Approach. Write the valve's force balance at the moment it opens (casing pressure on the exposed bellows area $A_b-A_p$, plus tubing pressure on the port area $A_p$, balanced against the nitrogen dome load on the full bellows area $A_b$); at closing the same balance applies with the tubing-pressure contribution removed (valve closes on casing pressure alone against the dome charge); then translate both depth pressures to surface using the given casing gas gradient.
(a) Bellows charge pressure. Opening force balance: $P_{vo}(A_b-A_p)+P_tA_p=P_bA_b\Rightarrow P_b=P_{vo}(1-R)+P_tR$. $P_b=962(1-0.1534)+680(0.1534)=814.6+104.3$. $\boxed{P_b=918.7\ \text{psi}}$ (at valve depth and temperature).
(b) Valve closing pressure at depth. At closing, casing pressure alone balances the dome load across the exposed area $(1-R)$: $P_{vc}(1-R)=P_b\Rightarrow P_{vc}=P_b/(1-R)=918.7/0.8466$. $\boxed{P_{vc}=1085.2\ \text{psi}}$.
(c) Surface opening and closing pressures. Subtracting the casing gas column over the 3300 ft to surface at 0.02 psi/ft: $P_{so}=P_{vo}-0.02\times3300=962-66$. $\boxed{P_{so}=896.0\ \text{psi}}$. $P_{sc}=P_{vc}-0.02\times3300=1085.2-66$. $\boxed{P_{sc}=1019.2\ \text{psi}}$.
Fig. 5 — Force balance on a casing-pressure-operated (IPO) gas-lift valve: casing pressure acts on the annular bellows area $(A_b-A_p)$, tubing pressure acts on the port area $A_p$, opposed by the nitrogen dome charge over the full bellows area $A_b$.
