ADVERTISEMENTS:

In this article we will discuss about the potential and problems of ground water flow.

**Ground Water Flow Potential****: **

ADVERTISEMENTS:

The total energy (or head, h) at any point in the ground water flow field per unit weight of water is given by-

h = z + p/γ_{w} + V^{2}/2g

Where z = elevation of the point (above a chosen datum); P/γ_{w} = pressure head and V^{2}/2g velocity head.

Since the ground water flow velocities are usually very small, V^{2}/2g is neglected, and

h = z + p/γ_{w} = piezometric head, at the point …(4.22)

ADVERTISEMENTS:

From Darcy’s law, V = Ki

Where K = coefficient of permeability of the formation;

And i = hydraulic gradient = grad (z + p/γ_{w}) = dh/ds.

ADVERTISEMENTS:

Therefore, V = – K (dh/ds) …(4.23)

The negative sign indicates the ground water flow in the direction of decreasing head, i.e., in the direction s.

It is convenient to introduce a velocity potential ɸ, defined as a scalar function of time and space, such that the velocity components in the x^{–}, y^{–} and z^{–} directions are given by-

u = – ∂ɸ/∂x, v = – ∂ɸ/∂y, w =∂ɸ/∂z …(4.24)

And generally, V = ∂ɸ/∂s …(4.25)

Where V is the velocity in the flow direction s.

From Eqs. (4.23) and (4.25),

dɸ/ds = K (dh/ds) or ɸ = Kh + C

Where C = a constant which disappears during differentiation and is usually taken as zero (since it does not alter the flow pattern in any manner).

ADVERTISEMENTS:

Therefore, ɸ = Kh …(4.26)

Since ground water can be considered incompressible for all practical purposes, i.e., ρ = constant, the equation of continuity for steady ground water flow is obtained by putting-

∂ρ/∂t = 0 in Eq. (4.61), as

∂u/∂x + ∂v/∂y + ∂w/∂z = 0 …(4.27)

From Eq. (4.24),

∂^{2}ɸ/∂x^{2} + ∂^{2}ɸ/∂y^{2} + ∂^{2}ɸ/∂z^{2} = 0 …(4.28)

Or, V^{2}ɸ = 0 …(4.29)

Which is the Laplace equation.

For a homogeneous and isotropic medium (formation), from Eqs. (4.24) and (4.26)-

u = – K (∂v/∂x), v = – K (∂h/∂y), w = – K (∂h/∂w)

From Eq. (4.27),

∂^{2}h/∂x^{2} + ∂^{2}h/∂y^{2 }+ ∂^{2}h/∂z^{2} = 0 …(4.30)

Or V^{2}h = 0 …(4.31)

i.e., for steady ground water flow in homogeneous isotropic medium, Laplace equation is satisfied, and a flow net can be constructed.

In anisotropic formation, if the permeabilities in the three principle directions are K_{x}, K_{y} and K_{z}, the velocity components are given by Darcy’s law as-

u = – K_{x} ∂h/∂x, v = – K_{y} ∂h/∂y, w = – K_{z} ∂h/∂z …(4.32)

From the continuity Eq. (4.27), for steady flow,

K_{x} ∂^{2}h/∂x^{2}, + K_{y} ∂^{2}h/∂y^{2}, + K_{z} ∂^{2}h/∂z^{2 }= 0 …(4.33)

**If a transformation could be made such that: **

X = x/√K_{x }, Y = _{}y/√K_{y} , Z = z/√K_{z}

Eq. (4.33) becomes ∂^{2}h/∂x^{2}, + ∂^{2}h/∂y^{2}, + K_{z} ∂^{2}h/∂z^{2 }= 0 …(4.34)

And hence Laplace equation is satisfied.

For a two-dimensional anisotropic systemic. K_{x} > K_{y}, the horizontal dimensions are reduced by multiplying by the factor √K_{y} /K_{x} , and a flownet is drawn in the transformed section. This creates an isotropic medium having an equivalent K = √K_{x}K_{y} . The ground water flow is computed by substituting K = √K_{x}K_{y} in the discharge equation and then the flownet is transferred back to the original section, which of course will not be an orthogonal network, Fig. 4.12.

**Ground Water Flow Problems****:**

Potential flow theory may be applied to obtain discharge and draw-down for various conditions of ground water flow, which are illustrated in the following examples.

**(a) ****Steady flow in a homogeneous aquifer, Fig. 4.13: **

Discharge per unit width, assuming flow to be horizontal and uniform everywhere in a vertical section-

q = V (h.1) = – (∂ɸ/∂s) h = – Kh (dh/dx)

Which is shown in Fig. 4.13 and also the water table actually observed in piezometers or determined by drawing a flownet. The actual water table is higher than that computed by the Dupuits equation, the departure being large in the neighbourhood of the well. The discrepancy is due to the fact that Dupuit flows are all assumed horizontal, whereas the actual velocities of the same magnitude have a drownward vertical component so that a greater saturated thickness is required for the same discharge; the assumption that dh/ds

≈ dh/dx is not correct in the vicinity of the well. The watertable approaches the well face tangentially and forms a seepage face because no consistent flow pattern can connect a watertable directly to a downstream free water surface. However, Eq. (4.36) yields correct values of q or K.

The vertical component of velocity (w) can be determined by considering the continuity equation in two dimensions-

The vertical velocity component increases with elevation and at the phreatic surface, z = h,

w = – q^{2}/kh^{2} …(4.38)

The negative sign indicates the downward direction.

**(b) Flow into horizontal galleries dug down to the impervious soil layer, Fig. 4.14****: **

Integrating, qx = Kh^{2}/2 + c_{1}

When x = 0, h = h_{1}; c_{1} = – Kh_{1}^{2}/2

q = K/2x (h^{2 }– h_{1}^{2}) …(4.39)

Which is parabolic. Putting h = H when x – L

q = K/2L (H^{2} – h_{1}^{2}) …(4.40)

From Eqs. (4.39) and (4.40), the equation to the phreatic line is-

h = √h_{1}^{2 }+ x/L (H^{2}-h_{1}^{2}) …(4.41)

The quantity of water flowing into the gallery of length I from both sides

Q = 2ql = Kl /L (H^{2} – h_{1}^{2}) …(4.42)

When the watertable in the gallery drops, i.e., h_{1} decreases, L increases; upon increasing h_{1} L decreases.

As an example, given H = 7 m, h_{1} = 2 m, L = 400 m, K = 60 m/day, I – 200 m

Flow from one side q = K/2L (H^{2} – h_{1}^{2}) = 60/2 × 400 (7^{2} – 2^{2}) = 3.375 m^{3}/day per m

(i) Flow into the gallery Q = 2ql = 2 × 3.375 × 200 = 1350 m^{3}/day

(ii) Equation to the phreatic surface

The GWT at 100 m from the face of the gallery is-

h = √4 + 0.1125 × 100 = 3.78 m

Or drawdown s = H – h = 7 – 3.78 = 3.22 m

(iii) The maximum drawdown at the gallery is

s_{max} = H – h_{1} = 7- 2 = 5m

**(c) Aquifer with Recharge: **

If the aquifer is being recharged by rainfall and if the steady infiltration rate be P (in suitable units), Fig. 4.15,

**Example:**

Rainfall at the rate 10 mm/hr falls on a strip of land 1 km wide laying between two parallel canals with 2 m difference in their water surface levels, Fig. 4.16. It is underlain by a horizontal impermeable stratum at 10 m below the water surface of the lower canal. Assuming a permeability of 12 m/day with vertical boundaries and all the rainfall infilters into the soil, compute the discharge per metre length into both the canals.

**Solution:**

So, there is a discharge of 120 m^{3}/day into both the canals, for each metre length of the strip of land.

**Note: **

(i) Recharge into land = Discharge into both the canals

= (0.01 × 24) (1000 × 1) = 120 + 120 = 240 m^{3}/d/m

= 240 m^{3}/d/m

**Note: **

(ii) If there were no recharge by rain, seepage from the upper canal to the lower canal, Eq. (4.36),

**Note:**

(iii) With recharge, the equation to phreatic surface, Eq. (4.45),

**(d) Steady flow in a confined aquifer of constant thickness, Fig. 4.17 Discharge per unit width: **

**(e) Steady flow in a confined aquifer of variable thickness, Fig. 4.18 Discharge per unit width: **