Build-up in a bounded reservoir and average reservoir pressure

Quick refresher

In a closed area, a build-up flattens out with time and reaches the average pressure $\bar p$ , the pressure that the area had at the moment of shut-in. We must wait as long as the well needs after start-up to reach pseudo-steady-state flow, often weeks.
The continuation of the Horner line, $p^*$ , here equals neither $p_i$ nor $\bar p$ . Usually $p^* > \bar p$ , but for a well near a boundary it can be the other way round.
The MBH method: $\bar p = p^* - (m/2.303)\,p_{DMBH}(t_{pDA})$ . The function $p_{DMBH}$ depends only on the shape of the area and the position of the well. It is calculated in advance (MBH charts).
If before shut-in the well had reached pseudo-steady-state flow, then $p_{DMBH} = \ln(C_At_{pDA})$ and $\bar p = p^* - m\lg(C_At_{pDA})$ . The formulas of Dietz and Ramey–Cobb give the same.
If the well had reached pseudo-steady-state flow before shut-in, the producing time that we put into the Horner function does not affect $\bar p$ , as long as $p^*$ and $t_{pDA}$ are calculated with the same time. The line, however, stays straight much longer if we take the time at which $t_{pDA} \approx 0.08$ .
Without the area and the shape of the region, we cannot make the correction. We find them from the first long test, the geology and the layout of the wells, and then we monitor the pressure with short build-ups.

Key relationships

p_{DMBH} = \dfrac{4\pi kh\,(p^* - \bar p)}{qB\mu} = \dfrac{2.303\,(p^* - \bar p)}{m}

dimensionless MBH correction

\bar p = p^* - m\,\lg\left(C_A\,t_{pDA}\right)

the well was in pseudo-steady-state flow before shut-in

t_{pDA} = \dfrac{3.553\cdot10^{-3}\,k\,t_p}{\phi\mu c_t A}

metric units: mD, h, mPa·s, 1/MPa, m²

\Delta t_{\bar p} = \dfrac{\phi\mu c_t A}{3.553\cdot10^{-3}\,k\,C_A}

Dietz time, h: on the MDH line (Miller–Dyes–Hutchinson: pressure against $\lg\Delta t$ ), at this moment the pressure equals $\bar p$

\bar p = p_i - \dfrac{B\,N_p}{c_t V_p}

material balance of a closed area (single-phase oil)

Common pitfalls

Taking $p^*$ for the average pressure in a developed accumulation.
Using the shape factor of a circle, 31.6, for the correction when the well stands near a fault or in a long block.
Putting in $\ln(C_At_{pDA})$ when the well had not yet reached pseudo-steady-state flow before shut-in. Then we need the MBH chart.
Thinking that the pressure has “almost equalised” because it rose slowly in the last hours. On a semilog scale, the rise can go on for weeks.
Applying a long build-up to a well among working neighbours without allowing for their influence.

After this lesson you can

explain why in a closed area the pressure after shut-in equalises at the average reservoir pressure and not at the initial one, and how long this takes
show why the continuation of the Horner line, $p^*$ , in a bounded area equals neither the initial nor the average pressure
calculate the average pressure by the MBH method from the dimensionless time $t_{pDA}$ and the shape of the area, and for a well that has reached pseudo-steady-state flow, by the formulas of Dietz and Ramey–Cobb
choose the producing time for the Horner function, and estimate how it affects the length of the straight line and the result
find the average pressure from the late points of a long build-up, and build the material balance of the area from the average pressures of repeated tests

Which pressure we call average

While the well produces, the pressure in the drainage area is not the same everywhere. Near the wellbore there is a drawdown cone (a funnel of low pressure), and far away the pressure is higher. The average reservoir pressure $\bar p$ is the average over the pore volume. If the area is closed and we stop the production, fluid flows from places of high pressure to places of low pressure, and the pressure equalises. For a constant compressibility, the equalised value is exactly $\bar p$ . The fluid that expands where the pressure falls is balanced by the fluid that is compressed where the pressure rises, so the total amount of fluid in the area does not change.

What the “average” pressure is for, and which one we need

Task: Productivity index, the inflow performance relationship of a well (the curve of rate against bottomhole pressure)
Which pressure we need: the average over the drainage area of this well at the moment of shut-in; the methods of the lesson give it

Task: Material balance of an accumulation or a block
Which pressure we need: the average over the whole connected pore volume; in pseudo-steady-state flow it is the average of the pressures of the areas, weighted by the rates (several wells)

Task: Comparison with a reservoir simulation model (a computer model of the reservoir)
Which pressure we need: the average over the cell with the well; we convert it from the build-up pressure, allowing for the cell size (Well model in a grid block)

Task: Inflow from an aquifer (a water-bearing formation)
Which pressure we need: the pressure at the contour of the accumulation, not the average over it (Aquifer models: water influx)

Task	Which pressure we need
Productivity index, the inflow performance relationship of a well (the curve of rate against bottomhole pressure)	the average over the drainage area of this well at the moment of shut-in; the methods of the lesson give it
Material balance of an accumulation or a block	the average over the whole connected pore volume; in pseudo-steady-state flow it is the average of the pressures of the areas, weighted by the rates (several wells)
Comparison with a reservoir simulation model (a computer model of the reservoir)	the average over the cell with the well; we convert it from the build-up pressure, allowing for the cell size (Well model in a grid block)
Inflow from an aquifer (a water-bearing formation)	the pressure at the contour of the accumulation, not the average over it (Aquifer models: water influx)

In a developed accumulation (one with many producing wells), shutting in one well does not stop its neighbours. The pressure in its area first rises, and then starts to fall together with the whole accumulation. So we relate a build-up to the drainage area of the well as it was at the moment of shut-in. The MBH method uses only the radial-flow line, the early part of the build-up, which the neighbours have not yet distorted. So it can be used in a developed accumulation too. For long build-ups among working wells, we must allow for the neighbours (Testing in a developed reservoir: Slider's method). The examples of the lesson are simpler: there is one well in a closed block.

CalculatedCalculated pressure in the formation along the long axis of the block of the lesson's examples (1000 × 500 m, the well 250 m from the short side), at the moment of shut-in after 4000 h of production and 24, 200 and 1000 h after it. The dotted line is the average pressure of the area. The pressure drop across the skin is not shown: in the wellbore at the moment of shut-in the pressure is another 0.8 MPa lower, 15.7 MPa. The drawdown cone near the well fills in a day, and the far end of the block equalises with the rest over weeks.

Pressure build-up in a closed area

We describe a build-up by superposition (adding up the effects of separate events): a shut-in is equal to starting an imaginary injection well at the same point, with the same rate. In a bounded area, we put into this sum the pressure function of the closed area, and not that of an infinite formation:

A build-up in a closed area as the sum of two rates

p_i - p_{ws}(\Delta t) = \frac{qB\mu}{2\pi kh}\Big[p_D(t_p + \Delta t) - p_D(\Delta t)\Big]

where:

$p_i$ initial reservoir pressure, Pa
$p_{ws}(\Delta t)$ bottomhole pressure a time $\Delta t$ after shut-in, Pa
$p_D(t)$ dimensionless pressure drop at the wall of a well that produces at a constant rate for a time $t$ in a closed area, without skin (the skin enters both terms in the same way and cancels), —
$t_p$ producing time at the rate $q$ before shut-in, s
$q,\ B,\ \mu$ rate at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s
$k,\ h$ permeability and net thickness, m², m

Units: SI (any consistent system of units)

Holds when

a single-phase slightly compressible fluid, a homogeneous formation, a closed area
a constant rate before shut-in (for a changing rate, the equivalent time $N_p/q$ )
we ignore wellbore storage (the well acting as a small tank): the well is shut in at the bottom, or the early points are not used

It differs from the formula of the lesson Pressure build-up analysis only by the function $p_D$ . While the disturbances have not reached the boundaries, this is a logarithm, and we get the Horner equation. Then $p_D$ starts to grow linearly with time, because the area is being depleted.

When both the producing time and the shut-in time are longer than the time to reach pseudo-steady-state flow, both values of $p_D$ equal $2\pi t_{DA} + \tfrac12\ln\big(4A/(\gamma C_Ar_w^2)\big)$ (bottomhole pressure through the initial pressure). The constants cancel, and only the depletion remains:

The pressure to which a build-up in a closed area levels off

p_{ws}(\Delta t \to \infty) = p_i - 2\pi t_{pDA}\,\frac{qB\mu}{2\pi kh} = p_i - \frac{qB\,t_p}{c_t V_p} = \bar p

where:

$t_{pDA}$ producing time referred to the area: $kt_p/(\phi\mu c_tA)$ , —
$A$ area of the drainage region, m²
$V_p = \phi hA$ pore volume of the area, m³
$c_t$ total compressibility of the rocks and fluids, 1/Pa
$\bar p$ average pressure of the area at the moment of shut-in, Pa

Units: SI; in metric units the pressure drop equals $qBt_p/(24\,c_tV_p)$ for $q$ in m³/d, $t_p$ in hours and $c_t$ in 1/MPa

Holds when

$\Delta t_{DA} = k\Delta t/(\phi\mu c_tA)$ is larger than the time to reach pseudo-steady-state flow for the given shape of the area
the area is closed, the neighbouring wells do not work, and the fluid is single-phase

The build-up reaches the pressure that the material balance gives to the area at the moment of shut-in: the average reservoir pressure. We must wait as long as the well needed after start-up to reach pseudo-steady-state flow, because the same $p_D(\Delta t)$ takes part in the equality. For a compact area with the well at the centre, this is $t_{DA}$ of about 0.1–0.15. For an elongated area, or a well near the edge, it is about one.

The build-up of the lesson's examples: how the pressure approaches the average (calculated by the model)

$\Delta t$ , h: 24
$\Delta t$ , d: 1.0
$p_{ws}$ , MPa: 19.515
$\bar p - p_{ws}$ , MPa: 0.625

$\Delta t$ , h: 72
$\Delta t$ , d: 3.0
$p_{ws}$ , MPa: 19.683
$\bar p - p_{ws}$ , MPa: 0.458

$\Delta t$ , h: 200
$\Delta t$ , d: 8.3
$p_{ws}$ , MPa: 19.831
$\bar p - p_{ws}$ , MPa: 0.310

$\Delta t$ , h: 400
$\Delta t$ , d: 17
$p_{ws}$ , MPa: 19.962
$\bar p - p_{ws}$ , MPa: 0.179

$\Delta t$ , h: 800
$\Delta t$ , d: 33
$p_{ws}$ , MPa: 20.081
$\bar p - p_{ws}$ , MPa: 0.060

$\Delta t$ , h: 1600
$\Delta t$ , d: 67
$p_{ws}$ , MPa: 20.134
$\bar p - p_{ws}$ , MPa: 0.007

$\Delta t$ , h	$\Delta t$ , d	$p_{ws}$ , MPa	$\bar p - p_{ws}$ , MPa
24	1.0	19.515	0.625
72	3.0	19.683	0.458
200	8.3	19.831	0.310
400	17	19.962	0.179
800	33	20.081	0.060
1600	67	20.134	0.007

A block of 1000 × 500 m, the well on the long axis 250 m from the short side, 4000 h of production at 45 m³/d. The average pressure of the area at the moment of shut-in is 20.141 MPa, and the initial pressure is 26.0 MPa. The data are assumed for illustration.

On a Horner plot, such a build-up looks like this. After the period of wellbore storage, there is the radial-flow straight line. Then the points leave it downward and level off towards $\bar p$ . The continuation of the line to a Horner ratio of 1 (an infinitely long shut-in) gives $p^*$ . Here it equals neither the initial pressure (that was long gone from the area) nor the average one. The Horner equation is built for an infinite formation, where an infinitely long shut-in would return the pressure to $p_i$ . In a closed area this is not so.

CalculatedHorner plot of the build-up of the lesson's examples: the points are the readings of the first 72 h, the thin line is the continuation of the same build-up by the model up to 3000 h, the dashed line is the radial-flow line and its continuation to

p^*

, and the dotted line is the average pressure. At 72 h the pressure is still almost 0.5 MPa below the average.

The MBH method: the correction from p* to the average pressure

The radial-flow line on a build-up in a closed area stays: in the first hours after shut-in, the pressure rises almost as in an infinite formation (the caveat is in the section on the producing time). So from the line we still find $kh$ , the skin and $p^*$ . What remains is to link $p^*$ with the average pressure. The difference between them depends only on the shape of the area, the position of the well in it and the dimensionless producing time. It is convenient to write it in dimensionless form and to calculate it once for each geometry. This is the MBH function (MBH method):

The dimensionless MBH correction

p_{DMBH}(t_{pDA}) = \frac{4\pi kh}{qB\mu}\,(p^* - \bar p) = \frac{2.303\,(p^* - \bar p)}{m}

where:

$p^*$ pressure on the continuation of the Horner line, drawn with the actual producing time $t_p$ , at a Horner ratio of 1, Pa
$\bar p$ average pressure of the area at the moment of shut-in, Pa
$m$ slope of the Horner line per decimal cycle (positive), Pa
$t_{pDA}$ producing time referred to the area: $kt_p/(\phi\mu c_tA)$ , —
$q,\ B,\ \mu,\ k,\ h$ rate before shut-in, formation volume factor, viscosity, permeability, thickness, m³/s, m³/m³, Pa·s, m², m

Units: SI; the second form is right in any units, if $p^*$ , $\bar p$ and $m$ are in the same pressure units

Holds when

the build-up has a radial-flow line (we confirm it with the pressure derivative)
a homogeneous formation, a single-phase fluid, a closed area, and one well in it (or the area is a part of a block, separated from the neighbouring wells by dividing lines of flow, the lines between the drainage areas of neighbouring wells)
the rate was constant before shut-in; on the stretch of the line $\Delta t \ll t_p$

From this comes the working formula: $\bar p = p^* - (m/2.303)\,p_{DMBH}(t_{pDA})$ . We take $p_{DMBH}$ from the MBH chart for a suitable shape of the area and position of the well.

In other unit systems (2)

Metric oilfield units

p^* - \bar p = 0.933\,\frac{q\,B\,\mu}{k\,h}\,p_{DMBH}

$q$ in m³/d at standard conditions, $\mu$ in mPa·s, $k$ in mD, $h$ in m; pressure in MPa.

$0.933 = 1/(4\pi\cdot0.0853) = 2.149/2.303$ is the slope per unit of natural logarithm (where the constants come from).

Oilfield units

p^* - \bar p = 70.6\,\frac{q\,B\,\mu}{k\,h}\,p_{DMBH}

$q$ in STB/d, $B$ in rb/STB, $\mu$ in cP, $k$ in mD, $h$ in ft; pressure in psi.

$70.6 = 162.6/2.303$ .

Superposition shows where this function comes from (the derivation is in the advanced sections):

The MBH function through the solution for a producing well

p_{DMBH}(t_{pDA}) = 4\pi t_{pDA} - 2\Big[p_D(t_{pD}) - \tfrac12\big(\ln t_{pD} + 0.809\big)\Big]

where:

$p_D(t_{pD})$ pressure function of a well in a closed area at a constant rate (without skin), at the moment of shut-in, —
$\tfrac12(\ln t_{pD} + 0.809)$ the same for an infinite formation (Line-source solution and the pressure function), —
$t_{pD}$ producing time in dimensionless form, $kt_p/(\phi\mu c_tr_w^2)$ , —

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

Holds when

the same as for the MBH function; $p_D$ is calculated for a specific shape of the area and position of the well.

The first term is the depletion by the material balance, $p_i - \bar p$ in units of $qB\mu/(4\pi kh)$ . The second is how much the boundaries have managed to lower the bottomhole pressure by the moment of shut-in, compared with an infinite formation. The pressure $p^*$ is lower than $p_i$ by exactly the same amount. If the boundaries have not yet been felt, the second term is zero, $p^* = p_i$ , and $p_{DMBH} = 4\pi t_{pDA}$ . The correction reduces to the material balance. In pseudo-steady-state flow, $p_{DMBH} = \ln(C_At_{pDA})$ , a straight line on a semilog plot. At $t_{pDA} = 1$ its ordinate equals $\ln C_A$ .

CalculatedCalculated MBH functions for a well in a closed square and in rectangles (the model of the lesson). The orange curve is the block of the lesson's examples (2:1, the well at a quarter of the length). The dashed curve is

4\pi t_{pDA}

: the boundaries have not yet influenced the well. With time, each curve settles on its own straight line

\ln(C_At_{pDA})

. The more elongated the area and the closer the well to its boundary, the lower the curve, and the later it settles on the line.

The MBH function for some closed areas (calculated by the model)

Shape and position of the well: Square, well at the centre
$C_A$: 30.88
$\ln(C_At_{pDA})$ applies when $t_{pDA} >$: 0.11
$p_{DMBH}$ at $t_{pDA}$ = 0.1: 1.15
at $t_{pDA}$ = 0.3: 2.23

Shape and position of the well: Rectangle 2:1, well at the centre
$C_A$: 21.84
$\ln(C_At_{pDA})$ applies when $t_{pDA} >$: 0.21
$p_{DMBH}$ at $t_{pDA}$ = 0.1: 0.96
at $t_{pDA}$ = 0.3: 1.88

Shape and position of the well: Rectangle 2:1, well on the axis at 1/4 of the length from the short side
$C_A$: 4.52
$\ln(C_At_{pDA})$ applies when $t_{pDA} >$: 0.98
$p_{DMBH}$ at $t_{pDA}$ = 0.1: 0.76
at $t_{pDA}$ = 0.3: 0.88

Shape and position of the well: Rectangle 2:1, well on the axis at 1/8 of the length from the short side
$C_A$: 0.58
$\ln(C_At_{pDA})$ applies when $t_{pDA} >$: 1.09
$p_{DMBH}$ at $t_{pDA}$ = 0.1: −0.10
at $t_{pDA}$ = 0.3: −0.75

Shape and position of the well: Rectangle 4:1, well at the centre
$C_A$: 5.38
$\ln(C_At_{pDA})$ applies when $t_{pDA} >$: 0.49
$p_{DMBH}$ at $t_{pDA}$ = 0.1: 0.34
at $t_{pDA}$ = 0.3: 0.61

Shape and position of the well: Rectangle 5:1, well at the centre
$C_A$: 2.36
$\ln(C_At_{pDA})$ applies when $t_{pDA} >$: 0.64
$p_{DMBH}$ at $t_{pDA}$ = 0.1: 0.04
at $t_{pDA}$ = 0.3: −0.05

Shape and position of the well	$C_A$	$\ln(C_At_{pDA})$ applies when $t_{pDA} >$	$p_{DMBH}$ at $t_{pDA}$ = 0.1	at $t_{pDA}$ = 0.3
Square, well at the centre	30.88	0.11	1.15	2.23
Rectangle 2:1, well at the centre	21.84	0.21	0.96	1.88
Rectangle 2:1, well on the axis at 1/4 of the length from the short side	4.52	0.98	0.76	0.88
Rectangle 2:1, well on the axis at 1/8 of the length from the short side	0.58	1.09	−0.10	−0.75
Rectangle 4:1, well at the centre	5.38	0.49	0.34	0.61
Rectangle 5:1, well at the centre	2.36	0.64	0.04	−0.05

The well is a line source in a homogeneous closed rectangle. “Applies” follows the criterion of the lesson: the difference from $\ln(C_At_{pDA})$ is less than 0.02 (the error of $\bar p$ is less than $0.01\,m$ ). The classical tabulated $C_A$ agree with these within 0.4%. Negative values mean that $\bar p > p^*$ .

Order of calculation by the MBH method

1
Horner line
From the points of radial flow we find the slope $m$ , the flow capacity $kh$ and $p^*$ , as in the lesson Pressure build-up analysis. In the Horner function we put the producing time $t_p$ , or $N_p/q$ .
2
Area and shape
From the map of faults, seismic data, the layout of the wells or the first long test, we set the area $A$ , its shape and the position of the well. In a developed accumulation we divide the total area among the wells in proportion to their rates.
3
Dimensionless time
$t_{pDA} = 3.553\cdot10^{-3}\,kt_p/(\phi\mu c_tA)$ for $t_p$ in hours, $k$ in mD, $\mu$ in mPa·s, $c_t$ in 1/MPa and $A$ in m².
4
Correction
From the MBH chart for a suitable shape we read $p_{DMBH}(t_{pDA})$ . If $t_{pDA}$ is larger than the time from which the curve coincides with its line, we take $\ln(C_At_{pDA})$ .
5
Average pressure
$\bar p = p^* - (m/2.303)\,p_{DMBH}$ . We check the result against the material balance and the previous tests.

Build-up of a producing well in a closed block (assumed data)

$\Delta t$ , h: 0.25
Horner ratio $(t_p + \Delta t)/\Delta t$: 16,001
$p_{ws}$ , MPa: 18.592

$\Delta t$ , h: 0.5
Horner ratio $(t_p + \Delta t)/\Delta t$: 8001
$p_{ws}$ , MPa: 18.737

$\Delta t$ , h: 1
Horner ratio $(t_p + \Delta t)/\Delta t$: 4001
$p_{ws}$ , MPa: 18.882

$\Delta t$ , h: 2
Horner ratio $(t_p + \Delta t)/\Delta t$: 2001
$p_{ws}$ , MPa: 19.026

$\Delta t$ , h: 4
Horner ratio $(t_p + \Delta t)/\Delta t$: 1001
$p_{ws}$ , MPa: 19.169

$\Delta t$ , h: 8
Horner ratio $(t_p + \Delta t)/\Delta t$: 501.0
$p_{ws}$ , MPa: 19.308

$\Delta t$ , h: 12
Horner ratio $(t_p + \Delta t)/\Delta t$: 334.3
$p_{ws}$ , MPa: 19.388

$\Delta t$ , h: 24
Horner ratio $(t_p + \Delta t)/\Delta t$: 167.7
$p_{ws}$ , MPa: 19.515

$\Delta t$ , h: 48
Horner ratio $(t_p + \Delta t)/\Delta t$: 84.3
$p_{ws}$ , MPa: 19.627

$\Delta t$ , h: 72
Horner ratio $(t_p + \Delta t)/\Delta t$: 56.6
$p_{ws}$ , MPa: 19.683

$\Delta t$ , h	Horner ratio $(t_p + \Delta t)/\Delta t$	$p_{ws}$ , MPa
0.25	16,001	18.592
0.5	8001	18.737
1	4001	18.882
2	2001	19.026
4	1001	19.169
8	501.0	19.308
12	334.3	19.388
24	167.7	19.515
48	84.3	19.627
72	56.6	19.683

The producing time $t_p$ is 4000 h at 45 m³/d; the well was shut in at the bottom, and wellbore storage ended in the first minutes. The data are calculated from a model of a homogeneous closed block of 1000 × 500 m and rounded to 0.001 MPa.

Worked exampleAverage pressure from a build-up by the MBH method

Given (assumed data)

Well and block: the only well in a closed tectonic block of 1000 × 500 m ( $A$ = 50 ha; 1 ha = 10,000 m²), on the long axis 250 m from the short side; $C_A$ = 4.52
Production before shut-in: 45 m³/d of oil; 7500 m³ produced, $t_p = N_p/q$ = 4000 h
$B$ ; $\mu$ ; $h$ ; $\phi$ ; $c_t$: 1.25 m³/m³; 1.2 mPa·s; 10 m; 0.20; $1.6\cdot10^{-3}$ MPa⁻¹
Initial pressure: 26.0 MPa
Build-up: the table above; the line is drawn through the readings at 0.5 and 2 h

Solution

Horner abscissas: $\lg(4000.5/0.5) = 3.90314$ and $\lg(4002/2) = 3.30125$ . Slope: $m = \dfrac{19.026 - 18.737}{3.90314 - 3.30125} = 0.4802$ MPa per cycle.
Flow capacity and permeability: $kh = 2.149\cdot\dfrac{45\cdot1.25\cdot1.2}{0.4802} = 302$ mD·m, and $k = 30.2$ mD.
Extrapolated pressure: $p^* = 19.026 + 0.4802\cdot3.30125 = 20.611$ MPa.
Dimensionless producing time: $t_{pDA} = \dfrac{3.553\cdot10^{-3}\cdot30.2\cdot4000}{0.20\cdot1.2\cdot1.6\cdot10^{-3}\cdot500{,}000} = 2.235$ . For this shape, the MBH function coincides with its line at $t_{pDA} > 0.98$ (the table), so $p_{DMBH} = \ln(C_At_{pDA})$ .
Correction: $C_At_{pDA} = 4.52\cdot2.235 = 10.10$ , $\lg 10.10 = 1.0043$ , and $m\lg(C_At_{pDA}) = 0.4823$ MPa.
Average pressure: $\bar p = 20.611 - 0.4823 = 20.129$ MPa.
Check by the material balance: the pore volume of the block is $500{,}000\cdot10\cdot0.20 = 10^6$ m³, and $\bar p = 26.0 - \dfrac{1.25\cdot7500}{1.6\cdot10^{-3}\cdot10^6} = 20.141$ MPa.

Answer: The average pressure in the block at the moment of shut-in is about 20.129 MPa. The pressure

p^*

is almost half a megapascal above it.

What it means. The MBH method gave the average pressure with an error of 0.012 MPa, about 0.2% of the pressure decline in the block (5.86 MPa). If we had taken $p^*$ for the average, the error would have been 0.47 MPa, and the material balance would have shown less depletion, and so overstated the hydrocarbons in place. We cannot wait for the equalisation: at 72 h the build-up is still 0.46 MPa below the average pressure, and to get to a difference of 0.05 MPa we must wait more than a month (the table above). The data in this example are assumed for illustration: they are calculated from a model of a homogeneous block.

If the well reached pseudo-steady-state flow: the formulas of Dietz and Ramey–Cobb

Producing wells usually work for months before shut-in, and their $t_{pDA}$ is much larger than the time to reach pseudo-steady-state flow. Then we do not need the MBH chart:

Average pressure after production in pseudo-steady-state flow

\bar p = p^* - m\,\lg\left(C_A\,t_{pDA}\right)

where:

$C_A$ shape factor of the area with the given position of the well, —
$t_{pDA}$ producing time referred to the area, —
$p^*,\ m$ continuation of the Horner line at the ratio 1, and its slope per decimal cycle, MPa

Units: any consistent units: $p^*$ , $\bar p$ and $m$ in the same pressure units

Holds when

before shut-in the well had reached pseudo-steady-state flow: $t_{pDA}$ is larger than the time from which the MBH function coincides with $\ln(C_At_{pDA})$ (the table above)
$p^*$ and $t_{pDA}$ are calculated with the same producing time

When $C_At_{pDA} > 1$ , the average pressure is below $p^*$ . When $C_At_{pDA} < 1$ , it is above. The second case happens for a well near a boundary (a small $C_A$ ) after a not too long production.

Two graphical forms of the same formula spare us the subtraction of the correction:

The Dietz method. On the MDH plot (pressure against $\lg\Delta t$ ), we continue the radial-flow line to the time $\Delta t_{\bar p}$ (the formula below). The pressure on the line at this moment is $\bar p$ . We do not need to know the producing time, but it must be much larger than $\Delta t_{\bar p}$ .
The Ramey–Cobb method. On the Horner plot, we read the pressure on the line at the Horner ratio $(t_p + \Delta t)/\Delta t = C_At_{pDA}$ .

Dietz time

\Delta t_{\bar p} = \frac{\phi\mu c_t A}{k\,C_A}

where:

$\Delta t_{\bar p}$ time on the continuation of the MDH line at which the line passes through $\bar p$ , s
$A,\ C_A$ area of the region and shape factor, m², —
$k,\ \phi,\ \mu,\ c_t$ permeability, porosity, viscosity, total compressibility, m², —, Pa·s, 1/Pa

Units: SI

Holds when

before shut-in the well worked in pseudo-steady-state flow, and the producing time is much larger than $\Delta t_{\bar p}$
the MDH line is drawn through the points of radial flow

In dimensionless form, $\Delta t_{DA} = 1/C_A$ . This is not the time in which the well will really reach the average pressure (that is several times longer). It is the point where the continuation of the line crosses the level $\bar p$ .

In other unit systems (2)

Metric oilfield units

\Delta t_{\bar p}\,[\text{h}] = \frac{\phi\;\mu\,[\text{mPa·s}]\;c_t\,[\text{MPa}^{-1}]\;A\,[\text{m}^2]}{3.553\cdot10^{-3}\,k\,[\text{mD}]\;C_A}

$\mu$ in mPa·s, $c_t$ in 1/MPa, $A$ in m², $k$ in mD; time in hours.

$3.553\cdot10^{-3}$ is the multiplier of the dimensionless time in these units (time to reach pseudo-steady-state flow).

Oilfield units

\Delta t_{\bar p}\,[\text{h}] = \frac{\phi\;\mu\,[\text{cP}]\;c_t\,[\text{psi}^{-1}]\;A\,[\text{ft}^2]}{2.637\cdot10^{-4}\,k\,[\text{mD}]\;C_A}

$\mu$ in cP, $c_t$ in 1/psi, $A$ in square feet, $k$ in mD; time in hours.

$2.637\cdot10^{-4}$ is the multiplier of the dimensionless time in oilfield units.

Which producing time to put into the Horner function

In the Horner function we put the producing time $t_p$ , and for a changing rate the equivalent time $N_p/q$ (the cumulative production divided by the last rate; the MDH method and equivalent time). In a bounded area, $p^*$ depends on this choice. If we replace $t_p$ by $t_p'$ , the points shift along the horizontal axis, and the continuation of the line changes by $m\lg(t_p'/t_p)$ . But the correction $m\lg(C_At_{pDA})$ changes by the same amount. So the average pressure from the formula of the previous section stays the same, if the well reached pseudo-steady-state flow and both values are calculated with the same time.

The producing time affects something else: how long the points lie on the line. After long production in pseudo-steady-state flow, the continuing depletion from the imaginary “producing” well of the superposition is mixed into the pressure rise after shut-in. So the slope on a Horner plot drawn with the actual $t_p$ gets smaller long before the boundaries are felt:

The local slope of a Horner plot after production in pseudo-steady-state flow

\frac{\text{local slope}}{m} = \left(1 + \frac{\Delta t}{t_p'}\right)\left(1 - 4\pi\,\Delta t_{DA}\right)

where:

$t_p'$ producing time put into the Horner function (not necessarily the actual one), h
$\Delta t_{DA}$ shut-in time referred to the area: $k\Delta t/(\phi\mu c_tA)$ , —
$m$ slope of radial flow, MPa

Units: dimensionless; $\Delta t$ and $t_p'$ in the same units

Holds when

before shut-in the well worked in pseudo-steady-state flow
after shut-in there is still radial flow around the well: the disturbance from the shut-in has not reached the boundaries

With the actual long $t_p$ , the first bracket is almost 1, and the slope falls by 5% already at $\Delta t_{DA} \approx 0.004$ . But if we put in the time at which $t_{pDA}' = 1/(4\pi) \approx 0.08$ , the first-order distortions cancel each other, and the line holds several times longer: about four times longer until a 5% deviation, and ten times until 1%. So after long production it is convenient to draw the Horner plot with $t_{pDA}' \approx 0.08$ , and to calculate the average pressure by the same formula with $C_At_{pDA}'$ .

Worked exampleThe same data: another time in the Horner function, and the Dietz method

Given (assumed data)

Build-up and formation: from the previous example; $k$ = 30.2 mD, $C_A$ = 4.52, $\phi\mu c_tA$ = 192 mPa·s·m²/MPa
Time for the Horner function: such that $t_{pDA}' = 0.08$
Line: through the readings at 2 h (19.026 MPa) and 12 h (19.388 MPa)

Solution

Horner time: $t_p' = \dfrac{0.08\cdot192}{3.553\cdot10^{-3}\cdot30.2} = 143.1$ h.
Abscissas: $\lg\dfrac{143.1 + 2}{2} = 1.86064$ , $\lg\dfrac{143.1 + 12}{12} = 1.11143$ . Slope: $m' = \dfrac{19.388 - 19.026}{1.86064 - 1.11143} = 0.4832$ MPa per cycle.
Continuation of the line: $p^{*\prime} = 19.388 + 0.4832\cdot1.11143 = 19.925$ MPa. This is 0.686 MPa lower than with the actual time.
Average pressure: $C_At_{pDA}' = 4.52\cdot0.08 = 0.3616$ , $\lg 0.3616 = -0.4418$ ; $\bar p = 19.925 - 0.4832\cdot(-0.4418) = 20.138$ MPa.
The Dietz method: $\Delta t_{\bar p} = \dfrac{192}{3.553\cdot10^{-3}\cdot30.2\cdot4.52} = 396$ h. The MDH line through the readings at 0.5 and 2 h has the slope $0.289/\lg 4 = 0.4800$ MPa per cycle. On it, at 396 h, the pressure is $19.026 + 0.4800\cdot\lg\dfrac{396}{2} = 19.026 + 0.4800\cdot2.2967 = 20.128$ MPa.

Answer: The short Horner time gives 20.138 MPa, the Dietz method gives 20.128 MPa, the MBH method in the previous example gave 20.129 MPa, and the material balance gives 20.141 MPa.

What it means. The differences are hundredths of a megapascal. The variant with the short Horner time is the most accurate: the line through the readings at 2–12 h lies on a stretch where the slope is distorted by less than 0.7%, while on the plot with the actual producing time the slope is already 8% smaller by 12 h. So with the actual time we had to draw the line through the first two hours only. The Dietz time, about 17 days, is not the moment when the well really reaches the average pressure: by the model, at 396 h the build-up is still 0.18 MPa below it. The data in this example are assumed for illustration.

The lesson Equivalent producing time and synthetic flowing time covers the generalisation of this idea: to put into the Horner function not the actual time, but a time at which the continuation of the line passes through the pressure at the boundary of the area. It also covers the cases with several wells in one block. If the production before shut-in was short, the slope is distorted not by the depletion but by the superposition itself. That case is covered in the lesson Effect of producing time on build-up analysis.

The late points of a build-up: the approach to the average pressure

If the build-up is long after all, we can find the average pressure without the shape of the area. This happens when the well was shut in for a long time for other reasons, when it is an observation well (a well used only to watch the pressure), or when the formation is very permeable. At the end of the build-up, the remainder $\bar p - p_{ws}$ decays exponentially. The pressure equalises in different directions and on different scales at once. But it is slowest along the longest dimension of the area, and this “mode” (a pattern of unevenness in the pressure, like one tone of a vibrating string) is the one that remains last.

A late build-up in a closed area

\bar p - p_{ws}(\Delta t) \approx a\,\e^{-\lambda\,\Delta t},\qquad \lambda \approx \frac{\pi^2\eta}{L^2}

where:

$a$ a constant (it depends on the shape of the area and the production history), Pa
$\lambda$ decay exponent, 1/s
$\eta = k/(\phi\mu c_t)$ hydraulic diffusivity of the formation, m²/s
$L$ length of the rectangular area (the long side), m

Units: SI; in metric units $\eta = 3.553\cdot10^{-3}\,k/(\phi\mu c_t)$ m²/h, and $\lambda$ comes out in 1/h

Holds when

a late period: the other, faster “modes” have already decayed (for a rectangle this is $\Delta t_{DA}$ of about 0.2–0.5 and more)
the formula for $\lambda$ is for a rectangular area and a well that is not at the middle of its length. At the middle, this mode does not appear, and the decay is faster. For a well at the centre of a circle, $\lambda = 14.7\,\eta/r_e^2$
one well in a closed area, and the neighbours do not work

From a few late readings we find $\lambda$ and $a$ , and with them $\bar p$ . Two long-standing methods are based on this. One is to choose $\bar p$ so that $\lg(\bar p - p_{ws})$ against $\Delta t$ lies on a straight line (Muskat's method for build-ups; do not confuse it with the method of the same name for forecasting solution gas drive, the drive in which gas that comes out of the oil pushes it to the wells, Performance prediction for solution gas drive reservoirs). The other is to plot the rate of pressure growth $\dd p_{ws}/\dd\Delta t$ against the pressure itself: this is a straight line that crosses the axis at $p_{ws} = \bar p$ (the Arps–Smith method).

Worked exampleAverage pressure from three late readings

Given (assumed data)

Well: the same, but shut in for a long time (the continuation of the build-up by the model)
Readings: 400 h — 19.962 MPa; 600 h — 20.037 MPa; 800 h — 20.081 MPa
Hydraulic diffusivity; length of the block: $\eta$ = 277.6 m²/h; 1000 m

Solution

Increments over equal intervals of 200 h: $d_1 = 20.037 - 19.962 = 0.075$ MPa, $d_2 = 20.081 - 20.037 = 0.044$ MPa. Their ratio is $q_r = d_2/d_1 = 0.5867$ .
Decay exponent: $\lambda = -\ln 0.5867/200 = 0.00267$ h⁻¹. By the formula, $\pi^2\eta/L^2 = 9.87\cdot277.6/1000^2 = 0.00274$ h⁻¹, which is close.
The rest of the rise after the last reading is the sum of a decreasing geometric progression: $d_2\,\dfrac{q_r}{1 - q_r} = 0.044\cdot\dfrac{0.5867}{1 - 0.5867} = 0.062$ MPa.
Average pressure: $\bar p = 20.081 + 0.062 = 20.143$ MPa.

Answer: From the late points,

\bar p \approx 20.143

MPa (the material balance gives 20.141 MPa).

What it means. The method works only when one mode is left. From the readings at 100, 200 and 300 h (19.727, 19.831 and 19.905 MPa), the same arithmetic would give 20.088 MPa, which is 0.05 MPa too low. The method also needs a shut-in of more than a month. For a producing well this is almost always unacceptable. For an observation well, or a well that has been idle for a long time, it is quite possible. The data in this example are assumed for illustration.

Repeated tests and the material balance

It is worth making the first test of a new well long. From the late part of the record we find the boundaries, the area and the shape of the region (Reservoir limit test, A sealing fault near the well: the method of images), and so $C_A$ . After that, short shut-ins are enough, only until the radial-flow line appears. The pressure $p^*$ and the MBH correction give the average pressure for each date, and the well is idle for hours, not weeks.

When we compare the skin from short build-ups, we use the same slope every time: the one from the flow capacity of the first, reliable test (if the flow capacity does not change noticeably). A short line gives the slope only roughly, and its scatter would turn into false changes of the skin. The average pressures of repeated tests, together with the cumulative production, give the material balance of the area:

Average pressure of a closed area and cumulative production

\bar p = p_i - \frac{B}{c_t V_p}\,N_p

where:

$N_p$ cumulative oil production at standard conditions at the moment of shut-in, m³
$V_p$ pore volume of the area connected to the well, m³
$B,\ c_t$ oil formation volume factor and total compressibility, m³/m³, 1/MPa
$p_i$ initial reservoir pressure, MPa

Units: any consistent units (for example, m³, 1/MPa and MPa)

Holds when

a closed area with no water inflow; single-phase oil above the bubble point pressure (the pressure below which gas comes out of the oil); $B$ and $c_t$ are constant
there is one well in the area, or $N_p$ and $V_p$ refer to the whole closed block

The average pressures against cumulative production lie on a straight line. Its slope gives $c_tV_p$ , and its continuation to $N_p = 0$ must return the initial pressure: this is a check. The oil in place of the area equals $V_p(1 - S_{wi})/B$ , where $S_{wi}$ is the initial water saturation, and $B$ is taken at the initial pressure (General material balance equation for oil reservoirs).

Worked examplePore volume from three tests

Given (assumed data)

Tests of the same well: after 2000, 3000 and 4000 h of production at 45 m³/d ( $N_p$ = 3750, 5625 and 7500 m³)
Horner lines: $m$ = 0.480 MPa per cycle in all three; $p^*$ = 23.399, 22.016 and 20.611 MPa
$C_A$ ; $B$ ; $c_t$: 4.52; 1.25; $1.6\cdot10^{-3}$ MPa⁻¹; $t_{pDA}$ at 4000 h is 2.235 (the first example)

Solution

Dimensionless time at 2000 and 3000 h: $2.235\cdot2000/4000 = 1.118$ and $2.235\cdot3000/4000 = 1.676$ . Both are larger than 0.98, so $\ln(C_At_{pDA})$ applies.
Average pressures: $23.399 - 0.480\cdot\lg(4.52\cdot1.118) = 23.399 - 0.480\cdot0.7036 = 23.061$ MPa; $22.016 - 0.480\cdot0.8794 = 21.594$ MPa; $20.611 - 0.480\cdot1.0044 = 20.129$ MPa.
Slope from the first and the last: for a production of $1.25\cdot(7500 - 3750) = 4687.5$ m³ at reservoir conditions, the average pressure fell by $23.061 - 20.129 = 2.932$ MPa. So $c_tV_p = 4687.5/2.932 = 1599$ m³/MPa, and $V_p = 1599/(1.6\cdot10^{-3}) = 999$ thousand m³.
Check: the continuation to $N_p = 0$ is $23.061 + 2.932 = 25.993$ MPa. It agrees with the initial pressure of 26.0 MPa.
If we take the $p^*$ values themselves instead of the average pressures: $c_tV_p = 4687.5/(23.399 - 20.611) = 1681$ m³/MPa. The volume is 5% larger, and the continuation to $N_p = 0$ gives 26.19 MPa, above the initial pressure.

Answer: The pore volume connected to the well is about 999 thousand m³ (in the model, 1000 thousand m³). Without the MBH correction it is overstated by 5%.

What it means. The difference $p^* - \bar p = m\lg(C_At_{pDA})$ grows with the producing time. So the series of $p^*$ falls more slowly than the average pressure. This gives the overstated volume and an “initial pressure” above the real one. The compressibility affects the result just as strongly: $V_p$ is inversely proportional to $c_t$ , and the pore compressibility is the least well known (Rock compressibility and the effect of stress on core properties). With water inflow the points deviate from the line upward, and a simple balance does not apply (Aquifer models: water influx). The data in this example are assumed for illustration.

The limits of the methods, and practice

Ways to find the average pressure from a build-up

Method: MBH from a chart
What we need to know: the area, the shape, the position of the well, $t_p$
When it applies: at any producing time; a Horner line is needed
The main risk: a wrong area or shape

Method: The formulas of Dietz and Ramey–Cobb
What we need to know: the area and $C_A$
When it applies: the well reached pseudo-steady-state flow before shut-in
The main risk: a producing time shorter than the time to reach pseudo-steady-state flow

Method: Late points (exponential)
What we need to know: only the readings themselves
When it applies: a shut-in to $\Delta t_{DA}$ of about 0.3 and more
The main risk: the duration; work of neighbouring wells

Method: Material balance
What we need to know: $V_p$ , $c_t$ , cumulative production
When it applies: a closed area, single-phase fluid
The main risk: water inflow, a wrong $c_t$

Method	What we need to know	When it applies	The main risk
MBH from a chart	the area, the shape, the position of the well, $t_p$	at any producing time; a Horner line is needed	a wrong area or shape
The formulas of Dietz and Ramey–Cobb	the area and $C_A$	the well reached pseudo-steady-state flow before shut-in	a producing time shorter than the time to reach pseudo-steady-state flow
Late points (exponential)	only the readings themselves	a shut-in to $\Delta t_{DA}$ of about 0.3 and more	the duration; work of neighbouring wells
Material balance	$V_p$ , $c_t$ , cumulative production	a closed area, single-phase fluid	water inflow, a wrong $c_t$

A layered formation. Layers of different permeability are depleted differently, and between them, in a shut-in well, there is crossflow through the wellbore (fluid flows from one layer to another). The pressure of such a well is not the average of any single layer. We measure the formation pressure layer by layer with a formation tester (Pressure measurements in producing reservoirs: barriers and accuracy).
Gas. The compressibility of gas depends on pressure, so we apply the methods in pseudo-pressures (pseudo-pressure), and build the material balance from $p/Z$ (Material balance for gas reservoirs).
Pressure below the bubble point pressure. There is free gas in the formation, and the compressibility and the mobility (permeability divided by viscosity) change with the pressure. The corrections become approximate.
Support from outside. If water approaches the area, or there is a gas cap (free gas above the oil), the area is not closed, and after shut-in the pressure does not level off at the balance of a closed area (Aquifer models: water influx).
Neighbouring wells. During a long shut-in, their work changes both the rise of the pressure and the level to which it tends (Testing in a developed reservoir: Slider's method).

Terms and symbols

Average reservoir pressure $\bar p$: The pressure of the drainage area averaged over the pore volume. In a closed area, the pressure reaches it after an infinitely long shut-in.
MBH function $p_{DMBH}$: The dimensionless difference $p^* - \bar p$ in units of $qB\mu/(4\pi kh)$ . It depends on the shape of the area, the position of the well and $t_{pDA}$ .
$t_{pDA}$: The producing time before shut-in, referred to the area: $kt_p/(\phi\mu c_tA)$ .
Dietz time $\Delta t_{\bar p}$: The time on the continuation of the MDH line at which it passes through $\bar p$ . In dimensionless form, $\Delta t_{DA} = 1/C_A$ .
Horner time $t_p'$: The producing time put into the Horner function. It may differ from the actual one, if $t_{pDA}$ in the correction is calculated with the same time.
Repeated tests: Short build-ups to monitor the reservoir pressure and the skin after the first, long test that established the boundaries of the area.

Common mistakes and misconceptions

Wrong

“The pressure has hardly grown in the last hours, so the build-up is over, and the last reading equals the average pressure.”

In fact

On a semilog scale the rise can go on for weeks: in the example of the lesson, at 72 h the pressure is 0.46 MPa below the average. The last reading is a lower limit of the average pressure, not the pressure itself.

Wrong

“ $p^*$ is always above the average pressure.”

In fact

Usually it is. But for a well near a boundary (a small $C_A$ ), after a not too long production, the MBH function is negative, and $\bar p > p^*$ .

Wrong

“For the correction it is enough to take $C_A$ = 31.6.”

In fact

31.6 is for a circle with the well at the centre. In the example of the lesson $C_A$ = 4.52, and with 31.6 the average pressure would come out $0.48\cdot\lg(31.6/4.52) \approx 0.41$ MPa lower.

Wrong

“Since the producing time is not known exactly, the MBH method is useless.”

In fact

After pseudo-steady-state flow is reached, the result does not depend on the time we choose, if $p^*$ and $t_{pDA}$ are calculated with the same time.

Wrong

“The Dietz time is how long we must keep the well shut in.”

In fact

It is a point on the continuation of the line. The real equalisation takes several times longer.

In depth: advanced detail

Derivation of the MBH function and its limiting forms

We write the build-up by superposition in dimensionless form: $p_{Ds} = 2\pi kh(p_i - p_{ws})/(qB\mu) = p_D(t_p + \Delta t) - p_D(\Delta t)$ . On the stretch of the Horner line after shut-in, there is radial flow around the well, $p_D(\Delta t) = \tfrac12(\ln\Delta t_D + 0.809)$ , and $\Delta t \ll t_p$ , so that $p_D(t_p + \Delta t) \approx p_D(t_p)$ . We add and subtract $\tfrac12\ln t_{pD}$ , and get $p_{Ds} = \tfrac12\ln(t_p/\Delta t) + p_D(t_{pD}) - \tfrac12(\ln t_{pD} + 0.809)$ . This is a Horner line with the slope $qB\mu/(4\pi kh)$ per unit of natural logarithm. Its continuation ( $t_p/\Delta t \to 1$ ) equals $p^* = p_i - \frac{qB\mu}{2\pi kh}\big[p_D(t_{pD}) - \tfrac12(\ln t_{pD} + 0.809)\big]$ .

The average pressure at the moment of shut-in comes from the material balance: $\frac{2\pi kh}{qB\mu}(p_i - \bar p) = 2\pi t_{pDA}$ . Subtracting, we get $\frac{4\pi kh}{qB\mu}(p^* - \bar p) = 4\pi t_{pDA} - 2\big[p_D(t_{pD}) - \tfrac12(\ln t_{pD} + 0.809)\big]$ , the formula of the main section. The well radius and the skin do not enter it: they shift $p_D$ and its logarithmic approximation in the same way.

In pseudo-steady-state flow, $p_D(t_{pD}) = 2\pi t_{pDA} + \tfrac12\ln\frac{4A}{\gamma C_Ar_w^2}$ , and $0.809 = \ln(4/\gamma)$ and $t_{pD}\,r_w^2/A = t_{pDA}$ . Substitution gives $p_{DMBH} = \ln(C_At_{pDA})$ . From this come both the Ramey–Cobb formula (on the Horner line $\ln[(t_p + \Delta t)/\Delta t] = \ln(C_At_{pDA})$ ) and the Dietz time. On the MDH line, $p = p^* - \frac{qB\mu}{4\pi kh}\ln(t_p/\Delta t)$ equals $\bar p$ at $t_p/\Delta t = C_At_{pDA}$ , that is, at $\Delta t_{DA} = 1/C_A$ .

The same arithmetic shows how long to wait for the equalisation. If the production was in pseudo-steady-state flow, then $\bar p - p_{ws}(\Delta t) = \frac{qB\mu}{4\pi kh}\big[p_{DMBH}(\Delta t_{DA}) - \ln(C_A\Delta t_{DA})\big]$ . The build-up approaches the average pressure in the same way as the MBH function, calculated for the shut-in time, approaches its line. In the example of the lesson, at 400 h of shut-in ( $\Delta t_{DA} = 0.22$ ) this difference is about 0.85, that is, 0.18 MPa.

Why the Horner line is shorter after long production

If before shut-in the well worked in pseudo-steady-state flow, then $p_D(t_p + \Delta t) = p_D(t_p) + 2\pi\Delta t_{DA}$ : while the well “keeps producing” in the superposition, the area is depleted further. After shut-in there is radial flow around the well: $p_D(\Delta t) = \tfrac12\ln\Delta t_D + \text{const}$ . So the pressure in the shut-in well is $p_{ws} = \text{const} + m_e\ln\Delta t - 2m_e\cdot2\pi\Delta t_{DA}$ , where $m_e = qB\mu/(4\pi kh)$ .

The Horner abscissa with the time $t_p'$ is $X = \ln[(t_p' + \Delta t)/\Delta t]$ , and $\dd X/\dd\Delta t = -t_p'/[\Delta t(t_p' + \Delta t)]$ . The derivative of the pressure against $\Delta t$ is $m_e(1 - 4\pi\Delta t_{DA})/\Delta t$ . We divide one by the other and get the slope $-\dd p_{ws}/\dd X = m_e(1 + \Delta t/t_p')(1 - 4\pi\Delta t_{DA})$ . When we expand the brackets, we see the first-order correction $\Delta t_{DA}(1/t_{pDA}' - 4\pi)$ . It vanishes at $t_{pDA}' = 1/(4\pi) = 0.0796$ . A calculation with the exact model of the lesson confirms the formula to the third digit, as long as the disturbance from the shut-in has not reached the boundaries.

How the curves of the lesson are calculated

The well is a line source in a homogeneous closed rectangle $L\times W$ . The Green's function of such an area (its response to a point source) is the product of the Green's functions of two “strips” with sealing edges: along $x$ , of width $L$ , and along $y$ , of width $W$ . For small times, it is convenient to calculate each as a sum of reflected sources (image wells), and for large times as a series of cosines. The dimensionless pressure at the wall of the well is the integral over time of their product. The contribution of the well itself is the familiar $\tfrac12E_1(r_D^2/(4t_D))$ , and the contribution of the boundaries is everything else.

In pseudo-steady-state flow, the ratio of the slope in ordinary coordinates to its limit equals $\Pi_x\Pi_y$ , where $\Pi_x = 1 + 2\sum_n \e^{-n^2\pi^2 t_{DA}W/L}\cos^2(n\pi x_D)$ , and $\Pi_y$ is the same with $L/W$ and $y_D$ . From this we see that the equalisation is a sum of decaying exponentials, and the slowest of them is along the long side. The calculation was checked in two independent ways: by direct summation of the exponential integrals of the reflected wells (the agreement is to $10^{-13}$ ), and against the classical tables of $C_A$ (within 0.4%). The average pressure over the area in the model at any moment equals $2\pi t_{DA}$ : the material balance holds.

Connections to other topics

Pressure build-up analysis — the Horner line, $p^*$ and the skin are the input data of the correction.
Pseudo-steady-state flow and the shape factor — the shape factor, $t_{DA}$ and the time to reach pseudo-steady-state flow.
Reservoir limit test — how to find the pore volume and the shape of the area from a long production.
Superposition principle — a build-up as the sum of two rates, and image wells.
Diagnosis with the pressure derivative — why at the end of a build-up in a closed area the derivative falls.
General material balance equation for oil reservoirs — where the average pressures go: the material balance equation.
Material balance for gas reservoirs — the same for gas: the $p/Z$ line.
Testing in a developed reservoir: Slider's method — a build-up among working wells.
Equivalent producing time and synthetic flowing time — which producing time to use for a changing rate and several wells.
Effect of producing time on build-up analysis — distortions of a build-up after a short production.
Well model in a grid block — comparing the pressure from a build-up with the pressure of a model cell.

Check yourself

1
A well produced for three months in a closed tectonic block, and then it was shut in. To what pressure will the build-up tend with time?
1. A.The extrapolated pressure $p^*$
2. B.The average pressure of the block at the moment of shut-in
3. C.The pressure at the boundary of the block at the moment of shut-in
4. D.The initial reservoir pressure
Show the answer
Correct answer: B. The average pressure of the block at the moment of shut-in
After shut-in the fluid is only redistributed inside the block. For a constant compressibility, the pressure equalises at the value averaged over the pore volume, the one that the material balance gives at the moment of shut-in. The initial pressure cannot be reached, because part of the fluid has been produced. $p^*$ is the continuation of a line, not a level of equalisation. The pressure at the boundary at the moment of shut-in is above the average.
2
A well at the centre of a closed rectangular area of 2:1 ( $C_A$ = 21.8) produced until shut-in to $t_{pDA}$ = 0.9. The Horner line has $m$ = 0.30 MPa per cycle and $p^*$ = 18.40 MPa. Find the average pressure.
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For a central well in a 2:1 area, the MBH function coincides with $\ln(C_At_{pDA})$ from $t_{pDA} \approx 0.2$ . So $\bar p = 18.40 - 0.30\cdot\lg(21.8\cdot0.9) = 18.40 - 0.30\cdot\lg 19.6 = 18.40 - 0.30\cdot1.292 \approx 18.01$ MPa.
3
What must we know, besides the build-up itself, to apply the MBH method?
1. A.The bubble point pressure of the oil
2. B.The area of the drainage region, its shape and the position of the well
3. C.The wellbore storage coefficient
4. D.Only the slope of the line and $p^*$
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Correct answer: B. The area of the drainage region, its shape and the position of the well
The correction $p_{DMBH}(t_{pDA})$ depends on the shape of the area and the position of the well, and the dimensionless time depends on the area. The slope and $p^*$ come from the points of the build-up. The bubble point pressure and wellbore storage matter for choosing the stretch of the line, but not for the correction.
4
Why, after long production, is it better to draw the Horner line with the time at which $t_{pDA}' \approx 0.08$ , and why does the average pressure not change from such a replacement?
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With the actual long time, the local slope is $m(1 + \Delta t/t_p)(1 - 4\pi\Delta t_{DA}) \approx m(1 - 4\pi\Delta t_{DA})$ , and it falls by 5% already at $\Delta t_{DA} \approx 0.004$ . At $t_{pDA}' = 1/(4\pi)$ , the first-order corrections cancel, and the line holds several times longer. A replacement of the time shifts $p^*$ by $m\lg(t_p'/t_p)$ , and $m\lg(C_At_{pDA})$ by the same amount. So $\bar p = p^* - m\lg(C_At_{pDA})$ does not change, if the well had reached pseudo-steady-state flow before shut-in.
5
A well has long been producing at the centre of a square area of 40 ha ( $C_A$ = 30.9); $k$ = 80 mD, $\phi\mu c_t = 4\cdot10^{-4}$ MPa⁻¹·mPa·s. At what shut-in time must we read the pressure on the continuation of the MDH line to get the average pressure?
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By the Dietz formula, $\Delta t_{\bar p} = \phi\mu c_tA/(3.553\cdot10^{-3}\,kC_A) = 4\cdot10^{-4}\cdot400{,}000/(3.553\cdot10^{-3}\cdot80\cdot30.9) = 160/8.78 \approx 18$ h. The well itself will not reach the average pressure by this time: this is only a point on the continuation of the line.
6
In a material balance calculation of the hydrocarbons in place, the values of $p^*$ were used instead of the average pressures of repeated build-ups. How does this affect the pore volume that we find?
1. A.Only the initial pressure will change
2. B.The volume will be overstated
3. C.It has no effect
4. D.The volume will be understated
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Correct answer: B. The volume will be overstated
$p^* - \bar p = m\lg(C_At_{pDA})$ grows with the producing time, so the series of $p^*$ falls more slowly than the average pressure. A smaller fall of the pressure for the same production means a larger $c_tV_p$ . In the example of the lesson the volume is overstated by 5%, and the continuation to zero production gives a pressure above the initial one.

Updated: 7 October 2026