Fracturing fluids and tip-screenout fracturing

Quick refresher

A fracturing fluid must create width, carry proppant to the end of the fracture, hold back leak-off, and avoid high friction in the pipes. After injection it must break down and flow out without leaving damage. These requirements contradict each other, and every fluid is a compromise.
Polymer solutions thin out under shear: the faster they are sheared, the thinner they get. By the power-law model the apparent viscosity is $\mu_a = K\dot\gamma^{\,n-1}$ . So the same fluid in the pipes is several times thinner than in a widened fracture, and more than a hundred times thinner than around a slowly settling grain.
The leak-off coefficient is made up of three resistances in series: the filter cake (a layer of polymer and solids) on the wall ( $C_w$ ), the zone invaded by the filtrate ( $C_v$ ), and the compression of the reservoir fluid ( $C_c$ ). With crosslinked gels the cake usually decides, and with slickwater (water with a friction reducer) the reservoir does.
In water a 20/40 grain (a grain between 0.42 and 0.84 mm) settles at about 0.1 m/s, in a linear gel at millimetres per second, and in a crosslinked gel at hundredths of a millimetre per second. So water carries proppant only by its flow and drops it in a bank (a heap on the bottom of the fracture), while a crosslinked gel holds the grains in suspension.
In a tip screen-out the proppant at the end of the fracture blocks it, the length no longer grows, and all further pumping goes into width and leak-off. The net pressure rises steeply, and this is how the screen-out is recognised. The width is limited by how well the fracture stays within the reservoir in height.
The pad for the screen-out is calculated for the moment when the fracture reaches the needed length: it must go into the reservoir exactly by that time. After the screen-out the leak-off goes through an unchanged area and slows down.

Key relationships

\mu_a = K\dot\gamma^{\,n-1},\quad \dot\gamma_{\text{slot}} = \dfrac{2n+1}{3n}\cdot\dfrac{6v}{w}

apparent viscosity and the shear rate at the wall of a slot

\dfrac{V}{A} = S_p + 2C_w\sqrt t

fluid-loss test: spurt loss and the cake coefficient

C_v = \sqrt{\dfrac{k\phi\Delta p}{2\mu_f}},\quad C_c = \Delta p\sqrt{\dfrac{k\phi c_t}{\pi\mu_r}}

leak-off limited by the filtrate and by the compression of the reservoir

v_s = \left[\dfrac{g\Delta\rho\,d^{\,n+1}}{18\cdot 3^{\,n-1}K}\right]^{1/n}

settling of a grain in a power-law fluid

\Delta V_L = 2A_lC_L\sqrt{t_{\text{so}}}\left[g(\delta) - \dfrac{\pi}{2}\right]

leak-off after the screen-out of the fracture tip

Common pitfalls

Describing a fluid by one viscosity. Without the shear rate the number is meaningless: in the pipes, in the fracture and around a grain the viscosity of one and the same fluid differs by orders of magnitude.
Carrying the laboratory leak-off coefficient over to the reservoir without calibration. The reservoir has natural fractures and interlayers, and we refine the coefficient from a calibration injection.
Taking slickwater for a permeable reservoir. The leak-off is huge, the efficiency is almost zero, and the proppant settles near the wellbore.
Expecting a screen-out with an excessive pad. The fracture will simply grow longer and thinner, and the needed width will not come.
Raising the net pressure after the screen-out without looking at the stress contrast. The fracture will go in height into the shales or towards water.

After this lesson you can

list the requirements for a fracturing fluid, and compare the main fluids (water with a friction reducer, linear and crosslinked gels, viscoelastic surfactants, foams) and the breakers (chemicals that thin the gel after pumping)
find the parameters of the power-law model (a simple formula for fluids that get thinner the faster they are sheared) from two points of the rheology curve (stress against shear rate), and calculate the apparent viscosity in the pipes, in the fracture and around a settling grain
split the leak-off coefficient into the filter cake, the filtrate (the liquid that passes through the cake into the rock) and the compressibility of the reservoir, find the cake coefficient and the spurt loss (the fluid that escapes at once, before the cake forms) from a laboratory test, and combine the mechanisms
estimate the settling velocity of proppant in water, in a linear gel and in a crosslinked gel, and explain how the fluid carries the proppant
calculate hydraulic fracturing with a tip screen-out: the pad and the time of the screen-out, the volume pumped to widen the fracture, the growth of the net pressure and the concentration of the slurry
make a plan for quality control of the fluid and the proppant before and during injection

What a fracturing fluid must do

How a fluid opens a fracture, and why the width depends on its viscosity only to the power ¼, is explained in the lessons on rock mechanics and on fracture design. There the fluid enters the calculation as two numbers, the apparent viscosity and the leak-off coefficient. Here we look at where these numbers come from and how the fluid itself is chosen.

In the pipes the fluid must flow easily. At rates of several cubic metres per minute the friction in the tubing is from a few to tens of megapascals, and a fluid that is thick in the pipes can make injection impossible. The shear in the pipes must not destroy its structure.
In the fracture it must be the opposite, and be viscous. The viscosity gives the net pressure and a width that is enough for the grains. The filter cake on the walls and low leak-off give the length. There the fluid must also carry the proppant to the end of the fracture and hold it in suspension until the fracture closes. Otherwise the grains settle to the bottom, and only part of the fracture will be propped.
After injection it must disappear. The viscosity must fall almost to that of water, so that the fluid flows out of the fracture. Polymer residue reduces the conductivity of the pack (conductivity in the reservoir), and the filtrate must not damage the reservoir at the walls.
All the time it must keep its properties at reservoir temperature, must not react with the rock and the reservoir fluids to their harm, and must be safe to handle and cheap.

The requirements argue with each other. Viscosity is needed in the fracture, but it gets in the way in the pipes. The cake reduces leak-off, but it stays on the walls. Stability at high temperature makes it harder to break the fluid down later. So fracturing fluids are systems: a base, a thickener, a crosslinker, a breaker of viscosity, and additives. They are chosen so that each property switches on in the right place and at the right time.

Types of fluids and additives

Most fracturing fluids are water-based. The thickener is the natural polymer guar or its derivatives (hydroxypropyl guar and its carboxymethyl derivative), and less often cellulose derivatives. Its loading is usually given in kilograms per cubic metre, from 2 to 5 kg/m³. In field practice it is given in pounds per 1000 gallons: 1 pound per 1000 gallons is 0.1198 kg/m³, so 30 pounds is about 3.6 kg/m³. A polymer solution is a linear gel. If we add a crosslinker, the polymer molecules join each other, and the viscosity grows tens of times. The result is a crosslinked gel.

The main fracturing fluids

Fluid: Slickwater
What it is made of: water with a friction reducer, about a litre of polyacrylamide per cubic metre
Where it is used: tight and shale rocks, large volumes at high rates
Limitations: hardly holds proppant; large leak-off

Fluid: Linear gel
What it is made of: a solution of guar or its derivatives
Where it is used: flushes, low proppant concentrations, low temperatures
Limitations: moderate viscosity that falls with temperature

Fluid: Borate crosslinked gel
What it is made of: guar and borates at pH 9–10
Where it is used: the main fluid at moderate temperatures
Limitations: the crosslinking is reversible: it is broken by shear and recovers

Fluid: Crosslinked gel with zirconium and titanium salts
What it is made of: guar derivatives and metal salts
Where it is used: high temperatures
Limitations: the crosslinking is irreversible: shear in the pipes breaks it

Fluid: Viscoelastic surfactants
What it is made of: a surfactant in brine, no polymer
Where it is used: permeable reservoirs where a clean pack matters
Limitations: large leak-off, a temperature limit, more expensive

Fluid: Foams and fluids with nitrogen or CO₂
What it is made of: gel and gas
Where it is used: depleted and water-sensitive reservoirs, fast flowback (return of the fluid to the surface)
Limitations: harder to run; the density changes with pressure

Fluid: Thickened hydrocarbons
What it is made of: oil or diesel fuel with a thickener
Where it is used: rocks sensitive to water
Limitations: fire hazard, cost

Fluid	What it is made of	Where it is used	Limitations
Slickwater	water with a friction reducer, about a litre of polyacrylamide per cubic metre	tight and shale rocks, large volumes at high rates	hardly holds proppant; large leak-off
Linear gel	a solution of guar or its derivatives	flushes, low proppant concentrations, low temperatures	moderate viscosity that falls with temperature
Borate crosslinked gel	guar and borates at pH 9–10	the main fluid at moderate temperatures	the crosslinking is reversible: it is broken by shear and recovers
Crosslinked gel with zirconium and titanium salts	guar derivatives and metal salts	high temperatures	the crosslinking is irreversible: shear in the pipes breaks it
Viscoelastic surfactants	a surfactant in brine, no polymer	permeable reservoirs where a clean pack matters	large leak-off, a temperature limit, more expensive
Foams and fluids with nitrogen or CO₂	gel and gas	depleted and water-sensitive reservoirs, fast flowback (return of the fluid to the surface)	harder to run; the density changes with pressure
Thickened hydrocarbons	oil or diesel fuel with a thickener	rocks sensitive to water	fire hazard, cost

The limits of use are approximate and depend on the composition of the particular system. Crosslinkers are often added with a delay, so that the gel crosslinks closer to the bottom of the well. In the pipes the fluid stays liquid, and the friction and the damage by shear are smaller.

A breaker is an additive that breaks down the polymer after injection. Oxidisers (persulfates) work at raised temperature. So that they do not break the gel too early, they are put in a shell that dissolves slowly. Enzymes break the polymer at moderate temperatures and in a narrow range of pH. For a borate gel the viscosity is also removed by lowering the pH, and for viscoelastic surfactants by contact with hydrocarbons or a solvent. The dose is a compromise. Too little breaker leaves polymer residue in the pack. Too much, and the gel breaks in the fracture before it closes, and the proppant settles.

The other additives solve particular problems. A biocide stops bacteria from breaking down the polymer in the tanks and from infecting the reservoir. A clay stabiliser (potassium chloride, ammonium salts) keeps the clays from swelling. A buffer holds the pH of crosslinking. Non-emulsifying surfactants stop emulsions from forming. Fluid-loss additives, which are fine particles and starch, cut the leak-off. Stabilisers protect the polymer at high temperature. Each additive is checked for compatibility with the others. A breaker put in the wrong place, or a biocide that disturbs crosslinking, can ruin the treatment.

Rheology: the power-law model and the shear rate

Polymer solutions are non-Newtonian: their viscosity falls as the shear rate (how fast the layers of fluid slide past one another) grows. In the range that matters for hydraulic fracturing they are described by the power-law model $\tau = K\dot\gamma^{\,n}$ (power-law model), with a flow behaviour index $n < 1$ . The rheology of a fracturing fluid is measured not with a simple viscometer but with a rheometer under pressure, at reservoir temperature and over the whole injection time, because crosslinked gels change with time and temperature. We find the parameters from two points of the curve, or from the whole curve by least squares:

Parameters of the power-law model from two points, and the apparent viscosity

n = \frac{\ln(\tau_2/\tau_1)}{\ln(\dot\gamma_2/\dot\gamma_1)},\qquad K = \frac{\tau_1}{\dot\gamma_1^{\,n}},\qquad \mu_a = K\dot\gamma^{\,n-1}

where:

$\tau_1,\ \tau_2$ shear stresses at the shear rates $\dot\gamma_1$ and $\dot\gamma_2$ , Pa
$\dot\gamma$ shear rate, s⁻¹
$n,\ K$ flow behaviour index and consistency index, —, Pa·sⁿ
$\mu_a$ apparent viscosity at the shear rate $\dot\gamma$ , Pa·s

Units: SI; $K$ in Pa·sⁿ (in oilfield units lbf·sⁿ/ft², with 1 lbf·sⁿ/ft² = 47.88 Pa·sⁿ)

Holds when

The power-law model holds in the range of the measurements. At very low shear rates the viscosity of a real fluid stops growing, and extrapolation overstates it.
The points are taken at reservoir temperature, after a hold that matches the injection time.

The smaller $n$ is, the stronger the thinning under shear. At $n = 0.45$ a tenfold rise of the shear rate lowers the viscosity by a factor of $10^{0.55} \approx 3.5$ .

To take the viscosity “in the right place”, we must know the shear rate there. For a slot of width $w$ and a pipe of diameter $d$ , the shear rate at the wall at an average velocity $v$ is:

Shear rate at the wall of a slot and of a pipe for a power-law fluid

\dot\gamma_{\text{slot}} = \frac{2n+1}{3n}\cdot\frac{6v}{w},\qquad \dot\gamma_{\text{pipe}} = \frac{3n+1}{4n}\cdot\frac{8v}{d}

where:

$\dot\gamma_{\text{slot}},\ \dot\gamma_{\text{pipe}}$ shear rate at the wall of a slot (a fracture) and of a pipe, s⁻¹
$v$ average velocity of the fluid: the rate divided by the area of the cross-section, m/s
$w,\ d$ width of the slot and inside diameter of the pipe, m
$n$ flow behaviour index, —

Units: SI

Holds when

The flow is laminar. The slot is two parallel walls (the cross-section of a fracture is really an ellipse, and the formula gives a characteristic shear rate).
At $n = 1$ we get the Newtonian $6v/w$ and $8v/d$ (shear rate in a pipe). The factors take into account the flatter velocity profile of a thinning fluid.

Each wing of the fracture (one half of it, on one side of the well) gets half of the rate. A narrow fracture at a high rate means hundreds of reciprocal seconds, and a widened one tens of them. In the pipes it is hundreds and thousands.

Worked exampleCrosslinked gel: the viscosity in the pipes and in the fracture

Given (assumed data)

Rheometer at reservoir temperature: at 40 s⁻¹ the stress is 26.0 Pa, at 170 s⁻¹ it is 50.0 Pa
Injection: 2.4 m³/min (0.04 m³/s) through tubing with an inside diameter of 100.5 mm
Fracture (the screen-out example below): height 20 m; average width before the screen-out 3.22 mm, at the end of injection 21.33 mm

Solution

The parameters: $n = \dfrac{\ln(50.0/26.0)}{\ln(170/40)} = 0.452$ , $K = 26.0/40^{0.452} = 4.91$ Pa·sⁿ.
In the pipes: $v = \dfrac{0.04}{\pi\cdot 0.1005^2/4} = 5.042$ m/s, $\dot\gamma = \dfrac{3\cdot0.452 + 1}{4\cdot0.452}\cdot\dfrac{8\cdot5.042}{0.1005} = 523$ s⁻¹, $\mu_a = 4.91\cdot523^{0.452 - 1} = 0.159$ Pa·s.
In the fracture before the screen-out: each wing gets 0.02 m³/s, $v = \dfrac{0.02}{20\cdot 0.00322} = 0.3106$ m/s, $\dot\gamma = \dfrac{2\cdot0.452 + 1}{3\cdot0.452}\cdot\dfrac{6\cdot0.3106}{0.00322} = 813$ s⁻¹, $\mu_a = 0.125$ Pa·s.
In the fracture at the end of injection: $v = \dfrac{0.02}{20\cdot 0.02133} = 0.0469$ m/s, $\dot\gamma = 18.5$ s⁻¹, $\mu_a = 0.99$ Pa·s.

Answer: The parameters of the gel are

n

= 0.452 and

K

= 4.91 Pa·sⁿ. The apparent viscosity is 0.159 Pa·s in the pipes, 0.125 Pa·s in the narrow fracture and 0.99 Pa·s in the widened fracture.

What it means. In a narrow, short fracture at a high rate the shear rate is even higher than in the pipes, and the viscosity is lower. This viscosity, 0.125 Pa·s, is the one used to calculate the width before the screen-out. When the fracture has widened, the shear rate fell by a factor of 44, and the viscosity rose by a factor of 7.9. The gel became thicker exactly where it must hold a lot of proppant. A Newtonian fluid with a viscosity of 1 Pa·s would give several times more friction in the pipes. Thinning under shear and a delay of crosslinking make injection possible. For the pipes the calculation is an estimate: the generalised Reynolds number of the gel (Metzner–Reed; a number that tells whether the flow is smooth or turbulent), $\rho v d\big/\big(\mu_a\tfrac{3n+1}{4n}\big)$ at a density of 1020 kg/m³, is here about 2500. That is at the edge of laminar (smooth) flow, for which the formula for the shear rate is written. We take the friction of a crosslinked gel in pipes from bench measurements for that fluid. The data in this example are assumed for illustration.

CalculatedApparent viscosity of a crosslinked and a linear gel against the shear rate (power-law model; below 1 s⁻¹ it is an extrapolation, dashed) and the viscosity of water. The points on the curve of the crosslinked gel are the places from the example: around a settling grain, in the fracture after the screen-out, in the pipes, and in the narrow fracture before the screen-out. The data are assumed for illustration.

Leak-off: what the leak-off coefficient is made of

Carter's law (volume balance with leak-off) describes leak-off with one leak-off coefficient $C_L$ : the filtration rate through the wall is $C_L/\sqrt{t}$ . The square root of time appears when the resistance to filtration grows with the volume of the fluid that has gone through. There are three such resistances, and they are in series: the polymer cake on the wall, the zone of the reservoir filled with filtrate (the liquid that has passed through the cake), and the reservoir fluid behind it, which has to be compressed or pushed back.

Cake ( $C_w$ , wall-building): The polymer and the particles of the fluid-loss additive settle on the wall. The cake grows with the volume of filtrate, and its resistance grows with it. Before the cake forms, a spurt loss $S_p$ goes through. We find the coefficient from a fluid-loss test.
Filtrate ( $C_v$ ): The liquid that has passed through the cake pushes back the reservoir fluid. The deeper the filtrate zone, the greater its resistance. It matters when the filtrate is viscous (viscoelastic surfactants, thickened hydrocarbons) or when there is no cake.
Reservoir ( $C_c$ ): The filtrate compresses the reservoir fluid and pushes it deeper. This resistance is set by the permeability, the compressibility and the viscosity of the reservoir fluid. It matters for fluids without a cake, in reservoirs with a barely compressible viscous fluid.

Fluid-loss test: the cake coefficient and the spurt loss

\frac{V}{A} = S_p + 2C_w\sqrt t,\qquad C_w = \frac{m}{2A},\qquad S_p = \frac{b}{A}

where:

$V$ volume of filtrate by the time $t$ , m³ (in the test, cm³)
$A$ filtration area: the end face of the core sample, m² (cm²)
$m,\ b$ slope and intercept of the line “filtrate volume against square root of time”, m³/s $^{1/2}$ , m³
$C_w$ leak-off coefficient set by the cake, m/s $^{1/2}$
$S_p$ spurt loss per unit area, m³/m² = m

Units: any consistent units: $C_w$ comes out in units of length per square root of the unit of time

Holds when

The leak-off rate through the cake is $C_w/\sqrt t$ . This gives the factor 2 on integration: $\int_0^t C_w\,dt'/\sqrt{t'} = 2C_w\sqrt t$ .
The pressure drop in the test is close to that in the reservoir (otherwise the coefficient is recalculated; for polymer cakes it grows with the pressure drop more slowly than the square root of it). The sample is of the same rock and permeability.

A straight line in the coordinates “volume against square root of time” means that the cake holds the leak-off back. The intercept on the volume axis is what went through before the cake formed. The standard filtration test of a drilling fluid is built in the same way (filtration of a mud).

Leak-off coefficients limited by the filtrate and by the compression of the reservoir, and their combination

C_v = \sqrt{\frac{k\,\phi\,\Delta p}{2\mu_f}},\qquad C_c = \Delta p\sqrt{\frac{k\,\phi\,c_t}{\pi\,\mu_r}},\qquad C_L = \frac{2C_cC_wC_v}{C_wC_v + \sqrt{C_w^2C_v^2 + 4C_c^2\left(C_w^2 + C_v^2\right)}}

where:

$C_v,\ C_c,\ C_L$ leak-off coefficients: by the filtrate, by the compression of the reservoir, and the overall one, the coefficient of Carter's law, m/s $^{1/2}$
$k,\ \phi$ permeability and porosity of the reservoir, m², —
$\Delta p$ pressure difference between the fracture and the reservoir, Pa
$\mu_f,\ \mu_r$ viscosity of the filtrate and of the reservoir fluid, Pa·s
$c_t$ total compressibility of the reservoir fluid and the pores, 1/Pa

Units: SI ( $k$ in m²: 1 mD = $0.987\cdot10^{-15}$ m²). In oilfield units the coefficients are more often given in ft/min $^{1/2}$ : 1 m/s $^{1/2}$ = 25.4 ft/min $^{1/2}$

Holds when

The filtration into the reservoir is linear (one-dimensional) at a constant pressure drop. The filtrate displaces the reservoir fluid like a piston.
The overall coefficient is an exact combination of three resistances in series at one common pressure drop, if the coefficients of the cake and of the filtrate grow as the square root of their share of the pressure drop, and the compression coefficient grows in proportion to it. The approximation $1/C_L = 1/C_w + 1/C_v + 1/C_c$ is also often used. It always gives a smaller value (in the example below, by 15%).
The derivation of the formulas is in the advanced section.

The smallest of the three numbers decides. The cake usually holds the leak-off of a crosslinked gel back most strongly. Slickwater has no cake, and the leak-off is limited by the reservoir, through the compression of the fluid. The higher the permeability, the larger $C_v$ and $C_c$ , and the more the cake matters.

Worked exampleLeak-off coefficient of a crosslinked gel in a permeable sandstone

Given (assumed data)

Fluid-loss test: a sandstone core disc, area 5.0 cm², pressure drop 10 MPa; the filtrate after 1, 4, 9, 16, 25 and 36 min: 4.9, 8.7, 12.6, 16.5, 20.4 and 24.2 cm³
Reservoir: $k$ = 300 mD, $\phi$ = 0.28, $c_t = 1.5\cdot10^{-3}$ 1/MPa, oil viscosity 3.0 mPa·s; filtrate viscosity 1.0 mPa·s; pressure difference between the fracture and the reservoir 10 MPa

Solution

A line through the points at 4 and 36 min (the others lie on it to within 0.1 cm³): the slope is $m = \dfrac{24.2 - 8.7}{6 - 2} = 3.875$ cm³/min $^{1/2}$ , and the intercept is $b = 8.7 - 2\cdot3.875 = 0.95$ cm³.
Cake: $C_w = \dfrac{3.875}{2\cdot 5.0} = 0.3875$ cm/min $^{1/2}$ $= \dfrac{0.3875\cdot10^{-2}}{\sqrt{60}} = 5.00\cdot10^{-4}$ m/s $^{1/2}$ (0.0127 ft/min $^{1/2}$ ). The spurt loss is $S_p = 0.95/5.0 = 0.19$ cm, which is 1.9 litres per square metre of wall.
Filtrate: $C_v = \sqrt{\dfrac{300\cdot 0.987\cdot10^{-15}\cdot 0.28\cdot 10\cdot10^{6}}{2\cdot 1.0\cdot10^{-3}}} = 2.04\cdot10^{-2}$ m/s $^{1/2}$ .
Reservoir: $C_c = 10\cdot10^{6}\sqrt{\dfrac{300\cdot 0.987\cdot10^{-15}\cdot 0.28\cdot 1.5\cdot10^{-9}}{\pi\cdot 3.0\cdot10^{-3}}} = 1.15\cdot10^{-3}$ m/s $^{1/2}$ .
Together: $C_wC_v = 1.020\cdot10^{-5}$ , the root $\sqrt{C_w^2C_v^2 + 4C_c^2(C_w^2 + C_v^2)} = 4.803\cdot10^{-5}$ , and $C_L = \dfrac{2\cdot1.15\cdot10^{-3}\cdot5.00\cdot10^{-4}\cdot2.04\cdot10^{-2}}{(1.020 + 4.803)\cdot10^{-5}} = 4.03\cdot10^{-4}$ m/s $^{1/2}$ (0.0102 ft/min $^{1/2}$ ). The approximation by the sum of reciprocals gives 3.43·10⁻⁴.
Slickwater (no cake): $\dfrac{2C_vC_c}{C_v + \sqrt{C_v^2 + 4C_c^2}} = 1.15\cdot10^{-3}$ m/s $^{1/2}$ , which is 2.9 times larger.
In 10 min, through a square metre of wall (one side), the gel loses $S_p + 2C_L\sqrt{600} = 21.6$ L, and water loses $2\cdot1.15\cdot10^{-3}\cdot\sqrt{600} = 56.3$ L.

Answer: The overall leak-off coefficient of the gel is about

4.03\cdot10^{-4}

m/s

^{1/2}

, and the spurt loss is 1.9 L/m². The cake holds the leak-off back. Water without a cake would go into the reservoir 2.6 times faster.

What it means. The cake gives the smallest of the three numbers, and the overall coefficient is only 19% less than the cake's own. For a permeable reservoir this is a lot: about eight times the coefficient from the calibration injection in a millidarcy sandstone (example of the lesson on design). So we do not pump slickwater into such reservoirs, and we add fluid-loss additives to the gel. A laboratory test gives an estimate. We refine the working value with a calibration injection, which includes all the interlayers and natural fractures. The screen-out calculation below uses $C_L = 4.0\cdot10^{-4}$ m/s $^{1/2}$ and $S_p$ = 1.9 L/m². The data in this example are assumed for illustration.

CalculatedThe fluid-loss test of the example: the volume of filtrate against the square root of time. The points are measurements, and the line goes through the points at 4 and 36 min. Its slope gives the cake coefficient, and its intercept on the volume axis gives the spurt loss. The data are assumed for illustration.

Proppant transport

A proppant grain is 2.5–3.5 times heavier than the fluid, and it settles while the fracture is open. If it settles before the fracture closes, the proppant collects at the bottom of the fracture, the top stays unpropped, and the working height is less than the created one. How fast a grain settles depends on the viscosity of the fluid, but at the shear rate that the settling itself creates, and that is very low. For a Newtonian fluid at low Reynolds numbers Stokes' law applies. For a power-law fluid we generalise it by estimating the shear rate around the grain as $3v_s/d$ :

Settling velocity of a grain in a Newtonian and in a power-law fluid

v_s = \frac{g\,\Delta\rho\,d^2}{18\,\mu},\qquad v_s = \left[\frac{g\,\Delta\rho\,d^{\,n+1}}{18\cdot 3^{\,n-1}K}\right]^{1/n}

where:

$v_s$ settling velocity of a single grain, m/s
$g$ acceleration of gravity, m/s²
$\Delta\rho$ difference between the densities of the grain and of the fluid, kg/m³
$d$ diameter of the grain, m
$\mu$ viscosity of a Newtonian fluid, Pa·s
$n,\ K$ parameters of the power-law model, —, Pa·sⁿ

Units: SI

Holds when

A single spherical grain at a low Reynolds number, $\rho_f v_s d/\mu \ll 1$ . The walls of the fracture are far away, and the fluid is at rest.
The second formula is Stokes' law with the apparent viscosity at the shear rate $3v_s/d$ . At $n = 1$ it coincides with the first. It is an estimate: at very low shear rates the real viscosity stops growing, and the formula understates the settling velocity.
In a concentrated slurry the grains hinder each other and settle more slowly. By the empirical law of Richardson and Zaki, at low Reynolds numbers the velocity is multiplied by about $(1 - c_v)^{4.65}$ , where $c_v$ is the volume fraction of proppant.

The settling velocity grows as the square of the grain diameter and in proportion to the density difference: a large heavy proppant settles much faster than a small light one. For a power-law fluid, a small $n$ makes the dependence on size even stronger.

Worked exampleHow fast a grain settles in water, in a linear gel and in a crosslinked gel

Given (assumed data)

Grain: 20/40 proppant, assumed diameter 0.6 mm, density 2700 kg/m³; fluid 1020 kg/m³; $\Delta\rho$ = 1680 kg/m³
Fluids: slickwater, 1.0 mPa·s; linear gel with $n$ = 0.60 and $K$ = 0.36 Pa·sⁿ; crosslinked gel with $n$ = 0.452 and $K$ = 4.91 Pa·sⁿ (the rheology example)

Solution

Water by Stokes: $v_s = \dfrac{9.81\cdot 1680\cdot(0.6\cdot10^{-3})^2}{18\cdot 1.0\cdot10^{-3}} = 0.330$ m/s. But the Reynolds number is $1020\cdot0.330\cdot 0.0006/0.001 = 202$ , so Stokes' law does not apply here.
With a correction of the drag for inertia (the Schiller–Naumann correlation $C_D = \dfrac{24}{Re}(1 + 0.15\,Re^{0.687})$ , solved by trial and error): $v_s = 0.0955$ m/s at $Re = 58$ and $C_D = 1.42$ .
Linear gel: $v_s = \left[\dfrac{9.81\cdot 1680\cdot(0.6\cdot10^{-3})^{1.6}}{18\cdot 3^{-0.4}\cdot 0.36}\right]^{1/0.6} = 2.52\cdot10^{-3}$ m/s. The shear rate around the grain is $3v_s/d = 12.6$ s⁻¹, and the apparent viscosity is 0.13 Pa·s.
Crosslinked gel: $v_s = \left[\dfrac{9.81\cdot 1680\cdot(0.6\cdot10^{-3})^{1.452}}{18\cdot 3^{-0.548}\cdot4.91}\right]^{1/0.452} = 1.79\cdot10^{-5}$ m/s. The shear rate around the grain is 0.09 s⁻¹, and the apparent viscosity is about 18 Pa·s.
In the 30 min from injection to closure of the fracture, a grain in the linear gel sinks by 4.5 m, and in the crosslinked gel by 3.2 cm. In water it covers 10 m (half the height of a fracture of 20 m) in 105 s. At a flow velocity in the fracture of 0.3 m/s, the flow carries it 32 m from the wellbore in that time.

Answer: A 20/40 grain settles at about 0.0955 m/s in water, at 2.52 mm/s in the linear gel, and at about 0.018 mm/s in the crosslinked gel. The crosslinked gel holds the proppant until the fracture closes.

What it means. Water carries proppant only by its flow. The grains drop out tens of metres from the wellbore and form a bank, which the flow pushes and builds up. So slickwater is pumped at high rates and with low concentrations of fine proppant. A linear gel is good for short injections and low concentrations. For the crosslinked gel the estimate is optimistic: at a shear rate of 0.09 s⁻¹ the power-law model is outside the range of the measurements and overstates the viscosity. But the real settling in it is still orders of magnitude slower than in the linear gel. In a concentrated slurry the grains hinder each other: at a proppant volume fraction of 0.2 the settling is about 2.8 times slower. The data in this example are assumed for illustration.

Convection. A slurry with proppant is heavier than clean fluid, and it flows down the fracture as a whole, carrying the grains. This is faster than the settling of single grains, especially in tall fractures.
Walls. When the width of the fracture is only a few times the size of a grain, the walls slow the grain and hinder its passage. At the entrance to a narrow fracture the grains form arches. This is how a screen-out arises.
Elasticity. Crosslinked gels are elastic, and this helps hold the grains better than viscosity alone shows. So proppant transport is also checked on test rigs with narrow slots.
Time. The viscosity falls with temperature and with the action of the breaker. What matters is the viscosity not at the start but at the end of injection and until the fracture closes.

Stopping the growth of the fracture: why and how

In a permeable reservoir the optimal fracture is short and very wide. In a 100 mD reservoir the same proppant that gives a fracture of 150 m with a width of about 2 mm in a millidarcy reservoir must go into a fracture of about 17 m with a width of about 2 cm (proppant number). Ordinary injection does not give such a width: the width grows only as the fourth root of the length and the viscosity, and the net pressure is limited by how well the fracture stays within the reservoir in height. We get the width in another way, by stopping the growth of the fracture in length (tip screen-out).

The pad is made small, so that all of it goes into the reservoir exactly when the fracture reaches the needed length. The first slurry with proppant reaches the end of the fracture and loses its fluid into the reservoir there, and the dehydrated proppant blocks the end of the fracture like a plug. The pressure at the tip no longer rises, and the fracture stops lengthening. Injection continues, but the area of the walls no longer grows. All the fluid goes into width and leak-off, and the fracture inflates and is packed with proppant. By the end of injection the slurry in the fracture is dehydrated almost to a dense pack, and after closure a wide, densely filled fracture remains.

The sign of the screen-out is the rise of the net pressure. While the fracture grows in length, the net pressure rises slowly. After the screen-out the fracture works like a vessel of constant area that is being pumped up, and the pressure rises steeply: on log–log axes with a slope of about one or more (reading the net pressure). Two limits restrict the growth. One is the stress contrast with the shales: a net pressure that approaches it pushes the fracture out of the reservoir in height (equilibrium height). The other is the allowed pressure at the wellhead and on the casing.

A short wide fracture is needed where what decides is not the length but the conductivity at the wellbore and the area of inflow. It changes the inflow in two ways:

Velocity at the wall. The fluid enters the fracture along its whole wall, and does not converge on a few perforation holes. So the filtration velocity is small. There is less force tearing off the grains of a weak sandstone and the fine particles, and less drawdown at the same rate. In a high-rate gas well a wide dense pack reduces the inertial losses (inertial losses in the pack).
Connection with the reservoir. Even a fracture only ten metres long goes beyond the damaged zone at the wellbore, whatever created it (choosing between acid and a fracture), and in height it crosses a thin alternation of sands and shales. So it connects to the wellbore also the layers that the perforations missed.

The first of these properties made the tip screen-out the basis of the frac-pack (Sand control methods), hydraulic fracturing combined with a gravel pack. The fracture reduces the inflow velocity at the wellbore, and the gravel pack around the screen (a filter pipe) holds back the sand that does start to move.

Calculating hydraulic fracturing with a tip screen-out

The calculation goes in two stages. Before the screen-out the fracture grows as in the lesson on design: the width by the PKN model, the leak-off by Carter's law. In a permeable reservoir the leak-off is strong, so we take the limiting case of strong leak-off: the factor of the width at the wellbore is 2.68, the average width is 0.567 of it, and the leak-off factor is $\kappa = \pi/2$ (PKN width). We find the pad by Nolte's rule (pad), but for the moment of the screen-out and not for the end of injection. After the screen-out the area of the walls is constant. Leak-off continues, but it slows down: every part of the wall filters more and more slowly, and no new parts appear.

Leak-off and the volume of the fracture after its growth has stopped

\Delta V_L = 2A_lC_L\sqrt{t_{\text{so}}}\left[g(\delta) - \frac{\pi}{2}\right],\quad g(\delta) = \sqrt\delta + (1 + \delta)\arcsin\frac{1}{\sqrt{1 + \delta}},\quad \delta = \frac{t - t_{\text{so}}}{t_{\text{so}}}

where:

$\Delta V_L$ leak-off during the time after the screen-out, m³
$A_l$ area of the wall of the fracture against the permeable layers (one side, both wings), m²
$C_L$ leak-off coefficient, m/s $^{1/2}$
$t_{\text{so}}$ time from the start of injection to the screen-out of the fracture, s
$g(\delta)$ leak-off function; $g(0) = \pi/2$ , —

Units: SI

Holds when

Before the screen-out the area of the fracture grew as the square root of time (the limit of strong leak-off), and every part of the wall filters by Carter's law from the moment it opened.
After the screen-out the area does not change, and the spurt loss all happened before the screen-out.
The same function describes the leak-off after the pumps stop in a calibration injection: $g(\delta) - \pi/2 = \tfrac{\pi}{4}G(\delta)$ , where $G$ is the second limiting form of Nolte's function, for a fracture whose area grew as the square root of time (the G function is written there for growth in proportion to time). The derivation is in the advanced section.

The volume of the fracture after the screen-out grows as $V_f(t) = V_{f,\text{so}} + q_i(t - t_{\text{so}}) - \Delta V_L$ . Since the pressure along a short, wide fracture is almost the same everywhere, the width increases evenly along the length. The greatest width at the wellbore grows by $\Delta\bar w/(\pi/4)$ , and the net pressure grows in proportion to it: $p_{net} = E'w_w/(2h_f)$ .

Worked exampleFrac-pack: the pad, the screen-out time and the packing of the fracture

Given (assumed data)

Reservoir: weak sandstone, $k$ = 300 mD; sand layers with a total thickness of 15 m in an interval of 20 m between thick shales; stress contrast 6.0 MPa; $E'$ = 3000 MPa
Fluid: the crosslinked gel from the examples above: $\mu_a \approx 0.125$ Pa·s in a growing fracture, $C_L = 4.0\cdot10^{-4}$ m/s $^{1/2}$ , spurt loss 1.9 L/m²
Injection: 2.4 m³/min (0.04 m³/s); fracture height 20 m
Aim: half-length 12 m; propped width 16 mm; lightweight ceramic 2700 kg/m³, pack porosity 0.38; the first slurry is 120 kg of proppant per cubic metre of slurry; by the end of injection the slurry in the fracture is dehydrated to 75% of a dense pack (assumed)

Solution

The width at the time of the screen-out: $w_w = 2.68\left(\dfrac{0.125\cdot 0.04\cdot 12}{3000\cdot10^{6}}\right)^{1/4} = 5.67$ mm, and the average is $\bar w = 0.5674\cdot5.67 = 3.22$ mm. The volume of the fracture is $2\cdot12\cdot20\cdot3.22\cdot10^{-3} = 1.546$ m³.
Leak-off: the area against the sands is $A_l = 2\cdot12\cdot15 = 360$ m². The factor is $2A_l\cdot\tfrac{\pi}{2}\cdot C_L = 0.452$ m³/s $^{1/2}$ , and the spurt loss is $2\cdot360\cdot0.0019 = 1.368$ m³.
The balance $0.04\,t = 1.546 + 1.368 + 0.452\sqrt t$ : the discriminant is $0.452^2 + 4\cdot0.04\cdot(1.546 + 1.368) = 0.6705$ , $\sqrt t = (0.452 + \sqrt{0.6705})/0.08 = 15.89$ , and $t_{\text{so}} = 252$ s (4.20 min). The pumped volume is 10.08 m³, and the efficiency is $\eta = 1.546/10.08 = 0.153$ .
The pad: $\varepsilon = (1 - 0.153)/(1 + 0.153) = 0.735$ , $V_{pad} = 0.735\cdot10.08 = 7.41$ m³. The other 2.67 m³ before the screen-out is the first slurry, which carries $2.67\cdot120 = 320$ kg of proppant.
The aim: the loading is $C_p = 0.016\cdot0.62\cdot2700 = 26.8$ kg/m², and the proppant in the fracture is $480\cdot26.8 = 12{,}864$ kg. The average width at the end of injection is $16/0.75 = 21.33$ mm, and the volume of the fracture is $480\cdot21.33\cdot10^{-3} = 10.24$ m³.
After the screen-out: $2A_lC_L\sqrt{t_{\text{so}}} = 2\cdot360\cdot4.0\cdot10^{-4}\cdot\sqrt{252} = 4.572$ m³. By trial and error, the packing time is $\Delta t$ = 371.4 s (6.19 min): $\delta = 1.474$ , $g = 2.9187$ , the leak-off is $4.572\cdot(2.9187 - 1.5708) = 6.163$ m³, the pumped volume is $0.04\cdot371.4 = 14.856$ m³, and the volume of the fracture is $1.546 + 14.856 - 6.163 = 10.239$ m³.
The net pressure: at the screen-out it is $3000\cdot5.67\cdot10^{-3}/(2\cdot20) = 0.425$ MPa. The width at the wellbore grows by $(21.33 - 3.22)/0.7854 = 23.06$ mm to 28.73 mm, and $p_{net} = 2.15$ MPa. The equilibrium height at this pressure is $20/\cos(\pi\cdot2.15/12.0) = 23.6$ m.
After the screen-out the slurry carries $12{,}864 - 320$ kg in 14.856 m³: on average 844 kg per cubic metre of slurry, or 1228 kg per cubic metre of fluid, about 10.2 pounds per gallon.

Answer: The pad is 7.41 m³. The growth of the fracture stops at 4.20 min. Then about 14.9 m³ of slurry, with an average concentration of 844 kg/m³, packs the fracture in 6.19 min. In all 24.94 m³ is pumped in 10.4 min, and the net pressure rises from 0.425 to 2.15 MPa.

What it means. The efficiency at the screen-out is only 0.153: almost all of the fluid pumped before the screen-out goes into the reservoir. So the pad is three quarters of the volume before the screen-out. Without the screen-out the same injection would go on lengthening the fracture, and the width would hardly grow. By the end of injection 41% of the pumped volume is in the fracture: packing is more efficient than growth. The net pressure of 2.15 MPa is more than a third of the stress contrast, and the fracture goes into the shales by 1.8 m on each side. The growth in height will reduce the width and the pressure a little, compared with the calculation at constant height. The result is a fracture with a dimensionless conductivity of about 0.67 at a pack permeability of 150 D, and an equivalent skin of about −2.8. The productivity is 1.67 times higher than that of an undamaged well without a fracture. The PKN model for a fracture whose length is about its height is an estimate. Such a fracture is stiffer than PKN assumes, so the real net pressure at the same width is higher than calculated, and the entry into the shales is larger. In practice the calculation is run with a model calibrated against the data of the calibration injection, and the course of packing is read from the net pressure. The data in this example are assumed for illustration.

CalculatedThe net pressure at the wellbore during the injection of the example (logarithmic axes): slow growth while the fracture lengthens, and steep growth after its growth stops at 4.2 min. The dashed line is the same injection without a screen-out: the fracture goes on lengthening. Calculated with the lesson's model.

CalculatedThe width of the fracture averaged over the height, along the wing, in the example: at the moment the growth stops (at the tip the fracture narrows to nothing) and at the end of injection, when the fracture has widened almost evenly along its length. The dashed line is the propped width of 16 mm after closure. Calculated with the lesson's model.

A pad with a margin is the enemy of the screen-out. If the pad is larger than calculated, the proppant reaches the end of the fracture later, the fracture grows longer, and the packing time and the width are less. So we base the pad on a calibration injection run with the same fluid.
A screen-out at the wellbore. If the width at the wellbore is small (a tortuous entrance of the fracture, perforations not in the direction of the fracture), the slurry will get stuck at the entrance before it reaches the end. The pressure then rises even more steeply than at a screen-out at the end. We tell them apart by the time: a screen-out at the end comes at the calculated moment, when the first slurry has reached the end of the fracture.
The rate at the end. During packing the rate is often lowered in steps to control the rise of pressure, and the injection ends with the most concentrated slurry. In a frac-pack the ring around the screen is packed next (packing).

Quality control of the fluid and the proppant

The design is calculated for particular properties of the fluid and the proppant. If they are different, the whole chain falls apart. So we check the quality before injection, during it and after it.

What we check

1
Water
Temperature, pH, and the content of iron, salts and bacteria. They change the crosslinking and the breaking of the gel. We check the water in every tank.
2
Base gel
The viscosity of the linear gel on a viscometer (viscometer) after the polymer has hydrated. It shows whether the thickener is dosed correctly.
3
Crosslinking and breaking
A sample with the crosslinker: the time of crosslinking and the look of the gel. A sample with the breaker, held at reservoir temperature: after how many hours the gel breaks. A rheometer: the viscosity at reservoir temperature over the injection time.
4
Proppant
Sieve analysis (sieve analysis), crush strength, roundness and sphericity, and turbidity (the share of dust). Samples from every batch.
5
During injection
Flowmeters and densitometers (instruments that measure the density): the proppant concentration from the density of the slurry, the rates of all the additives, and the pressure. Samples of the slurry from the blender.
6
After injection
A comparison of what was pumped with the design, by volumes and by the mass of proppant. Samples of the fluid that flows back when the well is brought on production: has the gel broken. The pressure record, to fit the model.

Proppant concentration from the density of the slurry

c_v = \frac{\rho_{\text{sl}} - \rho_{\text{fl}}}{\rho_p - \rho_{\text{fl}}},\qquad c = c_v\,\rho_p,\qquad c_{\text{fl}} = \frac{c}{1 - c_v}

where:

$c_v$ volume fraction of proppant in the slurry, —
$\rho_{\text{sl}},\ \rho_{\text{fl}},\ \rho_p$ density of the slurry (from the densitometer), of the clean fluid, and of the proppant grains, kg/m³
$c,\ c_{\text{fl}}$ mass of proppant per cubic metre of slurry and per cubic metre of clean fluid, kg/m³

Units: SI. The concentration per cubic metre of fluid is given in field practice in pounds per gallon: 1 lb/gal = 119.8 kg/m³

Holds when

The density of the fluid is constant. Gas in the slurry (foams, nitrogen) is taken into account separately.

The densitometer on the high-pressure line is the main tool for controlling the concentration. It shows that the blender holds to the schedule and how much proppant has gone into the well.

Worked exampleHow much proppant is in the slurry, by the densitometer

Given (assumed data)

Densitometer: 1450 kg/m³ while pumping a stage designed for 950 kg of proppant per cubic metre of fluid
Fluid and proppant: 1020 and 2700 kg/m³

Solution

Volume fraction: $c_v = \dfrac{1450 - 1020}{2700 - 1020} = 0.2560$ .
Per cubic metre of slurry: $0.2560\cdot2700 = 691$ kg/m³. Per cubic metre of fluid: $691/(1 - 0.2560) = 929$ kg/m³, about 7.75 pounds per gallon.
The deviation from the schedule: $929/950 - 1 = -2.2$ %.

Answer: The slurry has 929 kg of proppant per cubic metre of fluid, which is 2.2% less than the designed 950.

What it means. A deviation of a few per cent is usually acceptable. A systematic shortfall means that less proppant will go into the fracture than calculated, and sharp jumps of density mean that the blender or the proppant feed is running unevenly. The data in this example are assumed for illustration.

Pressure, chemicals and rules

The Rules for ensuring industrial safety at hazardous production facilities of the oil and gas industries (order of the Minister for Investment and Development of 30 December 2014, No. 355) require stimulation to be run by a work organisation plan. The plan lists the reagents, their maximum permissible concentrations and their hazard classes (paragraphs 733–734). The lines must be pressure tested (paragraph 737), and a danger zone with a radius of at least 50 m must be set up (paragraph 739). In hydraulic fracturing, packers that protect the casing are used (paragraph 759).
The Unified Rules for the Rational and Integrated Use of Subsoil (order of the Minister of Energy of 15 June 2018, No. 239) require that the use of any stimulation method at each new field is preceded by experimental studies. These studies must justify the process parameters under which the casing and the cement sheath are preserved (paragraph 425). Hydraulic fracturing of accumulations with a thin seal (cap rock) is allowed only if an analysis by a specialised organisation scientifically proves that the risk of loss of tightness of the accumulations is excluded (paragraph 318). A set of measurements on the well is taken before and after hydraulic fracturing (paragraph 336).
The description in the lesson is for teaching, and it is not a work procedure. Hydraulic fracturing is run by trained crews under an approved plan, with pressure-tested equipment and a plan of action for complications.

Terms and symbols

Fracturing fluid: The fluid that opens the fracture and carries the proppant. It sets the width, the leak-off and the transport.
Linear gel: A polymer solution (guar or its derivatives) without a crosslinker.
Crosslinked gel: A polymer solution whose molecules are joined by a crosslinker, borates or metal salts. The viscosity is tens of times higher.
Slickwater: Water with a small addition of polyacrylamide that reduces friction. It hardly holds proppant.
Breaker: An additive that breaks down the polymer after injection, so that the fluid flows out of the fracture.
Parameters $n$ and $K$: The flow behaviour index and the consistency index of the power-law model $\tau = K\dot\gamma^{\,n}$ .
Coefficients $C_w$ , $C_v$ , $C_c$: Leak-off limited by the cake, by the filtrate and by the compression of the reservoir. The overall coefficient is their combination in series.
Spurt loss $S_p$: The volume that goes through a unit area of the wall before the cake forms.
Tip screen-out (TSO): Deliberate plugging of the end of the fracture by proppant, after which injection widens the fracture.
Packing of the fracture: The stage after the growth stops, when the fracture widens and fills with dehydrated slurry.

Common mistakes and misconceptions

Wrong

“The more viscous the fluid, the better: a wider fracture and more reliable transport.”

In fact

The width grows only as the fourth root of the viscosity, but the friction in the pipes, the net pressure and the growth of the fracture in height all grow, and polymer residue spoils the pack. Viscosity is needed where it works, in the fracture and around the grains. For this we have thinning under shear and a delay of crosslinking.

Wrong

“The leak-off coefficient is a property of the fluid, and it can be taken from the data sheet.”

In fact

It depends also on the reservoir (permeability, fluid, pressure drop) and on natural fractures. The laboratory cake coefficient is only one of three components. We calibrate the working value with a calibration injection.

Wrong

“Slickwater is good for any reservoir: it is cheap and clean.”

In fact

In a permeable reservoir it has no cake, the leak-off is several times larger, and the efficiency is almost zero. It carries proppant only by its flow. Its place is tight rocks, large volumes and high rates.

Wrong

“A sharp rise of pressure during injection always means an emergency screen-out.”

In fact

In a tip screen-out a sharp rise of pressure is the aim: the fracture has stopped lengthening and is widening. Time helps tell it from a screen-out at the wellbore: a screen-out at the end comes at the calculated moment, when the first slurry has reached the end of the fracture.

Wrong

“For a short wide fracture it is enough to halve the pad.”

In fact

We calculate the pad from the fluid efficiency at the moment the fracture reaches the needed length. A mistake in either direction ruins the treatment: too little gives a screen-out at the wellbore, and too much gives a long thin fracture.

In depth: advanced detail

Where the leak-off coefficient formulas come from

Filtrate. Let the filtrate of viscosity $\mu_f$ fill a zone of depth $L$ , and let the whole pressure drop $\Delta p$ be spent across it. By Darcy's law the filtration velocity is $u = k\Delta p/(\mu_f L)$ , and the depth grows as $\phi\,dL/dt = u$ . From this $L\,dL = k\Delta p\,dt/(\mu_f\phi)$ , $L = \sqrt{2k\Delta p t/(\mu_f\phi)}$ and $u = \sqrt{k\phi\Delta p/(2\mu_f)}/\sqrt t$ . This is $C_v/\sqrt t$ .

Compression of the reservoir. If we raise the pressure at the wall by $\Delta p$ and hold it, a one-dimensional transient filtration appears in a semi-infinite reservoir, with the hydraulic diffusivity $k/(\phi\mu_r c_t)$ (Diffusivity equation). Its solution gives the flow through the wall $u = \Delta p\sqrt{k\phi c_t/(\pi\mu_r)}/\sqrt t$ . This is $C_c/\sqrt t$ .

Combination. All the mechanisms are in series: the flow is one, and the pressure drops add up. The cake and the filtrate give a coefficient proportional to the square root of their share of the pressure drop, and the compression of the reservoir gives one proportional to the share itself. From the equality of the flows and the sum of the pressure drops we get a quadratic equation for the overall coefficient. Its positive root is the formula of the lesson. If there is no cake, what remains is $C_{vc} = 2C_vC_c/(C_v + \sqrt{C_v^2 + 4C_c^2})$ . If one component is much smaller than the others, the overall coefficient is close to it.

The leak-off function after the screen-out

Let the area of one wall grow before the screen-out as $A(\tau) = A_l\sqrt{\tau/t_{\text{so}}}$ . This is how a fracture grows with strong leak-off. A part that opened at the moment $\tau$ has filtered $2C_L\sqrt{t - \tau}$ per unit area by the moment $t$ , and twice as much from both sides. If we add over the area ( $u = A/A_l$ , $\tau = t_{\text{so}}u^2$ ), we get the leak-off volume $V_L = 4C_LA_l\int_0^1\sqrt{t - t_{\text{so}}u^2}\,du$ , that is, $V_L = 2A_lC_L\sqrt{t_{\text{so}}}\,g(\delta)$ , where the integral can be taken exactly: $g(\delta) = \sqrt\delta + (1 + \delta)\arcsin(1/\sqrt{1 + \delta})$ .

At $\delta = 0$ the function equals $\pi/2$ . This is the factor $\kappa$ of strong leak-off from the volume balance of the lesson on design. At large $\delta$ it grows as $2\sqrt\delta$ : parts of the wall that opened long ago filter almost equally. The same function describes the leak-off after the pumps stop in a calibration injection, and the difference $g(\delta) - \pi/2$ , up to the factor $\pi/4$ , is the $G$ function of this form. If the fracture grew more slowly before the screen-out (weak leak-off), the function is different, but for permeable reservoirs the limit of strong leak-off is closer to the truth.

Connections to other topics

Fracture design: size and conductivity — proppant, the proppant number, the geometry models, the volume balance and the injection schedule.
Rock mechanics for hydraulic fracturing — the net pressure, the growth of a fracture in height, and the calibration injection.
Hydraulic fracturing: purpose and well selection — what a fracture gives, and how hydraulic fracturing is run.
Inflow to a hydraulically fractured well — inflow to a fracture, and choking of the fracture at the wellbore.
Sand control methods — hydraulic fracturing with a gravel pack in weak sandstones.
Drilling fluid rheology and flow regimes — the power-law model and the shear rate in pipes.
Acid treatment of carbonates and special wells — acid fracturing, viscoelastic surfactants and diversion.
Formation damage and damage skin — damage to the fracture walls by filtrate, and polymer residue.

Check yourself

1
A rheometer gave 18 Pa at 50 s⁻¹ and 36 Pa at 200 s⁻¹ for a gel. Find $n$ and $K$ and the apparent viscosity at 100 s⁻¹.
Show the answer
$n = \ln(36/18)/\ln(200/50) = \ln 2/\ln 4 = 0.5$ ; $K = 18/50^{0.5} = 2.55$ Pa·sⁿ; $\mu_a = 2.55\cdot100^{-0.5} = 0.255$ Pa·s. A check: at 200 s⁻¹ $\mu_a = 2.55/\sqrt{200} = 0.18$ Pa·s, and $0.18\cdot200 = 36$ Pa.
2
Which fluid is best suited for hydraulic fracturing of a permeable weak sandstone with a gravel pack?
1. A.Fresh water without additives
2. B.A linear gel of low concentration
3. C.Slickwater
4. D.A crosslinked gel or viscoelastic surfactants with leak-off control
Show the answer
Correct answer: D. A crosslinked gel or viscoelastic surfactants with leak-off control
We need viscosity to carry high concentrations of proppant, a cake or a viscous filtrate to hold back leak-off, and clean breaking. In a permeable reservoir slickwater will go almost entirely into the reservoir and will not carry the proppant. Fresh water, besides, can make clays swell and move.
3
In a fluid-loss test through a disc of area 10 cm², 6.0 cm³ was collected in 4 min and 12.0 cm³ in 16 min. Find the cake coefficient and the spurt loss.
Show the answer
The slope against the square root of time is $(12.0 - 6.0)/(4 - 2) = 3.0$ cm³/min $^{1/2}$ , and the intercept is $6.0 - 2\cdot3.0 = 0$ . $C_w = 3.0/(2\cdot10) = 0.15$ cm/min $^{1/2}$ $= 0.15\cdot10^{-2}/\sqrt{60} \approx 1.9\cdot10^{-4}$ m/s $^{1/2}$ . There is no spurt loss: the cake formed at once.
4
How many times faster does a grain of diameter 0.8 mm settle than a grain of 0.4 mm in the same Newtonian fluid (at low Reynolds numbers)?
1. A.2 times
2. B.4 times
3. C.8 times
4. D.The same
Show the answer
Correct answer: B. 4 times
By Stokes' law the velocity is proportional to the square of the diameter: $(0.8/0.4)^2 = 4$ . This is why a large proppant, which gives a higher conductivity, is harder to carry to the end of the fracture.
5
Five minutes after the start of injection the pressure in the fracture began to rise steeply, with a slope of about one on log–log axes. By the calculation, the first slurry should have reached the end of the fracture at the 5th minute. What has happened, and what should we do?
Show the answer
This is the expected screen-out at the end of the fracture. The time matches the calculated one, and the rise of pressure is the sign that the fracture is no longer lengthening and is widening. We continue the packing by the schedule, making sure that the net pressure does not approach the stress contrast with the shales and does not exceed the limits on the wellhead pressure. If necessary, we lower the rate. If the rise had started clearly earlier than calculated, it would have been a screen-out at the wellbore.
6
After the growth of the fracture has stopped, as much time has passed as before it ( $\delta = 1$ ). How many times less is the leak-off in this time than if the whole area had just opened?
Show the answer
$g(1) = 1 + 2\arcsin(1/\sqrt2) = 1 + 2\cdot\pi/4 = 2.571$ , so $g(1) - \pi/2 = 1.0$ , and the leak-off is $2A_lC_L\sqrt{t_{\text{so}}}\cdot1.0$ . A fresh wall would have filtered $2A_l\cdot2C_L\sqrt{t_{\text{so}}}$ in the same time, twice as much. The old parts of the wall are already covered with cake and filter more slowly. This is why the efficiency of injection grows after the screen-out.

Updated: 19 October 2026