.

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 .