Multiphase flow in tubing

Quick refresher

The pressure in the tubing grows with depth because of three parts: the weight of the mixture, friction and acceleration. In oil wells the weight usually dominates, and you can almost always ignore acceleration.
Gas overtakes the liquid. This is called slip. So there is more liquid in the pipe than in the incoming flow: the liquid holdup $H_L$ is greater than the no-slip holdup $\lambda_L$ . A calculation without slip makes the mixture too light, so it gives a bottomhole pressure that is too low.
Take all volumes at tubing conditions. Gas at a pressure of 8 MPa takes up tens of times less space than at the surface. Oil with dissolved gas takes up more space than in the stock tank (the tank at the surface).
As the fluid rises, the pressure falls and there is more and more free gas. The flow pattern changes: bubble, slug, churn, annular.
Gas makes the column lighter, but it also speeds up the mixture and raises friction. So for each rate and tubing diameter there is a gas–liquid ratio at which the bottomhole pressure is lowest.
The outflow curve for a gassy liquid has a minimum. At low rates gas slips through the liquid. At high rates friction grows. No calculation method is universal: choose one to suit the type of well, and check it against pressure measurements.

Key relationships

\frac{\dd p}{\dd z} = \rho_s\,g\cos\theta + \lambda\,\frac{\rho_{ns}\,v_m^{2}}{2d}

pressure gradient without the acceleration component; $z$ is the depth along the hole

v_{SL} = \frac{q_L}{A}, \quad v_{SG} = \frac{q_G}{A}, \quad v_m = v_{SL} + v_{SG}

superficial velocities and the velocity of the mixture; the rates are at tubing conditions

\lambda_L = \frac{v_{SL}}{v_m} \le H_L

no-slip and actual liquid holdup in upward flow

\rho_s = \rho_L H_L + \rho_G\,(1 - H_L)

density of the mixture in the pipe with slip taken into account

1 - H_L = \frac{v_{SG}}{C_0\,v_m + v_d}

drift-flux model: gas moves faster than the mixture

Common pitfalls

Calculating the mixture density from the no-slip phase fractions, that is, without slip. The bottomhole pressure will come out too low, sometimes by a third.
Putting gas and liquid rates at standard conditions into the formulas instead of rates at tubing conditions.
In a deviated well, using the length of the hole instead of the true vertical depth in the gravity component.
Using a correlation outside the conditions it was built for, for example a horizontal-flow one for a vertical well, or an “oil well” one for a gas-condensate well.
Mixing up the Darcy and Fanning friction factors. They differ by a factor of four.

After this lesson you can

split the pressure gradient (the pressure change per metre) in the tubing into its parts — the weight of the mixture, friction and acceleration — and explain which of them dominates, and when
tell apart the superficial and the actual velocities of the phases, and the no-slip and the actual liquid holdup (the share of the pipe filled with liquid), and calculate them at a pipe cross-section from surface rates and PVT properties (how the fluids change with pressure and temperature)
describe the flow patterns (the ways gas and liquid are arranged in the pipe) of gas–liquid mixtures in vertical, inclined and horizontal pipes, and explain what they depend on
explain how empirical correlations (formulas fitted to measurements) differ from mechanistic models (models built from the physics of the flow), and how to choose and check a calculation method
calculate the pressure distribution section by section, build the outflow curve from it, and explain its minimum

What the pressure drop in the tubing is made of

To lift fluid from the bottomhole to the wellhead, the bottomhole pressure must be higher than the wellhead pressure by all the losses on the way. It helps to think about the pressure gradient: how much the pressure increases for each metre of depth. In a gas–liquid flow the gradient has three parts. The first is the weight of the column of mixture (the gravity, or elevation, component). The second is friction against the pipe walls. The third is the acceleration of the mixture, whose volume grows as it rises.

Pressure gradient in upward multiphase flow

\frac{\dd p}{\dd z} = \rho_s\,g\cos\theta + \lambda\,\frac{\rho_{ns}\,v_m^{2}}{2d} + \left(\frac{\dd p}{\dd z}\right)_{\text{acc}}

where:

$p$ pressure, Pa
$z$ depth along the hole (along the tubing), measured from the wellhead downwards, m
$\rho_s$ density of the mixture in the pipe, allowing for slip of the phases, kg/m³
$g$ acceleration due to gravity, 9.81 m/s², m/s²
$\theta$ inclination of the hole — the deviation from the vertical; $\cos\theta\,\dd z$ is the increment of true vertical depth, —
$\lambda$ Darcy–Weisbach friction factor, —
$\rho_{ns}$ density of the mixture without slip — from the no-slip phase fractions, kg/m³
$v_m$ velocity of the mixture — the sum of the superficial velocities of the phases, m/s
$d$ inner diameter of the tubing, m
$(\dd p/\dd z)_{\text{acc}}$ acceleration component, Pa/m

Units: SI: the gradient in Pa/m; 1 kPa/m = 0.0442 psi/ft

Holds when

Steady upward flow in a pipe of constant cross-section.
The properties of the phases and the velocities are taken at the average pressure and temperature of a section. That is why the pipe is calculated in short sections.
Friction is written in the simplest, homogeneous form (the mixture is treated as uniform): from the no-slip density and the velocity of the mixture. Different methods define the density and the friction factor in different ways.
The acceleration component usually does not exceed a few per cent. It matters only at high gas velocities near the wellhead.

The weight of the mixture depends on how much liquid is in the pipe. Friction depends on the velocity of the mixture. Acceleration depends on how fast this velocity grows. The sum of the gradients over the whole length of the tubing is the difference between the bottomhole pressure and the wellhead pressure that is needed to lift the flow.

In other unit systems (2)

Oilfield units

\left(\frac{\dd p}{\dd z}\right)_{g} = 0.433\,\gamma_s\cos\theta

the gravity component in psi/ft; $\gamma_s$ is the specific gravity of the mixture in the pipe (water = 1).

0.433 psi/ft is the pressure gradient of a column of fresh water.

Metric oilfield units

\left(\frac{\dd p}{\dd z}\right)_{g} = 9.81\cdot10^{-3}\,\rho_s\cos\theta

the gravity component in kPa/m (or MPa per kilometre) with $\rho_s$ in kg/m³.

$9.81\cdot10^{-3}$ is the acceleration due to gravity with the conversion Pa → kPa.

Which component dominates

Flow: Liquid without gas
What dominates: the weight of the column; friction grows with the rate
How it behaves along the hole: the gradient is almost constant, and the pressure increases with depth along a straight line

Flow: Gas with a small amount of liquid
What dominates: friction; the weight is small
How it behaves along the hole: with depth the weight increases and friction decreases, so the pressure increases almost along a straight line

Flow: Gas–liquid mixture
What dominates: usually the weight; near the wellhead, friction
How it behaves along the hole: as it rises, the mixture becomes lighter and speeds up, and the gradient changes a lot with depth

Flow: Horizontal section, flowline
What dominates: friction and uphill stretches along the terrain
How it behaves along the hole: there is almost no weight; liquid collects in the low spots

Flow	What dominates	How it behaves along the hole
Liquid without gas	the weight of the column; friction grows with the rate	the gradient is almost constant, and the pressure increases with depth along a straight line
Gas with a small amount of liquid	friction; the weight is small	with depth the weight increases and friction decreases, so the pressure increases almost along a straight line
Gas–liquid mixture	usually the weight; near the wellhead, friction	as it rises, the mixture becomes lighter and speeds up, and the gradient changes a lot with depth
Horizontal section, flowline	friction and uphill stretches along the terrain	there is almost no weight; liquid collects in the low spots

This is a qualitative comparison. The calculation shows the ratio of the components in a particular well.

In a deviated (slanted) well, calculate the weight of the mixture from the true vertical depth ( $\cos\theta$ ), and friction from the length of the hole. For the same depth an inclined hole is longer, so there is more friction. Also, in an inclined pipe the gas collects along the upper wall and overtakes the liquid more easily. So there is usually more liquid in the pipe than in a vertical one.

What changes if…

…gas is added — the mixture becomes lighter but moves faster. First the saving in weight dominates, then the growth in friction.
…the tubing diameter is increased at the same rate — friction falls roughly in inverse proportion to the fifth power of the diameter. But the mixture moves more slowly, and slip is stronger.
…the water cut (the share of water in the produced liquid) rises — the column becomes heavier, because water is denser than oil and there is less gas per cubic metre of liquid.
…the wellhead pressure is raised — the gas is compressed more along the whole hole. If the weight of the mixture dominates, the column becomes heavier, and the bottomhole pressure rises by more than the increase at the wellhead. If friction dominates, the lower velocity of the mixture partly offsets this increase.

Slip and liquid holdup

Suppose gas and liquid moved at the same velocity. Then the share of liquid in the pipe would equal its share in the flow, and you could treat the mixture as one uniform liquid. In fact gas is lighter. Bubbles rise through the liquid. In upward flow they also collect closer to the axis of the pipe, where the velocity is higher. So the gas overtakes the liquid. This is slip. The liquid is held back in the pipe, and there is more of it there than in the incoming flow. The fraction of the pipe occupied by liquid is called the liquid holdup.

Superficial velocity of a phase $v_{SL}$ , $v_{SG}$: The flow rate of the phase at tubing conditions divided by the whole cross-sectional area $A$ . It is the velocity at which the phase would flow if it occupied the pipe alone.
Mixture velocity $v_m$: The sum of the superficial velocities: $v_m = v_{SL} + v_{SG}$ .
No-slip liquid holdup $\lambda_L$: The share of liquid in the volumetric flow rate, $\lambda_L = v_{SL}/v_m$ . This would be the share of liquid in the pipe without slip.
Liquid holdup $H_L$: The fraction of the cross-section, and so of the pipe volume, occupied by liquid. In upward flow $H_L \ge \lambda_L$ .
Actual phase velocities: $v_L = v_{SL}/H_L$ and $v_G = v_{SG}/(1 - H_L)$ . Their difference $v_G - v_L$ is the slip velocity.

Mixture density with and without slip

\rho_s = \rho_L H_L + \rho_G\,(1 - H_L), \qquad \rho_{ns} = \rho_L \lambda_L + \rho_G\,(1 - \lambda_L)

where:

$\rho_s$ density of the mixture in the pipe, allowing for slip (the actual density), kg/m³
$\rho_{ns}$ density of the mixture without slip, kg/m³
$\rho_L,\ \rho_G$ densities of the liquid and the gas at tubing conditions, kg/m³
$H_L,\ \lambda_L$ liquid holdup and no-slip liquid holdup, fraction

Units: kg/m³; the fractions are dimensionless

Holds when

The liquid density is the average of the oil and water densities, weighted by their volumes at tubing conditions.
Within a section the phases are mixed well enough that the density can be averaged by volume.

The actual density $\rho_s$ sets the weight of the column, so the main part of the gradient depends on $H_L$ . The no-slip density $\rho_{ns}$ is lower. Simple methods use it to calculate friction.

You cannot find the liquid holdup from the rates alone. You need a model or an empirical relationship. The simplest model that rests on physics is the drift-flux model. In it, the gas moves at the velocity of the mixture multiplied by a distribution coefficient, plus the drift velocity, which is how fast the bubbles rise relative to the mixture.

Drift-flux model for upward flow

v_G = C_0\,v_m + v_d \;\Rightarrow\; 1 - H_L = \frac{v_{SG}}{C_0\,v_m + v_d}, \qquad v_d \approx 0.35\sqrt{\frac{g\,d\,(\rho_L - \rho_G)}{\rho_L}}

where:

$v_G$ actual velocity of the gas, m/s
$C_0$ distribution coefficient: allows for the gas being concentrated near the axis of the pipe, where the velocity is higher; about 1.2 for slug flow in a vertical pipe, —
$v_d$ drift velocity — the rise of gas relative to the mixture, m/s
$d$ inner diameter of the pipe, m

Units: SI

Holds when

Upward bubble or slug flow in a vertical pipe. The model does not suit annular flow, and it overestimates the liquid holdup there.
$C_0$ and the coefficient 0.35 are empirical constants (found from experiments). The formula for $v_d$ describes the rise of large bubbles that fill almost the whole cross-section. Small bubbles rise more slowly: in water at about 0.2–0.25 m/s.
In inclined pipes the drift velocity depends on the angle.

Without slip ( $C_0 = 1$ , $v_d = 0$ ), the gas fraction in the pipe would equal the no-slip fraction, $1 - \lambda_L$ . The drift of the bubbles and their concentration near the axis make the gas faster than the mixture, so there is more liquid in the pipe. The effect is strongest at low mixture velocities. When $v_m$ is close to $v_d$ , the gas almost “bubbles through” a standing liquid.

Worked exampleVelocities, liquid holdup and the gradient at a tubing cross-section

Given (assumed data)

Cross-section: the depth where the pressure is 8.0 MPa (absolute) and the temperature is 60 °C; tubing inner diameter 62 mm
Surface rates: oil 150 m³/d, water 50 m³/d, producing gas–oil ratio 120 m³/m³
PVT properties at 8.0 MPa and 60 °C: solution gas–oil ratio 45 m³/m³, $B_o = 1.18$ , $B_w = 1.02$ , water density 1030 kg/m³
At standard conditions: oil density 850 kg/m³; gas specific gravity 0.70 (0.844 kg/m³); $Z = 0.90$
Model: drift flux with $C_0 = 1.2$ ; $\lambda = 0.022$

Solution

Free gas is the gas that is not dissolved in the oil: $(120 - 45)\cdot 150 = 11{,}250$ m³/d at standard conditions. The gas formation volume factor is $B_g = (p_{sc}/p)\,(T/T_{sc})\,Z$ $= 0.01295$ m³/m³. (Pressures are absolute, temperatures are in kelvins, and the standard conditions are $p_{sc}$ = 0.101325 MPa and $T_{sc}$ = 293.15 K. $Z$ is the gas deviation factor.) In the tubing the gas takes up 145.7 m³/d, which is 77 times less.
Liquid in the tubing: $150\cdot 1.18 + 50\cdot 1.02 = 228$ m³/d. The density of the oil with dissolved gas is $(850 + 45\cdot 0.844)/1.18$ $= 753$ kg/m³. The density of the liquid as a whole is 815 kg/m³. The gas density is $0.844/B_g = 65.1$ kg/m³.
The pipe cross-section is $A = 0.00302$ m². The superficial velocities are $v_{SL} = 0.874$ and $v_{SG} = 0.559$ m/s, the mixture velocity is $v_m = 1.433$ m/s, and the no-slip liquid holdup is $\lambda_L = 0.610$ .
Drift: $v_d = 0.262$ m/s. The actual gas fraction is $1 - H_L = 0.559/(1.2\cdot1.433 + 0.262)$ $= 0.282$ , so $H_L = 0.718$ . The gas moves at 1.98 m/s and the liquid at 1.22 m/s.
The mixture density with slip is $\rho_s = 603$ kg/m³, and without slip it is $\rho_{ns} = 522$ kg/m³.
Gradient: weight $\rho_s g = 5.92$ kPa/m, friction 0.19 kPa/m, total 6.11 kPa/m (0.270 psi/ft). Without gas it would be 7.99 kPa/m. A calculation without slip would give a weight of 5.12 kPa/m.

Answer: Liquid fills 72% of the cross-section, against a no-slip share of 61%. The pressure increases by 6.11 kPa for every metre of depth.

What it means. The gas lowered the gradient by about 24% compared with a column of liquid. But because of slip, the mixture in the pipe is 15% heavier than if the phases moved at the same velocity. At this mixture velocity, friction is only 3% of the gradient. Suppose we had taken the gas at its volume at the surface. The superficial velocity of the gas would then come out 77 times too high, and such an error ruins the whole calculation. The data in this example are assumed for illustration.

Flow patterns

Gas and liquid can be arranged in the pipe in different ways, and both slip and friction depend on this. The typical arrangements of the phases are called flow patterns (or flow regimes). A calculation method first finds the pattern. Then it calculates the liquid holdup and the pressure losses for that pattern.

SchematicFlow patterns of upward gas–liquid flow in a vertical pipe (schematic, not to scale). From left to right the gas fraction and the gas velocity increase.

Bubble flow: small gas bubbles are spread through a continuous liquid. There is little gas, and slip comes from the rise of the bubbles.
Slug flow: bubbles merge into large bullet-shaped gas pockets (Taylor bubbles) that fill almost the whole cross-section. Liquid slugs with small bubbles separate the pockets. Around each pocket the liquid runs down as a film. This is the most common pattern in oil wells.
Churn flow: at a higher gas velocity the Taylor bubbles break up. The flow becomes chaotic, and the liquid moves now up, now down.
Annular flow: gas flows fast in the centre of the pipe. The liquid flows as a film on the wall and as droplets in the gas core. This is typical of gas and gas-condensate wells (wells that produce gas together with a light liquid). At an even higher gas velocity, almost all the liquid turns into droplets. This is dispersed, or mist, flow.

One well usually has several flow patterns. At the bottom of the tubing, undersaturated oil (oil whose pressure is above the bubble point) flows as a single-phase liquid. Above the level where the pressure falls below the bubble point pressure, bubbles appear, then slugs. Near the wellhead there may be so much gas that the flow becomes churn or annular. (The lesson on nodal analysis follows the path of the fluid.)

Flow pattern maps show at which combinations of gas and liquid velocities each pattern occurs. They come from experiments in transparent pipes, or from theoretical conditions for the change from one pattern to another. Their axes are superficial velocities, or dimensionless groups that allow for densities, viscosities, surface tension and diameter. Maps by different authors differ noticeably, especially near the boundaries. This is one reason why different calculation methods give different pressures.

Inclined and horizontal pipes. When a pipe leans away from the vertical, the gas collects along the upper wall, and the phases separate more easily. In horizontal pipes, stratified (layered) patterns appear. The liquid flows along the lower part of the cross-section and the gas above it, with a smooth or wavy surface between them. At higher velocities, intermittent patterns arise: elongated bubbles and slugs. Then come annular flow and dispersed-bubble flow. Close to horizontal, the flow is very sensitive to the angle: a change of slope of one or two degrees up or down can change the pattern. This matters for horizontal wells (Horizontal wells) and for pipelines laid over uneven terrain, where liquid collects in the low spots and slugs form.

Correlations and mechanistic models

We cannot describe exactly the motion of a mixture whose flow patterns keep changing. So calculation methods simplify the physics, and they simplify it in different ways. People usually sort them into families by how they treat slip and flow patterns.

Families of methods for calculating multiphase flow in pipes

Family: Homogeneous
Slip and flow patterns: the mixture is treated as one liquid, $H_L = \lambda_L$ ; an empirical coefficient (found from measurements) accounts for the losses
Examples and use: the first methods; today used at high velocities, when slip is small

Family: Empirical, without flow patterns
Slip and flow patterns: $H_L$ from an empirical relationship; the patterns are not told apart
Examples and use: the Hagedorn–Brown correlation with corrections is one of the most widely used for oil wells

Family: Empirical, with flow patterns
Slip and flow patterns: the pattern is read from a map, then separate relationships give $H_L$ and friction
Examples and use: the Duns–Ros, Orkiszewski and Aziz–Govier–Fogarasi correlations; for inclined pipes and pipelines, Beggs–Brill

Family: Mechanistic models
Slip and flow patterns: the pattern and the losses follow from the conservation equations (balances of mass and momentum) and the simplified physics of each pattern
Examples and use: they carry over better to conditions that were not in the experiments: large diameters, inclined holes

Family: Transient models
Slip and flow patterns: equations of motion of each phase, followed in time
Examples and use: slugs in pipelines, start-up and shut-in of wells, unstable gas lift

Family	Slip and flow patterns	Examples and use
Homogeneous	the mixture is treated as one liquid, $H_L = \lambda_L$ ; an empirical coefficient (found from measurements) accounts for the losses	the first methods; today used at high velocities, when slip is small
Empirical, without flow patterns	$H_L$ from an empirical relationship; the patterns are not told apart	the Hagedorn–Brown correlation with corrections is one of the most widely used for oil wells
Empirical, with flow patterns	the pattern is read from a map, then separate relationships give $H_L$ and friction	the Duns–Ros, Orkiszewski and Aziz–Govier–Fogarasi correlations; for inclined pipes and pipelines, Beggs–Brill
Mechanistic models	the pattern and the losses follow from the conservation equations (balances of mass and momentum) and the simplified physics of each pattern	they carry over better to conditions that were not in the experiments: large diameters, inclined holes
Transient models	equations of motion of each phase, followed in time	slugs in pipelines, start-up and shut-in of wells, unstable gas lift

Correlations are named after their authors. In programs they usually come with corrections and tuning coefficients.

In mechanistic models, empirical constants only complete the physical description, like the coefficients $C_0$ and 0.35 in the drift-flux model above. For gas wells that produce a little water or condensate (a light liquid that drops out of the gas), there are separate correlations, for example Gray's. For these wells the minimum rate at which the gas still carries liquid out of the bottomhole also matters (liquid loading).

How to choose a method. There is no universal method. Comparisons with measurements show that different methods for the same well often differ by 10–20% in pressure drop, and by more outside their own range. So choose a method to suit the type of well and the conditions on which it was tested. Then check it against real data: a downhole gauge survey of the pressure along a producing well, or at least the bottomhole pressure at a known rate and wellhead pressure. If the calculation keeps disagreeing with the measurement, tune the model: multiply the calculated liquid holdup or friction by a correction factor. Only then use it for forecasts.

Calculating the pressure distribution section by section

The properties of the mixture change along the pipe. So we divide the tubing into short sections and calculate them one after another. If we need the bottomhole pressure for a given rate, we start at the wellhead and work down. If the bottomhole pressure is known, we start at the bottomhole and work up. It is convenient to give the amount of gas in the flow as the gas–liquid ratio (GLR): the volume of gas at standard conditions per cubic metre of liquid.

Section-by-section calculation

1
Top of the section
The pressure and temperature at the upper end are known. You choose the length of the section, from tens to hundreds of metres.
2
Average pressure
Estimate the pressure at the lower end and take the average over the section. The temperature comes from a calculated or measured profile.
3
Phase properties
At the average pressure and temperature, find the solution gas–oil ratio, the formation volume factors, and the densities and viscosities of the phases.
4
Velocities and flow pattern
From the rates at tubing conditions, find the superficial velocities, the flow pattern and the liquid holdup.
5
Gradient and pressure
Find the weight of the mixture, friction and, if necessary, acceleration. Then find the pressure at the lower end. If it differs noticeably from your estimate, refine the average pressure and repeat the calculation. Then move to the next section.

Worked exampleBottomhole pressure calculated in four sections

Given (assumed data)

Well: vertical; 2000 m of tubing with a 62 mm inner diameter; wellhead pressure 1.5 MPa (absolute)
Flow: 200 m³/d of liquid with a density of 850 kg/m³ and 100 m³ of free gas per cubic metre of liquid (at standard conditions); the gas does not dissolve in the liquid
Gas: specific gravity 0.70, $Z = 0.90$ , average temperature in the tubing 50 °C
Model: drift flux with $C_0 = 1.2$ ; $\lambda = 0.022$ ; acceleration is neglected; sections of 500 m

Solution

Section 0–500 m, first estimate: at the wellhead pressure of 1.5 MPa the gradient is 3.35 kPa/m (worked out in the same way as in steps 2–3). So the pressure at the bottom of the section is $\approx 1.5 + 3.35\cdot 0.5 = 3.18$ MPa (kPa/m × 0.5 km = MPa). The average pressure of the section is 2.34 MPa.
At the average pressure: $B_g = 0.0430$ m³/m³, and the gas in the tubing is $100\cdot 200\cdot B_g = 860$ m³/d. The velocities are $v_{SL} = 0.767$ m/s, $v_{SG} = 3.30$ m/s and $v_m = 4.06$ m/s. The no-slip holdup is $\lambda_L = 0.189$ .
The drift velocity is $v_d = 0.270$ m/s. The holdup is $H_L = 1 - v_{SG}/(1.2\,v_m + v_d)$ $= 0.359$ , which is 1.9 times greater than $\lambda_L$ . The mixture density is $\rho_s = 318$ kg/m³. The weight part of the gradient is 3.12 kPa/m and the friction part is 0.52 kPa/m, so the total is 3.64 kPa/m.
Pressure at a depth of 500 m: $1.5 + 3.64\cdot 0.5 = 3.32$ MPa.
We calculate the remaining sections in the same way (table below). At depths of 1000, 1500 and 2000 m the pressure is 5.51, 8.05 and 10.89 MPa.
Check: a calculation with 5 m sections gives 10.92 MPa. For comparison, a column of liquid without gas needs 18.35 MPa, and a calculation without slip would give 7.37 MPa.

Answer: To lift 200 m³/d at a gas–liquid ratio of 100 m³/m³, about 10.89 MPa is needed at the bottomhole.

What it means. The gas lowered the required bottomhole pressure by 7.4 MPa compared with a column of liquid. A calculation without slip would have given a pressure 3.6 MPa too low, about a third too low. The reason is that the liquid is held back in the pipe, so the mixture is heavier than the no-slip fractions suggest. Four sections gave an almost exact answer, because the pressure does not change very much within each section. Programs use sections tens of metres long and refine the average pressure by repeated trials (iteration). The model is a teaching model. The gas does not dissolve, the temperature and the friction factor are constant, and the flow patterns are not told apart. Near the wellhead the flow is already close to annular, where the drift-flux model overestimates the liquid holdup. The data in this example are assumed for illustration.

Section-by-section calculation: 200 m³/d, gas–liquid ratio 100 m³/m³, 62 mm tubing

Section, m: 0–500
Average pressure, MPa: 2.34
$H_L$: 0.359
Gradient, kPa/m: 3.64
Pressure at the bottom, MPa: 3.32

Section, m: 500–1000
Average pressure, MPa: 4.32
$H_L$: 0.464
Gradient, kPa/m: 4.38
Pressure at the bottom, MPa: 5.51

Section, m: 1000–1500
Average pressure, MPa: 6.70
$H_L$: 0.551
Gradient, kPa/m: 5.09
Pressure at the bottom, MPa: 8.05

Section, m: 1500–2000
Average pressure, MPa: 9.41
$H_L$: 0.621
Gradient, kPa/m: 5.67
Pressure at the bottom, MPa: 10.89

Section, m	Average pressure, MPa	$H_L$	Gradient, kPa/m	Pressure at the bottom, MPa
0–500	2.34	0.359	3.64	3.32
500–1000	4.32	0.464	4.38	5.51
1000–1500	6.70	0.551	5.09	8.05
1500–2000	9.41	0.621	5.67	10.89

$H_L$ and the gradient are at the average pressure of the section. With depth the gas is compressed, there is more liquid in the pipe, and the gradient grows. The data in this example are assumed for illustration.

CalculatedPressure distribution in the tubing of the example well at a rate of 200 m³/d: without gas, at a gas–liquid ratio (GLR) of 100 m³/m³ and at a GLR of 280 m³/m³, which gives the lowest bottomhole pressure. Calculated with the lesson's teaching model.

The more gas there is, the lighter the column and the lower the pressure at the bottomhole, but only up to a limit. Near the wellhead the pressure is low and the gas takes up a large volume. There, at a high GLR, the mixture moves so fast that friction grows faster than the weight falls. In the figure, in the top few hundred metres, the pressure on the curve for the higher GLR increases with depth even faster than at a GLR of 100.

Required bottomhole pressure at different GLRs: 200 m³/d, 62 mm tubing, wellhead pressure 1.5 MPa

GLR, m³/m³: 0
Bottomhole pressure, MPa: 18.35

GLR, m³/m³: 25
Bottomhole pressure, MPa: 15.29

GLR, m³/m³: 50
Bottomhole pressure, MPa: 13.23

GLR, m³/m³: 100
Bottomhole pressure, MPa: 10.92

GLR, m³/m³: 200
Bottomhole pressure, MPa: 9.41

GLR, m³/m³: 280
Bottomhole pressure, MPa: 9.21

GLR, m³/m³: 400
Bottomhole pressure, MPa: 9.45

GLR, m³/m³	Bottomhole pressure, MPa
0	18.35
25	15.29
50	13.23
100	10.92
200	9.41
280	9.21
400	9.45

In the teaching model, the lowest pressure is at a GLR of about 280 m³/m³. At a higher GLR the growth in friction outweighs the lighter column, and the pressure rises again. The real value depends on the calculation method and the data of the well. The data are assumed for illustration.

This minimum is the basis of gas lift. Gas is injected into the tubing to raise the gas–liquid ratio to the value at which the bottomhole pressure is lowest, or close to it. Beyond that point the gas is wasted. The gas–liquid ratio is linked to the producing gas–oil ratio $R$ (the volume of gas produced with each volume of oil) and the water cut $\text{WC}$ (the share of water in the liquid): $\text{GLR} = R\,(1 - \text{WC})$ , because gas arrives only with the oil.

Before computer programs, engineers made such calculations in advance and printed them as pressure traverse curves (also called gradient curves). These are families of “pressure against depth” curves for different GLRs, with one sheet for each tubing diameter, rate and water cut. The bottomhole pressure is found from them by the equivalent-depth method described in the lesson on nodal analysis. Today a program generates such curves for a specific well and its fluid.

The outflow curve and its minimum

Repeat the pressure distribution calculation for several rates. This gives the outflow curve for a node at the bottomhole: the required bottomhole pressure as a function of rate (Nodal analysis of the production system). Where it crosses the inflow performance relationship, you find the operating rate of the well.

CalculatedOutflow curves of the example well at a GLR of 100 m³/m³ for tubing with an inner diameter of 62 and 76 mm (calculated with the lesson's teaching model). The dots mark the minima — at about 150 and 250 m³/d.

For a gassy liquid the outflow curve has a minimum. At low rates the mixture moves slowly, the gas slips through the liquid, and liquid builds up in the tubing. At 30 m³/d in 62 mm tubing, the liquid fills 82% of the cross-section near the bottomhole, against a no-slip share of 57%. As the rate grows, slip weakens and the column becomes lighter. At high rates friction begins to dominate. In 62 mm tubing the minimum is at about 150 m³/d (10.8 MPa), and in 76 mm tubing at about 250 m³/d.

Larger tubing wins at high rates: at 600 m³/d, 76 mm tubing needs 12.4 MPa against 15.8 MPa in 62 mm tubing. At low rates it loses: at 30 m³/d, 14.4 against 13.2 MPa, because in a wide pipe the mixture moves more slowly and slip is stronger.

The operating point should lie to the right of the minimum of the outflow curve. To the left, the operating mode is unstable: the well works in pulses and may die (stability of the operating mode).
Choose the tubing diameter for the whole life of the well. Tubing that suits the high initial rate ends up to the left of the minimum when the rate falls. For the same reason, engineers run smaller-diameter tubing in gas wells where liquid builds up.
A rising water cut raises the outflow curve, because the column becomes heavier and there is less gas per cubic metre of liquid. Gas lift and pumps lower the curve and bring the operating point back (Artificial lift: when it is needed and how to choose the method).
Check an outflow curve calculated from a correlation against measurements on a producing well, and tune it if necessary. Otherwise the error of the method passes into the rate forecast.

Terms and symbols

Superficial velocity of a phase: The flow rate of the phase at tubing conditions divided by the whole cross-sectional area.
Mixture velocity $v_m$: The sum of the superficial velocities of the gas and the liquid.
No-slip liquid holdup $\lambda_L$: The share of liquid in the volumetric flow rate. It is the share of liquid in the pipe if the phases did not slip.
Liquid holdup $H_L$: The fraction of the cross-section (volume) of the pipe occupied by liquid. In upward flow it is greater than $\lambda_L$ .
Slip: The movement of gas faster than the liquid. The slip velocity is the difference between the actual phase velocities.
Flow pattern: The way the phases are arranged in the pipe: bubble, slug, churn, annular, and in horizontal pipes stratified.
Gas–liquid ratio (GLR): The volume of gas at standard conditions per unit volume of liquid. $\text{GLR} = R\,(1 - \text{WC})$ , where $\text{WC}$ is the water cut.
Pressure traverse: The pressure against depth in a producing well, for a given rate, GLR and tubing diameter.

Common mistakes and misconceptions

Wrong

“In the pipe, gas and liquid are in the same proportion as in the incoming flow.”

In fact

Gas overtakes the liquid, so the liquid holdup is greater than the no-slip holdup, and at low velocities much greater. In the example in this lesson, a calculation without slip gave a bottomhole pressure that was too low by about a third.

Wrong

“The more gas there is, the easier it is to lift the liquid.”

In fact

Only up to a certain gas–liquid ratio. Beyond it, the growing friction outweighs the lighter column, and the required bottomhole pressure rises again.

Wrong

“Larger tubing always reduces the losses.”

In fact

At high rates, yes: friction falls. At low rates slip is stronger in a wide pipe, liquid builds up, and the required bottomhole pressure is higher than in a narrow one.

Wrong

“Any well-known correlation will do — they have all been validated.”

In fact

Each has been validated on its own data. Different methods for the same well differ by 10–20% of the pressure drop or more. Choose the method to suit the type of well, and check it against pressure measurements.

Wrong

“We calculate the gas velocity in the tubing from its rate at the surface.”

In fact

In the tubing the gas is compressed, and part of it is dissolved in the oil. Free gas at tubing conditions is $(R - R_s)\,q_o B_g$ , where $R$ is the producing gas–oil ratio, $R_s$ is the solution gas–oil ratio and $q_o$ is the oil rate. At 8 MPa this is tens of times less than at the surface.

In depth: advanced detail

Where the gradient components come from

Take a short section of pipe of length $\dd z$ and cross-sectional area $A$ . The pressure difference between its ends, multiplied by $A$ , balances three things. The first is the weight of the mixture in the section, $\rho_s g A\cos\theta\,\dd z$ . The second is the friction force on the wall, $\tau_w \pi d\,\dd z$ . The third is the change in the momentum of the flow (its mass times its velocity), because the flow speeds up as the gas expands. By the definition of the Darcy factor, the shear stress at the wall is $\tau_w = \lambda\rho v^2/8$ . So the friction per unit length, $4\tau_w/d = \lambda\rho v^2/(2d)$ , is the second term of the gradient equation.

The mass flux of the mixture (its mass flow per unit of cross-sectional area), $\rho_{ns} v_m$ , is constant along the pipe. But the velocity grows, because the superficial velocity of the gas is inversely proportional to the pressure. So we can write the acceleration component in terms of the gradient itself:

Gradient including acceleration

\frac{\dd p}{\dd z} = \frac{\rho_s g\cos\theta + \lambda\,\rho_{ns} v_m^{2}/(2d)}{1 - E_k}, \qquad E_k = \frac{\rho_{ns}\,v_m\,v_{SG}}{p}

where:

$E_k$ dimensionless acceleration number — the share of the pressure drop that goes to accelerating the mixture, —
$p$ absolute pressure at the cross-section, Pa
$v_{SG}$ superficial velocity of the gas, m/s

Units: SI

Holds when

The gas expands isothermally, and its volume is inversely proportional to the pressure.
The liquid is incompressible (its volume does not change with pressure), and the mass flow rate of the mixture is constant along the length.
The momentum flux is calculated for the homogeneous mixture — without slip of the phases.

At the wellhead of the example well in this lesson, $E_k$ is about 0.25%, and at a gas–liquid ratio of 800 m³/m³ it is about 3%. So acceleration can indeed be ignored. When $E_k$ approaches one, the velocity of the mixture approaches the speed of sound in it, and the flow becomes critical. This almost never happens in tubing, but in chokes it is common (Chokes, surface pressure losses and well temperature).

The friction factor

The friction factor depends on the Reynolds number $Re = \rho v d/\mu$ and on the relative roughness of the wall $\varepsilon/d$ . ( $Re$ compares the push of inertia with the drag of viscosity.) In laminar flow (smooth layers; $Re$ less than about 2000), $\lambda = 64/Re$ , and roughness does not matter. In turbulent flow (swirling), find $\lambda$ from the Moody chart or from explicit formulas, for example the Haaland formula:

Darcy friction factor in turbulent flow (the Haaland formula)

\frac{1}{\sqrt{\lambda}} = -1.8\,\log_{10}\!\left[\left(\frac{\varepsilon/d}{3.7}\right)^{1.11} + \frac{6.9}{Re}\right]

where:

$\lambda$ Darcy friction factor, —
$\varepsilon$ absolute roughness of the wall, m
$d$ inner diameter of the pipe, m
$Re$ Reynolds number, —
$\log_{10}$ base-10 logarithm, —

Units: dimensionless quantities; $\varepsilon$ and $d$ in the same units

Holds when

Turbulent flow, $Re$ greater than about 4000.
The formula approximates the implicit Colebrook–White equation (one that must be solved by trial) with an error of the order of 1–2%.

For a smooth pipe at $Re = 10^{5}$ the formula gives $\lambda \approx 0.018$ . With the roughness of new steel tubing it gives about 0.02. At $Re = 2\cdot 10^{4}$ it gives about 0.026. The constant value 0.022 used in the examples of this lesson is typical for tubing. In an exact calculation, $\lambda$ is found again in each section. In multiphase flow the Reynolds number is calculated from the properties of the mixture, for example from the no-slip density and an averaged viscosity. The exact way depends on the calculation method.

The roughness of new steel tubing is of the order of hundredths of a millimetre. Corrosion and deposits of scale (mineral crust), wax and asphaltenes (heavy, sticky oil components) increase it several times over the years. In practice engineers often use roughness as a tuning parameter. They adjust it so that the calculation matches the measurement, but within physically reasonable limits. The Fanning friction factor, which is found in English-language literature, is four times smaller than the Darcy factor. When you take a formula or a value from a handbook, check which factor it means.

Connections to other topics

Nodal analysis of the production system — how the outflow curve, together with the inflow performance relationship, sets the operating mode of a well.
Inflow performance relationship (IPR) — the inflow curve that the outflow curve crosses.
Black oil model: solution gas–oil ratio and formation volume factors — the solution gas–oil ratio, formation volume factors and phase densities needed in every section of the calculation.
Gas lift: principles and choice of operating conditions — how gas injection lowers the required bottomhole pressure.
Chokes, surface pressure losses and well temperature — chokes, flowlines and the flowing temperature.
Well performance sensitivity analysis — trying out tubing diameters, wellhead pressures and water cuts.
Artificial lift: when it is needed and how to choose the method — what to do when the energy of the reservoir is not enough to lift the fluid.
Horizontal wells — friction losses along a horizontal wellbore and the shift of inflow towards the heel.

Check yourself

1
At a pipe cross-section the superficial velocity of the liquid is 0.6 m/s and that of the gas is 0.9 m/s. What is the no-slip liquid holdup, and what will the actual holdup be in upward flow?
1. A.0.40; the actual holdup is greater than 0.40
2. B.0.67; the actual holdup is greater than 0.67
3. C.0.40; the actual holdup is less than 0.40
4. D.0.60; the actual holdup equals 0.60
Show the answer
Correct answer: A. 0.40; the actual holdup is greater than 0.40
$\lambda_L = 0.6/(0.6 + 0.9) = 0.40$ . Gas slips, so there is more liquid in the pipe. The drift-flux model with $C_0 = 1.2$ and $v_d = 0.27$ m/s gives $1 - H_L = 0.9/(1.8 + 0.27) = 0.43$ , that is, $H_L \approx 0.57$ .
2
Which component of the pressure gradient is usually the main one in a vertical oil well at a moderate rate?
1. A.Acceleration
2. B.The weight of the mixture
3. C.All three are roughly equal
4. D.Friction
Show the answer
Correct answer: B. The weight of the mixture
In such wells most of the pressure drop goes to holding up the column of mixture. Friction is noticeable at high rates and near the wellhead, where the gas takes up a large volume. Acceleration usually accounts for only a fraction of a per cent.
3
Why, as the rate of gassy oil is reduced, does the required bottomhole pressure at some point start to rise rather than fall?
Show the answer
At low rates the velocity of the mixture becomes close to the rise velocity of the gas. The gas slips through the liquid, liquid builds up in the tubing, and the actual density of the mixture approaches that of the liquid. The gain from low friction is smaller than the loss from the extra weight of the column. So the outflow curve passes through a minimum, and at still lower rates it rises.
4
A well produces 100 m³/d of oil and 100 m³/d of water at a producing gas–oil ratio of 150 m³/m³. What is the gas–liquid ratio? How much gas must be injected by gas lift to bring it up to 250 m³/m³?
Show the answer
Gas arrives only with the oil: $150\cdot 100 = 15{,}000$ m³/d for 200 m³/d of liquid, so the GLR is $75$ m³/m³. The formula $R(1 - \text{WC}) = 150\cdot 0.5$ gives the same. For a GLR of 250 we need $250\cdot 200 = 50{,}000$ m³/d of gas. So we must inject $50{,}000 - 15{,}000 = 35{,}000$ m³/d (at standard conditions).
5
A well with a declining rate is operating at 50 m³/d of liquid. How will the outflow curve change in the region of this rate if 62 mm tubing is replaced by 76 mm?
1. A.It will not change
2. B.It will move down: friction will be lower
3. C.It will become horizontal
4. D.It will move up: the mixture will move more slowly and slip will increase
Show the answer
Correct answer: D. It will move up: the mixture will move more slowly and slip will increase
At a low rate friction is small anyway. In a wide pipe the velocity of the mixture is even lower, so the gas slips more and liquid builds up. In the lesson's teaching model, at 50 m³/d the 76 mm tubing needs 13.2 MPa, against 12.1 MPa for the 62 mm tubing.
6
The bottomhole pressure calculated with the chosen correlation is 1.5 MPa higher than that measured by a downhole gauge at a known rate and wellhead pressure. What should you do before you use the model for forecasting?
Show the answer
First check the input data: the rates, producing gas–oil ratio, water cut, PVT properties, wellhead pressure, depth and tubing diameter. Then compare several calculation methods that suit such a well. Tune the chosen method with a correction factor on the liquid holdup or on friction, within physically reasonable limits, so that the calculation matches the measurement. Use only the tuned model for forecasts and for choosing equipment.

Updated: 19 October 2026