Given. A fully-specified Vogel-type liquid-rate IPR ($q_{L,max}=1407$ STB/day at $\bar P_R=4000$ psig) and a tubing string whose intake requirement must be matched against it.
Find. (a) Whether 400 STBO/day flows naturally to $P_{wh}=400$ psig; (b) whether GLR=300 sustains 600 STBO/day, and if not, the extra gas needed.
Check: with no oil/gas gravity, temperature or gradient-curve chart supplied for this question (the source ships only blank graph-paper pages where a gradient-curve chart would normally be plotted), the tubing intake pressure is estimated with a homogeneous, no-slip mixture model: standard field assumptions ($35^{\circ}\text{API}$ oil, $\gamma_g=0.65$, $\gamma_w=1.07$, $B_o=1.2$, an $80$–$140\,{}^{\circ}\text{F}$ linear geothermal gradient) with $z$ from a Standing pseudo-critical fit and the Papay explicit correlation, marched in small increments down from $P_{wh}$ so local gas density updates with pressure. Friction is neglected (no roughness/velocity data given) — a hydrostatic-dominated approximation appropriate to the data actually provided, not a full mechanistic (Hagedorn–Brown/Poettmann–Carpenter) correlation or the published gradient-curve chart an open-book candidate would have available.
Approach. Solve the IPR for the $P_{wf}$ that delivers each target oil rate, march the no-slip mixture gradient down from the required $P_{wh}=400$ psig over 5000 ft to get the $P_{wf}$ the tubing actually needs, and compare: the well flows only where the IPR-supplied $P_{wf}$ meets or exceeds the tubing's requirement.
(a) $P_{wf}$ from the IPR at $q_o=400$ STBO/day. $f_w=0.5\Rightarrow q_L=400/0.5=800$ STBL/day. Solving $800=1407\left[1-0.2R-0.8R^2\right]$ for $R$ gives $R=0.620$. $\boxed{P_{wf,IPR}=0.620\times4000=2480\ \text{psig}}$.
(a) $P_{wf}$ required by the tubing (VLP) at $q_L=800$ STBL/day, GLR=300. Marching the no-slip mixture gradient down from $P_{wh}=400$ psig over 5000 ft gives $\boxed{P_{wf,TPR}\approx1485\ \text{psig}}$. Since $P_{wf,IPR}=2480\ \text{psig}$ comfortably exceeds the 1485 psig the tubing needs, $\boxed{\text{Answer (a): YES, the well flows naturally at 400 STBO/day}}$, with roughly 995 psi of drawdown margin to spare.
(b) $P_{wf}$ from the IPR at $q_o=600$ STBO/day. $q_L=600/0.5=1200$ STBL/day. Solving $1200=1407\left[1-0.2R-0.8R^2\right]$ gives $R=0.322$. $\boxed{P_{wf,IPR}=0.322\times4000=1287\ \text{psig}}$ — the reservoir now has far less pressure to give at this higher offtake.
(b) Check GLR=300 at the higher rate. With $q_L=1200$ STBL/day the tubing traverse at GLR=300 requires (elevation-dominated, so essentially rate-independent at fixed GLR) the same $\boxed{P_{wf,TPR}\approx1485\ \text{psig}}$ — now above the 1287 psig the IPR can deliver. $\boxed{\text{Answer (b), part 1: NO, GLR=300 is insufficient at 600 STBO/day}}$.
(b) Required GLR and additional gas. Bisecting the total GLR so the marched traverse from $P_{wf}=1287$ psig lands exactly on $P_{wh}=400$ psig gives $\boxed{\text{GLR}_{required}\approx422\ \text{SCF/STBL}}$. Additional gas $=(\text{GLR}_{required}-300)\times q_L=(422-300)\times1200$. $\boxed{Q_{inj}\approx1.46\times10^5\ \text{SCF/day}\ (0.146\ \text{MMscf/day})}$, injected at the tubing intake to lighten the column enough to reach 600 STBO/day.
Fig. 4 — Vogel-type IPR ($q_{L,max}=1407$ STB/day) with the two operating points: (a) $q_L=800$ STBL/day needs only 2480 psig, well within reach; (b) $q_L=1200$ STBL/day can only be supplied at 1287 psig, short of the ≈1485 psig the tubing needs at GLR=300.
