Generalization of the Darcy-Buckingham Law: Optimality and Water Flow in Unsaturated Media

Abstract

The Darcy-Buckingham law extends Darcy’s law from single phase flow in the subsurface to multiphase flow and was discovered by Edgar Buckingham (1867–1940). It is valid under the local-equilibrium condition that, however, does not always hold especially when fingering flow occurs. This chapter is devoted to generalizing the Darcy-Buckingham law by relaxing the local-equilibrium condition. The new development is based on an optimality principle that water flow in unsaturated media self-organizes in such a way that the resistance to water flow is minimized. Unlike the traditional form of the Darcy-Buckingham law, the relative permeability in the generalization is also a function of water flux, which is a direct result of the non-equilibrium flow behavior. The generalized Darcy-Buckingham law has been shown to be consistent with laboratory observations and field data for gravitational fingering flow in unsaturated soils. It is also the theoretical foundation for the active fracture model, the key constitutive relationship for modeling flow and transport in the unsaturated zone of Yucca Mountain, Nevada, that is the national geological disposal site for high-level nuclear waste in USA.

Appendix: An Alternative Derivation of Eq. 2.48 Without Using the Lagrange Multiplier

Equation 2.48 was derived in Sect. 2.4 based on the calculus of variations involving a Lagrange multiplier (Eq. 2.46). For the convenience of readers who are not familiar with the mathematical background of the Lagrange multiplier, this appendix provides an alternative derivation of Eq. 2.48 without using the Lagrange multiplier.

From Eq. 2.46, the Lagrangian (without using the Lagrange multiplier) for the given problem is

$$L = - K_{un} S_{*}$$

(2.A1)

where S _* is given as:

$$S_{*} \equiv \left( {\frac{\partial E}{\partial x}} \right)^{2} + \left( {\frac{\partial E}{\partial y}} \right)^{2} + \left( {\frac{\partial E}{\partial z}} \right)^{2}$$

(2.44-5)

Equation 2.48 is then derived by solving the following Euler-Lagrangian equation with w = E:

$$\frac{\partial L}{\partial w} - \frac{\partial }{\partial x}\left( {\frac{\partial L}{{\partial w_{x} }}} \right) - \frac{\partial }{\partial y}\left( {\frac{\partial L}{{\partial w_{y} }}} \right) - \frac{\partial }{\partial z}\left( {\frac{\partial L}{{\partial w_{z} }}} \right) = 0$$

(2.39)

Because E = h + z, we have

$$\frac{\partial L}{\partial w} = \frac{\partial L}{\partial E} = \frac{\partial L}{\partial h}\frac{\partial h}{\partial E} = \frac{\partial L}{\partial h} = - \frac{{\partial \left( {K_{un} S{}_{*}} \right)}}{\partial h}$$

(2.A2)

For $w_{x} = \frac{\partial E}{\partial x}$, we also have

$$\frac{\partial L}{{\partial w_{x} }} = - \frac{{\partial \left( {K_{un} S_{*} } \right)}}{{\partial S_{*} }}\frac{{\partial S{}_{*}}}{{\partial w_{x} }}$$

(2.A3-1)

$$\frac{{\partial \left( {K_{un} S_{*} } \right)}}{{\partial S_{*} }} = K_{un} + \frac{{\partial K_{un} }}{{\partial \left( {\log S_{*} } \right)}}$$

(2.A3-2)

$$\frac{{\partial S_{*} }}{{\partial w_{x} }} = 2\left( {\frac{\partial E}{\partial x}} \right)$$

(2.A3-3)

Combining Eqs. 2.A3-1 to 2A3-3 yields:

$$\frac{\partial L}{{\partial w_{x} }} = - 2K_{un} \frac{\partial E}{\partial x} - 2\frac{{\partial K_{un} }}{{\partial \left( {\log S_{*} } \right)}}\frac{\partial E}{\partial x}$$

(2.A4)

In a similar procedure, we obtain

$$\frac{\partial L}{{\partial w_{y} }} = - 2K_{un} \frac{\partial E}{\partial y} - 2\frac{{\partial K_{un} }}{{\partial \left( {\log S_{*} } \right)}}\frac{\partial E}{\partial y}$$

(2.A5)

$$\frac{\partial L}{{\partial w_{z} }} = - 2K_{un} \frac{\partial E}{\partial z} - 2\frac{{\partial K_{un} }}{{\partial \left( {\log S_{*} } \right)}}\frac{\partial E}{\partial z}$$

(2.A6)

Inserting Eqs. 2.A2 and 2.A4–2.A6 into Eq. 2.39 and using the following form of water mass balance equation;

$$\frac{{\partial \left( {K_{un} \frac{\partial E}{\partial x}} \right)}}{\partial x} + \frac{{\partial \left( {K_{un} \frac{\partial E}{\partial y}} \right)}}{\partial y} + \frac{{\partial \left( {K_{un} \frac{\partial E}{\partial z}} \right)}}{\partial z} = 0$$

(2.A7)

we finally obtain

$$\frac{{\partial \left( {\frac{{\partial K_{un} }}{{\partial (\log S_{*} )}}\frac{\partial E}{\partial x}} \right)}}{\partial x} + \frac{{\partial \left( {\frac{{\partial K_{un} }}{{\partial (\log S_{*} )}}\frac{\partial E}{\partial y}} \right)}}{\partial y} + \frac{{\partial \left( {\frac{{\partial K_{un} }}{{\partial (\log S_{*} )}}\frac{\partial E}{\partial z}} \right)}}{\partial z} = \frac{{S_{*} }}{2}\frac{{\partial K_{un} }}{\partial h}$$

(2.48)