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In this article we will discuss about:- 1. Well Flow near Aquifer Boundaries—Image Wells 2. Method of Images 3. Method of Images for Particular Cases.

**Well Flow near Aquifer Boundaries—Image Wells****: **

When a well is located close to a barrier (impermeable) or recharge boundary, marked deviations from a radial flow system occur. Solutions in such cases are readily obtainable by application of the method of images. An image, which may be an imaginary discharging or recharging well, creates a hydraulic system which is equivalent to the effects of a known physical boundary on the flow system. In essence, images enable an aquifer of finite extent to be transformed to one of infinite extent. This enables the radial flow equations to be applied to the modified system.

A common field problem is that of identifying and locating aquifer boundaries. Pumping test data can provide quantitative answers after application of the method of images and non- equilibrium equation. If a pumping well (PW) and an observation well (OW) are located near an unknown impermeable aquifer boundary (B), an image pumping well IPW will furnish the equivalent hydraulic flow system. From the law of times-

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r^{2}/t_{p} = r_{i}^{2}/t_{i} …(5.40)

Where r = distance of the observation well from the real well; r_{i} = distance of the observation well from the image well; t = time since pumping began to any selected drawdown before the boundary affects the drawdown, Fig. 5.24 and t_{i} = time since pumping began where the divergence of the drawdown curve from the type curve equals the selected drawdown, Fig. 5.24.

Half the value of r_{i} obtained from Eq. (5.40) may be taken approximately as the distance of the impermeable aquifer boundary from the discharging well. Alternatively, if the time (t_{0}) for zero drawdown is selected from the Jacob’s time-drawdown curve and t is the time at which a change of slope (divergence) is indicated, the distance of the aquifer boundary from the discharging well is obtained from the equation-

r^{2}/t_{0} = r_{i}^{2}/t …(5.41)

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Which is the same as the Eq. 5.15 used earlier.

The distance r_{i} obtained from Eq. (5.40) or (5.41) gives an arc on which the image well lies. Data from two or more observation wells are required to locate the image well from the intersection of the three arcs, Fig. 5.25. The aquifer boundary is midway and perpendicular to the line joining the pumped well and the image well.

**Example 1:**

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The coefficients of transmissibility and storage of a non-leaky artesian aquifer are 5 × 10^{5} lpd/m and 2 × 10^{-4}, respectively. The aquifer is bounded on one side by a barrier boundary. A fully penetrating production well has been discharging at a constant rate of 1200 lpm. The drawdown in an observation well 50 m from the pumped well due to the effects of the barrier boundary for a pumping period of 2 hours is 2 m. Compute the distance from the observation well to the discharging image well associated with the barrier boundary (Fig. 5.22).

**Solution:**

The time since pumping began to cause a drawdown 2 m in the observation well before the boundary affects the drawdown is given by-

**Example 2:**

The drawdown is 3 m in an observation well 10 m away from the pumping well (drilled in an artesian aquifer) after 10 min of pumping. What is the time since pumping started, for the same drawdown in another observation well 20 m away from the pumping well?

**Solution:**

Law of times for equal drawdown-

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r_{1}^{2}/t_{1} = r_{2}^{2}/t_{2}

10^{2}/10 = 20^{2}/t_{3}, t_{2} = 40 min

**Example 3:**

The drawdowns in an observation well 12 m away from the pumping well, are 2.6 m and 2.9 m, after 8 and 80 min, since pumping started. What are the corresponding drawdowns in an observation well 120 m away from the pumping well?

**Solution:**

Drawdown per log-cycle of distance is twice the drawdown per log-cycle of time. From the observation well data at r = 12 m,

Drawdown per log-cycle of time (t = 8 – 80 min)

∆s = 2.9 – 2.6 = 0.3 m

Drawdown per log-cycle of distance (r = 12 – 120 m)

∆s’ = 2 ∆s = 2 × 0.3 = 0.6 m

.** ^{.}**. Drawdown at r = 120 m, at t – 8 min, is = 2.6 – 0.6 = 2.0 m

Drawdown at r = 120, at t = 80 min, is = 2.0 + 0.3 = 2.3 m.

Which can also be obtained from the drawdown at r = 12 m at t = 120 min, as 2.9 – 0.6 = 2.3 m.

**Method of Images****: **

Aquifers bounded by an impermeable boundary (aquiclude) and by a recharge boundary (surface water body) are shown in Fig. 5.26 and 5.27.

These conditions can be modelled mathematically by imagining an equal well placed symmetrically about the image plan, which is the plane of the boundary itself. For a barrier boundary the image well is a discharge well and for a recharge boundary the image well is a recharge well. The drawdown at any point is the algebraic sum of the drawdowns resulting from the real (pumping) and image wells. For convenience, the static water level (s.w.l) in the real well is taken as the datum.

**Parallel Boundaries: **

In the event of two parallel boundaries, two image planes are considered which leads to an infinite set of images, Fig. 5.28. However, in practice, pairs of image wells are added until the next pair has negligible influence on the sum of all image well effect out to the point. The image well theory may be applied to such cases by taking into consideration the successive reflections on the boundaries.

It is convenient to remember ‘ROBS’, i.e., ‘Recharge boundary-Opposite, Barrier boundary—Same’ meaning that the image well is of opposite nature to that of the real (pumping) well in the case of a recharge boundary and of the same nature as the real well in the case of a barrier boundary; also when working the images through one image plane, the other image plane may be masked.

For example, consider working out the image wells (I) of the real (pumping) well PW due to two parallel boundaries— one barrier B and one recharge R, in Fig. 5.29 (a). I_{1} and I_{2} are the images of PW due to the boundaries R and B, respectively; mask B and find I_{5} the image of I_{2} due to R; mask R and find I_{4}, I_{5} the images of I_{1} I_{3}, respectively due to B; mask B and find I_{6}, I_{7} the images of I_{4}, I_{5 }respectively due to R and so on; also I_{1}, I_{3}, I_{4} and I_{5} are recharging image wells and I_{2}, I_{6} and I_{7 }are discharging image wells. Actually corrections need to be made only for a few of the nearest of the images, the number depending upon the accuracy desired. The image wells (I) of the real (pumping) well (PW) due to two parallel barrier boundaries (B) is shown in Fig. 5.29 (b).

**Rectangular and Wedge Shaped Boundaries: **

Image well systems for rectangular and wedge shaped boundaries are shown in Fig. 5.30. Each primary image will produce an unbalanced effect at the opposite boundary. Secondary image wells must be added at appropriate positions until the effects of both real and image wells are balanced at both boundaries.

If θ is the angle included between the boundaries, the exact number of image wells.

n_{i} = 360°/θ – 1 …(5.42)

For the wedge shaped boundary in Fig. 5.30 (b), θ = 45°,

n_{i} = 360°/45 – 1 = 7

The image wells are positioned in a circle whose centre is at the apex of the wedge and whose radius is equal to the distance from the real (pumped) well to the wedge apex. The image wells, I, are numbered in the sequence in which they are considered and located.

**Multiple Boundaries: **

The above procedures can be extended to analyse the pumping effects and locate multiple aquifer boundaries. The complexity of the solution increases with the number, irregularity and different types of boundaries present.

**Method of Images for Particular Cases: **

**1. Well Adjacent to a Stream or Line Source, Fig. 5.35: **

For confined non-leaky aquifer with no recharge, applying unsteady state equation, (Jacob’s) the drawdown at any time t,

When the unsteady equation is used, the time (t) drops out of the equation, and the drawdown is the same as that given by the Dupuit’s equation for the steady state. This shows that the presence of recharge boundary in the aquifer makes the flow steady and the drawdown at the well face or any observation well may be calculated by the Dupuit’s equation by putting an appropriate value for the radius of influence, which is R = 2a in the above case.

(b) Drawdown in the observation wells—1, 2, 3 and 4 applying the Dupuit’s equation (neglecting the well radius r_{w}),

**Note:**

r_{i} = Distance of the observation well from the image well.

The above results can be obtained by applying the unsteady state equation (Jacob’s), when the time factor drops out.

The steady-state drawdown at any point (x, y) is given by-

**Example 4:**

A 30 cm well is pumped at the rate of 1000 1pm. The transmissibility of the aquifer is 0.015 m^{2}/s. If the well is located at a distance of 120 m from a stream, what should be the drawdown?

(i) In the pumping well-

(ii) In an observation well 80 m away from the pumping well on the side opposite to the stream?

(iii) In an observation well 80 m away from the pumping well, on a line parallel to the stream?

**Solution:**

**2. Well in a Strip of Land Bounded by Two Streams, Fig. 5.36: **

The Image wells extended to infinity, but for practical purposes it is enough if only a pairs of image wells are considered for estimating drawdown.

The drawdown at the well face, due to the pumping well and the image wells I_{1}, I_{2}, I_{3}, I_{4}

and I_{5} is-

**Example 5:**

Water is pumped at the rate of 3000 1pm from a 30 cm well fully penetrating a strip aquifer 1.6 km wide bounded by two field ditches.

**Assuming the transmissibility of the aquifer as 0.02 m ^{2}/s, what is the drawdown at the well face if:**

(i) The well is at the centre of the strip

(ii) The well is at 400 m from one ditch

What is the drawdown in an observation well in case- (i), located at (300, 400 m) with reference to the pumping well?

**Solution: **

(i) Dupuit’s formula with-

To find the drawdown in the observation well for case (i), with pumping well as the origin of coordinates, from Eq. (5.61)

**3. Well in an Aquifer Bounded by Two streams at Right angles, Fig. 5.37: **

(a) The drawdown at the well face due to the pumping well and three image wells I_{1}, I_{2 }and I_{3}, applying the Dupuit’s equation-

(b) The drawdown in the observation well due to the pumping well and the three image and I_{3} applying the Dupuit’s equation.

**4. Well in an Aquifer Bounded by a Stream on One Side and Impermeable Barrier at 90°, Fig. 5.38: **

The drawdown at the well face due to the pumping well and the three image wells I_{1}, I_{2} and I_{3} by applying Dupuit’s equation-

The drawdown at the observation well can be found as in the previous case.

**5. Aquifer Bounded by a Recharge Boundary: **

If there is a recharge boundary adjacent to an aquifer, the flow into the well becomes steady, and the Dupuits equation can be used to find the drawdown at the face of the pumping well by substituting the appropriate value for the radius of influence R. The values of R in the Dupuit’s equation, are given for a number of boundary conditions, in Fig. 5.39.

**6. Well in an Aquifer Bounded by Two Impermeable Barriers at, Right Angles, Fig. 5.40: **

Since there is no recharge boundary, the flow into the well is not steady and the Dupuit’s equation cannot be used.

**Applying the Unsteady-State Equation (Jacob’s):**

Knowing the transmissibility and the storage coefficient of the aquifer, the drawdown at pumping well PW (X, Y) and the observation well OW(x, y ) can be determined at any time ‘t’ after pumping started.