Primary cementing · MunaiHub

Quick refresher

The aim is a continuous cement sheath from the shoe up to the planned height. Conductors and surface casing are cemented to the surface. Intermediate and production strings are cemented with an overlap of the previous shoe, or above all the formations that must be isolated. Liners are cemented over their whole length.
Single-stage cementing: spacers, the bottom plug, the cement slurry, the top plug, and displacement until the bump on the landing ring. The float valve (a one-way valve in the float collar) keeps the cement in the annulus. Long strings above weak formations are cemented in two stages, liners through drill pipe, and large-diameter strings through an inner string.
The main reason for poor cementing is mud that has not been displaced, especially on the narrow side of an eccentric annulus (one where the string lies off-centre). These help: standoff, a conditioned mud (mud that was circulated and treated before the job), spacers with a hierarchy of density and friction (each fluid is heavier and has more friction than the one before it), a high pump rate, and moving the string.
The volume of slurry has three parts. The first is the annulus from the top of cement down to the shoe (the open hole is taken from the caliper log, a log that measures the hole diameter, with a margin). The other two are the rathole (the hole below the shoe) and the shoe track (the pipes between the landing ring and the shoe). The volume of displacement is the inside volume of the string down to the landing ring.
While there is cement in the string, it is heavier than the mud in the annulus, so it flows down by itself (free fall). The fluid in the string sinks faster than the pumps supply it, and a void, a vacuum, opens at the top of the string. The returns (the flow coming out of the annulus at the surface) are then larger than the pump rate. At the end of displacement the annulus is heavier. The pump pressure then rises to the difference of the hydrostatic pressures plus friction. After the pumps stop, the float valve holds this difference.
The pressure at the bottom must stay above the pore pressure and below the fracture pressure, with friction included. The thickening time must be longer than the time the slurry spends in the well, with a margin.

Key relationships

V_{\text{sl}} = \sum_i A_{\text{ann }i}\,L_i\,(1 + x_i) + V_{\text{rat}} + V_{\text{st}}

volume of cement slurry: the annulus by intervals with a margin $x_i$ , the rathole and the shoe track

V_{\text{disp}} = \sum_j A_{\text{in }j}\,L_j

volume of displacement: the inside volume of the string down to the landing ring (plus the piping)

\Delta p_{\text{U}} = 0.00981\left(\sum \rho_{\text{ann}} h_{\text{ann}} - \sum \rho_{\text{str}} h_{\text{str}}\right)

excess of the annulus columns over the columns in the string, MPa (g/cm³, m); at the end of displacement this is the pump pressure without friction

\text{SO} = \dfrac{w_{\min}}{(D_{\text{hole}} - d)/2}

standoff: the smallest gap between the string and the wall, divided by the gap with perfect centring

Common pitfalls

Taking the volume of the open hole from the bit diameter and not from the caliper log with a margin.
Going on with displacement if the plug has not landed at the calculated volume. The cement from the shoe track goes into the annulus, and mud ends up at the shoe.
Taking returns lower than the pump rate in the middle of displacement for lost circulation, and a pump pressure near zero for a fault. This is how the pumps fill the void after free fall.
Checking the pressure on weak formations by hydrostatic pressure only. At the end of displacement the friction of the cement slurries in the annulus is added.
Forgetting that a margin of slurry will lift the cement above the plan if the hole is not as wide as assumed.

After this lesson you can

name the tasks of primary cementing and choose a method: single-stage, two-stage with a stage collar, liner, through an inner string, reverse circulation
describe the casing equipment and the surface equipment: the shoe, the float collar (a one-way valve with a landing ring), the plugs (rubber wipers that keep the fluids apart in the string), the cementing head (the fitting on top of the string for the cement), the centralisers and the pumping units
calculate the standoff of a string (how well it is centred in the hole) by the standard definition, and explain how standoff, the preparation of the mud, spacers (fluids that separate mud from cement), the flow regime and the movement of the string decide how completely the mud is displaced
calculate the volumes of slurry (with a margin), spacers and displacement (pushing the cement down with another fluid), the pressure at the end of the displacement, the bump (the pressure rise when the top plug lands), and the pressure difference across the float collar
check the pressure at the bottom and at weak rocks during the job against the pore pressure (the pressure of the fluid in the rock) and the fracture pressure (the pressure at which the rock breaks), and explain the free fall of fluid in the string (the fluid sinks faster than the pumps supply it)
tie the time of the job to the thickening time (how long the slurry stays pumpable), and describe waiting on cement (the pause while the cement hardens) and the control of the job by pressure, rate, volumes and hook load (the weight that the hook carries)

Aims and methods of cementing

The requirements for the cement sheath (to separate the formations, to hold the string, to protect it from outside) are explained in the lesson on oilwell cements (what a well needs from cement). Here we look at how to get it. The first decision is the height of cement (top of cement, TOC). Conductors and surface casing are cemented to the surface: they carry the weight of the wellhead and the preventers and protect fresh water. Intermediate and production strings are cemented with an overlap of the shoe of the previous string, or at least above all the permeable, gas-bearing and water-bearing formations that must be isolated. A liner (a string that hangs inside the previous one and does not reach the surface) is cemented over its whole length, together with the overlap. The project for building the well, and the safety rules, set the exact height and the overlap.

Methods of primary cementing

Method: Single-stage
How it is done: the slurry is pumped into the string between two plugs and displaced through the shoe into the annulus
When it is used: most strings

Method: Two-stage, with a stage collar
How it is done: first the lower interval is cemented through the shoe, then the ports of the collar are opened and the interval above it is cemented
When it is used: long strings above weak formations: the pressure of the whole column of cement would fracture them

Method: Liner
How it is done: the slurry is pumped through drill pipe; the drill pipe plug latches onto the liner plug and lands on the landing collar
When it is used: liners; narrow gaps

Method: Through an inner string (stab-in)
How it is done: drill pipe is stabbed into a seal in the shoe or the valve; the cement is pumped through it, with no plugs
When it is used: large-diameter conductors and surface casing

Method: Reverse circulation
How it is done: the slurry is pumped into the annulus from the surface, and the displaced fluid comes out through the string
When it is used: very weak formations: the friction of lifting the cement does not load them

Method	How it is done	When it is used
Single-stage	the slurry is pumped into the string between two plugs and displaced through the shoe into the annulus	most strings
Two-stage, with a stage collar	first the lower interval is cemented through the shoe, then the ports of the collar are opened and the interval above it is cemented	long strings above weak formations: the pressure of the whole column of cement would fracture them
Liner	the slurry is pumped through drill pipe; the drill pipe plug latches onto the liner plug and lands on the landing collar	liners; narrow gaps
Through an inner string (stab-in)	drill pipe is stabbed into a seal in the shoe or the valve; the cement is pumped through it, with no plugs	large-diameter conductors and surface casing
Reverse circulation	the slurry is pumped into the annulus from the surface, and the displaced fluid comes out through the string	very weak formations: the friction of lifting the cement does not load them

The special methods are described in more detail in the last section of the lesson.

Casing equipment and surface equipment

The string is run already fitted for cementing (parts of a string, preparing to run). From the bottom up:

Shoe (guide shoe, float shoe): Guides the string down the hole. A float shoe has a float valve (a one-way valve) built in, which is the second line of defence against the cement flowing back.
Float collar (float valve and landing ring): It sits one or two pipes above the shoe. The float valve does not let the heavy cement return into the string, and the plugs land on the ring. In Russian equipment, “throttling check valves” are common: self-filling (auto-fill) float valves. While the string is run they let mud into the string, and afterwards they work as an ordinary float valve and a “stop” ring.
Shoe track: The pipes between the float valve and the shoe. The top plug that follows the cement scrapes the leftover mud and cement off the wall of the string, and contaminated cement builds up in front of it: the last portion. When the plug lands on the ring, this portion stays in the shoe track, and clean cement goes out through the shoe. The shoe track also gives a margin for errors in the displacement volume. It is drilled out together with the float valve and the shoe.
Bottom plug (wiper plug): A rubber plug with a cavity closed by a diaphragm. It goes ahead of the cement, cleans the wall and separates the cement from the spacer. When it lands on the ring, the pressure bursts the diaphragm, and the cement passes through the plug.
Top plug (displacement plug): A solid plug. It goes behind the cement and separates it from the displacement fluid. When it lands on the bottom plug, the pressure rises sharply. This is the “bump”: the displacement is over.
Cementing head: It is screwed onto the string. The plugs are loaded in it and are released one after the other without taking the piping apart. The lines from the pumping units connect to it.
Centralisers, scratchers, turbulators: Centralisers hold the string in the middle of the hole (the next section). Scratchers scrape the mudcake (the layer of mud solids) off the wall. Turbulators are blades on the pipe that swirl the flow in important sections.
Stage collar: It is placed in the string at the boundary of the stages. Plugs or pressure open and close its ports.

SchematicSingle-stage cementing: washing; pumping the spacer and the cement between the plugs; displacement, where the bottom plug has landed on the ring and been burst, and the cement rises up the annulus; the bump, where the top plug has landed on the bottom plug and cement is left in the shoe track. Not to scale.

At the surface, cement is stored and fed by compressed air (pneumatically) from silos or cement mixing units. It is mixed with water in a jet or recirculating mixer and pumped by cementing units: high-pressure pumps on a truck chassis or a skid. (In the post-Soviet countries, for example, the TsA-320 units are common; the name is a transliteration of the Russian designation.) The main quality indicator of mixing is density: a densitometer (a density meter) at the outlet of the mixer measures it all the time (density and yield). To get a slurry of exactly the set density, it is sometimes prepared in advance in an averaging tank (a batch tank). The crew displaces with the units or with the rig mud pumps. Density, rate, volume and pressure are recorded over the whole job.

Centring the string

If the string is pressed against one wall, the annulus is eccentric: wide on one side and narrow on the other. The pressure drop along the length is the same on both sides, but the resistance of the narrow slot is much larger. So the cement goes along the wide side, while on the narrow side the drilling mud hardly moves, and a through channel of mud stays along the string. To prevent this, we centre the string. The measure of centring is set by the standard for centralisers (API Spec 10D, ISO 10427-1):

Standoff ratio of a string

\text{SO} = \frac{w_{\min}}{(D_{\text{hole}} - d)/2}\cdot 100\ \%

where:

$\text{SO}$ standoff ratio, the ratio of the smallest gap to the nominal gap, %
$w_{\min}$ the smallest distance between the outer surface of the string and the wall of the hole (standoff), mm
$D_{\text{hole}},\ d$ diameter of the hole and outside diameter of the string; $(D_{\text{hole}} - d)/2$ is the gap with perfect centring (annular clearance), mm

Units: any consistent units of length

Holds when

The cross-section of the hole is round. In a washout (a place where the hole has been washed wider than the bit made it), $D_{\text{hole}}$ is its diameter.
100% means that the string is exactly in the middle, and 0% means that it touches the wall.

At a centraliser the standoff depends on its design and on the side force on it. Between centralisers it also depends on how much the pipe sags. The standard sets the minimum restoring force of a bow-spring centraliser at a standoff of 67%, and it says that this is a condition for testing the centralisers themselves, not a norm for centring a string in a well. Cementing programmes often require not less than 70%, and more in critical intervals.

Rigid centralisers, solid or with blades, give a constant smallest gap and withstand large side forces. But in a washout the string can move off the axis by the full width of the washout. Bow-spring centralisers, with bowed strips, pass through narrow places and spring open in washouts. But the side force presses them in, so in an inclined hole and in bends we choose them by their restoring force (the force with which the springs push the string back to the centre). How many centralisers to run, and where, is calculated by standard methods (API RP 10D-2, ISO 10427-2) or by the manufacturers' programs: from the side force on each centraliser and the sag of the pipe between them.

Ratio of the speeds on the narrow and the wide side of an eccentric annulus

\frac{v_{\text{nar}}}{v_{\text{wid}}} = \left(\frac{\text{SO}}{2 - \text{SO}}\right)^{(n+1)/n}

where:

$v_{\text{nar}},\ v_{\text{wid}}$ average speed of the fluid in the narrowest and in the widest part of the annulus, m/s
$\text{SO}$ standoff, as a fraction, —
$n$ flow behaviour index of the power-law model of the fluid, —

Units: dimensionless

Holds when

Each part of the circumference of the annulus is treated as a narrow flat slot with its own width, and the pressure gradient along the length is the same for all parts (the narrow-slot approximation).
The flow is laminar, and the fluid is a power-law fluid with no yield stress. In a mud with a yield stress, the flow on the narrow side can stop altogether.
The width of the slot on the narrow side is proportional to $\text{SO}$ , and on the wide side to $2 - \text{SO}$ (as fractions of the nominal gap).

From the formula for losses in a slot (laminar flow), the speed at a given gradient grows as the width of the slot to the power $(n+1)/n$ . At $n \approx 0.4$ this power is about 3.5. So even a moderate eccentricity slows the narrow side sharply.

SchematicStandoff: on the left, the cross-section of the hole with the string moved to one wall, with the smallest gap and the nominal gap; on the right, a string between two centralisers in an inclined hole sags towards the low wall, and in the middle of the span the gap is smallest. Not to scale, the sag is exaggerated.

Worked exampleStandoff at the shoe: rigid or bow-spring centralisers, and how many pipes apart

Given (assumed data)

String at the shoe: 244.5 × 11.99 mm; weight in mud of 1.09 g/cm³ is 590.5 N/m; bending stiffness 12.28 MN·m²
Hole: 311.1 mm by the bit, with some washouts up to 340 mm; the hole deviates from vertical by up to 3°
Centralisers: rigid, with an outside diameter of 304.8 mm (12″); bow-spring, with a restoring force at 67% standoff of 9 kN by the data sheet
Requirement of the programme; lead slurry: standoff not less than 70%; $n$ = 0.407

Solution

Nominal gap in the 311.1 mm hole: $(311.1 - 244.5)/2 = 33.30$ mm. The rigid centraliser leaves not less than $(304.8 - 244.5)/2 = 30.15$ mm, so the standoff at the centraliser is $30.15/33.30 = 90.5$ %.
In a 340 mm washout the gap is 47.75 mm, and the same centraliser gives only $30.15/47.75 = 63.1$ %, which is less than 70%.
Between the centralisers the pipe is bent by the side component of its weight, $590.5\cdot\sin 3° = 30.9$ N/m. The sag in the middle of the span of a long pipe on equally spaced supports is $\delta = w_n L^4/(384\,EI)$ . With a centraliser every two pipes ( $L$ = 24 m) it is $30.9\cdot 24^4/(384\cdot 12{,}280{,}000) = 0.00217$ m, that is 2.17 mm. Every three pipes ( $L$ = 36 m) it is 11.0 mm.
Standoff in the middle of the span, with rigid centralisers in the 311.1 mm hole: $(30.15 - \delta)/33.30$ is 84.0% for every two pipes, and 57.5% for every three.
A bow-spring centraliser every two pipes carries $30.9\cdot 24 = 742$ N. With a linear spring characteristic, its standoff is $1 - 0.33\cdot 742/9000 = 97.3$ %, and in a washout the springs open out to the wall.
The ratio of the speeds of the lead slurry on the narrow and wide sides: at a standoff of 84.0% it is $(0.840/1.160)^{3.457} = 0.33$ , and at 57.5% it is $(0.575/1.425)^{3.457} = 0.043$ .

Answer: With rigid centralisers every two pipes, the standoff in the middle of the span is about 84.0%, and every three pipes only 57.5%. In a washout a rigid centraliser does not give even 70%, so in washout intervals the crew runs bow-spring ones.

What it means. Placing centralisers is a calculation, not a rule of “every pipe”. The sag grows as the fourth power of the span, and a step one and a half times longer gave a sag five times larger. At a standoff of 57.5%, the lead slurry on the narrow side moves more than twenty times slower than on the wide side, and that is where a channel of drilling mud stays. The sag model does not include the tension of the string, which straightens it. Near the shoe the tension is small, and the estimate is close to reality. Higher up the string the estimate is on the safe side. The standard methods also include the forces in curved sections, where tension presses the string against the wall. The data in this example are assumed for illustration.

Displacing the drilling mud

Cement does not wash away a thickened drilling mud by itself. If the mud in the annulus has stood for a long time, has built up a structure (gel strength) and has left a cake on the walls, the cement will take the easiest path and go round it. Good displacement comes from four measures.

Conditioning the mud. After running, the crew circulates until the mud at the outlet is uniform, usually for at least the volume of the annulus. The crew also lowers the yield point and the gel strength (how thick the mud is while it flows and while it stands still). A mobile mud is easier to displace, and the circulation shows whether the whole length of the annulus can be passed.
Spacers (spacer, and wash). A low-viscosity wash in turbulent flow (a swirling flow) washes away the mud and the cake. A heavier (weighted) spacer separates the cement from the drilling mud, because their mixture often thickens sharply and sets badly (contamination of cement). It also pushes the mud out like a piston. The compatibility of the fluids is checked in the laboratory.
The hierarchy of densities and friction. Each following fluid must be heavier than the previous one (in practice, by not less than 10%), and must give a larger friction gradient (more pressure lost per metre of annulus) at the same rate (by about 20% or more). Then it does not break through the displaced fluid in a tongue. The volume of spacer is enough for a height of not less than 150–250 m of annulus, or for 10 min of contact with the wall at the displacement rate. These are practical rules, not laws, and they are checked by calculation for each well.
Pump rate and movement of the string. The higher the rate, the better the displacement, as long as the equivalent circulating density (ECD) stays below the leak-off pressure. The ECD is the pressure at a depth while the fluid is moving, written as the density of a mud that would give the same pressure. The leak-off pressure is the pressure at which the rock starts to take fluid. Rotating the string (if the connections and the torque allow) or moving it up and down tears the stagnant mud off the narrow side. Moving it up and down also causes pressure surges.

Fluids of the example and the friction hierarchy (annulus in a 320 mm hole, 1.6 m³/min)

Fluid: Drilling mud
Density, g/cm³: 1.09
Readings at 100 / 3 rpm: 12 / 3
$n$ ; $K$ , Pa·sⁿ: 0.395; 0.804
Gradient, Pa/m: 340

Fluid: Wash fluid
Density, g/cm³: 1.00
Readings at 100 / 3 rpm: viscosity 1.5 mPa·s
$n$ ; $K$ , Pa·sⁿ: 1; 0.0015
Gradient, Pa/m: 92

Fluid: Spacer
Density, g/cm³: 1.25
Readings at 100 / 3 rpm: 30 / 8
$n$ ; $K$ , Pa·sⁿ: 0.377; 2.211
Gradient, Pa/m: 857

Fluid: Lead slurry
Density, g/cm³: 1.45
Readings at 100 / 3 rpm: 50 / 12
$n$ ; $K$ , Pa·sⁿ: 0.407; 3.157
Gradient, Pa/m: 1410

Fluid: Tail slurry
Density, g/cm³: 1.90
Readings at 100 / 3 rpm: 90 / 15
$n$ ; $K$ , Pa·sⁿ: 0.511; 3.331
Gradient, Pa/m: 2412

Fluid	Density, g/cm³	Readings at 100 / 3 rpm	$n$ ; $K$ , Pa·sⁿ	Gradient, Pa/m
Drilling mud	1.09	12 / 3	0.395; 0.804	340
Wash fluid	1.00	viscosity 1.5 mPa·s	1; 0.0015	92
Spacer	1.25	30 / 8	0.377; 2.211	857
Lead slurry	1.45	50 / 12	0.407; 3.157	1410
Tail slurry	1.90	90 / 15	0.511; 3.331	2412

The data are assumed for illustration. The gradients are from the slot formulas of the lesson on pressure losses. The flow of the wash is turbulent, so it works by washing away and does not follow the friction hierarchy. The density of each following viscous fluid is 15–31% higher than that of the previous one, and the gradient is 1.6–2.5 times higher.

Calculating volumes and materials

We find the volume of cement slurry by adding the annulus by intervals and what stays in the string and below it. The most uncertain part is the open hole. We take it from the caliper log (a measurement of the hole diameter) with a margin, and with no log from the bit diameter with a large margin.

Volume of cement slurry

V_{\text{sl}} = \sum_i A_{\text{ann }i}\,L_i\,(1 + x_i) + V_{\text{rat}} + V_{\text{st}}, \qquad A_{\text{ann}} = \frac{\pi}{4}\left(D^2 - d^2\right)

where:

$V_{\text{sl}}$ volume of cement slurry (one portion or all), m³
$A_{\text{ann }i},\ L_i$ area of the annulus and the length of the $i$ -th interval (in a cased interval, $D$ is the inside diameter of the previous string), m², m
$x_i$ margin: zero for a cased interval, and for the open hole taken from the caliper log and the experience of the area, fraction
$V_{\text{rat}},\ V_{\text{st}}$ volume of the rathole (the hole below the shoe) and of the shoe track, m³

Units: SI

Holds when

The volumes of the slurry do not change during pumping (we neglect compressibility and shrinkage).
The margin covers the inaccuracy of the hole volume. If the hole turns out smaller, the cement will rise higher than planned.

We find the mass of cement by dividing the volume by the slurry yield, and the water and the additives from the recipe (slurry yield).

Displacement volume

V_{\text{disp}} = \sum_j \frac{\pi}{4}\,d_{\text{in }j}^2\,L_j + V_{\text{pip}}

where:

$V_{\text{disp}}$ volume of displacement fluid up to the bump, m³
$d_{\text{in }j},\ L_j$ inside diameter and length of the $j$ -th section of the string from the wellhead to the landing ring, m
$V_{\text{pip}}$ volume of the head and the piping from the place where the top plug is released to the string, m³

Units: SI

Holds when

Nominal inside diameters. The real volume differs by fractions of a per cent, so the crew expects the bump near the calculated volume.
Pumps are counted by strokes with a volumetric efficiency, or by a flow meter. The compression of the mud and the string under pressure adds a little volume.

If the plug has not landed at the calculated volume, the crew pumps no more than half the volume of the shoe track. Beyond that the cement from the shoe track goes into the annulus, and displacement fluid ends up around the shoe.

Worked exampleVolumes, materials and time for cementing the 244.5 mm string

Given (assumed data)

Well and string: the 244.5 mm string by the sections of the lesson on string design, shoe at 2600 m, the “stop” landing ring at 2576 m (the shoe track is two pipes), bottom at 2605 m; surface casing 339.7 × 12.19 mm to 700 m
Hole by the caliper log: 320 mm from 700 to 2100 m, 316 mm deeper; a margin of 10% over the caliper log
Lead slurry: 1.45 g/cm³, yield 1.305 m³/t (the lightweight slurry with microspheres); from 500 m, which is 200 m above the surface casing shoe, to 2100 m
Tail slurry: class G cement at W/C 0.44: 1.90 g/cm³, yield 0.757 m³/t; from 2100 m to the shoe, plus the rathole and the shoe track
Spacers and displacement: wash 4 m³, spacer 10 m³; displacement with drilling mud; piping 0.2 m³; pump stroke 16.2 L, volumetric efficiency 0.95
Rates: spacer and lead slurry 1.0 m³/min, tail slurry 0.8; 5 min to release the top plug; displacement 1.6 m³/min, the last 3 m³ at 0.5 m³/min

Solution

Lead slurry: in the surface casing $200\cdot\frac{\pi}{4}(0.31532^2 - 0.2445^2) = 200\cdot 0.03114 = 6.23$ m³; in the open hole $1400\cdot 0.03347\cdot 1.10 = 51.54$ m³; in total 57.77 m³.
Tail slurry: in the annulus $500\cdot 0.03148\cdot 1.10 = 17.31$ m³, in the rathole $5\cdot\frac{\pi}{4}\cdot 0.316^2\cdot 1.10 = 0.43$ m³, in the shoe track $24\cdot\frac{\pi}{4}\cdot 0.22052^2 = 0.92$ m³; in total 18.66 m³.
Materials: cement for the lead slurry $57.77/1.305 = 44.3$ t (microspheres 6.3 t, water 33.2 m³), for the tail slurry $18.66/0.757 = 24.6$ t (water 10.8 m³); in total 68.9 t.
Displacement: the string down to the ring is $300\cdot 0.03819 + 1800\cdot 0.03885 + 476\cdot 0.03819 = 99.57$ m³, and with the piping 99.77 m³. The number of pump strokes is $99.77/(0.0162\cdot 0.95) = 6483$ . If the plug does not land, the crew pumps no more than half of the shoe track: 0.46 m³, or 30 strokes.
Time: $4 + 10 + 57.77/1.0 + 18.66/0.8 + 5 + (99.77 - 3)/1.6 + 3/0.5 = 167$ min, about 2.8 h. The lead slurry will spend 153 min in the well, and the tail slurry 95 min. With a margin of 1 h, their thickening time at the circulating temperature must be not less than 3.5 and 2.6 h.

Answer: We need 57.77 m³ of lead slurry and 18.66 m³ of tail slurry, which is 68.9 t of cement. The displacement is 99.77 m³, or 6483 pump strokes by the stroke counter. The job will take about 2.8 h.

What it means. The lead slurry is three quarters of the volume. A lightweight slurry is cheap per cubic metre and does not overload weak rocks. But its thickening time is the longest requirement of the job, because it is pumped first. The margin of 10% assumes that the hole is wider than the caliper log shows. If the hole is exactly as the caliper log shows, the cement will rise higher than planned: the tail slurry to 2040 m, and the lead slurry to 285 m. The pressure on the rocks will then be higher, and this is checked (the section on free fall). Materials are taken to the rig with a margin for losses on mixing. The data in this example are assumed for illustration.

How single-stage cementing goes

Order of work

1
Circulation
The crew circulates the string at depth and conditions the mud. The crew checks the returns (the flow that comes back out of the annulus) and the hook load, and if possible moves the string up and down or rotates it.
2
Piping
The crew installs the cementing head with the plugs loaded and pressure-tests the lines. The crew then checks the programme once more: volumes, rates, densities, allowable pressures.
3
Spacers and the bottom plug
The crew pumps the wash and the spacer and releases the bottom plug.
4
Cement slurry
The crew mixes and pumps the lead slurry and then the tail slurry, keeping the density steady all the time. The crew takes samples to check how the cement sets.
5
Top plug
The crew releases the top plug and checks on the indicator of the head that the plug has left.
6
Displacement
The crew displaces with mud at the set rate and counts the volume. The crew watches the pressure, the returns and the hook load, and lowers the rate before the calculated volume is reached.
7
The bump and the pressure test
When the plug has landed, the crew raises the pressure above the displacement pressure and then, if the programme says so, pressure-tests the string.
8
Checking the float valve
The crew releases the pressure smoothly and measures the volume that comes back. If it equals the compression volume (the fluid that the pressure squeezed in), the float valve holds. If fluid keeps coming, the valve does not hold: the crew closes the string and holds pressure in it until the cement sets.
9
Waiting on cement
The crew lands the string on the hanger (landing the string), installs the wellhead and waits until the cement gains strength (WOC).

Pressures during the job: hydrostatic, friction, bump

The string and the annulus are joined at the shoe, like the legs of a U-tube (communicating vessels). The pump pressure is what the column in the string lacks to balance the annulus, plus friction on the way. At the start of the job the string side is heavier (cement), and at the end the annulus side is. The pressure on the bottom of the hole and on its walls comes from the hydrostatic pressure of the columns in the annulus and from the friction in the annulus above the depth in question, as in drilling (ECD along the hole).

Imbalance of the columns and the pump pressure

\Delta p_{\text{U}} = 0.00981\left(\sum \rho_{\text{ann}} h_{\text{ann}} - \sum \rho_{\text{str}} h_{\text{str}}\right), \qquad p_{\text{pump}} = \Delta p_{\text{U}} + \Delta p_{\text{fr.ann}} + \Delta p_{\text{fr.str}}

where:

$\Delta p_{\text{U}}$ excess of the annulus columns over the columns in the string at the depth of the shoe; negative means that the string is heavier, MPa
$\rho_{\text{ann}},\ h_{\text{ann}}$ density and height of the columns of fluids in the annulus, g/cm³, m
$\rho_{\text{str}},\ h_{\text{str}}$ the same in the string, g/cm³, m
$p_{\text{pump}}$ pump pressure (at the cementing head), MPa
$\Delta p_{\text{fr.ann}},\ \Delta p_{\text{fr.str}}$ friction losses in the annulus and in the string at the given rate, MPa

Units: metric: MPa, g/cm³, m; 0.00981 MPa is the pressure of a column of 1 m of fluid with a density of 1 g/cm³

Holds when

The heights are vertical. The columns are counted from the wellhead to the depth of the shoe.
If $p_{\text{pump}}$ comes out negative by the formula, there is no pressure at the pumps: the fluid in the string goes down faster than it is pumped (free fall, the next section).

After the bump and the stop of the pumps, the friction disappears, and $\Delta p_{\text{U}}$ becomes the pressure difference that the float valve holds. If the valve does not hold, the cement will return into the string.

Friction in the annulus at the end of displacement (0.5 m³/min)

Fluid: Mud
Interval, m: 0–50
Speed, m/s: 0.268
Gradient, Pa/m: 242
Loss, MPa: 0.012

Fluid: Wash fluid
Interval, m: 50–179
Speed, m/s: 0.268
Gradient, Pa/m: 15
Loss, MPa: 0.002

Fluid: Spacer
Interval, m: 179–500
Speed, m/s: 0.268
Gradient, Pa/m: 620
Loss, MPa: 0.199

Fluid: Lead slurry
Interval, m: 500–700
Speed, m/s: 0.268
Gradient, Pa/m: 989
Loss, MPa: 0.198

Fluid: Lead slurry
Interval, m: 700–2100
Speed, m/s: 0.226
Gradient, Pa/m: 751
Loss, MPa: 1.051

Fluid: Tail slurry
Interval, m: 2100–2600
Speed, m/s: 0.241
Gradient, Pa/m: 1251
Loss, MPa: 0.626

Fluid	Interval, m	Speed, m/s	Gradient, Pa/m	Loss, MPa
Mud	0–50	0.268	242	0.012
Wash fluid	50–179	0.268	15	0.002
Spacer	179–500	0.268	620	0.199
Lead slurry	500–700	0.268	989	0.198
Lead slurry	700–2100	0.226	751	1.051
Tail slurry	2100–2600	0.241	1251	0.626

Calculated by the slot formulas. The open hole uses design diameters of 326.6 and 322.3 mm: such a hole holds the volume by the caliper log plus 10%. The flow is laminar everywhere except for the wash. The data are assumed for illustration.

Worked examplePressures at the end of displacement and after the bump

Given (assumed data)

After displacement (the plan): behind the string from top to bottom: mud 1.09 g/cm³ for 50.4 m, wash 1.00 for 128.5 m, spacer 1.25 for 321.1 m, lead slurry 1.45 for 1600 m, tail slurry 1.90 for 500 m; in the string, mud down to the ring (2576 m), and in the shoe track the tail slurry
Formations: pore pressure at the shoe 1.05 g/cm³; leak-off pressure under the surface casing shoe 1.50 g/cm³; fracture pressure at the shoe 1.784 g/cm³
Friction: the table above; in the string at 0.5 m³/min, 0.10 MPa
Bump and pressure test: by the programme, 3.5 MPa above the displacement pressure; then the string is pressure-tested at 30 MPa at the wellhead

Solution

Hydrostatic pressure of the annulus at the shoe: $0.00981\,(1.09\cdot 50.4 + 1.00\cdot 128.5 + 1.25\cdot 321.1 + 1.45\cdot 1600 + 1.90\cdot 500) = 0.00981\cdot 3854.81 = 37.82$ MPa, an equivalent mud weight of 1.483 g/cm³. At a depth of 700 m it is $0.00981\cdot 874.81 = 8.58$ MPa, or 1.250 g/cm³.
In the string: $0.00981\,(1.09\cdot 2576 + 1.90\cdot 24) = 27.99$ MPa. The excess of the annulus is $37.82 - 27.99 = 9.83$ MPa. This is what the pumps would show without friction, and what the float valve holds after the bump.
Pump pressure at the end of displacement: $9.83 + 2.09 + 0.10 = 12.02$ MPa. The bump is at about 15.5 MPa.
At the bottom at the end of displacement: $37.82 + 2.09 = 39.91$ MPa, an ECD of 1.565 g/cm³, which leaves 0.22 g/cm³ to the fracture pressure at the shoe. Under the surface casing shoe, with the friction above it of 0.41 MPa: $(8.58 + 0.41)/(0.00981\cdot 700) = 1.309$ g/cm³, and the margin to 1.50 is 0.19 g/cm³. At a rate of 1.6 m³/min it would be 1.619 and 1.346 g/cm³.
Overbalance (the excess of the mud pressure over the formation pressure) on the formations at the shoe after the stop: $37.82 - 0.00981\cdot 1.05\cdot 2600 = 37.82 - 26.78 = 11.04$ MPa. The wash in the annulus took from the overbalance of the mud (1.020 MPa) $0.00981\cdot(1.09 - 1.00)$ MPa for each metre of its column. The overbalance would disappear at a column of 1155 m, which is about 39 m³ in the 320 mm hole, and 4 m³ was pumped.
Pressure test at 30 MPa: at the wellhead this is an internal excess pressure, against the internal yield pressure of the top pipes of 47.4 MPa. At the ring, inside it is $30 + 0.00981\cdot 1.09\cdot 2576 = 57.54$ MPa, and outside it is 37.37 MPa, a difference of 20.17 MPa.

Answer: At the end of displacement the pumps will show about 12.0 MPa, and the bump will be at about 15.5 MPa. The float valve will hold 9.83 MPa. The ECD at the end of displacement is 1.565 g/cm³ at the bottom and 1.309 g/cm³ under the surface casing shoe, below the leak-off pressure.

What it means. The pump pressure at the end of displacement is almost entirely the excess of the heavy annulus. Friction adds about a fifth. So a rise of pressure towards the end of displacement is a normal sign that the cement is rising in the annulus, and a comparison with the calculation shows how much of it is there. The most dangerous moment for the rocks is the end of displacement, when the cement is highest, so the crew lowers the rate. At 1.6 m³/min the ECD under the surface casing shoe would be 0.037 g/cm³ higher. The string is pressure-tested at 30 MPa right after the bump, while the cement is liquid. If it is done after the cement has set, the string, which has expanded under the pressure, will later pull away from the set cement and leave a microannulus (a very thin gap). The data in this example are assumed for illustration.

Free fall: a vacuum forms in the string

While there is cement in the string and still drilling mud in the annulus, the string is heavier than the annulus. If the excess is larger than the friction at the set rate, the pump pressure falls to zero, and the fluid in the string goes down faster than it is pumped in. The string “goes on vacuum” (free fall, also called U-tubing, because the fluids behave like those in a U-shaped tube). A void forms at the top of the string, the fluid being pumped falls into it, and more comes out of the annulus than the pumps deliver. When the heavy fluids have gone out into the annulus, the excess becomes smaller, and the fall stops. The pumps then first have to fill the void. At this time the returns are smaller than the pump rate, and the pump pressure is near zero. Only then does the pressure begin to rise to the excess of the annulus.

Worked exampleFree fall while cementing the 244.5 mm string

Given (assumed data)

String when the top plug is released, if it had stayed full: from top to bottom: tail slurry down to 485 m, lead slurry down to 1973 m, spacer down to 2232 m, wash down to 2337 m, and mud below; in the annulus there is mud of 1.09 g/cm³
Friction: of the mud in the annulus at the lead slurry rate of 1.0 m³/min, 0.69 MPa
Model: the lesson's calculation by pumping steps (the advanced section): the rates and the volumes are from the example “Volumes, materials and time for cementing the 244.5 mm string”

Solution

Columns in the string: $1.90\cdot 485.4 + 1.45\cdot 1487.3 + 1.25\cdot 259.6 + 1.00\cdot 104.7 + 1.09\cdot 263.0 = 3794.7$ g/cm³·m; in the annulus $1.09\cdot 2600 = 2834$ .
The excess of the string is $0.00981\cdot(3794.7 - 2834) = 9.42$ MPa, which is $9.42/0.69 = 13.7$ times the friction. The string cannot stay full.
By the model, a vacuum starts to form in the string when 20 m³ have been pumped, which is 6 m³ of the lead slurry in all. By the time the top plug is released, the void at the top of the string is 27 m³, about 710 m.
The returns from the annulus are larger than the pump rate, up to 1.7 m³/min with the tail slurry pumped at 0.8 m³/min. In displacement the first 60 m³ go to filling the void: the returns are about 0.9 m³/min at a pump rate of 1.6, and the pump pressure is near zero.
Then the pressure rises to the excess of the annulus (the example “Pressures at the end of displacement and after the bump”). The greatest ECD at the bottom by the model is 1.59 g/cm³ before the rate is lowered, and the smallest is 1.12 g/cm³ at the start, while there is mud in the annulus. Under the surface casing shoe it is not more than 1.32 g/cm³.

Answer: For almost two hours there is a vacuum in the string: the void at the top reaches 710 m, the returns at the start are up to two times larger than the pump rate, and in the first 60 m³ of displacement they are almost half of it.

What it means. Free fall is not an accident but a property of working with a slurry heavier than the drilling mud, and we must expect it. A zero pump pressure and returns larger than the pump rate at the start, and returns smaller than the pump rate in the middle of displacement, are according to plan. Only a comparison with the calculation tells them from lost circulation. While the string is on vacuum, the fluids fall and mix at the boundaries, so plugs and spacers are especially needed. Suppose the hole is not wider than the caliper log and the cement rises higher than planned. Then the ECD under the surface casing shoe at the end of displacement is 1.45 g/cm³ by the same model, and the margin to 1.50 shrinks to 0.05 g/cm³. This is why the rate at the end of displacement is lowered. The data in this example are assumed for illustration.

CalculatedThe calculated pump pressure during the job for the example of the lesson: almost at once after the lead slurry starts, a vacuum forms in the string (zero pressure). After the void is filled, the pressure in the displacement rises to the excess of the annulus plus friction, about 12 MPa at the end. The dip before the end is the lowering of the rate to 0.5 m³/min.

CalculatedThe pump rate and the calculated returns from the annulus: in free fall the returns are larger than the pump rate, and in displacement, while the pumps fill the void, they are smaller. After that the two are equal.

CalculatedThe calculated equivalent mud weight during the job (the ECD while the pumps run) at the bottom (2600 m) and under the surface casing shoe (700 m), with the limits: the pore pressure and the fracture pressure at the bottom, and the leak-off pressure under the surface casing shoe. Both curves rise towards the end of displacement, when the cement is highest. When the rate is lowered, they fall for a time.

After displacement: waiting on cement and control

While the cement is liquid, it passes the full hydrostatic pressure to the formations. As it thickens, it gains gel strength, leans on the walls and the string, and stops passing the weight of the column downward. At the same time it loses volume: filtrate leaves, and chemical shrinkage goes on. The pressure on the formations at the shoe falls. In the interval when the slurry no longer passes pressure but is still permeable, gas may enter it and make a channel (gas migration). A simple estimate shows how large the risk is.

Worked exampleHow much pressure a thickening cement can “hold up”

Given (assumed data)

Cement column: from 500 m to the shoe at 2600 m: in the surface casing 200 m ( $D - d$ = 70.8 mm), in the 320 mm hole 1400 m (75.5 mm), in the 316 mm hole 500 m (71.5 mm)
Overbalance after displacement: 11.04 MPa at the shoe (the example “Pressures at the end of displacement and after the bump”)
Gel strength: about 48 Pa: the slurry starts to “hang up”; about 239 Pa: gas can no longer pass

Solution

The sum over the intervals: $\sum L/(D - d) = 200/0.0708 + 1400/0.0755 + 500/0.0715 = 28{,}361$ .
A structure with a stress $\tau$ holds a pressure difference of $4\tau\sum L/(D - d)$ , which is the same force balance as in flow: at 48 Pa, 5.45 MPa; at 239 Pa, 27.1 MPa.
The overbalance of 11.04 MPa will disappear when $4\tau\cdot 28{,}361 = 11.04\cdot 10^6$ Pa, that is, at $\tau = 97$ Pa, between 48 and 239 Pa.

Answer: In the transition period, the pressure on the formations at the shoe may fall below the pore pressure. If the sandstones near the top of the zone of abnormally high formation pressure (overpressure) carry gas, the gas may enter the cement that has not yet set.

What it means. The estimate is rough and on the safe side: the column builds up its structure unevenly, with the hot bottom earlier than the cold top, and some pressure is passed through the pore water. But it shows why, against gas-bearing formations, the crew uses a slurry with a short transition time, low fluid loss and no free fluid, and does not make the light lead slurry above them longer than necessary. The data in this example are assumed for illustration.

Waiting on cement (WOC) lasts until the cement at the shoe gains the strength needed to drill out the float collar and the shoe and to test below the shoe, often about 3.5 MPa in compression (cement tests). The hot bottom hardens faster than the cold top, so the tail slurry at the shoe sets the WOC, while the light lead slurry at the top may take days to gain strength. During the WOC, the crew lands the string at the wellhead and installs the spool and the preventers (landing and the next spool). Where the cement actually is, and how tightly it fills the annulus, is checked with temperature surveys and acoustic methods. That is the subject of the lesson Remedial cementing, cement plugs and cement evaluation.

What can be seen during cementing

What is seen: Pump pressure near zero, returns larger than the pump rate
Probable cause: free fall: a vacuum in the string
What the crew does: continue by the programme, keep the volumes by the stroke counter

What is seen: Pressure near zero, returns smaller than the pump rate
Probable cause: the pumps are filling the void after free fall
What the crew does: compare the volumes with the calculation; this is not lost circulation

What is seen: There is pressure, but it is lower than calculated or stops rising; returns smaller than the pump rate
Probable cause: lost circulation: the rock has started to take fluid
What the crew does: lower the rate; then follow the programme and the procedures

What is seen: Pressure higher than calculated and rising
Probable cause: a narrowing of the annulus, dehydration or thickening of the cement
What the crew does: lower the rate, without exceeding the allowable pressure

What is seen: The plug has not landed at the calculated volume
Probable cause: the plug has not left, or has slipped past; the volume of the string is inexact
What the crew does: pump no more than half of the shoe track

What is seen: The plug has landed before the calculated volume
Probable cause: the plug has landed on debris, or the cement has thickened in the string
What the crew does: compare volumes: some cement may have stayed in the string

What is seen: After the pressure is released, fluid flows back out of the string
Probable cause: the float valve does not hold
What the crew does: close the string and hold the pressure until the cement sets

What is seen	Probable cause	What the crew does
Pump pressure near zero, returns larger than the pump rate	free fall: a vacuum in the string	continue by the programme, keep the volumes by the stroke counter
Pressure near zero, returns smaller than the pump rate	the pumps are filling the void after free fall	compare the volumes with the calculation; this is not lost circulation
There is pressure, but it is lower than calculated or stops rising; returns smaller than the pump rate	lost circulation: the rock has started to take fluid	lower the rate; then follow the programme and the procedures
Pressure higher than calculated and rising	a narrowing of the annulus, dehydration or thickening of the cement	lower the rate, without exceeding the allowable pressure
The plug has not landed at the calculated volume	the plug has not left, or has slipped past; the volume of the string is inexact	pump no more than half of the shoe track
The plug has landed before the calculated volume	the plug has landed on debris, or the cement has thickened in the string	compare volumes: some cement may have stayed in the string
After the pressure is released, fluid flows back out of the string	the float valve does not hold	close the string and hold the pressure until the cement sets

A diagram for understanding the links. On a real well, the cementing programme and the approved procedures set the actions.

Another instrument is the hook load. While there is cement in the string, it gets heavier. When the cement goes out into the annulus, the heavy column outside pushes the string up, and the weight falls below the weight in mud: for the lesson's string by a third by the end of displacement (weight at landing). A difference from the calculation points to an error in the densities or the volumes.

Two-stage cementing, liners, an inner string, reverse circulation

Two-stage cementing. If the hydrostatic pressure of the whole column of cement, and the friction in displacement, would exceed the leak-off pressure of weak rocks, the column is divided into two stages. A stage collar goes into the string, usually above the weak rocks or near the shoe of the previous string. The crew cements the first stage through the shoe and displaces it with a plug that passes through the collar. Then the crew opens the ports of the collar, with a plug dropped into the string and pressure, or with pressure alone. The crew circulates through the ports and cements the second stage, and its displacement plug closes the ports. Below the collar there is often a packer-cup or a basket, so that the cement of the second stage does not fall down. The price is extra time, drilling out the collar, and the fact that after the first stage the string can no longer be moved up and down.

A liner is run on drill pipe and hung in the previous string (the liner hanger). The cement is pumped through the drill pipe, and behind it the crew sends a small drill pipe displacement plug. When it reaches the liner, it latches onto the liner plug and shears it off, and both go to the landing ring. Above the liner the cement is lifted with a margin. Then the crew sets the packer at the top of the liner, lifts the tool and washes the extra cement out by reverse circulation. The gaps around liners are small, so friction and ECD are large, centring is difficult, and the fluid loss of the cement must be especially low. Rotating the liner during displacement helps.

Through an inner string (stab-in) the crew cements large-diameter conductors and surface casing. The shoe or the float valve of the string has a built-in seat with a seal. The crew stabs drill pipe into it and pumps cement through the pipe until it comes out at the surface along the annulus. No plugs are needed: the displacement volume is the volume of the pipe, not of the string, almost no cement stays in the string, and the job is shorter. The crew usually “under-displaces” the cement a little, so that the displacement fluid does not go past the shoe. Then the crew checks the float valve and pulls the pipe.

In reverse circulation, the slurry is pumped from the surface directly into the annulus, and the displaced fluid comes out through the string. A float valve is not run, or a special one is used. The weak rocks at the bottom are not loaded by the friction of lifting the cement, and the pumping time is shorter. The main difficulty is to know when the cement has reached the shoe, so as not to push it into the string. The crew watches the volume and the density of the fluid coming out of the string.

SchematicSpecial methods: two-stage cementing with a collar and a packer-cup below it; a liner hung in the previous string, with cement in the overlap; cementing of a surface casing through drill pipe stabbed into the seal in the shoe. Not to scale.

Terms and symbols

Top of cement (TOC): The depth to which the cement slurry is lifted behind the string.
Float collar: Equipment one or two pipes above the shoe: it keeps the cement in the annulus and is the seat for the plugs.
Shoe track: The pipes between the landing ring and the shoe, in which contaminated cement stays.
Bottom and top plugs: The wiper plug with a diaphragm ahead of the cement, and the solid displacement plug behind it.
The bump: The rise of pressure when the top plug lands on the bottom plug: the displacement is over.
Standoff ratio: The smallest gap between the string and the wall divided by the gap with perfect centring, in per cent.
Spacer: A fluid between the drilling mud and the cement slurry: a low-viscosity wash, or a weighted spacer.
Hierarchy of density and friction: Each following fluid is heavier than the previous one and gives a larger friction gradient, so it pushes the previous one out like a piston.
Imbalance of the columns: The difference of the hydrostatic pressures of the annulus and the string at the depth of the shoe. At the end of displacement the float valve holds it.
Free fall: A heavy fluid in the string goes down faster than the pumps deliver. A vacuum forms in the string, and the returns are larger than the pump rate.
Stage collar: A collar with ports that can be closed, for cementing the second stage.

Common mistakes and misconceptions

Wrong

“We calculate the volume of the open hole from the bit diameter.”

In fact

The hole is almost always wider than the bit, especially in shales. We take the caliper log with a margin, and with no log, the bit diameter with a large margin. Then we check what happens if the hole is smaller and the cement rises higher than planned.

Wrong

“The pump pressure in the middle of displacement is zero: the pumps are faulty or there are losses.”

In fact

Most likely the pumps are filling the void left after free fall: the returns at this time are smaller than the pump rate. A calculation made in advance, which shows when the pressure should appear, helps to tell this from lost circulation.

Wrong

“The plug has not landed, so we must keep pumping.”

In fact

The crew pumps no more than half of the volume of the shoe track. Otherwise displacement fluid, not cement, will be at the shoe.

Wrong

“Cement is heavier than mud, so it will displace the mud from the annulus by itself.”

In fact

Cement takes the path of least resistance and goes round the thickened mud on the narrow side of an eccentric annulus. We need centring, conditioning of the mud, spacers and pump rate.

Wrong

“If the hydrostatic pressure at the end does not exceed the leak-off pressure, the job is safe.”

In fact

At the end of displacement the friction of the cement slurries in the annulus is added to the hydrostatic pressure. In the example of the lesson it is about 0.06 g/cm³ under the surface casing shoe, even at a lowered rate.

In depth: advanced detail

How the free fall is calculated

The lesson's model runs the pumping in small steps. The fluids move along the string and the annulus as a piston, without mixing. The position of each boundary is set by the volume that has left the shoe. At each step, the model checks whether the pump needs pressure at the current rate. It uses the formula for the imbalance of the columns, with friction by sections (the hydrostatic pressure and the losses by the formulas of the lesson on pressure losses, for each fluid by its own parameters).

Flow rate out of the shoe during free fall

p_{\text{str}}(z_{\text{top}}) - \Delta p_{\text{fr.str}}(Q) = p_{\text{ann}} + \Delta p_{\text{fr.ann}}(Q), \qquad \frac{\dd V_{\text{void}}}{\dd t} = Q - Q_{\text{pump}}

where:

$p_{\text{str}}(z_{\text{top}})$ the hydrostatic pressure of the column in the string at the shoe, when its top has gone down to the depth $z_{\text{top}}$ (above it there is a void), MPa
$p_{\text{ann}}$ the hydrostatic pressure of the columns in the annulus at the shoe, MPa
$\Delta p_{\text{fr.str}},\ \Delta p_{\text{fr.ann}}$ friction losses in the string and in the annulus at the rate $Q$ , MPa
$Q,\ Q_{\text{pump}}$ the flow rate out of the shoe (which is also the returns at the wellhead) and the pump rate, m³/min
$V_{\text{void}}$ the volume of the void at the top of the string, m³

Units: metric

Holds when

The pressure in the void is taken as atmospheric. In reality there is a partial vacuum there, which holds the fall back a little: the void is slightly smaller than calculated.
We ignore the inertia of the fluids: at each moment the flow rate is the one at which the heads balance.
The fluid being pumped lies at once on top of the column. The rathole is filled with mud and does not take part in the flow. The volume of the head and the piping (0.2 m³) is not included: it only shifts the displacement by 0.2 m³.

While the void grows, the top of the column in the string goes down, its pressure falls, and the outflow slows by itself. When the heavy fluids pass into the annulus, the outflow becomes smaller than the pump rate, and the void is filled. When it disappears, the ordinary formula for the pump pressure works again.

The calculation also shows what is hard to estimate by hand: how much the returns grow, and with them the friction in the annulus, during the fall. In the example of the lesson this is little, because the annulus is wide. But in narrow gaps a sharp rise of returns can raise the ECD at weak rocks higher than at the end of displacement. Specialised cementing programs add inertia, mixing at the boundaries, and a thermal calculation.

Connections to other topics

Oilwell cements and additives — cement classes, the density and yield of a slurry, thickening time and gas migration: the recipes of the lesson's lead and tail slurries.
Wellheads and running casing — the same well and the same string: running, fill-up, landing and the hook load after cementing.
Casing: purpose, types, properties and standards — the shoe, the float collar, the centralisers and the stage collar as parts of a string.
Casing design — the test pressure at the bump, and the loads during and after cementing.
Circulating pressure losses — the formulas for friction in the annulus and for ECD, from which the pressures during cementing are calculated.
Fracture gradient and leak-off testing — the leak-off pressure under the shoe and the fracture pressure, the upper limit of the pressures during cementing.
Well control: primary barrier and kick indicators — communicating vessels, the barriers of the well and gas migration as a threat to control.

Check yourself

1
Why is the float collar placed not in the shoe but one or two pipes higher?
1. A.To reduce the displacement volume
2. B.To make the string lighter
3. C.The centralisers require it
4. D.So that the contaminated cement that the plug collects from the walls stays in the shoe track, and clean cement comes out to the shoe
Show the answer
Correct answer: D. So that the contaminated cement that the plug collects from the walls stays in the shoe track, and clean cement comes out to the shoe
The top plug, which goes behind the cement, scrapes the leftover drilling mud and cement off the wall of the string, and contaminated cement builds up in front of it: the last portion. When the plug lands on the ring, this portion stays in the pipes between the ring and the shoe and is drilled out with them. Around the shoe, the most loaded place, there is clean cement. The shoe track also gives a margin for errors in the displacement volume.
2
A 177.8 mm string with a wall of 10.36 mm is cemented in 600 m of open hole by a caliper log of 225 mm with a margin of 15%; the shoe track is 20 m. What volume of cement slurry is needed, without the rathole?
Show the answer
Annulus: $\frac{\pi}{4}(0.225^2 - 0.1778^2) = 0.014932$ m²/m; $600\cdot 0.014932\cdot 1.15 \approx 10.30$ m³. Inside diameter: $177.8 - 2\cdot 10.36 = 157.08$ mm, and the shoe track is $20\cdot\frac{\pi}{4}\cdot 0.15708^2 \approx 0.39$ m³. In total, about 10.7 m³.
3
In the middle of displacement, the returns from the annulus are one and a half times smaller than the pump rate, and the pump pressure is near zero. What is this most likely?
1. A.The float valve has closed
2. B.Lost circulation: the rock is fractured
3. C.The plug has landed too early
4. D.The pumps are filling the void left in the string after free fall
Show the answer
Correct answer: D. The pumps are filling the void left in the string after free fall
In lost circulation there is pressure at the pumps, but it is usually lower than calculated or stops rising, and the returns fall. A zero pressure with returns smaller than the pump rate is a sign that the string was on vacuum and the pumps are filling the void at the top. A calculation shows in advance when the void will be filled and the pressure will appear. If it does not appear on time, the crew looks further for the cause.
4
The calculated displacement volume has been used up, and the bump has not come. What should the crew do, and why?
Show the answer
The crew can pump no more than half the volume of the shoe track (in the example of the lesson, about 0.46 m³), and then stop. If the crew goes on, the cement from the shoe track will go into the annulus, and displacement fluid will be around the shoe. The causes may be that the plug has not left the head or has slipped past, or that the volume of the string or the pump rate differs from the calculation. The crew finds out after the job.
5
A 244.5 mm string in a 311.1 mm hole rests on a rigid centraliser of 300 mm diameter. What is the standoff at the centraliser?
1. A.About 67%
2. B.About 83%
3. C.About 56%
4. D.100%
Show the answer
Correct answer: B. About 83%
The nominal gap is $(311.1 - 244.5)/2 = 33.3$ mm, and the smallest gap is $(300 - 244.5)/2 = 27.75$ mm. The standoff is $27.75/33.3 \approx 0.83$ , that is, 83%. Between the centralisers it is smaller, because of the sag of the pipe.
6
Suppose the leak-off test under the surface casing shoe in the lesson's well had given 1.30 g/cm³, not 1.50. Could the crew cement by the same programme? At the end of displacement, under the surface casing shoe, the static value is 1.25 g/cm³, and with friction at 0.5 m³/min it is 1.31 g/cm³.
Show the answer
No: 1.31 g/cm³ is more than 1.30, so at the end of displacement the rock under the shoe would start to take fluid, although statically there is a margin. The options: displace even more slowly at the end (less friction), make the lead slurry lighter or lower its height above the surface casing shoe, and if there is still no margin, cement in two stages with the collar below the surface casing shoe. Each option is checked by calculation again, including the worst case, when the cement rises higher than planned.

Updated: 16 October 2026