Steady-State Radial Flow — Wellbore and Average Reservoir Pressure: 24-Pet-A3 Fundamental Reservoir Engineering, May 2016, Q7

Question 7: Steady-State Radial Flow — Wellbore and Average Reservoir 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.

Oil rate, $q$	800 STBD
Drainage area, $A$	200 acres
Pressure at $r_e$, $p_e$	2000 psia
Oil viscosity, $\mu$	2 cP
Oil FVF, $B_o$	1.2 bbl/STB
Wellbore radius, $r_w$	0.33 ft
Pay-zone thickness, $h$	20 ft
Permeability, $k$	200 mD

Find. Wellbore flowing pressure, $p_{wf}$, and average reservoir pressure, $\bar p$ (no skin given, $s=0$).

Fig. 3 — Circular drainage area of external radius $r_e$; the known boundary condition is the pressure AT $r_e$ (steady state), from which both $p_{wf}$ and the pseudosteady-state average pressure $\bar p$ are found.

Approach. Convert the drainage area to an equivalent external radius, then use the steady-state radial-flow relation (external pressure $p_e$ known at $r_e$) to get $p_{wf}$; the average reservoir pressure follows from the standard pseudosteady-state relation, which differs from the external-boundary relation only by the constant $-3/4$ inside the log term.

Equivalent drainage radius. $A=\pi r_e^2\Rightarrow r_e=\sqrt{A/\pi}=\sqrt{(200)(43{,}560)/\pi}$, so $\boxed{r_e=1665\ \text{ft}}$.
Wellbore flowing pressure. With $k=200\ \text{mD}=0.2$ Darcy and $s=0$, steady state gives $q=\dfrac{7.08kh(p_e-p_{wf})}{\mu B_o\ln(r_e/r_w)}$, so $p_e-p_{wf}=\dfrac{q\mu B_o\ln(r_e/r_w)}{7.08kh}=\dfrac{800(2)(1.2)\ln(1665/0.33)}{7.08(0.2)(20)}=\dfrac{1920(8.526)}{28.32}=578.1$ psi. $\boxed{p_{wf}=2000-578.1=1421.9\ \text{psia}}$.
Average reservoir pressure. The pseudosteady-state form uses $\ln(r_e/r_w)-3/4$ in place of $\ln(r_e/r_w)$ for $\bar p-p_{wf}$, so $\bar p$ sits below $p_e$ by exactly the $-3/4$ term's contribution: $p_e-\bar p=\dfrac{q\mu B_o(3/4)}{7.08kh}=\dfrac{1920(0.75)}{28.32}=50.8$ psi. $\boxed{\bar p=2000-50.8=1949.2\ \text{psia}}$.

Quantity	Value
Equivalent drainage radius, $r_e$	1665 ft
Pressure drop, $p_e-p_{wf}$	578.1 psi
Wellbore flowing pressure, $p_{wf}$	1421.9 psia
Average reservoir pressure, $\bar p$	1949.2 psia

Topic: Steady-state radial (Darcy) flow and the relationship between external-boundary, average, and wellbore pressures.

Key relations: $q=\dfrac{7.08kh(p_e-p_{wf})}{\mu B_o[\ln(r_e/r_w)+s]}$ (external pressure known, $k$ in Darcy); $q=\dfrac{7.08kh(\bar p-p_{wf})}{\mu B_o[\ln(r_e/r_w)-3/4+s]}$ (average-pressure/pseudosteady-state form).

Why this works: Both relations describe the same radial pressure profile; the average reservoir pressure (the volumetrically-averaged pressure over the whole drainage area) is always slightly below the true boundary pressure $p_e$ because more of the drainage area's volume sits at pressures closer to $p_e$ than to $p_{wf}$ — the constant $3/4$ is the exact result of that volume-weighted average for a circular drainage area at pseudosteady state.

Common pitfall: Treating the given "reservoir pressure at the external radius" as if it were already the average reservoir pressure, and skipping the $-3/4$ correction entirely.

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 8 (steady and pseudosteady radial flow); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 6.

Question 7 of 7: Steady-State Radial Flow — Wellbore and Average Reservoir Pressure

Question 7: Steady-State Radial Flow — Wellbore and Average Reservoir Pressure (20 marks)