Question 1 of 5

Question 1 (10 Marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A closed (no-influx), undersaturated (single liquid phase) reservoir producing entirely by rock-and-fluid expansion.

Find. The current average reservoir pressure $\bar p$.

Approach. For a closed, undersaturated liquid reservoir, cumulative reservoir-barrel withdrawal is balanced entirely by rock+fluid expansion, giving a direct depletion material balance.

1. Cumulative production. $$N_p=q\times t=10{,}000\times(365\times6)=21{,}900{,}000\text{ STB} = 21.9\text{ MMSTB}$$
2. Compressibility-drive material balance. For an undersaturated, volumetric reservoir, $$N_pB_o=c_tV_p\,\Delta p\ \Rightarrow\ \Delta p=\frac{N_pB_o}{c_tV_p}$$ $$\Delta p=\frac{(21.9\times10^6)(1.4)}{(1\times10^{-5})(1\times10^9)}$$ $$\boxed{\Delta p\approx 3{,}066\text{ psi}}$$
3. Current average reservoir pressure. $$\bar p=p_i-\Delta p=5{,}000-3{,}066$$ $$\boxed{\bar p\approx 1{,}934\text{ psi}}$$ This large pressure drop for a comparatively modest withdrawal (about 2.2% of the pore volume, in STB terms) illustrates why undersaturated (single-liquid-phase) reservoirs are so pressure-sensitive: with no free gas or aquifer to buffer voidage, every barrel withdrawn must come from a very small total-compressibility expansion.