Recharge Boundary of a Well (With Diagram) | Geography

Under recharge boundary conditions, water level in the observation well will drawdown initially under the influence of the production well only. When the cone of depression spreads to the recharge boundary, the time-rate of drawdown will change, due to the influence of the image well. The time-rate of drawdown will thereafter continually decrease till eventually equilibrium conditions are reached, when recharge balances discharge.

Under equilibrium conditions in a non-leaky artesian aquifer-

s_r = s – s_i …(5.43)

Where s_r = drawdown in the observation well near a recharge boundary and s_i = build-up due to image well.

If the distance between the pumped well and the recharge boundary is a and ɸ is the angle between the line joining the pumping and image wells, and the line joining the pumping and observation wells, Fig. 5.31, then Eq. (5.47) can be expressed as-

Which when compared with Eq. (5.7) shows that in the first approximation this drawdown is equal to that at the face of a well in a circular island aquifer of radius 2a.

The percentage of pumped water diverted from a source of recharge depends upon the hydraulic properties of the aquifer, the distance of the recharge boundary from the pumping well and the pumping period. If Q_s is the volume of water drawn from the recharge source, the fraction of the well discharge Q_s/Q drawn is given by the dimensionless plot of ‘Q_s/Q versus a√4 K b t/S’ as shown in Fig. 5.32. The same result applies to an unconfined aquifer by replacing the storage coefficient with the specific yield, provided the drawdown is small compared to the saturated thickness of the aquifer.

The time required to reach approximate equilibrium conditions may be computed with the equation-

Example:

From the pumping tests in the area irrigated by the Ganga canal, UP, S = 0.11 and T = 1.5 × 10⁶ lpd/m. A well is to be sited to intercept the seepage and prevent waterlogging in the reach of the canal where the seepage losses are found to be 105 lps/km.

What should be the distance of the tubewell from the canal so that it does not induce extra seepage from the canal, and the discharge and spacing of tubewells in this area?

If a 30 cm tubewell is located at 250 m from the canal and pumped at the rate of 5000 lpm, determine:

(a) The time required to reach approximate equilibrium conditions.

(b) The percentage of pumped water furnished by the canal after pumping for one month.

(c) The resulting drawdown in the tubewell.

Solution:

The distance of the tubewell from the canal can be determined by trial by assuming a = 300 m and estimating r for s = 0.1 m along the perpendicular from the tubewell to the canal, i.e., ɸ = 0. For such a case Eq. (5.48) becomes-

s = 2.3 Q/2π T log₁₀ 2a – r/r

Assuming a minimum spacing of tubewells along a line parallel to the canal as 5a, the seepage loss on one side of the canal-

105 (5 × 300)/ 3 × 1000 = 78.8 lps. or 0.079 m³/sec

Assuming that this seepage loss is equal to the discharge from the well-

That is at a distance of 297.6 m from the tubewell towards the canal (or at 2.4 m from the canal towards the tubewell) the drawdown will be just 1 cm. The radius of influence is within the critical distance a, so that the pumping well does not induce extra seepage from the canal. Actually a fraction of the pumpage (as can be determined by constructing a flow net for the discharging well) comes from the landward side and the canal seepage intercepted is less than 78.8 lps. The radius of influence is at a safe margin from the canal and the well does not induce extra seepage from the canal.

The distance r for s = 0.1 m along a line parallel to the canal and on either side of the pumping well can be determined by putting ɸ = 90° or 270° in the Eq. (5.48)-

The distance r for s = 0.1 m along a line perpendicular to the canal and on the other side of the pumping well can be determined by putting ɸ = 180° in the Eq. (5.48)-

The distances r from the pumping well in the four directions corresponding to a given drawdown can be determined as shown in Table 5.17 and a flownet can be drawn as shown in Fig. 5.33.

As shown in Fig. 5.33, the tubewell is located at 300 m from the canal and the lines of equal drawdown (equipotential lines) are drawn by plotting the distances on the four sides of the well. The flow lines (stream lines) are then drawn by trial and error to form an orthogonal network called flownet. It can be seen from the flownet that the length of the canal which comes under the influence of pumping, with no extra seepage induced, is 1600 m. This distance normally varies from 4a to 6a.

The seepage from the canal in this reach at 105 lps/km is 105 × 1.6 = 168 lps, and from one side of the canal it is 84 lps. From the flownet, the fraction of the pumpage coming from the landward side = 79 × 5/20 = 20 lps. The balance of pumpage 79 – 20 = 59 lps comes from the canal side against a seepage of 84 lps from the canal. The well is not inducing extra seepage. If the pumpage can be increased to 110 lps, the water coming from landward side = 110 × 5/20 = 27.5 lps and the balance of 110 – 27.5 = 82.5 lps is drawn from canal seepage. Still, the well will not milk the canal with a narrow margin of 84 – 82.4 = 1.5 lps.

This can be confirmed by redrawing the flownet with the increased pumpage of 110 lps. Alternatively, the distance of the tubewell from the canal may be slightly reduced, instead of increasing the pumpage.

Spacing of Tubewells:

The tubewells should be so spaced that there is no interference of their cones of influence and the spacing between any two tubewells should be more than two times the radius of influence with a safe margin. In the above example, tubewells may be located on a line parallel to the canal with a spacing of more than 2r = 2 × 1050 = 2100 m, say 3 km apart, with an average pumpage of 100 lps.

(a) If a 30 cm tubewell is located at 250 m from the canal, the time required to reach approximate equilibrium is-

Aquifer Test—Recharge Boundary:

The distance-drawdown data in several observation wells parallel to the recharge boundary at the end of the test, when water levels are stabilised, are plotted (i.e., s_r vs. r) on a semi-log paper and the coefficient of transmissibility is determined from the Jacob’s method as-

T = 2.3Q/2π ∆s_r … (5.52)

The computed value of T and other known data for the observation well are substituted in Eq. (5.48) and the distance of the pumped well from the recharge boundary a is determined. Since the drawdowns are affected by recharge, the storage coefficient S cannot be computed from the distance-drawdown field data graph. After T and a are computed, the image well may be located and the coefficient of storage can be estimated by a process of trial and error with the Eqs. (5.44) through (5.46). Several values of S are assumed and s_r is determined. The value of S, for which s_r corresponds to the observed value, is taken as the coefficient of storage.