Pumping Well in a Leaky Artesian Aquifer | Ground Water | Geography

In this article we will discuss about:- 1. Leaky Artesian Aquifer 2. Unsteady Radial Flow—Leaky Artesian Aquifer 3. Steady Radial Flow—Leaky Artesian Aquifer 4. Hantush Inflection Point Method—Unsteady Radial Flow- Leaky Artesian Aquifer 5. Finite Number of Wells in a Leaky Artesian Aquifer in a Line 6. Circular Battery of Wells in a Leaky Artesian Aquifer.

Leaky Artesian Aquifer:

Aquifers which are overlain or underlain by semi-permeable strata are referred to as leaky aquifers. In such aquifers a significant portion of the yield may be derived by vertical leakage or seepage through the semi-confining formations into the aquifer, Fig. 5.14.

The velocity (v) of the downward vertical flow through the semi-confining layer (leakage rate, i.e., vertical discharge per unit area) is proportional to the difference between the ground water table and the piezometric head in the confined aquifer and is given by Darcy’s law.

V = K’ (Δh/b’) …(5.16)

And the total vertical leakage-

Q_c = v A_c

or Q_c= (K’/b’) Δh A_c …(5.17)

Where,

Q_c = leakage through the confining bed

K’ = coefficient of vertical permeability of confining bed

b’ = thickness of confining bed through which leakage occurs

A_c = area of confining bed through which leakage occurs

Δh = difference between the piezometric head in the aquifer and in the source bed above confining bed (GWT in Fig. 5.14).

The leakage (recharge) is proportional to the difference in head Δh and is more near the pumping well since the piezometric surface is very low. The ratio K’/b’ is called leakance and its reciprocal b’/K’ is called the hydraulic resistance ‘c’ of the confining layer and has the dimension of time. √Tc is called the leakage factor ‘B’ and has the dimension of length; Tis the transmissibility (= Kb) of the confined aquifer.

Example 1:

From the pumping tests of a semi confined aquifer of thickness 30 m and permeability 20 m/d, it is estimated that the recharge rate from an overlying unconfined aquifer through an aquitard of thickness 2 m, is 50 mm/year. The average piezometric surface in the semi confined aquifer is 16 m below the water table in the unconfined aquifer. Determine the hydraulic characteristics of the aquitard (semi-confining layer) and the aquifer.

If a well drilled in the aquifer is pumped at the rate of 5000 m³/day, how many square kilometers of recharge area are required to sustain the flow at the estimated recharge of 50 mm/year?

Solution:

Recharge through aquitard per unit area (1 m²)

Unsteady Radial Flow—Leaky Artesian Aquifer:

Hantush and Jacob (1965) showed that the drawdown of the piezometric head in a leaky confined aquifer is given by the equation-

Which are very similar to Eqs. (5.8) and (5.9) except that the well function contains the additional term r/B. Values of W (u, r/B) for various values of u and r/B are given by Hantush (1956) in Table 5.13. A family of leaky artesian type curves where values of W (u, r/B) versus 1/u for various values of r/B are plotted on log-log paper by Walton (1962) as shown in Fig. 5.15.

The B observed time-drawdown data s versus t is plotted on a transparent log-log paper of the same scale. The time-drawdown plot is superimposed on the type curves with the respective axes parallel and a match point is selected after the field-data curve has been matched with one of the curves of r/B in the family of type curves.

The coordinates W (u, r/B), 1/u, s and t at the match point and the value of r/B for the curve matched (by interpolation if the curve does not exactly coincide) are used to calculate T, S and K’ in the Eqs. (5.18), (5.8) and (5.19). It may be noted that as K → 0, r/B → 0 indicating that for an impermeable confining layer Eq. (5.18) reduces to that of non-leaky aquifer (Theis).

Example 2:

The time-drawdown data for unsteady flow in an observation well at 30 m from the pumped well are given in Table 5.14. The well was pumped at a constant discharge of 800 lpm. The thickness of the aquifer and the top semi-confining layer are 12 m and 4 m, respectively. Determine the aquifer constants.

Solution:

The time-drawdown data is plotted on a transparent log-log paper of the same scale as the Walton’s type curves and superimposed on the family of leaky artesian type curves, Fig. 5.16. The time-drawdown curve coincides with r/B = 0.2 type curve and the coordinates of the match-point P can be read as-

Steady Radial Flow—Leaky Artesian Aquifer:

When the time-drawdown data fall on the flat portions of the family of leaky artesian type curves indicating that the well discharge is balanced by leakage from the semi-confining layer, an equilibrium stage has been reached and the cone of depression is given by the formula-

This steady flow is maintained so long as the water level in the formation supplying leakage remains constant. Steady-state leaky artesian type curve is shown in Fig. 5.17 where values of ‘K_o(r/B) versus r/B’ are plotted on log-log paper. Field data on several observation wells (under steady state conditions),’s versus r,’ are plotted on transparent log-log paper of the same scale as the type curve.

The distance-drawdown field plot is superimposed on the type curves keeping the axes parallel and a match point is selected. The match point coordinates K₀(r/B), r/B, s and r are substituted in the Eqs. (5.20) and (5.21) to determine T and K’. The storage coefficient cannot be computed, under such conditions of flow, since the entire yield from the well is derived from leakage sources only. Values of K₀(r/B) for various values of r/B are given in Table 5.15.

Hantush (1956, 1964) found that if r/B < 0.05, Eq. (5.20) can be approximated by-

s = 2.3 Q/2π T log₁₀ 1.12 B/r …(5.22)

To determine T, a distance-drawdown plot ‘s versus log r,’ is drawn on semi-log paper, which will be a straight line for r/B < 0.05, and T is obtained from Eq. (4.82) as-

T = 2.3 Q/2π Δs

Where Δs is the drawdown per log cycle of distance r. If the straight line is extended to give an intercept r₀ on the abscissa where s = 0, then-

1.12 B/r₀ = 1, from Eq. (5.22)

Or B = r₀/1.12 …(5.23)

Since B = √Tc, the hydraulic resistance c is given by-

c = r₀²/1.25 T …(5.24)

Alternatively, if the coordinates of any point on the straight line plot are s and r, substitution of these values in Eq. (5.22) determines B, and c can be obtained from the relation B = √Tc.

Hantush Inflection Point Method—Unsteady Radial Flow- Leaky Artesian Aquifer:

Hantush (1956) developed a method for determining T, S and c from the time-drawdown data by reading on the plot of ‘s versus log t’ the values of s_i, t_i and Δs_i, where the subscript i refers to the inflection point I, i.e., the point where the drawdown (s_i) is one-half of the final or equilibrium drawdown given by the Eq. (5.20), i.e.,

Another solution by Hantush (1956) is for the time-drawdown data for several observation wells. The drawdown per log cycle of time (∆s) for the straight line portion of each curve (for each well at a known r) is determined. Then a plot ‘r versus log ∆s’ is drawn on a semi-log paper and a straight line of best-fit is drawn, and the slope of this line, i.e., change ∆r per log cycle of ∆s is determined. The straight line is extended to give the intercept ∆s₀ on the abscissa where r = 0 and, B and T are determined as-

B = ∆r/2.3 …(5.30)

T = 2.3 Q/4π ∆s₀

And c is obtained from the relation B = √Tc . To determine S, the values of Q, T and K₀ (r/B) for the ratio r/B, are substituted in Eq. (5.25) and s_i is calculated. The corresponding value of t_i is obtained from the time-drawdown plot and S is determined from Eq. (5.26).

A third method by Hantush (1964) is based on a simplified solution of Eq. (5.18) which is valid if t > 4t_i and Tt/SB > 2r.

Finite Number of Wells in a Leaky Artesian Aquifer in a Line:

This is illustrated in the following example:

Example 3:

A leaky artesian aquifer 30 m thick having a permeability of 20 m/day is situated above an impervious base and overlain by a semi-confining layer with a resistance of 2000 days against vertical leakage. On the top of this semi-confining layer, a homogeneous aquifer with constant water table is present. From the leaky artesian aquifer, ground water is abstracted by a series of 7 wells at intervals of 110 m in a straight line. Each well has a diameter of 40 cm, a screened length of 21 m in the middle of the aquifer, and is pumped at a constant rate of 600 m³/day. What is the maximum drawdown in any well?

Solution:

Fig. 5.46.

The interference of neighbouring wells is maximum at the central-well (No. 4) and the drawdown is maximum in the central well.

The drawdown in the central well is due to three components:

(i) Its own drawdown, assuming full penetration, s_w

(ii) Additional drawdown due to its partial penetration alone, *∆s_w

(iii) Drawdown (Σs) caused by the neighbouring wells at distances of

r = 110, 220, 330 m

Thus the total drawdown in the central well-

Circular Battery of Wells in a Leaky Artesian Aquifer:

If ‘n’ wells of equal capacity, spaced at ‘a’ in a circular array of radius R give a total discharge Q, Fig. 5.47, the drawdown at any point P may be obtained by the method of superposition as-

Where r_i = distance from P to the centre of the various wells. When R << B, the drawdown inside the circle is given by-

Neglecting the influence of point abstraction, the drawdown inside the circle is thus constant, equal to the drawdown in the centre C i.e.-

When the wells partially penetrate the aquifer, the drawdown at the well face will be augmented (∆s_w)_pp.

Outside the battery, at a distance V from its centre, the flow conditions are determined by the equation for a single well, i.e.-