Reservoir pressure and temperature

Quick refresher

If the water in a reservoir is connected to the surface through the rock, its pressure equals the pressure of its own column of water. This is normal (hydrostatic) pressure. A lasting departure from this line is abnormal pressure: abnormally high (overpressure) or abnormally low (subnormal pressure).
In every continuous phase (a connected body of one fluid) the pressure grows with depth at the rate $\rho g$ , which is the gradient. It is about 10–11 kPa/m in water, 6–8.5 kPa/m in oil, and 1–3 kPa/m in gas.
The pressure lines of different phases cross at the contacts. From measurements with a wireline formation tester (a tool that measures the pressure in the rock at chosen depths) we can find the gas–oil contact and the free water level (the level where the water pressure equals the pressure in the oil), even if no well has gone through them.
The water sets the pressure at the base of an accumulation. Going up, the pressure falls along the gradient of the lighter fluid, so at the top the pressure is higher than in the water at the same depth. The seal (the rock above that stops the oil and gas rising) holds this excess back, and drillers must allow for it.
We estimate reservoir temperature from the geothermal gradient — usually 2–4 °C per 100 m — and from measurements in the well. Measurements soon after drilling are too low, so we extrapolate them to the stabilised temperature.
During development we regard the reservoir as isothermal (at a constant temperature), except in zones where cold water or steam is injected.

Key relationships

p_w(z) = p_{\text{atm}} + \rho_w\,g\,z + C

pressure in a water-bearing layer: $C = 0$ is normal pressure, $C > 0$ overpressure, $C < 0$ subnormal pressure

\dfrac{\dd p}{\dd z} = \rho\,g

pressure gradient in a continuous phase; $\rho$ is the density at reservoir conditions

\Delta p = (\rho_w - \rho_{\text{hc}})\,g\,h

excess of the oil or gas pressure over the water pressure at a height $h$ above the free water level

T(z) = T_0 + G_T\,z

reservoir temperature from the geothermal gradient

Common pitfalls

Mixing absolute pressure (development, fluid properties) with gauge pressure (drilling, wellhead gauges).
Comparing pressures measured at different depths without bringing them to one reference depth (datum).
Extending a gradient measured over a few tens of metres all the way to the surface. A small measurement error turns into a large pressure error.
Taking the free water level found from pressure for the oil–water contact seen on the logs. The log contact lies higher, by a height that matches the entry pressure of the rock (the capillary pressure that oil needs to enter the pores).
Taking the reservoir temperature from the first measurement after drilling: it is usually 10–20 °C too low.

After this lesson you can

tell initial, current, static and average reservoir pressure apart, and do not confuse absolute pressure (measured from zero) with gauge pressure (read above atmospheric pressure)
draw the line of normal hydrostatic pressure (the pressure of a column of water) and judge from a measurement whether the pressure in the accumulation is normal
calculate the pressure gradients (how fast pressure grows with depth) of water, oil and gas from their densities in the reservoir, and find the positions of the contacts from pressure measurements at different depths
estimate the pressure at the top of an accumulation and explain why it is higher than the pressure in the aquifer water at the same depth
describe how reservoir pressure is measured and how it is brought to one reference depth (the datum)
estimate reservoir temperature from the geothermal gradient (how fast the temperature rises with depth) and from measurements in the well after drilling

What is meant by reservoir pressure

The fluids in the pores of a reservoir are under pressure. Its main source is the weight of a column of water that is connected to the surface through permeable rocks. Where there is no such connection, part of the weight of the rocks and other causes of abnormal pressure add to it (pore and overburden pressure). In field development we distinguish several “reservoir pressures”:

Initial reservoir pressure $p_i$: The pressure in the accumulation before development starts. Together with temperature it sets the initial state of the fluids and the hydrocarbons in place.
Current reservoir pressure: The pressure in the reservoir at a given moment of development. It falls as fluids are withdrawn, unless injection maintains it.
Static pressure: The pressure that settles in a shut-in well (a closed well). We use it to judge the pressure in the well's drainage area (the part of the reservoir that the well drains).
Average reservoir pressure $\bar p$: The average over the pore volume of the accumulation or of the drainage area (average reservoir pressure). It enters the material balance.
Bottomhole flowing pressure $p_{wf}$: The pressure at the bottom of a producing well; lower than the reservoir pressure by the drawdown (bottomhole flowing pressure).

Normal pressure and abnormal pressure

Suppose the pores of a reservoir are connected to the surface through permeable rocks. Then the pressure in the water at depth $z$ equals the pressure of the column of water above that point. We call this pressure normal, or hydrostatic. Its gradient is set by the density of the formation water, that is, by its salinity (the amount of dissolved salt): from 9.8 kPa/m for fresh water to 11 kPa/m and more for strong brines.

Pressure in a water-bearing layer

p_w(z) = p_{\text{atm}} + \rho_w\,g\,z + C

where:

$p_w$ absolute pressure in the water at depth $z$ , Pa
$p_{\text{atm}}$ atmospheric pressure, 0.101 MPa, Pa
$\rho_w$ density of the formation water, kg/m³
$g$ acceleration due to gravity, 9.81 m/s², m/s²
$z$ true vertical depth from the level where the water column begins (on land roughly from the ground surface, offshore from sea level), m
$C$ constant of the pressure anomaly (the departure from normal pressure) for a hydrodynamically connected system, Pa

Units: SI, absolute pressure

Holds when

The water is at rest and the column is continuous. The water density is constant. (If it changes with depth, we add up the pressure interval by interval.)
$C$ is the same throughout one hydrodynamically connected system (a system in which pressure is passed on freely).

$C = 0$ is normal pressure, $C > 0$ is abnormally high pressure (overpressure), and $C < 0$ is abnormally low pressure (subnormal pressure). Draw a straight line through the measurements in the water-bearing layers of an area and extend it up to zero depth. It reaches atmospheric pressure when the pressure is normal, and $p_{\text{atm}} + C$ when it is abnormal.

In other unit systems (2)

Metric units

p_w\,[\text{MPa}] = 0.101 + 0.00981\,\rho_w\,[\text{g/cm}^3]\;z\,[\text{m}] + C

$\rho_w$ in g/cm³, $z$ in m, $p_w$ and $C$ in MPa.

0.00981 MPa/m is the pressure of a column of liquid of density 1 g/cm³ and height 1 m.

Oilfield units

p_w\,[\text{psia}] = 14.7 + 0.433\,\gamma_w\;z\,[\text{ft}] + C

$\gamma_w$ is the specific gravity of the water (relative to fresh water), $z$ in feet, $p_w$ and $C$ in psia.

0.433 psi/ft is the gradient of fresh water.

Abnormal pressure lasts only under a seal that shuts the reservoir off from the surrounding waters. The topic Abnormal formation pressures: origin, prediction and recognition covers where it comes from and how to predict and recognise it while drilling. For the reservoir engineer something else matters. An accumulation with overpressure holds an extra store of pressure. Its undercompacted rocks (rocks whose pores have not yet been squeezed by the weight above them) are noticeably compressed as the pressure falls, and this compaction also pushes fluids out towards the wells (Reservoir drive mechanisms).

If the water in a water-bearing layer flows, its pressure differs from hydrostatic. An example is flow from a recharge area, where the layer reaches the surface, to a discharge area. Then the contacts of accumulations may be tilted in the direction of flow. For most accumulations the aquifer water moves very slowly, and we take the pressure distribution in it as hydrostatic.

Pressure gradients of oil and gas

Inside an accumulation the pores are filled with oil or gas. The pressure in each continuous phase changes with depth according to the hydrostatics of its density, with the gradient $\rho g$ . Oil and gas are lighter than water, so their pressure grows with depth more slowly. On a pressure–depth plot their pressure lines run closer to the vertical.

Pressure gradient in a continuous phase

\frac{\dd p}{\dd z} = \rho\,g

where:

$\dd p/\dd z$ pressure gradient — the increase in pressure per unit of true vertical depth, Pa/m
$\rho$ fluid density at reservoir conditions, kg/m³
$g$ acceleration due to gravity, m/s²

Units: SI

Holds when

The fluid is at rest (static equilibrium) and forms a continuous phase between the measurement points.
The density is constant with height — a good approximation for water and oil; for gas see below.

Gradient and density go together one to one, so a measured gradient tells us which fluid is in the reservoir. About 2 kPa/m means gas, about 7 kPa/m means oil, and about 10.5 kPa/m means water.

In other unit systems (2)

Metric units

\frac{\dd p}{\dd z}\,[\text{kPa/m}] = 9.81\,\rho\,[\text{g/cm}^3]

$\rho$ in g/cm³, gradient in kPa/m.

Oilfield units

\frac{\dd p}{\dd z}\,[\text{psi/ft}] = 0.433\,\rho\,[\text{g/cm}^3]

$\rho$ in g/cm³ (numerically the specific gravity relative to fresh water), gradient in psi/ft; 1 psi/ft = 22.62 kPa/m.

Fluid densities in the reservoir and pressure gradients

Fluid in the reservoir: Formation water
Density, g/cm³: 1.00–1.20
Gradient, kPa/m: 9.8–11.8
Gradient, psi/ft: 0.43–0.52

Fluid in the reservoir: Oil with dissolved gas
Density, g/cm³: 0.55–0.90
Gradient, kPa/m: 5.4–8.8
Gradient, psi/ft: 0.24–0.39

Fluid in the reservoir: Gas-condensate mixture
Density, g/cm³: 0.25–0.50
Gradient, kPa/m: 2.5–4.9
Gradient, psi/ft: 0.11–0.22

Fluid in the reservoir: Gas
Density, g/cm³: 0.05–0.30
Gradient, kPa/m: 0.5–2.9
Gradient, psi/ft: 0.02–0.13

Fluid in the reservoir	Density, g/cm³	Gradient, kPa/m	Gradient, psi/ft
Formation water	1.00–1.20	9.8–11.8	0.43–0.52
Oil with dissolved gas	0.55–0.90	5.4–8.8	0.24–0.39
Gas-condensate mixture	0.25–0.50	2.5–4.9	0.11–0.22
Gas	0.05–0.30	0.5–2.9	0.02–0.13

These are approximate ranges. The densities are at reservoir conditions. Oil in the reservoir is lighter than at the surface because of dissolved gas and heating (Black oil model: solution gas–oil ratio and formation volume factors). Gas in the reservoir is hundreds of times denser than at standard conditions (Gas formation volume factor, compressibility and viscosity).

The density of gas grows with pressure, so its gradient increases slightly with depth. For accumulations tens or hundreds of metres high, this change is a few per cent, and we take the gas pressure line as straight. For tall columns of gas, for example in the wellbore of a gas well, we calculate the pressure with the compression of the gas included (advanced section).

The water sets the pressure at the base of the accumulation. Going up through the accumulation, the pressure falls more slowly than it would in water. So at the top, the pressure of the oil or gas is higher than the pressure in the aquifer water at the same depth. This excess is the buoyancy of the hydrocarbons (the force that pushes the lighter fluid upwards). The seal must hold it back (capillary barrier of the seal).

Excess of the pressure at the top of the accumulation over the water pressure

\Delta p = (\rho_w - \rho_o)\,g\,h_o + (\rho_w - \rho_g)\,g\,h_g

where:

$\Delta p$ difference between the hydrocarbon pressure at the top and the water pressure extended along its gradient to the same depth, Pa
$\rho_w,\ \rho_o,\ \rho_g$ densities of water, oil and gas at reservoir conditions, kg/m³
$h_o,\ h_g$ heights of the oil part (from the free water level to the gas–oil contact) and of the gas part (from the contact to the top), m

Units: SI

Holds when

The accumulation is in hydrostatic equilibrium and the phases are continuous.
If there is no gas cap, $h_g = 0$ ; for a purely gas accumulation $h_o = 0$ .

The excess grows with the height of the accumulation, and it grows faster when the fluid is lighter. Over gas it is two to three times greater than over oil of the same height. The seal holds the accumulation as long as this excess is less than the seal's entry pressure, and the pressure at the top is itself less than the pressure at which fractures open in the seal.

Fluid contacts from pressure measurements

If pressure has been measured at several depths in each zone, the points lie on straight lines. The slope of each line matches the density of the fluid. The point where the oil and water lines cross is the free water level (FWL), where the pressures of the two phases are equal. The point where the gas and oil lines cross is the gas–oil contact (GOC). This is how we find the position of a contact even if no well has gone through it. Measurements in the oil zone are enough, together with a water pressure line built from measurements in a neighbouring well that has penetrated the same water-bearing layer.

How to find a contact from pressure

1
Measurements
A wireline formation tester measures the pressure in permeable layers through the whole section (Pressure measurement with the wireline formation tester). We reject measurements with a poor seal, and measurements affected by supercharging (extra pressure that mud filtrate, the liquid part of the drilling mud, has forced into the rock around the well).
2
Gradients
Draw a straight line through the points of each zone (with many points, by least squares, which means the line that fits the points best). The slope gives the density of the fluid. Check that this density is plausible.
3
Intersections
Find the depths where the lines cross. These are the gas–oil contact and the free water level.
4
Check
Compare the result with logs, core and samples. The oil–water contact seen on the logs lies above the free water level, by a height that matches the entry capillary pressure of the rock. The transition zone (where the water saturation changes gradually) begins above it (contact and transition zone).

Worked exampleGas–oil contact and free water level from formation tester measurements

Given (assumed data)

Gas cap: 2400 m — 27.296 MPa; 2440 m — 27.375 MPa
Oil rim: 2480 m — 27.530 MPa; 2525 m — 27.848 MPa
Water-bearing part: 2560 m — 28.163 MPa; 2600 m — 28.583 MPa
Measurement conditions: pressures absolute, true vertical depths from the surface

Solution

Gradients: gas $(27.375 - 27.296)\cdot 1000/40 = 1.975$ kPa/m, oil $(27.848 - 27.530)\cdot 1000/45 = 7.067$ kPa/m, water $(28.583 - 28.163)\cdot 1000/40 = 10.50$ kPa/m. The densities $\rho = (\dd p/\dd z)/g$ are 0.20 g/cm³ for gas, 0.72 g/cm³ for oil and 1.07 g/cm³ for water.
Gas–oil contact. We extend the oil line upward to 2440 m: $27.530 - 0.007067\cdot 40 = 27.2473$ MPa. The gas pressure there is 0.1277 MPa higher than this. With every metre of depth the two lines get closer by $7.067 - 1.975 = 5.092$ kPa/m, so they meet 25.1 m deeper. The GOC is at a depth of 2465.1 m.
Free water level. We extend the water line upward to 2525 m: $28.163 - 0.0105\cdot 35 = 27.7955$ MPa. The oil pressure there is 0.0525 MPa higher than this. With every metre of depth the two lines get closer by $10.50 - 7.067 = 3.433$ kPa/m, so they meet 15.3 m deeper. The FWL is at a depth of 2540.3 m.

Answer: The GOC is at about 2465 m, the free water level at about 2540 m; the height of the oil rim is about 75 m.

What it means. The densities confirm the fluids: gas, oil with dissolved gas, and saline water. The oil–water contact seen on the logs will lie above the level we found, by a height that matches the entry pressure of the rock. This height is a fraction of a metre in high-permeability sandstones, and metres or tens of metres in tight rocks. With two points per zone, we locate the contact to within about a metre: a measurement error of 1 kPa shifts the crossing point by 0.2–0.3 m, and an error of 1 m in true vertical depth gives a pressure error of 10 kPa in water. In practice there are more points, and we check the depths of the measurements against the directional survey (the record of the well's path). The data in this example are assumed for illustration.

CalculatedThe pressure measurements of the example (dots), the gas, oil and water lines through them, the contacts and the line of normal hydrostatic pressure of the area (water density 1.07 g/cm³).

Worked examplePressure regime and the pressure at the top of the accumulation

Given (assumed data)

Measurements and contacts: as in the previous example: water — 28.163 MPa at 2560 m; GOC — 2465 m, FWL — 2540 m; gas gradient 1.975 kPa/m
Normal pressure of the area: a column of formation water of density 1.07 g/cm³ from the surface
Top of the accumulation: 2385 m

Solution

Normal pressure at a depth of 2560 m: $0.101 + 0.00981\cdot 1.07\cdot 2560 = 26.973$ MPa. The measured pressure is larger by $C = 1.190$ MPa, so the pressure is abnormally high.
Pressure at the top, along the gas line: $27.296 - 0.001975\cdot 15 = 27.266$ MPa.
The aquifer water pressure at the depth of the top (the water line extended upward): $28.163 - 0.00981\cdot 1.07\cdot 175 = 26.326$ MPa. The excess at the top is $0.94$ MPa. We check it with the buoyancy formula: 75 m of oil and 80 m of gas give $0.94$ MPa.
Normal pressure at the depth of the top is 25.136 MPa. The pressure at the top is 2.13 MPa higher than this. As an equivalent mud weight (the density of a mud whose column gives this pressure, worked out from the gauge pressure): $(27.266 - 0.101)/(0.00981\cdot 2385) = 1.161$ g/cm³, against 1.07 for normal pressure.

Answer: The accumulation lies in an overpressure zone (

C \approx 1.2

MPa); the pressure at the top is 27.266 MPa, 0.94 MPa higher than the aquifer water pressure at the same depth and 2.13 MPa higher than normal.

What it means. A mud of density 1.12 g/cm³ balances normal pressure with a margin. At the top of this accumulation it would give only about 26.3 MPa of absolute pressure. That is less than the reservoir pressure, so a kick (an unwanted inflow of formation fluid into the well) is possible when the gas cap is penetrated. The seal, for its part, must hold the excess of 0.94 MPa over the water pressure. We should not estimate $C$ by extending to the surface a water gradient measured over only 40 m. A measurement error of 1 kPa would give an error in $C$ of the order of 0.06 MPa, so we take the water density from analyses of samples. The data in this example are assumed for illustration.

How reservoir pressure is measured

Ways of determining reservoir pressure

Method: Wireline formation tester
Where and when: open hole (before the steel casing is run)
What it gives: pressure at tens of points, gradients, samples (more)

Method: Drill stem test (DST)
Where and when: during or after drilling
What it gives: pressure and inflow from the interval

Method: Downhole pressure gauge
Where and when: shut-in wells
What it gives: static pressure; from the pressure build-up curve — reservoir pressure

Method: Permanent downhole gauges
Where and when: instrumented wells
What it gives: continuous pressure record

Method: Calculation from the wellhead or from the fluid level
Where and when: gas wells; wells with pumps
What it gives: from the wellhead pressure (gas) or the static fluid level

Method	Where and when	What it gives
Wireline formation tester	open hole (before the steel casing is run)	pressure at tens of points, gradients, samples (more)
Drill stem test (DST)	during or after drilling	pressure and inflow from the interval
Downhole pressure gauge	shut-in wells	static pressure; from the pressure build-up curve — reservoir pressure
Permanent downhole gauges	instrumented wells	continuous pressure record
Calculation from the wellhead or from the fluid level	gas wells; wells with pumps	from the wellhead pressure (gas) or the static fluid level

In gas wells we often calculate the reservoir pressure from the wellhead pressure of the shut-in well, adding the weight of the gas column (advanced section). In wells with pumps we calculate it from the static fluid level, which an echo sounder measures, and from the density of the liquid in the wellbore.

Measurements in wells drilled during development show how the different layers are being depleted. Suppose impermeable interbeds separate the layers. Then each layer loses pressure in its own way. On the pressure–depth profile we see steps, and segments shifted to the left of the initial line. From such profiles we find barriers and crossflow (flow between layers), see which layers the development drains, and tune the reservoir simulation model (History matching).

We compare pressures measured in different wells at different depths only after bringing them to a single datum depth (a reference depth). The datum is usually the middle of the accumulation by height, or the initial oil–water contact. From the corrected pressures we build isobar maps (maps of lines of equal pressure) and calculate the material balance.

Bringing pressure to a datum

p_{\text{datum}} = p_{\text{meas}} + \rho\,g\,(z_{\text{datum}} - z_{\text{meas}})

where:

$p_{\text{datum}}$ pressure brought to the datum depth, Pa
$p_{\text{meas}}$ measured pressure, Pa
$z_{\text{datum}},\ z_{\text{meas}}$ true vertical depths of the datum and of the measurement point, m
$\rho$ density of the fluid that fills the reservoir between these depths, at reservoir conditions, kg/m³

Units: SI

Holds when

The reservoir between the two depths is hydrodynamically connected and the fluid is at rest.
If a contact lies between the depths, we make the conversion section by section, each with its own density.

If the measurement point is deeper than the datum, the depth difference is negative, so the corrected pressure is less than the measured one. If the point is shallower than the datum, the opposite holds.

Reservoir temperature

Temperature in the earth's crust grows with depth, because heat flows from the interior to the surface. The increase in temperature per unit of depth is the geothermal gradient. In most sedimentary basins (large areas where thick layers of sediment have collected) it is 2–4 °C per 100 m (1.1–2.2 °F per 100 ft). Russian-language literature also uses the reciprocal quantity, the geothermal step. It is the depth over which the temperature rises by 1 °C. For a gradient of 3 °C per 100 m it is about 33 m.

Temperature at depth from the geothermal gradient

T(z) = T_0 + G_T\,z

where:

$T(z)$ temperature of the rocks at depth $z$ , °C
$T_0$ mean annual surface temperature (offshore, the temperature at the sea floor), °C
$G_T$ geothermal gradient, °C/m
$z$ true vertical depth, m

Units: °C and m (the gradient is often given in °C/100 m)

Holds when

We take the gradient as constant through the section. This is a rough approximation, because the gradient depends on the thermal conductivity of the rocks (how well they conduct heat) (advanced section).
The temperature is the stabilised one, undisturbed by drilling and development.

The formula gives a first estimate of the reservoir temperature before drilling. For calculating fluid properties, we replace it with the measured reservoir temperature.

In other unit systems (1)

Oilfield units

T\,[{}^{\circ}\text{F}] = T_0\,[{}^{\circ}\text{F}] + G_T\,[{}^{\circ}\text{F}/100\ \text{ft}]\cdot\frac{z\,[\text{ft}]}{100}

temperatures in °F, depth in feet; 1 °C/100 m = 0.549 °F/100 ft.

The worse the rocks conduct heat, the larger the gradient. With the same heat flow, the gradient is higher in shales than in sandstones and carbonates, and much higher than in rock salt, which conducts heat two to four times better than shales. So under a thick salt sequence, as in the subsalt deposits of the Pre-Caspian Basin, a reservoir at the same depth is colder than where there is no salt.

The mass of the rocks and their heat capacity (the heat they can store) are enormous, while the fluids move slowly. So during development we regard the reservoir as isothermal: its temperature hardly changes. The exceptions are the zones around injection wells when cold water or steam is injected, the near-wellbore zones of gas wells, where the gas cools as it expands, and thermal enhanced oil recovery methods.

The most reliable reservoir temperature comes from measurements in a well that has stood idle for a long time, and from the temperature of the inflow during a test. Logging, however, takes place soon after drilling, when the circulation of the drilling fluid is still having an effect: at the bottom of the hole the fluid cools the rocks. So the bottomhole temperature (BHT) recorded by a maximum thermometer on a logging sonde (a thermometer that keeps the highest reading) is usually 10–20 °C too low. The less time has passed since circulation stopped, the larger the error. We extrapolate the temperature to the stabilised value from several measurements made at different times.

Extrapolation of the bottomhole temperature to the stabilised temperature (Horner method)

T_{\text{bh}}(\Delta t) = T_{\text{res}} - m\,\lg\frac{t_c + \Delta t}{\Delta t}

where:

$T_{\text{bh}}(\Delta t)$ bottomhole temperature measured a time $\Delta t$ after circulation stopped, °C
$T_{\text{res}}$ stabilised reservoir temperature — the value at $\lg(\ldots) = 0$ , that is, as $\Delta t \to \infty$ , °C
$t_c$ duration of circulation at this depth before it stopped, h
$m$ slope of the line (per logarithmic cycle), °C

Units: time in the same units (hours); the logarithm is decimal (lg)

Holds when

This is an empirical method, by analogy with the pressure build-up curve. It describes the recovery of temperature around the wellbore with a line heat source model (heat released along a line, the wellbore).
We know the duration of circulation only roughly. We need at least three measurements, preferably from different runs to the same depth.

On a plot of temperature against $\lg\left((t_c + \Delta t)/\Delta t\right)$ , the measurements lie roughly on a straight line. The point where the line meets the temperature axis (its intercept) gives the reservoir temperature.

Worked exampleReservoir temperature from three bottomhole measurements

Given (assumed data)

Depth of the measurements: 2900 m; circulation before shutdown $t_c = 5$ h
Measurements: after 6 h — 85.2 °C; after 11 h — 91.2 °C; after 20 h — 95.2 °C
Mean annual surface temperature: 10 °C

Solution

Values on the horizontal axis: $\lg(11/6) = 0.2632$ , $\lg(16/11) = 0.1627$ , $\lg(25/20) = 0.0969$ .
Line through the first and third measurements: slope $m = \dfrac{95.2 - 85.2}{0.2632 - 0.0969} = 60.1$ °C per cycle. Reservoir temperature $T_{\text{res}} = 95.2 + 60.1\cdot 0.0969 = 101.0$ °C.
Check with the second measurement. On the line at $\lg(16/11) = 0.1627$ we get $101.0 - 60.1\cdot 0.1627 = 91.2$ °C, and 91.2 °C was measured. The points agree.
Geothermal gradient: $(101.0 - 10)/29 = 3.14$ °C/100 m. From the first measurement it would come out as only 2.59 °C/100 m.

Answer: The reservoir temperature is about 101 °C, 16 °C higher than the first measurement.

What it means. Suppose we take the first measurement for a gas accumulation, instead of the stabilised temperature. Then the $B_g$ of the gas in the examples of the topic Gas formation volume factor, compressibility and viscosity at 20 MPa comes out 6.7% too low (both the temperature and $Z$ are lower), and the gas in place is overstated by 7.2%. For oil, a temperature that is too low distorts the bubble point pressure and the viscosity. The data in this example are assumed for illustration.

The topic Formation water and formation temperature describes how log interpretation uses reservoir temperature, for converting the resistivity of formation water.

Terms and symbols

Initial reservoir pressure $p_i$: The pressure in the accumulation before development starts.
Normal (hydrostatic) pressure: The pressure of a column of formation water from the surface to the given depth.
Pressure gradient: The increase in pressure per metre of true vertical depth. In a continuous phase it equals $\rho g$ .
Free water level (FWL): The depth where the oil (gas) and water pressures are equal. We find it where their pressure lines cross.
Datum depth: The depth to which we bring all pressure measurements so that we can compare them.
Geothermal gradient and geothermal step: The gradient is the increase in temperature per unit of depth (°C/100 m). The step is its reciprocal: the depth over which the temperature rises by 1 °C.
BHT: The bottomhole temperature recorded during logging soon after drilling. It is too low, because circulation cools the hole.

Common mistakes and misconceptions

Wrong

“The pressure at the top of a gas accumulation equals the hydrostatic pressure for that depth.”

In fact

It is higher than the pressure in the aquifer water by the buoyancy $(\rho_w - \rho_g)gh$ . Under overpressure it is higher than the hydrostatic pressure also by the anomaly $C$ .

Wrong

“The contact from pressure and the oil–water contact from the logs are the same thing.”

In fact

The crossing of the pressure lines gives the free water level. The contact seen on the logs is higher, by a height that matches the entry capillary pressure of the rock.

Wrong

“The first temperature measurement on a log is the reservoir temperature.”

In fact

Circulation of the mud has cooled the bottom of the hole, so the measurement is usually 10–20 °C too low. We need to extrapolate from several measurements, or to measure in a well that has been idle for a long time.

Wrong

“The pressures in two wells can be compared directly.”

In fact

First we bring them to one datum with the fluid gradient. A depth difference of 50 m is 0.35 MPa in oil and 0.5 MPa in water.

In depth: advanced detail

Heat flow and layers of different thermal conductivity

Heat from the interior passes to the surface through all the layers of rock. In the steady state (when nothing changes with time), the heat flow is the same in every layer, if we neglect the heat that radioactive elements release in the rocks themselves. By Fourier's law, the temperature gradient in a layer is inversely proportional to its thermal conductivity:

Geothermal gradient from heat flow

G_T = \frac{q}{\lambda}

where:

$G_T$ geothermal gradient in the layer, K/m
$q$ heat flow density (the heat passing through each square metre) from the interior; in sedimentary basins usually 40–80 mW/m², W/m²
$\lambda$ thermal conductivity of the rocks of the layer: shales 1–2, sandstones and carbonates 2–4, rock salt about 5–6 W/(m·K), W/(m·K)

Units: SI

Holds when

Steady one-dimensional vertical heat flow.
The layers are horizontal. Above and below salt domes, the heat flow shifts sideways and the picture is more complicated.

Salt conducts heat well, so it “passes” the heat on with a small gradient. So the reservoir below the salt is colder. Above salt domes, by contrast, the rocks are warmer than their surroundings.

Worked exampleTemperature of a subsalt reservoir

Given (assumed data)

Heat flow $q$: 55 mW/m²; surface temperature 10 °C
0–1200 m: shales and sandstones, $\lambda = 1.8$ W/(m·K)
1200–3700 m: rock salt, $\lambda = 5.0$ W/(m·K)
3700–4300 m: subsalt carbonates, $\lambda = 2.8$ W/(m·K); the reservoir is at a depth of 4300 m

Solution

Gradients $q/\lambda$ : above the salt 3.06 °C/100 m, in the salt 1.10 °C/100 m, below the salt 1.96 °C/100 m.
Temperature at the top of the salt: $10 + 3.06\cdot 12 = 46.7$ °C; at the base of the salt: $46.7 + 1.10\cdot 25 = 74.2$ °C.
Reservoir temperature: $74.2 + 1.96\cdot 6 = 86.0$ °C.
If the gradient of the upper part of the section continued down to 4300 m, we would get $10 + 3.06\cdot 43 = 141$ °C.

Answer: The temperature of the subsalt reservoir is about 86 °C, 55 °C lower than the estimate from the gradient of the upper part of the section.

What it means. An error of several tens of degrees changes the forecast of the phase state of the fluid (gas, liquid or both), its viscosity and bubble point pressure, and the choice of materials and cements. So for subsalt accumulations we forecast the temperature from the heat flow and the thermal conductivity of the rocks, and then check it by measurements. The data in this example are assumed for illustration.

CalculatedTemperature through the section of the example: in the salt the gradient is small, and the subsalt reservoir at 4300 m is colder than the continuation of the gradient of the upper part of the section would give (dashed line).

Pressure in a column of gas: converting from the wellhead to the bottom of the hole

In a shut-in gas well the wellbore is filled with gas, so we can calculate the bottomhole pressure from the wellhead pressure. The density of gas is proportional to pressure. So pressure grows with depth exponentially, not linearly. We take $\dd p/\dd z = \rho_g g$ and $\rho_g = pM/(ZRT)$ , with an average $Z$ and an average temperature, and get:

Pressure in a static column of gas

p_{\text{bh}} = p_{\text{wh}}\,\exp\left(\frac{M\,g\,H}{\bar Z\,R\,\bar T}\right)

where:

$p_{\text{bh}},\ p_{\text{wh}}$ absolute pressures at the bottom and at the wellhead of the shut-in well, MPa
$M$ molar mass of the gas, $28.97\,\gamma_g$ , kg/kmol
$H$ true vertical depth of the bottom (the middle of the reservoir), m
$\bar Z,\ \bar T$ gas deviation factor at the average pressure, and the average absolute temperature in the wellbore, —, K
$R$ gas constant, 8314 J/(kmol·K), J/(kmol·K)

Units: SI ( $g = 9.81$ m/s²)

Holds when

The gas is at rest and there is no liquid in the wellbore.
We replace $Z$ and $T$ by their averages over the wellbore. We find $\bar Z$ at the average pressure, so we repeat the calculation once or twice.

The exponent is the ratio of the weight of the gas column to the pressure. The heavier the gas and the deeper the well, the more the bottomhole pressure exceeds the wellhead pressure. It is usually 10–30% higher.

In other unit systems (2)

Metric units

p_{\text{bh}} = p_{\text{wh}}\,\exp\left(\frac{0.03417\,\gamma_g\,H\,[\text{m}]}{\bar Z\,\bar T\,[\text{K}]}\right)

$H$ in m, $\bar T$ in K.

$0.03417 = 28.97\cdot 9.807/8314$ K/m. With $g$ = 9.81 m/s² it comes to 0.03418, and this is how the lesson on gas lift writes the constant.

Oilfield units

p_{\text{bh}} = p_{\text{wh}}\,\exp\left(\frac{0.01875\,\gamma_g\,H\,[\text{ft}]}{\bar Z\,\bar T\,[{}^{\circ}\text{R}]}\right)

$H$ in feet, $\bar T$ in degrees Rankine.

$0.01875 = 28.97/(144\cdot 10.73)$ °R/ft.

Worked exampleReservoir pressure from the wellhead pressure of a gas well

Given (assumed data)

Gas: as in the examples of the topic Ideal and real gas: the gas deviation factor (Z): $\gamma_g = 0.65$ , $T_{pc} = 207.8$ K, $p_{pc} = 4.63$ MPa
Wellhead pressure of the shut-in well: 20.5 MPa (absolute)
Depth of the middle of the reservoir: 2400 m; average temperature in the wellbore 60 °C

Solution

First approximation: $Z$ at the wellhead pressure and 60 °C is 0.8341. The exponent is $0.03417\cdot 0.65\cdot 2400/(0.8341\cdot 333.15) = 0.1918$ . The bottomhole pressure is $20.5\,\e^{0.1918} = 24.83$ MPa.
Refinement: the average pressure is $(20.5 + 24.83)/2 = 22.67$ MPa and $\bar Z = 0.8472$ . The exponent is 0.1889 and the bottomhole pressure is 24.76 MPa.
A further iteration changes the result by less than 0.01 MPa.

Answer: The reservoir pressure is about 24.76 MPa, 21% higher than the wellhead pressure.

What it means. A column of gas 2400 m high creates more than 4 MPa, so we cannot neglect it. If there is liquid (water or condensate) in the wellbore, a calculation for gas alone gives a bottomhole pressure that is too low. So we run a downhole gauge from time to time as a check. The data in this example are assumed for illustration.

Connections to other topics

Pore pressure and overburden pressure — hydrostatic pressure, gradients and equivalent mud weight in drilling.
Abnormal formation pressures: origin, prediction and recognition — where overpressure and subnormal pressure come from and how they are recognised while drilling.
Pressure gradients, fluid contacts and capillary equilibrium — pressure gradients, contacts and capillary equilibrium in detail.
Capillary pressure and the transition zone — why the oil–water contact lies above the free water level.
Gas formation volume factor, compressibility and viscosity — gas density and formation volume factor at reservoir pressure and temperature.
Reservoir drive mechanisms — how the energy of reservoir pressure displaces fluids towards the wells.

Check yourself

1
A line drawn through the pressure measurements in the water-bearing layers of an area and extended to zero depth reaches 3.0 MPa (absolute pressure). What does this mean?
1. A.The pressure in the water-bearing system is abnormally high, about 2.9 MPa above normal
2. B.The water is fresh
3. C.The pressure is normal
4. D.The pressure is abnormally low
Show the answer
Correct answer: A. The pressure in the water-bearing system is abnormally high, about 2.9 MPa above normal
With normal pressure the water column begins at the surface, so the line reaches atmospheric pressure, 0.1 MPa. An intercept of 3.0 MPa means a constant addition $C \approx 2.9$ MPa for the whole water-bearing complex. This is abnormally high pressure, and a seal holds it.
2
The oil density in the reservoir is 0.78 g/cm³. Find the pressure gradient in the oil zone in kPa/m and psi/ft.
Show the answer
$9.81\cdot 0.78 = 7.65$ kPa/m; $0.433\cdot 0.78 = 0.338$ psi/ft.
3
In the oil zone at a depth of 2300 m the pressure is 24.50 MPa and the oil gradient 7.0 kPa/m. In the water-bearing part of the reservoir, in a neighbouring well at a depth of 2400 m, the pressure is 25.10 MPa and the water gradient 10.3 kPa/m. Where is the free water level?
Show the answer
The oil line is $p = 24.50 + 0.0070\,(z - 2300)$ . The water line is $p = 25.10 + 0.0103\,(z - 2400)$ . We set them equal: $0.0033\,z = 8.02$ , so $z \approx 2430$ m, and the pressure there is 25.41 MPa. No well has penetrated this level, but we found its position from pressure, provided the layers are hydrodynamically connected.
4
Where is the excess of the hydrocarbon pressure over the aquifer water pressure at the same depth the greatest?
1. A.The same over the whole height
2. B.At the top of the accumulation
3. C.At the gas–oil contact
4. D.At the free water level
Show the answer
Correct answer: B. At the top of the accumulation
At the free water level the pressures are equal. Going up, the hydrocarbon pressure falls more slowly than the water pressure, so the difference grows with height and reaches its maximum at the top.
5
The first bottomhole temperature measurement after drilling is 78 °C. What can we say about the reservoir temperature?
1. A.It is above 78 °C: the circulation of the mud has cooled the bottom of the hole
2. B.It is below 78 °C: the mud has heated the bottom of the hole
3. C.It is equal to 78 °C
4. D.We can say nothing from a bottomhole measurement
Show the answer
Correct answer: A. It is above 78 °C: the circulation of the mud has cooled the bottom of the hole
At the bottom of the hole the circulating mud is colder than the rocks and cools them. After circulation stops the temperature recovers. The first measurement is usually 10–20 °C too low, and we find the stabilised temperature by extrapolating several measurements.
6
A pressure of 21.40 MPa was measured in the oil part of the reservoir at a depth of 2520 m. The oil density in the reservoir is 0.75 g/cm³. Bring the pressure to a datum at a depth of 2480 m.
Show the answer
$p_{\text{datum}} = 21.40 + 0.00981\cdot 0.75\cdot (2480 - 2520) = 21.40 - 0.29 = 21.11$ MPa. The datum is above the measurement point, so the corrected pressure is smaller.

Updated: 26 October 2026