Gas formation volume factor, compressibility and viscosity

Quick refresher

$B_g$ is the volume that the gas takes up in the reservoir for each standard cubic metre. In the reservoir the gas is squeezed tens to hundreds of times, so $B_g \ll 1$ (much less than 1). The inverse, $E = 1/B_g$ , is tens to hundreds.
$B_g$ is proportional to $Z/p$ . When the pressure is halved, the gas in the reservoir takes up about twice the volume, with a correction for the change in $Z$ .
The compressibility of an ideal gas is $1/p$ . For a real gas a correction is added that depends on the slope of $Z(p)$ . At reservoir pressures a gas is tens to hundreds of times more compressible than oil and water.
$Z$ and $c_g$ are different quantities. The first has no units and describes volume. The second is measured in MPa⁻¹ and describes how the volume changes.
Gas viscosity (resistance to flow) is 0.01–0.03 mPa·s, tens of times lower than that of water and oil. It rises with pressure. With temperature it rises at low pressure and falls at high pressure.
Always state the standard conditions for a gas. Going from 20 °C to 15 °C changes the standard volume by 1.7%.

Key relationships

B_g = \dfrac{p_{\text{sc}}}{p}\cdot\dfrac{Z\,T}{T_{\text{sc}}},\qquad E = \dfrac{1}{B_g}

gas formation volume factor and gas expansion factor ( $Z_{\text{sc}} = 1$ , pressures and temperatures absolute)

\rho_g = \dfrac{\rho_{g,\text{sc}}}{B_g}

gas density in the reservoir: mass is conserved

c_g = \dfrac{1}{p} - \dfrac{1}{Z}\left(\dfrac{\partial Z}{\partial p}\right)_T

isothermal compressibility of a real gas

Common pitfalls

Confusing $B_g$ in m³/m³ (or ft³/scf) with $B_g$ in rb/scf. The numbers for the same gas differ by a factor of 5.6.
Converting gas in place to 20 °C but production to 15 °C or 60 °F. For gas the discrepancy is 1.5–1.7%.
Confusing the compressibility $c_g$ with the gas deviation factor $Z$ .
Using the gas viscosity at atmospheric pressure to calculate inflow (the flow of gas into the well). At reservoir pressures of 20–30 MPa the viscosity is 1.5–2 times higher.

After this lesson you can

calculate the gas formation volume factor $B_g$ and its inverse $E = 1/B_g$ in metric and oilfield units, and use them to convert gas in place and gas volumes
explain how $B_g$ changes with pressure and why the standard conditions must always be stated for a gas
calculate the isothermal compressibility of a gas from the slope of $Z$ against pressure and compare it with the compressibility of oil, water and rock
explain why the viscosity of a gas rises with temperature at low pressure and falls at high pressure, and estimate it with a correlation
understand where these properties are used: estimating gas in place, material balance, inflow to a gas well, well testing

Gas formation volume factor

When gas rises from the reservoir to the surface, it expands. The pressure falls hundreds of times, and the temperature falls by tens of degrees. So the same gas takes up very different volumes in the reservoir and at the surface. To go from one to the other we need a multiplier: the gas formation volume factor. It is the volume of gas at reservoir conditions divided by the volume of the same amount of gas at standard conditions (formation volume factor).

We get it from the equation of state of a real gas (Ideal and real gas: the gas deviation factor (Z)). The same amount of gas $n$ takes up $V_{\text{res}} = ZnRT/p$ in the reservoir, and $V_{\text{sc}} = nRT_{\text{sc}}/p_{\text{sc}}$ at standard conditions. There the gas is practically ideal, so $Z_{\text{sc}} = 1$ . The ratio of these volumes does not depend on $n$ or $R$ :

Gas formation volume factor

B_g = \frac{V_{\text{res}}}{V_{\text{sc}}} = \frac{p_{\text{sc}}}{p}\cdot\frac{Z\,T}{T_{\text{sc}}}

where:

$B_g$ volume of gas in the reservoir per unit of its volume at standard conditions, m³/m³
$p,\ T$ reservoir pressure and temperature — absolute, MPa, K
$Z$ gas deviation factor at $p$ and $T$ , —
$p_{\text{sc}},\ T_{\text{sc}}$ standard conditions: 0.101325 MPa and 293.15 K (20 °C), MPa, K

Units: absolute pressures and temperatures in consistent units

Holds when

The gas is single-phase in the reservoir and remains a gas all the way to standard conditions. For wet gas (gas rich in heavier hydrocarbons) and gas condensate (gas that drops a light liquid), part of the gas turns into liquid at the surface, and the conversion is made from the composition (Types of reservoir fluids).
$Z_{\text{sc}} = 1$ . For natural gas at standard conditions, $Z$ differs from 1 by only fractions of a per cent.

The inverse, $E = 1/B_g$ , is the gas expansion factor. It shows how many standard cubic metres one cubic metre of gas in the reservoir contains. This number is easier to remember: for the lesson gas at 20 MPa and 80 °C it is about 190. In Russian-language practice of volumetric estimation (calculating gas in place from the size of the accumulation), the same factor is written as a product of three terms. These are the pressure ratio, a correction $\alpha = 1/Z$ for the departure from the Boyle–Mariotte law, and a temperature correction $f = T_{\text{sc}}/T$ .

In other unit systems (2)

Metric units, standard conditions of 20 °C

B_g = 3.456\cdot 10^{-4}\,\frac{Z\,T}{p}

$T$ — K, $p$ — MPa (absolute), $B_g$ — m³/m³.

$3.456\cdot 10^{-4} = 0.101325/293.15$ MPa/K. For standard conditions of 15 °C it is $3.516\cdot 10^{-4}$ ( $0.101325/288.15$ ).

Oilfield units

B_g\,[\text{ft}^3/\text{scf}] = 0.02828\,\frac{Z\,T}{p},\qquad B_g\,[\text{rb/scf}] = 0.00504\,\frac{Z\,T}{p}

$T$ — degrees Rankine, $p$ — psia. rb is a reservoir barrel (a barrel of gas at reservoir conditions), and scf is a standard cubic foot (a cubic foot at 14.696 psia and 60 °F).

$0.02828 = 14.696/519.67$ ; dividing by 5.615 ft³/bbl gives $0.00504$ .

CalculatedGas formation volume factor (top) and gas expansion factor (bottom) of the lesson gas at 80 °C, against reservoir pressure. The dashed line is the calculation for an ideal gas. The initial pressure of the example, 20 MPa, is marked.

Because $B_g$ is proportional to $Z/p$ , when the pressure is halved the gas in the reservoir takes up almost twice the volume. The hotter the reservoir, the larger $B_g$ is at the same pressure, because a hot gas is less dense. The departure from an ideal gas shows in the expansion factor. As long as $Z < 1$ , more gas fits into the reservoir than the ideal gas model predicts.

The mass of the gas does not change as it rises to the surface. So the same factor links the density of the gas in the reservoir to its density at standard conditions. We use the density in the reservoir to find the pressure gradient in the gas zone, which is the rise of pressure per metre of depth (Reservoir pressure and temperature).

Gas density in the reservoir

\rho_g = \frac{\rho_{g,\text{sc}}}{B_g},\qquad \rho_{g,\text{sc}} = 1.204\,\gamma_g

where:

$\rho_g$ gas density at reservoir conditions, kg/m³
$\rho_{g,\text{sc}}$ gas density at standard conditions (0.101325 MPa, 20 °C), kg/m³
$\gamma_g$ gas specific gravity relative to air; 1.204 kg/m³ is the density of air at 20 °C, —

Units: kg/m³; $B_g$ is for the same standard conditions as $\rho_{g,\text{sc}}$

Holds when

The gas is single-phase both at the surface and in the reservoir (all its mass remains gas).

This gives the same result as the equation of state $\rho_g = pM/(ZRT)$ . For example, for the lesson gas, $0.65\cdot 1.204/0.00530 \approx 148$ kg/m³.

Worked exampleGas formation volume factor and gas in place of a gas accumulation

Given (assumed data)

Gas: the lesson gas: $\gamma_g = 0.65$ ; at 20 MPa and 80 °C $Z = 0.869$ (example of the calculation of Z)
Initial reservoir pressure and temperature: 20.0 MPa (absolute), 80 °C
Accumulation: area 12 km², net thickness 18 m (the thickness of productive rock), porosity 0.16 (the fraction of the rock that is pore space), initial water saturation 0.25 (the fraction of the pore space filled with water)

Solution

Formation volume factor at standard conditions of 20 °C: $B_{gi} = \dfrac{0.101325}{20.0}\cdot\dfrac{0.869\cdot 353.15}{293.15} = 0.00530$ m³/m³; expansion factor $E = 188.6$ .
Gas-filled pore volume: $12\cdot 10^6\cdot 18\cdot 0.16\cdot(1 - 0.25) = 25.92$ million m³.
Gas initially in place: $G = 25.92\cdot 10^6/0.00530 = 4.89$ billion m³ (at 20 °C).
In oilfield units: $T = 635.67$ °R, $p = 2901$ psia, $B_{gi} = 0.00504\cdot 0.869\cdot 635.67/2901 = 0.000959$ rb/scf. The gas-filled pore volume is 163.0 million bbl, and the gas in place is $G = 170$ billion scf.
If we refer the gas in place to 15 °C instead: $B_{gi} = 0.00539$ m³/m³ and $G = 4.81$ billion m³. This is 1.7% less, although there is just as much gas in the reservoir.

Answer:

B_{gi} = 0.00530

m³/m³ (

E = 189

). Gas initially in place is about 4.89 billion m³ at 20 °C (170 billion scf at 60 °F).

What it means. Each cubic metre of pores filled with gas holds about 189 m³ of gas at standard conditions. An error in $Z$ passes straight into the gas in place. If we took the gas as ideal ( $Z = 1$ ), the gas in place would come out 13% lower. Gas in place and production must be converted to the same standard conditions. How much of this gas can be recovered depends on how far the pressure in the accumulation can be lowered (Material balance for gas reservoirs). The data in this example are assumed for illustration.

Isothermal compressibility of a gas

When the pressure in the accumulation falls, the gas expands and pushes itself towards the wells. This is the main source of energy of a gas accumulation. The measure of this expansion is the isothermal compressibility: the relative change of volume (the fraction by which the volume changes) for each unit change of pressure, at constant temperature.

Isothermal compressibility of a gas

c_g = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T = \frac{1}{p} - \frac{1}{Z}\left(\frac{\partial Z}{\partial p}\right)_T

where:

$c_g$ isothermal compressibility of the gas, MPa⁻¹
$V$ volume of a given amount of gas, m³
$p$ absolute pressure, MPa
$\left(\partial Z/\partial p\right)_T$ slope of $Z$ against pressure at constant temperature, MPa⁻¹

Units: any consistent units: MPa and MPa⁻¹ or psi and psi⁻¹

Holds when

The temperature is constant: we treat the reservoir as isothermal.
The gas is single-phase. The second form comes from substituting $V = ZnRT/p$ (the derivation is in the advanced section).

The minus sign makes the compressibility positive, because the volume decreases as the pressure rises. For an ideal gas, $Z = 1$ and $c_g = 1/p$ : at 1 MPa it is 1 MPa⁻¹, and at 20 MPa it is 0.05 MPa⁻¹. For a real gas, $c_g > 1/p$ as long as $Z$ falls with pressure (to the left of the minimum of the isotherm). When $Z$ rises, $c_g < 1/p$ .

The law of corresponding states lets us use a dimensionless quantity, the pseudo-reduced compressibility $c_{pr} = c_g\,p_{pc}$ . It follows from the slope of the generalised $Z$ chart at constant $T_{pr}$ :

Pseudo-reduced compressibility of a gas

c_{pr} = c_g\,p_{pc} = \frac{1}{p_{pr}} - \frac{1}{Z}\left(\frac{\partial Z}{\partial p_{pr}}\right)_{T_{pr}}

where:

$c_{pr}$ pseudo-reduced compressibility, —
$p_{pc},\ p_{pr}$ pseudo-critical and pseudo-reduced pressure, MPa, —

Units: dimensionless

Holds when

We take the slope from the same generalised chart (or its approximation) that gives $Z$ .

In this way we can find the compressibility of any natural gas from a single chart or equation. Get $c_{pr}$ and divide by the $p_{pc}$ of your gas.

CalculatedThe compressibility of the lesson gas (80 °C) divided by the compressibility of an ideal gas,

1/p

. The ratio is above 1 where

Z

falls with pressure and below 1 where

Z

rises.

Orders of magnitude of compressibilities in the reservoir, MPa⁻¹

What is compressed: Gas at 5 MPa
Compressibility: ≈ 0.2
Note: close to $1/p$

What is compressed: Gas at 20 MPa
Compressibility: ≈ 0.05
Note: decreases as the pressure rises

What is compressed: Undersaturated oil
Compressibility: $(0.7\text{–}3)\cdot 10^{-3}$
Note: above the bubble point pressure, so with no free gas

What is compressed: Formation water
Compressibility: $(0.4\text{–}0.6)\cdot 10^{-3}$
Note: depends weakly on pressure

What is compressed: Pore space of the rock
Compressibility: $(0.4\text{–}1.5)\cdot 10^{-3}$
Note: larger for unconsolidated sands

What is compressed	Compressibility	Note
Gas at 5 MPa	≈ 0.2	close to $1/p$
Gas at 20 MPa	≈ 0.05	decreases as the pressure rises
Undersaturated oil	$(0.7\text{–}3)\cdot 10^{-3}$	above the bubble point pressure, so with no free gas
Formation water	$(0.4\text{–}0.6)\cdot 10^{-3}$	depends weakly on pressure
Pore space of the rock	$(0.4\text{–}1.5)\cdot 10^{-3}$	larger for unconsolidated sands

Approximate values. The compressibilities of gas and oil are PVT properties (properties of the fluid). The compressibility of the pores is a mechanical property of the rock.

Worked exampleGas compressibility from the slope of the Z curve

Given (assumed data)

Lesson gas at 80 °C: $Z = 0.8633$ at 18 MPa, $0.8687$ at 20 MPa, $0.8778$ at 22 MPa
Reservoir: gas saturation 0.75, water saturation 0.25; compressibility of water $0.45\cdot 10^{-3}$ , of the pores $0.8\cdot 10^{-3}$ MPa⁻¹

Solution

Slope of the $Z$ curve near 20 MPa: $\dfrac{\partial Z}{\partial p} \approx \dfrac{0.8778 - 0.8633}{22 - 18} = 0.00363$ MPa⁻¹.
Compressibility: $c_g = \dfrac{1}{20} - \dfrac{0.00363}{0.8687} = 0.0500 - 0.0042 = 0.0458$ MPa⁻¹. This is 8% less than for an ideal gas.
Total compressibility of the reservoir: $c_t = S_g c_g + S_w c_w + c_f = 0.75\cdot 0.0458 + 0.25\cdot 0.00045 + 0.0008 = 0.0353$ MPa⁻¹. The gas accounts for 97% of the sum.

Answer:

c_g \approx 0.0458

MPa⁻¹. The total compressibility of the gas-saturated reservoir is 0.0353 MPa⁻¹, almost all of it from the gas.

What it means. At 20 MPa the lesson gas is just to the right of the minimum of the isotherm ( $Z$ is already rising), so its compressibility is slightly below $1/p$ . It is 31 times the compressibility of a typical undersaturated oil ( $1.5\cdot 10^{-3}$ MPa⁻¹). This is why people usually neglect the expansion of water and pores in a gas accumulation, but not in an oil accumulation. The total compressibility $c_t$ enters the interpretation of well tests (Well testing: purposes and types of test). The data in this example are assumed for illustration.

Gas viscosity

Dynamic viscosity $\mu$: The shear stress divided by the velocity gradient in a flow (how much the speed changes from one layer of the flow to the next). In SI it is measured in Pa·s. In practice it is measured in mPa·s, which has the same number as the centipoise (cP). This is the viscosity that enters Darcy's law.
Kinematic viscosity $\nu = \mu/\rho$: The dynamic viscosity divided by the density. It is measured in mm²/s, which has the same number as the centistokes (cSt). For a gas at atmospheric pressure it is larger than for water: a gas has a low viscosity, but it is very light.

Gas viscosity is about 0.01–0.03 mPa·s, tens of times lower than that of formation water and oil. This is one reason why gas moves so easily (has high mobility). With the same permeability and the same pressure difference, a gas flows through the rock much faster than a liquid.

Gas viscosity has a different mechanism from liquid viscosity. In a rarefied gas the molecules fly freely between neighbouring layers of the flow and carry momentum (their mass times their speed) with them. Fast molecules are slowed in a slow layer, and slow molecules are speeded up in a fast layer. The higher the temperature, the faster the molecules and the stronger this exchange. So at low pressure the viscosity of a gas rises with temperature and hardly depends on pressure. At high pressure the density of the gas comes close to that of a liquid. Each molecule is tightly surrounded by neighbours, and the viscosity depends on how they interact. Then viscosity rises with pressure and falls with temperature, because heating at the same pressure makes the gas less dense.

CalculatedViscosity of the lesson gas against pressure at 40, 80 and 120 °C, calculated with the Lee–Gonzalez–Eakin correlation. Between 13 and 19 MPa the curves cross in pairs. At lower pressure the hot gas is more viscous than the cold gas. At higher pressure it is the other way round.

Gas viscosity at reservoir conditions is rarely measured. It is a difficult experiment, and the correlations for natural gas are accurate to a few per cent. More often it is calculated in one of these ways:

From charts (the Carr–Kobayashi–Burrows correlation, 1950s). First find the viscosity at atmospheric pressure from the molar mass of the gas and the temperature. Add corrections for nitrogen, carbon dioxide and hydrogen sulfide, which raise the viscosity. Then multiply by the ratio $\mu/\mu_1$ (the viscosity at reservoir pressure divided by the viscosity at atmospheric pressure), which you read from the pseudo-reduced pressure and temperature.
From the gas density (the Lee–Gonzalez–Eakin correlation, 1960s). The viscosity is written in terms of the molar mass, the temperature and the density of the gas in the reservoir. This method suits software and is used in most PVT calculations.
From the composition at low pressure, using the mixing rule for the viscosities of the components (advanced section). A correction for pressure follows.

Viscosity of natural gas (Lee–Gonzalez–Eakin correlation)

\begin{gathered}\mu_g = 10^{-4}\,K\,\exp\left(X\rho_g^{\,Y}\right),\qquad K = \frac{(9.4 + 0.02\,M)\,T^{1.5}}{209 + 19\,M + T},\\ X = 3.5 + \frac{986}{T} + 0.01\,M,\qquad Y = 2.4 - 0.2\,X\end{gathered}

where:

$\mu_g$ gas viscosity, mPa·s (cP)
$T$ absolute temperature in degrees Rankine, $T\,[{}^{\circ}\text{R}] = 1.8\,T\,[\text{K}]$ , °R
$M$ molar mass of the gas, $28.97\,\gamma_g$ , g/mol
$\rho_g$ gas density at the pressure and temperature of the calculation, $pM/(ZRT)$ , g/cm³

Units: empirical: $T$ in °R, $\rho_g$ in g/cm³, $\mu_g$ in mPa·s

Holds when

It was built from measurements on natural gases at temperatures from about 40 to 170 °C and pressures up to 55 MPa. The error is a few per cent.
For sour gases, gases with a lot of nitrogen, and gas condensates the accuracy is lower. Calculate the gas density with a $Z$ that includes the correction for the sour components.

As $\rho_g \to 0$ , the exponential tends to 1, and $\mu_1 = 10^{-4}K$ is the viscosity at low pressure. It rises with temperature. The factor $\exp(X\rho_g^Y)$ describes how viscosity grows with density. In this way one formula covers both mechanisms.

Worked exampleGas viscosity in the reservoir

Given (assumed data)

Lesson gas: $\gamma_g = 0.65$ , $M = 18.83$ g/mol
Reservoir conditions: 20 MPa, 80 °C; gas density $147.6$ kg/m³ (example in the lesson on real gas behaviour)

Solution

Temperature in degrees Rankine: $T = 1.8\cdot 353.15 = 635.7$ °R; density $\rho_g = 0.1476$ g/cm³.
Numerator of $K$ : $(9.4 + 0.02\cdot 18.83)\cdot 635.7^{1.5} = 156{,}688$ . Denominator: $209 + 19\cdot 18.83 + 635.7 = 1202.4$ . So $K = 130.3$ . Viscosity at low pressure: $\mu_1 = 10^{-4}K = 0.0130$ mPa·s.
$X = 3.5 + 986/635.7 + 0.01\cdot 18.83 = 5.239$ and $Y = 2.4 - 0.2\cdot 5.239 = 1.352$ .
$\mu_g = 0.0130\cdot\exp\left(5.239\cdot 0.1476^{1.352}\right) = 0.0193$ mPa·s.

Answer: The gas viscosity in the reservoir is about 0.0193 mPa·s, which is 1.48 times the viscosity at atmospheric pressure.

What it means. Water at 80 °C has a viscosity of about 0.36 mPa·s, which is roughly 19 times higher. If we used the viscosity at atmospheric pressure to calculate inflow, we would overestimate the rate of the gas well by almost one and a half times. Viscosity changes with pressure, so gas inflow calculations account for it together with $Z$ (see the advanced section on pseudo-pressure). The data in this example are assumed for illustration.

Where the properties of a gas are used

Gas properties in engineering tasks

Task: Volumetric estimation of gas in place
Properties: $B_{gi}$ (through $Z$ )
Details: Volumetric estimation of hydrocarbons in place

Task: Material balance of the accumulation
Properties: $B_g$ , or $p/Z$
Details: Material balance for gas reservoirs

Task: Inflow to a gas well
Properties: $\mu_g$ and $Z$ (pseudo-pressure)
Details: Steady-state radial flow

Task: Well testing
Properties: $\mu_g$ , $c_g$ (through $c_t$ )
Details: Well testing: purposes and types of test

Task: Flow in the well and the pipeline
Properties: $\rho_g$ , $\mu_g$ , $Z$
Details: Multiphase flow in tubing

Task	Properties	Details
Volumetric estimation of gas in place	$B_{gi}$ (through $Z$ )	Volumetric estimation of hydrocarbons in place
Material balance of the accumulation	$B_g$ , or $p/Z$	Material balance for gas reservoirs
Inflow to a gas well	$\mu_g$ and $Z$ (pseudo-pressure)	Steady-state radial flow
Well testing	$\mu_g$ , $c_g$ (through $c_t$ )	Well testing: purposes and types of test
Flow in the well and the pipeline	$\rho_g$ , $\mu_g$ , $Z$	Multiphase flow in tubing

In all these calculations the gas properties change with pressure. So we give them as tables or correlations, not as a single number. In reservoir simulation models (computer models of the reservoir) we enter for the gas a table of $B_g$ (or $Z$ ) and $\mu_g$ against pressure at the reservoir temperature. In the black oil model the same table describes the gas of the gas cap (the free gas above the oil) and the gas that has come out of the oil (Black oil model: solution gas–oil ratio and formation volume factors).

Terms and symbols

Gas formation volume factor $B_g$: The volume of gas at reservoir conditions per unit of its volume at standard conditions. It is measured in m³/m³, ft³/scf or rb/scf.
Gas expansion factor $E$: The inverse of $B_g$ : how many standard cubic metres one cubic metre of reservoir gas contains.
Isothermal compressibility $c_g$: The relative decrease in the volume of a gas for each unit increase in pressure at constant temperature. It is measured in MPa⁻¹. For an ideal gas it is $1/p$ .
Pseudo-reduced compressibility $c_{pr}$: The dimensionless compressibility $c_g\,p_{pc}$ , found from the slope of the generalised $Z$ chart.
Total compressibility $c_t$: The compressibility of the reservoir together with its fluids: the sum of the compressibilities of the phases weighted by saturation, plus the pore compressibility.
Dynamic and kinematic viscosity: The dynamic viscosity $\mu$ is in mPa·s (cP). The kinematic viscosity $\nu = \mu/\rho$ is in mm²/s (cSt).

Common mistakes and misconceptions

Wrong

“ $B_g = 0.005$ means that from a cubic metre of reservoir we get 0.005 m³ of gas.”

In fact

It is the other way round. A cubic metre of gas at the surface takes up 0.005 m³ in the reservoir. A cubic metre of reservoir gas gives $1/B_g = 200$ m³ at the surface.

Wrong

“The compressibility of a gas is a constant, as it is for oil.”

In fact

It is roughly equal to $1/p$ , so it grows more than four times when the pressure falls from 20 to 5 MPa. That is why the balance of a gas accumulation uses $B_g(p)$ or $p/Z$ , not a constant compressibility.

Wrong

“The viscosity of a gas does not depend on pressure.”

In fact

It is independent of pressure only at low pressure. In the reservoir at 20–30 MPa it is 1.5–2 times higher than in the separator.

Wrong

“On heating, the viscosity of a gas falls, like that of any liquid.”

In fact

At low pressure the viscosity of a gas rises with temperature. It falls only at high pressures, when the density of the gas comes close to that of a liquid.

In depth: advanced detail

Derivation of the compressibility of a real gas

For $n$ moles of gas at constant temperature, $V = ZnRT/p$ , where $Z$ depends on pressure. We differentiate with respect to pressure (find the rate of change), using the rule for a quotient: $\left(\dfrac{\partial V}{\partial p}\right)_T = nRT\,\dfrac{p\,(\partial Z/\partial p) - Z}{p^2}$ .

Now divide by $-V = -ZnRT/p$ : $c_g = -\dfrac{p\,(\partial Z/\partial p) - Z}{Z\,p} = \dfrac{1}{p} - \dfrac{1}{Z}\dfrac{\partial Z}{\partial p}$ . For an ideal gas the derivative is zero, and $1/p$ remains. Substitute $p = p_{pc}\,p_{pr}$ . Then $\partial Z/\partial p = (\partial Z/\partial p_{pr})/p_{pc}$ , and after multiplying by $p_{pc}$ we get the formula for the pseudo-reduced compressibility.

Pseudo-pressure: viscosity and Z in the gas inflow formula

In Darcy's law for gas (the law that links flow rate to pressure difference), the flow rate in the reservoir changes along the flow path. The gas expands, and its viscosity and $Z$ depend on pressure. It is convenient to collect these changes in one function, the real gas pseudo-pressure (pseudo-pressure):

Real gas pseudo-pressure

m(p) = 2\int_{p_0}^{p}\frac{p'}{\mu_g(p')\,Z(p')}\,\dd p'

where:

$m(p)$ pseudo-pressure, MPa²/(mPa·s)
$p_0$ any reference pressure (usually zero or atmospheric); only differences of $m$ enter the calculation, MPa
$\mu_g,\ Z$ viscosity and gas deviation factor at pressure $p'$ and reservoir temperature, mPa·s, —

Units: consistent; in oilfield units psi²/cP

Holds when

The reservoir is isothermal, and the gas is single-phase.

With pseudo-pressure, the gas inflow equation is written in the same way as for a liquid. The only change is that the pressure difference is replaced by the difference of $m(p)$ . For the lesson gas, the product $\mu_g Z$ is nearly constant up to about 10 MPa. There $m(p)$ is proportional to $p^2$ , and we calculate the inflow from the difference of the squared pressures. Above about 25 MPa the ratio $p/(\mu_g Z)$ is nearly constant. There $m(p)$ grows linearly, and we work with the pressure difference, as for a liquid. In between, we need the exact integral.

Viscosity of a gas mixture at low pressure

If we know the composition of the gas and the viscosities of the components at the temperature we need and atmospheric pressure, we can estimate the viscosity of the mixture with the Herning–Zipperer mixing rule:

Viscosity of a gas mixture at low pressure

\mu_{\text{mix}} = \frac{\sum_i \mu_i\,y_i\sqrt{M_i}}{\sum_i y_i\sqrt{M_i}}

where:

$\mu_i$ viscosity of component $i$ at the same temperature and low pressure, mPa·s
$y_i,\ M_i$ mole fraction and molar mass of the component, —, g/mol

Units: viscosities in the same units

Holds when

The pressure is low (of the order of atmospheric) and the gas is close to ideal.
For reservoir pressures, multiply the result by the ratio $\mu/\mu_1$ from a chart, or recalculate it from the density.

The weight of a component is proportional to $y_i\sqrt{M_i}$ , so heavy components count for more than their mole fraction suggests. Nitrogen, carbon dioxide and hydrogen sulfide are more viscous than methane, so even a small amount of them noticeably raises the viscosity of the gas.

Connections to other topics

Ideal and real gas: the gas deviation factor (Z) — the gas deviation factor $Z$ , on which all the gas properties are based.
Volumetric estimation of hydrocarbons in place — estimating gas initially in place with the formation volume factor $B_{gi}$ .
Black oil model: solution gas–oil ratio and formation volume factors — gas in the black oil model: free gas, solution gas–oil ratio and underground withdrawal.
Material balance for gas reservoirs — how the decline of $p/Z$ is used to estimate gas in place and water influx into the accumulation.
Steady-state radial flow — gas inflow to a well and the real gas pseudo-pressure.
Reservoir pressure and temperature — where the reservoir pressure and temperature, at which the gas properties are calculated, come from.

Check yourself

1
The pressure in a gas accumulation has fallen from 20 to 10 MPa at constant temperature; $Z$ has changed little. How has the gas formation volume factor changed?
1. A.It has hardly changed
2. B.It has roughly doubled
3. C.It has increased fourfold
4. D.It has roughly halved
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Correct answer: B. It has roughly doubled
$B_g$ is proportional to $Z/p$ . When the pressure is halved and $Z$ is almost constant, $B_g$ roughly doubles. Each standard cubic metre of gas takes up twice as much room in the reservoir.
2
The gas in place has been calculated at standard conditions of 20 °C: 10.0 billion m³. What is it at standard conditions of 15 °C and the same pressure?
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At constant pressure the volume of an ideal gas is proportional to the absolute temperature: $10.0\cdot 288.15/293.15 = 9.83$ billion m³. The amount of gas in the accumulation is the same. Only the number is smaller, because a cubic metre at 15 °C holds more molecules.
3
What is the compressibility of an ideal gas at a pressure of 10 MPa?
1. A.0.01 MPa⁻¹
2. B.0.1 MPa⁻¹
3. C.Zero: an ideal gas is incompressible
4. D.10 MPa⁻¹
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Correct answer: B. 0.1 MPa⁻¹
For an ideal gas, $c_g = 1/p = 1/10 = 0.1$ MPa⁻¹. When the pressure falls by 0.1 MPa, the volume grows by about 1%.
4
How does the viscosity of a gas change on heating if the pressure is close to atmospheric?
1. A.It falls, as in liquids
2. B.It first rises, then falls
3. C.It rises
4. D.It does not change
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Correct answer: C. It rises
In a rarefied gas, viscosity comes from the transfer of momentum by flying molecules. When the gas is heated, they move faster, and the viscosity rises. Heating lowers the viscosity only at high pressures, when the gas is as dense as a liquid.
5
For the lesson gas at 80 °C $Z = 0.890$ at 24 MPa, $0.897$ at 25 MPa and $0.905$ at 26 MPa. Find the compressibility of the gas at 25 MPa and compare it with $1/p$ .
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Slope: $(0.905 - 0.890)/2 = 0.0075$ MPa⁻¹. $c_g = 1/25 - 0.0075/0.897 = 0.0400 - 0.0084 = 0.0316$ MPa⁻¹. This is 21% less than for an ideal gas, because at 25 MPa $Z$ rises with pressure and the gas resists compression more than an ideal gas does.
6
The density of a gas at standard conditions (20 °C) is 0.80 kg/m³, and its formation volume factor in the reservoir is 0.0050 m³/m³. Find the gas density in the reservoir and the pressure gradient in the gas zone.
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$\rho_g = 0.80/0.0050 = 160$ kg/m³. The gradient is $\rho_g g = 160\cdot 9.81 \approx 1570$ Pa/m $\approx 1.6$ kPa/m. This is more than six times lower than in water (Reservoir pressure and temperature).

Updated: 1 October 2026