Measurement while drilling

Quick refresher

An MWD system has sensors (for direction, gamma ray, pressure and vibration), electronics with memory, power from batteries or from a turbine alternator (a small turbine that the flow of mud spins), and a transmitter. At the surface it has a pressure sensor on the standpipe (the pipe that feeds mud to the drill string) and a decoder (a computer that reads the signal).
We take a survey of the path while the string is at rest. The module sends it after the pumps start again. During sliding (drilling with the string not rotating), the toolface arrives every few seconds.
Mud-pulse telemetry sends only a few bits per second (here a bit is the smallest unit of data, a 0 or a 1, not the drill bit). The signal fades more when the frequency is higher, the string is longer and the mud is thicker (more viscous). Gas in the mud almost blocks it.
Electromagnetic telemetry does not need circulation (the flow of mud), but it weakens in rocks of low resistivity. Wired drill pipe sends thousands of times more data, but it is expensive.
The sensors sit metres or tens of metres behind the bit. So a sensor learns that the well has left the reservoir layer later than the bit left it.
Real-time data are for steering, and data from memory are for interpretation. We check every survey against the field of the Earth.

Key relationships

p(L) = p_0\,e^{-\alpha L}, \qquad \alpha = \dfrac{2}{d\,c}\sqrt{\dfrac{\pi f\mu}{\rho}}

attenuation of a pressure pulse in the pipe (Newtonian fluid, an estimate)

n = \lceil\log_2(\Delta/\epsilon)\rceil, \qquad t = b/R

bits per quantity and the time to send a message

h_{\text{skin}} = \sqrt{\dfrac{2\rho_{\text{rock}}}{\omega\mu_0}} \approx 503\sqrt{\rho_{\text{rock}}/f}

skin depth: the scale of the attenuation of an electromagnetic signal

\tan\alpha_{\text{bed}} = \dfrac{H_2 - H_1}{x_2 - x_1}

apparent dip of the boundary of a layer from two crossings of it

Common pitfalls

Expecting mud-pulse telemetry to give data as detailed as wireline logging (logging with tools lowered into the well on a cable).
Forgetting that a sensor 15 m behind the bit reports the exit from the reservoir layer late.
Using real-time curves for quantitative interpretation when memory data are available.
Pumping coarse bridging agent (particles added to the mud to plug leaks) without agreeing it with the MWD contractor. The pulser and the turbine can clog.
Not checking the distances from the bit to each sensor against the real assembly.

After this lesson you can

describe the parts of an MWD/LWD system and the place of the sensors in the assembly
explain how a survey is taken at a connection and how the toolface is sent while drilling
compare mud-pulse telemetry (positive and negative pulses, continuous wave) with electromagnetic telemetry and wired drill pipe, and estimate the attenuation (fading) of the signal and the time needed to send data
choose a power source and estimate the life of the batteries
use LWD data to find the angle between the well and the reservoir layer, and decide how to bring the well back into the layer
list what we check in MWD/LWD data before we trust it

What an MWD system consists of

An MWD system has three parts: a downhole module in the bottom hole assembly, a communication channel and equipment at the surface. The module is built into non-magnetic drill collars. They stand above the bit, above the downhole motor (a motor in the string that turns the bit) or above the rotary steerable system (RSS, a steering tool that works while the string rotates). Inside it are the sensors, electronics with memory and a clock, a power source and a transmitter, which is called a pulser (it makes pressure pulses in the mud). At the surface a pressure sensor on the standpipe receives the signal, and a computer decodes it. The driller sees the result on a display at the console, and at the same time the data go to the office.

SchematicThe MWD system (diagram, not to scale): the module with sensors, power and pulser in the assembly above the motor, the pressure pulses that travel through the mud inside the pipe, the pressure sensor on the standpipe and the processing at the surface. The distances from the bit are only rough values.

What the system measures while drilling

Group: Direction
Quantities: inclination, azimuth, toolface
What for: steering the well (Survey instruments: magnetic, gyroscopic and inertial)

Group: Drilling mechanics
Quantities: weight and torque at the bit, rotation speed, shocks and vibration
What for: choice of the drilling parameters, protection of the tools (Bit performance and cost per metre)

Group: Pressure
Quantities: pressure in the annulus (the space between the string and the wall of the hole) and inside the pipe
What for: equivalent circulating density (ECD, the pressure of the moving mud written as a density), hole cleaning, signs of an influx (formation fluid entering the well) (Circulating pressure losses)

Group: Rocks (LWD)
Quantities: gamma ray (GR), resistivity, density, porosity, acoustic
What for: lithology (rock type), saturation (the share of the pores filled with oil, gas or water), geosteering (steering the well by the rocks it meets)

Group: State of the module
Quantities: temperature, supply voltage, flow
What for: monitoring how the system works

Group	Quantities	What for
Direction	inclination, azimuth, toolface	steering the well (Survey instruments: magnetic, gyroscopic and inertial)
Drilling mechanics	weight and torque at the bit, rotation speed, shocks and vibration	choice of the drilling parameters, protection of the tools (Bit performance and cost per metre)
Pressure	pressure in the annulus (the space between the string and the wall of the hole) and inside the pipe	equivalent circulating density (ECD, the pressure of the moving mud written as a density), hole cleaning, signs of an influx (formation fluid entering the well) (Circulating pressure losses)
Rocks (LWD)	gamma ray (GR), resistivity, density, porosity, acoustic	lithology (rock type), saturation (the share of the pores filled with oil, gas or water), geosteering (steering the well by the rocks it meets)
State of the module	temperature, supply voltage, flow	monitoring how the system works

The sensors are spread along the assembly. A direction survey describes the well at the depth of the sensors, which is usually 10–20 m above the bit. LWD tools in a long assembly sit even higher. To reduce this lag, we increasingly measure gamma ray and inclination right at the bit. A near-bit module sends them past the downhole motor to the MWD module over a short wireless link. The direction is measured by the same three accelerometers and three magnetometers as in magnetic survey instruments (how the angles are calculated from them).

A survey at a connection and the toolface while drilling

How a survey of the path is taken

1
Stop
Before a connection, drilling stops. The string is lifted a little above the bottom and its rotation stops. The accelerometers need a tool at rest. Otherwise they would feel the accelerations and the vibration of the string.
2
Survey
The module learns that the pumps have stopped from the loss of flow (a flow sensor, the turbine or the pressure shows it). After a set time it reads the six direction sensors, and the temperature with them.
3
Connection
While a pipe or a stand (several pipes screwed together) is added, the survey is kept in the memory of the module.
4
Transmission
After the pumps start, the module sends the survey first. Then it goes back to its main sequence of data: toolface, gamma ray, pressure.
5
Processing
At the surface we check the survey against the field of the Earth (how). Then we apply the declination and the grid convergence, calculate the coordinates (Directional survey calculations), and predict the position of the bit from how the assembly behaves.

While we drill in sliding mode, the toolface is the main quantity. It is sent every few seconds. The driller turns the string to hold the deflection tool in the right position against the reaction torque of the motor, the twist that the motor passes back to the string (more). When the string rotates, the toolface has no meaning. Then the tool sends continuous inclination and azimuth instead. They are less accurate than a survey at rest, but good enough to show the trend between surveys.

Communication also works in the other direction. To change the sequence of the data sent, the rate of transmission, or the task of the rotary steerable system, the crew sends a downlink command to the bottom of the well. This is an agreed pattern of changes in the flow rate of the mud or in the rotation speed of the string, which the module recognises. Such a command takes minutes, so it is planned ahead.

Mud-pulse telemetry

The most common channel is the drilling mud itself inside the drill pipe (mud-pulse telemetry). A valve in the module briefly changes the resistance to the flow. A pressure pulse then runs up the column of mud at the speed of sound in mud enclosed in a steel pipe, usually 1200–1500 m/s. A pressure sensor on the standpipe catches it. There are three ways to make pulses:

Positive pulses: a valve partly closes the passage inside the module, and the standpipe pressure rises for a moment.
Negative pulses: a valve opens an outlet from the pipe into the annulus for a moment, and the pressure falls.
Continuous wave: a rotating shutter, a “siren”, blocks the flow again and again and makes a pressure wave of a set frequency. The module sends data by shifting the phase of the wave (moving its crests a little earlier or later) or by changing its frequency.

The pulses themselves do not carry the information. Their timing does. In pulse systems a value is often coded by the “window” in which the next pulse falls after a reference pulse (a first pulse that starts the count). Eight windows give three bits per pulse. In continuous-wave systems a bit is sent by a phase shift at the next interval, or by no shift.

SchematicHow pressure pulses carry data (diagram). Top: a pulse system. After the reference pulse, the value is the number of the window in which the next pulse falls. Eight windows give three bits. Bottom: a continuous wave. Each interval, several periods long, carries one bit. A one is a change of the phase of the wave at the start of the interval.

Attenuation limits the rate of transmission. Pressure oscillations lose energy by friction against the walls of the pipe. The higher the frequency of the signal, the faster it dies out. A short pulse or a fast wave does not reach the surface from a great depth. For a Newtonian fluid (one whose viscosity does not depend on the rate of flow) in a rigid pipe, the losses in a thin layer at the wall give an exponential attenuation (the amplitude drops by the same share over each metre of pipe):

Attenuation of a pressure pulse in drill pipe

p(L) = p_0\,e^{-\alpha L}, \qquad \alpha = \frac{2}{d\,c}\sqrt{\frac{\pi f\,\mu}{\rho}}

where:

$p_0,\ p(L)$ amplitude of the pressure oscillation at the pulser and after $L$ metres of pipe, Pa
$L$ length of the string from the module to the standpipe, m
$\alpha$ attenuation coefficient, 1/m
$d$ inner diameter of the drill pipe, m
$c$ speed of the pulse in the mud inside the pipe, m/s
$f$ frequency of the signal, Hz
$\mu,\ \rho$ viscosity and density of the mud, Pa·s, kg/m³

Units: SI

Holds when

The fluid is Newtonian. Drilling mud is not Newtonian, so we put an effective viscosity in place of $\mu$ . The estimate is rough.
The losses come only from viscous friction in the layer at the wall, which is thin compared with the radius of the pipe. Turbulence of the flow and free gas add losses, so the calculation more likely underestimates the attenuation.
The pipes are the same along their whole length. There are no reflections from narrowings, the downhole motor, pumps and pulsation dampeners (devices that smooth out the pressure surges of the pumps). In practice such reflections change the signal strongly at some frequencies.

The attenuation grows as the square root of the frequency and of the viscosity, and it is inversely proportional to the diameter of the pipe. In a deep well with viscous mud and narrow pipes we have to use low frequencies, and with them a lower rate of transmission. A continuous wave usually needs several periods for one bit.

Worked exampleA pressure pulse from a depth of 3500 m

Given (assumed data)

String: 3500 m of pipe with an inner diameter of 108.6 mm
Mud: density 1250 kg/m³, effective viscosity 25 mPa·s
Pulse speed: 1300 m/s
Signal frequencies: 1, 4 and 12 Hz

Solution

Travel time to the standpipe: $3500/1300 = 2.7$ s. The reading “arrives late” by these seconds, which does not matter for drilling.
Attenuation at 1 Hz: $\alpha = \dfrac{2}{0.1086\cdot 1300}\sqrt{\dfrac{\pi\cdot 1\cdot 0.025}{1250}} = 1.123\cdot 10^{-4}$ 1/m. Then $\alpha L = 0.393$ , and the share of the amplitude that reaches the standpipe is $e^{-0.393} = 0.675$ .
At 4 and 12 Hz, $\alpha$ is larger by a factor of $\sqrt 4 = 2$ and $\sqrt{12} = 3.464$ . Then $\alpha L = 0.786$ and 1.361, and the shares of the amplitude that arrive are 0.456 and 0.256.
In decibels (a logarithmic unit for comparing the sizes of signals): $20\log_{10} 0.675 = -3.4$ dB at 1 Hz and $20\log_{10} 0.256 = -11.8$ dB at 12 Hz.

Answer: The pulse reaches the standpipe in 2.7 s. At 1 Hz about 68% of the amplitude arrives, and at 12 Hz about 26%.

What it means. A signal that is a quarter of its original size can still be picked out. But the noise of the pumps and pulsation dampeners arises at the surface and does not weaken with the depth of the well. The longer the string, the lower the working frequency we must choose, and the slower the transmission. The model is rough. Real mud is not Newtonian, and reflections from narrowings and from the motor cause dips at some frequencies. Its use is to show the trends. The data in this example are assumed for illustration.

CalculatedCalculated share of the pulse amplitude that reaches the standpipe, against the length of the string, for frequencies of 1, 4, 12 and 24 Hz. The pipe and the mud are as in the example, with the model of viscous attenuation in a rigid pipe. The scale of the shares is logarithmic.

How fast the channel sends data depends on the system and the conditions. Pulse systems send from a fraction of a bit to a few bits per second. Continuous-wave systems in good conditions send a little over ten bits per second. The fastest mud-pulse systems send up to about twenty. Gas in the mud harms the channel most: it sharply lowers the speed of sound and weakens the signal (see the advanced section). Aerated and foam muds (muds with air or foam in them), a low flow rate and the noise of the pumps also harm it. While the pumps are off, the mud-pulse channel is silent.

What the channel can send

A channel of a few bits per second forces us to save every bit. We send a quantity with the accuracy that is needed, not with the accuracy the sensors give: $n$ bits tell apart $2^n$ values. We add service bits to the data: a start marker, a message number and check bits that help to find errors. Instead of raw readings, the module usually sends angles that are already calculated, because that is shorter. The lesson on logging shows how the speed of the channel limits the detail of logging curves. Here we deal with surveys and the toolface.

Bits per quantity and the time to send a message

n = \left\lceil \log_2\frac{\Delta}{\epsilon} \right\rceil, \qquad t = \frac{b}{R}

where:

$n$ number of bits for one quantity (rounded up), bit
$\Delta,\ \epsilon$ range of the quantity and the step that is needed (resolution), units of the quantity
$b$ length of the message: data plus service bits, bit
$R$ useful rate of the channel, bit/s
$t$ time to send the message, s

Units: bits and seconds

Holds when

The quantity is quantised evenly over its whole range (it is split into equal steps).
Data compression and error-correcting codes are not included.
$R$ is the rate of useful data after decoding, not the frequency of the signal.

Every extra quantity and every extra bit of accuracy makes the transmission cycle longer, and all other data are updated less often.

Worked exampleTime to send a survey and the update rate of the toolface

Given (assumed data)

Accuracy needed: inclination 0.05° (0–180°), azimuth 0.1° (0–360°), toolface 64 positions around the circle, gamma ray 0.5 API units (0–255)
Service bits (assumed): a survey message has 10 bits and two check values of 10 bits each; a short message has 4 bits
Channel rate: 1.5 bit/s (deep well) and 6 bit/s (shallow well)
Rate of penetration (drilling speed) while sliding: 12 m/h

Solution

Inclination: $180/0.05 = 3600$ steps, so $n = 12$ bits ( $2^{12} = 4096$ ), with a step of $180/4096 = 0.044^\circ$ . Azimuth: $360/0.1 = 3600$ values, 12 bits. Toolface: 64 positions, 6 bits, a step of 5.6°. Gamma ray: $255/0.5 = 510$ steps, 9 bits.
Survey message: $10 + 12 + 12 + 2\cdot 10 = 54$ bits. At 1.5 bit/s this takes 36 s, and at 6 bit/s it takes 9 s.
Toolface: $4 + 6 = 10$ bits, updated every $10/1.5 = 6.7$ s. If the gamma ray is sent after each toolface ( $4 + 9 = 13$ bits), the toolface is updated once in $(10 + 13)/1.5 = 15.3$ s.
In this time the drill bit, drilling at 12 m/h, advances $12\cdot 15.3/3600 = 0.05$ m.

Answer: A survey is sent in 36 s in a deep well and in 9 s in a shallow one. At 1.5 bit/s the toolface is updated once in 6.7 s, and once in 15.3 s when the gamma ray is sent with it.

What it means. Each new curve in the sequence slows the update of the others. So we change the sequence to fit the task: the toolface more often while we orient the deflection tool, more logging curves in the productive layer. Five centimetres of drilling between updates are not a problem. But the reaction torque can turn the deflection tool faster than the next value arrives, and the driller has to act ahead of it. The data in this example are assumed for illustration.

Electromagnetic telemetry and wired drill pipe

Electromagnetic telemetry (EM telemetry) needs no mud. Above the module the string has an insulating sub, a short joint that does not conduct electricity. The transmitter applies a low-frequency alternating voltage, of a few hertz, across its two halves, and current spreads through the string and the surrounding rocks. At the surface a receiver measures the voltage between the string at the wellhead and an electrode that is grounded (stuck into the ground) some distance from the rig. The link can work in both directions, and it works without circulation. So it can be used when we drill with air, with foam or with aerated mud, where mud-pulse telemetry cannot work. A survey can also be sent while the pumps are off for a connection.

The limit of electromagnetic telemetry is attenuation in the rocks. An alternating electromagnetic field in a conducting medium weakens by a factor of about $e$ (2.72) over each skin depth:

Skin depth in conducting rocks

h_{\text{skin}} = \sqrt{\frac{2\rho_{\text{rock}}}{\omega\,\mu_0}}

where:

$h_{\text{skin}}$ skin depth: the distance over which the amplitude of the field falls by a factor of $e$ , m
$\rho_{\text{rock}}$ electrical resistivity of the rocks, Ω·m
$\omega$ angular frequency, $2\pi f$ , 1/s
$\mu_0$ magnetic constant, $4\pi\cdot 10^{-7}$ , H/m

Units: SI

Holds when

A uniform conducting medium and a plane wave. This estimate gives only the order of magnitude, that is, the right power of ten. In reality the current spreads from the sub along the string and through layers of different resistivity. The signal also weakens with distance where the skin effect is small, and casing and conducting mud add losses. The real range is set by the ratio of signal to noise at the receiver.
Displacement currents (the currents through insulating material) are negligible. At frequencies of a few hertz this is true.

The lower the frequency and the higher the resistivity of the rocks, the deeper the signal goes. That is why EM telemetry works at a few hertz. In sections (the sequences of rock layers that a well crosses) with low-resistivity shales and beds of salt water its range is clearly shorter than in high-resistivity sections.

In other unit systems (1)

In terms of the frequency in hertz

h_{\text{skin}} = 503.3\sqrt{\frac{\rho_{\text{rock}}}{f}}

$\rho_{\text{rock}}$ in Ω·m, $f$ in Hz, $h_{\text{skin}}$ in m.

$503.3 = 1/\sqrt{\pi\mu_0}$ , that is $1/\sqrt{4\pi^2\cdot 10^{-7}}$ .

Worked exampleHow deep an electromagnetic signal goes

Given (assumed data)

Depth of the transmitter: 2000 m
Frequency: 2 and 8 Hz
Resistivity of the rocks of the section: 1, 5 and 20 Ω·m

Solution

Skin depth at 2 Hz: $503.3\sqrt{1/2} = 356$ m, $503.3\sqrt{5/2} = 796$ m, $503.3\sqrt{20/2} = 1592$ m.
Over 2000 m there fit $2000/356 = 5.62$ , $2000/796 = 2.51$ and $2000/1592 = 1.26$ skin depths. The skin effect weakens the field to $e^{-5.62} \approx 0.36$ %, $e^{-2.51} \approx 8$ % and $e^{-1.26} \approx 28$ %.
At 8 Hz in rocks of 5 Ω·m: $h_{\text{skin}} = 503.3\sqrt{5/8} = 398$ m and $2000/398 = 5.03$ , so about 0.7% is left.

Answer: At 2 Hz the skin effect weakens a signal from 2000 m to 28% in a 20 Ω·m section and to 0.36% in a 1 Ω·m section. Raising the frequency four times in a 5 Ω·m section lowers the share from 8% to 0.7%.

What it means. The difference between the sections is tens of times. So we judge whether EM telemetry can be used from the resistivities in neighbouring wells. In low-resistivity sections we choose mud-pulse telemetry, or a system that can work with both channels. The plane-wave model gives only the order of magnitude. The data in this example are assumed for illustration.

Wired drill pipe is drill pipe with a cable built into the wall of the pipe and an induction coil in the end of every tool joint (the thick threaded end of a drill pipe). When two joints are screwed together, the coils of neighbouring pipes face each other and pass the signal without electrical contact. Booster subs (repeaters) are placed every few hundred metres. Such a network sends tens of thousands of bits per second in both directions, thousands of times more than mud-pulse telemetry. It also lets us put sensors along the whole string. But it is expensive and every joint needs careful handling, and the failure of one connection cuts the link. So it is used less often.

Communication channels of MWD

Channel: Mud-pulse: pulses
Rate, bit/s: from a fraction to a few
Needs circulation: yes
Main limit: attenuation with depth, gas in the mud

Channel: Mud-pulse: continuous wave
Rate, bit/s: up to a little over ten
Needs circulation: yes
Main limit: attenuation at high frequencies

Channel: Electromagnetic
Rate, bit/s: a few to tens, lower with depth
Needs circulation: no
Main limit: attenuation in low-resistivity rocks

Channel: Wired drill pipe
Rate, bit/s: tens of thousands
Needs circulation: no
Main limit: cost, reliability of the connections

Channel	Rate, bit/s	Needs circulation	Main limit
Mud-pulse: pulses	from a fraction to a few	yes	attenuation with depth, gas in the mud
Mud-pulse: continuous wave	up to a little over ten	yes	attenuation at high frequencies
Electromagnetic	a few to tens, lower with depth	no	attenuation in low-resistivity rocks
Wired drill pipe	tens of thousands	no	cost, reliability of the connections

Rough values, as a guide: the real rate depends on the depth, the mud, the rocks and the particular system.

Power for the downhole module

There is no cable to the bottom of the hole. So the module carries its energy with it or makes it on the spot.

Lithium batteries, usually primary (non-rechargeable) lithium thionyl chloride cells. They are compact and work without circulation. But the store of energy is limited, and at high temperature and high current the cells do not give all of it. A common limit is about 150 °C, and special cells go higher. Batteries are not recharged. We change a used pack when we pull out the assembly.
A turbine alternator: the flow of mud turns a turbine, and a generator makes the current. The power is enough for continuous-wave pulsers and for powerful LWD tools, and the working time is not limited. But the generator works only when mud circulates within a set range of flow rates. Screens keep debris out of it, and it wears out in abrasive mud. A turbine is often combined with a rechargeable battery or a battery pack, so that the module does not lose power when the pumps stop.

Worked exampleWill the batteries last for the run, and how much does the turbine take

Given (assumed data)

Battery pack: two parallel strings of 7 cells, each cell 3.6 V and 26 A·h; at downhole temperature 70% of the stored energy is delivered (assumed)
Average power used: 8 W; with a gamma ray tool and more frequent transmission, 12 W
Planned run: 90 h
Turbine alternator: pressure drop across the turbine 0.35 MPa at a flow rate of 35 L/s; the module needs 100 W

Solution

Voltage of a string: $7\cdot 3.6 = 25.2$ V. Capacity of two strings: $2\cdot 26 = 52$ A·h. Stored energy: $25.2\cdot 52 = 1310$ W·h. Delivered: $0.7\cdot 1310 = 917$ W·h.
Life at 8 W: $917/8 = 115$ h, so the run is covered. At 12 W: $917/12 = 76$ h, so the batteries run out before the planned trip out of the hole.
Hydraulic power of the flow through the turbine: $0.35\cdot 10^6\cdot 0.035 = 12{,}250$ W. The module needs 100 W, which is 0.8% of this power.

Answer: The batteries last 115 h at 8 W and only 76 h at 12 W. For 100 W the turbine needs less than one per cent of the power of the flow.

What it means. Batteries limit the time and the power. The turbine limits the working conditions: the flow rate, the cleanliness of the mud and wear. When we choose, we do not count average watts in general. We count the working hours of the run, with spare time for unplanned delays, because changing batteries means pulling out the assembly. The data in this example are assumed for illustration.

Logging while drilling: what each method gives

LWD tools measure the same physical properties as wireline tools. Their design and interpretation are covered in the lessons on logging methods. The lesson on open-hole logging explains how LWD differs from wireline logging. Here the point is what each method gives while we drill.

LWD methods and their role in drilling

Method: Gamma ray, including azimuthal
What it gives while drilling: lithology, correlation, approach to the top or the base of a layer
More: Gamma ray log and natural gamma ray spectrometry

Method: Resistivity from the attenuation and the phase shift of a wave
What it gives while drilling: saturation measured before the mud filtrate has invaded deeply; several depths of investigation
More: Resistivity logs: electrical and induction tools

Method: Azimuthal deep resistivity
What it gives while drilling: distance and direction to the boundaries of a layer, metres away from the well
More: the section on geosteering

Method: Density and neutron, borehole images
What it gives while drilling: porosity, gas; from an image, the dip of the layers relative to the well
More: Density log and photoelectric factor (Pe), Neutron log

Method: Acoustic, nuclear magnetic resonance
What it gives while drilling: porosity, rock properties, permeability
More: Sonic log, Nuclear magnetic resonance (NMR) logging

Method: Formation testing while drilling
What it gives while drilling: formation pressure and its gradient
More: Pressure measurement with the wireline formation tester

Method: Annular pressure while drilling
What it gives while drilling: ECD, hole cleaning, signs of an influx or of lost circulation
More: Circulating pressure losses

Method	What it gives while drilling	More
Gamma ray, including azimuthal	lithology, correlation, approach to the top or the base of a layer	Gamma ray log and natural gamma ray spectrometry
Resistivity from the attenuation and the phase shift of a wave	saturation measured before the mud filtrate has invaded deeply; several depths of investigation	Resistivity logs: electrical and induction tools
Azimuthal deep resistivity	distance and direction to the boundaries of a layer, metres away from the well	the section on geosteering
Density and neutron, borehole images	porosity, gas; from an image, the dip of the layers relative to the well	Density log and photoelectric factor (Pe), Neutron log
Acoustic, nuclear magnetic resonance	porosity, rock properties, permeability	Sonic log, Nuclear magnetic resonance (NMR) logging
Formation testing while drilling	formation pressure and its gradient	Pressure measurement with the wireline formation tester
Annular pressure while drilling	ECD, hole cleaning, signs of an influx or of lost circulation	Circulating pressure losses

Two features of LWD matter while drilling. Each sensor sits at its own distance from the bit, and it sees the rock some time after the bit has opened it (how we take this into account). And only a part of the data is sent in real time, with wide gaps between the points. We read the full curves from the memory of the tool after we pull it out. The annular pressure in real time is also important for well control. A rise or a fall of the equivalent mud weight that the drilling parameters do not explain is a reason to check for an influx or for lost circulation (Well control: primary barrier and kick indicators).

Geosteering: how to keep the well in the reservoir layer

Geosteering means steering the well by the geology, not only by the planned geometry. A horizontal well is designed from a map of the top of the reservoir layer, its upper boundary (Horizontal wells). But between wells the map is wrong by metres, and the layer can be only a few metres thick. So we correct the path as drilling goes on. From the gamma ray and the resistivity we see in which part of the layer the well runs and whether it is approaching the top or the base. From the angle between the well and the layer we decide how to change the inclination.

The main difficulty is the lag. The gamma ray sensor sits metres behind the bit, and when it sees the gamma activity rise, the bit has already passed the boundary. Three methods help: sensors closer to the bit; azimuthal measurements (readings taken separately in different directions around the well), which show on which side of the well the shale lies, above or below; and deep azimuthal resistivity, which notices a boundary metres before the well reaches it. We find the dip (the slope) of the layer along the well from two crossings of the same boundary:

Apparent dip of the boundary of a layer along the well

\tan\alpha_{\text{bed}} = \frac{H_2 - H_1}{x_2 - x_1}

where:

$\alpha_{\text{bed}}$ apparent dip of the boundary in the vertical plane of the well; positive if the boundary goes down in the direction of drilling, degrees
$H_1,\ H_2$ true vertical depths at which the sensor crossed the same boundary, m
$x_1,\ x_2$ horizontal positions of these points in the direction of the well, m

Units: metres; the angle in degrees

Holds when

The boundary is a plane, and the well runs in one vertical plane.
The angle is apparent, in the direction of the well. The true dip is larger if the well does not run along the dip.
The depths and positions are for the point of the sensor, not the bit, and come from the calculation of the path.

The well runs parallel to the boundary at an inclination of $90^\circ - \alpha_{\text{bed}}$ . If the layer goes down in the direction of drilling, the angle is less than 90°. If it rises, the angle is more than 90°.

CalculatedSection along the well of the example (the vertical scale is greatly stretched). The top and the base of the sandstone go down at 0.29°. The horizontal well comes up to the top, and the gamma ray sensor, 14 m behind the bit, notices this later than the bit. The well is brought down at 89° and then runs parallel to the top.

Worked exampleThe well has come up to the top: where is the bit, and which way to turn

Given (assumed data)

Layer: sandstone 5 m thick; the task is to run 1.5 m below the top
Landing (entering the layer): the gamma ray sensor crossed the top at a true vertical depth of 2500.0 m; the well was brought to 90° at a depth of 2501.5 m
After 300 m horizontally: the gamma ray rose to shale values, repeating in reverse order the pattern of the curve above the top at landing; the well ran horizontally at a depth of 2501.5 m all the time
Gamma ray sensor: 14 m behind the bit
Decision: reduce the inclination to 89.0°; the assembly changes the angle by 3° per 30 m

Solution

Apparent dip of the top: $\tan\alpha_{\text{bed}} = (2501.5 - 2500.0)/300 = 0.0050$ , so $\alpha_{\text{bed}} = 0.29^\circ$ . The top goes down in the direction of drilling. The well runs parallel to it at $90 - 0.29 = 89.71^\circ$ .
The bit is 14 m ahead of the sensor, at the same depth of 2501.5 m. There the top is deeper by $14\cdot 0.0050 = 0.07$ m. So the bit is already in the shale, 0.07 m above the top.
At an inclination of 89.0° the well goes deeper by $\tan 1^\circ = 0.01746$ m per metre, and the top goes deeper by 0.0050 m per metre. So the well goes below the top by $0.01746 - 0.0050 = 0.01246$ m per metre. To reach 1.5 m below the top we need $(1.5 + 0.07)/0.01246 = 126$ m horizontally.
After that we hold the well at 89.71°. The change from 90° to 89° at 3° per 30 m takes about 10 m of hole, so in reality the distance is a little longer.

Answer: The layer goes down at 0.29°. The bit has left the layer and is 0.07 m above the top, in the shale. At 89° the well returns to 1.5 m below the top after about 126 m, and from there it must run at 89.71°.

What it means. The angle between the well and the layer is only 0.29°, but over 300 m it used up the whole margin of 1.5 m. In a thin layer, tenths of a degree decide how much of the well ends up in the reservoir. The gamma ray alone cannot tell whether the well leaves through the top or through the base. Here we can see it from the pattern of the curve, which repeats the section above the top. An azimuthal gamma ray or azimuthal resistivity would answer at once: the readings from the upper side of the well would rise before those from the lower side. The data in this example are assumed for illustration.

Reliability and checking the data

An MWD module works in one of the hardest places in the well, just above the bit. It faces shocks and vibration, including twisting oscillation of the string, in which the bit sticks and then suddenly breaks free (stick-slip), temperatures of up to 150 °C and more, pressure of tens of megapascals, and abrasive mud with bridging agents. Failures are most often caused by vibration (it harms the electronics and the connectors), overheating, clogging of the pulser or the turbine by bridging materials, and wear. So the module reports the levels of shocks and vibration, and we choose the drilling parameters to lower them (Bit performance and cost per metre). We pump coarse bridging agent only by agreement with the MWD contractor, and we keep a spare module at the rig.

An assembly of many tools works like a chain: a run without a failure needs every tool to work. Say the chance of failure during a run is 3% for the pulser, 2% for the power supply, 1.5% for the direction module, 1% for the gamma ray and 2% for the resistivity tool. The chance of a run without failure is then $0.97\cdot 0.98\cdot 0.985\cdot 0.99\cdot 0.98 \approx 0.908$ . That is about one failed run in ten. Without the resistivity tool it is $0.927$ . The numbers are assumed, but the conclusion is general: every tool in the assembly pays for its data with a share of reliability, and a failed run means pulling out the assembly and losing rig hours (Drilling time and cost).

What we check in MWD and LWD data

1
Path surveys
Each survey is checked against the field of the Earth. We check the field model, the date, the declination and the grid convergence, the correction for the interference of the string if one was applied, and the agreement of repeat surveys (more).
2
Distances from the bit
The distance to each sensor must match the real assembly. An error of one metre shifts all the depths of surveys and curves.
3
Comparison with independent data
MWD surveys are compared with gyroscopic ones inside casing and at tie-in points (places where a new survey is tied to a trusted earlier one). LWD curves are compared with wireline logs, where there are any, and with neighbouring wells.
4
Quality of transmission
We look at the share of messages that failed the error check and at the gaps while the pumps are stopped. Doubtful values are not used until they are checked against memory.
5
Memory data
After the tool is pulled out, its memory is downloaded and matched to the real-time data by the clocks (more). For interpretation we use the memory data.
6
State of the module
Records of temperature, shocks and vibration, and supply voltage explain failures and warn of a failure on the next run.

Terms and symbols

MWD, LWD: Measurement of the direction and of the drilling parameters, and logging, by sensors in the assembly while drilling.
Pulser: The valve of the module that makes pressure pulses: positive, negative or a continuous wave.
Mud-pulse telemetry: Sending data by pressure pulses in the mud inside the pipe. It sends a few bits per second.
Electromagnetic telemetry: Sending data by low-frequency currents through the rocks. It does not need circulation.
Wired drill pipe: Drill pipe with a built-in cable and induction coils in the tool joints. It sends tens of thousands of bits per second.
Turbine alternator: A power source for the module, turned by the flow of mud.
Downlink: Sending commands to the module by an agreed change of the flow rate or of the rotation speed.
Skin depth $h_{\text{skin}}$: The distance over which an alternating field in a conducting medium weakens by a factor of $e$ .
Azimuthal gamma ray: A gamma ray log with the readings split by the sides of the well.
Geosteering: Steering the well by what LWD data show about the layer.

Common mistakes and misconceptions

Wrong

“MWD sends everything it measures at once.”

In fact

A channel of a few bits per second sends only a chosen sequence of data, with wide gaps between the points. The full data are kept in the memory of the module, and we can read them after we pull it out.

Wrong

“The gamma ray has risen, so the well is leaving the layer right now.”

In fact

The sensor sits metres behind the bit, and the bit left the layer earlier. And the gamma ray alone does not say whether the exit is through the top or the base. We need an azimuthal measurement or the pattern of the curve.

Wrong

“If mud-pulse telemetry works badly, raise the frequency to send faster.”

In fact

Attenuation grows with frequency. In a deep well with viscous mud we lower the frequency instead. The rate of transmission is limited by the ratio of signal to noise.

Wrong

“Electromagnetic telemetry is always better than mud-pulse telemetry.”

In fact

It does not depend on the mud or on circulation, but in low-resistivity sections and at great depth the signal is lost. The choice depends on the section, the depth and the method of drilling.

Wrong

“A turbine alternator is more reliable than batteries in everything.”

In fact

It has no time limit, but it does not work without circulation in the right range of flow rates, and it is vulnerable to debris and wear. That is why it is often combined with a rechargeable battery or a battery pack.

In depth: advanced detail

Speed of a pulse in the pipe and gas in the mud

The speed of sound in a liquid is $\sqrt{K/\rho}$ , where $K$ is the bulk modulus (how strongly the liquid resists compression). In a pipe, part of the compression “goes” into stretching the walls, so the effective modulus is smaller. For a thin-walled pipe the Korteweg correction takes this into account. It is known from calculations of water hammer (the pressure surge when a flow is stopped suddenly). For mud with free gas (gas that has not dissolved in the mud) we find the modulus of the mixture by Wood's rule, adding the compressibilities of the components (how easily each one is squeezed) in proportion to their volume fractions.

Speed of a pressure pulse in mud inside a pipe

c = \sqrt{\frac{K_{\text{eff}}}{\rho}}, \qquad \frac{1}{K_{\text{eff}}} = \frac{1}{K} + \frac{d}{E\,e}, \qquad \frac{1}{K} = \frac{1 - x}{K_{\text{m}}} + \frac{x}{p}

where:

$c$ speed of the pulse, m/s
$K_{\text{eff}},\ K$ effective bulk modulus, allowing for the pipe, and the bulk modulus of the liquid itself, Pa
$K_{\text{m}}$ bulk modulus of the mud without gas, Pa
$\rho$ density of the mud (with gas, the average by volume), kg/m³
$d,\ e$ inner diameter and wall thickness of the pipe, m
$E$ Young's modulus of steel, Pa
$x,\ p$ volume fraction of free gas and the pressure; the bulk modulus of the gas is taken equal to the pressure, —, Pa

Units: SI

Holds when

The pipe is thin-walled and free to move in the axial direction.
The gas is in small bubbles that are compressed isothermally (at constant temperature; the bulk modulus ≈ the pressure). In adiabatic compression (with no exchange of heat) the modulus is 1.2–1.4 times larger, and the speed is a little higher.
The wavelength is much larger than the diameter of the pipe and the size of the bubbles.

Gas adds almost no mass, but it greatly increases the compressibility. Even 1% of gas lowers the speed noticeably, and at a pressure of a few megapascals it lowers it several times. As the speed falls, the attenuation also grows, and mud-pulse telemetry stops working.

Worked exampleSpeed of a pulse in clean mud and with 1% gas

Given (assumed data)

Mud: $K_{\text{m}} = 2.4$ GPa, $\rho = 1250$ kg/m³
Pipe: 127 mm with a wall of 9.19 mm (inner diameter 108.6 mm), $E = 207$ GPa
Gas: 1% by volume at 20 MPa (gas density 150 kg/m³) and at 2 MPa (20 kg/m³)

Solution

Without the pipe: $\sqrt{2.4\cdot 10^9/1250} = 1385.6$ m/s. Correction for the pipe: $K d/(E e) = 2.4\cdot 10^9\cdot 0.1086/(2.07\cdot 10^{11}\cdot 0.00919) = 0.1370$ , so $c = 1385.6/\sqrt{1 + 0.1370} = 1299$ m/s. This is where the 1300 m/s of the example above comes from.
1% gas at 20 MPa: $1/K = 0.99/(2.4\cdot 10^9) + 0.01/(20\cdot 10^6)$ , so $K = 1.096$ GPa. The density is 1239.0 kg/m³, and the speed in the pipe is 912 m/s.
1% gas at 2 MPa: $K = 0.185$ GPa, the density is 1237.7 kg/m³, and the speed in the pipe is 384 m/s.

Answer: Clean mud carries the pulse at about 1299 m/s. 1% of gas lowers this to 912 m/s at 20 MPa and to 384 m/s at 2 MPa.

What it means. The pressure in the pipe is lowest at the top of the string, where it is close to the standpipe pressure. So gas in the mud that is pumped in changes the conditions for the pulse most strongly there, and the lower the standpipe pressure, the stronger the change. That is why mud-pulse telemetry loses the link on gassed, aerated and foam muds, and electromagnetic telemetry does not. The data in this example are assumed for illustration.

Connections to other topics

Survey instruments: magnetic, gyroscopic and inertial — the direction sensors, checking surveys against the field of the Earth, and non-magnetic drill collars.
Deflection tools and downhole motors — how a steerable motor is steered by the toolface, and what controls an RSS.
Open-hole logging: why, how and when — LWD and wireline logging, and the step of real-time curves.
Log acquisition, processing and presentation — depth and time of LWD data, memory data.
Horizontal wells — why horizontal wells are drilled and why they are steered along the layer.
Circulating pressure losses — the equivalent circulating density, which the annular pressure sensor monitors.

Check yourself

1
We drill with aerated mud. Which communication channel should we choose for MWD?
1. A.Mud-pulse with a continuous wave
2. B.Mud-pulse with positive pulses
3. C.Any: the mud does not affect the transmission
4. D.Electromagnetic
Show the answer
Correct answer: D. Electromagnetic
Gas in the mud sharply lowers the speed of sound and weakens the pressure pulses, so mud-pulse telemetry is unreliable on aerated mud. Electromagnetic telemetry sends the signal through the rocks and hardly depends on the mud. Its limits are the resistivity of the rocks and the depth.
2
Why, in a deep well, is a continuous-wave system switched to a lower frequency, even though this lowers the rate of transmission?
Show the answer
Attenuation grows as the square root of the frequency: $\alpha \propto \sqrt f$ . As the string gets longer, the share of the amplitude that arrives, $e^{-\alpha L}$ , falls faster the higher the frequency is, while the noise at the standpipe stays the same. At a low frequency the signal can be picked out reliably, although one bit takes more time. It is better to be slow than to make errors.
3
The channel gives 2 bit/s. The toolface message is 10 bits long, and the annular pressure message is 14 bits. How often is the toolface updated if the messages alternate, and if the pressure is sent only every second time?
Show the answer
If the messages alternate, the cycle is $10 + 14 = 24$ bits, which is 12 s. If the pressure is sent every second time, there is one pressure message for two toolface messages: $2\cdot 10 + 14 = 34$ bits in 17 s. The toolface is updated every 8.5 s on average, and the pressure every 17 s.
4
The gamma ray sensor is 18 m behind the bit. The well runs at an angle of 0.5° to the top of the layer and is approaching it. The sensor has just crossed the top. Where is the bit?
1. A.In the seal (the rock above the layer), about 0.16 m above the top
2. B.In the seal (the rock above the layer), 18 m above the top
3. C.Still in the layer
4. D.At the top
Show the answer
Correct answer: A. In the seal (the rock above the layer), about 0.16 m above the top
The bit is 18 m ahead along the well, and every metre the well rises relative to the top by $\tan 0.5^\circ \approx 0.0087$ m: $18\cdot 0.0087 \approx 0.16$ m. The bit left the layer 18 m of hole earlier, and the sensor notices it only now.
5
A battery pack stores 1500 W·h and, at downhole temperature, delivers 75% of it. The module uses 9 W. Will the pack last for a run of 100 h, and what changes if a tool of 4 W is connected?
Show the answer
The pack delivers $0.75\cdot 1500 = 1125$ W·h, so the life is $1125/9 = 125$ h. A run of 100 h is covered, with a margin. With the extra tool the life is $1125/13 \approx 86.5$ h, which is less than the run. We must change the batteries earlier, take a second pack, or switch to a turbine alternator.
6
How do LWD data from memory differ from real-time data, and which should we use to calculate hydrocarbons in place?
Show the answer
In real time, only some of the curves come through the channel, with a coarse step, tied to the depth at the moment of reception. The memory holds all the curves with a fine step, tied to depth by the clock of the tool. For interpretation and for calculating hydrocarbons in place we use the memory data, after we match the clocks and the depths. Real-time data serve for quick decisions.

Updated: 18 October 2026