Well Pressure via Real Gas Pseudopressure (Transient Flow): 24-Pet-B2 Oil and Gas Evaluation and Economics, December 2014, Q5

Question 5: Well Pressure via Real Gas Pseudopressure (Transient Flow) (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.

It also used a coarse reading of the $\psi(p)$ chart. Both are corrected below.

Check: the source data table does not list a gas viscosity, which the formula sheet’s $\eta=6.33k/(\phi\mu_ic_{ti})$ requires. Assumed $\mu_i=0.02$ cp, matching the value this same exam supplies explicitly for the sibling transient-flow problem in Question 6 (same gas-well-testing context, comparable reservoir temperature).

Given. $p_i=2000$ psia; $T=580^{\circ}$R; $h=39$ ft; $\phi=0.15$; $k=20$ mD; $r_w=0.4$ ft; $c_i=0.00053$ psi$^{-1}$; $q_g=7$ MMSCFD $=7000$ MSCFD; $t=36$ hr; $\mu_i=0.02$ cp (assumed, see the check note). The $\psi(p)$ function is given graphically; read from the exam chart: $p=$ 0, 500, 1000, 1500, 1750, 1850, 2000, 2500 psia $\to\psi=$ 0, 0.212, 0.910, 1.956, 2.598, 2.876, 3.325, 5.00 ($\times10^8$ psia$^2$/cp).

Find. $p_{wf}$ after 36 hours of constant-rate production.

Approach. Compute the dimensionless time $t_D$, apply the line-source $p_D$ solution (log approximation, since $t_D\gg100$), get $\Delta\psi$ from the transient-flow equation, subtract from $\psi(p_i)$ read off the given chart, then invert the same chart to recover $p_{wf}$.

Hydraulic diffusivity and dimensionless time. $\eta=6.33k/(\phi\mu_ic_i)=6.33(0.02)/(0.15\times0.02\times0.00053)=79{,}623\ \text{ft}^2/\text{day}$. $t=36/24=1.5$ day, so $t_D=\eta t/r_w^2=79{,}623(1.5)/0.4^2=\boxed{t_D=7.46\times10^5}$.
Dimensionless pressure. Since $t_D>100$: $p_D=\tfrac12(\ln t_D+0.809)=\tfrac12(\ln(7.46\times10^5)+0.809)$: $\boxed{p_D=7.166}$.
Pseudopressure at $r_w$. $\psi(r_w,t)=\psi_i-\dfrac{1.422\,q_{sc}T}{kh}p_D$. From the chart, $\psi(p_i=2000)=3.325\times10^8$ psia$^2$/cp. The formula sheet defines $k$ in Darcy for this equation too (the same units as in $\eta$), so $k=20$ mD $=0.020$ D, with $q_{sc}$ in MSCFD: $\Delta\psi=\dfrac{1.422(7000)(580)}{0.020(39)}(7.166)=\boxed{\Delta\psi=5.304\times10^7\ \text{psia}^2/\text{cp}}$ (identical to the textbook field form $1422\,q_{sc}T/(k_{mD}h)$). $\psi(r_w,36\text{hr})=3.325\times10^8-0.530\times10^8=\boxed{\psi_{wf}=2.795\times10^8\ \text{psia}^2/\text{cp}}$.
Invert the chart. $\psi=2.795\times10^8$ falls between the digitized points (1750 psia, $2.598\times10^8$) and (1850 psia, $2.876\times10^8$): $p_{wf}=1750+100\dfrac{2.795-2.598}{2.876-2.598}$, so $\boxed{p_{wf}\approx1821\ \text{psia}}$. The drawdown is about 180 psia after 36 hr (chart-reading precision about $\pm10$ psia).

[Figure not reproduced: Fig. 2 — Real gas pseudopressure function $\psi(p)$ read from the exam chart, with $\psi_i$ at 2000 psia and the 36-hr flowing point ($\Delta\psi=5.30\times10^7$, $p_{wf}\approx1821$ psia) marked. The same $\psi(p)$ function is reused for Question 6. See the official exam paper.]

Quantity	Value
Dimensionless time, $t_D$	$7.46\times10^5$
Dimensionless pressure, $p_D$	7.166
Pseudopressure drop, $\Delta\psi$	$5.30\times10^7$ psia$^2$/cp
Well pressure after 36 hr, $p_{wf}$	≈1821 psia

Topic: Transient radial flow of a real gas to a well, solved with the real-gas pseudopressure line-source method.

Key relations: $\psi(r,t)=\psi_i-\dfrac{1.422q_{sc}T}{kh}p_D$; $t_D=\eta t/r_w^2$, $\eta=6.33k/(\phi\mu_ic_t)$ ($k$ in Darcy); $p_D=\tfrac12(\ln t_D+0.809)$ for $t_D>100$.

Why this works: The pseudopressure transform linearizes the real-gas diffusivity equation exactly the way $p$ itself linearizes the liquid diffusivity equation, so the same line-source (Ei-function/log-approximation) solutions developed for oil wells carry over directly, provided $\psi$ (not $p$) is used throughout.

Common pitfall: Mixing permeability units. The formula sheet’s 1.422 and 6.33 constants both take $k$ in Darcy, and entering mD in one of them shifts $\Delta\psi$ by 1000×. A second trap is converting $\Delta\psi$ back to $\Delta p$ using a constant local slope evaluated at the wrong pressure — the $\psi$-$p$ relationship is markedly nonlinear (steepest near high pressure here), so a chart inversion (or the exact functional relationship) must be used, not a single global average slope.

Source: Lee & Wattenbarger, Gas Reservoir Engineering, Ch. 2 & 4; formula sheet (this exam).

Question 5 of 7: Well Pressure via Real Gas Pseudopressure (Transient Flow)

Question 5: Well Pressure via Real Gas Pseudopressure (Transient Flow) (20 marks)