Drilling rig: power and hoisting systems

Quick refresher

The load $W$ is the weight of the string on the hook plus the weight of the travelling equipment (the travelling block, hook and top drive). It is carried by $n$ working lines, the parts of the rope between the two blocks. Without friction in the sheaves (the pulleys), each line would carry $W/n$ . With friction, the fast line, which the drawworks drum pulls, carries more tension when hoisting: $F_{\text{fl}} = W/(\eta n)$ .
The block efficiency $\eta$ falls as the number of lines grows. With the API RP 9B calculation factor for sheaves on roller bearings (API RP 9B is a standard of the American Petroleum Institute), it is 0.874 with 6 lines, 0.811 with 10 and 0.755 with 14. Each extra pair of lines reduces the fast-line tension by less and less.
The derrick carries not only $W$ but also both ends of the line: $F_{\text{der}} = W + F_{\text{fl}} + F_{\text{dl}}$ . With 10 lines this is 22% more than the hook load (the weight hanging on the hook). So the permissible hook load is stated together with the number of lines strung (the reeving).
We check the line with the design factor: the breaking strength divided by the fast-line tension. By API RP 9B it must be at least 3 in ordinary work, and at least 2 when running casing and working stuck pipe (moving it up and down to free it).
The power at the drawworks drum is $N_{\text{drum}} = Wv/\eta$ . For the same power, the drawworks lifts a heavy string more slowly. While drilling, the pumps take the most power, and they take it all the time. While the string is pulled, the drawworks takes the most.
The line wears unevenly, at a few critical points. We count its work in tonne-kilometres (ton-miles in API practice). By a set programme, the crew slips the line (pulls a fresh length through the system) and cuts off the worn end.

Key relationships

\eta = \dfrac{K^n - 1}{K^n\,n\,(K - 1)}

block efficiency: $K$ = 1.04 for sheaves on roller bearings, 1.09 for plain bearings

F_{\text{fl}} = \dfrac{W}{\eta\,n}, \qquad F_{\text{dl}} = \dfrac{W}{n}, \qquad F_{\text{der}} = W + F_{\text{fl}} + F_{\text{dl}}

tension in the fast line and the dead line, and the derrick load

k_{\text{df}} = \dfrac{R}{F_{\text{fl}}}, \qquad N_{\text{drum}} = F_{\text{fl}}\,v_{\text{fl}} = \dfrac{W v}{\eta}

design factor of the line and power at the drawworks drum

T_{\text{rt}} = \dfrac{D\,(L_{\text{st}} + D)\,w + 4D\,(W_{\text{trav}} + C/2)}{9807}

line work for a full round trip, t·km ( $D$ , $L_{\text{st}}$ in m; $w$ in kN/m; $W_{\text{trav}}$ , $C$ in kN)

Common pitfalls

Dividing the load by the number of lines without the efficiency. With 6–14 lines the fast-line tension then comes out 13–25% too low, and the design factor of the line comes out too high.
Taking the derrick load as equal to the hook load. Both ends of the line also pull the derrick down.
Forgetting the travelling equipment. The travelling block with the hook and top drive weighs tens of tonnes and is part of the load on the lines.
Comparing only the weight of the string with the capacity of the rig. The travelling equipment and the overpull in a stuck-pipe case must go into the calculation too.
Deciding whether to slip the line from its look alone. Wear is concentrated at a few points, so the work of the line has to be counted.

After this lesson you can

name the elements of the power and hoisting systems and explain how the energy of the engines reaches the hook
compare mechanical drive, diesel-electric drive (DC and AC) and grid power, and estimate the power a rig needs and the fuel it burns
calculate the block efficiency, the tension in the fast line and in the dead line of the drilling line, and the design factor of the line
find the drawworks power, the hoisting speed of a string and the derrick load, and choose a rig class from the permissible hook load
estimate the work of the drilling line in tonne-kilometres for a round trip, drilling, running casing and a short trip, and explain why the line is slipped and cut

The power system: where the rig gets its energy

Rigs often work far from power lines, so they make their own energy. Diesel engines (prime movers) turn generators or drive machinery directly. If a power line is nearby, an electric rig can take its power from the grid. The energy goes to four groups of consumers: the drawworks, the mud pumps, the top drive or rotary table, and the auxiliary equipment. The auxiliary equipment includes mud cleaning, agitators (mixers), compressors, lighting, heating and the camp (housing for the crew). The general layout of a drilling rig is in the lesson on the drilling process.

The way energy passes from the engines to the machines is called the drive. On old rigs the drive was mechanical. Each diesel engine turned “its own” machine through chain or belt transmissions, clutches and torque converters. Then diesel-electric rigs appeared: the engines turn generators, and electric motors drive the machines. Since the 1970s some rigs have had all their generators feeding a common AC (alternating current) network. On these rigs, silicon controlled rectifiers (SCR) supply DC (direct current) motors. The latest generation is the AC drive: induction motors with variable frequency drives (VFD).

Types of rig drive

Drive: Mechanical
How it works: diesel engines turn the drawworks, pumps and rotary table through chain or belt drives, clutches and torque converters
Features: simple and self-contained, but each engine is tied to its own machines, so power cannot be moved from one machine to another; the running is less smooth

Drive: Diesel-electric DC (SCR)
How it works: generators feed a common network, and silicon controlled rectifiers supply DC motors
Features: any generator can supply any consumer; the number of generators running follows the load; smooth control

Drive: AC with variable frequency drives
How it works: induction motors with adjustable speed
Features: high torque (turning force) at low speed and at standstill, precise control of the drawworks, braking by the motors themselves

Drive: From the grid
How it works: an electric rig is supplied from a power line
Features: no fuel consumption or exhaust at the site; a power line must be nearby

Drive	How it works	Features
Mechanical	diesel engines turn the drawworks, pumps and rotary table through chain or belt drives, clutches and torque converters	simple and self-contained, but each engine is tied to its own machines, so power cannot be moved from one machine to another; the running is less smooth
Diesel-electric DC (SCR)	generators feed a common network, and silicon controlled rectifiers supply DC motors	any generator can supply any consumer; the number of generators running follows the load; smooth control
AC with variable frequency drives	induction motors with adjustable speed	high torque (turning force) at low speed and at standstill, precise control of the drawworks, braking by the motors themselves
From the grid	an electric rig is supplied from a power line	no fuel consumption or exhaust at the site; a power line must be nearby

The move to electric drive gave two things. The power of any generator can go to any machine. And the torque and speed of the drawworks, pumps and rotary table change gradually, so the string and the line are not shocked.

Data sheets often give the power of pumps, drawworks and engines in imperial horsepower: 1 hp = 0.7457 kW (the metric horsepower is 0.7355 kW). Pumps are the main continuous consumer during drilling. They give the mud the hydraulic power $pQ$ (MPa × L/s = kW). They take more than that from the shaft, because their efficiency is below 1 (Rig circulating and rotary systems). The drawworks gives short but high peaks when the string is pulled. The drawworks and the pumps rarely work at full power at the same time. So we choose the number of generators for the largest realistic combination of loads, and add a standby generator. Stopping circulation because a generator fails is dangerous for both the hole and well control.

Required electric power and fuel consumption

\begin{gathered} N_{\text{el}} = \frac{1}{\eta_{\text{drive}}}\left(\frac{pQ}{\eta_{\text{mech}}} + M\omega\right) + N_{\text{aux}},\\ G_{\text{f}} = g_{\text{f}}\,N_{\text{el}}\,t \end{gathered}

where:

$N_{\text{el}}$ electric power that the generators must produce, kW
$p,\ Q$ discharge pressure and pump rate, MPa, L/s
$\eta_{\text{mech}}$ mechanical efficiency of the pumps, —
$M,\ \omega$ torque and angular speed of the top drive or rotary table ( $\omega = 2\pi n_{\text{rot}}/60$ at a rotary speed $n_{\text{rot}}$ rpm), kN·m, rad/s
$\eta_{\text{drive}}$ efficiency of the electric drive, meaning the motors with their converters, —
$N_{\text{aux}}$ auxiliary consumers, kW
$G_{\text{f}},\ g_{\text{f}},\ t$ fuel consumption, specific consumption of the generators and running time, kg, kg/(kW·h), h

Units: kW, MPa × L/s = kW, kN·m × rad/s = kW

Holds when

The loads are constant during the calculation time. In reality they change with connections, circulating and trips, so the daily consumption is calculated from a load schedule.
The specific fuel consumption (the fuel burned per kilowatt-hour) depends on the generator load. For modern diesel generators near the rated load it is about 0.2 kg/(kW·h), and noticeably more at a light load.

The hydraulic power of the pumps and the shaft power of the top drive are divided by the efficiency to get the power at the input to the machines. That is then divided by the drive efficiency to get the power that the generators must produce.

Worked exampleHow much electricity and fuel a rig needs while drilling

Given (assumed data)

Pumps: 28 L/s at a pressure of 20 MPa, mechanical efficiency 0.90
Top drive: 18 kN·m at 120 rpm
Efficiency of the electric drive: 0.93
Auxiliary consumers: 300 kW
Generators: 3 × 1000 kW; specific consumption 0.22 kg/(kW·h), fuel density 0.84 kg/L
Drawworks when pulling the string: up to 1000 kW at the input shaft

Solution

Hydraulic power of the pumps: $pQ = 20\cdot 28 = 560$ kW. At the pump shaft: $560/0.90 = 622.2$ kW.
Top drive: $M\omega = 18\cdot 2\pi\cdot 120/60 = 226.2$ kW.
Generators: $N_{\text{el}} = (622.2 + 226.2)/0.93 + 300 = 1212$ kW. The pumps take $622.2/0.93 = 669$ kW of this, which is 55%.
Two generators run, each loaded to 61%. The third is standby.
Fuel per day: $0.22\cdot 1212\cdot 24 = 6399$ kg, or $6399/0.84 = 7618$ L.
When the string is pulled, the pumps are stopped and the drawworks takes 1000 kW at peak: $1000/0.93 + 300 = 1375$ kW.

Answer: While drilling, the rig uses about 1212 kW and burns about 7.6 m³ of diesel fuel a day. The peak while the string is pulled is about 1375 kW.

What it means. More than half of the energy during drilling goes to the pumps. So the pump rate and pressure are the main levers on fuel consumption. The peak when pulling is higher than the load while drilling, but two generators are enough for it too. The third is a standby, so that the failure of one generator does not stop circulation. The actual daily consumption is lower than this estimate, because part of the day goes on connections, circulating and trips at a different load. The data in this example are assumed for illustration.

The hoisting system: drawworks, block and tackle, and derrick

The hoisting system is a large block and tackle (a system of pulleys). The drawworks is a drum onto which the line is wound, with its drive and brakes. From the drum the line goes up to the crown block, a row of stationary sheaves at the top of the derrick. Then it goes down to the travelling block, which has movable sheaves, and so passes between the two blocks several times. Below the travelling block hang the hook and the top drive. The bails (steel links) carry the elevator, the clamp that grips the pipe. The end of the line that goes to the drum is the fast line. The other end is fixed at the base of the derrick, on the dead line anchor. When the line is slipped, new line is fed into the system from a spare reel next to the derrick.

SchematicThe block and tackle with 8 working lines, unfolded into one plane. The dead line runs from the anchor, which holds the weight sensor, over the stationary sheave of the crown block. Then the line goes alternately round the sheaves of the travelling block and the crown block. It leaves as the fast line to the drawworks drum. The load

W

is carried by 8 lines. The derrick receives

W

F_{\text{fl}}

and

F_{\text{dl}}

through the crown block. Not to scale.

If the line passes between the blocks $n$ times, the travelling block hangs on $n$ working lines. This number is called the number of lines strung, or the reeving. To raise the block by 1 m, the drum must wind in $n$ metres of line. So the speed of the fast line is $n$ times the speed of the block. In return, the force with which the drawworks pulls is, ideally, $n$ times smaller than the load. Rigs are usually reeved with 6 to 12 lines, and heavy rigs with more. The crew changes the reeving to suit the expected load.

Elements of the hoisting system

Element: Drawworks
What it does: winds in and pays out the fast line; the main brake is a band or disc brake, and the auxiliary brake (hydrodynamic or electromagnetic) absorbs energy when heavy strings are lowered; on an AC drive the motors themselves also brake

Element: Drilling line
What it does: double-lay steel wire rope, usually 6 strands of 19–25 wires twisted around a steel core; carries the whole load

Element: Crown block and travelling block
What it does: the stationary and movable sheaves of the block and tackle; below the travelling block are the hook, bails and elevator, and the top drive

Element: Dead line anchor
What it does: holds the dead line of the rope; the tension sensor of the weight indicator is mounted on it

Element: Derrick or mast with substructure
What it does: carries the crown block and the whole load; the substructure (the base) carries the rig floor; in the derrick are the monkeyboard (the derrickman's platform) and the fingerboard, which holds the stands (lengths of three joints)

Element	What it does
Drawworks	winds in and pays out the fast line; the main brake is a band or disc brake, and the auxiliary brake (hydrodynamic or electromagnetic) absorbs energy when heavy strings are lowered; on an AC drive the motors themselves also brake
Drilling line	double-lay steel wire rope, usually 6 strands of 19–25 wires twisted around a steel core; carries the whole load
Crown block and travelling block	the stationary and movable sheaves of the block and tackle; below the travelling block are the hook, bails and elevator, and the top drive
Dead line anchor	holds the dead line of the rope; the tension sensor of the weight indicator is mounted on it
Derrick or mast with substructure	carries the crown block and the whole load; the substructure (the base) carries the rig floor; in the derrick are the monkeyboard (the derrickman's platform) and the fingerboard, which holds the stands (lengths of three joints)

A tower-type derrick is assembled on site from parts. A mast is raised whole from a horizontal position, so it is quicker to erect and easier to transport.

Line tension and block efficiency

If the sheaves turned without friction, the tension in all the lines would be the same, $W/n$ . In fact, as the rope goes round a sheave, it has to overcome friction in the bearing and its own resistance to bending. So the tension is greater on the side of the sheave towards which the rope is pulled. When hoisting, the tension grows from line to line towards the fast line, and the fast line has the highest tension. The API RP 9B calculation factor for the ratio of tensions at one sheave is $K = 1.04$ for sheaves on roller bearings and $1.09$ for plain bearings. The manufacturer may recommend its own value.

Block efficiency of the reeving

\eta = \frac{W}{n\,F_{\text{fl}}} = \frac{K^n - 1}{K^n\,n\,(K - 1)}

where:

$\eta$ block efficiency — the ratio of the load to the ideal force $nF_{\text{fl}}$ , —
$W$ load on the lines: the weight of the string on the hook plus the weight of the travelling equipment, kN
$n$ number of working lines, —
$F_{\text{fl}}$ tension in the fast line when hoisting, kN
$K$ ratio of the tensions on the two sides of a rotating sheave: 1.04 for roller bearings, 1.09 for plain bearings, —

Units: dimensionless

Holds when

This is the ordinary rig reeving: the dead line passes over the stationary sheave of the crown block, and the rope goes round as many rotating sheaves as there are lines strung. If there are extra rotating sheaves in the path of the rope, their total number $S$ replaces $n$ in the exponent of the denominator, giving $K^S$ .
The block moves upward at constant speed. Acceleration, braking and jerks add dynamic loads that the formula does not include.
$K$ is a calculated value. It covers friction in the bearings and the loss in bending the rope. The sheave manufacturer may give its own value.

The more lines, the more sheaves are in the path of the rope, and the lower the efficiency. With 6 lines about 13% of the ideal gain in force is lost, and with 14 lines about 25%.

Tension in the fast line and the dead line

F_{\text{fl}} = \frac{W}{\eta\,n}, \qquad F_{\text{dl}} = \frac{W}{n}

where:

$F_{\text{fl}}$ tension in the fast line, the line with the highest tension when hoisting, kN
$F_{\text{dl}}$ tension in the dead line, taken equal to the static value, kN
$W,\ n,\ \eta$ load on the lines, number of lines and block efficiency, kN, —, —

Units: any units of force

Holds when

The dead line does not move, so its tension is taken as static, $W/n$ . When the block moves, friction in the sheaves makes the tension smaller when hoisting and larger when lowering (see the advanced section). For the derrick load when hoisting, $W/n$ errs on the safe side.
The ratio $F_{\text{fl}}/W = 1/(\eta n)$ is called the fast line factor.

The fast-line tension decides the design factor of the rope and the force that the drawworks must develop. The weight indicator measures the dead-line tension.

In other unit systems (1)

Oilfield units

F_{\text{fl}}\,[\text{lbf}] = \frac{W\,[\text{lbf}]}{\eta\,n}

the formula is the same in any units of force; 1 kN = 224.8 lbf, 1 tf (tonne-force) = 9.807 kN.

Block efficiency and loads on the line and the derrick for different numbers of lines

Lines $n$: 6
$\eta$ (roller): 0.874
$F_{\text{fl}}/W$: 0.1908
$F_{\text{der}}/W$: 1.357
$\eta$ (plain): 0.748

Lines $n$: 8
$\eta$ (roller): 0.842
$F_{\text{fl}}/W$: 0.1485
$F_{\text{der}}/W$: 1.274
$\eta$ (plain): 0.692

Lines $n$: 10
$\eta$ (roller): 0.811
$F_{\text{fl}}/W$: 0.1233
$F_{\text{der}}/W$: 1.223
$\eta$ (plain): 0.642

Lines $n$: 12
$\eta$ (roller): 0.782
$F_{\text{fl}}/W$: 0.1066
$F_{\text{der}}/W$: 1.190
$\eta$ (plain): 0.597

Lines $n$: 14
$\eta$ (roller): 0.755
$F_{\text{fl}}/W$: 0.0947
$F_{\text{der}}/W$: 1.166
$\eta$ (plain): 0.556

Lines $n$	$\eta$ (roller)	$F_{\text{fl}}/W$	$F_{\text{der}}/W$	$\eta$ (plain)
6	0.874	0.1908	1.357	0.748
8	0.842	0.1485	1.274	0.692
10	0.811	0.1233	1.223	0.642
12	0.782	0.1066	1.190	0.597
14	0.755	0.0947	1.166	0.556

Calculated with the efficiency formula, with $K$ = 1.04 (roller bearings) and $F_{\text{der}}/W = 1 + 1/(\eta n) + 1/n$ . The last column uses $K$ = 1.09. The values of $\eta$ and $F_{\text{fl}}/W$ agree with the API RP 9B tables for ordinary rig reeving, to within rounding.

CalculatedMechanical advantage (the gain in force)

W/F_{\text{fl}} = \eta n

against the number of lines. Without friction it equals

n

. With friction it grows more and more slowly and tends to the limit

1/(K-1)

: 25 for roller bearings and about 11 for plain bearings.

We check the line with the design factor: the nominal breaking strength of the line divided by the fast-line tension. The higher it is, the longer the line lasts. Small factors speed up wear and fatigue of the wires.

Design factor of the drilling line

k_{\text{df}} = \frac{R}{F_{\text{fl}}}

where:

$k_{\text{df}}$ design factor, —
$R$ nominal breaking strength of the line according to the standard or certificate, kN
$F_{\text{fl}}$ fast-line tension allowing for the block efficiency, kN

Units: any identical units of force

Holds when

The minimum values by API RP 9B are: 3 for the drilling line while drilling and tripping; 2 when running casing, when working stuck pipe and in other rare operations; and 2.5 for the line used to raise the mast.
For a worn rope and under shock loads, the actual factor is lower than the calculated one. Work near the minimum is possible only with a sound line and equipment, and without jerks.

This checks the weakest link, the line. The sheaves, hook, bails and derrick are checked against their own permissible loads.

Worked exampleFast-line tension and design factor for a 4800 m string

Given (assumed data)

String: vertical well of 4800 m: hook load off bottom (with the bit lifted off the bottom) 1463.1 kN, margin of overpull (the extra pull the string can safely take) 519.9 kN (string design)
Travelling equipment: travelling block with hook, top drive, bails and elevator — 300 kN
Reeving: 10 lines, sheaves on roller bearings ( $K$ = 1.04)
Line: 35 mm (1⅜″), EIPS steel (extra improved plow steel, a high-strength rope steel), nominal breaking strength 192,000 lbf = 854 kN

Solution

Block efficiency: $1.04^{10} = 1.4802$ , $\eta = \dfrac{1.4802 - 1}{1.4802\cdot 10\cdot 0.04} = 0.811$ .
Hoisting off bottom: $W = 1463.1 + 300 = 1763.1$ kN. Then $F_{\text{fl}} = 1763.1/(0.811\cdot 10) = 217.4$ kN and $F_{\text{dl}} = 1763.1/10 = 176.3$ kN.
Design factor: $k_{\text{df}} = 854/217.4 = 3.93$ .
Working the string (moving it up and down) with an overpull equal to the whole margin, 519.9 kN: $W = 2283.0$ kN, $F_{\text{fl}} = 2283.0/8.11 = 281.5$ kN, $k_{\text{df}} = 854/281.5 = 3.03$ .
For comparison, with 8 lines ( $\eta = 0.842$ , $\eta n = 6.736$ ): $k_{\text{df}} = 854\cdot 6.736/1763.1 = 3.26$ and $854\cdot 6.736/2283.0 = 2.52$ .

Answer: With 10 lines the fast line carries 217.4 kN when hoisting off bottom and 281.5 kN at the greatest overpull. The design factors, 3.93 and 3.03, are above the minimums of 3 and 2.

What it means. With 8 lines the factor in ordinary hoisting is still above 3. But when working the string it falls to 2.52. That is acceptable for a rare operation, but at such factors the line wears faster. If heavy operations are expected on a well, such as working stuck pipe or running long casing strings, the crew reeves more lines in advance. We check the line against the fast-line tension, not against $W/n$ . Without the efficiency, the tension with 10 lines would come out 19% too low. The data in this example are assumed for illustration. The breaking strength is the API RP 9B value for a line of this size.

Drawworks power and hoisting speed

The drawworks pulls the fast line with the force $F_{\text{fl}}$ at the speed $v_{\text{fl}} = nv$ , where $v$ is the speed of the travelling block. The power at the drum is the product of the two. Only a part of it, $Wv$ , goes into raising the load, and the rest is lost in the sheaves. A simple consequence follows: the power of the drawworks decides not what load it can raise but how fast. Other things limit the greatest load: the strength of the derrick, blocks and line, the brakes, and the greatest fast-line tension for which the drum and transmissions of the drawworks itself are designed.

Power at the drawworks drum and hoisting speed

\begin{gathered} N_{\text{drum}} = F_{\text{fl}}\,v_{\text{fl}} = \frac{W v}{\eta}, \qquad v_{\text{fl}} = n\,v,\\ N_{\text{in}} = \frac{N_{\text{drum}}}{\eta_{\text{dw}}} \end{gathered}

where:

$N_{\text{drum}}$ power at the drawworks drum, kW
$v,\ v_{\text{fl}}$ speed of the travelling block and speed of the fast line, m/s
$W,\ F_{\text{fl}},\ \eta,\ n$ load on the lines, fast-line tension, block efficiency, number of lines, kN, kN, —, —
$N_{\text{in}},\ \eta_{\text{dw}}$ power at the input shaft of the drawworks and the efficiency of its transmissions, kW, —

Units: kN × m/s = kW

Holds when

The motion is steady. Accelerating a string of hundreds of tonnes needs extra power, so acceleration and braking are made smooth.
When lowering, energy flows the other way: the load itself pulls the line, and the brakes or the motors in braking mode absorb the power $Wv$ (less friction).

At a given power, the hoisting speed is inversely proportional to the load. A full string is pulled slowly, and as it gets lighter towards the end of the trip it is pulled faster. This goes on until the maximum drawworks speed, or the speed limit set by swabbing (the pressure drop caused by pulling the string), is reached.

In other unit systems (1)

Oilfield units

N_{\text{drum}}\,[\text{hp}] = \frac{F_{\text{fl}}\,[\text{lbf}]\;v_{\text{fl}}\,[\text{ft/min}]}{33{,}000}

force in pounds-force, fast-line speed in feet per minute, power in imperial horsepower.

33,000 ft·lbf per minute is the definition of the imperial horsepower (745.7 W).

Worked exampleHow fast the drawworks will hoist a 4800 m string

Given (assumed data)

Load on the lines: 1763.1 kN (the string off bottom and the travelling equipment, example above)
Reeving: 10 lines, $\eta$ = 0.811
Drawworks: 1000 kW at the input shaft, transmission efficiency 0.90
Stand: three joints of 9.45 m, 28.35 m in all

Solution

Power at the drum: $N_{\text{drum}} = 0.90\cdot 1000 = 900$ kW.
Speed of the travelling block: $v = N_{\text{drum}}\,\eta/W = 900\cdot 0.811/1763.1 = 0.414$ m/s. Speed of the fast line: $10v = 4.14$ m/s.
Power at the hook: $Wv = 1763.1\cdot 0.414 = 730$ kW. The rest, $900 - 730 = 170$ kW, is lost in the sheaves.
A stand is raised in $28.35/0.414 = 68$ s.

Answer: The drawworks hoists the full string at about 0.414 m/s. That is about 68 s per stand, not counting the time for setting the slips (the wedges that hold the pipe) and unscrewing the stand.

What it means. Each stand pulled lightens the string, and the same power allows faster hoisting. In open hole, swabbing usually limits the hoisting speed, not the drawworks (pressure surge and swab). During acceleration and braking, dynamic loads limit it. GOST 16293-89 (the Soviet and Russian national standard for drilling rigs) requires that the hook can be raised at 0.1–0.25 m/s when working the string. A low speed at a large force is needed for pulling with overpull. The data in this example are assumed for illustration.

Derrick load and rig capacity

The crown block hangs at the top of the derrick, and all the lines pull down through it. The $n$ working lines carry the load $W$ . Besides them there are the fast line, which goes to the drawworks, and the dead line, which goes to the anchor. Both of these forces also pass to the derrick (see the first figure of the lesson).

Derrick load

F_{\text{der}} = W + F_{\text{fl}} + F_{\text{dl}} = W\,\frac{1 + \eta + \eta\,n}{\eta\,n}

where:

$F_{\text{der}}$ derrick load — the vertical force on the crown block, kN
$W,\ F_{\text{fl}},\ F_{\text{dl}}$ load on the lines, tension in the fast line and in the dead line, kN
$\eta,\ n$ block efficiency and number of lines, —

Units: any identical units of force

Holds when

The fast line and the dead line run down from the crown block almost vertically. Slanting lines and wind add horizontal forces, which are allowed for in the design of the derrick itself.
The dead line is taken as static ( $W/n$ ).
The load is shared unevenly among the legs of the derrick (see the advanced section).

With 10 lines the derrick carries 22% more than the hook load, and with 6 lines 36% more. The fewer the lines, the larger the “extra” load from the ends of the rope.

The capacity of a rig depends on the strength of several links, and each one is checked. The manufacturer designs and rates the derrick or mast for the greatest static hook load at a given number of lines. The load from the ends of the line has already been allowed for. The manufacturer also rates it for wind, with no stands and with stands in the fingerboard (API Spec 4F). The travelling block, hook, bails, elevators and swivels (the rotating joints that let mud flow into a turning string) have their own rated capacity (API Spec 8C, ISO 13535). The line is checked against the design factor.

For rigs of Soviet and Russian manufacture the main parameter is the permissible hook load. Under GOST 16293-89 (introduced in 1990), rigs are divided into 12 classes. Each class has a nominal drilling depth. This is the depth at which a drill string of 30 kg per metre loads the hook to half the permissible load. That leaves the rig a margin for overpull in a stuck-pipe case, for dynamic loads and for heavier casing strings. The standard allows the weight of the drill string to reach 0.6 of the permissible load. Then the design depth of a particular rig may differ from the nominal one.

Drilling rig classes under GOST 16293-89

Class: 1
Permissible hook load, kN (tf): 800 (82)
Nominal drilling depth, m: 1250

Class: 2
Permissible hook load, kN (tf): 1000 (102)
Nominal drilling depth, m: 1600

Class: 3
Permissible hook load, kN (tf): 1250 (127)
Nominal drilling depth, m: 2000

Class: 4
Permissible hook load, kN (tf): 1600 (163)
Nominal drilling depth, m: 2500

Class: 5
Permissible hook load, kN (tf): 2000 (204)
Nominal drilling depth, m: 3200

Class: 6
Permissible hook load, kN (tf): 2500 (255)
Nominal drilling depth, m: 4000

Class: 7
Permissible hook load, kN (tf): 3200 (326)
Nominal drilling depth, m: 5000

Class: 8
Permissible hook load, kN (tf): 4000 (408)
Nominal drilling depth, m: 6500

Class: 9
Permissible hook load, kN (tf): 5000 (510)
Nominal drilling depth, m: 8000

Class: 10
Permissible hook load, kN (tf): 6300 (642)
Nominal drilling depth, m: 10,000

Class: 11
Permissible hook load, kN (tf): 8000 (816)
Nominal drilling depth, m: 12,500

Class: 12
Permissible hook load, kN (tf): 10,000 (1020)
Nominal drilling depth, m: 16,000

Class	Permissible hook load, kN (tf)	Nominal drilling depth, m
1	800 (82)	1250
2	1000 (102)	1600
3	1250 (127)	2000
4	1600 (163)	2500
5	2000 (204)	3200
6	2500 (255)	4000
7	3200 (326)	5000
8	4000 (408)	6500
9	5000 (510)	8000
10	6300 (642)	10,000
11	8000 (816)	12,500
12	10,000 (1020)	16,000

In classes 3, 8, 9 and 11 the standard also permits rigs with loads of 1400, 4500, 5800 and 9000 kN. The tonnes-force in brackets are recalculated (1 tf = 9.807 kN); in data sheets they are often rounded.

Worked exampleDerrick load and rig class for a 4800 m well

Given (assumed data)

Load on the lines: 1763.1 kN when hoisting off bottom, 2283.0 kN with an overpull of 519.9 kN (the example on the design factor)
Tension in the ends with 10 lines: fast line 217.4 and 281.5 kN, dead line 176.3 and 228.3 kN
Mass of the string in air: 181.5 t — from the linear masses and section lengths in the string design

Solution

Derrick load when hoisting off bottom: $F_{\text{der}} = 1763.1 + 217.4 + 176.3 = 2156.8$ kN — 1.22 times the load on the hook.
At the greatest overpull: $F_{\text{der}} = 2283.0 + 281.5 + 228.3 = 2792.8$ kN.
Weight of the string in air: $181.5\cdot 9.81 = 1781$ kN. It must not exceed 0.6 of the permissible load, so we need a rig with a permissible load of at least $1781/0.6 = 2968$ kN. This is class 7 (3200 kN). Class 6 (2500 kN) has a limit of $0.6\cdot 2500 = 1500$ kN.
Check: the greatest hook load with overpull is $2283.0/3200 = 71$ % of the permissible.

Answer: A class 7 rig is needed (permissible hook load 3200 kN). When the string is worked with the full overpull, the derrick is loaded to 2792.8 kN.

What it means. The nominal depth of class 7 is 5000 m, more than ours. But we cannot choose a rig by depth alone. The nominal depth refers to a string of 30 kg/m, while the average mass of our string is 37.8 kg/m, because of the heavy drill collars (thick, heavy pipes at the bottom of the string) and the pipe with tool joints (the thick threaded ends of the pipes). The comparison by weight in air errs on the safe side, because in mud the string is lighter (1463.1 kN). The derrick is checked against its data sheet for 10 lines. A casing string can be heavier than a drill string. A 244.5 mm string weighs 1.46 MN in mud, almost as much as our drill string at 4800 m, already at a depth of 2600 m (hook load of a casing string). It is checked in the same way, but with the smallest design factor of the line, 2. The data in this example are assumed for illustration.

A lesson, not an operating procedure

The hoisting system holds hundreds of tonnes above people's heads. The hook load is never brought up to the limit of the derrick, blocks, bails and elevators given in their data sheets and markings. When stuck pipe is worked, two things limit the overpull: the margin of overpull of the string and the capacity of the rig.
The weight indicator must be in working order and calibrated. The block-height limiter stops the travelling block from being pulled into the crown block. The drawworks brakes are checked before work. During overpull and jarring (blows from the jar, a tool that hammers stuck pipe free), people are moved out of the danger zone.
The equipment data sheets, the project and programme of work, the industrial safety requirements and the procedures approved for the rig set the order of work, the permissible loads, the reeving, and the inspection and slipping of the line. Trained personnel carry out the work.

The drilling line: design factor, wear and work records

The drilling line is a double-lay steel wire rope. Wires are twisted into strands, and the strands are twisted around a core. Rigs usually use 6 strands of 19–25 wires, with an independent wire rope core (IWRC). The steel is high-strength: IPS or EIPS (improved and extra improved plow steel). API RP 9B gives these sizes: about 26–29 mm (1–1⅛″) for shallow wells, 29–32 mm for medium ones, and 32–45 mm (1¼–1¾″) for deep ones.

The line wears unevenly. It wears most in two places. The first is where it goes round the sheaves of the crown block and the travelling block at the moment when the pipe is picked up from the slips and the load is at its maximum. The second is at the crossover points of the wraps on the drum, where the neighbouring wraps crush the rope. Bending over the sheaves tires the wires (fatigue), so the sheaves are made large. For 6 × 19 Seale ropes, API RP 9B recommends a sheave diameter, measured at the bottom of the groove, of not less than 51 rope diameters where bending is the main cause of wear. It allows not less than 30 when part of the rope life is sacrificed to save weight and make the equipment easier to transport.

To spread the wear along the length, the crew periodically slips the line. It loosens the dead line anchor and pulls a new length of rope through the system from the spare reel, while the worn rope goes onto the drawworks drum. The length of the slip is chosen so that a length of rope from one critical point does not land on another critical point. When surplus rope has built up on the drum, the crew cuts it off. The length of the cut is set in drum wraps, and includes a half or a quarter of a wrap so that the crossover points shift. It is not made a multiple of the wrap length or of the distance between the critical points. The decision to slip and cut is made from the work of the line: the sum of the products of load and distance travelled.

Work of the drilling line in a full round trip

T_{\text{rt}} = \frac{D\,(L_{\text{st}} + D)\,w + 4D\,(W_{\text{trav}} + C/2)}{9807}

where:

$T_{\text{rt}}$ work of the line for hoisting and lowering a string to depth $D$ , t·km
$D$ depth (length of the string along the hole), m
$L_{\text{st}}$ length of a stand, m
$w$ weight of a metre of drill pipe in mud, kN/m
$W_{\text{trav}}$ weight of the travelling equipment: travelling block, hook, top drive, bails, elevator, kN
$C$ excess weight of the drill collars and heavy-weight pipe in mud over the weight of drill pipe of the same length, kN

Units: kN·m divided by 9807 kN·m per t·km (tonne-force × kilometre)

Holds when

The drill pipe is taken to reach the bottom, and the excess weight $C$ accounts for the heavy bottom part.
For each stand, the travelling equipment moves up and down over the length of the stand when hoisting, and the same when lowering. This gives $4D$ in total per round trip.
Friction is the same when lowering and when hoisting. Accelerations, vibrations and drag in a deviated hole are not included. This is a comparative measure of wear, not exact work.

The first term grows as the square of the depth, because at a greater depth the string is heavier and has to be pulled further. The second term grows linearly. With a heavy top drive it is comparable with the first.

In other unit systems (1)

Oilfield units, API RP 9B form

T_{\text{rt}}\,[\text{ton-mile}] = \frac{D\,(L_{\text{st}} + D)\,w}{10{,}560{,}000} + \frac{D\,(W_{\text{trav}} + C/2)}{2{,}640{,}000}

$D$ , $L_{\text{st}}$ in feet; $w$ in pounds per foot in mud; $W_{\text{trav}}$ , $C$ in pounds; work in ton-miles (2000 pounds × 1 mile).

$10{,}560{,}000 = 2000\cdot 5280$ is the number of foot-pounds in a ton-mile. $2{,}640{,}000$ is a quarter of it, so the second term is the same $4D(W_{\text{trav}} + C/2)$ . 1 ton-mile = 1.460 t·km = 14.32 MJ.

CalculatedLine work for a full round trip against depth, for the string and rig of the example below. The dashed curve is the drill pipe term, which grows as the square of the depth. The gap up to the solid curve is the term for the travelling equipment and the collars, which grows linearly. At a depth of 4800 m the two terms are comparable: 646 and 731 t·km.

Line work in drilling, running casing and a short trip

\begin{gathered} T_{\text{drill}} = 3\left[T_{\text{rt}}(D_2) - T_{\text{rt}}(D_1)\right],\\ T_{\text{short}} = T_{\text{rt}}(D_{\text{low}}) - T_{\text{rt}}(D_{\text{up}}),\\ T_{\text{csg}} = \frac{D\,(L_{\text{jt}} + D)\,w_{\text{csg}} + 4D\,W_{\text{trav}}}{2\cdot 9807} \end{gathered}

where:

$T_{\text{drill}}$ work in drilling from depth $D_1$ to $D_2$ , t·km
$T_{\text{short}}$ work in a short trip between the lower depth $D_{\text{low}}$ and the upper $D_{\text{up}}$ (for example, to the casing shoe and back), t·km
$T_{\text{csg}}$ work in running casing to depth $D$ , t·km
$L_{\text{jt}},\ w_{\text{csg}}$ length of a casing joint and weight of a metre of casing in mud, m, kN/m

Units: t·km; depths and lengths in m, weights in kN and kN/m

Holds when

Drilling goes in a cycle: drill the length of a joint or stand, pick up off bottom, ream (work the bit through the new section again), pick up for the connection, make the connection, and lower. In this cycle the line does work equal to three full round trips over the increase in depth. If the string is not reamed or lifted before the connection, it is two.
In a short trip, the whole string is raised and lowered over the interval between two depths. The work is exactly the difference between the work of full round trips at those depths.
Casing is only run in, so the work is half that of a full round trip. There are no collars in a casing string.

All the operations are expressed through the work of a full round trip, so they are easy to add up in the drilling line log.

Worked exampleLine work in a bit run and when to slip the line

Given (assumed data)

String: 4800 m: drill pipe 0.2733 kN/m in mud (G-105; slightly heavier upper S-135 pipe would add less than 0.1% of the work); drill collars 132.3 m at 1.101 kN/m; HWDP (heavy-weight drill pipe) 113.4 m at 0.6046 kN/m
Rig: travelling equipment 300 kN, stands of 28.35 m
Bit run: the bit drills from 4500 to 4800 m, then is pulled for replacement and a new one is run; if necessary, a short trip to the shoe at 3500 m
Slip programme (assumed): a slip of 7.5 m every 500 t·km; at every fourth slip, that is every 2000 t·km, 30 m is cut off

Solution

Excess weight of the bottom of the string: $C = 132.3\cdot(1.101 - 0.2733) + 113.4\cdot(0.6046 - 0.2733) = 147.1$ kN.
Round trip at 4800 m, the pipe term: $4800\cdot 4828.35\cdot 0.2733 = 6{,}334{,}023$ kN·m. The term for the travelling equipment and collars: $4\cdot 4800\cdot 373.55 = 7{,}172{,}160$ kN·m, where $373.55 = 300 + 147.1/2$ .
Work of the round trip: $T_{\text{rt}} = (6{,}334{,}023 + 7{,}172{,}160)/9807 = 1377.2$ t·km. The travelling equipment alone gives $4\cdot 4800\cdot 300/9807 = 587.3$ t·km, which is 43%.
A round trip at 4500 m, worked out in the same way, gives 1253.5 t·km. Drilling from 4500 to 4800 m: $T_{\text{drill}} = 3\,(1377.2 - 1253.5) = 371.1$ t·km.
Changing the bit at 4800 m is one more full round trip. In all, the bit run takes $371.1 + 1377.2 = 1748.3$ t·km, or 1197 ton-miles.
A short trip to 3500 m ( $T_{\text{rt}}$ = 877.4 t·km): $T_{\text{short}} = 1377.2 - 877.4 = 499.8$ t·km.
Together, $1748.3 + 499.8 = 2248.1$ t·km. This is more than 2000 t·km, so in that time the line is slipped 4 times, and at the fourth slip the worn end is cut off.

Answer: A bit run at a depth of 4500–4800 m gives the line about 1748 t·km of work. With a short trip it is about 2248 t·km. At such a depth, by the programme of the example, the line is cut after roughly every bit run.

What it means. Almost 43% of the work of a full round trip comes from the travelling equipment, because the heavy top drive goes up and down for every stand. The work grows with depth faster than linearly, so in deep intervals the line is slipped more often. The target of the programme is the work between cuts. We choose it from the recommendations of API RP 9B and the rope manufacturer. It depends on the rope diameter, the height of the derrick, the nature of the drilling and the design factor. We refine it from the condition of the cut-off ends. Inspection of the line is always more important than calculation. The data in this example are assumed for illustration, including the slip programme.

Terms and symbols

Drawworks: A drum with its drive and brakes that winds in the fast line of the drilling line.
Block and tackle, reeving: The crown block, the travelling block and the drilling line passed between their sheaves. The number of working lines $n$ is the number of lines strung.
Fast line and dead line: The fast line is the end of the line that goes to the drawworks drum ( $F_{\text{fl}}$ ). The dead line is the end fixed at the base of the derrick ( $F_{\text{dl}}$ ), where the weight indicator sensor is mounted.
Block efficiency $\eta$: The ratio of the load to the ideal force $nF_{\text{fl}}$ . It allows for friction and for bending of the line in the sheaves, and it falls as the number of lines grows.
Derrick load $F_{\text{der}}$: The load on the hook plus the tension in both ends of the line.
Permissible hook load: The greatest load on the hook for which a rig is designed. Rigs are divided into classes by it (GOST 16293-89). For a derrick it is stated together with the number of lines.
Design factor of the line $k_{\text{df}}$: The ratio of the breaking strength of the line to the fast-line tension. It must be at least 3 in drilling, and 2 when running casing and working stuck pipe.
Work of the line, t·km: The sum of the products of load and distance. It measures the wear of the line. 1 ton-mile = 1.46 t·km. It is the basis for slipping and cutting the line.
Slipping and cutting the line: Pulling a new length of line through the system and removing the worn end from the drum, by a programme based on the work of the line.

Common mistakes and misconceptions

Wrong

“The tension in the line is the hook load divided by the number of lines.”

In fact

That is only the tension of the dead line at rest. When hoisting, the fast line has more tension, by the factor of the block efficiency: with 10 lines it is 23% more than $W/n$ . We calculate the design factor of the line from the fast-line tension.

Wrong

“The derrick carries what the weight indicator shows.”

In fact

Both ends of the line also pull the derrick down. With 10 lines the derrick load is 22% more than the hook load, and one of the legs carries more than the others.

Wrong

“The more powerful the drawworks, the heavier the string the rig can lift.”

In fact

Power decides the hoisting speed. The derrick, blocks, hook, elevators and line limit the greatest load, through their permissible loads. In the drawworks itself, the greatest fast-line tension limits it.

Wrong

“The more lines, the better.”

In fact

The gain in force grows more and more slowly, the block moves more slowly, and more rope goes onto the drum. We choose the number of lines for the heaviest expected operation.

Wrong

“The line looks sound — it is too early to slip it.”

In fact

Wear is concentrated at a few points on the sheaves and the drum, and it builds up with the work of the line. Slipping and cutting follow a programme. Inspection supplements the programme, and overrides it if there is damage.

In depth: advanced detail

Where the efficiency formula comes from and what the weight indicator shows in motion

Number the working lines from the dead line ( $i = 1$ ) to the fast line ( $i = n$ ). Between neighbouring lines the rope goes round a rotating sheave. When hoisting, the tension of each next line is $K$ times that of the previous one. The last sheave of the crown block separates line $n$ from the fast line. So $T_i = F_{\text{fl}}/K^{\,n-i+1}$ , and the whole load is $W = \sum T_i = F_{\text{fl}}\left(K^{-1} + K^{-2} + \dots + K^{-n}\right)$ . The sum of this geometric progression (a series in which each term is a fixed multiple of the one before) gives $W = F_{\text{fl}}\,(K^n - 1)/[K^n(K - 1)]$ . From this we get $\eta = W/(nF_{\text{fl}})$ of the main section.

The stationary sheave of the crown block does not rotate, so the tension in the dead line is equal to the tension in the first line. When hoisting, it is the lowest of all: $F_{\text{dl}} = F_{\text{fl}}/K^n < W/n$ . When lowering, the drawworks does not pull but holds the rope back, and the friction changes sign. The tension now grows from the fast line towards the dead line. So the dead line has the highest tension, $F_{\text{dl}} > W/n$ . The weight indicator multiplies the dead-line tension by the number of lines. It therefore understates the hook load when hoisting and overstates it when lowering.

How large this difference is on a particular rig depends on the condition of the sheaves and bearings, and not only on the API calculation factor. We find it by calibration, comparing the readings during slow hoisting and lowering of the same string. Some modern systems apply a correction automatically. This matters for analysis of torque and drag. The difference between the hoisting and lowering weights comes not only from friction of the string against the walls, but partly from friction in the sheaves.

Uneven load on the legs of the derrick

The load from the working lines, $W$ , passes through the centre of the crown block. It is shared equally among the four legs of a tower derrick, $W/4$ each. But the ends of the line are attached asymmetrically. Suppose the dead line is anchored at leg A, and the drawworks stands between legs C and D. Then leg A carries an additional $F_{\text{dl}}$ , legs C and D carry $F_{\text{fl}}/2$ each, and leg B carries nothing beyond $W/4$ . Since $\eta > 0.5$ , $F_{\text{dl}} = W/n$ is greater than $F_{\text{fl}}/2 = W/(2\eta n)$ . So leg A is the most heavily loaded.

Equivalent derrick load and its utilisation factor

\begin{gathered} F_{\text{eq}} = 4\left(\frac{W}{4} + \frac{W}{n}\right) = W\,\frac{n + 4}{n},\\ E_{\text{der}} = \frac{F_{\text{der}}}{F_{\text{eq}}} = \frac{1 + \eta + \eta\,n}{\eta\,(n + 4)} \end{gathered}

where:

$F_{\text{eq}}$ equivalent load, which is the load the derrick would carry if all the legs were loaded as heavily as the most heavily loaded one, kN
$E_{\text{der}}$ derrick utilisation factor — the ratio of the actual load to the equivalent one, —
$W,\ n,\ \eta,\ F_{\text{der}}$ as in the main section, kN, —, —, kN

Units: any identical units of force

Holds when

The derrick has four legs, and the load from the working lines is shared equally among them.
The dead line is anchored at one leg and the drawworks is midway between two others. For another arrangement the distribution is different, and masts have their own scheme.

A derrick designed for an even load cannot be loaded to the full sum. With 10 lines the most heavily loaded leg carries as much as if the whole derrick carried 14% more than the actual load.

Worked exampleLoads on the derrick legs when hoisting a 4800 m string

Given (assumed data)

Load on the lines: $W$ = 1763.1 kN, 10 lines
Tension in the ends: $F_{\text{fl}}$ = 217.4 kN, $F_{\text{dl}}$ = 176.3 kN; $F_{\text{der}}$ = 2156.8 kN

Solution

Each leg's share of the working lines: $W/4 = 440.8$ kN.
Leg A (dead line): $440.8 + 176.3 = 617.1$ kN. Legs C and D (drawworks): $440.8 + 217.4/2 = 549.5$ kN. Leg B: 440.8 kN.
Equivalent load: $4\cdot 617.1 = 2468.4$ kN. Utilisation factor: $2156.8/2468.4 = 0.874$ .

Answer: The most heavily loaded leg carries 617.1 kN, which is 14% more than the average. The derrick must be checked for the equivalent load of 2468.4 kN.

What it means. The formula $W(n+4)/n$ gives the same result to within rounding: $1763.1\cdot 14/10 = 2468.3$ kN. This is why the derrick manufacturer decides where the drawworks and the dead line anchor go, and why the data sheet states the load for a particular reeving. The data in this example are assumed for illustration.

How the round-trip work formula is obtained

The work of the line is the load multiplied by the distance over which it acts. For a round trip we add three parts. The first is the travelling equipment, of weight $W_{\text{trav}}$ . For every stand it goes up and down about the length of a stand, both when the string is pulled and when it is run. That is $4D$ in all per round trip, and the work is $4DW_{\text{trav}}$ .

Drill pipe. When running in, the string grows from one stand to $D/L_{\text{st}}$ stands. On average the line carries half the sum of the initial and final weight, $\frac{1}{2}(wL_{\text{st}} + wD)$ , over the distance $D$ . If the friction is the same, pulling takes the same work. So for the round trip the work is $wD(L_{\text{st}} + D)$ . Drill collars and HWDP are heavier than pipe by $C$ . This excess travels the distance $D$ down and up, which adds $2CD$ . The sum $D(L_{\text{st}} + D)w + 4DW_{\text{trav}} + 2CD$ is the numerator of the formula of the lesson. Dividing by 9807 converts kN·m into t·km.

The formula is an agreed measure, not exact work. We need it to compare operations and keep records. So it is applied in the same way on all rigs, and the target of the slip programme is chosen from experience for a particular rig, rope and drilling conditions.

Connections to other topics

How a well is drilled: the process and the people involved — the systems of a drilling rig and how they work together.
Drill string design — the weight of the string in mud, the margin of overpull and the hook load: the starting loads for the hoisting system.
Wellheads and running casing — the hook load of a casing string when running and landing it.
Rig circulating and rotary systems — pumps, the circulating system, the rotary table and the top drive — the main consumers of power while drilling.
Circulating pressure losses — why the speed of running and pulling is limited by pressure surges.
Drilling time and cost — what makes up the time and the cost of a well.

Check yourself

1
The load on the lines is 1500 kN, the reeving is 12 lines, the sheaves are on roller bearings ( $\eta$ = 0.782). What is the fast-line tension, approximately?
1. A.160 kN
2. B.125 kN
3. C.1173 kN
4. D.1918 kN
Show the answer
Correct answer: A. 160 kN
$F_{\text{fl}} = W/(\eta n) = 1500/(0.782\cdot 12) = 1500/9.384 \approx 160$ kN. 125 kN is $W/n$ , without friction. The dead line has this tension at rest, while the fast line has 28% more tension when hoisting.
2
Why is the capacity of a derrick stated together with the number of lines in the reeving?
1. A.Because the weight of the travelling block depends on the number of lines
2. B.Because the ends of the line also load the derrick, and with fewer lines their share is larger
3. C.Because with more lines the rope wears faster
4. D.The number of lines is needed only for setting up the weight indicator
Show the answer
Correct answer: B. Because the ends of the line also load the derrick, and with fewer lines their share is larger
The derrick load is $F_{\text{der}} = W(1 + \eta + \eta n)/(\eta n)$ . With 6 lines it is 36% more than the hook load, and with 10 lines 22% more. So the fewer the lines, the smaller the hook load that the same derrick permits.
3
The string weighs 1200 kN in mud, the travelling equipment 250 kN, the rope has a breaking strength of 700 kN, and the reeving is 10 lines ( $\eta$ = 0.811). Is the design factor of the line sufficient to work a stuck string with an overpull of 400 kN?
Show the answer
$W = 1200 + 250 + 400 = 1850$ kN; $F_{\text{fl}} = 1850/8.11 \approx 228$ kN; $k_{\text{df}} = 700/228 \approx 3.07$ . For working stuck pipe, 2 is enough, so the line is not the limit. But we must also check the string (its margin of overpull) and the permissible load of the derrick and the travelling equipment.
4
The drawworks delivers 700 kW to the drum. At what speed will it raise a load of 1400 kN with a reeving of 8 lines ( $\eta$ = 0.842)?
Show the answer
$v = N_{\text{drum}}\eta/W = 700\cdot 0.842/1400 \approx 0.42$ m/s. The fast line moves 8 times faster, about 3.4 m/s. The drawworks would raise a load half as heavy twice as fast, if its maximum speed allows.
5
Why is the line work in a full round trip at 4000 m more than twice the work of a trip at 2000 m?
Show the answer
The drill pipe term $D(L_{\text{st}} + D)w$ grows almost as the square of the depth: at twice the depth the string is twice as heavy and has to be pulled twice as far. Only the term for the travelling equipment and the collars grows linearly. So in deep intervals the line is slipped and cut more often.
6
Sheaves on roller bearings ( $K$ = 1.04) have been replaced by sheaves on plain bearings ( $K$ = 1.09) with the same reeving of 10 lines. How will the fast-line tension change?
1. A.It will double
2. B.It will rise by about 26%
3. C.It will fall, because the rope slides less
4. D.It will not change: the load is still carried by 10 lines
Show the answer
Correct answer: B. It will rise by about 26%
The efficiency falls from 0.811 to 0.642, and the fast-line tension is inversely proportional to the efficiency: $0.811/0.642 \approx 1.26$ . The design factor of the line will fall in the same proportion.

Updated: 17 October 2026