Fracture design: size and conductivity

Quick refresher

The proppant that has reached the reservoir is the budget of the fracture, which is divided between length and width. Its dimensionless measure is the proppant number $N_{prop}$ : twice the product of the permeability of the pack and its volume in the reservoir, divided by the product of the permeability of the reservoir and the volume of the drainage area.
For $N_{prop}$ up to 0.1 the optimal dimensionless conductivity is about 1.6–1.7, and $1/J_{D,\max} \approx 0.99 - \tfrac12\ln N_{prop}$ : doubling the proppant adds only 10–20% to the productivity. At a larger proppant number the optimum shifts to a higher conductivity, and the fracture runs into the boundaries of the drainage area.
The calculation uses the conductivity in the reservoir, not the data-sheet value. The laboratory conductivity is taken at the greatest load $p_c - p_{wf,\min}$ and is reduced for polymer residue and embedment of the grains. For gas wells it is reduced further for inertial losses in the pack.
The width of a fracture depends on the viscosity, the rate and the modulus of the rock only to the power ¼: by the PKN model $w_w \approx 2.75\,(\mu q_i x_f/E')^{1/4}$ . We can hardly change the width with the fluid, while the modulus from soft rocks to stiff ones changes tens of times, and the width by 2–3 times. The net pressure grows as the fracture gets longer.
Leak-off sets the fluid efficiency $\eta$ : the share of the pumped volume that stays in the fracture. We calibrate it, and the leak-off coefficient, from the time the fracture takes to close after a calibration injection.
The pad is about $(1 - \eta)/(1 + \eta)$ of the pumped volume. We raise the proppant concentration by the law $\tau^{\varepsilon}$ with the same exponent, so that by the end of pumping it is even along the whole fracture.

Key relationships

N_{prop} = I_x^2 F_{CD} = \dfrac{2k_f V_{prop}}{k\,V_{res}}

the proppant number; $V_{prop}$ is the volume of the pack of both wings inside the reservoir

\dfrac{1}{J_{D,\max}} \approx 0.99 - \dfrac12\ln N_{prop},\quad F_{CD,opt} \approx 1.6\text{–}1.7

small proppant number, up to 0.1

\sigma_{\text{eff}} = p_c - p_{wf}

load on the proppant at the wellbore

w_w \approx 2.75\left(\dfrac{\mu q_i x_f}{E'}\right)^{1/4},\quad p_{net} = \dfrac{E' w_w}{2h_f}

the PKN model, consistent SI units

q_i t = A_f\bar w + 2A_l\kappa C_L\sqrt t,\quad \eta = \dfrac{A_f\bar w}{q_i t}

volume balance with leak-off, and the fluid efficiency

\varepsilon = \dfrac{1 - \eta}{1 + \eta},\quad c = c_e\,\tau^{\varepsilon}

pad fraction and the law of growth of the proppant concentration

Common pitfalls

Calculating the conductivity from the proppant data sheet at the load of the first months. When the bottomhole pressure is lowered, the load grows, the sand starts to crush, and the conductivity falls several times.
Counting all of the proppant as being in the reservoir when part of it has gone into the shales above and below. Only the part that is in the reservoir contributes to the productivity.
Sticking to the rule “ $F_{CD}$ = 1.6” at a large proppant number. There the optimum is higher, and the length of the fracture is limited by the size of the drainage area.
Raising the viscosity to get more width. The width grows only as the fourth root, the net pressure grows with it, and the fracture leaves the reservoir in height.
Taking a pad “with a margin”. Extra fluid lengthens the unpropped part of the fracture and increases leak-off and damage to the reservoir.

After this lesson you can

build the design chain: from the amount of proppant and the properties of the reservoir to the optimal half-length and width, then to the fluid volume, the pad (the first part of the fluid, pumped without proppant) and the schedule of the proppant concentration
calculate the conductivity of the proppant pack in the reservoir, taking into account the load, embedment of the grains (grains pressing into the fracture faces), polymer residue and inertial losses, and choose the proppant for the greatest expected load
calculate the proppant number (it compares the amount of proppant with the size of the drainage area), and from it find the optimal dimensionless conductivity, the half-length, the width and the greatest dimensionless productivity index
compare the PKN, KGD and radial geometry models (three simple ways to calculate the shape of a fracture), and estimate the width of the fracture and the net pressure (the pressure in the fracture above the closure pressure) during injection with the PKN model
make a volume balance with leak-off, find the fluid efficiency (the share of the pumped fluid that stays in the fracture), also from a calibration injection, and from it calculate the pad and how the proppant concentration should increase

The design chain

The lesson Hydraulic fracturing: purpose and well selection showed how a fracture of given length and conductivity changes the productivity of a well. The lesson Inflow to a hydraulically fractured well showed how to calculate the inflow to it in a bounded area and over time. The lesson Rock mechanics for hydraulic fracturing showed how the stresses in the rock set the closure pressure (the pressure at which the walls of the fracture come together), the width and the height of the fracture. Here this knowledge is joined in a design: what fracture to create, and how to do it.

We build the design from the reservoir to the injection, and not the other way round. First we decide what fracture is needed. A given amount of proppant can be laid in different ways, and there is one combination of length and width at which the productivity is greatest. Then we check whether such a fracture can be made. Is the width enough to let the proppant in? Will the fracture grow into the neighbouring layers? How much fluid will leak off into the reservoir? If the needed fracture cannot be made, we change the proppant, the fluid, the rate or the technology, and calculate again.

The order of the design calculation

1
Input data
Permeability and thickness of the reservoir, size of the drainage area (the well pattern), closure pressure and stress contrast with the barriers (how much higher the stress is in the layers above and below), modulus of the rock, leak-off coefficient from a calibration injection (lesson on rock mechanics), and the lowest planned bottomhole pressure.
2
Proppant
The greatest load on the proppant, the choice of type and size, and the conductivity of the pack in the reservoir.
3
Fracture dimensions
The proppant number for the chosen amount of proppant, the optimal dimensionless conductivity, the half-length and the width.
4
Geometry during injection
A fracture growth model, the width at the wellbore and the net pressure. A check that the fracture stays within the reservoir in height and that the width will let the grains through.
5
Volumes
Volume balance with leak-off: the injection time, the volume of slurry (the fluid with proppant in it) and the fluid efficiency.
6
Pad and injection schedule
The pad fraction, how the proppant concentration grows, the stages and the volume of the flush (the fluid that pushes the last proppant out of the well).
7
Size of the treatment
The same calculation for several amounts of proppant, and the choice by net present value, NPV (Net present value (NPV)).
8
Check
The pressure record of the main injection, a well test and the production after hydraulic fracturing. The input data are refined for the next wells.

Proppant and conductivity at reservoir conditions

After the fracture closes, its walls press on the proppant with a force that is set by the minimum stress in the rock. The proppant must carry this load without crushing, and it must keep the permeability of the pack (the packed layer of grains in the fracture). Three main types are used. Quartz sand is cheap, but under large loads the grains split, and the fragments plug the pores. Resin-coated sand has a resin coating that spreads the load and holds the fragments. A coating that cures in the reservoir also glues the grains together and stops them from being carried into the well. Ceramic is granules sintered (baked at high temperature) from kaolin or bauxite. It is strong and round, but more expensive.

Main types of proppant

Proppant: Quartz sand
Grain density, g/cm³: 2.65
Load at which it is usually used: up to 30–40 MPa
Price: low

Proppant: Resin-coated sand
Grain density, g/cm³: 2.5–2.6
Load at which it is usually used: up to 40–55 MPa
Price: medium

Proppant: Lightweight ceramic
Grain density, g/cm³: 2.6–2.8
Load at which it is usually used: up to 55–70 MPa
Price: higher

Proppant: Medium- and high-strength ceramic (bauxite)
Grain density, g/cm³: 3.2–3.6
Load at which it is usually used: 70 MPa and above
Price: high

Proppant	Grain density, g/cm³	Load at which it is usually used	Price
Quartz sand	2.65	up to 30–40 MPa	low
Resin-coated sand	2.5–2.6	up to 40–55 MPa	medium
Lightweight ceramic	2.6–2.8	up to 55–70 MPa	higher
Medium- and high-strength ceramic (bauxite)	3.2–3.6	70 MPa and above	high

Approximate limits for grains of size 20/40. They are checked by tests of the particular proppant at reservoir temperature and under long-term load. Heavy proppants settle faster in the fluid, and they are harder to carry to the end of the fracture.

Grain size is given, as for gravel, by a range of sieve numbers: 20/40 means grains that passed sieve No. 20 (mesh opening 0.84 mm) and were held by sieve No. 40 (0.42 mm). Large grains give a more permeable pack: the permeability grows roughly as the square of the grain size (Rock permeability). But they crush earlier, are carried less well by the fluid, and need a wider fracture. Rounded, spherical grains are stronger and give a more permeable pack than angular ones.

The load on the proppant is not the whole closure pressure. The fluid in the pores of the pack supports the grains with its pressure, so what loads the grain framework of the pack is the difference between the closure pressure and the pressure in the fracture. Near the wellbore the pressure in the fracture is almost equal to the bottomhole pressure, and the load there is greatest:

Effective stress on the proppant

\sigma_{\text{eff}} = p_c - p_{wf}

where:

$\sigma_{\text{eff}}$ effective closure stress on the proppant, MPa
$p_c$ closure pressure of the fracture, the minimum horizontal stress in the reservoir, MPa
$p_{wf}$ bottomhole pressure while the well is working, MPa

Units: any identical units of pressure

Holds when

Near the wellbore the pressure in the fracture equals the bottomhole pressure. Farther along the fracture it is higher, and the load is smaller.
As the reservoir is depleted (its pressure falls), the closure pressure falls with it, by $A\,\Delta\bar p$ , where $A$ is the poroelastic coefficient (Rock mechanics for hydraulic fracturing). This is usually less than the fall of the bottomhole pressure, so we calculate the design load from the initial closure pressure and the lowest planned bottomhole pressure. This is on the safe side.

The load grows when the bottomhole pressure is lowered, by putting the well on a pump or as the reservoir is depleted. So we choose the proppant not for the conditions of the first months, but for the greatest load over the whole life of the well.

Crushing. Fragments of grains plug the pores of the pack. For sand this starts at loads of about 30–40 MPa, and for ceramic clearly higher.
Embedment. The grains are pressed into the walls of the fracture, and the width decreases. The effect is stronger in soft rocks (shaly sandstones, chalks) and when there are few layers of grains.
Polymer residue. The thickener of the fracturing fluid leaves a filter cake (a layer of solids) on the walls and a gel residue in the pores of the pack. The share of conductivity that is retained is usually from a half to 90%, and it is worse at a low proppant loading. More breaker makes the pack cleaner, but if the viscosity drops too early, the proppant will settle before it reaches the end of the fracture.
Fine particles from the reservoir are carried into the pack during production. Load cycles at shut-ins and start-ups of the well break the grains further.
Inertial losses and multiphase flow in the pack reduce its effective permeability, especially in gas wells (see the advanced block below).

The width of the propped fracture is set by the amount of proppant per unit area of its wall, the loading $C_p$ , kg/m². The more layers of grains, the smaller the share of the width that is lost to embedment and filter cake at the walls:

Width and conductivity of a propped fracture

w_p = \frac{C_p}{(1 - \phi_p)\,\rho_p},\qquad (k_f w)_{\text{res}} = k_{f,\text{lab}}(\sigma_{\text{eff}})\,R\,(w_p - 2e)

where:

$w_p$ width of the proppant pack after closure, m
$C_p$ loading: the mass of proppant per 1 m² of one wall of the fracture, kg/m²
$\phi_p,\ \rho_p$ porosity of the pack and density of the proppant grains, —, kg/m³
$k_{f,\text{lab}}$ permeability of the pack from laboratory tests at the given load, mD
$R$ the share of the conductivity retained after the effect of the fracturing fluid (retained conductivity), —
$e$ depth of embedment of the grains into each wall, m

Units: conductivity in mD·m with $k$ in mD and the width in m

Holds when

The proppant lay evenly, without settling to the bottom of the fracture.
The factors $R$ and $e$ are found by tests with the fracturing fluid and with samples of the reservoir rock. Without tests, we choose cautious values.

The calculation of inflow and the proppant number use the conductivity in the reservoir, not the laboratory one. It is convenient to express it as an effective permeability of the pack: the conductivity divided by the full width $w_p$ .

CalculatedPack permeability of three 20/40 proppants against effective stress (the lesson's assumed curves, similar in shape to laboratory test curves). The vertical dashed line is the design load of the example, 31.8 MPa.

Worked exampleSand or ceramic: the conductivity in the reservoir

Given (assumed data)

Load: closure pressure 41.8 MPa (the calibration injection of the lesson on rock mechanics); the lowest planned bottomhole pressure is 10 MPa, and in the first months 25 MPa
Proppants 20/40: sand: 2650 kg/m³, pack porosity 0.36; lightweight ceramic: 2700 kg/m³, 0.38
Laboratory pack permeability (figure above): at 31.8 MPa: sand 76 D, ceramic 229 D; at 16.8 MPa sand is 174 D
Loading and losses: 4 kg/m²; retained share after polymer residue 0.7; embedment of 0.08 mm into each wall

Solution

The greatest load: $\sigma_{\text{eff}} = 41.8 - 10 = 31.8$ MPa (4612 psi). In the first months it is $41.8 - 25 = 16.8$ MPa.
The width of the pack: for ceramic $w_p = \dfrac{4}{(1 - 0.38)\cdot2700} = 2.39$ mm, for sand $\dfrac{4}{(1 - 0.36)\cdot2650} = 2.36$ mm. After subtracting the embedment $2\cdot0.08 = 0.16$ mm, 2.23 and 2.20 mm remain.
The conductivity in the reservoir at 31.8 MPa: ceramic $229\cdot10^3\cdot0.7\cdot2.23\cdot10^{-3} = 357$ mD·m, sand $76\cdot10^3\cdot0.7\cdot2.20\cdot10^{-3} = 117$ mD·m, which is 3.1 times less.
The effective permeability of the pack in the reservoir: for ceramic $357/2.39 = 149$ D; for sand $117/2.36 = 50$ D.
Sand in the first months: $174\cdot10^3\cdot0.7\cdot2.20\cdot10^{-3} = 268$ mD·m. At first sight this is enough, but when the bottomhole pressure falls to 10 MPa, the sand loses 56% of this conductivity.

Answer: At the design load, ceramic will keep about 357 mD·m in the reservoir and sand 117 mD·m. For the design calculation we take the permeability of the ceramic pack in the reservoir as 150 D.

What it means. Sand seems to pass under the conditions of the first months, and this is a trap. When the well is put on a pump and the bottomhole pressure is lowered, the conductivity of the fracture will fall by more than half. For a fracture of half-length 150 m in a 1 mD reservoir, this gives $F_{CD}$ = 2.4 for ceramic and 0.78 for sand. With sand the fracture ends up where the conductivity strongly limits the gain (effective wellbore radius at any conductivity). Whether the more expensive ceramic pays off is decided by the calculation of productivity and by economics, and we come back to it in the next section. The laboratory permeability here is assumed. Real curves are taken from tests of the particular proppant under long-term load. The data in this example are assumed for illustration.

More detail:Inertial losses in the proppant pack

Near the wellbore the whole inflow of a wing passes through the narrow cross-section of the fracture, and the filtration velocity in the pack (the flow rate per unit area) is high. The Forchheimer equation (Inflow at high rates: non-Darcy flow) adds an inertial term $\beta\rho v^2$ to the viscous resistance, and the pack behaves as if it were less permeable. If the inflow into the fracture is uniform along its length, the velocity in the pack falls linearly from the wellbore to the tip of the fracture. Integrating the losses along the length of the wing, we get the effective permeability:

Effective permeability of a pack with inertial losses

k_{f,\text{eff}} = \frac{k_f}{1 + \tfrac23\,\text{Fo}_w},\qquad \text{Fo}_w = \frac{\beta_f\,k_f\,\rho\,v_w}{\mu},\qquad v_w = \frac{qB/2}{h_p\,w_p}

where:

$k_{f,\text{eff}}$ permeability of the pack that, under Darcy's law, gives the same losses along the wing, m²
$\text{Fo}_w$ Forchheimer number in the pack at the wellbore, —
$\beta_f$ inertial resistance coefficient of the pack (from tests), m⁻¹
$\rho,\ \mu$ density and viscosity of the fluid at reservoir conditions, kg/m³, Pa·s
$v_w$ filtration velocity in the pack at the wellbore, m/s
$qB$ rate at reservoir conditions; each wing gets half, m³/s
$h_p,\ w_p$ height and width of the propped part of the fracture through which the flow goes, m

Units: SI

Holds when

The inflow into the fracture is uniform along the length, which is how a fracture of high conductivity behaves. At low conductivity the inflow is concentrated near the wellbore, and the losses are closer to the value at the wellbore, $k_f/(1 + \text{Fo}_w)$ .
The $\beta_f$ of a pack grows when grains crush, with gel residue, and when a second phase (for example water) is present.

The factor ⅔ comes from the integral. The viscous losses along the wing are proportional to $\int_0^1(1 - \xi)\,d\xi = 1/2$ , and the inertial ones to $\int_0^1(1 - \xi)^2\,d\xi = 1/3$ . For gas, the density in the reservoir is hundreds of kg/m³, several times less than for oil, and the viscosity is tens of times less. So at the same velocity $\text{Fo}_w$ is about ten times larger. With the large volume rates of gas in the reservoir, it is tens of times larger (example below).

Worked exampleHow much the inertial losses “eat” the conductivity of the fracture of a gas well

Given (assumed data)

Fracture: over the whole thickness of the reservoir, 20 m; $w_p$ = 4 mm; the permeability of the pack in the reservoir is 150 D; $\beta_f = 1\cdot10^5$ m⁻¹ (from tests of the pack)
Gas: 100 thousand m³/d; formation volume factor 0.005 m³/m³; density at standard conditions 0.78 kg/m³; viscosity in the reservoir 0.025 mPa·s
For comparison, the oil well of the design example: 60 m³/d, $B$ = 1.25, $\rho$ = 750 kg/m³, $\mu$ = 1.5 mPa·s, $h_p$ = 24 m, $w_p$ = 2.28 mm

Solution

The gas rate in the reservoir: $100{,}000\cdot0.005 = 500$ m³/d $= 5.787\cdot10^{-3}$ m³/s. The gas density in the reservoir is $0.78/0.005 = 156$ kg/m³.
The velocity in the pack at the wellbore: $v_w = \dfrac{5.787\cdot10^{-3}/2}{20\cdot0.004} = 0.0362$ m/s.
The Forchheimer number: $k_f = 150$ D $= 1.48\cdot10^{-10}$ m², and $\beta_f k_f\rho v_w = 10^5\cdot1.48\cdot10^{-10}\cdot156\cdot0.0362$ $= 8.36\cdot10^{-5}$ Pa·s. We divide by the viscosity $2.5\cdot10^{-5}$ Pa·s: $\text{Fo}_w = 3.34$ .
The effective permeability: $k_{f,\text{eff}} = \dfrac{150}{1 + \tfrac23\cdot3.34} = 46$ D, which is 31% of the Darcy value. At the wellbore itself it is $150/(1 + 3.34) = 35$ D.
The oil well: $v_w$ = 0.0079 m/s, $\text{Fo}_w$ = 0.059, and the loss of conductivity is about 4%.

Answer: In the gas well, inertial losses leave about 31% of the conductivity of the fracture. In the oil well they take away almost nothing.

What it means. For gas wells we design the conductivity of the fracture with a safety margin: a larger and stronger proppant, a higher loading, and $F_{CD}$ calculated from the effective conductivity, not the Darcy one. The effect depends on the rate: at half the rate, $\text{Fo}_w$ is half as large. The estimate with uniform inflow is optimistic, because at low conductivity the losses are closer to the value at the wellbore. The data in this example are assumed for illustration.

The proppant number and the optimal dimensionless conductivity

Suppose we have managed to lay a given volume of proppant pack in the reservoir. We can stretch it into a long thin fracture or gather it into a short wide one. A long fracture covers a larger part of the drainage area, but it carries fluid poorly. A short one carries fluid well, but it covers little. The lesson Hydraulic fracturing: purpose and well selection showed with an example that, for a fracture that is small compared with the area, the best result comes from $F_{CD}$ of about 1.6. To extend this result to fractures of any length, we use the dimensionless productivity index $J_D$ of a fracture in a closed area and the penetration ratio $I_x = 2x_f/x_e$ (a fracture in a bounded area).

The volume of the pack of both wings in the reservoir is $2x_f w h$ , and $F_{CD} = k_f w/(k x_f)$ . If we multiply $I_x^2$ by $F_{CD}$ , we get a quantity that contains only the volume of the pack, and not how it was divided between length and width:

The proppant number

N_{prop} = I_x^2\,F_{CD} = \frac{4x_f^2}{x_e^2}\cdot\frac{k_f w}{k\,x_f} = \frac{2\,k_f\,V_{prop}}{k\,V_{res}}

where:

$N_{prop}$ proppant number, —
$I_x$ penetration ratio of the fracture, $2x_f/x_e$ , —
$F_{CD}$ dimensionless conductivity of the fracture, —
$k_f,\ k$ permeability of the pack in the reservoir and permeability of the reservoir, mD
$V_{prop}$ volume of the proppant pack of both wings within the thickness of the reservoir, $2x_f w h$ , m³
$V_{res}$ volume of the drainage area, $x_e^2 h$ for a square area, m³

Units: dimensionless; $k_f$ and $k$ in the same units

Holds when

A vertical fracture in the centre of a square drainage area, parallel to its side.
Only the proppant within the reservoir goes into $V_{prop}$ . If the fracture is taller than the reservoir and the proppant is spread evenly over the height, this is the share $h/h_f$ of all the proppant.
$k_f$ is the permeability of the pack in the reservoir, taking into account the load, polymer residue and embedment (previous section).

The proppant number compares the conductivity that all the proppant can give with the conductivity of the drainage area. This is the “budget” of the fracture. At a given $N_{prop}$ , the choice of the half-length fixes the width and $F_{CD}$ uniquely, and the design problem becomes a search for the best way to divide the budget.

CalculatedDimensionless productivity index of a fracture in the centre of a closed square area against its dimensionless conductivity at a constant proppant number (the model of the lesson on inflow to a fractured well, pseudo-steady-state regime). The numbers at the curves are the proppant number. The curves 0.3, 0.82 and 3 stop on the left where the fracture reaches the boundaries (

I_x

= 1), and the curve 0.03 reaches the edge of the graph. The points are the maxima. The curve 0.82 is the design example below.

The optimal fracture in a square area (the lesson's model)

$N_{prop}$: 0.01
$F_{CD,opt}$: 1.69
$I_{x,opt}$: 0.08
$J_{D,\max}$: 0.304

$N_{prop}$: 0.03
$F_{CD,opt}$: 1.71
$I_{x,opt}$: 0.13
$J_{D,\max}$: 0.364

$N_{prop}$: 0.1
$F_{CD,opt}$: 1.75
$I_{x,opt}$: 0.24
$J_{D,\max}$: 0.464

$N_{prop}$: 0.3
$F_{CD,opt}$: 1.89
$I_{x,opt}$: 0.40
$J_{D,\max}$: 0.611

$N_{prop}$: 1
$F_{CD,opt}$: 2.43
$I_{x,opt}$: 0.64
$J_{D,\max}$: 0.881

$N_{prop}$: 3
$F_{CD,opt}$: 4.17
$I_{x,opt}$: 0.85
$J_{D,\max}$: 1.241

$N_{prop}$: 10
$F_{CD,opt}$: 10.96
$I_{x,opt}$: 0.96
$J_{D,\max}$: 1.609

$N_{prop}$	$F_{CD,opt}$	$I_{x,opt}$	$J_{D,\max}$
0.01	1.69	0.08	0.304
0.03	1.71	0.13	0.364
0.1	1.75	0.24	0.464
0.3	1.89	0.40	0.611
1	2.43	0.64	0.881
3	4.17	0.85	1.241
10	10.96	0.96	1.609

Calculated by searching for the maximum of $J_D$ with the model of a fracture of finite conductivity in a closed square. Published approximations of the same calculation give close values, within 1–2% in $J_D$ . For comparison, for a vertical well without skin in a square with a side of 500 m, $J_D$ = 0.14.

The table shows three ranges. At a small proppant number, less than 0.1, which is typical for reservoirs of medium and high permeability, the fracture is small compared with the area, and the optimum hardly depends on $N_{prop}$ : $F_{CD}$ is about 1.7 (the literature uses the rounded 1.6; the maximum is so flat that the difference in $J_D$ is hundredths of a per cent). For $N_{prop}$ from 0.1 to a few units, the optimal conductivity grows. The fracture approaches the boundaries, and more length gives less than more width. At large $N_{prop}$ , in tight reservoirs with big treatments, the fracture almost crosses the area ( $I_x \to 1$ ), and $J_D$ approaches the limit $6/\pi \approx 1.91$ .

Greatest productivity at a small proppant number

\frac{1}{J_{D,\max}} \approx 0.99 - \frac12\ln N_{prop},\qquad F_{CD,opt} \approx 1.6\text{–}1.7

where:

$J_{D,\max}$ the greatest dimensionless productivity index at the given proppant number, —
$N_{prop}$ proppant number, —

Units: dimensionless; natural logarithm

Holds when

$N_{prop}$ is up to 0.1. At 0.1 the formula overstates $J_{D,\max}$ by 0.7%, and at 0.3 already by 3%.
The area is a square and the regime is pseudo-steady-state. The derivation is in the advanced section.

The productivity grows with the amount of proppant logarithmically: doubling $N_{prop}$ reduces $1/J_D$ by only $\tfrac12\ln2 \approx 0.35$ , which means that it raises $J_D$ by 10–20% (the smaller $N_{prop}$ is, the smaller the gain). So we choose the size of the treatment by economics, not by “as much as will fit”.

Fracture dimensions from the proppant number and the chosen dimensionless conductivity

x_f = \sqrt{\frac{k_f\,V_{prop}}{2F_{CD}\,k\,h}},\qquad w = \sqrt{\frac{F_{CD}\,k\,V_{prop}}{2k_f\,h}}

where:

$x_f,\ w$ half-length and width of the propped fracture in the reservoir, m
$V_{prop}$ volume of the pack of both wings in the reservoir, m³
$k_f,\ k,\ h$ permeability of the pack and of the reservoir, thickness of the reservoir, mD, mD, m

Units: consistent; $k_f$ and $k$ in the same units

Holds when

It follows from $V_{prop} = 2x_f w h$ and $F_{CD} = k_f w/(k x_f)$ . An equivalent form is $x_f = I_x x_e/2$ with $I_x = \sqrt{N_{prop}/F_{CD}}$ .

At a small $N_{prop}$ the half-length grows as the square root of the ratio of the permeabilities of the pack and the reservoir. In a reservoir 100 times more permeable, the same proppant gives a fracture 10 times shorter and 10 times wider.

Worked exampleOptimal fracture dimensions for 40 t of ceramic

Given (assumed data)

Reservoir: $k$ = 1 mD, $h$ = 24 m, the sandstone of the lesson on rock mechanics; a square area of 500 × 500 m, $r_w$ = 0.108 m; oil with $\mu$ = 1.5 mPa·s, $B$ = 1.25
Proppant: 40 t of lightweight ceramic, 2700 kg/m³, pack porosity 0.38; the permeability of the pack in the reservoir is 150 D (previous example)
Fracture: height 35 m (checked in the example on injection); the proppant is spread evenly over the height

Solution

The volume of the pack in the reservoir: $V_{prop} = \dfrac{40{,}000\cdot24/35}{2700\cdot(1 - 0.38)} = 16.39$ m³.
The proppant number: $N_{prop} = \dfrac{2\cdot150{,}000\cdot16.39}{1\cdot500^2\cdot24} = 0.82$ .
The optimum by the model (the figure and the table above): $F_{CD} = 2.28$ , $I_x = 0.60$ , $J_{D,\max} = 0.826$ .
The dimensions: $x_f = I_x x_e/2 = 0.60\cdot250 = 150$ m. The width in the reservoir is $w = \dfrac{16.39}{2\cdot150\cdot24} = 2.28$ mm. The conductivity is $k_f w = 150{,}000\cdot2.28\cdot10^{-3} = 342$ mD·m.
Without a fracture ( $S = 0$ ): $J_D = \dfrac{1}{\ln(500/0.108) - 1.311} = 0.140$ . The fracture raises the productivity by a factor of 5.9: $J = \dfrac{0.536\cdot1\cdot24}{1.5\cdot1.25}\cdot0.826 = 5.67$ m³/(d·MPa).
The same proppant used differently: a fracture of 250 m, up to the boundaries ( $I_x$ = 1, $F_{CD}$ = 0.82), gives $J_D = 0.730$ . A fracture of 60 m ( $F_{CD}$ = 14.2) gives $J_D = 0.616$ .

Answer: The optimal fracture has a half-length of about 150 m, a width of 2.28 mm and a conductivity of 342 mD·m (

F_{CD}

= 2.28). The productivity of the well rises by a factor of 5.9.

What it means. The optimum is flat: fractures of 100 and 200 m with the same proppant give $J_D$ of 0.767 and 0.792, which is 7% and 4% less. The extreme options lose more: the long thin one loses 12%, and the short thick one 25%. The same 40 t of sand, with a pack permeability of 50 D, would give $N_{prop}$ = 0.27 and $J_{D,\max}$ = 0.59, which is 28% less than ceramic. Whether the difference pays for the price is decided by economics. A width of 2.3 mm is about four layers of 20/40 grains. It is not worth making the fracture thinner, because embedment and gel residue would take away a large share of the conductivity. The data in this example are assumed for illustration.

The optimum is flat, and the conductivity is uncertain

The optimum in $F_{CD}$ is flat: in the design example a fracture with an $F_{CD}$ twice as high or twice as low as the optimal one loses about 5–6% of the productivity. But the conductivity in the reservoir is known only to within a factor of several, and with time it only falls: through crushing, plugging, and the growth of the load with depletion. So it is safer to make the error towards a higher conductivity. A fracture designed exactly at the optimum will be to the left of it after a few years.
In the first months, while the inflow is bilinear (the early flow regime of a fracture), it is set by the conductivity, not by the length (flow regimes after start-up). In gas wells, inertial losses eat part of the conductivity. So in practice we design $F_{CD}$ somewhat above the optimum for the proppant number.

A real design is also limited by technical conditions. A fracture cannot be longer than the drainage area ( $I_x \le 1$ ), and with closely spaced wells we must not let the fractures of neighbouring wells meet. The width is limited by the stiffness of the rock and by the net pressure at which the fracture stays in the reservoir (next section). A pack that is too thin, less than three or four layers of grains, loses a large share of its conductivity. In permeable reservoirs the optimum needs a width of centimetres. We get it by deliberately stopping the growth of the fracture in length (Fracturing fluids and tip-screenout fracturing), and in weak sandstones we combine such a fracture with a gravel pack (a layer of gravel that holds back the sand; Sand control methods).

How much proppant: the optimal fracture for different amounts of ceramic (the reservoir of the design example)

Proppant, t: 10
$N_{prop}$: 0.20
$x_f$ , m: 84
$w$ , mm: 1.02
$J_{D,\max}$: 0.55

Proppant, t: 20
$N_{prop}$: 0.41
$x_f$ , m: 114
$w$ , mm: 1.50
$J_{D,\max}$: 0.67

Proppant, t: 40
$N_{prop}$: 0.82
$x_f$ , m: 150
$w$ , mm: 2.28
$J_{D,\max}$: 0.83

Proppant, t: 80
$N_{prop}$: 1.64
$x_f$ , m: 186
$w$ , mm: 3.67
$J_{D,\max}$: 1.03

Proppant, t: 160
$N_{prop}$: 3.28
$x_f$ , m: 215
$w$ , mm: 6.35
$J_{D,\max}$: 1.27

Proppant, t	$N_{prop}$	$x_f$ , m	$w$ , mm	$J_{D,\max}$
10	0.20	84	1.02	0.55
20	0.41	114	1.50	0.67
40	0.82	150	2.28	0.83
80	1.64	186	3.67	1.03
160	3.28	215	6.35	1.27

The height of the fracture is 35 m and the pack permeability 150 D in all options. In reality a bigger treatment grows more in height, so a smaller share of the proppant gets into the reservoir and $J_D$ is lower than in the table. On the other hand, a thick pack loses a smaller share of its width to embedment, which raises $J_D$ a little. Each doubling of the proppant adds less and less to $J_D$ . For a well without a fracture $J_D$ = 0.14.

The amount of proppant to pump is decided by economics. For several sizes of treatment we build a production forecast (first transient inflow, then stabilised), convert it to discounted revenue, and subtract the cost of the treatment. The cost grows faster than linearly, because a larger fracture needs disproportionately more fluid. We choose the size with the highest NPV (Net present value (NPV)), and we check the sensitivity to the oil price and to the uncertainty of the conductivity with scenarios (Sensitivity analysis and scenarios).

Geometry models: PKN, KGD and radial

The fracture dimensions found from the proppant number are the target. To reach it, we must know how the fracture grows during injection. Fracture growth models solve three problems together. The elastic one: what width the net pressure creates. The hydraulic one: what pressure is needed to push a viscous fluid along a narrow slit. And the volume balance: how much fluid stays in the fracture and how much goes into the reservoir. The simplest, two-dimensional models are PKN and KGD, named after the first letters of the surnames of their authors (history of the models), and the radial one. They fix the height of the fracture (or treat the fracture as round) and give the length and the width. Pseudo-three-dimensional and three-dimensional models also calculate the growth in height, but they need the stress profile and the properties of all the layers.

SchematicThree two-dimensional models of a fracture (schematic, not to scale). Left: the view onto the plane of the fracture. Right: the vertical cross-section at the wellbore. PKN is a long fracture in a reservoir between barriers, with an elliptical cross-section of height

h_f

. KGD is a short, tall fracture, with the same width over the height and a width that falls towards the tip. Radial is a circle of radius

R

around the injection interval.

Two-dimensional fracture models

Model: PKN
Where it applies: a long fracture whose height is held by barriers ( $2x_f$ is much larger than $h_f$ )
Width at net pressure: $w = 2p_{net}h_f/E'$
Net pressure as the fracture grows: rises

Model: KGD
Where it applies: a short fracture, the height is greater than the length
Width at net pressure: $w = 4p_{net}x_f/E'$
Net pressure as the fracture grows: falls

Model: Radial
Where it applies: no barriers; the fracture grows from a point in all directions
Width at net pressure: $w = 8p_{net}R/(\pi E')$
Net pressure as the fracture grows: falls

Model	Where it applies	Width at net pressure	Net pressure as the fracture grows
PKN	a long fracture whose height is held by barriers ( $2x_f$ is much larger than $h_f$ )	$w = 2p_{net}h_f/E'$	rises
KGD	a short fracture, the height is greater than the length	$w = 4p_{net}x_f/E'$	falls
Radial	no barriers; the fracture grows from a point in all directions	$w = 8p_{net}R/(\pi E')$	falls

The width is the greatest one, in the middle of the smallest dimension of the fracture, at a uniform net pressure. $E'$ is the plane-strain modulus. For the names of the models and their history, see the lesson on rock mechanics.

Why does the width depend so weakly on the fluid? The width is proportional to the net pressure divided by the stiffness of the rock. The pressure is needed to push the fluid through the slit, and the pressure drop along a slit is inversely proportional to the cube of its width. A wide slit needs less pressure, and a small pressure creates a narrow slit. The balance is reached at a width proportional to the fourth root of the product of viscosity, rate and length, divided by the modulus of the rock. A fluid twice as viscous widens the fracture by only 19%, while the modulus from soft rocks to stiff ones changes tens of times, and the width by 2–3 times.

Width of the fracture at the wellbore and the net pressure by the PKN model

w_w \approx 2.75\left(\frac{\mu\,q_i\,x_f}{E'}\right)^{1/4},\qquad \bar w \approx 0.58\,w_w,\qquad p_{net} = \frac{E'\,w_w}{2h_f}

where:

$w_w$ the largest width of the fracture at the wellbore during injection, m
$\bar w$ average width over the whole fracture, m
$\mu$ apparent viscosity of the fluid in the fracture, Pa·s
$q_i$ injection rate, into both wings together, m³/s
$x_f$ half-length of the created fracture, m
$E'$ plane-strain modulus of the rock, $E/(1 - \nu^2)$ , Pa
$p_{net},\ h_f$ net pressure at the wellbore and the height of the fracture, Pa, m

Units: SI (any consistent system of units)

Holds when

The fracture is much longer than its height, the height is constant, and every vertical cross-section is an ellipse. The fluid is Newtonian and the flow is laminar.
The factor 2.75 and the ratio 0.58 are the midpoint between the solution without leak-off (2.82 and 0.59) and the limiting case of strong leak-off (2.68 and 0.57). The derivation is in the advanced section. For fluids whose viscosity falls as the shear rate grows, we take the apparent viscosity at the shear rate in the fracture.
The factor refers to the total rate into both wings. If we use the rate into one wing instead, the factor is $2^{1/4} \approx 1.19$ times larger, about 3.27. The formula is often given in this form, so when comparing formulas, check which rate the factor refers to.

The width grows with the length of the fracture, so the net pressure grows too. This is the sign by which a PKN-type fracture is recognised during injection. For the fracture to let the proppant through, the width at the wellbore must be at least about three diameters of the largest grains. Otherwise the grains form an arch across the fracture, and a screen-out (the proppant blocks the fracture) stops the injection.

In other unit systems (1)

Oilfield units

w_w\,[\text{in}] \approx 0.356\left(\frac{\mu\,[\text{cP}]\;q_i\,[\text{bbl/min}]\;x_f\,[\text{ft}]}{E'\,[\text{psi}]}\right)^{1/4}

$\mu$ in cP, $q_i$ in barrels per minute into both wings, $x_f$ in feet, $E'$ in psi.

$0.356 = 2.75\cdot0.00329/0.0254$ , where $0.00329 = (1.171\cdot10^{-10})^{1/4}$ . The factor $1.171\cdot10^{-10}$ converts cP·(bbl/min)·ft/psi to Pa·s·(m³/s)·m/Pa, and $0.0254$ converts metres to inches. With the shear modulus $G$ the formula is often written as $0.3\,[\mu q_i (1 - \nu) x_f/G]^{1/4}$ . This is the same factor: $(1 - \nu)/G = 2/E'$ and $0.3\cdot2^{1/4} \approx 0.357$ .

The KGD model describes short fractures: the height is greater than the length, and the elastic cross-section is in the horizontal plane, so the width is proportional to the net pressure and to the length of the fracture. The same logic gives a width that grows as $(\mu q_i x_f^2/(E' h_f))^{1/4}$ , and the net pressure falls as the fracture lengthens. The radial model suits fractures in a uniform rock mass without barriers, and horizontal wells with transverse fractures at the early stage of growth. In practice we use pseudo-three-dimensional models, which calculate the height themselves from the stress profile. Two-dimensional models are needed for understanding, for estimates, and for checking a computer calculation.

Volume balance, leak-off and fluid efficiency

While the fracture is open, fluid filters through its walls into the reservoir. On a freshly opened wall the leak-off is fast. Then it is slowed by the filter cake of the polymer, by the zone invaded by the filtrate (the liquid that enters the rock), and by the compression of the reservoir fluid. By Carter's law the leak-off rate per unit area of wall falls as $C_L/\sqrt{t - \tau}$ , where $\tau$ is the moment when this part of the wall opened, and $C_L$ is the leak-off coefficient. The same leak-off set the pressure decline after the calibration injection (the G function). The pumped volume divides into the volume of the fracture and the volume that has gone into the reservoir:

Volume balance during injection with leak-off

q_i t = A_f\,\bar w + 2A_l\left(\kappa\,C_L\sqrt t + S_p\right),\qquad \eta = \frac{A_f\,\bar w}{q_i t}

where:

$q_i,\ t$ injection rate of the slurry and the injection time, m³/s, s
$A_f$ area of one wall of both wings, $2x_f h_f$ , m²
$\bar w$ average width of the fracture, m
$A_l$ the part of this area against the permeable reservoir, $2x_f h_l$ , m²
$\kappa$ a factor that takes into account that parts of the wall opened at different times: from 1.42 (PKN without leak-off) to $\pi/2 \approx 1.57$ (strong leak-off), —
$C_L$ leak-off coefficient, m/s $^{1/2}$
$S_p$ spurt loss (instant leak-off before the filter cake forms), per unit area, m
$\eta$ fluid efficiency: the share of the pumped volume that is in the fracture at the end of injection, —

Units: SI (any consistent system of units)

Holds when

Carter's law: the leak-off rate is inversely proportional to the square root of the time since the part of the wall opened. It goes through both walls, hence the factor 2.
Leak-off goes only into the permeable reservoir. The proppant is part of the pumped volume, but it does not leak off.
$\kappa$ comes from averaging $2\sqrt{t-\tau}$ over the area, when the area grows by a power law. In the examples of the lesson $\kappa$ = 1.5, and spurt loss is not counted.

The first term grows slowly with time, because the width and the length grow as fractional powers. The second term grows faster. So the efficiency falls as injection goes on: big treatments and slow injections lose a larger share of the fluid into the reservoir. The efficiency is higher at a lower leak-off coefficient, a higher rate and a wider fracture, that is, in soft rocks and with a more viscous fluid.

The leak-off coefficient is hard to calculate in advance. It depends on the permeability, on the fluid, and on which layers the fracture has crossed. So we find it from the calibration injection. After shut-in, the area of the fracture no longer grows, the fluid goes out through the walls, and the volume of the fracture falls in proportion to the $G$ function. If we treat the fracture as a PKN fracture, its volume is linked with the net pressure through the compliance $\pi h_f/(2E')$ (the average width per unit of net pressure), and the slope of the line “pressure against $G$ ” gives the leak-off coefficient:

Efficiency and leak-off coefficient from a calibration injection

\eta = \frac{\frac{3\pi}{16}G_c}{1 + \frac{3\pi}{16}G_c},\qquad C_L = \frac{m\,h_f^2}{E'\,h_l\sqrt{t_p}}

where:

$G_c$ value of the $G$ function at the moment the fracture closes, —
$m$ slope (absolute value) of the line “pressure against $G$ ” before closure, Pa
$h_f,\ h_l$ height of the fracture during the calibration injection, and the height through which the leak-off goes (the permeable reservoir), m
$E'$ plane-strain modulus, Pa
$t_p$ duration of injection, s

Units: SI; $C_L$ in m/s $^{1/2}$

Holds when

The $G$ function has the form used in the lesson on rock mechanics: the area of the fracture grew in proportion to time during injection. With the second limiting form the efficiency comes out a few per cent less.
After shut-in the area does not change, there is no spurt loss, and by the time of closure there is no fluid left in the fracture.
The compliance is that of a PKN fracture. For short fractures we need the compliance of the KGD model, and the leak-off coefficient will be different.

The first formula follows from the fact that the leak-off after shut-in equals the volume of the fracture at shut-in, and the leak-off before shut-in equals the rest of the pumped volume. Their ratio is expressed through $G_c$ . The later the fracture closes, the larger $G_c$ and the higher the efficiency.

Worked exampleLeak-off and efficiency from a calibration injection

Given (assumed data)

Calibration injection (lesson on rock mechanics): 8 min of injection, closure 40 min after shut-in: $G_c$ = 4.27, slope $m$ = 0.609 MPa per unit of $G$
Fracture and rock: height during the calibration injection 31.2 m, leak-off into a reservoir 24 m thick; $E'$ = 23,100 MPa

Solution

Efficiency: $0.589\cdot4.27 = 2.52$ (here $0.589 = 3\pi/16$ ), and $\eta = 2.52/(1 + 2.52) = 0.72$ .
Leak-off coefficient: $C_L = \dfrac{0.609\cdot31.2^2}{23{,}100\cdot24\cdot\sqrt{480}} = 4.88\cdot10^{-5}$ m/s $^{1/2}$ . This is 1.24·10⁻³ ft/min $^{1/2}$ in oilfield units.
If we assume that the fluid leaks off through the whole height of the fracture, 31.2 m, and not only through the reservoir, $C_L$ would come out $31.2/24 = 1.3$ times smaller: 3.75·10⁻⁵ m/s $^{1/2}$ .

Answer: In 8 min about 72% of the pumped volume stayed in the fracture. The leak-off coefficient is about 4.9·10⁻⁵ m/s

^{1/2}

What it means. The efficiency refers to the calibration injection. For the main injection it will be lower, because the main injection is longer, and over its time the leak-off grows faster than the volume of the fracture (next example). The leak-off coefficient, on the contrary, is a property of the pair “reservoir and fluid”, and we carry it over to the calculation of the main injection if the calibration was done with the same fluid. The assumption about the height of the leak-off changes the answer by almost a quarter, so the design must state it. The data in this example are assumed for illustration.

Worked exampleHow much fluid is needed for a fracture of 150 m

Given (assumed data)

Aim: $x_f$ = 150 m, 40 t of ceramic (the example on the proppant number)
Injection: slurry rate 4.8 m³/min (0.08 m³/s), apparent viscosity of the fluid in the fracture 0.25 Pa·s
Rock: $E'$ = 23,100 MPa, stress contrast with the shales 5.9 MPa, reservoir 24 m; fracture height 35 m
Leak-off: $C_L = 4.9\cdot10^{-5}$ m/s $^{1/2}$ (previous example), $\kappa$ = 1.5, leak-off into the reservoir

Solution

The width at the wellbore: $w_w = 2.75\left(\dfrac{0.25\cdot0.08\cdot150}{23{,}100\cdot10^6}\right)^{1/4} = 9.28$ mm. The average is $\bar w = 0.58\cdot9.28 = 5.38$ mm.
The net pressure at the wellbore: $p_{net} = \dfrac{23{,}100\cdot9.28\cdot10^{-3}}{2\cdot35} = 3.06$ MPa. A check of containment: the equilibrium height is $24/\cos(\pi\cdot3.06/11.8) = 35.0$ m, which agrees with the assumed one.
The proppant will pass: 9.28 mm at the wellbore is more than three diameters of the largest 20/40 grains ( $3\cdot0.84 = 2.5$ mm).
The volume of the fracture: $V_f = 2\cdot150\cdot35\cdot5.38\cdot10^{-3} = 56.5$ m³.
Leak-off: the area against the reservoir is $A_l = 2\cdot150\cdot24 = 7200$ m², and the factor is $2A_l\kappa C_L = 2\cdot7200\cdot1.5\cdot4.9\cdot10^{-5} = 1.058$ m³/s $^{1/2}$ .
The balance $0.08\,t = 56.5 + 1.058\sqrt t$ is a quadratic equation in $\sqrt t$ . The discriminant is $1.058^2 + 4\cdot0.08\cdot56.5 = 19.20$ , and its root is 4.382. So $\sqrt t = (1.058 + 4.382)/0.16 = 34.0$ s $^{1/2}$ , and $t = 34.0^2 = 1156$ s $= 19.27$ min.
The pumped volume is $V_i = 4.8\cdot19.27 = 92.5$ m³ of slurry, and 56.5 m³ is in the fracture. The efficiency is $\eta = 0.611$ .

Answer: A fracture of 150 m will be made by injecting about 92.5 m³ of slurry in 19.27 min at an efficiency of 0.611. The net pressure of 3.06 MPa holds the fracture within 35 m.

What it means. The efficiency of the main injection is lower than that of the calibration injection (0.72), because it lasts longer. The volume of slurry includes 14.8 m³ of the proppant itself, so the clean fluid is about 78 m³. If we take a fluid twice as viscous, the width grows by only 19%, but the net pressure rises to 3.6 MPa, at which the equilibrium height is already 42 m. The proppant will go into the shales, and the fracture may reach the neighbouring layers. The PKN model with a constant height is an estimate. We run the design calculation with a pseudo-three-dimensional model calibrated against the data of the calibration injection. The data in this example are assumed for illustration.

CalculatedCalculated distribution of the pumped volume in the example: how much slurry has been pumped, how much has stayed in the fracture and how much has gone into the reservoir at each moment of injection (the PKN model with leak-off by Carter's law). The points on the curve of the fracture volume are the 5th and the 10th minute, and the labels give the half-length at that moment.

The pad and the proppant injection schedule

Injection starts with a pad: fluid without proppant. It opens the fracture and creates the width into which the grains will go. But its main job is to take up the leak-off. The fluid that goes first ends up at the end of the fracture, and it leaks into the reservoir the longest. If the pad is too small, the proppant-carrying fluid loses its water at the end of the fracture too early. The proppant then forms a plug there, the growth of the fracture stops, and the pressure rises sharply. This is a screen-out before all of the proppant has been pumped. If the pad is too large, then by the end of pumping a part without proppant remains at the end of the fracture, and it will close, while the extra fluid only damages the reservoir.

We raise the proppant concentration gradually. Slurry that was pumped earlier goes farther along the fracture and loses fluid for longer, so its concentration in the fracture rises more strongly. If we start with a dilute slurry and bring the concentration up to the maximum by the end of pumping, then by the moment of shut-in the proppant will be spread evenly along the fracture. The rule for the pad fraction and the law of growth of the concentration was found by K. Nolte in the mid-1980s:

Pad fraction and the law of growth of the proppant concentration

\varepsilon = \frac{1 - \eta}{1 + \eta},\qquad V_{pad} = \varepsilon V_i,\qquad c(t) = c_e\left(\frac{t - t_{pad}}{t_e - t_{pad}}\right)^{\varepsilon},\qquad c_e = \frac{M}{\eta V_i}

where:

$\varepsilon$ exponent of the law of growth of the concentration; it is also the pad fraction of the pumped volume, —
$\eta$ fluid efficiency at the end of injection, —
$V_{pad},\ V_i$ volume of the pad and the whole pumped volume, m³
$c(t)$ proppant concentration in the injected slurry, kg/m³ of slurry
$t_{pad},\ t_e$ end of the pad and end of injection, min
$c_e$ concentration at the end of injection, which is also the concentration in the whole fracture at shut-in, kg/m³
$M$ mass of proppant, kg

Units: consistent

Holds when

Leak-off follows Carter's law, and the area of the fracture grows during injection. The proppant does not settle and moves together with the fluid.
The concentration is the mass of proppant per unit volume of slurry. In field practice it is given per volume of clean fluid: $c_{\text{fl}} = c/(1 - c/\rho_p)$ .
The efficiency is taken from the calculation of the main injection, calibrated against the calibration injection. When in doubt, we make the pad a little larger: a screen-out costs more than extra fluid.

An analysis of leak-off by Carter's law shows that the pad fraction and the exponent of the law of growth coincide. Then the exponent itself follows from the balance of mass. The slurry of volume $(1 - \varepsilon)V_i$ with the concentration $c_e\tau^{\varepsilon}$ carries $c_e(1 - \varepsilon)V_i/(1 + \varepsilon)$ of proppant, and by shut-in all of it must fit into the volume of the fracture $\eta V_i$ at the concentration $c_e$ . The equality holds exactly at $\varepsilon = (1 - \eta)/(1 + \eta)$ . The lower the efficiency, the larger the pad and the closer the growth of the concentration to linear. At $\eta \to 1$ no pad is needed, and the slurry is pumped at once at the final concentration.

Pad fraction by the fluid efficiency

Efficiency $\eta$: Pad $\varepsilon$ , % of the volume
0.2: 67
0.4: 43
0.6: 25
0.8: 11

Efficiency $\eta$	0.2	0.4	0.6	0.8
Pad $\varepsilon$ , % of the volume	67	43	25	11

At low efficiency, most of the fluid is pumped without proppant.

The width of the propped fracture follows from the same balance. By shut-in the fracture of average width $\bar w$ is filled with slurry at the concentration $c_e$ , so the loading is $C_p = c_e\bar w$ , and after closure the width of the pack is $w_p = c_e\bar w/((1 - \phi_p)\rho_p)$ . The ratio of the propped width to the created width equals the ratio of the share of grains in the slurry to their share in the pack. To prop more width we need a more concentrated slurry at the end of injection, and it is limited by what the pumps can still pump and the fluid can still carry.

Worked exampleThe pad and the concentration stages

Given (assumed data)

Injection (previous example): $V_i$ = 92.5 m³ of slurry, $\eta$ = 0.611, volume of the fracture 56.5 m³, average width 5.38 mm, rate 4.8 m³/min
Proppant: 40 t of ceramic, 2700 kg/m³, pack porosity 0.38; five stages of slurry of equal volume

Solution

The exponent and the pad: $\varepsilon = \dfrac{1 - 0.611}{1 + 0.611} = 0.241$ . The pad is $0.241\cdot92.5 = 22.3$ m³, about 5 min of injection.
The concentration at the end of injection, and in the fracture at shut-in: $c_e = \dfrac{40{,}000}{56.5} = 708$ kg/m³ of slurry, or 960 kg per 1 m³ of clean fluid.
The law of growth: $c = 708\,\tau^{0.241}$ , where $\tau$ is the fraction of the slurry injection time. The average concentrations of the five stages of 14.0 m³ each are in the table below. Together they carry 40.0 t of proppant.
The loading after closure: $C_p = c_e\bar w = 708\cdot5.38\cdot10^{-3} = 3.81$ kg/m². The width of the pack is $\dfrac{3.81}{0.62\cdot2700} = 2.28$ mm, which is 42% of the average created width.

Answer: The pad is about 22.3 m³. We raise the concentration from zero to 708 kg/m³ of slurry by the law

\tau^{0.241}

, and all 40 t go into the fracture with a loading of about 3.81 kg/m².

What it means. The width of the pack in the reservoir came out the same as the proppant-number calculation required. This is as it should be: the design is self-consistent. We finish the injection with a flush: clean fluid in a volume equal to that of the casing down to the perforated interval. We must not flush too far. The part of the fracture near the wellbore would be left without proppant and would close, and the choking would take away most of the gain (choking of the fracture). Flushing too little would leave proppant in the wellbore, and it would have to be washed out. The data in this example are assumed for illustration.

Slurry stages (example)

Stage: 1
Slurry volume, m³: 14.0
Concentration, kg/m³ of slurry: 387
Same, kg/m³ of fluid: 452
Proppant, t: 5.43

Stage: 2
Slurry volume, m³: 14.0
Concentration, kg/m³ of slurry: 528
Same, kg/m³ of fluid: 656
Proppant, t: 7.41

Stage: 3
Slurry volume, m³: 14.0
Concentration, kg/m³ of slurry: 598
Same, kg/m³ of fluid: 769
Proppant, t: 8.40

Stage: 4
Slurry volume, m³: 14.0
Concentration, kg/m³ of slurry: 649
Same, kg/m³ of fluid: 855
Proppant, t: 9.12

Stage: 5
Slurry volume, m³: 14.0
Concentration, kg/m³ of slurry: 690
Same, kg/m³ of fluid: 927
Proppant, t: 9.69

Stage	Slurry volume, m³	Concentration, kg/m³ of slurry	Same, kg/m³ of fluid	Proppant, t
1	14.0	387	452	5.43
2	14.0	528	656	7.41
3	14.0	598	769	8.40
4	14.0	649	855	9.12
5	14.0	690	927	9.69

The average concentration of each stage is the average of the law $c_e\tau^{\varepsilon}$ over its interval. In practice the first stage is made with a small constant concentration, and all stages are rounded to values that are easy to hold on the blender (the machine that mixes proppant into the fluid).

CalculatedProppant concentration in the injected slurry in the example: the pad, the continuous law of growth

c_e\tau^{\varepsilon}

, and the five stages that replace it in practice.

Checking the design: before, during and after injection

Before injection. The calibration injection gives the closure pressure, the net pressure, the leak-off coefficient and the efficiency. A stepwise reduction of the rate separates the losses in the perforations (the holes shot through the casing) from those in the tortuous (twisting) start of the fracture (Rock mechanics for hydraulic fracturing). With these data we recalculate the model and the design of the main injection: the pad, the rate and the concentrations.
During injection. The net pressure on a log–log plot shows whether the fracture is growing in length, in height, or has been blocked by proppant at its end. A sharp rise of pressure earlier than calculated is a sign of an approaching screen-out. Then we lower the concentration, or stop feeding the proppant and flush the slurry out.
After injection. We fit the pressure record with the model, and so refine the height, the leak-off and the modulus. A well test after the well has been brought on production, from the bilinear and linear parts of the record, gives the conductivity and the half-length of the fracture that really work (Inflow to a hydraulically fractured well, Diagnosis with the pressure derivative). We compare them with the design.
From production. We compare production with the forecast, remembering that the inflow is transient in the first months. Systematic differences are a reason to revise the input data for the next wells, and not only the result of one treatment.

Terms and symbols

Proppant number $N_{prop}$: It is $I_x^2F_{CD} = 2k_fV_{prop}/(kV_{res})$ , a dimensionless measure of the amount of proppant in the reservoir. It sets the greatest possible productivity of the fracture.
Optimal dimensionless conductivity: The $F_{CD}$ at which a given volume of proppant gives the greatest $J_D$ . It is about 1.6–1.7 at a small $N_{prop}$ , and higher at a large one.
Effective closure stress on the proppant: The closure pressure minus the pressure in the fracture. It is greatest at the wellbore at the lowest bottomhole pressure.
Proppant loading $C_p$: The mass of proppant per square metre of one wall of the fracture, kg/m². It sets the width of the pack.
Retained conductivity: The share of the conductivity of the pack that remains after the effect of the fracturing fluid.
Proppant embedment: Grains pressing into the walls of the fracture under load, which reduces its width.
PKN, KGD and radial models: Two-dimensional models of fracture growth: a long fracture with a held height, a short tall one, and a round one.
Leak-off coefficient $C_L$: The constant of Carter's law: the leak-off rate through the wall is $C_L/\sqrt{t-\tau}$ .
Fluid efficiency $\eta$: The share of the pumped volume that is in the fracture at the end of injection.
Pad: Fluid without proppant at the start of injection. It creates the width and takes up the leak-off.

Common mistakes and misconceptions

Wrong

“In a low-permeability reservoir the fracture should be as long as possible.”

In fact

For a given amount of proppant, lengthening makes the fracture thinner, and beyond the optimum the productivity falls. In the lesson's example a fracture of 250 m, up to the boundaries of the area, gives a $J_D$ that is 12% less than that of the optimal 150 m. The length is also limited by the size of the drainage area.

Wrong

“The conductivity of a fracture is taken from the proppant data sheet.”

In fact

The data-sheet tests correspond to a certain load, loading and clean fluid. In the reservoir the conductivity is less: the load grows when the bottomhole pressure falls, polymer residue and embedment of the grains take away part of it, and in gas wells there are inertial losses. In the lesson's example, 229 D on the laboratory curve became 150 D in the reservoir.

Wrong

“To get width, it is enough to raise the viscosity of the fluid.”

In fact

The width grows only as the fourth root of the viscosity: a fluid twice as viscous gives +19%. The net pressure grows with the width, and the fracture starts to grow in height. In a stiff rock we get width from the proppant concentration at the end of injection, or by deliberately stopping the growth of the fracture in length.

Wrong

“Efficiency is a property of the fracturing fluid.”

In fact

It depends on both the reservoir and the injection: the same fluid in the lesson's example gave 0.72 in 8 min of calibration injection, and 0.61 in 19 min of the main injection. The property of the pair “fluid and reservoir” is the leak-off coefficient, and the efficiency is recalculated for each injection.

Wrong

“It is safer to take the pad with a large margin.”

In fact

An extra pad leaves an unpropped part at the end of the fracture, which will close. It also increases leak-off and damage to the reservoir. We calculate the pad from the efficiency calibrated against the calibration injection, and allow only a moderate margin.

In depth: advanced detail

Where the PKN width formula comes from

In the PKN model every vertical cross-section of the fracture is an ellipse of height $h_f$ , opened by the net pressure: the greatest width is $w = 2p_{net}h_f/E'$ , as in the lesson on rock mechanics. The flow of a Newtonian fluid (one whose viscosity does not depend on the flow) along an elliptical channel with the narrow axis $w$ is $q = \dfrac{\pi h_f w^3}{64\mu}\left(-\dfrac{\partial p}{\partial x}\right)$ . If we express the pressure through the width, we get $q = -\dfrac{\pi E'}{512\mu}\dfrac{\partial (w^4)}{\partial x}$ .

If we take the rate to be the same along the whole wing and equal to $q_i/2$ , integration from the wellbore to the tip of the fracture, where the width is zero, gives $w_w^4 = \dfrac{256}{\pi}\cdot\dfrac{\mu q_i x_f}{E'}$ . That is the factor $(256/\pi)^{1/4} \approx 3.00$ . This is what Perkins and Kern did. The profile of the width along the length is $w_w(1 - x/x_f)^{1/4}$ , and the average width is $(\pi/4)\cdot(4/5) = \pi/5 \approx 0.63$ of $w_w$ .

In reality part of the pumped fluid stays in the fracture along the way, and the rate falls towards the tip. Nordgren added the accumulation of fluid to the equation. Without leak-off the problem has a self-similar solution. The numerical solution for this lesson gives the factor 2.82 and the ratio of the average width to the greatest 0.59. With strong leak-off the fluid is lost along the length so that the rate falls as $(2/\pi)\arccos(x/x_f)$ . Then the factor is 2.68 and the ratio is 0.57. Real injections lie between these limits, so the lesson takes 2.75 and 0.58. The error from this choice is less than 3% in width, much less than the uncertainty of the modulus and the viscosity. The Perkins and Kern solution, 3.00, overstates the width by 6–12%. If the factor refers to the rate into one wing, all the values are larger by $2^{1/4} \approx 1.19$ : 3.57 for Perkins and Kern, 3.35 without leak-off, and 3.27 for the value taken in the lesson. The last one coincides with the often-quoted factor that comes from the rounded coefficients of the Nordgren solution. The full numerical solution of the PKN model with leak-off by Carter's law, for the lesson's example injection, gives at the wellbore a factor of 2.76 and a ratio of the average width to the greatest of 0.58.

For the KGD model the elastic cross-section is horizontal: $w = 4p_{net}x_f/E'$ , and the fluid flows along a slit of height $h_f$ and constant width. If we repeat the argument, we get a width proportional to $(\mu q_i x_f^2/(E'h_f))^{1/4}$ . The pressure falls as such a fracture lengthens, because the width grows faster than the length of the channel along which the fluid is pushed. The numerical factors of the KGD and radial models depend on how the flow at the very tip of the fracture is described, and different authors give different values. For estimates the powers matter more than the factors.

Why the optimum is about 1.6–1.7

If the fracture is small compared with the area, we can replace it by a well of effective wellbore radius $r_w'$ , and for a square with a side $x_e$ we have $1/J_D = \ln(x_e/r_w') - 1.311$ (Inflow to a hydraulically fractured well). We write $x_e/r_w' = (2/I_x)\cdot(x_f/r_w')$ and substitute $I_x^2 = N_{prop}/F_{CD}$ :

Productivity of a short fracture through the proppant number

\frac{1}{J_D} = \ln2 - 1.311 - \frac12\ln N_{prop} + \left[\frac12\ln F_{CD} + \ln\frac{x_f}{r_w'}\left(F_{CD}\right)\right]

where:

$\ln(x_f/r_w')(F_{CD})$ the universal curve of the effective wellbore radius of a fracture of finite conductivity, —
$N_{prop},\ F_{CD}$ proppant number and dimensionless conductivity, —

Units: dimensionless; natural logarithms

Holds when

The fracture is small compared with the area ( $I_x$ is below about 0.3), the regime is pseudo-steady-state, and the area is a square.

At a constant $N_{prop}$ , only the expression in square brackets depends on the choice of the fracture. The first term grows with $F_{CD}$ (the fracture is shorter), and the second falls (the fracture carries fluid better). The sum has a minimum. It does not depend on $N_{prop}$ . This is why all short fractures have the same optimum.

The expression in square brackets by the model of the lesson on inflow to a fractured well

$F_{CD}$: 0.50
$\ln(x_f/r_w')$: 2.151
Sum in the brackets: 1.804

$F_{CD}$: 1.00
$\ln(x_f/r_w')$: 1.645
Sum in the brackets: 1.645

$F_{CD}$: 1.69
$\ln(x_f/r_w')$: 1.344
Sum in the brackets: 1.606

$F_{CD}$: 3.00
$\ln(x_f/r_w')$: 1.102
Sum in the brackets: 1.651

$F_{CD}$: 10.00
$\ln(x_f/r_w')$: 0.831
Sum in the brackets: 1.982

$F_{CD}$	$\ln(x_f/r_w')$	Sum in the brackets
0.50	2.151	1.804
1.00	1.645	1.645
1.69	1.344	1.606
3.00	1.102	1.651
10.00	0.831	1.982

The minimum of the sum is about 1.606 at $F_{CD} \approx 1.69$ . Together with $\ln2 - 1.311 = -0.618$ this gives the constant 0.989 in the formula $1/J_{D,\max} = 0.99 - \tfrac12\ln N_{prop}$ . Near the optimum the sum changes by hundredths: the optimum is flat.

Connections to other topics

Hydraulic fracturing: purpose and well selection — dimensionless conductivity, equivalent skin and the choice of wells for hydraulic fracturing.
Inflow to a hydraulically fractured well — inflow to a fracture in a bounded area and over time, the basis of the calculation by the proppant number.
Rock mechanics for hydraulic fracturing — closure pressure, net pressure, the G function and holding the fracture in height.
Fracturing fluids and tip-screenout fracturing — fracturing fluids and deliberate stopping of fracture growth for short, wide fractures.
Inflow at high rates: non-Darcy flow — the Forchheimer equation and the inertial resistance coefficient.
Sand control methods — hydraulic fracturing with a gravel pack in weak permeable sandstones.
Net present value (NPV) — choosing the size of a treatment by net present value.
Diagnosis with the pressure derivative — how to find the working length and conductivity of a fracture from a test after hydraulic fracturing.

Check yourself

1
At a proppant number of 0.02, we decided to double the amount of proppant (the same pack permeability, in the same reservoir). How will the greatest dimensionless productivity index change?
1. A.It will rise by about 13%
2. B.It will not change: the optimal conductivity is the same
3. C.It will rise by a factor of $\sqrt2$
4. D.It will double
Show the answer
Correct answer: A. It will rise by about 13%
At a small proppant number $1/J_{D,\max} = 0.99 - \tfrac12\ln N_{prop}$ . At 0.02 this is $0.99 + 1.956 = 2.946$ , and $J_D = 0.339$ . At 0.04 it is $0.99 + 1.609 = 2.599$ , and $J_D = 0.385$ . The gain is about 13%: the productivity grows logarithmically, and the size of the treatment is chosen by economics.
2
The closure pressure in a reservoir is 48 MPa. For the first two years the well will flow naturally at a bottomhole pressure of 30 MPa, and then it will be put on a pump with a bottomhole pressure of 12 MPa. By then the reservoir pressure will have fallen by 6 MPa, and the poroelastic coefficient is 0.6. For what load should we choose the proppant, and will quartz sand do?
Show the answer
In the first years the load is $48 - 30 = 18$ MPa. After depletion the closure pressure will fall by $0.6\cdot6 = 3.6$ MPa, to 44.4 MPa, and at a bottomhole pressure of 12 MPa the load becomes $44.4 - 12 = 32.4$ MPa, almost twice as much. We choose the proppant for this greatest load. Sand starts to crush at 30–40 MPa, so we need resin-coated sand or ceramic. The choice is decided by comparing the productivity and the price.
3
A reservoir 15 m thick lies between thick shales with a stress contrast of 6 MPa. The fracture must grow to a half-length of 120 m. Which two-dimensional model fits, and how will the net pressure behave while the fracture grows normally?
1. A.PKN; the pressure will rise slowly
2. B.Any: all the models give the same
3. C.Radial; the pressure will be constant
4. D.KGD; the pressure will fall
Show the answer
Correct answer: A. PKN; the pressure will rise slowly
The fracture is much longer than its height, and the height is held by barriers. These are the conditions of the PKN model. The width at the wellbore grows with the length as $x_f^{1/4}$ , so the net pressure grows too. A fall of pressure in such a fracture would show that a barrier has been broken through, and a sharp rise would show that the proppant has run into the end of the fracture.
4
After a calibration injection the fracture closed at $G_c = 2$ . Estimate the efficiency of this injection and the pad fraction, if the main injection will have the same efficiency. Why in practice do we take a larger pad?
Show the answer
Since $0.589\cdot2 = 1.178$ , the efficiency is $\eta = 1.178/2.178 \approx 0.54$ , and the pad is $\varepsilon = 0.46/1.54 \approx 0.30$ , about 30% of the volume. The main injection is longer than the calibration one, and its efficiency is lower, and a lower efficiency calls for a larger pad. So we recalculate the efficiency of the main injection with a model that uses the leak-off coefficient found from the calibration injection, and do not carry it over directly.
5
The apparent viscosity of the fluid in the fracture was doubled at the same rate. How will the width of the fracture at the wellbore and the net pressure change by the PKN model?
1. A.The width by 41%, the pressure will not change
2. B.The width by 19%, the pressure by 19%
3. C.The width by a factor of two, the pressure by a factor of two
4. D.The width will not change, the pressure by a factor of two
Show the answer
Correct answer: B. The width by 19%, the pressure by 19%
$w_w \propto (\mu q_i x_f/E')^{1/4}$ , so the width grows by a factor of $2^{1/4} = 1.19$ . The net pressure $E'w_w/(2h_f)$ is proportional to the width and grows in the same way. The rise of pressure can take the fracture out of the reservoir in height, so viscosity is a poor way to get width.
6
The same volume of proppant as in the lesson's example is going to be pumped into a 100 mD reservoir of the same thickness and with the same well pattern. What will the optimal fracture be, and why is it hard to make by ordinary injection?
Show the answer
The proppant number falls 100 times, to 0.008. At this value the optimum is $F_{CD} \approx 1.7$ . By the formula for the dimensions, the expression under the root for the half-length is $150{,}000\cdot16.4 = 2.46\cdot10^6$ , divided by $2\cdot1.7\cdot100\cdot24 = 8160$ , that is, about 300 m², and $x_f \approx 17$ m. The width in the reservoir is $16.4/(2\cdot17\cdot24) \approx 0.02$ m, about 2 cm. Ordinary injection does not make such a width. We get it by stopping the growth of the fracture in length with proppant at its end and continuing to pump to widen it. This is the technique of short, wide fractures (Fracturing fluids and tip-screenout fracturing), and in weak sandstones it is combined with a gravel pack.

Updated: 10 October 2026