Reservoir Engineering: Rock and Fluid Properties, Flow in Porous Media, Drive Mechanisms, Material Balance, Reserves, Geomechanics and Simulation

Section 4 of the GATE Petroleum Engineering (PE) paper, the paper’s widest and most numerical section, in one chapter and the syllabus’s order: petrophysical properties of reservoir rocks; coring and core analysis; reservoir fluid properties; the phase behaviour of hydrocarbon systems; single-phase and multiphase flow through porous media; water and gas coning; reservoir pressure measurement; reservoir drives, drive mechanisms and recovery factors; the theory of material balance; reserve estimation and its techniques; field development and reservoir management; reservoir geomechanics — stress analysis and fracturing; and the basics of reservoir simulation — mathematical models, finite-difference discretisation, Cartesian and radial grids, black-oil simulation and history matching. Units are stated with every formula: darcy units (cm³/s, darcy, cm², atm, cP, cm) or oilfield units (STB/d, md, ft, psi, cP, bbl/STB), never mixed.

1. Petrophysical properties of reservoir rocks, and coring and core analysis

Porosity φ is pore volume over bulk volume; total porosity counts all pores, effective porosity only the interconnected ones that can deliver fluid. Saturation Sₒ, S_w, S_g is the fraction of pore volume each fluid fills, and they sum to one; the connate or irreducible water saturation S_wi is the water that cannot be displaced. Permeability k is defined by Darcy’s law for the laminar flow of a single fluid that fills the pores: in darcy units q = kAΔp/(μL), with q in cm³/s, k in darcy, A in cm², Δp in atm, μ in cP and L in cm; in oilfield units q = 0.001127kAΔp/(μL), with q in bbl/d, k in md, A in ft², Δp in psi and L in ft. Gas permeability measured at low pressure is higher than the liquid value because gas molecules slip at the pore walls (the Klinkenberg effect); extrapolating k_gas against 1/p̄ to infinite pressure gives the equivalent liquid permeability.

Layers in parallel (flow along the beds) average arithmetically by thickness, k̄ = Σkᵢhᵢ/Σhᵢ; beds in series (flow across them) average harmonically, k̄ = L/Σ(Lᵢ/kᵢ), so the tightest bed dominates. Wettability is the tendency of one fluid to spread on the rock in the presence of another, measured by the contact angle or by the Amott and USBM tests; most sandstones are water-wet and many carbonates oil-wet or mixed-wet. Capillary pressure is the pressure difference across a curved interface, Pc = p_non-wetting − p_wetting = 2σ cos θ/r for a capillary of radius r, and in a reservoir it equals Δρ g h above the free-water level, which is why saturation varies through the transition zone. Relative permeability k_r is the fraction of absolute permeability available to a phase when several flow together; each curve ends at its residual (immobile) saturation, and the curves cross at a water saturation above 0.5 in a water-wet rock.

Coring recovers rock for direct measurement: conventional cores cut with a core barrel and annular core bit give continuous full-diameter samples; sidewall cores (percussion or rotary) sample chosen depths after logging. Routine core analysis measures porosity (helium porosimeter by Boyle’s law, or summation of fluids), air permeability (corrected for Klinkenberg slip) and fluid saturations (Dean-Stark distillation or retort). Special core analysis (SCAL) measures capillary pressure (porous-plate, centrifuge, mercury injection), relative permeability (steady-state and unsteady-state displacements), wettability, rock compressibility and electrical properties for Archie’s constants. Rock compressibility c_f = (1/V_p)(dV_p/dp) and the fluid compressibilities combine into the total compressibility cₜ = Sₒcₒ + S_wc_w + S_gc_g + c_f that appears in every transient equation.

⚠️ Parallel is arithmetic, series is harmonic
Layers of 10 ft at 200 md, 20 ft at 50 md and 30 ft at 10 md carrying flow along the beds average to (2000 + 1000 + 300)/60 = 55 md. The same three as beds in series would average far lower, because a harmonic mean is pulled toward its smallest term. Read the flow direction before choosing the mean.

2. Reservoir fluid properties and phase behaviour

Oil is described at stock-tank conditions by its API gravity, °API = 141.5/SG − 131.5, where SG is the specific gravity at 60 °F; water is 10 °API and lighter oils have higher values. In the reservoir, oil carries dissolved gas: the solution gas-oil ratio Rs (scf/STB) rises with pressure up to the bubble-point pressure p_b and is constant above it. The oil formation volume factor Bo (reservoir bbl/STB) is the reservoir volume of oil plus its dissolved gas per stock-tank barrel; it rises slightly as pressure falls toward p_b (the liquid expands) and then falls below p_b as gas comes out of solution. The gas formation volume factor from the real-gas law is Bg = 0.02827zT/p ft³/scf with T in °R and p in psia (equivalently 0.00504zT/p bbl/scf), where z is the gas compressibility factor read from the Standing-Katz chart at the pseudo-reduced pressure and temperature. Oil viscosity falls with pressure down to p_b as gas dissolves and then rises below it.

The pressure-temperature phase envelope of a reservoir fluid is bounded by the bubble-point curve and the dew-point curve, which meet at the critical point. Its highest pressure is the cricondenbar and its highest temperature the cricondentherm. Where the reservoir temperature falls relative to the envelope classifies the fluid: to the left of the critical point it is oil (black oil far to the left, volatile oil close to it); between the critical temperature and the cricondentherm it is a retrograde gas condensate, in which liquid condenses in the reservoir as pressure falls below the upper dew point — the opposite of normal behaviour — and much of it stays trapped; beyond the cricondentherm it is a wet gas (liquids form only at the separator) or a dry gas (none form at all). PVT laboratories measure these properties by flash liberation (gas stays in contact with its oil, as in the separator) and differential liberation (gas removed as it forms, as in the reservoir below p_b).

The five reservoir fluids
FluidReservoir temperature relative to the envelopeBehaviour on depletion
Black oilWell below the critical temperatureGas evolves below the bubble point; low shrinkage
Volatile oilBelow but close to the critical temperatureLarge gas release and shrinkage just below the bubble point
Retrograde gas condensateBetween the critical temperature and the cricondenthermLiquid drops out in the reservoir below the dew point
Wet gasAbove the cricondenthermSingle phase in the reservoir; liquid at separator conditions
Dry gasAbove the cricondenthermSingle phase in the reservoir and at the surface

3. Single- and multiphase flow through porous media, coning and pressure measurement

For radial flow of oil to a well of radius r_w from an outer radius r_e, integrating Darcy’s law gives, in oilfield units, the steady-state rate q = 0.00708kh(p_e − p_wf)/[μBo ln(r_e/r_w)], with q in STB/d, k in md, h in ft, p in psi and μ in cP. A bounded reservoir that has reached pseudo-steady state (every point declining at the same rate) uses the average pressure p̄ and ln(r_e/r_w) − 0.75, and a skin s adds to the logarithm in both. Before the boundaries are felt the flow is transient and obeys the diffusivity equation, whose line-source solution is the basis of well testing. The productivity index J = q/(p̄ − p_wf) summarises a well’s inflow in STB/d/psi.

When water displaces oil, the fractional flow of water at a point, neglecting gravity and capillarity in horizontal flow, is f_w = 1/[1 + (k_ro/k_rw)(μ_w/μ_o)]. It depends on saturation through the relative permeabilities and on the viscosity ratio, so a viscous oil gives a high water cut early. The Buckley-Leverett frontal-advance theory moves each saturation at a velocity proportional to df_w/dS_w, which produces a shock front; the Welge tangent from the initial saturation to the f_w curve gives the front saturation and the average saturation behind it, hence the displacement efficiency at breakthrough.

Coning is the rise of an underlying water (or the fall of an overlying gas cap) toward the perforations of a producing well, because the pressure drawdown near the well exceeds the gravity head that holds the contact flat. Below a critical rate the cone is stable and the well produces clean oil; above it the cone breaks through, the water cut or gas-oil ratio jumps, and oil recovery is lost. Remedies: produce below the critical rate, perforate away from the contact, use horizontal wells (whose smaller drawdown per unit length cresting instead of coning), inject gel or cement barriers, or use downhole water separation. Reservoir pressure is measured with permanent downhole gauges, bottom-hole gauges run during tests, and wireline formation testers (RFT/MDT) that take a pressure at many depths in one trip; plotting those pressures against depth gives fluid gradients (about 0.43-0.45 psi/ft for water, lower for oil, much lower for gas) whose intersections locate the gas-oil and oil-water contacts.

🎯 Why a viscous oil waters out early
At a saturation where k_rw/k_ro = 0.5, a 5 cP oil displaced by 0.5 cP water gives f_w = 1/(1 + 2 × 0.1) = 0.83: the well already produces five parts water to one of oil. The same rock with a 0.5 cP oil would give f_w = 1/(1 + 2) = 0.33. Lowering the mobility of the displacing water is the whole idea of polymer flooding.

4. Drive mechanisms, recovery factors and material balance

Primary recovery is driven by the energy stored in the reservoir. In a solution-gas (depletion) drive the oil and its gas expand; above the bubble point only rock and liquid compressibility work and pressure falls fast, and below it gas comes out of solution, the gas-oil ratio climbs steeply and recovery is low. A gas-cap drive is pushed by an expanding cap of free gas, pressure falls more slowly and recovery is higher, especially with gravity segregation. A water drive is supported by influx from an aquifer, pressure is maintained, water cut rises steadily and recovery is the highest of the primary drives. Gravity drainage in thick, steeply dipping, permeable reservoirs can recover more still, slowly. Compaction drive comes from the collapse of pore volume in weak rock, with surface subsidence. Most reservoirs are a combination. Typical primary recovery factors are often quoted as roughly 5-30% of OOIP for solution-gas drive, 20-40% for gas-cap drive and 35-75% for water drive, with the actual value set by rock, fluid and rate.

The material balance equation treats the reservoir as a tank: the reservoir volume of fluids produced equals the expansion of what remains plus water influx. In the Havlena-Odeh form, F = N(Eₒ + mE_g + E_f,w) + W_e, where F = Nₚ[Bₒ + (Rₚ − R_s)B_g] + WₚB_w is the underground withdrawal, Eₒ = (Bₒ − Bₒᵢ) + (R_sᵢ − R_s)B_g the expansion of oil and its original gas, E_g = Bₒᵢ(B_g/B_gᵢ − 1) the expansion of a gas cap of relative size m, E_f,w = Bₒᵢ(1 + m)(c_wS_wi + c_f)Δp/(1 − S_wi) the expansion of connate water and shrinkage of pore volume, and W_e the water influx. It is straightened for interpretation: with no gas cap and no influx, F against Eₒ + E_f,w is a line through the origin of slope N. For an undersaturated oil with no influx the balance reduces to NₚBₒ = NBₒᵢcₑΔp, with effective compressibility cₑ = (cₒSₒ + c_wS_wi + c_f)/(1 − S_wi). For a volumetric gas reservoir it is the famous straight line p/z = (pᵢ/zᵢ)(1 − G_p/G), so the gas initially in place G is where the p/z line reaches zero.

Drive mechanisms at a glance
DrivePressureProducing GORWater production
Solution gasDeclines rapidlyLow, then rises sharply, then fallsLittle or none
Gas capDeclines slowlyRises steadily in up-dip wellsLittle or none
Water driveStays highStays lowEarly and rising
Gravity drainageDeclinesLow in down-dip wellsLittle

5. Reserve estimation, decline curves, and field development and reservoir management

Reserves are estimated by four techniques in rough order of the data they need. Volumetric: in oilfield units the oil initially in place is N = 7758Ahφ(1 − S_wi)/Bₒᵢ STB, with A in acres, h in ft, 7758 bbl per acre-ft, and gas in place G = 43 560Ahφ(1 − S_wi)/B_gᵢ scf with B_gᵢ in ft³/scf; multiplied by a recovery factor they give reserves. Material balance needs production and pressure history. Decline-curve analysis extrapolates a producing well’s rate. Simulation combines everything. Arps’ decline family is q = qᵢ/(1 + bDᵢt)^(1/b): exponential decline (b = 0) is q = qᵢe^(−Dt), a straight line on a semilog plot of rate against time, with cumulative Nₚ = (qᵢ − q)/D; hyperbolic decline has 0 < b < 1; harmonic decline (b = 1) is q = qᵢ/(1 + Dᵢt) with Nₚ = (qᵢ/Dᵢ)ln(qᵢ/q). Exponential decline is the most conservative of the three.

Field development turns a discovery into a producing plan: the number, type and spacing of wells, the drive to rely on and whether to support it by water or gas injection, the surface facilities and export route, and the production profile — build-up, plateau and decline — that the economics are run against. Reservoir management is the continuing loop of data acquisition (pressures, logs, tests, production), model updating and decisions (infill drilling, workovers, injection changes, EOR) that aims to maximise economic recovery over the field’s life rather than rate in any one year.

🧠 Recognise the decline from the plot that is straight
Exponential decline is straight on log q against t and on q against Nₚ (Cartesian); harmonic decline is straight on log q against Nₚ. A hyperbolic decline is straight on none of them, and b has to be fitted.

6. Reservoir geomechanics, and the basics of reservoir simulation

Three principal stresses act at depth: the vertical (overburden) stress σ_v, the weight of the rock above — often near 1 psi/ft — and the maximum and minimum horizontal stresses σ_H and σ_h. Rock responds to effective stress, σ′ = σ − αp, where p is pore pressure and α the Biot coefficient (close to 1 for weak, porous rock). Under the common assumption of lateral constraint (uniaxial strain), σ_h = [ν/(1 − ν)](σ_v − αp) + αp, with ν Poisson’s ratio. A hydraulic fracture opens against the least principal stress and so propagates perpendicular to σ_h: vertical fractures in most reservoirs, whose azimuth follows σ_H. The fracture gradient is essentially σ_h per foot of depth. Production lowers p and raises effective stress, compacting the reservoir, closing natural fractures and, in weak rock, producing sand; injection does the opposite and can reopen faults. Wellbore stability is the same analysis around a hole: too little mud weight collapses it (shear failure), too much fractures it (tensile failure).

A reservoir simulator solves the equations of mass conservation for each component, Darcy’s law for each phase, and PVT relations, on a grid. The mathematical model is a set of non-linear partial differential equations; finite-difference discretisation replaces derivatives by differences between grid blocks (the second-derivative formula of the Taylor series), in space and in time. Explicit time stepping is simple but stable only for small steps; implicit stepping is unconditionally stable and needs a linear solve every step; IMPES (implicit pressure, explicit saturation) is the classic compromise. Cartesian grids of rectangular blocks model fields; radial (r-θ-z) grids, with block radii spaced logarithmically outward from the well, model near-well behaviour such as coning. The black-oil model has three phases (oil, gas, water) and three pseudo-components (stock-tank oil, surface gas, water), with gas allowed to dissolve in oil through R_s and all properties functions of pressure — the workhorse for most oil fields; compositional models track individual hydrocarbon components where composition changes matter. History matching adjusts the uncertain inputs (permeability, aquifer size, relative permeability) until the model reproduces the observed pressures, rates and water cuts; only then is it trusted to forecast. A history match is not unique, which is why forecasts are reported with ranges.

⚠️ Fractures open against the least stress
A fracture’s plane is normal to σ_min. With σ_v the largest stress at depth (normal-faulting regime) σ_h is the least, so fractures are vertical and strike along σ_H. Only at shallow depth or in strongly compressive settings, where σ_v becomes the least stress, do hydraulic fractures turn horizontal.

Key takeaways

  • Darcy: q = kAΔp/(μL) in darcy units, 0.001127kAΔp/(μL) oilfield; radial steady state q = 0.00708khΔp/[μBₒ ln(r_e/r_w)]; parallel layers average arithmetically, series harmonically.
  • Bₒ rises to the bubble point then falls; B_g = 0.02827zT/p ft³/scf; retrograde condensates lie between the critical temperature and the cricondentherm.
  • f_w = 1/[1 + (k_ro/k_rw)(μ_w/μ_o)]; coning is controlled by a critical rate; water drive keeps pressure and recovers most, solution-gas drive least.
  • N = 7758Ahφ(1 − S_wi)/Bₒᵢ; MBE F = N(Eₒ + mE_g + E_f,w) + W_e; gas p/z falls linearly to zero at G_p = G; Arps decline q = qᵢ/(1 + bDᵢt)^(1/b).
  • Effective stress σ − αp; fractures run perpendicular to σ_h; black-oil simulators solve three phases on Cartesian or radial finite-difference grids and are trusted only after history matching.

Practice questions (17)

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.

  1. Brine of viscosity 2 cP flows through a core of length 5 cm and cross-section 10 cm² at 0.5 cm³/s under a pressure difference of 1 atm. The permeability of the core is ______ md.

    Numerical answer — type the value.

    Show answer

    Answer: 500

    In darcy units k = qμL/(AΔp) = (0.5 × 2 × 5)/(10 × 1) = 5/10 = 0.5 darcy = 500 md. Check the other way: with k = 0.5 D, q = 0.5 × 10 × 1/(2 × 5) = 0.5 cm³/s, as given.
  2. Three horizontal layers are produced together with flow parallel to the bedding: 10 ft of 200 md, 20 ft of 50 md and 30 ft of 10 md. The average permeability is ______ md.

    Numerical answer — type the value.

    Show answer

    Answer: 55

    Parallel flow averages by thickness: k̄ = Σkh/Σh = (200 × 10 + 50 × 20 + 10 × 30)/(10 + 20 + 30) = (2000 + 1000 + 300)/60 = 3300/60 = 55 md. The harmonic (series) mean of the same beds, 60/(10/200 + 20/50 + 30/10) = 60/3.45 = 17.4 md, is the trap option.
  3. A water-wet capillary of radius 10 μm contains oil and water with interfacial tension 30 dyn/cm and contact angle 0°. The capillary pressure is ______ kPa.

    Numerical answer — type the value.

    Show answer

    Answer: 6

    Pc = 2σ cos θ/r = 2 × 30 × 1/(10 × 10⁻⁴ cm) = 60/0.001 = 60 000 dyn/cm². Since 1 Pa = 10 dyn/cm², that is 6000 Pa = 6 kPa. In SI directly: 2 × 0.030 N/m / 10⁻⁵ m = 6000 Pa.
  4. The permeability of a core measured with air at low mean pressure is higher than its permeability to a non-reactive liquid. This is because of

    1. gas slippage at the pore walls (the Klinkenberg effect)
    2. turbulence of the gas at high velocity
    3. capillary end effects
    4. swelling of clays by air
    Show answer

    Answer: A — gas slippage at the pore walls (the Klinkenberg effect)

    At low pressure the gas mean free path approaches the pore size and molecules slip along the walls, so more gas flows than Darcy’s law with the liquid permeability predicts. Plotting k_gas against 1/p̄ and extrapolating to 1/p̄ = 0 recovers the liquid-equivalent value. Turbulence (non-Darcy flow) would reduce apparent permeability, not raise it.
  5. A gas at 2000 psia and 660 °R has z = 0.90. Its formation volume factor is ______ ft³/scf (to four decimal places). (Use B_g = 0.02827zT/p ft³/scf.)

    Numerical answer — type the value.

    Show answer

    Answer: 0.0084

    B_g = 0.02827 × 0.90 × 660/2000 = 0.02827 × 0.297 = 0.008396 ft³/scf, which is 0.0084. Check from the ideal-gas ratio: at 14.7 psia and 520 °R, B_g = z(p_sc/p)(T/T_sc) = 0.9 × (14.7/2000) × (660/520) = 0.9 × 0.00735 × 1.2692 = 0.008396.
  6. A reservoir whose temperature lies between the critical temperature and the cricondentherm of its fluid, at a pressure above the dew-point curve, contains a

    1. retrograde gas condensate
    2. black oil
    3. dry gas
    4. undersaturated volatile oil
    Show answer

    Answer: A — retrograde gas condensate

    That temperature window is the retrograde region: the fluid is a single-phase gas at initial pressure, and as pressure drops below the upper dew point liquid condenses in the reservoir. Oils lie to the left of the critical point; wet and dry gases lie beyond the cricondentherm.
  7. Oil flows radially at steady state to a well with k = 100 md, h = 30 ft, p_e − p_wf = 500 psi, μ = 2 cP, Bₒ = 1.2 bbl/STB, r_e = 1000 ft and r_w = 0.5 ft. Using q = 0.00708khΔp/[μBₒ ln(r_e/r_w)], the rate is ______ STB/d (to the nearest whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 582

    Numerator 0.00708 × 100 × 30 × 500 = 10 620. ln(1000/0.5) = ln 2000 = 7.6009, so the denominator is 2 × 1.2 × 7.6009 = 18.242, and q = 10 620/18.242 = 582.2 STB/d. Check through the productivity index: J = 0.00708 × 3000/18.242 = 1.1644 STB/d/psi, and 1.1644 × 500 = 582.2.
  8. In horizontal waterflooding, at a certain saturation k_rw/k_ro = 0.5. The oil viscosity is 5 cP and the water viscosity 0.5 cP. Neglecting capillary and gravity effects, the fractional flow of water is ______ (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.83

    f_w = 1/[1 + (k_ro/k_rw)(μ_w/μ_o)] = 1/[1 + 2 × (0.5/5)] = 1/(1 + 0.2) = 0.833, which is 0.83. Equivalently f_w = λ_w/(λ_w + λ_o) with mobilities in the ratio (0.5/0.5) : (1/5) = 1 : 0.2, giving 1/1.2.
  9. Water coning in an oil well is best controlled by

    1. producing below the critical rate
    2. perforating immediately above the oil-water contact
    3. increasing the drawdown
    4. lowering the oil viscosity by dilution with water
    Show answer

    Answer: A — producing below the critical rate

    A cone is stable when the drawdown that lifts it is balanced by the gravity difference between water and oil; that balance defines the critical rate. Perforating close to the contact or raising drawdown both make coning worse.
  10. An oil reservoir covers 640 acres with 25 ft net pay, porosity 0.20, connate water saturation 0.25 and initial oil formation volume factor 1.25 bbl/STB. Using N = 7758Ahφ(1 − S_wi)/Bₒᵢ, the original oil in place is ______ million STB (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 14.9

    Bulk volume 640 × 25 = 16 000 acre-ft; 7758 × 16 000 = 124.13 million bbl; × 0.20 = 24.83 million bbl of pore space; × 0.75 = 18.62 million reservoir bbl of oil; ÷ 1.25 = 14.90 million STB. In cubic feet as a check: 16 000 × 43 560 = 6.970 × 10⁸ ft³, and ÷ 5.6146 gives 1.2413 × 10⁸ bbl, the same.
  11. An undersaturated oil reservoir with no water influx is depleted from its initial pressure by 1500 psi, still above the bubble point. Bₒᵢ = 1.25 bbl/STB, Bₒ at the new pressure = 1.27 bbl/STB, and the effective compressibility cₑ = 1.8 × 10⁻⁵ psi⁻¹. Using NₚBₒ = NBₒᵢcₑΔp, the recovery factor is ______ % (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.66

    Nₚ/N = BₒᵢcₑΔp/Bₒ = 1.25 × 1.8 × 10⁻⁵ × 1500/1.27. The numerator is 1.25 × 0.027 = 0.03375, and 0.03375/1.27 = 0.02657, i.e. 2.66%. Recovery above the bubble point is small because only rock and liquid compressibility drive it.
  12. A volumetric dry-gas reservoir has an initial p/z of 5000 psia. After 20 Bscf has been produced, p/z is 4000 psia. The original gas in place is ______ Bscf.

    Numerical answer — type the value.

    Show answer

    Answer: 100

    p/z = (pᵢ/zᵢ)(1 − G_p/G), so 4000/5000 = 1 − 20/G, giving 20/G = 0.2 and G = 100 Bscf. Graphically the line from (0, 5000) through (20, 4000) falls 50 psia per Bscf and reaches zero at 5000/50 = 100 Bscf.
  13. A reservoir shows a slowly declining pressure, a producing GOR that stays low, and a water cut that appears early and rises steadily. The dominant drive is most likely

    1. water drive
    2. solution-gas drive
    3. gas-cap drive
    4. compaction drive
    Show answer

    Answer: A — water drive

    Aquifer influx replaces the voidage, so pressure holds up; since pressure stays high little gas comes out of solution, so the GOR stays low; and the aquifer water itself is produced early. Solution-gas drive would show a rapidly falling pressure and a sharply rising GOR; a gas cap would raise the GOR.
  14. A well declines exponentially from 1000 STB/d with a nominal decline rate of 0.2 per year. Its rate after 3 years is ______ STB/d (to the nearest whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 549

    q = qᵢe^(−Dt) = 1000 × e^(−0.6) = 1000 × 0.5488 = 548.8 STB/d, which is 549. The cumulative production over the three years is (qᵢ − q)/D = (1000 − 548.8)/0.2 = 2256 STB/d·yr, or about 823 000 STB.
  15. A well follows hyperbolic decline with qᵢ = 1000 STB/d, initial decline rate Dᵢ = 0.5 per year and b = 0.5. Its rate after 2 years is ______ STB/d (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 444.4

    q = qᵢ/(1 + bDᵢt)^(1/b) = 1000/(1 + 0.5 × 0.5 × 2)^(1/0.5) = 1000/(1.5)² = 1000/2.25 = 444.4 STB/d. An exponential decline at the same initial rate would give 1000e^(−1) = 367.9, lower, which is why exponential extrapolation is the conservative one.
  16. At 10 000 ft the overburden stress is 10 000 psi and the pore pressure 4650 psi. For a laterally constrained formation with Poisson’s ratio 0.25 and Biot coefficient 1, σ_h = [ν/(1 − ν)](σ_v − αp) + αp. The minimum horizontal stress is ______ psi (to the nearest whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 6433

    ν/(1 − ν) = 0.25/0.75 = 1/3. Effective vertical stress 10 000 − 4650 = 5350 psi; one third of it is 1783.3 psi; adding the pore pressure back gives σ_h = 1783.3 + 4650 = 6433.3 psi, which is 6433 — a fracture gradient of about 0.643 psi/ft.
  17. Which statements about reservoir simulation are correct? (More than one option may be correct.)

    1. A black-oil model has three phases, with gas allowed to dissolve in oil through the solution gas-oil ratio
    2. IMPES solves pressure implicitly and saturation explicitly
    3. Radial grids usually space block boundaries logarithmically away from the well
    4. A successful history match proves that the model is unique
    Show answer

    Answer: A — A black-oil model has three phases, with gas allowed to dissolve in oil through the solution gas-oil ratio; B — IMPES solves pressure implicitly and saturation explicitly; C — Radial grids usually space block boundaries logarithmically away from the well

    The first three are the standard definitions: three phases with dissolved gas via R_s; implicit pressure with explicit saturation; logarithmic radial spacing because pressure varies with ln r near a well. History matching is an inverse problem with many solutions, so a match shows consistency, not uniqueness.