Formation water and formation temperature

Quick refresher

The ions (charged particles) of the salts in the formation water carry the current in a rock. The higher the salinity and the temperature, the lower the resistivity of the water, $R_w$ , and of the rock.
We express salinity in g/L (the mass of salt in a litre of solution) or in ppm (milligrams of salt per kilogram of solution). The two differ by the density of the solution. Log interpretation uses the equivalent NaCl concentration (the concentration of an NaCl solution that conducts as well as the water).
When a solution is heated from 20 to 100 °C, its resistivity falls almost threefold. We bring all the resistivities ( $R_w$ , $R_{mf}$ , $R_m$ ) to formation temperature.
Temperature usually increases with depth by 2–4 °C per 100 m. Between the surface and the bottom of the hole we usually interpolate it linearly (along a straight line).
Circulating mud cools the bottom of the hole, so a temperature measured soon after circulation stops is below the formation temperature. If we take several readings over time, we can extrapolate (extend the trend) to the formation temperature.

Key relationships

R_2 = R_1\,\dfrac{T_1 + 21.5}{T_2 + 21.5}

conversion of the resistivity of an NaCl solution from temperature $T_1$ to $T_2$ (°C), the Arps formula

T(z) = T_0 + G_T\,z

linear increase of temperature with depth; $G_T$ is the geothermal gradient

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

recovery of the bottom-hole temperature after circulation stops (the Horner method)

Common pitfalls

Taking the resistivity of a sample at laboratory temperature as the formation $R_w$ .
Treating the bottom-hole temperature measured on the first tool run as the formation temperature.
Confusing g/L and ppm: for brines the difference reaches 17%.
Estimating $R_w$ from the total salinity of a water of complex composition as if it were a pure NaCl solution.

After this lesson you can

tell apart the water in the pores, the water bound to clays, and the water of crystallisation (water built into mineral crystals), and explain which of them determines $R_w$
express salinity in g/L, in ppm (parts per million) and as an equivalent NaCl concentration, and convert from one unit to another
estimate the resistivity of water from its salinity, and convert resistivity from one temperature to another
find the formation temperature from the geothermal gradient, and from several bottom-hole readings corrected for the effect of circulation
estimate how errors in temperature and salinity lead to errors in $R_w$ and in water saturation

Water in the rock as the interpreter sees it

Formation evaluation comes down to three quantities. Porosity tells how much fluid the rock holds. Saturation tells what share of the pores water, oil and gas occupy. Permeability tells how easily the fluids move to the well. Logging measures none of them directly. We calculate them from the physical properties of the rock, and for that we need to know three “external” quantities: what the mineral matrix (the solid skeleton of the rock) is made of, what water is in the pores, and at what temperature the water is.

In interpretation we describe the mineral matrix by its chemical composition, not by grain size, because the composition determines the tool readings. That is why quartz siltstone and chert behave on logs like sandstone, chalk behaves like limestone, and gypsum differs sharply from anhydrite.

Rock classes in log interpretation

Class: Sandstone
Main mineral: quartz SiO₂
What belongs to it: also quartz siltstone, chert, conglomerate

Class: Limestone
Main mineral: calcite CaCO₃
What belongs to it: also chalk

Class: Dolomite
Main mineral: dolomite CaMg(CO₃)₂
What belongs to it: the readings differ noticeably from limestone

Class: Salt
Main mineral: halite NaCl
What belongs to it: rock salt, salt-bearing sequences

Class: Anhydrite, gypsum
Main mineral: CaSO₄; CaSO₄·2H₂O
What belongs to it: the water in gypsum crystals changes its readings sharply

Class: Shale
Main mineral: clay minerals
What belongs to it: clays and claystones, without telling the minerals apart

Class	Main mineral	What belongs to it
Sandstone	quartz SiO₂	also quartz siltstone, chert, conglomerate
Limestone	calcite CaCO₃	also chalk
Dolomite	dolomite CaMg(CO₃)₂	the readings differ noticeably from limestone
Salt	halite NaCl	rock salt, salt-bearing sequences
Anhydrite, gypsum	CaSO₄; CaSO₄·2H₂O	the water in gypsum crystals changes its readings sharply
Shale	clay minerals	clays and claystones, without telling the minerals apart

This is a simplified division, used to choose the parameters of the mineral matrix. The geological name of the rock may differ.

We must also tell the kinds of water in the rock apart. For logging they are different “waters”.

Free formation water: It fills the pores, can move and conducts current. Its resistivity at formation temperature is $R_w$ .
Capillary-bound water: The same formation water, but held in small pores and in films on the grains. It is immobile, yet it conducts current in the same way and counts towards water saturation (Capillary pressure and the transition zone).
Clay-bound water: The surface of clay particles holds it, together with the exchangeable cations (positive ions that can be swapped). It adds a conductivity that Archie's equation for clean rocks does not include (Water saturation in shaly sands).
Water of crystallisation: It is part of the crystal lattice of minerals, such as gypsum. It does not conduct current, but it contains hydrogen, so the neutron log sees it.

Salinity of formation waters

Formation waters are solutions of salts. Most of the ions in them are usually sodium Na⁺ and chloride Cl⁻. There are also noticeable amounts of calcium Ca²⁺, magnesium Mg²⁺, potassium K⁺, sulfate SO₄²⁻ and bicarbonate HCO₃⁻. The total content of dissolved salts is called salinity. In Russian-language practice we express it in grams per litre of solution (g/L). In international practice it is often given in ppm (parts per million): milligrams of salt per kilogram of solution. A litre of brine weighs more than a kilogram, so the number in g/L is larger than the number in g/kg.

Converting salinity from g/L to ppm

C_\text{ppm} = \frac{1000\,C_{\text{g/L}}}{\rho}

where:

$C_\text{ppm}$ salinity in milligrams of salt per kilogram of solution, ppm = mg/kg
$C_{\text{g/L}}$ salinity in grams of salt per litre of solution, g/L
$\rho$ density of the solution; for NaCl solutions near 20 °C, approximately $\rho \approx 1 + 0.73\,w$ , where $w$ is the mass fraction of salt, g/cm³ (= kg/L)

Units: as indicated; $C_{\text{g/L}}/\rho$ is grams of salt per kilogram of solution

Holds when

The density and the salinity refer to the same temperature.
The density formula is an approximation for NaCl solutions at room temperature. For waters of complex composition we measure the density.

For fresh waters the two numbers almost coincide. For brines they differ: 150 g/L at a density of 1.10 g/cm³ is about 136,000 ppm, which is 9% less. Relationships fitted to measurements, and charts, for NaCl solutions are usually built in ppm.

Waters by salinity

Water: Fresh
Salinity: up to 1 g/L
Example: surface waters and shallow groundwater

Water: Brackish
Salinity: 1–10 g/L
Example: near the recharge areas of the formations

Water: Saline
Salinity: 10–50 g/L
Example: seawater — about 35 g/L

Water: Brine
Salinity: more than 50 g/L
Example: deep horizons, sections with salt-bearing sequences

Water	Salinity	Example
Fresh	up to 1 g/L	surface waters and shallow groundwater
Brackish	1–10 g/L	near the recharge areas of the formations
Saline	10–50 g/L	seawater — about 35 g/L
Brine	more than 50 g/L	deep horizons, sections with salt-bearing sequences

This is one of the common classifications. A saturated NaCl solution (one that cannot dissolve any more salt) at room temperature contains about 26% salt by mass (about 264,000 ppm, or 316 g/L). Solubility increases only slightly with heating, up to 28% at 100 °C.

Salinity usually increases with depth. Near the surface, rainwater (meteoric water) flushes the formations. In deep, stagnant horizons the water stays in contact with the rocks for a long time. The rule is not absolute. Near the outcrops of formations (where they reach the surface) and in recharge zones (where surface water enters the formation), the water is freshened to a great depth. Near salt-bearing sequences there are brines close to saturation. In some deep horizons salinity decreases again, for example because compacting shales release relatively fresh water. By V. A. Sulin's classification, the waters of oil and gas fields most often belong to the calcium chloride and sodium bicarbonate types.

Log charts and formulas are built for NaCl solutions. For the same mass, other ions conduct current differently: bicarbonate and sulfate are noticeably worse than chloride, and magnesium is better than sodium. So for a water of complex composition we convert the salinity into an equivalent NaCl concentration. This is the sum of the ion concentrations, each multiplied by a weighting coefficient that depends on the total salinity. A more reliable way is to measure the resistivity of the sample directly.

We get the resistivity of the formation water from analyses of samples (formation tests, production), from catalogues of the waters of the area, and from logs: from the SP (Spontaneous potential (SP) log) and from water-bearing zones. We compare the methods with one another (Determining formation water resistivity).

Formation temperature and the geothermal gradient

Heat from the Earth's interior constantly flows towards the surface, so temperature increases with depth. Daily and seasonal changes of air temperature die out within the first few metres. At a depth of 15–40 m there lies a neutral layer with a constant temperature, close to the mean annual air temperature. We take this temperature as the “surface” temperature $T_0$ in calculations. Offshore, we take $T_0$ as the temperature at the seabed. In permafrost regions (ground that stays frozen all year) the upper tens or hundreds of metres of the section have a negative temperature, so linear interpolation from the surface does not work there.

Temperature at depth for a constant gradient

T(z) = T_0 + G_T\,z,\qquad G_T = \frac{T_{TD} - T_0}{z_{TD}}

where:

$T(z)$ temperature of the rocks at depth $z$ , °C
$T_0$ mean annual temperature of the surface (neutral layer) or of the seabed, °C
$G_T$ geothermal gradient, °C/m (in practice °C/100 m)
$z$ vertical depth from the surface (for deviated wells, measured vertically, not along the hole), m
$T_{TD},\ z_{TD}$ formation temperature at total depth, and the total depth, °C, m

Units: °C and metres; $1\ ^\circ\text{C}/100\ \text{m} = 0.549\ ^\circ\text{F}/100\ \text{ft}$

Holds when

The gradient is constant over the section. This is a good approximation for a uniform sedimentary sequence without thick salt layers and without active groundwater movement.
$T_{TD}$ is the stabilised formation temperature, not the temperature measured immediately after circulation.

In Russian-language literature, the reciprocal $1/G_T$ is called the geothermal step. It is the number of metres we have to go down for the temperature to rise by 1 °C. At $G_T = 3\ ^\circ\text{C}/100\ \text{m}$ the step is about 33 m.

In sedimentary basins the gradient is usually 2–4 °C per 100 m. In ancient, “cold” platforms it is at the lower end of this range or below. In rifts (zones where the crust is stretching) and in young basins with a high heat flow it is 4–6 °C per 100 m or more. Two quantities determine it: the heat flow $q$ from the Earth's interior, and the thermal conductivity of the rocks, $\lambda$ (how well they conduct heat). By Fourier's law, $G_T = q/\lambda$ . With a typical heat flow of about 60 mW/m² and a thermal conductivity of shales of about 1.5–2 W/(m·K), the gradient comes out at 3–4 °C per 100 m.

Salt conducts heat 2–4 times better than shales. So in a thick salt sequence the gradient is small, and below it the rocks are colder than the average gradient would give. Above salt domes, on the contrary, it is warmer.
Groundwater movement evens out the temperature: descending flows cool the section, and flows rising along faults warm it.
Rapid sedimentation does not give recently buried sequences time to warm up, so the gradient in them is temporarily lower.

We need temperature for more than resistivities. The SP coefficient, the properties of oil, gas and water, the maturation of source rocks (Source rocks, maturation and migration), and the choice of mud, cement and equipment all depend on it. For more on reservoir conditions, see the topic Reservoir pressure and temperature.

Bottom-hole temperature and the correction for circulation

On every run of logging tools we record the bottom-hole temperature (BHT). In the past a maximum-reading thermometer attached to the tool did this; now the tool's own sensors do it. We write the BHT in the log heading together with the depth, the time circulation ended, and the time of the measurement.

But by the time of logging, the well has not yet come into thermal equilibrium with the rocks. While the hole is circulated, the mud brings the cold of the upper intervals down to the bottom and carries heat upwards. Near the bottom the rocks cool down, and higher up the hole they warm up. After circulation stops, the temperature near the bottom recovers slowly, over hours and days. For this reason the temperature measured on the first run is usually below the formation temperature, often by 10 °C or more. The shorter the time since circulation stopped, the larger the shortfall. If we build a gradient from this temperature, all the temperatures over the section come out too low.

If there have been several runs to the same depth, we can extrapolate the temperature from them to infinite time after circulation stops. Temperature recovery follows the same law as pressure recovery after a well is shut in. This is the Horner method (Pressure build-up analysis):

Recovery of the bottom-hole temperature after circulation stops

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

where:

$T_{\text{bh}}(\Delta t)$ temperature at the bottom of the hole a time $\Delta t$ after circulation stops, °C
$T_{\text{res}}$ formation (stabilised) temperature at this depth, °C
$m$ slope of the line: the drop in temperature per logarithmic cycle (one tenfold change of the ratio). It depends on the cooling power and on the thermal conductivity of the rocks, °C
$t_c$ duration of circulation at this depth before it stopped, h
$\Delta t$ time from the end of circulation to the measurement, h

Units: °C; $t_c$ and $\Delta t$ in the same units; $\lg$ is the decimal logarithm

Holds when

The cooling of the bottom of the hole is like the removal of heat by a line source of constant power (a thin line that takes away heat) during $t_c$ .
The readings are taken at one depth, and the time $\Delta t$ is not too short (usually more than a few hours).
We need at least three readings (two points cannot show whether the data lie on a line), recorded together with the time circulation stopped.

On a plot of “temperature against $\lg\,[(t_c + \Delta t)/\Delta t]$ ”, the points lie on a straight line. The value of the line at a ratio equal to one ( $\Delta t \to \infty$ ) is an estimate of the formation temperature. This is an approximation. When $t_c$ is not known accurately and the waiting times are short, the error can be several degrees. The most reliable temperature is one measured in a well that has stood for weeks without circulation, or during tests and in production.

Worked exampleFormation temperature from three bottom-hole readings, and the gradient

Given (assumed data)

Total depth (vertical): 3500 m
Duration of circulation before pulling out $t_c$: 4 h
Bottom-hole readings 6, 12 and 20 h after circulation stopped: 109.5, 113.5 and 115.0 °C
Mean annual surface temperature $T_0$: 10 °C
Depth of the formation of interest: 3080 m

Solution

Abscissae (the horizontal values of the plot): $\lg(10/6) = 0.2218$ , $\lg(16/12) = 0.1249$ , $\lg(24/20) = 0.0792$ .
Slope from the outer points: $m = \dfrac{115.0 - 109.5}{0.2218 - 0.0792} = 38.6$ °C per cycle. The middle reading deviates from this line by only 0.3 °C, so the points do lie on a straight line.
Extrapolation to $\Delta t \to \infty$ : $T_{\text{res}} = 115.0 + 38.6\cdot 0.0792 = 118.1$ °C. A least-squares line through all three points (the line that fits them best) gives 118.2 °C.
Gradient: $G_T = (118.1 - 10)/35 = 3.09$ °C per 100 m (a depth of 3500 m is 35 hundreds of metres; 1.69 °F per 100 ft). The geothermal step is 32 m.
Temperature of the formation at 3080 m: $T = 10 + 3.09\cdot 30.8 = 105$ °C. If we had built the gradient from the first reading of 109.5 °C, we would have got 97.6 °C, which is 7.5 °C lower.

Answer: The formation temperature at total depth (the bottom of the well) is about 118.1 °C, the gradient is 3.09 °C per 100 m, and the temperature of the formation of interest is about 105 °C.

What it means. The first reading put the formation temperature at total depth too low by 8.6 °C. This temperature goes into the conversion of the resistivities of water and filtrate (the example below), into the SP coefficient, and into calculations of fluid properties. The extrapolation is an estimate. If the circulation time is not known accurately, we check the result against measurements made during a formation test. The data in this example are assumed for illustration.

CalculatedThe bottom-hole temperature readings of the example in Horner coordinates. The horizontal axis is the ratio

(t_c + \Delta t)/\Delta t

on a logarithmic scale (time increases to the right). The line is drawn through the outer points and extended to a ratio equal to one.

Water resistivity: salinity and temperature

Ions create the electrical conductivity of water. The more salt is dissolved, the more charge carriers there are and the lower the resistivity, though not in proportion: in strong solutions the ions get in each other's way. When water is heated, it becomes less viscous, the ions move faster, and near 25 °C the conductivity rises by about 2% per degree. From 20 to 100 °C the resistivity of a solution decreases almost threefold.

Resistivity of an NaCl solution at 24 °C from its concentration

R_{24} \approx 0.0123 + \frac{3647.5}{C_\text{ppm}^{\,0.955}}

where:

$R_{24}$ resistivity of the solution at 24 °C (75 °F), Ω·m
$C_\text{ppm}$ concentration of NaCl (or equivalent NaCl concentration), ppm

Units: empirical: $R$ in Ω·m, $C$ in ppm

Holds when

An empirical approximation (a formula fitted to tables of measurements) for NaCl solutions, from hundreds of ppm to nearly saturated. The error is usually within a few per cent, and larger for nearly saturated brines.
For a water of another composition, $C$ is the equivalent NaCl concentration.

Seawater (about 35,000 ppm) has about 0.18 Ω·m at 24 °C, a 100,000 ppm brine about 0.07 Ω·m, and a saturated solution about 0.04 Ω·m. Resistivity falls more slowly than concentration rises: a tenfold increase in salinity reduces $R$ by about a factor of eight.

Converting the resistivity of a solution to another temperature (the Arps formula)

R_2 = R_1\,\frac{T_1 + 21.5}{T_2 + 21.5}

where:

$R_1,\ R_2$ resistivity of the solution at temperatures $T_1$ and $T_2$ , Ω·m
$T_1,\ T_2$ temperatures, °C

Units: °C; the resistivities in any identical units

Holds when

An empirical approximation for NaCl solutions in the usual range of formation temperatures.
In logging we apply it to all aqueous solutions: formation water, filtrate and drilling mud. For mudcake and for waters of complex composition the accuracy is lower.

Resistivity is inversely proportional to the sum $T + 21.5$ : as if conductivity grew linearly with temperature and vanished at −21.5 °C. The formula does not depend on salinity, so the same water, when heated from 20 to 100 °C, becomes $121.5/41.5 \approx 2.9$ times more conductive.

In other unit systems (2)

Temperature in °F

R_2 = R_1\,\frac{T_1 + 6.77}{T_2 + 6.77}

$T_1$ , $T_2$ in degrees Fahrenheit.

The sum $T + 6.77$ vanishes at −6.77 °F — the same point as −21.5 °C: $1.8\cdot(-21.5) + 32 \approx -6.7$ .

Form with a temperature coefficient

\rho_\text{w}(t) = \frac{\rho_\text{w}(18)}{1 + \alpha\,(t - 18)}

$t$ in °C, $\rho_\text{w}(18)$ is the resistivity of the water at 18 °C; this is how the correction is often written in Russian-language literature.

At $\alpha = 1/39.5 \approx 0.025\ ^\circ\text{C}^{-1}$ the formula coincides with the Arps formula: $18 + 21.5 = 39.5$ .

CalculatedResistivity of NaCl solutions of three concentrations against temperature, calculated with the two formulas above. The middle curve is the water of the example below; the points are its resistivity at 20 °C in the laboratory and at the formation temperature of 105 °C.

Worked exampleFormation water resistivity from an analysis of a sample

Given (assumed data)

Salinity of the sample (almost entirely NaCl): 62 g/L
Density of the sample at 20 °C: 1.043 g/cm³
Resistivity of the sample measured in the laboratory: 0.123 Ω·m at 20 °C
Formation temperature: 105 °C (from the previous example)

Solution

Concentration in ppm: $C = 1000\cdot 62/1.043 = 59{,}444$ ppm (59.4 g/kg). Density check: $1 + 0.73\cdot 0.0594 = 1.043$ g/cm³ — consistent with the measurement.
Resistivity at 24 °C from the concentration: $R_{24} = 0.0123 + 3647.5/{59{,}444}^{0.955} = 0.113$ Ω·m.
The laboratory measurement converted to 24 °C: $0.123\cdot 41.5/45.5 = 0.112$ Ω·m — a discrepancy of 0.7%: the sample really is close to an NaCl solution.
At formation temperature: $R_w = 0.123\cdot\dfrac{20 + 21.5}{105 + 21.5} = 0.040$ Ω·m; from the concentration it comes out as 0.041 Ω·m.
If we take the uncorrected temperature of 97.6 °C from the previous example, $R_w = 0.043$ Ω·m — 6% more.

Answer: The resistivity of the formation water at formation temperature is about 0.040 Ω·m, three times lower than the laboratory value.

What it means. By Archie's equation, for $n = 2$ , the water saturation is proportional to $\sqrt{R_w}$ (Archie's equation and saturation of clean reservoirs). If we put in the laboratory value of 0.123 Ω·m, $S_w$ comes out too high by a factor of 1.75. A formation with a true water saturation of 30% would then appear to be 52% water-saturated. If the temperature is too low by 7.5 °C and nobody notices, the cost is smaller: about 3% in $S_w$ . Even that should not be neglected. The data in this example are assumed for illustration.

Terms and symbols

Salinity: The total content of dissolved salts in water, in g/L or ppm.
ppm: Milligrams of salt per kilogram of solution. It equals $1000\,C_{\text{g/L}}/\rho$ .
Equivalent NaCl concentration: The concentration of an NaCl solution with the same electrical conductivity as a water of complex composition.
Formation water resistivity $R_w$: The resistivity of the water in the pores at formation temperature.
Geothermal gradient $G_T$: The increase of temperature per unit of depth, in °C/100 m. Its reciprocal is the geothermal step, in m/°C.
Neutral layer: A layer at a depth of 15–40 m with a constant temperature, close to the mean annual one. It gives the datum for the “surface” temperature.
Bottom-hole temperature (BHT): The temperature measured by a tool at the bottom of the hole. Because circulation cools the bottom, it is usually below the formation temperature.
Circulation time $t_c$ and time since it stopped $\Delta t$: The input data for correcting the bottom-hole temperature by the Horner method.

Common mistakes and misconceptions

Wrong

“The bottom-hole temperature from the first tool run is the formation temperature.”

In fact

Circulation cools the bottom of the hole, and the first readings are below the formation temperature, often by 10 °C or more. We get the formation temperature by extrapolating several readings, by regional corrections, or from measurements in wells that have stood idle.

Wrong

“The resistivity of a sample measured in the laboratory is $R_w$ .”

In fact

That is the resistivity at laboratory temperature. At a formation temperature of 100 °C it is almost three times lower. Without conversion, the water saturation is overestimated by a factor of about 1.7.

Wrong

“A gram per litre and ppm are practically the same thing.”

In fact

For fresh waters they are; for brines they are not. The number in ppm is smaller than the number in mg/L by the factor by which the density of the solution exceeds 1 g/cm³: for brines, by 3–17%.

Wrong

“The temperature gradient is the same throughout the section.”

In fact

It is inversely proportional to the thermal conductivity of the rocks: smaller in salt, larger in shales. Water movement and rapid sedimentation change it too. Linear interpolation is an approximation for uniform sequences.

Wrong

“The $R_w$ of any water can be found from its total salinity with the NaCl formulas.”

In fact

Only for waters close to an NaCl solution. For waters with a large share of calcium, magnesium, sulfates and bicarbonates, we need the equivalent NaCl concentration or a direct measurement of the resistivity of the sample.

In depth: advanced detail

Where the logarithm in the circulation correction comes from

Imagine that circulation removes heat from the rocks at a constant power $Q$ per metre of borehole for a time $t_c$ , and then stops. The conduction of heat towards the borehole follows the same diffusion equation as fluid flow towards a well. So the solution repeats the line-source solution (Line-source solution and the pressure function). After the stop, the temperature equals the formation temperature, minus the cooling from a “sink” that has worked since time 0, plus the heating from an equal “source” switched on at the moment of the stop. This is the principle of superposition (adding the effects of separate causes) (Superposition principle).

At long times the exponential integrals are replaced by logarithms, and the difference of the two solutions gives $T_{\text{res}} - T_{\text{bh}} = \dfrac{Q}{4\pi\lambda}\ln\dfrac{t_c + \Delta t}{\Delta t}$ , where $\lambda$ is the thermal conductivity of the rocks. If we change to the decimal logarithm, we get the slope $m = 2.303\,Q/(4\pi\lambda)$ . This is exactly the slope used in the example of the lesson. The approximation works when $\Delta t$ is much longer than the time $r_w^2/(4a)$ , where $a \approx 10^{-6}$ m²/s is the thermal diffusivity of the rocks (how fast a temperature change spreads through them). For a hole of radius 0.11 m this time is less than an hour.

Connections to other topics

Spontaneous potential (SP) log — the SP coefficient depends on temperature, and $R_w$ from the SP requires $R_{mf}$ at formation temperature.
Archie's equation and saturation of clean reservoirs — how an error in $R_w$ carries over into an error in water saturation.
Determining formation water resistivity — all the ways of finding $R_w$ and how they are cross-checked.
Borehole environment: invaded zone and mudcake — the resistivities of the mud, filtrate and mudcake in the log heading.
Resistivity logs: electrical and induction tools — why the resistivity of rocks depends on the water in the pores.
Reservoir pressure and temperature — reservoir pressure and temperature in reservoir engineering.
Pressure build-up analysis — the same Horner method for pressure build-up.
Source rocks, maturation and migration — how the temperature of the subsurface controls the maturation of organic matter.

Check yourself

1
The mean annual surface temperature is 12 °C, and the stabilised temperature at a depth of 2500 m is 87 °C. What is the geothermal gradient?
1. A.2.5 °C/100 m
2. B.3.0 °C/100 m
3. C.3.5 °C/100 m
4. D.4.0 °C/100 m
Show the answer
Correct answer: B. 3.0 °C/100 m
$G_T = (87 - 12)/2500 = 0.030$ °C/m, that is 3.0 °C per 100 m; the geothermal step is about 33 m per degree.
2
The resistivity of a formation water sample is 0.20 Ω·m at 25 °C. What will it be at a formation temperature of 90 °C?
Show the answer
By the Arps formula: $R = 0.20\cdot(25 + 21.5)/(90 + 21.5) = 0.20\cdot 46.5/111.5 \approx 0.083$ Ω·m — 2.4 times lower.
3
On three successive tool runs the bottom-hole temperature rose: 96, 101 and 103 °C. What does this mean?
1. A.The tools are faulty
2. B.The bottom of the hole is warming up after circulation stopped, and the formation temperature is above 103 °C
3. C.The formation is cooling over time
4. D.The formation temperature is the average of the three readings, 100 °C
Show the answer
Correct answer: B. The bottom of the hole is warming up after circulation stopped, and the formation temperature is above 103 °C
Circulation cooled the rocks near the bottom. After it stopped, the temperature has been recovering towards the formation temperature. The last reading has not yet reached equilibrium. We find the formation temperature by extrapolation in Horner coordinates.
4
Brine analysis: 150 g/L, density 1.10 g/cm³. How much is that in ppm, and why does the difference matter for charts?
Show the answer
$C = 1000\cdot 150/1.10 \approx 136{,}000$ ppm. Charts and empirical formulas for NaCl are built in ppm (mass fractions). If we put in 150,000, $R_w$ comes out too low, and for a brine an error of 10% in concentration is noticeable.
5
In which section, for the same heat flow, will the geothermal gradient be smallest?
1. A.In a thick shale sequence
2. B.In a thick sequence of rock salt
3. C.In water-bearing sandstones
4. D.The gradient does not depend on the rocks
Show the answer
Correct answer: B. In a thick sequence of rock salt
By Fourier's law $G_T = q/\lambda$ . The thermal conductivity of salt is 2–4 times higher than that of shales, so in a salt-bearing sequence temperature increases with depth more slowly.
6
Water saturation was calculated by Archie's equation with an $R_w$ measured at 20 °C, although the formation is at 110 °C. What are the sign and the approximate scale of the error?
Show the answer
The true $R_w$ is lower than the laboratory value by a factor of $(110 + 21.5)/(20 + 21.5) \approx 3.2$ . At $n = 2$ water saturation is proportional to $\sqrt{R_w}$ , so it is overestimated by a factor of about $\sqrt{3.2} \approx 1.8$ — a productive bed may appear to be water-bearing.

Updated: 26 October 2026