Well Pressure via Real Gas Pseudopressure — Two-Rate Superposition: 24-Pet-B2 Oil and Gas Evaluation and Economics, May 2015, Q5

Question 5: Well Pressure via Real Gas Pseudopressure — Two-Rate Superposition (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. $p_i=2000$ psia; $T=580^{\circ}$R; $h=39$ ft; $\phi=0.15$; $k=20$ mD; $r_w=0.4$ ft; $c_t=0.00053$ psi$^{-1}$; $\mu=0.0158$ cp; $q_1=7$ MMSCFD $=7000$ MSCFD for $0\le t<36$ hr, then $q_2=21$ MMSCFD $=21{,}000$ MSCFD for $36\le t\le108$ hr. The $\psi(p)$ function is given graphically (points read at $p=0,500,1000,\ldots,2500$ psia: $\psi=0,\,0.25,\,1.05,\,2.1,\,3.4,\,5.0\ (\times10^8)$ psia$^2$/cp).

Find. $p_{wf}$ after 108 hours of two-rate production.

Approach. The rate changed partway through, so the constant-rate pseudopressure-drop equation cannot be applied directly — use superposition in time: treat the history as two rate increments ($\Delta q_1=q_1$ at $t=0$, and $\Delta q_2=q_2-q_1$ at $t=36$ hr), and sum each increment’s own pseudopressure-drop contribution evaluated at its own elapsed time since it began.

Diffusivity/time-group. $\eta=6.33k/(\phi\mu c_t)=6.33(0.020)/(0.15\times0.0158\times0.00053)$: $\eta=1.008\times10^5\ \text{ft}^2/\text{day}$ ($k=20$ mD $=0.020$ Darcy, per the formula sheet’s stated units).
Dimensionless times for each rate increment. $\Delta q_1$ has been flowing for the full $t=108$ hr $=4.5$ day: $t_{D,1}=\eta t/r_w^2=1.008\times10^5(4.5)/0.4^2=2.835\times10^6$. $\Delta q_2$ has been flowing for $108-36=72$ hr $=3.0$ day: $t_{D,2}=1.008\times10^5(3.0)/0.4^2=1.890\times10^6$. Both exceed 100, so the log-approximation $p_D=0.5(\ln t_D+0.809)$ applies to each: $p_{D,1}=0.5(\ln(2.835\times10^6)+0.809)=7.833$; $p_{D,2}=0.5(\ln(1.890\times10^6)+0.809)=7.630$.
Superposition sum for the total pseudopressure drop. $\psi_i-\psi_{wf}=\dfrac{1.422T}{kh}\left[\Delta q_1\,p_{D,1}+\Delta q_2\,p_{D,2}\right]=\dfrac{1.422(580)}{20(39)}\left[7000(7.833)+14{,}000(7.630)\right]$: $\boxed{\Delta\psi=1.709\times10^5\ \text{psia}^2/\text{cp}}$.
Convert to pressure via the $\psi(p)$ curve. $\psi_i=\psi(p_i=2000)=3.400\times10^8$ psia$^2$/cp (from the given graph). $\psi_{wf}=\psi_i-\Delta\psi=3.400\times10^8-1.709\times10^5=3.39829\times10^8$ psia$^2$/cp. Reading the inverse of the $\psi(p)$ curve at this value: $\boxed{p_{wf}=1999.3\ \text{psia}}$.

[Figure not reproduced: Fig. 2 — Real gas pseudopressure $\psi$ vs. pressure $p$ (source chart), with $\psi_i$ at the initial reservoir pressure and $\psi_{wf}$ after 108 hr of two-rate production marked — the two points are nearly coincident because the total pseudopressure drop is small relative to $\psi_i$. See the official exam paper.]

Quantity	Value
Diffusivity group, $\eta$	$1.008\times10^5$ ft$^2$/day
$t_{D,1}$ (full 108 hr), $t_{D,2}$ (last 72 hr)	$2.835\times10^6$, $1.890\times10^6$
Total pseudopressure drop, $\Delta\psi$	$1.709\times10^5$ psia$^2$/cp
$\psi_{wf}$ at $t=108$ hr	$3.398\times10^8$ psia$^2$/cp
Well pressure, $p_{wf}$ at $t=108$ hr	1999.3 psia

the $\psi(p)$ curve depends only on gas composition and $T$ (not on $k,h,\phi,r_w,c_t,\mu$), both of which match, and the axis ranges of the two charts are identical.

Topic: Real-gas pseudopressure transient-flow analysis with a rate change (superposition in time).

Key relations: $\psi(r,t)=\psi_i-\dfrac{1.422q_{sc}T}{kh}p_D$; superposition for a multi-rate history: $\psi_i-\psi_{wf}(t)=\dfrac{1.422T}{kh}\sum_j\Delta q_j\,p_D(t_{D,j})$, each increment evaluated at its own elapsed time; $p_D=0.5(\ln t_D+0.809)$ for $t_D>100$.

Why this works: The diffusivity equation is linear, so the pressure response to an arbitrary rate history is the sum of the responses to a sequence of constant-rate step changes, each starting at its own moment and evaluated using its own elapsed (not total) time since that step began.

Common pitfall: Using the total flowing time (108 hr) for both rate increments, or using only the final rate (21 MMSCFD) for the whole 108 hr — both skip the superposition principle and give a materially different (and wrong) pseudopressure drop.

Source: Lee & Wattenbarger, Gas Reservoir Engineering, Ch. 4 (pseudopressure, superposition); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 6.

Question 5 of 7: Well Pressure via Real Gas Pseudopressure — Two-Rate Superposition

Question 5: Well Pressure via Real Gas Pseudopressure — Two-Rate Superposition (20 marks)