Linear Flow of a Slightly Compressible Oil — Downstream Pressure and Rate: 24-Pet-A3 Fundamental Reservoir Engineering, Undated paper, Q3

Question 3: Linear Flow of a Slightly Compressible Oil — Downstream Pressure and Rate (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.

Quantity	Value
Length, $L$	1,000 ft
Cross-sectional area, $A$	800 ft²
Permeability, $k$	250 md
Oil viscosity, $\mu$	2.5 cp
Oil compressibility, $c$	$68\times10^{-6}$ psi$^{-1}$
Upstream pressure, $p_1$	2,500 psia
Flow rate at $p_1$, $q_1$	50 res bbl/day

Find. (1) The downstream pressure $p_2$; (2) the flow rate $q_2$ evaluated at $p_2$.

Approach. Above the bubble point the oil is a single-phase slightly compressible liquid, so mass (not volume) is conserved along the sand body; apply the formula sheet's Linear Flow of Slightly Compressible Fluids equation twice, once with the reference pressure set to the known upstream point $p_1$ (to solve for $p_2$) and once with it set to $p_2$ (to get $q_2$).

Set up the governing equation. $q_R=\dfrac{0.001127\,kA}{\mu L c}\ln\!\left[\dfrac{1+c(p_R-p_2)}{1+c(p_R-p_1)}\right]$. The coefficient is $K=\dfrac{0.001127\,kA}{\mu Lc}=\dfrac{0.001127(250)(800)}{2.5(1{,}000)(68\times10^{-6})}=\dfrac{225.4}{0.17}$, so $\boxed{K=1{,}325.9\ \text{bbl/day}}$.
Solve for $p_2$ using $p_R=p_1$. With $p_R=p_1$, the denominator term vanishes ($1+c(p_1-p_1)=1$) and $q_1=K\ln[1+c(p_1-p_2)]$. Rearranging: $1+c(p_1-p_2)=e^{q_1/K}=e^{50/1325.9}=e^{0.03771}=1.03844$, so $p_1-p_2=0.03844/c=0.03844/(68\times10^{-6})=565.3$ psi. Thus $\boxed{p_2=2{,}500-565.3=1{,}934.7\ \text{psia}}$.
Solve for $q_2$ using $p_R=p_2$. Now the numerator term vanishes: $q_2=K\ln\!\left[\dfrac{1}{1+c(p_2-p_1)}\right]=1{,}325.9\ln\!\left[\dfrac{1}{1-0.03844}\right]=1{,}325.9\ln(1.03998)$, so $\boxed{q_2=52.0\ \text{res bbl/day}}$ — slightly higher than $q_1$, consistent with the lower-pressure oil downstream being less compressed and occupying more reservoir volume per unit mass.

Quantity	Value
Downstream pressure, $p_2$	1,934.7 psia
Flow rate at $p_2$, $q_2$	52.0 res bbl/day

Topic: Steady-state linear Darcy flow of a slightly compressible liquid, where volumetric rate varies along the flow path even though mass rate is constant.

Key relations: $q_R=\dfrac{0.001127\,kA}{\mu Lc}\ln\!\left[\dfrac{1+c(p_R-p_2)}{1+c(p_R-p_1)}\right]$, the field-unit integrated form of Darcy's law combined with $\rho(p)=\rho_{ref}[1+c(p-p_{ref})]$ for a slightly compressible fluid under steady mass-conservation.

Why this works: unlike an incompressible fluid, a slightly compressible liquid's density (and hence its reservoir-barrel volume per unit mass) changes measurably with pressure; conserving mass rather than volume along the sand body means the volumetric rate quoted at one end of the system is not simply repeated at the other end — it must be referenced to the local pressure, which is exactly what setting $p_R$ equal to each end pressure in turn accomplishes.

Common pitfall: treating $q_1=q_2$ as if the fluid were incompressible, or forgetting that the "reference pressure" in the formula must match the pressure at which the given/asked-for flow rate is actually measured.

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 3 (flow of compressible fluids); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 6.

Question 3 of 7: Linear Flow of a Slightly Compressible Oil — Downstream Pressure and Rate

Question 3: Linear Flow of a Slightly Compressible Oil — Downstream Pressure and Rate (20 marks)