Thermal-Method Screening and Wet-Steam Volume: 24-Pet-A7 Secondary and Enhanced Oil Recovery, May 2014, Q3

Problem 3: Thermal-Method Screening and Wet-Steam Volume (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; a steam flood at 300 psia requires 5 bbl of (cold) water flashed to 80% quality steam to displace 1 bbl of oil; at 300 psia, $v_{liq}=0.0189$ ft³/lbm, $v_{vap}=1.543$ ft³/lbm; unit conversions 1 bbl $=0.159$ m³, 1 kg $=2.205$ lb, 1 ft $=0.305$ m.

Find. (a) Why high porosity and low pressure favour thermal recovery; (b) effect of porosity on thermal-front propagation rate; (c) the volume of wet (80% quality) steam at 300 psia required to displace 1.0 bbl of oil.

Approach. (a)-(b) are qualitative heat-balance arguments from the Marx-Langenheim heated-zone model; (c) converts the mass of 5 bbl of cold water into the specific volume it occupies once flashed to 80%-quality wet steam at 300 psia, using $v_{mix}=v_{liq}+x(v_{vap}-v_{liq})$.

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 (about $577^{\circ}$F / $303^{\circ}$C at 1300 psia, versus about $417^{\circ}$F / $214^{\circ}$C at 300 psia) — 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. Thermal fronts lag the injected-fluid front because part of the heat carried by the injected fluid is left behind in the rock matrix it passes through, and porosity sets how large that rock heat sink is relative to the fluid. For a hot-fluid (condensed-steam/hot-water) front, a heat balance on a unit bulk volume gives the thermal-front velocity as a fraction of the fluid-front velocity, $$\frac{v_T}{v_f}=\left[1+\frac{(1-\phi)\,\rho_sC_s}{\phi\,\rho_wC_w}\right]^{-1}.$$ With grain $\rho_sC_s\approx2.2\times10^6$ J/m³·°C and water $\rho_wC_w\approx4.18\times10^6$ J/m³·°C, $v_T/v_f\approx0.25$ at $\phi=0.15$ but $\approx0.45$ at $\phi=0.30$. Higher porosity therefore speeds up the thermal front (per pore volume injected): less rock per unit pore volume has to be heated before the front can move on, so less of the injected heat is left behind in the matrix. Low porosity slows it: the front falls far behind the injected fluid, the heated zone grows slowly, and more heat is lost to the over- and underburden meanwhile, which is one reason the screening criterion in Part a asks for $\phi>0.25$. (At a fixed volumetric injection rate the absolute front speed is $u\rho_wC_w/M$, with $M=(1-\phi)\rho_sC_s+\phi\rho_wC_w$; that changes only weakly with porosity because the pore water is itself heat-hungry, but the pore volume heated per unit of heat injected still rises with $\phi$.)

c. Wet-steam volume to displace 1.0 bbl oil.

Convert the 5 bbl of water to a mass. Using the given conversions, $5\text{ bbl}=5\times0.159=0.795\ \text{m}^3$ of liquid water. At standard water density $\rho\approx1000\ \text{kg/m}^3$, $$m=0.795\times1000=795.0\ \text{kg}=795.0\times2.205=\boxed{1753.0\ \text{lbm}}.$$
Specific volume of the 80%-quality wet steam at 300 psia. $$v_{mix}=v_{liq}+x(v_{vap}-v_{liq})=0.0189+0.80(1.543-0.0189)=0.0189+0.80(1.5241)=\boxed{1.2382\ \text{ft}^3/\text{lbm}}.$$
Total wet-steam volume. $$V_{steam}=m\,v_{mix}=1753.0\times1.2382=2170.5\ \text{ft}^3.$$ Converting back to barrels using the same given factors ($1\ \text{ft}^3=1/(0.305)^3\times0.159^{-1}$... more directly, $1\ \text{bbl}=0.159\ \text{m}^3=0.159/0.305^3=5.604\ \text{ft}^3$), $$V_{steam}=\frac{2170.5}{5.604}=\boxed{387.3\ \text{bbl of wet steam (at 300 psia, 80\% quality) to displace 1.0 bbl of oil.}}$$

Quantity	Value
Mass of 5 bbl water	1753.0 lbm
Specific volume of 80%-quality wet steam @ 300 psia	1.2382 ft³/lbm
Total wet-steam volume	2170.5 ft³
Wet-steam volume per 1.0 bbl oil displaced	387.3 bbl

Topic: Thermal-method screening criteria, the Marx-Langenheim heated-zone growth model, and wet-steam specific volume.

Key relations: volumetric heat capacity $M=(1-\phi)\rho_rc_r+\phi\sum S_j\rho_jc_j$; wet-steam (two-phase) specific volume $v_{mix}=v_{liq}+x(v_{vap}-v_{liq})$, a mass-weighted mixing rule identical in form to the in-situ-quality relation used elsewhere in this subject.

Why this works: once water is flashed to a given quality at fixed pressure, its specific volume balloons almost entirely because of the vapor fraction's very large specific volume relative to the liquid's ($v_{vap}/v_{liq}\approx82$ at 300 psia here) — even a "wet" 80%-quality mixture occupies roughly 65 times the volume of the liquid water it was generated from, which is why steam-flood injectivity and surface generating capacity, not just heat content, are major field design constraints.

Common pitfall: converting the 5 bbl of water using the wrong reference density (e.g. the saturated-liquid density AT 300 psia, which differs slightly from standard cold-water density) instead of the intended "5 bbl of water" measured before flashing; forgetting to convert the final wet-steam volume back into the same barrel units the question asks for (an easy place to leave the answer in ft³ and misreport a much larger apparent number).

Source: Prats, Thermal Recovery, SPE Monograph Vol. 7, Ch. 2-3 (Marx-Langenheim model, steam properties); Green & Willhite, Enhanced Oil Recovery, Ch. 10 (steam flooding).

Question 3 of 4: Thermal-Method Screening and Wet-Steam Volume

Problem 3: Thermal-Method Screening and Wet-Steam Volume (25 marks)