Well performance sensitivity analysis

Quick refresher

Sensitivity analysis is a nodal analysis repeated many times. We change a parameter (or a combination of parameters), find the new operating point and compare the rates. Before that we tune the model to measurements. Otherwise the model's own error gets mixed up with the effects.
Shifting either curve by $\delta$ MPa changes the rate by about $\delta/(1/J + s_{\text{out}})$ , where $s_{\text{out}}$ is the slope of the outflow curve at the operating point. If the outflow curve is flat compared with the inflow curve, the reservoir decides the rate. If it is steep, the lift system decides.
The best tubing diameter grows with productivity. A low-rate well does better with narrow tubing (the gas slips less through the liquid). A high-rate well does better with wide tubing (there is less friction). We choose the tubing for the whole period of natural flow, not for the first year.
Rising water cut makes the liquid column heavier and lowers the gas–liquid ratio, and depletion lowers the inflow curve. Together they bring the well to its natural flow limit, beyond which artificial lift is needed. We plan the change ahead of time, because near the limit the operating mode is unstable.
We compare the effects of the parameters by their elasticities (the relative change of the rate for a relative change of the parameter) and by a tornado chart for plausible ranges. The uncertainty of the model of flow in the tubing can be as large as the effect of the wellhead pressure or of the water cut.

Key relationships

\Delta q \approx \dfrac{\delta_{\text{in}} + \delta_{\text{out}}}{1/J + s_{\text{out}}}

linear estimate: the inflow curve rises, or the outflow curve falls, by $\delta$ MPa

E_x = \dfrac{\Delta q/q}{\Delta x/x}

elasticity of the rate with respect to the parameter $x$

\text{GLR} = R\,(1 - \text{WC}),\qquad \rho_L = (1 - \text{WC})\,\rho_o + \text{WC}\,\rho_w

gas–liquid ratio and the density of the liquid at a water cut $\text{WC}$

J = \dfrac{0.536\,k h}{\mu B\left[\ln(r_e/r_w) - 0.75 + S\right]}

productivity index with the skin factor (metric oilfield units)

Common pitfalls

Calculating sensitivity on a model that is not tuned. The error of the correlation for flow in the tubing is as big as many of the effects.
Adding up the gains from interventions that were calculated one by one. They interact, and the sum can be bigger or smaller.
Choosing tubing for today's rate without checking where the operating point will be in a few years.
Waiting until the well dies. Near the natural flow limit the operating mode is unstable, so artificial lift is prepared in advance.
Comparing parameters by a step of “one unit”: 1 MPa, 1% of water cut, 1 mm of diameter. The scales are different. Compare elasticities or plausible ranges instead.

After this lesson you can

find the operating point of the base case, and use the slopes of the inflow and outflow curves (the pressure the reservoir can give, and the pressure the well needs, at each rate) to decide what limits the rate more
calculate and explain how the rate depends on the tubing diameter, wellhead pressure, water cut, gas–oil ratio, skin factor and reservoir pressure
compare the effects of the parameters on one scale, using elasticities (the percentage change of the rate for a 1% change of a parameter) and a tornado chart (a chart of bars that show how far the rate moves for each parameter), and tell the parameters we can control from the uncertain ones
use a forecast of reservoir pressure and water cut to find when the well will stop flowing naturally, and estimate the gain from artificial lift
set up the calculation of the cases in software: tune the model to measurements, build a matrix of cases and check that the operating mode is stable

The base case, and what limits the rate

We start the analysis with the base case: a model of the well that reproduces today's operating mode. In this lesson it is the lesson's well from the lesson on chokes and surface losses. The well is vertical and 2000 m deep. Its tubing is 73 mm (62 mm inside). The average reservoir pressure is $\bar p_R = 22$ MPa, and the bubble point pressure (the pressure at which gas starts to come out of the oil) is 12 MPa. The productivity index (the rate per unit of pressure drawdown) $J$ is about 30 m³/(d·MPa) at a skin factor of 3. The water cut is 10%, and the producing gas–oil ratio (the volume of gas produced with each volume of oil) is 90 m³/m³. There is no choke (a narrow opening that holds back the flow). The wellhead pressure of 1.5 MPa is set by the gathering network. The inflow curve is composite (composite inflow performance relationship: a straight line above the bubble point and a curve below it). The outflow curve comes from the drift-flux model of the lesson Multiphase flow in tubing. In this model the gas moves faster than the liquid, by a drift velocity. All the gas is free, and the temperature in the tubing is constant. All the data are assumed.

The first question is what limits the rate: the reservoir or the lift system. The slopes of the curves at the operating point give the answer (what shifts the curves). With the node at the bottomhole, the slope of the inflow curve is $-1/J$ . Each extra cubic metre per day costs $1/J$ MPa of bottomhole pressure. The slope of the outflow curve, $s_{\text{out}}$ , shows how much more bottomhole pressure the lift system needs for each extra cubic metre per day. Suppose we raise the inflow curve by $\delta$ MPa (the reservoir pressure has grown), or lower the outflow curve by $\delta$ (we have lowered the wellhead pressure). Then the operating point moves by $\Delta q \approx \delta/(1/J + s_{\text{out}})$ . The derivation is in the advanced section.

Worked exampleThe base operating mode and a diagnosis from the slopes of the curves

Given (assumed data)

Well: the lesson's well: $\bar p_R = 22$ MPa, $p_b = 12$ MPa, $J = 29.85$ m³/(d·MPa), tubing of 62 mm inside diameter, $\text{WC} = 10$ %, $R = 90$ m³/m³
Wellhead pressure $p_{wh}$: 1.5 MPa (absolute)

Solution

Operating point: $q = 287.1$ m³/d and $p_{wf} = 12.38$ MPa. This is still above the bubble point pressure, on the straight part of the inflow performance relationship.
Where the 22 MPa go: the drawdown (the pressure drop in the reservoir) takes 9.62 MPa (44%), the lift up the tubing $p_{wf} - p_{wh}$ takes 10.88 MPa (49%), and the wellhead pressure takes 1.5 MPa (7%).
Slopes: for the inflow, $1/J = 1/29.85 = 0.03350$ MPa per m³/d. For the outflow (from two neighbouring rates, ±10 m³/d), $s_{\text{out}} = 0.00793$ MPa per m³/d. That is 4.2 times smaller.
The response to a shift of the curves is $1/(0.03350 + 0.00793) = 24.1$ m³/d for each megapascal.
A check. With the reservoir pressure 1 MPa higher, the estimate is 24.1 and the exact calculation gives 23.9 m³/d. With a wellhead pressure of 1.0 MPa instead of 1.5, the outflow curve falls by 0.797 MPa. This is more than 0.5, because the gas in the tubing is less compressed and the column gets lighter. The estimate is $0.797\cdot 24.1 = 19.2$ , and the exact value is 18.3 m³/d.
The linear estimate does not catch big changes. Doubling $J$ (a well treatment) gives 462.9 m³/d (+61%). Tubing of 76 mm instead of 62 mm gives 303.5 m³/d (+5.7%).

Answer: The well works at 287.1 m³/d. At the operating point the outflow curve is four times flatter than the inflow curve, so the reservoir mainly decides the rate.

What it means. A flat outflow curve means that the lift system “lets through” almost everything the reservoir gives. The gain will come from work on the reservoir, such as treating the near-wellbore zone and maintaining the pressure. Changing the tubing will add little. The linear estimate is good for small changes. For decisions we always calculate the whole new operating point. The data in this example are assumed for illustration.

Tubing diameter: the best one depends on the productivity and on the year

The tubing diameter acts on the outflow curve in two opposite ways (the outflow curve and its minimum). The wider the tubing, the lower the velocity of the mixture and the friction. This is a gain at high rates. But at a lower velocity the gas slips through the liquid more, liquid collects in the well (liquid loading), and the column gets heavier. This is a loss at low rates. So every well has its own best diameter, and it grows with the productivity.

CalculatedThe operating rate of the lesson's well against the inside diameter of the tubing, at four levels of productivity (permeability 0.25, 0.5, 1 and 2 times the base value). The dots mark the highest rates within the range of the graph. Calculated with the lesson's teaching models.

Operating rate, m³/d, for different tubing and productivity (the lesson's well, calculated)

$J$ , m³/(d·MPa): 7.5
Tubing 60 mm: 75
Tubing 73 mm: 70
Tubing 89 mm: 61
Tubing 114 mm: 46

$J$ , m³/(d·MPa): 14.9
Tubing 60 mm: 148
Tubing 73 mm: 152
Tubing 89 mm: 145
Tubing 114 mm: 122

$J$ , m³/(d·MPa): 29.9
Tubing 60 mm: 248
Tubing 73 mm: 287
Tubing 89 mm: 303
Tubing 114 mm: 290

$J$ , m³/(d·MPa): 59.7
Tubing 60 mm: 354
Tubing 73 mm: 463
Tubing 89 mm: 551
Tubing 114 mm: 607

$J$ , m³/(d·MPa)	Tubing 60 mm	Tubing 73 mm	Tubing 89 mm	Tubing 114 mm
7.5	75	70	61	46
14.9	148	152	145	122
29.9	248	287	303	290
59.7	354	463	551	607

The inside diameters are 50, 62, 76 and 100 mm. The productivity was changed through the permeability; the other data are the base values. The data are assumed.

For the base well the best inside diameter is about 80 mm (the 89 mm tubing has an inside diameter of 76 mm). But the gain of the 89 mm tubing over the 73 mm tubing is small: 303 against 287 m³/d. Something else matters more. The rate falls with time, and tubing that is the best today turns out to be too wide after a few years. The calculation with the forecast shows this.

Rate by year of the forecast for different tubing, m³/d (0 means the well does not flow)

Year: $\bar p_R$ , $\text{WC}$: 0: 22.0 MPa, 10%
60 mm: 248
73 mm: 287
89 mm: 303
114 mm: 290

Year: $\bar p_R$ , $\text{WC}$: 1: 21.0 MPa, 18%
60 mm: 220
73 mm: 249
89 mm: 257
114 mm: 233

Year: $\bar p_R$ , $\text{WC}$: 2: 20.0 MPa, 27%
60 mm: 187
73 mm: 205
89 mm: 201
114 mm: 162

Year: $\bar p_R$ , $\text{WC}$: 3: 19.0 MPa, 36%
60 mm: 148
73 mm: 152
89 mm: 133
114 mm: 0

Year: $\bar p_R$ , $\text{WC}$: 4: 18.2 MPa, 45%
60 mm: 103
73 mm: 87
89 mm: 0
114 mm: 0

Year: $\bar p_R$ , $\text{WC}$: 5: 17.5 MPa, 53%
60 mm: 0
73 mm: 0
89 mm: 0
114 mm: 0

Year: $\bar p_R$ , $\text{WC}$	60 mm	73 mm	89 mm	114 mm
0: 22.0 MPa, 10%	248	287	303	290
1: 21.0 MPa, 18%	220	249	257	233
2: 20.0 MPa, 27%	187	205	201	162
3: 19.0 MPa, 36%	148	152	133	0
4: 18.2 MPa, 45%	103	87	0	0
5: 17.5 MPa, 53%	0	0	0	0

The forecast of the reservoir pressure and water cut is assumed, as if a reservoir model had produced it. The other data are the base values.

Worked exampleTubing for the whole period of natural flow

Given (assumed data)

Well: the lesson's well, $p_{wh} = 1.5$ MPa
Forecast: the table above: the pressure falls from 22 to 17.5 MPa and the water cut grows from 10 to 53% over five years
Options: tubing of 60, 73, 89 and 114 mm

Solution

Year 0: the 89 mm tubing is the best, with 303 m³/d against 287 for 73 mm and 248 for 60 mm.
Year 2: the 89 mm tubing is already behind the 73 mm tubing, 201 against 205. The 114 mm tubing gives 162.
Year 3: 73 mm gives 152, 60 mm gives 148 and 89 mm gives 133. The 114 mm tubing no longer flows naturally.
Year 4: the 89 mm tubing has stopped. The 73 mm tubing gives 87 and the 60 mm tubing gives 103 m³/d. In year 5 no option flows.
A rough total for years 0–4 is the sum of the rates multiplied by 365 days: 60 mm gives 331, 73 mm gives 358, 89 mm gives 326 and 114 mm gives 250 thousand m³ of liquid.

Answer: For the whole period of natural flow, the 73 mm tubing is the best (about 358 thousand m³). The 89 mm tubing wins in the first two years and loses later. The 60 mm tubing keeps the well flowing longer, but gives little at the start.

What it means. A total based on the rates “at the start of each year” overstates the production of all the options, because the rate also falls within the year. But it can still be used to compare the options. We make the decision on tubing together with the decision on artificial lift. If a pump will be installed after three or four years anyway, the choice of diameter will also depend on the size of the pump and on its delivery. Running a velocity string (narrower tubing put in during the late period, so that the gas slips less) prolongs natural flow, but here only for a short time: in year 5 even the 60 mm tubing does not flow. The data in this example are assumed for illustration.

Wellhead pressure, water cut and gas–oil ratio

The wellhead pressure is a back pressure on the whole well. It is set by the choke, the flowline and the separator (flowline and gathering network). For a mixture with gas, each megapascal at the wellhead costs more than a megapascal at the bottom. At a higher pressure the gas in the whole column is compressed more, and the column is heavier. On the lesson's well, lowering the wellhead pressure from 1.5 to 1.0 MPa lowers the outflow curve by 0.8 MPa.

Operating rate at different wellhead pressures (the lesson's well)

$p_{wh}$ , MPa: 1.0
Rate, m³/d: 305.4

$p_{wh}$ , MPa: 1.5
Rate, m³/d: 287.1

$p_{wh}$ , MPa: 2.0
Rate, m³/d: 266.7

$p_{wh}$ , MPa: 2.5
Rate, m³/d: 245.3

$p_{wh}$ , MPa: 3.0
Rate, m³/d: 223.3

$p_{wh}$ , MPa	Rate, m³/d
1.0	305.4
1.5	287.1
2.0	266.7
2.5	245.3
3.0	223.3

About 41 m³/d for each megapascal of wellhead pressure. The data are assumed.

Water cut acts on the outflow twice. Water is heavier than oil, and gas comes only with the oil, so the gas–liquid ratio (the volume of gas per volume of liquid) $\text{GLR} = R\,(1 - \text{WC})$ falls. In the lesson's model, water cut does not affect the inflow: the productivity index for liquid is taken as constant. In reality it depends on the phase permeabilities and viscosities (how easily each fluid moves through the rock), and it can fall or grow. Which way the simplification errs depends on the reservoir. The main consequence is seen in the oil: the liquid rate falls moderately, but the oil rate falls much faster.

Water cut: the rates of liquid and of oil, and the natural flow limit (the lesson's well,

\bar p_R = 22

MPa)

$\text{WC}$ , %: 0
Liquid, m³/d: 300.3
Oil, m³/d: 300.3
Flows naturally when $\bar p_R$ is at least, MPa: 14.14

$\text{WC}$ , %: 20
Liquid, m³/d: 271.3
Oil, m³/d: 217.0
Flows naturally when $\bar p_R$ is at least, MPa: 15.62

$\text{WC}$ , %: 40
Liquid, m³/d: 229.4
Oil, m³/d: 137.6
Flows naturally when $\bar p_R$ is at least, MPa: 17.32

$\text{WC}$ , %: 60
Liquid, m³/d: 167.7
Oil, m³/d: 67.1
Flows naturally when $\bar p_R$ is at least, MPa: 19.19

$\text{WC}$ , %: 80
Liquid, m³/d: 78.7
Oil, m³/d: 15.7
Flows naturally when $\bar p_R$ is at least, MPa: 21.06

$\text{WC}$ , %	Liquid, m³/d	Oil, m³/d	Flows naturally when $\bar p_R$ is at least, MPa
0	300.3	300.3	14.14
20	271.3	217.0	15.62
40	229.4	137.6	17.32
60	167.7	67.1	19.19
80	78.7	15.7	21.06

The natural flow limit is the lowest reservoir pressure at which the inflow and outflow curves still cross. The data are assumed.

The gas–oil ratio makes the column lighter, but it also makes the mixture move faster, so friction grows. The rate therefore has a flat maximum against it (the lowest bottomhole pressure at a certain GLR). On the lesson's well the rate grows from 196 m³/d at $R$ = 30 m³/m³ to 314 at 250, and at 400 m³/m³ it falls again to 300. Gas lift (Gas lift: principles and choice of operating conditions) is based on the same maximum. The injected gas raises the GLR to the value at which the bottomhole pressure is lowest. In the calculation the gas–oil ratio is not linked to the bubble point pressure. This is a simplification: in reality a higher gas–oil ratio also means a higher bubble point pressure and a different inflow performance relationship.

Gas–oil ratio and operating rate (the lesson's well,

\text{WC} = 10

$R$ , m³/m³: 30
GLR, m³/m³: 27
Rate, m³/d: 195.5

$R$ , m³/m³: 60
GLR, m³/m³: 54
Rate, m³/d: 255.6

$R$ , m³/m³: 90
GLR, m³/m³: 81
Rate, m³/d: 287.1

$R$ , m³/m³: 150
GLR, m³/m³: 135
Rate, m³/d: 310.9

$R$ , m³/m³: 250
GLR, m³/m³: 225
Rate, m³/d: 314.0

$R$ , m³/m³: 400
GLR, m³/m³: 360
Rate, m³/d: 300.5

$R$ , m³/m³	GLR, m³/m³	Rate, m³/d
30	27	195.5
60	54	255.6
90	81	287.1
150	135	310.9
250	225	314.0
400	360	300.5

The data are assumed.

Skin factor and well stimulation: the gain depends on the lift system

The skin factor is part of the productivity index (the inflow formula with skin). Lowering the skin turns the inflow curve about the point $q = 0$ , $p_{wf} = \bar p_R$ . The outflow curve decides how much the rate grows. If it is flat, the rate grows almost in proportion to $J$ . If it is steep, the rate grows only until it runs into the friction in the tubing. So we estimate the gain from treating the near-wellbore zone, or from hydraulic fracturing, together with the tubing that the well will use afterwards. Perforation also acts through skin (Effect of completion on inflow).

CalculatedThe operating rate of the lesson's well against the skin factor, with 89 mm tubing (upper curve, dashed), 73 mm (middle curve, solid) and 60 mm (lower curve). The dots mark the cases of the example: skin 6 and 0 with 73 mm tubing, and skin 0 with 89 mm tubing. Calculated with the lesson's teaching models.

Worked exampleTreating the near-wellbore zone, changing the tubing, or both

Given (assumed data)

Well: the lesson's well, but a well test showed a skin factor of 6 (assumed data)
Interventions: an acid treatment that lowers the skin to 0; replacing the 73 mm tubing with 89 mm; both together; hydraulic fracturing with a skin of $-4$
Reservoir: $k = 80$ mD, $h = 12$ m, $\mu = 1.4$ mPa·s, $B = 1.2$ , $\ln(r_e/r_w) = \ln 3000 = 8.006$

Solution

Productivity index: $J = 0.5358\cdot 80\cdot 12/\left(1.4\cdot 1.2\cdot(8.006 - 0.75 + S)\right)$ (the constant 0.536 taken with four digits, as 0.5358). At $S = 6$ it is 23.1, at $S = 0$ it is 42.2 and at $S = -4$ it is 94.0 m³/(d·MPa).
The rate now ( $S = 6$ , 73 mm tubing): 230.8 m³/d.
Treatment only: 372.0 m³/d (+141.2 m³/d). New tubing only: 234.5 m³/d (+3.7). Both interventions: 416.7 m³/d (+185.9). This is 41.0 m³/d more than the sum of the separate gains.
Hydraulic fracturing with a skin of $-4$ : 580 m³/d with 73 mm tubing and 746 with 89 mm tubing.

Answer: The treatment gives +141.2 m³/d, the new tubing alone gives +3.7, and together they give +185.9 m³/d.

What it means. At a skin of 6 the reservoir limits the well, and wider tubing gives almost nothing. After the treatment the rate has grown, the friction in the tubing has become noticeable, and now the same tubing adds tens of cubic metres per day. The stronger the stimulation, the more the lift system matters: after hydraulic fracturing the 89 mm tubing gives 29% more than the 73 mm tubing. The order of decisions is first the treatment, then a check of the lift system against the new inflow curve. It is the same as in the lesson on nodal analysis, and here it is confirmed on oil with dissolved gas. The data in this example are assumed for illustration.

What matters more: elasticities and the tornado chart

The parameters are measured in different units, so it makes no sense to compare them by a step of “one unit”. It is more convenient to use the elasticity: by how many per cent the rate changes when the parameter changes by one per cent. We find it with two calculations, one with the parameter a little above the base value and one a little below.

Elasticity of the rate with respect to a parameter

E_x = \frac{\Delta q/q}{\Delta x/x} \approx \frac{q(x + \Delta x) - q(x - \Delta x)}{2\,q}\cdot\frac{x}{\Delta x}

where:

$E_x$ elasticity of the operating rate with respect to the parameter $x$ , —
$q$ operating rate of the base case, m³/d
$x,\ \Delta x$ base value of the parameter and its small change (±5% in the lesson), units of the parameter

Units: dimensionless

Holds when

small changes: the elasticity describes the response near the base case and does not replace a calculation of big changes
for fractions and for quantities near zero (water cut, skin factor) a relative step is inconvenient; their effect is shown in absolute steps or in ranges

An elasticity of 1.85 for the reservoir pressure (see the table below) means that a pressure 1% higher gives a rate about 1.85% higher. The sign gives the direction: a rise of the wellhead pressure lowers the rate.

Elasticities of the operating rate of the lesson's well (calculated)

Parameter: Reservoir pressure $\bar p_R$
Elasticity: 1.85
What it means: the strongest: the pressure is spent both on the inflow and on the lift

Parameter: Permeability $k$
Elasticity: 0.81
What it means: almost proportional: the well is limited by the inflow

Parameter: Tubing diameter $d$
Elasticity: 0.48
What it means: noticeable, but less than permeability

Parameter: Skin factor $S$
Elasticity: −0.24
What it means: at $S = 3$ ; the bigger the skin, the stronger the effect

Parameter: Gas–oil ratio $R$
Elasticity: 0.23
What it means: moderate: gas still helps, but less and less; the maximum rate is near $R$ = 200 m³/m³

Parameter: Wellhead pressure $p_{wh}$
Elasticity: −0.20
What it means: weak in percentage terms, but 1 MPa at the wellhead is worth about 40 m³/d

Parameter	Elasticity	What it means
Reservoir pressure $\bar p_R$	1.85	the strongest: the pressure is spent both on the inflow and on the lift
Permeability $k$	0.81	almost proportional: the well is limited by the inflow
Tubing diameter $d$	0.48	noticeable, but less than permeability
Skin factor $S$	−0.24	at $S = 3$ ; the bigger the skin, the stronger the effect
Gas–oil ratio $R$	0.23	moderate: gas still helps, but less and less; the maximum rate is near $R$ = 200 m³/m³
Wellhead pressure $p_{wh}$	−0.20	weak in percentage terms, but 1 MPa at the wellhead is worth about 40 m³/d

The step is ±5% of the base value. The data are assumed.

The elasticity does not tell us how uncertain the data themselves are. A tornado chart shows this (spider and tornado charts). For each parameter we calculate the rate at the ends of its plausible range and draw it as a bar. The bars are arranged in order of falling length. The ranges include both the uncertainty of the data (permeability and skin from well tests, reservoir pressure, gas–oil ratio) and the uncertainty of the model itself, for example of the distribution coefficient $C_0$ in the model of flow in the tubing.

CalculatedTornado chart of the lesson's well: the operating rate at the ends of the assumed ranges of the parameters. The vertical line is the base case. Calculated with the lesson's teaching models.

How to read the tornado chart. The two top bars are the skin factor and the permeability. The rate depends on them most, and both come from well test data. So before choosing an intervention, it is worth checking them with a well test (Pressure build-up analysis). The bar of the tubing flow model (271–307 m³/d) is almost as long as the bars of the wellhead pressure and of the water cut. If the model is not tuned to measurements, we cannot trust a difference of two dozen cubic metres between the cases.

Depletion, rising water cut and when artificial lift is needed

A fall of the reservoir pressure lowers the inflow curve, and a rise of the water cut raises the outflow curve. At some combination of the two, the curves stop crossing and the well stops flowing. This is the natural flow limit. For each water cut we can find the lowest reservoir pressure at which the well still works on its own. If we join such points, we get a boundary on the plane “water cut against reservoir pressure”. The forecast of the reservoir model lies on the same plane as a line through the years.

CalculatedThe natural flow limit of the lesson's well for 60 mm tubing (grey line), 73 mm (solid) and 89 mm (dashed). Above the line the well flows naturally. The dots are the assumed forecast of reservoir pressure and water cut for years 0–5, from left to right. Calculated with the lesson's teaching models.

Worked exampleWhen the well will need a pump

Given (assumed data)

Well: the lesson's well, 73 mm tubing, $p_{wh} = 1.5$ MPa
Forecast of the reservoir model (assumed): years 0–5: $\bar p_R$ = 22.0, 21.0, 20.0, 19.0, 18.2, 17.5 MPa; $\text{WC}$ = 10, 18, 27, 36, 45, 53%
Artificial lift for comparison: a pump that holds the bottomhole pressure at 9.0 MPa (assumed)

Solution

The natural flow rate by year is 287, 249, 205, 152, 87 and 0 m³/d. The oil rate is 258, 204, 150, 97, 48 and 0 m³/d.
The natural flow limit at the water cut of each year is 14.85, 15.46, 16.19, 16.96, 17.77 and 18.52 MPa. The margin is the difference between the reservoir pressure and the limit: 2.04 MPa in year 3, 0.43 MPa in year 4 and −1.02 MPa in year 5.
Year 5: the margin is negative, so the well does not flow. Year 4: formally the well flows, but the operating point (87 m³/d) lies to the left of the minimum of the outflow curve (187 m³/d). This is in the region of strong gas slippage, where the flow pulses.
With a pump holding the bottomhole pressure at 9 MPa (below the bubble point pressure, so the rate comes from the composite curve): year 3 gives 289, year 4 gives 265 and year 5 gives 244 m³/d of liquid.

Answer: Natural flow will end between years 4 and 5. Artificial lift must be prepared for year 3 or 4, when a pump would give 3.0 times more than natural flow.

What it means. The pump is needed before the well dies. Already in year 3 it would give almost twice as much liquid. In year 4 the natural flow is unstable. Any upset can “kill” it, such as a stop for repair or a rise of the pressure in the gathering header (the main pipe that collects the production of several wells). The preparation includes choosing the method (Artificial lift: when it is needed and how to choose the method), the equipment, and the power supply or the gas for gas lift (Gas lift: principles and choice of operating conditions, Electric submersible pumps (ESP)). It takes months. So we set the date from the forecast, not after the well has stopped. While the reservoir pressure is above the bubble point pressure, as in this forecast, the gas–oil ratio is constant. Below the bubble point pressure the inflow curve shrinks faster (forecast of the inflow performance relationship). The data in this example are assumed for illustration.

The map of the natural flow limit also answers the question about tubing: narrow tubing moves the limit down and wide tubing moves it up. The shift is small, and it gets smaller as the water cut grows. Between 60 mm and 73 mm tubing it is 0.9 MPa at a water cut of 10% and 0.8 MPa at 50%. The forecast crosses the 89 mm line a year earlier than the 73 mm line, and it crosses the 60 mm line in the same year as the 73 mm line. So in the late period, a change to narrower tubing is a short delay, not a replacement for artificial lift.

How calculation software does this

Well-modelling software does the same as this lesson, only faster and in more detail. It divides the wellbore and the flowline into short sections. For each section it calculates the properties of the phases, the flow regime, the pressure and the temperature (calculation by sections, temperature of the flow). It finds the operating point and runs through the cases of a given matrix. The value of the result depends not on the software, but on how well the model is prepared.

The order of a sensitivity analysis

1
Data
the well design and the directional survey (the measured path of the well), PVT properties (bubble point pressure, solution gas–oil ratio, viscosities), the inflow performance relationship from well tests, the pressures at the wellhead and in the gathering network, and the temperature.
2
Tuning
using a measurement of the rate and the bottomhole pressure (with a downhole gauge, or by recalculation), we choose the correlation for flow in the tubing. We adjust, within reasonable limits, the corrections to the liquid holdup (the share of the pipe volume filled with liquid) or to the friction. We take the productivity index from a well test. Without tuning, the cases cannot be compared.
3
Matrix of cases
the parameters we control (tubing, choke, wellhead pressure, treatment, lift method) and the uncertain ones (reservoir pressure, water cut, gas–oil ratio, skin) over their plausible ranges. First one at a time, then the real combinations.
4
Checking the modes
the operating point lies to the right of the minimum of the outflow curve, or at least is stable. The limits are respected: the drawdown (for sand production and coning, which is water or gas pulled up towards the well), the velocities (for erosion), the pressures (for the equipment), and the temperatures (for hydrates, which are ice-like solids that plug lines, and for paraffin wax).
5
Over time
a forecast of reservoir pressure and water cut from the reservoir model; the year of the change to artificial lift, and the choice of its method.
6
Updating
after each new measurement we compare the model with it and retune it if needed. This is how we monitor the well.

Terms and symbols

Sensitivity analysis: A scan of the input data of a well model, with the operating rate calculated for each case.
Base case: A well model tuned to today's measured operating mode. All the changes are counted from it.
Inflow limit and lift limit: The reservoir mainly decides the rate if the outflow curve at the operating point is flat compared with the inflow curve. The lift system decides if it is steep.
Elasticity $E_x$: The relative change of the rate divided by the relative change of the parameter.
Tornado chart: Bars showing the change of the rate at the ends of the plausible ranges of the parameters, arranged in order of falling length.
Natural flow limit: The combination of reservoir pressure and water cut (or of other parameters) at which the inflow and outflow curves stop crossing.
Velocity string: Narrower tubing put in during the late period, to reduce gas slippage and prolong natural flow.

Common mistakes and misconceptions

Wrong

“The well is limited either by the reservoir or by the tubing; we only need to find out which.”

In fact

Usually both the reservoir and the tubing limit it, in different proportions. The slopes of the curves at the operating point show the proportion. After any intervention the proportion changes. In the lesson's example, after the reservoir treatment the gain from new tubing also became noticeable.

Wrong

“The wider the tubing, the higher the rate.”

In fact

Only for wells that are productive enough. At $J$ of about 7.5 m³/(d·MPa), 60 mm tubing gives 75 m³/d and 114 mm tubing gives 46: in wide tubing the gas slips through the liquid, and the column gets heavier. Even in a productive well, wide tubing stops flowing earlier.

Wrong

“The treatment gives +140, the new tubing +4, so together they give +144 m³/d.”

In fact

The effects interact: in the lesson's example they gave about +186 m³/d together. We calculate the operating point for the combination of interventions that we plan to carry out.

Wrong

“While the well flows naturally, it is too early to think about a pump.”

In fact

Near the natural flow limit the operating mode is unstable, and preparing artificial lift takes months. We set the date of the change from the forecast of reservoir pressure and water cut. In the lesson's example it is a year before natural flow stops.

Wrong

“The model of flow in the tubing is exact, so a difference of 20 m³/d between the cases is real.”

In fact

The uncertainty of the correlation is of the same order. First we tune the model to a measurement of the bottomhole pressure and the rate, and only then compare the cases.

In depth: advanced detail

Where the linear estimate of the response comes from

With the node at the bottomhole, near the operating point $(q_0, p_0)$ we write the curves as straight lines. The inflow is $p = p_0 - (q - q_0)/J + \delta_{\text{in}}$ and the outflow is $p = p_0 + s_{\text{out}}(q - q_0) - \delta_{\text{out}}$ . Here $\delta_{\text{in}}$ is the rise of the inflow curve and $\delta_{\text{out}}$ is the fall of the outflow curve when a parameter changes. If we set them equal, we get the new operating point.

Linear estimate of the change of the operating rate

\Delta q \approx \frac{\delta_{\text{in}} + \delta_{\text{out}}}{1/J + s_{\text{out}}}

where:

$\Delta q$ change of the operating rate, m³/d
$\delta_{\text{in}}$ rise of the inflow curve at the same rate (for example, a growth of the reservoir pressure), MPa
$\delta_{\text{out}}$ fall of the outflow curve at the same rate (for example, from a lower wellhead pressure), MPa
$J$ productivity index; $1/J$ is the magnitude of the slope of the inflow curve, m³/(d·MPa)
$s_{\text{out}}$ slope of the outflow curve at the operating point, MPa/(m³/d)

Units: metric oilfield units (MPa, m³/d)

Holds when

small changes, for which the curves near the operating point can be replaced by straight lines
a change of $J$ also changes the slope of the inflow; for big changes (a reservoir treatment, a change of tubing) the whole new point is calculated

The denominator is the “stiffness” of the system. If $s_{\text{out}} \ll 1/J$ , the rate responds to a shift of either curve almost as $J\delta$ , and the reservoir limits the well. If $s_{\text{out}} \gg 1/J$ , the response is about $\delta/s_{\text{out}}$ and hardly depends on $J$ , and the lift system limits the well. On the falling branch of the outflow curve, $s_{\text{out}} < 0$ , so the denominator is small, and small disturbances change the rate strongly. This is the instability near the natural flow limit.

Operation near the natural flow limit

In year 4 of the forecast the lesson's well works at 87 m³/d, and the minimum of its outflow curve is at about 187 m³/d at 14.7 MPa. The operating point lies on the falling branch. The outflow curve falls there, but it is flatter than the inflow curve, so for slow changes (in a quasi-static sense) the operating mode is stable (stability of the operating mode). In practice it is not reliable: the velocity of the mixture is low, the gas slips, and the flow comes in slugs (plugs of liquid separated by gas). The denominator of the linear estimate is 0.019 here against 0.041 in year 0, and the response to disturbances is twice as strong. A rise of the wellhead pressure by 0.2 MPa lowers the rate from 87 to 63 m³/d, and a rise by 0.3 MPa stops the well. In year 0 the same rise of 0.2 MPa would cost 3% of the rate.

So we set the date of the change to artificial lift with a margin. It is not the time when the curves stop crossing, but the time when the operating point comes close to the minimum of the outflow curve, or when the oil rate becomes lower than a pump or gas lift would give after counting the costs. We use the same sign, an operating point to the left of the minimum, when we choose tubing for the late period.

Connections to other topics

Nodal analysis of the production system — the inflow and outflow curves, the operating point and stability: the whole analysis is built on them.
Chokes, surface pressure losses and well temperature — the same well at the surface: chokes, the flowline, the separator and the temperature of the flow.
Inflow performance relationship (IPR) — the composite inflow curve, the skin factor and the forecast of the curve as the reservoir is depleted.
Multiphase flow in tubing — the outflow curve of liquid with gas, its minimum and the role of the tubing diameter.
Artificial lift: when it is needed and how to choose the method — what to do when a well stops flowing naturally, and how to choose a lift method.
Gas lift: principles and choice of operating conditions — how gas injection uses the maximum in the dependence of the rate on the gas–liquid ratio.
Sensitivity analysis and scenarios — the same tools (spider and tornado charts, ranges) in the economics of a project.
Pressure build-up analysis — how permeability and skin factor, the most influential parameters, are found more exactly.

Check yourself

1
A low-productivity well ( $J$ of about 7.5 m³/(d·MPa)) that produces oil with dissolved gas has 89 mm tubing and a rate of 61 m³/d. Which is more likely to give more after a workover: 114 mm tubing or 73 mm tubing?
1. A.73 mm: less gas slippage
2. B.114 mm: less friction
3. C.The only option is to lower the pressure in the separator
4. D.The diameter does not affect the rate of a low-productivity well
Show the answer
Correct answer: A. 73 mm: less gas slippage
At a low rate the friction in the tubing is small, while in wide tubing the gas slips through the slow liquid and the column gets heavier. In the lesson's model, at this productivity the 73 mm tubing gives 70 m³/d, the 89 mm tubing gives 61 and the 114 mm tubing gives 46.
2
A well has a productivity index of 25 m³/(d·MPa), and the slope of its outflow curve at the operating point is 0.012 MPa per m³/d. Lowering the wellhead pressure lowers the outflow curve by 0.6 MPa. Estimate the gain in rate. What limits the well more?
Show the answer
$1/J = 0.04$ MPa per m³/d, and $\Delta q \approx 0.6/(0.04 + 0.012) \approx 11.5$ m³/d. The slope of the outflow is three times smaller than the slope of the inflow, so the reservoir mainly limits the well. A treatment of the near-wellbore zone that increases $J$ will give more than a fight for a lower wellhead pressure.
3
On a tornado chart the longest bars are for the skin factor and the permeability, and the bar of the tubing flow model is as long as the bar of the wellhead pressure. What should we do first?
1. A.Lower the wellhead pressure at once, since it is a parameter we control
2. B.Double the ranges of all the parameters
3. C.Nothing: tornado charts are not used for wells
4. D.Find the skin and permeability more exactly with a well test and tune the flow model to a measurement, and only then choose an intervention
Show the answer
Correct answer: D. Find the skin and permeability more exactly with a well test and tune the flow model to a measurement, and only then choose an intervention
Long bars of data show that the decision depends most on their uncertainty, so we make them more exact first. The bar of the model is comparable to the effect of the wellhead pressure. While the model is not tuned, a difference between cases of this size is not proved.
4
In the forecast the reservoir pressure in years 3, 4 and 5 is 19.0, 18.2 and 17.5 MPa. The natural flow limit at the forecast water cut is 17.0, 17.8 and 18.5 MPa. When will the well stop flowing naturally, and when should artificial lift be prepared?
Show the answer
The margin, $\bar p_R$ minus the limit, is 2.0, 0.4 and −1.0 MPa. Natural flow will stop between years 4 and 5. The lift should be ready by year 3 or 4. In year 4 the margin is small, the operating point is already to the left of the minimum of the outflow curve, and the operating mode is not reliable. Preparing the equipment takes months.
5
After an acid treatment the rate grew by 140 m³/d. A calculation showed that new tubing before the treatment would have given +4 m³/d. What should we expect from changing the tubing after the treatment?
1. A.More: after the treatment the tubing limits the well more
2. B.Less: the treatment has already used up the possibilities
3. C.The same +4 m³/d
4. D.The rate will fall
Show the answer
Correct answer: A. More: after the treatment the tubing limits the well more
The treatment turned the inflow curve, the rate grew, and the friction in the tubing became a noticeable part of the losses: the outflow curve is steeper at the new operating point. Now wider tubing gives tens of cubic metres per day. In the lesson's example it is about 45 m³/d.

Updated: 19 October 2026