Capillary Pressure & Saturation-Height

How the curved interface between oil and water in a pore creates a pressure jump - and a transition zone above the free-water level.

Petroleum EngineeringMultiphase FlowFree preview
⏱️ About 16 min
Capillary Pressure & Saturation-Height — illustration
Illustrative image (AI-generated).

Why does oil float above water in a reservoir, with a fuzzy transition zone between them? The answer is capillary pressure.

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The big idea: Capillary pressure is the pressure jump across the curved oil-water interface in a pore; it scales as $2\sigma\cos\theta/r$ and builds the transition zone above the free-water level.
🎯 By the end, you'll be able to
  • Define capillary pressure as the pressure difference across a curved oil-water interface
  • Apply $P_c = 2\sigma\cos\theta/r$ and explain why smaller pores give higher capillary pressure
  • Distinguish drainage from imbibition and describe the capillary-pressure hysteresis loop
  • Use the saturation-height relation $h = P_c/(\Delta\rho\, g)$ to locate the transition zone above free water
📎 Helpful to know first

Capillary Pressure in a Pore

At a curved oil-water interface inside a pore, the pressure is not the same on both sides. Capillary pressure is that pressure difference, $P_c$, defined as the pressure in the non-wetting phase minus the pressure in the wetting phase. In a water-wet oil-water system oil is the non-wetting phase, so $P_c$ is positive. For an idealized cylindrical pore throat of radius $r$, the Young-Laplace result gives $P_c = 2\sigma\cos\theta / r$, where $\sigma$ is the oil-water interfacial tension and $\theta$ is the contact angle. The key dependence is on $r$: smaller pores have higher capillary pressure, which is why capillary forces dominate in tight rock.

Drainage, Imbibition, and Hysteresis

Saturation changes by two opposite processes. In drainage, the non-wetting phase (oil) displaces the wetting phase (water) - saturation moves toward lower water saturation, and capillary pressure rises. In imbibition, the wetting phase (water) spontaneously displaces oil, and capillary pressure falls. The two paths do not retrace each other: the $P_c$-versus-saturation curve traced during drainage sits above the one traced during imbibition. This path dependence is called hysteresis, and it means a rock's capillary behaviour depends on its saturation history.

✨ Saturation-Height and the Leverett J-Function

Capillary pressure has a height equivalent. Balancing capillary pressure against the density contrast between water and oil gives the saturation-height relation $h = P_c/(\Delta\rho\, g)$, where $\Delta\rho$ is the water-minus-oil density contrast and $g$ is gravity. It says how far above the free-water level a given capillary pressure lifts the oil-water interface - and so it defines the transition zone, the interval over which water saturation grades from 100% toward connate. To compare capillary pressures across rocks of different porosity and permeability, the Leverett J-function normalises $P_c$ into a dimensionless curve - a standard way to average capillary-pressure data across a field.

📝 Worked example: (a) An oil-water interface has interfacial tension $\sigma = 0.025$ N/m, contact angle $\theta = 0$ degrees (so $\cos\theta = 1$), and pore radius $r = 5$ microns ($5 \times 10^{-6}$ m). Compute the capillary pressure $P_c = 2\sigma\cos\theta/r$. (b) At a point in the transition zone the capillary pressure is $P_c = 20000$ Pa (20 kPa). With water density $1000$ kg/m$^3$, oil density $800$ kg/m$^3$ (so $\Delta\rho = 200$ kg/m$^3$) and $g = 9.81$ m/s$^2$, find the height $h = P_c/(\Delta\rho\, g)$ above the free-water level.
  1. (a) Capillary pressure: $P_c = 2\sigma\cos\theta/r = 2 \times 0.025 \times 1 / (5 \times 10^{-6})$.
  2. Numerator: $2 \times 0.025 \times 1 = 0.05$.
  3. Divide: $P_c = 0.05 / (5 \times 10^{-6}) = 10000$ Pa $= 10$ kPa.
  4. (b) Saturation-height: $h = P_c/(\Delta\rho\, g) = 20000/(200 \times 9.81)$.
  5. Denominator: $200 \times 9.81 = 1962$.
  6. Divide: $h = 20000/1962 = 10.2$ m above the free-water level.
✓ (a) $P_c = 10000$ Pa $= 10$ kPa. (b) The point sits about $10.2$ m above the free-water level, within the oil-water transition zone.

Check your understanding

1. The capillary pressure in a single tube is $P_c = 2\sigma\cos\theta/r$. With everything else fixed, how does capillary pressure change as the pore radius $r$ gets smaller?
$P_c$ is inversely proportional to $r$, so smaller pores give a higher capillary pressure - which is why capillary forces dominate in tight rock.
2. An oil-water transition zone has $P_c = 20000$ Pa, water at $1000$ kg/m$^3$, oil at $800$ kg/m$^3$ ($\Delta\rho = 200$ kg/m$^3$), and $g = 9.81$ m/s$^2$. Using $h = P_c/(\Delta\rho\, g)$, what is the height above the free-water level?
$h = 20000/(200 \times 9.81) = 20000/1962 = 10.2$ m above the free-water level.
✅ Key takeaways
  • Capillary pressure is the pressure jump across a curved oil-water interface: $P_c = 2\sigma\cos\theta/r$, so smaller pores mean higher $P_c$.
  • Drainage (oil entering) and imbibition (water entering) trace a hysteresis loop - the curve depends on saturation history.
  • Saturation-height $h = P_c/(\Delta\rho\, g)$ places the transition zone above the free-water level; the Leverett J-function averages $P_c$ across a field.
➡️ Capillary pressure fixes the saturation at which each phase can flow - which is exactly what relative permeability curves describe next.