Exercises in Hydrogeology

By Bill Scott

Division of Science and Engineering
School of Environmental Science
Murdoch University
Perth, Western Australia

These exercises and following solutions and .html codes were produced with the symbolic manipulator Maple. They are available in an active form as solving.mws; those who have Maple are free to use this sheet in their study, along with the passive version solving.html. Please feel free to copy and use the material. However, the author retains the copywrite. He request that he be acknowledged, should the material be passed along in any form. Permission of the author is required before copies are passed along for profit.

Note that some lines contain Maple input. These may seem cryptic but are mostly english. Specifically, the subs( ) command substitutes known equalities into an equation or expression. The % is a 'ditto' operation that simply 'brings down' the calculation last done (usually from the previous line). The %% 'brings down' the result two statements removed. The solve( ) command solves for the required unknown or unknowns.

This material is available at

Exercise A1. Horizontal Groundwater Flow

Groundwater is flowing horizontally along a profile that is uniformly 2.1 metres deep. Two piezometers are located 230 metres apart in the profile; the northern one reads 10.421 metres AHD; the southern one reads 11.682 metres AHD. What is the direction of groundwater flow? What is the hydraulic gradient? If the hydraulic conductivity is 2 metres/day, what is the Darcy velocity? With a porosity of 0.4, how long do you expect it takes for the groundwater to flow between the two points?

1. Recognition

The lectures and readings should show links with other such problems, and Darcy's law. The first 'answer' is an orientation problem, simply looking at the numbers gives the 'answer', to where the groundwater is moving. The other 'answers' are quantitative and require numerical digestion.

Right here it is best to simply regurgitate some of the ordinary specifics. Nearly all the problems in groundwater hydrology relate to Darcy's law:

v = K*i = K*(H1-H2)/L = K*(h[1]+z[1]-(h[2]+z[2]))/L

2. Interpretation

The two total head values are measured relative to AHD (Australian Height Datum) and relate to a reference level (Sea Level). The definition is such that the 'upstream' value is given the position 1 and the 'downstream' value is given the position 2. It doesn't really matter if, in reality, we choose wrongly, the calculated velocity will simply become negative. In this case, if we choose position 1 as the south point and position 2 as the north point, the subtraction produces a positive number and the flow is positive. That means, immediately, that the horizontal flow is from south to north.

This gives some orientation and we draw a picture. the picture should show the dimensions of interest; in this case the horizontal, the direction of flow; and the vertical, in which the pressure gauges, the piezometers, are placed. The height of water in the piezometers above the given absolute position indicates the pressure that exists at that point. The view presented is a 'vertical crossection', a most common presentation of groundwater flow.

Include the data that are given, as water levels in tubes, and include the reference level, sea level. Try to make it accurate, so as not to be confused. Include an arrow to indicate the flow direction.

3. Related Information

Darcy's law is a description of the flow rate through a soil normalised for the area of flow. The so-called Darcy velocity is simply this flow rate m^3/day divided by the area of flow m^2 The net result has units of m/day ; though it has units of velocity, it is not a velocity. The flow rate per day is measured and divided by the area; this is the 'Darcy velocity'. It is precisely v = Q/A , a volumetric flux or flow/unit area. The 'real velocity' is much larger than this; the expected velocity is the pore velocity vp = v/n , the velocity of a non-reactive dye or tracer. Indeed, for a small porosity n around 0.2, the 'real velocity' may be 5 times the 'Darcy velocity'. Simply; groundwater must 'scream' through the pores to 'make up' for the space taken up by the grains.

4. Focus and Do

The hydraulic gradient i is the fall over the run.

> i = (H[1]-H[2])/L;

i = (H[1]-H[2])/L

> subs(H[1]=11.682*metres,H[2]=10.421*metres,
L= 230*metres,%);

i = .5482608696e-2

This is the hydraulic gradient. A further substitution gives the Darcy velocity

> v = K*i; subs(%%,K = 2*metres/day,%);

v = K*i

v = .1096521739e-1*metres/day

A further substitution produces the pore velocity or 'real velocity'. The porosity n = .4 cares for the soil matrix; the real velocity is 250% greater than the Darcy velocity.

> vp = v/n; subs(%%,n = .4,%);

vp = v/n

vp = .2741304348e-1*metres/day

5. Solve with units

The solution and units are included above. This gives an immediate check on the correctness of the 'answer'. Still, the 'accuracy' can be no more than the number of digits; round the answer off to 3 digits. Note that later we will do the calulations with the full 10 digit default, to remove any possibility of adding any 'round-off' error in this calculation. The second operation, here, assigns a name to the equality.

> evalf(%,3); vp[subs]:=%:

vp = .274e-1*metres/day

6. Check the answer

We haven't finished. Rethink what a velocity is. Look at your old Physics books, how one uses velocity during a trip, to estimate arrival times.

> vp = distance/time; time = solve(%,time); subs(vp[subs],distance = 230*metres,%); evalf(rhs(%)*years/(365*day),2);

vp = distance/time

time = distance/vp

time = 8394.160583*day


The gradients seem OK. Groundwater flows slowly and 23 years is a reasonable time to travel 230 metres. Remember that the hydraulic gradient is small.

Exercise A2. Vertical Groundwater Flow

Nearly quicksand conditions are being experienced in a swale. A hole is drilled and a nested set of piezometers placed in the profile; piezometer A, at 5 metres; piezometer B, at 2 metres. Is the groundwater flowing upward? Which piezometer is expected to show the highest total head? The difference in total heads between the two piezometers is 2.2 metres. What is the vertical hydraulic gradient? If the vertical hydraulic conductivity is 0.5 metres/day, what is the vertical Darcy velocity? If the area of concern is 0.21 hectares, how much water is expected to be flowing? If there are no other sources or sinks, this would be the flow of water up and onto the surface. State your answer in m^3/day .

1. Recognition

A problem like this one needs to be treated as an exercise on the imagination. Firstly, because groundwater problems are invariably 3D and, secondly, this problem, particularly, is revealing; is some depth emerging in your understanding of groundwater flow?--the flow is vertical and the measurement of tendency for flow, the piezometers, are also vertical. Can you sort this out? Are you comfortable with the concepts?

Also, a clear understanding of terms in groundwater hydrology or hydrogeology is necessary. That is, a 'nest of piezometers' like a 'nest of birds' consists of several piezometers in a single hole. Piezometers are no more or no less than 'pressure metres'; these are located at different heights in the same hole. But the piezometers are, in reality, tubes filled with water, also in the vertical. This becomes a jumbled mess, if indeed, we put all the devices on a single drawing. Instead, we 'pretend' that the piezometers occupy several holes which are displaced slightly in the horizontal. But remember, all the piezometers are really 'in the same hole'.

Darcy's law still applies. Write it down. To minimise confusion we use large A and large B instead of 1 and 2 for the positions. The idea is that the vertical is different from the horizontal, and a slightly different nomenclature may increase clarity. If, however, this change in nomenclature is difficult and causes you stress, use 1 and 2. The symbols are there for your purposes and to make things clear; use them to your advantage.

v = K*i = K*(H[A]-H[B])/L = K*(h[A]+z[A]-(h[B]+z[B]))/L

Of course, position A and position B are now in the vertical. We chose them so that, hopefully A is upstream of B. That is, the total pressure at position A is greated than at position B; if, in fact we calculate that the pressure at position B is greater than the pressure at position A

Going over the points in our formatted approach:

2. Interpretation

Restating the verbal description in my own words, it seems piezometers are placed in the vertical flow to measure the upward flow in a swale. Knowing the values for the primary parameters (head differences, lengths, the gradient), we can calculate and quantify the groundwater flows. An appropriate diagram, slightly distorted to allow insight, reveals the heart of the problem:

There are no new 'formulas' except that we need to emphase the vertical, perhaps using a focused form of Darcy's law

v = Q/A = K*i = K*(H[A]-H[B])/L

3. Related Information

There is a need to subtract the two vertical positions to get the distance between the piezometers, the L value in Darcy's law. This is done on the graph and should clarify the problem. Notice that L = 3 . It is also presumed that the shape of the swale is such that the concave features do not destroy any reference level one may be using. Remember that, in reality, both piezometers are in the same hole and must have a common reference. Notionally, this reference could be A; the measurements shown, 5m and 3m, would then correspond to the total head values H[A] and H[B]

4. Focus and Do

Quicksand indicates that the water is flowing upward so the lowest level, 5 metres below the ground (the measurement with the piezometer at point A in the figure), has the highest reading. Please accept that this is an inherent result from knowing that there is quicksand, it is not a feature of the depth of the reading . If there had been recharge or downward flow in the region (as is usual in a swale in the winter), the upper piezometer, 2 metres below the ground (the measurement in the piezometer at point B in the figure) would definitely register the highest reading. There are no if, ands or buts about this--please accept that the total head (as measured by the absolute height of water in a hole) directly predicts the way the water is flowing, as well as the amount of flow. That is Darcy's law .

First the hydraulic gradient:

> i=(H[A]-H[B])/L;

> subs(H[A]-H[B]=2.2,L=3,%);

i = (H[A]-H[B])/L

i = .7333333333

Next, the Darcy velocity or flow per unit area.

> v=K*i;

> subs(K=0.5*metres/day,%%,%);

v = K*i

v = .3666666666*metres/day

Third, the flow rate Q .

> v=Q/A;

v = Q/A

> subs(%%,A=0.21*hectare,%);

.3666666666*metres/day = 4.761904762*Q/hectare

> Q=solve(%,Q);

Q = .7699999998e-1*metres*hectare/day

5. Solve with units

Here, this means, 'fix the units!'. They are there, but mixed. We also round off to give a better idea of accuracy and make the answer look nicer.

> subs(hectare=(100*metres)^2,%):

> evalf(%,3);

Q = 770.*metres^3/day

6. Check the answer

There is little one may do to 'check this' other than redo the calculations. Here we consider presentating the results in a little different way. There is a known relation between the vertical gradient and the development of quicksand (Bouwer, pg 329)

> i[q] = (1-n)*rho[p]+theta-1;

i[q] = (1-n)*rho[p]+theta-1

where the particle density of sand rho[p] is approximately 2.6 and the water contant theta is approximately 0.3 and the porosity n is also about 0.3 . (We should be critical here and note that Bouwer ignores the units for density.) That means that

> subs(rho[p]=2.6,n=.4,theta=.4,%);

i[q] = .96

It seems the given hydraulic gradient is not quite enough to cause quicksand, but then, we don't know the specific properties of the sand; certainly the result is in the right ball-park.

A feel for the actual flow may be obtained by considering that the area is one-fifth of a hectare, which is about half an acre.

> 0.21*hectare*(1*acre/(0.4047*hectare));


For interest's sake, we consider the acre-foot as a measure of water volume; we all seem to come back to acres and feet, one way or another. In addition acre-feet are standard, imperial measures of water supply volume. Work it out. It dates back to the time serfs were given land, notionally 10 acres, formed by division of a mile into 8 pieces. With a mile equal to 5280 feet, we can stumble through the conversion.

> 8*8*10*acre*ft=1*mile*1*mile*ft;

640*acre*ft = mile^2*ft

> subs(1*mile=5280*ft,%);

640*acre*ft = 27878400*ft^3

But we need to convert the ft^3 into m^3 . Try an inch conversion first.

> algsubs(1*ft^3=(12*inches)^3,%);

640*acre*ft = 48173875200*inches^3

Next, we recollect the conversion between metres and inches.

> isolate(1*metre=39.36*inches,inches);

inches = .2540650406e-1*metre

> subs(%,%%);

640*acre*ft = 790034.9670*metre^3

> evalf(isolate(%,acre*ft),4);

acre*ft = 1234.*metre^3

This seems a bit frivilous, just for a conversion factor. The point is, the quantity of water rising in this system would be a little more than half this in a day.

> evalf(770/1234)*acre*ft: evalf(%,2);


The Darcy velocity calculation suggest that the actual rise of water would be

> .3666666666*metre/day*(3.28*ft/metre);


This is enough to go over your 'wellys' and give you wet feet! This calculation presumes that there is no runoff or horizontal flow, which would ordinarily take over. Though the Darcy velocity is, in reality, a flow rate per unit area, it is also the 'velocity' that would occur, if there were no intervening soil matrix. Here the pore velocity or 'real velocity' within the porous soil matrix, is given by

> vp=%/0.4;

vp = 3.006666665*ft/day

The 'velocity' that would be seen within the soil matrix or the 'velocity' of movement of an inert, passive dye.

Look closely at the following comparisons: The amount of water buildup on the surface ( 1.2*ft in a day), is less in magnitude than that perceived as acre-feet ( .62*acre*ft ). This is because acre*ft and m^3 are volumes; the area over which this volume is observed is .21*hectare or about half an acre. In a day, the given volume of .62*acre*ft is concentrated on half an acre and the height to which it rises is about twice this magnitude or 1.2*ft


Exercise A3. Groundwater Supply

Murdoch lies on the edge of the Jandakot Groundwater Mound, south of Perth. A rough value of the hydraulic gradient through the campus is 5 metres/kilometre. The notional depth of the 'unconfined' aquifer system is about 20 metres. Consider that a series of production wells may be used to harvest half of the groundwater flowing underneath Murdoch, roughly a flow area 1 kilometre by 10 metres. If the hydraulic conductivity is 10 metres/day, what is the estimated Darcy velocity? How much water could theoretically be collected from such a scheme?

1. Recognition

The acre-foot, as presented above, is usually used in reference to water supply. Water supply is the objective of the 'rough' calculation in this exercise. It is another horizontal flow problem, like Exercise A1, but where the hydraulic gradient is given directly. The gradient i is presumed approximately constant in the horizontal, through the whole scene. That means there is a steady, constant flow of groundwater through, a 'through flow'. A portion (half) is presumed to be available for water supply to the campus. The solution estimates the Darcy velocity or flow rate per unit area, then the total flow through a crossection of aquifer; Half of this flow is the final estimate of possible water supply.

Again, Darcy's law is required, given here in alternate forms:

Q = A*v

Q = A*K*i

2. Interpretation

The drawing shows the steady gradient in terms of the water flow. The overall flow is a product of the Darcy velocity and the crossection of flow.


3. Related Information

Note that the depth of the aquifer is constant. This type of calculation relies on either

The term transmissivity T, with units of m^2/day is used for the product of the hydraulic conductivity K and the aquifer depth d. In terms of T, Darcy's law may be written:

Q = A*K*i = w*d*K*i = w*T*i

Notice that the water is drawn off from 10 metres of depth, only half of the whole depth of the aquifer, 20 metres. Hence the total depth is 20 metres and the transmissivity is 200*m^2/day but Q is based on only a 10 m depth.


4. Focus and Do

> v=K*i;

> subs(K=10*m/day,i=5*m/(1000*m),%);

The estimated Darcy velocity.

v = K*i

v = 1/20*m/day

> subs(%,A=1000*m*10*m,Q=A*v);

Q = 500*m^3/day

The rough amount of water available to the campus.


5. Solve with units

The units, here, would probably be kilolitres in Perth, a standard used by the Water Corporation.

> subs(m^3=kilolitre,%);

Q = 500*kilolitre/day

Elsewhere, m^3 would be convenient. In the U.S, following the definition of acre-feet above, this is equivalent to

> isolate(1234*m^3=acre*ft,m^3);

m^3 = 1/1234*acre*ft

> subs(%,%%%): evalf(%,2);

Q = .41*acre*ft/day


6. Check the answer

Try working out the problem by first converting the crossection to acres. The use of different units can make an independent cross-check to your workings.

> A=10*m*1000*m;

A = 10000*m^2

> A=1*hectare/(.4047*hectare/acre);

A = 2.470966148*acre

> v=.05*m/day*3.28*ft/m;

v = .1640*ft/day

> subs(%,%%,Q=A*v): evalf(%,2);

Q = .41*acre*ft/day


Exercise A4. Horizontal Flow Affected by Salinity

In the wheatbelt, there is often 'patchiness' in the brackishness of the water; some is higher in salinity than the sea; some is good, potable water. In every case, piezometric measurements should be coupled with salinity measurements. The point water head measured in a saline bore is less than what would be measured in a bore containing fresh water.

Two bores are placed 540 metres apart to measure horizontal flow within a confined aquifer with reasonably uniform geological features. Piezometer A is placed at 210 m AHD and contains fresh water; piezometer B is at the same level but contains saline water with a density of 1.15*gms/(cm^3) A is north of B. The water level in A rises to 222.421 m AHD; the water level in B rises to 221.090 m AHD. What is the total fresh water head (relative to AHD) in piezometer A? What is the total fresh water head (relative to AHD) in piezometer B? If A is north of B, which direction is the water flowing?

1. Recognition

Several features appear, including a confined aquifer, saline water of density 1.15*gms/(cm^3) , and the use of AHD (Australian Height Datum) values directly for measurements of total head. One important feature is absent here; a water table. That is, since the aquifer is confined, and has no unconfined upper surface, there is no water table. That being the case, if we need to represent the pressures as measured in the confined aquifer, below, we could form an artificial surface, the surface that would be formed by a multitude of piezometers, drilled at all points within the confined aquifer. That surface is called the 'piezometric surface' and it indicates the groundwater flow magnitude and direction; it is dashed in the figure below.

This simplest of concepts becomes corrupted, however, if the piezometric fluid density varies. It is a tricky business because the water in our piezometers is saline. Remember that the holes, the piezometers, are no more than pressure metres, where the measure of pressure is the height water rises above the given level. Less water rise will be experienced if the water in the hole is salty, and the measurement of pressure is corrupted. Imagine if mercury were use as the manometer or pressure guage fluid. It does occur naturally, you know, and in places it has been reported that 'swimming holes' of mercury occur. In those 'swimming holes' you could literally 'walk on water'! If, indeed, mercury were in the piezometer as a fluid, the water would rise only

> evalf(1/13.6*100,2)*percent;


of the 'fresh water' value. That is, the pressure measurement would be in error by a factor of 13.6

It is necessary to correct the pressure head measurements to their 'fresh water' values, making them 'fresh water heads', before either the flow direction or the flow magnitude can be assertained; indeed, before the piezometric surface can be properly drawn!


2. Interpretation

The drawing shows the two piezometers. The real value of the measured head, the 'point water head', is higher in A than in B. But don't let that deceive you! Note the uncorrected piezometric surface, which doesn't correspond to a water table or the proper piezometric surface! The two confining surfaces could be of any depth and even above the piezometric surface. The thicknesses of the confining layers are unknown, but it is presumed that, in the least, the transmissivity of the confined aquifer is constant, if indeed, the thickness varies. Note that the drawing is not done to scale, except for preserving the relative point water head values of the two holes.

Again, as in Example A1, we have horizontal flow and we need to worry about pressure head and elevation head separately. the form of Darcy's law likely to be of most use is

v = K*(h[A]+z[A]-(h[B]+z[B]))/L


3. Related Information

Pressure is directly proportional to density, that is the basis of the 'hydrostatic equation'. Pressure can also be considered as a potential energy per unit volume. Head, similarily, can be considered potential energy per unit weight. In primary units this all becomes a little clearer:

> pascal=kg*m/sec^2/m^2;

pascal = kg/(m*sec^2)

> rhs(%)/(kg/m^3*m/sec^2);


The denominator is the weight per unit volume. This produces what we consider 'metres of head'.

An alternate view is that the formula p = rho*g*h applies; this should be clear if you are a diver or from your basic physics or atmospheric science or any scientific discipline.

It is then clear (see Fetter, pg 139 or the Practical Manual, Darcy's Law 2) that the fresh water head values are obtained from the point water heads by multiplying by the density (or specific gravity or density ratio)

`fresh water head` = `point water head`*rho[`point ...

Of course, in units of gms/(cm^3) , fresh water has a density of 1. This means we just multiply by the density in gms/(cm^3) .

Importantly, do not apply the density correction to the elevation head. It is already 'normalised' and in 'per unit weight units'. The effect of density is only on the pressure measured and a distortion of the length that would be measured in the manometric height in the piezometer, above the specific position.


4. Focus and Do

first we calculate the 'fresh water head' values in both piezometers. Piezometer A has fresh water in it so the measured value is the appropriate fresh water head value.

> h[A]=12.421*m;

> h[B]=11.090*m*(1.15*gms/cm^3/(1*gms/cm^3));

h[A] = 12.421*m

h[B] = 12.75350*m

All the known values are now substituted into the hydraulic gradient. This gives the direction of flow.

> values:=%%,z[A]=210*m,%,z[B]=210*m,L=540*m;

values := h[A] = 12.421*m, z[A] = 210*m, h[B] = 12....

> i = ((h[A]+z[A])-(h[B]+z[B]))/L;

> subs(values,%);

i = (h[A]+z[A]-h[B]-z[B])/L

i = -.6157407408e-3

This negative value means that the flow is not from A to B. When corrected for density, to produce 'fresh water heads', the flow is from B to A. The flow is from the south to the north.

5. Solve with units

Again, the units are properly cared for. As an exercise, however, we now work out the real pressures (in Pascals) at points A and B. To avoid confusion, the measured heights at A and B are entered directly into the equation for the pressure p = rho*g*h . First, however, we go through a simplest of units conversion, to be sure we don't make errors.

> {rho[A]=1*gms/cm^3,rho[B]=1.15*gms/cm^3}:

> solve({1*kg=1000*gms,m=100*cm},{gms,cm}):

> subs(%,%%);

{rho[B] = 1150.00*kg/(m^3), rho[A] = 1000*kg/(m^3)}...

Now we substitute values into the hydrostatic or pressure equation. To keep down confusion about the true value of the pressure heads, the measured heights are inserted directly into the equation.

> subs(%,

> subs(g=9.806*m/sec^2,%);

{p[B] = 125060.8210*kg/(m*sec^2), p[A] = 121800.326...

A conversion to Pascals gives the expected units for pressure at the specific positions A and B.

> subs(kg/(m*sec^2)=Pascal,%);

{p[A] = 121800.3260*Pascal, p[B] = 125060.8210*Pasc...

The vertical locations of both piezometers are the same so it is clear that the flow is from B to A, or south to north. Calculating the gradient requires dividing by the weight of fresh water per m^3 . The density of fresh water is 1000*kg/m^3; gravity is 9.806*m/sec^2; the product is the force exerted by gravity per unit volume of fresh water, the weight per unit volume.

> subs(%%,(p[A]-p[B])/(1000*kg/m^3*9.806*m/sec^2));


This is the height difference (the pressure head difference) between A and B. The hydraulic gradient is this value divided by the distance between the two piezometers.

> i=%/(540*m);

i = -.6157407409e-3

6. Check the answer

The above is a check, but we can go further. For instance, what would be the height of water in piezometer A if it were brackish, the 'point water head'? Even though this information is useless for calculation of the actual flow, it should give an indication of direction. The height of water, written in a simplest way, would be

> 12.421*m/1.15;


A value smaller than the 'point water value', a value that would be measured, were piezometer A filled with brackish water. Subtract the comparable level in piezometer B. The negative value varifies that the flow should be from south to north.

> %-11.090*m;


Exercise A5. Flow Through an Aquitard

An upper, unconfined aquifer overlies a confined aquifer. The confining layer in between is 0.5 metres thick. An exploration hole reveals that the water table is 4.32 metres above the confining layer. The hole is developed into a piezometer with a PVC casing installed just above the confining layer; when equilibrated, the piezometric measurement is 3.82 metres above the confining layer. A further redevelopment of the hole (removal of the casing and redrilling) places the piezometer just below the confining layer; the equilibrated piezometric measurement is 2.51 m, with reference to the bottom of the confining layer. If the water is all fresh and the K[z] of the unconfined aquifer is 1 metre/day, what is the recharge rate into the confined aquifer from the upper unconfined aquifer? What is the hydraulic conductivity of the confining layer?

1. Recognition

Another vertical flow problem, as might emerge from the usual, multiple aquifer systems or simply an investigation of seepage flow into lakes and rivers. In lakes and rivers, the piezometers might be installed directly below the water body, and the upper water table measurement is made directly on the lake, and is the lake surface. Steady flow is being considered, with no changes in time. Also, only a single flow direction is considered, the vertical, using the Dupuit-Forcheimer approximation.

The flow within the upper, confined aquifer is calculated first, from the phreatic or piezometric measurements. This flow must pass through the confining layer below. Since there is fresh water everywhere, it is clear that the flow is downward. The upper measurement, at the water table, is greater in an absolute sense than the lower measurement, just above the confining layer. Hence, we designate the upper piezometer 1 and the lower piezometer 2. In reality, there is but one hole, but our 'model' requires a piezometer at the water table, another one just above the confining layer as well as a third one, piezometer 3, just below the confining layer. This last piezometer allows an estimate of the hydraulic conductivity of the confining layer.


2. Interpretation

In this case, we write Darcy's law twice, once between positions 1 and 2 and, separately, between 2 and 3.

v = K*(h[1]+z[1]-(h[2]+z[2]))/L

v = K*(h[2]+z[2]-(h[3]+z[3]))/L

It is important to get the references on the same basis, which we will take to be the bottom of the confining layer.


3. Related Information

If we were investigating the 'hydraulic resistance', d/K [=] days, the effective confining layer would be at the bottom of the lake, otherwise the equation set would be the same. The use of a known medium to investigate the unknown is a standard approach in science, and in hydrogeology.

Often, such problems are algebraic, with two or more unknowns that must be solved for simultaneously (see, e. g., Bouwer, pg 53); this should not cause any stress, but allow you to revise your algebra. Again, the use of a symbolic manipulator can easily solve some of the more complex problems of this sort.

Were we investigating the 'negative pressures' within a soil, a similar method might apply, except the hydraulic conductivity would be a function of history and water content. All these calculations are strickly valid only in equilibrium flow, that is, steady flow that doesn't change in time or kind. Still, the solutions are 'first guesses' and can provide valuable insight into environmental hydrology.


4. Focus and Do

Plug the data into the two Darcy equations

> v = K[z]*(h[1]+z[1]-(h[2]+z[2]))/L;

v = K[z]*(h[1]+z[1]-h[2]-z[2])/L

> subs(K[z]=1*m/day,L=4.322*m,

v = .1161499306*m/day

The Darcy velocity in the second equation produces the wanted solution. Be careful with the measurements; follow the diagram carefully. Remember the 3rd piezometer is measured relative to the bottom of the confining layer, so z[3] = 0 .

> v = K*(h[2]+z[2]-(h[3]+z[3]))/L;

v = K*(h[2]+z[2]-h[3]-z[3])/L

> subs(%%, L=0.5*m,

> h[2]=3.82*m,z[2]=.5*m,h[3]=2.51*m,z[3]=0,%);

.1161499306*m/day = 3.620000000*K

> K[`confining layer`]=solve(%,K);

K[`confining layer`] = .3208561619e-1*m/day


5. Solve with units

Round off properly. The units are better in cm/day .

> subs(m=100*cm,%): evalf(%,2);

K[`confining layer`] = 3.2*cm/day


6. Check the answer

The confining layer has a K value only 3% of the upper aquifer, and this is expected. However, this is not a check of the workings. Try calculating the Darcy velocity using the calculated K and data from positions 2 and 3.

> v = K*(h[2]+z[2]-(h[3]+z[3]))/L;

> subs(%%, L=0.5*m,K=.032*m/day,

> h[2]=3.82*m,z[2]=.5*m,h[3]=2.51*m,z[3]=0,%);

v = K*(h[2]+z[2]-h[3]-z[3])/L

v = .1158400000*m/day

Further back calculate to get z[1] . This could indirectly locate the height of the water table, provided there were equilibrium flow. Note that L = z[1]-.5*m

> v = K*(h[1]+z[1]-(h[2]+z[2]))/(z[1]-.5*m);

v = K*(h[1]+z[1]-h[2]-z[2])/(z[1]-.5*m)

> subs(v=0.1161*m/day, K=1*m/day,

> h[1]=0,h[2]=3.82*m,z[2]=.5*m,%);

.1161*m/day = m*(z[1]-4.32*m)/(day*(z[1]-.5*m))

> solve(%,z[1]);


The height measured in piezometer 1 is

> %-.5*m: evalf(%,4);