Water Resources Engineering
Prof.

R. Srivastava

Indian Institute of Technology, Kanpur
Lecture

No. # 08

In the previous lecture we had looked at the groundwater which is the water in the sub
surface and we saw that if we want to utilise the groundwater, we have to know the
amount of water present, the amount available for withdrawal and how fast we can
withdraw it. The other thing which we should look at is how deep the water table is
because that will affect the cost of pumping. So the groundwater table has let us say
horizontal level initially but when we start pumping it. It will start to go down with time
and we should know how this water table will change with time so that we know how we
have to lower the pump if the water level goes down very much and we had looked at a
mass balance or continuity equation in which we take an element of the aquifer.
(Refer Slide Time: 01:14)

We look at the mass of water owing into this element and the change of mass within
this element and the continuity equation. This tells us that these should be balanced and
looking at that mass balance will allow us to determine how the water level or the
pressure in the aquifer is changing at that point. So in order to look at the mass balance
we should know the amount coming in and the amount going out. If we take an
elementary volume like this, we have delta X delta y and delta 2 then we can nd out the
storage of water within this volume and how it is changing with time. So we need to
know two things. One is the amount of water coming in, amount of water going out. This
will depend on the velocity which can be obtained from the Darcy's
law and therefore it
depends on the hydraulic conductivity K and the hydraulic gradient i. Q also depends on
the area of cross section which in this case will be delta X delta y or delta y delta 2,
depending on the phase we are considering. So combining Darcy law and the area,
velocity term, will give us an idea about Q. The other thing which we need to look at is

the storage within this volume, the amount of water that comes out of this storage for a
unit drop of head. We shall look into these things today. Let us first look at the hydraulic
conductivity K. We had discussed briey that this K is a function of both the uid
property and the medium property. If we take the analogy with ow in circular pipes, in
which let us say we have a drop of head equal to delta h in a length of L then for laminar
ow, we know the well known equation for the head loss in which mu is the, dynamic
viscosity of the uid V is an average velocity, gamma is the specific weight and D is the
diameter of pipe. If we want to compare this with ow through porous media, we can
think of the D as being the diameter of the particles because that will be related to the
diameter of the pores. So in the porous media the equation which we have is the Darcys
law which says that V = K delta h over L. So if we compare these equations, the K will
come out to be dependent on the medium property, the diameter of the particle and it will
also depend on the uid property, specific weight and the dynamic viscosity as gamma
over mu. This is uid property and d square denotes the porous medium property. So
typically gamma over mu is taken out of this equation.
(Refer Slide Time: 06:53)

The hydraulic conductivity K is written as gamma over mu times some intrinsic
permeability or specic permeability which is only a property of the medium. This K is
some constant into the grain size square based on the pipe ow analogy. For different
porous medium for example for sand and clay, the value of C may be different and there
are a lot of empirical equations which are used to correlate the intrinsic permeability with
grain size. Gamma over mu varies with the uid. If we have ow of water, this value
will be different, for oil it will be different, but in general, for groundwater ow, we have
ow of water. Even for water, the value of gamma and mu may be different at different
temperatures. In general in groundwater ow, we will assume that the temperature
variation is very small and water is the owing uid. Therefore K is commonly used and
not the intrinsic permeability. Commonly used term is the hydraulic conductivity since
water is the owing uid and temperature can be assumedto be constant. If not then it is
better to use the intrinsic permeability becausethat will not change with uid properties.
The second thing which we want to see is the amount of water that can come out of the
aquifer for drop of head.

(Refer Slide Time: 08:56)

So for a given drop of head, supposethere is this volume which has some solid particles
and the rest filled with water. We have already discussed that we will only consider
saturated ow conditions. So there is no air inside this. Now this voluIne is subjected to
some pressure and we will start with the conned aquifer case because that is easier to
derive. If we put a tube here, a piezometer here, water will rise up to a certain level,
which may be below the ground level. It may even be above the ground level. In this
case it is called a owing aquifer. Now suppose the piezometeric head is here, what is
required to be seen is the amount of water that will come out, if we lower the
piezometeric head by certain amount delta h. This lowering of the head may be because
of pumping or it may be because of head drop in some other area where this aquifer is
connected. But let us take this delta h as the head drop and if we take an element of the
aquifer here, let us say we take some element here; we want to see the amount of water
that can come out of this aquifer for head drop of delta h.
Now let us look at the mechanism of how the water comes out for unconfined aquifer. It
is quiet straight forward. This is the groundwater table bedrock. If we lower the
groundwater table by some amount delta h, water will come out because of drainage of
this volume. But in unconfined and in conned aquifers there is no change or not much
change in volume. There is little change which we will see.But there is not much change
in the volume. Therefore the water which comes out is because of the compressibility
and when we talk about compressibility, it would be compressibility of water as well as
the compressibility of the medium. So we can call that aquifer or formation
compressibility. When we lower the piezometeric head, the water from this element,
supposecomes out of the element, this water is under some pressure. It has not been kept
constant but now it has been lowered because of this lowering delta h. When we reduce
the pressure,water will expand and therefore it will come out of this element.
The other thing which happensis becauseof this lowering of pressure in the water, since
the total stressremains same, the over burden pressure remains same. Lowering of water
pressure means increase on the grain pressure. If there is some grain pressure, let us call
the pressure sigma. This acts on the grains and will increase if we lower the value of

(Refer Slide Time: 13:09). So the pressure in the water is let us say p, the pressure on the
grains is sigma, now sigma + p will be the total pressure which remains constant. By
lowering the piezometeric head, we are lowering p and therefore sigma will increase.
Due to this, the aquifer or the formation will be compressed (becauseof this increase in
stress) and that compression will lead to further release of water from this element. We
shall look at ways to derive these terms which represent release of water due to metrics
compressibility or the formation compressibility, release of water due to water
compressibility. Let us define a term which is known as the storage coefficient.
(Refer Slide Time: 14:07)

This is denoted by S. S is the volume of water released from the formation from a prism
of unit cross section area with unit drop of piezometric head so that 3 since here that S is
the volume and it is from a prism of unit cross section area for a unit drop of piezometric
head. So if we take this conned aquifer of height B and considered a unit area of (Refer
Slide Time: 16:05), this area is 1 and this has some piezometric head which we will call
h, the datum can be anywhere. The datum is taken at the base of the aquifer that is at the
impermeable rock.
Now this is some value of piezometric head and now if we lower this by a unit amount,
we determine the amount of water that will come out of storage. So that term is denoted
by the storage coefcient S. Now the mass of water which is present inside this prism M
can be related to the volume and the density is rho. Rho is the density of water into the
volume which will be (since the area of cross section has been taken as unity); the
volume will be equal to B into porosity. The mass of water contained within this
confined aquifer, the prism of base area 1 can be written as rho p eta which is the
porosity and therefore if this mass is changing, it will change because of change in all
these parameters. Suppose we have some change here, delta B, and some change here
delta rho, then when we are lowering the piezometeric head, the aquifer thickness B will
also change. As we have seenthat, by lowering p we are increasing the effective pressure
on the grains and therefore the aquifer will be compressed. B will reduce the porosity.
Similarly it will also change. So all these three parameters change and their net effect
will cause some change in mass delta M. Our aim is to find out this delta M becausewe

can then relate the storage coefficient S with delta M. Let us look at how to find out this
delta M for certain change in pressure or the piezometeric head h.
(Refer Slide Time: 19:01)

We now start with our original mass. Since all these three can change, we can write this
as (Refer Slide Time: 19:17), so change in porosity will cause some mass change in
formation thickness. So this is the effect of change in porosity formation thickness and
density. Total change in mass occurs because of change in porosity, change in the
formation thickness and change in the density of water. We need to evaluate these
changes and for that we introduce a property which is known as compressibility. This is
inverse of the modulus of velocity and is dened as change in volume per unit volume
divided by a change in pressure. This is basically a strain over stress. For water, we can
define a compressibility which we can say is represented by beta. In general a positive
change in pressure will cause a negative change in volume and therefore we put in
negative sign here, where Vw is the volume of water and delta Vw is the change in this
volume. Delta p of course is the change in pressure. As we know, the piezometeric head
h is p over gama + z, where p is the pressure and z is the elevation. Therefore a change in
pressure, delta p can be written as gama into change in piezometeric head, delta h. So we
will be using this relation later to nd out the storage coefcient. So let us look at the
water compressibility equation.

(Refer Slide Time: 22:14)

Now Vw is the volume of water and therefore if we write rho Vw, this will be the mass of

water, Vw will be the volume within the aquifer. Now if you look at this mass of water
rho, Vw continuity tells us that matter cannot be created or destroyed. Therefore this
mass should remain constant. So we can relate a change in volume of water with a
change in density. So if this is not changing, then the sum of these two changesshould be
equal to 0 and from here we can find out the term delta Vw over Vw will be equal to
delta rho over rho. We can write this as delta rho over rho delta p and therefore the
change in density of water can be written in terms of its compressibility as delta rho, beta
rho delta. So using this equation, we have related the change in the density of water with
compressibility of water, its density and the change in the pressure.The next thing which
we look at is compressibility of the formation of the aquifer material. We define the
formation compressibility or aquifer compressibility and use a symbol alpha. In the
same way as we did for water, we now denote the volume of the formation or the aquifer.
So in this case since we are taking unit area we can say that V will be equal to B into 1
and delta V is the change in V, delta p is the pressure.
Since we are using p as a symbol of pressure in the water, we can use a different symbol
here and we can write pf as delta p in the formation or we can denote this term also by
sigma. We can use delta pf or delta sigma for this term. We can nd out delta b from
here, delta rho from here. If we look at the equation for delta M, we had delta rho, delta
B and delta eta. So we need these three terms to find out the change in mass using this
equation. Delta rho can be related with the water compressibility. Delta B can be
obtained from here over B delta sigma. We have also seen earlier that sigma + p is
constant therefore delta sigma will be equal to delta and therefore we can also write
delta B from here as alpha B delta. The third term which we need is how the porosity is
changing with change in pressure and for that we can assume what is generally a very
good assumption.

(Refer Slide Time: 28:02)

The volume of solids is constant, so if we have an aquifer material here, no matter how
much the pressure is changing in this aquifer, water pressure and the grain pressure, the
amount of solid material present in this area, volume of solids which we denote as Vs,
remains the same becausethe grains are assumedto be incompressible. Now Vs can be
written as 1

eta. If the thickness as we have taken is B, and cross section area is 1, then

the amount of solid volume present inside this unit area and height B will be 1 eta
times B, becausetotal volume is B. Out of that, eta B is the volume of the liquid water in
this case. Therefore total volume of solids Vs will be equal to 1- eta B and if this is a
constant, then we can use the same for formulation

as we use before, that delta Vs will be

equal to 0 and this would imply that B times minus (Refer Slide Time: 30:29) equal to 0.
From here we can obtain delta eta which is the change in porosity. We can relate it with
the change in the formation thickness B.
Now we have all three parts of the equation, formulated delta eta from here which is a
function of delta B, delta B we have from here and delta rho, we have from this equation
so putting all these three in this equation which tells us total change in mass can be
obtained as delta M in terms of the compressibility of the water, alpha and the
compressibility of the aquifer formation beta. So we write delta M in terms of the aquifer
thickness B into alpha + the porosity into beta multiplied by the specific weight gamma.
Now we will need this because we have said that we want to find out a storage
coefcient S which is the volume releasedper unit change in h. This h is the piezometric
head and is given by pl gama + Z. We have related all the changeswith delta p, so delta p
can be replaced by gama delta h. We would write in terms of delta p and then write that
delta p as gama delta h, which is nothing but delta p. Then to nd out a storage
coefcient, we know this is the volume and we have written our equation in terms of
mass. S will be equal to the mass released divide by rho per unit change in the
piezometeric head. Therefore S will turn out to be gama alpha + eta beta into B. This
equation gives us the storage coefficient and representsthe volume of water released for
a prism of unit cross section area over the entire thickness B.

(Refer Slide Time: 34: 10)

Similar to the storage coefficient S, we have a term, which is known as specic storage
S5 and this is the volume of water released from unit volume. so if we compare it with
the denition of the storage coefficient, if we divide S by B, we would get SSbecauseS
was the volume released from the entire thickness B and therefore if you take unit
volume, the amount of water released will be given by S over B and then we can write
this too in terms of the compressibility and the porosity as gama alpha + eta beta.Ss will
be used when we derive our continuity equation for a conned aquifer case. For
unconfined aquifer, the storage coefcient S is equal to specific yield although there is a
small component which comes from the compressibility. It can be ignored with the

respect to Sy. So compressibility terms are generally negligible and therefore for
unconfinedaquifer,the storagecoefcient S is takenas Sy.Using this storagecoefcient
S, specific storage S5, we can derive the continuity equation and that gives us the
relation between the head, the peizometeric head in the aquifer and it relates with the
amount of water we are pumping out of the aquifer. As we have seen, the storage
coefcient can be defined for unconned and confined aquifers. The specic storage for
confined aquifers can also be defined and we will use that, to derive the equation of
motion for a conned aquifer.

(Refer Slide Time: 36:57)

This is the conning layer and at the bottom we have this bedrock. So in this confined
aquifer, we can look at an elementary volume and see how the head change affects the
amount of storage and the amount of water which is coIr1ing in and then by balancing
these two, we can derive an equation which will tell us how the head is changing with
time and space. So this spatial and temporal variation of the head can be obtained by the
continuity equation. So if we take this element, let us enlarge it. We will be using the
Cartesian coordinate system but sometimes it is better to use a radial coordinate system.
For example if you are pumping from a well, then the piezometeric head will be
symmetric about the well in the radial direction and therefore, it would be convenient
using a radial coordinate system. But let us start with the Cartesian coordinate system.
We take this elementary volume which has size delta X, delta y and delta 2. Of course our
X, y and 2 directions are like this.
Let us consider the ow in the X direction. Same thing can be done in y and z direction
also. But let us start with the X direction and say that there is some velocity qX which is
the Darcy velocity in this case and not the seapagevelocity. Because of this velocity qX,
the mass which is coming in the mass uX will be rho into qX. What we want to see is
what is the net mass owing into this control volume. But the mass which is going out
can be written by using rho qX which is the mass coming in from the left phase + the rate
of change multiplied by the distance delta X. Sometimes we assumethat this whole term
rho is inside the differential here. So when we take del by delta X, we put rho inside the
derivative but the variation of rho is very small with respect to the other term which is
del qX by del X. If we write rho del qX/del X and write qX, then compared to this term, it
can be neglected and therefore generally we write this term only as rho del qX/del X. We
will be making an assumption that the change in density can be ignored and therefore it
can be taken out of the derivative and we can write as rho del qX/del X.

(Refer Slide Time: 41 :38)

Therefore the net inow would be rho qX which is the inow from the left phase minus
the outow from the right phase which is rho qX plus and using our assumption, we have
taken rho out of the derivative del/del X of qX into delta X and this will result in minus
rho. Now this is the mass uX rate per unit area. We have multiplied it by the area. We
can say that the net inow per unit area is given by this and therefore inow is in the X
direction. Remember that we are talking only about the X direction. Let us do other
directions. Inow in X direction can be written as minus rho multiplied by the area which
is in this case, delta y delta 2. As you can see that this phase has an area delta y delta 2,
this term gives us the mass coIning in the element in the X direction. We can now use
Darcy law to relate it with the head.
As we know qX now, we will make another assumption here. We have a medium which
is isotropic. If we make that assumption, then we do not need to use this subscript X. For
isotropic medium, we can remove this X because K will be same in all direction.
Sometimes the K is not same in all directions. For eXample if you have a layered
formation, then K will be different for ow along the layers and it will be different for
ow perpendicular to the layers. But here we assume that the medium is isotropic and
therefore we will ignore this subscript X. Using Darcy law, we can write qX as K
delh/del

X and

therefore

from

this

we

can write

the inow

in the X direction

as rho

K.

This term representsthe mass of water owing in the X direction. Net inow of mass into
this control volume delta X, delta y, delta 2 and similar eXpressioncan be derived for y
and 2 directions.

10

(Refer Slide Time: 45:31)

All of them can be added together to get the total or the net mass inow into the system
as rho K. Delta X, delta y, delta 2 can be combined as delta V. This is the Volume of the
element into; we have the contribution from the X direction, del two h/del X square and
similar terms for y and z direction. Now the continuity equation says that this net mass
inow should be equal to the change of storage within the control Volume. Now if we
use the storage coefficient S, we have seenthat S representsthe Volume releasedper unit
cross section area using a prism but if you want to use elements with area delta X delta y
and delta 2 we would need to use specific storage rather than a storage coefcient. So we
use S5 which is the specic storage that represents the Volume released from a unit
Volume. So Volume of water released from unit aquifer Volume for unit head drop.
If you want to find out the Volume of water released from this elemental Volume for a
head drop of delta h, we can write an equation which tells us that mass released per unit
time from this elementary Volume can be written as rho Ss delta V delta h oVer delta t and
the way we write this eXpressionis that Ss is from a unit Volume, therefore we have to
multiply with delta h. Ss gives us the Volume and therefore we multiply with rho to get
the mass. The change of mass or unit time in the elementary Volume delta In/delta t can
be related with the specic storage Ss the density delta V delta h and delta t and in the
limit as delta t tends to 0, we can write (Refer Slide Time: 49: 39) and the other limit

which we will take is, when this element reduces to a point that delta V tends to 0. So
when we equate net mass inow which is given by this eXpressionwith change of mass
within the control Volume. So these two terms should be equal and therefore delta V will
cancel out, rho will also cancel out and we should note that rho will cancel out because

we have assumedthat change of rho is Very small or negligible. If we do not ignore that
term, then rho will also be included in the equations.

11

(Refer Slide Time: 50:33)

If we equate these two, we nally get an equation which can be written as del square h.
This equation representsthat we have made some assumptions. So I will write them here.
First assumption is that it is a conned aquifer. Second assumption which we have made
is that the aquifer is isotropic. That is the reason why we have only K. Otherwise we will
have Kx, Ky, Kz, three different terms which will be included in this. The term del
square can be expanded in Cartesian coordinates. It has a very simple form and if you
have some so situation where you have radial and X is symmetric ow. Symmetric means
that theta will not be coming into the equation. Then you get a term in which there is del
h/del theta = 0. So that term has not been considered as del 2h/del theta square term
which has been neglected becauseh is not a function of theta. Using this equation we can
obtain the variation of h for a given condition.
We can also write this equation in a little different form by noting that the storage
coefcient S is specic storage into the depth of aquifer and we have already defined the
transmissivity as K into B. So if we use this S and T, we can also write this equation in
terms of the storage coefcient and the transmissivity. So either we can use specic
storage and hydraulic conductivity or we can use storage coefficient and transIr1issivity
and solve this equation for given conditions. Since this is second order equation space,
rst order in time, we need one initial condition and two boundary conditions to solve
this equation.

12

(Refer Slide Time: 54:22)

For example if we consider the case of one dimensional ow, we have a confining layer
ground level and suppose we are looking at a case where ow is occurring between two
water bodies, one of them here is such that the height of water is h1 and the height of
water here is hz. The piezometeric head in the confined aquifer may be like this, or like
this or it may be a straight line like this. So in order to nd out which is the variation in
the confined aquifer, we need to solve the equation of motion which is again written here
as this. Now if we make the assumption that the ow is steady then del h by del t
becausesteady means the parametersnot changing with time so del h by del t will be = 0
and therefore we will get a simple equation which is known as the Laplace equation.
This equation has a form similar to the head diffusion equation therefore this is also
called the diffusion equation. Now let us say that this is a steady state ow condition
therefore

we will

need

to solve

this

and let us also

assume

that

this

is a one dimensional

ow. So the ow is taking place only in X direction. So we can write since the head is
not changing in other directions, this equation simplies to del 2 h/del X square = 0. The
solution of which is h = C1 X + C2. C1 and C2 are constants which will be obtained by
the boundary conditions. In this case, considering the length of the aquifer to be L, the
boundary conditions are at X = 0, h = h1 and at X = L, h = hz. So using these two
boundary conditions, we can obtain the solution of this equation which tells us that h is
linear. In this case if we have a ow between these two water bodies, the piezometeric
head in the aquifer will vary linearly between h1 and hz and the variation can be obtained
directly from applying these two boundary conditions from which we can write h = h1
(Refer Slide Time: 58:35). So this is obtained by applying these two boundary conditions
in this equation.

13

(Refer Slide Time: 58:55)

If we want to nd out the ow rate, we can use the Darcys law which gives us q = Ki, i
in this case is del h/del X, because we are considering one dimensional ow which is
simply h1 hz over L. Since the piezometeric head is linear, i is constant throughout the
length and is given by the delta h which is h1h2 divided by L and this will give us the
apparent velocity or Darcy velocity using which we can nd out q into area. If we
consider unit width, then area will be equal to the thickness of the aquifer which is b. So
we can write this as K h1 hz over L into B. We have seen today how to obtain the
equation of motion for combined aquifer, how to solve it for given boundary conditions
and the same thing can be done for unconned aquifer. The thing is that unconned
aquifer is a little more complicated because the water level itself determines the
thickness of the aquifer. Thickness of the aquifer is not constant but is varying from
place to place. So we need to make some simplifying assumptions in order to derive that
equation. Similarly for conned aquifers also we have solved the equation for a one
dimensional steady state ow conditions.
If we go for unsteady ow, it will be a little more complicated or if we go for radial
coordinate system, the solution will be a little more complicated. So we will look at these
solutions for conned aquifer and we will also look at how to derive the equation of
motion for unconfined aquifer in the next lecture.

14