# Box 6 Adding Recharge to the Unconfined Aquifer System

Consider the addition of a flux boundary condition on the top of the model. This could represent uniform recharge to the upper surface between the constant-head boundaries (Figure Box 6-1). Now the flow rate will vary from one x-location to another. Depending on the relative values of h1, h2, recharge, W, hydraulic conductivity, K, and boundary heads, the flow may be entirely to the right, to the left, or a groundwater divide may form within the land mass at which the flow, q’, is zero with increasing flow volume in opposite directions from the divide to the boundaries.

A few schematics of possible configurations are shown in Figure Box 6-2 to illustrate the substantial variations that can occur in this simple one-dimensional system. Flow directions and velocities, as well as the shape of the water table can be drastically different.

To incorporate surficial recharge into the analytical solution, first consider a small prism of the aquifer, as shown in Figure Box 6-3.

Flow through the left face of Figure Box 6-3 is shown in Equation Box 6-1.

 (Box 6-1)

where:

 q’x = flow per unit width (L2/T) dy = width of the face into the page (L)

Flow through the right face of Figure Box 6-3 is shown in Equation Box 6-2.

 (Box 6-2)

The value of is different on each side of the element, so the value of Q(q’ dy) is different on each side. The difference between flow on each side of the element must equal the amount of flow entering as recharge from the top of the system. The difference in flow can be expressed in terms of the one-dimensional, unconfined, steady-state groundwater flow equation, and equated to the amount of flow that enters the element from the top between the two sides (W dx dy) as in Equation Box 6-3.

 (Box 6-3)

where:

 W = recharge rate from the surface (L/T)

Gathering the derivative terms and dividing both sides by –K provides Equation Box 6-4.

 (Box 6-4)

Integrating Equation Box 6-4 yields Equation Box 6-5.

 (Box 6-5)

where:

 C1, C2 = constants of integration (L, L2, respectively)

By applying boundary conditions as done for the case without recharge (h = h1 at x = 0 and h = h2 at x = L), the constants can be determined. Solving for the constants provides Equation Box 6-6. Solving Equation Box 6-6 for h at x results in Equation Box 6-7. See Figures Box 6-1 and Box 6-3 for terms.

 (Box 6-6)
 (Box 6-7)

where:

 hx = head at x some distance from the origin, x=0 (L) x = distance from the origin (L) h1 = head at the origin (L) h2 = head at L (L) L = distance from the origin to h2 K = hydraulic conductivity (L/T) W = recharge rate (L/T)

To determine q’ as a function of x, Equation Box 6-6 can be differentiated with respect to x resulting in Equation Box 6-8.

 (Box 6-8)

Multiplying Equation Box 6-8 by –K provides Equation Box 6-9.

 (Box 6-9)

Substituting q’x for the left-hand side of Equation Box 6-9 provides Equation Box 6-10.

 (Box 6-10)

The next task is to determine the location of the groundwater divide. An expression for the location of the divide, d, can be determined by setting q’x equal to zero in Equation Box 6-10 and solving for the x position of the divide (d) as shown in Equation Box 6-11.

 (Box 6-11)

where:

 d = x location of the groundwater divide (L)

The maximum head (or minimum in the case of a negative value of W) can be determined by substituting d for x in Equation Box 6-7 resulting in Equation Box 6-12.

 (Box 6-12)

where:

 hmax/min = maximum head for positive W and minimum for negative W (L)

The spreadsheet provided at this link allows you to explore the nature of this flow system for various input parameter values.