A permeable water-bearing sand is drilled with water-based mud kept above the pore pressure. Which zone lies immediately beyond the mudcake, with most of its original pore fluid displaced by mud filtrate?

The flushed zone, whose resistivity is Rxo

The uninvaded zone, whose true resistivity is Rt

The flushed zone, whose resistivity is Rxo

The transition zone, whose pores hold only the original formation water

The mudcake itself, whose resistivity is Rmc

A thick clean permeable sand contains formation water of resistivity 0.08 Ω·m and is drilled with mud whose filtrate resistivity is 0.8 Ω·m. Relative to the shale baseline, the SP opposite the sand is:

a negative (leftward) deflection of about K mV, because Rmf exceeds Rw by a factor of ten

a positive (rightward) deflection of about K mV, because the saline formation water conducts better than the filtrate

no deflection, because the SP responds to permeability and is unaffected by the salinity contrast

a deflection whose sign depends on how permeable the sand is, with its size fixed by the mud pressure

Which statement about the gamma-ray log is correct?

An arkosic sand rich in K-feldspar can read high without being shaly, so gamma-ray shale volume overestimates its clay

A clean limestone reads higher than shale because carbonate rock concentrates uranium and thorium in its lattice

The log cannot be run in cased holes, because the gamma rays from the formation cannot penetrate steel casing at all

An oil-bearing sand reads lower than a water-bearing sand of the same clay content, because oil contains no potassium-40 or other radioactive atoms

An arkosic sand rich in K-feldspar can read high without being shaly, so gamma-ray shale volume overestimates its clay

A well is drilled with non-conductive oil-based mud. Which tool is suited to measuring the deep formation resistivity, and why?

The induction tool, because its coils induce eddy currents magnetically and need no conducting path through the mud

The laterolog, because its focusing currents flow most easily through resistive mud into the rock

The normal device, because its current electrodes need no contact with a conducting fluid in the borehole

The induction tool, because its coils induce eddy currents magnetically and need no conducting path through the mud

The SP device, because the potential it records between its two electrodes is directly proportional to the formation resistivity

Opposite a bed, the micro-normal curve reads consistently higher than the micro-inverse curve (positive separation). What does this indicate?

A permeable bed carrying a mudcake, with the micro-inverse more affected by the low-resistivity cake

A shale bed, because the clay is more conductive than the mudcake

A permeable bed carrying a mudcake, with the micro-inverse more affected by the low-resistivity cake

A hydrocarbon-bearing zone, because oil raises the resistivity seen by the shallower micro-inverse curve

A hole washout, because the larger diameter makes the micro-normal read the mud resistivity

In a clean sandstone the density porosity, computed with ρma = 2.65 g/cm³ and ρf = 1.0 g/cm³, is 24%, and the neutron porosity, correctly scaled for sandstone, is 11%. Which explanation is most likely?

Gas fills part of the pores: its low density inflates density porosity and its low hydrogen index depresses neutron porosity

Thin shale laminae holding clay-bound water, which lower the hydrogen index seen by the neutron tool and so depress its reading

Pyrite and other heavy minerals raise the bulk density, which makes the density porosity read well above the true porosity

The pores are completely flushed with fresh mud filtrate, so both tools read the true porosity and differ only by random noise

In a carbonate with large isolated vugs, the density–neutron porosity is 18% while the Wyllie sonic porosity, with correct matrix and fluid transit times, is 11%. Which statement best explains the difference?

The first arrival takes the faster matrix path around isolated vugs, so sonic porosity is mainly primary; the gap estimates vuggy porosity

Every vug slows the first arrival wave, so the sonic value is the true total porosity and the density–neutron value is an overestimate

Wyllie’s equation always underestimates porosity in carbonates because the matrix transit time assumed for limestone is set far too high

The first arrival takes the faster matrix path around isolated vugs, so sonic porosity is mainly primary; the gap estimates vuggy porosity

Isolated vugs contain no fluid, so only the density and neutron tools can detect them, while the sonic tool cannot respond to them at all

In a CPMG sequence, what is the purpose of the train of 180° pulses that follows the initial 90° pulse?

To refocus spins dephased by static-field inhomogeneity into spin echoes, so that echo amplitudes decay with T₂

To repolarise the protons between echoes, so that the longitudinal relaxation time T₁ is measured directly

To raise the Larmor frequency, so that more hydrogen nuclei fall inside the sensitive volume of the tool

To saturate the hydrogen in the rock matrix, so that only the pore-fluid hydrogen contributes to the measured signal

To refocus spins dephased by static-field inhomogeneity into spin echoes, so that echo amplitudes decay with T₂

Which combination of CBL and VDL observations indicates cement well bonded to both the casing and the formation?

Low CBL amplitude, with weak casing arrivals and strong wavy formation arrivals

Moderate CBL amplitude, with strong casing arrivals and no formation arrivals

Low CBL amplitude, with weak casing arrivals and strong wavy formation arrivals

High CBL amplitude, with weak casing arrivals and weak formation arrivals

Which of the following statements about the SP, caliper, temperature and dipmeter logs are correct?

The SP cannot be recorded in oil-based mud, because no conducting path links the electrode to the formation

Across a permeable bed with mudcake, the caliper reads a diameter smaller than the bit size

In a vertical hole, the dip azimuth on a dipmeter points toward the pad that records the bed boundary at the greater depth

The SP cannot be recorded in oil-based mud, because no conducting path links the electrode to the formation

Across a permeable bed with mudcake, the caliper reads a diameter smaller than the bit size

A temperature log run an hour after circulation stops records the undisturbed geothermal gradient

In a vertical hole, the dip azimuth on a dipmeter points toward the pad that records the bed boundary at the greater depth

Which of the following statements about NMR logging are correct?

Water in small pores has a shorter T₂ than water in large pores, because surface relaxation dominates in small pores

NMR porosity needs no matrix density or matrix transit time, because only hydrogen in the pore fluids contributes

Water in small pores has a shorter T₂ than water in large pores, because surface relaxation dominates in small pores

NMR porosity needs no matrix density or matrix transit time, because only hydrogen in the pore fluids contributes

For fluids in rock, T₂ is always longer than or equal to T₁

The part of the T₂ distribution lying above the cut-off is the bound fluid, BVI

Which of the following statements about cement-bond logging are correct?

The CBL amplitude is read at the 3 ft receiver, while the VDL waveform is displayed from the 5 ft receiver

A microannulus can give a high CBL amplitude opposite sound cement, and logging under pressure often reduces it

The CBL amplitude is read at the 3 ft receiver, while the VDL waveform is displayed from the 5 ft receiver

A microannulus can give a high CBL amplitude opposite sound cement, and logging under pressure often reduces it

Light foamed cement attenuates the casing signal more than normal cement, so it always shows a lower amplitude

An off-centre tool raises the CBL amplitude, so a well-bonded section can be mistaken for free pipe

Geophysical Well Logging: SP, Resistivity, Radioactivity, Sonic, NMR and Cement-Bond Logs, and Quantitative Formation Evaluation

This chapter of Part B2 of the Geology and Geophysics (GG) paper answers the heading “Geophysical Well Logging”: “Principles and techniques of geophysical well-logging, SP, resistivity, induction, gamma ray, neutron, density, sonic, temperature, dip meter, caliper, nuclear magnetic resonance - longitudinal and transverse relaxation, CPMG sequence, porosity characterization, cement bond logging, micro-logs, Quantitative evaluation of formations from well logs.” It teaches the borehole environment and the invasion profile, the SP, gamma-ray, caliper, temperature and dipmeter logs, the resistivity, induction and micro-logs with Archie’s law, the density, neutron and sonic porosity logs, NMR logging, cement-bond logging, and the workflow of quantitative evaluation. The numericals are Archie water saturation, density and sonic porosity, shale volume from the gamma-ray index, Rw from the SP, NMR bound-fluid arithmetic and bulk volume water.

1. Principles: the borehole environment and the family of logs

A well log is a continuous record of a physical property against depth, made by lowering a sonde on a wireline. Drilling mud is kept at a pressure above that of the pores (overbalance), so in a permeable bed the mud filtrate invades the rock and the solids build a mudcake on the wall. Three zones result: the flushed zone next to the hole (resistivity Rxo, water saturation Sxo), a transition zone, and the uninvaded zone with true resistivity Rt and saturation Sw. The resistivities of mud (Rm), filtrate (Rmf), cake (Rmc) and formation water (Rw) all fall as temperature rises, and each tool has its own depth of investigation, so comparing shallow and deep readings reveals invasion and hence permeability.

The logs and what each is used for
Log	Measures	Main uses
SP	Natural potential, mV	Permeable beds, Rw, shale volume
Gamma ray	Natural radioactivity, API	Shale volume, correlation
Caliper	Hole diameter	Washouts, mudcake, corrections
Resistivity, induction, micro	Resistivities Rt, Ri, Rxo	Saturation, invasion, thin beds
Density, neutron, sonic	Bulk density, hydrogen index, transit time	Porosity, lithology, gas
Temperature, dipmeter	Fluid temperature; bed dip	Cement top, gas entry; structure
NMR, cement bond	Hydrogen relaxation; casing amplitude	Pore sizes, bound fluid; cement quality

2. SP, gamma ray, caliper, temperature and dipmeter logs

The SP log records the natural potential between an electrode in the hole and one at the surface. It arises from the electrochemical effect (membrane potential across shale plus the liquid-junction potential) and a small electrokinetic part. In a thick clean permeable bed the deflection from the shale baseline reaches the static SP, SSP = −K log₁₀(Rmf/Rw), where K is about 65 + 0.24·T (T in °C, mV per decade; about 71 mV at 25 °C). With fresh mud (Rmf > Rw) the sand shows a negative (leftward) deflection; if Rmf < Rw it is positive; if Rmf = Rw there is none. SP locates permeable beds and their boundaries (at the inflection points) and gives Rw, but not the magnitude of permeability, and it needs a conductive mud, so it fails in oil-based mud and in cased holes.

The gamma-ray log measures natural radioactivity from potassium-40 and the uranium and thorium series, in API units, with a scintillation detector. Shale is high, while clean sandstone, limestone, anhydrite and salt are low. The gamma-ray index is I_GR = (GR − GR_min)/(GR_max − GR_min); the linear estimate V_sh = I_GR is an upper bound, and Larionov’s non-linear corrections are V_sh = 0.083(2^3.7·I_GR − 1) for younger (Tertiary) rocks and 0.33(2^2·I_GR − 1) for older rocks. Arkosic (K-feldspar) or micaceous sands, glauconite and potash can read high without being shaly, so a spectral log separating K, Th and U is safer. The log works in cased holes and in any mud.

The caliper log gives the hole diameter from spring-loaded arms: it reads below bit size against permeable beds (mudcake) and above it against washouts in shale, salt or coal; it corrects other logs and gives the cement volume. The temperature log shows the geothermal gradient (typically 25 to 30 °C per km) and local anomalies: cooling where gas enters, a warm anomaly at the hydrating cement top, flow into or out of the hole. A log run soon after circulation shows mud-disturbed values. Fluid resistivities are brought to formation temperature with Arps’ relation R₂ = R₁(T₁ + 21.5)/(T₂ + 21.5) (T in °C). The dipmeter has pads with microresistivity buttons at 90° apart and an orientation package (inclinometer and compass). In a vertical hole of diameter d, a planar boundary seen at depths differing by Δz on opposite pads has tan(dip) = Δz/d, and the bed dips toward the pad that sees it deeper.

⚠️ High gamma ray is not always shale, and the SP is not permeability

The linear form V_sh = I_GR overestimates shale volume in most rocks, and a radioactive but clean sand (arkose, mica, glauconite) pushes the estimate higher still because the log reads high without any clay. The SP size depends on the salinity contrast and shale content, not on how permeable the sand is: a tight sand and a very permeable one can show the same deflection.

3. Resistivity, induction and micro-logs; Archie’s law

Dry rock is an insulator, so the current is carried by the brine in the pores. Archie’s law relates a fully water-saturated rock to its water: the formation factor F = Ro/Rw = a/φ^m, where φ is porosity, m the cementation exponent (about 2 for consolidated sandstone, lower in loose sand), and a a constant near 1. When hydrocarbons are present the resistivity index is I = Rt/Ro = Sw^(−n), with saturation exponent n about 2, so Sw = [a·Rw/(φ^m·Rt)]^(1/n). Hydrocarbons do not conduct, so Rt rises as Sw falls. Archie assumes a clean formation: clay conducts too, so in shaly sand Archie gives Sw too high.

Normal and lateral tools pass current between electrodes and read a potential, and are limited by mud and bed-thickness effects. The laterolog forces the current by guard electrodes into a thin horizontal sheet, so it works well in saline mud and high formation resistivity, and it needs a conductive mud. The induction tool needs no current path through the mud: a transmitter coil at about 20 kHz induces eddy-current loops in the formation, whose magnetic field induces a signal in the receiver proportional to formation conductivity (σ = 1/R). It suits fresh, oil-based mud and air-filled holes, and is best at low to moderate resistivity. Dual tools give shallow, medium and deep readings (Rxo, Ri, Rt).

Micro-logs use closely spaced electrodes on a pad pressed against the wall. The micro-normal (2-inch) investigates slightly deeper than the micro-inverse (1 × 1 inch). Where a mudcake of low resistivity coats the wall, the micro-inverse is more affected by it and reads lower, so positive separation (micro-normal above micro-inverse) marks a permeable bed, while shale shows little or negative separation. Micro-logs also give Rxo, thin-bed detail and hole-wall quality.

⚠️ Match the tool to the mud

Laterologs need a conductive path, so they suit salty mud; induction logs need none, so they suit fresh or oil-based mud. A question that names the mud type is really asking which tool can work at all.

4. Density, neutron and sonic porosity logs

The density tool presses a pad with a gamma-ray source (caesium-137) against the wall. The gamma rays lose energy by Compton scattering in proportion to the electron density, which tracks bulk density ρb; two detectors correct for mudcake. For a liquid-filled rock φ_D = (ρma − ρb)/(ρma − ρf). Typical matrix densities are about 2.65 g/cm³ for quartz sandstone, 2.71 for limestone and 2.87 for dolomite, and ρf is about 1.0 for fresh filtrate. Gas has low density, so ρb is too low and φ_D too high. The photoelectric factor Pe separates lithologies (about 1.8 quartz, 5.1 calcite, 3.1 dolomite).

The neutron tool emits fast neutrons from an americium–beryllium source. Hydrogen, with nearly the neutron’s mass, slows them most, so the count rate measures the hydrogen index, which in a clean rock is the liquid-filled porosity; the log is scaled in limestone porosity units. Gas has a low hydrogen index, so the neutron porosity reads too low, while bound water in shale makes it read high. Plotted on compatible scales, a clean gas sand shows crossover, with neutron porosity below density porosity, whereas shale shows the neutron above the density. Neutron and density together fix porosity and lithology.

The sonic tool measures the compressional-wave transit time Δt (μs/ft) over a receiver spacing; two transmitters and receivers (borehole compensation) cancel hole-size effects. The Wyllie time-average equation, Δt = φ·Δt_f + (1 − φ)·Δt_ma, gives φ_S = (Δt − Δt_ma)/(Δt_f − Δt_ma), with Δt_ma about 55 μs/ft for sandstone, 47 for limestone and 43 for dolomite and Δt_f about 189 for water. It holds for consolidated, clean, liquid-filled rock; in loose sands the answer is divided by a compaction factor, and gas raises Δt. The first arrival travels the fast matrix path around isolated vugs and fractures, so the sonic porosity is primary porosity, and the neutron–density porosity minus φ_S estimates secondary porosity. Cycle skipping gives spurious high Δt spikes. Sonic also ties wells to seismic and supplies elastic constants.

5. Nuclear magnetic resonance logging: T1, T2, CPMG and porosity

Hydrogen nuclei in pore fluids behave as small magnets. In a static field B₀ they align and precess at the Larmor frequency f = (γ/2π)B₀, which is 42.58 MHz per tesla for hydrogen. The tool polarises the protons for a wait time, and the magnetisation grows with the longitudinal (spin–lattice) relaxation time T₁: M_z = M₀(1 − e^(−t/T₁)), about 95% after 3T₁. A 90° radio-frequency pulse then tips the magnetisation into the transverse plane, where it decays with the transverse (spin–spin) relaxation time T₂; T₂ ≤ T₁. Only fluid hydrogen is seen, so the signal is independent of the rock matrix.

Static-field inhomogeneity dephases the spins faster than T₂ alone would. The CPMG sequence (Carr–Purcell–Meiboom–Gill) follows the 90° pulse by a train of 180° pulses at echo spacing TE; each refocuses the spins into a spin echo, and the echo amplitudes decay as e^(−t/T₂). The decay is a sum of exponentials, and inversion converts it into a T₂ distribution whose total area is the NMR porosity and whose bins are porosity in each pore size. Relaxation is mainly at the pore surface, 1/T₂ ≈ 1/T₂,bulk + ρ₂(S/V), so small pores (clay-bound and capillary-bound water) have short T₂ and large pores long T₂.

A T₂ cut-off splits the distribution: porosity below it is bound fluid, BVI (bulk volume irreducible), and above it free fluid, FFI, so MPHI = BVI + FFI. Common defaults are about 33 ms for sandstone and about 92 ms for carbonate, ideally calibrated on core. Then Swirr = BVI/MPHI. Clay-bound water (T₂ below about 3 ms) belongs to the total NMR porosity but is left out of the effective porosity that many logs report, so state which porosity MPHI means. Permeability is estimated with the Timur–Coates (k ∝ φ⁴(FFI/BVI)²) and SDR (k ∝ φ⁴·T₂,lm²) models, and hydrocarbons are typed from their different T₁, T₂ and diffusion contrasts with water.

🎯 Why NMR is different from the other porosity logs

Density, neutron and sonic porosities all need a matrix value (ρma, Δt_ma or a limestone scale), and a wrong lithology biases them. NMR responds only to hydrogen in fluids and needs no matrix, and, uniquely, its T₂ spectrum also tells how the porosity is divided between small and large pores.

6. Cement-bond logging

After casing is cemented, the cement must seal each zone from the next (zonal isolation). A cement-bond log (CBL) is a sonic tool run in the cased hole: a transmitter at about 20 kHz and two receivers, 3 ft and 5 ft away. The CBL amplitude is the height of the first casing arrival at the 3 ft receiver (in mV). Where casing is free, it rings and the amplitude is high; where cement is bonded to the casing, energy leaks into the cement and the amplitude is low. The attenuation can be written as 20·log₁₀(A_free/A) in dB, divided by the 3 ft spacing for dB/ft, and a bond index is the attenuation in the zone divided by that in a fully bonded section.

The VDL (variable-density log) displays the full waveform at the 5 ft receiver as light and dark bands. Free pipe gives strong, straight, parallel casing arrivals and weak formation signal. A good bond to both casing and formation gives weak casing arrivals and strong, wavy formation arrivals. Pitfalls: a thin microannulus (a gap between casing and cement from pressure or temperature change) raises the amplitude despite good cement, and re-running the log under pressure usually lowers it; an eccentred tool lowers the amplitude; very fast formations arrive before the casing signal; and light or foamed cement attenuates less. Ultrasonic pulse-echo tools map the cement all around the casing and show channels that the average CBL misses.

⚠️ High CBL amplitude does not always mean no cement

Read the amplitude together with the VDL and the transit time. A microannulus can raise the amplitude behind a sound cement sheath, an off-centre tool can lower it falsely, and a fast formation can arrive before the casing signal, so the log is run centralised and, if a microannulus is suspected, repeated under pressure before the job is judged to have failed.

7. Quantitative evaluation of formations from logs

The standard workflow is:
(1) find permeable beds (SP, gamma ray, caliper, micro-log);
(2) estimate V_sh from the gamma ray;
(3) get porosity from density, sonic or neutron–density, corrected for shale and gas;
(4) get Rw from the SP (Rw = Rmf·10^(SSP/K)), from a nearby water zone (Rw = Ro·φ^m/a) or from a catalogue;
(5) read Rt from the deep curve;
(6) compute Sw by Archie;
(7) report the hydrocarbon saturation Sh = 1 − Sw, the bulk volume water BVW = φ·Sw, and the hydrocarbon pore volume per unit bulk volume φ(1 − Sw).

Worked example. A sand reads ρb = 2.40 g/cm³ with ρma = 2.65 and ρf = 1.00, so φ = 0.25/1.65 = 0.152. With Rt = 30 Ω·m, Rw = 0.04 Ω·m and a = 1, m = n = 2: Sw = [0.04/(0.152² × 30)]^(1/2) = 0.24, so Sh = 0.76, BVW = 0.152 × 0.24 ≈ 0.0365 and the hydrocarbon pore volume is 0.152 × 0.76 ≈ 0.1155. For m = n = 2, BVW reduces to (a·Rw/Rt)^(1/2), independent of porosity, which is why a constant BVW over an interval signals a zone at irreducible water saturation.

Comparing Sxo (from Rxo and Rmf, with the same Archie form) with Sw gives the movable hydrocarbon, Sxo − Sw, because a large difference means the filtrate has flushed out oil. Shaly-sand models (Simandoux, Indonesian, dual-water) lower Sw relative to Archie. Net pay is the thickness that passes cut-offs on porosity, shale volume and Sw. Together with cased-hole logs (cement bond, temperature) the evaluation also checks completion integrity.

Key takeaways

Mud filtrate invades permeable beds and leaves mudcake; the flushed zone has Rxo, the uninvaded zone Rt. The SP gives SSP = −K log(Rmf/Rw) (negative against sand when Rmf > Rw), locates permeable beds and gives Rw, but needs conductive mud and not permeability.
Gamma ray gives I_GR and V_sh (linear is an upper bound; Larionov corrects), but K-feldspar, mica and glauconite mislead. Caliper, temperature and dipmeter (tan dip = Δz/d) give hole size, thermal anomalies and bed dip.
Archie: F = a/φ^m and Sw = [aRw/(φ^m Rt)]^(1/n). Laterologs need conductive mud; induction suits fresh or oil-based mud; positive micro-log separation marks permeable beds.
φ_D = (ρma − ρb)/(ρma − ρf); Wyllie φ_S = (Δt − Δt_ma)/(Δt_f − Δt_ma). Gas makes density porosity high and neutron porosity low (crossover); sonic sees primary porosity only.
NMR: T₁ builds the magnetisation, T₂ decays in the CPMG echo train, short T₂ means small pores; a T₂ cut-off (about 33 ms in sandstone) splits BVI from FFI, and NMR needs no matrix value.
CBL amplitude is low for bonded cement and high for free pipe, read with the VDL (casing against formation arrivals); microannulus is the classic trap. Evaluation runs φ, Rw, Sw, then Sh = 1 − Sw, BVW = φ·Sw and Sxo − Sw for movable hydrocarbon.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A permeable water-bearing sand is drilled with water-based mud kept above the pore pressure. Which zone lies immediately beyond the mudcake, with most of its original pore fluid displaced by mud filtrate?
1. The uninvaded zone, whose true resistivity is Rt
2. The flushed zone, whose resistivity is Rxo
3. The transition zone, whose pores hold only the original formation water
4. The mudcake itself, whose resistivity is Rmc
Show answer
Answer: B — The flushed zone, whose resistivity is Rxo
The zone next to the hole that filtrate has largely flushed is the flushed zone, with resistivity Rxo and water saturation Sxo. The uninvaded zone lies beyond the invasion front and keeps its original fluids (Rt, Sw). The transition zone is only partly invaded, so its pores hold a mixture, not formation water alone, and the mudcake is the deposited solids on the wall, not a zone of the rock.
A thick clean permeable sand contains formation water of resistivity 0.08 Ω·m and is drilled with mud whose filtrate resistivity is 0.8 Ω·m. Relative to the shale baseline, the SP opposite the sand is:
1. a negative (leftward) deflection of about K mV, because Rmf exceeds Rw by a factor of ten
2. a positive (rightward) deflection of about K mV, because the saline formation water conducts better than the filtrate
3. no deflection, because the SP responds to permeability and is unaffected by the salinity contrast
4. a deflection whose sign depends on how permeable the sand is, with its size fixed by the mud pressure
Show answer
Answer: A — a negative (leftward) deflection of about K mV, because Rmf exceeds Rw by a factor of ten
With SSP = −K log₁₀(Rmf/Rw) = −K log₁₀(0.8/0.08) = −K, the deflection is negative and about K mV (K is about 70 mV per decade near 25 °C). The sign is set by which of Rmf and Rw is larger, not by the permeability or the mud pressure, and a positive deflection would need Rmf < Rw, which is the reverse of the stated case.
Which statement about the gamma-ray log is correct?
1. A clean limestone reads higher than shale because carbonate rock concentrates uranium and thorium in its lattice
2. The log cannot be run in cased holes, because the gamma rays from the formation cannot penetrate steel casing at all
3. An oil-bearing sand reads lower than a water-bearing sand of the same clay content, because oil contains no potassium-40 or other radioactive atoms
4. An arkosic sand rich in K-feldspar can read high without being shaly, so gamma-ray shale volume overestimates its clay
Show answer
Answer: D — An arkosic sand rich in K-feldspar can read high without being shaly, so gamma-ray shale volume overestimates its clay
Potassium-40 in K-feldspar (and in mica or glauconite) makes a clean but radioactive sand read like shale, so I_GR overstates the shale volume there; a spectral log separating K, Th and U exposes the cause. Carbonates are usually low in gamma ray, the log works in cased holes because high-energy gamma rays pass through steel with only some attenuation, and the fluid in the pores does not change the formation’s radioactivity appreciably.
A well is drilled with non-conductive oil-based mud. Which tool is suited to measuring the deep formation resistivity, and why?
1. The laterolog, because its focusing currents flow most easily through resistive mud into the rock
2. The normal device, because its current electrodes need no contact with a conducting fluid in the borehole
3. The induction tool, because its coils induce eddy currents magnetically and need no conducting path through the mud
4. The SP device, because the potential it records between its two electrodes is directly proportional to the formation resistivity
Show answer
Answer: C — The induction tool, because its coils induce eddy currents magnetically and need no conducting path through the mud
Electrode tools (normal, lateral, laterolog) must inject current through the mud, which a non-conductive oil-based mud blocks. The induction tool’s transmitter coil induces eddy-current loops in the formation and its receiver senses their magnetic field, which is proportional to conductivity, so no conducting path is needed. The SP is a natural potential that also needs a conductive mud and does not give resistivity.
Opposite a bed, the micro-normal curve reads consistently higher than the micro-inverse curve (positive separation). What does this indicate?
1. A shale bed, because the clay is more conductive than the mudcake
2. A permeable bed carrying a mudcake, with the micro-inverse more affected by the low-resistivity cake
3. A hydrocarbon-bearing zone, because oil raises the resistivity seen by the shallower micro-inverse curve
4. A hole washout, because the larger diameter makes the micro-normal read the mud resistivity
Show answer
Answer: B — A permeable bed carrying a mudcake, with the micro-inverse more affected by the low-resistivity cake
The micro-inverse (1 × 1 inch) is shallower, so a thin mudcake of low resistivity pulls its reading down more than that of the micro-normal (2 inch), which looks slightly deeper into the invaded rock. Mudcake forms only against permeable beds, so positive separation marks permeability; shale shows little or negative separation. Hydrocarbon content is not what drives this separation, and a washout would drag both curves towards the mud value.
In a clean sandstone the density porosity, computed with ρma = 2.65 g/cm³ and ρf = 1.0 g/cm³, is 24%, and the neutron porosity, correctly scaled for sandstone, is 11%. Which explanation is most likely?
1. Gas fills part of the pores: its low density inflates density porosity and its low hydrogen index depresses neutron porosity
2. Thin shale laminae holding clay-bound water, which lower the hydrogen index seen by the neutron tool and so depress its reading
3. Pyrite and other heavy minerals raise the bulk density, which makes the density porosity read well above the true porosity
4. The pores are completely flushed with fresh mud filtrate, so both tools read the true porosity and differ only by random noise
Show answer
Answer: A — Gas fills part of the pores: its low density inflates density porosity and its low hydrogen index depresses neutron porosity
Here the density porosity exceeds the neutron porosity by a wide margin, the crossover signature of gas: gas lowers ρb, so φ_D computed with a liquid ρf is too high, and its low hydrogen index makes φ_N too low. Clay-bound water contains hydrogen and raises, not lowers, the neutron reading, so shale laminae push the neutron above the density rather than below it. Heavy minerals raise ρb and so lower φ_D, again the opposite direction, and full flushing would make the two porosities agree.
In a carbonate with large isolated vugs, the density–neutron porosity is 18% while the Wyllie sonic porosity, with correct matrix and fluid transit times, is 11%. Which statement best explains the difference?
1. Every vug slows the first arrival wave, so the sonic value is the true total porosity and the density–neutron value is an overestimate
2. Wyllie’s equation always underestimates porosity in carbonates because the matrix transit time assumed for limestone is set far too high
3. The first arrival takes the faster matrix path around isolated vugs, so sonic porosity is mainly primary; the gap estimates vuggy porosity
4. Isolated vugs contain no fluid, so only the density and neutron tools can detect them, while the sonic tool cannot respond to them at all
Show answer
Answer: C — The first arrival takes the faster matrix path around isolated vugs, so sonic porosity is mainly primary; the gap estimates vuggy porosity
A compressional first arrival follows the fastest route, which in a rock with isolated vugs is the connected matrix around them, so the sonic porosity records primary (intergranular) porosity. Density and neutron respond to total pore volume, so (φ_DN − φ_S) = 18 − 11 = 7 porosity units estimates the secondary vuggy porosity. The vugs hold fluid like other pores, and the stem states that the matrix and fluid transit times are correct, so a wrong limestone matrix value cannot be the cause of the gap.
In a CPMG sequence, what is the purpose of the train of 180° pulses that follows the initial 90° pulse?
1. To repolarise the protons between echoes, so that the longitudinal relaxation time T₁ is measured directly
2. To raise the Larmor frequency, so that more hydrogen nuclei fall inside the sensitive volume of the tool
3. To saturate the hydrogen in the rock matrix, so that only the pore-fluid hydrogen contributes to the measured signal
4. To refocus spins dephased by static-field inhomogeneity into spin echoes, so that echo amplitudes decay with T₂
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Answer: D — To refocus spins dephased by static-field inhomogeneity into spin echoes, so that echo amplitudes decay with T₂
After the 90° pulse the transverse magnetisation fades quickly because spins in slightly different fields dephase. Each 180° pulse reverses that dephasing and forms a spin echo, so the train of echo amplitudes decays only through true transverse relaxation, e^(−t/T₂). The pulses do not measure T₁, which needs a recovery or a separate polarisation sequence, they do not change the Larmor frequency (set by B₀), and the matrix has no mobile hydrogen to saturate.
Which combination of CBL and VDL observations indicates cement well bonded to both the casing and the formation?
1. High CBL amplitude, with strong straight parallel casing arrivals and weak formation arrivals
2. Moderate CBL amplitude, with strong casing arrivals and no formation arrivals
3. Low CBL amplitude, with weak casing arrivals and strong wavy formation arrivals
4. High CBL amplitude, with weak casing arrivals and weak formation arrivals
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Answer: C — Low CBL amplitude, with weak casing arrivals and strong wavy formation arrivals
Good bonding to the casing drains energy from the pipe, giving a low CBL amplitude and weak casing arrivals, and good bonding to the formation passes the energy on, so the formation arrivals on the VDL are strong and wavy. High amplitude with straight casing bands is free pipe. Moderate amplitude with strong casing arrivals and no formation signal means the cement is not bonded to the formation, and a high amplitude cannot occur together with weak casing arrivals.
Which of the following statements about the SP, caliper, temperature and dipmeter logs are correct?
1. The SP cannot be recorded in oil-based mud, because no conducting path links the electrode to the formation
2. Across a permeable bed with mudcake, the caliper reads a diameter smaller than the bit size
3. A temperature log run an hour after circulation stops records the undisturbed geothermal gradient
4. In a vertical hole, the dip azimuth on a dipmeter points toward the pad that records the bed boundary at the greater depth
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Answer: A — The SP cannot be recorded in oil-based mud, because no conducting path links the electrode to the formation; B — Across a permeable bed with mudcake, the caliper reads a diameter smaller than the bit size; D — In a vertical hole, the dip azimuth on a dipmeter points toward the pad that records the bed boundary at the greater depth
Without a conductive mud the SP electrode has no electrical contact with the rock, so the log fails. Mudcake builds up opposite permeable beds, and the caliper reads under bit size there. A downdip pad meets a dipping boundary deeper, so the dip is toward the deeper-reading pad. A log run soon after circulation still shows the mud’s cooling, not the undisturbed gradient, which needs a long shut-in.
Which of the following statements about NMR logging are correct?
1. Water in small pores has a shorter T₂ than water in large pores, because surface relaxation dominates in small pores
2. NMR porosity needs no matrix density or matrix transit time, because only hydrogen in the pore fluids contributes
3. For fluids in rock, T₂ is always longer than or equal to T₁
4. The part of the T₂ distribution lying above the cut-off is the bound fluid, BVI
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Answer: A — Water in small pores has a shorter T₂ than water in large pores, because surface relaxation dominates in small pores; B — NMR porosity needs no matrix density or matrix transit time, because only hydrogen in the pore fluids contributes
In small pores the fluid is close to the grain surface, where relaxation is fast, so T₂ is short; large pores give long T₂. Only fluid hydrogen is seen, so no matrix value is needed. In fact T₂ is never longer than T₁, and the part of the distribution above the cut-off is the free fluid FFI, while the part below it is bound fluid.
Which of the following statements about cement-bond logging are correct?
1. The CBL amplitude is read at the 3 ft receiver, while the VDL waveform is displayed from the 5 ft receiver
2. A microannulus can give a high CBL amplitude opposite sound cement, and logging under pressure often reduces it
3. Light foamed cement attenuates the casing signal more than normal cement, so it always shows a lower amplitude
4. An off-centre tool raises the CBL amplitude, so a well-bonded section can be mistaken for free pipe
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Answer: A — The CBL amplitude is read at the 3 ft receiver, while the VDL waveform is displayed from the 5 ft receiver; B — A microannulus can give a high CBL amplitude opposite sound cement, and logging under pressure often reduces it
The 3 ft receiver gives the CBL amplitude and the 5 ft receiver the VDL waveform. A microannulus lets the pipe ring despite good cement and a pressured pass closes the gap. Foamed or light cement attenuates less (lower acoustic impedance), so it shows a higher amplitude than its quality deserves, and an off-centre tool lowers the amplitude, making a free pipe look better bonded rather than worse.
A clean water-wet sandstone has porosity 0.20, true resistivity Rt = 20 Ω·m and formation-water resistivity Rw = 0.05 Ω·m. Take Archie’s constants a = 1, m = 2 and n = 2, with Sw = [a·Rw/(φ^m·Rt)]^(1/n). The water saturation Sw, as a fraction, is ____ to two decimal places.
Numerical answer — type the value.
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Answer: 0.25
φ^m = 0.20² = 0.04, so φ^m·Rt = 0.04 × 20 = 0.8 and a·Rw = 0.05. The ratio is 0.05/0.8 = 0.0625, and Sw = 0.0625^(1/2) = 0.25. Check through the formation factor: F = 1/0.04 = 25, Ro = 25 × 0.05 = 1.25 Ω·m, I = Rt/Ro = 16, and Sw = I^(−1/2) = 0.25.
In a clean sandstone the density log reads ρb = 2.32 g/cm³; take ρma = 2.65 g/cm³ and ρf = 1.00 g/cm³. The deep resistivity is Rt = 40 Ω·m and Rw = 0.05 Ω·m, with Archie’s a = 1, m = 2, n = 2. The bulk volume water BVW = φ·Sw, expressed as a percentage of bulk volume, is ____ to one decimal place.
Numerical answer — type the value.
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Answer: 3.5
Density porosity φ = (2.65 − 2.32)/(2.65 − 1.00) = 0.33/1.65 = 0.20. Then Sw = [0.05/(0.20² × 40)]^(1/2) = [0.05/1.6]^(1/2) = (0.03125)^(1/2) = 0.1768. BVW = 0.20 × 0.1768 = 0.03536, which is 3.5% of bulk volume. Check: for m = n = 2, BVW = (a·Rw/Rt)^(1/2) = (0.05/40)^(1/2) = 0.03536, the same.
A sonic log in a clean, consolidated, water-filled sandstone reads Δt = 94 μs/ft. Take Δt_ma = 55 μs/ft and Δt_f = 185 μs/ft and use the Wyllie time-average equation. The porosity, in percent, is ____.
Numerical answer — type the value.
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Answer: 30
From Δt = φ·Δt_f + (1 − φ)·Δt_ma, φ = (Δt − Δt_ma)/(Δt_f − Δt_ma) = (94 − 55)/(185 − 55) = 39/130 = 0.30, which is 30%. Check: 0.30 × 185 + 0.70 × 55 = 55.5 + 38.5 = 94 μs/ft.
In an older consolidated rock, a gamma-ray log reads GR = 70 API in a bed, with GR_min = 20 API in clean sand and GR_max = 120 API in shale. Use Larionov’s relation for older rocks, V_sh = 0.33(2^(2·I_GR) − 1). The shale volume V_sh, as a fraction, is ____ to two decimal places.
Numerical answer — type the value.
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Answer: 0.33
I_GR = (70 − 20)/(120 − 20) = 0.5. Then V_sh = 0.33(2^(2 × 0.5) − 1) = 0.33(2 − 1) = 0.33. The linear estimate V_sh = I_GR = 0.50 is the tempting wrong answer: it is an upper bound, and Larionov’s non-linear relation gives a smaller value.
An NMR T₂ distribution in a sandstone has partial porosities of 3 porosity units (pu) with T₂ below 3 ms, 5 pu with T₂ between 3 and 33 ms, and 10 pu with T₂ above 33 ms. Treat all porosity with T₂ below the 33 ms cut-off, clay-bound water included, as bound fluid (BVI), and take MPHI as the total of the three partial porosities. The irreducible water saturation Swirr = BVI/MPHI, as a percentage rounded to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 44
MPHI = 3 + 5 + 10 = 18 pu. BVI is the porosity with T₂ below the cut-off: 3 + 5 = 8 pu, and FFI = 10 pu (check: 8 + 10 = 18). So Swirr = 8/18 = 0.444, which is 44%. Taking only the 3–33 ms bin as bound (5/18 = 28%) forgets the clay-bound water that also lies below the cut-off.
The static SP is given by SSP = −K·log₁₀(Rmf/Rw) with K = 70 mV per decade. A thick clean water sand shows a deflection of −35 mV from the shale baseline, and the mud filtrate resistivity at formation temperature is Rmf = 0.40 Ω·m. The formation-water resistivity Rw is ____ Ω·m, to two decimal places.
Numerical answer — type the value.
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Answer: 0.13
From −35 = −70·log₁₀(0.40/Rw), log₁₀(0.40/Rw) = 0.5, so 0.40/Rw = 10^0.5 = 3.162 and Rw = 0.40/3.162 = 0.1265, which is 0.13 Ω·m to two decimals. Equivalently Rw = Rmf·10^(SSP/K) = 0.40 × 10^(−0.5). Rw comes out smaller than Rmf, as it must for a negative deflection.