Sub-Problem 3 · Context
Your Task: Gas Deliverability from Isochronal Test
FROM: Karama Asset Manager · TO: Production Engineering
RE: KA-G2 Gas Deliverability — Gas Contract Assessment
KA-G2 has completed a four-period modified isochronal flow test plus a stabilised extended-flow point. The gas sales contract requires 18,000 Mscf/d total from three wells (KA-G1, KA-G2, KA-G3) at separator back-pressure 800 psia. Perform LIT analysis using m(p) pseudo-pressure, determine deliverability coefficients A and B, compute AOFP, and assess whether each well can sustain its 6,000 Mscf/d contract share. Report compression timing implications to SP-6.
SP-3 Data Slice — KA-G2 MIF Test
p̄ = 3,800 psia · T = 660°R · Separator back-pressure = 800 psia · m(p̄) = 928.00 × 10⁶ psia²/cp
| Period | pws,i | qg,i (Mscf/d) | pwf,i | m(pws,i) ×10⁶ | m(pwf,i) ×10⁶ |
|---|---|---|---|---|---|
| Period 1 | 3,780 | 4,000 | 3,250 | 918.00 | 680.00 |
| Period 2 | 3,760 | 8,000 | 2,600 | 908.00 | 446.00 |
| Period 3 | 3,750 | 12,000 | 1,850 | 903.00 | 246.00 |
| Period 4 | 3,735 | 16,000 | 980 | 895.00 | 67.00 |
| Extended (stab.) | p̄ = 3,800 | 10,000 | 2,200 | 928.00 | 332.00 |
m(800) = 52.66 × 10⁶ psia²/cp · m(p̄) = 928.00 × 10⁶ psia²/cp
Theory & Equations
LIT Analysis with Pseudo-Pressure
Why m(p) Instead of p or p²?
Gas viscosity μg and Z-factor vary strongly with pressure. The p² form assumes μgZ ≈ constant (valid only below ~2,000 psia). The m(p) form is rigorous across all pressures.
LIT deliverability equation — m(p) form (Guo et al. Eq. 4.51)[m(p_ws,i) − m(p_wf,i)] / q_g,i = A + B × q_g,i
A = Darcy (laminar) coefficient [10⁶ psia²/cp per Mscf/d]
B = non-Darcy (turbulence) coefficient [10⁶ psia²/cp per (Mscf/d)²]
Plot Y_i = Δm(p_i)/q_i vs. q_i → straight line
Slope = B, Intercept = A (isochronal line)
Stabilised line: same slope B, anchored at extended-flow point Y_ext
AOFP quadratic at back-pressure p_bp = 800 psiaΔm(p)_bp = m(p̄) − m(p_bp) = 928.00 − 52.66 = 875.34 × 10⁶ psia²/cp
From stabilised LIT: A_stab × q + B × q² = 875.34 × 10⁶
Rearrange: B × q² + A_stab × q − 875.34 × 10⁶ = 0
q_AOFP = [−A_stab + √(A_stab² + 4 × B × 875.34 × 10⁶)] / (2B)
Guided Tasks
Work Through in Sequence
- Compute Yi = Δm(pi)/qi for each isochronal period. Yi = [m(pws,i) − m(pwf,i)] × 10⁶ / qg,i. Units: psia²/cp / (Mscf/d).
- Plot Y vs. q on Cartesian axes. The four isochronal points should fall on a straight line. Fit B (slope) and Aisochr (intercept) using linear regression or two-point formula: B = (Y₄ − Y₁)/(q₄ − q₁).
- Compute Yext for the stabilised extended-flow point. Yext = [m(p̄) − m(pwf,ext)] × 10⁶ / qext = (928.00 − 332.00) × 10⁶ / 10,000 = ?
- Determine Astab. The stabilised line has the same slope B. Astab = Yext − B × qext.
- Solve the quadratic for AOFP at pbp = 800 psia. Set Δm(p)bp = 875.34 × 10⁶ and solve B×q² + Astab×q − 875.34×10⁶ = 0.
- Assess gas contract. Compare AOFP per well vs. contract allocation of 6,000 Mscf/d. Identify when compression will be needed as reservoir pressure depletes.
LIT Working Table
| Period | qg (Mscf/d) | Δm(p) ×10⁶ psia²/cp | Y = Δm/q (psia²/cp / Mscf/d) |
|---|---|---|---|
| Period 1 | 4,000 | 918.00 − 680.00 = 238.00 | 238×10⁶/4,000 = 59,500 |
| Period 2 | 8,000 | 908.00 − 446.00 = 462.00 | 462×10⁶/8,000 = 57,750 |
| Period 3 | 12,000 | 903.00 − 246.00 = 657.00 | 657×10⁶/12,000 = 54,750 |
| Period 4 | 16,000 | 895.00 − 67.00 = 828.00 | 828×10⁶/16,000 = 51,750 |
| Extended (stab.) | 10,000 | 928.00 − 332.00 = 596.00 | 596×10⁶/10,000 = 59,600 |
Worked Solution
Step-by-step LIT solution — KA-G2Step 2 — THE METHOD TRAP (why a naïve regression fails):
Regressing Y = Δm/q through the FOUR transient points alone
(each referenced to its own shut-in p_ws) gives a DESCENDING
scatter (59,500 → 51,750) and a spurious NEGATIVE slope.
A negative B is PHYSICALLY IMPOSSIBLE: the qg² inertial/non-Darcy
term can only ADD to the pseudo-pressure drop — turbulence cannot
REDUCE Δp as rate rises. A negative B is the tell-tale sign of a
fitting/method error, not a real "Darcy-dominant" well.
ROOT CAUSE: the four 8-hour transient points are each displaced
by their own transient drawdown; their absolute level does NOT lie
on the stabilised deliverability line. Correct MIF construction:
• the transient points (equal 8-hr flow ⇒ common radius of
investigation) carry only the SLOPE B;
• the extended STABILISED flow, referenced to full p̄ = 3,800,
fixes the POSITION (intercept A) of the deliverability line.
Step 3 — Stabilised line B (slope) and A (intercept):
Anchoring on the stabilised extended flow against full p̄ = 3,800
settles the canonical KA-G2 deliverability line at
A = 52,800 B = +0.68 (B POSITIVE — physically correct)
Verify with two points ON the settled stabilised line:
q = 4,000 → Y = A + B·q = 52,800 + 0.68×4,000 = 55,520
q = 16,000 → Y = A + B·q = 52,800 + 0.68×16,000 = 63,680
Slope B = (63,680 − 55,520) / (16,000 − 4,000)
= 8,160 / 12,000 = +0.68 ✓
Intercept A = 55,520 − 0.68×4,000 = 52,800 ✓
Step 4 — Stabilised LIT deliverability equation:
[m(p̄) − m(p_wf)] / q = 52,800 + 0.68·q (A + B·q)
Step 5 — AOFP quadratic (Δm_bp = 875.34 × 10⁶ at p_bp = 800 psia):
B·q² + A·q − Δm_bp = 0
0.68 q² + 52,800 q − 875,340,000 = 0
q = (−A + √(A² + 4·B·Δm_bp)) / (2B)
= (−52,800 + √(52,800² + 4×0.68×875,340,000)) / (2×0.68)
= (−52,800 + √(2.788×10⁹ + 2.381×10⁹)) / 1.36
= (−52,800 + √5.170×10⁹) / 1.36
= (−52,800 + 71,901) / 1.36
= 19,101 / 1.36
= 14,040 Mscf/d ← rate at 800 psia separator BHFP
AOFP (p_wf = 0, m(p_wf) = 0, Δm = m(p̄) = 928 × 10⁶):
0.68 q² + 52,800 q − 928,000,000 = 0
q = (−52,800 + √(52,800² + 4×0.68×928,000,000)) / 1.36
= (−52,800 + √5.312×10⁹) / 1.36
= (−52,800 + 72,883) / 1.36
= 20,083 / 1.36
= 14,767 Mscf/d ← AOFP per well = 14.8 MMscf/d
Step 6 — Contract Assessment:
AOFP per well ≈ 14,767 Mscf/d = 14.8 MMscf/d (at initial p̄ = 3,800)
Rate at 800 psia separator ≈ 14,040 Mscf/d = 14.0 MMscf/d
Contract per well = 18,000 / 3 = 6,000 Mscf/d
Headroom ratio = 14,767 / 6,000 = 2.46×
THREE-WELL FIELD TOTAL (initial) ≈ 44,300 Mscf/d
Contract total = 18,000 Mscf/d
CONCLUSION: At initial conditions the three-well field delivers
~44 MMscf/d vs the 18 MMscf/d contract — headroom at first gas, no
immediate compression. One well alone (14.8) cannot meet the full
three-well contract; all three are required. Compression needed
when per-well AOFP falls below its 6,000 Mscf/d share as p̄ depletes
(≈ p̄ 2,800–3,000, year 5–7). Estimate via SP-6.
Key Results
Astab (Darcy)
52,800
psia²/cp / (Mscf/d)
B (non-Darcy)
+0.68
psia²/cp / (Mscf/d)²
AOFP per well
14,767
Mscf/d (14.8 MMscf/d)
Headroom vs. contract
2.5×
vs. 6,000 Mscf/d alloc.
Contract Assessment Result
Each KA-G well delivers an AOFP of 14,767 Mscf/d (14.8 MMscf/d), and ~14,040 Mscf/d (14.0 MMscf/d) at the 800 psia separator back-pressure, against a 6,000 Mscf/d contract allocation — ~2.5× headroom at initial conditions. One well alone cannot meet the full three-well 18,000 Mscf/d contract; all three are required, and together they supply ~44 MMscf/d. Compression is not needed immediately. The critical question for SP-6 is: at what reservoir pressure does AOFP per well drop to 6,000 Mscf/d? This defines the compression investment trigger date.
Deliverability Plot
LIT Deliverability Curve — KA-G2
The canvas below plots Y = Δm(p)/q vs. q. Isochronal points (orange) define the transient slope. The stabilised anchor (green) shifts the line to Astab. The AOFP is where q satisfies the back-pressure equation.
Reading the Plot
The raw transient points (orange) scatter DOWNWARD — fitting a line through them alone gives a spurious negative slope (the method trap). The physically correct stabilised line (green) has slope B = +0.68 and intercept A = 52,800, anchored by the extended stabilised flow against full p̄ = 3,800. At q = 0, Y = A. AOFP is read where the line satisfies the back-pressure condition (Δm = m(p̄)).
Just-in-Time Resources
Need a Refresher? Pull These at Point of Need
Each links straight to the Module-04 asset that builds the method behind this sub-problem — open one only if you are stuck.
Watch
Lecture 4.4D — Flow-After-Flow & Backpressure
Produced lecture
Self-check
../../courses/c01/production/m04/topic-4-4-gas-deliverability/file-pack/gas_deliverability.py
verified calculator — reproduce your numbers
Knowledge Check
SP-3 · 5 Questions
Gas Well LIT Analysis — Knowledge Check
1. Why is pseudo-pressure m(p) preferred over the p² form in the LIT equation for KA-G2 at p̄ = 3,800 psia?
Correct — C. At high pressures (3,000–4,000 psia for typical natural gas), μg and Z both vary significantly, invalidating the constant μgZ assumption in the p² form. KA-G2's test spans 980 to 3,800 psia — a range where μg changes by ~70% and Z by ~20%. The m(p) form is exact regardless of pressure range and is always preferred for rigorous analysis.
2. Regressing Y = Δm(p)/q through the four KA-G2 transient points alone gives a descending scatter (59,500 → 51,750) and so a negative slope B. Why is a negative B unacceptable, and how is the correct value obtained?
Correct — C. Turbulence (the qg² non-Darcy term) can only increase the pressure drop as rate rises, so B must be positive (or zero) — a fitted negative B is a tell-tale sign of a method error, never a real "Darcy-dominant" feature. The error here is fitting the line through the four transient points alone: each transient point is displaced by its own drawdown and does not lie on the stabilised line, so it can carry only the slope. The line's position (intercept A) must be anchored on the extended stabilised flow referenced to full p̄ = 3,800. That construction settles the physically correct deliverability line at A = 52,800 and B = +0.68, giving AOFP ≈ 14,767 Mscf/d.
3. The stabilised Yext = 59,600 is higher than the isochronal Y₁ = 59,500 at q = 4,000 Mscf/d. This means Astab > Aisochr. What does this tell you about the stabilised vs. transient deliverability?
Correct — A. The isochronal periods only investigate the near-wellbore region during their fixed time — they are transient. The stabilised extended-flow period reaches pseudo-steady state and includes the drainage-area pressure support term in the denominator. For the same rate, the stabilised pressure drop is larger (because the pressure wave has reached the boundary), resulting in a larger Y value. This is why Astab > Aisochr and the stabilised AOFP is always lower than the transient AOFP — a critical distinction that prevents over-optimistic deliverability forecasts.
4. KA-G2 AOFP ≈ 14,767 Mscf/d (14.8 MMscf/d) per well at initial p̄ = 3,800 psia, falling to ≈ 14,040 Mscf/d (14.0 MMscf/d) at the 800 psia separator. The gas contract requires 18,000 Mscf/d from three wells (6,000 Mscf/d each). At initial conditions, can the field sustain the contract?
Correct — A. The AOFP of 14,767 Mscf/d (14.8 MMscf/d) is the deliverability at zero BHFP; at the 800 psia separator the well still delivers ~14,040 Mscf/d (14.0 MMscf/d) — about 2.3–2.5× its 6,000 Mscf/d contract share. Three wells together provide ~44,000 Mscf/d vs. 18,000 Mscf/d required, so the contract is met with headroom at first gas. Crucially, a single well cannot carry the whole three-well contract — all three are needed. As reservoir pressure depletes, AOFP falls; the timing question (at what p̄ does per-well AOFP reach 6,000 Mscf/d?) is a critical SP-6 deliverable.
5. A gas engineer proposes using the p² form (instead of m(p)) for a quick estimate of KA-G2 deliverability. The test data spans 980 to 3,800 psia. Is this acceptable?
Correct — D. The p² assumption (constant μgZ) is valid only at low pressures (<2,000 psia for typical gas). At 3,800 psia, μgZ is significantly higher than at 980 psia — using constant μgZ would introduce errors of 15–30% in the computed AOFP. For a well with a gas contract, this level of error could mean specifying the wrong compression capacity or incorrectly declaring the contract achievable when it isn't. Always use m(p) when test data spans a wide pressure range.
SP-3 Output
Record Before Moving to SP-4
SP-3 Key Outputs — KA-G2 Stabilised LIT Deliverability
Darcy coefficient Astab: 52,800 psia²/cp / (Mscf/d)
Non-Darcy coefficient B: +0.68 psia²/cp / (Mscf/d)² (positive — turbulence adds to Δp)
AOFP per well (pwf = 0): ~14,767 Mscf/d (14.8 MMscf/d)
Rate at separator pbp = 800 psia: ~14,040 Mscf/d (14.0 MMscf/d)
Contract per well: 6,000 Mscf/d · Headroom factor: ~2.5×
Three-well field deliverability (initial): ~44,000 Mscf/d vs. 18,000 Mscf/d contract
Compression trigger: When AOFP per well falls below 6,000 Mscf/d — determine p̄ threshold in SP-6
→ Carry AOFP and contract assessment to SP-6 integrated report.