24-Pet-A6 Well Logging and Formation Evaluation, May 2017, Q3

Question 3 (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 five-layer column with permeability and depth as tabulated, a family of lab capillary pressure–saturation curves (one per layer, keyed by permeability), and the reservoir's water-oil contact.

Given data
$\rho_w$	66 lb/ft³	$\rho_o$	35 lb/ft³
WOC	6,070 ft	Depth asked	6,030 ft

Fig. Q3 – five-layer column with depth, permeability, and the WOC/FWL datum.

Find. (a) the FWL; (b) $S_w$ at 6,030 ft (which falls in the K=100 md layer, 6,025–6,040 ft).

Approach. Use the universal (rock-independent) $P_c$–height relation to relate FWL and WOC, then convert the height above FWL at 6,030 ft into a $P_c$ value and read the corresponding $S_w$ off the K=100 md curve.

Capillary-pressure gradient. $$P_c=\frac{\rho_w-\rho_o}{144}\,h=\frac{66-35}{144}\,h=0.2153\,h\ \text{(psi, }h\text{ in ft above FWL)}$$
Free Water Level. By definition $P_c=0$ exactly at $S_w=100\%$ (no oil–water interface curvature once the pore space is fully water-saturated) – that is precisely the definition of the FWL. The given WOC (6,070 ft) is the depth at which the well shows $S_w=100\%$, so the two coincide here: $$\boxed{FWL=6{,}070\text{ ft}}$$ (In general FWL sits below an apparent contact picked at some $S_w\lt100\%$, e.g. a "last oil show" pick in a tight streak – but this problem defines WOC via full water saturation, so no offset applies.)
Height above FWL at 6,030 ft and the corresponding $P_c$. $$h=FWL-6{,}030=6{,}070-6{,}030=40\text{ ft}$$ $$P_c=0.2153\times40\approx 8.61\text{ psi}$$
Read $S_w$ from the applicable layer curve. Depth 6,030 ft lies in the 6,025–6,040 ft interval, i.e. the K=100 md layer. Reading the K=100 md capillary curve at $P_c\approx8.6$ psi: $$\boxed{S_w(6{,}030\text{ ft})\approx 40\%}$$

Question 3 – final results
Quantity	Value
$P_c$ gradient	0.2153 psi/ft
Free Water Level	6,070 ft
$P_c$ at 6,030 ft	8.61 psi
$S_w$ at 6,030 ft (K=100 md curve)	≈ 40%

Check – the $S_w$ reading in step 4 is taken graphically off the printed lab chart (K=100 md curve at $P_c\approx8.6$ psi); ordinary chart-reading tolerance of a few percentage points applies, consistent with the resolution of the original grid.

Topic: Free water level and the capillary transition zone in a layered reservoir.

Key relations: $P_c=(\rho_w-\rho_o)h/144$ (universal, rock-independent height relation); $S_w$ at any depth comes from the LOCAL rock-type's own $P_c$–$S_w$ curve evaluated at that depth's $P_c$.

Why this works: The oil–water hydrostatic gradient sets $P_c$ purely as a function of height above FWL, regardless of rock type; only the mapping from $P_c$ to $S_w$ is rock-dependent (through pore-throat size, i.e. permeability). Tighter rock (lower K) has a larger irreducible water saturation and a taller transition zone for the same $P_c$.

Common pitfall: Applying one layer's $P_c$–$S_w$ curve to a different layer's depth interval, or confusing an apparent (log-resolution) WOC with the true FWL when they are not defined at exactly $S_w=100\%$.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 4 (capillary pressure and fluid distribution); Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 2.

Question 3 of 8

Question 3 (10 Marks)