The Productivity Index J collapses the entire radial inflow equation into one number that directly connects reservoir deliverability to well design decisions — drawdown targets, tubing sizing, artificial lift timing, and stimulation economics. It is the single most universally used metric in daily production engineering.
▶
Lecture 2.2: The Productivity Index — Concept, Measurement, and Application
18:40
Derives J from the PSS radial inflow equation, showing how every reservoir and fluid variable collapses into a single slope. Demonstrates a two-rate PI test on a live dataset from the Karama Field, constructs the full IPR curve, identifies the AOFP, plots the NODAL operating point, and uses Flow Efficiency to screen a stimulation candidate. Three worked field cases included.
In Topic 2.1 you derived the radial inflow equation from first principles, a relationship between flow rate Q, average reservoir pressure P̄, wellbore flowing pressure Pwf, and the rock and fluid properties k, h, re, rw, μ, B, and skin S. That equation works, but it requires knowing every one of those quantities. In the field, the engineer often does not have reliable values for all of them and even when they do, re-evaluating the full equation every time they want to predict a rate is cumbersome.
The Productivity Index J solves both problems at once. It packages all the reservoir and fluid terms into a single constant: J = 0.00708kh / [μB(ln(re/rw) − 0.75 + S)]. Once J is known, either calculated from reservoir data or measured directly from a well test, rate prediction, pressure targeting, stimulation screening, and depletion planning all reduce to simple arithmetic. J is the language that connects reservoir engineering to production engineering, and fluency in it is expected of every completions and production engineer.
This topic derives J, explains its physical meaning, shows how to measure it from field data, and covers its five most important engineering applications. The interactive simulators on Section 7 let you build intuition by exploring how each reservoir and fluid variable changes J and the IPR curve, essential preparation for the Karama Field KRM-4 problem set.
LEARNING OBJECTIVES
After completing this topic, you will be able to:
1. Define J and state its units (STB/day/psi); derive it from the PSS radial inflow equation. 2. Identify which reservoir and fluid variables drive J high or low, and which the engineer can control. 3. Construct a complete IPR from a single PI measurement: compute AOFP, tabulate Q vs Pwf, plot the straight line. 4. Predict rate at any target Pwf, or calculate required Pwf to achieve a target rate. 5. Perform a two-rate PI test: compute J from ΔQ/ΔPwf, validate against P̄, and infer skin S from the intercept. 6. Calculate Flow Efficiency (FE) and use it to quantify damage and estimate stimulation uplift in STB/day. 7. Explain when the linear PI model is valid (single-phase, above bubble point) and why it breaks down at and below Pb. 8. Apply J to depletion planning: predict how rate at fixed Pwf declines as P̄ falls, and identify the artificial lift trigger pressure.
PREREQUISITE
Topic 2.1 — Darcy’s Law for Radial Flow is the direct prerequisite. You must be comfortable with the PSS radial inflow equation, the skin factor S, and the log-drainage-ratio geometry term ln(re/rw). The PI derivation on Section 1 is a two-line factorisation of that equation. If the derivation feels unfamiliar, revisit Topic 2.1 Sections 3 – 4 before proceeding.
PBL CONNECTION — KRM-4 PROBLEM SET
The Karama Field KRM-4 well problem set (Module 02) requires you to:
Sub-Problem 2: Compute J from a two-rate well test (Q⊂1 = 350 STB/day at Pwf = 4,267 psia; Q⊂2 = 620 STB/day at Pwf = 3,817 psia; P̄ = 4,850 psia). Compare measured J to the theoretical J from reservoir data. Calculate implied skin S.
Sub-Problem 3: Construct the full KRM-4 IPR, identify AOFP, assess whether the well meets the 1,200 STB/day target at current separator back-pressure (Pwf = 3,100 psia), and determine at what P̄ artificial lift becomes necessary.
Section 1 of 7
Deriving and Defining the Productivity Index
J emerges naturally from the PSS radial inflow equation by a two-line factorisation. Understanding that derivation makes every downstream application intuitive.
1.1 Factorising the PSS Radial Inflow Equation
The PSS radial inflow equation from Topic 2.1 is:
Q = (0.00708 · k · h) / (μ · B · [ln(rₑ/rₗ) − 0.75 + S]) · (P̄ − Pₓᵤ)Q = flow rate (STB/day) | k = permeability (mD) | h = net pay (ft)
μ = viscosity (cp) | B = FVF (RB/STB) | rₑ/rₗ = drainage/wellbore radius ratio
S = skin factor (dimensionless) | P̄ = avg reservoir pressure | Pₓᵤ = BHP (psia)
For a given well at a given point in its life, every quantity in the denominator except (P̄−Pwf) is approximately constant for single-phase oil above the bubble point. So the entire pre-multiplier of (P̄−Pwf) is a constant, we call it J, the Productivity Index.
J = 0.00708 · k · h / (μ · B · [ln(rₑ/rₗ) − 0.75 + S])Units: STB/day/psi. Numerator: kh product (mD·ft), reservoir transmissibility.
Denominator: fluid resistance × geometry × (1 + damage). J increases when kh increases, when μ or B decrease, or when S decreases.
And by direct inversion:
J = Q / (P̄ − Pₓᵤ)This is the measurement form. J is the ratio of flow rate to drawdown at any stabilised operating condition.
1.2 Physical Meaning of J
J is the additional STB/day produced for every additional psi of drawdown applied. A well with J = 5 STB/day/psi produces 500 STB/day more if wellbore pressure is reduced by 100 psi. A well with J = 0.5 STB/day/psi requires 1,000 psi of additional drawdown to achieve the same incremental production.
This linear proportionality is what makes J so powerful. Once you know J, rate prediction, pressure targeting, and uplift calculations all become one-line algebra. The entire complex machinery of radial flow, Darcy’s law, log-radius geometry, skin, condenses into a single measurable constant that any field engineer can use directly.
High J Well (>5 STB/d/psi)
Large kh, low viscosity, low skin. Produces prolifically at modest drawdown. Artificial lift deferred. Example: Brent Group sandstone, GoA turbidite.
Moderate J (1–5 STB/d/psi)
Typical of many North Sea producers. Stimulation vs artificial lift is the key engineering decision. PI testing determines priority action.
Low J Well (<1 STB/d/psi)
Tight formation, high viscosity, or severe damage. Hydraulic fracturing or horizontal drilling may be needed just to achieve commercial rates.
1.3 Units and Sign Convention
J has units of STB/day/psi in oilfield units. It is always a positive number. If your calculation yields a negative J, the error is in the sign of the pressures, drawdown is always P̄−Pwf, never the reverse. In SI units J would be m³/(s·Pa), but oilfield units are universal in practice for oil wells.
1.4 Worked Derivation — KRM-4 Baseline
Reservoir data for Karama Field KRM-4: k = 18 mD, h = 95 ft, re = 1,320 ft, rw = 0.354 ft, μ = 1.4 cp, Bo = 1.25 RB/STB, S = 0 (undamaged baseline).
WORKED EXAMPLE 1Theoretical J for KRM-4 (Undamaged)
Note: The actual measured J on KRM-4 from the well test is 0.60 STB/day/psi, implying skin S ≈ +4 (taken as +5 in the canonical dataset). Sub-Problem 2 of the problem set works through this exactly.
1.5 Sensitivity: What Changes J?
Decomposing J into its terms shows immediately which factors dominate:
Term
Role in J
Can Engineer Change?
Typical Range of Influence
k · h
Numerator — transmissibility
No (rock property)
1× to 1000× across formations
μ
Denominator — fluid resistance
No (fluid property)
0.5 cp (light) to 10,000 cp (bitumen)
Bo
Denominator — shrinkage
No
1.05–1.6 RB/STB (minor effect)
ln(rₑ/rₗ)
Denominator — geometry
Mostly fixed
6–9 for typical wells
S
Denominator — damage/stimulation
YES, primary lever
−7 (fracture) to +50 (severe damage)
Engineering implication: The only term in the J equation that the production engineer controls directly is skin S. Reducing S via acid stimulation or removing damage restores lost productivity. Hydraulic fracturing creates a large negative equivalent skin S′ that can multiply J by 5–20×. Everything else is set by geology and fluid properties.
The constant 0.00708 is a unit conversion factor that arises when combining Darcy’s law in field units. In its SI form, Darcy’s radial inflow equation contains no conversion constant. In field units:
0.00708 = 1/(141.2 × π / (2π)) × (1/1.127) × ..., the exact derivation combines: 1 mD = 9.869×10⁻¹^6 m²; 1 psi = 6,895 Pa; 1 STB/day = 1.840×10⁻^6 m³/s; 1 cp = 10⁻³ Pa·s; 1 ft = 0.3048 m. When all these conversions are collected, the leading coefficient evaluates to 0.007082, conventionally rounded to 0.00708. It appears in every oilfield inflow equation in the same form.
For gas wells, viscosity and compressibility factor Z both vary strongly with pressure, so a simple linear PI does not apply. The rigorous approach uses the real-gas pseudo-pressure function m(P):
Q₉ = k·h·[m(P̄)−m(Pₓᵤ)] / [1,424·T·(ln(rₑ/rₗ)−0.75+S)]Q₉ in Mscf/day | T in °R | m(P) in psi²/cp, defined as ∫(2P/μz)dP
A pseudo-PI can be defined as J₉ = Q₉ / [m(P̄)−m(Pₓᵤ)]. For a first approximation at low pressures, the squared-pressure form Q₉ = J₉·(P̄²−Pwf²) is used. Gas well IPR is covered in Module 04.
Section 2 of 7
The IPR — The Straight-Line Inflow Performance Relationship
The IPR is the graphical expression of Q = J·(P̄−Pwf). It displays a well's complete deliverability envelope at a given reservoir pressure and is the starting point for every production engineering design decision.
2.1 Constructing the IPR Straight Line
Since Q = J(P̄−Pwf), plotting Q on the x-axis against Pwf on the y-axis gives a straight line with slope −1/J (i.e., a steeper line means a smaller J). The line runs from the shut-in point (Q=0, Pwf=P̄) to the AOFP (Q=J×P̄, Pwf=0).
Figure 2.2.1 — Anatomy of the Straight-Line IPR
Key Points on the IPR
Shut-in (Q=0): Pwf = P̄ — no drawdown, no flow.
AOFP (Pwf=0): Q = J×P̄, theoretical maximum at zero wellbore pressure.
Operating point: Set by separator back-pressure + hydrostatic + friction losses. The well actually runs here.
Slope = −1/J: Steeper line = smaller J = less productive well.
Effect of Stimulation on IPR
Reducing skin S increases J → the IPR line rotates anti-clockwise around the shut-in point P̄. Slope becomes less steep. AOFP increases. The operating point moves right (higher rate at same Pwf) or upward (higher Pwf at same rate).
Depletion: As P̄ falls, the line shifts downward, same slope, lower AOFP. Rate at fixed Pwf declines linearly.
2.2 The Absolute Open Flow Potential (AOFP)
AOFP = J × P̄Theoretical maximum rate if P_wf could be reduced to zero (atmospheric). Used as an upper bound on deliverability and as a common inter-well comparison metric. Measured in STB/day. The AOFP is never achievable in practice (tubing can't sustain P_wf = 0) but it sets the ceiling of the IPR.
AOFP VALIDITY CAVEAT
The linear IPR AOFP is valid only if the entire Pwf range from P̄ to zero remains above the bubble point Pb. If Pb is close to P̄, as Pwf drops below Pb, free gas evolves near the wellbore, ko drops with rising gas saturation, and the real IPR curves below the straight-line extrapolation. In those cases, the linear AOFP over-estimates deliverability. Always verify: Is the entire production drawdown range above Pb? If not, use the Vogel or composite IPR (Topic 4.2).
2.3 Worked Example — KRM-4 IPR Construction and Target Assessment
Required P_wf for 1,200 STB/day:
P_wf = P̄ − Q_target/J = 4,850 − 1,200/0.926 = 4,850 − 1,296 = 3,554 psia
At current P_wf = 3,100 psia:
Q = 0.926 × (4,850 − 3,100) = 0.926 × 1,750 = 1,621 STB/day ✓ exceeds target
Maximum P_wf that still meets target:
3,554 psia — leaving 454 psi margin above current back-pressure.
With measured J = 0.60 STB/d/psi (skin S≈+5):
Q at P_wf=3,100 = 0.60 × 1,750 = 1,050 STB/day ✗ MISSES target by 150 STB/day
Required P_wf = 4,850 − 1,200/0.60 = 4,850 − 2,000 = 2,850 psia
→ Well cannot meet target at current back-pressure without stimulation or lift.
2.4 IPR Family Across Depletion
As the reservoir depletes (P̄ declines), a family of IPR lines is generated, each parallel (same slope = same J), but with a progressively lower P̄ intercept. At fixed Pwf, rate decreases linearly as P̄ declines.
Figure 2.2.2 — IPR Family at Successive Reservoir Pressures (Depletion)
2.5 The IPR and NODAL Analysis — Preview
The IPR alone does not determine what the well actually produces. Production rate is set by the intersection of the IPR (inflow: reservoir to sandface) with the Vertical Lift Performance (VLP) or Tubing Performance Relationship (TPR) curve (outflow: sandface to surface). This intersection is the NODAL operating point. Changing tubing size, wellhead pressure, or installing artificial lift shifts the VLP, moving the operating point. Stimulation rotates the IPR. Optimising well deliverability requires managing both curves simultaneously, the domain of Module 07 NODAL Analysis.
Key insight: The IPR is the reservoir’s supply curve. The VLP is the tubing’s demand curve. They intersect at the single stable operating point. Stimulation (higher J) rotates the IPR left; reducing wellhead pressure lowers the VLP; artificial lift creates an additional ΔP that lowers effective Pwf. All three change the operating point. NODAL Analysis (Module 07) optimises all three together.
Section 3 of 7
Measuring PI in the Field
The PI can be determined directly from well test data without knowing k, h, S, or fluid properties individually. This section covers the two-rate PI test, stabilisation requirements, and how to infer skin from the intercept method.
▶
Field Demonstration: Running a Two-Rate PI Test on KRM-4
12:30
Live walkthrough of a two-rate PI test sequence on Karama Field KRM-4. Shows how to identify stabilisation from BHP recorder output, apply the slope and intercept methods to compute J and infer skin, and validate the result against the theoretical PI from radial inflow equation parameters. Includes discussion of common field errors (early shut-in, rate instability, phase segregation).
3.1 The Two-Rate PI Test
A PI test measures J directly from rates and pressures without needing reservoir properties. The simplest version uses two stabilised flow points. At each rate, the well must reach pseudo-steady-state so that Pwf is at its true stabilised value for that rate and reservoir pressure.
Establish baseline P̄ independently: Run a pressure build-up (Horner plot, MDH plot) or use material balance to determine the current average reservoir pressure P̄. This is the y-intercept of the IPR.
Produce at rate Q₁ until stabilised: Set the choke to give rate Q₁ and hold until bottomhole pressure Pwf stops changing (flatlines on BHPG). In a 100 mD reservoir this may take hours; in a 1 mD reservoir, days. Record stabilised Pwf,1.
Change rate to Q₂ and re-stabilise: Change the choke to give rate Q₂ (either higher or lower, higher is preferred for accuracy). Hold until Pwf again stabilises. Record stabilised Pwf,2.
Calculate J, Two-Rate (Slope) Method:
J = (Q₂ − Q₁) / (Pwf,1 − Pwf,2)This is the slope of the IPR line between the two test points. Note: P_wf,1 is at the LOWER rate (higher BHP), so P_wf,1 > P_wf,2 and the difference is positive.
Validate using P̄ (Intercept Method): If J is correct, then Q = J(P̄−Pwf) should hold at both test points. Check both. Discrepancy >10% usually means Pwf is not fully stabilised, or P̄ is wrong.
TWO-RATE PI TEST SUMMARY
J = ΔQ / ΔPwf = (Q₂−Q₁) / (Pwf,1−Pwf,2)Does not require knowing P̄. Measures the IPR slope directly from two stabilised points. Works even if P̄ is uncertain, as long as P̄ has not changed between the two flow periods.
J = Q / (P̄ − Pwf)Single-point method (requires known P̄). Average the two J values from each test point as a consistency check. If J values differ by >15%, re-evaluate stabilisation.
WORKED EXAMPLE 3KRM-4 Two-Rate PI Test Calculation
Test data: P̄ = 4,850 psia (from prior build-up); Q₁ = 350 STB/day at Pwf,1 = 4,267 psia; Q₂ = 620 STB/day at Pwf,2 = 3,817 psia.
From J definition: J = 0.00708kh / [μB·(ln(rₑ/rₗ) − 0.75 + S)]
Rearrange: ln(rₑ/rₗ) − 0.75 + S = 0.00708kh / (J·μ·B)
= 12.11 / (0.60 × 1.4 × 1.25)
= 12.11 / 1.050 = 11.533
S = 11.533 − (8.224 − 0.75) = 11.533 − 7.474 = +4.06 ≈ +4 (taken as +5 in the canonical dataset)
Interpretation: Skin S = +4–5 represents moderate formation damage.
Flow Efficiency = J_measured/J_ideal = 0.60/0.926 = 0.648 (≈ 0.65, 65% of potential).
3.2 Stabilisation Time — A Critical Constraint
The PI test is only valid at pseudo-steady-state (PSS) conditions, after the transient pressure pulse from the rate change has reached the drainage boundary. If you read Pwf before stabilisation, the flowing pressure appears lower than PSS, and J is under-estimated. Stabilisation time tPSS can be estimated from:
Tight reservoirs: If tPSS is days to weeks, a stabilised PI test is impractical. Use transient pressure analysis (PTA, Module 30) instead, this extracts k, S, and P̄ mathematically from the transient data, allowing J to be calculated without waiting for stabilisation. For routine production monitoring of established wells, flowing material balance or reservoir simulation provide updates to J over time.
3.3 Common PI Test Errors — Field Pitfalls
If Pwf is read before PSS is reached, the apparent drawdown is larger than the true PSS drawdown (because the transient effect adds extra pressure drop near the wellbore). The computed J = Q/ΔP will therefore be smaller than the true PSS J. The well appears worse than it is. This is the most common PI test error. Solution: use a BHPG logger with continuous P vs time output and only accept the reading when dP/dt < 0.01 psi/hr.
The average reservoir pressure P̄ required for the intercept method is the fully stabilised static pressure, not the pressure read 1–2 hours after shut-in. A quick shut-in reads only the near-wellbore pressure recovery; P̄ requires either a Horner extrapolation to infinite shut-in time, or a material balance calculation. Using an under-estimated P̄ gives an over-estimated J and an erroneous IPR.
PI test rates must be at stable, measured surface conditions and converted to reservoir conditions using Bo. Unstable wellhead choke settings, multiphase slug flow, or inaccurate test separators all introduce rate uncertainty that propagates directly into J. A 10% error in Q gives a 10% error in J. Always confirm rate stability with at least 2 hours of consistent separator readings at each test point before accepting Pwf as stabilised.
Section 4 of 7
Engineering Applications of the Productivity Index
The PI bridges reservoir characterisation and production engineering. This section covers five core applications: rate prediction, pressure targeting, Flow Efficiency and stimulation screening, depletion planning, and inter-well comparison for field development.
4.1 Rate Prediction at a Given Pwf
The most fundamental application: given J (from well test or calculation) and current P̄ (from material balance or build-up), predict rate at any target operating pressure.
Q = J × (P̄ − Pwf)P_wf is the BHFP at the sand-face. It is related to wellhead pressure by: P_wf = P_wh + ρgH − friction losses. In practice, P_wf at the perforations is what the choke and tubing system sets, covered in Module 07.
4.2 Required Pwf for a Target Rate
Pwf,required = P̄ − Qtarget / JTells the engineer the maximum BHP at which the well can still meet Q_target. If P_wf,required is below the minimum achievable BHP from the tubing system, artificial lift is needed.
WORKED EXAMPLE 4Artificial Lift Trigger — KRM-4 Depletion Planning
Problem: Jmeasured = 0.60 STB/d/psi. Target = 800 STB/day. Tubing analysis shows the minimum achievable BHP without lift is Pwf,min = 2,900 psia. At what average reservoir pressure P̄ does the well fail to meet target under natural flow?
At target Q = 800 STB/day, required P_wf:
P_wf = P̄ − 800/0.60 = P̄ − 1,333 psi
Natural flow holds while P_wf,required ≥ P_wf,min:
P̄ − 1,333 ≥ 2,900
P̄ ≥ 4,233 psia
→ When P̄ falls below 4,233 psia, the well cannot produce 800 STB/day naturally.
With gas lift reducing effective P_wf to 1,500 psia (at P̄ = 3,800 psia):
Q_AL = 0.60 × (3,800 − 1,500) = 0.60 × 2,300 = 1,380 STB/day
Artificial lift installed early (when P̄ = 4,233 psia) extends plateau production
by the time taken for P̄ to drop from 4,850 to 4,233 psia, potentially 1–2 years
depending on reservoir volume and offtake rate.
Engineering decision: Plan lift installation when P̄ reaches ~4,400 psia (safety margin above the 4,233 psia trigger) to avoid a production dip.
4.3 Flow Efficiency — Quantifying Damage and Stimulation Uplift
Flow Efficiency (FE) compares the actual well performance to its ideal undamaged performance. It is the engineer’s primary tool for deciding whether to stimulate a well.
FLOW EFFICIENCY DEFINITION
FE = Jactual / JidealJ_ideal = J calculated with S = 0 (undamaged, from reservoir properties)
J_actual = J measured from well test (includes all damage effects)
Equivalently: FE = (ln(rₑ/rₗ)−0.75) / (ln(rₑ/rₗ)−0.75+S)
FE = 1.00: Well performing at undamaged ideal, no remedial action needed. FE < 1.00: Damaged. FE = 0.50 means half the potential PI lost to damage. FE > 1.00: Stimulated (negative skin). FE = 1.50 means 50% more productive than undamaged natural matrix.
WORKED EXAMPLE 5KRM-4 Stimulation Screening via FE
4.4 Inter-Well Comparison and Field Development Ranking
Comparing J and FE across wells in a field identifies which wells are underperforming relative to their reservoir quality, and therefore which are the highest-value stimulation candidates. The specific productivity index J/h (covered in Topic 2.3) enables comparison across wells with different pay thicknesses.
Well
k (mD)
h (ft)
S (inferred)
J_ideal
J_measured
FE
Action
KRM-1
22
110
+2
1.35
1.18
0.87
Monitor
KRM-2
18
95
+14
0.926
0.41
0.44
Priority acid job
KRM-3
25
130
−1
1.68
1.75
1.04
Best in field — model well
KRM-4
18
95
+5
0.926
0.60
0.65
Acid stimulation candidate
KRM-5
12
75
+9
0.575
0.18
0.31
Urgent priority acid job
Decision rule: FE < 0.50 is a near-universal acid stimulation trigger if reservoir properties justify commercial rates after treatment.
FE < 0.75 warrants detailed economic evaluation.
FE > 1.0 indicates successful stimulation or natural fracturing, investigate for additional well locations in the same area.
4.5 Depletion Planning and Reserves Estimation
If J is approximately constant above the bubble point (which it is for single-phase oil where μ and B change only modestly with pressure), then the production decline at fixed Pwf is linear in P̄. Integrating this over the reservoir material balance gives a first-order production forecast:
Q(t) = J × [P̄(t) − Pwf]P̄(t) is tracked via material balance (e.g. p/z plot for gas, volumetric balance for oil above bubble point).
When P̄(t) approaches P_wf, Q → 0 (economic limit reached).
This is the drive-mechanism depletion model, more sophisticated decline analysis is in Module 09.
For a volumetric closed reservoir (no aquifer, no gas cap), the material balance for undersaturated oil above Pb is:
Np = N · ct · (Pi − P̄)
where N = OOIP (STB), ct = total compressibility (psi⁻¹), Pi = initial pressure, P̄ = current pressure. Combined with Q = J(P̄−Pwf), this gives P̄ as a function of cumulative production Np, and therefore Q as a function of Np. This is the basis for Reservoir Simulation initial depletion modelling. For typical North Sea chalk: N = 50 MMSTB, ct = 15×10⁻6; psi⁻¹ → 1 psi of reservoir depletion recovers ~750 STB from reservoir compressibility alone.
Section 5 of 7
Factors Affecting the Productivity Index
J is not fixed for the life of a well. Understanding what drives it up or down, and what the engineer can and cannot control, is the foundation for production optimisation strategy.
5.1 Skin Factor S — The Engineer's Primary Lever
Skin appears additively with ln(re/rw) in the denominator of J. Because ln(re/rw) is typically 7–9, a large positive skin (+20, +30) can be the dominant term in the denominator, severely suppressing J. Conversely, a large negative skin from hydraulic fracturing (equivalent skin S' = −6 to −8) can reduce the denominator to <2, multiplying J by 4–10×. The following table uses KRM-4 parameters:
Skin S
Denom. Factor (ln(re/rw)−0.75+S)
J (STB/d/psi)
vs S=0
Cause
+30
37.47
0.185
−80%
Severe scale/asphaltene plugging
+20
27.47
0.252
−73%
Heavy mud invasion, perforation damage
+10
17.47
0.396
−57%
Moderate formation damage
+5
12.47
0.555
−40%
Mild damage (KRM-4 current state)
0
7.47
0.926
Baseline (undamaged)
Clean completion, no damage
−2
5.47
1.27
+37%
Mild acid stimulation
−5
2.47
2.80
+202%
Effective matrix acid job
−7
0.47
14.6
+1,477%
Hydraulic fracture (equiv. skin)
Key insight: Moving from S=−5 to S=−7 (2 units of additional negative skin from deepening a hydraulic fracture) gives a further 5× increase in J. This diminishing-return pattern explains why hydraulic fracture design focuses on extending fracture half-length to maximise equivalent negative skin.
5.2 kh — The Reservoir Quality Factor (Uncontrollable)
J is linearly proportional to the kh product. Doubling k doubles J; doubling h doubles J. However, kh is a rock property set by depositional and diagenetic history, it cannot be changed by engineering. Pre-drill kh estimates from seismic analogue typically carry ±50–200% uncertainty. This is why PI testing the first well in a new field is so critical: it collapses kh uncertainty from ±200% to ±10–20%.
This 7× spread in J maps directly to a 7× spread in forecast production rate, the dominant source of production uncertainty before the well is drilled.
Near-Wellbore k vs Bulk k
Damage affects the effective k in the near-wellbore zone (r < rs), not the bulk reservoir. Skin S is an efficient way to represent this, it lumps the damaged-zone transmissibility loss into one number without needing to resolve the radial damage profile. Matrix acidising restores kdamaged back toward kreservoir, reducing S. The bulk k (from core or well test analysis) is unchanged.
5.3 Fluid Properties μ and Bo (Uncontrollable)
Viscosity μ is in the denominator of J: higher viscosity = lower J. This is the fundamental reason why heavy oil requires so much more drawdown (or artificial lift, or thermal stimulation) to achieve the same rate as light oil.
Oil Type
API°
μ at reservoir (cp)
Bo (RB/STB)
J relative to 35°API light oil
Light crude (North Sea)
35–45
0.4–1.5
1.15–1.40
1.0× (baseline)
Medium crude
25–35
2–10
1.05–1.20
0.15–0.5×
Heavy crude
15–25
10–100
1.01–1.10
0.015–0.15×
Bitumen / extra-heavy
<15
1,000–10,000+
≈1.0
<0.01×
5.4 Effect of Pressure Depletion on J — Three Regimes
① Above Bubble Point
Single-phase oil. μ and Bo change slowly with pressure. J approximately constant. Rate at fixed Pwf declines linearly as P̄ falls. The linear PI model is valid.
② Near Bubble Point (P̄ approaching Pb)
Gas begins to exsolve near the wellbore where P is lowest. ko starts to decrease. J begins to decline even before P̄ reaches Pb. Transition zone — composite IPR applies.
③ Below Bubble Point
Two-phase flow in reservoir. ko = k × kro(Sg), declining with rising gas saturation. J is no longer constant. IPR curves away from straight line (Vogel shape). Linear PI invalid. Use Vogel IPR (Topic 4.1).
Below the bubble point, gas exsolves from the oil and occupies pore space. The effective permeability to oil becomes ko = k × kro(Sg), where kro is the relative permeability to oil, a function that falls from 1.0 at Sg = 0 toward zero at high gas saturations. The apparent PI therefore becomes:
Japparent = 0.00708 · k · kro(Sg) · h / [μo(P) · Bo(P) · (ln(rₑ/rₗ)−0.75+S)]
As Sg builds up, kro decreases and so does J. The IPR curve is no longer linear. This is precisely what Vogel’s empirical correlation captures, it models the curving IPR below Pb without requiring a full relative permeability dataset. Topic 2.4 covers this in detail.
At high flow velocities near the wellbore (especially in gas wells and high-rate oil wells), Darcy’s law underestimates pressure drop because inertial (non-Darcy) effects become significant. The effective skin becomes:
Stotal = Smechanical + D · QD = non-Darcy flow coefficient (day/STB or day/Mscf) | Q = flow rate (STB/day or Mscf/day)
This means J is rate-dependent, higher rates give higher effective skin and lower apparent J. The PI test must be run at rates representative of normal operating conditions. If J from a low-rate test is applied to predict high-rate performance, it will over-estimate the deliverability. Non-Darcy effects are most significant in gas wells and are covered in Module 04.
Section 6 of 7 — Interactive
Interactive Simulators — PI and IPR
Three tools: (1) a PI Calculator that builds J from reservoir parameters and plots the full IPR; (2) a Two-Rate PI Test Calculator that mirrors the Sub-Problem 2 calculation; (3) a Flow Efficiency and Stimulation Value tool for stimulation screening. Use these before tackling the KRM-4 problem set.
Simulator 1 — PI Calculator and IPR Curve Builder INTERACTIVE
Initialising…
Engineering experiments to try:
• Set S = +15, then reduce to 0 — watch J and operating Q recover
• Increase k from 18 to 200 mD — observe the 11× J improvement
• Move Pwf up until Q drops to exactly 1,200 STB/day — that is your maximum back-pressure
• Set Pb = P̄ − 200 psia (barely undersaturated) — note the validity warning
• Increase μ from 1.4 to 15 cp (heavier oil) — observe J collapse by >10×
Simulator 2 — Two-Rate PI Test Calculator INTERACTIVE
Initialising…
Default values above are the Sub-Problem 2 test data for KRM-4. The simulator computes J by the slope method, validates it against P̄ at both test points, and infers skin S. Adjust sliders to explore sensitivity.
Simulator 3 — Flow Efficiency and Stimulation Value INTERACTIVE
Initialising…
Default values are KRM-4 parameters. Try setting Jactual = 0.41 (KRM-2 in the field table) — observe that FE = 0.44 generates >$20MM/year uplift from a successful acid job.
PBL EXERCISE — PREPARE FOR SUB-PROBLEMS 2 AND 3
Using Simulator 2, enter the KRM-4 test data: P̄ = 4,850 psia; Q₁ = 350 STB/day at Pwf,1 = 4,267 psia; Q₂ = 620 STB/day at Pwf,2 = 3,817 psia; kh = 1,710 mD·ft; μ×B = 1.75; ln(re/rw) = 8.22.
Record from Simulator 2: (a) Jmeasured by slope method; (b) Jmeasured by intercept method at each point; (c) implied skin S.
Then using Simulator 1 with k=18, h=95, P̄=4,850, Pb=3,650, μ=1.4, Bo=1.25, rₑ/rₗ=3,729, and your inferred S: Record from Simulator 1: (d) Jtheoretical; (e) Q at Pwf=3,100 psia; (f) whether the 1,200 STB/day target is met; (g) maximum Pwf that still meets target.
These values are the answers to Sub-Problems 2 and 3 of the KRM-4 problem set.
Assessment — Section 7 of 7
Knowledge Check
Ten questions at problem-set difficulty. Covers PI definition, IPR construction, two-rate test, Flow Efficiency, depletion planning, and validity limits. Immediate feedback with worked explanations provided.
1. A well produces 1,560 STB/day with a measured BHFP of 3,420 psia. The current average reservoir pressure P̄ = 4,980 psia. What is the Productivity Index J?
J = Q / (P̄ − Pwf) = 1,560 / (4,980 − 3,420) = 1,560 / 1,560 = 1.00 STB/day/psi. Note that both the numerator and the drawdown happen to equal 1,560 here — a coincidence that makes the arithmetic clean. The common error is dividing by P̄ = 4,980 psia rather than the drawdown, which would give 0.31. Always use (P̄ − Pwf), never P̄ alone.
2. Using J = 1.00 STB/day/psi from Q1 and P̄ = 4,980 psia, what is the AOFP, and what physical assumption must hold for this to be valid?
AOFP = J × P̄ = 1.00 × 4,980 = 4,980 STB/day. The critical validity assumption is that the IPR is linear throughout the entire Pwf range from P̄ to zero. This holds only if single-phase oil exists at all pressures — meaning Pwf never drops below the bubble point Pb. If Pb is, say, 4,500 psia, the AOFP from the linear IPR is grossly over-estimated and the real IPR curves below it below Pb. Option D (closed boundary) relates to the flow regime (PSS), not the AOFP validity.
3. A two-rate PI test gives: Q₁ = 500 STB/day at Pwf,1 = 4,600 psia; Q₂ = 1,200 STB/day at Pwf,2 = 3,950 psia. What is J by the slope method?
J = (Q₂−Q₁)/(Pwf,1−Pwf,2) = (1,200−500)/(4,600−3,950) = 700/650 = 1.077 ≈ 1.08 STB/day/psi. Key points: (1) It is Pwf,1−Pwf,2 (lower-rate minus higher-rate pressure, giving a positive number) divided into the rate increase; (2) this method does not require knowing P̄ — it computes the IPR slope directly from two stabilised points; (3) 1.077 rounds to 1.08, closest to option D.
4. Using J = 1.077 STB/day/psi from Q3, and P̄ = 5,200 psia, what Pwf is needed to produce 2,000 STB/day? Can a well with minimum achievable BHP of 3,100 psia meet this target?
Pwf,required = P̄ − Q/J = 5,200 − 2,000/1.077 = 5,200 − 1,857 = 3,343 psia. For the target to be met under natural flow, the actual Pwf must be below 3,343 psia (lower BHP = more drawdown = more production). Since 3,100 psia < 3,343 psia, the tubing can deliver the required BHP (with 243 psi of margin to spare). So yes, the target is achievable. Option A is correct — note that the logic is: Pwf,actual < Pwf,required means target is achievable (more drawdown than needed), not less.
5. A well has Jideal (S=0) = 2.40 STB/day/psi and Jmeasured = 0.72 STB/day/psi. What is the Flow Efficiency and what does it imply for stimulation strategy?
FE = Jmeasured/Jideal = 0.72/2.40 = 0.30. The well operates at only 30% of its undamaged potential — 70% of deliverability is suppressed by formation damage. This is a high-priority stimulation candidate. FE values below 0.50 almost always justify a detailed acid stimulation evaluation. Option C is wrong: FE = 0.30 is not within normal tolerance — it represents severe damage that costs the operator significant daily revenue. Option A has the ratio inverted (Jideal/Jactual = 3.33 is not Flow Efficiency).
6. Reservoir pressure P̄ declines from 4,800 psia to 4,100 psia over one year. J = 1.2 STB/day/psi (constant). Pwf is held fixed at 2,800 psia by a surface choke. By what percentage does the daily production rate decline over this year?
Qyear0 = 1.2 × (4,800 − 2,800) = 1.2 × 2,000 = 2,400 STB/day. Qyear1 = 1.2 × (4,100 − 2,800) = 1.2 × 1,300 = 1,560 STB/day. Decline = (2,400 − 1,560)/2,400 = 840/2,400 = 35%. This is natural depletion decline: even with constant J and constant Pwf, rate falls as P̄ depletes because the drawdown (P̄−Pwf) shrinks. This is why reservoir management and artificial lift are so important for maintaining plateau production.
7. What is the critical assumption that makes J a TRUE CONSTANT (not varying with rate or time) in the single-phase PI model?
J is the constant of proportionality between Q and (P̄−Pwf). For this to be truly constant, every term in J = 0.00708kh/[μB(ln(re/rw)−0.75+S)] must be constant. This requires: (1) constant k — no changing saturation (so no free gas → Pwf above Pb); (2) constant μ and B — approximately true for small pressure changes above Pb; (3) constant S — no progressing damage; (4) PSS or SS flow regime (not transient). When gas evolves below Pb, kro changes with gas saturation and J becomes rate-dependent — the IPR curves downward (Vogel behaviour).
8. A well has J = 0.80 STB/day/psi with skin S = +8, and ln(re/rw) = 7.90. What would J be after a successful acid stimulation that reduces S to −2?
Current denom. = ln(re/rw) − 0.75 + S = 7.90 − 0.75 + 8 = 15.15. New denom. = 7.90 − 0.75 + (−2) = 5.15. Since J × denom = constant (= 0.00708kh/(μB)): Jnew = Jold × (old denom/new denom) = 0.80 × 15.15/5.15 = 0.80 × 2.942 = 2.35 ≈ 2.36. The closest answer is B (2.42 with slight rounding). The key calculation: reducing skin from +8 to −2 (a change of 10 skin units) reduces the denominator from 15.15 to 5.15 — a factor of 2.94×. J nearly triples. Option A is wrong: skin directly modifies J because it appears in J's denominator.
9. Why would the AOFP calculated from a straight-line IPR be unreliable if the current P̄ is only 200 psia above the bubble point pressure Pb?
When P̄ is only slightly above Pb, a modest drawdown will push Pwf below Pb. Below Pb, gas exsolves from the oil and occupies pore space, reducing the relative permeability to oil: ko = k×kro(Sg). As gas saturation rises, kro falls, Japparent decreases, and the IPR curves below the linear prediction. Extrapolating the linear IPR all the way to Pwf = 0 therefore over-estimates the AOFP. In practice: if (P̄−Pb) < 20% of total drawdown range, use the composite Vogel IPR. Option D (surface equipment) is a real practical constraint but is separate from the physical validity issue asked about here.
10. An acid stimulation job costing $1.2 MM is expected to increase J from 0.45 to 1.80 STB/day/psi. At P̄ = 4,500 psia, Pwf = 3,000 psia, oil price $80/STB, variable opex uplift $15/STB: what is the approximate daily incremental revenue and simple payback period?
ΔP = 4,500 − 3,000 = 1,500 psi. Qbefore = 0.45 × 1,500 = 675 STB/day. Qafter = 1.80 × 1,500 = 2,700 STB/day. ΔQ = 2,700 − 675 = 2,025 STB/day. Net revenue per STB = $80 − $15 = $65. Daily incremental revenue = 2,025 × $65 = $131,625 ≈ $132,000/day. Payback = $1,200,000 / $131,625 = 9.1 days. Answer A is correct ($132,000/day, ~9 days). This extraordinarily fast payback (less than 2 weeks) is realistic for a successful acid stimulation in a high-value reservoir — the capital cost is tiny relative to the sustained production uplift. It illustrates why stimulation economics are often the highest-return intervention available to the production engineer.
TOPIC COMPLETE — NEXT STEPS
You have completed Topic 2.2: The Productivity Index. You are ready to proceed to:
Topic 2.3 — Specific Productivity Index: Normalise J against net pay thickness (J/h) to enable honest inter-well comparison when pay thickness varies. Essential for multi-well field development ranking and identifying reservoir quality vs damage issues.
Topic 2.4 — Limitations of the Linear PI: Understand precisely when the straight-line model breaks down — below the bubble point, at high rates with non-Darcy effects, and for gas wells — and which replacement models to apply (Vogel, composite IPR, back-pressure equation).
KRM-4 Problem Set Sub-Problems 2 & 3: Apply the two-rate PI test calculation, construct the full IPR, and make the production target assessment and artificial lift recommendation for the Karama Field development team.