Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2025 / Paper 3 / 1

Let $S_g$ be the fraction of the pore space occupied by gas, and let $\varphi$ be the porosity. At fixed porosity and unchanged rock framework, the bulk density isReplacing water with much less dense gas therefore decreases the bulk density approximately linearly, as in Figure 1. The liquid and gas densities here are those at reservoir conditions.

A pore fluid has negligible static shear modulus, so the rock framework chiefly controls the effective shear modulus $\mu$. If $\mu$ changes little, $V_s=\sqrt{\mu /\rho }$ increases gently as the density falls. This does not imply that S waves travel through the pore gas itself: their restoring force comes from the connected solid framework.

The P wave speed depends on both bulk modulus and shear modulus:Gas is much more compressible than water. Even a small connected gas fraction can strongly reduce the effective fluid bulk modulus, approximately through the compliance average ${\kappa }_f^{-1}\simeq (1-S_g)/{\kappa }_w+S_g/{\kappa }_g$ for a pressure-equilibrated mixture. The saturated rock's effective bulk modulus consequently falls rapidly at first, overwhelming the smaller density reduction and causing the sharp initial fall in $V_p$. Once the fluid is already very compliant, further gas replacement changes the stiffness less; the continuing density decrease can then make $V_p$ rise again. Figure 1 assumes a particular saturation response: patchy fluids, changing effective pressure, and frequency-dependent fluid motion can alter it. The plotted $V_s$ is only nearly constant, so the plot should not be interpreted as an exact fixed-$\mu$ constitutive model.

For incidence from cap rock (1) into reservoir (2), use the conventional normal-incidence stress-amplitude reflection coefficientHere acoustic impedance is the P wave impedance; $\rho V_s$ is the corresponding shear-wave impedance. Positive $R$ preserves the chosen stress-wave seismic reflection polarity; negative $R$ reverses it. These are amplitude coefficients, not reflected energy fractions. At normal incidence in the ideal isotropic model there is no P-to-S mode conversion. The cap-rock impedances are

The following are approximate readings of the supplied Figure 1 PNG, checked against the matching source PDF, rather than exact tabulated parameters. At zero gas, use $\rho \simeq 1.75\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, $V_p\simeq 1.75\mathrm{km}{\mathrm{s}}^{-1}$, and $V_s\simeq 0.60\mathrm{km}{\mathrm{s}}^{-1}$; at full gas use $\rho \simeq 1.25\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, $V_p\simeq 1.15\mathrm{km}{\mathrm{s}}^{-1}$, and $V_s\simeq 0.65\mathrm{km}{\mathrm{s}}^{-1}$. The left-hand $V_p$ endpoint is crowded against the axis and the density curve; readings differing by roughly $0.05$ in the plotted units are reasonable. The resulting coefficients deserve about two significant figures, not the apparent precision of the intermediate arithmetic.

Both end members change seismic reflection polarity, but the P wave change is much larger: approximately $+0.15$ to $-0.22$, compared with $+0.08$ to $-0.05$ for S waves. Thus a calibrated P wave survey is better for distinguishing these two saturation end members and should give a stronger reflection from the fully gas-filled reservoir. A negative P wave reflection relative to the water-filled prediction is a useful gas indicator in this stipulated model. In a real seismic reflection survey, lithology, porosity, effective pressure, and seismic wavelet phase also affect the response, so a bright or reversed reflection alone does not prove the presence of gas.

For the fully water-filled reservoir, the approximate densities and wave speeds giveand henceBoth are positive: the reservoir has larger P wave and shear-wave impedances than the cap rock, and neither reflection reverses the chosen stress-wave seismic reflection polarity.

For the fully gas-filled reservoir,soBoth impedances are now smaller than those of the cap rock. Both reflections reverse seismic reflection polarity, with a substantially larger amplitude for the P wave reflection. Reading the right-hand $V_p$ endpoint instead as $1.20\mathrm{km}{\mathrm{s}}^{-1}$ would give $R_p\simeq -0.20$, preserving the interpretation.

The P wave source is normally the more useful depth source for the end members and for the stated approximately 50% gas case, because the predicted $|R_p|$ exceeds $|R_s|$, making the reservoir-top arrival easier to identify, assuming comparable source and recording quality. For a horizontal reflector and vertical two-way travel time $t$ entirely through the homogeneous cap rock,or $h=V_{s1}{t}_s/2=300t_s\mathrm{m}$ for an identified S wave arrival, with times in seconds. The reservoir's own wave speed is not the overburden wave speed used in this conversion. Nonzero-offset data require moveout correction and an appropriate overburden velocity model; no numerical depth can be obtained without a measured time.

At approximately 50% gas, Figure 1 gives $\rho \simeq 1.5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, $V_p\simeq 1.05\mathrm{km}{\mathrm{s}}^{-1}$, and $V_s\simeq 0.62$–$0.63\mathrm{km}{\mathrm{s}}^{-1}$. ThusThe S wave reflector is then weak because its impedance is nearly matched to the cap rock. An exact impedance match, ${\rho }_2{V}_{s2}={\rho }_1{V}_{s1}$, gives $R_s=0$: a reservoir top can exist without producing an observable normal-incidence S wave reflection. Rounding the 50% $V_s$ to $0.60$ would suggest exactly that cancellation, but the plotted line sits slightly higher. The continuous $Z_s$ curve crosses the cap-rock value at a somewhat greater saturation; the coarse figure does not justify a precise percentage.

There is also a smaller-saturation P wave impedance match: $R_p$ changes sign between zero and modest gas saturation, so it must pass through zero. Thus neither source guarantees a detectable reflector at every saturation. Joint P wave and S wave observations help distinguish a weak impedance contrast from an absent reservoir; a slower S wave speed alone is no reason to prefer it when its reflected amplitude is near zero.