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PEC EPE Petro-Gas practice questions

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Question 1Reservoir Engineering

Which primary drive mechanism typically provides the highest oil recovery factor in a reservoir that has a strong, active aquifer and good vertical communication between the oil zone and the water leg?

  1. Water drive Correct answer.
  2. Gas cap expansion drive
  3. Solution gas drive
  4. Rock and fluid expansion drive

Explanation

An active aquifer supplies both pressure support and a displacing front that sweeps oil toward the producers, so water-drive reservoirs commonly recover roughly 30–50% of OOIP. Solution-gas and purely compaction-driven reservoirs generally deliver far lower primary recovery because the available energy per barrel of oil is much smaller.

Question 2Reservoir Engineering

In a strongly water-wet sandstone core at irreducible water saturation, the relative permeability to water is closest to which value?

  1. 0.5
  2. 1.0
  3. 0 Correct answer.
  4. 0.2

Explanation

At irreducible water saturation the water occupies the smallest pores and is held in place by capillary forces, so it cannot flow and its relative permeability is zero. Water relative permeability rises only as water saturation increases above Swi and reaches 1.0 when the oil saturation falls to residual.

Question 3Reservoir Engineering

The oil formation volume factor, Bo, is defined as the reservoir volume occupied at reservoir conditions by which quantity?

  1. One stock-tank barrel of oil plus its dissolved gas Correct answer.
  2. One standard cubic foot of separator gas at base conditions
  3. One pound-mole of gas at standard conditions
  4. One barrel of water at reservoir conditions

Explanation

Bo converts a stock-tank barrel of saleable oil, together with the gas dissolved in it at reservoir pressure, into the larger volume that barrel occupies in the reservoir. Because oil shrinks as dissolved gas is released, Bo is greater than one for most live oils and decreases toward unity as pressure falls below the bubble point.

Question 4Reservoir Engineering

A drainage area covers 640 acres with an average net pay of 30 ft, porosity of 0.22, and connate water saturation of 0.25. The oil formation volume factor is 1.30 rb/stb. Using the 7,758 constant for barrels per acre-foot, what is the original oil in place?

  1. 18.9 MMstb Correct answer.
  2. 9.5 MMstb
  3. 14.2 MMstb
  4. 25.2 MMstb

Explanation

Volumetric OOIP is 7,758 × A × h × φ × (1 − Sw) / Bo, giving 7,758 × 640 × 30 × 0.22 × 0.75 / 1.30 ≈ 18.9 million stock-tank barrels. The 7,758 factor converts acre-feet to barrels so that area in acres and thickness in feet can be used directly.

Question 5Reservoir Engineering

A well drains a circular reservoir with a drainage radius of 1,000 ft and a wellbore radius of 0.33 ft. Permeability is 50 mD, net pay is 40 ft, and the pressure at the drainage boundary is 3,000 psia while the flowing bottomhole pressure is 2,000 psia. With oil viscosity of 1.5 cp and Bo of 1.2 rb/stb, what is the steady-state oil rate using the radial flow equation?

  1. 981 stb/d Correct answer.
  2. 490 stb/d
  3. 1,960 stb/d
  4. 3,920 stb/d

Explanation

For steady-state radial flow in field units, q = 0.00708 k h (Pe − Pw) / [μ Bo ln(re/rw)]. Substituting gives 0.00708 × 50 × 40 × 1,000 divided by 1.5 × 1.2 × ln(3,030), which evaluates to about 981 stb/d.

Question 6Reservoir Engineering

At a given water saturation in a waterflood, the oil relative permeability is 0.20 and the water relative permeability is 0.08. Oil viscosity is 2.0 cp and water viscosity is 0.5 cp, and capillary effects are neglected. What is the fractional flow of water at that saturation?

  1. 0.62 Correct answer.
  2. 0.50
  3. 0.55
  4. 0.38

Explanation

Fractional flow uses mobilities, so fw = (krw/μw) / [(krw/μw) + (kro/μo)]. Here the water mobility is 0.16 and the oil mobility is 0.10, giving fw = 0.16/0.26 ≈ 0.62.

Question 7Reservoir Engineering

A reservoir was discovered at 3,200 psia, and laboratory analysis gives a bubble point pressure of 2,500 psia for the crude. The reservoir is currently at 2,300 psia with no water influx. Which statement correctly describes the reservoir fluids at current conditions?

  1. The oil is undersaturated and no free gas is present in the reservoir
  2. The oil is saturated and free gas is present in the reservoir Correct answer.
  3. The oil shows retrograde behaviour and drops out condensate in the pores
  4. The oil is saturated but all released gas remains dissolved at these conditions

Explanation

Once reservoir pressure falls below the bubble point, the crude can no longer hold all its dissolved gas, so the oil is saturated and a free gas phase exists in the pore space. Free gas then competes with oil for flow and becomes the main source of reservoir energy through its expansion.

Question 8Reservoir Engineering

A p/z versus cumulative gas production plot for a volumetric dry gas reservoir gives a straight line. Extrapolating that line to p/z equal to zero provides which quantity?

  1. The original gas in place Correct answer.
  2. The abandonment reservoir pressure
  3. The cumulative water influx
  4. The absolute open flow potential

Explanation

The gas material balance in the form p/z = (pi/zi)(1 − Gp/G) is linear in cumulative production, so the extrapolated intercept at p/z = 0 corresponds to Gp = G. The slope itself is related to the initial p/z divided by the original gas in place.

Question 9Reservoir Engineering

A volumetric dry gas reservoir had an initial pressure of 4,000 psia with z of 0.90. Current pressure is 3,000 psia with z of 0.85, and cumulative production to date is 20 Bcf. Assuming no water influx and no rock compaction, what is the original gas in place?

  1. 130 Bcf
  2. 150 Bcf
  3. 52 Bcf
  4. 97 Bcf Correct answer.

Explanation

The gas material balance gives G = Gp / [1 − (p/z)/(pi/zi)]. The initial p/z is 4,444 psia and the current p/z is 3,529 psia, so the production fraction is 0.206 and G = 20/0.206 ≈ 97 Bcf.

Question 10Reservoir Engineering

A solution-gas-drive well produces 800 stb/d at a flowing bottomhole pressure of 2,000 psia when the average reservoir pressure is 3,000 psia. Applying Vogel's inflow performance relationship, what is the maximum oil rate at a flowing pressure of zero?

  1. 1,250 stb/d
  2. 1,565 stb/d Correct answer.
  3. 1,410 stb/d
  4. 1,750 stb/d

Explanation

Vogel's relation is qo/qmax = 1 − 0.2(Pwf/Pr) − 0.8(Pwf/Pr)². With Pwf/Pr = 0.667 the bracket evaluates to 0.511, so qmax = 800/0.511 ≈ 1,565 stb/d.

Question 11Reservoir Engineering

Which term describes the fraction of bulk rock volume that consists of interconnected pore space available for fluid flow?

  1. Effective porosity Correct answer.
  2. Absolute permeability
  3. Connate water saturation
  4. Total porosity

Explanation

Effective porosity counts only the connected pores that can store and transmit fluids. Total porosity also includes isolated pores that do not contribute to flow, so the two are not interchangeable.

Question 12Reservoir Engineering

The pressure at which the first liquid droplet condenses from a hydrocarbon gas during isothermal pressure reduction is called the:

  1. Critical pressure
  2. Hydrostatic pressure
  3. Bubble-point pressure
  4. Dew-point pressure Correct answer.

Explanation

The dew point marks the transition from a single-phase gas to a gas with an initial liquid phase. Bubble-point pressure describes the opposite transition, where the first gas bubble forms from a liquid.

Question 13Reservoir Engineering

Stock-tank oil volumes in reservoir engineering are reported at which reference conditions?

  1. Separator pressure and separator temperature
  2. Reservoir temperature and reservoir pressure
  3. Bubble-point pressure and reservoir temperature
  4. Standard surface temperature and pressure Correct answer.

Explanation

Stock-tank oil is the oil remaining after dissolved gas has been released and the oil is measured at standard surface conditions. Separator conditions are part of the surface processing train but are not the stock-tank reference.

Question 14Reservoir Engineering

A water-wet oil reservoir has a water-oil contact with zero capillary pressure. At 100 ft above that contact, water density is 1.03 g/cm3 and oil density is 0.80 g/cm3. What is the capillary pressure at that elevation?

  1. 20.0 psi
  2. 1.0 psi
  3. 10.0 psi Correct answer.
  4. 5.0 psi

Explanation

Capillary pressure is the buoyancy difference over the height: Pc = h(ρw - ρo)g. The density difference of 0.23 g/cm3 gives about 0.100 psi/ft, so 100 ft corresponds to about 10 psi.

Question 15Reservoir Engineering

In a solution-gas-drive reservoir with no gas cap, no water influx, and negligible rock and water compressibility, which drive index is approximately 1.0?

  1. Rock-compaction drive index
  2. Solution-gas drive index Correct answer.
  3. Water-drive index
  4. Gas-cap drive index

Explanation

Drive indices sum to about 1.0 when all active mechanisms are included. With no gas cap, no water influx, and negligible compaction, the solution-gas mechanism accounts for essentially all of the reservoir energy.

Question 16Reservoir Engineering

In the radial-flow equation, a positive skin factor most directly represents:

  1. The pressure drop across the drainage boundary
  2. A stimulation effect that increases productivity
  3. Turbulence in the tubing string
  4. Near-wellbore damage that adds pressure loss Correct answer.

Explanation

Positive skin adds an extra pressure drawdown near the wellbore, commonly caused by mud damage, fines, or completion restrictions. Negative skin, by contrast, indicates stimulation or improved near-wellbore flow.

Question 17Reservoir Engineering

Why is the real-gas pseudo-pressure, m(p), used in gas well deliverability analysis?

  1. It accounts for viscosity and z-factor variation with pressure Correct answer.
  2. It replaces relative permeability in the flow equation
  3. It removes the need to know reservoir temperature
  4. It converts a gas reservoir into a two-phase oil-water system

Explanation

The pseudo-pressure integral incorporates the pressure-dependent gas viscosity and z-factor. This makes the gas flow equation more nearly linear than a formulation based on pressure alone.

Question 18Reservoir Engineering

In one-dimensional Buckley-Leverett waterflood theory, the shock-front water saturation is found by:

  1. Connecting the initial oil saturation to the residual oil saturation
  2. Using the water saturation at which oil relative permeability is zero
  3. A tangent from initial water saturation to the fractional-flow curve Correct answer.
  4. Taking the maximum water saturation on the fractional-flow curve

Explanation

The Buckley-Leverett front saturation is determined by the tangent construction from the initial water saturation to the fractional-flow curve. The slope of that tangent gives the frontal advance velocity of the shock front.

Question 19Reservoir Engineering

In a gas-condensate reservoir produced below the dew point, condensate banking near the wellbore reduces gas productivity mainly because:

  1. The liquid dropout reduces gas relative permeability Correct answer.
  2. The gas viscosity increases sharply at lower pressure
  3. The absolute permeability of the rock increases
  4. The water saturation decreases below connate water

Explanation

Below the dew point, heavier components condense and accumulate near the wellbore as a liquid bank. This liquid occupies pore space and lowers the relative permeability to gas, reducing deliverability.

Question 20Reservoir Engineering

Before reservoir pressure falls below the bubble point in an undersaturated oil reservoir with no gas cap and no water influx, the active drive mechanism is:

  1. Gas-cap expansion drive mechanism
  2. Rock and fluid expansion drive Correct answer.
  3. Active water-drive support mechanism
  4. Solution-gas expansion drive mechanism

Explanation

Above the bubble point, no free gas has evolved and no gas cap or aquifer is present. Production is supported by expansion of the oil and connate water plus reduction of pore volume as pressure declines.

Question 21Reservoir Engineering

In reservoir engineering, which term is defined as the ratio of effective permeability to a fluid at a given saturation to the viscosity of that fluid?

  1. Transmissibility
  2. Mobility Correct answer.
  3. Storativity
  4. Hydraulic diffusivity

Explanation

Mobility is λ = k/μ, where k is the effective permeability to the flowing phase and μ is its viscosity. It measures how easily a fluid moves through rock under a unit pressure gradient. Transmissibility adds thickness, while diffusivity and storativity include compressibility and porosity terms.

Question 22Reservoir Engineering

The solution gas-oil ratio, Rs, of a crude oil is defined as which quantity?

  1. The reservoir barrels of gas released from one stock-tank barrel of oil at surface
  2. The ratio of produced gas rate to produced oil rate at the separator
  3. The reservoir volume of oil divided by the stock-tank volume of oil produced at surface
  4. The standard cubic feet of gas dissolved in one stock-tank barrel of oil Correct answer.

Explanation

Rs is the quantity of dissolved gas associated with one stock-tank barrel of oil at the prevailing reservoir pressure and temperature. It is reported in scf/STB and is determined from laboratory differential or flash liberation tests. Produced GOR is a surface measurement and is not the same as Rs.

Question 23Reservoir Engineering

The real-gas deviation factor z approaches a value of 1.0 under which set of reservoir conditions?

  1. Low pressure and high temperature, far from the phase envelope Correct answer.
  2. High pressure and low temperature, near the hydrocarbon dew point
  3. High pressure and high temperature, well above the critical point
  4. Low pressure and low temperature, where condensate forms

Explanation

At low pressure and high temperature, gas molecules are far apart and interact weakly, so behavior approaches that of an ideal gas and z is close to 1. Near the critical point or dew point, attractive forces reduce z well below 1. At high pressure, repulsive forces can push z above 1.

Question 24Reservoir Engineering

A waterflood displacement has a mobility ratio M = 2.5, calculated as water mobility divided by oil mobility. Which interpretation is correct?

  1. The displacement is unfavorable because oil will finger through water.
  2. The displacement is favorable because water is less mobile than oil.
  3. The mobility ratio has no effect on areal sweep efficiency.
  4. The displacement is unfavorable because water is more mobile than oil. Correct answer.

Explanation

M greater than 1 means the displacing water is more mobile than the displaced oil, which promotes viscous fingering and early water breakthrough. This reduces volumetric sweep efficiency. A favorable waterflood generally requires M ≤ 1.

Question 25Reservoir Engineering

The Leverett J-function is used to scale capillary pressure between a core sample and a reservoir. If the J-function value is the same in both systems, the capillary pressure Pc is proportional to which group?

  1. σ sinθ (k/φ)
  2. σ cosθ √(φ/k) Correct answer.
  3. μ cosθ √(k/φ)
  4. σ cosθ (φ k)

Explanation

The Leverett J-function is J(Sw) = (Pc/(σ cosθ)) √(k/φ). Rearranging shows that, at equal J and wettability, Pc scales as σ cosθ √(φ/k). Porosity and permeability enter through the square root of their ratio.

Question 26Reservoir Engineering

A pot aquifer has an initial water volume of 5.0 × 10^8 bbl and a water compressibility of 3.0 × 10^-6 psi^-1. If reservoir pressure drops by 500 psi, what water influx is estimated, neglecting rock compressibility?

  1. 2,500,000 bbl
  2. 750,000 bbl Correct answer.
  3. 1,500,000 bbl
  4. 250,000 bbl

Explanation

For a pot aquifer, We ≈ cw Vi Δp = (3.0 × 10^-6)(5.0 × 10^8)(500) = 750,000 bbl. Rock compressibility would add a small additional term. The pot model assumes rapid pressure communication and a fixed aquifer volume.

Question 27Reservoir Engineering

In a gas well deliverability equation, the rate-dependent skin term Dq represents which effect?

  1. The pressure loss caused by water coning at the perforations
  2. The skin caused by mud filtrate invasion around the wellbore
  3. The skin caused by partial penetration of the perforated interval
  4. Additional pressure loss from high-velocity turbulent flow near the wellbore Correct answer.

Explanation

High gas velocity near a gas well creates inertial and turbulent losses that increase with rate. These are modeled as a rate-dependent skin Dq, where D is the non-Darcy or turbulence coefficient. It is separate from mechanical skin factors such as partial penetration or mud damage.

Question 28Reservoir Engineering

For a water-wet reservoir, how does the imbibition capillary pressure curve compare with the drainage curve at the same water saturation?

  1. Imbibition capillary pressure is always zero.
  2. Imbibition capillary pressure is always positive and larger than drainage.
  3. Imbibition capillary pressure is lower than drainage at the same water saturation. Correct answer.
  4. Imbibition capillary pressure equals drainage because hysteresis is negligible.

Explanation

Capillary pressure exhibits hysteresis because the contact angle and pore-scale fluid configuration differ between drainage and imbibition. For a water-wet system, the imbibition curve generally lies below the drainage curve at the same saturation. The difference is important when modeling waterflood and imbibition processes.

Question 29Reservoir Engineering

A rich gas-condensate reservoir is produced below its dew point. As reservoir pressure declines and condensate drops out in the formation, what is the expected trend in the produced separator gas-oil ratio?

  1. The gas-oil ratio decreases because more condensate is produced from the formation.
  2. The gas-oil ratio decreases because the gas becomes richer near the wellbore.
  3. The gas-oil ratio remains constant because the condensate is immobile.
  4. The gas-oil ratio increases because the produced gas becomes leaner. Correct answer.

Explanation

Below the dew point, heavier components condense and are left behind as an immobile liquid phase in the reservoir. The gas reaching the well is therefore leaner, so the separator produces less condensate per unit of gas and the gas-oil ratio rises. This behavior contrasts with an oil reservoir, where solution gas evolves and GOR usually falls before rising.

Question 30Reservoir Engineering

A dry gas reservoir with an active water drive is analyzed with a p/z versus cumulative production plot. How does the data typically deviate from the straight-line volumetric depletion trend?

  1. The data plot above the volumetric line because water influx supports reservoir pressure. Correct answer.
  2. The data plot below the volumetric line because gas dissolves in the invading water.
  3. The data plot below the volumetric line because water blocks gas flow.
  4. The data plot above the volumetric line because water influx adds gas to the reservoir.

Explanation

For volumetric depletion, p/z declines linearly with cumulative gas production. Water influx reduces the pressure decline for a given amount of gas produced, so the measured p/z remains higher than the volumetric line and plots above it. This deviation is used to estimate water influx and aquifer support.

Question 31Drilling Engineering

A vertical well section is filled with static water-based mud of density 1,200 kg/m3 to a true vertical depth of 2,500 m. Using g = 9.81 m/s2, what is the hydrostatic pressure at that depth?

  1. 24.5 MPa
  2. 29.4 MPa Correct answer.
  3. 31.8 MPa
  4. 30.6 MPa

Explanation

Hydrostatic pressure in a static vertical column is P = ρ g TVD. Substituting 1,200 kg/m3 × 9.81 m/s2 × 2,500 m gives about 29.4 MPa. The calculation uses true vertical depth, not measured depth.

Question 32Drilling Engineering

Which casing string is normally the first to be run and cemented in a well and is used to isolate shallow unconsolidated formations and support the wellhead?

  1. Conductor casing Correct answer.
  2. Intermediate casing
  3. Production casing
  4. Liner

Explanation

Conductor casing is the shallowest casing string, run and cemented before surface casing. It stabilizes the top hole, isolates shallow formations, and provides a foundation for the wellhead.

Question 33Drilling Engineering

While drilling, a flow check confirms that the well is flowing with the pumps off. What is the correct immediate well-control action?

  1. Circulate the influx at the current pump rate
  2. Shut in the well and record stabilized pressures Correct answer.
  3. Pull the drill string to surface to inspect the bit
  4. Increase the mud weight and keep drilling

Explanation

Once a kick is confirmed, the well must be shut in promptly to stop further influx and allow stabilized pressures to be recorded. Those shut-in pressures are needed to determine the kill mud weight and method. Continuing to drill or circulating before shut-in can enlarge the influx.

Question 34Drilling Engineering

A well is being drilled at 3,000 m TVD with mud density 1,450 kg/m3. The annular friction loss at the current flow rate is 1.2 MPa. What is the equivalent circulating density at that depth? Use g = 9.81 m/s2.

  1. 1,491 kg/m3 Correct answer.
  2. 1,475 kg/m3
  3. 1,450 kg/m3
  4. 1,410 kg/m3

Explanation

ECD adds the annular pressure loss, expressed as an equivalent mud density, to the static mud density. The conversion is Δρ = ΔP/(g × TVD), so 1.2 MPa/(9.81 × 3,000 m) equals about 41 kg/m3. Thus ECD ≈ 1,450 + 41 = 1,491 kg/m3.

Question 35Drilling Engineering

A leak-off test at a 2,000 m TVD shoe gives a formation integrity equivalent mud weight of 1.68 sg. The well is currently drilled with 1.15 sg mud. What maximum surface pressure can be applied at the shoe before reaching the tested limit? Use g = 9.81 m/s2 and 1 sg = 1,000 kg/m3.

  1. 12.8 MPa
  2. 10.4 MPa Correct answer.
  3. 15.0 MPa
  4. 8.2 MPa

Explanation

The allowable surface pressure is the difference between the tested equivalent mud weight and the current mud weight, converted to pressure over true vertical depth. Using ΔP = (1.68 − 1.15) × 1,000 × 9.81 × 2,000 gives about 10.4 MPa. Measured depth would overstate the limit.

Question 36Drilling Engineering

While drilling a naturally fractured carbonate reservoir, the mud level in the active pits falls, pump pressure decreases, and returns at the shakers decrease. Which is the most likely explanation?

  1. Mud is escaping into a fractured or vuggy formation Correct answer.
  2. The bit nozzles are partially plugged and restricting flow
  3. A kick is entering the well and pushing mud out of the hole
  4. The mud weight is too low to balance formation pore pressure

Explanation

Falling pit level, reduced pump pressure, and reduced returns are classic signs of lost circulation. In a naturally fractured carbonate, mud can escape into fractures or vugs faster than it is pumped. A kick would usually raise pit volume, while plugged nozzles would raise pump pressure.

Question 37Drilling Engineering

A rig crew reports that the Marsh funnel reading of the active mud is 48 s/qt. Which property does this measurement primarily represent?

  1. Gel strength after 10 minutes static aging
  2. Funnel viscosity, a field consistency check Correct answer.
  3. Yield point in lb/100 ft2
  4. Plastic viscosity in centipoise

Explanation

The Marsh funnel measures the time for one quart of mud to flow through a standard orifice, reported as funnel viscosity in seconds per quart. It is a quick field consistency check and does not separate plastic viscosity from yield point. Those parameters require a rotational viscometer.

Question 38Drilling Engineering

A directional survey over a 45 m course length shows a dogleg angle of 2.5°. What is the dogleg severity for this interval?

  1. 1.67°/30 m Correct answer.
  2. 2.50°/30 m
  3. 0.83°/30 m
  4. 1.25°/30 m

Explanation

Dogleg severity normalizes the measured dogleg angle to a standard 30 m course length: DLS = dogleg angle × 30 / course length. Here 2.5° × 30/45 = 1.67°/30 m. Reporting 2.5° directly would describe the angle over 45 m, not the severity.

Question 39Drilling Engineering

A well is shut in on a kick at 3,600 m TVD. Current mud density is 1,320 kg/m3. SIDPP is 4.0 MPa and SICP is 6.3 MPa. What kill mud density is required, and what does the SICP–SIDPP difference indicate?

  1. 1,548 kg/m3; the difference is annular friction loss
  2. 1,320 kg/m3; the difference shows the choke is closed
  3. 1,433 kg/m3; the difference is the fracture gradient
  4. 1,433 kg/m3; the difference reflects influx pressure Correct answer.

Explanation

Kill mud density equals current density plus SIDPP converted over TVD: 1,320 + 4.0 MPa/(9.81 × 3,600 m) ≈ 1,433 kg/m3. In a shut-in well, SICP minus SIDPP represents the pressure deficit caused by the lighter influx in the annulus. Annular friction is not present because there is no circulation.

Question 40Drilling Engineering

During a wait-and-weight kill, the pump is brought to kill rate and the choke is adjusted so drillpipe pressure follows the calculated schedule. Why is the casing shoe pressure monitored continuously?

  1. To measure the weight of the kill mud as it is pumped down the drill pipe
  2. To confirm that the choke is fully open before kill mud reaches the bit
  3. To keep shoe pressure below the formation fracture pressure Correct answer.
  4. To calculate the new slow circulating pressure after each stand

Explanation

The shoe is commonly the weakest exposed point in the annulus, so the kill must not exceed its fracture pressure. The drillpipe gauge is used to hold bottomhole pressure constant, while the casing gauge tells the choke operator whether the shoe is being protected. If shoe pressure approaches the limit, the pumping schedule or choke adjustment must be reviewed.

Question 41Drilling Engineering

During a well-control event with the drill string in the hole, which BOP element is designed to close and seal around a specific drill pipe outside diameter?

  1. Shear rams
  2. Annular preventer
  3. Pipe rams Correct answer.
  4. Blind rams

Explanation

Pipe rams carry elastomer seals sized to the drill pipe outside diameter in use, so they close and seal around that pipe body while leaving the pipe bore open. Blind rams close an open hole, shear rams cut pipe, and an annular preventer seals on a range of shapes rather than a specific pipe size.

Question 42Drilling Engineering

Which casing string is normally run below surface casing to isolate unstable or pressure-transition formations and to provide a seat for the next liner or production casing?

  1. Production casing
  2. Intermediate casing Correct answer.
  3. Conductor casing
  4. Surface casing

Explanation

Intermediate casing is set below surface casing to isolate problem formations and to provide a competent seat for the next hole section, liner, or production string. Conductor casing is the shallowest string, surface casing is set above it in the sequence, and production casing is the final wellbore conduit.

Question 43Drilling Engineering

Which bit type removes rock primarily by a shearing action and is commonly selected for high penetration rates in homogeneous shale?

  1. Impregnated diamond bit
  2. Down-the-hole hammer bit
  3. PDC bit Correct answer.
  4. Roller-cone bit

Explanation

Polycrystalline diamond compact cutters shear the rock as the bit rotates, which gives high rates of penetration in uniform, low-abrasion formations such as shale. Roller-cone bits crush and gouge, impregnated diamond bits grind hard abrasive rock, and hammer bits rely on percussion.

Question 44Drilling Engineering

What is the primary duty of a decanting centrifuge in a weighted water-based mud system?

  1. Degas the mud before it enters the mud pump
  2. Increase the mud weight by adding barite to the active system
  3. Remove large drilled cuttings from the shale shaker underflow
  4. Separate fine low-gravity solids from the weighted mud so the barite-rich stream can be returned Correct answer.

Explanation

A decanting centrifuge spins the mud at high speed so fine low-gravity solids are separated while the denser barite-rich stream is retained and returned to the active system. It is a fine-solids removal device, not a degasser or a weighting device.

Question 45Drilling Engineering

Which operational change will most directly reduce surge pressure when running casing into a well?

  1. Increasing the running speed of the casing string
  2. Increasing the yield point of the mud
  3. Reducing the number of centralizers
  4. Reducing the running speed of the casing string Correct answer.

Explanation

Surge pressure is generated by the casing acting like a piston and by mud adhering to the moving pipe, and it increases with running speed and mud rheology. Slowing the running speed lowers the induced pressure, while higher yield point or tighter annular restrictions tend to increase it.

Question 46Drilling Engineering

In primary cementing, what is the main purpose of using bottom and top cementing plugs?

  1. To separate the cement slurry from the mud ahead and behind it and to wipe the casing interior Correct answer.
  2. To centralize the casing in the borehole
  3. To test the casing shoe after cementing
  4. To increase the density of the cement slurry

Explanation

The bottom plug runs ahead of the cement slurry and separates it from the mud, while the top plug follows the slurry and separates it from the displacement fluid. Both plugs also wipe the inside of the casing as they travel down, reducing mud contamination of the cement.

Question 47Drilling Engineering

In a 60° inclined hole section, which change most improves cuttings transport when the rate of penetration is high?

  1. Reducing the flow rate while increasing weight on bit
  2. Decreasing the mud yield point and stopping pipe rotation
  3. Raising annular velocity and maintaining adequate low-shear-rate viscosity Correct answer.
  4. Lowering the annular velocity and reducing mud low-shear viscosity

Explanation

In inclined holes, cuttings tend to settle and form beds, so transport depends on sufficient annular velocity and low-shear-rate viscosity to keep cuttings moving and suspended. Pipe rotation also helps disturb beds, while lower flow, lower viscosity, or stopping rotation worsens hole cleaning.

Question 48Drilling Engineering

While slide drilling with a mud motor in a build-and-hold section, the directional driller wants to increase inclination without changing azimuth. Where should the toolface be oriented?

  1. High side of the hole Correct answer.
  2. Left side of the hole
  3. Low side of the hole
  4. Right side of the hole

Explanation

A high-side toolface directs the motor bit to build inclination, while a low-side toolface tends to drop inclination. Right and left toolface orientations are used to turn the wellpath to the right or left rather than to build angle without azimuth change.

Question 49Drilling Engineering

While drilling a permeable zone, the differential pressure between the mud column and the formation is 5.0 MPa. The projected contact area between a drill collar and the borehole wall is 0.20 m², and the friction coefficient between the collar and the mud cake is 0.25. What is the approximate differential sticking force?

  1. 500 kN
  2. 1,000 kN
  3. 125 kN
  4. 250 kN Correct answer.

Explanation

Differential sticking force is estimated as differential pressure multiplied by contact area multiplied by the friction coefficient. Using 5.0×10^6 Pa × 0.20 m² × 0.25 gives 250,000 N, or about 250 kN.

Question 50Drilling Engineering

Two 12/32 in nozzles and one 14/32 in nozzle are installed in a drill bit. What is the total flow area (TFA) of the three nozzles?

  1. 0.221 in²
  2. 0.371 in² Correct answer.
  3. 0.150 in²
  4. 0.742 in²

Explanation

Nozzle area is calculated with the circle area formula using the nozzle diameter in inches. Two 12/32 in nozzles contribute about 0.221 in², the 14/32 in nozzle contributes about 0.150 in², and the total is approximately 0.371 in².