Question 3 of 4: Thermal-Method Screening and In-Situ Steam Quality

Problem 3: Thermal-Method Screening and In-Situ Steam Quality (25 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. Screening thresholds $\phi>0.25$, $p\lt1300$ psia; typical injected steam quality 0.7; in a swept zone at $p=1.0$ MPa: $S_w=0.25$, $S_o=0.35$, $S_{steam}=0.40$; $\rho_{liq}=885$ kg/m³, $\rho_{vap}=5.31$ kg/m³.

Find. (a) Why high porosity and low pressure favour thermal recovery; (b) effect of porosity on thermal-front propagation rate; (c) in-situ steam quality, and confirmation it sits just inside the saturated-liquid line.

Approach. (a)-(b) are qualitative heat-balance arguments from the Marx-Langenheim heated-zone model; (c) is a mass-fraction (quality) calculation from the two water-bearing phase saturations and their known densities at 1.0 MPa saturation conditions.

a. Why high porosity and low pressure favour thermal methods. Thermal EOR works by injecting (or generating in-situ) heat, and its efficiency is set by how much of that heat ends up raising the temperature of the oil versus being "wasted" heating the rock matrix or lost to the surroundings. Higher porosity means a larger fraction of the swept bulk volume is pore fluid rather than solid rock, so per unit of oil-bearing rock, less rock mass has to be heated to the same temperature before the oil itself gets hot — this raises thermal (volumetric heating) efficiency and reduces the fuel/steam requirement per barrel of oil contacted. Lower reservoir pressure matters because the saturation temperature of steam falls with pressure ($T_{sat}\approx182\,{}^\circ\text{C}$ at 1300 psia vs. much lower at typical thermal-project pressures) — a low-pressure reservoir needs a lower steam temperature (hence less energy per unit mass of steam generated, and less wellbore heat loss getting it downhole) to still achieve a large viscosity-reducing temperature rise; low pressure also generally signals a reservoir that has already lost most of its primary depletion energy, making it a good economic candidate for the very energy-intensive investment thermal recovery requires.

b. Effect of porosity on thermal-front propagation rate. For a fixed rate of heat injection, the rate at which the heated (steam) zone advances is inversely proportional to the volumetric heat capacity of the rock-plus-fluid system that must be raised to steam temperature, $M=(1-\phi)\rho_r c_r+\phi(S_o\rho_oc_o+S_w\rho_wc_w+\cdots)$. Water's volumetric heat capacity ($\rho c\approx4.18\times10^6$ J/m³·°C) is roughly double a typical sandstone's ($\rho c\approx2.1\times10^6$ J/m³·°C), so higher porosity increases $M$ (more of the heat-hungry, high-heat-capacity pore fluid per unit bulk volume) and slows the thermal front for a given heat injection rate; conversely, lower porosity (more low-heat-capacity rock, less fluid to heat) lets the same heat input advance the front faster. This is the opposite sense from Part a's porosity effect on overall thermal efficiency — higher porosity is more heat-efficient per barrel of oil contacted, but the heated zone itself grows more slowly.

c. In-situ steam quality.

1. Set up the quality definition. In-situ (local) steam quality is a mass fraction of the water-only system (liquid water + steam), excluding the immiscible oil phase: $$x=\frac{m_{vapor}}{m_{vapor}+m_{liquid}}=\frac{\rho_{vap}\,S_{steam}}{\rho_{vap}\,S_{steam}+\rho_{liq}\,S_w}$$ (porosity and bulk volume cancel between numerator and denominator).
2. Substitute. $$x=\frac{5.31\times0.40}{5.31\times0.40+885\times0.25}=\frac{2.124}{2.124+221.25}=\frac{2.124}{223.37}=\boxed{0.0095\ (0.95\%).}$$
3. Confirm "just barely inside the saturated liquid line." On the H-P diagram, a saturated liquid-vapor mixture at fixed pressure sits at a horizontal position between the saturated-liquid boundary ($x=0$) and the saturated-vapor boundary ($x=1$), at fractional position $x$. Here $x=0.0095$ is only about 1% of the way from the saturated-liquid line toward the saturated-vapor line — i.e. the state point is almost entirely liquid, sitting just barely to the two-phase side of the saturated-liquid boundary, confirming the given statement quantitatively. This is far below the 0.7 quality typical of the steam as injected: nearly all of the latent heat carried by the injected steam has already condensed into the liquid phase (giving up heat to the rock and oil) by the time it reaches this point in the reservoir, leaving only a trace of the water mass still as vapor even though both phases remain present and saturated.