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: Write the linear IPR equation:
Since the IPR is linear, the flow rate q and the flowing bottomhole pressure $p_{wf}$ are related by a straight line of the form $q = a + b\,p_{wf}$, where a and b are constants to be found from the two given data points.
Step 2: Find the slope using the two data points:
The two points are $(p_{wf} = 4000, q = 200)$ and $(p_{wf} = 1000, q = 600)$. The slope is $b = \frac{q_2 - q_1}{p_{wf2} - p_{wf1}} = \frac{600 - 200}{1000 - 4000} = \frac{400}{-3000} = -0.13333$ STB/day per psi.
Step 3: Write the line equation using one known point:
Using the point $(4000, 200)$, the line is $q = 200 + b\,(p_{wf} - 4000) = 200 - 0.13333\,(p_{wf} - 4000)$.
Step 4: Substitute the target pressure:
At $p_{wf} = 2875$ psi, $p_{wf} - 4000 = 2875 - 4000 = -1125$. So $q = 200 - 0.13333 \times (-1125) = 200 + 150 = 350$.
Step 5: Check with the other point:
As a check, at $p_{wf} = 1000$, $q = 200 - 0.13333 \times (1000 - 4000) = 200 - 0.13333 \times (-3000) = 200 + 400 = 600$, which matches the given second data point, confirming the line equation is correct.
Final Answer:
\[ \boxed{350 \text{ STB/day}} \]