathrm{IGV}}

where:

$\mathrm{CEPL}$ cementational porosity loss, % of the original rock volume
$C$ intergranular cement, % of the present volume

Units: percent

Holds when

The same as for COPL.
Cement fills the pores without pushing the grains apart.

The cement occupies $C$ % of the present rock volume; the factor converts this share into percent of the original volume. If COPL is greater than CEPL, porosity was taken mainly by compaction; if it is smaller, by cement.

Worked exampleCompaction or cement: two samples of one sandstone

Given (assumed data)

Samples: one sandstone from nearby depths of about 3 km; the grains are the same in size and sorting
Sample A: grains coated with chlorite; intergranular pores 21%, intergranular cement 3%, no matrix
Sample B: no coats; intergranular pores 9%, cement 16% (mostly quartz), no matrix
Initial porosity $P_i$: 40% (well-sorted sand); 45% as a check

Solution

Intergranular volume: for A $21 + 3 = 24$ %, for B $9 + 16 = 25$ %. The grains are packed almost equally tightly.
Loss by compaction: for A $\mathrm{COPL} = 100\cdot(40 - 24)/(100 - 24) = 21.05$ %, for B $100\cdot(40 - 25)/(100 - 25) = 20.00$ %.
Loss due to cement: for A $\mathrm{CEPL} = 3\cdot 60/76 = 2.37$ %, for B $16\cdot 60/75 = 12.80$ % (here $60 = 100 - P_i$ , and 76 and 75 are $100 - \mathrm{IGV}$ ).
Balance check. Of the initial 40%, sample A has $40 - 21.05 - 2.37 = 16.58$ % of the original volume left. The rock itself has shrunk to $100\cdot 60/76 = 78.95$ % of the original volume, and the present 21% of pores is $0.21\cdot 78.95 = 16.58$ % of the original: it agrees. For B: $40 - 20.00 - 12.80 = 7.20$ and $0.09\cdot 80.00 = 7.20$ .
The share of compaction in the porosity loss: for A $21.05/(21.05 + 2.37) = 0.90$ ; for B, 0.61. At $P_i$ = 45%, 0.93 and 0.69.

Answer: Compaction took almost the same from both samples — 21.05 and 20.00% of the original volume. The difference was made by cement: 2.37% in the sample with chlorite coats and 12.80% in the sample without them.

What it means. In both samples compaction took more porosity than cement did — this is typical of sandstones. But the difference in quality between them is entirely the work of cement: where the grains are coated with chlorite, quartz had nothing to grow on. For prediction, it follows that at this depth what decides is not so much the depth itself as where the coats occur. We look for them in core and tie them to facies. The choice of initial porosity changes the numbers but not the conclusion. The calculation assumes that the grains did not dissolve. If pores left by dissolved grains are wrongly included in the intergranular volume, compaction will be underestimated. The data in this example are assumed for illustration.

CalculatedIntergranular volume and cement of the two samples of the example. The “equal losses” line is where COPL = CEPL at an initial porosity of 40%; below it porosity was taken mainly by compaction, above it by cement. The dashed lines are lines of equal present intergranular porosity.

The porosity–permeability relationship

Logs give porosity in every well. Permeability comes only from core and well tests. So in models, permeability is almost always worked out from porosity, using a relationship built on core samples. For sandstones of one type, such a relationship is often close to a straight line in semilog coordinates (permeability on a logarithmic axis, porosity on an ordinary one). When porosity changes by a few percent, permeability changes by a factor of several.

Semilog relationship between porosity and permeability

\lg k = a + b\,\phi

where:

$k$ permeability, mD
$\phi$ porosity, fraction
$a,\ b$ constants fitted to the samples; $b$ is the number of decimal orders of magnitude by which permeability grows when porosity grows by one unit, —

Units: empirical: $k$ in mD, $\phi$ as a fraction, $\lg$ is the decimal logarithm

Holds when

The samples are of one rock type and one scale. Permeability is brought to the same conditions: reservoir stress, and liquid flow (measured with liquid, or corrected for gas slip with the Klinkenberg correction).
The relationship has not been checked outside the porosity range of the samples.
$a$ and $b$ are found by least squares, regressing $\lg k$ on $\phi$ . This is the right way when permeability is predicted from porosity. A regression of $\phi$ on $\lg k$ gives a different, steeper line.

The relationship is empirical (fitted to data). Its physical basis is the Kozeny–Carman model, in which, for unchanged pore geometry, permeability grows as $\phi^3/(1 - \phi)^2$ (see the section on rock types). Over a narrow range of porosity this curve is almost a straight line on a semilog plot.

Such a relationship has two traps. The first is mixing rock types. At the same porosity, samples of different texture and with different clays differ in permeability by orders of magnitude. The common line then passes between the clouds of points and describes none of them. The second trap is the meaning of a point on the line. A regression on the logarithm gives the mean of the logarithm. That is the median (the middle value), not the arithmetic mean. For a lognormal scatter (the logarithms of permeability scatter in a bell-shaped way), it equals the geometric mean. For flow along the layers, exactly the arithmetic mean is what we need (means of lognormal permeability):

Arithmetic mean for a lognormal scatter about the relationship

\bar k = k_G\,e^{\sigma^2/2},\qquad \sigma = s\,\ln 10

where:

$\bar k$ arithmetic mean permeability at the given porosity, mD
$k_G$ the value on the regression line — the median, which is also the geometric mean, mD
$\sigma$ standard deviation of the natural logarithm of permeability about the line, —
$s$ the same in decimal logarithms — the residual standard deviation of the regression, —

Units: any units of permeability

Holds when

The scatter about the line is lognormal and the same at any porosity.
If the samples are a mixture of several rock types, the scatter is not lognormal, and the formula gives only the order of magnitude of the correction.

This is the power mean of a lognormal permeability at $\omega = 1$ . The correction grows quickly with the scatter: at $s$ = 0.3 it is about 1.3 times, and at $s$ = 0.85 almost seven times.

Twelve core plugs from one interval

No.: 1
$\phi$ , fraction: 0.186
$k$ , mD: 250
Type from thin sections: 1

No.: 2
$\phi$ , fraction: 0.214
$k$ , mD: 500
Type from thin sections: 1

No.: 3
$\phi$ , fraction: 0.238
$k$ , mD: 570
Type from thin sections: 1

No.: 4
$\phi$ , fraction: 0.262
$k$ , mD: 940
Type from thin sections: 1

No.: 5
$\phi$ , fraction: 0.158
$k$ , mD: 18
Type from thin sections: 2

No.: 6
$\phi$ , fraction: 0.192
$k$ , mD: 25
Type from thin sections: 2

No.: 7
$\phi$ , fraction: 0.221
$k$ , mD: 65
Type from thin sections: 2

No.: 8
$\phi$ , fraction: 0.247
$k$ , mD: 69
Type from thin sections: 2

No.: 9
$\phi$ , fraction: 0.141
$k$ , mD: 0.96
Type from thin sections: 3

No.: 10
$\phi$ , fraction: 0.172
$k$ , mD: 1.5
Type from thin sections: 3

No.: 11
$\phi$ , fraction: 0.205
$k$ , mD: 3.7
Type from thin sections: 3

No.: 12
$\phi$ , fraction: 0.233
$k$ , mD: 4.2
Type from thin sections: 3

No.	$\phi$ , fraction	$k$ , mD	Type from thin sections
1	0.186	250	1
2	0.214	500	1
3	0.238	570	1
4	0.262	940	1
5	0.158	18	2
6	0.192	25	2
7	0.221	65	2
8	0.247	69	2
9	0.141	0.96	3
10	0.172	1.5	3
11	0.205	3.7	3
12	0.233	4.2	3

Type 1 is a clean medium- to coarse-grained sandstone of channel bars; type 2 is a fine-grained laminated sandstone with kaolinite in the pores; type 3 is a fine-grained sandstone with fibrous illite. Permeability to liquid at reservoir stress. The data are assumed for illustration.

Worked exampleOne relationship for the whole interval

Given (assumed data)

Samples: the 12 plugs of the table above, all together, without splitting by type
Task: permeability at a porosity of 0.20, and its scatter

Solution

Means: $\bar\phi = 0.2058$ , the mean $\lg k$ is 1.540.
Sums over the plugs: $S_{\phi\phi} = \sum(\phi - \bar\phi)^2 = 0.01514$ and $S_{\phi k} = \sum(\phi - \bar\phi)(\lg k - \overline{\lg k}) = 0.2671$ .
Slope and intercept: $b = 0.2671/0.01514 = 17.64$ ; $a = 1.540 - 17.64\cdot 0.2058 = -2.090$ .
The residual standard deviation is $s = 0.85$ (in decimal orders of magnitude), the coefficient of determination is $R^2 = 0.39$ : the line explains less than half of the scatter of $\lg k$ .
At $\phi = 0.20$ : $\lg k = -2.090 + 17.64\cdot 0.20 = 1.438$ , that is, $k = 27.4$ mD. The plugs nearest in porosity are No. 6 (0.192) — 25 mD, No. 11 (0.205) — 3.7 mD and No. 2 (0.214) — 500 mD. At almost the same porosity, No. 2 is $500/27.4 \approx 18$ times more permeable than the value on the line, and No. 11 is $27.4/3.7 \approx 7$ times less permeable.
If the scatter is taken as lognormal: $\sigma = 0.85\cdot 2.303 = 1.96$ , the correction is $e^{\sigma^2/2} = 6.83$ , and the arithmetic mean at this porosity is $27.4\cdot 6.83 \approx 187$ mD.

Answer: At a porosity of 0.20 the common relationship gives about 27.4 mD. But plugs of almost the same porosity differ from this value by a factor of 7–18. If the scatter is taken as lognormal, the arithmetic mean at such a porosity is 6.83 times the value on the line.

What it means. The line passes between the clouds of the three rock types and describes none of them. In a model it will blur the differences in permeability. The channel sandstones will get a value that is too low and the illite-bearing rocks one that is too high. The forecast of flow rates and of water breakthrough (the moment water reaches the producing well) will be wrong in both directions. The correction to the arithmetic mean can be treated here only as an estimate of the order of magnitude: the scatter is made up of three groups and is not lognormal. The way out is to split the plugs into rock types, each with its own relationship (next section). The data in this example are assumed for illustration.

Rock types and the flow zone indicator

A rock type (also called a petrophysical type) is a group of rocks with a similar pore structure. Their pores and throats have similar sizes. So they have a similar porosity–permeability relationship, similar capillary pressure curves (how hard it is for one fluid to push another out of the pores) and similar relative permeabilities (how well each fluid flows when two fluids share the pores). A useful type meets three conditions. It differs in flow properties. It can be explained geologically, by facies and burial history. And it can be recognised on logs; otherwise we cannot extend it to wells without core.

One common way to identify types is the flow zone indicator (FZI). It follows from the Kozeny–Carman model. For the same pore geometry, the quantity $\sqrt{k/\phi}$ is proportional to the ratio of pore volume to grain volume. This quantity is the characteristic size of the channels. (The derivation is in the advanced section.)

Reservoir quality index and flow zone indicator

\mathrm{RQI} = 0.0314\sqrt{\frac{k}{\phi}},\qquad \phi_z = \frac{\phi}{1 - \phi},\qquad \mathrm{FZI} = \frac{\mathrm{RQI}}{\phi_z}

where:

$\mathrm{RQI}$ reservoir quality index, µm
$k$ permeability, mD
$\phi$ effective porosity, fraction
$\phi_z$ ratio of pore volume to grain volume (normalised porosity), —
$\mathrm{FZI}$ flow zone indicator, µm

Units: $k$ in mD, $\phi$ as a fraction; RQI and FZI in µm

Holds when

$0.0314 = \sqrt{0.9869\cdot 10^{-3}}$ converts millidarcy to square micrometres under the root.
Permeability is single-phase (to liquid or with the Klinkenberg correction), porosity is effective, and the test conditions are the same for all samples.
The basis is the model of a bundle of winding (tortuous) capillaries. FZI gathers the specific surface of the pores (their surface area per unit of volume), their tortuosity and shape into one number.

RQI is the characteristic size of the channels (up to a constant factor): of the order of a micrometre or more in good reservoirs, hundredths of a micrometre in tight ones. FZI separates the geometry of the pores from their quantity: samples of one type have nearly the same FZI at different porosities. On a plot of $\lg\mathrm{RQI}$ against $\lg\phi_z$ , each type lies on its own straight line with a slope of 1, and the position of the line gives the FZI.

Porosity–permeability relationship for a rock type

k = 1014\,\mathrm{FZI}^2\,\frac{\phi^3}{(1 - \phi)^2}

where:

$k$ permeability, mD
$\mathrm{FZI}$ flow zone indicator of the type — usually the geometric mean over its samples, µm
$\phi$ porosity, fraction

Units: $k$ in mD, FZI in µm

Holds when

$1014 = 1/0.0314^2$ .
The formula is just the definitions of RQI and FZI solved for $k$ .

On a semilog plot this is not a straight line but a gently curved one, concave downwards: its slope decreases as porosity grows. When porosity rises from 0.15 to 0.25, the permeability of a rock of one type grows by about a factor of 6.

Throat radius from Winland's equation

\lg r_{35} = 0.732 + 0.588\,\lg k - 0.864\,\lg \phi

where:

$r_{35}$ radius of the pore throats at which mercury, forced into the rock in a capillary pressure test, has filled 35% of the pore volume, µm
$k$ air permeability without the Klinkenberg correction, mD
$\phi$ porosity, % (not a fraction)

Units: empirical: $r_{35}$ in µm, $k$ in mD, $\phi$ in percent

Holds when

It is a regression on samples for which mercury curves, porosity and permeability were all available. For other rocks, especially tight ones, check it against your own data.
Porosity is substituted in percent.

The 35% level was chosen because the radius at this saturation proved to be most closely related to permeability. A common explanation is that by this point mercury has already filled the connected network of the widest channels, which carries the main flow. Like FZI, $r_{35}$ divides rocks into groups by throat size. Mercury curves are covered in the lesson on capillary pressure.

Worked exampleThree rock types by FZI

Given (assumed data)

Samples: the same 12 plugs (the table in the previous section) with types from thin sections
Comparison: permeability at $\phi = 0.20$ ; the common relationship gives 27.4 mD

Solution

Plug No. 2 ( $\phi$ = 0.214, $k$ = 500 mD): $\mathrm{RQI} = 0.0314\cdot\sqrt{500/0.214} = 0.0314\cdot 48.34 = 1.518$ µm; $\phi_z = 0.214/0.786 = 0.2723$ ; $\mathrm{FZI} = 1.518/0.2723 = 5.57$ µm.
Plug No. 11 ( $\phi$ = 0.205, $k$ = 3.7 mD): $\mathrm{RQI} = 0.0314\cdot 4.248 = 0.1334$ µm; $\phi_z = 0.205/0.795 = 0.2579$ ; $\mathrm{FZI} = 0.517$ µm — at almost the same porosity, 11 times smaller.
We do the same for all the plugs (table below). For type 1, FZI is from 4.92 to 5.57 µm, for type 2 from 1.51 to 1.90, and for type 3 from 0.44 to 0.52. These are three tight groups, and each coincides with the type from thin sections.
The geometric means of FZI of the types: 5.20, 1.69 and 0.474 µm. The boundaries between the types are halfway on a logarithmic scale: $\sqrt{5.20\cdot 1.69} = 2.96$ and $\sqrt{1.69\cdot 0.474} = 0.90$ µm.
Permeability at $\phi = 0.20$ : $\phi^3/(1 - \phi)^2 = 0.008/0.64 = 0.0125$ ; type 1 — $1014\cdot {5.20}^2\cdot 0.0125 = 343$ mD, type 2 — 36.2 mD, type 3 — 2.85 mD.
For comparison, here is the throat radius from Winland's equation: for plug No. 2 $r_{35} \approx 14.8$ µm, for No. 11 about 0.86 µm. The equation was built on air permeability without corrections. Here we substituted the smaller permeability to liquid at reservoir stress. So both radii, especially that of the tight plug, are somewhat underestimated.

Answer: Each type is described by its own curve: at a porosity of 0.20 type 1 gives about 343 mD, type 2 36.2 mD and type 3 2.85 mD, while the common relationship gives 27.4 mD for all of them.

What it means. Within a type, FZI varies by ten or twenty percent, not several-fold. So the relationship by type predicts the permeability of the plugs with an accuracy that the common line cannot reach. Here the types coincided with the thin sections because the example was built to be clean. In practice we check the FZI boundaries against the geology, and we make sure that the type can be recognised on logs — by shale volume, facies and nuclear magnetic resonance (NMR) logs (Nuclear magnetic resonance (NMR) logging). In a model, we first spread the rock types through the grid, then porosity. Permeability then comes from the relationship of each cell's own type. The data in this example are assumed for illustration.

RQI,

\phi_z

and FZI of the plugs

No.: 1
Type: 1
RQI, µm: 1.151
$\phi_z$: 0.229
FZI, µm: 5.04

No.: 2
Type: 1
RQI, µm: 1.518
$\phi_z$: 0.272
FZI, µm: 5.57

No.: 3
Type: 1
RQI, µm: 1.537
$\phi_z$: 0.312
FZI, µm: 4.92

No.: 4
Type: 1
RQI, µm: 1.881
$\phi_z$: 0.355
FZI, µm: 5.30

No.: 5
Type: 2
RQI, µm: 0.335
$\phi_z$: 0.188
FZI, µm: 1.79

No.: 6
Type: 2
RQI, µm: 0.358
$\phi_z$: 0.238
FZI, µm: 1.51

No.: 7
Type: 2
RQI, µm: 0.539
$\phi_z$: 0.284
FZI, µm: 1.90

No.: 8
Type: 2
RQI, µm: 0.525
$\phi_z$: 0.328
FZI, µm: 1.60

No.: 9
Type: 3
RQI, µm: 0.082
$\phi_z$: 0.164
FZI, µm: 0.50

No.: 10
Type: 3
RQI, µm: 0.093
$\phi_z$: 0.208
FZI, µm: 0.45

No.: 11
Type: 3
RQI, µm: 0.133
$\phi_z$: 0.258
FZI, µm: 0.52

No.: 12
Type: 3
RQI, µm: 0.133
$\phi_z$: 0.304
FZI, µm: 0.44

No.	Type	RQI, µm	$\phi_z$	FZI, µm
1	1	1.151	0.229	5.04
2	1	1.518	0.272	5.57
3	1	1.537	0.312	4.92
4	1	1.881	0.355	5.30
5	2	0.335	0.188	1.79
6	2	0.358	0.238	1.51
7	2	0.539	0.284	1.90
8	2	0.525	0.328	1.60
9	3	0.082	0.164	0.50
10	3	0.093	0.208	0.45
11	3	0.133	0.258	0.52
12	3	0.133	0.304	0.44

Calculated by the formulas above from the porosity and permeability in the table of plugs.

CalculatedThe plugs of the example on a porosity–permeability plot (permeability on a logarithmic scale). The three curves are the relationships of the three rock types, from their mean FZI. The dashed straight line is the common relationship through all the plugs.

In carbonates (limestones and dolomites) the approach is the same, but we identify the types first of all by pore type and particle size (petrophysics of carbonates). Vugs (large irregular cavities) and micropores break the link between FZI and the geometry of the channels, so FZI alone is usually not enough to separate the rocks.

Predicting quality away from wells

Between wells and on new structures, reservoir quality has to be predicted. A reliable prediction rests on causes, not on average values. First comes where the facies are and what the framework is made of. Then comes what burial has done to them.

What a prediction is made of

1
Facies model
We reconstruct the depositional environments from core, logs and correlation, and build facies maps: where the channels and bars are, where the floodplain is, where the poorly sorted deposits are. Each facies is matched with rock types and their proportions (Facies, facies successions and sedimentary logs, Well correlation).
2
Burial and temperature
From burial history curves and the geotherm (how temperature rises with depth), we estimate the depths and temperatures the rocks have been through (Traps and timing). We build porosity trends separately for facies and compositions. Be careful when you carry a trend from shallower depths into the intervals of active quartz cementation.
3
Seismic data
Porosity and lithology change acoustic impedance (rock density times sound speed). So we use seismic attributes and inversion (a method that turns seismic data into rock properties) to map porosity and sandstone thickness between wells. We calibrate the relationship at the wells (Seismic interpretation).
4
Geostatistics
Rock types, and then porosity, are spread among the cells of the model. The method takes account of wells, trends and correlation lengths (the distances over which values stay related). Permeability is assigned from the relationship of its own type (Spatial correlation and the variogram, Permeability upscaling for single-phase flow).
5
Checking
We compare the prediction with new wells, tests and production history. Discrepancies suggest which factor has been underestimated (core, logs and tests).

Worked exampleWill the permeability be enough on a neighbouring, deeper structure?

Given (assumed data)

Discovery: type 2 sandstone at a depth of 2.2 km, mean porosity 0.21
Neighbouring structure: the same sandstone at a depth of 3.1 km
Regional compaction trend of the sandstones: $\phi(z) = \phi_1\,e^{-c\,(z - z_1)}$ , $c$ = 0.27 km⁻¹
Geotherm: 15 °C at the surface, 30 °C per kilometre
Permeability cut-off for economic flow: 5 mD
FZI of the types: type 1 — 5.20; type 2 — 1.69; type 3 — 0.474 µm (previous example)

Solution

Porosity from the trend: $\phi = 0.21\cdot e^{-0.27\cdot 0.9} = 0.21\cdot 0.7843 = 0.165$ .
Permeability of type 2: $\phi^3/(1 - \phi)^2 = 0.00644$ , $k = 1014\cdot {1.69}^2\cdot 0.00644 = 18.7$ mD — above the cut-off. If a facies of type 1 turns out to be there, the permeability is 177 mD. For type 3 it is 1.47 mD, which is below the cut-off.
Temperature: at the discovery $15 + 30\cdot 2.2 = 81$ °C, on the structure $15 + 30\cdot 3.1 = 108$ °C. The sandstone on the structure has gone deeper into the zone of active quartz cementation than all the wells from which the trend was built.
If quartz adds another 3% of cement: $\phi = 0.165 - 0.03 = 0.135$ , $\phi^3/(1 - \phi)^2 = 0.00329$ , $k = 1014\cdot {1.69}^2\cdot 0.00329 = 9.5$ mD.

Answer: We expect the type 2 sandstone on the structure to have a porosity of about 0.165 and a permeability of about 18.7 mD. With additional quartz cement, the permeability is about 9.5 mD. In a type 3 facies it will not reach 5 mD in either case.

What it means. The main uncertainty here is not the trend but the facies and the cement. A trend built from wells at a temperature of about 81 °C knows nothing about cementation at 108 °C, so it most likely overestimates porosity. The calculation that subtracts cement is also optimistic. It leaves FZI unchanged, but cement in the throats lowers FZI too, so the permeability may fall by more than half. Before drilling, we check the core of the discovery for protective coats on the grains. We also use the seismic data and the facies model to see which facies to expect on the structure. The result is given as a range, not as a single number. The data in this example are assumed for illustration.

Terms and symbols

Reservoir quality, storage and flow properties: The porosity and permeability of a rock and the structure of the pore space behind them.
Initial porosity and permeability: The properties of the sediment immediately after deposition, before compaction and cementation.
Authigenic clay minerals: Clay minerals that grew in the pores after burial: grain coats, pore fill, bridges.
Intergranular volume (IGV): Pores, cement and matrix between the framework grains, % of the rock volume; a measure of compaction.
COPL and CEPL: Porosity losses by compaction and due to cement, converted to the original rock volume.
RQI: The reservoir quality index $0.0314\sqrt{k/\phi}$ , µm: the characteristic size of the channels.
FZI: The flow zone indicator: RQI divided by $\phi/(1 - \phi)$ ; it characterises the pore geometry and is almost constant within a rock type.
Rock type: A group of rocks with a similar pore structure and its own porosity–permeability relationship; a hydraulic flow unit.
$r_{35}$: The throat radius at 35% mercury saturation; estimated from porosity and permeability by Winland's equation.

Common mistakes and misconceptions

Wrong

“The plugs have the same porosity, so their permeability is similar.”

In fact

Permeability is controlled by the throats, not by the pore volume. At a porosity of 0.20 the three rock types of the lesson's example differ in permeability by more than a factor of a hundred.

Wrong

“The value on the regression line of $\lg k$ on porosity is the mean permeability of the layer.”

In fact

It is the median. For flow along the layers the arithmetic mean is needed, which for a lognormal scatter is larger than the median by a factor of $e^{\sigma^2/2}$ — several times when the scatter is large.

Wrong

“The thin section shows 21% pores against an initial 40%, so compaction and cement together took 19% of the rock volume.”

In fact

The present percentages refer to the compacted, smaller volume. In the lesson's example such a sample has about 16.6% of the original volume left as pores, and the losses are about 23.4%.

Wrong

“The sandstone has only 3–5% of clay, so it does not affect permeability.”

In fact

What matters is the mineral and the way it occurs: a few percent of fibrous illite in the throats reduces permeability by an order of magnitude or more, while the same amount of structural clay is almost harmless.

Wrong

“The porosity trend from the wells of a field can be extended to a neighbouring, deeper structure.”

In fact

Only if the same facies and the same processes are there. Deeper and hotter, in the interval of quartz cementation, porosity falls faster than the trend built from shallower wells, and carrying it over overestimates the quality.

In depth: advanced detail

Where FZI comes from and what it says about the pores

The Kozeny–Carman model pictures the pores as a bundle of winding (tortuous) channels. In its general form it links permeability to porosity, to the specific surface of the pores, and to their tortuosity (how winding they are) and shape:

The Kozeny–Carman model in terms of specific surface

k = \frac{\phi^3}{(1 - \phi)^2}\cdot\frac{1}{F_s\,\tau^2\,S_{gv}^2}

where:

$k$ permeability, µm²
$\phi$ porosity, fraction
$S_{gv}$ specific surface of the pores — the area of the pore walls per unit volume of grains, µm⁻¹
$\tau$ tortuosity — the ratio of the length of the fluid path to the length of the sample, —
$F_s$ shape factor of the channel cross-section, —

Units: consistent: $k$ in µm² when $S_{gv}$ is in µm⁻¹

Holds when

The pores are a network of tortuous channels, and the flow is laminar (smooth, not turbulent).
For grain packings the product $F_s\tau^2$ is close to 5. For cemented rocks and rocks with clays it is unknown and is included in FZI.

Divide both sides by $\phi$ and take the square root: $\sqrt{k/\phi} = \phi_z/(\sqrt{F_s}\,\tau\,S_{gv})$ . On the left is RQI (the factor 0.0314 merely converts millidarcy to µm²), on the right is $\phi_z$ multiplied by a quantity that depends only on the pore geometry. That quantity is what is called $\mathrm{FZI} = 1/(\sqrt{F_s}\,\tau\,S_{gv})$ .

For a packing of identical spheres of diameter $d$ , the specific surface is $S_{gv} = 6/d$ . With $F_s\tau^2 = 5$ we obtain the formula of the permeability lesson, $k = d^2\phi^3/[180\,(1 - \phi)^2]$ (the Kozeny–Carman model). So the FZI of spheres is $d/\sqrt{180}$ .

Worked exampleHow much “surface” compaction and clays have added

Given (assumed data)

Packing of spheres: diameter 0.25 mm = 250 µm
Rock types of the example: FZI 5.20, 1.69 and 0.474 µm

Solution

FZI of the packing of spheres: $250/\sqrt{180} = 250/13.42 = 18.6$ µm.
How many times larger the product $\sqrt{F_s}\,\tau\,S_{gv}$ is for the types than for the spheres: $18.6/5.20 = 3.6$ ; $18.6/1.69 = 11.0$ ; $18.6/0.474 = 39$ .
At equal porosity, permeability is proportional to the square of FZI: type 3 is lower than the packing of spheres by a factor of ${39}^2 = 1521$ .

Answer: Allowing for tortuosity, the “effective surface” of the clean type 1 sandstone is 3.6 times larger than that of 0.25 mm spheres, and that of the illite sandstone is 39 times larger.

What it means. Type 1 is close to an ideal packing: its FZI is lowered by compaction, by the irregular shape of the grains and by a little clay. In type 3, the illite fibres have increased the area of the walls that slow the flow many times over. They have also lengthened the path. That is why the permeability is three orders of magnitude (more than a thousand times) lower than that of a packing of spheres of the same porosity. FZI cannot separate the contributions of surface and tortuosity: it is one number for both effects. The data in this example are assumed for illustration.

Where the porosity-loss formulas come from

Take 100 units of volume of fresh sediment with a porosity of $P_i$ %. It contains $100 - P_i$ units of grains. Suppose the grains neither dissolve nor are added. After compaction, the same grains occupy $100 - \mathrm{IGV}$ % of the present rock volume, so the whole present volume is $V = 100\,(100 - P_i)/(100 - \mathrm{IGV})$ original units.

The intergranular space of the present rock is $\mathrm{IGV}\cdot V/100$ original units. Compaction reduced it from $P_i$ to this value. The difference is $P_i - \mathrm{IGV}\,(100 - P_i)/(100 - \mathrm{IGV})$ . Reduce it to a common denominator and it becomes $100\,(P_i - \mathrm{IGV})/(100 - \mathrm{IGV})$ — the COPL formula.

Cement occupies $C$ % of the present volume, that is, $C\cdot V/100 = C\,(100 - P_i)/(100 - \mathrm{IGV})$ original units — this is CEPL. In a clean sandstone without matrix, the remaining pores $P_{ig}\cdot V/100$ are equal to $P_i - \mathrm{COPL} - \mathrm{CEPL}$ : this is what the example checked. The formulas do not account for pores left by dissolution or for grains replaced by cement. These are counted separately in the thin section.

Connections to other topics

Sediment texture: grain size, sorting and shape — grain size, sorting and packing, and the initial permeability of loose sand.
Porosity — types of porosity, secondary porosity and the exponential compaction trend.
Rock permeability — Darcy's law, the Kozeny–Carman model and averaging over layers.
Core analysis — how the porosity and permeability of core are measured and why clay-bearing plugs are dried gently.
Reservoir heterogeneity and permeability averaging — the scatter of permeability, its means and the ratio $k_v/k_h$ .
Reservoir architecture and performance — how bodies of different quality make up a reservoir and communicate with one another.
Carbonate sediments, environments and carbonate porosity — pore types of carbonates and their petrophysical classes.
Water saturation in shaly sands — how clay distorts logs and the estimate of saturation.
Capillary pressure and the transition zone — mercury curves, channel sizes and the Leverett J-function by rock type.
Nuclear magnetic resonance (NMR) logging — relaxation time distributions and permeability from NMR logging — signs of rock types in wells.

Check yourself

1
Two sandstones with a porosity of 18% contain 5% of authigenic clay each: in the first it is booklets of kaolinite in separate pores, in the second fibrous illite. Which of them is more likely to be the more permeable?
1. A.The first
2. B.The second
3. C.The same: the clay content is equal
4. D.The one in which the clay grew earlier
Show the answer
Correct answer: A. The first
Kaolinite occupies separate pores and leaves the throats largely free, but illite fibres block the pores and throats across the flow. At the same porosity and the same clay content, the rock with illite is usually an order of magnitude or more less permeable, and its irreducible water saturation is higher.
2
From a thin section: intergranular pores 18%, intergranular cement 12%, no matrix. Take the initial porosity as 40%. What took more porosity — compaction or cement?
Show the answer
$\mathrm{IGV} = 18 + 12 = 30$ %. $\mathrm{COPL} = 100\cdot(40 - 30)/(100 - 30) \approx 14.3$ %, $\mathrm{CEPL} = 12\cdot 60/70 \approx 10.3$ % of the original volume: compaction took slightly more. Check: $40 - 14.3 - 10.3 = 15.4$ %, and the present 18% of pores in a volume compressed to $100\cdot 60/70 \approx 85.7$ % of the original is $0.18\cdot 85.7 \approx 15.4$ %.
3
A straight line was drawn by least squares through a cloud of plugs on a plot of porosity against $\lg k$ . What does a point on the line give at a given porosity?
1. A.Roughly the median, that is, the geometric mean of permeability
2. B.The highest permeability
3. C.The arithmetic mean of permeability
4. D.The harmonic mean
Show the answer
Correct answer: A. Roughly the median, that is, the geometric mean of permeability
A regression on the logarithm averages logarithms. The mean of the logarithm is the logarithm of the geometric mean. For a scatter that is symmetric in logarithms (lognormal), the geometric mean coincides with the median. The arithmetic mean is larger by a factor of $e^{\sigma^2/2}$ .
4
A plug: $\phi$ = 0.20, $k$ = 100 mD. Find RQI and FZI and decide which type of the lesson's example it is closer to (the boundaries are 2.96 and 0.90 µm).
Show the answer
$\mathrm{RQI} = 0.0314\sqrt{100/0.20} = 0.0314\cdot 22.36 \approx 0.702$ µm; $\phi_z = 0.20/0.80 = 0.25$ ; $\mathrm{FZI} \approx 0.702/0.25 \approx 2.81$ µm. This lies between 0.90 and 2.96, so it is type 2, but right at the boundary with type 1. Such a plug should be looked at in thin section before it is assigned to a type.
5
Why do sandstones with continuous chlorite coats on the grains often keep a high porosity at great depths?
1. A.Chlorite dissolves quartz cement
2. B.The coats raise the pore pressure
3. C.The coats stop quartz overgrowths from starting to grow on the grains
4. D.Chlorite makes the grains stronger, so there is no compaction
Show the answer
Correct answer: C. The coats stop quartz overgrowths from starting to grow on the grains
Quartz cement grows on the surface of quartz grains, continuing their crystal lattice. A continuous coat covers this surface, so the cement can start growing only in gaps in the coat. Coats do not stop compaction, so the intergranular volume of such rocks is usually the same as that of their neighbours. What is preserved is exactly what cement has not filled.
6
A porosity trend was built from wells that penetrated the reservoir at depths of 1.8–2.4 km (about 70–87 °C) and gives 0.19 at a depth of 4 km (about 135 °C). In which direction is the prediction most likely to be wrong, and why?
Show the answer
It is most likely too high (an overestimate). The wells of the trend reached the reservoir only at the start of the interval of active quartz cementation (which becomes noticeable above about 70–80 °C). At 4 km the rock has gone through most of that interval. Quartz cement will take porosity beyond what mechanical compaction takes and, in the throats, permeability as well. The exceptions are sandstones with protective coats on the grains or with abnormally high pore pressure. We look for their signs in core and in pressure measurements.

Updated: 26 October 2026