Simplified water influx models: Fetkovich and Carter–Tracy

Quick refresher

Superposition keeps the whole pressure history. The Fetkovich method keeps one number, the average pressure of the aquifer $\bar p_a$ . The Carter–Tracy method keeps the cumulative influx $W_{e,n-1}$ . That is why they are fast, and also why they have an error.
The Fetkovich method treats the aquifer as a tank behind a resistance: the water leaves the tank through a narrow “pipe”. Take a step in which the pressure at the contact (the boundary between the accumulation and the aquifer) stays constant. During that step the tank pressure moves towards the contact pressure along an exponential curve (fast at first, then more and more slowly). Its time constant is $\tau = c_tW/J$ (the bigger it is, the slower the tank pressure follows the contact pressure). The influx during the step is the loss of water stored in the tank. This is the exact solution of the pseudo-steady-state model (a model in which the pressure in a closed aquifer falls at the same rate everywhere), so the error lies in the model itself. The model has no transient period (the first years, before the pressure drop has reached the outer edge of the aquifer). So it underestimates the influx from a large aquifer in those years.
The formula for the productivity index $J$ of a radial aquifer (a ring around the accumulation), with $\ln r_{aD} - 3/4$ in the denominator, is correct only for $r_{aD} \gg 1$ . Here $r_{aD}$ is the ratio of the outer radius of the aquifer to the radius of the accumulation. At $r_{aD} \approx 2$ the denominator is negative, and we need the exact formula for a ring.
The Carter–Tracy method uses the influence function $p_D(t_D)$ (the solution for a constant influx rate) and its derivative (its slope). It is exact if the influx rate has been constant from the very start. If not, the method replaces the past with “an instant influx at the start plus a constant rate”. It is then wrong by a few per cent, even with very short steps.
Rules that use the average pressure of the step lag behind by half a step if the aquifer reacts in less time than one step takes. These are superposition with half drops (each step takes half of the pressure fall in the interval before it and half of the fall in the interval after it) and the Fetkovich method. The Carter–Tracy method takes the pressure at the end of the step, so it has no such lag.
A shorter step removes only the step error. Superposition gets closer and closer to the exact solution. The Fetkovich method gets closer to its own pseudo-steady-state model, and the Carter–Tracy method to its own approximation.

Key relationships

\Delta W_{e,n} = c_tW\,\big(\bar p_{a,n-1} - \bar p_n\big)\big(1 - \e^{-\Delta t_n/\tau}\big),\qquad \tau = \frac{c_tW}{J}

the Fetkovich step; $\bar p_n = (p_{n-1} + p_n)/2$

\bar p_{a,n} = p_i - \frac{W_{e,n}}{c_tW}

average aquifer pressure after the step

J = \frac{0.536\,f\,k\,h}{\mu_w\,(\ln r_{aD} - 3/4)}

pseudo-steady-state productivity index of a radial aquifer, in m³/(d·MPa), for $r_{aD} \gg 1$

W_{e,n} = W_{e,n-1} + \big(t_{Dn} - t_{D,n-1}\big)\,\frac{U\,(p_i - p_n) - W_{e,n-1}\,p_D'(t_{Dn})}{p_D(t_{Dn}) - t_{D,n-1}\,p_D'(t_{Dn})}

the Carter–Tracy step

p_D \approx \frac{2\,t_D}{r_{aD}^2 - 1} + \ln r_{aD} - \frac34

influence function of a closed aquifer after the pressure drop has reached its outer boundary ( $r_{aD} \gg 1$ )

Common pitfalls

Using the Fetkovich method for a large aquifer without a correction for the transient period. The influx of the first years comes out too low by tens of per cent. The method cannot be used for an infinite aquifer.
Using the formula for $J$ with $\ln r_{aD} - 3/4$ for a small aquifer ( $r_{aD} \lesssim 3$ ).
Expecting a shorter step to make the Carter–Tracy method exact.
Putting the pressure drop of the last step into the Carter–Tracy formula instead of the full pressure decline $p_i - p_n$ , or using the influence function of the wrong aquifer.
Mixing up the influence function $p_D$ (pressure for a constant influx rate) with the dimensionless influx $W_D$ (influx for a constant pressure).

After this lesson you can

explain how a step-by-step method differs from superposition and what “state” of the aquifer it carries from one step to the next
calculate the storage capacity (the water the aquifer gives up per megapascal of pressure drop), the maximum encroachable water (the water it would give up if its pressure fell to zero), the productivity index and the time constant of an aquifer with the Fetkovich method, and carry out a step-by-step influx calculation
explain what the influence function $p_D(t_D)$ is (the pressure drop at the contact when water flows in at a constant rate, written without units), and carry out a step-by-step calculation with the Carter–Tracy formula
compare both methods with the exact solution on one pressure history, and separate the error of the model from the error of the step
choose a method and a step length for a small, a large and an infinite aquifer

Why simplify superposition

The van Everdingen–Hurst method is exact, but wasteful (superposition of pressure steps). Every past pressure step keeps producing influx. So on each new date we must add up the whole sum again. After $n$ steps the sum has $n$ terms, and over the whole history it needs $n(n+1)/2$ values of the dimensionless influx (the influx caused by one unit of pressure drop). In a forecast we repeat the sum for every trial pressure. In a reservoir simulation model the aquifer is connected to dozens of boundary cells, and there are hundreds or thousands of time steps. At every step the program also improves the pressure by repeated trials (iterations).

A step-by-step (recursive) method builds each step on the result of the one before. It carries a small “state” of the aquifer from one step to the next. From this state, the pressure at the contact and the length of the step, it finds the influx for the step at once. The amount of calculation grows in proportion to the number of steps, not to its square. The only question is how much information about the past is lost when the history is dropped.

What each method remembers about the past

Method: van Everdingen–Hurst
What is carried from step to step: the whole history of pressure steps
Aquifer model: the diffusivity equation (how a pressure change spreads through the rock)
Exact if…: the steps are small

Method: Fetkovich
What is carried from step to step: the average aquifer pressure $\bar p_a$
Aquifer model: a tank behind a resistance
Exact if…: the aquifer is in the pseudo-steady-state regime

Method: Carter–Tracy
What is carried from step to step: the cumulative influx $W_{e,n-1}$
Aquifer model: the diffusivity equation, through the influence function
Exact if…: the influx rate has been constant from the start

Method	What is carried from step to step	Aquifer model	Exact if…
van Everdingen–Hurst	the whole history of pressure steps	the diffusivity equation (how a pressure change spreads through the rock)	the steps are small
Fetkovich	the average aquifer pressure $\bar p_a$	a tank behind a resistance	the aquifer is in the pseudo-steady-state regime
Carter–Tracy	the cumulative influx $W_{e,n-1}$	the diffusivity equation, through the influence function	the influx rate has been constant from the start

The Fetkovich method changes the aquifer model. The Carter–Tracy method keeps the model and changes only how it accounts for the past.

The Fetkovich method: the aquifer as a tank behind a resistance

You already know the pseudo-steady-state model from the lesson on aquifer models. We represent a finite aquifer (one of limited size) by a tank. The pressure in the tank is the average aquifer pressure $\bar p_a$ . The link between the tank and the accumulation is a resistance, like the resistance of a well in the pseudo-steady-state regime (Pseudo-steady-state flow and the shape factor). Water flows into the accumulation at a rate proportional to the difference $\bar p_a - p$ , and the tank empties according to its own material balance (the water that leaves it lowers its pressure). The Fetkovich method is a way to calculate with this model when the pressure at the contact changes, without superposition.

The pseudo-steady-state aquifer model in the notation of the Fetkovich method

\frac{\dd W_e}{\dd t} = J\,\big(\bar p_a - p\big),\qquad W_e = c_tW\,\big(p_i - \bar p_a\big) = W_{ei}\left(1 - \frac{\bar p_a}{p_i}\right),\qquad W_{ei} = c_tW\,p_i

where:

$W_e$ cumulative water influx at reservoir conditions, m³
$J$ productivity index of the aquifer in pseudo-steady-state flow (the inflow for each unit of pressure difference), m³/(d·MPa)
$\bar p_a$ average pressure in the aquifer, MPa
$p$ pressure at the contact (the average pressure of the accumulation), MPa
$c_tW$ storage capacity of the aquifer (the water it gives up per megapascal of pressure drop): the compressibility $c_w + c_f$ times its pore volume $W$ , m³/MPa
$W_{ei}$ maximum encroachable water, m³
$p_i$ initial pressure (of the aquifer and the accumulation), MPa

Units: m³, days, MPa

Holds when

The aquifer is closed. After the transient period it drains so that the pressure falls at the same rate everywhere in it (the pseudo-steady-state regime).
Compressibility and permeability are constant. The pressure at the contact equals the average pressure of the accumulation.

$W_{ei}$ is a reference quantity: it is the volume of water that the aquifer would give up if its pressure fell to zero. This limit can never be reached. The real limit for a pressure decline of $\Delta p$ is $c_tW\Delta p$ , the elastic store (the most water that the aquifer can give up by expanding). The ratio $W_{ei}/p_i$ is simply the storage capacity $c_tW$ , so the formulas can be written without $W_{ei}$ . Both forms are found in software and in the literature.

Suppose that during a step of length $\Delta t_n$ the pressure at the contact is constant and equal to $\bar p_n$ , and at the start of the step the tank pressure is $\bar p_{a,n-1}$ . Then the difference $\bar p_a - \bar p_n$ falls exponentially: it shrinks by the same fraction in every equal interval of time, and the time constant $\tau = c_tW/J$ sets how fast (the exponential of the pseudo-steady-state model). The influx during the step equals the loss of water stored in the tank. The whole past of the aquifer is reduced to one number: the tank pressure at the start of the step.

The Fetkovich step

\Delta W_{e,n} = c_tW\,\big(\bar p_{a,n-1} - \bar p_n\big)\left(1 - \e^{-\Delta t_n/\tau}\right),\qquad \bar p_{a,n} = p_i - \frac{W_{e,n-1} + \Delta W_{e,n}}{c_tW},\qquad \bar p_n = \frac{p_{n-1} + p_n}{2}

where:

$\Delta W_{e,n}$ influx during step $n$ , m³
$\bar p_{a,n-1},\ \bar p_{a,n}$ average aquifer pressure at the start and at the end of the step; $\bar p_{a,0} = p_i$ , MPa
$\bar p_n$ average pressure at the contact during the step: the mean of the readings at its two ends, MPa
$\Delta t_n$ length of the step, d
$\tau$ time constant of the aquifer, $c_tW/J$ , d
$W_{e,n-1}$ cumulative influx at the start of the step, m³

Units: m³, days, MPa

Holds when

Within the step, the pressure at the contact is replaced by its average value.
The pseudo-steady-state aquifer model (the equations above).

If the step is short compared with $\tau$ , the factor in brackets is close to $\Delta t_n/\tau$ . Then the influx during the step is $J(\bar p_{a,n-1} - \bar p_n)\Delta t_n$ , as in the steady-state model, except that the pressure difference that drives the flow is measured from the average aquifer pressure, not from the initial one. If the step is long, the tank has time to reach the pressure at the contact, and the aquifer behaves like a small “tank”. In the form with $W_{ei}$ , the factor in front of the brackets is $W_{ei}/p_i$ and the exponent is $-Jp_i\Delta t_n/W_{ei}$ . The two forms are the same.

Productivity index of the aquifer for the Fetkovich method

J_{\text{rad}} = \frac{2\pi f\,k\,h}{\mu_w\left(\ln r_{aD} - \dfrac34\right)},\qquad J_{\text{lin}} = \frac{3\,k\,w\,h}{\mu_w\,L}

where:

$J_{\text{rad}}$ radial aquifer on a fraction $f$ of the perimeter, closed at the radius $r_a$ ; $r_{aD} = r_a/r_o$ , m³/(s·Pa)
$J_{\text{lin}}$ linear aquifer of width $w$ along the contact and length $L$ , closed at the far end, m³/(s·Pa)
$k,\ h,\ \mu_w$ permeability and thickness of the aquifer, viscosity of water, m², m, Pa·s

Units: SI

Holds when

Pseudo-steady-state regime: the pressure difference that drives the flow is the average aquifer pressure minus the pressure at the contact.
The formula for a radial aquifer is correct for $r_{aD} \gg 1$ . For small $r_{aD}$ use the exact expression for a ring (see the advanced section).

These are the formulas for inflow to a well in the pseudo-steady-state regime, with the accumulation playing the role of the well. The term $3/4$ and the factor 3 appear because the pressure difference that drives the flow is measured from the average pressure of a closed aquifer. If the pressure at the outer boundary is held constant (an aquifer with recharge, fed with water from outside), the difference is measured from the initial pressure. Then the formulas keep only $\ln r_{aD}$ and $kwh/(\mu_wL)$ , the tank becomes infinite, and the method turns into the Schilthuis steady-state model (steady-state model).

In other unit systems (2)

Metric oilfield units

J_{\text{rad}} = \frac{0.536\,f\,k\,h}{\mu_w\left(\ln r_{aD} - 0.75\right)},\qquad J_{\text{lin}} = \frac{0.256\,k\,w\,h}{\mu_w\,L}

$J$ in m³/(d·MPa) at reservoir conditions; $k$ in mD; $h$ , $w$ and $L$ in m; $\mu_w$ in mPa·s.

$0.536 = 2\pi\cdot0.0853$ and $0.256 = 3\cdot0.0853$ , where 0.0853 is the constant of Darcy's law in these units (Rock permeability).

Oilfield units

J_{\text{rad}} = \frac{7.08\cdot10^{-3}\,f\,k\,h}{\mu_w\left(\ln r_{aD} - 0.75\right)},\qquad J_{\text{lin}} = \frac{3.381\cdot10^{-3}\,k\,w\,h}{\mu_w\,L}

$J$ in reservoir barrels per day per psi; $k$ in mD; $h$ , $w$ and $L$ in ft; $\mu_w$ in cP.

$7.08\cdot10^{-3} = 2\pi\cdot1.127\cdot10^{-3}$ and $3.381\cdot10^{-3} = 3\cdot1.127\cdot10^{-3}$ .

Worked exampleWater influx by the Fetkovich method

Given (assumed data)

Aquifer: the large aquifer of the lesson Aquifer models: water influx: it surrounds the whole accumulation, which has a radius $r_o = 1595.8$ m, and it is closed at a radius of 16 km ( $r_{aD}$ = 10.0265); thickness 10 m, porosity 0.20, permeability 50 mD, water viscosity 0.4 mPa·s, $c_t = 1.25\cdot10^{-3}$ MPa⁻¹
Pressure history: the yearly readings of the lesson Unsteady-state water influx: the van Everdingen–Hurst method: 30.00 MPa at the start, then 28.47, 26.52, 24.79, 23.73, 22.87, 22.37, 21.89 and 21.44 MPa at the end of the first to the eighth year

Solution

Pore volume of the aquifer: $W = \pi\,(16{,}000^2 - 1595.8^2)\cdot10\cdot0.20 = 1592.5$ million m³. Storage capacity: $c_tW = 1990.6$ thousand m³/MPa. Maximum encroachable water: $W_{ei} = c_tWp_i = 59.72$ million m³.
Productivity index: $\ln 10.0265 - 0.75 = 1.5552$ , so $J = 0.536\cdot1\cdot50\cdot10/(0.4\cdot1.5552) = 430.8$ m³/(d·MPa). Time constant: $\tau = 1990.6\cdot10^3/430.8 = 4621$ d, or 12.65 years. For a yearly step, $1 - \e^{-365.25/4621} = 0.0760$ .
First year: the average pressure at the contact is $\bar p_1 = (30 + 28.47)/2 = 29.235$ MPa, and the aquifer is still untouched ( $\bar p_{a0} = 30$ ). Influx: $\Delta W_{e,1} = 1990.6\cdot(30 - 29.235)\cdot0.0760 = 115.7$ thousand m³. Aquifer pressure: $\bar p_{a1} = 30 - 115.7/1990.6 = 29.942$ MPa.
Second year: $\bar p_2 = (28.47 + 26.52)/2 = 27.495$ MPa, and $\Delta W_{e,2} = 1990.6\cdot(29.942 - 27.495)\cdot0.0760 = 370.2$ thousand m³. The total is 485.9 thousand m³, and $\bar p_{a2} = 30 - 485.9/1990.6 = 29.756$ MPa.
Continue the same way, step by step. At the end of the eighth year $W_e$ = 5492.3 thousand m³, and the aquifer pressure is 27.24 MPa when the pressure at the contact is 21.44 MPa.

Answer: By the Fetkovich method the aquifer gave up 5492.3 thousand m³ of water by the end of the eighth year, which is 3.4% less than the exact solution. In the first year it gave 115.7 thousand m³, which is 39% less.

What it means. During the whole calculation the method kept one number, the pressure of the “tank”, and never went back to earlier steps. By the eighth year the aquifer pressure had fallen only to 27.24 MPa. The large aquifer is far from exhausted, and the pressure difference of 5.8 MPa between it and the contact drives water into the accumulation. But in the first years the method greatly understates the influx. The model of a tank behind a resistance does not include the transient period. In that period the parts of the aquifer nearest to the accumulation give up the water. The section that compares the methods shows this in detail. The data in this example are assumed for illustration.

The solution for a constant influx rate: the influence function

The Carter–Tracy method needs a second problem for a radial aquifer, the companion of the van Everdingen–Hurst problem. Suppose that water enters the accumulation at a constant rate $q_w$ . How does the pressure at the contact fall? This is the problem with a constant rate (constant terminal rate), known from well testing (the well pressure function). The accumulation again plays the role of the well, and there is no skin factor at the contact (the skin factor is a number that shows how much the zone right around a well helps or hinders flow). Its dimensionless answer is called the aquifer influence function.

Influence function of a radial aquifer

p_D(t_D) = \frac{2\pi f k h\,\big(p_i - p_o(t)\big)}{q_w\,\mu_w};\qquad p_D \approx 2\sqrt{t_D/\pi}\ \ (t_D \ll 1),\qquad p_D \approx \tfrac12\big(\ln t_D + 0.809\big)\ \ \text{(infinite)},\qquad p_D \approx \frac{2t_D}{r_{aD}^2 - 1} + \ln r_{aD} - \frac34\ \ \text{(closed, late)}

where:

$p_o(t)$ pressure at the contact for a constant influx rate, Pa
$q_w$ constant water influx rate at reservoir conditions, m³/s
$t_D$ dimensionless time of the aquifer, as in the van Everdingen–Hurst method, —
$f,\ k,\ h,\ \mu_w$ fraction of the perimeter, permeability, thickness of the aquifer, viscosity of water, —, m², m, Pa·s

Units: SI; a dimensionless function

Holds when

A uniform radial aquifer. The influx rate is constant from the moment $t = 0$ . There is no skin at the contact.
The logarithmic form holds for $t_D$ above about 25, while the aquifer still behaves as an infinite one. The last form holds after the pressure drop has reached the closed boundary, for $r_{aD} \gg 1$ (for small $r_{aD}$ , use the exact expression for a ring instead of $\ln r_{aD} - 3/4$ ).

The pressure drop at the contact grows at first like the square root of time, then like the logarithm of time. For a closed aquifer, after the pressure drop reaches its boundary, it grows linearly: the aquifer empties like a tank. The term $2t_D/(r_{aD}^2 - 1)$ is the fall of its average pressure, and the constant term is the pressure drop across the resistance. The Fetkovich model reproduces exactly this late straight line: a tank with the storage capacity $c_tW$ behind a resistance equal to the constant term. So it is correct when the aquifer works in the pseudo-steady-state regime, and it is wrong in the transient period.

CalculatedThe influence function

p_D(t_D)

, the pressure decline at the contact for a constant influx rate, for closed aquifers with

r_{aD}

= 2, 5 and 10 and for an infinite aquifer (the time axis is logarithmic; calculated from the diffusivity equation).

The two problems are linked. For any influx history, the pressure decline at the contact is a convolution (a sum of responses) of the influx rate with the derivative of the influence function. Each change in the rate gives its own response, and the responses add up, as when the rate of a well changes (the Duhamel integral). So if we know $p_D(t_D)$ , we can recover the influx from the pressure history. The Carter–Tracy method does this approximately, step by step.

Influence function of the large aquifer of the lesson (

r_{aD}

≈ 10.03) at the dates of the yearly readings

Year: 1
$t_D$: 6.115
$p_D$: 1.4438
$p_D' = \dd p_D/\dd t_D$: 0.06693

Year: 2
$t_D$: 12.231
$p_D$: 1.7391
$p_D' = \dd p_D/\dd t_D$: 0.03622

Year: 3
$t_D$: 18.346
$p_D$: 1.9253
$p_D' = \dd p_D/\dd t_D$: 0.02629

Year: 4
$t_D$: 24.462
$p_D$: 2.0727
$p_D' = \dd p_D/\dd t_D$: 0.02250

Year: 5
$t_D$: 30.577
$p_D$: 2.2051
$p_D' = \dd p_D/\dd t_D$: 0.02103

Year: 6
$t_D$: 36.693
$p_D$: 2.3317
$p_D' = \dd p_D/\dd t_D$: 0.02046

Year: 7
$t_D$: 42.808
$p_D$: 2.4560
$p_D' = \dd p_D/\dd t_D$: 0.02024

Year: 8
$t_D$: 48.923
$p_D$: 2.5795
$p_D' = \dd p_D/\dd t_D$: 0.02015

Year	$t_D$	$p_D$	$p_D' = \dd p_D/\dd t_D$
1	6.115	1.4438	0.06693
2	12.231	1.7391	0.03622
3	18.346	1.9253	0.02629
4	24.462	2.0727	0.02250
5	30.577	2.2051	0.02103
6	36.693	2.3317	0.02046
7	42.808	2.4560	0.02024
8	48.923	2.5795	0.02015

Calculated by solving the diffusivity equation with a constant influx rate at the contact and a closed outer boundary. By the eighth year the derivative is almost equal to $2/(r_{aD}^2 - 1) = 0.0201$ : the aquifer has reached the pseudo-steady-state regime.

The Carter–Tracy method

This is the idea of the Carter–Tracy method, proposed in 1960. During the current step, treat the influx rate as constant, and take the pressure at the contact at the end of the step. Choose the rate so that the pressure decline at the end of the step, calculated from the influence function, equals the measured decline. To avoid adding up the whole history, replace it too with a simple one. Imagine that the water has flowed at the same constant rate from the very beginning, and that at the initial moment an instant influx was added on top. Choose that instant influx so that exactly $W_{e,n-1}$ has accumulated by the start of the step.

The Carter–Tracy step

W_{e,n} = W_{e,n-1} + \big(t_{Dn} - t_{D,n-1}\big)\,\frac{U\,\Delta p_n - W_{e,n-1}\,p_D'(t_{Dn})}{p_D(t_{Dn}) - t_{D,n-1}\,p_D'(t_{Dn})},\qquad \Delta p_n = p_i - p_n

where:

$W_{e,n},\ W_{e,n-1}$ cumulative influx at the end of the current and the previous step, m³
$t_{Dn},\ t_{D,n-1}$ dimensionless time at the end of the current and the previous step, —
$\Delta p_n$ total pressure decline at the contact from the initial pressure to the end of the step (not the drop over one step), MPa
$U$ aquifer constant, $2\pi f\phi c_thr_o^2$ , m³/MPa
$p_D,\ p_D'$ influence function and its derivative with respect to $t_D$ at the end of the step, —

Units: m³, MPa; time is dimensionless

Holds when

The influx rate is constant within the step.
The whole history before the step is replaced by an equivalent one: an instant influx at the initial moment plus the constant rate of the current step from the very beginning (see the advanced section).
The influence function matches the aquifer: a closed one with its own $r_{aD}$ , or an infinite one.

The fraction in the formula is the influx rate during the step, in m³ per unit of dimensionless time. The numerator is the measured pressure decline multiplied by $U$ , minus the response to the instant influx $W_{e,n-1}$ . The denominator is the response to a unit constant rate from the very beginning, minus the water that this rate would have given before the start of the step. For the first step ( $W_{e,0} = 0$ , $t_{D0} = 0$ ) the formula gives $W_{e,1} = U\Delta p_1t_{D1}/p_D(t_{D1})$ . This is the influx at a constant rate that lowered the pressure by $\Delta p_1$ by the end of the step. If the influx rate really has been constant from the very beginning, the formula is exact for any step length.

Worked exampleWater influx by the Carter–Tracy method

Given (assumed data)

Aquifer: the same one: $U = 40{,}000$ m³/MPa, $t_D$ = 6.115 per year, closed at $r_{aD}$ ≈ 10.03; the influence function is in the table above
Pressure history: the same one, yearly steps; $\Delta p_n = 30 - p_n$

Solution

First year: $t_{D1}$ = 6.1154, $p_D = 1.4438$ , and there is no cumulative influx yet. $W_{e,1} = 6.1154\cdot40{,}000\cdot1.53/1.4438 = 259.2\cdot10^3$ m³.
Second year: $t_{D2}$ = 12.231, $p_D = 1.7391$ , $p_D' = 0.03622$ . Numerator: $40{,}000\cdot3.48 - 259.2\cdot10^3\cdot0.03622 = 129.81\cdot10^3$ m³. Denominator: $1.7391 - 6.1154\cdot0.03622 = 1.5176$ . Rate: $a_2 = 129.81\cdot10^3/1.5176 \approx 85.54\cdot10^3$ m³ per unit of $t_D$ .
Influx at the end of the second year: $W_{e,2} = 259.2 + 85.54\cdot6.1154 = 782.3$ thousand m³.
Continue in the same way. At the end of the eighth year $W_e$ = 5754.2 thousand m³. The influx rate of the last years is about 2379 m³/d.

Answer: By the Carter–Tracy method with a yearly step, the aquifer gave up 5754.2 thousand m³ by the end of the eighth year, which is 1.2% more than the exact solution. In the first year it gave 259.2 thousand m³, which is 36% more.

What it means. This method also kept one number, the cumulative influx. From the solution for the aquifer it took only the influence function and its derivative at the current moment. In the first step it assumed that water flowed at a constant rate for the whole year, a rate at which the pressure at the contact falls by 1.53 MPa by the end of the year. In reality the pressure fell steadily, so the influx at the start of the year was weaker, and the method overstated the influx of the first year by a third. The influx rate of the last years is almost equal to the underground withdrawal per day (the volume of oil, gas and water taken out of the accumulation, measured at reservoir conditions). The pressure in the accumulation is now changing slowly. The data in this example are assumed for illustration.

CalculatedWhat the Carter–Tracy method puts in place of the real history at the fourth yearly step: a straight line through its own influx at the end of the third year, with a slope equal to the rate of the current step. At the initial moment it starts with an instant “negative” influx of about −0.9 million m³. The curve is the exact solution for the same pressure history, and the dots are the results of the method year by year.

Comparison with the exact solution on one history

Let us compare three calculations on one pressure history and one aquifer. The exact solution comes from the diffusivity equation for the aquifer, with the pressure at the contact changing along a straight line between the readings. In the illustrative history of this lesson it agrees with the influx found from the material balance (checking against the balance). All three calculations get only the yearly readings.

Large aquifer: the exact influx and the errors of the calculations with a yearly step

Year: 1
exact influx, thousand m³: 190.5
superposition, %: −16.2
Fetkovich, %: −39.2
Carter–Tracy, %: 36.1

Year: 2
exact influx, thousand m³: 667.1
superposition, %: −6.3
Fetkovich, %: −27.2
Carter–Tracy, %: 17.3

Year: 3
exact influx, thousand m³: 1372.8
superposition, %: −2.9
Fetkovich, %: −19.4
Carter–Tracy, %: 9.2

Year: 4
exact influx, thousand m³: 2188.2
superposition, %: −1.2
Fetkovich, %: −13.6
Carter–Tracy, %: 5.0

Year: 5
exact influx, thousand m³: 3057.4
superposition, %: −0.8
Fetkovich, %: −9.7
Carter–Tracy, %: 3.2

Year: 6
exact influx, thousand m³: 3935.2
superposition, %: −0.4
Fetkovich, %: −6.8
Carter–Tracy, %: 2.1

Year: 7
exact influx, thousand m³: 4810.5
superposition, %: −0.3
Fetkovich, %: −4.8
Carter–Tracy, %: 1.6

Year: 8
exact influx, thousand m³: 5685.7
superposition, %: −0.3
Fetkovich, %: −3.4
Carter–Tracy, %: 1.2

Year	exact influx, thousand m³	superposition, %	Fetkovich, %	Carter–Tracy, %
1	190.5	−16.2	−39.2	36.1
2	667.1	−6.3	−27.2	17.3
3	1372.8	−2.9	−19.4	9.2
4	2188.2	−1.2	−13.6	5.0
5	3057.4	−0.8	−9.7	3.2
6	3935.2	−0.4	−6.8	2.1
7	4810.5	−0.3	−4.8	1.6
8	5685.7	−0.3	−3.4	1.2

The error is the deviation from the exact solution. Superposition uses half drops: each step takes half of the pressure fall in the interval before it and half of the fall in the interval after it. The Fetkovich method uses the index $J$ from the formula with $\ln r_{aD} - 3/4$ .

CalculatedThe error of the cumulative influx against the exact solution for the history of this lesson: superposition, the Fetkovich method and the Carter–Tracy method with a yearly step (solid lines), and the last two methods with a monthly step (dashed). Superposition with a monthly step agrees with the exact solution within 0.4%.

Error with a yearly, monthly and weekly step, % of the exact solution

Calculation: superposition, yearly step
Year 1: −16.2
Year 2: −6.3
Year 4: −1.2
Year 8: −0.3

Calculation: superposition, monthly step
Year 1: −0.4
Year 2: −0.2
Year 4: 0.0
Year 8: 0.0

Calculation: Fetkovich, yearly step
Year 1: −39.2
Year 2: −27.2
Year 4: −13.6
Year 8: −3.4

Calculation: Fetkovich, monthly step
Year 1: −38.4
Year 2: −26.7
Year 4: −13.3
Year 8: −3.3

Calculation: Carter–Tracy, yearly step
Year 1: 36.1
Year 2: 17.3
Year 4: 5.0
Year 8: 1.2

Calculation: Carter–Tracy, monthly step
Year 1: −4.5
Year 2: −7.3
Year 4: −6.0
Year 8: −2.3

Calculation: Carter–Tracy, weekly step
Year 1: −7.4
Year 2: −9.0
Year 4: −6.8
Year 8: −2.6

Calculation	Year 1	Year 2	Year 4	Year 8
superposition, yearly step	−16.2	−6.3	−1.2	−0.3
superposition, monthly step	−0.4	−0.2	0.0	0.0
Fetkovich, yearly step	−39.2	−27.2	−13.6	−3.4
Fetkovich, monthly step	−38.4	−26.7	−13.3	−3.3
Carter–Tracy, yearly step	36.1	17.3	5.0	1.2
Carter–Tracy, monthly step	−4.5	−7.3	−6.0	−2.3
Carter–Tracy, weekly step	−7.4	−9.0	−6.8	−2.6

The monthly and weekly (1/52 of a year) steps are built from the same yearly readings (inside a year, the pressure follows a straight line).

Superposition with half drops (each step takes half of the pressure fall in the interval before it and half of the fall in the interval after it) underestimates the influx only because of the step. As the step gets smaller, it comes closer and closer to the exact solution, because its aquifer model is complete.
The Fetkovich method underestimates the influx in the first years by almost the same amount with any step. This is a model error. From the very start, a tank behind a resistance gives up water the way a real aquifer does only after the pressure drop has reached its boundary. At the start, a large aquifer works like an infinite one, and its parts near the accumulation give up water much faster. The error gets smaller as the aquifer reaches the pseudo-steady-state regime. Here it falls to 3% by the eighth year. For an aquifer that will not “feel” its boundary for a long time, the error stays large, and for an infinite aquifer the method cannot be used.
The Carter–Tracy method with a yearly step overestimates the influx of the first years. It gives the whole first year the rate that is needed to lower the pressure by the end of the year, but in reality the influx at the start of the year was weaker. With a monthly step the sign of the error changes (the method now understates the influx), and with a weekly step the error is larger still in size. When the step gets even smaller, the calculation does not approach the exact solution but the solution of its own formula. For the history of this lesson this is about −8 to −10% in the first three years and −2.6% by the eighth (a daily step gives the same). The formula replaces the past with an equivalent history, and this approximation does not go away when the step gets smaller. It becomes exact only when the influx rate stays unchanged for a long time (see the advanced section).

Worked exampleThe same methods for a small aquifer

Given (assumed data)

Aquifer: the small aquifer of the lesson Aquifer models: water influx: it surrounds the whole accumulation and is closed at a radius of 3.2 km ( $r_{aD}$ = 2.005); the other properties are the same
Pressure history: the same; the step is one year

Solution

Storage capacity: $c_tW = 60.4$ thousand m³/MPa, 33 times less than for the large aquifer. $U$ is the same, 40,000 m³/MPa.
The formula with 3/4 cannot be used: $\ln 2.005 - 0.75 = -0.054$ . The exact expression for a ring gives 0.3171 instead of this number, so $J = 0.536\cdot50\cdot10/(0.4\cdot0.3171) = 2113$ m³/(d·MPa) and $\tau = 60.4\cdot10^3/2113 = 28.6$ d. The aquifer responds within a month.
The exact influx at the end of the first and the eighth year is 85.2 and 515.1 thousand m³. Superposition with a yearly step gives 46% less in the first year and 2.2% less by the eighth. The Fetkovich method gives the same.
The Fetkovich method with a monthly step: the error is −0.8% in the first year. The same yearly step, but with the pressure changing along a straight line inside the step (see the advanced section): 0.00%.
The Carter–Tracy method with a yearly step: 0.6% in the first year and 0.00% by the eighth.

Answer: For the small aquifer the Carter–Tracy method is accurate already with a yearly step, while superposition with half drops and the Fetkovich method with the same step underestimate the influx of the first year by almost a half.

What it means. A small aquifer catches up with the pressure at the contact within a month, and it gives up water almost at once in response to the current contact pressure. Rules that use the average pressure of the step lag by half a step. In the first year they “see” an average pressure decline of 0.77 MPa instead of the 1.53 MPa at the end of the year, so the influx comes out almost half as large. The Carter–Tracy method takes the pressure at the end of the step and does not have this problem. The Fetkovich model is correct for such an aquifer. Only the step needs to be fixed: make it shorter, or allow for the steady change of pressure inside the step. The data in this example are assumed for illustration.

Which method to choose

Where each method works

Aquifer: small, responds within weeks
Superposition: the step is shorter than the response time
Fetkovich: exact $J$ and a short step
Carter–Tracy: good

Aquifer: large, closed
Superposition: the reference
Fetkovich: underestimates the first years
Carter–Tracy: error of a few per cent

Aquifer: very large or infinite
Superposition: the reference
Fetkovich: cannot be used
Carter–Tracy: with the influence function of an infinite aquifer

Aquifer	Superposition	Fetkovich	Carter–Tracy
small, responds within weeks	the step is shorter than the response time	exact $J$ and a short step	good
large, closed	the reference	underestimates the first years	error of a few per cent
very large or infinite	the reference	cannot be used	with the influence function of an infinite aquifer

Superposition with short steps is the reference. It is worth checking the step-by-step methods against it at least once on a real pressure history.

The error of the Fetkovich method in the transient period can be avoided. Start the calculation with superposition, and switch to the Fetkovich method when the pressure drop has reached the boundary of the aquifer. At the moment of the switch, take the “tank” pressure from the cumulative influx, $\bar p_a = p_i - W_e/(c_tW)$ . The example shows how much the moment of the switch matters.

Worked exampleSuperposition at the start, the Fetkovich method afterwards

Given (assumed data)

Aquifer and history: the large aquifer and the history of this lesson; before the switch, superposition with quarterly steps, after it, Fetkovich steps one year long
Time for the pressure drop to reach the boundary: about 3.3 years (the example in the lesson on aquifer models)

Solution

Switch after the fourth year: the tank pressure is $\bar p_a = 30 - 2185.0/1990.6 = 28.902$ MPa. The error in the fifth to the eighth year is −0.8, −0.4, 0.0 and 0.4%.
Switch after the second year, when the aquifer still works as an infinite one: the error in the third year is −7.6%, and by the eighth year −1.5%.
For comparison, the Fetkovich method from the very start gives −9.7% in the fifth year and −3.4% in the eighth.

Answer: If we switch to the Fetkovich method after the pressure drop has reached the boundary, the error does not exceed 1%. A switch that is too early brings back the error of the transient period.

What it means. The Fetkovich method is correct where its model is correct, after the transient period. Superposition is needed only in the first years, while it is still cheap to calculate. After that the calculation goes step by step. The time to reach the boundary is estimated from the hydraulic diffusivity (how fast a pressure change spreads through the rock) and the size of the aquifer, $t_a \approx (r_a - r_o)^2/(4\eta)$ . The data in this example are assumed for illustration.

Terms and symbols

Step-by-step (recursive) method: A calculation of the influx during a step from a small “state” of the aquifer taken from the previous step, without adding up the whole pressure history.
Maximum encroachable water $W_{ei}$: It equals $c_tWp_i$ : the influx if the aquifer pressure fell to zero. It is the reference quantity of the Fetkovich method, and $W_{ei}/p_i = c_tW$ .
Time constant $\tau$: It equals $c_tW/J$ . It is the time in which a tank behind a resistance gives up about 63% of its elastic store when the pressure at the contact is constant.
Influence function $p_D(t_D)$: The dimensionless pressure decline at the contact for a constant influx rate. It is the basis of the Carter–Tracy method.
Equivalent history: What the Carter–Tracy method puts in place of the past: an instant influx at the initial moment and the constant rate of the current step.
Model error and step error: The first does not go away when the step gets smaller (the Fetkovich method in the transient period, the Carter–Tracy method). The second does (superposition).

Common mistakes and misconceptions

Wrong

“The shorter the step, the more accurate any method.”

In fact

Only superposition gets closer and closer to the exact solution. With a monthly step the Fetkovich method is wrong by almost as much as with a yearly step. In the example of this lesson the Carter–Tracy method changes the sign of its error: +36% with a yearly step, −4 to −7% with a monthly step and −7 to −9% with a weekly step in the first years. The error does not go away as the step gets smaller.

Wrong

“The Fetkovich method is just a fast form of the van Everdingen–Hurst method, and the results are always close.”

In fact

They are close after the transient period. For the large aquifer of this lesson, the method underestimates the influx of the first year by 39% and of the fourth year by 14%. Switching to it after the pressure drop has reached the boundary gives an error of less than 1%.

Wrong

“In the Carter–Tracy formula, you use the pressure drop over the step.”

In fact

You use the full pressure decline at the contact from the initial pressure, $p_i - p_n$ . The formula compares it with the response of the whole equivalent influx history.

Wrong

“ $W_{ei}$ is the water that the aquifer will give up.”

In fact

$W_{ei} = c_tWp_i$ is the influx if the pressure fell to zero; it is a reference quantity. The real limit for a pressure decline of $\Delta p$ is $c_tW\Delta p$ . For the large aquifer of this lesson, at a decline of 8.6 MPa it is 17 million m³ out of the 59.7 million m³ of “maximum encroachable water”.

In depth: advanced detail

Where the Carter–Tracy formula comes from, and why it is approximate

For any influx history, the pressure decline at the contact is the sum of the responses to each small increment of the influx (superposition in the constant-rate solution). If the cumulative influx $W_e(\lambda)$ is given as a function of dimensionless time, then

Pressure at the contact as a convolution of the influx history

U\,\big(p_i - p(t_D)\big) = \int_0^{t_D}\frac{\dd W_e}{\dd\lambda}\;p_D'(t_D - \lambda)\dd\lambda

where:

$W_e(\lambda)$ cumulative influx up to the dimensionless moment $\lambda$ , m³
$p_D'$ derivative of the influence function, —
$U$ aquifer constant, m³/MPa

Units: m³ (both sides)

Holds when

The problem is linear. Before the start of development the pressure in the aquifer equals $p_i$ everywhere.
An instant influx of volume $V$ at the moment $\lambda$ gives the response $V\,p_D'(t_D - \lambda)$ , the derivative of the response to a constant rate.

For a constant rate $a$ from the very beginning, the integral equals $a\,p_D(t_D)$ , which is the definition of the influence function. The inverse operation, finding the influx from the pressure history, is the van Everdingen–Hurst problem. In Laplace transforms (a mathematical tool that turns this kind of integral into a product) the solutions of the two problems are linked by the relation $\bar q_D(s)\,\bar p_D(s) = 1/s^2$ , where $\bar q_D$ is the transform of the dimensionless influx rate at constant pressure.

The Carter–Tracy method puts into this integral not the real history but an equivalent one: a constant rate $a_n$ from the very beginning plus an instant influx $W_{e,n-1} - a_nt_{D,n-1}$ at the moment $\lambda = 0$ . It agrees with the real history in two ways: $W_{e,n-1}$ has accumulated by the start of the step, and the rate during the step is $a_n$ . Putting it into the integral, we get:

Pressure decline from the equivalent influx history

U\,\Delta p_n = a_n\,p_D(t_{Dn}) + \big(W_{e,n-1} - a_n\,t_{D,n-1}\big)\,p_D'(t_{Dn})

where:

$a_n$ constant influx rate of the current step, m³ per unit of $t_D$
$W_{e,n-1} - a_n\,t_{D,n-1}$ instant influx of the equivalent history at the initial moment, m³
$\Delta p_n$ total pressure decline at the contact by the end of the step, MPa

Units: m³ (both sides)

Holds when

The equivalent history instead of the real one; the influence function is for the given aquifer.

Solving this equation for $a_n$ , we get the formula of the method. In the figure in the section on the method, the instant influx of the equivalent history is negative: the rate is growing, and the present constant rate “from the very beginning” would have given too much water.

If we make the steps as small as we like, the rate of the step $a_n$ becomes the derivative $\dd W_e/\dd t_D$ , and $W_{e,n-1}$ and $t_{D,n-1}$ become the current $W_e$ and $t_D$ . The recursive formula turns into a differential equation. It contains only the values of the influence function at the current moment, not the whole convolution:

The limit of the Carter–Tracy formula for very short steps

\big(p_D - t_D\,p_D'\big)\,\frac{\dd W_e}{\dd t_D} + p_D'\,W_e = U\,\big(p_i - p(t_D)\big)

where:

$W_e$ cumulative influx by the method, m³
$t_D$ dimensionless time of the aquifer, —
$p_D,\ p_D'$ influence function and its derivative at the moment $t_D$ , —
$U$ aquifer constant, m³/MPa
$p_i - p(t_D)$ pressure decline at the contact, MPa

Units: m³ (both sides); time is dimensionless

Holds when

The steps tend to zero. The aquifer and the influence function are the same as in the formula of the method.

The solution of this equation agrees with the exact one only if the influx rate is constant. In the early period ( $p_D \approx 2\sqrt{t_D/\pi}$ ), for a constant pressure decline $\Delta p$ , it gives $W_e = \tfrac23\sqrt{\pi t_D}\,U\Delta p$ instead of the exact $2\sqrt{t_D/\pi}\,U\Delta p$ . This is larger by a factor of $\pi/3$ , that is by 4.7%. For a steady pressure decline at a rate $g$ (per unit of $t_D$ ) we get $W_e = \tfrac25\sqrt\pi\,Ug\,t_D^{3/2}$ instead of the exact $\tfrac{4}{3\sqrt\pi}\,Ug\,t_D^{3/2}$ . This is $3\pi/10 \approx 0.94$ of the exact value, that is 6% less. So in the history of this lesson, where the pressure falls almost steadily, the method with short steps underestimates the influx.

The error of the method does not go away when the step gets smaller. It is set by how far the influx history differs from a constant rate, not by the length of the step.

The exact resistance of a ring-shaped aquifer

The productivity index $J$ of the Fetkovich method is the ratio of the influx rate to the difference between the average aquifer pressure and the pressure at the contact in the pseudo-steady-state regime. For a ring $1 \le r_D \le r_{aD}$ with an impermeable outer boundary, this difference can be found exactly by solving the diffusivity equation for the case where the pressure falls at the same rate at every point:

Dimensionless resistance of a ring-shaped aquifer in the pseudo-steady-state regime

b(r_{aD}) = \frac{4\,r_{aD}^4\ln r_{aD} - 3\,r_{aD}^4 + 4\,r_{aD}^2 - 1}{4\,\big(r_{aD}^2 - 1\big)^2}\;\xrightarrow{\;r_{aD}\gg1\;}\;\ln r_{aD} - \frac34,\qquad J = \frac{2\pi f k h}{\mu_w\,b(r_{aD})}

where:

$b$ dimensionless difference between the average aquifer pressure and the pressure at the contact, in units of $q_w\mu_w/(2\pi fkh)$ , —
$r_{aD}$ ratio of the outer radius of the aquifer to the radius of the accumulation, —

Units: dimensionless; $J$ in SI (in metric oilfield units $2\pi \to 0.536$ )

Holds when

A uniform ring. The outer boundary is impermeable. The flow has settled into the pseudo-steady-state regime.

For $r_{aD} = 10$ the exact value is 1.59, against 1.55 from the simplified formula. $J$ differs by 3%, which is small compared with the error of the transient period. For $r_{aD} = 2$ the exact value is 0.32, while the simplified one is negative. The same $b$ appears in the late-time form of the influence function (the straight line that it approaches): $p_D \approx 2t_D/(r_{aD}^2 - 1) + b(r_{aD})$ .

The Fetkovich step when the pressure at the contact changes

The rule of the average pressure of the step is accurate while the step is short compared with $\tau$ . If the pressure at the contact changes along a straight line from $p_{n-1}$ to $p_n$ inside the step, the tank equations can be solved exactly even for a long step:

The Fetkovich step when the pressure at the contact changes linearly

\Delta W_{e,n} = c_tW\left[\big(\bar p_{a,n-1} - p_{n-1}\big)\big(1 - \e^{-\Delta t_n/\tau}\big) + \big(p_{n-1} - p_n\big)\left(1 - \frac{\tau}{\Delta t_n}\big(1 - \e^{-\Delta t_n/\tau}\big)\right)\right]

where:

$p_{n-1},\ p_n$ pressure at the contact at the start and at the end of the step, MPa
$\bar p_{a,n-1}$ average aquifer pressure at the start of the step, MPa
$\tau,\ \Delta t_n$ time constant of the aquifer and length of the step, d

Units: m³, MPa, days

Holds when

The pseudo-steady-state aquifer model. The pressure at the contact changes along a straight line inside the step.

For $\Delta t_n \ll \tau$ the formula agrees with the average-pressure rule. For $\Delta t_n \gg \tau$ the second term comes close to the whole pressure drop of the step. The tank keeps up with the contact and lags behind it by only the time $\tau$ , while the average-pressure rule lags by half a step. For the small aquifer of this lesson with a yearly step, this formula gives the influx practically without error.

Connections to other topics

Unsteady-state water influx: the van Everdingen–Hurst method — the exact solution and superposition, against which the simplified methods are compared.
Aquifer models: water influx — the pseudo-steady-state aquifer model and its time constant.
Line-source solution and the pressure function — the well pressure function is the same influence function when the accumulation plays the role of the well.
Pseudo-steady-state flow and the shape factor — the pseudo-steady-state regime, and where the term 3/4 in the productivity formula comes from.
Superposition principle — the Duhamel integral, which the Carter–Tracy method replaces with an approximation.
History matching of aquifer parameters — matching the aquifer parameters to the history for the chosen method.
Straight-line material balance (Havlena–Odeh method) — water influx in the straight-line form of the material balance.

Check yourself

1
What “state” of the aquifer does the Fetkovich method carry from step to step?
1. A.Only the pressure at the contact
2. B.The average pressure of the aquifer
3. C.The whole history of pressure steps
4. D.The influence function and its derivative
Show the answer
Correct answer: B. The average pressure of the aquifer
A tank behind a resistance is fully described by the tank pressure. If we know it at the start of the step, and the pressure at the contact, we find the influx during the step from an exponential with the time constant $\tau = c_tW/J$ , and the new tank pressure from its balance. So the past steps are not needed.
2
The storage capacity of an aquifer is $c_tW$ = 500 thousand m³/MPa, $J$ = 200 m³/(d·MPa), and the initial pressure is 25 MPa. At the start of a half-year step (182.6 d) the aquifer pressure is 24.60 MPa, and the pressure at the contact falls during the step from 23.8 to 23.2 MPa. Find the influx during the step and the aquifer pressure at the end of the step.
Show the answer
$\tau = 500{,}000/200 = 2500$ d; $1 - \e^{-182.6/2500} = 1 - \e^{-0.0730} = 0.0704$ . The average pressure at the contact during the step is $(23.8 + 23.2)/2 = 23.5$ MPa. The influx is $\Delta W_e = 500{,}000\cdot(24.60 - 23.5)\cdot0.0704 \approx 38.7$ thousand m³. The aquifer pressure falls by $38{,}700/500{,}000 = 0.077$ MPa, to 24.52 MPa.
3
When does the Carter–Tracy formula give an exact result for any length of step?
1. A.When the influx rate has been constant since the start of development
2. B.When the pressure at the contact is constant
3. C.When the aquifer is infinite
4. D.When the step is shorter than a day
Show the answer
Correct answer: A. When the influx rate has been constant since the start of development
The formula replaces the past with the constant rate of the current step from the very beginning, plus an instant influx at the initial moment. If the rate really is constant, the instant influx is zero and the replacement is exact. When the pressure at the contact is constant, the influx rate falls, and even with very small steps the method overstates the influx: by 4.7% in the early period.
4
An aquifer with a radius of 2.4 km surrounds an accumulation with a radius of 1.5 km. Can we take $J$ for the Fetkovich method from the formula with $\ln r_{aD} - 3/4$ ?
1. A.No: the Fetkovich method works only for linear aquifers
2. B.No: $\ln 1.6 - 0.75 < 0$ , so $J$ comes out negative; the exact expression for a ring is needed
3. C.Yes, the formula works for any aquifer
4. D.Yes, but the time is multiplied by 3/4
Show the answer
Correct answer: B. No: $\ln 1.6 - 0.75 < 0$ , so $J$ comes out negative; the exact expression for a ring is needed
$r_{aD} = 2.4/1.5 = 1.6$ , and $\ln 1.6 = 0.47$ , which is less than 0.75. The simplified formula was derived for aquifers many times larger than the accumulation. The exact resistance of the ring at $r_{aD} = 1.6$ is positive and small (about 0.2). Such an aquifer responds fast, almost like a small “tank”.
5
A calculation by the Fetkovich method for a large aquifer underestimates the influx of the first years by tens of per cent. Will shorter steps help? What should we do?
Show the answer
No. This is an error of the model, not of the step. A tank behind a resistance gives up water from the very start at the rate $J\Delta p$ . A large aquifer in the transient period behaves like an infinite one and gives up water faster: the influx grows in proportion to the square root of time. Calculate the early period by superposition (or by the Carter–Tracy method), and switch to the Fetkovich method after the pressure drop has reached the boundary. In the example of this lesson this gives an error of less than 1%.
6
A calculation by the Carter–Tracy method with a yearly step gave an influx 3% above the exact one, and with a monthly step 5% below it. Is there a mistake in the calculations?
1. A.No: the method has two errors, one from the length of the step and one from the approximation itself. When the step gets smaller, the second one stays, and it can have the opposite sign
2. B.No: a monthly step always underestimates the influx
3. C.Yes: as the step gets smaller, the error should only get smaller
4. D.Yes: the influence function was taken wrongly
Show the answer
Correct answer: A. No: the method has two errors, one from the length of the step and one from the approximation itself. When the step gets smaller, the second one stays, and it can have the opposite sign
With a long step the method gives the whole step the rate that is needed for the pressure at its end. If the pressure inside the step falls steadily, the influx of the first steps comes out too high. With a short step the error that comes from the equivalent history remains, and for a steady pressure decline it is negative. In the example of this lesson: +5% and −6% by the fourth year.

Updated: 7 October 2026