The Workflow

Part 4, Part 4: Gassmann and Fluid Substitution

Learning objectives

Explain why a logged reservoir gives you the saturated rock, not the dry frame Gassmann's forward equation wants
Run fluid substitution as three steps: strip the in-situ fluid out, hold the frame fixed, put the new fluid in
Recover the dry-frame bulk modulus with the Gassmann inverse and confirm the round trip is exact
Follow K through the three-panel pipeline as the fluid is swapped, starting always from the brine leg

You Never Have the Dry Frame

The equation of Part 4.1 asks for the dry frame, and the subsurface refuses to hand it over. A well log measures the rock as it is, full of its in-situ fluid: it gives you $V_P$ P, $V_S$ S, and density for the saturated rock, and from those with the porosity you can back out the saturated bulk modulus $K_{sat}$ sat and the shear modulus $\mu$ . What you do not have is $K_{dry}$ dry. Gassmann's forward equation is written the inconvenient way for this: it takes $K_{dry}$ dry in and gives $K_{sat}$ sat out. So the first practical move is to run it backward.

Three Steps

Every fluid substitution is the same three-step move. One, strip the in-situ fluid out. The Gassmann equation solved for $K_{dry}$ dry is its exact algebraic inverse; feed that inverse the in-situ $K_{sat}$ sat, the mineral $K_{min}$ min, the in-situ fluid modulus $K_{fl}$ fl, and the porosity, and it returns the one dry-frame bulk modulus consistent with them. For the sand, the in-situ brine $K_{sat} = 12.35$ sat=12.35 GPa inverts cleanly to $K_{dry} = 6.0$ dry=6.0 GPa. Two, hold the frame fixed. The recovered $K_{dry}$ dry and the shear modulus $\mu$ are properties of the solid skeleton, and swapping the pore fluid changes neither, so freeze them (the shear modulus never moved in the first place). Three, put the new fluid in. Run the forward equation with the new fluid's modulus, then rebuild the density with the new fluid and recompute the velocities. Gas gives $K_{sat} = 6.12$ sat=6.12 GPa; oil, a middle fluid, gives 8.26. That whole trip, saturated to dry to re-saturated, is fluid substitution.

The Round Trip Is Exact

There is a clean check that the machinery is sound: substitute a fluid back for itself. Take the brine rock, strip it to the dry frame at $K_{dry} = 6.0$ dry=6.0, then put brine straight back in. $K_{sat}$ sat returns to 12.35 exactly, because the forward step and the inverse step are one equation solved two ways. The workflow adds no error of its own; every uncertainty it carries comes in through its inputs, which is the subject of Part 4.4. And the leg it starts from is almost always brine. Most reservoirs are water-wet, so the brine-saturated rock is the natural in-situ state (Part 3.2), and the study asks what a hydrocarbon would do to it. The dry frame is an intermediate you pass through, never a state the rock is actually in.

Run the pipeline in the figure and watch the bulk modulus move: 12.35 with brine, down to 6.0 stripped dry, and back up to whatever the new fluid supplies, 6.12 for gas, 8.26 for oil. The bulk modulus is doing all the visible work here, but a real substitution moves the density and both velocities too, and they do not move together. The next section reads the full anatomy of the change: what moves, what stays put, and why the pattern is the signature that finds gas.

References

Smith, T. M., Sondergeld, C. H., & Rai, C. S. (2003). Gassmann fluid substitutions: A tutorial. Geophysics, 68(2), 430-440.
Mavko, G., Mukerji, T., & Dvorkin, J. (2009). The Rock Physics Handbook (2nd ed.). Cambridge University Press.