Given. A well loaded to the desired oil rate with a straight-line IPR, full well/completion data, and the surface casing operating pressure (no proprietary gradient-curve chart is legible on the source page — solved instead from a static gas-column formula and a homogeneous no-slip mixture model, per the method below).
Well depth
8000 ft
Productivity index, $J$
1 bbl/day/psi
Average reservoir pressure, $\bar P_R$
3000 psi
Desired oil rate, $q_o$
1000 STB/day
Oil gravity / gas gravity
$35^\circ$API / 0.65
Formation GLR
100 SCF/STB
Surface / bottomhole temperature
$100^\circ$F / $200^\circ$F
Wellhead pressure, $P_{wh}$
100 psi
Tubing ID
2.441 in.
Differential pressure across operating valve
100 psi
Surface gas operating injection pressure, $P_{so}$
1200 psi
Find. (a) The depth of the point of gas injection; (b) the required injection gas rate.
Check: no PVT table or gradient-curve chart is provided for this well (pages 7–10 of the paper are blank grid/redacted pages meant for the exam's own proprietary gradient charts), so the injected/produced gas is modelled as fully "free" gas in a homogeneous, no-slip mixture (constant $Z=0.85$, $B_o\approx1$) — the standard simplification for gas-lift design questions when no PVT/chart data is supplied.
Approach. (1) Get $P_{wf}$ from the straight-line IPR. (2) March the tubing pressure with the formation GLR only, from $P_{wf}$ at 8000 ft up toward surface — this is the well's natural (un-lifted) flowing traverse. (3) Compute the casing gas-column pressure (static gas gradient) from the surface operating pressure. (4) The point of gas injection is where the casing pressure, less the valve differential, first drops to meet the natural tubing traverse (marching up from bottom). (5) Above that point, march the tubing pressure again with the total GLR (formation + injected), and bisect the injected GLR so that the traverse arrives at the given $P_{wh}$ at the surface.
Natural tubing traverse (formation GLR = 100 SCF/STB). Marching the homogeneous no-slip mixture gradient $dP/dD=\rho_m(P,T)/144$ upward from $(D,P)=(8000\ \text{ft},\,2000\ \text{psi})$, using the linear temperature profile $T(D)=100+0.0125D\ ^\circ$F, the traverse reaches only $\approx21$ psi at the surface — far below $P_{wh}=100$ psi, confirming the well cannot flow naturally on formation gas alone and needs lift assistance.
Casing gas-column pressure. $P_{cas}(D)=P_{so}\exp\!\left[\dfrac{0.01875\,\gamma_g D}{\bar T_R(D)\,Z}\right]$ (absolute pressures internally, $Z=0.85$, $\bar T_R(D)$ the average absolute temperature from surface to $D$), starting from $P_{so}=1200$ psi. This rises only gently with depth (gas is light), from 1200 psi at surface to $\approx1351$ psi at 8000 ft.
Locate the point of injection. The valve opens where $P_{cas}(D)-\Delta P_{valve}$ first equals the natural tubing traverse, searching from bottom to top. The two curves cross at $\boxed{D_{inj}\approx5753\ \text{ft}}$, where both sides equal $\approx1280$ psi.
Required total GLR above the injection point. Above $D_{inj}$, re-march the mixture traverse from $(5753\ \text{ft},\,1280\ \text{psi})$ up to the surface, now solving (by bisection) for the total GLR that lands exactly on $P_{wh}=100$ psi. This gives $\boxed{GLR_{total}=164.5\ \text{SCF/STB}}$ — more gas than the formation alone supplies, which is exactly what continuous gas lift adds.
Injected GLR and gas rate. $GLR_{inj}=GLR_{total}-GLR_{formation}=164.5-100=64.5$ SCF/STB. $q_{gas,inj}=GLR_{inj}\times q_o=64.5\times1000$. $\boxed{q_{gas,inj}\approx64{,}500\ \text{SCF/day}\ (0.064\ \text{MMSCF/day})}$.
Fig. 3 — Casing operating-pressure traverse (green, less the 100-psi valve differential) crossing the well's natural formation-GLR tubing traverse (red) at the point of injection, $D_{inj}\approx5753$ ft; above that point the lifted tubing traverse (blue, total GLR = 164.5 SCF/STB) carries the flow to the required $P_{wh}=100$ psi.
