Nodal analysis of the production system

Quick refresher

Pressure falls from the average reservoir pressure $\bar p_R$ to the separator pressure $p_{sep}$ . At any given moment both ends are fixed. The total of the losses between them depends on the rate, and it equals $\bar p_R - p_{sep}$ at only one rate: the operating rate.
At the node, inflow equals outflow, and there is a single pressure. So the operating point is where two curves cross. The inflow curve shows the pressure at the node that the part of the system before the node can provide. The outflow curve shows the pressure needed to carry the flow on to the separator.
Where you put the node does not change the answer. A node at the bottomhole (the bottom of the well) separates the reservoir, described by its inflow performance relationship (the curve of rate against bottomhole pressure), from the lift system. A node at the wellhead separates the well from the surface facilities.
In a vertical oil well with a moderate rate, most of the pressure goes to hold up the weight of the fluid column in the tubing. Friction grows roughly as the square of the rate, so it limits wells with high rates.
Improvements to single parts of the system do not simply add up. The gain from treating the reservoir depends on the lift system. The gain from larger tubing depends on the inflow.

Key relationships

\bar p_R - \Delta p_{\text{up}}(q) = p_{node} = p_{sep} + \Delta p_{\text{down}}(q)

condition at the node: at the operating rate, inflow and outflow give the same pressure

q = J\,(\bar p_R - p_{wf})

straight-line inflow performance relationship: the inflow curve for a node at the bottomhole (single-phase inflow, that is, oil or water alone)

p_{wf} = p_{wh} + \rho g H + \lambda\,\dfrac{H}{d}\,\dfrac{\rho v^2}{2}

outflow curve for a node at the bottomhole, for liquid without gas in vertical tubing, SI

Common pitfalls

Expecting a higher productivity index (more flow for each unit of pressure drop) to raise the rate in proportion. Friction in the tubing takes part of the gain.
Choosing the tubing for the current operating mode alone. As the rate falls and the water cut rises, tubing that is too large starts to fill up with liquid.
Taking the crossing with the falling (left) branch of the outflow curve as the operating mode. The operation is unstable there.
Mixing volumes at standard conditions (a fixed pressure and temperature at the surface) with volumes at tubing conditions (the real pressure and temperature in the tubing) when calculating the flow velocity in the tubing.

After this lesson you can

list the elements of the production system, from the reservoir to the separator (the surface vessel that splits oil, gas and water), and explain what the pressure loss in each of them depends on
state the conditions at a node and draw the inflow and outflow curves for a chosen node
find the operating point, where the reservoir and the lift system (the tubing that carries the fluid up the well) work together, both on a graph and by calculation
explain how reservoir pressure, skin factor (a number that shows how much the zone right around the well helps or hinders flow), tubing diameter, wellhead pressure and water cut (the share of water in the produced liquid) shift the operating point
choose where to put the node for a given engineering question, and find out what limits the rate: the reservoir or the lift system

The production system: from the reservoir to the separator

Before it reaches the separator, the fluid travels a long way. First it flows through the reservoir rock towards the well. Then it passes the near-wellbore zone and the perforation tunnels (holes made through the casing, the steel lining of the well, into the rock), or the gravel pack (a layer of gravel that holds back sand). It rises up the tubing (the pipe inside the well that carries the fluid up), past the downhole equipment, for example a subsurface safety valve (SSSV, a valve that shuts the well in an emergency). At the wellhead (the top of the well at the surface) it passes the choke, a narrow opening that limits the flow. Then it goes along the flowline (the pipe from the well to the plant) into the separator, where the gas is separated from the liquid. The pressure falls in every section.

SchematicThe production system and its main nodes: 1 — the reservoir (average reservoir pressure

\bar p_R

), 2 — the sandface (the wall of the well), 3 — the bottomhole, at the tubing intake, 4 — the wellhead, just before the choke, 5 — the separator. Arrows show the direction of flow.

At any moment the two ends of the system are fixed. At the start is the average reservoir pressure $\bar p_R$ in the drainage area of the well (the part of the reservoir that feeds the well). It changes slowly, over months and years of field development. At the end is the separator pressure $p_{sep}$ . A pressure controller holds it steady, because the whole processing plant depends on it. Everything between the two ends must fit into this pressure difference.

Pressure budget of the production system

\bar p_R - p_{sep} = \Delta p_{\text{res}} + \Delta p_{\text{comp}} + \Delta p_{\text{tbg}} + \Delta p_{\text{choke}} + \Delta p_{\text{line}}

where:

$\bar p_R$ average reservoir pressure in the drainage area, converted to the depth of the middle of the perforated interval (the part of the casing with holes), MPa
$p_{sep}$ separator pressure, MPa
$\Delta p_{\text{res}}$ pressure loss in the reservoir: $\bar p_R - p_{wfs}$ , where $p_{wfs}$ is the flowing pressure at the sandface (the wall of the well), MPa
$\Delta p_{\text{comp}}$ loss in the near-wellbore zone and the completion (the perforations and gravel pack that connect the well to the rock): $p_{wfs} - p_{wf}$ , where $p_{wf}$ is the bottomhole flowing pressure, MPa
$\Delta p_{\text{tbg}}$ loss in lifting through the tubing, including the weight of the fluid column and the pressure lost in the downhole equipment: $p_{wf} - p_{wh}$ , where $p_{wh}$ is the wellhead pressure, MPa
$\Delta p_{\text{choke}}$ loss across the choke: $p_{wh} - p_{ch}$ , where $p_{ch}$ is the pressure after the choke, MPa
$\Delta p_{\text{line}}$ loss in the flowline and the gathering system: $p_{ch} - p_{sep}$ , MPa

Units: any pressure unit, the same for all terms

Holds when

Steady state (nothing changes with time): the same flow passes through every element, with the volumes given at the same conditions.
Each term depends on the rate and on the fluid properties. So the equality holds at only one rate: the operating rate.
The weight of the fluid column counts as part of the tubing loss. In a vertical well it is usually the largest term.

The pressure difference between the reservoir and the separator is the pressure “budget”. Each element spends its share. If one element spends more, less is left for the others, and the rate falls.

What the pressure loss in each element of the system depends on

Element: Reservoir
The loss is greater if…: permeability (how easily the rock lets fluid through) and thickness are lower, or viscosity (how thick the oil is) is higher; also below the bubble point pressure (the pressure at which the first gas bubbles appear), because gas comes out of solution
How it depends on the rate: in proportion to the rate (Darcy's law); faster than that below the bubble point pressure

Element: Near-wellbore zone and completion
The loss is greater if…: the near-wellbore zone is damaged, the perforation tunnels are few or short, the gravel pack is plugged
How it depends on the rate: in proportion to the rate, plus a part caused by the inertia of fast flow that grows roughly as the square of the rate

Element: Tubing
The loss is greater if…: the well is deeper, the fluid is heavier (more water, less gas), the pipe diameter is smaller
How it depends on the rate: the weight of the column depends little on the rate; friction grows roughly as the square of the rate

Element: Choke
The loss is greater if…: the flow area is smaller
How it depends on the rate: rises sharply with the rate; in critical flow (the fluid moves at the speed of sound in the choke) the rate no longer depends on the pressure after the choke

Element: Flowline
The loss is greater if…: the pipeline is longer and narrower, the route has uphill sections
How it depends on the rate: friction grows roughly as the square of the rate

Element	The loss is greater if…	How it depends on the rate
Reservoir	permeability (how easily the rock lets fluid through) and thickness are lower, or viscosity (how thick the oil is) is higher; also below the bubble point pressure (the pressure at which the first gas bubbles appear), because gas comes out of solution	in proportion to the rate (Darcy's law); faster than that below the bubble point pressure
Near-wellbore zone and completion	the near-wellbore zone is damaged, the perforation tunnels are few or short, the gravel pack is plugged	in proportion to the rate, plus a part caused by the inertia of fast flow that grows roughly as the square of the rate
Tubing	the well is deeper, the fluid is heavier (more water, less gas), the pipe diameter is smaller	the weight of the column depends little on the rate; friction grows roughly as the square of the rate
Choke	the flow area is smaller	rises sharply with the rate; in critical flow (the fluid moves at the speed of sound in the choke) the rate no longer depends on the pressure after the choke
Flowline	the pipeline is longer and narrower, the route has uphill sections	friction grows roughly as the square of the rate

The losses in the reservoir and in the near-wellbore zone together are described by the well's inflow performance relationship.

The shares of the losses differ greatly from well to well. In a vertical oil well with a moderate rate, most of the pressure usually goes to lifting the fluid up the tubing. Almost all of that holds up the weight of the column, and little goes to friction. In wells with high rates, in horizontal wells, and in subsea wells with long flowlines, friction takes the main share. There are no universal “right” proportions. That is why the system is calculated as a whole.

The node, inflow and outflow

Choose a point in the system: the node. The elements before the node (upstream), from the reservoir to the node, form the inflow side. The elements after the node (downstream), from the node to the separator, form the outflow side. In steady state, two conditions hold at the node. As much fluid leaves the node as arrives at it. And the node has a single pressure.

Start from the reservoir pressure and subtract the losses up to the node. This gives the pressure at the node as seen from the inflow side. Now start from the separator pressure and add the losses after the node. This gives the pressure that the node needs, as seen from the outflow side. Both depend on the rate.

Inflow curve to the node

p_{node} = \bar p_R - \Delta p_{\text{up}}(q)

where:

$p_{node}$ pressure at the node, MPa
$\bar p_R$ average reservoir pressure, MPa
$\Delta p_{\text{up}}(q)$ sum of the pressure losses in all elements from the reservoir to the node at rate $q$ , MPa
$q$ flow rate, usually at standard conditions, m³/d

Units: any consistent units of pressure and rate

Holds when

Steady state: the reservoir pressure does not change during the calculation.
All rates are given at the same conditions.

The pressure at the node that the part of the system before the node can provide at each rate. The higher the rate, the bigger the losses before the node, so the inflow curve falls.

Outflow curve from the node

p_{node} = p_{sep} + \Delta p_{\text{down}}(q)

where:

$p_{sep}$ separator pressure, MPa
$\Delta p_{\text{down}}(q)$ sum of the pressure losses in all elements from the node to the separator at rate $q$ , MPa

Units: any consistent units of pressure and rate

Holds when

The separator pressure is held constant.
The losses in each element are worked out for the real composition of the flow: gas, oil and water.

The pressure needed at the node to carry the rate $q$ through all the elements to the separator. The outflow curve usually rises with the rate.

The point where the two curves cross is the operating point of the reservoir and the lift system. It is the one rate at which the part of the system before the node delivers exactly as much as the part after the node can pass, at the same pressure at the node. On a graph, it is the crossing point of the curves. In a calculation, it is the root (the solution) of the equation $\bar p_R - \Delta p_{\text{up}}(q) = p_{sep} + \Delta p_{\text{down}}(q)$ .

Inflow and outflow curves for a node at the bottomhole

Most often the node is put at the bottomhole, at the tubing intake (where the fluid enters the tubing). The inflow curve is then the well's inflow performance relationship (IPR): how the rate depends on the bottomhole flowing pressure $p_{wf}$ , the pressure at the bottom of the well while it produces. For oil or water alone (single phase), the curve is close to the straight line $q = J(\bar p_R - p_{wf})$ . Here $J$ is the productivity index, the rate per unit of pressure drop. The difference $\bar p_R - p_{wf}$ is the drawdown, the pressure drop that pushes fluid into the well. When the bottomhole pressure is below the bubble point pressure, gas comes out of solution from the oil and the curve bends.

For a node at the bottomhole, the outflow curve shows the bottomhole pressure needed to lift the rate $q$ to the wellhead, for a given wellhead pressure $p_{wh}$ (in Russian-language practice, the “buffer pressure”). This pressure has three parts: the wellhead pressure, the weight of the fluid column in the tubing, and the friction losses. The pressure loss from speeding up the flow is usually ignored in wells. For a liquid without free gas (gas that has come out of the oil), all three parts are easy to calculate.

Outflow curve for liquid without gas in vertical tubing

p_{wf} = p_{wh} + \rho g H + \lambda\,\frac{H}{d}\,\frac{\rho v^2}{2},\qquad v = \frac{4q}{\pi d^2}

where:

$p_{wf}$ bottomhole flowing pressure needed to lift the rate $q$ , Pa
$p_{wh}$ wellhead pressure, Pa
$\rho$ density of the liquid in the tubing, kg/m³
$g$ acceleration due to gravity, 9.81 m/s², m/s²
$H$ true vertical depth of the node; in a vertical well equal to the tubing length, m
$\lambda$ Darcy–Weisbach friction factor; in turbulent flow in steel pipes usually 0.015–0.03, —
$d$ inner diameter of the tubing, m
$v$ mean velocity of the liquid in the tubing, m/s
$q$ volumetric flow rate of the liquid at tubing conditions, m³/s

Units: SI

Holds when

Single-phase incompressible liquid (its volume does not change with pressure): the pressure along the whole length of the tubing is above the bubble point pressure, and there is no free gas.
Vertical tubing of constant diameter. In a deviated (slanted) well, the weight of the column uses the true vertical depth, and friction uses the length along the hole.
The density and the friction factor are constant. Speeding up of the flow is ignored.
The flow is turbulent (swirling, not smooth): the Reynolds number $Re = \rho v d/\mu$ is above a few thousand. $\lambda$ depends on $Re$ and on how rough the pipe wall is. English-language literature also uses the Fanning friction factor, which is four times smaller: $f_F = \lambda/4$ .

The weight of the column does not depend on the rate. Friction grows as the square of the velocity. So the outflow curve starts at the level $p_{wh} + \rho g H$ and bends upwards. At the same rate, friction is inversely proportional to the fifth power of the diameter. If the tubing diameter is 1.2 times smaller, the friction loss is greater by a factor of $1.2^5 \approx 2.5$ .

In other unit systems (2)

Metric oilfield units

p_{wf} = p_{wh} + 9.81\cdot10^{-6}\,\rho H + 0.1086\,\frac{\lambda\,\rho\,H\,q^2}{d^5}

$p$ in MPa, $\rho$ in kg/m³, $H$ in m, $q$ in m³/d (at tubing conditions), $d$ in mm.

$9.81\cdot10^{-6}$ is the acceleration due to gravity with the conversion Pa → MPa. In SI the friction loss is $8\lambda\rho H q^2/(\pi^2 d^5)$ ; $0.1086 = 8\cdot10^{9}/(\pi^2\cdot 86{,}400^2)$ converts m³/d → m³/s, mm → m and Pa → MPa.

Oilfield units

p_{wf} = p_{wh} + 0.433\,\gamma H + 1.148\cdot10^{-5}\,\frac{\lambda\,\gamma\,H\,q^2}{d^5}

$p$ in psi, $\gamma$ is the specific gravity of the liquid (water = 1), $H$ in ft, $q$ in bbl/d, $d$ in inches.

$0.433$ psi/ft is the pressure gradient of fresh water; $1.148\cdot10^{-5}$ follows from the same expression $8\lambda\rho H q^2/(\pi^2 d^5)$ after converting the units.

For gassy oil (oil with gas dissolved in it), the calculation is more complicated. As the fluid rises, the pressure falls and gas comes out of solution. The mixture becomes lighter but moves faster, and the gas overtakes the liquid. The outflow curve is then calculated with multiphase flow correlations (formulas for gas and liquid flowing together), section by section along the length of the tubing. Its meaning is the same: the pressure needed at the bottomhole to lift a given rate.

Worked exampleOperating point of a flowing well (node at the bottomhole)

Given (assumed data)

Average reservoir pressure $\bar p_R$: 25.0 MPa
Productivity index $J$ (single-phase inflow): 150 m³/(d·MPa)
True vertical depth of the node $H$ (vertical well, tubing run to the bottomhole): 2000 m
Inner diameter of the tubing $d$: 62 mm (73 mm or 2⅞ in tubing)
Oil in the tubing: density $\rho$ , viscosity $\mu$: 850 kg/m³; 2.0 mPa·s
Wellhead pressure $p_{wh}$: 1.5 MPa
Friction factor $\lambda$: 0.022
Free gas in the tubing: none: the bubble point pressure is below the wellhead pressure; the formation volume factor is close to 1

Solution

Inflow curve to the node: $p_{wf} = \bar p_R - q/J = 25.0 - q/150$ (MPa, with $q$ in m³/d).
Weight of the column: $\rho g H = 850\cdot 9.81\cdot 2000$ Pa $= 16.68$ MPa. Even at zero rate, the pressure needed at the bottomhole is $p_{wh} + \rho g H = 18.18$ MPa.
Friction: the cross-section of the tubing is $A = \pi d^2/4 = 0.00302$ m². The velocity is $v = q/(86{,}400\,A)$ . The friction loss $\lambda\dfrac{H}{d}\dfrac{\rho v^2}{2}$ is proportional to $q^2$ . At $q = 1000$ m³/d it would be 4.43 MPa.
Outflow curve: $p_{wf} = 18.18 + 4.43\,(q/1000)^2$ MPa.
The operating point is the positive root of the quadratic equation $25.0 - q/150 = 18.18 + 4.43\,(q/1000)^2$ . We get $q = 699$ m³/d and $p_{wf} = 20.34$ MPa.
Check the type of flow: $v = 2.68$ m/s and $Re = \rho v d/\mu \approx 71{,}000$ . The flow is turbulent, so the assumed $\lambda$ is reasonable.
Where the 25 MPa of reservoir pressure goes: drawdown 4.66 MPa (18.6%), weight of the column 16.68 MPa (66.7%), friction 2.16 MPa (8.7%), wellhead pressure 1.5 MPa (6.0%).

Answer: The well produces about 699 m³/d at a bottomhole pressure of 20.34 MPa.

What it means. Two thirds of the reservoir pressure goes to holding up the column of oil in the tubing. Only 18.6% drives the inflow from the reservoir. Friction takes only 8.7% so far, but it grows as the square of the rate. The more productive the reservoir, the more the tubing limits the rate. A liquid with no gas in the tubing is a simplification, chosen so that you can work out the outflow curve by hand. In gassy oil the column is lighter, and the calculation uses multiphase flow correlations. The data in this example are assumed for illustration.

Where to put the node

Where you put the node does not change the answer: the rate and the pressures at all points of the system come out the same. You choose the node for convenience. Put it so that the element you study lies wholly on one side, and its effect shows up as a shift of a single curve.

Typical positions of the node

Node: Bottomhole, $p_{wf}$
Inflow side: reservoir and near-wellbore zone (inflow performance relationship)
Outflow side: tubing, wellhead, choke, flowline, separator
Questions it suits: tubing diameter, well stimulation (treating the reservoir to raise inflow) and hydraulic fracturing, depletion (falling reservoir pressure), rising water cut, choice of a pump or gas lift

Node: Sandface, $p_{wfs}$
Inflow side: the reservoir only
Outflow side: the completion and everything beyond it
Questions it suits: number and length of the perforation tunnels, gravel pack — see Effect of completion on inflow

Node: Wellhead, $p_{wh}$
Inflow side: reservoir and tubing
Outflow side: choke, flowline, separator
Questions it suits: pressure in the gathering system (the pipe network that collects fluid from the wells), diameter and length of the flowline, connecting new wells

Node: Separator, $p_{sep}$
Inflow side: the whole well and the flowline
Outflow side: — (the pressure is fixed)
Questions it suits: choice of separation pressure, the need for gas compression

Node: Choke or subsurface safety valve
Inflow side: everything up to the element
Outflow side: everything after the element
Questions it suits: the pressure drop across the element itself: the difference between the pressures of the two curves at each rate

Node	Inflow side	Outflow side	Questions it suits
Bottomhole, $p_{wf}$	reservoir and near-wellbore zone (inflow performance relationship)	tubing, wellhead, choke, flowline, separator	tubing diameter, well stimulation (treating the reservoir to raise inflow) and hydraulic fracturing, depletion (falling reservoir pressure), rising water cut, choice of a pump or gas lift
Sandface, $p_{wfs}$	the reservoir only	the completion and everything beyond it	number and length of the perforation tunnels, gravel pack — see Effect of completion on inflow
Wellhead, $p_{wh}$	reservoir and tubing	choke, flowline, separator	pressure in the gathering system (the pipe network that collects fluid from the wells), diameter and length of the flowline, connecting new wells
Separator, $p_{sep}$	the whole well and the flowline	— (the pressure is fixed)	choice of separation pressure, the need for gas compression
Choke or subsurface safety valve	everything up to the element	everything after the element	the pressure drop across the element itself: the difference between the pressures of the two curves at each rate

An element whose pressure drop is given as a function of the rate (a choke, a subsurface safety valve) is called a functional node.

For a node at the wellhead, the inflow curve is the pressure that is left in the fluid at the wellhead after it has been lifted up the tubing: $p_{wh}(q) = p_{wf}(q) - \Delta p_{\text{tbg}}(q)$ . Here $p_{wf}(q)$ is taken from the inflow performance relationship. This curve is called the tubing performance curve. For a liquid without gas, it falls more steeply than the inflow performance relationship, because friction in the tubing is added to the losses in the reservoir. The outflow curve is the pressure needed at the wellhead to carry the flow through the choke and the flowline to the separator. If the wellhead pressure is held constant, the outflow curve is a horizontal line. The operating rate is where the tubing performance curve crosses it.

The fluid on its way to the separator

Real oil almost always contains dissolved gas. On the way to the separator its state changes: it is liquid at first, then liquid plus gas. Follow the path of undersaturated oil, whose reservoir pressure is above the bubble point pressure $p_b$ , so that all its gas stays dissolved. In the reservoir it is single-phase (liquid only). Near the bottomhole the pressure is below the reservoir pressure. But as long as it stays above $p_b$ , the oil remains a liquid. The temperature hardly changes. As the fluid rises up the tubing, the pressure falls quickly and the fluid cools. At some depth the pressure drops below the bubble point pressure, and bubbles of gas begin to form. Higher up there is more and more gas, and the bubbles merge. The flow passes in turn through bubble flow (small gas bubbles in the liquid), slug flow (large gas pockets separated by liquid) and other flow patterns. In the separator the gas and the liquid are separated.

State of undersaturated oil on its way to the separator

Section: Reservoir, $p > p_b$
State of the fluid: single-phase liquid
What it means for the calculation: the inflow performance relationship is a straight line, the productivity index is constant

Section: Near-wellbore zone at $p_{wf} < p_b$
State of the fluid: gas comes out of the oil near the well
What it means for the calculation: the effective permeability to oil (how easily oil flows when gas is also present) falls, and the inflow performance relationship bends

Section: Tubing below the depth where $p = p_b$
State of the fluid: liquid
What it means for the calculation: the weight of the column is greatest

Section: Tubing above this depth
State of the fluid: gas–liquid mixture, with more and more gas towards the wellhead
What it means for the calculation: the mixture is lighter but faster, and friction grows; the gas overtakes the liquid (slip)

Section: Separator
State of the fluid: gas and liquid separated
What it means for the calculation: the amount of gas that comes out depends on the separation pressure and temperature

Section	State of the fluid	What it means for the calculation
Reservoir, $p > p_b$	single-phase liquid	the inflow performance relationship is a straight line, the productivity index is constant
Near-wellbore zone at $p_{wf} < p_b$	gas comes out of the oil near the well	the effective permeability to oil (how easily oil flows when gas is also present) falls, and the inflow performance relationship bends
Tubing below the depth where $p = p_b$	liquid	the weight of the column is greatest
Tubing above this depth	gas–liquid mixture, with more and more gas towards the wellhead	the mixture is lighter but faster, and friction grows; the gas overtakes the liquid (slip)
Separator	gas and liquid separated	the amount of gas that comes out depends on the separation pressure and temperature

An important consequence follows: the amount of gas in the flow at each point depends on the pressure, and therefore on the rate. That is why the inflow and outflow curves of gassy oil are calculated together with the fluid's PVT properties (how its volume and other properties change with pressure and temperature). These are the solution gas–oil ratio (how much gas the oil holds in solution), the formation volume factors, and the densities and viscosities of the phases at each pressure. The basics of phase behaviour are covered in the lessons Phase behaviour: pure substances and binary mixtures and Types of reservoir fluids.

What shifts the curves

The operating point moves only when at least one of the curves moves. It helps to remember which changes act on which curve, for a node at the bottomhole.

Effect of changes on the curves for a node at the bottomhole

Change: Fall in reservoir pressure
Curve: inflow
How it shifts: moves down; below the bubble point pressure it also shrinks (the rates get lower), because gas takes up a growing share of the pore space (the gas saturation grows)
What happens to the rate: falls; the well may stop flowing

Change: Damage to the near-wellbore zone (rise in skin factor)
Curve: inflow
How it shifts: rotates about the point $q = 0$ , $p_{wf} = \bar p_R$ : at the same bottomhole pressure the rate is lower
What happens to the rate: falls

Change: Treatment of the near-wellbore zone, hydraulic fracturing
Curve: inflow
How it shifts: rotates the other way: $J$ increases
What happens to the rate: rises, but by less than $J$ does if the lift system is the limit

Change: Larger tubing diameter
Curve: outflow
How it shifts: moves down at high rates (less friction); at low rates of gassy oil it may move up
What happens to the rate: rises in high-rate wells

Change: Lower wellhead or separator pressure
Curve: outflow
How it shifts: moves down
What happens to the rate: rises

Change: Rise in water cut
Curve: outflow (and inflow)
How it shifts: the outflow curve moves up: the column is heavier and there is less gas per unit of liquid; the inflow curve changes because of the relative permeabilities (how oil and water get in each other's way in the rock)
What happens to the rate: falls; the well may die (stop flowing)

Change: Gas lift or a pump
Curve: outflow
How it shifts: moves down: gas lightens the column, a pump adds pressure
What happens to the rate: an operating point appears where natural flow is impossible

Change	Curve	How it shifts	What happens to the rate
Fall in reservoir pressure	inflow	moves down; below the bubble point pressure it also shrinks (the rates get lower), because gas takes up a growing share of the pore space (the gas saturation grows)	falls; the well may stop flowing
Damage to the near-wellbore zone (rise in skin factor)	inflow	rotates about the point $q = 0$ , $p_{wf} = \bar p_R$ : at the same bottomhole pressure the rate is lower	falls
Treatment of the near-wellbore zone, hydraulic fracturing	inflow	rotates the other way: $J$ increases	rises, but by less than $J$ does if the lift system is the limit
Larger tubing diameter	outflow	moves down at high rates (less friction); at low rates of gassy oil it may move up	rises in high-rate wells
Lower wellhead or separator pressure	outflow	moves down	rises
Rise in water cut	outflow (and inflow)	the outflow curve moves up: the column is heavier and there is less gas per unit of liquid; the inflow curve changes because of the relative permeabilities (how oil and water get in each other's way in the rock)	falls; the well may die (stop flowing)
Gas lift or a pump	outflow	moves down: gas lightens the column, a pump adds pressure	an operating point appears where natural flow is impossible

How nodal analysis answers engineering questions

Calculation procedure

1
Input data
The reservoir pressure and the inflow performance relationship, found from well tests. The well design: depth, tubing diameter, equipment. The wellhead or separator pressure. The fluid properties: density, producing gas–oil ratio (the volume of gas produced with each volume of oil), water cut and PVT properties.
2
Choice of the node
Choose it so that the element you study lies wholly on one side of the node.
3
Curves
Draw the pressure at the node against the rate, from the inflow side and from the outflow side.
4
Operating point
The crossing point of the curves. Check that it lies on the stable branch of the outflow curve.
5
Sensitivity
Repeat the calculation for other tubing diameters, wellhead pressures, skin factors, water cuts and future reservoir pressures.

Trying out different options is called sensitivity analysis. The lesson Well performance sensitivity analysis covers it in a systematic way. The example below shows its main idea on the well from the previous example.

Worked exampleLarger tubing, a near-wellbore treatment or a lower wellhead pressure?

Given (assumed data)

Base case A: the data of the previous example: $J = 150$ m³/(d·MPa), $d = 62$ mm, $p_{wh} = 1.5$ MPa
Option B — larger tubing: $d = 76$ mm (89 mm or 3½ in tubing)
Option C — treatment of the near-wellbore zone: $J = 300$ m³/(d·MPa), the same tubing
Option D: treatment and larger tubing together
Option E — lower wellhead pressure: $p_{wh} = 0.8$ MPa, everything else as in case A

Solution

Base case (previous example): $q_{\text{A}} = 699$ m³/d.
Larger tubing: at the same rate, friction is lower by a factor of $(76/62)^5 = 2.77$ . The outflow curve is $p_{wf} = 18.18 + 1.60\,(q/1000)^2$ MPa. The operating point is $q_{\text{B}} = 850$ m³/d (+22%).
Treatment of the near-wellbore zone with the same tubing: $q_{\text{C}} = 920$ m³/d (+32%). The productivity index has doubled, but the rate has risen by only about a third. Friction in the tubing (3.76 MPa) is now greater than the drawdown (3.07 MPa).
Both improvements together: $q_{\text{D}} = 1271$ m³/d (+82%). This is more than the sum of the separate gains (+53%).
The lower wellhead pressure lowers the outflow curve by 0.7 MPa. The result is $q_{\text{E}} = 752$ m³/d (+8%). The gain of 53 m³/d is smaller than $J\cdot 0.7 = 105$ m³/d, because friction grows along with the rate.

Answer: Larger tubing gives about 850 m³/d, the treatment of the near-wellbore zone 920, both improvements together 1271, and the lower wellhead pressure 752 m³/d. The base case gives 699 m³/d.

What it means. Both the reservoir and the lift system limit this well. So each improvement alone gives little, but together they almost double the rate. The larger tubing lets the well use the higher productivity of the reservoir. This is the main lesson of the systems approach: evaluate the elements together. Economics decides which option to choose: the cost of the workover (a repair job on the well) and of the treatment, the risks, and the payback period (the time needed to earn back the cost). The data in this example are assumed for illustration.

CalculatedCurves calculated for the data of the examples: inflow for

J = 150

and

300

m³/(d·MPa), outflow for tubing with an inner diameter of 62 and 76 mm. A — base case, B — larger tubing, C — treatment of the near-wellbore zone, D — both improvements.

If the outflow curve lies above the inflow curve at every rate, the curves do not cross. The energy of the reservoir is then not enough to lift the liquid to the wellhead, and the well cannot flow on its own. This happens as the reservoir pressure falls and the water cut rises. Artificial lift is then needed. A pump or gas lift lowers the outflow curve, and an operating point appears again. By repeating the calculation with future reservoir pressures and water cuts, nodal analysis shows when this will happen, and it lets you compare ways of lifting the fluid.

The method also works the other way round. Engineers measure the rate and the pressures at the wellhead and at the bottomhole, and use them to tune the well model. They adjust the productivity index, or a correction in the formula that describes flow in the tubing. The tuned model is then used for forecasts and for optimisation (finding the best settings).

Terms and symbols

Node: A point in the system where the pressure from the inflow side is compared with the pressure from the outflow side. Its position is chosen to suit the question being solved.
Inflow curve: The pressure at the node as a function of rate, as seen from the reservoir side. For a node at the bottomhole, it is the inflow performance relationship.
Outflow curve (VLP, TPR): The pressure needed at the node to carry the rate to the separator. For a node at the bottomhole, it is the lift curve.
Operating point: The crossing point of the inflow and outflow curves: the operating rate and the pressure at the node.
Functional node: An element whose pressure drop is given as a function of rate, such as a choke or a subsurface safety valve.
Drawdown: The difference between the average reservoir pressure and the bottomhole pressure, $\bar p_R - p_{wf}$ .
Buffer pressure: The Russian-language name for the wellhead pressure measured in the Christmas tree (the valve assembly on top of the well), above the tubing. Its symbol is $p_{wh}$ .

Common mistakes and misconceptions

Wrong

“If we double the productivity index, we double the rate.”

In fact

The rate rises by less. A higher rate means more friction in the tubing, so part of the pressure saved in the reservoir is lost in the lift system. In the example in this lesson, doubling $J$ raised the rate by only about a third.

Wrong

“The larger the tubing diameter, the better.”

In fact

For a liquid without gas, friction is indeed lower. But in large tubing at low rates, gas slips up through the liquid, the column becomes heavier, and the well's operation becomes unstable. The diameter has to suit the whole producing life of the well.

Wrong

“The operating point depends on where the node is placed.”

In fact

It does not: the node only divides the system into two parts. The choice of node decides which curves you draw and how clearly you can see the effect of the element you study.

Wrong

“If we lower the wellhead pressure by 0.7 MPa, the rate will rise by $0.7J$ .”

In fact

The operating point moves along the inflow performance relationship, and friction grows with the rate, so the gain is smaller. In the example in this lesson it is about 53 m³/d instead of 105. For gassy oil, the amount of gas in the tubing also changes.

In depth: advanced detail

Stability of the operating mode: an outflow curve with a minimum

For gassy oil, the outflow curve for a node at the bottomhole usually has a minimum (it is called J-shaped). At low rates, gas slips up through the slowly rising liquid. Liquid builds up in the tubing, and the column becomes heavier, so the required bottomhole pressure is high. As the rate increases, slip decreases and the pressure falls to a minimum. Beyond that point, friction begins to dominate and the curve rises.

SchematicAn outflow curve with a minimum and two inflow curves (schematic, not to scale). The upper inflow line crosses the outflow curve twice: only the right-hand intersection can be the operating point. After depletion the inflow curve moves down and no longer crosses the outflow curve — the well stops flowing.

An inflow line can cross such a curve twice, and only the right-hand crossing can be the operating point. Check it. At the right-hand point, suppose the rate happens to rise slightly. The lift system will then need more pressure than the reservoir can provide, and the rate will fall back. If the rate falls slightly, the reservoir will provide more than the lift system needs, and the rate will rise again. At the left-hand point it is the other way round. A small deviation grows, and the rate either moves to the right-hand point or falls to zero: the well “dies”.

Condition for a stable operating point (quasi-static)

\frac{\dd p_{\text{out}}}{\dd q} > \frac{\dd p_{\text{in}}}{\dd q}

where:

$p_{\text{out}}(q)$ pressure at the node from the outflow curve, MPa
$p_{\text{in}}(q)$ pressure at the node from the inflow curve, MPa
$q$ flow rate, m³/d

Units: any consistent units

Holds when

A simplified quasi-static treatment (each state is taken as if the flow had time to settle): the inertia of the flow and the compressibility of the gas in the wellbore are ignored.
Real swings of the operating mode (for example, repeated surges of gas and liquid) are studied with dynamic models.

The inflow curve always falls, so any crossing with a rising branch of the outflow curve is stable. On a falling branch, the point is stable only if the outflow curve falls more gently than the inflow curve.

The practical conclusion: operating conditions to the left of the minimum of the outflow curve are unreliable. The well works in fits and starts, builds up liquid and may stop. So choose the tubing diameter so that the operating point stays to the right of the minimum, even at the end of the producing life. Tubing that is too large in a well with a falling rate is a common cause of an early shutdown. In gas wells the same effect is called liquid loading: below the critical velocity, the gas stops carrying out droplets of water and condensate.

How the outflow curve is built for gassy oil

Before computer programs, engineers built the outflow curve by hand from pressure traverse curves (also called gradient curves). These are ready-made charts of pressure against depth, for a given tubing diameter, rate, water cut and several gas–liquid ratios. The method is this. On the curve for the required gas–liquid ratio, find the depth that matches the wellhead pressure (the “equivalent depth”). Add the tubing length to it. Then read the pressure at the new depth on the same curve. This is the bottomhole pressure. Repeat this for several rates, and you get the outflow curve.

For a node at the wellhead the procedure is reversed. Take the bottomhole pressure from the inflow performance relationship and find its equivalent depth. Subtract the tubing length, and read the pressure that will remain at the wellhead. This is how the tubing performance curve is built.

Computer programs do the same numerically. They divide the tubing into short sections. In each section they find the properties of the phases, the flow pattern, the liquid holdup (the share of the pipe volume filled with liquid) and the pressure gradient (the pressure change per metre of length), using the average pressure and temperature of the section. Then they add up the losses of all the sections. The lesson Multiphase flow in tubing covers these calculations. The lesson Chokes, surface pressure losses and well temperature covers chokes, flowlines and flowing temperature.

Nodal analysis of a gas well

For a gas well the scheme is the same, and only the curves change. The inflow curve is built from an equation with the difference of the squares of the pressures (see the lesson on the inflow performance relationship). All pressures are absolute (measured from zero pressure, not from atmospheric pressure). In the outflow curve, the weight of the gas column is small, but friction is large and grows quickly with the rate. The gas density in the tubing changes with pressure, so the calculation takes into account the gas deviation factor (a number that corrects the ideal gas law for real gas) and the temperature. At low rates the main threat is liquid loading at the bottomhole (liquid collecting at the bottom of the well). At high rates it is friction in the tubing and inertial losses near the well (Inflow at high rates: non-Darcy flow).

Connections to other topics

Well inflow: the well performance diagram — the path of the fluid from the reservoir to the separator and the places where pressure is lost — an introductory overview.
Inflow performance relationship (IPR) — how to build the inflow curve: the productivity index, Vogel's equation, Fetkovich's method.
Multiphase flow in tubing — how to calculate the outflow curve for a gas–liquid mixture.
Chokes, surface pressure losses and well temperature — chokes, flowlines and temperature — the surface part of the system.
Well performance sensitivity analysis — trying out options in a systematic way: tubing, wellhead pressure, water cut, skin factor.
Skin factor — the additional resistance of the near-wellbore zone in the pressure budget.
Artificial lift: when it is needed and how to choose the method — what to do when the inflow and outflow curves do not intersect.

Check yourself

1
Where is it most convenient to put the node in order to compare tubing of different diameters?
1. A.At the separator
2. B.The node has to be placed at each point in turn
3. C.At the bottomhole of the well
4. D.At the sandface in front of the perforations
Show the answer
Correct answer: C. At the bottomhole of the well
The answer does not depend on where the node is. But with the node at the bottomhole, the tubing diameter changes only the outflow curve. The inflow curve (the inflow performance relationship) stays the same. So you only need to draw several outflow curves on one plot, together with a single inflow performance relationship.
2
For a node at the bottomhole, the well's inflow curve is $p_{wf} = 20 - q/100$ and the outflow curve is $p_{wf} = 14 + 2\cdot10^{-6}\,q^2$ (pressures in MPa, rate in m³/d). Find the operating point.
Show the answer
Set the two equal: $2\cdot10^{-6}q^2 + 0.01q - 6 = 0$ . The positive root is $q = \dfrac{-0.01 + \sqrt{10^{-4} + 4.8\cdot10^{-5}}}{4\cdot10^{-6}} \approx 541$ m³/d. The bottomhole pressure is $20 - 5.41 \approx 14.6$ MPa. Check against the outflow curve: $14 + 2\cdot10^{-6}\cdot 541^2 \approx 14.6$ MPa.
3
After an acid treatment the well's productivity index has doubled, but the rate has risen by only 15%. The most likely explanation is:
1. A.A measurement error: the rate must have doubled
2. B.The rate is limited by the lift system — friction in the tubing or the wellhead pressure
3. C.The productivity index does not depend on the treatment
4. D.The acid reduced the reservoir pressure
Show the answer
Correct answer: B. The rate is limited by the lift system — friction in the tubing or the wellhead pressure
The treatment rotated the inflow curve, but the operating point moved along the steep outflow curve. The drawdown saved in the reservoir went into friction in the tubing at the higher rate. To gain from the treatment, lower the outflow curve, for example with larger tubing or a lower wellhead pressure.
4
Why can the crossing of the inflow curve with the left-hand, falling branch of the outflow curve not be an operating mode?
Show the answer
At this point the outflow curve falls more steeply than the inflow curve. If the rate rises slightly, the pressure that the lift system needs falls by more than the pressure the reservoir provides, and the rate keeps rising, up to the right-hand crossing. If the rate falls slightly, the lift system needs more pressure than the reservoir provides, and the rate keeps falling until the well stops. Small deviations are not damped but grow. So this mode is unstable.
5
How will the outflow curve change as the water cut of the production rises, if the liquid rate and the oil's producing gas–oil ratio stay the same?
1. A.It will move down: water makes the flow lighter
2. B.It will become a horizontal line
3. C.It will move up: the column is heavier and there is less gas per unit of liquid
4. D.It will not change: the outflow curve depends only on the tubing diameter
Show the answer
Correct answer: C. It will move up: the column is heavier and there is less gas per unit of liquid
Water is heavier than oil, and gas arrives only with the oil, so the gas–liquid ratio falls. Both effects make the column heavier, so a higher bottomhole pressure is needed to lift the same rate. This is why wells with a rising water cut eventually stop flowing.
6
Why can you not find the rate of a well from the inflow performance relationship alone?
Show the answer
The inflow performance relationship says what rate the reservoir will deliver at each bottomhole pressure. But it does not set the bottomhole pressure itself. That depends on what lies beyond the bottomhole: the weight of the column and friction in the tubing, the wellhead pressure and the separator pressure. You get the rate only from a joint solution: the crossing of the inflow performance relationship with the outflow curve.

Updated: 22 September 2026