Pressure measurements in producing reservoirs: barriers and accuracy

Quick refresher

In a producing accumulation the pressure profile is a snapshot of how the formation has responded to production and injection. It carries two kinds of information: how far the pressure has fallen from its initial value, and what shape the profile has. We can interpret it only if we know the production and injection of each unit.
Vertical flow tilts the line. For upward flow the gradient is larger than the hydrostatic one, for downward flow it is smaller, by $\mu u_z/k_z$ . Vertical velocities are small. So the difference is visible only when the vertical permeability is low. Otherwise the unit is in vertical equilibrium and looks hydrostatic even while fluid flows through it.
A barrier shows up as a step in potential between units, but only if uneven depletion keeps it going. The steady-state step equals the difference between the own depletion rates of the units (how fast each unit would lose pressure on its own), multiplied by the time constant of the barrier. Units depleted in proportion to their pore volumes give no step, even if the shale is continuous.
To turn the step into the transmissibility of the barrier, we need the flow rates of the units (from production logging, a survey that measures how much each interval flows) and their pore volumes. They set the error of the result, not the pressure gauge.
A tight unit that gives its fluid to a neighbouring unit produces a parabolic potential profile. Its amplitude gives the vertical permeability. The probe of the tester sees the horizontal permeability. So a laminated unit (one made of many very thin layers) can give clean pretests even when its $k_z$ is very small.
Within one survey, the repeatability of the gauge matters. Between surveys and between wells, absolute accuracy and common depth matching (all depths matched to the same reference) matter. In a producing accumulation we reject supercharging only by mobility (the permeability divided by the viscosity, which the tester finds at each point). Repeat surveys give the equalisation time. A pulse from a neighbouring well reveals baffles that depletion does not show.

Key relationships

\dfrac{\dd p}{\dd z} = \rho g + \dfrac{\mu\,u_z}{k_z}

gradient with vertical flow; $u_z > 0$ is upward flow, $z$ is depth measured downward

\delta\Phi_b = \dfrac{\mu\,u_z\,b}{k_b}

potential step across a barrier of thickness $b$ and permeability $k_b$

\Delta_{\text{ss}} = (r_2 - r_1)\,\tau,\quad \tau = \dfrac{c_tV_1V_2}{(V_1 + V_2)\,T}

steady-state step between two units; $r_i = q_i/(c_tV_i)$ are the own depletion rates of the units

\Delta\Phi_{\max} = \dfrac{\mu\,u\,h}{2\,k_z}

potential difference across a tight unit of thickness $h$ that gives up its fluid to a neighbour

Common pitfalls

Taking a hydrostatic profile with no steps as proof of good vertical communication.
Taking a high point in a tight layer of a depleted accumulation for a less depleted layer, without checking the mobility.
Calculating the transmissibility of a barrier from the step without the flow rates of the units and their pore volumes, or without checking whether the step has had time to settle.
Comparing pressures from different years and different wells without allowing for the absolute accuracy of the gauges, the depth matching and the datum depth.
Reading a small deviation of the gradient from the hydrostatic one as flow, without checking the directional survey (the measured path of the well) and the fluid density.

After this lesson you can

explain how the pressure profile of a producing accumulation differs from the equilibrium one, and what production and injection data we need to interpret it
calculate the viscous component of the gradient for vertical flow, test whether it is significant given the errors, and estimate the effective vertical permeability
find the potential step across a barrier and, with the model of two units that exchange fluid (crossflow), link it to the difference in depletion rates, the transmissibility of the barrier and the equalisation time
recognise the parabolic profile of a tight unit that gives up its fluid to a neighbouring unit, and use it to estimate the vertical permeability of the unit
tell when the repeatability of the readings matters and when their absolute accuracy matters, and reject supercharged points without relying on straight lines
use repeat surveys, wells that cross faults, and pulses made by neighbouring wells to test barriers and the communication between blocks

A new well in a producing accumulation

Before production starts, the fluids have had geological time to come to equilibrium. In each hydraulically connected system, the pressure of the continuous phase (the phase that forms an unbroken path through the pores) lies on a straight line with the slope $\rho g$ . The task of the survey is to find these lines and the points where they cross (Pressure gradients, fluid contacts and capillary equilibrium). After the wells start to produce, the pressure at every point of the formation is set by the history of production and injection. It changes across the area and over time. A new well drilled into a producing accumulation sees this process at one place and at one moment, like an observation well.

An open hole (a well with no steel casing across the formations) is convenient for this. While the well is full of drilling mud, the thin layers do not exchange fluid through the hole, so the tester measures the pressure of each layer separately. This is impossible in a cased well (one lined with steel pipe) that is open to several formations. After shut-in the formations flow into each other through the wellbore, and the pressure in the well matches none of them (Build-up in a bounded reservoir and average reservoir pressure). The main problem in an open hole is supercharging of tight layers (extra pressure around the probe, caused by mud filtrate that has soaked into the rock). We come back to it below.

The profile holds two kinds of information. The pressure drop from the initial pressure shows how much fluid has been taken from a unit, compared with the pore volume that the unit drains. The shape of the profile — steps, slopes and bends — shows how the units are connected to each other. To compare points at different depths, we convert each pressure to one datum depth (a chosen reference depth), using the fluid density of the unit. We do it in the same way as when we bring readings to a datum. The result is the potential $\Phi = p + \rho g\,(z_{\text{datum}} - z)$ , which is the pressure that the fluid would have at the datum depth. In a unit where nothing flows, the potential is the same over the whole thickness. So the points of such a unit fall on a vertical line on a graph of potential against depth. Fluid flows from the higher potential to the lower one.

CalculatedFour typical potential profiles in an observation well, calculated with the models of the lesson's examples: a unit in vertical equilibrium; upward flow in a unit with low vertical permeability; two units separated by shale, with uneven depletion; a tight laminated unit above a permeable one. The horizontal axis is the potential relative to a reference value. Each graph has its own scale.

How to read a potential profile in a producing formation

Shape on the graph: A vertical line to the left of the initial level
What is happening in the formation: the unit is depleted and in vertical equilibrium
What to check: the drop against the withdrawals and the pore volume (see the section on repeat surveys)

Shape on the graph: A slope: the potential grows with depth
What is happening in the formation: upward flow in a unit with low vertical permeability
What to check: whether the slope is significant, the directional survey, the fluid density

Shape on the graph: A step between units
What is happening in the formation: a barrier, with uneven depletion
What to check: the flow rates of the units, whether the shale is continuous, supercharging

Shape on the graph: An arc inside a unit
What is happening in the formation: a tight unit gives its fluid to its neighbour
What to check: the mobility of the points, whether the top is impermeable

Shape on the graph: One point to the right of its neighbours
What is happening in the formation: most often supercharging
What to check: the mobility, a repeat pretest

Shape on the graph	What is happening in the formation	What to check
A vertical line to the left of the initial level	the unit is depleted and in vertical equilibrium	the drop against the withdrawals and the pore volume (see the section on repeat surveys)
A slope: the potential grows with depth	upward flow in a unit with low vertical permeability	whether the slope is significant, the directional survey, the fluid density
A step between units	a barrier, with uneven depletion	the flow rates of the units, whether the shale is continuous, supercharging
An arc inside a unit	a tight unit gives its fluid to its neighbour	the mobility of the points, whether the top is impermeable
One point to the right of its neighbours	most often supercharging	the mobility, a repeat pretest

All this applies to single-phase flow (only one fluid moves). If free gas has come out of solution in a unit, or if water and oil are moving, the gradient reflects both phases. Such profiles are covered in the lesson Two-phase flow and reservoir description from formation-tester data.

Vertical flow: the gradient is not hydrostatic

Vertical flow in a formation is common. Edge water (water from the aquifer around the accumulation) rises under a depleted oil rim (the thin ring of oil between the water below and the gas cap, the layer of free gas, above). Injection goes into one unit while production comes from another. Units that deplete unevenly pass fluid to each other. By Darcy's law, vertical flow is driven by a difference in potential, not by a difference in pressure. So the pressure gradient that the tester measures has two parts: a hydrostatic part and a viscous part. The viscous part is the pressure lost because the rock resists the flow.

Pressure gradient with vertical flow

\frac{\dd p}{\dd z} = \rho g + \frac{\mu\,u_z}{k_z}

where:

$\dd p/\dd z$ pressure gradient with depth, as given by the pretests, Pa/m
$z$ true vertical depth, measured downward, m
$\rho$ fluid density at reservoir conditions, kg/m³
$g$ acceleration of gravity, 9.81, m/s²
$u_z$ vertical Darcy velocity (the flow rate through a unit of horizontal area); positive for upward flow, m/s
$\mu$ fluid viscosity at reservoir conditions, Pa·s
$k_z$ effective vertical permeability to this fluid, over the measured interval and its surroundings; if the other phase is at residual saturation, the phase permeability, m²

Units: SI

Holds when

single-phase flow; the velocity $u_z$ and the permeability $k_z$ are constant over the measured interval, so the profile is a straight line
if $k_z$ changes with depth, the viscous component changes too: a thin tight layer gives a step (see the section on barriers)
depths are true vertical depths, and the fluid density is known independently (from the analysis of a sample)

The measured gradient minus the hydrostatic one is the viscous component $G_{\text{visc}} = \mu u_z/k_z$ . With upward flow the pressure grows with depth faster than the hydrostatic pressure, and with downward flow more slowly. On a graph of pressure against depth, the line for upward flow moves to the right of the hydrostatic line as depth increases.

In other unit systems (1)

Metric oilfield units

G\,[\text{MPa/m}] = 0.00981\,\rho\,[\text{g/cm}^3] + \frac{u_z\,[\text{m/d}]\;\mu\,[\text{mPa·s}]}{0.0853\,k_z\,[\text{mD}]}

$G$ is the gradient in MPa/m, $\rho$ is in g/cm³, $u_z$ is in m/d (cubic metres per day through a square metre), $\mu$ is in mPa·s, $k_z$ is in mD.

0.00981 MPa/m is the pressure of a one-metre column of liquid with a density of 1 g/cm³. 0.0853 is the constant of Darcy's law in metric oilfield units (Darcy's law).

A unit in which the viscous component is smaller than its error is said to be in vertical pressure equilibrium (vertical equilibrium). This does not mean that nothing flows. It also does not mean that water and oil separate easily under gravity: the phases need much higher vertical velocities to separate that way (Segregated flow and gravity displacement). But if we do find a slope, it gives a number that is almost impossible to get any other way: the effective vertical permeability on a scale of tens of metres vertically and hundreds of metres across the area. For this we need the vertical velocity $u_z$ . We take it from the material balance (a count of the fluid taken out and the fluid that has come in) or from the reservoir simulation model. We can check the movement of the contact with repeat pulsed-neutron logging in neighbouring wells (a log run in a cased well that shows where the oil, water and gas are).

Pretests of the water leg in a new well on the flank (six years after the start of production)

Depth, m: 2552.0
Pressure, MPa: 25.7265
Mobility, mD/(mPa·s): 34

Depth, m: 2558.0
Pressure, MPa: 25.7896
Mobility, mD/(mPa·s): 12

Depth, m: 2564.0
Pressure, MPa: 25.8555
Mobility, mD/(mPa·s): 58

Depth, m: 2570.0
Pressure, MPa: 25.9191
Mobility, mD/(mPa·s): 21

Depth, m: 2576.0
Pressure, MPa: 25.9854
Mobility, mD/(mPa·s): 9

Depth, m: 2582.0
Pressure, MPa: 26.0487
Mobility, mD/(mPa·s): 40

Depth, m: 2588.0
Pressure, MPa: 26.1128
Mobility, mD/(mPa·s): 27

Depth, m: 2594.0
Pressure, MPa: 26.1788
Mobility, mD/(mPa·s): 16

Depth, m	Pressure, MPa	Mobility, mD/(mPa·s)
2552.0	25.7265	34
2558.0	25.7896	12
2564.0	25.8555	58
2570.0	25.9191	21
2576.0	25.9854	9
2582.0	26.0487	40
2588.0	26.1128	27
2594.0	26.1788	16

The accumulation is the one from the lessons Reservoir pressure and temperature and Pressure gradients, fluid contacts and capillary equilibrium: a gas cap, an oil rim and edge water. The water leg is the water-filled part of the section below the oil. All the readings are valid (they passed the quality checks). The pressures are absolute and the depths are true vertical depths. The data are assumed for illustration.

Worked exampleUpward flow in the water leg and vertical permeability

Given (assumed data)

Pretests: eight valid readings from the table above, from 2552 to 2594 m
Formation water from the analysis: 1.092 g/cm³ at 20 °C, $B_w = 1.020$ ; density error ±0.005 g/cm³ (the water line from the analysis)
Water influx into the oil rim over the last year: 520 m³/d at reservoir conditions from the material balance (the error is assumed to be ±30%); area of the base of the rim 3.2 km²
Water viscosity at reservoir temperature: 0.45 mPa·s

Solution

A line through the eight points by the least-squares method, which finds the line that fits the points best (gradient from the points): $\bar z = 2573.0$ m, $\sum(z - \bar z)^2 = 1512$ m², $G = 10.770$ kPa/m, scatter $s = 0.80$ kPa, $s_G = 0.80/\sqrt{1512} = 0.021$ kPa/m.
Hydrostatic gradient of the water: $\rho_w = 1.092/1.020 = 1.0706$ g/cm³, $\rho_w g = 9.81\cdot 1.0706 = 10.503$ kPa/m, and its error is $9.81\cdot 0.005 = 0.049$ kPa/m.
Viscous component: $G_{\text{visc}} = 10.770 - 10.503 = 0.267$ kPa/m. Its error is $\sqrt{0.021^2 + 0.049^2} = 0.053$ kPa/m. The ratio is $0.267/0.053 \approx 5.0$ , so the upward flow is clearly detected.
Vertical velocity: $u_z = 520/(3.2\cdot 10^6) = 1.625\cdot 10^{-4}$ m/d. We assume here that the water enters the rim evenly through its base. If part of it comes from the flanks through the formation, $u_z$ at the well is smaller, and the $k_z$ found below is too high.
Vertical permeability from the metric form: $k_z = \dfrac{1.625\cdot 10^{-4}\cdot 0.45}{0.0853\cdot 0.000267} = 3.2$ mD.
Error: the viscous component is known to within $0.053/0.267 = 20$ %, and the inflow to ±30%. Together they give $\sqrt{20^2 + 30^2} = 36$ %, so $k_z$ lies between 2.1 and 4.4 mD.
Detection limit: two standard errors are 0.106 kPa/m. With the same inflow, a viscous component of this size would correspond to $k_z \approx 8$ mD. In a unit with higher vertical permeability the profile would look hydrostatic.

Answer: The water in the water leg is moving upward. The effective vertical permeability is about 3.2 ± 1.2 mD.

What it means. This is the vertical permeability of the whole water-bearing interval, including its thin shale layers. It is several times lower than the horizontal permeability of the sandstone. It decides how fast bottom water will rise towards the wells (Water and gas coning). The main uncertainty is the inflow from the material balance, not the readings. Before the calculation it is worth checking the directional survey. At an inclination of 25°, an error of 0.5° changes the vertical intervals by 0.41%, and the gradient by 0.044 kPa/m, which is 16% of the viscous component that we found. Supercharging does not get in the way here: all the points have mobilities of 9–58 mD/(mPa·s). The data in this example are assumed for illustration.

Accuracy: what matters in a producing accumulation

In an accumulation that is not producing, we need the error of the gradient to find the fluid density and the contact. Here it answers a different question: does the gradient differ from the hydrostatic one enough for us to say that fluid is flowing? The difference contains two independent errors. One is the error of the slope of the line through the pretests (standard error of the gradient). The other is the error of the hydrostatic gradient, which we calculate from the fluid density found by analysing a sample.

Significance of the deviation from the hydrostatic gradient

G_{\text{visc}} = G - \rho g,\qquad \sigma_{G_{\text{visc}}} = \sqrt{s_G^2 + (g\,\sigma_\rho)^2}

where:

$G,\ s_G$ gradient from the pretests of the unit, and its standard error, Pa/m
$\rho,\ \sigma_\rho$ fluid density in the formation from the analysis of a sample, and its standard error, kg/m³
$G_{\text{visc}}$ viscous component, an estimate of $\mu u_z/k_z$ , Pa/m
$\sigma_{G_{\text{visc}}}$ its standard error, Pa/m

Units: SI (in practice kPa/m)

Holds when

the errors of the slope and of the density are random and independent; the points of the line are valid and belong to one unit
a systematic scale error of the depths (a wrong directional survey) is not in the formula and is checked separately
we call the flow detected if $|G_{\text{visc}}|$ is larger than two or three of its errors. With few points the error of the slope is only a rough estimate, so the threshold is taken from Student's t coefficient (a factor from statistical tables that widens the threshold when there are few points)

A density error of 0.005 g/cm³ is 0.049 kPa/m. It is often larger than the error of a slope found from many points. So the water density from the analysis and the oil density from the sample set the detection limit for vertical flow.

We check the depth scale (whether the measured depths are stretched or squeezed compared with the true ones) separately. A constant shift of the depths does not tilt the line, but an error that grows with depth does. In an inclined well, an error $\delta\theta$ in the inclination angle $\theta$ changes the vertical intervals by a factor of about $1 - \delta\theta\,\tan\theta$ (Directional survey calculations). The gradient changes by the same share in the opposite direction. An error of half a degree at an angle of 25° is 0.4% of the gradient, about 0.04 kPa/m in water. That is comparable with the error of the viscous component itself. So we interpret the viscous component only after we are sure of the directional survey (depth accuracy).

Which accuracy matters for which conclusion

Conclusion: Gradient and viscous component
What is compared: points of one survey
What matters: repeatability of the gauge, relative depths, scale of the directional survey, fluid density
Size of the error: 0.02–0.05 kPa/m

Conclusion: Step between units
What is compared: points of one survey
What matters: repeatability, supercharging of tight points
Size of the error: about 1 kPa

Conclusion: Drop from the initial pressure
What is compared: surveys of different years
What matters: absolute accuracy of the gauges, depth matching, datum depth
Size of the error: 10–15 kPa

Conclusion: Block pressures from different wells
What is compared: wells at the same time
What matters: the same factors, and readings taken on the same date
Size of the error: 15–20 kPa

Conclusion	What is compared	What matters	Size of the error
Gradient and viscous component	points of one survey	repeatability of the gauge, relative depths, scale of the directional survey, fluid density	0.02–0.05 kPa/m
Step between units	points of one survey	repeatability, supercharging of tight points	about 1 kPa
Drop from the initial pressure	surveys of different years	absolute accuracy of the gauges, depth matching, datum depth	10–15 kPa
Block pressures from different wells	wells at the same time	the same factors, and readings taken on the same date	15–20 kPa

Approximate values for quartz gauges and good depth matching (pressure and depth accuracy, comparing wells). A constant error of the instrument does not change the gradients and steps within one survey, but it counts in full when surveys are compared.

Supercharging without straight lines. In an accumulation that is not producing, a supercharged point gives itself away by lying to the right of the line of valid points (pretest classes). In a producing accumulation there are no such straight lines. A high pressure in a thin layer can mean supercharging, or it can mean a less depleted layer. We cannot tell which from the point alone. So we reject points by mobility, using an absolute threshold that we have tested beforehand where straight lines do exist: in appraisal wells (wells drilled to size up a new discovery) before production started, or in units that we know to be in vertical equilibrium (how to recognise supercharging).

The danger of supercharging is higher in a producing accumulation. The density of the mud is chosen for the highest formation pressure in the section. So the overbalance on the depleted units (the amount by which the mud pressure exceeds the formation pressure) is larger than usual, by several megapascals. Filtration through the mud cake (the thin layer of mud solids on the wall of the hole) grows with the overbalance, and so does supercharging in the tight layers of the depleted units (model of filtrate injection).

Depleted and injection-supported units in one hole

In a producing accumulation the pressure can differ by several megapascals along the section. Some units are depleted. Others have kept their initial pressure or exceeded it because of injection. The mud must hold back the highest of them, so the depleted units are under a large overbalance. This raises the risk of lost circulation (mud flowing into the formation). It also raises the risk of differential sticking, where the pressure difference holds the drill string or the tester against the wall of the hole. The pressure profile of each new well gives a better idea of the pressures to expect in the next ones.
This is a teaching explanation, not an instruction. The mud density, the survey programme and the time spent at each test point are set by approved plans and well control procedures, and staff with up-to-date training carry them out (Well control: primary barrier and kick indicators).

A barrier in the section: the potential step

Take two permeable units, each in vertical equilibrium, separated by a thin tight layer. This can be a shale parting (in field practice, a shale break) or a cemented layer (barriers and baffles). If fluid crosses the parting, the profile is vertical in both units, and the potential changes by a step at the parting. The whole pressure difference that the crossflow needs falls on the tight layer. We find the step by extending the lines of both units to the middle of the parting and taking the difference of their pressures there. The standard error of the step is the square root of the sum of the squares of the errors of the two lines at that depth (error of a line far from the points).

Potential step across a barrier

\delta\Phi_b = \frac{\mu\,u_z\,b}{k_b}

where:

$\delta\Phi_b$ potential step: the potential of the unit that the crossflow comes from, minus the potential of the other unit, Pa
$u_z$ crossflow velocity through the barrier: the flow rate divided by the area of the barrier, m/s
$b$ thickness of the barrier (from the log), m
$k_b$ vertical permeability of the barrier, m²
$\mu$ viscosity of the fluid that flows across, Pa·s

Units: SI

Holds when

the barrier is uniform and continuous; its storage capacity (how much fluid it gives up as the pressure falls) is small, so the flow through it is almost steady-state
the units above and below the barrier are in vertical equilibrium and are filled with one fluid

The step is proportional to the crossflow. Without crossflow a barrier cannot be seen, however tight the shale is. Pressure readings show a barrier only when production forces fluid to flow through it.

In other unit systems (1)

Metric oilfield units

\delta\Phi_b\,[\text{MPa}] = \frac{u_z\,[\text{m/d}]\;\mu\,[\text{mPa·s}]\;b\,[\text{m}]}{0.0853\,k_b\,[\text{mD}]}

$u_z$ is in m/d, $\mu$ in mPa·s, $b$ in m, $k_b$ in mD.

0.0853 is the constant of Darcy's law in metric oilfield units.

This gives the main rule: an observed step confirms a barrier, but the lack of a step does not rule one out. There will be no step if the units deplete at the same rate (see the next section). There will be none if a barrier of huge area lets the needed crossflow through with a pressure difference smaller than the error of the readings. There will also be none if shut-in wells that are open to both units stand nearby, because the units equalise through their wellbores. Shale that shows on the logs of all the wells but gives no step on the profile is either discontinuous, with “windows” in it, or production has not created any crossflow through it.

The same pressure, different movement of water

A continuous shale, and a much tighter shale with a single “window”, can give the same pressures. Over weeks and months the pressure equalises through the window over the whole area. For waterflooding (pushing oil out with injected water) the two cases are different. Through the continuous shale, water seeps slowly everywhere. Through the shale with a window, it passes only where the front has arrived. Pressure spreads through the formation by diffusion, and quickly. Saturation is carried by the flow, and slowly. So a model that was tuned to the pressure is also checked against the movement of the water: against pulsed-neutron logging in cased wells, the water cut (the share of water in the produced liquid), and the order in which water breaks through in the units (History matching).

Differential depletion of two units

The most common cause of crossflow is differential depletion. The same wells drain the units, but the units give up fluid at different speeds relative to their pore volumes. They differ in productivity, in the thickness open to the well, and in injection. Picture each unit as a tank with the same potential over its whole area and thickness, and the barrier between them as a partition that lets fluid through (isolated blocks). Then the balance for each unit includes the withdrawal and the crossflow:

Material balance of two units with crossflow

\begin{aligned} c_tV_1\,\frac{\dd\Phi_1}{\dd t} &= -q_1 - q_{12}, \\ c_tV_2\,\frac{\dd\Phi_2}{\dd t} &= -q_2 + q_{12}, \qquad q_{12} = T\,(\Phi_1 - \Phi_2) \end{aligned}

where:

$\Phi_1,\ \Phi_2$ potentials of units 1 and 2, brought to one datum, MPa
$V_1,\ V_2$ pore volumes of the units in the area drained by the wells of the site, m³
$q_1,\ q_2$ net withdrawal from the units (production minus injection) at reservoir conditions, m³/d
$q_{12}$ crossflow from unit 1 to unit 2 through the barrier, m³/d
$T$ transmissibility of the barrier, $0.0853\,k_bA/(\mu b)$ , where $A$ is its area, m³/(d·MPa)
$c_t$ total compressibility of the rock and the fluids, 1/MPa

Units: metric oilfield: MPa, m³, m³/d, 1/MPa, days

Holds when

the potential is the same everywhere inside each unit: the unit is much more permeable than the barrier, and the drawdown around the wells (the extra pressure drop that drives fluid into a well) is ignored
the withdrawals are constant and the compressibility is constant; inflow from outside (edge water) is counted as a negative withdrawal
the storage capacity of the barrier itself is small; the flow is single-phase

The pressure of each unit falls at a rate that is proportional to its withdrawal and crossflow, divided by its storage capacity (capacitance) $c_tV$ , which is the volume of fluid that the unit gives up for each MPa of pressure drop. The crossflow grows with the potential difference, so in the end it evens out the rates of decline.

Steady-state step and time constant of the barrier

\begin{aligned} \Delta_{\text{ss}} &= (r_2 - r_1)\,\tau = \frac{q_2V_1 - q_1V_2}{T\,(V_1 + V_2)}, \\ r_i &= \frac{q_i}{c_t V_i}, \qquad \tau = \frac{c_t\,V_1V_2}{(V_1 + V_2)\,T} \end{aligned}

where:

$\Delta_{\text{ss}}$ potential difference $\Phi_1 - \Phi_2$ after the units have reached a common pseudo-steady-state regime (both losing pressure at the same rate), MPa
$r_1,\ r_2$ own depletion rates: how fast the potential of each unit would fall without crossflow, MPa/d
$\tau$ time constant of the barrier: the step grows as $1 - e^{-t/\tau}$ and reaches 95% of its steady-state value after $3\tau$ , d

Units: metric oilfield, as above

Holds when

constant withdrawals and equal initial potentials of the units
the compressibility is not in the steady-state step: it cancels in the right-hand form

The step equals the difference of the depletion rates multiplied by the equalisation time. The tighter the barrier and the more the rates differ, the larger the step. If the units give up fluid in proportion to their pore volumes ( $r_1 = r_2$ ), there is no step, whatever the barrier. The potential averaged over the pore volume does not depend on the crossflow, and it falls in the same way as in a single tank (see the advanced section).

The time $\tau$ is the same constant as for a block behind a baffle (time constant of a block). The only difference is that for two tanks of comparable size we put $V_1V_2/(V_1 + V_2)$ in place of $V_p$ . The table shows how shales of different permeability show up in the block of the example below.

A shale parting 3 m thick in a block of 3 × 2 km with uneven depletion of the units

$k_b$ , mD: 0.0002
$T$ , m³/(d·MPa): 28.4
$\tau$ , d: 728
Step after 2 years, kPa: 2082.1
Steady-state, kPa: 3289.8

$k_b$ , mD: 0.002
$T$ , m³/(d·MPa): 284
$\tau$ , d: 72.8
Step after 2 years, kPa: 329.0
Steady-state, kPa: 329.0

$k_b$ , mD: 0.02
$T$ , m³/(d·MPa): 2843
$\tau$ , d: 7.28
Step after 2 years, kPa: 32.9
Steady-state, kPa: 32.9

$k_b$ , mD: 0.2
$T$ , m³/(d·MPa): 28,433
$\tau$ , d: 0.73
Step after 2 years, kPa: 3.3
Steady-state, kPa: 3.3

$k_b$ , mD	$T$ , m³/(d·MPa)	$\tau$ , d	Step after 2 years, kPa	Steady-state, kPa
0.0002	28.4	728	2082.1	3289.8
0.002	284	72.8	329.0	329.0
0.02	2843	7.28	32.9	32.9
0.2	28,433	0.73	3.3	3.3

Calculated with the two-unit model and the data of the example below: own depletion rates 8.503 and 3.987 kPa/d, oil 1.2 mPa·s. The data are assumed for illustration.

CalculatedGrowth of the potential step between the units for shales of four permeabilities (two-unit model, data of the example below). At first all the curves run together, because the units deplete at their own rates. Then the crossflow stops the growth at the level

(r_2 - r_1)\tau

. The dashed line is 3 kPa, the approximate limit for detecting a step in one survey.

Pretests of the observation well after two years of production

Depth, m: 2703.0
Pressure, MPa: 25.2121
Mobility, mD/(mPa·s): 140
Unit: unit 2 (upper)

Depth, m: 2706.0
Pressure, MPa: 25.2324
Mobility, mD/(mPa·s): 95
Unit: unit 2 (upper)

Depth, m: 2709.5
Pressure, MPa: 25.2589
Mobility, mD/(mPa·s): 210
Unit: unit 2 (upper)

Depth, m: 2712.5
Pressure, MPa: 25.2800
Mobility, mD/(mPa·s): 120
Unit: unit 2 (upper)

Depth, m: 2715.0
Pressure, MPa: 25.2991
Mobility, mD/(mPa·s): 160
Unit: unit 2 (upper)

Depth, m: 2720.5
Pressure, MPa: 25.6670
Mobility, mD/(mPa·s): 60
Unit: unit 1 (lower)

Depth, m: 2724.0
Pressure, MPa: 25.6936
Mobility, mD/(mPa·s): 85
Unit: unit 1 (lower)

Depth, m: 2728.0
Pressure, MPa: 25.7210
Mobility, mD/(mPa·s): 44
Unit: unit 1 (lower)

Depth, m: 2732.0
Pressure, MPa: 25.7511
Mobility, mD/(mPa·s): 70
Unit: unit 1 (lower)

Depth, m: 2736.5
Pressure, MPa: 25.7844
Mobility, mD/(mPa·s): 38
Unit: unit 1 (lower)

Depth, m: 2740.0
Pressure, MPa: 25.8086
Mobility, mD/(mPa·s): 52
Unit: unit 1 (lower)

Depth, m	Pressure, MPa	Mobility, mD/(mPa·s)	Unit
2703.0	25.2121	140	unit 2 (upper)
2706.0	25.2324	95	unit 2 (upper)
2709.5	25.2589	210	unit 2 (upper)
2712.5	25.2800	120	unit 2 (upper)
2715.0	25.2991	160	unit 2 (upper)
2720.5	25.6670	60	unit 1 (lower)
2724.0	25.6936	85	unit 1 (lower)
2728.0	25.7210	44	unit 1 (lower)
2732.0	25.7511	70	unit 1 (lower)
2736.5	25.7844	38	unit 1 (lower)
2740.0	25.8086	52	unit 1 (lower)

The shale parting is at 2716–2719 m on the log, and it can be traced in all the wells of the block. Water-based mud. All the readings are valid, the pressures are absolute and the depths are true vertical depths. The data are assumed for illustration.

Worked exampleTransmissibility of a shale parting from the pressure step

Given (assumed data)

Block: 3 × 2 km; unit 2 (upper): 14 m, porosity 0.21; unit 1 (lower): 22 m, porosity 0.19
Net withdrawals from production logging: unit 2: 300 m³/d, unit 1: 200 m³/d at reservoir conditions; assumed errors: ±10% for each flow rate, ±15% for the pore volumes from the maps
Formation: total compressibility $2.0\cdot10^{-3}$ MPa⁻¹, undersaturated oil (the pressure is above the bubble point pressure, so there is no free gas), 1.2 mPa·s; production has gone on for two years
Survey: the table above

Solution

Lines of the units: upper $G = 7.264$ kPa/m, lower $G = 7.271$ kPa/m. Both correspond to oil with a density of about 0.74 g/cm³, so each unit is in vertical equilibrium. At the middle of the parting (2717.5 m) the lower line is to the right of the upper one by $\Delta = 328.7 \pm 1.0$ kPa.
Pore volumes: $V_2 = 6.0\cdot10^6\cdot 14\cdot 0.21 = 17.64\cdot 10^6$ m³, $V_1 = 25.08\cdot 10^6$ m³. Own depletion rates: $r_2 = 300/(0.002\cdot 17.64\cdot10^6) = 8.503$ kPa/d, $r_1 = 3.987$ kPa/d. The difference is 4.516 kPa/d.
Time constant: $\tau = \Delta/(r_2 - r_1) = 328.7/4.516 = 72.8$ d. This is much less than two years, so the step has settled.
Transmissibility: $V_1V_2/(V_1 + V_2) = 25.08\cdot 17.64/42.72 = 10.356\cdot 10^6$ m³, $T = c_tV_1V_2/\big((V_1 + V_2)\tau\big) = 0.002\cdot 10.356\cdot 10^6/72.8 = 284.5$ m³/(d·MPa).
Permeability of the shale: $k_b = T\mu b/(0.0853\,A) = 284.5\cdot 1.2\cdot 3/(0.0853\cdot 6.0\cdot 10^6) = 0.0020$ mD. Crossflow from the lower unit to the upper one: $T\Delta = 284.5\cdot 0.3287 = 93.5$ m³/d.
Error. Use $T = (q_2V_1 - q_1V_2)/\big(\Delta(V_1 + V_2)\big)$ , where $\Delta(V_1 + V_2) = 0.3287\cdot 42.72 = 14.042\cdot 10^6$ . The contribution of $q_2$ is $25.08/14.042\cdot 30 = 53.6$ . The contribution of $q_1$ is $17.64/14.042\cdot 20 = 25.1$ . The contribution of each pore volume is $V_1V_2(q_1 + q_2)/\big(\Delta(V_1 + V_2)^2\big)\cdot 0.15 = 55.3$ m³/(d·MPa). Together they give $\sqrt{53.6^2 + 25.1^2 + 2\cdot 55.3^2} = 98$ , that is ±34%. We treat the errors as independent here. An error of the area that is common to both units does not affect $T$ , because it cancels. So our estimate of the error is, if anything, on the cautious side (too large). The error of the step itself (1.0 kPa) is negligible.

Answer: The transmissibility of the shale parting is about 285 ± 98 m³/(d·MPa), and its vertical permeability is about 0.0020 mD. The parting is continuous and almost impermeable, although it lets 93.5 m³/d through.

What it means. The step is measured to within one per cent, but the transmissibility only to within a third. The information about the barrier lies in the flow rates of the units and their pore volumes, not in the pressure gauge. The compressibility did not enter the result, because it cancelled. The step exists only because the upper unit depletes 2.1 times faster than the lower one. If it were produced at 140.7 m³/d instead of 300, which is in proportion to its pore volume, there would be no step with the same shale. In the model we treat the units as separate, and the parting is given by a transmissibility that was also checked against the movement of the water. If the time constant were comparable with the production time, the step would still be growing, and a calculation from the steady-state step would give a time constant that is too short. Then we solve the full equation (see the advanced section). The data in this example are assumed for illustration.

A tight unit that gives its fluid to its neighbour

Another common case is a thick unit with low vertical permeability that lies on top of a permeable one. The wells are open to both, but mainly the permeable unit produces. The tight unit gives up its fluid by flowing down into it. If the top of the tight unit is impermeable, the flow inside it is only vertical. Once the two units have reached a common pseudo-steady-state regime, every layer of the tight unit loses pressure at the same rate. The flow velocity grows from zero at the top to its largest value at the boundary with the permeable unit. So the potential profile is not straight but curved. It is a parabola that reaches its largest value at the top.

Potential profile in a tight unit that flows into its neighbour

\begin{aligned} \Phi(\zeta) - \Phi_1 &= \Delta\Phi_{\max}\left[\frac{2\zeta}{h} - \left(\frac{\zeta}{h}\right)^2\right], \\ \Delta\Phi_{\max} &= \frac{\mu\,u\,h}{2\,k_z}, \qquad u = \frac{q}{A}\cdot\frac{V_2}{V_1 + V_2} \end{aligned}

where:

$\zeta$ height above the base of the tight unit, that is above the boundary with the permeable unit, m
$h$ thickness of the tight unit, m
$\Phi_1$ potential of the permeable unit (it is in vertical equilibrium), Pa
$u$ crossflow velocity at the boundary between the units: the share of the withdrawal that comes from the tight unit, divided by the area, m/s
$q,\ A$ net withdrawal of the site (all of it through the permeable unit) and its area, m³/s, m²
$V_1,\ V_2$ pore volumes of the permeable and the tight unit, m³
$k_z$ effective vertical permeability of the tight unit, m²

Units: SI

Holds when

the top of the tight unit is impermeable, the flow in it is vertical, and no fluid flows to the wells along the tight unit itself
common pseudo-steady-state regime: both units lose pressure at the same rate (the time criterion is given below)
single-phase flow, and $k_z$ is the same over the thickness of the unit

The crossflow through the boundary is the share of the total withdrawal that equals the share of the tight unit in the pore volume. With common depletion every cubic metre of pore space gives up the same amount. The factor one half in the formula for the amplitude is there because the flow velocity in the unit grows linearly from zero, so on average it is half the velocity at the boundary.

In other unit systems (1)

Metric oilfield units

\Delta\Phi_{\max}\,[\text{MPa}] = \frac{u\,[\text{m/d}]\;\mu\,[\text{mPa·s}]\;h\,[\text{m}]}{2\cdot 0.0853\,k_z\,[\text{mD}]}

$u$ is in m/d, $\mu$ in mPa·s, $h$ in m, $k_z$ in mD; $q$ is in m³/d, $A$ in m².

0.0853 is the constant of Darcy's law in metric oilfield units.

When the profile of the tight unit settles

t_{Dh} = \frac{k_z\,t}{\phi\,\mu\,c_t\,h^2} > 2

where:

$t_{Dh}$ dimensionless time of the vertical drainage of the unit, —
$t$ time since the start of withdrawal, s
$\phi,\ c_t,\ h$ porosity, total compressibility and thickness of the tight unit, —, 1/Pa, m

Units: SI

Holds when

the withdrawal from the permeable unit is roughly constant; all of it goes through that unit, and the crossflow from the tight unit builds up from zero
the limit holds for any ratio of the pore volumes of the units. If a constant rate were taken from the tight unit itself from the start, $t_{Dh} > 0.5$ would be enough (see the advanced section)

At $t_{Dh} = 2$ the potential difference across the unit reaches at least 99% of its steady-state value. At $t_{Dh} = 0.5$ in the lesson's example it reaches only 81%. The crossflow from the tight unit is small at first and builds up while the permeable unit leads it in pressure decline (see the advanced section). The time grows as the square of the thickness: a unit twice as thick takes four times as long to settle.

In other unit systems (1)

Metric oilfield units

t_{Dh} = \frac{0.0853\,k_z\,[\text{mD}]\;t\,[\text{d}]}{\phi\;\mu\,[\text{mPa·s}]\;c_t\,[\text{MPa}^{-1}]\;h^2\,[\text{m}^2]}

$k_z$ is in mD, $t$ in days, $\mu$ in mPa·s, $c_t$ in 1/MPa, $h$ in m.

0.0853 is the same constant of Darcy's law: it converts mD·MPa/(mPa·s) into m²/d.

The tester shows such a profile clearly, for an important reason: the probe and the tight unit “see” different permeabilities. The mobility from a pretest mostly reflects the horizontal permeability of a few centimetres of rock at the wall (which permeability the probe sees). Vertical flow through a unit with frequent thin shale layers is controlled by the harmonic mean across the layers. This average is dominated by the tightest layers, and it can be hundreds or thousands of times smaller than the horizontal permeability (Reservoir heterogeneity and permeability averaging). So a laminated unit gives clean pretests with no supercharging, and at the same time a strongly curved profile. But if the unit is tight in the horizontal direction too, its pretests are supercharged, and we must tell the bend of the profile from supercharging by the mobility.

Pretests of the observation well at a site with a laminated unit

Depth, m: 2749.0
Pressure, MPa: 26.4723
Mobility, mD/(mPa·s): 28
Unit: laminated unit

Depth, m: 2751.5
Pressure, MPa: 26.4837
Mobility, mD/(mPa·s): 21
Unit: laminated unit

Depth, m: 2754.0
Pressure, MPa: 26.4844
Mobility, mD/(mPa·s): 32
Unit: laminated unit

Depth, m: 2756.5
Pressure, MPa: 26.4754
Mobility, mD/(mPa·s): 18
Unit: laminated unit

Depth, m: 2759.0
Pressure, MPa: 26.4598
Mobility, mD/(mPa·s): 25
Unit: laminated unit

Depth, m: 2761.5
Pressure, MPa: 26.4614
Mobility, mD/(mPa·s): 120
Unit: permeable unit

Depth, m: 2766.0
Pressure, MPa: 26.4929
Mobility, mD/(mPa·s): 85
Unit: permeable unit

Depth, m: 2771.0
Pressure, MPa: 26.5303
Mobility, mD/(mPa·s): 150
Unit: permeable unit

Depth, m: 2776.0
Pressure, MPa: 26.5659
Mobility, mD/(mPa·s): 96
Unit: permeable unit

Depth, m	Pressure, MPa	Mobility, mD/(mPa·s)	Unit
2749.0	26.4723	28	laminated unit
2751.5	26.4837	21	laminated unit
2754.0	26.4844	32	laminated unit
2756.5	26.4754	18	laminated unit
2759.0	26.4598	25	laminated unit
2761.5	26.4614	120	permeable unit
2766.0	26.4929	85	permeable unit
2771.0	26.5303	150	permeable unit
2776.0	26.5659	96	permeable unit

The laminated unit is at 2748–2760 m, and a continuous shale covers it from above. The permeable unit is at 2760–2778 m. Low-fluid-loss mud (mud that loses little filtrate into the rock). The pressures are absolute and the depths are true vertical depths. The data are assumed for illustration.

Worked exampleVertical permeability of a laminated unit from the bend of the profile

Given (assumed data)

Laminated unit: 12 m, porosity 0.16; the top is impermeable
Permeable unit: 18 m, porosity 0.22; site area 4.0 km²; all the withdrawal, 300 m³/d at reservoir conditions, comes from this unit; assumed errors: ±10% for the withdrawal, ±15% for the share of the tight unit in the pore volume from the maps
Fluid and rock: oil 1.2 mPa·s, total compressibility $2.0\cdot10^{-3}$ MPa⁻¹; mud filtrate 0.40 mPa·s
Survey: the table above; production has gone on for more than a year

Solution

Line of the permeable unit through four points: $G = 7.236$ kPa/m, oil in vertical equilibrium. At the boundary of the units (2760 m) the pressure is 26.4502 MPa.
Excess of the points of the laminated unit over this line: 16.8, 50.5, 77.6, 95.0 and 101.7 kPa at heights of 1.0, 3.5, 6.0, 8.5 and 11.0 m above the boundary. Shape factors $2\zeta/h - (\zeta/h)^2$ : 0.160, 0.498, 0.750, 0.915, 0.993.
A least-squares fit with one parameter, the amplitude: $\Delta\Phi_{\max} = \sum f e/\sum f^2 = 274.0/2.6595 = 103.0$ kPa. The scatter of the points is $s = 0.68$ kPa, and the error is $0.68/\sqrt{2.6595} = 0.42$ kPa. The points fall on a parabola, so the profile has settled.
Crossflow velocity: the share of the tight unit is $1.92/(3.96 + 1.92) = 0.3265$ (pore volumes per square metre: $12\cdot 0.16$ and $18\cdot 0.22$ m). The crossflow is $300\cdot 0.3265 = 97.95$ m³/d, and $u = 97.95/(4.0\cdot 10^6) = 2.449\cdot 10^{-5}$ m/d.
Vertical permeability: $k_z = \dfrac{1.2\cdot 2.449\cdot 10^{-5}\cdot 12}{2\cdot 0.0853\cdot 0.1030} = 0.020$ mD. The error is $\sqrt{10^2 + 15^2} = 18$ %, which is about ±0.004 mD.
Checks. Time to settle ( $t_{Dh} = 2$ ): $2\cdot 0.16\cdot 1.2\cdot 0.002\cdot 144/(0.0853\cdot 0.020) = 65$ d, which is much less than the production time. Supercharging: by the supercharging model, for a mud whose filtration is five times lower than in the examples of the lesson Supercharging and permeability from formation-tester pretests, the excess for mobilities of 18–32 mD/(mPa·s) is from 0.49 to 0.85 kPa. Against a background of about a hundred kilopascals this is not important.

Answer: The effective vertical permeability of the laminated unit is about 0.020 ± 0.004 mD, and the horizontal permeability from the pretests is 7–13 mD. The ratio

k_v/k_h

is of the order of 0.002.

What it means. A unit with good reservoir properties according to the pretests gives up its oil only through its base, and slowly: its potential lags behind the permeable unit by 103.0 kPa. In waterflooding the water will go through the permeable unit and will displace oil from the laminated one mainly from below. This must be built into the model as separate layers with low vertical transmissibility (Displacement in heterogeneous reservoirs). If the points of the laminated unit had fallen on a vertical line, its vertical permeability would have been above the detection limit. If high points with low mobility had appeared near the parabola, we would have checked them for supercharging first. The data in this example are assumed for illustration.

CalculatedThe potential profile from the example: the points of the permeable unit fall on a vertical line, and the points of the laminated unit fall on a parabola fitted by least squares. The potential is measured from the potential of the permeable unit.

Repeat surveys, blocks and faults

A tester survey in an open hole is usually made once in each well, before the casing is run. It is repeated in the next new wells. In wells that are already cased, the pressure of the formations comes from permanent downhole gauges in wells that produce each formation separately, and from special testers for cased holes. A series of surveys over time gives what no single survey can give: the rates of change.

Pressure decline and drained volume. The potential of a site, averaged over the pore volume, falls by the total net withdrawal divided by $c_tV$ , as in a closed area, whatever the crossflows inside. If the pressure in one part of the accumulation falls faster than the withdrawals and the mapped volume imply, then the wells drain a smaller volume. A fault or a pinch-out (where the layer thins out to nothing) cuts off part of it. This is one of the most important uses of the surveys: checking the drainage areas. Here we compare surveys of different years and different wells. So the absolute accuracy of the instruments and a single datum depth are important.

Blocks and faults. Take two wells in neighbouring blocks, surveyed at the same time. Each block shows vertical equilibrium, but at different levels. That means the fault holds back the crossflow. Equal levels mean either good communication over the production time, or that the blocks deplete in proportion to their volumes. A fault between blocks is described by the same two-tank model as the shale parting. The only difference is that the barrier stands vertically, and its transmissibility is set by the fault zone (fault transmissibility in the model). Wells that cross a fault are especially valuable. In one survey, with one gauge, they measure both blocks. The pressure difference is then known to about a kilopascal, instead of 15–20 kPa when different wells are compared.

Worked exampleA fault between blocks from three surveys over time

Given (assumed data)

Blocks: west: pore volume from the maps 30 million m³, net withdrawal 350 m³/d; east: 45 million m³, 150 m³/d (injection makes up for most of the withdrawal)
Formation: total compressibility $2.0\cdot10^{-3}$ MPa⁻¹, oil 1.2 mPa·s; the fault is open to flow over a length of 2.5 km, with a formation height of 30 m
Surveys: three new wells crossed the fault 150, 300 and 600 days after the start of production. The pressure in the east block is higher than in the west block by 0.470, 0.728 and 0.948 MPa (the error of each step is assumed to be ±2 kPa)

Solution

The step grows as $\Delta(t) = \Delta_{\text{ss}}(1 - e^{-t/\tau})$ . For the times $t$ and $2t$ : $\Delta(2t)/\Delta(t) - 1 = e^{-t/\tau}$ . Here $x = 0.728/0.470 - 1 = 0.5489$ , and $\tau = 150/(-\ln 0.5489) = 150/0.5998 = 250.1$ d.
Steady-state step: $\Delta_{\text{ss}} = 0.470/(1 - 0.5489) = 0.470/0.4511 = 1.042$ MPa.
Check with the third well: $1.042\cdot(1 - e^{-600/250.1}) = 1.042\cdot 0.9092 = 0.947$ MPa against the measured 0.948. They agree within the error.
Transmissibility of the fault: $V_1V_2/(V_1 + V_2) = 30\cdot 45/75 = 18$ million m³, $T = 0.002\cdot 18\cdot 10^6/250.1 = 143.9$ m³/(d·MPa). The crossflow from east to west after the step settles is $143.9\cdot 1.042 = 150$ m³/d.
Check of the pore volumes: from the pressures, $\Delta_{\text{ss}}/\tau = 1.042/250.1 = 4.1663$ kPa/d. From the withdrawals and the maps, $350/(0.002\cdot 30\cdot 10^6) - 150/(0.002\cdot 45\cdot 10^6) = 5.8333 - 1.6667 = 4.1666$ kPa/d. The maps are confirmed.
Fault zone transmissibility: $k_f/l_f = T\mu/(0.0853\,A_f) = 143.9\cdot 1.2/(0.0853\cdot 75{,}000) = 0.027$ mD/m, where $A_f$ is the area of the fault that is open to flow.
Accuracy: step errors of ±2 kPa give $\tau$ an error of about ±6 d. If the steps were found by comparing different wells (±16 kPa), the error would grow to ±48 d.

Answer: The fault passes fluid. The time constant is about 250 d and the transmissibility is 143.9 m³/(d·MPa). After the step settles, the east block gives the west block about 150 m³/d.

What it means. A single survey would show only a step. It could not tell a fault that passes fluid from one that does not. With a sealing fault the step grows without limit, at a rate equal to the difference between the own depletion rates. With a fault that passes fluid it levels off. A series of surveys gives the equalisation time, and with it the transmissibility, without the block flow rates, which are needed only for a check. We compare this result with the estimate of the fault transmissibility from a well test of a neighbouring well (transmissibility from a well test). The well test sees the nearby part of the fault, and the surveys see the whole fault. The model assumes constant withdrawals. If the withdrawals changed, we compare the steps with a full model. The data in this example are assumed for illustration.

If there are more than two blocks or units, and the withdrawals change in time, we do not use formulas. We solve the same system of balances for many tanks. This is the simplified multi-tank material balance model (each block is one tank; Lumped multi-tank material balance model) or, more often, a reservoir simulation model. It also happens that a fault that held a pressure difference for a long time starts to pass fluid. The pressure in a depleted block suddenly rises, with no visible cause in the injection. So we check the fault transmissibility found from the early surveys against the later ones.

Neighbouring wells and pulse interference testing

A pressure profile is a snapshot of one point of a transient pressure field (a field that is still changing). Near a producing well the pressure is lower than the average of the unit by the drawdown of that well, and more so in the unit that produces better. Near an injection well it is higher. If a neighbouring well changes its rate a few days before the survey, it changes the pressures of the units in different ways, and it can create a step or hide one. So in the interpretation we note the distances to the working wells and the history of their rates. When we compare the survey with a model, we take the pressure of the model cells at the place of the observation well and on the day of the survey (which pressure to compare with which).

We can also use this effect on purpose. Before the survey we change the rate of a neighbouring well that is open to only one unit, for example we shut in an injection well. The unit of that well responds quickly. The adjacent unit receives the disturbance only through the barrier, and it lags behind. Such a pulse survey is a kind of interference test (a test in which we change the rate in one well and watch the pressure in another). The difference is that we read the response not from a record of pressure in time at the responding well, but from a profile through the whole thickness of the formation, taken in one day. Vertical interference with two probes of one tool (observation probe) sees centimetres and metres. A pulse survey sees hundreds of metres between wells.

Lag of the unit behind the barrier during a pulse

\tau_{\text{loc}} = \frac{\phi_1 h_1\,c_t\,\mu\,b}{k_b},\qquad \Delta p_2 - \Delta p_1 \approx \tau_{\text{loc}}\,\frac{\dd(\Delta p_2)}{\dd t}

where:

$\tau_{\text{loc}}$ local time constant: how long unit 1 behind the barrier takes to catch up with the pressure change of unit 2 near the observation well, s
$\phi_1,\ h_1$ porosity and thickness of unit 1 (its storage capacity per unit area is $\phi_1h_1c_t$ ), —, m
$b,\ k_b$ thickness and vertical permeability of the barrier, m, m²
$\Delta p_2,\ \Delta p_1$ pressure changes of the unit with the working well and of the unit behind the barrier, Pa

Units: SI

Holds when

the well that creates the pulse is open only to unit 2, and unit 1 receives the disturbance only through the barrier
lateral flow inside unit 1 is ignored; this overestimates the lag if unit 1 is permeable across the area
$\Delta p_2$ is the pressure of unit 2 at the observation well, including the crossflow. The barrier feeds fluid into unit 2 over the whole area that the pulse has reached. The leakage factor is $B = \sqrt{k_2h_2b/k_b}$ (the distance over which the shale can feed unit 2; $k_2$ and $h_2$ are the permeability and thickness of unit 2). If $B$ is smaller than the distance to the well that creates the pulse, the line-source solution for unit 2 alone strongly overestimates its response (see the example below). The line-source solution is the standard formula for the pressure around a well in an infinite reservoir
the approximate equality holds for an almost linear change of the pressure of unit 2 and for times much longer than $\tau_{\text{loc}}$ ; the exact relation is $\Delta p_2 - \Delta p_1 = \tau_{\text{loc}}\,\dd(\Delta p_1)/\dd t$

The pressure difference between the units during a pulse is the rate of change of pressure multiplied by the lag time. The principle is the same as for the steady-state step during depletion. The difference is that the rate here is the rate of change of the pressure of unit 2 at the observation well, not the difference of the depletion rates. Also, the lag time is set by the storage capacity of unit 1 alone, not by both units of the site, so it is longer.

In other unit systems (1)

Metric oilfield units

\tau_{\text{loc}}\,[\text{d}] = \frac{\phi_1\,h_1\,[\text{m}]\;c_t\,[\text{MPa}^{-1}]\;\mu\,[\text{mPa·s}]\;b\,[\text{m}]}{0.0853\,k_b\,[\text{mD}]}

$h_1$ and $b$ are in m, $c_t$ in 1/MPa, $\mu$ in mPa·s, $k_b$ in mD; $\tau_{\text{loc}}$ is in days.

0.0853 is the constant of Darcy's law in metric oilfield units.

Worked exampleA shale that depletion does not show, seen in a pulse survey

Given (assumed data)

Formation: the block from the example of the shale parting; upper unit 2: 150 mD, 14 m, porosity 0.21; lower unit 1: 22 m, porosity 0.19; oil 1.2 mPa·s, $c_t = 2.0\cdot10^{-3}$ MPa⁻¹
Shale between the units at this place: 3 m, permeability 0.2 mD
Pulse: an injection well that is open only to unit 2, 700 m from the observation well, was injecting 800 m³/d (at reservoir conditions); it was shut in 96 h before the survey

Solution

Without crossflow through the shale, unit 2 at the observation well would have fallen according to the line-source solution (Line-source solution and the pressure function, metric form): $x = 0.21\cdot 1.2\cdot 0.002\cdot 700^2/(0.01421\cdot 150\cdot 96) = 1.2069$ , $E_1(1.2069) = 0.1567$ , and $\Delta p_2 = 0.933\cdot 800\cdot 1.2/(150\cdot 14)\cdot 0.1567 = 0.4265\cdot 0.1567 = 0.0668$ MPa.
But the shale feeds unit 2. The local time constants are: for unit 1, $\tau_{\text{loc}} = 0.19\cdot 22\cdot 0.002\cdot 1.2\cdot 3/(0.0853\cdot 0.2) = 1.76$ d; for unit 2, $\tau_2 = 0.21\cdot 14\cdot 0.002\cdot 1.2\cdot 3/(0.0853\cdot 0.2) = 1.24$ d. The leakage factor is $B = \sqrt{150\cdot 14\cdot 3/0.2} = 177$ m. Both time constants are less than four days, and $B$ is much smaller than 700 m. During the pulse the shale has time to feed unit 2 over the whole area that the pulse has reached, so the solution for unit 2 alone does not apply here.
Calculation with the model of two units with crossflow (see the advanced section): by the time of the survey the pressure of unit 2 has fallen by 19.76 kPa, which is 3.4 times less than without crossflow, and the pressure of unit 1 has fallen by 9.13 kPa. The difference between the units is $19.76 - 9.13 = 10.63$ kPa.
The same shale during ordinary depletion of the block: $T = 0.0853\cdot 0.2\cdot 6.0\cdot 10^6/(1.2\cdot 3) = 28{,}433$ m³/(d·MPa), $\tau = 0.002\cdot 10.356\cdot 10^6/28{,}433 = 0.728$ d, and the step is $4.516\cdot 0.728 = 3.3$ kPa. This is at the detection limit.
For comparison: with a 2 mD shale, $\tau_{\text{loc}} = 0.176$ d, and the pressure of unit 2 at the well falls at the time of the survey at a rate of 6.1 kPa/d (by the same model). The estimate by the formula of the main section, $0.176\cdot 6.1 = 1.1$ kPa, is close to the calculation (1.0 kPa). For $t \gg \tau_{\text{loc}}$ the formula works. With a 0.002 mD shale, as in the example of the parting, unit 1 has hardly moved (0.4 kPa), and the difference between the units is 65 kPa, almost the whole fall of unit 2.

Answer: The pulse shows a difference of 10.6 kPa between the units. This is 3.2 times the step of 3.3 kPa that the same shale would give over years of depletion.

What it means. A pulse survey is more sensitive for two reasons. First, the lag time of unit 1 is set by its own storage capacity (1.76 d), not by the shorter equalisation time of the two units during depletion (0.728 d). Second, the pressure of unit 2 at the observation well falls faster at the time of the survey (9.0 kPa/d) than the own depletion rates of the units differ (4.516 kPa/d). On the other hand, the view of a pulse is limited to the neighbourhood of the pair of wells. It tests the barrier at one particular place, while depletion tests it on average over the block. So each complements the other, and together they help to find “windows”. We must also allow for the crossflow inside unit 2 itself: a calculation for unit 2 alone would overestimate its response by a factor of 3.4, and with it the difference between the units. We choose the time of the survey by calculation for the particular pair of wells. In the example the difference is still growing, and by day 8 it would have reached 19 kPa. The model ignores lateral flow in unit 1. If unit 1 is permeable across the area, the difference is smaller, so here the estimate is an upper limit. The data in this example are assumed for illustration.

CalculatedPressure fall at the observation well after the injection well is shut in (the data of the example): unit 2 and unit 1 behind the 0.2 mD shale, by the model of two units with crossflow. The dashed line is unit 2 without crossflow (line-source solution). The vertical line is the time of the survey, 96 h after shut-in.

Pressure profiles and the reservoir model

The simple models of this lesson explain the shapes of the profiles and give estimates. In a real accumulation there are more units. Vertical equilibrium does not hold everywhere. A tight unit also gives up fluid sideways towards the wells. The injection changes, and the flow has not become steady. So we use the profiles quantitatively when we tune a layered reservoir simulation model to fit the measurements (history matching). We adjust the vertical permeabilities of the units and the transmissibility multipliers of the partings (factors that scale the flow between model cells) until the calculated pressures in each observation well on the day of its survey agree with the measured ones (History matching).

Pressures and flow rates together. The two-unit model shows that the same pressure difference can come from different combinations of barrier transmissibility and distribution of withdrawals. So we match the model at the same time to the pressure profiles and to the distribution of inflow and injectivity (how much each unit takes) among the units. The latter comes from production logging (measurement of the flow of each unit in a working well) in producing and injection wells. A model fitted to the pressures alone can distribute the withdrawals wrongly.
Early, while it is simpler. Barriers show up best during the depletion period, before injection evens out the pressures. Then the pressure differences between the units are larger, and the mechanism is simpler. The model will take the partings found in advance into account when waterflooding or gas injection begins. They decide how gravity will separate water, oil and gas along the section (Segregated flow and gravity displacement).
Pressure and saturation. Pressure is an average over large areas of the formation, while water moves along particular paths. A model tuned to the pressure is checked against the movement of the water: against pulsed-neutron logging, the water cut and the order of breakthrough in the units (see the section on barriers).

Terms and symbols

Observation well: A well whose measurements are used to watch the formation. A new well in a producing accumulation, with an open hole, is an observation well for the time of the tester survey.
Potential $\Phi$: The pressure converted to the datum depth with the fluid density of the unit: $\Phi = p + \rho g(z_{\text{datum}} - z)$ . Fluid flows from the higher potential to the lower one.
Viscous component of the gradient $G_{\text{visc}}$: The difference between the measured and the hydrostatic gradients, $\mu u_z/k_z$ . It is positive for upward flow.
Vertical pressure equilibrium: The state of a unit in which the viscous component is smaller than its error. The profile is hydrostatic, although vertical flow may be going on.
Potential step across a barrier: The difference between the potentials of the units on the two sides of a tight parting. It appears only when fluid crosses the parting.
Own depletion rate $r$: The rate at which the potential of a unit would fall without crossflow: the net withdrawal divided by its storage capacity $c_tV$ .
Time constant of a barrier $\tau$: The equalisation time of two units or blocks through a barrier: $c_tV_1V_2/((V_1 + V_2)T)$ .
Pulse survey: A tester survey made soon after a neighbouring well that is open to one unit only has changed its rate. It reveals baffles that depletion does not show.

Common mistakes and misconceptions

Wrong

“The pressures of both units lie on one line, so the units communicate well.”

In fact

This means only that the crossflow keeps up with the uneven depletion, or that there is almost no depletion. A continuous shale gives no step if the units give up fluid in proportion to their volumes. We check communication from the flow rates of the units, with a pulse survey, and from the movement of the water.

Wrong

“The step of 330 kPa was measured accurately, so the transmissibility of the parting is known accurately too.”

In fact

The transmissibility is set by the difference of the depletion rates of the units, that is by their flow rates and pore volumes. Their errors give ±30–40%, while the error of the step is a fraction of a per cent.

Wrong

“A point in a tight layer of a depleted unit is higher than its neighbours, so the layer is isolated and less depleted.”

In fact

First look at the mobility. In depleted units the overbalance is large, and tight layers are supercharged more than usual. A separate pressure level is confirmed by several points with good mobility.

Wrong

“A water gradient of 10.77 kPa/m instead of the hydrostatic 10.50 is a measurement error.”

In fact

It may be upward flow. We compare the deviation with its error, which includes the error of the water density, and we check the directional survey. Only then do we decide between flow and error.

Wrong

“The pressure of the unit in the new well is 2 MPa below the initial pressure, so the difference from last year's survey in a neighbouring well is the decline over the year.”

In fact

Comparing different wells and years needs one datum depth, an absolute calibration of the gauges and depth matching. It also means allowing for the distance to the working wells: the drawdown of a neighbouring producing well is part of the reading.

In depth: advanced detail

Exact solution for two units

Add the balances of the two units. The crossflow cancels, and the potential averaged over the pore volume, $\bar\Phi = (V_1\Phi_1 + V_2\Phi_2)/(V_1 + V_2)$ , changes as in a single tank. Now subtract them, after dividing each by its storage capacity. The difference $\Delta = \Phi_1 - \Phi_2$ obeys the equation $\dd\Delta/\dd t = (r_2 - r_1) - \Delta/\tau$ with the same constant $\tau$ . For equal initial potentials the solution is:

Potentials of the two units in time

\begin{aligned} \bar\Phi(t) &= \Phi_0 - \frac{(q_1 + q_2)\,t}{c_t\,(V_1 + V_2)}, \qquad \Delta(t) = \Delta_{\text{ss}}\left(1 - e^{-t/\tau}\right), \\ \Phi_1 &= \bar\Phi + \frac{V_2}{V_1 + V_2}\,\Delta, \qquad \Phi_2 = \bar\Phi - \frac{V_1}{V_1 + V_2}\,\Delta \end{aligned}

where:

$\Phi_0$ initial potential, common to both units, MPa
$\bar\Phi$ potential of the site averaged over the pore volume, MPa
$\Delta,\ \Delta_{\text{ss}},\ \tau$ potential difference, its steady-state value and the time constant (main section), MPa, MPa, d

Units: metric oilfield, as in the main section

Holds when

constant withdrawals, equal initial potentials, and the other assumptions of the two-unit model.

The average potential does not depend on the barrier. We use it to check the total drained volume. The barrier only shares the decline out between the units. It is tempting to write that after the steady state is reached each unit falls from its initial value by “its own withdrawal with crossflow” divided by its own storage capacity. This is wrong. Such an expression loses the difference that built up while the step was settling, and it gives both units the same potential.

In the example of the shale parting, the potential averaged over the pore volume fell in 730 days by $500\cdot 730/(0.002\cdot 42.72\cdot 10^6) = 4.272$ MPa. The lower unit fell by 4.136 MPa and the upper unit by 4.465 MPa. If the time constant is comparable with the production time, we find the transmissibility from a single step $\Delta(t)$ by solving the equation $\Delta(t) = \Delta_{\text{ss}}(T)\,(1 - e^{-t/\tau(T)})$ for $T$ by trial. An estimate from the steady-state step then gives a time constant that is too short and a transmissibility that is too high.

Where the parabola and the settling criterion come from

In the common pseudo-steady-state regime the pressure at every point of the tight unit falls at the same rate. So each layer of the unit gives up the same volume per unit of thickness. Everything that the layers above it have given up, up to the impermeable top, passes through a section at height $\zeta$ . So the velocity is $u(\zeta) = u\,(1 - \zeta/h)$ . By Darcy's law $\dd\Phi/\dd\zeta = \mu u(\zeta)/k_z$ . Integrating from the boundary, we get $\Phi - \Phi_1 = (\mu u/k_z)(\zeta - \zeta^2/2h)$ , which is the formula of the main section. At the top, $\zeta = h$ , the difference is largest, $\mu u h/(2k_z)$ .

If a constant rate had been taken from a layer with an impermeable top from the very start, the fraction of the steady-state potential difference that has built up across the unit would be $1 - \sum_{n\ \text{odd}} \tfrac{8}{n^2\pi^2}\,e^{-n^2\pi^2 t_{Dh}}$ . At $t_{Dh}$ = 0.1 it is 70%, at 0.2 it is 89%, at 0.3 it is 96%, and at 0.5 it is 99.4%. But in the lesson's examples all the withdrawal goes through the permeable unit, and the crossflow from the tight unit builds up from zero. At the boundary of the layer the potential equals that of the permeable unit, and that potential falls faster the smaller the pore volume of the permeable unit is. Solving the same problem together with the balance of the permeable unit (a series in the eigenfunctions of the layer: the natural shapes in which the pressure evens out across it) gives, at $t_{Dh}$ = 0.5, 1 and 1.5, the shares 81%, 96% and 99%, if the pore volume of the permeable unit is twice as large, as in the example. If it is a hundred times larger, the shares are 70%, 91% and 98%. The 99% level is reached at $t_{Dh} = 2$ . This gives the criterion of the main section. For the laminated unit of the example, $t_{Dh} = 2$ is 65 days. With a vertical permeability ten times lower it is 648 days. In a very tight, thick unit the profile stays unsettled for years, its bend is smaller than the steady-state bend, and an estimate with the formula of the main section overestimates $k_z$ .

A pulse in two units with crossflow

In the example of the pulse survey, unit 2 carries the pulse across the area, and unit 1 exchanges fluid with it only through the shale. Per unit of area these are two storage capacities, $\phi_2h_2c_t$ and $\phi_1h_1c_t$ , linked by a crossflow of $0.0853\,k_b/(\mu b)$ per megapascal of pressure difference. Unit 2 also carries fluid across the area. The same equations describe a formation with dual porosity (a fractured rock in which the fractures carry the flow and the matrix blocks store the fluid): unit 2 plays the role of the fractures, and unit 1 the role of the matrix blocks. The solution for a well in unit 2 whose rate has changed by $q$ is easiest to write in the Laplace domain, that is, after a Laplace transform (a mathematical device that replaces the time $t$ by a new variable $s$ and makes the equations simpler).

Response of two units with crossflow to a change in the rate of a well in unit 2

\bar p_2(s) = \frac{2\cdot 0.933\,q\mu}{k_2h_2\,s}\,K_0\!\left(\frac{r}{B}\sqrt{\tau_2 s + \frac{\tau_{\text{loc}}\,s}{1 + \tau_{\text{loc}}\,s}}\right),\qquad \bar p_1(s) = \frac{\bar p_2(s)}{1 + \tau_{\text{loc}}\,s}

where:

$\bar p_2,\ \bar p_1$ Laplace transforms of the pressure changes of units 2 and 1 at a distance $r$ from the well, MPa·d
$s$ Laplace transform variable, 1/d
$q$ change in the rate of the well in unit 2; for a shut-in injection well, its former injection rate at reservoir conditions, m³/d
$k_2,\ h_2$ permeability and thickness of unit 2, mD, m
$\mu$ fluid viscosity, mPa·s
$r$ distance from the well that creates the pulse to the observation well, m
$B$ leakage factor $\sqrt{k_2h_2b/k_b}$ , m
$\tau_2,\ \tau_{\text{loc}}$ local time constants of units 2 and 1, $\phi_ih_ic_t\mu b/(0.0853\,k_b)$ , d
$K_0$ modified Bessel function of the second kind of order zero, —

Units: metric oilfield: MPa, m³/d, mD, m, mPa·s, days

Holds when

unit 1 does not carry flow across the area (lateral flow in it is ignored), the storage capacity of the shale is small, and the formation is uniform across the area and unbounded
the pressures in time are found by numerical inversion of the Laplace transform (turning the result in $s$ back into time; Stehfest algorithm)
without crossflow ( $k_b \to 0$ ) the formula turns into the line-source solution (Line-source solution and the pressure function)

The crossflow matters when the duration of the pulse is longer than the local constant $\tau_2$ and the leakage factor is smaller than the distance between the wells. Then the shale feeds unit 2 over the whole area that the pulse has reached, and the response of unit 2 is much smaller than the solution for one unit gives. The second equality is the same link between unit 1 and unit 2 as in the main section.

Connections to other topics

Pressure measurement with the wireline formation tester — pretests, reading classes, pressure and depth accuracy, the gradient from a set of points.
Pressure gradients, fluid contacts and capillary equilibrium — equilibrium profiles of an accumulation that is not producing, and separate fluid systems.
Supercharging and permeability from formation-tester pretests — supercharging of tight layers, and mobility as the criterion for rejecting points.
Reservoir architecture and performance — barriers, baffles, isolated blocks and the time constant of a block.
Partially communicating faults — fault transmissibility from a well test.
Displacement in heterogeneous reservoirs — how layering and crossflow affect the waterflooding of units.
History matching — matching a layered model to pressures and flow rates.
Pseudo-steady-state flow and the shape factor — pressure decline of a closed area, and the pore volume.
Build-up in a bounded reservoir and average reservoir pressure — why the pressure of a shut-in well with several layers is not the formation pressure.
Lumped multi-tank material balance model — multi-tank material balance for block pressures.
Two-phase flow and reservoir description from formation-tester data — pressure profiles when two phases move.

Check yourself

1
In a new well both units lie on one line of the oil gradient, and there is no step across the shale parting, although the shale shows on the logs of all the wells of the block. What can we say?
1. A.Without the flow rates of the units, nothing definite: with depletion in proportion to the pore volumes there is no step, even if the shale is continuous
2. B.The shale is permeable, since the pressure is the same
3. C.The step is hidden by supercharging
4. D.The shale is discontinuous, and the units communicate well
Show the answer
Correct answer: A. Without the flow rates of the units, nothing definite: with depletion in proportion to the pore volumes there is no step, even if the shale is continuous
Only uneven depletion keeps a step going. If the withdrawals from the units are proportional to their pore volumes, there is no reason for crossflow, and even an impermeable shale gives no step. We need the flow rates of the units. We can test the parting with a pulse survey or from the movement of the water. A window in the shale is only one of the possibilities.
2
In the water leg the gradient from the pretests is $10.62 \pm 0.03$ kPa/m, and the formation water density from the analysis is $1.065 \pm 0.005$ g/cm³. Is there vertical flow, and in which direction?
Show the answer
The hydrostatic gradient is $9.81\cdot 1.065 = 10.448$ kPa/m, and the difference is $10.62 - 10.448 = 0.172$ kPa/m. Its error is $\sqrt{0.03^2 + 0.049^2} = 0.057$ kPa/m, so the deviation is three errors: there is flow. The gradient is larger than the hydrostatic one, so the water flows upward. Before drawing the conclusion we check the directional survey: an error of 1.6% in the depth scale would give the same deviation.
3
Unit 1: pore volume 20 million m³, net withdrawal 150 m³/d. Unit 2: 10 million m³, 150 m³/d. The transmissibility of the parting is 500 m³/(d·MPa), and the total compressibility is $1.5\cdot10^{-3}$ MPa⁻¹. Find the steady-state step and the time it takes to settle.
Show the answer
Own depletion rates: $r_1 = 150/(0.0015\cdot 20\cdot10^6) = 5.0$ kPa/d and $r_2 = 10.0$ kPa/d. $V_1V_2/(V_1 + V_2) = 6.67\cdot10^6$ m³, and $\tau = 0.0015\cdot 6.67\cdot10^6/500 = 20$ d. The step is $(10.0 - 5.0)\cdot 20 = 100$ kPa, with unit 1 higher. Check: $(150\cdot 20 - 150\cdot 10)\cdot 10^6/(500\cdot 30\cdot 10^6) = 0.1$ MPa. After $3\tau$ = 60 d the step reaches 95% of this value.
4
A thin layer with a mobility of 0.3 mD/(mPa·s) in a depleted unit lies 60 kPa to the right of the line through its neighbouring points, which have mobilities of 50–200. What is the most likely explanation?
1. A.An isolated, less depleted layer
2. B.A depth error
3. C.Upward flow in the layer
4. D.Supercharging: the overbalance in the depleted unit is large, and the tight layer does not take the filtrate away
Show the answer
Correct answer: D. Supercharging: the overbalance in the depleted unit is large, and the tight layer does not take the filtrate away
In a producing accumulation we cannot tell supercharging from a separate pressure level by the position of the point, so the mobility decides. With a mobility of 0.3 mD/(mPa·s) and a large overbalance, supercharging of tens of kilopascals is to be expected. An isolated layer would be confirmed by several points with good mobility at their own level. A depth error would shift the point by 60 kPa only if it were about 8 m.
5
Why can a pulse survey after the shut-in of a neighbouring injection well reveal a shale parting that gave no noticeable step over years of depletion?
Show the answer
The difference between the units is the rate of change of pressure multiplied by the lag time of the unit behind the barrier. During depletion the rate is the difference of the own depletion rates of the units, a few kilopascals per day, and the time is the equalisation time of the two units. After the neighbouring well is shut in, the pressure at the observation well changes faster. The lag time is set by the storage capacity of the single unit behind the barrier, so it is longer. In the lesson's example a 0.2 mD shale gave about 3 kPa during depletion, but in the pulse after four days it gave about 11 kPa, and the difference was still growing. Such a pulse must be calculated for both units together, because the shale also feeds the unit with the working well.
6
To find how far the pressure of a unit has fallen since the discovery, we compare the survey in a new well with the first survey in the discovery well. What matters most for this comparison?
1. A.The mobility of the fluid
2. B.The number of pretests in the unit
3. C.The repeatability of the gauge at one test point
4. D.The absolute accuracy of both gauges, common depth matching and the datum depth
Show the answer
Correct answer: D. The absolute accuracy of both gauges, common depth matching and the datum depth
The decline is the difference between readings of different instruments in different wells and at different times. The constant errors of the gauges and of the depths do not cancel here, as they do within one survey, but enter in full: ±10–15 kPa. In addition, we bring the pressures to one datum depth and allow for the drawdown of the wells working nearby.

Updated: 28 October 2026