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%matplotlib notebook
import matplotlib.pyplot as plt
import numpy as np
Till now, we only considered storage at the water table and gave very simple, but practical examples largely ignoring spatial dimensions. Spatial dimensions will be dealt with later. In this section, we handle the physics of elastic storage and will give some interesting everyday examples that are sometimes easily overlooked.
Elastic storage is the only storage occurring in confined (and semi-confined) aquifers, i.e. in aquifers without a water table, aquifer that are completely filled with water from floor to ceiling. In such aquifers we have no lowering of the water table whatsoever, unless the head is lowered to beneath the ceiling of the aquifer, a case further ignored here.
Therefore, in confined aquifers storage can only result from compression of the water and depression of the aquifer. The compressibility of the water and the grains themselves is quite obvious, but often the less obvious storage is the most important part. This is the deformation of the soil skeleton, the bulk matrix or the (bulk) porous medium as it is called.
To analyze the physics of elastic storage, we start with noting that the total load at any depth is carried by the total (vertical pressure) $\sigma_{t}$ or $p$ N/m2. This total pressure must equal the sum of the vertical grain pressure (the so-called effective stress, $\sigma_{e}$) and the water pressure $\sigma_{w}$
$$ p=\sigma_{e}+\sigma_{w} $$This is indicated in figure [fig:The-weight-of-ground-supported-by-sigmaw+sigmae]. The brown beams and the springs are imaginary. They replace the volume $V_{0}$ (1 m3 say) that has been cut out of the aquifer. The two imaginary springs have the same properties as the water and the porous medium respectively. Let us see that happens when the pressure is increased by $\Delta p$. Then the volume (or height) $V_{0}$ is reduced by $\Delta V$ and the springs pressures are increased by $\Delta\sigma_{w}$ and $\Delta\sigma_{e}$ respectively. The springs have different stiffness, so $\Delta\sigma_{w}\ne\Delta\sigma_{e}$. However each string will always carry a fixed proportion of the total stress. Se we may write
$$ \frac{\Delta\sigma_{w}}{\Delta p}=LE $$where LE is this fixed proportion and is called the loading efficiency. The LE must obviously lie between 0 and 1 and is fixed for any particular porous medium. So if we put a weight, a layer of sand for instance, on ground surface, p will increase by $\Delta p$ equal to the weight of that sand per $m^{2}$. and $\Delta\sigma_{w}=LE\,\Delta p$. If we have a piezometer in the aquifer, the the water level (the head) will rise by
$$ \Delta\phi=\frac{LE}{\rho f}\Delta p $$Now assume that $\Delta p$ is caused by an change of the barometer pressure as is shown in the figure below. Then the same reasoning applies, as the subsoil does not know whether it is the barometer pressure of a layer of sand that causes the pressure increase.
Figure: The weight of the ground plus water is supported by two pressures, the water pressure $\sigma_{w}$ and the effective pressure $\sigma_{e}$
A nice and famous early example of loading efficiency is the impact of a train stopping at a station and leaving again some time later (figure above). The weight of the locomotive compresses the aquifer a bit, thus reducing its pore space. This in turn compresses the groundwater, which cannot readily escape. Hence, its pressure rises and it starts to flow sidewards, so that the pressure gradually decreases towards its original trend. When the train leaves, the opposite occurs. The removal of the load reduces the effective stress, which causes the aquifer to bounce back, providing more pore space to the water, which depressurizes and increases somewhat in volume. This reduced water pressure causes surrounding groundwater to flow towards to fill up the gap because of which the pressure gradually normalizes.
Figure: Water level fluctuation in a confined aquifer produced by a train stopping near an observation well [Todd (1959), Todd and Mays (2005)]
Q: Think of another way for the water to escape from a semi-confined aquifer.
The figure below, left, shows the situation where the pressure increase is caused by a load (of sand) on ground surface; the right hand picture show how the same pressure increase is caused by an increase of the barometer pressure. The question is, how does the barometer pressure change alter the head (water level) in the piezometer?
Figure: Effect on head in confined aquifer by a load $\Delta p$ on surface versus an increase of the barometric pressure.
There is no difference for the pressures in the aquifer between the two pictures. However there is a difference in the piezometer. The barometer pressure also works on the water level in the piezometer, which the sand layer does not. So on the one hand the water level in the piezometer wants rise by $\Delta\sigma_{w}/\rho g=LE\Delta p/\rho g$ as it was the case in the left picture, but at the same time it also wants to decline under the full barometer pressure that now works on its water surface, which is different from the left picture. To see how the head changes exactly, we only have to write out what the pressure change at the bottom of the piezometer is:
$$\rho g\Delta\phi+\Delta p = \Delta\sigma_{w}$$$$ = LE\,\Delta p$$Therefore we have
$$\Delta\phi=\frac{1}{\rho g}\left(LE-1\right)\Delta p$$and, because $LE-1<0$ we rather write this with $1-LE$, which is positive, as
$$\Delta\phi = -\frac{1}{\rho g}\left(1-LE\right)\Delta p$$$$= -\frac{1}{\rho g}BE\,\Delta p$$
Where $BE$ is the so-called barometric efficiency, which also is between 0 and 1. Hence
$$LE+BE=1$$This means that the water pressure always takes a fixed fraction LE of the total pressure increase $\Delta p$ whatever the cause of the pressure increase and the head would rise accordingly $\Delta\phi=LE\,\Delta p/LE$. But if the pressure increase is caused by the barometer pressure, the the head declines according to fixed portion of the barometer increase namely $\Delta\phi=-BE\,\Delta p/\rho g$. Plus we have $+LE=1$.
Figure: Example of a high degree (75%) of barometric efficiency [Todd (1959), Todd and Mays (2005)]. It shows the response of the head in a well penetrating a confined aquifer together with the barometric pressure. Note that the axes on the right is reversed to show the similarity of the two curves (head down when barometer pressure goes up and vice versa).
A famous example of the barometer efficiency was given by [Todd (1959), Todd and Mays (2005)], figure [fig:Todd-baro-graph]. This example is used here because its is famous as one of the first published. However, barometer effects are always seen in piezometers in confined aquifers. The barometer efficiency generally varies between 20% and 80%.
The barometer influence causes a noisy behavior of head time series form confined and semi-confined aquifers, because, even if the groundwater was in perfect rest, barometer fluctuations do affect both the head measured in piezometers as the pressure measured in pressure gauges. Only if heads are measured a short time intervals of hours rather than weeks, the noisy behavior of the head in confined aquifers actually shows its clear one-to-one relation with the course of the barometer pressure. Therefore, such a noisy time series behavior actually shows that a piezometer is in a (semi)-confined aquifer. Unless we have very thick unsaturated zones with substantial resistance against air flow, we will not see much if any barometer fluctuation in water table aquifers ([Rasmusen and Crawford (1997)]).
If the total pressure p is increased by $\Delta p$, the the porous medium is compressed together with the water that it contains. Clearly, the increase of the water pressure will also compress the individual grains. However, sand grains are about 50 times harder, or stiffer, or less compressible than the water. Therefore, the effect of the grains being compressed can be safely neglected. However, the porous medium itself is far less stiff than the grains. It is compressed due to some deformation of the grains et the expense of the porosity.
Hence the volume $V_{0}$ is compressed by $\Delta V$ when the pressure $p$ is increased by $\Delta p$. Because the aquifer is assumed of infinite lateral extent, the only possible movement is downward, meaning that $\Delta V=\Delta H$ the change of the thickness of the considered part of the layer that we replaced by the springs. This means both springs underlie the same compression.
Let the water have a compressibility $\alpha$ meaning that $a$ m3 of water would be compressed by the fraction $\alpha$ for each in crease of the water pressure by 1 N/m2. Let the porous medium have a compressibility of $\beta$, meaning that one m3 of the porous medium would be compressed by the factor $\beta$ for each N/m2 increase of effective $\sigma_{e}$. These compressibilities, therefore, have dimension [m3/m3/(N/m2)=m2/N.
Now consider that the soil was put under an extra total pressure of $\Delta p$ causing it to be compressed by the fraction $\Delta H/H_{0}=\Delta V/V_{0}$. Then the effective pressure increases by
$$ \Delta\sigma_{e}=-\frac{\Delta V/V_{0}}{\beta} $$Because the grains are incompressible, it follows that the change of pore volume equals the change of the total volume. Therefore, for the water we have a relative volume change ( = compression) of
$$\Delta V/\left(\epsilon V_{0}\right)$$This is because the water volume change equals the total change of the porous medium, but the total water volume equals porosity times the volume of the porous medium $V_{0w}=\epsilon V_{0}$. Therefore, the water pressure increase is
$$\Delta\sigma_{w} = \frac{\Delta V/\left(\epsilon V_{0}\right)}{\alpha}$$$$= \frac{\Delta V/V_{0}}{\epsilon\alpha}$$
And, therefore
$$\frac{\Delta\sigma_{e}}{\Delta\sigma_{w}} = \frac{\epsilon\alpha}{\beta}$$and so,
$$\Delta\sigma_{e} = \frac{\epsilon\alpha}{\beta}\Delta\sigma_{w}$$Then we can eliminate $\Delta\sigma_{e}$ from the pressure equation:
$$\Delta p = \Delta\sigma_{w}+\Delta\sigma_{e}$$$$\Delta p = \left(1+\frac{\epsilon\alpha}{\beta}\right)\Delta\sigma_{w}$$which finally yields the loading efficiency in terms of the compressibilities of the water and the porous medium.
$$LE=\frac{\Delta\sigma_{w}}{\Delta\sigma_{p}}=\frac{\beta}{\beta+\epsilon\alpha}$$And because $BE=1-LE$ we also have
$$BE=\frac{\epsilon\alpha}{\beta+\epsilon\alpha}$$Figure: Compression of the porous medium, while the volume of the grains remains unchanged because their compressibility is negligible compared to that of both the water and the porous medium.
When the aquifer is compressed, only the pore space is reduced because the grains are almost incompressible compared to the water and the porous medium. Therefore, the reduction of the pore space is the same as the reduction of the total volume of the porous medium, as it is conceptually shown in figure [fig:Compression-of-the-porous-medium].
Now consider the situation in which we lower the water pressure, for instance because by extracting water from the aquifer. The specific storage coefficient is
$$S_{s}=-\frac{\partial V/V_{0}}{\partial\phi}$$it is the volume of water released from the porous medium per m of lowering of the head $\phi$ (a negative $\Delta\phi$ yields a positive amount of water). It is also immediately clear that the dimension of $S_{s}$ is $\mbox{\ensuremath{\left(m^{3}/m^{3}\right)}/m=\ensuremath{m^{-1}}}$, the volume of water released per $\mbox{m}^{3}$ of the porous medium per m of head decline.
Lowering of the water pressure does not change the total pressure in any way. Therefore, $\Delta p=0$, which yields
$$0=\Delta\sigma_{w}+\Delta\sigma_{e}$$The amount of water squeezed out of the porous medium changes due to a positive change of the effective pressure equals the reduction of the pore volume. It amount is ($\Delta\sigma_{e}>$0, reduces the pore volume by $\Delta V$ due to which this volume is released form the porous medium).
$$\Delta V_{pm}=+V_{0}\beta\Delta\sigma_{e}$$Where pm means”porous medium”.
And the water, of which we have $\epsilon V_{0,}$, expands by
$$\Delta V_{w}=-\alpha\left(\epsilon V_{0}\right)\Delta\sigma_{w}$$(A negative value of $\Delta\sigma_{w}$ releases a positive volume of water).
The total amount of water released equals the volume squeezed out due to the reduction of the pore space plus the volume that is generated by expansion of the water due to the reduction of the water pressure:
$$\Delta V = -\alpha\left(\epsilon V_{0}\right)\Delta\sigma_{w}+V_{0}\beta\Delta\sigma_{e}$$and because $\Delta\sigma_{e}=-\Delta\sigma_{w}$ in this case (see equation [eq:deltaSigma_w+deltaSigma_e=0]) we have
$$\frac{\Delta V}{V_{0}}=-\left(\alpha\epsilon+\beta\right)\sigma_{w}$$and so
$$\frac{\Delta V/V_{0}}{\Delta\sigma_{w}}=-\left(\epsilon\alpha+\beta\right)$$using $\Delta\sigma_{w}=\rho g\Delta\phi$ yields
$$S_{s} = -\frac{\Delta V/V_{0}}{\Delta\phi}$$$$ = \rho g\left(\epsilon\alpha+\beta\right)$$which, considering that we reduce the $\Delta$ to infinitisimal small $\partial$, completes the proof (see equation [eq:Ss-definition]).
The compressibility of water is
$$\alpha = -\frac{1}{V_{w,0}}\frac{\partial V_{w}}{\partial\sigma_{w}}\,\,{\left[L^{2}/F\right]^{2}}$$$\alpha_{w}\approx\ 4.4\times 10^{-10}$ [m2/N]. Clearly, $\partial W_{w}/V_{w,0}$ is the relative change of the water volume. There is some dependency on dissolved components, water containing dissolved gas, may be up to three times more compressible than water without dissolved gas under normal pore pressure (Lyons, William C. (2010): Working Guide to Reservoir Engineering; Elsevier).
The compressibility of the porous medium is
$$\beta=-\frac{1}{V_{T,0}}\frac{\partial V_{T}}{\partial\sigma_{e}}$$where $\partial V_{T}/V_{T,0}$ is the relative change of the volume of the porous medium, that is, the total volume of the considered soil (including its pores). $\sigma_{e}$ is the effective stress, i.e. that part of the total stress $\sigma_{t}$ or p that is not carried by the water pressure $\sigma_{w}$. The total pressure equals the weight of the overburden, i.e. that of the overlying formations including the water in that it contains. Hence $\sigma_{e}=p-\sigma_{w}$.
The soil compressibility $\alpha$ is the gradient of a stress-strain curve (relative volume change versus applied effective stress) of a dry soil sample put under increased stress in the laboratory, such that sideward movement is prevented as it is the case in the soil under uniform vertical stress. Unlike water, the compressibility of soil is not necessarily a constant. If the soil is put under higher stress than it ever supported, then it consolidates, meaning that the change of volume is largely irreversible. But under lower than historic stress, a compressibility can be determined, and truly elastic behavior assumed. It is clear that this compressibility depends on porosity.
[Van der Gun (1980)] presented the following relation between the compressibility of aquifers and depth based on laboratory measurements from laboratory carried out by Van der Knaap (1959)
$$\beta=\epsilon\left(3\times10^{-11}+6.6\times10^{-11}z^{-0.7}\right)$$where $z$ is the depth below ground surface and $\epsilon$ is porosity. Then applying [eq:]
$$S_{s}=\rho g\left(\epsilon\alpha+\beta\right)$$With the relation of [Van der Gun (1980)] we obtain the graphs shown in [fig:Van der Gun-specific-storage]. As can be concluded from the graph, values in the order of 10e-5 m2/N are often found in practice, where we generally have porosities of around 35% in fluviatile and eolian sandy aquifers.
Question: Is it feasible that compressibility of the porous medium is proportional to porosity?
Figure: Computed specific storage coefficient $Ss=\rho g\left(\epsilon\alpha+\beta\right) \,{\left[m^{-1}\right]}$ as a function of depth below ground surface using the relation by [Van der Gun (1980)])
Explain what loading efficiency is.
What factors contribute to the elastic storage coefficient and what factor may be neglected?
If a load \Delta p is placed on top of a confined aquifer and the water pressure in the aquifer is increased to $\Delta\sigma_{w}=LE\,\Delta p$, then how much does the head in a piezometer in that aquifer change?
The same question for the situation where $\Delta p$ is caused by an increase of the barometer.
Assume we have a pressure gauge (pressure transducer or pressure sensor) in a piezometer in the confined aquifer that measures the absolute pressure (air pressure + water pressure). On day 1, the barometer rises by \Delta p and later with zero rise of the barometer, a load $\Delta p$ is placed on ground surface. What is the difference if any in the registration by the pressure gauge in the piezometer, and what is the difference in head?
The measurements of pressure gauges confined and semi-confined aquifers are corrected for barometer pressure changes by subtracting the barometer pressure from the pressure measurement. Does this mean that the fluctuations of the barometer pressure are eliminated by this correction? Of so explain why. If not also explain why.
What is actually the result of this correction of the registered pressures? What actually do we get by this correction?
How can we compute the specific yield from the measured barometric efficiency? Note: $BE = 1-\frac{\beta}{\epsilon\alpha+\beta}$, $Sy = \rho g\,(\epsilon\alpha+\beta)$. Think of what we can easily estimate and what we know, respectively what we don't know? Assume that porosity \epsilon can be reasonably well estimated.
Consider a confined aquifer and two situations. First there is a loading on ground surface with value $\Delta p$. The head is measured both in a piezometer and in a pressure gauge (which measures the absolute water pressure in the aquifer). What is the difference in the two measurements?
In the same location consider a barometer pressure increase of the same magnitude as the surface loading $\Delta p$ before, so $\Delta a=\Delta p$. What is the difference in the head measured with a piezometer and that measured with a pressure gauge?
What is the difference between the heads measured with the piezometer in the two cases?
What is the difference between the pressures measured with the pressure gauges in the two cases?
How much is the barometer effect in an unconfined aquifer.
If we want to register head in a (semi)confined aquifer by means of a pressure gauge, we have to correct for barometer variations. Is this also the case in a cold climate in which the water in the top of the piezometer is frozen and the ice completely blocks the piezometer from the ambient air?
If an infinitely extended surface load would remain constant. How will the head or pressure in a piezometer in a semi-confined aquifer behave?
What could we do with such behavior?
With two pressure transducers, one measuring the barometer pressure and the other the water pressure in some piezometer in a confined aquifer how can we compute the barometer efficiency? What parameter we still miss to obtain true numerical values?
How does the head in a water-table aquifer react to barometer fluctuations?
How large may the variation of the head due to barometer fluctuations become given a range of atmospheric pressure from variation between 970 to 1040 mbar (=cm head)?
What values do you expect for total elastic storage coefficients of aquifers in practice?
How could we measure the elastic storage coefficient in a confined aquifer below the sea bottom?
Does the value of the specific yield that we may derive from barometer efficiency, water storativity and porosity refer to the value of the measuring point or to the thickness of the entire aquifer?
How useful is it to measure local porosity at the screen position of the piezometer to compute the storage coefficient of the aquifer?