Capillary Phase Trapping

Abstract

After primary and secondary recovery, substantial amounts of hydrocarbon remain entrapped by capillary forces. This phenomenon, known as capillary trapping. Different factors affect the microscopic trapping mechanism such as pore structure, wettability, capillary pressure, interfacial tension, relative permeability, initial saturation, and other properties of the rock and fluids. Understanding the amount of trapped phase is vital for different applications such as enhance oil recovery (EOR), enhanced gas recovery (EGR), and carbon capture and storage (CCS). In EOR and EGR, the purpose is reducing residual oil saturation, while for CCS, it is opposite, i.e. the target is maximising the amount of trapped CO₂. There are two main capillary trapping mechanisms, snap-off and by-passing which are described in detail in this chapter. The laboratory methods and empirical mathematical models that are used to measure and predict the trapped phase saturation are discussed in this chapter. Furthermore, there are various techniques for the mobilization of the trapped hydrocarbon phase. Gas injection is one of the effective methods to remove the trapped phase. This chapter discusses some of the key aspects of phase trapping and mitigation methods.

Appendix: Derivation of Velocity Ratio Expression

The purpose of this Appendix is to drive the velocity ratio expression in the pore doublet model, which determine the interface position (Fig. 10.25).

Fig. 10.25

Schematic of the pore doublet structure

The magnitude of capillary pressure affect the velocity of the interface in both capillary tube. The static capillary pressure function (P_c equation) is defined as Eq. 10.23.

$$P_{c} = \frac{2\sigma \cos \theta }{R}$$

(10.24)

where σ is the interfacial tension between invading and invaded fluid, θ is the advancing contact angle and R is the radius of the capillary.

Furthermore, the volumetric flow rate in either path is given by Hagen-Poiseuille (H-P) equation:

$$q = \frac{{\pi R^{4} }}{8\mu }\frac{\Delta P}{L}$$

(10.25)

where R is the radius of the capillary, L is the length of the flow path of the doublet, assumed to be the same in both branches and μ is the viscosity, assuming that both fluids have the same viscosity. The magnitude of viscose pressure drop can be calculated using Eq. 10.24, if the flow rate was known. Equation 10.24, implies the assumptions of cylindrical capillaries of circular cross-section and of fully developed and steady laminar flows that none of which is likely to be met in reality.

The total volumetric flow rate through the doublet, if the water is supplied at the volume flow rate, q_w, to the doublet can be written as:

$$q_{w} = q_{1} + q_{2} = \frac{\pi }{8\mu L}\left( {R_{1}^{4} \Delta P_{1} + R_{2}^{4} \Delta P_{2} } \right)$$

(10.26)

where q_l is volume flow rates in capillary 1 and q₂ is volume flow rates in capillary 2. The flow direction from left to right are taken with a positive sign and in reverse direction with a negative sign. It is assumed that the interfaces advance simultaneously in both capillaries of pore doublet.

The pressure drop across each capillary is the sum of the capillary pressure and the viscous pressure drop. Since the two capillaries are parallel, the net pressure difference between the two branch points must be the same when taken over either of the two capillaries 1 and 2.

$$\Delta P_{1} - \,P_{c1} = \Delta P_{2} - P_{c2}$$

(10.27)

And so

$$\Delta P_{c} = P_{c2} - P_{c1} = \Delta P_{2} - \Delta P_{1}$$

(10.28)

In Eq. 10.26, the capillary pressure is positive for imbibition process and negative for drainage one.

Equation (10.27) can be rearranged as:

$$\Delta P_{1} = \Delta P_{2} - \Delta P_{c}$$

(10.29)

Substitute Eqs. (10.28) into (10.25):

$$q_{w} = \frac{\pi }{8\mu L}\left( {R_{1}^{4} \Delta P_{2} - R_{1}^{4} \Delta P_{c} + R_{2}^{4} \Delta P_{2} } \right)$$

(10.30)

So Eq. (10.29) can be rearranged as:

$$q_{w} = \frac{{\pi \Delta P_{2} }}{8\mu L}\left( {R_{1}^{4} + R_{2}^{4} } \right) - \frac{{\pi R_{1}^{4} }}{8\mu L}\Delta P_{c}$$

(10.31)

Then, $\Delta P_{2}$ can be found using Eq. (10.30).

$$\Delta P_{2} = \left( {q_{w} + \frac{{\pi R_{1}^{4} \Delta P_{c} }}{8\mu L}} \right)*\frac{8\mu L}{{\pi \left( {R_{1}^{4} + R_{2}^{4} } \right)}}$$

(10.32)

Equation (10.25) it can be rearranged as:

$$q_{1} = q_{w} - q_{2} = q_{w} - \frac{{\pi R_{2}^{4} }}{8\mu L}\Delta P_{2}$$

(10.33)

Then Eqs. (10.32) together with Eq. (10.31) results in:

$$q_{1} = q_{w} - \frac{{\pi R_{2}^{4} }}{8\mu L}*\left( {q_{w} + \frac{{\pi R_{1}^{4} \Delta P_{c} }}{8\mu L}} \right)*\frac{8\mu L}{{\pi \left( {R_{1}^{4} + R_{2}^{4} } \right)}}$$

(10.34)

Equation 10.33 can be simplified to:

$$q_{1} = q_{w} - \frac{{R_{2}^{4} }}{{\left( {R_{1}^{4} + R_{2}^{4} } \right)}}q_{w} - \left( {\frac{{\pi R_{1}^{4} R_{2}^{4} \Delta P_{c} }}{{8\mu L\left( {R_{1}^{4} + R_{2}^{4} } \right)}}} \right)\, = q_{w} \left( {\frac{{R_{1}^{4} }}{{R_{1}^{4} + R_{2}^{4} }}} \right) - \left( {\frac{{\pi R_{1}^{4} R_{2}^{4} \Delta P_{c} }}{{8\mu L\left( {R_{1}^{4} + R_{2}^{4} } \right)}}} \right)$$

(10.35)

Dividing this equation by R₁⁴ and multiplying by $\left( {R_{1}^{4} + R_{2}^{4} } \right)$:

$$\frac{{q_{1} \left( {R_{1}^{4} + R_{2}^{4} } \right)}}{{R_{1}^{4} }} = q_{w} - \left( {\frac{{\pi R_{2}^{4} \Delta P_{c} }}{8\mu L}} \right)$$

(10.36)

So the equation of q₁ and q₂ are as follows:

$$q_{1} = \frac{{q_{w} - \left( {\frac{{\pi R_{2}^{4} \Delta P_{c} }}{8\mu L}} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }} = \frac{{q_{w} - \frac{{\pi R_{2}^{4} \sigma \cos \theta }}{4\mu L}\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }}$$

(10.37)

$$q_{2} = \frac{{\left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} q_{w} + \left( {\frac{{\pi R_{2}^{4} \Delta P_{c} }}{8\mu L}} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }} = \frac{{\left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} q_{w} + \frac{{\pi R_{2}^{4} \sigma \cos \theta }}{4\mu L}\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }}$$

(10.38)

Then expressions of the velocities, v₁ and v_2, will be obtained by dividing each flow rate by its respective cross-sectional area ($\frac{{\pi D_{1}^{2} }}{4}$ and $\frac{{\pi D_{2}^{2} }}{4}$):

$$v_{1} = \frac{{q_{1} }}{{\pi R_{1}^{2} }} = \frac{{q*\frac{1}{{\pi R_{1}^{2} }} - \left( {\frac{{\pi R_{2}^{4} \Delta P_{c} }}{8\mu L}*\frac{1}{{\pi R_{1}^{2} }}} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }} = \frac{{\frac{q}{{\pi R_{1}^{2} }} - \frac{{R_{2}^{4} }}{{4\mu LR_{1}^{2} }}*\left( {\left( {\sigma \cos \theta } \right)*\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }}$$

(10.39)

$$v_{2} = \frac{{q_{2} }}{{\pi R_{2}^{2} }} = \frac{{\left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} q*\frac{1}{{\pi R_{2}^{2} }} + \left( {\frac{{\pi R_{2}^{4} \Delta P_{c} }}{8\mu L}*\frac{1}{{\pi R_{2}^{2} }}} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }} = \frac{{\frac{{R_{2}^{2} }}{{R_{1}^{4} }}*\frac{q}{\pi } + \frac{{R_{2}^{2} }}{4\mu L}*\left( {\left( {\sigma \cos \theta } \right)*\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)} \right)}}{{1 + \left( {\frac{{R_{2} }}{{R_{1} }}} \right)^{4} }}$$

(10.40)

And

So dividing these two velocity to find the ratio of velocity:

$$\frac{{v_{2} }}{{v_{1} }} = \frac{{\frac{{R_{2}^{2} }}{{R_{1}^{4} }}*\frac{q}{\pi } + \frac{{R_{2}^{2} }}{4\mu L}*\left( {\left( {\sigma \cos \theta } \right)*\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)} \right)}}{{\frac{q}{{\pi R_{1}^{2} }} - \frac{{R_{2}^{4} }}{{4\mu LR_{1}^{2} }}*\left( {\left( {\sigma \cos \theta } \right)*\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)} \right)}}$$

(10.41)

Dividing this expression by $R_{2}^{2} \sigma \cos \theta$ and multiplying by $4\mu LR_{1}$

$$\frac{{v_{2} }}{{v_{1} }} = \frac{{\frac{4\mu Lq}{{\pi R_{1}^{3} \sigma \cos \theta }} + \left( {\frac{{R_{1} }}{{R_{2} }} - 1} \right)}}{{\frac{4\mu Lq}{{\pi R_{1} R_{2}^{2} \sigma \cos \theta }} - \frac{{R_{2}^{2} }}{{R_{1} }}*\left( {\frac{1}{{R_{2} }} - \frac{1}{{R_{1} }}} \right)}} = \frac{{4N_{cap} + \left( {\frac{1}{\beta } - 1} \right)}}{{\frac{{4N_{cap} }}{{\beta^{2} }} - \beta^{2} *\left( {\frac{1}{\beta } - 1} \right)}}$$

(10.42)

Finally, Eq. 10.19 can be further simplified to:

$$\frac{{v_{2} }}{{v_{1} }} = \frac{{4N_{cap} + \left( {\frac{1}{\beta } - 1} \right)}}{{\frac{{4N_{cap} }}{{\beta^{2} }} - \beta^{2} *\left( {\frac{1}{\beta } - 1} \right)}}$$

(10.43)

In which

$$N_{cap} = \frac{\mu Lq}{{\pi R_{1}^{3} \sigma \cos \theta }}$$

(10.44)

And

$$\beta = \frac{{R_{2} }}{{R_{1} }}$$

(10.45)