Civil Engineering
Prof. P.K. Mohapatra
Indian Institute of Technology, Kanpur
Lecture

N o. # 27

In todays class, we will discuss about reservoirs. When we say reservoirs, the purpose may
be many folds. It may be used for ood control.
(Refer Slide Time: 00:32)

It may be used for irrigation, power generation, pisciculture or aquaculture, drinking water
supply, can be sometimes used for recreation. We should remember that why we may build a
reservoir due to one or more than one these reasons. Suppose the channel is owing with
water, we should know what the inow is to the system and then accordingly we can design
the reservoir based on our requirements. When we say requirement, we can call it demand
sometimes. Based on the two terms, one is the demand the other is the supply we will design
the reservoir.

(Refer Slide Time: 02:06)

When we say reservoir, our purpose is basically three folds. One is to calculate what the
demand is. What is the reservoir capacity and then how we should operate the reservoir?
We shall discuss about these two and we will solve some problems.
(Refer Slide Time: 02:34)

There is something called the water availability. What is water availability? Suppose I have
the data for 50 years, let us analyze the daily discharges for a stream for 50 years. What I
should do is accumulate these values per month or calculate monthly discharges. You sum
the daily discharge values for the whole month. Now you get 12 values for a year. For 50
years, you get 50 multiplied by 12, that many values. Suppose I want to know the water
availability so I can plot here, 0 percent. This is 100 percent and this is discharge and the
series I get, I can write them in descending order. Means, the highest should be at the top.
You can say Z to A. When I plot these values something like this, the highest value I should
assign is 0 percent and to the lowest value, I should assign is 100 percent. It means 100
2

percent of the times; this discharge is available to me. Suppose I want to find out what is 75
percent dependable ow is, then here this is 75 percent. Suppose I want to know what 90
percent is, then I go here and nd out the discharge. This is 90 percent dependable ow
(Refer Slide Time: 05:13). This is 75 percent dependable ow.
(Refer Slide Time: 05:15)

A thumb rule for deciding the reservoir is generally 75 percent dependable ow is used for
irrigation and 90 percent dependable ow is used for power supply and 100 percent
dependable ow is used for drinking water supply. Now let us go to the next thing to decide
how I can find out the reservoir capacity. As I said earlier, we need to know what the demand
or what is the water requirement is. The purpose can be many. For whatever is the purpose,
depending on that, the requirement of water and the supply of water will be.

(Refer Slide Time: 06:28)

Now I will introduce you two terms, one is the mass curve and later I shall tell you about the
differential mass curve. Let us see what mass curve is. In mass curve I plot here time and in
the other axis, I plot cumulative volume. It means the curve will look like this becausethis is
a cumulative volume. A value let us say here (Refer Slide Time: 07:12), cannot be less than
its previous value becausewhat is cumulative is, if I have a series, supposethis is t everyday
or at monthly or yearly values and I have here the ow volume. Supposethis is time period 1,
2, 3, 4 like this and here I have let us say 100, 110, 80, 30 maybe 120 like this, then this
cumulative volume will be 100 and then I keep on adding these numbers. That means 210
then 80, which is 290 then 30, 320 etc. Subsequentvalues are in this series. A value cannot be
less than the previous values. This is called the mass curve. So for the stream under
consideration for which we are trying to design the reservoir, we need to nd out the mass
curve of supply.

(Refer Slide Time: 08:41)

We nd out, we calculate from the given data mass curve, the supply and remember this is
mass curve for ow volume. We also need to prepare mass curve for demand and on the
same plot, we try to see these two curves. Let us say supply is like this and demand is let us
say like this. In this case, this is supply and this is demand. Then we know that at this place
supply is more than demand. Let me make another curve with dotted line so that it will be
clearer.

(Refer Slide Time: 09:55)

This is time this is mass curve or I can say Q, can be supply or demand. Let us say supply is
through a solid line. So supply is like this. This is S and let us say demand is a dotted line.
This is demand, so consider this portion from this to this. Let us say this supply curve is like
this, so here supply is more than demand. We do not require the reservoir volume to store
water because supply is already above demand. Consider this portion here, demand is more
than the supply and we need some reservoir. We need some storage so that we can store the
5

water and whenever required at this time, from this time onwards we can supply that water
from the reservoir. This is the supply due to ow of the stream. But we should reserve water
so that

we can fulll

this demand.

(Refer Slide Time: 11:37)

Let us see a systematic procedure to find out the reservoir capacity. First of all you prepare
the mass curve for supply and then you prepare the demand curve. Let us say wherever
demand is higher; you should concentrate on that location. For example here, demand is more
I should nd out the difference. Similarly here demand is more, I should find out what is the
difference. Similarly here demand is more and I should find out what is the difference. Here
demand is more I should find out what is the difference (Refer Slide Time: 12:27) and let us

say we have 1, 2, 3, 4 volumes at 4 different places. The demand is higher than the supply
and

we

need

to know

what

the

extra

amount

is.

We

should

store

to fulfill

the

demand.

I

should nd out maximum of these differences. Let us say this is sigma Q demand sigma Q
supply. Wherever you consider these and wherever this value is maximum, that will be the
reservoir capacity. This thing will be clearer when we solve one problem and let us go to the
next thing.

(Refer Slide Time: 13:32)

Just now I indicated that demand is through a constant curve like this. Do you think the
demand will be constant? What does this mean? This means in time, this is time; let us say
these are the values against each month. So for each month the demand is constant that is
why this is a straight line with constant slope, becausewhen I keep on adding, let us say this
is X, X, X, X, X always. Then when I keep on adding, this will give me a straight line and with
a constant slope. So sigma Q will be X, 2X, 3X, 4X and so on. We get a straight line. But do
you think in reality this is the case?
(Refer Slide Time: 14:34)

What we observe is the demand is also a variable in time. For eXample during summer we
need more water. Similarly during winter we may need less water, so this demand will again
be a variable thing. Let us say this is the variable demand, similarly this is the variable
supply. This is supply. Then I should consider at different times what could be the difference
and out of these differences I should accept for my design value or the reservoir capacity
7

should

be the maximum

one.

So I consider

the maximum

of these

differences.

Whether

it is a

constant demand curve or a variable demand curve, it does not matter. What we have to do is

plot the mass curve both for the demand and the supply and try to see where the maximum
difference is and based on that, we plot the maximum difference. We decide the reservoir
capacity. I will move on to the next thing.
(Refer Slide Time: 16: 19)

There is another technique to nd out the reservoir capacity. It is called the differential mass
curve. It is called the differential mass curve becausewe plotted the mass curve and then we
find out the difference. This technique that I am going to describe is called the sequent peak
algorithm. In this sequent peak algorithm, generally a computer programming is used and
please remember that in order to design the reservoir, to arrive at a number for the reservoir
capacity, what is done is, it not just 1 or 2 years generally, but we consider for the proposed
site, we consider 40-50 years discharge value. When we consider such a long series, you
know better to use a computer program. This sequent peak algorithm is a computer program.
It is an algorithm which systematically calculates the reservoir capacity. Let us look into
procedure in sequentpeak algorithm.

(Refer Slide Time: 17:50)

Let us say this is time and this is say supply and this is demand. What we do is in the
previous method, in the differential mass curve first we find out the mass curve for this and
then mass curve for this. We compare the mass curves then nd out what the capacity is,
but here what we do is first we find out the difference. We find out the difference and you can
analyze the difference between these two columns. You get numbers here. Suppose this is
supply, this is demand then you get a series.
(Refer Slide Time: 18:59)

Some numbers will be positive, some numbers will be negative. You get a series like this.
Remember if supply is more than demand, then supply minus demand will be positive. If
supply is less, then supply minus demand will be negative. In sequent peak algorithm we try
to find out the peaks. Suppose this is peaks remember, this is time and in y axis, I have the
fourth column. In the previous table I showed this fourth column. This is time, this is supply,
this
9

is demand

and this

is the difference

of these

two.

So this

column

is that

difference.

What

I do here is I consider the peak and then the next peak. Let us say that I neglect these two
peaks because these values are less compared to this peak value because this is called a
sequent peak algorithm. I have to search the next higher peak. Let us say this is another peak
and if the series continue then I should find out where the peak is and remember I do not
consider these peaks becausethey are less compared to this peak. So once I am here I should
search for the next higher peak compared to this peak. In this particular figure I have 1, 2, 3,
4 peaks. In between the peaks, let us say this is peak 1, this is peak 2, this is peak 3, this is
peak 4(Refer Slide Time: 21:22) and remember the decision how to find out the peaks. The
peak has to be higher than these values and these two should not be considered becausethese
are less compare to this. Here in between the two peaks, I should find out the minimum
trough. Supposethe 1, 2, 3, 4 troughs are present, out of these 4 troughs I should decide the
minimum one. Similarly in between these two peaks, there may be more than one trough on
mini

mass.

I should

find

out the minimum

one.

Similarly in between these two peaks, I have to nd out whichever is the minimum value. Let
us say this is T1, this is T2, and this is T3. For calculating the reservoir capacity, what I
should do is you just find out P1 T1 for the difference between these 2. Similarly P2 T 2
and P3 T3 and find out what is the maximum between these three. Do not forget that in
reality, in the series, it may be a very long series and in the series, your computer program
will do this operation. It will try to pick the numbers, the peaks and troughs and in between
peaks and troughs, it will try to find out where the maximum difference is, depending on
which the reservoir capacity will be decided. Now I will repeat once again. While calculating
these differences you have to be slightly careful because this will be with a minus sign and
this will be with a plus sign. When we say difference, basically we add these two magnitudes.
This is the peak; this is the trough so we add these magnitudes to find out the reservoir
capacity. These things will be clearer when we will solve some problems. So far we have
discussed about the estimation of reservoir capacity and about three different things. One is
the constant demand, the other is the variable demand and also we have discussed about the

algorithm which is sequent peak. Also you should remember that conversely, one can do the
problem, suppose a reservoir capacity is known, the reservoir volume is known, and then we
can determine the volume. The requirement of water is the demand that needs to be fullled
by these reservoirs. Can that be determined? Yes. So now we will see different problems.
These concepts should be clearer.

10

(Refer Slide Time: 25:07)

This is a problem one can calculate the minimum storage required to maintain a demand rate
of 40 meter cube per second. These are the ows given for 12 different months. In January, it
is 60 cumec. Please remember these two units cumec and metre cube per second are same.
These are the numbers given for different inows per year.
(Refer Slide Time: 25:44)

What we should do is we should first prepare the mass curve for ow. When I say ow it
means this is the supply. Then I should prepare the mass curve for demand and I must
compare these two to find out the difference, that is the reservoir capacity. Do not forget that
mass

11

curve

is time.

This

is cumulative

ow.

(Refer Slide Time: 26:27)

So here are the steps for different months. I have the supply. This number 60 indicates the
average daily discharge. I should multiply 31 because January has 31 days. This is the
monthly value. The 60 is the average daily discharge and for the month, this is the discharge
you can say this is the supply. Similarly for other months, we have these Q max per metre
cube per second as supply. Then this is the demand, this is a case of constant demand. The
water required is 40 Q max. Again these are the daily values, so for respective months, I
should find out the number, the monthly values. This is again in cumec. What needs to be
done? I should prepare the mass curve. Remember these numbers are only for the month. So I
must prepare the mass curve, only then I should compare the supply, the demand and the
required reservoir capacity.
(Refer Slide Time: 28:02)

This is the mass curve for supply. Remember the previous numbers. What should I do? I
should add the first number which will be 1860. I should add this to this and similarly I keep
12

on adding. I get the cumulative curve which gives me the mass curve for supply. This is the
mass curve for supply. In x axis, I will have the months here. I will have the months from
January to December and in this axis I will have the supply. Whatever is the number, this will
be the mass curve and similarly for demand I can get the mass curve by adding the previous
column values. I must plot this and this. So here it will not be like this becausethe demand at
some places will be more than the supply. Only then we require a reservoir. If the supply is
higher than the demand, then there is no need for the reservoir. So let us say this is something
like this then let me draw here one more figure. Let us say this is the demand because we
have a constant demand. It will be with a constant slope and let us say the supply initially for
example here 1860 is more than 1,240. Initially it is more but then gradually it will be less
than that and then towards the end it is again more the supply is more. If you examine the
critical months, it will be January, February, March. So look here supply is more compare to
the demand but look at this. These 3 months are the dry seasons,and these 3 months are
critical. We need a reservoir to cutter the requirement of water and these three periods will
indicate what the required capacity is.
SupposeI find out the difference here, for example here the demand is 7240 and the supply is
6080, this is 1160, is required. Let us say this is one point and this is another, so my intention
is to nd out the difference between these two plots at different points and for that I should
draw the tangents at different locations parallel to this line here. Depending on the maximum
difference, from the demand we determine the capacity and if you do this graphically we will
find out that this value will be slightly higher. Here the analytical number is 1160 but if you
do it graphically, you may nd it to be slightly higher. I believe the value will be around 1800
or 1900. It means the reservoir capacity will be that much. So what I have to do is, I plot this
curve and take the difference and wherever the difference is maximum, I consider that to be

the reservoir capacity. I think the case of constant demand is clear. Now let us see the second
case.

(Refer Slide Time: 32:53)

This caseis to calculate the demand that can be fulfilled by storage of 3600 cumec days. This
is the case of reverse problem. It means in the other problem demand was given and we have
to nd out the reservoir capacity.
13

Here reservoir capacity is given and we have to find out the demand. It can be satisfied
through this reservoir. The exercise will be pretty similar. I will come back to this once again.
(Refer Slide Time: 33:31)

These are the inow values so the procedure will be like this. I have to prepare the mass
curve for the supply. This is supply. So prepare the mass curve for the supply and now you do
not know the demand but you know the reservoir capacity. So what you have to do is you
guess some lower values in the mass curve. Suppose this is the mass curve for the supply,
then you can take this or this or this. You draw the tangents. Then let us say, my first guess
value is this. I draw a tangent here. Then add storage I know this storage and the problem
storage is given. So add the storage. Here through this you pass a line like this becauseyou
assume

that

the

demand

is still

constant.

This

mass

curve

for

demand

will

have

a constant

slope and if I get this line, what is to be done I know that this is the mass curve for demand.
Once this curve is known I can easily nd out the monthly values. For example here it will be
with a constant value. You can deduct the previous months value. For example here the
monthly value will be this minus this. You get the demand discharge, monthly demand
discharge. The objective is in this problem to solve the problem for the demand curve. We
know the reservoir capacity and through this exercise, through this problem we can know the
demand it can be fullled through this reservoir. Now let us see the next problem which will
be on variable demand. Pleaseremember that for the sake of explaining you I assumethat the
demand is constant. It need not be though. So let us see the next problem. This is for the
previous problem.

14

(Refer Slide Time: 36:36)

For a proposed reservoir the following data was calculated. Data will be given. Next, the
prior water rights required the release of natural ow of or ve cumec, whichever was less.
This is important to understand sometimes, supposeyou build a reservoir, but there might be
some rulings. You have to maintain some ow. You cannot tap all the water. So here in this
question, it is given that the release of natural ow of ve cumec or whichever is less, have to
be released for requirement. This, remember is not the demand. This is the water rights
requirement. Then assuming an average reservoir area of 20 kilometer square, the reservoir
area is given. Estimate the storage required to meet these demands. So let us seethe data.
(Refer Slide Time: 37:43)

This is the data. These are months again the data is for one year. The second column is ow.
Flow means ow in the stream and then this is the demand. This is the evaporation. Now
when water ows, there will be some loss due to evaporation. This is evaporation and this is
the rainfall. We have to be very careful because in this particular question, you know units
15

are given here by mistake, so in the data you will be provided with units. For example rainfall
may be in centimeter or in millimeter. Similarly evaporation will be in terms of some depth
and ow will be in cumec and this demand will be again in cumec. These are the datas. What
is the procedure we follow? We follow the sameprocedure.
(Refer Slide Time: 38:48)

Consider all inows and we call it a supply. For the given system you consider all the inows
and what will be the inows. Let us see this will be inow, this will be positive. This will
also be becauserainfall will add to the system so rainfall should be also considered as inow
to the system. Now let us see what the demands are. Consider all outows becausedemand is
something which we have to provide. The reservoir has to cutter so while calculating
demand, we must consider this. This is the water required and also evaporation becausethis
is a loss. For this system, this is a loss. So these two columns will give me the positive things.
For the inow to the system, these two columns will give me the negative things or loss or
from the system. You now prepare the mass curves as usual that means in the x axis you take
time or month. In y axis you take cumulative ow. Here there is a trick. You should be
consistent as far as your units are concerned. Here I will give you the clue. The clue is either
you use volume for all the components. For example rainfall, you should consider volume or
you should consider the depth as the unit. You are free to use either volume or depth as the
unit. While preparing the mass ow curve for the supply and for the demand you should be
vigilant that your unit should be consistent.

16

(Refer Slide Time: 41 :10)

If you are adding, for example here when you add ow, this is in cumec. This is in
centimeters. You try to write in one unit. If you are using volume then to this depth, you
multiply the reservoir area. You get the volume. Similarly, if you are interested in nding out
the depth, for this ow when you divide by the area, these two steps are over and I also know
how to prepare the mass curve. It is all same compared to both and as usual the maximum
difference; the maximum difference between the supply and the demand curve will give us
the reservoir capacity. The only difference in this problem is that the demand is not a straight
line.

(Refer Slide Time: 42:03)

In the earlier programs demand was plotted like this becausewe assumedthat the demand is
constant but here, it will not be constant because of variable demands. It does not matter. You

plot your supply and whatever the supply is, you plot the demand. Wherever it is maximum,
you consider it at all places wherever it is maximum let us say at this place it is maximum.
17

That will be the reservoir capacity. I think this is clear. The example explains you how to take
into account the variable demands. Let us see the next one which is on sequent peak
algorithm.
(Refer Slide Time: 42:55)

The average monthly ows in to a reservoir in a period of 2 years are given above. Please
remember while you are planning for a reservoir, suppose you have been asked to design a
reservoir, you have been asked to nd out the reservoir capacity then the data should be for
more than 10 years. Remember this, only for the sake of it; we plotted here the data for one
year. But in reality you have to use more than 10 years data and again in sequent peak
algorithm, it is very useful because you have a long series time data all right. This is the
average monthly ows into a reservoir in a period of 2 years. It is given below. If uniform
discharge at 90 cumec is desired from this reservoir, we calculate the minimum storage
capacity of the reservoir. Again you have the demand which is constant and you can also
solve this problem the way we did the first problem which is through differential mass curve
but here we are trying to solve this problem through sequentpeak algorithm. For June it is 20,
July 60 and likewise we have here data for 1 year.

18

(Refer Slide Time: 44:27)

Next we have data for another one year, so what is to be done.
(Refer Slide Time: 44:33)

We have data for 2 years; the method says you if you are within the periods data prepare the
data for two in periods. This is because we assume that the saIne ow will continue for the
next two years. For example here we have data for 2 years and we must prepare data for 4
years. Then where from will we get the data for the next 2 years? The same data will be used
for third and fourth year. Fourth year means we have 12 values. For the first year, twelve
values, for second year and the third year, these values will be used and for the fourth year,
these values will be used. Find net ow volume for each month. What you have to do is you
do not have to prepare the mass curve. In the other method we have to prepare the mass
curve. Suppose I have data for 24 months or may be 48 months. We use these 48 months to
prepare the mass curve. Here we are not doing that, remember this. We use on monthly basis
and we try to nd out what the difference is. What we have to do is we are not preparing the
19

mass curve but we will do it on monthly basis. We have to seethe difference between supply
and demand for each month. Means in the data sheet, we should consider row wise. So here

we have the data for inow and in the question, this is the 90 cumec for demand. We consider
row vise and in a particular row we should see what the demand and the supply is and the
difference. We have to prepare data this way.
(Refer Slide Time: 46:53)

This may be serial number. This will be time and in our case it will be months and the data.
For our case it is from June onwards and remember we have 48 values. This will be 1, this

will be 48 and we have supply for each month. We have been given a number, so that the
number has to be written here followed by the demand. Here you put 90 everywhere and then
here you prepare S D. This is supply; this is demand so you prepare supply demand.
Whenever it is rainy or in the so called non dry period, we will have to supply more than the
demand. Therefore, the number that comes will be positive. Sometimes it may be negative.
For

each

of these

rows

one has to find

out

the difference

between

these

two

and

we should

also remember the procedure we followed and what is the procedure? We nd net ow
volume.

For each month

and it is S

D.

In the computer program in fact you use the subscript i for this and this i is the index for the
respective month and for the respective month you nd out the difference. When I say
difference, it may be positive or it may be negative S D and then we nd the cumulative net
ow volume. Then in this column, you have to do something like a mass curve. In the earlier
differential mass curve technique, what we did was preparing the mass curve here. After
supply you prepare the mass curve for supply. After demand you prepare the mass curve for
demand and then compare these two but here individually you prepare the difference between
the 2 at each month. Then you prepare a cumulative series which means sigma S D. So here
there will be only one. Here it will be this plus this. Here it will be this plus this and we will
get some numbers.

20

(Refer Slide Time: 49:58)

This will be treated to nd out the peaks and the troughs and as we discussed. Now this is
your rst peak, you should see where the next higher peak is. This is a peak, this is a peak,
this is a peak (Refer Slide Time: 50:27) but these peaks are lesser compared to this. But this
might be higher than this so if this is peak 1, this is peak 2. Similarly in this series remember
only this is an example and in this example we have 48 rows but in reality it will be more
than that. You can guess if you have 50 years data, how big will the series be? The computer
program will search and then you find out where the sequential peak is, that means one peak
and the next peak should be higher than the rst peak. Similarly you find out all the peaks
and in between 2 peaks you will nd the minimum of all the troughs. These are all the
troughs. These 12 are also the troughs. But out of so many troughs, I will go for this one,
because this is the minimum one. Then you decide P T. If you have more than one peak,
then corresponding troughs in the gap of two peaks, you nd out the minimum troughs.
Similarly between P2 and P3, there might be so many troughs and you decide to go for the
minimum trough. So, P2 T2, similarly P3 T3 and you consider all these numbers and the
maximum number will be your reservoir capacity.

21

(Refer Slide Time: 52:18)

Once again I will repeat the procedure which says, use the sequent peak algorithm. It means
to find out where P1, P2 or T 1, T 2, T 3 is and then find the maximum of all the P

T values.

Out of all the P T series, you nd out the maximum one and the maximum value gives you
the reservoir capacity. This is a sequent peak algorithm. This is more useful when we have a
longer series and if you have lesser data then we may want to go for a differential mass curve.
When we compare the results, they give more or less the same results. We should get same
results. If one is using the differential mass curve and the other one is using the sequent peak
algorithm, one should get the same results. But the sequent peak algorithm has lesser errors.
So it is generally preferred. I think with the help of these four examples, the concept of
estimating the reservoir capacity is clear and in the next lecture we will study the design of
culvert.

22