Rig circulating and rotary systems

Quick refresher

The mud moves round a closed loop: tanks → charge pumps and mud pumps → discharge manifold → standpipe → rotary hose → swivel or top drive → string → bit → annulus → flowline → solids control → tanks.
A mud pump is a reciprocating positive-displacement pump. This means its pistons move back and forth, and each stroke pushes out a fixed volume almost regardless of the pressure. In field practice a stroke is one revolution of the shaft. In a single-acting triplex pump, three cylinders work per stroke. In a double-acting duplex pump, two cylinders work, each in both directions.
The actual output is smaller than the calculated one, by the factor of the volumetric efficiency, about 0.9–1.0. Worn valves and pistons, gas and poor filling of the cylinders lower it.
A smaller liner allows a higher pressure, because the rod load is the pressure acting on the piston area. But it gives less output per stroke. The power of the drive limits the product of pressure and flow rate.
The strokes to the bit are the internal volume of the string divided by the output per stroke. The strokes up the annulus are the volume of the annulus divided by the same output. The stroke counter shows a true volume only if the volumetric efficiency is right.
The rotary table turns the string through the kelly, and the string is lengthened one joint at a time. The top drive turns the string from above. With it the crew can drill with stands, and can rotate and circulate the string while pulling out. The power of rotation is $N_{\text{rot}} = M\omega$ .

Key relationships

V_{\text{stk}} = 3\cdot\frac{\pi}{4}\,d^2 L\,\eta_v, \qquad V_{\text{stk}} = \frac{\pi}{2}\left(2d^2 - d_{\text{rod}}^2\right) L\,\eta_v

output per stroke of a single-acting triplex pump and of a double-acting duplex pump

Q = \frac{V_{\text{stk}}\,n}{60}, \qquad N_{\text{hyd}} = p\,Q, \qquad p_{\max} = \frac{F_{\text{rod}}}{\pi d^2/4}

pump rate, hydraulic power (MPa × L/s = kW) and the allowable pressure of a liner

Z_{\text{bit}} = \frac{V_{\text{string}}}{V_{\text{stk}}}, \qquad Z_{\text{ann}} = \frac{V_{\text{ann}}}{V_{\text{stk}}}

strokes to the bit and up the annulus

N_{\text{rot}} = M\,\omega = \frac{2\pi\,M\,n_{\text{rot}}}{60}

power of rotating the string: kN·m × rad/s = kW

Common pitfalls

Calculating the flow rate from the strokes with the nameplate (rated) volumetric efficiency when the valves and pistons are worn. The volumes and the circulation times then “drift”.
Drilling the lower intervals with large liners. The pumps reach the allowable liner pressure before they reach the power limit.
Confusing a pump stroke with a piston stroke. In the field count, three cylinders of a triplex pump deliver in one stroke.
Taking the torque at the rotary table or the top drive (the turning force) for the torque at the bit. In a deviated well most of it goes to friction of the string against the walls.
Explaining a change in standpipe pressure by a single cause. The pressure falls with a washout in the string, with pump wear and with a kick.

After this lesson you can

describe the path of the mud through the surface circulating system and what each element is for: tanks, pumps, manifold, standpipe, rotary hose, swivel or top drive
explain how a reciprocating (piston) mud pump is built, how a single-acting triplex pump differs from a double-acting duplex pump, and what the liners, valves, pulsation dampener and charge pump do
calculate the output of a pump per stroke and per unit of time, using the volumetric efficiency; find the allowable pressure for the liners; and choose liners for an interval
calculate the strokes to the bit and up the annulus and the circulation time, and check the volumetric efficiency by measurement
compare the rotary table with a kelly and the top drive, and calculate the power needed to rotate the string
list the parameters the driller watches, explain how they are measured and what changes in them mean

The path of the mud and the surface circulating system

The mud circulates round a closed loop. Centrifugal charge pumps (pumps that spin the mud outward) take it from the suction tank and feed it to the intake of the mud pumps. The mud pumps force the mud under pressure into the discharge manifold, a system of pipes and valves. From there it goes on into the standpipe, a vertical pipe on the derrick. From the top of the standpipe, a flexible rotary hose carries the mud to the swivel or the top drive, and from there into the drill string. The mud leaves through the bit nozzles and rises up the annulus. At the wellhead it enters the flowline (the pipe or channel that carries the mud away from the wellhead), and it returns to the tanks through the solids control equipment. The lesson on pressure losses during circulation covers how the pump pressure is used up along this path.

The surface part of the circulating system

Element: Suction (active) tank
Purpose: the pumps take the mud from it; its volume is watched all the time, because a gain or a loss reveals a kick or lost circulation

Element: Reserve tanks, slug tank
Purpose: a reserve of mud, weighted mud, and viscous and other slugs (pills), which are small batches of special mud pumped separately

Element: Mixing unit
Purpose: a jet mixer (a hopper with an eductor), barite (a heavy powder that weights the mud) and bentonite (a clay powder) hoppers, a tank for chemicals

Element: Agitators and mud guns
Purpose: mixers that keep the weighting material and the solids in suspension and even out the properties of the mud between the tanks

Element: Charge pumps
Purpose: centrifugal pumps that fill the cylinders of the mud pumps under pressure

Element: Mud pumps, manifold, standpipe
Purpose: create the pressure and the flow; the manifold valves direct the mud to the standpipe, to the fill-up line or to a tank

Element: Rotary hose, swivel or top drive
Purpose: a flexible link to the moving travelling block, and the point where the mud enters the rotating string

Element: Flowline and solids control equipment
Purpose: shale shakers (vibrating screens), a degasser (takes gas out of the mud), hydrocyclones and centrifuges (spin solids out of the mud) (Solids control in drilling fluids)

Element: Trip tank
Purpose: a small tank that measures volumes accurately during trips (tripping)

Element	Purpose
Suction (active) tank	the pumps take the mud from it; its volume is watched all the time, because a gain or a loss reveals a kick or lost circulation
Reserve tanks, slug tank	a reserve of mud, weighted mud, and viscous and other slugs (pills), which are small batches of special mud pumped separately
Mixing unit	a jet mixer (a hopper with an eductor), barite (a heavy powder that weights the mud) and bentonite (a clay powder) hoppers, a tank for chemicals
Agitators and mud guns	mixers that keep the weighting material and the solids in suspension and even out the properties of the mud between the tanks
Charge pumps	centrifugal pumps that fill the cylinders of the mud pumps under pressure
Mud pumps, manifold, standpipe	create the pressure and the flow; the manifold valves direct the mud to the standpipe, to the fill-up line or to a tank
Rotary hose, swivel or top drive	a flexible link to the moving travelling block, and the point where the mud enters the rotating string
Flowline and solids control equipment	shale shakers (vibrating screens), a degasser (takes gas out of the mud), hydrocyclones and centrifuges (spin solids out of the mud) (Solids control in drilling fluids)
Trip tank	a small tank that measures volumes accurately during trips (tripping)

The design decides the number and volume of the tanks. There must be enough mud at the surface, with a margin, to fill the well and to make up for losses.

A rig usually has two or three pumps. They are connected to the manifold in parallel, so their outputs add up. The flow rate we need is not the only thing that decides how many to run. The power matters too. In the examples below, 30 L/s at 12.5 MPa is already beyond a single pump, and both pumps run at half the stroke rate that one pump would need. If one pump is enough, the second is kept ready to start. The failure of a pump must not interrupt circulation for long.

The mud pump: how it works

Mud pumps are reciprocating positive-displacement pumps. A piston moves back and forth in a cylinder. On the backward stroke the cylinder fills with mud through the suction valve. On the forward stroke the piston pushes the mud out through the discharge valve. The output hardly depends on the pressure. The pump displaces its volume even when the pressure rises, so with the line shut the pressure builds up until the relief valve opens. Such pumps suit high pressures and an abrasive fluid (one that wears metal) with solids. We can calculate their output from the number of strokes.

A pump has two parts. In the power end, the crankshaft, connecting rods and crossheads turn rotation into the back-and-forth motion of the rod. The fluid end holds the replaceable cylinder liners (the sleeves of the cylinders), the pistons on the rods, and the suction and discharge valves with their seats. The liners, pistons and valves wear fastest, and the crew changes them on the rig.

SchematicA cylinder of a single-acting triplex pump (left) and of a double-acting duplex pump (right) on the forward stroke of the piston. In the single-acting cylinder the mud is pushed out only in front of the piston. In the double-acting cylinder, mud is pushed out in front of the piston while the space behind it fills. On the return stroke the reverse happens, but the volume behind the piston is smaller by the volume of the rod. Not to scale.

Triplex and duplex mud pumps

: Action
Triplex (three-piston): single-acting: discharges on the forward stroke
Duplex (two-piston): double-acting: discharges in both directions

: Cylinders
Triplex (three-piston): 3, 120° apart
Duplex (two-piston): 2, 90° apart

: Stroke rate
Triplex (three-piston): higher
Duplex (two-piston): lower

: Flow
Triplex (three-piston): smoother
Duplex (two-piston): strongly pulsating

: Weight, maintenance
Triplex (three-piston): lighter, simpler and cheaper
Duplex (two-piston): heavier; the rod seal works under pressure

	Triplex (three-piston)	Duplex (two-piston)
Action	single-acting: discharges on the forward stroke	double-acting: discharges in both directions
Cylinders	3, 120° apart	2, 90° apart
Stroke rate	higher	lower
Flow	smoother	strongly pulsating
Weight, maintenance	lighter, simpler and cheaper	heavier; the rod seal works under pressure

Almost all modern rigs have triplex pumps. Duplex pumps are found on older rigs. The advanced section covers the flow pulsations.

Three important devices are fitted to a pump. The pulsation dampener on the discharge line is a vessel with an elastic diaphragm, filled with nitrogen. It smooths the pressure fluctuations caused by the thrusts of the pistons. The centrifugal charge pump fills the cylinders under pressure. Without it the mud has no time to enter the cylinder during the stroke, especially at a high stroke rate and with a thick mud, and the volumetric efficiency falls. The relief valve on the discharge line releases the pressure if it goes above what the liners and the piping allow, for example with a closed valve or a plugged bit.

Pump output per stroke and per unit of time

In field practice a pump stroke is one revolution of the crankshaft. In one stroke, each of the three pistons of a triplex pump moves forward once. Each of the three cylinders displaces its own volume $\frac{\pi}{4}d^2L$ . In a duplex pump, each of the two pistons delivers both forward (the full volume) and backward (the volume less the rod). The stroke rate of a pump is given in strokes per minute (spm). The stroke counters on the driller's console count the strokes from the start of pumping.

Output of a mud pump per stroke and per unit of time

\begin{gathered} V_{\text{stk}} = 3\cdot\frac{\pi}{4}\,d^2 L\,\eta_v \ \ \text{(triplex)}, \qquad V_{\text{stk}} = \frac{\pi}{2}\left(2d^2 - d_{\text{rod}}^2\right) L\,\eta_v \ \ \text{(duplex)},\\ Q = \frac{V_{\text{stk}}\,n}{60} \end{gathered}

where:

$V_{\text{stk}}$ output per stroke (revolution of the shaft), m³
$d,\ d_{\text{rod}}$ inside diameter of the liner (piston diameter) and diameter of the rod, m
$L$ length of the piston stroke, m
$\eta_v$ volumetric efficiency (fill factor) — the fraction of the calculated volume that is actually delivered, —
$n$ strokes per minute, strokes/min
$Q$ pump rate, m³/s

Units: SI. It is convenient to take $d$ and $L$ in decimetres, because then $V_{\text{stk}}$ comes out directly in litres

Holds when

The volume swept by the piston is calculated from the full stroke. In a duplex pump the return stroke displaces the volume of the ring-shaped space between the liner and the rod.
The volumetric efficiency accounts for leakage through the valves and pistons, incomplete filling of the cylinders and compression of gas in the mud. For a pump in good order with a charge pump it is close to 1. For a worn pump it falls to 0.9 and below.
With several pumps the outputs add up, and the total counter counts the strokes of all the pumps.

The output is proportional to the square of the liner diameter and to the stroke rate. Change the liners to change the output per stroke. Change the stroke rate to change the output per minute.

In other unit systems (1)

Oilfield units

V_{\text{stk}}\,[\text{bbl}] = 0.000243\,d^2 L\,\eta_v\ \ \text{(triplex)}, \qquad V_{\text{stk}}\,[\text{bbl}] = 0.000162\left(2d^2 - d_{\text{rod}}^2\right) L\,\eta_v

$d$ , $d_{\text{rod}}$ and $L$ are in inches, the output per stroke in barrels; the pump rate in gallons per minute is $Q = d^2 L\,\eta_v\,n/98.04$ (triplex) and $(2d^2 - d_{\text{rod}}^2)L\,\eta_v\,n/147.1$ (duplex).

$0.000243 = (3\pi/4)/9702$ and $0.000162 = (\pi/2)/9702$ , because a barrel holds 9702 cubic inches. $98.04 = 4\cdot 231/(3\pi)$ and $147.1 = 2\cdot 231/\pi$ , because a gallon holds 231 cubic inches.

Worked exampleOutput per stroke and stroke rate for 30 L/s

Given (assumed data)

Pumps: two triplex pumps, 160 mm liners, piston stroke 228.6 mm (9″), at most 140 strokes per minute
Volumetric efficiency: 0.95
Required pump rate: 30 L/s — as in the lessons on pressure losses and on bit hydraulics

Solution

Volume of one cylinder per stroke (dimensions in decimetres): $\frac{\pi}{4}\cdot 1.60^2\cdot 2.286 = 4.596$ L.
Three cylinders: $3\cdot 4.596 = 13.79$ L; with the volumetric efficiency: $13.79\cdot 0.95 = 13.10$ L per stroke.
For 30 L/s we need $30\cdot 60/13.10 = 137.4$ strokes/min in all, which is 68.7 strokes/min on each of the two pumps.
One pump would give 30 L/s at 137.4 strokes/min. But at a standpipe pressure of 12.5 MPa it would have to give the mud $12.5\cdot 30 = 375$ kW. That is more than its drive allows: $0.90\cdot 390 = 351$ kW.
In oilfield units: $13.10/158.99 = 0.0824$ bbl per stroke. The pump rate is $30/0.06309 = 476$ gpm.

Answer: The output per stroke is 13.10 L. Two pumps at 68.7 strokes/min each give 30 L/s.

What it means. We convert the stroke counter into a volume with this output per stroke, so the volumetric efficiency must be right. If the valves have worn and it has fallen, the actual pump rate will be lower than the counter shows. We check this by measurement (see the example below) or with a flowmeter. The oilfield formula gives the same result: $0.000243\cdot 6.299^2\cdot 9\cdot 0.95 = 0.0824$ bbl. The pump data are assumed for illustration. The pump rate of 30 L/s and the pump limits are the same as in the lessons on pressure losses and on bit hydraulics.

Pressure, power and the choice of liners

A pump is limited in three ways. The pressure is limited by the load on the rod. The pressure of the mud acts on the piston area and presses on the rod and the whole power end, and the manufacturer sets the largest rod load for the pump. So each liner has its own allowable pressure: the smaller the diameter, the higher the pressure. The flow rate with a given liner is limited by the highest stroke rate. The power is limited by the drive. The hydraulic power $pQ$ cannot exceed the drive power multiplied by the mechanical efficiency of the pump, which is about 0.9. In addition, the pressure must not exceed what the manifold, standpipe, hose and swivel allow, and the relief valve is set with a margin above the working pressure.

Allowable liner pressure and the power limit

p_{\max} = \frac{F_{\text{rod}}}{\pi d^2/4}, \qquad N_{\text{hyd}} = p\,Q \le \eta_{\text{mech}}\,N_{\text{drive}}

where:

$p_{\max}$ allowable discharge pressure for a liner of diameter $d$ , MPa
$F_{\text{rod}}$ largest rod load according to the pump data sheet, MN
$d$ liner diameter, m
$N_{\text{hyd}}$ hydraulic power that the pumps give to the mud, kW (MPa × L/s)
$\eta_{\text{mech}},\ N_{\text{drive}}$ mechanical efficiency of the pump and the power of its drive, —, kW

Units: MN/m² = MPa; MPa × L/s = kW

Holds when

The strength of the fluid end, and not only the rod load, may also limit the pressure. For small liners the rated pressure can be lower than the one calculated from the rod load.
Pump data sheets give the allowable pressure and the output at the highest stroke rate for each liner in a table. The formula explains where these numbers come from.

On a plot of flow rate against pressure, each liner gives its own operating envelope (the region where the pump can work). Its boundary has three parts: a horizontal line of allowable pressure, a power hyperbola that is common to all liners, and a vertical line of the largest flow rate. The lesson on bit hydraulics covers how the operating point for the bit is chosen within this envelope.

CalculatedOperating envelopes of the two pumps of the example, with 160, 170 and 180 mm liners (the numbers by the vertical lines). The envelopes are set by the allowable pressure from the rod load of 362 kN, the common power limit of 702 kW and the largest flow rate at 140 strokes/min. Point A is drilling the 215.9 mm hole at 30 L/s and 12.5 MPa. Point B is the same well with new nozzles, 30.8 L/s at 18.0 MPa. Point C is the upper interval, 65 L/s at 8 MPa.

Worked exampleWhich liners to put in the pumps

Given (assumed data)

Pumps: two, stroke 228.6 mm, at most 140 strokes per minute, rod load 362 kN, drive 390 kW each, mechanical efficiency 0.90, volumetric efficiency 0.95
Upper interval: 393.7 mm hole: 65 L/s is needed at a pressure of about 8 MPa (assumed)
Lower interval: 215.9 mm hole: the operating conditions chosen in the lesson on bit hydraulics — 30.8 L/s at 18.0 MPa

Solution

Power limit of the two pumps: $2\cdot 0.90\cdot 390 = 702$ kW. The upper interval needs $8\cdot 65 = 520$ kW and the lower one $18.0\cdot 30.8 = 554$ kW, so the power is enough in both.
Allowable pressure $p_{\max} = 0.362/(\pi d^2/4)$ : for 160 mm, $0.362/0.02011 = 18.0$ MPa; for 170 mm, 15.9 MPa; for 180 mm, 14.2 MPa.
Output per stroke of one pump: 160 mm, 13.10 L; 170 mm, 14.79 L; 180 mm, 16.58 L. The largest flow rate of the two pumps at 140 strokes/min is 61.1, 69.0 and 77.4 L/s.
Upper interval: 160 mm liners will not give 65 L/s (the limit is 61.1 L/s). 170 mm and 180 mm liners will, and the pressure of 8 MPa is below the allowable one for both. The stroke rate for 65 L/s is $65\cdot 60/(2\cdot 14.79) = 132$ with 170 mm liners and $65\cdot 60/(2\cdot 16.58) = 118$ with 180 mm liners.
Lower interval: only liners of 160 mm and smaller allow 18.0 MPa (for 170 mm the limit is 15.9 MPa). The flow rate of 30.8 L/s needs $30.8\cdot 60/(2\cdot 13.10) = 71$ strokes/min per pump.

Answer: In the upper interval the crew fits 180 mm liners, and the pumps run at 118 strokes/min instead of 132. Before drilling the 215.9 mm hole, it changes the liners to 160 mm, the largest size that allows 18 MPa.

What it means. We choose the largest liner whose allowable pressure covers the expected working pressure. With a large liner the required flow rate is reached at a lower stroke rate, so the valves, pistons and liners wear more slowly. Changing the liners when moving to the next interval is routine work. The 160 mm liners give exactly the limits of the lesson on bit hydraulics: 18.0 MPa and about 700 kW. The operating conditions of the lower interval lie right at the limit of their pressure (point B in the figure above). The relief valve is set no higher than the allowable pressure of the liners fitted. So such operating conditions leave almost no margin for pressure spikes. The data in this example are assumed for illustration.

Strokes and volumes: to the bit, up the annulus, a full circuit

A positive-displacement pump is also a flowmeter: each stroke is a known volume. So on the rig volumes are counted in strokes. How many strokes does a slug pumped from the surface need to reach the bit? How many to bring the mud from the bottom up to the wellhead (“bottoms up”)? How many for a full circuit of the well? We find the time by dividing the volume by the pump rate. This is also how the lag time of cuttings is calculated (the lesson on the drilling process).

Strokes to the bit and up the annulus

Z_{\text{bit}} = \frac{V_{\text{string}}}{V_{\text{stk}}}, \qquad Z_{\text{ann}} = \frac{V_{\text{ann}}}{V_{\text{stk}}}, \qquad t = \frac{V}{Q}

where:

$Z_{\text{bit}},\ Z_{\text{ann}}$ strokes to the bit and up the annulus from the bit to the wellhead, strokes
$V_{\text{string}}$ internal volume from the pumps to the bit: surface lines, drill pipe, drill collars and the rest of the assembly, m³
$V_{\text{ann}}$ volume of the annulus from the bit to the wellhead, by sections, m³
$V_{\text{stk}}$ actual output per stroke (with the volumetric efficiency); with several identical pumps, the output per stroke of any one of them, and the strokes are counted in total, m³
$t,\ V,\ Q$ time to pump a volume $V$ at pump rate $Q$ , s, m³, m³/s

Units: SI; the volume of the hole and of the pipe is $\frac{\pi}{4}(D^2 - d^2)$ per metre of length

Holds when

The pipe and the hole have their nominal diameters. Washouts increase the volume of the annulus, so the actual time is checked with a tracer.
The mud is incompressible and is not lost anywhere. With lost circulation or an influx the volumes stop adding up, and that is the warning sign.

The number of strokes does not depend on the pump rate; the time does. That is why pumping schedules for a kill and for circulating the hole are built in strokes (the pressure schedule for killing the well).

Worked exampleStrokes to the bit and “bottoms up” in a 3000 m well

Given (assumed data)

Well: from the lesson on pressure losses: the bit on the bottom at 3000 m; casing to 2300 m with an inside diameter of 222.4 mm, below it a 215.9 mm hole
String: drill pipe 127 × 108.6 mm, 150 m of 158.8 × 71.4 mm drill collars; surface lines — standpipe 30 m × 101.6 mm, hose 20 m and the bore of the top drive 8 m × 76.2 mm
Pumps: two, 13.10 L per stroke (160 mm liners, the example above), 30 L/s in total

Solution

Internal volume: surface lines 0.371 m³. Drill pipe: $2850\cdot\frac{\pi}{4}\cdot 0.1086^2 = 26.40$ m³. Drill collars: $150\cdot\frac{\pi}{4}\cdot 0.0714^2 = 0.601$ m³. In total 27.37 m³.
Strokes to the bit: $27.37/0.01310 = 2089$ . The time at 30 L/s is $27.37/0.030 = 912$ s, about 15 min.
Annulus: around the drill collars $150\cdot\frac{\pi}{4}(0.2159^2 - 0.1588^2) = 2.52$ m³. Around the drill pipe in open hole $550\cdot\frac{\pi}{4}(0.2159^2 - 0.127^2) = 13.17$ m³. In the casing $2300\cdot\frac{\pi}{4}(0.2224^2 - 0.127^2) = 60.21$ m³. In total 75.90 m³.
Strokes for “bottoms up”: $75.90/0.01310 = 5794$ . The time is $75.90/0.030 = 2530$ s, about 42 min.
A full circuit of the well takes $2089 + 5794 = 7883$ strokes, about 57 min.

Answer: The strokes to the bit are 2089 (about 15 min), “bottoms up” 5794 (about 42 min), a full circuit 7883.

What it means. The annulus holds almost three times as much as the string, so “bottoms up” is the longest part of the circuit. Before a trip, the crew circulates for at least this long to carry out the cuttings and gas (tripping). We need the strokes to the bit for the kill schedule and to place a slug exactly at the bit. The line from the pumps to the standpipe was not included in the volume. If the actual output per stroke is lower, the real mud lags behind the stroke count. The data in this example are assumed for illustration.

Worked exampleChecking the volumetric efficiency by measurement

Given (assumed data)

Pump: one, 160 mm liners, stroke 228.6 mm — calculated volume 13.79 L per stroke
Measurement: the pump takes mud from an isolated calibrated tank (a tank with a measured volume scale). In 250 strokes the level in it fell by a volume of 3.18 m³

Solution

Actual output per stroke: $3180/250 = 12.72$ L.
Volumetric efficiency: $12.72/13.79 = 0.922$ , instead of the assumed 0.95.
Two such pumps at a total of 137.4 strokes/min give a pump rate of $137.4\cdot 12.72/60 = 29.13$ L/s, not 30.
“Bottoms up” in the well of the example above: the number of strokes is $75.90/0.01272 = 5967$ instead of 5794. The time is $75.90/0.02913 = 2606$ s, about 43.4 min instead of 42.2 min.

Answer: The volumetric efficiency is about 0.922. The pump delivers 3% less than assumed in the calculations, and the strokes for “bottoms up” are in fact 173 more than calculated.

What it means. Such a drop is a reason to inspect the valves, seats and pistons, and to check the charge pump. Until the pump is repaired, we calculate volumes from the measured output per stroke. Otherwise slugs will end up in the wrong place, cuttings samples will be shifted in depth, and the kill pressure schedule will be built on the wrong number of strokes. The data in this example are assumed for illustration.

Rotating the string: the rotary table, swivel and top drive

The rotary table is a rotating table at the centre of the rig floor, driven by an engine. The master bushing is fitted in its opening, and the kelly bushing sits in the master bushing. Its rollers grip the square or hexagonal kelly (the top pipe of the string) and pass the rotation to it. They let the kelly slide down as the hole deepens. While drilling, the table turns to the right, which is clockwise seen from above. When the string is lengthened or pulled, the crew hangs it in the rotary table on slips (wedges that grip the pipe).

Above the kelly is the swivel. It hangs from the hook, carries the weight of the whole string and lets the string rotate. The mud passes from the rotary hose into the rotating string through the swivel's gooseneck (a curved pipe) and its packed washpipe (a pipe with seals). The lesson on the drill string covers how the kelly itself, its valves and its sub (a short connecting piece) are built.

The top drive combines the swivel and the motor. Electric motors (AC or DC) or hydraulic motors turn a hollow shaft through a gearbox, and the string is screwed onto this shaft. The drive hangs from the travelling block and moves along guides that take up the reaction torque (the turning force that pushes back on the drive). Beneath it is a pipe-handling mechanism: a rotating link hanger, link tilt, an elevator and a torque wrench for making up and breaking out connections. There are also ball valves for shutting off the string (valves in the string). The rotary table stays in place with a top drive, because the string is set on slips in it.

Rotary table with a kelly and a top drive

: What turns the string
Rotary table and kelly: the rotary table through the kelly bushing
Top drive: the drive motor through its own shaft

: Making a connection
Rotary table and kelly: one joint at a time, about 9–12 m: the kelly is unscrewed and screwed onto the next joint from the mousehole
Top drive: a stand of three joints, about 27–28 m: the drive is screwed onto the stand from above

: Tripping
Rotary table and kelly: the kelly is set back in the rathole; the string cannot be rotated or circulated unless the kelly is made up again
Top drive: the string can be connected to quickly at any height, and rotated and circulated while it is moving

: Features
Rotary table and kelly: simpler and cheaper
Top drive: heavier moving equipment; power and control lines have to run up the derrick

	Rotary table and kelly	Top drive
What turns the string	the rotary table through the kelly bushing	the drive motor through its own shaft
Making a connection	one joint at a time, about 9–12 m: the kelly is unscrewed and screwed onto the next joint from the mousehole	a stand of three joints, about 27–28 m: the drive is screwed onto the stand from above
Tripping	the kelly is set back in the rathole; the string cannot be rotated or circulated unless the kelly is made up again	the string can be connected to quickly at any height, and rotated and circulated while it is moving
Features	simpler and cheaper	heavier moving equipment; power and control lines have to run up the derrick

The mousehole is a sleeve in the rig floor that holds the next joint; during a trip the kelly with the swivel is set back in the neighbouring rathole.

The crew screws connections together, and unscrews them, with tongs (large wrenches): one holds the pipe and the other turns it. A pneumatic or hydraulic spinner spins the connection up first. On many rigs a single machine does all this work. It is called the iron roughneck. It moves up to the pipe at the rotary table, spins up the connection itself and tightens it to the specified torque. People do not have to stand with tongs next to the rotating pipe, and a consistent torque protects the threads. Slips also come with a pneumatic or hydraulic drive.

Power of rotating the string

N_{\text{rot}} = M\,\omega, \qquad \omega = \frac{2\pi\,n_{\text{rot}}}{60}

where:

$N_{\text{rot}}$ power at the shaft of the rotary table or top drive, kW
$M$ torque on the string at the surface, kN·m
$\omega,\ n_{\text{rot}}$ angular velocity and rotary speed, rad/s, rpm

Units: kN·m × rad/s = kW

Holds when

The rotation is steady. The torque at the surface is the sum of the torque at the bit and the friction of the string against the walls. In a deviated well the friction is usually the larger part (drag and torque).
The power at the input of the drive is higher by its efficiency.

A top drive gives its full torque at low speed. Above a certain speed it runs into the power limit, and the available torque falls in inverse proportion to the speed. The strength of the string also limits the torque. The limit is set below the make-up torque (the torque used to tighten a connection) of the weakest connection (the top of the string).

In other unit systems (1)

Oilfield units

N_{\text{rot}}\,[\text{hp}] = \frac{M\,[\text{ft·lbf}]\;n_{\text{rot}}\,[\text{rpm}]}{5252}

the torque is in foot-pounds force, the speed in revolutions per minute, the power in horsepower.

$5252 = 33{,}000/(2\pi)$ , because 1 hp = 33,000 ft·lbf per minute and in one revolution the shaft turns through $2\pi$ radians. 1 kN·m = 737.6 ft·lbf.

Worked exampleHow much power the rotation of the string takes

Given (assumed data)

Torque while drilling: 15.8 kN·m — the directional well from the lesson on drill string design; of this 4.0 kN·m is the torque at the bit
Rotary speed: 120 rpm
Top drive (assumed): continuous torque 45 kN·m, power 600 kW

Solution

Angular velocity: $\omega = 2\pi\cdot 120/60 = 12.57$ rad/s. Power: $N_{\text{rot}} = 15.8\cdot 12.57 = 198.6$ kW, or $198.6/0.7457 = 266$ hp.
Check in oilfield units: $15.8$ kN·m $= 11{,}653$ ft·lbf, and $11{,}653\cdot 120/5252 = 266$ hp.
The bit accounts for $4.0\cdot 12.57 = 50.3$ kW, which is 25%. The rest goes to friction of the string against the walls.
The corner speed of the drive (the speed up to which the torque stays constant): $60\cdot 600/(2\pi\cdot 45) = 127$ rpm. At 120 rpm the drive gives the full 45 kN·m, which is 2.8 times more than needed. At 180 rpm it gives only $600/(2\pi\cdot 180/60) = 31.8$ kN·m.

Answer: Rotating the string takes 198.6 kW, a third of the drive power. Only 50.3 kW of this reaches the bit.

What it means. The torque margin is needed not for average drilling but for the peaks. When the string rotates unstably or gets stuck, the torque rises for a short time to several times its normal value. So a torque limit is set on the drive, below the make-up torque of the weakest tool joint. The drive then stops before it over-tightens or twists off the string. The power at the bit is only part of the power of rotation. In deviated wells this part is the smaller, the greater the friction. The data in this example are assumed for illustration.

Monitoring the drilling parameters

We cannot see the well, so we judge what happens at the bottom from instruments. Their readings are brought together on the driller's console. The mud logging unit records them continuously, against time and against depth, and adds its own: gas, mud properties and a description of the cuttings. The values themselves matter less than their changes when the drilling parameters have not been changed.

What is measured on the rig

Parameter: Hook load, weight on bit
How it is measured: a tension sensor on the dead line of the drilling line; the weight on bit is the difference between the weight off bottom and on bottom
What for: drilling parameters, overpull and set-downs at tight spots (weight of the string)

Parameter: Depth, rate of penetration
How it is measured: a sensor of drawworks drum revolutions (the position of the block) and an “in slips” indicator
What for: footage drilled, change of rock, a sudden rise in the rate

Parameter: Rotary speed, torque
How it is measured: sensors on the drive of the rotary table or top drive; the torque from the motor current or from a sensor
What for: drilling parameters; tight spots, vibrations, unstable rotation

Parameter: Standpipe pressure
How it is measured: a pressure sensor on the standpipe manifold
What for: pressure losses in the system: a washout, a plugged nozzle, pump wear

Parameter: Pump strokes
How it is measured: stroke counters on each pump
What for: the flow rate in: the stroke rate multiplied by the output per stroke

Parameter: Flow out
How it is measured: a flow sensor in the flowline or a flowmeter
What for: comparison with the flow rate in: an influx or lost circulation

Parameter: Volume of mud
How it is measured: level sensors in the tanks, the total active volume, the trip tank
What for: a gain or a loss of mud

Parameter: Gas, hydrogen sulfide, mud density and temperature
How it is measured: gas detectors and sensors at the outlet and the inlet
What for: gas shows, contamination of the mud, danger to people

Parameter	How it is measured	What for
Hook load, weight on bit	a tension sensor on the dead line of the drilling line; the weight on bit is the difference between the weight off bottom and on bottom	drilling parameters, overpull and set-downs at tight spots (weight of the string)
Depth, rate of penetration	a sensor of drawworks drum revolutions (the position of the block) and an “in slips” indicator	footage drilled, change of rock, a sudden rise in the rate
Rotary speed, torque	sensors on the drive of the rotary table or top drive; the torque from the motor current or from a sensor	drilling parameters; tight spots, vibrations, unstable rotation
Standpipe pressure	a pressure sensor on the standpipe manifold	pressure losses in the system: a washout, a plugged nozzle, pump wear
Pump strokes	stroke counters on each pump	the flow rate in: the stroke rate multiplied by the output per stroke
Flow out	a flow sensor in the flowline or a flowmeter	comparison with the flow rate in: an influx or lost circulation
Volume of mud	level sensors in the tanks, the total active volume, the trip tank	a gain or a loss of mud
Gas, hydrogen sulfide, mud density and temperature	gas detectors and sensors at the outlet and the inlet	gas shows, contamination of the mud, danger to people

The design and the work programme define the set of sensors. On modern rigs the readings are also sent to the operator's office.

How to read the changes

What has changed: Standpipe pressure falls slowly at the same stroke rate
Possible causes: a washout in a tool joint or a pipe; pump wear (the volumetric efficiency falls); light fluid in the annulus during a kick

What has changed: Standpipe pressure rises sharply
Possible causes: a nozzle is plugged, the assembly is balled up (packed with sticky cuttings), the downhole motor has stalled (pressure drop across the bit)

What has changed: Torque rises and fluctuates
Possible causes: poor hole cleaning, a narrowing of the hole, unstable rotation of the string, a worn bit

What has changed: Hook load is higher than calculated when pulling out and lower when running in
Possible causes: drag (friction of the string against the wall of the hole) is increasing, because of cuttings, ledges or a narrowing of the hole (drag)

What has changed: Flow out rises, the volume in the tanks gains
Possible causes: a kick — signs of a kick

What has changed: Flow out falls, the volume decreases
Possible causes: mud is being lost to the formation

What has changed	Possible causes
Standpipe pressure falls slowly at the same stroke rate	a washout in a tool joint or a pipe; pump wear (the volumetric efficiency falls); light fluid in the annulus during a kick
Standpipe pressure rises sharply	a nozzle is plugged, the assembly is balled up (packed with sticky cuttings), the downhole motor has stalled (pressure drop across the bit)
Torque rises and fluctuates	poor hole cleaning, a narrowing of the hole, unstable rotation of the string, a worn bit
Hook load is higher than calculated when pulling out and lower when running in	drag (friction of the string against the wall of the hole) is increasing, because of cuttings, ledges or a narrowing of the hole (drag)
Flow out rises, the volume in the tanks gains	a kick — signs of a kick
Flow out falls, the volume decreases	mud is being lost to the formation

No single sign proves the cause on its own. We check it against the other parameters and against what the crew was doing. Transfers between tanks, filling the hole and adding water also change the volume in the tanks.

A lesson, not an operating procedure

The discharge lines, the standpipe and the hose work at pressures of tens of megapascals. The rupture of a line under pressure can be fatal. The relief valves of the pumps are never switched off or blocked. The rotary hose is secured with safety clamps. Nobody is kept at the lines during pressure tests and when the pumps are started.
The top drive, the rotary table and the tongs are equipment that rotates with a large torque. Work on them is done with the power isolated and locked out.
The signs of a kick call for action under approved well control procedures. The equipment data sheets, the design and the work programme, the industrial safety requirements and the rig's procedures set the sequence of work, the allowable pressures and the settings. Trained personnel carry out the work.

Terms and symbols

Mud pump: A reciprocating positive-displacement pump: a single-acting triplex or a double-acting duplex.
Pump liner: The replaceable sleeve of a cylinder. Its diameter decides the output per stroke and the allowable pressure.
Pump stroke, number of strokes: A stroke is one revolution of the crankshaft. The stroke rate of a pump is given in strokes per minute, and volumes are calculated from the stroke counters.
Volumetric efficiency $\eta_v$: The ratio of the actual output to the output calculated from the geometry. It is about 0.9–1.0, and it falls with wear.
Pulsation dampener: A vessel with nitrogen behind an elastic diaphragm on the discharge line. It smooths the pressure pulsations.
Strokes to the bit, “bottoms up”: The internal volume of the string and the volume of the annulus, expressed in pump strokes.
Rotary table, kelly, swivel: The rotating table; the many-sided pipe that it turns; and the device hung on the hook through which the mud enters the rotating string.
Top drive: A motor with a swivel on the travelling block that turns the string from above. It makes it possible to drill with stands and to rotate the string while pulling out.
Iron roughneck: A machine that screws connections together and unscrews them, with a metered torque.

Common mistakes and misconceptions

Wrong

“The pump output is the strokes multiplied by the volume of a cylinder.”

In fact

A triplex pump has three cylinders working per stroke. A duplex pump has two, each working in both directions. The actual output is lower than the calculated one, by the factor of the volumetric efficiency.

Wrong

“To raise the pressure, fit larger liners.”

In fact

It is the other way round. The smaller the liner, the higher the allowable pressure, because the rod load is the pressure multiplied by the piston area. Larger liners give a greater output at a lower pressure.

Wrong

“The standpipe pressure has fallen, so the mud has become lighter and the pump rate can be raised.”

In fact

A fall in pressure at the same stroke rate is a reason to look for the cause: a washout in the string, pump wear or a kick. First check the flow out, the volume in the tanks and the pumps.

Wrong

“The torque on the top drive is the torque at the bit.”

In fact

At the surface the torque is the sum of the torque at the bit and the friction of the string against the walls. In a deviated well the friction is often the larger part.

Wrong

“A top drive is needed only to drill faster.”

In fact

It also lets the crew connect to the string at any moment, and rotate and circulate it while pulling out. This matters with tight spots and with a kick during a trip.

In depth: advanced detail

Why a triplex pump delivers more smoothly

Over one revolution of the shaft, the piston speed varies almost like a sine wave. It is zero at the dead centres (the ends of the stroke) and greatest in the middle of the stroke. The output of a cylinder is proportional to the speed of its piston while the piston is discharging. In a single-acting triplex pump, three cylinders discharge 120° apart. When one piston slows down, the next speeds up, and the total flow varies only between 0.91 and 1.05 of the mean. In a double-acting duplex pump, two cylinders are 90° apart and each discharges in both directions, but the return stroke displaces less, by the volume of the rod. The fluctuations here are much greater: from 0.71 to 1.21 of the mean for a pump with 170 mm liners and a 70 mm rod.

CalculatedInstantaneous pump output relative to the mean over one revolution of the shaft: a single-acting triplex pump and a double-acting duplex pump (170 mm liners, 70 mm rod). Calculated for sinusoidal piston motion (the connecting rod is much longer than the crank).

Flow pulsations turn into pressure pulsations. These shake the lines, the hose and the connections, and they interfere with the signals of mud-pulse MWD tools (measurement-while-drilling tools, which send data up from the hole as pressure pulses in the mud). That is why a pulsation dampener is fitted on the discharge line of each pump and charged with nitrogen to the pressure the manufacturer recommends for the working pressure of the pump. The real picture also depends on the length of the connecting rod and on how the valves open. But the calculation shows the main thing correctly: with three single-acting cylinders the flow is several times smoother.

Where the duplex pump formula comes from

On the forward stroke the piston displaces the volume $\frac{\pi}{4}d^2L$ from the front chamber. On the return stroke it displaces mud from the rear chamber. But the rod occupies part of the cross-section of that chamber, so the volume is smaller: $\frac{\pi}{4}(d^2 - d_{\text{rod}}^2)L$ . In one revolution of the shaft one cylinder delivers the sum, $\frac{\pi}{4}(2d^2 - d_{\text{rod}}^2)L$ , and two cylinders deliver twice as much: $\frac{\pi}{2}(2d^2 - d_{\text{rod}}^2)L$ . So in the field count a duplex pump makes four discharges per stroke, and a triplex pump three.

The rod seal in the rear chamber works under the full discharge pressure and wears. This is one of the reasons why duplex pumps have given way to triplex pumps, in which the rod on the power end side is not exposed to the pressure of the mud.

Connections to other topics

How a well is drilled: the process and the people involved — the systems of a rig and the lag time of cuttings.
Circulating pressure losses — how the pump pressure is used up along the path of the mud.
Bit hydraulics optimisation — choosing the flow rate and the nozzles within the operating envelope of the pumps.
Solids control in drilling fluids — the solids control equipment of the circulating system.
Well control: primary barrier and kick indicators — kick indicators on the rig instruments and the trip tank.
Well kill methods — strokes to the bit in the kill schedule.
Drilling rig: power and hoisting systems — the power system, the block and tackle and the weight of the top drive on the hook.

Check yourself

1
Liners of 140 mm have been fitted in the triplex pumps of the example (piston stroke 228.6 mm); the volumetric efficiency is 0.95. What is the output of one pump per stroke?
1. A.20.1 L
2. B.10.56 L
3. C.3.34 L
4. D.10.03 L
Show the answer
Correct answer: D. 10.03 L
The volume of one cylinder is $\frac{\pi}{4}\cdot 1.4^2\cdot 2.286 \approx 3.519$ L (dimensions in decimetres). Three cylinders work per stroke, which gives 10.56 L. With the volumetric efficiency, $10.56\cdot 0.95 \approx 10.03$ L. 10.56 L is the calculated volume without losses, 3.34 L is only one cylinder with losses, and 20.1 L is two pumps.
2
Why do liners of a smaller diameter allow a higher discharge pressure?
1. A.They are thicker and stronger
2. B.A smaller piston moves faster
3. C.There is less leakage through them
4. D.The rod load equals the pressure multiplied by the piston area; with a smaller area the same force allows a higher pressure
Show the answer
Correct answer: D. The rod load equals the pressure multiplied by the piston area; with a smaller area the same force allows a higher pressure
The power end is designed for the largest rod load. With a smaller liner the same force corresponds to a higher pressure: $p_{\max} = F_{\text{rod}}/(\pi d^2/4)$ . In return the output per stroke is smaller, in proportion to $d^2$ .
3
The internal volume of a drill string is 24 m³, the pump output is 15 L per stroke, and the pump runs at 120 strokes per minute. How many strokes to the bit, and how many minutes is that?
Show the answer
$Z_{\text{bit}} = 24/0.015 = 1600$ strokes. At 120 strokes per minute this is $1600/120 \approx 13.3$ min. If the actual volumetric efficiency is lower than assumed, the slug will reach the bit later than the counter shows.
4
At a constant stroke rate the standpipe pressure fell from 18.0 to 16.5 MPa in an hour. What should be checked?
Show the answer
A fall in pressure at the same stroke rate means that the mud passes along its path more easily, or that less of it is being delivered. Possible causes are a washout in the string (the pressure falls gradually and keeps falling), pump wear (a lower volumetric efficiency) or a kick (light fluid in the annulus). First compare the flow out and the volume in the tanks. If they are normal, check the pumps. If a washout in the string is suspected, pull the string to find the washed-out pipe before it fails.
5
The rotary table turns the string at 90 rpm with a torque of 12 kN·m. What power is needed to rotate it?
Show the answer
$\omega = 2\pi\cdot 90/60 = 9.42$ rad/s, and $N_{\text{rot}} = 12\cdot 9.42 \approx 113$ kW (about 152 hp). The power at the input of the drive is higher by the efficiency of the rotary table transmission.
6
What advantages does a single-acting triplex pump have over a double-acting duplex pump?
1. A.Its output does not depend on the volumetric efficiency
2. B.It delivers more smoothly, is lighter, runs at a higher stroke rate and is simpler to maintain
3. C.It does not need a pulsation dampener
4. D.It delivers more per stroke with the same liners
Show the answer
Correct answer: B. It delivers more smoothly, is lighter, runs at a higher stroke rate and is simpler to maintain
Three cylinders 120° apart give an almost steady flow, the pump is lighter, it allows a higher stroke rate, and the rod seal does not work under pressure. It is still fitted with a pulsation dampener, and the volumetric efficiency is accounted for in the same way.

Updated: 28 October 2026