Bit performance and cost per metre

Quick refresher

The rate of penetration $R$ rises almost linearly with the weight on bit when the weight is above a threshold and the hole bottom is cleaned in time. Above the founder point, extra weight hardly speeds up drilling: balling (clay stuck to the bit), poor cleaning or vibrations “eat” it.
As the rotary speed rises, the rate grows more slowly than linearly, especially in hard rocks. A higher rotary speed speeds up drilling, but it lengthens the path of the cutters over the rock for each metre.
Drill-off test: we lock the brake of the drawworks (the winch that holds the string), and as the bit goes deeper it “eats up” the weight, while the string stretches. The advance in one step equals the stretch $\Delta G\sum L_i/(EA_i)$ , and the rate is the advance divided by the time of the step.
Mechanical specific energy (MSE) is the energy per unit volume of rock broken. When the bit works well, it is about three times the strength of the rock under the bit. A rise in MSE at the same operating parameters is a sign of balling, vibrations or wear.
The cost per metre $C_{\text{m}} = [C_{\text{b}} + C_{\text{h}}(t_{\text{d}} + t_{\text{conn}} + t_{\text{trip}})]/H$ takes account of both the speed and the length of the run. So we compare bits by it, and not by the speed or the footage alone.
A bit is worth pulling when the cost of the next metre equals the average cost per metre of the run. The minimum is flat, and at the rig the decision also weighs the risk of breakage, a change of rock and the nearness of the casing point (the depth at which the casing is to be run).

Key relationships

C_{\text{m}} = \dfrac{C_{\text{b}} + C_{\text{h}}\,(t_{\text{d}} + t_{\text{conn}} + t_{\text{trip}})}{H}, \qquad v_{\text{run}} = \dfrac{H}{t_{\text{d}} + t_{\text{trip}}}

cost per metre and trip-averaged rate of a run

\Delta L = \Delta G\sum \frac{L_i}{E A_i}, \qquad R = \frac{\Delta L}{\Delta t}

drill-off test: the advance in one step and the rate of penetration

\text{MSE} = \frac{G}{A} + \frac{2\pi N M}{A R}

mechanical specific energy: the work of the weight and of rotation per unit volume of rock

\frac{C_{\text{h}}}{R(t^*)} + c_{\text{conn}} = C_{\text{m}}(t^*)

time to pull the bit: the cost of the next metre equals the average cost per metre

Common pitfalls

Comparing bits by the rate of penetration or by the footage, and not by the cost per metre.
Keeping the weight above the founder point. The rate does not grow, while the cutting structure and the bottom hole assembly (the heavy lower part of the string) suffer from vibrations.
Calculating the specific energy from the torque at the top drive in a deviated well. The friction of the string against the walls overstates it several times.
Pulling a bit because the rate falls, without checking in the cuttings whether the rock has changed.
Comparing the cost per metre of runs with different trip depths, without bringing them to the same conditions.

After this lesson you can

explain how the rate of penetration depends on the weight on bit, the rotary speed, the cleaning of the hole bottom, the overbalance and wear, and what the threshold (the weight below which the bit only scratches the rock) and the founder point (the weight above which extra weight hardly speeds up drilling) are
run and process a drill-off test (a test in which the weight on the bit falls away as the bit drills): from the stretch of the string find the rate at each weight, the threshold and the working weight on bit
calculate the mechanical specific energy from the weight, torque (turning force), rotary speed and rate, compare it with the strength of the rock under the bit, and read its changes
calculate the cost per metre and the trip-averaged rate of a run, and compare the runs of offset wells with corrections for depth and for the reason for pulling
find the time to pull a bit from the minimum of the cost per metre and from the maximum of the trip-averaged rate, and explain why the decision at the rig is not made by calculation alone
describe the Bourgoyne–Young model of the rate of penetration and its limits

Rate of penetration, trip-averaged rate and cost per metre

The rate of penetration $R$ is the speed of deepening while the bit is on the bottom, in metres per hour or feet per hour. The mud logging unit (the unit that records the drilling parameters at the surface) records its value at each moment from the movement of the travelling block, the block that moves up and down in the derrick (the tower of the rig). We find the average over a run by dividing the footage $H$ by the drilling time $t_{\text{d}}$ . But a bit that drills fast and also wears fast makes us pull the string more often. Every pull and every run back in takes hours during which the rig makes no footage. So in Russian-language practice we also calculate the trip-averaged rate next to the rate of penetration. It is the footage divided by the drilling time together with the tripping time (the time of pulling the string out and running it back in).

Average rate of penetration and trip-averaged rate

\bar R = \frac{H}{t_{\text{d}}}, \qquad v_{\text{run}} = \frac{H}{t_{\text{d}} + t_{\text{trip}}}

where:

$\bar R$ average rate of penetration over the run, m/h
$v_{\text{run}}$ trip-averaged rate of the run (the footage divided by the drilling time and the tripping time together), m/h
$H$ footage of the bit run, m
$t_{\text{d}}$ time of drilling with the bit on the bottom, h
$t_{\text{trip}}$ time of the tripping operations of the run: pulling the bit, changing it and running in the new one, h

Units: m and h

Holds when

the trip time is charged to the run at the end of which the bit is pulled;
some calculations add the time of connections, circulating and reaming to the denominator. When we compare runs, we count it in the same way for all of them.

The trip-averaged rate is lower than the rate of penetration, and the gap grows as the run gets shorter and the well deeper. It counts time, but not the price of the bit. For cheap bits its maximum and the minimum of the cost per metre almost coincide. For expensive bits they differ (see the section on the time to pull the bit).

Money combines both factors. The cost per metre (in oilfield units, the cost per foot) is the price of the bit plus the cost of all the time of the run, divided by the metres drilled. The time of the run covers drilling, connections (adding a new stand to the string; a stand is three pipes joined together) and tripping. We take the full cost of an hour (the spread rate): the rig with its crew and all services that are paid by time, such as the mud, the MWD tool (the measurement-while-drilling tool that sends measurements up from the hole), mud logging and fuel.

Cost per metre

C_{\text{m}} = \frac{C_{\text{b}} + C_{\text{h}}\left(t_{\text{d}} + t_{\text{conn}} + t_{\text{trip}}\right)}{H}

where:

$C_{\text{m}}$ cost per metre of the run, $/m
$C_{\text{b}}$ price of the bit (less its residual value, if it will be run again), $
$C_{\text{h}}$ cost of an hour of rig work with all services that are paid by time, $/h
$t_{\text{d}},\ t_{\text{conn}},\ t_{\text{trip}}$ time of drilling, of connections and of tripping in the run, h
$H$ footage of the run, m

Units: any currency; time in hours, footage in metres

Holds when

the formula leaves out costs that do not depend on the bit or on time (casing, cement, mobilisation, which is moving the rig and crew to the site). It compares runs and does not estimate the cost of a well;
the time of problems and repairs is not charged to the bit, unless the bit is to blame;
the formula does not count risk (breakage, metal left on the bottom, an unsuccessful next run). We weigh it separately.

The rate of penetration enters through $t_{\text{d}} = H/\bar R$ . The length of the run enters through the number of metres over which the price of the bit and the tripping are divided. So a slow bit with a long run can turn out cheaper than a fast one with a short run.

In other unit systems (1)

Oilfield units

C_{\text{ft}} = \frac{C_{\text{b}} + C_{\text{h}}\left(t_{\text{d}} + t_{\text{conn}} + t_{\text{trip}}\right)}{H_{\text{ft}}} = \frac{C_{\text{m}}}{3.281}

cost per foot, footage in feet.

1 m = 3.281 ft.

Estimate of the tripping time and the connection time

t_{\text{trip}} \approx 2\,\frac{L}{L_{\text{st}}}\,\frac{\tau_{\text{st}}}{60} + t_{\text{fix}}, \qquad t_{\text{conn}} \approx \frac{H}{L_{\text{st}}}\,\frac{\tau_{\text{conn}}}{60}

where:

$L$ depth from which the bit is pulled: the length of the string along the hole (in a deviated well, more than the true vertical depth), m
$L_{\text{st}}$ length of a stand (three pipes joined together): 28.35 m, three pipes of 9.45 m, as in the lesson Drilling rig: power and hoisting systems, m
$\tau_{\text{st}}$ average time per stand when pulling or running in, including the work of the slips (wedges that grip the pipe) and the tongs (the tools that screw the pipes together), min
$t_{\text{fix}}$ time for changing the bit, handling the bottom hole assembly (BHA, the heavy lower part of the string) and checks, h
$\tau_{\text{conn}}$ time of one connection (adding a stand to the string) when drilling with stands, min

Units: m, min and h

Holds when

this is an estimate for planning. In reality the speed of running in and pulling out changes along the hole. In open hole (the part not yet lined with casing), the pressure surges that the moving pipe causes limit the speed (running and pulling pipe), and in cased hole it is higher;
the lesson's examples take $\tau_{\text{st}}$ = 2.5 min, $t_{\text{fix}}$ = 2.0 h and $\tau_{\text{conn}}$ = 5 min. These are assumptions, not norms.

With these assumptions $t_{\text{trip}} = L/340.2 + 2.0$ h. The string is pulled and run in at an average of 680 m an hour in each direction, and every 340 m of depth adds an hour of tripping to the run. A deep run costs more simply because its trip is longer.

Worked exampleThree runs of offset wells: rate, trip-averaged rate, cost per metre

Given (assumed data)

Runs: run 1: roller-cone bit 617, 3920–4105 m in 30.5 h, bit US$16,000; run 2: PDC M323, 4105–4540 m in 36.5 h, US$45,000; run 3: PDC M433, 4180–4800 m in 59.0 h, US$52,000 (the reports from the lesson Bit selection and dull grading)
Rig: US$2000 per hour with all services
Tripping and connections: from the estimate above: $t_{\text{trip}} = L/340.2 + 2.0$ h, $t_{\text{conn}} = H/340.2$ h

Solution

Tripping from the depth of pulling: run 1 — $4105/340.2 + 2.0 = 14.07$ h; run 2 — 15.35 h; run 3 — 16.11 h. Connections: 0.54, 1.28 and 1.82 h.
Average rate of penetration: 6.07, 11.92 and 10.51 m/h.
Trip-averaged rate: run 1 — $185/(30.5 + 14.07) = 4.15$ m/h; run 2 — $435/(36.5 + 15.35) = 8.39$ ; run 3 — $620/(59.0 + 16.11) = 8.25$ m/h.
Cost per metre: run 1 — $(16{,}000 + 2000\cdot 45.11)/185 = 574.2$ US$/m, where $45.11 = 30.5 + 0.54 + 14.07$ ; run 2 — $(45{,}000 + 2000\cdot 53.13)/435 = 347.7$ ; run 3 — $(52{,}000 + 2000\cdot 76.93)/620 = 332.0$ US$/m.
If the bit of run 3 had been pulled from the same depth as in run 2 (tripping 15.35 h): $(52{,}000 + 2000\cdot(59.0 + 1.82 + 15.35))/620 = 329.6$ US$/m.

Answer: The metre was cheapest in run 3, at 332.0 US$/m. Run 2 was faster (11.92 m/h) and had a higher trip-averaged rate (8.39 m/h), but a metre in it cost 347.7 US$. In run 1, with the roller-cone bit, it was 574.2 US$/m.

What it means. The order of the runs depends on the measure we use. By the rate of penetration and the trip-averaged rate, run 2 is ahead. By the footage and the cost per metre, run 3 is ahead. The cost per metre is the fairer measure, but we compare by it with reservations too. The bit of run 3 was pulled from a greater depth, and its trip cost more. In equal conditions its advantage would grow from 15.7 to 18.1 US$/m. It was pulled at the planned total depth, not because of wear, so this bit could do more than the record shows. Finally, the rocks of the two wells are a little different. How the cost per metre changes during a run, and when it pays to pull the bit, is in the section on the time to pull. The time and the cost of a well as a whole are the subject of the lesson Drilling time and cost. The data in this example are assumed for illustration.

What the rate of penetration depends on

A bit breaks rock by pressing in (indentation), cutting off (shearing) or grinding (how a bit breaks rock). The speed of deepening is the depth that the cutting structure penetrates in one revolution, multiplied by the number of revolutions, as long as the fragments get away from the bottom in time. So the main levers of the driller are the weight, the rotary speed and the flushing. The limits are set by the rock, the mud, wear and vibrations.

Weight on bit $G$ . While the weight is small, the cutting structure only scratches the rock. Noticeable drilling begins from some threshold. Above it the rate grows almost in proportion to the added weight, because each kilonewton presses the inserts or cutters deeper. Then the growth slows down. The bottom is not cleaned in time, the bit gets covered with clay, vibrations begin, or the cutters sink fully into the rock. The weight above which an increase hardly speeds up drilling is called the founder point.
Rotary speed $N$ . If the depth of penetration per revolution did not change, the rate would grow in proportion to the rotary speed. In fact, as the rotary speed rises, each element of the cutting structure passes over a point of the bottom faster and faster. A fragment has no time to break off and leave, part of the energy goes into friction and heat, and the advance per revolution decreases. So the rate grows more slowly than the rotary speed, and the harder the rock, the more noticeably. PDC bits run at a higher rotary speed than roller-cone bits, because every revolution of a roller-cone bit uses up some of the life of its bearings (bearings of the cones). With a downhole motor (a motor in the string that the mud turns), the speed of the bit is the sum of the speed of the string and the speed of the motor shaft (downhole motors).
Cleaning of the bottom. The bit grinds again any fragment that the pressure difference presses to the bottom. The force and the position of the jets decide how much energy goes into new fragments (what the jets do at the bottom). Poor cleaning moves the founder point towards lower weights.
Overbalance and mud. The higher the mud pressure above the pore pressure, the stronger the rock under the bit (confined strength) and the harder the fragments are pressed down. The solids and the viscosity of the mud also slow down drilling. PDC bits usually drill faster in oil-based muds, where clay sticks to the cutters less.
Wear of the cutting structure. As wear grows, the inserts and cutters become blunt, the wear flat (the flat worn surface) grows, and at the same weight the depth of penetration decreases. The rate falls from the beginning to the end of the run.
Vibrations. Stick-slip (the bit sticks and then jumps free), whirl (the bit rotating off its centre) and axial shocks carry energy away and break the cutting structure (shocks and vibrations). Their signs are swings of the torque and of the rotary speed at the top drive (the motor that turns the string from above).

The drilling rate in clays, corrected for the weight, the rotary speed and the bit diameter, is also used to recognise abnormally high pressures. This is the d-exponent. There we need not the rate itself, but how easily the rock drills when all other things are equal.

Drill-off test: how to find the working weight

The simplest way to learn how the rate depends on the weight in a given rock with a given bit is to run a test right on the bottom. The classic method is the drill-off test, proposed in 1958 by A. Lubinski. The driller puts the largest permitted weight on the bit and locks the brake of the drawworks (the winch that holds the string), so the top of the string no longer goes down. The bit goes on drilling. As it goes deeper, the string stretches, and the weight on the bit is “drilled off”: it decreases, and the hook load (the weight carried by the hook of the rig) rises by the same amount. We record the time in which the weight falls by each step.

How the test is done

1
Preparation
We choose a uniform unit of rock, set the flow rate and the rotary speed that will be held for the whole test, and check the hook load reading with the bit off the bottom.
2
Weight
We bring the weight smoothly up to the largest that is permitted for the bottom hole assembly and the bit (how much the assembly can give), and we lock the brake.
3
Record
While the weight falls, we record the time of each step, by hand or from the mud logging unit. The test ends when the rate falls almost to zero.
4
Processing
From the compliance of the string (how much it stretches per unit of load) we find the advance in a step, and from the time we find the rate. We plot the rate against the weight, and find the threshold and the founder point.
5
Repeat
If needed, we repeat the test at another rotary speed or flow rate. That also shows the best rotary speed.

Advance in a step and rate of penetration in a drill-off test

\Delta L = \Delta G\sum_i \frac{L_i}{E A_i}, \qquad R = \frac{\Delta L}{\Delta t}

where:

$\Delta L$ advance of the bit in a step: the stretch of the string, m
$\Delta G$ drop in the weight on bit in a step (the hook load rises by the same amount), N
$L_i,\ A_i$ length and cross-section area of the body of the $i$ -th part of the string: drill pipe, heavy-weight drill pipe (HWDP), drill collars, m, m²
$E$ modulus of elasticity of steel, 207,000 MPa, Pa
$\Delta t$ time of a step, s
$R$ rate of penetration at the mean weight of the step, m/s

Units: SI. In practical units the advance in mm equals the weight in kN multiplied by $\sum L_i\cdot 10^6/(E A_i)$ ( $L_i$ in m, $E$ in MPa, $A_i$ in mm²), and the rate in m/h is $3.6\,\Delta L/\Delta t$ ( $\Delta L$ in mm, $\Delta t$ in s)

Holds when

a vertical or nearly vertical well: a change of weight is passed along the whole string without losses to friction against the walls;
the top of the string is fixed: the stretch of the drilling line when the hook load rises is small compared with the stretch of the string;
the cross-section is the pipe body. Tool joints and upset ends (the thickened ends of the pipes) are stiffer, so the compliance and the rate are overstated a little, by a few per cent;
we assign the rate within a step to its mean weight. We take small steps, but such that each lasts at least tens of seconds, otherwise the error of timing is large.

Almost all the compliance comes from the long, thin-walled drill pipe. Drill collars hardly stretch. The longer the string, the greater the advance per kilonewton, and the more slowly the weight is drilled off. So in deep wells the test is more accurate.

Worked exampleDrill-off test at a depth of 4560 m

Given (assumed data)

String: as in the lesson on drill string design: drill pipe 127 × 9.19 mm — 4314.3 m, heavy-weight drill pipe 127 × 76.2 mm — 113.4 m, drill collars 165.1 × 71.4 mm — 132.3 m; $E$ = 207,000 MPa
Bit and parameters: PDC 215.9 mm in dolomite, 120 rpm; the brake is locked at 120 kN, the largest weight that the assembly is designed for
Times of the steps of 10 kN: 120 → 110 kN — 23 s; 110 → 100 — 23 s; 100 → 90 — 25 s; 90 → 80 — 29 s; 80 → 70 — 36 s; 70 → 60 — 46 s; 60 → 50 — 65 s; 50 → 40 — 111 s

Solution

Cross-section areas: drill pipe $\pi(127^2 - 108.62^2)/4 = 3401.3$ mm², HWDP 8107.3 mm², drill collars 17,404.5 mm².
Compliance per kilonewton: $4314.3\cdot 10^6/(207{,}000\cdot 3401.3) = 6.1277$ mm/kN, for the HWDP 0.0676, for the drill collars 0.0367. Together they are 6.2320 mm/kN, so in a step of 10 kN the bit advances 62.32 mm.
The rate is $R = 3.6\cdot 62.32/\Delta t = 224.35/\Delta t$ m/h at the mean weight of the step: 115 kN — 9.75; 105 — 9.75; 95 — 8.97; 85 — 7.74; 75 — 6.23; 65 — 4.88; 55 — 3.45; 45 kN — 2.02 m/h.
The six points from 45 to 95 kN lie on the straight line $R = 0.1399\,G - 4.246$ (fitted by the method of least squares, the usual way to draw the best straight line through points). It crosses the weight axis at $G = 4.246/0.1399 \approx 30$ kN. This is the threshold.
At 105 and 115 kN the line would give 10.44 and 11.84 m/h, but the measured rate is 9.75 at both. That is 0.69 and 2.09 m/h lower. The founder point lies between 95 and 105 kN.

Answer: The threshold is about 30 kN, and the linear part continues up to about 100 kN. It is reasonable to hold the working weight at about 95–100 kN, because more weight hardly speeds up drilling.

What it means. The calculation of the specific energy below will show that above 100 kN the energy rises too: the extra energy does not go into breaking the rock. We look for the cause of the founder point by its signs: swings of torque (vibrations), a rise in the standpipe pressure (the pump pressure read at the surface) with sticking of clay (balling), and the cuttings. If we remove the cause, the linear part will extend to higher weights. Here every step lasts more than 20 s, and a stopwatch or the record of the mud logging unit gives the rate with an accuracy of a few per cent. In soft rocks at 20–30 m/h a step would be drilled off in a few seconds, and instead of a test with a locked brake it is more convenient to use stepwise trials: several minutes of drilling at a constant weight on each step. The data in this example are assumed for illustration.

CalculatedThe drill-off test of the example: the rate of penetration of each step at its mean weight. The solid straight line is drawn through the six lower points and continued as a dashed line. The two upper points lie below it: the bit has passed the founder point. The line crosses the weight axis at the threshold.

Mechanical specific energy

To understand whether a bit works well, it helps to ask how much energy goes into each cubic centimetre of rock drilled out. R. Teale in 1965 defined the mechanical specific energy (MSE) as the work done by the weight and the rotation in a unit of time, divided by the volume of rock drilled out in the same time. The work of the weight is $G R$ , the work of rotation is $2\pi N M$ , and the volume is $A R$ , where $A$ is the area of the hole bottom.

Mechanical specific energy

\text{MSE} = \frac{G}{A} + \frac{2\pi N M}{A R}

where:

$\text{MSE}$ mechanical specific energy: the energy per unit volume of rock broken, Pa (J/m³)
$G$ weight on bit, N
$A$ area of the hole bottom, $\pi D^2/4$ ; $D$ is the bit diameter, m²
$N$ rotary speed of the bit, rev/s
$M$ torque at the bit, N·m
$R$ rate of penetration, m/s

Units: SI (any consistent system; $N$ and $R$ in the same units of time)

Holds when

the formula takes the weight and the torque at the bit itself. The torque at the surface includes the friction of the string against the walls and overstates the energy;
the energy of the mud jets is not counted: this is the mechanical energy of the bit;
the values are averaged over a time during which the rock and the parameters did not change.

The first term, the work of the weight, is usually a few megapascals and is small. Almost all the energy comes through rotation. Specific energy is the price of breaking the rock. The more of it goes into friction, impacts and regrinding of cuttings, the higher it is for the same rock.

In other unit systems (2)

Metric units

\text{MSE}\,[\text{MPa}] = \frac{4000\,G}{\pi D^2} + \frac{480{,}000\,N M}{D^2 R}

$G$ in kN, $D$ in mm, $N$ in rpm, $M$ in kN·m, $R$ in m/h.

$4000 = 1000\cdot 4$ : kN into N, and the area $\pi D^2/4$ in mm². $480{,}000 = 2\pi\cdot 60\cdot 10^6\cdot 4/(\pi\cdot 1000)$ : revolutions per minute into revolutions per hour, kN·m into N·mm, m/h into mm/h.

Oilfield units

\text{MSE}\,[\text{psi}] = \frac{4\,W}{\pi d^2} + \frac{480\,N T}{d^2 R}

$W$ is the weight in pounds-force; $d$ is the bit diameter in inches; $N$ in rpm; $T$ is the torque in ft·lbf; $R$ in ft/h.

$480 = 2\pi\cdot 60\cdot 12\cdot 4/(\pi\cdot 12)$ : revolutions per minute into revolutions per hour, ft·lbf into in·lbf, ft/h into in/h.

The smallest specific energy at which a rock can be broken is close to its compressive strength, and at the bottom of the hole, to the confined strength $C_{\text{conf}}$ . In practice even a well-working bit spends only about a third of the energy on breaking the rock. F. Dupriest and W. Koederitz introduced the control of specific energy at rigs in 2005. They took the mechanical efficiency of a bit (the share of the energy that really breaks the rock) at its peak as 30–40%, and they multiplied the specific energy by 0.35 to compare it directly with the strength of the rock. The ratio $C_{\text{conf}}/\text{MSE}$ is the estimate of this efficiency.

We measure the torque at the bit with sensors above the bit, if the assembly has them (Measurement while drilling). Otherwise we estimate it through the bit friction coefficient $\mu$ , which R. Pessier and M. Fear introduced in 1992. If the weight is spread evenly over the face of the bit, the torque is proportional to the weight and the radius:

Torque at the bit from the bit friction coefficient

M = \frac{\mu\,G\,D}{3}, \qquad \text{MSE} = \frac{G}{A}\left(1 + \frac{2\pi\mu N D}{3R}\right)

where:

$\mu$ bit friction coefficient: it covers both cutting and friction; for PDC bits it is usually larger than for roller-cone bits, —
$G,\ D,\ N,\ R,\ A$ weight, bit diameter, rotary speed, rate of penetration, area of the hole bottom, consistent units

Units: consistent units. In metric units $M = \mu G D/3000$ (torque in kN·m, weight in kN, diameter in mm), and the rotational part of the MSE is $160\,\mu N G/(D R)$ MPa ( $N$ in rpm, $R$ in m/h)

Holds when

the weight is spread evenly over the face of the bit, so the mean radius at which the friction force acts is a third of the diameter;
we find $\mu$ for a given bit and rock from torque measurements. It changes with wear, with balling and in another rock.

With a known $\mu$ , we can estimate the torque at the bit even without sensors above the bit. If in the same rock $\mu$ at the bottom rises while the rate falls, the cutters most probably rub on their worn flat surfaces (wear flats) or are covered with clay.

Worked exampleSpecific energy in the drill-off test

Given (assumed data)

Parameters at 95 kN: bit 215.9 mm, 120 rpm, rate of penetration 8.97 m/h (the test above)
Torque: the bit friction coefficient $\mu = 3\cdot 4.0/(100\cdot 0.2159) = 0.556$ , from a torque of 4.0 kN·m at 100 kN in the lesson Drill string design; at the top drive at 95 kN the torque is 7.1 kN·m
Rock: dolomite, strength under the bit 178.8 MPa (the dolomite example)

Solution

Torque at the bit: $M = 0.556\cdot 95\cdot 0.2159/3 = 3.80$ kN·m. Area of the hole bottom: $\pi\cdot 215.9^2/4 = 36{,}609.6$ mm².
Work of the weight: $95{,}000/36{,}609.6 = 2.59$ MPa. Work of rotation: $480{,}000\cdot 120\cdot 3.80/(215.9^2\cdot 8.97) = 523.5$ MPa. $\text{MSE} = 526.1$ MPa.
Efficiency of the bit: $178.8/526.1 = 34$ %.
From the torque at the top drive: the rotational part is $480{,}000\cdot 120\cdot 7.1/(215.9^2\cdot 8.97) = 978.1$ MPa, $\text{MSE} = 980.7$ MPa, and the “efficiency” is 18%.
At 115 kN ( $M$ = 4.60 kN·m, $R$ = 9.75 m/h): $\text{MSE} = 3.14 + 583.0 = 586.1$ MPa, which is 11% more. At 45 kN ( $M$ = 1.80, $R$ = 2.02): $1.23 + 1101.1 = 1102.3$ MPa.

Answer: At the working weight of 95 kN the specific energy is 526.1 MPa. This gives an efficiency of about 34%, like that of a well-working bit. Above the founder point the specific energy rises, and at a small weight it is twice as high.

What it means. The specific energy confirms the drill-off test: it is lowest where the rate still grows linearly. At 45 kN the bit hardly penetrates the rock and rubs against it, so the energy goes into heat and wear. From the torque at the top drive the efficiency would come out half as large. That torque contains the friction of the string against the walls, which the bit does not feel, and in a deviated well the difference is even greater. It is most useful to follow the specific energy as a trend. If in the same rock and with the same parameters it grows, we look for balling, vibrations or wear. The data in this example are assumed for illustration.

CalculatedMechanical specific energy at the steps of the drill-off test (the torque at the bit comes from the bit friction coefficient). At a small weight the cutters hardly penetrate, and the energy is high. On the linear part it is lowest. Beyond the founder point it rises again. The dashed line is the strength of dolomite under the bit: even at its minimum the specific energy is almost three times as large.

How to read changes in the specific energy

What we see: MSE rises, the rate falls, torque and rock are the same
Likely cause: wear of the cutting structure: the cutters rub with their wear flats

What we see: MSE and standpipe pressure rise, the rate falls in clays
Likely cause: balling: clay covers the cutting structure

What we see: MSE rises, torque and rotary speed swing
Likely cause: vibrations: stick-slip, whirl

What we see: MSE and the rate change in a step
Likely cause: a change of rock: we check it by the cuttings and by the logs recorded while drilling

What we see: MSE is high at a small weight and falls as the weight rises
Likely cause: the depth of cut is small: the weight can be raised

What we see	Likely cause
MSE rises, the rate falls, torque and rock are the same	wear of the cutting structure: the cutters rub with their wear flats
MSE and standpipe pressure rise, the rate falls in clays	balling: clay covers the cutting structure
MSE rises, torque and rotary speed swing	vibrations: stick-slip, whirl
MSE and the rate change in a step	a change of rock: we check it by the cuttings and by the logs recorded while drilling
MSE is high at a small weight and falls as the weight rises	the depth of cut is small: the weight can be raised

We compare the specific energy with itself, in the same rock and with the same bit. The absolute value depends on the torque from which it was calculated.

The Bourgoyne–Young model of the rate of penetration

The test shows how the rate depends on the weight here and now. To forecast the rate and to choose the parameters for a whole interval of the hole, we need models that link the rate to all factors at once. The best known is the Bourgoyne–Young model, built for roller-cone bits. In it the rate is the product of eight factors, and each factor stands for one influence.

Bourgoyne–Young model of the rate of penetration

R = f_1 f_2 f_3 f_4 f_5 f_6 f_7 f_8, \qquad f_5 = \left[\frac{w - w_{\text{th}}}{4 - w_{\text{th}}}\right]^{a_5}, \quad f_6 = \left(\frac{N}{60}\right)^{a_6}, \quad f_7 = e^{-a_7 h}

where:

$R$ rate of penetration, ft/h
$f_1$ drillability of the rock with the given type of bit and mud: the basic factor, ft/h
$f_2,\ f_3,\ f_4$ compaction of rocks with depth, the departure of the pore pressure from normal, and the overbalance, —
$w,\ w_{\text{th}}$ weight per inch of bit diameter and its threshold, in thousands of pounds-force per inch (1 kN/mm ≈ 5.71); 4 is the reference value that scales the ratio, about 0.70 kN/mm, thousand lbf/in
$N$ rotary speed, rpm
$h$ fraction of the cutting structure that is worn: 0 is new, 1 is completely worn, —
$f_8$ bit hydraulics, through the impact force of the jets, —
$a_5,\ a_6,\ a_7$ exponents that are fitted to the data of offset wells, —

Units: oilfield units of the original model

Holds when

the factors are independent: the influence of each does not depend on the others. This is a simplification;
we fit the coefficients $a_1 \ldots a_8$ by regression (a statistical fit) to the data of offset wells for each unit of rock. The authors suggested reasonable limits, for example $a_5$ from 0.5 to 2.0, $a_6$ from 0.4 to 1.0, and $a_7$ from 0.3 to 1.5;
the model was built for roller-cone bits. For PDC bits we use it with care, or use other models.

At $a_5 = 1$ the rate grows in proportion to the weight above the threshold. This is the linear part of the drill-off test. The case $a_6 < 1$ means that the rate grows more slowly than the rotary speed. The factor $f_7$ shows how the rate falls as wear grows. For a forecast, what matters is not so much the formulas as the quality of the data from which the coefficients are fitted.

Worked exampleWhat a higher rotary speed gives

Given (assumed data)

Parameters: 95 kN, 120 rpm, 8.97 m/h; specific energy 526.1 MPa, of which 523.5 MPa is rotation (the example above)
Change: the rotary speed is raised to 150 rpm at the same weight and the same torque
Exponent $a_6$: 0.6, assumed for the example within the limits 0.4–1.0

Solution

Rate factor: $(150/120)^{0.6} = 1.25^{0.6} = 1.1433$ . Rate: $8.97\cdot 1.1433 = 10.26$ m/h.
The rotational part of the specific energy is proportional to $N/R$ : the factor is $1.25/1.1433 = 1.0933$ . $\text{MSE} = 2.59 + 523.5\cdot 1.0933 = 574.9$ MPa, which is 9% more.
The path of a cutter over the rock per metre drilled is also proportional to $N/R$ , and it will grow by the same 9%.
For comparison: at $a_6$ = 0.4 the rate would grow by a factor of 1.093, and at $a_6$ = 1.0 by a factor of 1.25.

Answer: The rate will grow by about 14%, to 10.26 m/h. But every metre will cost 9% more energy, and the path of the cutters over the rock will grow by the same amount.

What it means. A higher rotary speed pays off if the rig time saved is worth more than the extra wear. With a long trip and a run that is not limited by wear, the answer is usually yes. If the run is limited by the wear of the cutters, the bit will drill less, and the cost per metre may rise. The answer comes from the cost per metre and from a test at several rotary speeds. The data in this example are assumed for illustration.

When to pull the bit

As the run goes on, the rate falls, and every next metre costs more, because it takes more rig hours. While the next metre is cheaper than the average cost per metre of the run, it lowers that average, and it pays to keep drilling. When the next metre becomes dearer than the average, every new metre raises it, and it pays to pull the bit. The minimum of the average cost is where the cost of the next metre equals the average.

Condition for the best time to pull the bit

\frac{C_{\text{h}}}{R(t^*)} + c_{\text{conn}} = C_{\text{m}}(t^*) \quad\Longleftrightarrow\quad R(t^*) = \frac{H(t^*)}{t^* + t_{\text{trip}} + C_{\text{b}}/C_{\text{h}}}

where:

$t^*$ drilling time with the bit after which it pays to pull it, h
$R(t^*)$ rate of penetration at this moment, m/h
$H(t^*)$ footage by this moment, m
$c_{\text{conn}}$ cost of connections per metre of footage, $C_{\text{h}}\tau_{\text{conn}}/(60 L_{\text{st}})$ , $/m
$C_{\text{h}},\ C_{\text{b}},\ t_{\text{trip}}$ cost of a rig hour, price of the bit, tripping time, $/h, $, h

Units: $, h, m

Holds when

the tripping time of a run is constant. In fact it grows with the depth of pulling, so the calculation tells us to pull a little too late;
the rock in the interval is uniform, and the rate falls only because of wear;
the risk of breakage and of losing the cutting structure is not counted. It makes us pull earlier.

The second form is a tangent (a straight line that touches a curve at one point). On the plot of footage against time, the straight line from the point $-(t_{\text{trip}} + C_{\text{b}}/C_{\text{h}})$ on the time axis touches the footage curve at the moment $t^*$ . The price of the bit, expressed in hours of rig time, moves this point to the left, and the touch point comes later: it pays to run an expensive bit longer. Without the price of the bit, the same line from $-t_{\text{trip}}$ gives the maximum of the trip-averaged rate, $R(t^*) = v_{\text{run}}(t^*)$ . The cost of connections per metre does not depend on time and does not shift the moment of pulling.

To find the moment in advance, we need a curve of how the rate falls. In the example below it is exponential (the rate loses the same fraction of itself every hour). This is what the Bourgoyne–Young model gives if the wear grows in proportion to time ( $h = t/\tau$ , and then $\lambda = a_7/\tau$ ):

Fall of the rate of penetration with wear (the model of the example)

R(t) = R_0\,e^{-\lambda t}, \qquad H(t) = \frac{R_0}{\lambda}\left(1 - e^{-\lambda t}\right)

where:

$R_0$ rate of a new bit, m/h
$\lambda$ rate of decline: the fraction of the speed lost per hour, 1/h
$t$ time of drilling with the bit, h
$H(t)$ footage by the moment $t$ , m

Units: m, h

Holds when

the rock, the weight and the rotary speed are constant; only the wear changes;
$R_0$ and $\lambda$ are fitted to the record of the rate in similar runs. With a change of rock the curve is different.

The larger $\lambda$ is, the earlier it pays to pull the bit. At $\lambda t = 1$ the rate falls to 37% of the initial one.

Worked exampleWhen it paid to pull the bit in run 2

Given (assumed data)

Run: PDC M323, 435 m in 36.5 h; the rate fell from about 18.75 to 7.0 m/h; pulled because the rate of penetration fell
Model: $R = 18.75\,e^{-0.027t}$ m/h, fitted to the record of the run
Costs: bit US$45,000; rig US$2000 per hour; tripping from 4540 m — 15.35 h; connections — US$5.88 per metre ( $2000/340.2$ )

Solution

Footage: $H(t) = (18.75/0.027)(1 - e^{-0.027t}) = 694.4\,(1 - e^{-0.027t})$ m. Check: $H(36.5) = 694.4\cdot(1 - 0.3733) = 435.2$ m, $R(36.5) = 18.75\cdot 0.3733 = 7.00$ m/h. The model reproduces the run.
The price of the bit in rig hours: $45{,}000/2000 = 22.5$ h. The point on the time axis: $-(15.35 + 22.5) = -37.85$ h.
The touch point: $e^{0.027t} - 1 = 0.027\,(37.85 + t)$ . Numerically, $t^* = 42.8$ h.
At this moment $H = 694.4\cdot(1 - 0.3149) = 475.7$ m and $R = 18.75\cdot 0.3149 = 5.90$ m/h. The cost per metre is $(45{,}000 + 2000\cdot(42.8 + 15.35))/475.7 + 5.88 = 345.0$ US$/m. The next metre costs $2000/5.90 + 5.88 = 344.9$ US$/m. They are equal to within rounding.
With the bit actually pulled at 36.5 h: $(45{,}000 + 2000\cdot 51.85)/435.2 + 5.88 = 347.6$ US$/m. This is 2.6 US$/m more, which is less than 1%.
If we use the maximum of the trip-averaged rate (without the price of the bit, from the point $-15.35$ h): $t = 29.3$ h, $H = 379.6$ m, $v_{\text{run}} = 379.6/(29.3 + 15.35) = 8.50$ m/h, and the cost per metre is 359.7 US$/m.
Check at the rig: at 36.5 h the next metre costs $2000/7.00 + 5.88 = 291.6$ US$, which is less than the average 347.6, so it was worth drilling further. At 45 h the rate is 5.56 m/h and the next metre costs 365.6 US$, which is already more than the average 345.2.

Answer: The minimum cost per metre is about 345.0 US$/m, with pulling after 42.8 h (475.7 m). Pulling at 36.5 h cost 2.6 US$/m more. The maximum of the trip-averaged rate would have called for pulling after only 29.3 h.

What it means. The minimum is flat: between 37 and 50 h the cost per metre differs from the lowest by less than one per cent (see the figure below). The criterion of the trip-averaged rate, which does not know the price of the bit, would tell us to pull the bit 13.5 h earlier, and the metre would cost 14.7 US$ more. For an expensive PDC bit the price of the bit matters. For a cheap roller-cone bit the two criteria almost coincide. The calculation itself is most useful before the run: it shows what each hour of work with a worn bit costs. At a particular moment at the rig, the decision also takes into account the points listed below. The data in this example are assumed for illustration.

CalculatedCost per metre of run 2 against the hours the bit would drill before it is pulled (solid), and the cost of the next metre (dashed). The curves cross at the minimum of the average cost. The actual pulling was 6 h earlier, and the metre cost only US$2.6 more.

CalculatedFootage of run 2 against the drilling time, and two tangents. The solid one is drawn from the point −37.85 h (tripping plus the price of the bit in rig hours) and touches the curve at 42.8 h: the minimum of the cost per metre. The dashed one is drawn from the point −15.35 h (tripping only) and touches the curve at 29.3 h: the maximum of the trip-averaged rate.

The next metre against the average. It is more convenient to watch not the average cost, but the cost of the next metre. We calculate it from the current rate, $C_{\text{h}}/R + c_{\text{conn}}$ , and compare it with the accumulated cost per metre of the run. When the first catches up with the second, it is time to end the run.
Wear or rock? The rate falls both with wear and when we enter a stronger thin layer, but pulling helps only in the first case. With wear, the specific energy and the bit friction coefficient grow in the same rock and with the same operating parameters. With a change of rock, the rate changes in a step, and the cuttings and the logs recorded while drilling show a different rock.
The cost of the opposite mistake. A worn cutting structure does not only drill slowly. In a PDC bit with the cutters of the shoulder worn halfway (the code 1-4 of run 2), the risk of a ring-shaped groove and of breakages grows. In a roller-cone bit, the risk of a bearing failure and of a lost cone grows (the code of run 1, examples of dull grading). Metal on the bottom costs more than several hours of drilling: it needs an extra run with a magnet or a milling tool.
How much is left to drill. A new run is a new bit plus the trip. In run 2 it is $45{,}000 + 2000\cdot 15.35 = 75{,}700$ US$. If about 40 m remain to the casing point (the depth at which the casing is to be run), it is cheaper to drill them with the worn bit. Even at 5 m/h that is 8 h, or US$16,000.

Terms and symbols

Rate of penetration, $R$: The speed of deepening while the bit is on the bottom. The average over a run is $H/t_{\text{d}}$ .
Trip-averaged rate, $v_{\text{run}}$: The footage divided by the drilling time together with the tripping time.
Cost per metre, $C_{\text{m}}$: The price of the bit plus the cost of the time of the run, divided by the footage. It is the main criterion for comparing bits.
Threshold weight: The weight below which the cutting structure hardly penetrates the rock. On the plot of rate against weight, it is the point where the straight line crosses the axis.
Founder point: The weight above which the rate stops growing linearly.
Drill-off test: A test with a locked brake. From the time of the fall in weight and the compliance of the string, we find the rate at different weights.
Mechanical specific energy (MSE): The energy of the weight and of rotation per unit volume of rock drilled out. We compare it with the strength of the rock under the bit.
Bit friction coefficient, $\mu$: The ratio that links the torque at the bit with the weight and the diameter: $M = \mu G D/3$ .
Bourgoyne–Young model: A model of the rate of penetration as a product of eight factors. The coefficients are fitted to the data of offset wells.
Cost of the next metre: It equals $C_{\text{h}}/R + c_{\text{conn}}$ and shows what the metre that the bit is drilling now costs. When it equals the average cost per metre of the run, it pays to pull the bit.

Common mistakes and misconceptions

Wrong

“The higher the rate of penetration, the cheaper the metre.”

In fact

The cost per metre also depends on the price of the bit and on the length of the run. In the lesson's example, the fastest bit gave a dearer metre than the slower one, which drilled more.

Wrong

“We should always hold the largest weight that the assembly can stand.”

In fact

Above the founder point extra weight hardly speeds up drilling, but the specific energy, the vibrations and the wear grow. We hold the working weight near the upper end of the linear part, and we raise it only after removing the cause of the founder point.

Wrong

“The specific energy from the torque at the top drive shows the work of the bit.”

In fact

The torque at the surface includes the friction of the string against the walls. In a deviated well it may be several times larger than the torque at the bit. We need the torque at the bit: from sensors above the bit or through the bit friction coefficient.

Wrong

“The rate has halved, so it is time to pull the bit.”

In fact

First we check whether the rock has changed (cuttings, logs, specific energy), and we compare the cost of the next metre with the average. It pays to run an expensive bit longer. The minimum of the cost is flat, and the decision also depends on the risk and on the nearness of the casing point.

Wrong

“The drill-off test suits any well.”

In fact

It assumes that a change of weight is passed along the string without friction. In a deviated well and with a downhole motor its readings are distorted. There we look for the working weight by stepwise trials and by the specific energy.

In depth: advanced detail

Where the condition for the best time to pull comes from

We write the cost per metre as a function of the drilling time: $C(t) = [C_{\text{b}} + C_{\text{h}}(t + t_{\text{trip}})]/H(t) + c_{\text{conn}}$ , where the cost of connections per metre $c_{\text{conn}}$ is constant. The derivative with respect to time is zero when $C_{\text{h}}H(t) = [C_{\text{b}} + C_{\text{h}}(t + t_{\text{trip}})]\,H'(t)$ . Since $H'(t) = R(t)$ , we get $R(t^*) = H(t^*)/(t^* + t_{\text{trip}} + C_{\text{b}}/C_{\text{h}})$ . The slope of the line from the point $-(t_{\text{trip}} + C_{\text{b}}/C_{\text{h}})$ to a point of the curve is equal to the slope of the curve itself, which means that the line is tangent to the curve. We substitute this into $C(t^*)$ and get $C(t^*) = C_{\text{h}}/R(t^*) + c_{\text{conn}}$ : the average cost equals the cost of the next metre.

A constant $c_{\text{conn}}$ is simply added to the cost per metre and does not change the moment of pulling. The price of the bit shifts the starting point. In the example of run 2, at a price of US$15,000 the touch point would come after about 34.7 h, and at US$90,000 after 51.5 h. The cost of a rig hour acts the other way: the higher it is, the smaller the price of the bit “in hours”, and the earlier it pays to pull. In the limit, when the bit costs almost nothing, we are left with the criterion of the trip-averaged rate.

In fact the tripping time grows with the depth of pulling, and so with the drilling time. Taking this into account adds a term to the condition that makes every extra hour of drilling a little dearer, and the optimum shifts a little earlier. For a run of a few hundred metres at a depth of 4–5 km, the shift is small compared with the flatness of the minimum.

Where the formula for specific energy comes from

In a time $\Delta t$ a bit that rotates at the speed $N$ under the torque $M$ does the work of rotation $2\pi N M\,\Delta t$ . The weight $G$ , going down with the bit by $R\,\Delta t$ , does the work $G R\,\Delta t$ . In the same time the volume $A R\,\Delta t$ is drilled out. We divide the work by the volume: $\text{MSE} = (G R + 2\pi N M)/(A R) = G/A + 2\pi N M/(A R)$ . The unit is the joule per cubic metre, which is the pascal. So the specific energy is measured in the same units as the strength of the rock.

If we express the torque through the bit friction coefficient, $M = \mu G D/3$ , the rotational part becomes $2\pi N\mu G D/(3 A R) = 8\mu N G/(3 D R)$ in consistent units. At a constant $\mu$ it is proportional to the ratio of the weight to the rate, that is, inversely proportional to the depth of penetration per unit of weight. So on the linear part of the drill-off test, where $R \approx b(G - G_{\text{th}})$ , the specific energy falls towards a constant value. Beyond the founder point, where the rate grows more slowly than the weight, it rises again.

Connections to other topics

Bit selection and dull grading — bit codes, the strength of the rock under the bit, bit records and the dull codes of the same three runs.
Drill bits: types and design — how different bits break the rock and why their cutting structure wears.
Drill string design — the 4800 m string, its compliance and the largest weight that the assembly gives.
Bit hydraulics optimisation — flow rate and nozzles: the cleaning of the bottom, on which the founder point depends.
Drilling rig: power and hoisting systems — stands, pulling the string and the work of the line in a run: where the tripping time comes from.
Rig circulating and rotary systems — what we measure at the rig: weight, rotary speed, torque, pressure; the power of rotation.
Abnormal formation pressures: origin, prediction and recognition — the d-exponent: the corrected rate of drilling as a sign of abnormal pressures.

Check yourself

1
Run 1: a bit that cost US$20,000 drilled 300 m in 30 h. Run 2: a bit that cost US$35,000 drilled 520 m in 60 h. Tripping is 14 h in both runs, the rig costs US$2000 per hour, and we do not count connections. Which run is cheaper?
1. A.It cannot be compared without the price of oil
2. B.Run 2: about 352 US$/m against 360
3. C.They are equal
4. D.Run 1: it is faster, 10 m/h against 8.7
Show the answer
Correct answer: B. Run 2: about 352 US$/m against 360
Run 1: $(20{,}000 + 2000\cdot 44)/300 = 360$ US$/m. Run 2: $(35{,}000 + 2000\cdot 74)/520 \approx 351.9$ US$/m. The slower and dearer bit gave a cheaper metre, because its price and the trip were divided by a larger footage.
2
At a depth of 3000 m the string has 2850 m of drill pipe 127 × 9.19 mm and 150 m of drill collars 165.1 × 71.4 mm. In a drill-off test the weight fell from 100 to 90 kN in 40 s. What is the rate of penetration?
Show the answer
Compliance: the pipe is $2850\cdot10^6/(207{,}000\cdot 3401.3) \approx 4.048$ mm/kN, the collars are $150\cdot10^6/(207{,}000\cdot 17{,}404.5) \approx 0.042$ mm/kN, and together they are about 4.09 mm/kN. In the step of 10 kN the bit advanced 40.9 mm, so the rate is $3.6\cdot 40.9/40 \approx 3.7$ m/h. The string is shorter than in the lesson's example, so at the same rate the same step is drilled off faster.
3
In a deviated well the specific energy calculated from the torque at the top drive is three times larger than that from the torque at the bit given by the MWD tool. What follows?
1. A.The MWD tool is faulty
2. B.Specific energy cannot be calculated in deviated wells
3. C.The bit works three times worse and must be changed
4. D.Most of the torque at the surface goes into the friction of the string against the walls; we judge the efficiency of the bit by the torque at the bit
Show the answer
Correct answer: D. Most of the torque at the surface goes into the friction of the string against the walls; we judge the efficiency of the bit by the torque at the bit
The torque at the surface is the sum of the torque at the bit and the friction of the string against the walls, which is large in a deviated well. The specific energy from the surface torque is useful as a trend. But the work of the bit itself is shown by the torque at the bit, or by its estimate through the bit friction coefficient.
4
Why is the calculated time to pull the bit only a guide, and what finally decides it at the rig?
Show the answer
The minimum of the cost per metre is flat: pulling a few hours earlier or later changes the cost per metre by fractions of a per cent (in the lesson's example, by US$2.6/m). At the rig the final decision weighs several things: the risk of breakage and of losing the cutting structure; a check, by the cuttings and the specific energy, of whether the rock has changed; the nearness of the casing point or the total depth; and, for roller-cone bits, the life of the bearings.
5
The cost of a rig hour has risen, while the price of the bit and the fall of the rate are the same. How does the best time to pull shift?
1. A.Earlier: the price of the bit, expressed in rig hours, has become smaller, and the point of the tangent has moved to the right
2. B.We must pull at once
3. C.Later: time has become dearer, so we must drill longer
4. D.It does not change
Show the answer
Correct answer: A. Earlier: the price of the bit, expressed in rig hours, has become smaller, and the point of the tangent has moved to the right
The tangent is drawn from the point $-(t_{\text{trip}} + C_{\text{b}}/C_{\text{h}})$ . As $C_{\text{h}}$ grows, the term $C_{\text{b}}/C_{\text{h}}$ gets smaller, the point comes closer to the origin, and the touching comes earlier. The optimum moves towards the maximum of the trip-averaged rate.
6
The rotary speed was raised from 100 to 140 rpm at the same weight and the same torque. The exponent $a_6$ = 0.7. How will the rate of penetration and the specific energy change?
Show the answer
The rate will grow by a factor of $1.4^{0.7} \approx 1.266$ , which is about 27%. The rotational part of the specific energy is proportional to $N/R$ : $1.4/1.266 \approx 1.106$ , which is 10–11% more. Drilling is faster, but each metre needs more energy and a longer path of the cutters over the rock. We check the gain by the cost per metre.

Updated: 28 October 2026