Pseudo-Steady-State Inflow — Flowing Wellbore Pressure: 24-Pet-A3 Fundamental Reservoir Engineering, December 2015, Q5

Question 5: Pseudo-Steady-State Inflow — Flowing Wellbore Pressure (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $q=500$ STBD, $A=200$ acres, $\bar p=2000$ psia, $k=200$ mD, $B_o=1.252$ bbl/STB, $\mu=1$ cP, $h=50$ ft, $s=-1$, $r_w=0.3$ ft.

Find. The flowing bottom-hole pressure $p_{wf}$.

Check: the bubble point pressure (1000 psia) is well below $\bar p=2000$ psia and the computed $p_{wf}$, so the well stays above $p_b$ and single-phase liquid (Darcy) flow applies throughout the drainage area — it is not needed in the calculation itself.

Approach. Convert the drainage area to an equivalent circular drainage radius, then apply the pseudo-steady-state radial inflow equation referenced to the given average reservoir pressure $\bar p$ (which uses the $-3/4$ shape-factor term, not $-1/2$, since $\bar p$ — not the boundary pressure $p_e$ — is what is given).

Equivalent drainage radius. $A=200\ \text{ac}\times43{,}560\ \text{ft}^2/\text{ac}=8{,}712{,}000\ \text{ft}^2=\pi r_e^2\ \Rightarrow\ r_e=\sqrt{A/\pi}=\boxed{1665.3\ \text{ft}}$.
Radial geometry term. $\ln(r_e/r_w)=\ln(1665.3/0.3)=\boxed{8.622}$.
Solve for $p_{wf}$. $q=\dfrac{0.00708\,kh(\bar p-p_{wf})}{\mu B_o\left[\ln(r_e/r_w)-\tfrac34+s\right]}$. The bracket is $8.622-0.75+(-1)=6.872$, so $\Delta p=\dfrac{q\,\mu B_o\,(6.872)}{0.00708\,kh}=\dfrac{(500)(1)(1.252)(6.872)}{0.00708(200)(50)}=\dfrac{4302}{70.8}=60.8$ psi. $\boxed{p_{wf}=\bar p-\Delta p=2000-60.8=1939.2\ \text{psia}}$.

Quantity	Value
Drainage radius, $r_e$	1665.3 ft
$\ln(r_e/r_w)$	8.622
Pressure drop, $\Delta p$	60.8 psi
Flowing bottom-hole pressure, $p_{wf}$	1939.2 psia

Concept & Theory

Topic: Pseudo-steady-state (PSS) radial inflow.

Key relations: $q=\dfrac{0.00708\,kh(\bar p-p_{wf})}{\mu B\left[\ln(r_e/r_w)-\tfrac34+s\right]}$ referenced to the volumetric average drainage-area pressure $\bar p$; the equivalent form referenced to the (constant) boundary pressure $p_e$ instead uses $\ln(r_e/r_w)-\tfrac12$.

Why this works: Once the reservoir reaches PSS, the pressure profile shape is fixed and the $-3/4$ constant is the exact geometric factor relating the volumetric average pressure of a circular drainage area to the pressure at its outer boundary; using $-\tfrac12$ here (the $p_e$-referenced form) would understate $\Delta p$ since $\bar p>p_e$ always.

Common pitfall: Mixing the $-3/4$ (average-pressure) and $-1/2$ (boundary-pressure) forms, or using $k$ in Darcy with the field-units 0.00708 constant (which already assumes $k$ in md).

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 1–2 (steady/pseudo-steady inflow performance); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 6.

Question 5 of 7: Pseudo-Steady-State Inflow — Flowing Wellbore Pressure

Question 5: Pseudo-Steady-State Inflow — Flowing Wellbore Pressure (20 marks)