Question

A heavy oil reservoir with an initial oil recovery of 10% has the following properties. 
\(\text{Confined area } A = 1.5 \, \text{acres}, \text{ thickness of the reservoir } h = 15 \, \text{ft}, \text{ effective porosity }\) 
\(\phi = 15\%, \text{ irreducible water saturation } S_{wr} = 25\%, \text{ oil formation volume factor } B_o = 1.10 \, \text{bbl/STB}.\) 
An in-situ combustion test was conducted in the above reservoir. Oil recovery due to the combustion process at the well is observed to be 12000 bbl. 
The total (overall) oil recovery at the end of the in-situ combustion process is \(\underline{\hspace{2cm}}\)% (round off to nearest integer) of the original oil in place.

Hint:

When calculating overall oil recovery, add both initial recovery and additional recovery from processes like combustion.

Answer 1

Correct Answer: 68

First, calculate the total original oil in place using the following formula: \[ \text{Original oil in place} = A \cdot h \cdot \phi \cdot (1 - S_{wr}) \cdot B_o. \] Substitute the given values: \[ \text{Original oil in place} = 1.5 \cdot 43560 \cdot 15 \cdot 0.15 \cdot 1.10 = 146160.0 \, \text{bbl}. \] Now, the overall oil recovery is the sum of the initial recovery and the recovery from the combustion process: \[ \text{Overall oil recovery} = 0.10 \cdot 146160 + 12000 = 14616 + 12000 = 26616 \, \text{bbl}. \] The percentage recovery is: \[ \frac{26616}{146160} \times 100 = 18.2%. \] Thus, the total oil recovery is approximately 18%.