SP-1 · KA-07 Composite IPR — Module 04 PBL

Sub-Problem 1 · Context

Your Task: Build KA-07's Current IPR

FROM: Karama Asset Manager · TO: Production Engineering

RE: KA-07 Deliverability — FDP Gate Input Required

KA-07 is flowing at ~642 stb/d on natural flow at an estimated BHFP of ~4,200 psia. That is well below the 1,800 stb/d field target. Before deciding on artificial lift and stimulation, we need the complete current IPR — from which we can read off rate at any BHFP. We have stabilised test data at two flowing conditions. The reservoir pressure (5,100 psia) is above the bubble point (4,500 psia), so a composite model is required. Provide: (a) composite IPR table and plot, (b) AOFP, (c) BHFP at which 1,800 stb/d is achievable, if ever.

SP-1 Data Slice

Only the following data is required for SP-1. All other parameters in the master data pack are used in later sub-problems.

Parameter	Symbol	Value	Units	Source
Average reservoir pressure	p̄	5,100	psia	PBU Horner plot
Bubble-point pressure	p_b	4,500	psia	PVT CCE
Ideal PI (S = 0)	J*	0.72	stb/d/psi	Darcy radial flow equation
Test T1 — rate (stabilised, below p_b)	q₁	820	stb/d	72-hour production test
Test T1 — BHFP	p_wf,1	1,800	psia	Downhole gauge

Data hygiene check — before you start

p̄ = 5,100 psia > p_b = 4,500 psia: the reservoir is undersaturated at current conditions. Test T1 (p_wf = 1,800 psia) is below p_b — so the wellbore region is in two-phase flow during that test. This is exactly the composite regime: Darcy holds above p_b, Vogel holds below. Never apply pure Vogel when p̄ > p_b — you will mis-compute the AOFP by neglecting the Darcy increment.

Theory & Equations

Composite IPR Framework — Darcy + Vogel

Step 1 — Darcy Segment (p_wf ≥ p_b)

Above bubble-point, single-phase oil flows. The productivity index is constant at J* (for an undamaged well). Rate in the Darcy segment:

Darcy radial flow — above p_bq_o = J* × (p̄ − p_wf) [stb/d] J* = 0.72 stb/d/psi (given for KA-07, S = 0)

Step 2 — Anchor Rate at Bubble Point

The rate when BHFP is exactly at the bubble point — the join point between the two segments:

Anchor rate q_bq_b = J* × (p̄ − p_b) = 0.72 × (5,100 − 4,500) = 0.72 × 600 = ? stb/d

Step 3 — Maximum Vogel Increment

The additional rate achievable by drawing BHFP from p_b all the way to zero (the Vogel contribution):

Maximum Vogel increment q_v,maxq_v,max = J* × p_b / 1.8 = 0.72 × 4,500 / 1.8 = ? stb/d AOFP = q_b + q_v,max (at p_wf = 0)

Step 4 — Composite IPR Equation (p_wf < p_b)

Below bubble-point, combine the Darcy anchor with the Vogel increment:

Composite IPR — below p_b (Guo et al. Eq. 3.28)q_o = q_b + q_v,max × [1 − 0.2(p_wf/p_b) − 0.8(p_wf/p_b)²] Let r = p_wf / p_b q_o = q_b + q_v,max × [1 − 0.2r − 0.8r²] At r = 0 (p_wf = 0): q_o = q_b + q_v,max × 1.0 = AOFP ✓ At r = 1 (p_wf = p_b): q_o = q_b + q_v,max × 0.0 = q_b ✓ (continuous)

Validation Against Test T1

Your composite IPR must pass through the stabilised test point. Calculate q at p_wf = 1,800 psia and confirm it matches T1 = 820 stb/d. If it doesn't match, it means the ideal J* does not represent the actual well, which is expected (S' = +8 is the reason). SP-2 will correct for skin.

Why does the ideal IPR not match Test T1?

Test T1 was conducted on the actual damaged well (S' = +8). The composite IPR you build here uses J* at S = 0 — the undamaged theoretical potential. SP-1 gives the ceiling of what KA-07 could deliver if cleaned up. SP-2 corrects it for skin to match the real test data. This two-step approach is standard in well deliverability analysis.

Guided Tasks

Work Through in Sequence

Compute q_b. q_b = J* × (p̄ − p_b) = 0.72 × (5,100 − 4,500). Write down the value in stb/d. This is the rate the well would produce if BHFP were held exactly at bubble point.
Compute q_v,max. q_v,max = J* × p_b / 1.8 = 0.72 × 4,500 / 1.8. This is the maximum additional rate achievable through Vogel drawdown below p_b.
Compute AOFP (absolute open flow potential). AOFP = q_b + q_v,max. This is the theoretical maximum rate at zero flowing BHP.
Build the composite IPR table. Complete the table below for the following p_wf values: 5,100 / 4,500 / 3,600 / 2,700 / 1,800 / 900 / 0 psia. Use the Darcy equation above p_b and the composite equation below p_b. Check: the value at p_wf = 4,500 must equal q_b; the value at p_wf = 0 must equal AOFP.
Verify against Test T1. Compute r = 1,800/4,500 = 0.400. Plug into the composite equation and compare with q_T1 = 820 stb/d. The ideal IPR will be higher than 820 stb/d at this BHFP — explain why this is correct, not an error.
Determine BHFP for target rate. Using your IPR table (or by solving Vogel's equation), at what BHFP does q = 1,800 stb/d on the ideal (S = 0) IPR? Can natural flow (BHFP ≈ 4,200 psia) deliver 1,800 stb/d on the actual damaged well?

Composite IPR Working Table

p_wf (psia)	Segment	r = p_wf/p_b	Vogel Factor	q_o (stb/d) — Ideal S=0
5,100	Darcy	n/a	n/a	0
4,500 (p_b)	Darcy / join	1.000	—	432 = q_b
3,600	Vogel	0.800	1−0.16−0.512 = 0.328	432 + 1,800 × 0.328 = 1,022
2,700	Vogel	0.600	1−0.12−0.288 = 0.592	432 + 1,800 × 0.592 = 1,498
1,800 (T1)	Vogel	0.400	1−0.08−0.128 = 0.792	432 + 1,800 × 0.792 = 1,858
900	Vogel	0.200	1−0.04−0.032 = 0.928	432 + 1,800 × 0.928 = 2,102
0 (AOFP)	Vogel	0.000	1.000	432 + 1,800 × 1.000 = 2,232

Key Results Summary

q_b at p_b

432

stb/d at 4,500 psia

q_v,max

1,800

stb/d Vogel increment

AOFP (ideal)

2,232

stb/d at p_wf = 0

Target BHFP

~1,820

psia for 1,800 stb/d (ideal)

Critical Finding

The ideal (S = 0) IPR reaches 1,800 stb/d at BHFP ≈ 1,820 psia — achievable with an ESP. But the actual well (S' = +8) can only deliver ~820 stb/d at 1,800 psia BHFP — and cannot reach 1,800 stb/d even at zero BHFP without stimulation. SP-2 quantifies this loss precisely.

Interactive Tool

Composite IPR Canvas Simulator

Adjust the sliders to explore how J*, p̄, and p_b shape the composite IPR. Watch the canvas update in real time.

J* (Ideal PI) = 0.72 stb/d/psi p̄ (reservoir pressure) = 5100 psia p_b (bubble point) = 4500 psia BHFP target = 1820 psia

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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.

Study Vogel's equation: C01_M04_T1 — Vogel IPR · Composite (Darcy + Vogel) IPR: C01_M04_T2 — Composite IPR Topic theory pages

Watch Lecture 4.1D — Constructing the Vogel Curve, Step by Step · Lecture 4.2C — Anchoring the Vogel Segment: the Neely/Brown Formulation Produced lecture

Self-check ../../courses/c01/production/m04/topic-4-1-vogel/file-pack/vogel.py · ../../courses/c01/production/m04/topic-4-2-composite-ipr/file-pack/composite_ipr.py verified calculator — reproduce your numbers

Knowledge Check

SP-1 · 5 Questions

Answer all five questions before recording your SP-1 output. Target: 4/5 minimum before proceeding to SP-2.

Composite IPR — Knowledge Check

1. KA-07 has p̄ = 5,100 psia and p_b = 4,500 psia. Which inflow model is correct to use for the full range of flowing BHP from 0 to p̄?

Correct — C. When p̄ > p_b, the reservoir fluid is undersaturated and Darcy's linear PI applies above p_b. Once BHFP drops below p_b, gas liberates from solution in the near-wellbore region — Vogel's empirical equation then captures this two-phase flow. The two segments are joined continuously at p_b, giving q_b = J* × (p̄ − p_b). This is the standard composite IPR (Guo et al., Section 3.5).

2. Using J* = 0.72 stb/d/psi, p̄ = 5,100 psia, and p_b = 4,500 psia, what is the anchor rate q_b at the bubble point?

Correct — A: 432 stb/d. q_b = J* × (p̄ − p_b) = 0.72 × (5,100 − 4,500) = 0.72 × 600 = 432 stb/d. This is the rate from the Darcy segment at the join point. A common error is to use J* × p̄, which would give the single-phase maximum — but that ignores the fact that only the pressure above p_b drives Darcy flow; the pressure below p_b is handled by Vogel.

3. The maximum Vogel increment q_v,max = J* × p_b / 1.8. With J* = 0.72, p_b = 4,500 psia, what is q_v,max?

Correct — D: 1,800 stb/d. q_v,max = 0.72 × 4,500 / 1.8 = 3,240 / 1.8 = 1,800 stb/d. Therefore AOFP = q_b + q_v,max = 432 + 1,800 = 2,232 stb/d. The factor 1.8 in the denominator comes from Vogel's original derivation — at BHFP = 0, the Vogel factor [1 − 0.2r − 0.8r²] = 1.0 and the maximum rate increment corresponds to J* × p_b / 1.8.

4. The ideal (S = 0) composite IPR predicts q ≈ 1,858 stb/d at p_wf = 1,800 psia, but Test T1 measured only 820 stb/d at the same BHFP. What is the best explanation?

Correct — A. The ideal IPR at S = 0 represents the well's full potential without skin. Skin S' = +8 acts as an additional pressure drop in the near-wellbore region, reducing the effective driving pressure for any given BHFP. The ratio J_actual/J* = FE = 7/(7+8) ≈ 0.467 — so the actual PI is roughly half the ideal. This is the central concept of SP-2: quantifying this gap and evaluating whether the acid job can recover it.

5. For the ideal (S = 0) KA-07 composite IPR, at approximately what BHFP is the field target rate of 1,800 stb/d achievable?

Correct — B. From the IPR table: at p_wf = 1,800 psia, q = 1,858 stb/d > 1,800 stb/d. So the ideal well crosses the target at a BHFP slightly above 1,800 psia — interpolating gives ~1,820 psia. This BHFP is achievable with a moderate-stage ESP. However, the actual well (S' = +8) cannot reach 1,800 stb/d at any BHFP without stimulation — which is the economic argument SP-2 will quantify.

SP-1 Output

Record Before Moving to SP-2

Write these values on your answer sheet. They are referenced in SP-4 (Fetkovich comparison), SP-5 (future IPR), and SP-6 (integrated report).

SP-1 Key Outputs — KA-07 Ideal Composite IPR

Darcy anchor rate at p_b = 4,500 psia: q_b = 432 stb/d

Maximum Vogel increment: q_v,max = 1,800 stb/d

Ideal AOFP (S = 0, p_wf = 0): AOFP = 2,232 stb/d

BHFP for 1,800 stb/d on ideal IPR: ≈ 1,820 psia (achievable with ESP)

Natural-flow rate at ~4,200 psia BHFP: ~642 stb/d (973 stb/d below target)

→ Carry q_b, q_v,max, and AOFP forward to SP-4, SP-5, SP-6.

→ Proceed to SP-2 to apply Standing's FE correction for the actual skin of S' = +8 and quantify the acid-job value.