Permeability & Darcy's Law

Permeability measures a rock's ability to transmit fluids, governed by Darcy's Law.

Petroleum EngineeringPetrophysicsFree preview
⏱️ About 14 min
Permeability & Darcy's Law — illustration
Illustrative image (AI-generated).

A rock can be highly porous, like a sponge, but if those pores are completely sealed off from one another, no fluid will ever flow. The property that dictates connectivity and flow is permeability.

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The big idea: Permeability ($k$) is the measure of a rock's ability to transmit fluids, quantified by Darcy's Law.
🎯 By the end, you'll be able to
  • Define absolute permeability and its standard units (darcy, millidarcy)
  • Apply Darcy's Law in lab units to solve for permeability
  • Explain the distinction between porosity and permeability
  • Recognize the role of unit constants in field-scale equations
📎 Helpful to know first

Defining Permeability

Permeability ($k$) is a property of the porous medium that measures the ease with which fluids can flow through its interconnected pore network. The standard unit is the darcy, though most reservoir rocks have permeabilities measured in millidarcies (md). A rock with high porosity does not necessarily have high permeability; a rock can be porous but impermeable if the pores are isolated.

✨ Darcy's Law in the Lab

In laboratory Darcy units, the proportionality constant is exactly 1. Darcy's Law is written linearly as $q = (k \cdot A \cdot \Delta p) / (\mu \cdot L)$, where $q$ is flow rate, $A$ is cross-sectional area, $\Delta p$ is pressure drop, $\mu$ is fluid viscosity, and $L$ is length. In field units, a conversion constant is required, but the fundamental relationship remains the same.

📝 Worked example: A core plug is tested in the lab. Water (viscosity $\mu = 1.0$ cp) flows through the plug at a rate $q = 1.0$ cm³/s. The core has a cross-sectional area $A = 2.0$ cm², length $L = 3.0$ cm, and experiences a pressure drop $\Delta p = 2.0$ atm. Calculate the absolute permeability in darcies.
  1. Rearrange Darcy's Law to solve for $k$: $k = (q \cdot \mu \cdot L) / (A \cdot \Delta p)$.
  2. Substitute the given lab values: $k = (1.0 \cdot 1.0 \cdot 3.0) / (2.0 \cdot 2.0)$.
  3. Calculate the numerator: $1.0 \times 1.0 \times 3.0 = 3.0$.
  4. Calculate the denominator: $2.0 \times 2.0 = 4.0$.
  5. Divide to find $k$: $k = 3.0 / 4.0 = 0.75$ darcy.
  6. Convert to millidarcies: $0.75 \times 1000 = 750$ md.
✓ The absolute permeability of the core is 0.75 darcy, or 750 md.

Check your understanding

1. Which statement correctly contrasts porosity and permeability?
Porosity is the void fraction (storage), whereas permeability is the connectivity of those voids allowing fluid transmission (flow).
2. Using Darcy's Law in lab units, if $q=1.0$ cm³/s, $A=2.0$ cm², $\Delta p=2.0$ atm, $\mu=1.0$ cp, and $L=3.0$ cm, what is $k$?
$k = (1.0 \times 1.0 \times 3.0) / (2.0 \times 2.0) = 0.75$ darcy, which equals 750 md.
✅ Key takeaways
  • Permeability quantifies the ability of a rock to transmit fluids through interconnected pores.
  • In lab Darcy units, $k = (q \cdot \mu \cdot L) / (A \cdot \Delta p)$ with a constant of 1.
  • A rock can have high porosity but zero effective permeability if pores are unconnected.
➡️ Having established how much fluid a rock can hold and how easily it flows, we now examine how different fluids share that pore space.