Oil viscosity and interfacial tension

Quick refresher

Dynamic viscosity $\mu$ is measured in mPa·s, and 1 mPa·s equals 1 centipoise (cP). Kinematic viscosity $\nu = \mu/\rho$ is measured in mm²/s, and 1 mm²/s equals 1 centistokes (cSt). Darcy's law uses the dynamic viscosity.
Oil viscosity falls very steeply when the oil is heated. It rises as the oil gets denser, falls as gas dissolves in it, and rises only weakly with pressure. That is why reservoir oil is least viscous at the bubble point.
Viscosity is calculated in three steps. First comes dead oil, which has lost all its dissolved gas: $\mu_{od}$ comes from the °API gravity (a density scale on which heavier oil has a lower number) and the temperature. Next comes oil at the bubble point: $\mu_{ob}$ comes from $\mu_{od}$ and the solution gas–oil ratio (the volume of gas dissolved in each unit volume of oil). Last comes undersaturated oil, which is above the bubble point: its viscosity comes from $\mu_{ob}$ and $p/p_b$ . The first step is the least reliable: at the same density, the viscosities of different oils can differ several-fold.
In the reservoir the gas–oil tension is a few mN/m. It rises as the pressure falls, and it disappears when the gas and oil mix completely (full miscibility). The oil–water tension is usually 10–35 mN/m. Surfactants (soap-like chemicals) lower it by orders of magnitude.
The parachor method: $\sigma^{1/4} = \sum_i \mathcal{P}_i\,(x_i\rho_L/M_L - y_i\rho_V/M_V)$ . The tension is proportional to the fourth power of the sum, so a 5% error in the parachors (constants of the pure substances) gives about a 20% error in the tension.
Capillary pressure is proportional to $\sigma\cos\theta$ , where $\theta$ is the contact angle. The capillary number $N_c = \mu v/\sigma$ compares the viscous forces with the capillary ones. Residual oil, the oil left in the rock after flooding, starts to decrease when $N_c$ rises to roughly $10^{-5}\text{–}10^{-4}$ .

Key relationships

\nu = \dfrac{\mu}{\rho},\qquad 1\ \text{mPa·s} = 1\ \text{cP},\quad 1\ \text{mm}^2/\text{s} = 1\ \text{cSt}

kinematic viscosity and units

\mu_{ob} = a\,\mu_{od}^{\,b}

correction for dissolved gas; $a$ and $b$ decrease with the solution gas–oil ratio (Beggs–Robinson correlation)

\mu_o = \mu_{ob}\left(\dfrac{p}{p_b}\right)^{m}

undersaturated oil (Vasquez–Beggs correlation)

\sigma^{1/4} = \sum_i \mathcal{P}_i\left(x_i\dfrac{\rho_L}{M_L} - y_i\dfrac{\rho_V}{M_V}\right)

parachor method: $\sigma$ in mN/m, $\rho$ in g/cm³, $M$ in g/mol

N_c = \dfrac{\mu\,v}{\sigma}

capillary number, SI

Common pitfalls

Putting the solution gas–oil ratio in m³/m³, the pressure in MPa and the temperature in °C into the correlations. They are built for scf/STB (cubic feet of gas per barrel of stock-tank oil), psia (pounds per square inch, absolute) and °F.
Taking the dead oil viscosity from a correlation when it could be measured. The error can be several-fold, and almost all of it carries over into the viscosity of the reservoir oil.
Confusing the kinematic viscosity of dead oil from analyses (mm²/s at 20 or 50 °C) with the dynamic viscosity of reservoir oil.
Treating the gas–oil tension as constant: as pressure falls from the bubble point to atmospheric, it rises more than sixfold for the lesson oil (the example oil used throughout this lesson).
Mixing parachors, densities and molar masses taken from different descriptions of the composition. The result depends on each of them to the fourth power.

After this lesson you can

distinguish dynamic from kinematic viscosity and convert their units
explain how oil viscosity depends on temperature, density, dissolved gas and pressure, and why it is lowest at the bubble point (the pressure at which gas starts to come out of the oil)
calculate oil viscosity with a chain of correlations (for dead, saturated and undersaturated oil) and estimate its error by comparing with laboratory data
calculate the gas–oil interfacial tension by the parachor method and know the typical values of the oil–water tension
convert capillary effects to a different interfacial tension and estimate the capillary number

Viscosity: definition, units and mechanism

Viscosity is the internal friction of a fluid: how strongly one layer resists the sliding of the next. In a Newtonian fluid (one whose viscosity does not depend on the shear rate) the shear stress is proportional to the shear rate. The constant of proportionality is the dynamic viscosity $\mu$ (shear stress and shear rate). This is the viscosity that appears in Darcy's law. For the same permeability and the same pressure difference, the flow rate is inversely proportional to viscosity (Darcy's law).

Dynamic and kinematic viscosity

\tau = \mu\,\frac{\dd v}{\dd y},\qquad \nu = \frac{\mu}{\rho}

where:

$\tau$ shear stress between the fluid layers, Pa
$\dd v/\dd y$ shear rate: the change of velocity across the flow, s⁻¹
$\mu$ dynamic viscosity, Pa·s
$\nu$ kinematic viscosity, m²/s
$\rho$ fluid density at the same conditions, kg/m³

Units: SI: Pa·s and m²/s

Holds when

The fluid is Newtonian: its viscosity does not depend on the shear rate. Oil above the wax appearance temperature (the temperature at which wax crystals start to form) usually is.
The flow is laminar (smooth, in layers, with no turbulence).

Dynamic viscosity sets how strongly a fluid resists flowing when a pressure difference pushes it. Kinematic viscosity sets how it flows under its own weight. Glass capillary viscometers, in which the oil runs out through a thin tube by itself, measure the kinematic viscosity. That is why analyses of dead oil (oil that has lost its dissolved gas) often report the kinematic viscosity at 20 and 50 °C, while inflow calculations (how much oil flows from the reservoir into the well) need the dynamic viscosity at reservoir conditions.

In other unit systems (1)

Practical units

1\ \text{mPa·s} = 1\ \text{cP} = 10^{-3}\ \text{Pa·s},\qquad 1\ \text{mm}^2/\text{s} = 1\ \text{cSt} = 10^{-6}\ \text{m}^2/\text{s}

cP is the centipoise, cSt the centistokes; in oilfield formulas viscosity is always in cP, that is, in mPa·s.

Converting kinematic viscosity to dynamic: $\mu\,[\text{mPa·s}] = \nu\,[\text{mm}^2/\text{s}]\cdot\rho\,[\text{g/cm}^3]$ .

In a liquid each molecule is tightly packed among its neighbours. For one layer to slide, its molecules must squeeze past the others and overcome the pull between them. Four main rules follow from this. Heating makes the molecules move faster, so viscosity falls quickly, almost exponentially. Large molecules (long-chain paraffins, which are wax-like, and resins and asphaltenes, which are heavy and tar-like) cling to one another, which is why heavy oils are viscous. Dissolved gas, made of small methane and ethane molecules, pushes the large molecules apart and thins the oil. Pressure pushes the molecules closer together, so viscosity rises slowly. A gas works differently, and at low pressure the effect of temperature is the opposite: its viscosity rises as it heats up (gas viscosity).

Worked exampleViscosity of the lesson oil in the reservoir and at the surface

Given (assumed data)

Dead oil at 85 °C and atmospheric pressure: $\mu_{od} = 5.53$ mPa·s (laboratory), density $850/1.05 = 809.5$ kg/m³
Reservoir oil at the bubble point pressure of 18 MPa: $\mu_{ob} = 1.00$ mPa·s, density 708 kg/m³ (Density and compressibility of reservoir oil)
Reservoir permeability: 50 mD

Solution

Kinematic viscosity of the dead oil at 85 °C: $\nu = 5.53/0.8095 = 6.83$ mm²/s; of the reservoir oil: $\nu = 1.00/0.708 = 1.41$ mm²/s.
Mobility $k/\mu$ (how easily the fluid moves through the rock): for the reservoir oil $50/1.00 = 50$ mD/(mPa·s), for the dead oil at the same temperature $50/5.53 = 9.0$ mD/(mPa·s).

Answer: Dissolved gas makes the lesson oil 5.5 times less viscous: 1.00 mPa·s in the reservoir against 5.53 mPa·s for the same oil without gas at the same temperature.

What it means. If the dead oil viscosity is used by mistake, the calculated well rate will be 5.5 times too low. The error is even larger if the kinematic viscosity from an analysis at 20 °C is used: that oil has no gas, and the temperature is 65 °C lower. The data in this example are assumed for illustration.

Dead oil: temperature and density

To find the viscosity of reservoir oil, we start with the dead oil viscosity $\mu_{od}$ : the viscosity of the oil without its dissolved gas, at reservoir temperature and atmospheric pressure. The laboratory measures it on the oil left at the end of a differential liberation test, which lowers the pressure step by step and removes the gas at each step (viscosity in PVT studies). It can also be measured on dead oil at several temperatures. For the lesson oil at 85 °C it is 5.53 mPa·s.

Temperature. The heavier the oil, the more steeply its viscosity falls when it is heated. Heating a light oil by 20 °C lowers its viscosity by a factor of about 1.5; for a heavy oil the fall is several-fold. To find the viscosity between the measured temperatures, engineers interpolate (fit a smooth curve) with empirical formulas. In one, the logarithm of viscosity is a straight line against the inverse of the absolute temperature (in kelvin), which is the Andrade equation. In the other, the logarithm of the logarithm of viscosity is a straight line against the logarithm of temperature.
Density. The heavier the oil (the lower its °API), the more viscous it is: from a few mPa·s for light oils at reservoir temperature to thousands for heavy ones.
Composition of the heavy part. Paraffins, naphthenes, aromatics, resins and asphaltenes (the families of hydrocarbons and tar-like compounds that make up the heavy part of oil) in different proportions give viscosities that differ several-fold at the same density. Density is only an indirect sign of composition.

Beggs–Robinson correlation for dead oil viscosity

\mu_{od} = 10^{x} - 1,\qquad x = 10^{\,3.0324 - 0.02023\,\mathrm{API}}\;T^{-1.163}

where:

$\mu_{od}$ dead oil viscosity at temperature $T$ and atmospheric pressure, cP (mPa·s)
$\mathrm{API}$ gravity of the dead oil in degrees API, °API
$T$ temperature, °F

Units: oilfield units: °F; the result is in cP, that is, mPa·s

Holds when

The data behind the correlation cover 16–58 °API and 70–295 °F (21–146 °C).
At low temperatures (near the lower limit of its data, and below) the correlation tends to overestimate the viscosity.
The oil is Newtonian, which holds above the wax appearance temperature.

The double exponential (a power of ten inside another power of ten) shows the main point: when the temperature or the °API rises, viscosity falls very quickly. But the correlation knows only the oil's density and temperature. Its error for a single oil can be several-fold, so engineers try to measure the dead oil viscosity. The formulas of other correlations are in the advanced section.

CalculatedDead oil viscosity against temperature, from the Beggs–Robinson correlation, for oils of 25, 34.2 and 45 °API (logarithmic scale). The dot is the measurement for the lesson oil (34.2 °API) at 85 °C: 5.53 mPa·s, against 2.20 from the correlation.

Worked exampleDead oil viscosity of the lesson oil from correlations and from measurement

Given (assumed data)

Lesson oil: 34.2 °API; reservoir temperature 85 °C = 185 °F
Laboratory: $\mu_{od} = 5.53$ mPa·s at 85 °C

Solution

Beggs–Robinson correlation: the exponent $3.0324 - 0.02023\cdot 34.2 = 2.3405$ ; $10^{2.3405} = 219.0$ ; $185^{-1.163} = 0.002308$ ; $x = 219.0\cdot 0.002308 = 0.5055$ .
$\mu_{od} = 10^{0.5055} - 1 = 2.20$ mPa·s, which is 60% below the measured value.
Three other correlations (formulas in the advanced section): Beal in Standing's form, 1.71; Glaso, 1.82; Egbogah–Ng, 2.42 mPa·s.

Answer: The correlations give 1.7–2.4 mPa·s against a measured 5.53: for the lesson oil they underestimate the dead oil viscosity by a factor of 2.3–3.2.

What it means. The correlations agree with one another, and all are wrong in the same direction: the lesson oil is much more viscous than an “average” oil of the same density. This happens with oils rich in resins and with high-wax oils. Over large data sets the average error of the correlations is tens of per cent, but for a single oil a several-fold difference is not rare. That is why the dead oil viscosity at reservoir temperature is the first thing to measure. Without a measurement, all the later steps inherit this error. The data in this example are assumed for illustration.

Dissolved gas: viscosity of saturated oil

Each cubic metre of gas dissolved in the oil lowers its viscosity, because the small gas molecules push the large oil molecules apart. So reservoir oil is least viscous at the bubble point, where the oil holds the most gas. Below the bubble point the gas leaves the oil, and the viscosity rises. Below the bubble point the viscosity curve is the mirror image of the curve of the solution gas–oil ratio $R_s(p)$ (solution gas–oil ratio): where $R_s$ falls quickly, viscosity rises quickly.

Beggs–Robinson correlation for the viscosity of oil with dissolved gas

\mu_{o} = a\,\mu_{od}^{\,b},\qquad a = 10.715\,(R_s + 100)^{-0.515},\qquad b = 5.44\,(R_s + 150)^{-0.338}

where:

$\mu_o$ viscosity of the gas-saturated oil at pressure $p \le p_b$ (at $p = p_b$ it is $\mu_{ob}$ ), cP (mPa·s)
$\mu_{od}$ viscosity of the same oil without gas at the same temperature, cP
$R_s$ solution gas–oil ratio of the oil at pressure $p$ , gas at 60 °F, scf/STB
$a,\ b$ multiplier and exponent; both are less than one and decrease with the solution gas–oil ratio, —

Units: oilfield units: $R_s$ in scf/STB (from m³/m³ with the gas at 20 °C, multiply by 5.53), viscosity in cP

Holds when

It is built on data for several hundred oils. At $R_s = 0$ it gives $a \approx b \approx 1$ , that is, $\mu_o \approx \mu_{od}$ .
$\mu_{od}$ is the viscosity of the same oil at the same temperature. A measured value is better than a calculated one.
Below the bubble point the formula takes the solution gas–oil ratio at the current pressure, that is, the gas that remains in the oil.

The gas correction makes the oil less viscous. At the same time it “squeezes” the error of the starting viscosity: because $b < 1$ , the relative error of $\mu_{od}$ passes into $\mu_{ob}$ in a weaker form, raised to the power $b$ . But the error does not disappear: a $\mu_{od}$ that is half the true value gives, at $b = 0.6$ , a $\mu_{ob}$ that is a third too low.

Worked exampleViscosity of the lesson oil at the bubble point and below

Given (assumed data)

Solution gas–oil ratio: 18 MPa: 110 m³/m³ (608 scf/STB); 12 MPa: 76.3 (422); 6 MPa: 41.0 m³/m³ (227 scf/STB)
Dead oil viscosity at 85 °C: measured 5.53 mPa·s; by the Beggs–Robinson correlation 2.20 mPa·s (example above)
Laboratory: 1.00 mPa·s at 18 MPa, 1.30 at 12 MPa, 1.96 at 6 MPa

Solution

At the bubble point: $a = 10.715\cdot 708^{-0.515} = 0.3649$ , $b = 5.44\cdot 758^{-0.338} = 0.5785$ ; $\mu_{ob} = 0.3649\cdot 5.53^{0.5785} = 0.981$ mPa·s (deviation from the laboratory value −1.9%).
At 12 MPa: $a = 0.4270$ , $b = 0.6362$ , $\mu_o = 1.268$ mPa·s (−2.5%); at 6 MPa: $a = 0.5433$ , $b = 0.7325$ , $\mu_o = 1.901$ mPa·s (−3.0%).
The same correction applied to the viscosity of 2.20 mPa·s from the correlation: $0.3649\cdot 2.20^{0.5785} = 0.576$ ; at 12 MPa, 0.705; at 6 MPa, 0.968 mPa·s; the deviations are −42, −46 and −51%.
The error passes on in a weaker form: $(2.20/5.53)^{0.5785} = 0.587$ . The dead oil viscosity is too low by 60%, and at the bubble point by 41%.

Answer: On the assumed data of the lesson, the Beggs–Robinson correction that starts from the measured dead oil viscosity is 2–3% away from the laboratory values. Starting from the viscosity given by the correlation, it is 42–51% too low.

What it means. The main point here is the first step. Without a measured dead oil viscosity the chain is wrong by almost a factor of two, and the gas correction only weakens that error. With a measured $\mu_{od}$ the result is close to the laboratory values, but this comes from the assumed data of the lesson. It does not show how accurate the correction is. For real oils, gas corrections are wrong by 10–30% on average, and by more for an individual oil. Another widely used correction, Chew–Connally, gives a viscosity one third higher at the bubble point for the same oil and the same $\mu_{od}$ (advanced section). Gas corrections, like bubble point correlations, should be checked against your own samples (error and tuning). The data in this example are assumed for illustration.

Above the bubble point

Above the bubble point the solution gas–oil ratio stays constant, so viscosity changes only because the oil is compressed. As the pressure rises, the molecules move closer together, and the viscosity rises almost in a straight line with pressure, usually by 10–20% for every 10 MPa. For the lesson oil it rises from 1.00 mPa·s at the bubble point to 1.20 mPa·s at the initial pressure of 30 MPa.

Vasquez–Beggs correlation for the viscosity of undersaturated oil

\mu_o = \mu_{ob}\left(\frac{p}{p_b}\right)^{m},\qquad m = 2.6\,p^{1.187}\exp\!\big(-11.513 - 8.98\cdot 10^{-5}\,p\big)

where:

$\mu_o,\ \mu_{ob}$ viscosity at pressure $p$ and at the bubble point, cP (mPa·s)
$p,\ p_b$ pressure and bubble point pressure, absolute, psia
$m$ exponent, increasing with pressure, —

Units: oilfield units: pressure in psia ( $p\,[\text{psia}] = p\,[\text{MPa}]/0.006895$ ), viscosity in cP

Holds when

The oil is undersaturated (above the bubble point), $p \ge p_b$ . $\mu_{ob}$ is the measured or calculated viscosity at the bubble point.
The database is the same as for the solution gas–oil ratio and compressibility correlations of the same authors (PVT correlations for oil and gas condensate, compressibility).

The ratio $\mu_o/\mu_{ob}$ depends only on the pressures: the correction is the same for all oils at the same $p$ and $p_b$ . This is a simplification. But above the bubble point viscosity changes little, and the error of the correction is usually smaller than the error of $\mu_{ob}$ itself.

Worked exampleViscosity of the lesson oil at 30 and 25 MPa

Given (assumed data)

Bubble point pressure: 18 MPa = 2610.7 psia; $\mu_{ob} = 1.00$ mPa·s (laboratory)
Laboratory: 1.20 mPa·s at 30 MPa, 1.12 mPa·s at 25 MPa

Solution

30 MPa = 4351.1 psia: $p^{1.187} = 20{,}846$ ; $\exp(-11.513 - 8.98\cdot 10^{-5}\cdot 4351.1) = 6.7651\cdot 10^{-6}$ ; $m = 2.6\cdot 20{,}846\cdot 6.7651\cdot 10^{-6} = 0.3667$ .
$\mu_o = 1.00\cdot(30/18)^{0.3667} = 1.206$ mPa·s, which is 0.5% above the laboratory value.
25 MPa = 3625.9 psia: $m = 0.3152$ , $\mu_o = 1.00\cdot(25/18)^{0.3152} = 1.109$ mPa·s, which is 1.0% below the laboratory value.
If, instead of the laboratory $\mu_{ob}$ , we take the value calculated from the measured dead oil viscosity, 0.981 mPa·s, then at 30 MPa we get $0.981\cdot 1.206 = 1.183$ mPa·s.

Answer: By the correlation the viscosity is 1.206 mPa·s at 30 MPa and 1.109 mPa·s at 25 MPa; the laboratory values are 1.20 and 1.12.

What it means. Above the bubble point, the Vasquez–Beggs correction is wrong by about one per cent on the assumed data of the lesson. For real oils the error of such corrections is usually a few per cent. The main uncertainty sits in $\mu_{ob}$ . So above the bubble point too, the quality of the result depends on the viscosity at the bubble point and, behind it, on the dead oil viscosity. The data in this example are assumed for illustration.

Laboratory data and correlations: what to use in the calculation

CalculatedViscosity of the lesson oil at 85 °C: laboratory points, and chains of the Beggs–Robinson correlation (below the bubble point) and the Vasquez–Beggs correlation (above it). One chain starts from the measured dead oil viscosity (solid line), the other from the viscosity calculated by the correlation (dashed line).

The data for the lesson oil are assumed, but the balance between the factors is typical. The dead oil viscosity sets the level of the viscosity curve, and its correlations are wrong by tens of per cent, and sometimes several-fold. The gas and pressure corrections capture the shape of the curve more reliably: a minimum at the bubble point, a slow rise above it and a fast rise below it. For real oils their error is usually 10–30% below the bubble point and a few per cent above it. So the working rule is: measure what sets the level, use the correlations for the shape, and check them against measurements on your own samples.

How to get the viscosity of reservoir oil for calculations

1
A full PVT study is available
Take the measured viscosities at reservoir temperature above and below the bubble point. Interpolate between the points, and above the last point continue with the pressure correction.
2
Only the dead oil has been measured
Use the measurements at two or three temperatures to find the dead oil viscosity at reservoir temperature, then apply the gas and pressure corrections. The error runs from a few per cent to tens of per cent, depending on the oil and the correction.
3
One reservoir oil measurement is available
Adjust the chain of correlations with one multiplier: the measured value at that point divided by the calculated one. Check that the curve stays smooth and has its minimum at the bubble point.
4
There are no measurements at all
Use correlations for all three steps. The result is only a rough estimate of the size of the number, and it can be wrong several-fold. Take a sample for a viscosity measurement at the first opportunity (Reservoir fluid sampling).

A reservoir simulation model (a computer model of how the fluids move in the reservoir) takes viscosity as a table against pressure, together with $R_s$ and $B_o$ (black oil model). Here $B_o$ is the oil formation volume factor: the volume the oil takes in the reservoir per unit volume at the surface. The table must be consistent. The viscosity minimum must sit at the bubble point, and below it the viscosity must rise faster where the solution gas–oil ratio falls faster.

Heavy and highly viscous oils

Oils of 10 to 20–22 °API (density roughly 920–1000 kg/m³) are usually called heavy. Extra-heavy oils are heavier than water (below 10 °API). Natural bitumens are oils with a reservoir viscosity above roughly 10,000 mPa·s. The boundaries differ between classifications (density classes). In the reservoir such oils have a viscosity from tens to tens of thousands of mPa·s. They hold little gas, so the reservoir viscosity is close to the dead oil viscosity at the same temperature.

Temperature decides everything. Heating by 100 °C lowers the viscosity of a heavy oil tens or hundreds of times. Thermal methods rely on this: steam injection and in-situ combustion (burning part of the oil underground) (thermal models).
Oil moves much less easily than water. The water–oil mobility ratio (how much more easily water moves than oil) is large, water breaks through to the wells early, and ordinary waterflooding is not very effective (effect of viscosity).
General correlations do not fit. Correlations built on average oils make their biggest errors for heavy oils. Separate formulas exist for heavy oils, but measuring at several temperatures is the most reliable way.
Non-Newtonian behaviour. Oils rich in asphaltenes, and waxy oils cooled below the wax appearance temperature, behave as non-Newtonian fluids. Their viscosity depends on the shear rate, and cooled waxy oils develop a yield stress, a minimum stress below which they do not flow at all (Retrograde condensation, wax, asphaltenes and high-pressure fluids).

In Kazakhstan, heavy, highly viscous oils are produced in the west of the country, on the Buzachi Peninsula, at the Karazhanbas and North Buzachi fields. Steam has been injected at Karazhanbas since the 1980s. The high-wax oils of Mangyshlak cause a different problem: their viscosity rises sharply as they cool in the well and in pipelines.

Gas–oil interfacial tension: the parachor method

The lesson on wettability explains what interfacial tension is, what units it is measured in and how it relates to wettability. Here we look at how to calculate it for reservoir oil and gas. For dead oil at room temperature, the tension at the boundary with air or gas is about 25–35 mN/m. In the reservoir, gas dissolves in the oil and the oil partly evaporates into the gas, so the two phases become more alike and the tension falls to a few mN/m. When the fluids mix completely (full miscibility), the boundary disappears and the tension is zero (miscible displacement).

Experiments show that, for a pure substance, the fourth root of the surface tension is proportional to the difference between the molar densities of the liquid and the vapour (the moles per unit volume), over a wide range of temperatures. The constant of proportionality is called the parachor. It is a constant of the substance that hardly changes with temperature, and it is found from measured tensions and densities. For mixtures, we add up the parachors of the components, weighted by their mole fractions (their shares of the moles) in each phase.

Parachor method: tension at the liquid–gas interface

\sigma^{1/4} = \sum_i \mathcal{P}_i\left(x_i\,\frac{\rho_L}{M_L} - y_i\,\frac{\rho_V}{M_V}\right)

where:

$\sigma$ interfacial (surface) tension, liquid–gas, mN/m
$\mathcal{P}_i$ parachor of component $i$ , (mN/m) $^{1/4}$ ·cm³/mol
$x_i,\ y_i$ mole fractions of the component in the liquid and in the gas, —
$\rho_L/M_L,\ \rho_V/M_V$ molar densities of the liquid and the gas: the density divided by the molar mass of the phase, mol/cm³

Units: the units must be exactly these: $\sigma$ in mN/m (dyn/cm), densities in g/cm³, molar masses in g/mol

Holds when

The phases are in equilibrium, and their compositions and densities are known, from an experiment or from an equation-of-state calculation (a thermodynamic model of the fluids) (Vapour–liquid equilibrium).
For a pure substance the sum reduces to a single term: $\sigma^{1/4} = \mathcal{P}(\rho_L - \rho_V)/M$ .
The parachors of heavy fractions have not been measured. They are estimated from the molar mass, for example $\mathcal{P} = -11.4 + 3.23M - 0.0022M^2$ ( $M$ in g/mol).
Near the critical point, and in systems close to it, the method is less accurate.

The contrast between the phases sets the tension: the more the liquid and the gas differ in molar density, the larger it is. The fourth power makes the method sensitive: a 5% error in the sum means about a 20% error in the tension.

Parachors of pure components

Component: Methane $\mathrm{C_1}$
Parachor: 77.0
Component: Isopentane
Parachor: 225.0

Component: Ethane $\mathrm{C_2}$
Parachor: 108.0
Component: Normal pentane
Parachor: 231.5

Component: Propane $\mathrm{C_3}$
Parachor: 150.3
Component: Normal hexane
Parachor: 271.0

Component: Isobutane
Parachor: 181.5
Component: Nitrogen $\mathrm{N_2}$
Parachor: 41.0

Component: Normal butane
Parachor: 189.9
Component: Carbon dioxide
Parachor: 78.0

Component	Parachor	Component	Parachor
Methane $\mathrm{C_1}$	77.0	Isopentane	225.0
Ethane $\mathrm{C_2}$	108.0	Normal pentane	231.5
Propane $\mathrm{C_3}$	150.3	Normal hexane	271.0
Isobutane	181.5	Nitrogen $\mathrm{N_2}$	41.0
Normal butane	189.9	Carbon dioxide	78.0

Parachors are in (mN/m) $^{1/4}$ ·cm³/mol; for heavy fractions, use the formula given in the assumptions above.

When the oil is described by the black oil model (a simple model that treats the fluid as stock-tank oil plus dissolved gas), the compositions of the phases are usually not known. The method is then applied to two pseudo-components, which stand in for the real mixture: the stock-tank oil and the separator gas. The liquid contains both. Per 1 m³ of stock-tank oil it holds $\rho_{o,\text{sc}}/M_o$ kmol of oil and $R_s/V_m$ kmol of gas, in a volume of $B_o$ . The gas phase is the gas alone (we neglect the evaporation of oil), and its molar density in the reservoir is $1/(V_m B_g)$ . Here $V_m$ is the molar volume of gas at standard conditions: 24.055 m³/kmol at 0.101325 MPa and 20 °C.

Worked exampleGas–oil tension of the lesson oil at 12 MPa

Given (assumed data)

Properties at 12 MPa: $R_s = 76.3$ m³/m³, $B_o = 1.258$ , $B_g = 0.00897$ m³/m³
Stock-tank oil: 850 kg/m³, molar mass 215 kg/kmol (assumed)
Separator gas (mole fractions, assumed): $\mathrm{N_2}$ 0.005; $\mathrm{CO_2}$ 0.010; $\mathrm{C_1}$ 0.740; $\mathrm{C_2}$ 0.090; $\mathrm{C_3}$ 0.080; isobutane 0.025; n-butane 0.040; n-pentane 0.010; molar mass 23.2 kg/kmol ( $\gamma_g = 0.80$ )

Solution

Parachors: for the gas, the sum of $y_i\mathcal{P}_i$ over its composition with the parachors from the table, $\mathcal{P}_g = 94.2$ (methane alone gives $0.740\cdot 77.0 = 57.0$ ); for the oil, $-11.4 + 3.23\cdot 215 - 0.0022\cdot 215^2 = 581.4$ .
Liquid per 1 m³ of stock-tank oil: oil $850/215 = 3.953$ kmol, gas $76.3/24.055 = 3.172$ kmol; the mole fraction of oil is $x_o = 3.953/7.125 = 0.5548$ and that of gas $0.4452$ ; the molar density is $7.125/1.258 = 5.664$ kmol/m³ = 0.005664 mol/cm³.
Gas phase: $1/(24.055\cdot 0.00897) = 4.634$ kmol/m³ = 0.004634 mol/cm³.
The sum: $581.4\cdot 0.5548\cdot 0.005664 + 94.2\cdot(0.4452\cdot 0.005664 - 0.004634) = 1.827 - 0.199 = 1.628$ .
$\sigma = 1.628^4 = 7.02$ mN/m. If the parachor of the oil is 5% higher or lower, the result is 8.7 or 5.6 mN/m.

Answer: The tension at the interface between the lesson oil and gas at 12 MPa is about 7 mN/m.

What it means. The gas term is negative: a unit volume of the gas phase holds more gas component than the same volume of liquid does, and this reduces the contrast between the phases. At the bubble point the same calculation gives 3.5 mN/m, at 6 MPa 13, and at atmospheric pressure about 23 mN/m (chart below). As the oil loses its gas, the phases become more different and the tension rises. A scatter of ±20%, from the uncertain parachor, is the usual accuracy of the method. For important calculations, measure the tension on a sample at reservoir conditions, for example from the shape of a hanging (pendant) drop, and adjust the parachors to match the measurement. The data in this example are assumed for illustration.

CalculatedGas–oil interfacial tension of the lesson oil below the bubble point, calculated by the parachor method for two pseudo-components. There is no free gas above the bubble point. An uncertainty of 5% in the parachor of the oil shifts the curve by about a fifth up or down.

Oil–water tension, capillary pressure and the capillary number

At the boundary between oil and water the tension is higher. Pure hydrocarbons against water give about 50 mN/m. Crude oils at atmospheric conditions usually give 20–40 mN/m, and in the reservoir 10–35 mN/m. Oil lowers the tension with its polar components (resins, asphaltenes and naphthenic acids, whose molecules have one end that attracts water), which collect at the boundary. So the tension depends on the composition of the oil and on the salinity (salt content) and acidity of the water. It falls as temperature rises and depends only weakly on pressure. Surfactants (surface-active agents, soap-like chemicals) lower it to hundredths and thousandths of a mN/m. Such ultra-low values are measured from the shape of a drop in a spinning tube.

The tension enters the capillary pressure as a multiplier, together with the contact angle (the angle at which the fluid boundary meets the rock surface): $p_c \propto \sigma\cos\theta$ for the same pore geometry (converting a laboratory curve). That is why the same rock gives different transition zones for different pairs of fluids. By the Leverett J-function, the height above the level of zero capillary pressure, at the same saturation (the same share of the pore space filled with the fluid), is proportional to $\sigma\cos\theta/\Delta\rho$ (Leverett J-function).

Worked exampleTransition zones at the gas–oil and oil–water contacts

Given (assumed data)

Oil–water: $\sigma = 25$ mN/m, $\theta = 30^\circ$ measured through the water; water 1040 kg/m³, oil 708 kg/m³
Gas–oil at the gas–oil contact (GOC), lesson oil at the bubble point: $\sigma = 3.5$ mN/m (parachor method, chart above), the oil wets the rock better than the gas does: $\cos\theta = 1$ ; gas $0.963/0.00598 = 161$ kg/m³
Rock: the same in both cases; the oil–water transition zone from the capillary curve is 15 m (assumed)

Solution

Wetting tensions ( $\sigma\cos\theta$ ): oil–water $25\cos 30^\circ = 21.7$ mN/m, gas–oil $3.5$ mN/m.
Density differences: water–oil $1040 - 708 = 332$ kg/m³, oil–gas $708 - 161 = 547$ kg/m³.
Ratio of the heights at the same saturation: $\dfrac{3.5/547}{21.7/332} = 0.098$ .
Transition zone at the GOC: $15\cdot 0.098 = 1.5$ m.

Answer: The transition zone at the gas–oil contact is about 10 times thinner than at the oil–water contact: about 1.5 m against 15 m.

What it means. The gas–oil tension in the reservoir is six times smaller than the oil–water wetting tension, and the oil–gas density difference is 1.6 times larger than the water–oil one. Both factors make the transition zone at the GOC thin. During depletion (production) the pressure falls and the gas–oil tension rises, so the capillary forces that hold the oil back when gas tries to displace it become stronger. The data in this example are assumed for illustration.

Tension also decides what happens to the residual oil. After waterflooding, part of the oil stays behind as droplets trapped in the pores. To push a droplet through a pore throat (a narrow gap between grains), the pressure difference across it must exceed the capillary pressure of its menisci, the curved ends of the droplet (how wettability determines residual oil). The capillary number describes the balance between the viscous forces that push the droplet and the capillary forces that hold it.

Capillary number

N_c = \frac{\mu\,v}{\sigma}

where:

$N_c$ capillary number, —
$\mu$ viscosity of the displacing phase (water), Pa·s
$v$ Darcy velocity (flow rate per unit area), m/s
$\sigma$ interfacial tension between the displacing and the displaced phase, N/m

Units: SI, dimensionless

Holds when

This is the simplest form. Definitions with $\cos\theta$ in the denominator, or with the pore velocity instead of the Darcy velocity, are also found. Only numbers from the same definition can be compared.
The thresholds that follow are rough guides. Residual oil starts to decrease from about $N_c \sim 10^{-5}\text{–}10^{-4}$ (higher in oil-wet rocks, where oil clings to the rock surface), and almost all of it is displaced at about $N_c \sim 10^{-2}$ .

To move trapped oil, the capillary number has to be raised by several orders of magnitude. The velocity and the viscosity of the water cannot be changed that much. What remains is to lower the tension, and surfactants do exactly that.

Worked exampleCapillary number in waterflooding and with a surfactant

Given (assumed data)

Waterflooding: Darcy velocity 0.3 m/d, water viscosity at reservoir temperature 0.4 mPa·s (assumed)
Oil–water tension: 25 mN/m

Solution

Velocity: $v = 0.3/86{,}400 = 3.47\cdot 10^{-6}$ m/s.
$N_c = \dfrac{0.4\cdot 10^{-3}\cdot 3.47\cdot 10^{-6}}{0.025} = 5.6\cdot 10^{-8}$ .
To get $N_c = 10^{-5}$ at the same velocity, the tension must fall to $0.4\cdot 10^{-3}\cdot 3.47\cdot 10^{-6}/10^{-5} = 1.39\cdot 10^{-4}$ N/m = 0.14 mN/m, which is 180 times lower; for $N_c = 10^{-2}$ , it must fall to1.4 × 10⁻⁴ mN/m.

Answer: In waterflooding

N_c \approx 6\cdot 10^{-8}

. To start displacing the residual oil, the tension must be lowered by a factor of about 180. For almost complete displacement it must fall to ten-thousandths of a mN/m.

What it means. Ordinary waterflooding works at capillary numbers two to three orders of magnitude below the threshold. That is why the residual oil hardly depends on how much water is pumped through. Surfactants that lower the tension to thousandths of a mN/m raise the capillary number to $10^{-3}$ and above, into the range where residual oil decreases noticeably. The data in this example are assumed for illustration.

Terms and symbols

Dynamic viscosity $\mu$: The ratio of shear stress to shear rate, measured in mPa·s (the same number as cP). It is the viscosity that enters Darcy's law.
Kinematic viscosity $\nu$: The dynamic viscosity divided by the density, measured in mm²/s (the same number as cSt). Glass capillary viscometers measure it.
Dead oil viscosity $\mu_{od}$: The viscosity of oil without dissolved gas, at reservoir temperature and atmospheric pressure. It is the starting point for calculating the viscosity of reservoir oil.
Oil viscosity at the bubble point $\mu_{ob}$: The lowest viscosity of reservoir oil. All the gas is dissolved, and compression has not yet begun.
Parachor $\mathcal{P}$: A constant of the substance in the method for calculating tension: $\sigma^{1/4} = \mathcal{P}(\rho_L - \rho_V)/M$ . For mixtures the parachors are added up, weighted by the mole fractions.
Capillary number $N_c$: The ratio of the viscous forces to the capillary forces, $\mu v/\sigma$ . In waterflooding it is of the order of $10^{-7}$ , and residual oil starts to decrease at about $10^{-5}\text{–}10^{-4}$ .

Common mistakes and misconceptions

Wrong

“The oil viscosity is 35 mm²/s: that is what the dead oil analysis at 20 °C says.”

In fact

That is the kinematic viscosity of oil without gas at surface temperature. The viscosity of reservoir oil is the dynamic one, at reservoir temperature and with dissolved gas, and it can be tens of times lower. For the lesson oil it is 5.53 mPa·s without gas at 85 °C and 1.00 mPa·s with gas at the bubble point.

Wrong

“The correlation matched the laboratory value at the bubble point, so it is right for this oil.”

In fact

A match at one point may be a coincidence. For the lesson oil, the Beggs–Robinson and Chew–Connally corrections applied to the same dead oil viscosity differ by a third. Check a correlation at several points and on several samples.

Wrong

“The higher the pressure, the lower the viscosity: more gas is dissolved in the oil.”

In fact

That holds only below the bubble point. Above it the solution gas–oil ratio is constant, and viscosity rises as pressure increases. The minimum is at the bubble point.

Wrong

“We will take the gas–oil tension from a reference table, as for the water–air pair.”

In fact

It depends strongly on pressure and composition. For the lesson oil it runs from 23 mN/m at atmospheric pressure to 3.5 mN/m at the bubble point. Calculate it for your own fluids, or measure it at reservoir conditions.

Wrong

“To recover residual oil, it is enough to inject water faster.”

In fact

The capillary number has to be raised by several orders of magnitude, and the injection rate cannot be increased that much. Trapped oil is set moving by lowering the interfacial tension or by changing the wettability.

In depth: advanced detail

Other viscosity correlations

The correlation equations used in the examples of the lesson are in oilfield units: temperature $T$ in °F, pressure in psia, solution gas–oil ratio in scf/STB at 60 °F and viscosity in cP. $\lg$ is the common logarithm (base 10). Do not put metric quantities into these equations.

Dead oil viscosity by Beal in Standing's form

\mu_{od} = \left(0.32 + \frac{1.8\cdot 10^{7}}{\mathrm{API}^{4.53}}\right)\left(\frac{360}{T + 200}\right)^{a},\qquad a = 10^{\,0.43 + 8.33/\mathrm{API}}

where:

$\mu_{od},\ \mathrm{API},\ T$ dead oil viscosity, gravity in degrees API and temperature, cP, °API, °F

Units: oilfield units

Holds when

It approximates a chart built for oils at 100 and 200 °F. At high temperatures it tends to underestimate the viscosity.

The first factor is the viscosity at 160 °F (71 °C), where the second factor equals one. The second factor describes how viscosity changes with temperature. The change is steeper the heavier the oil: the exponent $a$ grows as °API decreases.

Glaso correlation for dead oil viscosity

\mu_{od} = 3.141\cdot 10^{10}\,T^{-3.444}\,(\lg\mathrm{API})^{\,10.313\lg T - 36.447}

where:

$\mu_{od},\ T,\ \mathrm{API}$ as above, cP, °F, °API

Units: oilfield units

Holds when

It was built on North Sea oils, like the same author's bubble point correlation (correlation equations).

The effects of density and temperature on viscosity are tied together here: the exponent of $\lg\mathrm{API}$ depends on temperature.

Egbogah–Ng correlation for dead oil viscosity

\lg\lg(\mu_{od} + 1) = 1.8653 - 0.025086\,\mathrm{API} - 0.5644\,\lg T

where:

$\mu_{od},\ T,\ \mathrm{API}$ as above, cP, °F, °API

Units: oilfield units

Holds when

It has the same form as the Beggs–Robinson correlation, with coefficients fitted to a different database.

The double logarithm of viscosity is a straight line against the logarithm of temperature. This is the usual way to make the viscosity curves of crude oils and petroleum products plot as straight lines.

Chew–Connally correction for dissolved gas (Standing's form)

\mu_{o} = 10^{a}\,\mu_{od}^{\,b},\qquad a = R_s\,(2.2\cdot 10^{-7}R_s - 7.4\cdot 10^{-4}),\qquad b = \frac{0.68}{10^{\,8.62\cdot 10^{-5}R_s}} + \frac{0.25}{10^{\,1.1\cdot 10^{-3}R_s}} + \frac{0.062}{10^{\,3.74\cdot 10^{-3}R_s}}

where:

$\mu_o,\ \mu_{od}$ viscosity of the oil with and without gas at the same temperature, cP
$R_s$ solution gas–oil ratio, scf/STB

Units: oilfield units

Holds when

It started as a chart, and the formulas approximate that chart. At $R_s = 0$ it gives $\mu_o \approx \mu_{od}$ ( $a = 0$ , $b = 0.992$ ).

It has the same form $a\,\mu_{od}^{\,b}$ as the Beggs–Robinson correction, but $a$ and $b$ depend on the solution gas–oil ratio in a different way.

Undersaturated oil viscosity by Beal in Standing's form

\mu_o = \mu_{ob} + 0.001\,(p - p_b)\,\big(0.024\,\mu_{ob}^{1.6} + 0.038\,\mu_{ob}^{0.56}\big)

where:

$\mu_o,\ \mu_{ob}$ viscosity at pressure $p$ and at the bubble point, cP
$p,\ p_b$ pressure and bubble point pressure, psia

Units: oilfield units

Holds when

$p \ge p_b$ . Viscosity grows linearly with pressure.

The rise of viscosity per unit of pressure is larger for viscous oils, because of the $\mu_{ob}$ terms in the brackets.

Worked exampleOther correlations for the lesson oil

Given (assumed data)

Lesson oil: 34.2 °API, 185 °F, $R_{sb} = 608$ scf/STB, $\mu_{od} = 5.53$ cP (measured), $p_b = 2610.7$ psia, $\mu_{ob} = 1.00$ cP (measured)

Solution

Beal in Standing's form: $a = 10^{\,0.43 + 8.33/34.2} = 4.716$ ; $\mu_{od} = (0.32 + 1.8\cdot 10^7/34.2^{4.53})\,(360/385)^{4.716} = 1.71$ cP.
Glaso: $\mu_{od} = 1.82$ cP; Egbogah–Ng: $\mu_{od} = 2.42$ cP.
Chew–Connally at 608 scf/STB: $10^{a} = 0.4280$ , $b = 0.6566$ ; $\mu_{ob} = 0.4280\cdot 5.53^{0.6566} = 1.316$ cP, which is 32% above the laboratory value (Beggs–Robinson gave 0.981).
Beal for undersaturated oil at 30 MPa (4351.1 psia): $\mu_o = 1.00 + 0.001\cdot 1740.4\cdot(0.024 + 0.038) = 1.108$ cP, which is 7.7% below the laboratory value (Vasquez–Beggs gave 1.206).

Answer: For the dead oil of the lesson the correlations give 1.71–2.42 cP against a measured 5.53. The gas corrections that start from the same

\mu_{od}

differ by a third. Above the bubble point the differences are smaller, up to 8%.

What it means. The dead oil viscosity correlations agree with one another better than with the laboratory value. They share a common weakness: all they know about the oil is its density and temperature. The errors at the different steps are of different sizes: tens of per cent, and several-fold, for dead oil; tens of per cent for the gas correction; a few per cent for the pressure correction. The data in this example are assumed for illustration.

Why tension grows as the fourth power of the phase contrast

Tension is the excess energy of the molecules at the interface. A molecule at the surface of a liquid has fewer neighbours than one inside, so it receives less attraction than it would deeper in the liquid. The denser the liquid compared with the vapour, the bigger this shortfall and the higher the tension. At the critical point the liquid and the vapour become identical, and the tension vanishes. The empirical parachor rule links these quantities very steeply: tension grows roughly as the fourth power of the difference between the molar densities of the phases.

Two practical consequences follow. The first is high sensitivity. Uncertainties in the phase densities, in their compositions and in the parachors of the heavy fractions are amplified fourfold. So the method needs consistent data: compositions and densities from one equation-of-state calculation or from one experiment. The second is a rapid fall of tension as the phases come closer together. In gas condensate systems (gas that drops liquid out when the pressure falls) and in enriched gas injection (injecting gas enriched with heavier components), the tension falls to fractions of a mN/m. The capillary forces that hold the liquid in place weaken. The condensate that has dropped out in the reservoir, or the residual oil, then moves more easily than the curves measured at ordinary tensions predict. Near the critical point the exponent is somewhat smaller than four, and simple parachors are in greater error.

Connections to other topics

Laboratory PVT tests — how the viscosity of reservoir oil is measured in the laboratory.
PVT correlations for oil and gas condensate — PVT property correlations: error, selection and tuning against samples.
Density and compressibility of reservoir oil — density and compressibility of reservoir oil, two more of its properties in the black oil model.
Wettability and interfacial phenomena — interfacial tension, contact angle and Young's equation.
Capillary pressure and the transition zone — how tension enters the capillary pressure and the Leverett J-function.
Fractional flow function and mobility ratio — how the ratio of oil and water viscosities governs displacement.
Types of reservoir simulation model — thermal models for highly viscous oils.

Check yourself

1
At what pressure is the viscosity of reservoir oil lowest?
1. A.At the bubble point pressure
2. B.At the initial reservoir pressure
3. C.Oil viscosity does not depend on pressure
4. D.At atmospheric pressure
Show the answer
Correct answer: A. At the bubble point pressure
Above the bubble point the solution gas–oil ratio is constant, and viscosity rises with pressure because the oil is compressed. Below the bubble point the oil loses the gas that thinned it, and viscosity rises as the pressure falls. The minimum is at the bubble point, where the most gas is dissolved and the compression is least.
2
The dynamic viscosity of an oil is 3.2 mPa·s and its density is 760 kg/m³. Find the kinematic viscosity and express the dynamic viscosity in cP and Pa·s.
Show the answer
$\nu = 3.2/0.760 = 4.21$ mm²/s (cSt). $\mu = 3.2$ mPa·s $= 3.2$ cP $= 0.0032$ Pa·s.
3
The dead oil viscosity at reservoir temperature is 4.0 cP and the solution gas–oil ratio is 400 scf/STB. Estimate the viscosity of the saturated oil by the Beggs–Robinson correlation.
Show the answer
$a = 10.715\cdot 500^{-0.515} = 0.4365$ ; $b = 5.44\cdot 550^{-0.338} = 0.6447$ ; $\mu_{ob} = 0.4365\cdot 4.0^{0.6447} = 0.4365\cdot 2.444 = 1.07$ cP. The dissolved gas has lowered the viscosity almost fourfold.
4
The dead oil viscosity from a correlation turned out to be half the true value. By how much will the viscosity of the saturated oil be wrong if the exponent in the gas correction is $b = 0.6$ ?
1. A.It will not be wrong: the gas correction will remove the error
2. B.Also by a factor of two
3. C.By a factor of four
4. D.By about a third
Show the answer
Correct answer: D. By about a third
$\mu_{ob} = a\,\mu_{od}^{\,b}$ , so the ratio of the wrong value to the true value is $0.5^{0.6} = 0.66$ . The viscosity of the saturated oil is too low by about a third. The correction weakens the error but does not remove it.
5
How does the gas–oil interfacial tension change as the reservoir pressure falls below the bubble point?
1. A.It rises
2. B.It does not change
3. C.It falls
4. D.It becomes zero
Show the answer
Correct answer: A. It rises
Less gas stays dissolved in the oil, and the gas itself is less dense at lower pressure. The phases differ more and more, so the tension rises: for the lesson oil, from 3.5 mN/m at the bubble point to 7 mN/m at 12 MPa and 13 mN/m at 6 MPa.
6
Water displaces oil at a Darcy velocity of 0.5 m/d; the water viscosity is 0.6 mPa·s and the oil–water tension is 30 mN/m. Find the capillary number. What will it become if a surfactant lowers the tension to 0.01 mN/m?
Show the answer
$v = 0.5/86{,}400 = 5.79\cdot 10^{-6}$ m/s; $N_c = 0.6\cdot 10^{-3}\cdot 5.79\cdot 10^{-6}/0.030 = 1.16\cdot 10^{-7}$ . With the surfactant the tension is 3000 times lower: $N_c = 3.5\cdot 10^{-4}$ . This is above the threshold of $10^{-5}\text{–}10^{-4}$ , so residual oil starts to be displaced.

Updated: 16 October 2026