SP-5 · AOF, p/Z Analysis & Recommendation: GK-22

Sub-Problem 5 · The Integration Challenge

Assembling the Complete Deliverability Picture for the FEED Gate

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FEED Gate Deadline: All Five Carry-Forward Values Now Required

This sub-problem cannot begin until you have CF-1 through CF-4 in hand

The Gashaka FEED gate closes in four hours. The FPSO facilities team is waiting for your deliverability note. You now hold all five carry-forward values from SP-1 through SP-4, the inputs to the full gas deliverability equation. SP-5 is the integration step: assemble the equation, solve for AOF, derive all associated outputs, and write the recommendation.

SP-5 requires more time than the earlier sub-problems (~60 min) because it has five distinct tasks:

AOF Calculation: Solve the gas deliverability equation as a quadratic in q_g (the D·q non-Darcy term makes it implicit). Report the theoretical AOF at p_wf = 0 and the deliverability at the operating p_wf,min = 3,500 psia.
Condensate and Voidage at AOF: Convert AOF to condensate rate (STB/d) and reservoir voidage (res bbl/d).
p/Z Analysis and IGIP Estimate: Build the p/Z straight line from two data points and extrapolate to IGIP.
Six-Variable Sensitivity Study: Vary k, h, Z, r_e, r_w, and S by ±30% (±2 for S); calculate P10/P50/P90 AOF.
Recommendation Memo: Write the FEED-gate engineering note (max 500 words + table).

D·q Nonlinearity: Why You Solve a Quadratic, Not Iterate

The gas deliverability equation contains G_total(q) = ln(r_e/r_w) − ¾ + S + D·q in the denominator, so the rate q appears on both sides: the equation is implicit in q. GK-22 is a high-rate well and its non-Darcy coefficient (D·q = 4.340 at the 12,400 Mscfd DST, ⇒ D = 3.5×10⁻⁴ /Mscfd) is large enough that the non-Darcy term dominates at AOF rates. Fixed-point iteration (guess q → update D·q → recompute q) diverges here: it oscillates instead of settling. The correct method is to substitute G_total = G_basic + D·q into the equation and solve the resulting quadratic q·C·(G_basic + D·q) = N exactly in one step: see Task 5.1.

Carry-Forward Assembly

Your SP-1 to SP-4 Values: Assemble Before Calculating

Fill in the blank column from your own SP-1 through SP-4 results before checking the reference values. If any of your values differ by more than 5% from the reference, re-work the relevant sub-problem before continuing.

CF Code	Parameter	Reference Value	Units	Goes Into
CF-1d	Flow model	PSS gas deliverability: q ∝ k_effh(p̅²−p_wf²)/[1,424μZT·G]	—	Equation form
CF-2c	Z-factor at p̅	1.40	—	Numerator and denominator
CF-2d	B_g at p̅	0.000714 res bbl/scf	res bbl/scf	Voidage calc
CF-2h	Gas viscosity μ_g	0.0312	cp	Denominator
CF-3a	Effective permeability k_eff	35.7	mD	Numerator
CF-4b	G_basic	10.788	—	Denominator (base)
CF-4c	D·q at DST rate	4.340 (at 12,400 Mscfd)	—	Denominator (first estimate)
CF-4d	G_total at DST rate	15.128	—	Denominator (reference at DST rate)
—	h (net pay)	94	ft	Numerator
—	T (reservoir)	745	°R	Denominator
—	p̅ (initial)	7,350	psia	Numerator (p̅² term)
—	p_wf,min	3,500	psia	Numerator (p_wf² term)
—	R_v (condensate yield)	112	STB/MMscf	Condensate rate calc

Task 5.1

AOF Calculation: Quadratic Solution

Your result: Task 5.1 (AOF)

The Gas Deliverability Equation

PSS Gas Deliverability: Pressure-Squared Form q_g [Mscfd] = k_eff · h · (p̅² − p_wf²) ──────────────────────────────────────── 1,424 · μ_g · Z · T · G_total(q) where G_total(q) = ln(r_e/r_w) − 3/4 + S + D·q_g = 9.038 − 0.750 + 2.500 + D·q_g = 10.788 + 0.00035 · q_g This is nonlinear because G_total depends on q_g itself. Because the equation is implicit, you CANNOT just divide N by a fixed G_total, and for GK-22's large D, fixed-point iteration diverges. Solve it as a quadratic (below). Denominator constant (excluding G_total): C = 1,424 · μ_g · Z · T = 1,424 × 0.0312 × 1.40 × 745 = 1,424 × 32.54 = 46,339 (field-units constant 1,424 absorbs the unit factors) Numerator constant N = k_eff · h · (p̅² − p_wf²), two cases: • Theoretical AOF, p_wf = 0: N₀ = 35.7 × 94 × 7,350² = 35.7 × 94 × 54,022,500 = 1.8129 × 10¹¹ mD·ft·psia² • Operating deliverability, p_wf,min = 3,500 psia: N₃₅₀₀ = 35.7 × 94 × (7,350² − 3,500²) = 35.7 × 94 × 41,772,500 = 1.4018 × 10¹¹ mD·ft·psia²

Quadratic Solution: Exact, One Step

Solve the Implicit Equation as a Quadratic Substituting G_total = G_basic + D·q into q_g = N/(C·G_total) and clearing the denominator: q_g · C · (G_basic + D·q_g) = N (C·D)·q_g² + (C·G_basic)·q_g − N = 0 Coefficients (independent of p_wf): a = C·D = 46,339 × 3.5×10⁻⁴ = 16.219 b = C·G_basic = 46,339 × 10.788 = 499,908 Positive root: q_g = (−b + √(b² + 4·a·N)) / (2a) CASE 1 — Theoretical AOF (p_wf = 0), N₀ = 1.8129 × 10¹¹: 4·a·N₀ = 4 × 16.219 × 1.8129×10¹¹ = 1.1761 × 10¹³ b² = 2.4991 × 10¹¹ b² + 4aN₀ = 1.2011 × 10¹³ √(b² + 4aN₀) = 3,465,689 q_g = (−499,908 + 3,465,689) / (2 × 16.219) = 2,965,781 / 32.437 = 91,431 Mscfd ≈ 91.4 MMscfd ← AOF (theoretical, p_wf = 0) Check: D·q = 3.5×10⁻⁴ × 91,431 = 32.0, so non-Darcy dominates at AOF. CASE 2 — Deliverability at p_wf,min = 3,500 psia, N₃₅₀₀ = 1.4018 × 10¹¹: 4·a·N₃₅₀₀ = 4 × 16.219 × 1.4018×10¹¹ = 9.0942 × 10¹² b² + 4aN₃₅₀₀ = 2.4991×10¹¹ + 9.0942×10¹² = 9.3441 × 10¹² √(…) = 3,056,810 q_g = (−499,908 + 3,056,810) / 32.437 = 2,556,902 / 32.437 = 78,826 Mscfd ≈ 78.8 MMscfd ← deliverability at p_wf,min = 3,500 psia Check: D·q = 3.5×10⁻⁴ × 78,826 = 27.6.

AOF vs deliverability: do not conflate them. The Absolute Open Flow is the theoretical maximum at zero sandface pressure: 91.4 MMscfd. You cannot actually flow at p_wf = 0; against the real minimum sandface pressure p_wf,min = 3,500 psia the well delivers 78.8 MMscfd. Both come straight from the quadratic, only the numerator N changes. Note that non-Darcy (D·q ≈ 28–32) is several times larger than the Darcy group G_basic = 10.79 at these rates, which is why this is a turbulence-limited high-rate well and why iterating on D·q would never converge.

AOF (theoretical, p_wf=0)

91.4

MMscfd

Deliverability (p_wf,min=3,500)

78.8

MMscfd

D·q at AOF

32.0

apparent skin

p_wf,min

3,500

psia

Task 5.2

Condensate Rate and Reservoir Voidage at AOF

Your results, Task 5.2

Condensate rate (STB/d)

Reservoir voidage (res bbl/d)

Condensate Rate and Voidage at AOF — GK-22 AOF (theoretical, p_wf = 0): q_g,AOF = 91,431 Mscfd = 91.4 MMscfd Condensate rate at AOF: q_c,AOF = q_g,AOF [MMscfd] × R_v [STB/MMscf] = 91.4 × 112 = 10,237 STB/d ≈ 10,240 STB/d Reservoir gas voidage at AOF (using B_g at p̅ = 0.000714 res bbl/scf = 0.714 res bbl/Mscf): V_void,AOF = q_g,AOF [Mscfd] × B_g [res bbl/Mscf] = 91,431 × 0.714 = 65,282 res bbl/d ≈ 65,280 res bbl/d Note: B_g at p̅ = 7,350 psia is used here. As reservoir pressure declines during production, B_g changes — the voidage per unit surface production changes accordingly (see SP-2 p/Z table). Summary — GK-22 at AOF (theoretical, p_wf = 0): Gas rate: 91.4 MMscfd Condensate rate: 10,240 STB/d Reservoir voidage: 65,280 res bbl/d

These "at AOF" figures are the theoretical maxima at p_wf = 0. The AOF voidage of 65,280 res bbl/d is ~7.4× the DST voidage (8,850 res bbl/d at 12.4 MMscfd): the DST rate was far below open-flow potential. The facilities team should not design to the theoretical AOF. Size the condensate stabilisation train and separators to the recommended design rate (≈63 MMscfd at p_wf,min with a −20% margin: see the memo table), not to the 91.4 MMscfd theoretical ceiling.

Gas AOF (p_wf=0)

91.4

MMscfd

Condensate at AOF

10,240

STB/d

Reservoir Voidage at AOF

65,280

res bbl/d

Design Rate (–20% of deliv.)

63.0

MMscfd recommended

Task 5.3

p/Z Analysis and IGIP Estimate

Your result: Task 5.3 (IGIP)

p/Z Material Balance: IGIP Estimation For a volumetric gas reservoir: p/Z = (p_i/Z_i) × [1 − G_p/G] → Straight line: p/Z vs. G_p with slope = −(p_i/Z_i)/G Two data points: Point 1 (initial): G_p = 0, p/Z = 7,350/1.40 = 5,250 psia Point 2 (DST): G_p from cumulative production during 24h test q = 12.4 MMscfd × 1 day = 12.4 MMscf = 0.0124 Bscf p after test ≈ 7,300 psia (minimal depletion in 24h test) Z at 7,300 psia ≈ 1.39 → p/Z ≈ 5,252 psia (Insufficient depletion from a single DST for reliable IGIP from p/Z) Using reserve estimate approach: Pore volume = Aφh(1−S_wi) = (640 acres × 43,560 ft²/acre) × 0.21 × 94 × 0.82 = 27,878,400 × 0.21 × 94 × 0.82 = 450,970,349 ft³ ≈ 451 MMft³ IGIP = PV / B_gi = 451 × 10⁶ ft³ / 0.02829ZT/p = 451 × 10⁶ / (0.02829 × 1.40 × 745/7,350) [ft³/scf to scf to Bscf] B_gi = 0.02829 × 1.40 × 745 / 7,350 = 0.004015 ft³/scf IGIP = 451 × 10⁶ / 0.004015 = 1.123 × 10¹¹ scf = 112 Bscf Projected p/Z at key depletion points (from SP-2 table): p = 7,350 psia: Z=1.40, p/Z = 5,250 psia (initial) p = 6,200 psia: Z≈1.28, p/Z = 4,844 psia (dew-point threshold) p = 5,000 psia: Z≈1.09, p/Z = 4,587 psia (mid-depletion) p = 3,500 psia: Z≈0.88, p/Z = 3,977 psia (design abandonment) G_p at each point (from p/Z straight line through IGIP = 112 Bscf): G_p = IGIP × [1 − (p/Z)/(p_i/Z_i)] At p=6,200: G_p = 112 × [1 − 4,844/5,250] = 112 × 0.077 = 8.7 Bscf At p=5,000: G_p = 112 × [1 − 4,587/5,250] = 112 × 0.126 = 14.1 Bscf At p=3,500: G_p = 112 × [1 − 3,977/5,250] = 112 × 0.242 = 27.1 Bscf

The IGIP of ~112 Bscf is a volumetric estimate based on a single well and assumed drainage area. It is highly uncertain: IGIP uncertainty is typically ±50% at this appraisal stage. The p/Z plot will need additional production data points to constrain the straight line. A concave-upward p/Z would indicate aquifer support; linear behaviour confirms volumetric depletion.

Dew-Point Arrival from p/Z

The p/Z analysis shows that the reservoir reaches dew-point pressure (p = 6,200 psia) after approximately 8.7 Bscf of cumulative production. At the recommended design rate of ~63 MMscfd (= 23 Bscf/year), this corresponds to only ~0.4 years, roughly 4–5 months after first production. (At the full theoretical AOF of 91.4 MMscfd the dew point would be reached in under 3 months.) The high deliverability of this well means depletion is rapid: the FEED gate recommendation must explicitly flag that condensate banking risk activates within the first year of production, not several years out.

Task 5.4

Six-Variable Sensitivity Study

Your results, Task 5.4 (P10 / P90 AOF)

P10 AOF (MMscfd)

P90 AOF (MMscfd)

The base case is the theoretical AOF (p_wf = 0) of 91.4 MMscfd. Each variable is varied individually by ±30% (or ±15% for Z, ±2 units for skin S) while holding all others at base case, and the AOF is re-solved as a quadratic each time (the non-Darcy term must be re-evaluated). The facilities team uses the P10 for conservative design. Note: because non-Darcy D·q dominates at AOF rates, the numerator/denominator sensitivities are sub-linear: a ±30% input change in k or h moves AOF only ~±17%, not ±30%, because the q² turbulence term partly absorbs the change.

Variable	Base Value	Low Scenario (P10)	AOF Low (MMscfd)	High Scenario (P90)	AOF High (MMscfd)	AOF Swing (%)
Base Case	—	—	91.4	—	91.4	—
k_eff (mD)	35.7	−30%: 25.0 mD	74.4	+30%: 46.4 mD	106.1	±30% → −19%/+16% AOF (damped by D·q)
h (ft)	94	−30%: 66 ft	74.5	+30%: 122 ft	106.0	±30% → −19%/+16% AOF (damped by D·q)
Z-factor	1.40	−15%: 1.19	100.3	+15%: 1.61	84.4	lower Z → higher AOF (q ∝ 1/Z)
r_e (ft)	2,980	−30%: 2,086 ft	91.9	+30%: 3,874 ft	91.1	±30% → ±1% AOF (damped by ln & D·q)
r_w (ft)	0.354	−30%: 0.248 ft	91.0	+30%: 0.460 ft	91.8	±30% → ±1% AOF (inside ln)
Skin S	+2.5	S = +4.5	89.0	S = +0.5 (stim)	93.9	±2 units → ±3% AOF (damped by D·q)
Combined Low (k, h, S at P10)	—	k=25, h=66, S=4.5	58.1	—	—	P10 AOF = 58.1 MMscfd
Combined High (k, h, S at P90)	—	—	—	k=46.4, h=122, S=0.5	125.3	P90 AOF = 125.3 MMscfd

Key Sensitivity Insights

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What the Sensitivity Table Reveals

Four critical design insights for the FEED gate team

k and h are still the dominant uncertainties: a ±30% change in either moves AOF by about −19%/+16% (≈74–106 MMscfd). The swing is sub-linear, not ±30%, because the non-Darcy q² term partly absorbs any rate change. Since k from 12 core plugs has genuine ±40% uncertainty, the true AOF range is even wider.
Z-factor: lower Z gives higher AOF. Because q ∝ 1/Z, the −15% case (Z = 1.19) raises AOF to ~100 MMscfd while the +15% case (Z = 1.61) lowers it to ~84 MMscfd, a moderate but real effect worth carrying in the PVT uncertainty quantification.
r_e and r_w uncertainties are negligible for AOF: a 30% change in drainage radius moves AOF only ~1% — the natural log dampens them, and the dominant non-Darcy term dampens them further. Optimising well spacing does not meaningfully improve single-well deliverability.
Skin is the actionable lever, but its leverage is muted here: a successful stimulation reducing S from +2.5 to +0.5 raises AOF only ~3% (89.0 → 93.9 MMscfd) because non-Darcy turbulence (not mechanical skin) controls this well's deliverability. The real prize is reducing the non-Darcy coefficient (e.g., bigger perforation area / completion), not acidising for skin.
P10 AOF = 58.1 MMscfd: the conservative design case; the recommended design rate (≈63 MMscfd at p_wf,min with a −20% margin) sits close to this, so the facilities concept is robust to the downside.

KWL: Final Reflection

Complete the L Column for All Five Sub-Problems

Return to your Hub KWL table one final time. Complete the L column for all five sub-problems: this is the deliverable for the KWL component of the assessment. Use the reference structure below.

Your KWL: L column (all five sub-problems)

K: What You Knew

Gas-condensates have dew points, not bubble points
B_g = 0.005035ZT/p from real gas law
k_eff = k × k_rg(S_wi)
G = ln(r_e/r_w) − ¾ + S for PSS
Gas wells experience non-Darcy turbulence

W: What Was Uncertain

Flow model for single-phase gas above p_d
Z > 1 at high p_pr, physical reason
k_rg = 0.94 vs 1.0, principle vs. magnitude
D·q magnitude relative to mechanical skin
How banking impairs deliverability quantitatively
How to estimate IGIP from single p/Z point

L: What You Learned

SP-1: GK-22 is rich gas-condensate; Vogel is wrong; banking at 6,200 psia
SP-2: Z=1.40 correct; B_g=0.714 res bbl/Mscf; voidage 8,860 res bbl/d
SP-3: k_eff=35.7 mD; banking can reduce AOF to 37% of initial
SP-4: D·q=4.34, 74% larger than S; G_total=15.13; stim worth +20%
SP-5: AOF=91.4 MMscfd (p_wf=0), deliv=78.8 at 3,500 psia; cond=10,240 STB/d; IGIP≈112 Bscf; solve quadratic not iterate; non-Darcy dominates

Task 5.5: Final Deliverable

FEED-Gate Engineering Recommendation Memo: Template

Complete this memo using your SP-1 through SP-5 results. Maximum 500 words for the narrative sections; the results table is not word-limited. This is submitted to the facilitator for assessment against the SP-5 marking rubric.

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FEED-Gate Engineering Note: GK-22 · Gashaka Field

Prepared by: [Your Name] · Date: [Date] · Distribution: Field Development Manager, FPSO Facilities Team

1. Fluid Classification

Section 1: Fluid Classification

2. Deliverability Results

Parameter	Value	Units	Basis
Absolute Open Flow (AOF, theoretical)	91.4	MMscfd	PSS gas deliverability, quadratic solve, p_wf = 0
Deliverability at p_wf,min	78.8	MMscfd	Quadratic solve, p_wf,min = 3,500 psia
Condensate rate at AOF	10,240	STB/d	AOF × R_v = 91.4 × 112
Reservoir voidage at AOF	65,280	res bbl/d	AOF [Mscfd] × B_g = 91,431 × 0.714
IGIP estimate (volumetric)	~112	Bscf	Pore volume / B_gi, 640-acre spacing
Recommended design rate	63.0	MMscfd	Deliverability × 0.80 margin = 78.8 × 0.80
Recommended condensate design	~7,060	STB/d	Design gas rate × R_v = 63.0 × 112

3. Sensitivity Summary (P10/P50/P90)

Scenario	AOF (MMscfd)	Dominant Driver
P10 (pessimistic: low k, h, high S)	58.1	k=25 mD, h=66 ft, S=+4.5
P50 (base case AOF, p_wf=0)	91.4	k=35.7 mD, h=94 ft, S=+2.5
P90 (optimistic: high k, h, low S)	125.3	k=46.4 mD, h=122 ft, S=+0.5

4. Key Assumptions and Limitations

Section 4: Key Assumptions and Limitations

5. Condensate Banking Risk Flag: CRITICAL

Section 5: Condensate Banking Risk Flag

6. Recommendation

Section 6: Recommendation

SP-5 Assessment: What the Facilitator is Marking

Your recommendation memo is assessed on eight criteria: (1) correct fluid classification with justification; (2) correct AOF value (±10% tolerance); (3) condensate rate correctly derived from AOF; (4) reservoir voidage correctly derived; (5) IGIP estimate with method stated; (6) P10/P50/P90 sensitivity table with correct dominant variable identification; (7) banking risk quantified (not just mentioned); (8) recommendation distinguishes design rate from AOF and gives justification. A memo that omits the banking risk caveat or uses Vogel instead of the gas deliverability equation scores below 50%.

Just-in-Time Resources

This integration sub-problem draws on every strand of Module 01. Reach for these the moment you get stuck assembling the deliverability chain.

Study

Module 01 · Topic 4: Drainage Geometry & Radial Flow (the deliverability equation lives here)
Module 01 · Topic 1: PVT Behaviour & Fluid Classification (μ, Z, R_v)
Module 01 · Topic 2: Formation Volume Factor & Gas FVF (B_g) (voidage & p/Z IGIP)
Module 01 · Topic 3: Permeability & Relative Permeability (k·h)

Watch

Lecture 1.4A — Geometry Matters: How Boundary Shape Controls Deliverability
Lecture 1.4B — Radial Flow to Steady-State IPR: Step by Step
Lecture 1.1 — Expanding the Phase Envelope: From Gibbs to Reservoir Fluid Dynamics

Self-check

geometry_calculator.py, verified calculator; reproduce your radial-flow geometry numbers.
gas_deliverability.py, verified AOF/deliverability toolkit (in preparation; reproduce your AOF, condensate and voidage numbers once published).

Knowledge Check

SP-5: Five MCQ

Q1: Why Solve the AOF as a Quadratic

Why must the GK-22 AOF be obtained by solving a quadratic, rather than by dividing N by a fixed G_total or by fixed-point iteration on D·q?

A. Because the pressure terms (p̅² − p_wf²) change as the reservoir depletes: the equation must be updated at each time step

B. Because G_total = G_basic + D·q contains q_g itself, so the equation is implicit; substituting gives (C·D)q² + (C·G_basic)q − N = 0, an exact quadratic. With GK-22's large D, fixed-point iteration on D·q does not converge (it oscillates) so the quadratic is the correct method

C. Because μ_g and Z change significantly between p̅ = 7,350 and p_wf = 3,500 psia and must be evaluated at the mean pressure

D. A quadratic is not required, using G_total from the DST rate gives an adequate one-step answer

Explanation

The D·q term makes G_total a function of the very rate being solved for, so q = N/(C·G_total) is implicit. Multiplying through gives the exact quadratic (C·D)q² + (C·G_basic)q − N = 0, whose positive root is the answer (AOF = 91.4 MMscfd at p_wf = 0; 78.8 MMscfd at p_wf = 3,500). Fixed-point iteration (guess q, update D·q, recompute q) diverges for GK-22 because the non-Darcy term is so large (D·q ≈ 32 at AOF, three times G_basic): each update overshoots and the estimate oscillates instead of settling. Option D is wrong because a fixed DST G_total ignores that D·q grows steeply with rate.

Q2: Condensate Rate at AOF

At q_g,AOF = 91.4 MMscfd and condensate yield R_v = 112 STB/MMscf, the condensate surface rate at AOF is:

A. 112 STB/d: R_v has units STB/MMscf and the rate is directly 112

B. 1,390 STB/d: this is the DST condensate rate at 12.4 MMscfd, not at AOF

C. 10,240 STB/d, q_c = 91.4 MMscfd × 112 STB/MMscf = 10,237 STB/d ≈ 10,240

D. 8,830 STB/d, using the DST gas rate of 78.8 MMscfd instead of the AOF

Explanation

q_c = q_g [MMscfd] × R_v [STB/MMscf] = 91.4 × 112 = 10,237 STB/d ≈ 10,240 STB/d at the theoretical AOF. The yield R_v is assumed constant at initial conditions (GK-22 is above the dew-point, so condensate yield is the recombined separator value). Below the dew-point, R_v will change as heavier components preferentially condense in the reservoir. Note the facilities train is sized to the recommended design rate (~63 MMscfd ⇒ ~7,060 STB/d), not the theoretical AOF.

Q3: Dominant Sensitivity Parameter

From the six-variable sensitivity study, which parameter has the LEAST impact on AOF when varied ±30%?

A. k (permeability): core analysis uncertainty is large but reservoir geology controls k, not engineering

B. Skin S: stimulation can remove skin so it is irrelevant to the inherent uncertainty

C. r_e (drainage radius): a 30% change in r_e produces only ~1% AOF change, because r_e appears inside a natural logarithm which heavily dampens its influence (and the dominant non-Darcy term dampens it further)

D. Z-factor: the Z-factor is measured from the PVT report and carries no uncertainty

Explanation

r_e varying ±30% (2,086 to 3,874 ft) changes G_basic by only ~0.6 units, moving AOF from 91.9 to 91.1 MMscfd (about ±1%) because the natural log dampens it and the large non-Darcy term dampens it further still. By contrast k and h each move AOF by roughly −19%/+16% (74–106 MMscfd), Z (±15%) moves it ~±9% (100/84 MMscfd, with lower Z giving higher AOF), and skin S (±2 units) moves it ~±3% (89.0/93.9 MMscfd). Note all of these are sub-linear here because non-Darcy turbulence dominates GK-22's deliverability. Engineering implication: optimising drainage radius (well spacing) barely moves single-well AOF; reducing the non-Darcy coefficient (completion/perforation design) and controlling skin are the levers that matter.

Q4: p/Z IGIP Interpretation

The volumetric IGIP estimate for GK-22 is approximately 112 Bscf. During production monitoring, the p/Z plot is found to curve upward (p/Z declining more slowly than the straight-line extrapolation predicts). This indicates:

A. The IGIP is larger than 112 Bscf: additional reservoir volume is being drained beyond the 640-acre spacing

B. Aquifer or pressure support is maintaining reservoir pressure above the pure depletion trend: the reservoir is not volumetrically closed; this could be natural water influx from a connected aquifer or pressure diffusion from adjacent reservoir compartments

C. The gas is behaving more compressibly than expected: a revised Z-factor correlation should be applied

D. Condensate banking below the dew-point is causing p/Z to deviate from linearity by reducing the effective producing volume

Explanation

A straight p/Z plot is the signature of a volumetric (closed) gas reservoir, no energy source other than gas expansion. An upward-curving p/Z (pressure declining more slowly than expected for the cumulative production) means additional energy is being supplied, most commonly natural water influx from a connected aquifer. This is a diagnostic of critical importance: an aquifer-supported reservoir may seem to have excellent pressure maintenance early in life, but water eventually invades the reservoir and can trap gas below the encroaching water front, reducing ultimate recovery. Option A would manifest as a HIGHER straight-line IGIP estimate from the p/Z, not curvature.

Q5: Condensate Banking Inclusion in FEED Recommendation

The FEED recommendation memo should flag condensate banking risk because:

A. Banking will occur immediately at first production so the FEED separator must already account for a banked-reservoir k_rg = 0.35 from day one

B. Banking is a regulatory compliance issue: all deepwater HPHT wells must include banking risk in their FEED documentation regardless of reservoir conditions

C. Banking will reduce AOF to ~37% of initial within a few years; mitigating it requires pressure maintenance infrastructure (lean gas injection, compression, injection wells) that must be scoped now: the FEED lock-in window closes after this gate and retrofitting deepwater subsea architecture is prohibitively expensive

D. Banking only affects the condensate rate, not the gas rate: it is a surface production chemistry issue, not a reservoir engineering deliverability concern

Explanation

The SP-3 analysis showed that mature condensate banking reduces k_rg from 0.94 to ~0.35, cutting AOF to ~37% of initial, from 91.4 MMscfd to ~33.8 MMscfd. Because GK-22 is so deliverable, p̅ crosses p_d = 6,200 psia within roughly the first year of production (much sooner at higher offtake), so this decline arrives early. Addressing it after FEED (e.g., post-FID or during production operations) means retrofitting a subsea injection system, which for a deepwater FPSO-based development can cost hundreds of millions of dollars more than incorporating it at FEED. The economic case for evaluating lean gas cycling at FEED is therefore straightforward, even if the technical decision is ultimately deferred.

Facilitator Guide

Discussion Points and Common Errors

This section is for the GK-22 PBL debrief session.

Three Most Common Errors in This PBL

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Error 1: Applying Vogel IPR to a Single-Phase Gas Well

Engineers who missed SP-1's key conclusion sometimes apply Vogel, which underestimates performance for gas wells. The correct PSS gas deliverability model (quadratic solve) gives a theoretical AOF of 91.4 MMscfd (78.8 MMscfd at the operating p_wf,min = 3,500 psia). Use the debrief to reinforce that Vogel is not a universal gas-condensate model: it is specifically derived for solution-gas-drive oil reservoirs and is physically wrong for single-phase gas above the dew-point.

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Error 2: Assuming Z = 1 and Computing B_g Without the Z Correction

Engineers who assume Z = 1 compute B_g = 0.005035 × 1.0 × 745/7,350 = 0.510 res bbl/Mscf, 28% lower than the correct 0.714. This cascades into a 28% underestimate of voidage and a 28% overestimate of the numerator constant N (because Z appears in both B_g and the deliverability equation denominator: the denominator overestimate partially compensates). Net effect on AOF: approximately 13% overestimate. Emphasise that at p_pr = 11, Z > 1 is physically correct and must be used.

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Error 3: Omitting D·q and Treating Gas Deliverability as Linear

Engineers who omit D·q entirely (treating G_total = G_basic = 10.788, a Darcy-only linear solve) compute AOF = N₀/(C·G_basic) ≈ 363 MMscfd, nearly 4× the correct 91.4 MMscfd. For GK-22 the non-Darcy term is not a small correction: D·q ≈ 32 at AOF, three times larger than G_basic, so dropping it grossly overstates deliverability. This is the most consequential error for FEED gate sizing: it would massively oversize the separator. Emphasise that (a) D·q is dominant, not optional, for high-rate gas wells; (b) the implicit equation must be solved as a quadratic, never by fixed-point iteration (which diverges here); and (c) the Jones correlation D = 3.5×10⁻⁴ is an estimate that requires multi-rate well testing to refine.

Discussion Questions for Plenary Debrief (10 min)

What would the AOF calculation look like if GK-22 were a horizontal well with 600 ft of open hole? Which terms change and in which direction?
If the FEED gate decides NOT to include lean gas cycling, what production strategy would you recommend to extend the useful life of the well before banking becomes severe? What rate would you recommend?
The p/Z analysis gives IGIP ≈ 112 Bscf. The GK-18 discovery well (from which GK-22 was appraised) estimated IGIP at 180 Bscf. What explanations would you consider for the discrepancy, and what additional data would you acquire to resolve it?
A third appraisal well (GK-25) is planned in a structurally higher position. How would you design its PVT and SCAL programme differently, knowing what you now know from GK-22?

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