Question 2 of 8

Question 2 (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. Steady, incompressible linear (Darcy) flow updip through a tilted bed, with the outlet cross-section as the representative flow area.

Find. The outlet pressure $p_2$.

Approach. Apply Darcy's law for tilted linear flow, which combines a viscous pressure-drop term with a hydrostatic (gravity) term because the flow is updip.

1. Fluid specific gravity and flow area. $\gamma=\rho/62.4=44/62.4=0.7051$ (relative to water). Using the given outlet face as the (uniform) flow area, $A=W\times h=500\times30=15{,}000\text{ ft}^2$.
2. Darcy's law for tilted linear flow. $$q=\frac{1.127\times10^{-3}\,kA}{\mu L}\Big[(p_1-p_2)-0.433\gamma L\sin\theta\Big]$$ Rearranging for $p_2$: $$p_2=p_1-\underbrace{\frac{q\mu L}{1.127\times10^{-3}kA}}_{\text{viscous drop}}-\underbrace{0.433\gamma L\sin\theta}_{\text{elevation (gravity) term}}$$
3. Viscous pressure drop. $$\Delta p_{visc}=\frac{(5)(2)(2{,}500)}{1.127\times10^{-3}(60)(15{,}000)}\approx 24.6\text{ psi}$$
4. Elevation (gravity) term. $$\Delta p_{grav}=0.433(0.7051)(2{,}500)\sin(7^\circ)\approx 93.0\text{ psi}$$ This is the larger of the two terms – the well must overcome most of the pressure difference simply to lift the oil up the $7^\circ$ dip, since the flow rate (5 bbl/day through a 15,000 ft² face) is very slow.
5. Outlet pressure. $$p_2=2{,}100-24.6-93.0$$ $$\boxed{p_2\approx 1{,}982.3\text{ psi}}$$