Question 2 of 7: Capillary Pressure, Solution GOR, and Core Saturation Volumes

Question 2: Capillary Pressure, Solution GOR, and Core Saturation Volumes (20 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.

Find. (1) gas-phase pressure $P_g$; (2) initial solution gas-oil ratio $R_{si}$; (3) oil volume in the core; (4) core porosity from the mass-imbibition method.

Approach. Each sub-part is a direct application of a single governing relation: capillary pressure across a curved interface, the linear solubility of gas in an under-saturated oil below its bubble point, saturation-weighted pore volume, and the mass-balance definition of porosity.

1. (1) Capillary pressure and gas-phase pressure. Water is the wetting phase (small contact angle), so the meniscus curves toward the gas and $P_c=P_g-P_w=\dfrac{2\sigma\cos\theta}{r}$, with $r=d/2=5\times10^{-6}$ m: $P_c=\dfrac{2(0.072)\cos15^\circ}{5\times10^{-6}}=27{,}819$ Pa $=27.82$ kN/m². Then $\boxed{P_g=P_w+P_c=110+27.82=137.8\ \text{kN/m}^2}$.
2. (2) Initial solution GOR. Above the bubble point the oil is a single liquid phase, so $R_s$ is constant at its bubble-point value; taking the standard linear solubility model ($R_s=0$ at atmospheric pressure, rising linearly to $R_{sb}$ at $p_b$) with the given slope, $\boxed{R_{si}=R_{sb}=\left(\dfrac{dR_s}{dp}\right)p_b=0.25(2{,}500)=625\ \text{SCF/STB}}$.
3. (3) Oil volume in the core. With only oil and water present, $S_o=1-S_w=1-0.17=0.83$: $V_o=\phi\,V_b\,S_o=0.20(50)(0.83)$, so $\boxed{V_o=8.3\ \text{cm}^3}$.
4. (4) Porosity by mass imbibition. Pore volume equals the mass of water imbibed divided by water density: $V_p=\dfrac{m_{sat}-m_{dry}}{\rho_w}=\dfrac{167.7-153.0}{1.00}=14.7\ \text{cm}^3$. Bulk volume $V_b=\pi r^2 L=\pi(1.25)^2(15)=73.63\ \text{cm}^3$. So $\boxed{\phi=V_p/V_b=14.7/73.63=0.1996\approx20.0\%}$.