Oil and Gas Well Testing: Diffusivity Solutions, Superposition, Drawdown and Build-up, Derivatives and Type Curves, Wellbore Effects, Multiple-Well and Gas Well Tests
1. The diffusivity equation, its solutions, radius of investigation and superposition
Transient radial flow of a slightly compressible fluid obeys (1/r)∂/∂r(r ∂p/∂r) = (φμcₜ/0.0002637k)∂p/∂t in oilfield units. It is solved in dimensionless form: p_D = kh(pᵢ − p)/(141.2qBμ), t_D = 0.0002637kt/(φμcₜr_w²) and r_D = r/r_w. For a well producing at constant rate in an infinite reservoir, the line-source (exponential-integral) solution is p_D = −½Ei(−r_D²/(4t_D)). When t_D/r_D² exceeds about 25 — which at the wellbore happens within seconds to minutes — Ei(−x) ≈ ln x + 0.5772, and the solution becomes the semilog approximation p_D = ½[ln(t_D) + 0.80907]. That is the basis of every straight line in this chapter: during infinite-acting radial flow, pressure is linear in log t.
The radius of investigation is how far the pressure disturbance has travelled by time t: r_i = √(kt/(948φμcₜ)) ft, with t in hours. It grows as √t, so a test four times as long sees only twice as far, and it tells the tester how long to flow to see a suspected boundary. The diffusivity equation is linear, so solutions add: this is the principle of superposition. In space, the pressure drop at a point is the sum of the drops caused by every well — and a sealing fault is represented by an image well at the same distance on the far side, producing at the same rate. In time, a rate change is represented by a new well starting at the change with rate equal to the change; a build-up (shut-in after producing at q for t_p) is the producing well continuing plus an injector of rate q starting at shut-in.
2. Drawdown and build-up analysis, Horner’s approximation and drill-stem testing
In a drawdown test the well flows at constant rate from uniform initial pressure. During infinite-acting radial flow p_wf = pᵢ − m[log t + log(k/(φμcₜr_w²)) − 3.23 + 0.869s], a straight line on semilog paper with slope m = 162.6qBμ/(kh) psi per log cycle. So the permeability-thickness is kh = 162.6qBμ/m, and the skin follows from the pressure on the straight line at 1 hour: s = 1.151[(pᵢ − p_1hr)/m − log(k/(φμcₜr_w²)) + 3.23]. In a build-up test the well is shut in after producing for t_p; superposition gives the Horner equation p_ws = pᵢ − m log[(t_p + Δt)/Δt], a straight line of the same slope m against log of the Horner time ratio, which extrapolates at a ratio of 1 (infinite shut-in) to p* — the initial pressure for a new well, a value used to find the average pressure for a depleted one. For build-up the skin is s = 1.151[(p_1hr − p_wf)/m − log(k/(φμcₜr_w²)) + 3.23], with p_wf the flowing pressure at shut-in. Horner’s approximation is the replacement of a variable rate history by a single constant rate q for an equivalent producing time t_p = 24N_p/q hours, N_p being the cumulative production since the last stabilisation.
The skin converts to an extra pressure drop at the well, Δp_skin = 141.2qBμs/(kh) = 0.869ms, and to a flow efficiency — the ratio of the drawdown an undamaged well would need to the actual drawdown. The Miller-Dyes-Hutchinson (MDH) plot of p_ws against log Δt is a simpler build-up plot valid when Δt ≪ t_p. A drill-stem test (DST) is a temporary completion on drill pipe: packers isolate the zone, a downhole tester valve opens and closes it, and the usual sequence is a short initial flow (to relieve supercharging), an initial shut-in (for initial pressure), a longer main flow and a final shut-in (for kh, skin and p*), with fluid samples recovered in the pipe. It is the first measurement of a new discovery’s deliverability.
| Quantity | Formula |
|---|---|
| Slope of the straight line (psi/cycle) | m = 162.6qBμ/(kh) |
| Permeability | k = 162.6qBμ/(mh) |
| Skin, build-up | s = 1.151[(p_1hr − p_wf)/m − log(k/(φμcₜr_w²)) + 3.23], p_wf at shut-in |
| Pressure drop across the skin | Δp_skin = 0.869ms |
| Horner producing time (h) | t_p = 24N_p/q |
| Radius of investigation (ft) | r_i = √(kt/(948φμcₜ)) |
| Wellbore storage coefficient (bbl/psi) | C = qBΔt/(24Δp) on the unit-slope line |
3. Pressure-derivative analysis, type curves, wellbore effects and multilayer reservoirs
Immediately after a rate change the surface rate changes but the sandface rate does not, because the wellbore itself stores or releases fluid — wellbore storage, with coefficient C = V_w c_w (bbl/psi) for a wellbore full of fluid of compressibility c_w, or C = 144A_u/(5.615ρ) for a changing liquid level (A_u the wellbore area in ft², ρ in lbm/ft³). While storage dominates, Δp is proportional to Δt, a unit slope on log-log axes, and C = qBΔt/(24Δp) from any point on it. Other wellbore effects are afterflow in build-ups, phase redistribution (gas rising in a shut-in wellbore, which can make pressure hump upward) and the thermal and leak effects that masquerade as reservoir features. Storage hides the early radial-flow line, so the semilog straight line must be picked only after storage has ended.
The pressure derivative, Δp′ = dΔp/d(ln t) (the Bourdet derivative, computed with smoothing), plotted with Δp on log-log axes, turns each flow regime into a recognisable shape. Wellbore storage: Δp and Δp′ together on a unit slope. Transition: a hump in the derivative, larger for larger skin. Infinite-acting radial flow: a flat (horizontal) derivative at level 0.5 in dimensionless terms, i.e. Δp′ = m/2.303; its level gives kh without picking a semilog line. Linear flow (a fracture or channel): half slope. Bilinear flow (finite-conductivity fracture): quarter slope. A sealing fault: the derivative steps up to twice its radial level (the semilog slope doubles). Closed boundaries in drawdown: a late unit slope (pseudo-steady state). A constant-pressure boundary: the derivative falls away. Type curves are families of dimensionless solutions (for example p_D against t_D/C_D for various C_De^(2s)); matching field data to a curve by sliding on log-log axes gives kh from the pressure match, C from the time match and skin from the curve chosen, and the derivative makes the match far less ambiguous.
In multilayer reservoirs the layers may be commingled (communicating only through the wellbore) or in crossflow (communicating in the rock). A conventional test sees the total kh and an average skin; the layers separate only through their differing depletion, their late-time boundary responses, or layer-by-layer tests with production-logging rates measured alongside pressure. Early-time behaviour of a layered system with crossflow can resemble that of a dual-porosity (naturally fractured) reservoir, which shows a characteristic dip in the derivative as fluid moves from the matrix to the fractures.
4. Injection, multiple-well, gas-well and inflow testing
Injection-well testing mirrors production testing: an injectivity test injects at constant rate (the analogue of a drawdown) and a falloff test shuts the injector in (the analogue of a build-up), analysed with the same semilog and derivative methods for kh and skin, with care for the mobility contrast between the injected-fluid bank and the reservoir beyond it. A step-rate test injects at increasing rates; the break in the plot of stabilised pressure against rate marks the fracture-extension pressure. Multiple-well tests measure the reservoir between wells rather than near one: in an interference test an active well changes rate and one or more observation wells record the resulting small pressure change, analysed with the line-source solution (the Ei function) or type curves for the transmissivity kh/μ and the storativity φcₜh between the wells, and for directional permeability if several observation wells are used. A pulse test uses a sequence of short rate pulses, whose time lag and amplitude at the observation well give the same properties faster and with less interference from other field activity.
Gas-well testing must allow for gas properties that change strongly with pressure: the real-gas pseudo-pressure m(p) = 2∫(p/μz)dp linearises the diffusivity equation, with p² usable at low pressure and p at high pressure. Gas wells also show rate-dependent (non-Darcy) skin, s′ = s + Dq, which is separated by testing at several rates. Deliverability tests fit the empirical back-pressure equation q = C(p̄² − p_wf²)ⁿ, with 0.5 ≤ n ≤ 1, and extrapolate to the absolute open flow (AOF) at p_wf = 0 (or atmospheric): the flow-after-flow (back-pressure) test uses stabilised rates; the isochronal test uses equal flow periods each followed by shut-in to static pressure, for tight wells that take too long to stabilise; and the modified isochronal test uses equal flow and shut-in periods, with one extended stabilised point. Inflow testing covers both the flow periods that measure how a well delivers and the inflow (negative) test, in which pressure in a completed or suspended well is reduced below formation pressure and the well is watched for flow, to verify that a barrier — a cement plug, liner lap or packer — holds.
Key takeaways
- p_D = −½Ei(−r_D²/4t_D), and at late time p_D = ½(ln t_D + 0.80907): radial flow is a straight line against log t.
- m = 162.6qBμ/(kh); s = 1.151[(p_1hr − p_wf)/m − log(k/φμcₜr_w²) + 3.23]; Δp_skin = 0.869ms; t_p = 24N_p/q for Horner.
- r_i = √(kt/948φμcₜ) grows as √t; superposition in space gives image wells, in time gives build-up.
- Derivative signatures: unit slope storage, flat radial flow, half slope linear, doubling for a sealing fault, late unit slope for a closed drawdown.
- Interference and pulse tests measure kh/μ and φcₜh between wells; gas deliverability follows q = C(p̄² − p_wf²)ⁿ from flow-after-flow or isochronal tests.
Practice questions (14)
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 build-up test on an oil well that produced at 250 STB/d (Bₒ = 1.2 bbl/STB, μ = 0.8 cP, h = 25 ft) gives a Horner straight-line slope of 40 psi per log cycle. Using m = 162.6qBμ/(kh), the formation permeability is ______ md (to two decimal places).
Numerical answer — type the value.
Show answer
Answer: 39.02
k = 162.6qBμ/(mh) = 162.6 × 250 × 1.2 × 0.8/(40 × 25). The numerator is 162.6 × 240 = 39 024 and the denominator 1000, so k = 39.02 md. Check: with k = 39.02, m = 162.6 × 240/(39.02 × 25) = 39 024/975.5 = 40.0 psi/cycle.In a build-up test, m = 40 psi/cycle and p_1hr − p_wf = 400 psi. The formation has k = 40 md, φ = 0.20, μ = 0.8 cP, cₜ = 1.5 × 10⁻⁵ psi⁻¹ and r_w = 0.3 ft. Using s = 1.151[(p_1hr − p_wf)/m − log(k/(φμcₜr_w²)) + 3.23], the skin factor is ______ (to two decimal places).
Numerical answer — type the value.
Show answer
Answer: 5.71
φμcₜr_w² = 0.2 × 0.8 × 1.5 × 10⁻⁵ × 0.09 = 2.16 × 10⁻⁷, so k/(φμcₜr_w²) = 40/2.16 × 10⁻⁷ = 1.852 × 10⁸, whose log is 8.268. Then s = 1.151(400/40 − 8.268 + 3.23) = 1.151 × 4.962 = 5.71: a damaged well. Its skin pressure drop is 0.869 × 40 × 5.71 = 198 psi.A well test gives a semilog slope of 40 psi/cycle and a skin factor of 5. The additional pressure drop across the skin zone, Δp_skin = 0.869ms, is ______ psi (to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 173.8
Δp_skin = 0.869 × 40 × 5 = 173.8 psi. The factor 0.869 is 141.2/162.6, which converts the slope back to the 141.2qBμ/(kh) form: Δp_skin = 141.2qBμs/(kh) gives the same number.Before a build-up test a well produced 12 000 STB since its last shut-in, and its last stabilised rate was 400 STB/d. The Horner pseudo-producing time is ______ hours.
Numerical answer — type the value.
Show answer
Answer: 720
t_p = 24N_p/q = 24 × 12 000/400 = 24 × 30 = 720 hours — the time the well would have needed at its last rate to produce the same cumulative volume, which is what lets a variable history be treated as one constant-rate period.For k = 50 md, φ = 0.20, μ = 1 cP and cₜ = 1 × 10⁻⁵ psi⁻¹, the radius of investigation after 24 hours of flow, from r_i = √(kt/(948φμcₜ)) with t in hours, is ______ ft (to the nearest whole number).
Numerical answer — type the value.
Show answer
Answer: 796
kt = 50 × 24 = 1200; 948φμcₜ = 948 × 0.2 × 1 × 10⁻⁵ = 0.001896; the ratio is 632 911 ft² and its square root 795.6, which is 796 ft. Check the scaling: after 96 hours (four times as long) r_i = 2 × 795.6 = 1591 ft.During early build-up data on a unit-slope line, Δp = 100 psi at Δt = 0.1 h for a well that produced 500 STB/d with B = 1.2 bbl/STB. Using C = qBΔt/(24Δp), the wellbore storage coefficient is ______ bbl/psi.
Numerical answer — type the value.
Show answer
Answer: 0.025
C = 500 × 1.2 × 0.1/(24 × 100) = 60/2400 = 0.025 bbl/psi. The 24 converts the rate from per day to per hour: qB/24 = 25 bbl/h, and 25 × 0.1 h = 2.5 bbl stored for a 100 psi rise.On a log-log diagnostic plot, infinite-acting radial flow is identified by
Show answer
Answer: A — a flat (zero-slope) pressure derivative
In radial flow Δp is linear in ln t, so dΔp/d(ln t) is constant: a horizontal derivative whose level gives kh. Unit slope on both curves is wellbore storage, half slope is linear flow, and a falling derivative indicates a constant-pressure boundary.A single straight sealing fault near a tested well causes the late-time semilog slope, compared with the early radial-flow slope, to
Show answer
Answer: A — double
By superposition in space the fault acts as an image well producing at the same rate; once its pressure disturbance reaches the test well, two wells’ worth of drawdown accumulate, and the semilog slope becomes 2m. On the derivative plot the flat level steps up to twice its value.Which statements about wellbore effects in pressure-transient tests are correct? (More than one option may be correct.)
Show answer
Answer: A — While wellbore storage dominates, Δp and its derivative lie on a unit-slope line on log-log axes; B — Wellbore storage can hide the early part of the semilog straight line; C — Phase redistribution in a shut-in well can make the pressure rise and then hump
Storage makes sandface rate lag surface rate, giving Δp ∝ Δt (unit slope) and masking radial flow until it ends; gas segregating upward in a closed wellbore can raise and then lower the gauge pressure. Storage belongs to the wellbore volume and fluid compressibility, not to the reservoir.An interference test between an active well and an observation well primarily measures
Show answer
Answer: A — the transmissivity (kh/μ) and storativity (φcₜh) of the reservoir between the wells
The observation well is not flowing, so its own skin and storage hardly affect the signal; the line-source solution at distance r gives kh/μ from the pressure match and φcₜh from the time match. That inter-well storativity is what a single-well test cannot give.A step-rate injection test is run mainly to determine
Show answer
Answer: A — the fracture-extension pressure of the formation
Injecting at stepwise higher rates and plotting stabilised pressure against rate gives two straight segments; the break between them is where the formation starts to fracture and accept fluid more easily. The result sets the maximum safe injection pressure for a waterflood.Two stabilised points of a gas-well back-pressure test give q₁ = 1 MMscf/d at Δ(p²) = 1 × 10⁶ psia² and q₂ = 4 MMscf/d at Δ(p²) = 8 × 10⁶ psia². For q = C(Δp²)ⁿ, the exponent n is ______ (to two decimal places).
Numerical answer — type the value.
Show answer
Answer: 0.67
n = log(q₂/q₁)/log(Δ₂/Δ₁) = log 4/log 8 = 0.6021/0.9031 = 0.667, which is 0.67. Exactly, 4 = 2² and 8 = 2³, so n = 2/3. It lies between 0.5 (fully turbulent) and 1 (laminar), as it must.Which statements about gas-well and inflow testing are correct? (More than one option may be correct.)
Show answer
Answer: A — An isochronal test suits low-permeability gas wells that take a long time to stabilise; B — Real-gas pseudo-pressure accounts for the variation of μ and z with pressure; C — An inflow (negative) test checks whether a barrier holds when pressure above it is reduced
Isochronal tests avoid waiting for stabilisation at every rate; m(p) = 2∫p/(μz)dp linearises the gas diffusivity equation; the negative test lowers pressure to look for inflow past a plug, lap or packer. Non-Darcy skin is Dq — it grows with rate, which is why gas wells are tested at several rates.In a conventional drill-stem test, the purpose of the short initial flow period followed by an initial shut-in is mainly to
Show answer
Answer: A — relieve the supercharged pressure near the well and then record the initial reservoir pressure
Mud filtrate invasion leaves the near-well zone above reservoir pressure; a brief flow bleeds this off, and the shut-in that follows builds up to the true initial pressure. The longer second flow and final shut-in are then analysed for kh, skin and p*.