Inflow at high rates: non-Darcy flow

Quick refresher

Darcy's law is the limit for slow flow. At high velocities an inertial term, $\beta\rho v^2$ , adds to the viscous term. Their ratio is the Forchheimer number, $\text{Fo} = k\beta\rho v/\mu$ .
This is not turbulence. The flow in the pores stays laminar (smooth, with no chaotic swirling). The extra losses come from the fluid speeding up and slowing down in channels whose width keeps changing.
The viscous losses are spread evenly over the logarithm of the distance (each tenfold step in distance takes the same pressure drop). The inertial losses fall off as $1/r$ : half of them occur in the ring out to $2r_w$ , and 95% out to $20r_w$ . So they behave like a skin, but a skin that is proportional to the rate.
The apparent skin (the skin that a test at rate $q$ shows) is $S' = S + Dq$ , where $Dq$ is the Forchheimer number at the wall of the well. For gas, $D$ includes the gas density at standard conditions and the viscosity at the bottom of the well.
We separate $S$ and $D$ with tests at several rates. A straight line of $S'$ against $q$ has the intercept $S$ (where it meets the vertical axis) and the slope $D$ . A straight line of $\Delta m/q$ against $q$ gives the coefficients $a$ and $b$ of the two-term equation.
Damage to the near-wellbore zone and partial penetration make $D$ several times larger: $\beta$ grows as permeability falls, and the inertial losses grow as the square of the velocity at the wall.

Key relationships

-\dfrac{\dd p}{\dd x} = \dfrac{\mu}{k}\,v + \beta\rho\,v^2

Forchheimer equation, consistent units

S' = S + Dq,\qquad D = \dfrac{\beta\rho_{sc}k}{2\pi h\mu_w r_w}

apparent skin of a gas well, $q$ at standard conditions

m(\bar p) - m(p_{wf}) = a\,q + b\,q^2

two-term gas inflow equation in pseudo-pressures

a = U\left(\ln\dfrac{r_e}{r_w} - \dfrac34 + S\right),\quad b = UD,\quad U = 1.290\,\dfrac{T}{kh}

metric units: $q$ in thousand m³/d, $m$ in MPa²/(mPa·s)

Common pitfalls

Treating the whole skin measured at a high rate in a gas well as damage, and planning an acid treatment (pumping acid in to clear the rock around the well) on that basis.
Estimating the absolute open flow from a single rate with Darcy's formula. The inertial losses grow as the square of the rate, so the potential comes out too high.
Using the gas density at reservoir conditions in $D$ together with a rate at standard conditions, or using the viscosity at reservoir pressure instead of bottomhole pressure.
Plotting $\Delta m/q$ against $q$ with transient flow periods (stretches of production at one rate, while the pressure is still changing) of different lengths. Each flow period has its own coefficient $a$ .
Taking $\beta$ from a correlation and applying it to a particular reservoir without a check. The scatter of the correlations is an order of magnitude (a factor of ten).

After this lesson you can

explain where the inertial losses in a porous medium come from and why they are not turbulence; write the Forchheimer equation, and use the Forchheimer number to judge when the losses cannot be ignored
describe how we measure the inertial resistance coefficient $\beta$ on core (rock samples taken from a well) and estimate it from correlations (formulas fitted to measured data), and what it depends on
derive the steady-state inflow (flow whose pressures no longer change with time) with an inertial term, show that these losses are concentrated at the wall of the well, and so arrive at a skin that grows with the rate: $S' = S + Dq$
calculate the coefficient $D$ of a gas well in metric and in oilfield units, and write the two-term gas inflow equation using pseudo-pressure (a modified pressure that allows for the changing viscosity and compressibility of gas)
separate the skin from the inertial losses with a test at several rates, draw the stabilised inflow performance line, and find the absolute open flow (the rate the well would give at zero bottomhole pressure)
explain why damage to the near-wellbore zone, partial penetration (only part of the reservoir thickness is open) and perforation make the inertial losses many times larger

Why Darcy's law stops working

Darcy's law describes slow flow. All of the pressure drop goes into viscous friction (the fluid rubbing against itself and against the rock), so the drop is proportional to the Darcy velocity (the flow rate divided by the area it crosses; see Rock permeability). At high velocities the flow becomes non-Darcy. Pore channels are not straight tubes. The fluid squeezes through narrow throats, spreads out in the pores, goes round the grains and keeps changing direction. At every narrow point it has to speed up, and just after it has to slow down again. Part of the pressure drop is spent on this. That part is proportional not to the velocity but to its square, like the kinetic energy $\rho v^2/2$ . At low velocities this part is negligible. At high velocities it is comparable with the viscous part, and it can even exceed it.

Forchheimer equation

-\frac{\dd p}{\dd x} = \frac{\mu}{k}\,v + \beta\rho\,v^2

where:

$v$ Darcy velocity: the flow rate per unit cross-sectional area (not the velocity in the pores), m/s
$\mu$ dynamic viscosity of the fluid, Pa·s
$k$ permeability, m²
$\beta$ inertial resistance coefficient of the rock, m⁻¹
$\rho$ fluid density, kg/m³
$x$ coordinate along the flow, m

Units: SI (any consistent system of units)

Holds when

The flow is single-phase (one fluid) and isothermal (constant temperature). At every point the velocity points along the pressure gradient (the direction in which pressure changes fastest).
$\beta$ is a property of the rock, like $k$ . We find it by experiment (see below).
As $v \to 0$ , the second term vanishes and the equation becomes Darcy's law.

The first term is the viscous resistance and the second is the inertial resistance. In Russian-language literature this equation is also called the two-term law of filtration. There the second term is sometimes written as $\rho v^2/l$ , where $l = 1/\beta$ is called the macro-roughness coefficient.

The ratio of the second term to the first shows how large the inertial term is. This ratio is the Forchheimer number:

Forchheimer number

\text{Fo} = \frac{\beta\rho v^2}{\mu v/k} = \frac{k\beta\rho v}{\mu}

where:

$\text{Fo}$ ratio of the inertial resistance to the viscous, —
$k,\ \beta,\ \rho,\ v,\ \mu$ as in the Forchheimer equation, m², m⁻¹, kg/m³, m/s, Pa·s

Units: no units; the quantities are in consistent units

Holds when

This is an estimate at one point of the flow. Near a well, $v$ and Fo change with distance.

At Fo = 0.01 the pressure gradient is 1% larger than Darcy's law gives. At Fo = 0.1 it is 10% larger. At Fo > 1 the inertial resistance dominates. The same quantity is called the modified Reynolds number for a porous medium.

Why is it mainly gas that matters? While the correction is small, Darcy's law sets the velocity: $v \approx k|\dd p/\dd x|/\mu$ . We put this into the Forchheimer number. At a given pressure gradient, the number is then proportional to $\rho/\mu^2$ : the viscosity enters squared. Gas in the reservoir is tens of times less viscous than light oil, but only a few times less dense. So at the same pressure gradient, the Forchheimer number for gas is hundreds of times larger.

Gas and gas-condensate wells (gas-condensate is gas that drops out liquid as the pressure falls) — almost always. These are the wells where the inertial skin is measured in every test.
Oil and water wells — when permeability is high, viscosity is low and rates are high, especially if only part of the reservoir thickness is open.
Perforation tunnels (holes shot through the casing into the rock) and gravel packs (a ring of gravel that holds back sand) — the flow squeezes into a small area, so the velocities in them are much higher than in the reservoir (Effect of completion on inflow).
Hydraulic fractures with proppant (cracks made by pumping fluid at high pressure and held open by grains called proppant) — the inertial losses in the proppant pack reduce the effective conductivity of the fracture (Inflow to a hydraulically fractured well).
Gas injection wells and gas flow tests on core — the same physics with the flow in the opposite direction.

The inertial resistance coefficient

The inertial resistance coefficient $\beta$ (the beta factor) is a property of the rock, like permeability. It has units of one over length: m⁻¹ in SI and ft⁻¹ in oilfield units (1 ft⁻¹ = 3.281 m⁻¹). The more twisting, uneven and narrow the pore channels are, the larger $\beta$ is. So, on the whole, it grows as permeability falls.

We measure $\beta$ on core (rock samples taken from a well) by flowing gas through a sample at several rates. As the gas moves along the sample it expands, so the Darcy velocity changes. But the mass flux $G = \rho v$ — the mass of gas passing through a unit area per second — is the same in every cross-section. We write the Forchheimer equation using $G$ , multiply it by the gas density, $\rho = pM/(ZRT)$ , and integrate over the length of the sample (add up the pressure changes along it). The result is a straight line:

Gas test on core: the linear form of the Forchheimer equation

\frac{M\,(p_1^2 - p_2^2)}{2ZRT\,L\,\mu\,G} = \frac{1}{k} + \beta\,\frac{G}{\mu}

where:

$p_1,\ p_2$ absolute pressures (measured from zero, not from atmospheric pressure) at the inlet and outlet of the sample, Pa
$G$ mass flux: the mass flow rate of gas divided by the cross-sectional area of the sample, kg/(m²·s)
$M$ molar mass of the gas, kg/mol
$R$ universal gas constant, 8.314, J/(mol·K)
$T$ temperature of the test, K
$L$ length of the sample, m
$\mu,\ Z$ gas viscosity and gas deviation factor (the correction that makes the ideal gas law fit a real gas; at the pressures of the test $Z \approx 1$ ), Pa·s, —

Units: SI; pressures absolute

Holds when

The flow is isothermal and single-phase. $Z$ and $\mu$ do not change over the range of pressures in the test.
Gas slippage (the Klinkenberg correction: gas molecules slip along the pore walls, so the rock looks more permeable than it is) is small. To make sure of this, the test is run with back pressure (pressure held at the outlet), so that the mean pressure in the sample is about a megapascal. Otherwise slippage is allowed for separately (Rock permeability).

Plot the left-hand side (vertical axis) against $G/\mu$ (horizontal axis). The test points fall on a straight line. It meets the vertical axis at $1/k$ , and its slope is $\beta$ . We use the same device — plotting the pressure losses per unit of rate against the rate — in well tests too.

Worked examplePermeability and the coefficient

\beta

from a core test

Given (assumed data)

Sample: length 4.0 cm, diameter 2.54 cm
Gas: nitrogen: $M$ = 0.02801 kg/mol, $\mu$ = 0.0178 mPa·s at 25 °C (298.15 K), $Z = 1$
Outlet pressure $p_2$: 0.6000 MPa (absolute; held by a back-pressure regulator so that gas slippage is small)
Mass flux $G$ in four tests: 0.05; 0.15; 0.30; 0.45 kg/(m²·s)
Inlet pressure $p_1$: 0.6464; 0.7490; 0.9192; 1.1007 MPa (absolute)

Solution

For each test, calculate the horizontal value $X = G/\mu$ and the vertical value $Y$ — the left-hand side of the formula (pressures in pascals, $\mu = 1.78\cdot10^{-5}$ Pa·s). Test 1: $X_1 = 2809$ m⁻¹, $Y_1 = 9.178\cdot10^{12}$ m⁻². Test 4: $X_4 = 25{,}281$ m⁻¹, $Y_4 = 15.015\cdot10^{12}$ m⁻².
Tests 2 and 3 give $Y = 10.633\cdot10^{12}$ and $12.826\cdot10^{12}$ m⁻². These points lie on the same straight line.
Find the slope from the first and last tests: $\beta = (Y_4 - Y_1)/(X_4 - X_1)$ $= 2.60\cdot10^{8}$ m⁻¹.
Find the intercept on the vertical axis: $1/k = Y_1 - \beta X_1 = 8.448\cdot10^{12}$ m⁻². So $k = 1.184\cdot10^{-13}$ m² $= 120$ mD.
Now the Forchheimer number. In test 1, $\text{Fo} = k\beta G/\mu = 0.086$ . In test 4 it is 0.78, so the inertial resistance is 78% of the viscous resistance.

Answer: The permeability of the sample is about 120 mD, the inertial resistance coefficient about 2.60·10⁸ m⁻¹.

What it means. If we calculated the permeability by Darcy's law from test 4 alone, we would get $1/Y_4$ , which is 44% lower. The inertial losses would be “hidden” in a permeability that is too low. That is why gas permeability is measured at low rates, and $\beta$ is found from the slope over several rates. The correlation given below gives 2.4·10⁸ m⁻¹ for 120 mD, which is the same order of magnitude (the same power of ten). The data in this example are assumed for illustration.

When there is no core, we estimate $\beta$ from correlations (formulas fitted to measurements on many samples) with permeability, and sometimes with porosity as well. A widely used one for consolidated rocks (rocks whose grains are cemented together) is:

Correlation of the inertial resistance coefficient with permeability

\beta = \frac{7.64\cdot10^{10}}{k^{1.201}}

where:

$\beta$ inertial resistance coefficient, m⁻¹
$k$ permeability of the rock, mD

Units: empirical: $\beta$ in m⁻¹, $k$ in mD

Holds when

The rocks are consolidated and the gas flow is single-phase. Measurements scatter around the correlation by up to an order of magnitude (a factor of ten).
Irreducible water (water held in the rock that cannot be produced) or condensate that has dropped out narrows the channels. Then $\beta$ for gas increases, sometimes several times over.

$\beta$ is roughly proportional to $k^{-1.2}$ : a rock ten times less permeable has a $\beta$ 16 times larger. The correlation is good for a first estimate. For a project, we find $\beta$ (or, better still, the coefficient $D$ of the well) from core and well tests.

In other unit systems (1)

Oilfield units

\beta\,[\text{ft}^{-1}] = \frac{2.33\cdot10^{10}}{k\,[\text{mD}]^{1.201}}

$\beta$ in ft⁻¹, $k$ in mD.

$7.64\cdot10^{10} = 3.281\cdot2.33\cdot10^{10}$ , where 3.281 converts ft⁻¹ to m⁻¹.

The order of magnitude of

\beta

from the correlation

$k$ , mD: 1
$\beta$ , m⁻¹: $7.6\cdot10^{10}$
$\beta$ , ft⁻¹: $2.3\cdot10^{10}$

$k$ , mD: 10
$\beta$ , m⁻¹: $4.8\cdot10^{9}$
$\beta$ , ft⁻¹: $1.5\cdot10^{9}$

$k$ , mD: 100
$\beta$ , m⁻¹: $3.0\cdot10^{8}$
$\beta$ , ft⁻¹: $9.2\cdot10^{7}$

$k$ , mD: 1000
$\beta$ , m⁻¹: $1.9\cdot10^{7}$
$\beta$ , ft⁻¹: $5.8\cdot10^{6}$

$k$ , mD	$\beta$ , m⁻¹	$\beta$ , ft⁻¹
1	$7.6\cdot10^{10}$	$2.3\cdot10^{10}$
10	$4.8\cdot10^{9}$	$1.5\cdot10^{9}$
100	$3.0\cdot10^{8}$	$9.2\cdot10^{7}$
1000	$1.9\cdot10^{7}$	$5.8\cdot10^{6}$

The values are only a guide. At the same permeability, different rocks differ by a factor of several.

Inertial losses at the well: a skin that grows with the rate

Near a well the Darcy velocity increases towards the wellbore. The whole flow passes through the side surface of a cylinder of radius $r$ , so $v = q/(2\pi rh)$ . The viscous term of the Forchheimer equation grows towards the well as $1/r$ , and the inertial term grows as $1/r^2$ . We put this velocity into the equation and integrate from the wall of the well to the outer boundary, as in the derivation of the Dupuit equation (Steady-state radial flow):

Steady-state radial flow with inertial losses

p_e - p_w = \frac{q\mu}{2\pi kh}\ln\frac{r_e}{r_w} + \frac{\beta\rho q^2}{(2\pi h)^2}\left(\frac{1}{r_w} - \frac{1}{r_e}\right)

where:

$q$ flow rate at reservoir conditions; for a liquid $q = q_s B$ , m³/s
$p_e,\ p_w$ pressure at the outer boundary and at the wall of the well, Pa
$r_e,\ r_w$ radius of the outer boundary and radius of the well, m
$k,\ h,\ \mu$ permeability, reservoir thickness, viscosity, m², m, Pa·s
$\beta,\ \rho$ inertial resistance coefficient and fluid density, m⁻¹, kg/m³

Units: SI; $\ln$ is the natural logarithm

Holds when

The inflow is steady-state and single-phase, and the liquid is slightly compressible (we treat gas below, through the mass flow rate).
The reservoir is homogeneous (the same everywhere). The well is an open hole (not cased or perforated) over the whole thickness. $\beta$ and $\rho$ are constant.

The drawdown (the pressure difference that drives the flow) is no longer proportional to the rate. A term proportional to $q^2$ adds to the term proportional to $q$ (the Dupuit equation). If the rate doubles, the viscous losses double and the inertial losses grow fourfold.

What matters most is where the pressure is lost. Here are the shares of the viscous and the inertial losses that fall on the ring between the wall of the well and the radius $r$ :

Shares of viscous and inertial losses within radius

r

\delta_{\text{visc}}(r) = \frac{\ln(r/r_w)}{\ln(r_e/r_w)},\qquad \delta_{\text{inert}}(r) = \frac{1/r_w - 1/r}{1/r_w - 1/r_e} \approx 1 - \frac{r_w}{r}

where:

$\delta_{\text{visc}},\ \delta_{\text{inert}}$ the viscous and inertial pressure losses between $r_w$ and $r$ , each as a share of the corresponding losses between $r_w$ and $r_e$ , —
$r$ distance from the well axis, m

Units: no units; radii in the same units

Holds when

Steady-state radial flow, as in the previous formula. The approximate equality holds when $r_e \gg r_w$ .

Half of the inertial losses fall on the ring from $r_w$ to $2r_w$ , 90% within $10r_w$ and 95% within $20r_w$ . For a typical well, that is within two metres of the axis. The viscous losses are spread much more widely: only about a third of them fall on the same two metres.

CalculatedCalculated shares of the viscous and the inertial pressure losses within the distance

r

from the axis of the lesson's well (

r_w

= 0.108 m,

r_e

= 500 m). Almost all the inertial losses occur in the first metre.

The inertial losses are concentrated in a thin ring at the wellbore. So it is convenient to count them as part of the skin (Skin factor) — a skin that grows with the rate (the non-Darcy coefficient). We divide the inertial term by the “unit” of radial pressure drop, $q\mu/(2\pi kh)$ :

Rate-dependent skin

p_e - p_{wf} = \frac{q\mu}{2\pi kh}\left(\ln\frac{r_e}{r_w} + S + Dq\right),\qquad D = \frac{\beta\rho k}{2\pi h\mu r_w}

where:

$S$ skin that does not depend on the rate: damage to the near-wellbore zone, geometry of penetration, perforation, —
$D$ non-Darcy coefficient: the coefficient of the inertial skin, s/m³
$Dq$ inertial skin at the rate $q$ , —
$p_{wf}$ bottomhole flowing pressure of the producing well, Pa
$q,\ \mu,\ k,\ h,\ \beta,\ \rho,\ r_e,\ r_w$ as in the formula above; $q$ is at reservoir conditions, SI

Units: SI; $\ln$ is the natural logarithm

Holds when

$r_e \gg r_w$ , so the term $1/r_e$ is dropped.
The well is an open hole over the whole thickness, and $\beta$ at the wellbore is the same as in the reservoir. With damage and with partial penetration, $D$ is larger (see the last section).

The skin $S' = S + Dq$ that a test measures is called the apparent skin. The product $Dq$ is the Forchheimer number at the wall of the well: $Dq = k\beta\rho v_w/\mu$ , where $v_w = q/(2\pi r_w h)$ is the velocity at the wall. The inertial skin is large when the flow right at the wall is already clearly non-linear.

In other unit systems (2)

Metric oilfield units (liquid)

D\,[1/(\text{m}^3/\text{d})] = 1.818\cdot10^{-18}\,\frac{\beta\,\rho\,B\,k}{h\,\mu\,r_w}

$\beta$ in m⁻¹, $\rho$ in kg/m³, $B$ in m³/m³, $k$ in mD, $h$ and $r_w$ in m, $\mu$ in mPa·s; $q$ in m³/d at standard conditions.

$1.818\cdot10^{-18}$ is $0.9869\cdot10^{-15}$ divided by $86{,}400\cdot2\pi\cdot10^{-3}$ . It converts mD to m², days to seconds and mPa·s to Pa·s.

Oilfield units (liquid)

D\,[1/(\text{STB/d})] = 1.021\cdot10^{-14}\,\frac{\beta\,\rho\,B\,k}{h\,\mu\,r_w}

$\beta$ in ft⁻¹, $\rho$ in g/cm³, $B$ in rb/STB (reservoir barrels per stock-tank barrel), $k$ in mD, $h$ and $r_w$ in ft, $\mu$ in cP; $q$ in STB/d (stock-tank barrels per day).

The same unit conversion as in the SI formula.

For gas, the density and the volume flow rate both change along the flow path. But their product, the mass flow rate, is the same at all radii. So the formula for a gas well uses the rate at standard conditions and the gas density at the same conditions. The viscosity is taken at the bottom of the well, where the inertial losses are concentrated (the derivation is in the advanced section):

Non-Darcy coefficient of a gas well

D = \frac{\beta\,\rho_{sc}\,k}{2\pi h\,\mu_w\,r_w}

where:

$D$ non-Darcy coefficient per unit of gas rate at standard conditions, s/m³
$\rho_{sc}$ gas density at standard conditions: $1.204\,\gamma_g$ at 20 °C ( $\gamma_g$ is the gas specific gravity relative to air, that is, the gas density divided by the air density), kg/m³
$\mu_w$ gas viscosity at the bottomhole pressure and reservoir temperature, Pa·s
$\beta,\ k,\ h,\ r_w$ as above, m⁻¹, m², m, m

Units: SI

Holds when

The mass flow rate is the same at all radii, and the inflow is isothermal.
The inertial losses are concentrated at the wellbore, so the viscosity is taken at the bottomhole pressure.
The well is an open hole over the whole thickness, and the near-wellbore zone is undamaged.

The only quantity that depends on pressure is the viscosity at the bottom of the well. Over the range of rates in an ordinary test it changes by about 10%. So we take the $D$ of a gas well as constant, as long as the penetration and the condition of the near-wellbore zone do not change. Tests measure exactly this $D$ .

In other unit systems (2)

Metric oilfield units

D\,[1/(\text{thousand m}^3/\text{d})] = 1.818\cdot10^{-15}\,\frac{\beta\,\rho_{sc}\,k}{h\,\mu_w\,r_w}

$\beta$ in m⁻¹, $\rho_{sc}$ in kg/m³ (20 °C, 0.101325 MPa), $k$ in mD, $h$ and $r_w$ in m, $\mu_w$ in mPa·s; the rate is in thousand m³/d.

This is a thousand times the constant for a liquid, because the rate is in thousands of cubic metres.

Oilfield units

D\,[1/(\text{Mscf/d})] = 2.71\cdot10^{-15}\,\frac{\beta\,k\,M\,p_{sc}}{h\,\mu_w\,r_w\,T_{sc}}

$\beta$ in ft⁻¹, $k$ in mD, $M$ in g/mol, $p_{sc}$ in psia (pounds per square inch, absolute), $T_{sc}$ in °R (degrees Rankine, the absolute temperature scale that goes with °F), $h$ and $r_w$ in ft, $\mu_w$ in cP; the rate is in thousand standard cubic feet per day.

The density at standard conditions is written as $p_{sc}M/(RT_{sc})$ . The constant comes from converting the units.

Worked exampleInertial skin of a gas well from rock properties

Given (assumed data)

Reservoir: $k$ = 30 mD, $h$ = 12 m, the wellbore is open over the whole thickness, no damage
Well: $r_w$ = 0.108 m; radius of the drainage area 500 m
Gas: the gas from the lesson on gas properties: $\gamma_g$ = 0.65, reservoir temperature 80 °C
Rate: 600 thousand m³/d at a bottomhole pressure of about 17 MPa
$\beta$: from the correlation with permeability

Solution

$\beta = 7.64\cdot10^{10}/30^{1.201}$ $= 1.286\cdot10^{9}$ m⁻¹.
The gas density at standard conditions is $\rho_{sc} = 1.204\cdot0.65 = 0.7826$ kg/m³. The gas viscosity at 17 MPa and 80 °C is $\mu_w = 0.01794$ mPa·s.
The numerator is $\beta\rho_{sc}k = 1.286\cdot10^{9}\cdot0.7826\cdot30$ $= 3.019\cdot10^{10}$ , and the denominator is $h\mu_w r_w = 12\cdot0.01794\cdot0.108 = 0.02325$ . So $D = 1.818\cdot10^{-15}\cdot3.019\cdot10^{10}/0.02325$ $= 2.36\cdot10^{-3}$ (thousand m³/d)⁻¹.
The inertial skin is $Dq = 2.36\cdot10^{-3}\cdot600 = 1.42$ .
Check through the velocity at the wall. The mass flow rate is $5.43$ kg/s, so the mass flux is $G_w = 5.43/(2\pi\cdot0.108\cdot12) = 0.667$ kg/(m²·s), and $\text{Fo}_w = k\beta G_w/\mu_w = 1.42$ . This is the same number.
For an undamaged well in pseudo-steady state (the flow regime after the pressure disturbance has reached the boundaries), the logarithmic term is $\ln(500/0.108) - 0.75 = 7.690$ . The share of the inertial skin in the resistance is $1.42/(7.690 + 1.42) = 16$ %.

Answer:

D \approx 2.36\cdot10^{-3}

(thousand m³/d)⁻¹. At 600 thousand m³/d the inertial skin of 1.42 is about 16% of the resistance of an undamaged well.

What it means. At the wall itself, the inertial resistance is 1.42 times the viscous resistance. But the Forchheimer number falls as $1/r$ : at one metre from the axis it is about 0.15, and at ten metres about 0.015. This is an estimate for an open hole, made with a correlation. In real wells $D$ is usually larger. The flow squeezes into the perforation tunnels, the near-wellbore zone is damaged, and not the whole thickness works (see the last section). That is why $D$ is measured. The data in this example are assumed for illustration.

Gas inflow: the two-term equation in pseudo-pressures

The viscosity and the gas deviation factor $Z$ change with pressure. So, as with Darcy's law, we integrate the Forchheimer equation for gas in terms of the pseudo-pressure $m(p) = 2\int p\,\dd p/(\mu Z)$ (real gas pseudo-pressure; for its properties see Gas formation volume factor, compressibility and viscosity). Pseudo-pressure is a modified pressure that allows for these changes. The derivation is in the advanced section. Once the pressure disturbance has reached the boundaries of the drainage area (the area from which the well takes its fluid) and the inflow has become pseudo-steady-state (Pseudo-steady-state flow and the shape factor), the result is the two-term inflow equation:

Two-term gas inflow equation in pseudo-pressures

\begin{gathered} m(\bar p) - m(p_{wf}) = a\,q + b\,q^2\\ a = U\left(\ln\frac{r_e}{r_w} - \frac34 + S\right),\qquad b = U D\\ U = \frac{p_{sc}\,T}{\pi kh\,T_{sc}}\end{gathered}

where:

$m(p)$ gas pseudo-pressure, Pa²/(Pa·s)
$\bar p,\ p_{wf}$ average pressure in the drainage area and bottomhole flowing pressure, absolute, Pa
$q$ gas rate at standard conditions, m³/s
$U$ the “unit” of pseudo-pressure drop in radial inflow, Pa²/(Pa·s)/(m³/s)
$T,\ T_{sc},\ p_{sc}$ reservoir temperature, and temperature and pressure at standard conditions, K, K, Pa
$a,\ b$ viscous and inertial flow resistance coefficients, by the formulas
$S,\ D$ the rate-independent skin and the non-Darcy coefficient, —, s/m³

Units: SI; pressures absolute; $\ln$ is the natural logarithm

Holds when

The inflow is pseudo-steady-state, in a circular area with the well at the centre. For another shape of area we change the logarithmic term, as in the lesson on pseudo-steady-state flow.
Using $m(\bar p)$ instead of the volume-averaged pseudo-pressure is the usual approximation.
$D$ is taken with the gas viscosity at the bottom of the well. The gas in the reservoir is single-phase.

The coefficient $a$ is the resistance of the reservoir and the near-wellbore zone according to Darcy's law, and $b$ is the inertial resistance. In Russian-language practice they are known as the flow resistance coefficients. As long as $\mu Z$ is almost constant (for the gas of the lesson, up to about 10 MPa), $\Delta m \approx \Delta(p^2)/(\mu Z)$ . Then the equation becomes the two-term equation with squared pressures from the lesson Inflow performance relationship (IPR). Its coefficients $A$ and $B$ are $a$ and $b$ multiplied by $\mu Z$ .

In other unit systems (2)

Metric oilfield units

U = 1.290\,\frac{T}{kh}

$m$ in MPa²/(mPa·s), $q$ in thousand m³/d at 20 °C and 0.101325 MPa, $T$ in K, $k$ in mD, $h$ in m; $a$ in MPa²/(mPa·s) per thousand m³/d, $b$ per (thousand m³/d)².

$1.290$ is $1000\cdot0.101325$ divided by $\pi\cdot0.0853\cdot293.15$ . Here 0.0853 is the constant of Darcy's law in these units.

Oilfield units

U = 1422\,\frac{T}{kh}

$m$ in psi²/cP, $q$ in Mscf/d (thousand standard cubic feet per day; standard conditions are 14.7 psia and 60 °F), $T$ in °R, $k$ in mD, $h$ in ft.

The same constant 1422 as in the formula for steady-state gas inflow (Steady-state radial flow).

We solve the quadratic equation for the rate. The absolute open flow (AOF) is the conventional rate at a bottomhole pressure equal to atmospheric pressure. We find it from the same formula:

Rate of a gas well from the two-term equation

q = \frac{-a + \sqrt{a^2 + 4b\,\Delta m}}{2b},\qquad \Delta m = m(\bar p) - m(p_{wf})

where:

$q$ gas rate at standard conditions, thousand m³/d
$\Delta m$ pseudo-pressure drop between the drainage area and the bottom of the well, MPa²/(mPa·s)
$a,\ b$ coefficients of the two-term equation, by the formula above
$q_{max}$ absolute open flow: $q$ at $\Delta m_{max} = m(\bar p) - m(p_{\text{atm}})$ , thousand m³/d

Units: metric oilfield (or any units consistent with $a$ and $b$ )

Holds when

The inflow is stabilised (the pressures no longer change with time) and the coefficients $a$ and $b$ are known. For $b = 0$ the formula becomes $q = \Delta m/a$ .

Without the inertial term, the inflow performance line (with $q$ along the horizontal axis and $\Delta m$ along the vertical axis) would be a straight line. With the inertial term the line bends, and each further step of the drawdown gives a smaller increase in rate.

Tests at several rates: how to separate $S$ and $D$

A test at one rate gives one apparent skin, $S' = S + Dq$ , which is the sum of two unknowns. To separate them, we test the well at several rates that differ clearly, best of all by a factor of two or three. We process the results in one of two ways.

From the skins of the flow periods. We interpret the pressure record of each flow period in the usual way, but in pseudo-pressures (Pressure drawdown analysis, Pressure build-up analysis). The record is the drawdown while the well flows, or the build-up after shut-in. This gives the flow capacity $kh$ (permeability times thickness) and the apparent skin $S'_j$ . The points fall on a straight line:

Apparent skin of the flow periods

S'_j = S + D\,q_j

where:

$S'_j$ apparent skin of the $j$ -th flow period, from interpreting its pressure record, —
$q_j$ rate of the $j$ -th flow period (for a build-up, the rate before shut-in), thousand m³/d
$S,\ D$ intercept on the vertical axis and slope of the line, —, (thousand m³/d)⁻¹

Units: any; $D$ is in units that are the inverse of the rate units

Holds when

All flow periods are interpreted with one reservoir model, and the flow capacity $kh$ is the same for all of them.
The skin and $D$ do not change during the test (the well has already been cleaned up after completion).

The inertial pressure drop vanishes at once when the well is shut in, like any pressure drop across a skin. So the skin from a build-up curve contains the $Dq$ of the rate at which the well was flowing before shut-in.

From the inflow performance line. We divide the two-term equation by the rate: $\Delta m/q = a + bq$ . If we plot the points of the flow periods with $q$ on the horizontal axis and $\Delta m/q$ on the vertical axis, they fall on a straight line with slope $b$ and intercept $a$ . But this is true only if the coefficient $a$ is the same for all flow periods. While the inflow is transient (the pressure disturbance is still spreading), $a$ contains a logarithmic term that grows with time, instead of $\ln(r_e/r_w) - 3/4$ . So it is different for flow periods of different lengths. This gives three kinds of test:flow-after-flow, isochronal and modified isochronal.

Multi-rate tests of gas wells

Type of test: Flow-after-flow
How it is run: 3–5 flow periods one after another, each until the pressure stabilises
What it gives: $a$ and $b$ ; in a tight reservoir (one with very low permeability) it takes a very long time

Type of test: Isochronal
How it is run: flow periods of equal length, with the pressure building up fully between them, plus one long flow period
What it gives: $b$ from the short flow periods, $a$ from the long one

Type of test: Modified isochronal
How it is run: flow periods and shut-ins of equal length
What it gives: about the same; the pressure before the flow period is used instead of $\bar p$

Type of test	How it is run	What it gives
Flow-after-flow	3–5 flow periods one after another, each until the pressure stabilises	$a$ and $b$ ; in a tight reservoir (one with very low permeability) it takes a very long time
Isochronal	flow periods of equal length, with the pressure building up fully between them, plus one long flow period	$b$ from the short flow periods, $a$ from the long one
Modified isochronal	flow periods and shut-ins of equal length	about the same; the pressure before the flow period is used instead of $\bar p$

In Russian-language practice the first kind is called the method of steady-state withdrawals. Stabilisation means that the flow has reached the pseudo-steady-state regime. For a well at the centre of a circular area this happens at $kt/(\phi\mu c_tA) \approx 0.1$ . For the well of the lesson (30 mD, area radius 500 m) that is about three days. In a similar reservoir with a permeability of 3 mD it is about a month.

CalculatedAn isochronal test of the lesson's well, plotted as

\Delta m/q

against the rate (calculated with the lesson's model). The points of four 6-hour flow periods lie on a line with slope

b

. The stabilised line is parallel to it and lies higher by the difference of the logarithmic terms.

Worked exampleDamage skin and the coefficient

D

from an isochronal test

Given (assumed data)

Well: the lesson's well (see the example above): $k$ = 30 mD, $h$ = 12 m, $r_w$ = 0.108 m, the lesson gas, 80 °C
Test: four flow periods of 6 h each; after each one the well is shut in until the pressure recovers to 20 MPa
Rates $q_j$: 200; 350; 500; 650 thousand m³/d
Interpretation of the flow periods in pseudo-pressures: $kh$ = 360 mD·m for all flow periods; apparent skins 2.76; 3.48; 4.32; 5.22

Solution

The averages are $\bar q = 425$ thousand m³/d and $\bar S' = 3.945$ .
The slope from the first and last flow periods is $D \approx (5.22 - 2.76)/(650 - 200)$ $= 5.47\cdot10^{-3}$ . From all four flow periods (by least squares, the best-fit straight line) $D = 5.48\cdot10^{-3}$ (thousand m³/d)⁻¹.
The intercept is $S = \bar S' - D\bar q$ $= 3.945 - 5.48\cdot10^{-3}\cdot425$ $= 1.62$ .
The estimate of $D$ for an open, undamaged hole (the example above) is $2.36\cdot10^{-3}$ . The measured $D$ is 2.3 times larger.
At a rate of 650 thousand m³/d the inertial skin $Dq = 3.56$ is 68% of the apparent skin 5.22.

Answer: The rate-independent skin is about 1.62. The non-Darcy coefficient is

D \approx 5.48\cdot10^{-3}

(thousand m³/d)⁻¹.

What it means. Damage and perforation give a small skin, about 1.62. But the apparent skin of 5.22 at the highest flow period is more than half inertial. An acid treatment would reduce only $S$ . The measured $D$ is 2.3 times larger than the estimate for an open hole. This may point to a constriction of the inflow near the well: the flow squeezes into the perforation tunnels, and part of the interval may not be working. But the correlation for $\beta$ scatters widely, so such a difference is not yet proof. The working interval is checked by production logging (a survey run in the producing well that shows where the fluid enters). The scatter of the points about the line is the usual error of interpretation, so we take at least three or four flow periods. The data in this example are assumed for illustration.

CalculatedThe apparent skins of the four flow periods of the example, and the least-squares line (the best-fit straight line). The intercept on the vertical axis is the rate-independent skin

S

, and the slope is

D

. The dashed line shows the same skin with the

D

of an open, undamaged hole.

How to plan a multi-rate test

1
Clean up the well
The skin must stay constant during the test. After completion, flow the well until all the mud filtrate (the liquid that leaked into the rock from the drilling fluid) has been produced and the rate is stable.
2
Choose the rates
Use 3–4 flow periods, usually with the rate increasing from one to the next. The largest rate is two to three times the smallest, but it must not cause sand production or destroy the near-wellbore zone.
3
Choose the duration
If there is enough time for each rate to stabilise, run the flow periods one after another. If not, use an isochronal scheme: flow periods of equal length and one long flow period.
4
Record the pressure and the rate
Use bottomhole gauges and measure the rate in every flow period. Record the pressure at the end of each flow period and during the shut-ins.
5
Process in two ways
Plot the skins of the flow periods against the rate, and plot $\Delta m/q$ against the rate. Agreement between the two estimates of $D$ is a quality check.

The inflow performance line of a gas well and the absolute open flow

Once $S$ and $D$ are known, and $kh$ has been found from the same pressure records, we can calculate the stabilised inflow performance line without waiting for the flow to stabilise. The formula with $\ln(r_e/r_w) - 3/4$ , for the drainage area of the well, gives the coefficient $a$ , and the product $UD$ gives $b$ . We do the same in an isochronal test that has no long flow period.

Worked exampleStabilised inflow performance line of the lesson's well

Given (assumed data)

Test results (the example above): $S$ = 1.616 (before rounding), $D$ = 5.48·10⁻³ (thousand m³/d)⁻¹, $kh$ = 360 mD·m
Drainage area: circular, $r_e$ = 500 m; $r_w$ = 0.108 m; average pressure 20 MPa; 80 °C (353.15 K)
Pseudo-pressure of the lesson gas at 80 °C: $m$ (20 MPa) = 27,699; $m$ (15 MPa) = 16,573; $m$ (0.1013 MPa) = 0.8 MPa²/(mPa·s)

Solution

$U = 1.2903\cdot353.15/360 = 1.2657$ (the constant 1.290 with one more digit).
Then $\ln(500/0.108) = 8.440$ , $a = 1.2657\cdot(8.440 - 0.75 + 1.616)$ $= 11.779$ , and $b = UD = 6.936\cdot10^{-3}$ .
Take a bottomhole pressure of 15 MPa. Then $\Delta m = 27{,}699 - 16{,}573 = 11{,}126$ , and $a^2 + 4b\Delta m = 447.4$ , whose square root is 21.152. So $q = (21.152 - 11.779)/(2b)$ $= 676$ thousand m³/d. Without the inertial losses it would be $\Delta m/a = 945$ thousand m³/d.
For the absolute open flow, $\Delta m_{max} = 27{,}699 - 0.8 = 27{,}698$ . This gives $q_{max} = 1322$ thousand m³/d. Without the inertial losses it would be $2351$ .
At the 15 MPa flow period the inertial skin, $Dq = 3.70$ , is comparable with $a/U = \ln(r_e/r_w) - 0.75 + S = 9.31$ .
If we draw a straight line through this single flow period, as for a liquid ( $q$ proportional to $\Delta m$ ), we get $q_{max} = 27{,}698\cdot676/11{,}126$ $= 1683$ thousand m³/d.

Answer: At a bottomhole pressure of 15 MPa the well will give about 676 thousand m³/d. The absolute open flow is about 1322 thousand m³/d.

What it means. The inertial losses reduce the rate at a drawdown of 5 MPa by 28%, and they reduce the absolute open flow by 44%. The larger the drawdown, the larger their effect. A straight line through a single flow period would overestimate the absolute open flow by 27%. A calculation without inertial losses would overestimate it by a factor of 1.8. The absolute open flow is only a conventional measure of potential. The real rate is limited by the tubing, by sand production and by water cones (water drawn up towards the well in the shape of a cone; Nodal analysis of the production system). The data in this example are assumed for illustration.

CalculatedThe calculated stabilised inflow performance line of the lesson's well (average pressure 20 MPa), with inertial losses and without them (by Darcy's law). The points mark the rate at a bottomhole pressure of 15 MPa and the absolute open flows.

What increases the inertial losses at a well

The formulas for $D$ above apply to an open hole in an undamaged reservoir. Anything that raises the velocity at the wall, or raises $\beta$ where the velocity is highest, increases $D$ — usually several times over.

A damaged zone. The permeability in it is lower, so $\beta$ , which is roughly proportional to $k^{-1.2}$ , is higher. And the velocity is highest right at the wall. Suppose that in a ring of radius $r_s$ the permeability is reduced to $k_s$ and the inertial resistance coefficient has grown to $\beta_s$ . If we repeat the integration over the two rings, we get:

Apparent skin with a damaged zone

S' = \left(\frac{k}{k_s} - 1\right)\ln\frac{r_s}{r_w} + Dq,\qquad D = D_0\left[\frac{\beta_s}{\beta}\left(1 - \frac{r_w}{r_s}\right) + \frac{r_w}{r_s}\right]

where:

$k_s,\ \beta_s$ permeability and inertial resistance coefficient of the damaged zone, mD, m⁻¹
$r_s$ radius of the damaged zone, m
$D_0$ non-Darcy coefficient of the same well without damage, (thousand m³/d)⁻¹

Units: any consistent units; $D$ and $D_0$ in the same units

Holds when

The damaged zone is a cylinder around the wellbore, and the flow in it is steady-state. The first term is the Hawkins formula (Skin factor).
$r_e \gg r_s$ , and the hole is open over the whole thickness.

Suppose the permeability in a zone of radius 0.5 m around a well of radius 0.108 m is reduced fivefold. By the correlation, $\beta$ grows by a factor of $5^{1.2} \approx 6.9$ , and $D$ grows by a factor of about 5.6. The damage skin and the inertial skin grow together.

Partial penetration. Suppose only an interval $h_w$ out of the whole thickness $h$ is working. Then the velocity at the wall is $h/h_w$ times higher. The damage skin, referred to the whole thickness, grows $h/h_w$ times. The inertial losses grow as the square of the velocity, that is, $(h/h_w)^2$ times. A pseudo-skin of flow convergence (an extra skin that comes from the flow bending towards the open interval), $S_{pp}$ , adds to them (Partial penetration and well deviation):

Apparent skin with partial penetration

S' = S_d\,\frac{h}{h_w} + S_{pp} + D_0\left(\frac{h}{h_w}\right)^{2} q

where:

$S_d$ damage skin of the penetrated interval (the part that is open), by the Hawkins formula, —
$h,\ h_w$ reservoir thickness and the working (not merely perforated) interval, m
$S_{pp}$ pseudo-skin of partial penetration, —
$D_0$ non-Darcy coefficient of a well that penetrates the whole thickness, (thousand m³/d)⁻¹

Units: no units; $D_0 q$ in consistent units

Holds when

The damaged zone is thin compared with the interval, and the flow near the wellbore is radial towards the interval.
The inertial losses in the zone of flow convergence, farther from the wellbore, are small compared with the losses at the wall.

If a third of the thickness is open, the inertial skin is nine times larger. So a high $D$ together with a small $S$ often means that only part of the interval is working. Production logging in a producing well shows the working interval.

Perforation. The flow squeezes into individual tunnels, so the velocity in and around the tunnels is much higher than at an open hole. The inertial losses in the tunnels, especially those packed with gravel, are a noticeable part of $D$ in cased wells (wells lined with steel casing; see Effect of completion on inflow).
A hydraulic fracture increases the inflow area and makes the $D$ of the well several times smaller. But the inertial losses in the proppant pack reduce the effective conductivity of the fracture itself (Hydraulic fracturing: purpose and well selection).
Two-phase flow. Condensate that drops out of the gas, or water near the wellbore, narrows the channels for gas and increases both $S$ and $D$ .

Terms and symbols

Non-Darcy (inertial) flow: Flow in which an inertial resistance, proportional to the square of the velocity, adds to the viscous resistance of Darcy's law. It is noticeable in gas wells and high-rate wells.
Forchheimer equation: The two-term law of filtration, $-\dd p/\dd x = \mu v/k + \beta\rho v^2$ .
Inertial resistance coefficient $\beta$: A property of the rock, with units of one over length. It increases as permeability decreases, and we measure it on core at several flow rates.
Forchheimer number Fo: The ratio of the inertial resistance to the viscous resistance, $k\beta\rho v/\mu$ . At the wall of the well it equals the inertial skin $Dq$ .
Non-Darcy coefficient $D$: The inertial skin per unit of rate. For a gas well with an open, undamaged hole it is $\beta\rho_{sc}k/(2\pi h\mu_w r_w)$ .
Apparent skin $S'$: The skin measured in a test at a single rate: $S' = S + Dq$ .
Flow resistance coefficients $a$ and $b$: The coefficients of the two-term gas inflow equation $\Delta m = aq + bq^2$ : the viscous and the inertial resistance.
Isochronal test: A test with flow periods of equal length, with shut-ins between them and one long flow period. It is used when stabilisation takes too long.

Common mistakes and misconceptions

Wrong

“Non-Darcy skin is turbulence in the reservoir.”

In fact

The flow in the pores stays laminar. The losses come from the fluid speeding up and slowing down in channels whose width keeps changing. The word “turbulent” in the names is historical.

Wrong

“From the build-up curve after flow at a high rate, the skin of the gas well is +5 — the near-wellbore zone is badly damaged.”

In fact

This is the apparent skin $S + Dq$ . First we separate it by testing at several rates. In the lesson's example, the skin at the highest flow period is +5.2, but the rate-independent skin is about 1.6.

Wrong

“The absolute open flow can be estimated from a single rate by extending the inflow line according to Darcy's law.”

In fact

The inertial losses grow as the square of the rate, so such an estimate is too high. For the lesson's well, a straight line through the flow period at 15 MPa gives a quarter more, and a calculation with no inertial losses at all gives 1.8 times more. We need several flow periods and the two-term equation.

Wrong

“ $D$ is a constant of the reservoir: calculate it from the correlation and do not test.”

In fact

$D$ is a characteristic of the well. It depends on the perforation, the working interval and damage, and it changes with time. An estimate from the correlation for an open, undamaged hole usually comes out lower than the $D$ of a real well.

Wrong

“The reservoir density of the gas and the viscosity at reservoir pressure are put into the formula for $D$ of a gas well.”

In fact

The mass flow rate is constant. So we take the density at standard conditions together with the rate at standard conditions. We take the viscosity at the bottomhole pressure, where the inertial losses are concentrated.

In depth: advanced detail

Derivation of the two-term equation for gas

We write the Forchheimer equation for radial gas inflow using the mass flow rate $\dot m = \rho_{sc}q$ , which is the same at all radii. The Darcy velocity is $v = \dot m/(2\pi rh\rho)$ . We substitute it and multiply both sides by $\rho/\mu$ :

Forchheimer equation for gas in terms of the mass flow rate

\frac{\rho}{\mu}\,\frac{\dd p}{\dd r} = \frac{\dot m}{2\pi kh}\cdot\frac{1}{r} + \frac{\beta\,\dot m^2}{(2\pi h)^2\mu}\cdot\frac{1}{r^2}

where:

$\dot m$ mass flow rate of the gas, kg/s
$\rho = pM/(ZRT)$ gas density at pressure $p$ , kg/m³
$\mu$ gas viscosity at pressure $p$ , Pa·s

Units: SI

Holds when

The inflow is isothermal and steady-state, to a well that is an open hole over the whole thickness.

The left-hand side equals $\dfrac{M}{RT}\cdot\dfrac{p}{\mu Z}\,\dfrac{\dd p}{\dd r}$ , that is, $\dfrac{M}{2RT}\,\dfrac{\dd m}{\dd r}$ . This is where the pseudo-pressure comes from. In the second term on the right, the viscosity changes with pressure. But this term is concentrated at the wall, so we can take the viscosity in it at the bottomhole pressure. We integrate from $r_w$ to $r_e$ and use $\dot m = \rho_{sc}q$ and $\rho_{sc} = p_{sc}M/(RT_{sc})$ . This gives:

Steady-state gas inflow with inertial losses

m(p_e) - m(p_w) = \frac{p_{sc}T\,q}{\pi kh\,T_{sc}}\left(\ln\frac{r_e}{r_w} + \frac{\beta\rho_{sc}k}{2\pi h\mu_w r_w}\,q\right)

where:

$m(p_e),\ m(p_w)$ pseudo-pressure at the outer boundary and at the wall of the well, Pa²/(Pa·s)
$\mu_w$ gas viscosity at the bottomhole pressure, Pa·s

Units: SI; pressures absolute

Holds when

The term $1/r_e$ in the inertial part is dropped ( $r_e \gg r_w$ ).

The expression in brackets is the logarithmic term plus $Dq$ , with the coefficient $D$ of a gas well. For pseudo-steady-state inflow, we replace $\ln(r_e/r_w)$ by $\ln(r_e/r_w) - 3/4$ and add the skin $S$ . This gives the two-term equation of the main section.

Inertial skin with changing rates and in transient flow

The inertial pressure drop, like the pressure drop across a skin, appears and disappears together with the rate, because the thin ring at the wellbore does not store fluid. So when we superpose the flow periods (Superposition principle; we add up the effects of each rate change), the transient part is the sum of the contributions of all the rate changes. The skin and the inertial term relate only to the current rate $q_n$ :

Bottomhole pressure of a gas well after several rate changes

m(p_i) - m(p_{wf}) = U\left[\sum_{j=1}^{n}(q_j - q_{j-1})\,\tfrac12\left(\ln t_{D,j} + 0.809\right) + (S + Dq_n)\,q_n\right]

where:

$p_i$ initial reservoir pressure, MPa
$q_j$ rate of the $j$ -th flow period, $q_0 = 0$ , thousand m³/d
$t_{D,j}$ dimensionless time from the start of the $j$ -th flow period: $3.553\cdot10^{-3}\,k(t - t_{j-1})$ divided by $\phi\mu_i c_{ti}r_w^2$ , —
$U$ $1.290\,T/(kh)$ in metric units, MPa²/(mPa·s) per thousand m³/d
$\mu_i,\ c_{ti}$ gas viscosity and total compressibility at the initial pressure, mPa·s, MPa⁻¹

Units: metric oilfield; $t$ in h, $k$ in mD, $\phi$ as a fraction, $r_w$ in m

Holds when

The reservoir is infinite, and the flow is radial (the logarithmic approximation).
The product $\mu c_t$ is taken at the initial pressure. At large drawdowns, pseudo-time (a modified time that allows for the changing gas properties) is used instead of time.

The term $(S + Dq_n)q_n$ is non-linear in the rate (the rate appears squared), so we cannot split it over the flow periods as we do the logarithmic terms. Two practical consequences follow. First, the skin from each pressure record is the $S + Dq$ of the flow period in which it was recorded. Second, when the well is shut in, the bottomhole pseudo-pressure rises at once by $U(S + Dq_n)q_n$ .

The β coefficient of a packed bed of grains: gravel and proppant

The Ergun equation, known from chemical engineering, describes flow through a packed bed of grains — a gravel pack, proppant, loose sand — well. It has the form of the Forchheimer equation. If we compare the terms, we get the permeability and $\beta$ from the grain diameter $d_p$ and the porosity (the share of the volume that is pore space):

Permeability and the β coefficient of a packed bed of grains

k = \frac{\phi^3 d_p^2}{150\,(1 - \phi)^2},\qquad \beta = \frac{1.75\,(1 - \phi)}{\phi^3 d_p} = \frac{1.75\,d_p}{150\,(1 - \phi)\,k}

where:

$d_p$ mean diameter of the grains (by the ratio of volume to surface), m
$\phi$ porosity of the pack, fraction
$k,\ \beta$ permeability and inertial resistance coefficient, m², m⁻¹

Units: SI

Holds when

The pack is unconsolidated (loose), and the grains are of similar size. The constants 150 and 1.75 are empirical (found from experiments).

For a given pack, $\beta k$ is proportional to the grain size. For example, for gravel with 1 mm grains and a porosity of 0.35, the permeability is about 700 D and $\beta$ is about $3\cdot10^4$ m⁻¹. The velocities in perforation tunnels packed with gravel are so high that, even with such a small $\beta$ , the inertial losses in them are significant.

Connections to other topics

Skin factor — the skin and its components; the inertial component grows with the rate.
Steady-state radial flow — the Dupuit equation, gas inflow and pseudo-pressure, on which the derivation is built.
Gas formation volume factor, compressibility and viscosity — the viscosity, $Z$ and pseudo-pressure of the lesson gas.
Inflow performance relationship (IPR) — the two-term equation with squared pressures and the equation with an exponent.
Partial penetration and well deviation — the pseudo-skin of partial penetration and of an inclined wellbore, the term $S_{pp}$ . It is added to the inertial skin, which has grown by a factor of $(h/h_w)^2$ .
Pressure build-up analysis — how the skin of each flow period is found from the build-up curve.
Superposition principle — superposition of flow periods with a changing rate.
Effect of completion on inflow — inertial losses in perforation tunnels and gravel packs.
Nodal analysis of the production system — how the inflow performance line of a gas well, together with the lift (the flow up the well through the tubing), sets the rate.

Check yourself

1
In a test of a gas well at two rates, the apparent skin is 3.1 at 300 thousand m³/d and 4.9 at 600 thousand m³/d. What are the rate-independent skin $S$ and the coefficient $D$ ?
1. A. $S$ = 1.3; $D$ = 0.006 (thousand m³/d)⁻¹
2. B. $S$ = 3.1; $D$ = 0
3. C. $S$ = 4.9; $D$ = 0.003 (thousand m³/d)⁻¹
4. D. $S$ = 0; $D$ = 0.0082 (thousand m³/d)⁻¹
Show the answer
Correct answer: A. $S$ = 1.3; $D$ = 0.006 (thousand m³/d)⁻¹
The slope is $D = (4.9 - 3.1)/(600 - 300) = 0.006$ (thousand m³/d)⁻¹, and the intercept is $S = 3.1 - 0.006\cdot300 = 1.3$ . At 600 thousand m³/d the inertial skin is 3.6, which is more than the damage skin.
2
What share of the inertial pressure losses falls on the ring between the wall of the well and a distance of five well radii (the drainage area is much larger)?
1. A.about 95%
2. B.about 80%
3. C.about 50%
4. D.about 20%
Show the answer
Correct answer: B. about 80%
The share of the inertial losses within $r$ is about $1 - r_w/r = 1 - 1/5 = 0.8$ . The viscous losses in the same ring are only $\ln 5/\ln(r_e/r_w)$ , which is about 20% when $r_e/r_w$ is several thousand.
3
How will the coefficient $D$ and the logarithmic term of the inflow equation change if the well is drilled with a bit (the cutting tool) of twice the diameter and everything else stays the same?
1. A. $D$ will be halved, the logarithmic term will decrease by about 0.7
2. B.both will be halved
3. C. $D$ will not change, the logarithmic term will decrease by 0.7
4. D. $D$ will decrease fourfold
Show the answer
Correct answer: A. $D$ will be halved, the logarithmic term will decrease by about 0.7
$D$ is inversely proportional to $r_w$ : the velocity at a wall of twice the area is half as high. The logarithmic term decreases by $\ln 2 \approx 0.69$ . At a value of about 8, that is less than 10%. So the radius of the wellbore matters much more for the inertial losses than for the viscous ones.
4
Why do the gas density at standard conditions and the viscosity at the bottomhole pressure enter the formula for $D$ of a gas well?
Show the answer
The inertial term is proportional to $\rho v^2$ . For gas, $\rho$ and $v$ both change with pressure. But their product, the mass flux, depends only on the area for a given mass flow rate. So $\rho v^2 = (\rho v)^2/\rho$ . When we integrate through the pseudo-pressure, what remains is the mass flow rate $\rho_{sc}q$ , that is, the density at standard conditions together with the rate at standard conditions. The viscosity stays in the inertial term. That term is concentrated at the wall of the well, so we take the viscosity at the bottomhole pressure.
5
An oil well has $k$ = 500 mD, all 10 m of the thickness open, $\mu$ = 0.5 mPa·s, oil density in the reservoir 750 kg/m³, $B$ = 1.3, $r_w$ = 0.108 m and a rate of 2000 m³/d. Do we need to take the inertial losses into account? And what if only a quarter of the thickness were working?
Show the answer
By the correlation, $\beta = 7.64\cdot10^{10}/500^{1.201}$ $\approx 4.4\cdot10^7$ m⁻¹. By the metric formula for a liquid, the numerator is $1.818\cdot10^{-18}\,\beta\rho Bk \approx 3.9\cdot10^{-5}$ and the denominator is $h\mu r_w = 0.54$ . So $D \approx 7.2\cdot10^{-5}$ (m³/d)⁻¹ and $Dq \approx 0.14$ , which is about 2% of the logarithmic term. This is negligible. If a quarter of the thickness were working, $D$ would be 16 times larger and $Dq \approx 2.3$ . The pseudo-skin of partial penetration would be added to it, and that is already a noticeable part of the resistance.
6
Why are all the short flow periods of an isochronal test of equal length, and why do we need one long flow period?
Show the answer
While the inflow is transient, the coefficient $a$ contains a logarithmic term that grows with time. For flow periods of equal length that start from the same state (after the pressure has built up), this term is the same. So their points, plotted as $\Delta m/q$ against $q$ , lie on a straight line with the correct slope $b$ . The short flow periods do not give the stabilised coefficient $a$ . We find it from one long flow period that is carried to stabilisation: we draw a line through its point with the same slope. Or we calculate it from $kh$ and $S$ for the known drainage area.

Updated: 26 October 2026

Quick refresher

After this lesson you can

Why Darcy's law stops working

The inertial resistance coefficient

Inertial losses at the well: a skin that grows with the rate

Gas inflow: the two-term equation in pseudo-pressures

Tests at several rates: how to separate SSS and DDD

The inflow performance line of a gas well and the absolute open flow

What increases the inertial losses at a well

Terms and symbols

Common mistakes and misconceptions

In depth: advanced detail

Connections to other topics

Check yourself

Tests at several rates: how to separate $S$ and $D$