Question

The drainage oil-water capillary pressure data for a core retrieved from a homogeneous isotropic reservoir is listed in the table. The reservoir top is at 4000 ft from the surface and the water-oil contact (WOC) depth is at 4100 ft. \[ \begin{array}{|c|c|} \hline \textbf{Water saturation (\%)} & \textbf{Capillary pressure (psi)} \\ \hline 100.0 & 0.0 \\ \hline 100.0 & 5.5 \\ \hline 99.0 & 5.6 \\ \hline 89.2 & 6.4 \\ \hline 81.8 & 6.9 \\ \hline 44.2 & 11.2 \\ \hline 29.7 & 17.1 \\ \hline 25.1 & 36.0 \\ \hline \end{array} \] Assume the densities of water and oil at reservoir conditions are 1.04 g/cc and 0.84 g/cc, respectively. The acceleration due to gravity is 980 cm/s2. The interfacial tension between oil and water is 35 dynes/cm and the contact angle is 0 degree. The depth of free-water level (FWL) is at ________ ft (rounded off to one decimal place).

Hint:

Remember: Free water level (FWL) is above the WOC by the capillary rise distance corresponding to the measured Pc.

Answer 1

Step 1: Recall relation between capillary pressure and height above FWL.
\[ P_c = (\rho_w - \rho_o) g h \] where - \(P_c\) = capillary pressure at a given depth, - \(h\) = height above free water level (FWL), - \(\rho_w, \rho_o\) = densities of water and oil. 

Step 2: Convert densities to consistent units.
\[ \rho_w = 1.04 \, g/cc = 1.04 \times 10^3 \, kg/m^3 \] \[ \rho_o = 0.84 \, g/cc = 0.84 \times 10^3 \, kg/m^3 \] \[ \Delta \rho = \rho_w - \rho_o = 200 \, kg/m^3 \] 

Step 3: Convert capillary pressure.
Given \(P_c = 5.5 \, \text{psi}\) at 100% water saturation. 1 psi = 68950 dynes/cm\(^2\). \[ P_c = 5.5 \times 68950 = 379225 \, dynes/cm^2 \] Convert to SI: \(1 \, dyne/cm^2 = 0.1 \, Pa\). \[ P_c = 379225 \times 0.1 = 37922.5 \, Pa \] 

Step 4: Height corresponding to this capillary pressure.
\[ h = \frac{P_c}{\Delta \rho \, g} = \frac{37922.5}{200 \times 9.81} \] \[ h = \frac{37922.5}{1962} \approx 19.34 \, m \] Convert to feet: \[ h = 19.34 \times 3.28 = 63.5 \, ft \] 

Step 5: Locate FWL.
Water-oil contact (WOC) is at 4100 ft. Since capillary rise is \(h = 63.5 \, ft\), \[ FWL = WOC - h = 4100 - 63.5 = 4036.5 \, ft \] 

Step 6: Correction using interfacial tension.
Given interfacial tension = 35 dynes/cm, contact angle = 0. \[ P_c = \frac{2\sigma \cos \theta}{r} \] But the tabulated capillary data already accounts for lab measurement, so we rely on the Pc-height relation used above. 

Final Answer: \[ \boxed{4036.5 \, \text{ft}} \]