Gassmann fluid substitution: predicting the seismic response of fluid changes

The central insight

Replacing one pore fluid with another changes the rock’s saturated bulk modulus $K_{\mathrm{sat}}$: a stiff fluid such as brine resists compression and stiffens the rock, a soft one such as gas hardly does. The saturated shear modulus does not change. A fluid has no shear stiffness, so swapping one fluid for another leaves the rock’s resistance to shearing exactly as the dry frame sets it, $\mu_{\mathrm{sat}} = \mu_{\mathrm{dry}}$.

That second statement is what lets pre-stack seismic separate fluid from lithology. A change of fluid moves $K$ and $\rho$ but not $\mu$; a change of rock moves all three. Section 5.4 shows how the change of reflection amplitude with angle reads the difference.

Gassmann’s equation

For a rock of porosity $\phi$ whose solid has bulk modulus $K_{\mathrm{min}}$, whose empty (dry) frame has bulk modulus $K_{\mathrm{dry}}$, and whose pores hold a fluid of modulus $K_{\mathrm{fl}}$:

The saturated modulus is the dry frame’s plus a fluid term. The fluid term is large when $K_{\mathrm{fl}}$ is large and nearly vanishes for gas. Its numerator, $(1 - K_{\mathrm{dry}}/K_{\mathrm{min}})^{2}$, measures how compliant the frame is against its own mineral: a soft frame leaves the fluid much to do, a stiff one little.

The equation behaves correctly at its limits. A fluid with no stiffness gives back $K_{\mathrm{dry}}$; a fluid as stiff as the mineral, or a porosity of zero, gives $K_{\mathrm{min}}$; and a frame with no stiffness at all gives Wood’s equation for grains suspended in fluid. Figure 5.3’s implementation is checked against all four.

Density and velocity

Density is a volume average, $\rho = (1-\phi)\,\rho_{\mathrm{min}} + \phi\,\rho_{\mathrm{fl}}$, with $\rho_{\mathrm{fl}} = S_{\mathrm{w}}\rho_{\mathrm{b}} + (1 - S_{\mathrm{w}})\rho_{\mathrm{g}}$ for brine at water saturation $S_{\mathrm{w}}$ and gas in the rest of the pore space. The velocities follow from the moduli:

• $V_{\mathrm{P}}$ falls with gas. $K_{\mathrm{sat}}$ falls a long way and $\rho$ a little; the modulus wins.
• $V_{\mathrm{S}}$ rises with gas. $\mu$ is unchanged and $\rho$ falls, so $V_{\mathrm{S}}$ goes up by the square root of the density ratio, a few percent.
• $\rho$ falls in proportion to the gas, however the fluids are arranged.
• $V_{\mathrm{P}}/V_{\mathrm{S}}$ falls. Since $(V_{\mathrm{P}}/V_{\mathrm{S}})^{2} = K_{\mathrm{sat}}/\mu + \tfrac{4}{3}$, it follows $K_{\mathrm{sat}}$ alone; the density cancels. In the clean sand of Figure 5.3 it goes from 1.82 in brine to 1.49 in gas.

The fluids, and how they share the pores

Fluid moduli depend strongly on pressure and temperature. Figure 5.3 takes them from Batzle and Wang (1992). Some later sources print $-1820$ for the salinity-squared term of their brine velocity; the site uses the paper’s $-820$, which fits measured high-salinity data better. At 2 km of burial (20 MPa, 75 °C), brine of 100,000 ppm NaCl has $K_{\mathrm{b}} = 2.98$ GPa and $\rho_{\mathrm{b}} = 1.054$ g/cm³, and methane-rich gas of specific gravity 0.6 has $K_{\mathrm{g}} = 0.0406$ GPa and $\rho_{\mathrm{g}} = 0.132$ g/cm³. Gas is stiffer and denser at depth: its modulus is 0.017 GPa at 1 km and 0.066 GPa at 3 km.

When the two fluids share every pore finely enough for their pressures to equalise within a wave cycle, the rock feels the Wood (Reuss) average of the fluids. This is uniform saturation:

The soft gas dominates the sum. With 5 % gas, $1/K_{\mathrm{fl}} = 0.95/2.98 + 0.05/0.0406 = 0.32 + 1.23 = 1.55$ GPa$^{-1}$, so $K_{\mathrm{fl}}$ $= 0.65$ GPa, a fifth of brine’s modulus; with 10 % gas it is 0.36 GPa, an eighth.

When the gas sits instead in patches larger than the distance over which pore pressure can equalise during a wave cycle, each patch responds as a fully brine-saturated or fully gas-saturated rock. The stiffest the rock can then be at low frequency is the patchy limit: Gassmann applied to each patch, and the P-wave moduli $M = K_{\mathrm{sat}} + \tfrac{4}{3}\mu$ averaged harmonically, $1/M = S_{\mathrm{w}}/M_{\mathrm{b}} + (1 - S_{\mathrm{w}})/M_{\mathrm{g}}$ (Mavko and Mukerji 1998). Real rocks lie between the two limits. Brie et al. (1995) proposed an empirical law, $K_{\mathrm{fl}} = (K_{\mathrm{b}} - K_{\mathrm{g}})\,S_{\mathrm{w}}^{e} + K_{\mathrm{g}}$, often used with $e \approx 3$ for gas and water in sands, where in Figure 5.3 it stays within about 2 m/s of the band between the limits. It interpolates; it does not bound. At $e = 1$ it becomes the Voigt (arithmetic) average of the fluids, which makes the rock stiffer even than the patchy limit, and at large $e$ its fluid modulus falls below Wood’s average, softer than any mixture of brine and gas can be, so the rock it predicts is softer than uniform mixing allows.

Reading Figure 5.3

• The figure opens on the clean sand, all brine, with $V_{\mathrm{P}}$ $= 3236$ m/s. Plate (a) already draws the whole story: under uniform mixing the curve drops almost vertically as the first gas enters, while the patchy limit declines gently.
• At $S_{\mathrm{w}}$ $= 0.95$, 5 % gas mixed through every pore takes 418 m/s off $V_{\mathrm{P}}$, 90 % of the 463 m/s that full gas takes. At $S_{\mathrm{w}}$ $= 0.9$ the fall is 481 m/s, more than full gas: the fluid’s share of the modulus in plate (c) has nearly gone, but the brine still in the pores keeps the rock heavy. The lowest $V_{\mathrm{P}}$, 2718 m/s, comes at $S_{\mathrm{w}}$ $\approx 0.70$; past it the modulus has little left to lose (the P-wave modulus falls only another 1.8 %, the density 5.7 %), and $V_{\mathrm{P}}$ rises again to 2773 m/s.
• Held in patches, the same 10 % gas costs only 62 m/s.
• With plate (a) on $V_{\mathrm{S}}$ or $\rho$ there is one curve, the same for every mixing law. $V_{\mathrm{S}}$ rises 75 m/s (4 %) from brine to gas; the density falls 0.184 g/cm³ in a straight line.
• With plate (a) on $V_{\mathrm{P}}/V_{\mathrm{S}}$, most of its fall under uniform mixing happens in the first tenth of gas, from 1.82 to 1.54, against 1.49 in full gas; in patches the first tenth takes it only to 1.77.
• The shaley sand, whose frame is softer still, loses 600 m/s to full gas. The limestone loses 250 m/s, 5 % of its $V_{\mathrm{P}}$: brine adds about as much stiffness as in the clean sand (8.8 against 8.0 GPa), but its stiff frame and large shear term make that a small share of its P-wave modulus.

Why a little gas does so much

The fall in $V_{\mathrm{P}}$ follows the fluid modulus, not the saturation. The Reuss average is controlled by its softest member, so $K_{\mathrm{fl}}$ collapses with the first few percent of gas and Gassmann’s fluid term collapses with it, while the density follows the saturation in a straight line. A few percent of gas, too little to produce (fizz water), can therefore make an amplitude anomaly almost as strong as a commercial gas column: a classic false positive among direct hydrocarbon indicators. The density, which falls only 5 % of the way for 5 % gas, is the better witness to saturation, and estimating it needs reflections at far angles (Section 5.4).

Read the other way, a fall in $V_{\mathrm{P}}$ does not fix the saturation either. In the clean sand at 2 km, a fall of 400 m/s means about 4 % gas if the gas is finely mixed and about 82 % if it is in patches.

Gassmann’s assumptions

• Low frequency. Pore pressure must equalise through the pore space within a wave cycle. Seismic frequencies usually allow this; sonic logs are nearer the edge, and ultrasonic laboratory measurements (around 1 MHz) do not, so wave-induced flow makes rocks measured there stiffer than Gassmann predicts.
• Connected pores. Fluid must move freely between pores. Isolated vugs, moldic pores and crack-like pores break this, one reason carbonates are hard.
• One mineral modulus. The solid is treated as homogeneous, with a single $K_{\mathrm{min}}$. A mixed mineralogy needs an averaged value (the shaley sand of Figure 5.3 uses the Hill average of 80 % quartz and 20 % clay, 32.6 GPa), and the result is sensitive to it.
• Isotropy and a closed system. The rock is isotropic, no fluid flows into or out of the rock volume during a wave cycle, and the fluid neither softens nor reacts with the frame. Anisotropic rocks such as shales need the general form of Brown and Korringa (1975).
• A mixing law. Uniform mixing is assumed unless the patchy limit or an empirical law such as Brie’s replaces it, and the choice can matter as much as the fluid itself.

Common pitfalls

• The dry frame has to come from somewhere. In practice $K_{\mathrm{dry}}$ is found by inverting Gassmann from a logged brine-saturated rock ($V_{\mathrm{P}}$, $V_{\mathrm{S}}$, $\rho$ and an assumed brine modulus). Errors in the logs, in $K_{\mathrm{min}}$ or in the in-situ fluid properties pass straight into the prediction. The frames of Figure 5.3 are given dry.
• $K_{\mathrm{min}}$ varies with mineralogy. Quartz 36.6 GPa, calcite 76.8 GPa, dolomite 94.9 GPa, clay about 21 GPa (Mavko et al. 2009). Check the value you assume.
• Fluid properties vary with depth. The gas modulus changes several-fold between 10 and 30 MPa. Use Batzle and Wang at in-situ pressure and temperature rather than one textbook number.
• Uniform or patchy. Field data sometimes show a smaller fall in $V_{\mathrm{P}}$ than the uniform prediction; patchy saturation is one cause, and a small fall does not rule out a large gas saturation.

The QI workflow that uses Gassmann

A typical quantitative-interpretation pipeline:

1. Acquire pre-stack seismic data and extract intercept and gradient volumes by AVO analysis (Section 5.4).
2. At a well in the survey, read $V_{\mathrm{P}}$, $V_{\mathrm{S}}$ and $\rho$ in the brine-saturated reference state, and compute $K_{\mathrm{sat}}$ and $\mu$ from them.
3. Invert Gassmann for the dry-frame $K_{\mathrm{dry}}$; $\mu$ is already in hand.
4. Run Gassmann forward for each fluid scenario (full gas, oil, mixed saturations, uniform or patchy) on the same dry frame, with fluid properties at reservoir conditions, and predict $V_{\mathrm{P}}$, $V_{\mathrm{S}}$ and $\rho$ for each.
5. Model the AVO response of each scenario from those predictions.
6. Compare with the observed AVO, and keep the scenarios whose predicted response matches the data.

The loop is only as good as Gassmann’s prediction. Where the assumptions fail (poorly connected carbonate pores, strongly dispersive rocks, anisotropic shales), the interpretation is uncertain, and interpreters calibrate with more wells or turn to elastic inversion.