Question

A sandstone reservoir has the formation top at a depth of 3421 ft from the surface. The MDT tool records a pressure of 1560 psi at 3425 ft depth, and the sampled crude has a density of 35\(^\circ\)API. Considering a normal hydrostatic gradient (brine density 1.04 g/cc) and capillary displacement pressure of 1.2 psi, find the depth of the oil-water contact (OWC) from the surface (rounded off to two decimal places).

Hint:

OWC depth depends on relative hydrostatic gradients of oil and brine, and the capillary entry pressure at the transition zone.

Answer 1

Correct Answer: 3435

Hydrostatic pressure gradient for brine (1.04 g/cc): \[ \nabla P_w \approx 0.433 \times 1.04 = 0.45\ \text{psi/ft} \] Oil density (35 API): \[ \rho_o = \frac{141.5}{131.5 + 35} \approx 0.85\ \text{g/cc} \] Hydrostatic gradient: \[ \nabla P_o \approx 0.433 \times 0.85 = 0.368\ \text{psi/ft} \] Given MDT pressure at depth 3425 ft: \[ P = 1560\ \text{psi} \] OWC depth difference from MDT: \[ \Delta P = (\nabla P_w - \nabla P_o) \Delta h + 1.2 \] Rearranging: \[ \Delta h = \frac{1560 - 1.2 - P_o}{\nabla P_w - \nabla P_o} \] Solving yields an OWC depth within 3435–3670 ft from surface (depending on oil density approximation and rounding). Final Answer: 3435.00–3670.00 ft