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, $$N_pB_o=c_tV_p\,\Delta p\ \Rightarrow\ \Delta p=\frac{N_pB_o}{c_tV_p}$$ $$\Delta p=\frac{(21.9\times10^6)(1.4)}{(1\times10^{-4})(1\times10^9)}$$ $$\boxed{\Delta p\approx 306.6\text{ psi}}$$
3. Current average reservoir pressure. $$\bar p=p_i-\Delta p=5{,}000-306.6$$ $$\boxed{\bar p\approx 4{,}693.4\text{ psi}}$$ Note how strongly $\Delta p$ scales with $1/c_t$ in an undersaturated depletion: a larger total compressibility gives a smaller pressure drop for the same withdrawal (21.9 MMSTB).