# Mod-01 Lec-12 Steady Flow into Wells (Contd.); Unsteady Flow into Wells T
Welcome to this lecture number twelve so in which we will continue with our previous lectures
ahh ahhh article on steady flow through wells, So in lecture number eleven we discuss the
dteady through a confined into well through a confined aquifer So in which the confined
aquifer thickness was B and the well was penetrating the entire thickness of the confined aquifer
and the water was horizontal before the the pumping began which resulted in steady flow
into the aquifer into the well and the which also resulted in steady discharge of Q from
the well And we got the expression for the ahhh the discharge which is popularly known
as the yeah this is the expression And yeah the original expression in terms of the
hydraulic conductivity is Q=2 PIE and K into B H2 – H1 divided by natural log of
R2 by R1. So this is popularly known as themes equation and this when we simplify
and assume applying the boundary conditions at the extreme points that is the extreme
point when in the there is no drawdown and the radius is equal to radius of influence
R capital R And H the head is equal to the total this head or the piezometric head and
that so this is the ahhh the upstream most extreme the downstream most extreme is the
cylindrical where this S the drawdown is equal to drawdown in the well equal to SW R1is equal
to radius of the well that is RW and H1 the head is equal to the the depth of the water
in the well in the confined aquifer So we get this expression and these expression can
be used ahhh for with in case of pumping test to determine the hydraulic and aquifer parameters
such as the hydraulic conductivity and tranmissivity. So therefore so these expressions
can be used in steady flow pumping test for fully penetrating wells in confined aquifer
for determining the hydraulic conductivity K and transmissibility
or transmissibility T of the aquifer So here we get an expression for the hydraulic conductivity
K for the steady pumping test through the confined aquifer having a fully penetrating
well as ah well as Q divided by 2 PIE B into H2 – H1 into natural log off R2 by R1. And
the same thing can also be written as Q divided by 2 PIE B and in this case that is SW and
natural log off R divided by RW. Now here so this is the expression
for K in a steady flow pumping test through a fully penetrating well in
a confined aquifer Similarly the same expression using this equation
you can write down the expression for T as Q divided by 2 PI H2 – H1 into natural log
of R2 by R1 this can also be written as Q divided by 2 PI into SW into natural log of
R the radius of influence divided by RW the well radius. So this is the expression for
K comma T so this is the expression for K and this is expression for T. So in a steady
flow pumping test through a fully penetrating well in a confined aquifer So we need to know
to determine this hydraulic conductivity K in transmissivity T we need to know the steady
flow rate from this fully penetrating well in a confined aquifer. We need to know the
thickness of the confined aquifer we need to know either the depths between two the
observation wells the depths of water in the two observations wells that is H1 and H2 and
their radial with respect to the their radial distance with respect to the well access the
vertical well axis R1 and R2 And either we need to know this H1 and H2, R1 and R2 or
we need to know the well radius RW the drawdown in the value SW and the radius of influence
in the capital R and so in this case so we can easily determine easily estimate the value
of the two of the three aquifer parameters formation consents that is the hydraulic conductivity
K and transmissivity or transmissibility T. Now let us go to the case of steady flow
through a fully penetrating well in an unconfined aquifer
So far I have the seen the steady flow through a fully penetrating well in an confined aquifer
and now let us g to this steady flow through a fully penetrating well in a unconfined aquifer
So for this the diagram would be there is a well which is fully penetrating through
an unconfined aquifer. And in this case this is the water surface this is the water table
before the pumping started and after the pumping started so this water
table will show the steady draw down. So this is the well axis and this is the steady discharge
from this well in the unconfined aquifer and this is the lower impervious strata for the
unconfined aquifer And this is the radius of the well is RW and
the diameter is 2RW and this is the the radius of the influence or radius of the
circle of the influence is R and the head their measured above the bottom impervious
layer as capital H and at any two points and here what happens is so this is the we need to apply the dupuit’s assumptions.
And so this is the water table by Dupuit’s assumption and this is the true water table
And here this is a H1 and drawdown there is H1 and similarly for the second observation
well the head is H2 and the draw down there is S2. And obviously here the the Dupuit’s
assumptions which we discussed in the previous lecture or have considered that is applied.
So the Dupuit’s assumptions accordingly that is a say for small inclinations
of free surface the stream lines can be approximated as approximated or assumed
as horizontal And second assumption is the hydraulic grade
the hydraulic gradient is equal to slope of free surface
and is independent of depth is independent let me write here independent of depth. So
here the first assumption that is the streamlines are assumed to be horizontal so this is a
more or less true to accept close the well where this stream lines are significantly
deviating from this one but still with these assumption the that means the error in estimating
the discharge is a quite less So therefore because if we ahhh consider the ahh the inclination
of stream line near to the well s then we have to modify our expression and for the
discharge and it will make it more complicated. So therefore we are making the the to Dupuit’s
assumption and this one. So here the this one so let me show you the the flow that is
radially into the well and here it is a HW and this is a SW okay and it any general point
this is H at distance of R from the well axis the drawdown is S okay So now we will derive
the expression for this steady flow so we know that the the radially involved velocity
VR is given by K into DH by DR this is the Darcy’s Law and so this is a by Darcy’s
law So therefore the study flow rate through the well is given by this VR into the area
which contributes and that area which is ahh contributing in this case will be=2 PI R
which is the ahhh the circumference of the circle like that radial distance R multiplied
by H which is the ahhh depth of water at measured with respect to the horizontal impervious
flow multiplied by this is the area multiplied by the velocity that is VR and this is=2
PI RH into K into DH by DR So therefore we can write this as Q divided by 2 PI into K
into DR by R this is=HDH okay. So this expression needs to be integrated that is between the
limits R1 the lower limit R1 and the upper limit R2 and for the H the lower limit is
H1 the upper limit is H2. So therefore we get Q=PI K H2 square – H1 square divided
by natural log of R2 by R1. And in this case so this is the expression or the flow rate
so at the extreme points so here this case that is H2=H R2=R and
H1=HW, R1=R2 So therefore this expression can be re-written as Q=PI into K into H
square – HW square divided by natural log of R divide by RW. So these are the expressions
for the steady flow rate in terms of hydraulic conductivity KH which is the the depth of
water above the impervious layer the in the unconfined aquifer HW is the depth of water
in the well and R is the radius of the influence RW is the well radius and of course if you
express a same this for two observation wells in that case so this is the expression to
observation wells at a distance of R1 and R2 at a radial distance of R1 and R2 where
the depth of water is H1 and H2. So these are the this is the expression so now let
us so in this case it should be noted it is to be noted
that the time required for achieving a steady state is longer for unconfined aquifer And so these two expressions let us write
down the approximate forms approximate forms of equation the steady flow
rate Q. So when SW that is the drawdown in the well very small so in that case that is
the SW=H – HW is a small relative to H therefore H, H square – HW square you can
write this as H + HW into H – HW. So this H + HW is approximately=2H and H – HW
is approximately=SW So therefore the approximate expressions for the steady flow rate=2Pi
that is Pi into K into 2 H SW divided by natural log of R divided by RW and this can be written
as 2 Pi into T into SW divided by natural log of R divide by RW. So these are
the approxi this is one of approximate expression for Q when the SW is very small Similarly this we can also write this Q is approximately
=2 Pi into T into S1 – S2 divided by natural log of R2 by R1. So in this case that is so here that is H2 square – H1 square so
here that is assuming H2 square – H1 square=H2 + H1 into H2 – H1 and this H2 + H we
can take this to the approximately equal to that is 2T 2 that is H into and this S2 so
this S2 H2 – H1=S2 – S1 So therefore we get this expression that is so these two expression will give the approximate
expression for steady flow rate through this fully penetrating well in an unconfined aquifer
and obviously their almost same as the expression for steady slow through a fully penetrating
well in a confined aquifer. So now we have discussed the steady flow through a penetrating
well in an unconfined aquifer as well as a fully penetrating well in an confined And
now in this case so here we can also write down so therefore this T is approximately
approximated as K into H1 + H2 by 2. So therefore this K into H1 + H2=2T so this is a with
this assumption so we get this approximate expression for this Q. Now let us consider
the unconfined aquifer that is unconfined aquifer with uniform recharge. So in this
previous two cases we did not consider any recharge into the well so now let us consider
the unconfined aquifer with uniform recharge So in this case that is the this is the ground
level and this is the well in the unconfined aquifer
and this is the water table and of course here we are getting the the uniform recharge
and let us take this as a W recharge rate W and this is the this steady flow through
this well this Q, QW and here so this is the impermeable strata and let
us consider a strip wherein to the inflow is Q and outflow is Q + DQ So this DQ is the
one which gets contribution from this recharge the constant recharge and in this case let
us consider the water table and let us consider this steady height as
say H0 and the radius of influence as R0 and here so this radius is R and then this elemental
strip of radial distance DR okay So now we can write down
for this case we can write down the expression for the discharge DQ is given by the area
that is -2Pi R into DR into the uniform rate of recharge. So if we integrate this one
we get that is Q=- Pi into R square into W + constant of Integration and the boundary
conditions are at the well R tends to 0 and Q tends to QW. Therefore QW=C therefore
we can write down this Q=- Pi R square into W + QW okay If you substitute the expression
for Q which is – 2 Pi R K into H DH by DR so here this K into DH by DR is the velocity
as per Darcy’s law and 2 Pi R into H is the the area which contributes flow so this
is=- Pi R square into W + QW so if we integrate this expression integrating applying the the
boundary condition it is a H=H0 the upstream most boundary
condition and at R=R0 So therefore we can write down after integration we will get and
expression for the the drawdown curve H0 square – H square will be=this is W divided by
2 K into R square – R0 square + QW divided by Pi K into natural log of R0 by R. So this
is the so the expression for this draw down curve in an unconfined aquifer with uniform recharge
okay So if you know the value of the depth of water above the
lower impervious layer the uniform rate of recharge hydraulic conductivity and the radius
influence R0 QW which is the steady flow through the this well fully penetrating the unconfined
aquifer. So in that case this will be the expression for the drawdown And we know that when R=R0 Q=0 that means beyond the this
cone of influence there is no flow contribution. Therefore substituting this values
we get QW=Pi into R0 square into W in this case so that is a
so when R=R0 that case so this Q=0 so therefore if you apply this expression so in that case
this QW into Pi into R0 square into W So therefore so this is the expression for the the steady
flow through this well in an unconfined aquifer having a uniform
recharge and this is the so this is the expression for steady flow through a fully
penetration well in a confined aquifer in an unconfined aquifer with uniform recharge.
So it is fairly simple so simply the area of the circle of influence multiplied by the
the uniform rate of recharge so that will be given by that will give the expression
for the the steady flow through the well And now let us consider a well in a uniform flow uniform flow through and unconfined aquifer
so again this is a say fully penetrating well and uniform flow through. So far we have
seen we have consider the water table to be horizontal so that only after the pumping
starts the well starts getting contribution from the aquifer. Now instead of that suppose
we consider a well which is drilled through an unconfined aquifer
having a sloping water table below the sloping ground In this case we consider so this is
the sloping water table so with slope of I okay and let me denote
the well axis here and the the uniform discharge through this
well and here what happen is after the pumping starts then the water table will show a drawdown
curve like this. So this is the drawdown curve and in this case obviously here
so up to this point only there will be contribution to the well and obviously in this right side
it is fully flowing this one And if you draw the so in this case
let us I am sorry this is let us say this is the unconfined aquifer and here this is the top confining layer
and the confined aquifer thickness is B and now in let me so this represents the stagnation
this vertical line represent stagnation and let us also draw the top view for this If
this is the let me draw view in this case this is the well and let me so this represents
stagnation surface and within this. So these are the streamlines and so here at any
let us consider this as + YL and let us consider this as a +5 goes all the way up to here this
is – 5L and this is this distance this say – XL and obviously this is the
X and Y coordinate and beyond this the flow lines will be like this okay So now for this
case so there is an stagnation surface because
of original slope that is original hydraulic gradient due to an inclined water table So
in this case we can write the expression for the hydraulic conductivity as 2 Q k=2Q divided
by Pi into R into HU + HD into IU + ID and for here this HU is the
upstream saturated HD is the downstream saturated thickness
and this IU is the upstream water table slope and ID this is the downstream water table
slope So by super imposing a radial flow and 1D flow we get we can write down the expression
for this stagnation surface that is Y by – Y by X=Tan of 2Pi KBI divided by Q into Y.
So this is the expression so this B is the saturated that is the aquifer thickness and
Q is the steady discharge of course am not going into the derivation of this so therefore
and this Y L is given by + or – Q divided by 2KBI so this is the distance of the stagnation
surface at distance and okay And excel is given by – Q divided
by 2Pi KBI so we will stop here and we will continue in the next lecture thank you

## 5 thoughts on “Mod-01 Lec-12 Steady Flow into Wells (Contd.); Unsteady Flow into Wells”

1. molla berhe says:

it so fine

2. Ariel Cabrera Arandia says:

Fine.

3. Rahul Saikia says:

Thanks

4. mahenderreddy dasireddy says:

nice but please speed up ur explanation it is very slow more time consuming

5. bala murugan says:

sound is not enough ..have to improve ut