Question

A volumetric undersaturated solution gas drive reservoir (without gas cap, no water influx, and with no initial gas saturation) has an initial water saturation of $15%$ which remains unchanged during production. After the production of $10%$ of the initial oil (measured at surface conditions), the oil formation volume factor $B_o$ is reduced from its initial value of $1.4~\text{bbl/STB$ to $1.2~\text{bbl/STB}$. The final gas saturation in percentage is _________ (rounded to one decimal place).}

Hint:

For undersaturated solution-gas drive with constant $S_w$, use $N=\dfrac{S_o\,PV}{B_o}$. A given % oil produced at surface translates to a relation between $S_{o,f}$ and $B_{o,f}$, from which $S_g$ follows since $S_o+S_w+S_g=1$.

Answer 1

Data: $S_{w}=0.15$ (constant), initial $S_{g0}=0$, so $S_{o0}=1-0.15-0=0.85$. $B_{o,i}=1.4$, $B_{o,f}=1.2$. $N_p=0.10\,N_{0}$ (STB).
Step 1: Express initial and final stock-tank oil.
Initial STB in place: $N_0=\dfrac{S_{o0}\,PV}{B_{o,i}}=\dfrac{0.85\,PV}{1.4}$.
Final oil saturation $S_{o,f}=1-S_w-S_{g,f}=0.85-S_g$.
Final STB in place: $N_f=\dfrac{(0.85-S_g)\,PV}{1.2}$.
Step 2: Use the given $10%$ production.
$N_p=N_0-N_f=0.10\,N_0$ \Rightarrow\ $N_f=0.90\,N_0$.
Thus, \[ \dfrac{(0.85-S_g)\,PV}{1.2}=0.90\left(\dfrac{0.85\,PV}{1.4}\right) \Rightarrow 0.85-S_g=1.2\times 0.90 \times \dfrac{0.85}{1.4}=0.655714\dots \] \[ \Rightarrow\ S_g=0.85-0.655714\dots=0.194286\dots \] Step 3: Convert to percentage.
$S_g(%)=0.194286\times 100=19.4286%\ \Rightarrow\ 19.4%$ (one decimal place).