Inflow to a hydraulically fractured well

Quick refresher

The quantity $\ln(x_f/r_w')$ , which is $S_f + \ln(x_f/r_w)$ , depends only on $F_{CD}$ . So there is one curve for all fractures. At high conductivity $r_w' \to x_f/2$ . At low conductivity $r_w' \to 0.28\,k_f w/k$ , and the length of the fracture stops mattering.
If the fracture is comparable with the drainage area, the effective wellbore radius overstates the productivity. Then we need the dimensionless productivity index $J_D$ of a closed area. In a square it does not exceed $6/\pi \approx 1.91$ . This is the limit of linear flow to a fracture across the whole width of the area. For a vertical well without skin it is about 0.14.
After start-up the flow regimes follow one another. First comes bilinear flow (the fluid flows in a straight line in the fracture and in the rock at the same time), where the pressure drop grows as $t^{1/4}$ . Then linear flow (the pressure drop grows as $\sqrt t$ ; this happens only at very high conductivity). Then pseudo-radial flow (from $t_{Dxf} \approx 3$ –5), where the flow looks like the flow to an ordinary well. Last comes pseudo-steady-state flow, when the boundary of the drainage area is felt.
In a tight reservoir (one with very low permeability) with a long fracture, the boundary of the area makes itself felt before pseudo-radial flow begins. In a square area this already happens when the penetration ratio $I_x$ (the fracture length as a share of the size of the area) is more than about 0.2–0.25.
While the flow is transient (still changing with time), the productivity is higher than the stabilised one. If we compare the first weeks after hydraulic fracturing with the stabilised rate before it, we overstate the gain several times over.
Choking of the fracture at the wellbore and damage to its faces both add a positive skin. Choking is more dangerous, because the whole flow of one wing (one half of the fracture, on one side of the well) passes through a short stretch.

Key relationships

t_{Dxf} = \dfrac{k\,t}{\phi\mu c_t x_f^2}

dimensionless time based on the half-length (SI; in metric oilfield units the factor is $3.553\cdot10^{-3}$ with $t$ in hours)

r_w' \to x_f/2

fracture of high conductivity, $F_{CD}$ from 30

r_w' \to 0.2807\,k_f w/k

fracture of low conductivity, $F_{CD}$ up to 0.1: the length has no effect

J_D = \dfrac{qB\mu}{2\pi kh\,(\bar p - p_{wf})},\qquad I_x = \dfrac{2x_f}{x_e}

dimensionless productivity index and penetration ratio of the fracture

p_{wD} = 2.45\,\dfrac{t_{Dxf}^{1/4}}{\sqrt{F_{CD}}},\qquad p_{wD} = \sqrt{\pi\,t_{Dxf}}

bilinear and linear flow

s_{ch} \approx \dfrac{\pi k\,x_s}{k_{f,ch}w},\qquad s_{fs} = \dfrac{\pi b_s}{2x_f}\left(\dfrac{k}{k_s} - 1\right)

choke skin of the fracture and skin of its faces

Common pitfalls

Taking $r_w' = x_f/2$ for a fracture whose half-length is comparable with half the side of the drainage area. The productivity will be overstated, and for a fracture across the whole width of the area it will be overstated by a factor of two or more.
Judging the effect of hydraulic fracturing by the rate of the first weeks, and comparing it with the stabilised rate before the treatment.
Assuming that a longer fracture always increases the inflow. When $F_{CD}$ is below 0.1, the extra length hardly works.
Looking for the derivative plateau (the flat part of the pressure derivative curve) to find $kh$ for a well with a long fracture in a tight reservoir. There may be no pseudo-radial period.
Confusing the penetration ratio $I_x$ of the fracture with the degree of penetration of the reservoir $h_w/h$ .

After this lesson you can

plot the effective wellbore radius against the dimensionless conductivity over the whole range, explain both of its asymptotes (the lines the curve approaches at its two ends) and the limits of the approximation
calculate the dimensionless productivity index of a fractured well in a closed drainage area, and explain why it cannot exceed the limit of linear flow
recognise the bilinear, linear and pseudo-radial flow regimes, estimate how long each lasts, and understand when there is no pseudo-radial regime at all
explain and estimate why the rates of the first weeks and months after hydraulic fracturing overstate the long-term gain
calculate the skin (a number that shows how much the zone right around the well helps or hinders flow) from choking of the fracture at the wellbore (a short stretch next to the wellbore where the fracture carries fluid much less easily than designed), and the skin of its faces, and estimate when they take the gain away from the treatment

What the equivalent skin leaves out

The lesson Hydraulic fracturing: purpose and well selection gave the main tool. Some months after start-up, a well with a fracture behaves like an ordinary well of radius $r_w'$ . The dimensionless conductivity $F_{CD} = k_f w/(k x_f)$ shows what part of the fracture length works. This is enough for a quick estimate, but four questions remain:

how the effective wellbore radius behaves at very low and very high conductivity, and where the accuracy of its approximation ends
what to do when the fracture is comparable with the drainage area, for example with closely spaced wells in a low-permeability reservoir
how long the transient period lasts, and why the first rates after hydraulic fracturing are misleading
what happens if the fracture is poorly connected to the wellbore, or if its faces are damaged by the fracturing fluid

It is convenient to write the solutions for a fractured well in dimensionless variables (variables with the units removed, so that one curve fits all cases), with the half-length of the fracture as the length scale. The dimensionless pressure drop is the same as in the lessons on transient flow: $p_D = 2\pi kh\,\Delta p/(qB\mu)$ , without skin (Line-source solution and the pressure function). The negative skin of the fracture does not appear in it separately, because the solution is built for the fracture itself and not for the wellbore. We measure time in dimensionless units based on the half-length:

Dimensionless time based on the fracture half-length

t_{Dxf} = \frac{k\,t}{\phi\,\mu\,c_t\,x_f^2}

where:

$t_{Dxf}$ dimensionless time, scaled by the square of the fracture half-length, —
$k$ permeability of the reservoir, m²
$t$ time since the well started at a constant rate, s
$\phi,\ \mu,\ c_t$ porosity, viscosity of the fluid at reservoir conditions, total compressibility, —, Pa·s, Pa⁻¹
$x_f$ propped half-length of the fracture, m

Units: SI (any consistent system of units)

Holds when

A single-phase, slightly compressible fluid. The hydraulic diffusivity $k/(\phi\mu c_t)$ is the same as in the diffusivity equation (Diffusivity equation).

At $t_{Dxf} \approx 1$ the pressure disturbance has moved away from the fracture by a distance close to its half-length. The duration of each regime grows as $x_f^2/k$ . A fracture twice as long in a reservoir ten times less permeable “lives” in the transient regime forty times longer.

In other unit systems (2)

Metric oilfield units

t_{Dxf} = 3.553\cdot10^{-3}\,\frac{k\,t}{\phi\,\mu\,c_t\,x_f^2}

$k$ in mD, $t$ in h, $\mu$ in mPa·s, $c_t$ in MPa⁻¹, $x_f$ in m.

$3.553\cdot10^{-3}$ is the same factor as for hydraulic diffusivity in the lesson Diffusivity equation.

Oilfield units

t_{Dxf} = 2.637\cdot10^{-4}\,\frac{k\,t}{\phi\,\mu\,c_t\,x_f^2}

$k$ in mD, $t$ in h, $\mu$ in cP, $c_t$ in psi⁻¹, $x_f$ in ft.

The lesson's model

All the curves and examples are calculated for one model. It has a vertical fracture over the whole thickness of the reservoir, with two identical wings. The reservoir is homogeneous and isotropic (the same properties in all directions). The fluid is single-phase and slightly compressible. The flow follows Darcy's law both in the reservoir and in the fracture, and the conductivity is constant along the length (unless choking is stated). The fracture is divided into segments with uniform inflow, and a system of equations is solved for the inflow into each segment. This is a semi-analytical solution of the classical problem of a fracture of finite conductivity (an advanced section at the end of the lesson). Real fractures are not uniform in height and length. So the model results are a guide, not an exact forecast.

Effective wellbore radius at any conductivity

The reservoir gives fluid to the fracture along its whole length, and the fracture carries the accumulated flow to the wellbore. Suppose the fracture carries fluid much more easily than the reservoir supplies it ( $F_{CD}$ is large). Then the pressure along it is almost constant, and the whole fracture works as an extension of the wellbore. Suppose it carries fluid less easily ( $F_{CD}$ is small). Then the pressure along the fracture rises quickly from the wellbore to its tip. The reservoir gives fluid mainly to the part of the fracture next to the wellbore, and the far part hardly works.

The dimensionless form of the problem gives an important result: at late times the quantity $\ln(x_f/r_w')$ depends only on $F_{CD}$ . So one curve describes any fracture in any reservoir. We only need to calculate the dimensionless conductivity. Then we subtract to get the equivalent skin (the skin that would give the same inflow as the fracture):

Equivalent skin of a fracture from the universal curve

S_f = \ln\frac{x_f}{r_w'}\bigg|_{F_{CD}} - \ln\frac{x_f}{r_w}

where:

$S_f$ equivalent skin of a well with a fracture, —
$\ln(x_f/r_w')\big|_{F_{CD}}$ value of the universal curve at the given dimensionless conductivity: from $\ln 2 \approx 0.693$ (infinite conductivity) to several units (low conductivity), —
$x_f,\ r_w,\ r_w'$ fracture half-length, wellbore radius and effective wellbore radius, m

Units: dimensionless; natural logarithms

Holds when

The flow is pseudo-radial: the pressure disturbance has moved away from the fracture by several of its half-lengths.
The fracture is small compared with the drainage area (the penetration ratio is below about 0.3, see the next section).
The conductivity is uniform along the length. Choking at the wellbore and damage to the faces are accounted for by separate skins (last section).

The first term describes the fracture, and the second describes the wellbore to which it is “attached”. So the smaller the wellbore radius, the larger the skin that the same fracture gives (larger in absolute value, that is, ignoring the sign). For $x_f$ = 50 m and $r_w$ = 0.108 m, the second term is $\ln(x_f/r_w) = 6.14$ .

The curve has two asymptotes. At high conductivity $r_w' \to x_f/2$ , the limit used in the lesson on hydraulic fracturing. At low conductivity the effective wellbore radius no longer depends on the length of the fracture:

Effective wellbore radius of a fracture of low conductivity

r_w' \approx \frac{1}{2\gamma}\,\frac{k_f w}{k} = 0.2807\,\frac{k_f w}{k}

where:

$r_w'$ effective wellbore radius of a well with a fracture, m
$k_f w$ fracture conductivity, mD·m
$k$ permeability of the reservoir, mD
$\gamma$ $\e^{0.5772} = 1.781$ , the exponential of Euler's constant, as in the inflow formulas; $1/(2\gamma) = 0.2807$ , —

Units: $k_f w/k$ has the dimension of length: with $k_f w$ in mD·m and $k$ in mD, the radius comes out in metres

Holds when

$F_{CD} \lesssim 0.1$ . At $F_{CD}$ = 0.1 the exact curve lies 3.3% below the asymptote, and at 0.05 it lies 1.6% below.
The flow is pseudo-radial. The fracture is longer than the part that really works (lengths of about $k_f w/k$ ).

A fracture of low conductivity works only along the stretch next to the wellbore. The ratio of the fracture and reservoir conductivities sets the length of this stretch. Everything beyond it hardly gives any fluid. So it is useless to lengthen such a fracture. Doubling the conductivity doubles $r_w'$ and reduces the skin by $\ln 2 \approx 0.69$ . The constant 0.2807 was found with the lesson's model: as $F_{CD}$ decreases, the model values converge to $1/(2\gamma)$ . This is the same value that the exact solution for an infinitely long fracture gives.

CalculatedCalculated effective wellbore radius against dimensionless conductivity (the lesson's model, double-logarithmic scale) and its asymptotes:

r_w'/x_f = 0.2807\,F_{CD}

at low conductivity and

r_w'/x_f = 0.5

at high conductivity. The transition between them lies roughly between

F_{CD}

= 0.3 and 30.

The approximation from the hydraulic fracturing lesson against the lesson's model

$F_{CD}$: 0.01
$r_w'/x_f$ , model: 0.0028
$r_w'/x_f$ , approximation: 0.0045
Deviation of $r_w'$: +60.7%

$F_{CD}$: 0.05
$r_w'/x_f$ , model: 0.0138
$r_w'/x_f$ , approximation: 0.0169
Deviation of $r_w'$: +22.6%

$F_{CD}$: 0.1
$r_w'/x_f$ , model: 0.0271
$r_w'/x_f$ , approximation: 0.0303
Deviation of $r_w'$: +11.7%

$F_{CD}$: 0.3
$r_w'/x_f$ , model: 0.0755
$r_w'/x_f$ , approximation: 0.0780
Deviation of $r_w'$: +3.2%

$F_{CD}$: 1
$r_w'/x_f$ , model: 0.193
$r_w'/x_f$ , approximation: 0.192
Deviation of $r_w'$: −0.5%

$F_{CD}$: 10
$r_w'/x_f$ , model: 0.436
$r_w'/x_f$ , approximation: 0.435
Deviation of $r_w'$: −0.1%

$F_{CD}$: 100
$r_w'/x_f$ , model: 0.493
$r_w'/x_f$ , approximation: 0.493
Deviation of $r_w'$: +0.0%

$F_{CD}$: 1000
$r_w'/x_f$ , model: 0.499
$r_w'/x_f$ , approximation: 0.492
Deviation of $r_w'$: −1.3%

$F_{CD}$	$r_w'/x_f$ , model	$r_w'/x_f$ , approximation	Deviation of $r_w'$
0.01	0.0028	0.0045	+60.7%
0.05	0.0138	0.0169	+22.6%
0.1	0.0271	0.0303	+11.7%
0.3	0.0755	0.0780	+3.2%
1	0.193	0.192	−0.5%
10	0.436	0.435	−0.1%
100	0.493	0.493	+0.0%
1000	0.499	0.492	−1.3%

The approximation is accurate to within 1.5% in $r_w'$ from $F_{CD} \approx 0.5$ up to thousands. At smaller values it overstates $r_w'$ more and more, and there it is better to use the asymptote $0.2807\,k_f w/k$ or the model. An error of $\varepsilon$ % in $r_w'$ changes the skin by about $\varepsilon/100$ .

Worked exampleA long fracture of low conductivity in a permeable reservoir

Given (assumed data)

Reservoir: $k$ = 40 mD, $r_w$ = 0.108 m
Fracture: $x_f$ = 70 m, $k_f w$ = 140 mD·m (little proppant)
Options: the same fracture half as long; the same length with twice the conductivity

Solution

Dimensionless conductivity: $F_{CD} = \dfrac{140}{40\cdot 70} = 0.05$ , so we are on the low-conductivity branch.
Asymptote: $r_w' \approx 0.2807\cdot\dfrac{140}{40} = 0.98$ m. By the model, $r_w' = 0.966$ m and $S_f = -\ln\dfrac{0.966}{0.108} = -2.19$ .
The approximation from the hydraulic fracturing lesson would give $r_w' = 1.185$ m and $S_f = -2.40$ . The effective wellbore radius is overstated by 23%.
A fracture half as long, $x_f$ = 35 m: $F_{CD} = 0.10$ . By the model $r_w' = 0.949$ m and $S_f = -2.17$ , almost the same.
A conductivity of 280 mD·m at $x_f$ = 70 m: $F_{CD} = 0.10$ , $r_w' = 1.899$ m, $S_f = -2.87$ .

Answer: The fracture works like a well of radius about 0.966 m (

S_f \approx

−2.19). Halving the length changes almost nothing. Doubling the conductivity improves the skin by 0.68.

What it means. Here it is conductivity that is missing, not length. Half of the fracture, and with it half of the fluid and proppant, is wasted. A permeable reservoir needs a short, wide fracture. The data in this example are assumed for illustration.

A fracture in a bounded drainage area

The effective wellbore radius describes the fracture as the flow “sees” it from far away. Suppose the half-length is comparable with the size of the drainage area. With closely spaced wells in a low-permeability reservoir, this is common. Then there is no “far away”. Fluid goes from the boundaries of the area straight to the plane of the fracture, and a large part of the flow is linear (along parallel lines). In this case we calculate the productivity of the well in a closed area directly. We compare the length of the fracture with the size of the area through the penetration ratio:

Penetration ratio of a fracture

I_x = \frac{2x_f}{x_e}

where:

$I_x$ penetration ratio of the fracture: the share of the side of the area that the fracture covers, —
$x_f$ propped half-length of the fracture, m
$x_e$ side of the square drainage area (for a rectangular area, the side parallel to the fracture), m

Units: dimensionless

Holds when

The fracture passes through the centre of the area, parallel to its side. At $I_x$ = 1 the ends of the fracture reach the boundaries.

Do not confuse it with the degree of penetration of the reservoir $h_w/h$ (Partial penetration and well deviation). That compares the opened thickness with the full thickness. Here we compare the length of the fracture with the size of the area in plan view.

We express productivity as the dimensionless productivity index. It is convenient because it lets us compare wells and fractures in reservoirs with any $k$ , $h$ and fluid properties:

Dimensionless productivity index

J_D = \frac{qB\mu}{2\pi kh\,(\bar p - p_{wf})},\qquad J = \frac{2\pi kh}{B\mu}\,J_D

where:

$J_D$ dimensionless productivity index, —
$J$ productivity index with respect to the average reservoir pressure, m³/(s·Pa)
$q,\ B,\ \mu$ rate at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s
$k,\ h$ permeability and thickness of the reservoir, m², m
$\bar p,\ p_{wf}$ average pressure in the drainage area and bottomhole flowing pressure, Pa

Units: SI (any consistent system of units)

Holds when

The regime is pseudo-steady-state in a closed area (Pseudo-steady-state flow and the shape factor). In the transient regime $J_D$ depends on time (section on the first rates).
The inflow is single-phase. For a vertical well in a square, $1/J_D$ equals $\ln(x_e/r_w) - 1.311 + S$ . Here $-1.311$ is $\tfrac12\ln\big(4/(\gamma C_A)\big)$ at $C_A$ = 30.88, where $C_A$ is the Dietz shape factor (a number that depends on the shape of the area and the position of the well).

$J_D$ is the inverse of the dimensionless drawdown. For a vertical well without skin in a square with a side of 300–500 m, $J_D \approx 0.14$ –0.15. For a short fracture the same formula works, with $r_w$ replaced by $r_w'$ and the skin equal to zero.

In other unit systems (1)

Metric oilfield units

J\,[\text{m}^3/(\text{d}\cdot\text{MPa})] = \frac{0.536\,k\,h}{\mu\,B}\,J_D

$k$ in mD, $h$ in m, $\mu$ in mPa·s, $B$ in m³/m³; the rate is at standard conditions.

$0.536 = 2\pi\cdot0.0853$ is the constant of the Dupuit equation in these units.

The longer the fracture, the larger $J_D$ , but the growth has a limit. A fracture of infinite conductivity across the whole width of a square ( $I_x$ = 1) turns the inflow into linear flow. The fluid from each half of the area goes to the plane of the fracture along parallel lines. A well with one fracture cannot work better than this in that area:

Upper limit of the dimensionless productivity index

J_D \le J_{D,\max} = \frac{6}{\pi}\cdot\frac{x_e}{y_e}

where:

$J_{D,\max}$ dimensionless productivity index for linear inflow to a fracture of infinite conductivity along the whole length of the area, —
$x_e$ side of the area along which the fracture runs, m
$y_e$ the other side of the area (across the fracture), m

Units: dimensionless

Holds when

The area is closed and rectangular, the fracture runs through the centre parallel to the side $x_e$ , and the regime is pseudo-steady-state.
There is one vertical fracture over the whole thickness of the reservoir. The derivation is in the advanced section.

For a square the limit is $6/\pi \approx 1.91$ , about 13 times that of an undamaged vertical well. In an area elongated along the fracture the limit is higher, because the fluid has a shorter way to the fracture. In an area elongated across it the limit is lower, because the way is longer.

CalculatedCalculated dimensionless productivity index of a fracture in the centre of a closed square area against the penetration ratio, for three values of

F_{CD}

(the lesson's model, pseudo-steady-state regime). The dashed line is the estimate through the effective wellbore radius for a fracture of infinite conductivity. The horizontal line is the limit

6/\pi

Worked exampleThree fracture half-lengths in a tight reservoir on a 400 × 400 m well pattern

Given (assumed data)

Reservoir: $k$ = 0.2 mD, $h$ = 12 m, $\mu$ = 1.0 mPa·s, $B$ = 1.2 m³/m³
Drainage area: a 400 × 400 m square, the well in the centre, $r_w$ = 0.108 m
Before fracturing: skin +1
Fractures: $x_f$ = 40, 100 and 200 m; all have $k_f w$ = 200 mD·m

Solution

Before fracturing: $J_D = \dfrac{1}{\ln(400/0.108) - 1.311 + 1} = \dfrac{1}{8.217 - 1.311 + 1} = 0.1265$ ; $J = \dfrac{0.536\cdot0.2\cdot12}{1.0\cdot1.2}\,J_D = 1.072\cdot J_D = 0.136$ m³/(d·MPa).
The penetration ratios $I_x = 2x_f/400$ are 0.2, 0.5 and 1.0. The dimensionless conductivities $F_{CD} = 200/(0.2\,x_f)$ are 25, 10 and 5.
$J_D$ from the closed-square model: 0.569, 0.999 and 1.397. The productivity $J$ is 0.61, 1.07 and 1.50 m³/(d·MPa), and the increase factors are 4.5, 7.9 and 11.0.
The estimate through the effective wellbore radius ( $r_w'$ from the model at the same $F_{CD}$ , the formula for a vertical well in a square): $J_D$ = 0.574, 1.103 and 2.966. It overstates the result by 1, 10 and 112%.
The limit for this area: $J_{D,\max} = 6/\pi = 1.91$ , which means $J = 2.05$ m³/(d·MPa).

Answer: Fractures of 40, 100 and 200 m increase the productivity by factors of 4.5, 7.9 and 11.0. The effective wellbore radius is fine for the first, overstates the second by 10%, and overstates the third by more than a factor of two.

What it means. Doubling the length from 100 to 200 m at the same conductivity increased the productivity by only a factor of 1.40. The fracture runs into the boundaries of the area, and its $F_{CD}$ falls. We assess a long fracture in a closely spaced pattern only with the closed-area model. Choosing the length and conductivity for a given amount of proppant is a design problem (Fracture design: size and conductivity). The data in this example are assumed for illustration.

Flow regimes after the well starts

The formulas with the effective wellbore radius and with $J_D$ describe stabilised inflow. Right after start-up the picture is different. The pressure disturbance is only starting to spread from the fracture, and the geometry of the flow changes several times. On the diagnostic plot (a log–log graph of the pressure change and its derivative against time; Diagnosis with the pressure derivative), each regime is recognised by its slope.

SchematicFlow regimes towards a well with a hydraulic fracture (plan view, schematic, not to scale). Bilinear: flow along the fracture and towards it from the part of the reservoir near the wellbore. Linear: parallel streamlines to the whole plane of the fracture. Pseudo-radial: the flow from far away converges on the fracture as on a well of large radius. Pseudo-steady-state: the whole closed area is drained.

Linear flow in the fracture itself. It lasts seconds to minutes. It is almost always hidden by wellbore storage (the effect of fluid stored in the wellbore itself; Wellbore storage).
Bilinear flow. The pressure in the fracture falls towards the wellbore, and the fluid flows along the fracture. In the reservoir it flows towards the plane of the fracture, but only in the part near the wellbore, because the disturbance has not yet reached the end of the fracture. Two linear flows at the same time make the pressure drop grow as $t^{1/4}$ . This is a slope of ¼ on a log–log graph.
Linear flow. If the conductivity is very high ( $F_{CD}$ above about 300), the pressure along the fracture evens out before the disturbance has gone far from it. The reservoir then gives fluid across the whole plane of the fracture. The pressure drop grows as $\sqrt t$ , a slope of ½. At the usual $F_{CD}$ from 1 to 100 this regime is almost absent.
Transition period: the streamlines bend as they go round the ends of the fracture.
Pseudo-radial flow. The disturbance has moved away from the fracture by several of its lengths, and the flow converges on it as on a well of radius $r_w'$ . The derivative reaches the plateau of radial flow. From the plateau we find $kh$ , and from the pressure we find the skin $S_f$ .
Pseudo-steady-state regime: the disturbance has reached the boundaries of the closed area. The productivity is set by $J_D$ from the previous section.

Bilinear flow to a fracture of finite conductivity

p_{wD} = \frac{\pi}{\Gamma(5/4)\sqrt2}\,\frac{t_{Dxf}^{1/4}}{\sqrt{F_{CD}}} = 2.45\,\frac{t_{Dxf}^{1/4}}{\sqrt{F_{CD}}}

where:

$p_{wD}$ dimensionless bottomhole pressure drop, $2\pi kh\,\Delta p/(qB\mu)$ , —
$t_{Dxf}$ dimensionless time based on the half-length, —
$F_{CD}$ dimensionless conductivity of the fracture, —
$\Gamma(5/4)$ gamma function, 0.9064, —

Units: dimensionless

Holds when

These are early times. For $F_{CD} \ge 3$ the regime lasts until about $t_{Dxf} \approx 0.1/F_{CD}^2$ . For $F_{CD}$ from 1.6 to 3 it lasts longer than this formula gives (at $F_{CD}$ = 1.6, up to $t_{Dxf} \approx 0.7$ ). For $F_{CD} < 1$ it lasts less: only a short stretch of the fracture near the wellbore works, and already at $t_{Dxf}$ of about $F_{CD}^2/400$ the inflow starts to turn into pseudo-radial flow, skipping the linear regime.
The fracture is uniform and is not choked at the wellbore. The effect of the wellbore has ended.

The length of the fracture has dropped out of the formula: in this regime the flow does not yet “know” where the fracture ends. If we write out $p_{wD}$ , $t_{Dxf}$ and $F_{CD}$ in full, we get a pressure drop that depends on the conductivity $k_f w$ , not on $x_f$ . On a graph of the pressure drop against $t^{1/4}$ this is a straight line. Its slope gives $k_f w$ , and a positive intercept on the pressure axis points to choking of the fracture at the wellbore.

In other unit systems (2)

Metric oilfield units

\Delta p\,[\text{MPa}] = 1.117\,\frac{qB\mu}{h\sqrt{k_f w}}\left(\frac{t}{\phi\mu c_t k}\right)^{1/4}

$q$ in m³/d, $\mu$ in mPa·s, $h$ in m, $k_f w$ in mD·m, $t$ in h, $c_t$ in MPa⁻¹, $k$ in mD.

$1.117 = 2.4508\cdot(3.553\cdot10^{-3})^{1/4}/0.5358$ (exact values of the coefficients; $0.5358 = 2\pi\cdot0.08527$ ).

Oilfield units

\Delta p\,[\text{psi}] = 44.1\,\frac{qB\mu}{h\sqrt{k_f w}}\left(\frac{t}{\phi\mu c_t k}\right)^{1/4}

$q$ in STB/d, $\mu$ in cP, $h$ in ft, $k_f w$ in mD·ft, $t$ in h, $c_t$ in psi⁻¹, $k$ in mD.

$44.1 = 141.2\cdot2.45\cdot(2.637\cdot10^{-4})^{1/4}$ .

Linear flow to a fracture of high conductivity

p_{wD} = \sqrt{\pi\,t_{Dxf}}\quad\Longleftrightarrow\quad \Delta p = \frac{qB\mu}{2hx_f}\sqrt{\frac{t}{\pi k\phi\mu c_t}}

where:

$\Delta p$ bottomhole pressure drop after start-up at a constant rate, Pa
$q,\ B,\ \mu$ rate at standard conditions, formation volume factor, viscosity, m³/s, m³/m³, Pa·s
$h,\ x_f$ thickness of the reservoir and half-length of the fracture, m
$k,\ \phi,\ c_t,\ t$ permeability, porosity, total compressibility, time, m², —, Pa⁻¹, s

Units: SI (any consistent system of units)

Holds when

$F_{CD}$ is above about 300. The regime lasts until $t_{Dxf} \approx 0.016$ . At lower conductivity bilinear flow takes its place.
The fracture runs over the whole thickness of the reservoir. If the fracture is less tall than the reservoir, the first part of the record is distorted.

From the slope of the straight line “pressure drop against $\sqrt t$ ” we find the product $x_f\sqrt k$ . If the permeability is known from a test before fracturing, we find the half-length. The skin of the fracture faces adds a constant to the pressure drop, so the line does not pass through the origin (Diagnosis with the pressure derivative).

In other unit systems (2)

Metric oilfield units

\Delta p\,[\text{MPa}] = 0.1972\,\frac{qB\mu}{h\,x_f}\sqrt{\frac{t}{k\phi\mu c_t}}

$q$ in m³/d, $\mu$ in mPa·s, $h$ and $x_f$ in m, $t$ in h, $k$ in mD, $c_t$ in MPa⁻¹.

$0.1972 = \sqrt{\pi\cdot3.553\cdot10^{-3}}/0.5358$ .

Oilfield units

\Delta p\,[\text{psi}] = 4.064\,\frac{qB\mu}{h\,x_f}\sqrt{\frac{t}{k\phi\mu c_t}}

$q$ in STB/d, $\mu$ in cP, $h$ and $x_f$ in ft, $t$ in h, $k$ in mD, $c_t$ in psi⁻¹.

$4.064 = 141.2\cdot\sqrt{\pi\cdot2.637\cdot10^{-4}}$ .

Pseudo-radial flow to a fracture

p_{wD} = \frac12\left(\ln t_{Dxf} + 0.809\right) + \ln\frac{x_f}{r_w'}

where:

$p_{wD},\ t_{Dxf}$ dimensionless pressure drop and dimensionless time based on the half-length, —
$\ln(x_f/r_w')$ value of the universal curve at the given $F_{CD}$ : $\ln 2$ for infinite conductivity, —
$0.809$ $\ln(4/\gamma)$ , the constant of the logarithmic approximation, —

Units: dimensionless; natural logarithms

Holds when

The derivative reaches the plateau 0.5 to within 7–8% at $t_{Dxf} = 3$ and to within 5% at $t_{Dxf} = 5$ , for any $F_{CD}$ from 1 to infinity.
The reservoir is infinite: the disturbance has not yet reached the boundaries of the area.

If we change from $t_{Dxf}$ to the dimensionless time based on the wellbore radius, we get the usual semilog formula (pressure against the logarithm of time) with the skin $S_f$ (Pressure drawdown analysis). So in the pseudo-radial period the fracture is seen as a negative skin, and only there.

Boundaries of the flow regimes (the lesson's model)

Regime: Bilinear
Derivative: slope ¼, 4 times lower than the pressure drop
When: until $t_{Dxf} \approx 0.1/F_{CD}^2$ (for $F_{CD} \ge 3$ )
What it gives: $k_f w$ ; choking

Regime: Linear
Derivative: slope ½, half the pressure drop
When: for $F_{CD} \gtrsim 300$ , until $t_{Dxf} \approx 0.016$
What it gives: $x_f\sqrt k$ ; face skin

Regime: Pseudo-radial
Derivative: plateau 0.5
When: from $t_{Dxf} \approx 3$ –5
What it gives: $kh$ , $S_f$

Regime: Boundary of the area
Derivative: the derivative rises
When: in a square, from $t_{DA} \approx 0.03$ –0.05
What it gives: area, $J_D$

Regime	Derivative	When	What it gives
Bilinear	slope ¼, 4 times lower than the pressure drop	until $t_{Dxf} \approx 0.1/F_{CD}^2$ (for $F_{CD} \ge 3$ )	$k_f w$ ; choking
Linear	slope ½, half the pressure drop	for $F_{CD} \gtrsim 300$ , until $t_{Dxf} \approx 0.016$	$x_f\sqrt k$ ; face skin
Pseudo-radial	plateau 0.5	from $t_{Dxf} \approx 3$ –5	$kh$ , $S_f$
Boundary of the area	the derivative rises	in a square, from $t_{DA} \approx 0.03$ –0.05	area, $J_D$

$t_{DA} = kt/(\phi\mu c_tA)$ is the dimensionless time based on the area (Pseudo-steady-state flow and the shape factor). In a square $t_{DA} = t_{Dxf}\,I_x^2/4$ . The boundaries show up on the derivative noticeably (by a few per cent) earlier than for a vertical well, because the ends of the fracture are closer to the boundaries.

CalculatedCalculated dimensionless pressure derivatives of a well with a fracture in an infinite reservoir, for three values of

F_{CD}

(the lesson's model, without wellbore storage and skin). On the left is the bilinear part with a slope of ¼. For a fracture of very high conductivity it is the linear part with a slope of ½. On the right all the curves reach the plateau 0.5 of pseudo-radial flow.

Worked exampleHow long the regimes last in a tight reservoir

Given (assumed data)

Reservoir and area: from the previous example: $k$ = 0.2 mD, $\phi$ = 0.12, $\mu$ = 1.0 mPa·s, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹, a 400 × 400 m square
Fractures: $x_f$ = 40 m ( $F_{CD}$ = 25) and $x_f$ = 100 m ( $F_{CD}$ = 10)

Solution

Hydraulic diffusivity: $\eta = 3.553\cdot10^{-3}\cdot\dfrac{0.2}{0.12\cdot1.0\cdot1.5\cdot10^{-3}} = 3.95$ m²/h. In one hour $t_{Dxf}$ grows by $\eta/x_f^2$ : 0.002467 and 0.0003948.
The end of bilinear flow, $t_{Dxf} = 0.1/F_{CD}^2$ , is 0.00016 and 0.001. That is after 0.06 and 2.5 h, usually still within the effect of the wellbore.
The start of pseudo-radial flow, $t_{Dxf} \approx 3$ , comes after 51 and 317 d.
The dimensionless time based on the area at that moment: $t_{DA} = 3\,I_x^2/4$ = 0.03 and 0.19. The boundary of the area becomes visible on the derivative at $t_{DA} \approx 0.03$ –0.05, which means after $t_{DA}A/(24\eta)$ = 51–84 d ( $A$ = 160,000 m²).

Answer: For the 40 m fracture the derivative barely has time to approach the plateau after 51 d, and almost at once the boundary starts to show. For the 100 m fracture there is no pseudo-radial regime: the boundary arrives earlier.

What it means. At the start of the pseudo-radial regime, $t_{DA} = 0.75\,I_x^2$ . From this we get a rule: in a square area this regime has time to form only if $I_x$ is smaller than about 0.2–0.25. In tight reservoirs with long fractures we cannot find $kh$ from the derivative plateau. We determine it before fracturing or by matching with calculated curves, and we forecast the inflow with a model that includes both the fracture and the boundaries. The data in this example are assumed for illustration.

Why the first rates overstate the effect

Right after start-up the whole drawdown falls on a thin layer of the reservoir along the huge surface of the fracture. The pressure gradients at its faces are large, and the well produces a lot. As the disturbance spreads, the same drawdown is spread over a larger and larger volume. The gradients fall, and the rate declines towards the stabilised value. A well before fracturing has usually been producing at a stabilised rate for a long time. If we compare the first weeks after the treatment with the stabilised rate before it, the gain will be overstated.

The transient productivity index shows this in numbers. It is the ratio of the rate to the difference between the average reservoir pressure and the bottomhole pressure at time $t$ . In dimensionless form, at a constant rate the average pressure in a closed area falls by $2\pi t_{DA}$ (in units of $p_D$ ; Reservoir limit test), so:

Transient productivity index and apparent gain factor

J_D(t) = \frac{1}{p_{wD}(t) - 2\pi t_{DA}},\qquad n_{\text{app}}(t) = \frac{J_D(t)}{J_{D,\text{pre}}}

where:

$J_D(t)$ dimensionless productivity index of the fractured well at time $t$ (constant rate), —
$p_{wD}(t)$ dimensionless bottomhole pressure drop since the start of production, —
$t_{DA}$ dimensionless time based on the area; $2\pi t_{DA}$ is the average fall of pressure in the area, —
$n_{\text{app}}(t)$ apparent factor of productivity increase at time $t$ , —
$J_{D,\text{pre}}$ stabilised $J_D$ of the same well before hydraulic fracturing, —

Units: dimensionless

Holds when

The area is closed and homogeneous, and the inflow is single-phase. Before the treatment the well worked in the pseudo-steady-state regime.
At constant bottomhole pressure the rate changes with time in a similar way, but not exactly the same. This formula is a measure, not a forecast of the rate day by day.

At large times $p_{wD} - 2\pi t_{DA}$ tends to $1/J_D$ of the pseudo-steady-state regime, and the apparent factor tends to the true one. While the regime is transient, the denominator is smaller, and the factor is larger.

CalculatedApparent factor of productivity increase after hydraulic fracturing against production time (the lesson's model): the reservoir and area from the examples, fractures of 40 and 100 m. The dashed horizontal lines are the stabilised values, which the curves approach from above.

Worked exampleWhat the first weeks after hydraulic fracturing will show

Given (assumed data)

Well: from the examples above: $k$ = 0.2 mD, a 400 × 400 m square, skin before the treatment +1
Fracture: $x_f$ = 100 m, $F_{CD}$ = 10
Comparison: the productivity after the treatment against the stabilised productivity before it

Solution

The stabilised gain factor, from the example of the three half-lengths: 7.90.
After 7 d: $t_{Dxf}$ = 0.066, the apparent gain factor is 16.0, which is 2.0 times the stabilised one.
After 30 d: $t_{Dxf}$ = 0.28, the gain factor is 10.25 (1.3 times larger).
After 90 and 365 d: 8.3 and 7.9.

Answer: The first week would show a gain by a factor of 16.0, the first month by 10.25, and the long-term gain is 7.90.

What it means. Anyone who judges the treatment by the first month overrates it by 30% and will be disappointed half a year later, even if the fracture has kept its conductivity. In reality the rate also falls because the reservoir pressure and the fracture conductivity fall. We compare the results of treatments by the stabilised productivity, or by a model that includes the transient regime. The data in this example are assumed for illustration.

Choking of the fracture and damage to its faces

A fracture can lose part of its gain in two ways. Choking (a choked fracture) is a stretch of length $x_s$ right next to the wellbore, where the conductivity $k_{f,ch}w$ is much lower than designed. The proppant may have been pushed too far away from the wellbore, or carried out of the fracture when the well was first brought on production. The grains near the wellbore may have crushed. Or the fracture may be joined to the wellbore through a narrow, winding channel. Fracture-face damage is a thin layer of the reservoir, of thickness $b_s$ , along the surface of the fracture. Its permeability $k_s$ has been reduced by the filtrate (the liquid that leaks out of the fracturing fluid into the rock) and by the polymer of the fracturing fluid.

Choke skin of a fracture

s_{ch} \approx \pi k\,x_s\left(\frac{1}{k_{f,ch}w} - \frac{1}{k_f w}\right) \approx \frac{\pi k\,x_s}{k_{f,ch}w}

where:

$s_{ch}$ additional skin from choking, —
$k$ permeability of the reservoir, mD
$x_s$ length of the choked stretch from the wellbore, m
$k_{f,ch}w,\ k_f w$ conductivity of the choked stretch and of the rest of the fracture, mD·m

Units: dimensionless; permeabilities and conductivities in consistent units (mD and mD·m)

Holds when

The whole flow of a wing passes along the fracture through the choked stretch, so each wing carries half of the rate.
This is an upper estimate. If $k_{f,ch}w$ is not much larger than $k\,x_s$ , part of the fluid bypasses the choked stretch through the reservoir, and the actual skin is smaller (in the example below, by about 40%).

Derivation: by Darcy's law, the pressure drop across the choked stretch is $\Delta p = (q/2)\,\mu x_s/(k_{f,ch}w\,h)$ (at reservoir conditions). Dividing it by $qB\mu/(2\pi kh)$ gives the skin. A short stretch of low conductivity is “expensive”, because the whole flow goes through it. A stretch of 0.3 m with a conductivity of 150 mD·m in a 300 mD reservoir gives about +1.8.

Skin of the fracture faces

s_{fs} = \frac{\pi b_s}{2x_f}\left(\frac{k}{k_s} - 1\right)

where:

$s_{fs}$ skin of the fracture faces, —
$b_s$ thickness of the damaged layer from the face of the fracture into the reservoir, m
$k_s$ permeability of the damaged layer, mD
$x_f,\ k$ half-length of the fracture and permeability of the reservoir, m, mD

Units: dimensionless

Holds when

The layer is thin, $b_s \ll x_f$ , and the same along the whole fracture. The flow through it is perpendicular to the face.
In the linear regime $s_{fs}$ is added to $p_{wD}$ exactly. In the pseudo-radial regime the total skin grows by about $s_{fs}$ , and never by less. The model gives 5–20% more at $s_{fs}$ from 0.5 to 1, and relatively even more at small $s_{fs}$ , because the inflow is redistributed along the fracture.

This is the Hawkins formula for the skin of a damaged zone (Skin factor), applied to the plane of the fracture. The same flow passes through the area of four faces (two sides of each wing), $4x_fh$ , instead of $2\pi r_wh$ . The area is huge, so even strong damage of a long fracture gives a small skin. The skin becomes noticeable for short fractures in permeable reservoirs.

Worked exampleA short, wide fracture in a permeable reservoir: what choking and face damage took away

Given (assumed data)

Reservoir: $k$ = 300 mD, $r_w$ = 0.108 m
Fracture as designed: $x_f$ = 6 m, $k_f w$ = 3000 mD·m
Choking: at the wellbore, $x_s$ = 0.3 m with $k_{f,ch}w$ = 150 mD·m (the proppant was pushed away from the wellbore)
Faces: damaged layer $b_s$ = 0.05 m, $k/k_s$ = 30

Solution

As designed: $F_{CD} = \dfrac{3000}{300\cdot6} = 1.67$ . By the model $\ln(x_f/r_w') = 1.351$ , and $S_f = 1.351 - \ln\dfrac{6}{0.108} = -2.67$ .
Choking: $s_{ch} \approx \pi\cdot300\cdot0.3\left(\dfrac{1}{150} - \dfrac{1}{3000}\right) = 1.79$ .
Faces: $s_{fs} = \dfrac{\pi\cdot0.05}{2\cdot6}\,(30 - 1) = 0.38$ .
The sum by the formulas: $-2.67 + 1.79 + 0.38 = -0.50$ . The model with both defects gives $S = -0.67$ .
The productivity relative to the design, at $X = \ln(r_e/r_w) - 0.75$ = 7.179 ( $r_e$ = 300 m): $\dfrac{X + S_f}{X + S} = \dfrac{7.179 - 2.67}{7.179 - 0.67} = 0.69$ .

Answer: Instead of the designed skin −2.67 the well has about −0.67, and it gives 69% of the designed rate.

What it means. The main damage was done by the 30 cm at the wellbore, not by the damage on the whole surface of the faces. The choke formula overstates the effect here. The model gives an addition of 1.06 instead of 1.79, because part of the fluid bypasses the choked stretch through the reservoir. But the conclusion is the same. For short fractures in permeable reservoirs, the main thing is a reliable connection of the fracture to the wellbore, and the proppant must not be pushed away from the wellbore (Fracture design: size and conductivity). The data in this example are assumed for illustration.

How to assess inflow to a well with a fracture

1
Dimensionless conductivity
Use the conductivity of the fracture at reservoir conditions, not the proppant data sheet. For gas wells, include the inertial losses in the pack.
2
A short or a long fracture
If $I_x$ is below 0.2–0.3, take the effective wellbore radius from the universal curve (at $F_{CD} < 0.1$ , from the asymptote $0.28\,k_f w/k$ ). If it is larger, take $J_D$ from the closed-area model, not above $6/\pi$ for a square.
3
Time scale
Find $t_{Dxf}$ per day and $t_{DA}$ : how long the transient inflow will last, and whether there will be a pseudo-radial regime before the boundaries arrive.
4
Fracture defects
Take the choke skin and the face skin as upper estimates. For short fractures in permeable reservoirs they decide the outcome.
5
Forecast
Transient inflow in the first months, then stabilised inflow. Check that the tubing and the lift equipment can carry the higher rate (Nodal analysis of the production system).
6
Check after the treatment
Diagnosis from the pressure derivative: slopes ¼ and ½, conductivity and half-length from the early stretches, and a comparison of the stabilised productivity with the calculation.

Terms and symbols

Dimensionless time based on the half-length $t_{Dxf}$: It equals $kt/(\phi\mu c_t x_f^2)$ . At $t_{Dxf} \approx 1$ the pressure disturbance has moved away from the fracture by roughly its half-length.
Bilinear flow: The early regime of a fracture of finite conductivity: linear flow in the fracture and towards it at the same time. The pressure drop grows as $t^{1/4}$ .
Linear flow: Flow along parallel lines to the plane of a fracture of high conductivity. The pressure drop grows as $\sqrt t$ .
Pseudo-radial flow: The late regime in which the flow from far away converges on the fracture as on a well of radius $r_w'$ .
Penetration ratio $I_x$: It equals $2x_f/x_e$ : the share of the side of the drainage area that the fracture covers.
Dimensionless productivity index $J_D$: It equals $qB\mu/(2\pi kh(\bar p - p_{wf}))$ . It lets us compare wells and fractures in different reservoirs.
Choking of the fracture: A stretch of low conductivity right next to the wellbore, through which the whole flow of a wing passes. It gives the skin $s_{ch}$ .
Fracture-face skin $s_{fs}$: The extra resistance of a thin damaged layer of the reservoir along the surface of the fracture.

Common mistakes and misconceptions

Wrong

“The effective wellbore radius $x_f/2$ is fine for any fracture of high conductivity.”

In fact

Only for a fracture that is small compared with the drainage area. In the lesson's example, at $I_x$ = 0.5 the estimate through the effective wellbore radius overstates the productivity by 10%, and at $I_x$ = 1 by more than a factor of two. For a square, $J_D$ cannot exceed $6/\pi$ .

Wrong

“If we double the length of the fracture, the effective wellbore radius doubles.”

In fact

This is true only at high conductivity ( $F_{CD}$ from 30). At $F_{CD}$ below 0.1 the effective wellbore radius is about $0.28\,k_f w/k$ and does not depend on the length. Doubling the length at the same conductivity also halves $F_{CD}$ .

Wrong

“The rate of the first month after hydraulic fracturing is its result.”

In fact

In a tight reservoir the first month falls in the transient regime, and the productivity at that time is higher than the stabilised one. In the lesson's example, the apparent gain factor over the first month was almost a third larger than the true one.

Wrong

“If the test is long, the derivative plateau will certainly appear.”

In fact

For a long fracture in a tight reservoir, the boundary of the area makes itself felt before pseudo-radial flow begins: in a square, when $I_x$ is more than about 0.2–0.25.

Wrong

“Damage to the fracture faces is the most dangerous.”

In fact

It is spread over a huge area, and for long fractures it gives a small skin. Choking at the wellbore is more dangerous: the whole flow passes through a short stretch.

In depth: advanced detail

Where the limit 6/π comes from

Let a fracture of infinite conductivity divide the closed area $x_e \times y_e$ in half along the side $x_e$ . Each half is a block of reservoir that is $L = y_e/2$ long in the direction of flow, with an impermeable outer face. The fluid flows to the fracture linearly, through the area $x_e h$ on one side. In the pseudo-steady-state regime, only the fluid released beyond a cross-section at a distance $y$ from the fracture passes through that cross-section. That is the share $1 - y/L$ of the flow $q/2$ .

Integrating Darcy's law, we get the pressure difference between the point $y$ and the fracture, $\dfrac{q\mu}{2kx_eh}\left(y - \dfrac{y^2}{2L}\right)$ (at reservoir conditions, $B$ = 1). The average value of this difference over the layer is $\dfrac{q\mu L}{6kx_eh}$ . From this, $\bar p - p_{wf} = \dfrac{q\mu\,y_e}{12\,kx_eh}$ and $J_D = \dfrac{q\mu}{2\pi kh(\bar p - p_{wf})} = \dfrac{6}{\pi}\cdot\dfrac{x_e}{y_e}$ .

The same logic explains the terms $-1/2$ and $-3/4$ in the radial formula (Pseudo-steady-state flow and the shape factor). The difference is only in the geometry. A fracture of finite conductivity that does not reach the boundaries always gives a smaller $J_D$ . The lesson's model reproduces the limit $6/\pi$ at $I_x$ = 1 and infinite conductivity to the fifth digit.

How the lesson's curves were calculated

A wing of the fracture is divided into 16–60 segments (up to 240 at very low conductivity), with shorter segments near the wellbore and near the tip of the fracture. On each segment the inflow from the reservoir is taken as uniform but unknown. The pressure in the reservoir at the middle of each segment is expressed through all the inflows with the influence function of a line source. The pressure drop along the fracture is expressed through Darcy's law for the flow in it. The condition “the pressure in the fracture equals the pressure in the reservoir at its face” at the middle of the segments, together with the given total rate, gives a system of linear equations.

Infinite reservoir, transient inflow: the influence function is the integral of the modified Bessel function $K_0$ in the Laplace space. We return to real time by numerical inversion with the Stehfest method (Stehfest algorithm), as in the lessons on transient flow.
The pseudo-radial constant $\ln(x_f/r_w')$ : the limit of the same problem at large times, where $K_0$ is replaced by a logarithm. At infinite conductivity the solution gives exactly $\ln 2$ .
Closed square: the influence function is written as a series of cosines across the area. The slowly converging part of the series is summed exactly through the Clausen function. The model was checked against a finite-difference solution and against the limit $6/\pi$ .

Checks of the model: the bilinear asymptote $2.4508\,t^{1/4}/\sqrt{F_{CD}}$ and the linear one $\sqrt{\pi t}$ are reproduced at early times to within tenths of a per cent. As $F_{CD}$ decreases, the effective wellbore radius converges to $0.2807\,k_f w/k$ . The transient solution for the square coincides at the start with the solution for an infinite reservoir, and at the end it reaches the pseudo-steady-state $J_D$ . The error of the calculations themselves is tenths of a per cent, much less than the uncertainty of the real parameters of a fracture.

Fracture half-length from the linear-flow part of the record

If the conductivity of the fracture is very high, the early part of the pressure record after start-up or shut-in is linear flow, and we find the half-length from it. For this we plot the pressure drop against $\sqrt t$ and check the slope ½ on the diagnostic plot. Then, from the slope $m_l$ of the straight line, we calculate $x_f$ with the metric form of the linear-flow formula.

Worked exampleHalf-length from the slope of the line “pressure drop against the square root of time”

Given (assumed data)

Production conditions: $q$ = 15 m³/d, $B$ = 1.2, $\mu$ = 1.0 mPa·s, $h$ = 12 m
Reservoir: $k$ = 0.05 mD (from the test before fracturing), $\phi$ = 0.12, $c_t = 1.5\cdot10^{-3}$ MPa⁻¹
Pressure record: slope ½ on the derivative from 6 to 60 h; the slope of the line $m_l$ = 1.23 MPa per $\sqrt{\text{h}}$ , and the line passes through zero

Solution

From the linear-flow formula: $x_f = \dfrac{0.1972\,qB\mu}{h\,m_l\sqrt{k\phi\mu c_t}}$ .
$\sqrt{0.05\cdot0.12\cdot1.0\cdot1.5\cdot10^{-3}} = 0.0030$ .
$x_f = \dfrac{0.1972\cdot15\cdot1.2\cdot1.0}{12\cdot1.23\cdot0.0030} = 80$ m.
Check of the regime: linear flow ends at $t_{Dxf} \approx 0.016$ , which means at $t = \dfrac{0.016\,\phi\mu c_t x_f^2}{3.553\cdot10^{-3}\,k} = 104$ h. The interval 6–60 h lies inside the regime.

Answer: The half-length of the fracture is about 80 m. The skin of the faces is close to zero, since the line passes through the origin.

What it means. The result depends directly on the permeability. What we found is the product $x_f\sqrt k$ , so an error of a factor of two in $k$ gives an error of a factor of $\sqrt2$ in $x_f$ . If the conductivity is not very high, the early part is bilinear or transitional, and the linear-flow formula will understate $x_f$ . Then we use matching with calculated curves. The data in this example are assumed for illustration.

In the same way, from the slope of the line “pressure drop against $t^{1/4}$ ” on the bilinear part, we find the conductivity: $k_f w = \left(\dfrac{1.117\,qB\mu}{h\,m_b}\right)^2\dfrac{1}{\sqrt{\phi\mu c_t k}}$ , where $m_b$ is the slope in MPa per $\text{h}^{1/4}$ . Both estimates need the fracture to cover the whole thickness of the reservoir, and the effect of the wellbore to have ended by that time.

Connections to other topics

Hydraulic fracturing: purpose and well selection — dimensionless conductivity, equivalent skin and the gain factor (the folds of increase) are the basis of this lesson.
Skin factor — the effective wellbore radius and the Hawkins formula on which the skin of the fracture faces is built.
Pseudo-steady-state flow and the shape factor — the shape factor and stabilised inflow in a closed area.
Diagnosis with the pressure derivative — the diagnostic plot on which bilinear, linear and pseudo-radial flow are recognised.
Fracture design: size and conductivity — how to choose the length and conductivity of a fracture for a given amount of proppant.
Inflow at high rates: non-Darcy flow — inertial losses in the proppant pack reduce the effective conductivity of a fracture in a gas well.
Nodal analysis of the production system — whether the tubing and lift equipment can carry the rate of the well after hydraulic fracturing.
Effect of completion on inflow — the resistance of the perforations and of the gravel pack, for wells where the fracture is connected to the wellbore through perforations.

Check yourself

1
A fracture with a half-length of 80 m and a conductivity of 100 mD·m has been made in a 25 mD reservoir. What will doubling its length give, at the same conductivity?
1. A.The productivity will double
2. B.The effective wellbore radius will double, and the skin will fall by 0.69
3. C.Almost nothing
4. D.The effective wellbore radius will halve
Show the answer
Correct answer: C. Almost nothing
$F_{CD} = 100/(25\cdot80) = 0.05$ , which is the low-conductivity branch. The effective wellbore radius is $\approx 0.28\cdot100/25 \approx 1.1$ m and does not depend on the length: only the stretch next to the wellbore works. Raising the conductivity would help.
2
A well sits in the centre of a square area of 300 × 300 m. The fracture has infinite conductivity and a half-length of 150 m. An engineer estimated the productivity through $r_w' = x_f/2$ . How big was the error?
Show the answer
$I_x = 2\cdot150/300 = 1$ : the fracture runs across the whole width, and $J_D = 6/\pi \approx 1.91$ . The estimate through the effective wellbore radius gives $r_w' = 75$ m, $\ln(300/75) = 1.386$ , and $J_D = 1/(1.386 - 1.311) \approx 13$ . This overstates the productivity almost 7 times. The effective wellbore radius describes only fractures that are small compared with the area.
3
On the diagnostic plot after hydraulic fracturing, in the first hours the derivative runs parallel to the pressure drop and 4 times below it. What regime is this, and what can we find from it?
1. A.Pseudo-radial flow; the permeability of the reservoir
2. B.Bilinear flow; the conductivity of the fracture
3. C.Wellbore storage; the wellbore storage coefficient
4. D.Linear flow; the half-length of the fracture
Show the answer
Correct answer: B. Bilinear flow; the conductivity of the fracture
If the pressure drop grows as $t^n$ , the derivative equals $n\,\Delta p$ . A ratio of ¼ means $n$ = ¼, which is bilinear flow to a fracture of finite conductivity. From the slope of the line “pressure drop against $t^{1/4}$ ” we find $k_f w$ , and an intercept on the pressure axis points to choking. For linear flow the ratio would be ½, and for wellbore storage it would be 1.
4
Why might a test of a well with a fracture of 150 m half-length in a tight reservoir, with 400 m between wells, never show a derivative plateau?
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Pseudo-radial flow starts at $t_{Dxf} \approx 3$ , and by that time $t_{DA} = 3\,I_x^2/4$ . Here $I_x = 0.75$ and $t_{DA} \approx 0.42$ . The boundary of the area made itself felt long ago (it is noticeable already from $t_{DA}$ of about 0.03–0.05). The flow passes from the bilinear or transition regime straight into depletion of the area. We find $kh$ before fracturing or by matching with calculated curves.
5
In the first month after hydraulic fracturing, a well produced 14 m³/d. Before the treatment it produced 1.5 m³/d, stabilised, at the same drawdown. Can we say that the productivity rose 9 times?
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No. The first month in a tight reservoir is the transient regime, and the productivity at that time is higher than the stabilised one. Before the treatment the well was already working in the stabilised regime. The true factor comes from stabilised data or from a calculation with a model of transient inflow to a fracture. In the lesson's example, under similar conditions, the first month overstated the gain by almost a third. The rate will decline even if the fracture keeps its conductivity.
6
A short fracture in a 400 mD reservoir has a stretch of 0.2 m next to the wellbore with a conductivity of 100 mD·m. About how large is the choke skin?
1. A.2.5
2. B.0.03
3. C.25
4. D.0.25
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Correct answer: A. 2.5
$s_{ch} \approx \pi k x_s/(k_{f,ch}w) = \pi\cdot400\cdot0.2/100 \approx 2.5$ . This is an upper estimate, because part of the flow will bypass the stretch through the reservoir. For a short fracture with a designed skin of about −3, this is the loss of most of the gain.

Updated: 28 October 2026