Question

The Inflow Performance Relationship (IPR) for a vertical well in a single-phase oil reservoir was found to be linear. The flowing bottomhole pressures are 4000 psi and 1000 psi at oil flow rates of 200 STB/day and 600 STB/day, respectively.
At the flowing bottomhole pressure of 2875 psi, the value of flow rate (in STB/day) is __________. (Answer in integer)
[STB: Stock Tank Barrel]

Hint:

Fit a straight line between the two given flow rate and pressure pairs, then read off the flow rate at 2875 psi.

Answer 1

Correct Answer: 350

Step 1: Define the productivity index for a linear IPR:
For a linear inflow performance relationship, the productivity index J is defined as the change in flow rate per unit drop in flowing bottomhole pressure, $J = \frac{\Delta q}{\Delta p_{wf} \text{(drop)}} = \frac{q_2 - q_1}{p_{wf1} - p_{wf2}}$, using the pressure drop as a positive quantity since flow rate increases as pressure decreases.
Step 2: Compute J from the given data:
Using $q_1 = 200$ STB/day at $p_{wf1} = 4000$ psi and $q_2 = 600$ STB/day at $p_{wf2} = 1000$ psi, $J = \frac{600 - 200}{4000 - 1000} = \frac{400}{3000} = 0.13333$ STB/day per psi.
Step 3: Write the IPR using the productivity index form:
The linear IPR can be written as $q = q_1 + J\,(p_{wf1} - p_{wf})$, which says the flow rate increases above the reference rate $q_1$ in proportion to how much the flowing pressure has dropped below the reference pressure $p_{wf1}$.
Step 4: Substitute the numbers for the target pressure:
At $p_{wf} = 2875$ psi, the pressure drop from the reference is $p_{wf1} - p_{wf} = 4000 - 2875 = 1125$ psi. So $q = 200 + 0.13333 \times 1125 = 200 + 150 = 350$ STB/day.
Step 5: Confirm using the second reference point:
Using $q_2 = 600$ at $p_{wf2} = 1000$ instead, $q = 600 + J\,(p_{wf2} - p_{wf}) = 600 + 0.13333\,(1000 - 2875) = 600 + 0.13333 \times (-1875) = 600 - 250 = 350$ STB/day, which agrees with the earlier result.
Final Answer:
\[ \boxed{350 \text{ STB/day}} \]