Lecture 29: Well Hydraulics-2

Lecture 29: Well Hydraulics-2


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Yeah. In lecture number 29 ah we are going to ah discuss the ah Well Hydraulics related
to Confined Aquifer ok. So, in the previous lecture, ah we derived equation of ah water
flow to well, in case of unconfined aquifers now we focus on ah flow to well in confined
aquifer ah . So, ah in case of confined aquifer we have
we have seen the previous ah lecture. So, the the first thing is the static water level.
So, that ah it is also called peizometric surface, because this is this is a confined
aquifer ok this is confined aquifer. And, ah this is a pumping water level ok.
So, this is pumping water level this h w and r w is the the ah radius of the well, and
this is a drawdown curve and d h by d r ok. And so, this is a static water level and R
is a your radius of influence and b is the thickness of the aquifer and water is ah flowing
horizontally to the well screen. And, the well is complete fully penetrated
to the aquifer ok. So, in this case so, how to ah derive equation for finding out the
ah the the pumping rate in case of confined, confined I mean in case of wells ok.
So, let us let us see ah the derivation which is similar to in case of unconfined aquifer.
So, here the some of the assumptions are aquifer is homogeneous and isotropic having uniform
thickness. So, ah and then ah ; that means, the K is not varying ah in the direction as
well as with space ok. And, the pumped well penetrates entirely thickness
of aquifer fully penetrating well. And, flow well is in steady state of course, and top
bottom a top and bottom of aquifer ah bounded by bedrock or impervious layer, because it
is a confined aquifer. So, the top and bottom or impervious layers ah the well is pumped,
at constant rate of course, it is similar to in case of unconfined aquifer, aquifer
is seemingly infinite in areal extension. So, as a areal extent is it ah ; that means,
if you this aquifer right. So, . So, ah the aquifer is seemingly infinite aerial extent
so; that means it is going somewhere. So, here the infinite extends as d h by d r which
is equal to 0 . So, ah using Darcy’s ah law. So, Q is equal to K i into a.
So, Q is equal to K . So, i is d h by d r and here ah the a area of cross section . So,
this is the 2 impervious layers. So, this is a ah confined aquifer and this is a screen
let us say. So, water is flowing to the ah well horizontally and with this thickness
ok. So, now, you know the surface area of this is so, 2 pi r into b ok.
So, the surface area. So, because because the water is going ah flowing through the
surface right water is flowing through the surface so, the 2 pi r b. So, this is area
into K into i. So, ah the similar to ah unconfined aquifer we are going to integrate I mean solve
this . So, let us see. So, d h it take the d h out.
So, then Q by 2 pi ah b into K into d r by r. And, the boundary conditions here are when
r equal to r w is nearer to the well. So, h is equal to h w that is the pumping water
level when r equal to r the extreme points. So, if you take these the extreme right . So,
the extreme point and R equal to r . So, then H is equal to capital H ok so, same thing.
So, with these boundary conditions put the boundary conditions in this equation right.
And finally, you get h minus h w which is equal to Q by K b right ah l n of R by r w,
because this is d r by r. So, integration you will get l n of r ok so, then limits r
w into R u get l n R V r w . So, the next Q is equal. So, put the values
ah arrange the values and Q is equal to 2 pi K b H minus h w into l n R by r w ok . So,
this equation is called Dupuit Thiem equation. So, here the ah definition of each variable
is given Q K ah b thickness R radius of influence, r w radius of well h w pumping water level
ok . So, then ah let us say ah similar to in a
unconfined aquifer, we are we can also ah use the same equation to find out the ah ah
aquifer properties in a K and t. So, here we ah going to put one observation well, and
second observation well at r 1 distance and r 2 distances at r 1 distance 1 observation
well and r 2 distance second observation well. Ok.
Similarly, so, after ah pumping we observe ah h 1 this ah water level here and h 2 water
level here and knowing the h value. So, you can find out the S 1 which is equal to capital
H minus h 1 . So, similarly S 2 so, that is a draw down which is equal to capital H minus
h 2 ok . So, from 2 observation points we get to draw
down points. So, with the 2 draw down points ah we are going to ah get an equation for
K as well as T . So bring the equation back. So, that is a
this is a original I mean the flow to confine a well in confined aquifer, 2 pi K b into
H minus h w or by ln r 2 by r 1 ok. So, where l n r 2 by r 1 so, for this ah what happen.
So, ah 2 r from 2 wells 2 wells you get ah you know different ah S 1 S 2 S ah values.
So, so, if you have a number you know observation wells number of observations wells. So, you
get in our different draw down points . So, put the ah ah put the ah distances right.
So, from the centre of well to different observation wells that is r 1 r 2 r 3 r 4 so, for every
distance. And, the corresponding you know the observation
so, the draw down points right . So, radius the distance on x axis and draw down on y
axis and you get ah a curve like this right and for one log cycle. So, find out what is
the ah difference in draw down ok for one log cycle .
So, here for one log cycle ln this is the log axis in l n this is the log axis for one
log cycle it is the value is 2.30 ok. So, then substitute that here. So, for one log
cycle is 2.30 and delta S. So, delta s is the the correspond the difference in the draw
down corresponding to the one log cycle. So, ah so, if you have this kind of curve
it is easier to find the value of ah you know the T ok. So, this is known right and this
is known Q is known and you can estimate T ok. From this graph from this graph delta
S is known and Q is known and you can find out the T. So, this is the graphically you
can estimate. And, you can also ah use equations to solve you know K and T .
So, that is ah that is similar to in case of ah confined aquifer ah unconfined aquifer
ok. So, let us see here an example ah there is a 10 centimetre diameter well penetrates
10 meter ah thick confined aquifer. So, the study state draw down were ah found
to be 2.5 and 0.05 meter at a distance of 10 metre and 40 meter respectively ok from
the centre of the well. So, when the well was operated with a constant discharge rate
of 125 litre per minute for 12 hours. So, using Dupuit Thiem equation, then calculate
the transmissibility and hydraulic conductivity. So, here if you see this I mean example. So,
the Q is given r 1 10 metre and r 2 40 metre there are 2 observation wells installed near
to the main well, and the corresponding drawdown are 2.5 and 0.05 ok.
So, here if you see so, the pumping water level. ah . So, this is a pumping water level
let us say this is the pumping water level ok this is going out. So, ah at a distance
so, here one observation well there is a another observation well. So, here one this is ah
10 metre and let us say another observation well at 40 metre.
So, draw down draw down at this point. So, this is 2.5 and at this point this is 0.05,
that is what given ok um And, so, let us solve so, in that case what is the ah value of hydraulic
conductivity and transmissibility? So, the so, this equation if you see the original
equation so, Q is equal to right 2 pi T into s 1 minus s 2 divided by l n of r 2 by r 1
ok. So, this is original equation. So, original equation also I mean if you see the ah equation
to well in confined aquifer. So, that is 2 pi T ah h minus h w divided by l n of r 2
by r 1 . So, this was the equation. So, since H is
ah sorry ah so, since s 1 is equal to H minus h 1 let us say right s 2 is equal to H minus
h 2 ok. So, this can be written as ah I mean the same equation the same equation can be
written like Q is equal to 2 pi T into h 2 minus h 1 divided by l n of r 2 by r 1 ok
. Now, if you substitute if you if you substitute
this this value right ah h 2. So, h 1 is equal to h 1 is equal to in this H minus s 1, h
2 is equal to H minus s 2. So, substitute this right h minus s 2 minus h h ah s minus
h 1 minus s 1 ok. So, h h gets cancelled and you get s 1 minus s 2 divided by l n r 2 by
r 1 ok. So, that is equation you see here right.
So, T is equal to 0 point. So, substitute the values substitute the values is so, so
the Q is given r 2 r 1 is given and s 1 s 2 is given. So, finally, T is this right and
if you can multiply generally K. So, K is equal to ah sorry T is equal to K into b.
So, we know ah we know b right. So, multiply and divided by so, K is equal to T by b.
So, transmissibility divided by the thickness you get the K value ok.
So, there is another example. So, there is a confined aquifer has a source of recharge
as shown in figure. So, this is a recharge right, this is the confined aquifer because
this is a impervious layers and this is also an impervious layer and this is the confined
aquifer . So, all from here to here this is confined aquifer and the recharge is taking
place from this end ok. So, the hydraulic conductivity of the aquifer
is 50 meter per day ok. The hydraulic conductivity is aquifer is 50 meter per day and the porosity
is point 2 right 20 percent right the piezometric head in 2 wells. So, well 1 and well 2 ah
thousand meter apart is 55 meter ah and ah 50 meter respectively ok.
So, one is ah the piezometric heads in 2 wells. So, this is 1000 meters. So, they have a 1000
meter apart right. So, in one well they found 55 metre and other well the 50 meter respectively
from the common datum ok from the common datum suppose suppose this is the datum line right.
So, from here let us say ah 55 and here ah sorry here you found sorry this is the common
datum right. So, so, at this point this is 55 so, at this point this is 50. So, there
is a difference of 5 metre here and the common datum the average thickness of the 30 meter.
So, 30 meter is the average thickness that is 30 meter right.
And, then the average of width is 5 kilometre. So, average width ah the thickness is ah 30
meter ok and width is So, width is 5 kilo meter. So, 5 kilometre means this is ah I
mean ah passing through the board you can say ok. So, that is the thickness thickness
goes like this right and ah. So, the question is determine the rate of flow through aquifer
and then the second determine the time of travel from the head of the aquifer to 0.4
kilometre downstream assuming no ah depression or diffusion ok.
So, 4 kilo meter downstream. What is the head of aquifer and then determine the rate of
flow through the aquifer so, these 2 points. So, one is rate of flow through the aquifer
right. So, this is Q and second is the determine the time travel from head of the aquifer.
So, let us see if this is the head of aquifer. So, from here if you take 4 kilo metre and
downright so, how long it is going to take from water to reach one extreme to 4 kilo
4 kilometre down ok . So, ah let us see here so, area of cross section
of the floor. So, 30 metre thickness and 5 kilometre is the width. So, so, here the ah
aquifer . So, this is 5 kilo metre and this is the thickness 30 metre. So, water is entering
through these ok . So and hydraulic gradient h 1 minus h 2 by L so, that is 55 50 by 1000
right. So, 1000 is the length. So, that will be this is hydraulic gradient
I see is equal to 5 into 10 power minus 3 then the rate of flow Q, which is Darcy’s
equation K i A. So, Q is equal to so, K is given 50 metre per day right and 15 into 10
power 4 and 5 so, this is area. And, this is ah ah hydraulic gradient now this is sorry
this is the area and this is the hydraulic gradient. And finally, you get ah Q that is
37500 metre cube per day ok. And, then we are going to use the Darcy’s velocity. So,
that is Q by A. So, using Darcy’s velocity you get ah 0.25
metre per day and the seepage velocity. So, which is so, this is Darcy’s velocity so,
but the I mean the the aquifer has porosity right, it has porosity so; that means, the
0.2 ah 0.2; that means, 20 percent. So, 20 percent is really effective. So, effective
porosity so, out of 100 percent only 20 percent is participating in flow process. So, we need
to consider that ah . So, so then so, the V divided by 0.2. So, you get 1.2 metre per
day and time to travel 4 kilometre downstream. So, you knowing the velocity right this is
the seepage velocity and the distance you get 3200 days or 8.7 7 years. So, this this
shows clearly the velocity of flow in aquifer is very slow you can see. So, for 4 kilometre
so, it is taking 8.77 years ok . So and there is another example like theme
equation using theme equation? So, 1 metre ah diameter of well right a well diameter
is a 1 metre. And, constant flow ah rate 1 1 3 metre cube per hour, and b is a thickness
of aquifer h not is static water level, and there are 2 observation points h 1 is 38.2
metre. So, that is located at r 1 of 15 metre and h 2 39.5 metre located at ah ah 50 meter.
So, find head in the main ah main pumping well and draw down the corresponding draw
down. So, the question here is here is ah the ground surface right. So, there is the
2 impervious layer . This is impervious layer 1 and imperious layer
2 and let us say this is the well right. And, this is the confined aquifer and it got ah
a thickness of ah 30 metre . So, let us say initially this is the static water level ok.
So, the static water level . So, let us say this is a drawdown curve for that there are
2 observation points. So, observation point here and observation
point here ok this is O 1 O 2 . So, ah this is located at this is located at 15 metre
and other one is located at 50 metre 5 0 50 metre and the corresponding ah h 1 h 2. So,
here from here to here so, the h 1 so, this is ah let us say h 1 and here this is h 2
energy end this is h 0 ok And so, find out what is ah ah what is you
know h w so; that means, pumping water level at this point. So, what is h w the pumping
water level? And, r w, so 1 metre diameter, so r w is given. So, the here r w that is
radius of well is ah 1 metre divided by like 0.5 metre here point metre diameter of the
well. So, question is what is h w and S w S w is this right; this is S w or s w, which
is equal to h 0 minus h w ok. Since, h 0 is given if you find out h w you
can find S w ok right. So, so this is the equation for ah estimating transmissivity
in ah in case of confined aquifer, we are going to use this equation to find out the
first transmissivity. So, the T is equal to 1 1 3 right, ah divided
by 2 pi into 1.8 minus 0.5 l n 50 by. So, this is the transmissivity you get the transmissivities
ah you get T is equal to 16.66. So, ah then once you know T now you use h is equal to
h 0 plus Q by 2 pi T l n r by R this is same equation, if you see Q is equal to 2 pi T
into ah h minus h 0 divided by l n r by capital R right.
So, ok so, Q into 2 pi T h minus h 0 right, ah yeah so, from this equation. So, the same
equation if we use so, h is equal to h 0 plus Q by 2 pi T l n into so, T is known. So, then
you can similarly. So, then if h is equal to h w right you can use as h 1 or h 2 so,
h 2 plus this one. So, from this equation you get h w is equal
to 39.5 h 2 is given 1 1 3 by 2 pi 16.2 this one . Now, h w is equal to 34.5 metre right.
So, knowing h w, h w is known now h 0 is known. So, S w is 40 minus 34.5 and you get S w.
So, that is drawdown in pumping well. So, that is 5.5 metre ok .
So, then next is well loss. So, well loss ah . So, basically if you see the draw down
curve. So, the draw down curve, so what happens so, draw down is here ending is here, but
the pumping water level is showing here ok. So, this much ah really it is a loss due to
some losses. So, the losses could be happens in 2 cases; one is due to the turbulence in
the well, like the water flow turbulence in the well, and also the resistance a flow in
the aquifer system ok. So, the 2 are really ah causing the head loss
right or head loss or in the water level variation in the pumping water level. So, so I supposed
I mean ah if there is no well loss. So, I supposed to get my draw down curve like this
right like this, but in theoretically it is I mean the practically you will have the draw
down curve here and the pumping water level here. So, this difference is due to the well
loss ok ah . So, let us see ah how to estimate the well loss in case of ah aquifer?
So, here in a pumping ah artesian well right ah the total drawdown at the well S w can
be considered to be made up of 3 parts ok. So, S w whatever you are getting the the draw
down curve ok. So, that S w this is S w, and this is the pumping as well as this is the
ah let us say h naught and this is h w ok. So, the h s w is equal to h 0 minus S w ok.
So, S w ah really the S w is causing the ah due to the 2 parts made up of 3 3 parts.
So, one is head loss required to cause laminar porous ah media ah called formation or aquifer.
So, that is aquifer loss. So, here the water is flowing through the ah horizontally to
the screen pipe ok to the screen pipe . So, during that process there is a head loss ok.
So, that also causing the ah I mean that is causing the draw down here. And, the other
one is drop of piezometric head required to a sustain turbulent flow in the region nearest
to the well where the Reynolds number may be larger than unity ok. So, here the other
case is so, because of the turbulence, which is present inside the pumping well or pumping
well. So, the turbulence; that means, the it is not laminar flow. So, once it is coming
flow which is adding into the system, ah and and also because the pumping because the pumping.
So, so here the water is flowing ah laminar and coming very slowly right, at the same
time the pumping really ah doing aggressive ah things.
So, that really creates the turbulence inside the ah well. So, that definitely Reynolds
number ah definitely more than one. So, that also causing the um draw down then head loss
through the well screen. So, this is head loss through the well screen
also ah another thing. So, the S w L, which is the laminar porous media flow or the aquifer
due to aquifer ok, well loss due to aquifer. So, that is proportional to Q and all and
and this one the turbulent flow or the flow through screen and the turbulence. So, that
is the square root of Q ok. So, the well loss ah can be expressed like this ah . So, the
draw down in the pumping well which is equal to C 1 into Q C 2 into Q square C 1 into Q
this is ah due to aquifer and C 2 Q square this is ah due to the well ok.
So, C 1 Q 1 the formation the aquifer loss and C 2 Q square this is a well loss ok there
are 2 things you have to consider. So, the draw down ah we are seeing in the pumping
well is equal to the well loss due to aquifer and well loss due to well ok yeah. So, ah
so, in this lecture we focused on ah ah flowing with water flow to a ah well installed in
a confined aquifer. So, we derived the equation steady state equation for estimating the flow
rate ah or flow to well in confined aquifer. And, also the procedure to find out the aquifer
parameters or aquifer parameters like transmissivity and hydraulic conductivity right.
Ah. Using you know a installing series of observation wells and then you get a graph
ah ah drawing you know distances on x axis and draw downs on y axis and from the graph,
you can you know ah use in the main equation and find out the transmissivity and the corresponding
hydraulic conductivity, if you know the thickness of aquifer. So, then we ah solve some problems
and then also we ah solved I mean we formulated in equation for you know well loss.
So, well loss is happening basically I mean 3. So, one is due to the head due to the aquifer
ok. So, the other one is head due to the screen openings and third is the head due to ah head
cross due to turbulence in the well. So, combining these 3 ah 3 we will definitely cause the
head loss or drawdown in the ah pumping well ok so.
Thank you .

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