2.2 Non-hydrostatic pore pressure

Pore pressure is not hydrostatic everywhere. In fact, many times pore pressure is an unknown! In a system of “connected pores” under hydrostatic equilibrium (water does not move), pore pressure increases hydrostatically. Non-hydrostatic variations of pore pressure are usually located adjacent to low permeability barriers (e.g., shale formations) that do not allow pore pressure to achieve hydrostatic equilibrium fast enough compared to the rate of sedimentation and pore pressure relief. Hence, pore pressure gets locked-in. In the example in Figure 2.9, pore pressure is hydrostatic until $z \sim 8,500$ ft. Overpressure develops from $z \sim 8,500$ ft to $z \sim 11,000$ ft (likely due to a low permeability mudrock). Pore pressure below $z \sim 11,000$ ft is quite different from hydrostatic!

Figure 2.9: Overpressure example in the Monte Cristo field (Image credit: [Zoback 2013]).

A convenient parameter to relate pore pressure and total vertical stress is the dimensionless overpressure parameter $\lambda_p$:

$$\lambda_p(z) = \frac{P_p(z)}{S_v(z)}$$ (2.14)

In stationary conditions $P_p$ cannot be larger than $S_v$, hence, $\lambda_p \leq 1$. In onshore conditions $\lambda_p \sim 0.44$ means hydrostatic pore pressure ( $\lambda_p \neq 0.44$ in hydrostatic conditions offshore, why?). Reservoir overpressure is good for hydrocarbon recovery (more energy to flow to the wellbore), however, it may cause geomechanical challenges for drilling (kicks). Parameter $\lambda_p \rightarrow 1$ means little effective stress. We will see later that rocks have effective stress-dependent strength. Hence, overpressure leads to weak rocks, especially if they are not well cemented, difficult to drill.

2.2.1 Mechanisms of overpressure

There are several mechanisms that may contribute to overpressure $P_p > P_p^{\text{hydrostatic}}$. First, hydrocarbon accumulations create overpressure due to buoyancy. Overpressure $\Delta P_p$ is proportional to hydrocarbon column height $h$ and difference in mass density of pore fluids $\rho_{brine} - \rho_{hydrocarbon}$

$$\Delta P = (\rho_{brine} - \rho_{hydrocarbon}) g h$$ (2.15)

where $h$ is measured from the hydrocarbon-brine contact line upwards. A connected pore structure is needed throughout the buoyant phase.

Figure 2.10: Mechanisms of overpressure: Hydrocarbon column.

Second, changes in temperature cause fluids to dilate. If the fluids cannot escape quickly enough, then pore pressure increases. Third, clay diagenesis can expell water molecules. For example, when montmorillonite converts to illite at high pressure and temperature, previously bound water molecules get “added” to the pore space. Under constant pore volume conditions, such addition will result in increases of pore pressure. Fourth, hydrocarbon generation also induces overpressure. With hydrocarbon generation, the original organic compounds transform in another phase which occupies more volume at the same pressure conditions. Overpressure in organic-rich shales is a good indicator of hydrocarbon presence.

Figure 2.11: Other mechanisms of overpressure

2.2.2 Disequilibrium compaction and excess pore pressure

Changes of vertical and horizontal stresses can induce pore pressure changes. Pore pressure increases when a rock/sediment is compressed (such that the pore volume decreases) under conditions in which the fluid cannot escape quickly enough. Figure 2.12 shows a schematic representation of this concept.

• (LEFT) A weight is added on a receptacle with a moving top lid. The top has a hole so that water can escape through the tube when the valve is not closed. The valve is closed now, so the water takes the weight $W$ and pore pressure increases a magnitude $\Delta P = W/A$, where $A$ is the area of the top lid.
• (MIDDLE) Somebody opens the valve. The fluid starts to drain out. When the top touches the sediment grains, the weight $W$ starts to transfer to the sediment, so that the water takes now just a fraction of the weight and pore pressure reduces accordingly.
• (RIGHT) The sediment takes all the load $W$ so now the vertical effective stress on the top is $\sigma_v = W/A$. The fluid does not support the weight $W$ anymore $\Delta P = 0$. The time it takes to arrive to this scenario depends on the tube and valve hydraulic conductivity and the overpressure generated by the weight $W$.

Figure 2.12: Schematic of the consolidation problem.

Imagine now a sediment layer saturated with water. There is an impervious layer at the bottom. Water cannot escape from the sides either. Water can only escape from the top.

Figure 2.13: Disequilibrium compaction: initial conditions

An overburden is placed quickly on the sediment so that it compacts an amount $\Delta z$. Initially, the pore pressure increases everywhere the same amount. The pore pressure decreases preferentially at the top boundary (where it can flow) and the rate of pore pressure change is proportional to the hydraulic diffusivity parameter

$$D_h = \frac{M k} {\mu}$$ (2.16)

where $M$ is the constrained rock “stiffness” (inverse of 1D compressibility $C_{bp}$), $k$ is the sediment (vertical) permeability, and $\mu$ is the fluid viscosity. The one-dimensional equation to this problem is

$$\frac{\partial P_p}{\partial t} = D_h \frac{d^2P_p}{dz^2}$$ (2.17)

Figure 2.14: Disequilibrium compaction: load application

The solution of the partial differential equation above give us a characteristic time $T_{ch}$ for which $\sim 2/3$ of the pore pressure is relieved,

$$T_{ch} = \frac{L^2}{D_h}$$ (2.18)

where $L$ is the characteristic distance of drainage. In our example $L$ is the thickness of the sediment layer, the longest straight path to a draining boundary.

EXAMPLE 2.5: Calculate the characteristic time of pore pressure diffusion for a 100 m thick sediment with top drainage considering
(a) a sand with $k =$ 100 mD and $M =$ 1 GPa, and
(b) a mudrock with $k =$ 100 nD and $M =$ 1 GPa.
The water viscosity is 1 mPa s.

SOLUTION
(a) Sand: $T_{ch} \sim$ 1.1 day.
(b) Mudrock: $T_{ch} \sim$ 3000 years. $\: \: \blacksquare$

The example in Figure 2.16 is a measurement of pore pressure $P_p$ that utilizes measurements of porosity in mudrocks (Track 3) to predict overpressure (Track 4). The concept is simple: the porosity of clay-rich rock decreases with effective stress (Figure 2.15). Let us assume the following equation for such dependence

$$\phi = \phi_o \exp (-\beta \sigma_v)$$ (2.19)

Figure 2.15: Schematic example of shale porosity as a function of effective vertical stress. Effective stress (rather than total stress) causes rock mechanical compaction.

Under hydrostatic pore pressure conditions, vertical effective stress will always increase with depth. However, in the presence of overpressure, effective stress may increase less steeply or even decrease with depth. Hence, mudrocks with porosity higher than the porosity expected at that depth (in hydrostatic conditions) indicate overpressured sediment intervals (Figure 2.17).

Figure 2.16: Example of overpressure in the Gulf of Mexico.

Figure 2.17: Schematic representation of overpressure due to disequilibrium compaction. Overpressure develops when the rate of sedimentation exceeds the rate of pore pressure diffusion, i.e., water does not have enough time to escape when squeezed within the rock.

Figure 2.18 shows maps of pressure gradients in the Gulf of Mexico. Data from offshore locations show that overpressure (gradient $>$ 0.44 psi/ft ) starts to develop at 2 to 3 km of depth below seafloor.

Figure 2.18: Pore pressure gradients in the Gulf of Mexico (GOM). Overpressure (gradient $>$ 0.44 psi/ft) in GOM is mostly due to disequilibrium compaction and often increases with depth (depth increasing from Pleistocene to Miocene). Source: Rose et al. 2020, https://doi.org/10.1190/INT-2019-0019.1.

EXAMPLE 2.6: Calculate the pore pressure $P_p$ and overpressure parameter $\lambda_p$ in a muddy sediment located offshore with porosity $\phi = 0.298$. The total depth is 2000 m and the water depth is 500 m (assume a lithostatic gradient of 22 MPa/km below the seafloor). Laboratory tests indicate a compaction curve with parameters $\beta = 3.2 \times 10^{-2}$ MPa$^{-1}$ and $\phi_0 = 0.38$. How much higher than expected hydrostatic value is the pore pressure?

SOLUTION
First, calculate total vertical stress:

$$S_v = 10 \times 0.5 + 22 \times 1.5 = 38 .$$

Second, calculate effective vertical stress from using the measured porosity and Equation 2.19:

$$\sigma_v = - \frac{\ln \left( \phi / \phi_o \right)}{\beta} = 7.6 .$$

Third, calculate pore pressure from $S_v = \sigma_v + P_p$, so

$$P_p = S_v - \sigma_v = 38 - 7.6 = 30.40 .$$

Hence, the overpressure parameter is

$$\lambda_p = \frac{P_p}{S_v} = \frac{30.40 \text{ MPa}}{38 \text{ MPa}} = 0.8. \: \: \blacksquare$$

2.2.3 Reservoir depletion

Opposite to overpressure, “underpressure” occurs when pore pressure is lower than hydrostatic. The most common reason of underpressure is reservoir depletion. In compartmentalized reservoirs with poor water recharge drive, pore pressure may stay low for long periods of time. Reservoir depletion usually brings along lower total horizontal stresses which lower the fracture gradient and make drilling problematic because of decreased difficulty to create open-mode fractures.

Figure 2.19: Example of decreased fracture pressure (between 5,100 ft and 5,400 ft) due to depletion of sandy intervals.