SOIL MECHANICS - CHAPTER 11 doc

FLOW TOWARDS WELLSFor the theoretical analysis of groundwater flow several computational methods are available, analytical or numerical.. Studying groundwater flow is of great importance

FLOW TOWARDS WELLS

For the theoretical analysis of groundwater flow several computational methods are available, analytical or numerical Studying groundwater flow

is of great importance for soil mechanics problems, because the influence of the groundwater on the behavior of a soil structure is very large Many

Q 0

. h 0

. H .

.

.

.. .

Figure 11.1: Single well in aquifer.

dramatic accidents have been caused by higher pore water pres-sures than expected For this reason the study of groundwater flow requires special attention, much more than given in the few chapters of this book In this chapter one more example will be presented: the flow caused by wells Direct applications include the drainage of a building pit, or the production of drinking water by a system of wells.

The solutions to be given here apply to a homogeneous sand layer, confined between two impermeable clay layers, see Fig-ure 11.1 This is denoted as a confined aquifer , assuming that the pressure in the groundwater is sufficiently large to ensure complete saturation in the sand layer.

In this case the groundwater flows in a horizontal plane In this plane the cartesian coordinate axes are denoted as x and y The groundwater flow is described by Darcy’s law in the horizontal plane,

(11.1)

and the continuity equation for an element in the horizontal plane,

68

It now follows, if it is assumed that the hydraulic conductivity k is constant, that the partial differential equation governing the flow is

2 h

This is again Laplace’s equation, but this time in a horizontal plane.

The problem to be considered concerns the flow in a circular region, having a radius R, to a well in the center of the circle This is an important basic problem of groundwater mechanics The boundary conditions are that at the outer boundary (for r = R) the groundwater head

It is postulated that the solution of this problem is

r

outer boundary (r = R), and r is a polar coordinate,

That the expression (11.4) indeed satisfies the differential equation (11.3) can be verified by substitution of this solution into the differential equation The solution also satisfies the boundary condition at the outer boundary, because for r = R the value of the logarithm is 0 (ln(1) = 0) The boundary condition at the inner boundary can be verified by first differentiating the solution (11.4) with respect to r This gives

dh

This means that the specific discharge in r-direction is, using Darcy’s law,

area 2πrH of such a cylinder,

This quantity appears to be constant, independent of r, which is in agreement with the continuity principle It appears that through every

flowing towards the center of the circle That is precisely the required boundary condition, and it can be conclude that the solution satisfies all conditions, and therefore must be correct.

The flow rate very close to the center is very large, because there the discharge Q 0 must flow through a very small surface area At the outer boundary the available is very large, so that there the flow rate will be very small, and therefore the gradient will also be small This makes it plausible that the precise form of the outer boundary is not so important The solution (11.4) can also be used, at least as a first approximation, for a well in a region that is not precisely circular, for instance a square Such a square can then be approximated by a circle, taking care that the total circumference is equal to the circumference of the square.

expected This confirms that by pumping the groundwater head will indeed be lowered.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Figure 11.2: Sink and source.

It is important to note that the differential equation (11.3) is linear, which means that solutions can be added This is the superposition principle Using this principle solutions can be obtained for a system of many wells, for instance for a drainage system All wells should be operating near the center of a large area, the outer boundary of which is schematized to a circle

of radius R For a system of n wells the solution is

n

X

j=1

r j

influence of all wells has simply been added to obtain the solution The

approximately equal to R, the radius of the area, provided that the wells are all located in the vicinity of the center of that area Then all logarithms are

at least approximately.

In Figure 11.2 the potential lines and the stream lines have been drawn for the case of a system of a single well and a single recharge well in an infinite field, assuming that the discharges of the well and the recharge well are equal In mathematical physics these singularities are often denoted as a sink and a source.

Problems

11.1 For a system of air conditioning water is extracted from a layer of 10 m thickness, having a hydraulic conductivity of 1 m/d The discharge is

50 m3/d At a distance of 100 m the water is being injected into the same layer by a recharge well What is the influence on the groundwater head in the

point just between the two wells?

11.2 A well in a circular area of radius 1000 m appears to lead to a lowering of the groundwater table (a drawdown) of 1 m at a distance of 10 m from the well What is the drawdown at a distance of 100 m?

11.3 Make a sketch of the solution (11.4) for values of r/R from 0.001 to 1 The value 0.001 applies to the value r = rw, where rw is the radius of the tube through which the water is being produced Assume that h0= 20 m, H = 10 m, and Q0/2πkH = 1 m What is the limiting value of the head when the radius of the tube is very small, rw→ 0?

11.4 If R → ∞ the solution (11.4) can not be used because ln(0) = −∞ Does this mean that in a very large island (Australia) no groundwater can be produced?