Real-Gas Pseudopressure Drawdown: 24-Pet-B2 Oil and Gas Evaluation and Economics, May 2016, Q6

Question 6: Real-Gas Pseudopressure Drawdown (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.

Check: this question omits reservoir temperature entirely, and the required real-gas pseudopressure formula needs it.

Given.

Quantity	Value
Flow rate, $q_{sc}$	7,000 MSCFD (constant)
Initial pressure, $p_i$	2,000 psia
Temperature, $T$	580 °R (see the check note)
Thickness, $h$	39 ft
Viscosity, $\mu$	0.0158 cP
Porosity, $\phi$	0.15
Permeability, $k$	20 mD
Well radius, $r_w$	0.4 ft
Compressibility, $c_t$	0.00053 psi$^{-1}$
Time	36 hr

Find. Flowing bottomhole pressure $p_{wf}$ after 36 hr.

Approach. Compute dimensionless time, check whether the semilog (Ei-function) approximation applies, get the dimensionless pressure, convert to a real-gas pseudopressure drop via the transient flow equation, then invert through the digitized $\psi(p)$-vs-$p$ chart to recover $p_{wf}$.

Dimensionless time. Using the formula sheet's diffusivity grouping ($k$ in mD, $t$ in days): $$t_D=\frac{6.33\,k\,t}{\phi\mu c_t r_w^2}=\frac{6.33(20)(36/24)}{0.15(0.0158)(0.00053)(0.4)^2}$$ $$t_D=9.45\times10^5 \gg 100 \Rightarrow \text{semilog (log) approximation applies}$$
Dimensionless pressure. $$p_D=\tfrac{1}{2}\left(\ln t_D+0.809\right)=\tfrac{1}{2}\left(\ln(9.45\times10^5)+0.809\right)$$ $$\boxed{p_D=7.284}$$
Pseudopressure drop. $$\Delta\psi=\frac{1.422\,q_{sc}T}{kh}\,p_D=\frac{1.422(7000)(580)}{20(39)}(7.284)$$ $$\boxed{\Delta\psi=53{,}900\ \text{psia}^2/\text{cp}}$$
Invert through the $\psi(p)$ chart. Reading $\psi_i$ at $p_i=2000$ psia off the digitized chart gives $\psi_i=3.400\times10^8$; subtracting the drawdown and reading back: $$\psi_{wf}=\psi_i-\Delta\psi=3.400\times10^8-5.39\times10^4=3.39946\times10^8$$ $$\boxed{p_{wf}\approx1999.8\ \text{psia}}$$ The pressure drop after only 36 hr is tiny (<1 psi) — consistent with this well's very high $\psi$-vs-$p$ sensitivity near $p_i$ (the chart's slope is steep there) combined with a moderate rate and only a day and a half of production.

Digitized $\psi(p)$-vs-$p$ chart used to convert the computed pseudopressure drop back to a flowing bottomhole pressure.

Quantity	Result
$t_D$	$9.45\times10^5$
$p_D$ (log approximation)	7.284
Pseudopressure drop, $\Delta\psi$	53,900 psia$^2$/cp
Flowing bottomhole pressure, $p_{wf}$	1,999.8 psia

Topic: Real-gas pseudopressure transient well-test analysis.

Key relations: $t_D=6.33kt/(\phi\mu c_tr_w^2)$; $p_D=\tfrac12(\ln t_D+0.809)$ for $t_D>100$; $\Delta\psi=1.422q_{sc}T\,p_D/(kh)$.

Why this works: The pseudopressure transform $\psi(p)=2\int p/(\mu Z)\,dp$ linearizes the nonlinear real-gas diffusivity equation, so the well-known liquid-flow $p_D$-$t_D$ solutions (here the log/Ei-function approximation, valid once $t_D>100$) apply directly to $\Delta\psi$ instead of $\Delta p$ — only the final inversion through the gas-specific $\psi(p)$ curve is nonlinear.

Common pitfall: Applying $\Delta p=141.2q\mu B\,p_D/(kh)$ (the liquid form) directly to a gas well instead of working in pseudopressure — at high pressure the two are nearly equivalent, but the gap grows as pressure (and $\mu Z$ variation) departs from initial conditions.

Source: Lee & Wattenbarger, Gas Reservoir Engineering, Ch. 4 (real-gas pseudopressure well testing); this exam’s own formula sheet.

Question 6 of 7: Real-Gas Pseudopressure Drawdown

Question 6: Real-Gas Pseudopressure Drawdown (20 marks)