UNIT 16 STORAGE AND DISTRIBUTION

RESERVOIRS
Structure
16.1 Introduction
Objectives
16.2 Distribution Reservoirs
16.3 Types of Distribution Reservoirs
16.3.1 SurfaceReservoirs
16.3.2 Elevated Resewoirs
16.3.3 Types with Respect to Construction Materials
16.4 Position and Elevation of Reservoirs
16.5 Storage Capacity of Distribution Reservoirs
16.5.1 Balancing Storage
16.5.2 Contingency Storage
16.5.3 Fire Storage
16.6 Design for Storage
16.6.1 Mass Curve Method
16.6.2 Analytical Method
16.7 Guidelines for Design
16.7.1 Economical Depth
16.7.2 Roofing
16.7.3 Walls for Concrete Reservoirs
16.7.4 Floors of Reservoirs
16.7.5 Pipings
16.7.6 Testing for Watertightness
16.8 Elevated Reservoirs
16.8.1 Intz Tanks
16.8.2 Standpipes
16.9 Summary
16.10 Key Words
16.11 Answers to SAQs

16.1 INTRODUCTION
A water supply project consists of taking of raw water from source (river, stream or
reservoir) conveying it to purification units, and finally distribution to the consumers.
Thus, a water supply project consists of
(i) works for collection of water,
(ii) works for conveyance ofwater,
(iii) works for purification and treatment, and
(iv) works for distribution of water to the consumers.
The proper planning consists of an economical efficiently functioning units with
minimum recurring and operational expenditure and trouble. It has been discussed in
earlier units that water demand is not constant. There are several factors, which cause
fluctuations in demand and well planned water supply scheme has to supply watet as and
when required depending upon the demand. Earlier it has been discussed that maximum
daily demand goes as high as 1.8 times the average daily demand, of course only at peak
hours, i.e., for certain hours in the day. We cannot expect purification units to be made to
purify water at double of the average demand at peak periods. Hence, in the system one
or more reservoirs are inserted to furnish elasticity to the distribution system so that
different units to work at a constant average rate.
Distribution reservoirs also called seyice reservoirs are the storage reservoirs, which
-A_-- ---r-J L-- d? l..f-- ...-
r-- A..-'----,l. L
,
..
-, -,A ,
-,
-
,,
:
, .I,.., A..A,-
Water Distribution fire, breakdowns, repairs etc. Thus, helping in absorbing hourly fluctuations in the normal
water demand.

Objectives
The unit will deal with distribution or service reservoirs. After going through this unit,
you will know about
purpose and use of distribution reservoirs,
its suitable position in the system,
different kinds of service or distribution reservoirs,
procedure and guidelines for desi n, and
8
estimation of designed capacity of a distribution reservoir.

16.2 DISTRIBUTION RESERVOIRS

In a water supply scheme, an independence of action is desirable from the standpoint of
economy and safety and in many cases is of importance with respect to the quality of the
water. For example where water is brought from the source through a long conduit, a
distributing or equalising reservoir enables the conduit to be operated at a comparatively
uniform rate and hence, can be made of minimum size. Similarly, such a reservoir makes
it possible to reduce the capacity of purpps, filters or other similar units and to operate
them more uniformly and economically. In a *all works, the pumps will operate at full
capacity for a portion for the day only, reducing the running cost of the pumps, to get
relief from power failures or load shedding, which has become common in our country.
In case of a ground water supply, a small reservoir increases the capacity of the source by
making the demand more uniform. In a large distributing system, several reservoirs
placed at different points or locations may effect considerable economy in the size of the
pipe system and other units including the pumping plant. As a measure of safety against
the interruption of the supply from accidents to conduit or machinery, distributing
reservoirs are of great value. It gives cheaply additional safety against interruption when
units may require duplication. With respect to quality also these reservoirs are of great
advantage as they provide an additional opportunity for sedimentation and making it
possible to avoid taking water from streams during period of great turbidity.
Small reservoirs may also be provided for convenience in operation. A receiving reservoir
at the terminals of conduits, small reservoirs for regulating pressure at intermediate points
and similar reservoirs on air chambers at pumping stations, may be provided for

Elevated distribution

ELEVATION

.--PLAN

Figure 16.1 :Working of a Distribution Tank

I equalising the action of pumps, The function of a balancing or dishibution reservoir is Storage And
DIstribwon Reservoirs

/ explained in Figure 16.1. When the demand for water exceeds the rate of supply, the
water £lows into the distribution system both from the elevated distribution reservoir as
well as water through direct pumping by means of a bypass loop (as shown in Figure
1
.
16.1) by closing valve at A but keeping valves B. C and D open.
When demand for water is less than the rate of supply, then the required demand is met
by water pumped through the bypass loop, while the balance pumped water fills up the
balancing reservoir by closing the valve D but keeping valves A, B and C open to
regulate the flow. '

The main functions served by the distribution reservoirs are summarised below :
(i) They absorb the hourly variations in demand and allow the water treatment
units and pumps to operate at a constant rate. This reduces the running,
maintenance and operational (RMO)costs and improves efficiency.
(ii) They help in maintaining constant pressure in the distribution mains. In their
absence, the pressure falls down as the demand of water increases.
(iii) The pumping of water in shifts is made possible by them without affecting the
supply. Thus, 8 to 12 hours of pumping can be carried out so as to supply the
whole day demand.
(iv) The water stored in these reservoirs can be supplied during emergencies such as
breakdown of pumps, heavy fire demand etc.
(v) They provide an overall economy by reducing the sizes of pumps, pipe lines
and treatment unit and this is the most important aspect of service or
distribution reservoirs.

16.3 TYPES OF DISTRIBUTION RESERVOIRS

Depending upon their elevation with respect to the ground, reservoirs may be classified
as
(a) Surface reservoirs, and
(b) elevated reservoirs.

16.3.1 Surface Reservoirs

A surface reservoir may be constructed at ground level or below the ground level. Hence,
they are also known as ground reservoirs. They are generally constructed at higher points
of the area to be served. If the area to be served is undulating having more than one high
points, more than one distribution reservoir may be provided. In that case the area is
divided into several zones and a separate reservoir is provided in each zone for proper
distribution of water.
In a gravitation type of distribution system, water is stored in the ground distribution
reservoir and then directly sent from there to distribution mains, which are situated at
lower elevations. In a combined gravity and pumping system, the treated water is first
stored in a ground reservoir and then pumped to an elevated service reservoir from which
water is supplied into the distribution mains.

16.3.2 Elevated Reservoirs

Elevated reservoirs are mostly rectangular or circular overhead tanks erected at a suitable
elevation above the ground level and supported on towers. They are constructed where
the pressure requirements require considerable elevation above the ground surface to
serve multi-storyed buildings, They are provided in the area where the combined gravity
and pumping system for water distribution is adopted. Water is pumped into these
elevated tanks from filter units and then supplied to the consumers.
Stand Pipes
Stand pipes are a kind of elevated tanks without any supporting towers for resting
the tank body. They are tall cylindrical shell resting directly on ground. Stand pipes
are generally 15 to 20 m high and 10 to 15 m in diameter. Figure 16.8 shows
typical cross-section of a stand pipe.
Water Distribution 16.3.3 Types with Respect to Construction Materials
With respect to form of construction i.e. construction materials, reservoirs may be
classified as :
(i) Earthen,
(ii) Masonry,
(iii) Reinforced concrete, and
(iv) Steel.
When the reservoir is not required to be elevated above the natural ground level, the
capacity is to be larger than the most economical form and the usual one is the open
reservoirs with earthen embankments. The storage of surface water in such reservoirs
does not usually affect their quality if they have previously been stored in large
impounding reservoirs. But in the case of ground water or filtered surface water, it is
desirable that they are stored in closed tanks or reservoirs. Covered reservoirs are mostly
built with masonry walls and covers with RCC slabs. To economise the construction, it is
better to put them partly in excavation and partly above the surface. Elevated reservoirs
are provided when a reservoir requires to be considerably elevated.

16.4 POSITION AND ELEVATION OF RESERVOIRS

If the service storage is to be of maximum value as a safeguard to the undertaking against
breakdown then it should be positioned as near as possible to the area of demand. From
the service storage, the distribution system should spread directly with such a
ramification of mains that no single breakage could cause a severe interruption to the
continuity of the supply. There should be sufficient interconnection between the
distribution mains so that if a breakdown of any one main occurs, the supply may be
maintained by re-routing the water. It is not always possible to have a high point, which is
in the centre of the distribution area and in such circumstances, the best possible should
be done. If the high point is remote from the area of demand, the area is kept to feed the
demand by two major mains from the service reservoir. If there is some high ground,
which is not quite high enough then a water tower or several water towers may be
provided to meet the requirement. It has to be seen that service reservoir is located at
such an elevation that a steady pressure is available at all points of the distribution
system, sufficient to reach the topmost storey of three or four storey buildings.
The elevation at which it is desirable to position a service or distribution reservoir
depends upon the distance of the reservoir from the distribution area and the elevation of
the highest buildings to be served. If the distribution area varies widely in elevation, it
may be necessary to use two or more service reservoirs at different levels so that the
lower areas do not receive an unduly high pressure. Generally 45 to 75 m static pressure
is that which suits best to the domestic distribution system. It results in distribution pipes
of moderate size and thickness and wastage from leaks is not excessive at this pressure.
Pressure below 95 m may cause trouble in supplying extensive distribution areas and
pressure above 120 m may be too high to result in excessive leakage losses.

Figure 16.2 :Location of a Distribution Reservoir

In a pumping system, the pumps are usually located near one side of the city and for the
best results, the reservoir should be located at some point on the opposite side of the
distributing system. In this way, full benefit of the reservoir is secured in equalising pump
action, providing reserve supply, reducing average head on the pumps and making the Storage And
pressure more uniform in the system. Distribution Reservoirs

Figure 16.2 illustrates this condition : P is a pumping station, B is the reservoir and LM
is the area to be served. PX is the pumping head. During night, water flows into the
reservoir and the hydraulic gradient will be line XB. During day when consumption is
more than the pumpage, the reservoir supplies the deficiency and water flows to the
intermediate area from both directions giving pressure line XYB. With no reservoir or one
located at L, the gradient would have been line XY', steeper than XY, if the size of pipes
remain the same. A reservoir placed near the pumps at L will serve to equalise the pump
load only and will provide reserve supply but it will not have effect on the distribution
system. Hence, better location for the reservoir is at B.
If at all a service reservoir is placed at I,, it will be a small stand pipe or a small elevated
tank. The proper elevation of a reservoir depends on the required pressure in the mains.
Where more than one zone of pressure is employed, a site is selected to serve all but the
highest zone. The highest zone is then operated without a reservoir or with a tank or a
stand pipe.

16.5 STORAGE CAPACITY OF DISTRIBUTION

RESERVOIRS
It has been discussed that a distribution reservoir has mainly two functions :
(i) to balance the fluctuating demand from the distribution system against the
output from the source.
(ii) to maintain continuous supply in case of any breakdown at the source or in the
main trunk lines.

6.5.1 Balancing Storage

The first one is also commonly known as balancing storage or equalising or operating
storage while the second one is popularly known as breakdown storage. In addition to
these two types of storage, some storage of water has to be done for tackling the
out-break of fire in the city or the distribution area and the storage for that is known as
fire storage.
The source itself may give fluctuating output in step with the demand. But it is not
economical to follow this policy. If the peak flow rate is twice the average flow then the
source units are to be made of capacity to deliver twice the average demand and the
delivery pipeline, coagulation units and filters should also be capable of delivering twice
the average demand. Thus, the size of units has to be made double of the average
demand, which will not be used continuously hence, there will be only 50% usage of the
capital spent on these works. Therefore, it is to be seen that source and treatment works
including the main trunk line to work as much as possible to their maximum capacity and
the change in demand to be balanced by a service or distribution reservoir.

HOURS -
Figure 163 :Typical Variation in Consumption Rates for a Small Area
Water Distribution Figure 16.3 shows two consecutive days consumption into three 8-hour periods. Since
Sunday is a holiday in most of the places, maximum day demand of a week has been
. taken into consideration.
From midnight until 8 a.m. flow is less than average (% loss in the figure shown), from
8 a.m. until 4 p.m. flow is greater than average (% greater in the figure shown) and from
4 p.m. until midnight, the flow is about the average. Now it can be deduced that during
the period 8 a.m. to 4 p.m. excess flow will have to be from the storage (distribution
reservoir). This amount of water will have to be put back into the reservoir during the
following night in order to fill the reservoir ready for the next day. The maximum
demand figure varies 15% to 40% in excess of the average for the whole day depending
upon the size of the area considered. The lower percentage 'is applicable to large
industrial area of a population one lac or more and the higher percentage applies to a
smaller urban area of population under 20,000.
In the above calculation, minimum storage required to average out the flow of one
particular day has been taken. In practice, storage more than this will be required due to
following reasons :
(i) There will be variations from one day to the next day.
(ii) It may not be always possible to have the service reservoir full every morning at
8.00 a.m.
If double the theoretical requirement is provided then the contents of the service reservoir
above or below the 50% full line can easily be fluctuated and this will give greater
flexibility. Supposing thht an 8 hour maximum flow rate is 30% in excess of the average
flow for the day, then the amount of storage required for the distribution or service
reservoir
= 2 x 0.30 x (1 /3) (day's total consumption) x 125%
= 25% of day's total consumption
i.e., one quarter of the day's consumption. This storage will meet the hourly fluctuation of
flow.
But this much amount of storage may not be sufficient to meet all contingencies.

16.5.2 Contingency Storage

Contingency storage is also known as breakdown storage. This storage is required to
meet breakdowns at source, repair of pipe bursts on Mains. The storage depends upon the
nature of the source, the layout of Mains and safety precautions to be taken.
Bore hole pumping stations or boosting stations can have four to five hours interruption
of supply due to electricity supply failure at the same time such units should be designed
for not more than 22 hours working out of 24 hours to take into account of routine
maintenance. Water treatment plants also required to be shut down for four to five hours
for necessary repairs. River intake sources sometimes present hazards due to sudden
pollution, but this problem is tackled by providing raw water storage at the intake
because polluted water is not allowed to pass to the treatment plants. For repair of major
Mains, six to eight hours time for repair and two to four hours for refilling the system is
allowed when the Main has been repaired. The loss of water from burst is also substantial.
A breakdown may not conveniently occur just when the reservoir is full. It may occur
during the maximum demand period when the overdraw may rise to 25% of the day's
supply. In addition to these, some allowance is to be made for "bottom water". Bottom
water has to be kept in the reservoir to prevent its complete emptying.

16.5.3 Fire Storage

This provision takes care of the requirement of water for extinguishing fires. Although
this requirement seems to be high but when analysed, it comes to around 1 to 5 litres per
day per person depending upon type of city. Hence, sometimes, storage for fire-fighting is
clubbed with contingency storage.
Only for balancing flows to an average sized distribution system, about one quarter of
day's supply is stored and thought to be sufficient. But this is not sufficient to run the
supply against breakdown and fire-fighting. The minimum storage for safeguarding the
continuance of a supply during breakdowns is one day's supply. But where daily and
seasonal fluctuations are large (chances of more fire in summer) and trunk Mains are not
duplicated, it is desirable to provide three days capacity to the distribution reservoirs.
 Storage And
16.6 DESIGN FOR STORAGE DistributionReservoirs

The storage capacity of balancing or service reservoirs is worked out with the help of
hydrographs of inflow and outflow by mass curve method or by analytical tabular
solution.

16.6.1 Mass Curve Method

A mass diagram is the plot of accumulated supply or demand versus time. The supply is
also known as inflow and demand as outflow. First mass curve of supply, known as
supply line is drawn and over this demand curve is superimposed. The amount of
balancing storage is determined by adding the maximum ordinates between the demand
and supply lines. First hourly demand for all 24 hours from the day of maximum
requirement is determined. Cumulative demand is plotted against time, which is known
as mass curve of demand. Next cumulative supply against time is plotted, which is a
straight line if the supply is constant. The storage required is calculated as the sum of the
two maximum ordinates between demand and supply lines. (Reference to Example 16.3
and Figure 16.2 will make the procedure more clear.)

16.6.2 Analytical Method

In this method cumulative hourly demand and cumulative hourly supply are tabulated for
all the 24 hours. The hourly excess demand and hourly excess supply are worked out. The
summation of maximum of the excess of demand and the maximum of excess of supply
gives the required storage capacity. The method will be more clear with the following
example :
Example 16.1
A city has population of 1.5 lakhs and it has to be supplied water at the rate of
200 litres per person per day. The hourly variation in demand is given in the table.
Find out the capacity of the distribution reservoir to be provided for balancing the
variable demand against a constant rate of pumping :
(i) when the pumping is done for all the 24 hours.
(ii) when pumping is done from 6 a.m. to 11 a.m. and *en 2 p.m. to 9 p.m.

Periodofdays Oto4 4to8 8to12 12to16 16to20 20to24

in hours

% of average 16 70 190 88 166 70

hourly flow .

Solution
Average daily supply = 200 x 15,000
= 30 x 106 litres = 30 million litres
Average hourly demand = 30 / 24 = 1.25 million litres.

Period Demand Cum. Cum. Const. Cum. Excess of Excess of

in hrs in units of demand in demand in pumping pumping demand 10 supply in
Avg. hrly. terms of million rate in in ML ML= Co1.(4) ML= Col.
demand Avg. hrly. Itrs. (Co13 MLmr - CoI.(6) 6 - Co1.4 .
demand x 1.25 MI,) (+ ve values) (+ ve
values)

1 2 3 4 5 6 7 8

1 0.16 0.16 0.20 1.25 1.25 - 1.05

2 0.16 0.32 0.40 1.25 2.50 - 2.10

3 0.16 0.48 0.60 1.25 3.75 - 3.15

4 0.16 0.64 0.80 1.25 5.00 - 4.20

5 0.70 1.34 1.675 1.25 6.25 - 4.575

Water Distribution
Period Demand Cum. Cum. Const Cum. Excess of Excess of
in hrs in units of demand in demand in pumping pumping demand 10 supply in
Avg. hrly. terms of million rate in in ML ML= Co1.(4) ML= Col.
demand Avg. hrly. Itrs. (COW MWhr - Co1.(6) 6 - Co1.4
demand x 1.25 ML) (+ ve values) (+ ve
values)

1 2 3 4 5 6 7 8
-
-
6
-
0.70 2.04
-2.59
-
1.25 7.50 4.95

7 0.70 2.74 3.425 1.25 8.75 -

8 0.70 3.44 4.30 1.25 10.00 -
9 1.90 5.34 6.675 1.25 11.25 - 4.575

~~-~~
10 1.90 7.24 9.05 1.25 12.50 - 3A5

11 1.90 9.14 11.425 1.25

-
13.75 -
1.25 15.00 - 1.20

13 0.88 11.92 14.90

14 0.88 12.80 16.00

'15 0.88 13.68 17.10

16 0.88 14.56 18.20 1.25 20.00 - 1.80

17 1.66 16.22 20.275 1.25 21.25 - 0.975

18 1.66 17.88 22.35 1.25 22.50 - 0.150

19 1.66 19.54 24.425 1.25 23.75 0.675 -

pp--ppP-

20 1.66 21.20 26.50 1.25 25.00 1.500 -

21 0.70 21.90 27.375 1.25 26.25 1.125 -
22 0.70 22.60 28.25 1.25 27.50 0.75 -
23 0.70 23.30 29.125 1.25 28.75 0.375 -
24 0.70 24.00 30.00 1.25 30.00 - -
From above table, it is observed that
(a) the maximum excess of demand = 1.5 million litres
(b) the maximum excess of supply = 5.70 million litres
Hence, total storage required = (a) + (b) = 1.5 + 5.7 = 7.2 million litres
Case (ii) when pumping is done for limited period.
When pumping is done from 6 a.m. to 11 p.m. and 2 p.m. to 9 p.m. i.e., 12 hrs. i.e.,
Rate of supply = 30112 = 2.5 million litreshr.

Perlod Demand Cum. Cum. Cmt. Cum. Excess of Exeesl of

in hrs in units of demand in demand in pumping pumping demand in supply in
Avg. hrly. terms of million rate in inML ML= Cd.(4) ML= Cd.
demand Avg. hrly. Itrs. (Co1.3 MWhr - Cd.(6) -
6 CoL4
demand x 1.25 ML) (+ ve values) (+ ve
values)
-------
1 2 3 4 5 6 7

1 0.16 0.16 0.20 - - 0.20

2 0.16 0.32 0.40 - - 0.40 -

 Storage And
Period Demand Cum. Cum. Const. Cum. Excess of Excess of Distribution Reservoirs
in hrs in units of demand in demand in pumping pumping demand 10 supply in
Avg. hrly. terms of Wion rate in in ML ML= Co1.(4) ML= CoL
demand Avg. hrly. lta-s. (Co13 MUhr - CoL(6) 6 - ColA
demand x 1.25 ML) (+ ve values) (+ ve
values)

1 2 3 4 5 6 7 8

3 0.16 0.48 0.60 - - 0.60 -

4 0.70 0.64 0.80 - - 0.80 -
5 0.70 1.34 1.675 - - 1.675 -
6 0.70 2.04 2.55 - - 2.55 -
7 0.70 2.74 3.425 2.5 2.5 0.925 -
8 0.70 3.44 4.30 2.5 5.0 - 0.70

9 1.90
- -
10 1.90 7.24 9.05 2.5 10.0 - 0.950

11 1.90 9.14 11.425 2.5 12.50 - 1.075

12 1.90 11.04 13.80 - 12.50 1.30 -

13 0.88 11.92 14.90 - 12.50 2.40 -

14 0.88 12.80 16.00 - 12.50 3.50 -

15 0.88 13.68 17.10 2.5 15.00 2.10 -
16 0.88 14.56 18.20 2.5 17.50 0.70 -
17 1.66 16.22 20.275 2.5 20.00 0.275 -
18 1.66 17.88 22.35 2.5 22.50 - 0.150

19 1.66 19.54 24.425 2.5 25.00 - 0.575

20 1.66 2 1.20 26.50 2.5 27.50 - 1.00

21 0.70 21.90 27.375 2.5 30.00 - 2.625

22 0.70 22.60 28.25 - 30.00 - 1.75

23 0.70 23.30 29.125 - 30.00 - 0.875

24 0.70 24.00 30.00 - 30.00 - -

From the above table, the maximum excess of demand = 3.50 million litres
the maximum excess of supply = 2.625 million litres
:. Total storage required = 3.5 + 2.625 = 6.125 million litres (Ans.)
Example 16.2
Calculate the storage required to supply the demand shown in the following table if
the inflow of water to the reservoir is maintained at a uniform rate throughout 24
hours.

The 00-04 04-08 08-12 12-16 16-20 20-24

Demand in 0.48 0.87 1.33 1.00 0.82 0.54

million Litres
Water Distribution Solution
Total demand during the day = 0.48 + 0.87 + 1.33 + 1.0 + 0.82 + 0.54
= 5.04 million litres
Total supply during the day = Total demand = 5.04 million litres
.-. Constant hourly supply = 5.04/24 = 0.21 M1
.-. 4 hourly supply = 0.21 x 4 = 0.84 M1

Timein Demand Cum. Pumping Cum. Excess of Excess of

hrs in Demand In Ml pumping demand (Co1.3 supply (Co1.5 -
million in Ml in Ml - Co1.5) (+ ve CoU) (+ ve
litres values only) values only)

(1) (2) (3) (4) (5) (6) (7)

0-4 0.48 0.48 0.84 0.84 - 0.36

4-8 0.87 1.35 0.84 1.68 . - 0.36

8-12 1.33 2.68 0.84 2.52 0.16 -

12 - 16 1O
.O 3.68 0.84 3.36 0.32 -
16- 20 0.82 4.50 0.84 4.40 0.10 -
20 - 24 0.54 5.04 0.84 5.04 - -

From above table, it is observed that the maximum of excess of demand = 0.32 M1
and the maximum of excess of supply = 0.36 M1.
Therefore, total storage required = 0.32 + 0.36 = 0.69 M1= 680,000 litres (Ans.)
Example 16.3
A town has population of one lakh. It is to be supplied with water at the rate of
200 litres per head per day. The variation in demand is as follows :
6 a.m. to 9 a.m. 40% of total
9 a.m. to 12 noon 10% of total
12 noon to 3 p.m. 10% of total
3 p.m. to 6 p.m. 15% of total
6 p.m. to 9 p.m. 25% of total
Determine the capacity of the service reservoir when the pumping is at a uniform
rate from 6 a.m. to 6 p.m.
Solution
. :. Total daily requirement = 1,00,000 x 200
= 20 x lo6 = 20 MI
Calculation for cumulative demand

Demand and supply lines have been drawn in firm and dotted lines respectively in
Figure 16.4. Two maximum ordinates enclosed between demand and supply lines have
been red out as 5 million litres and 3 million litres respectively.
 Storage And
DistributionReservoirs

TIME IN HOURS

Figure 16.4
-
.: Total Storage required = 5 + 3 = 8 million litres (Ans.)

16.7 GUIDELINES FOR DESIGN

For economical design of service or distribution reservoirs, site condition is very
important. The reduced level of sites gives an idea whether the reservoir will be on
ground or overhead tank supported on columns at higher elevation. The distribution
reservoirs may be made of RCC or masonry. Depending upon their elevation with respect
to the ground, they are classified as :
(a) surface reservoirs, and
(b) elevated reservoirs.
Surface Reservoirs
They are circular, square or rectangular tanks constructed at ground level or below
the ground level. Therefore, they are also sometimes known as ground reservoirs.
They are mostly constructed at highest point in the city.
A circular tank is geometrically the most economic in shape giving the least
amount of walling for a given capacity. But for circular tank, flat ground should be
available which may not be always feasible. To make best use of available land,
rectangular tanks are more economical. In fact, shape df the land available is often
determining factor in built up areas. A rectangular tank or reservoir with ratio of
sides 1.2 : 1.5 shows a benefit over a square tank when an internal well is provided
to divide the reservoir into two compartments.
When size of the reservoir is very big earthen reservoirs are constructed. But
distribution or service reservoirs should not be provided without cover because
they contain treated water, which is directly supplied to the consumers. Hence, it is
always preferable to make a masonry or concrete tank with cover.
When not limited by other considerations, the locations and elevation of the
bottom are so chosen as to secure the most economical relation between excavation
and filling. Where a town or locality is served by a single service reservoir, it is
desirable to divide the reservoir into two or more basins for convenience of
cleaning and repair.

16.7.1 Economical Depth

The most economical depth can be determined by trial. However, considering various
elements, most economical depth may be determined approximately by analysis.
Let us assume the reservoir to be a square one having side length as x and depth = h,
Q = given capacity and C = cost per unit area of all portion whose cost is proportional to
Water Distribution the area such as land, reservoir lining, cover etc. The cost of exterior wall per metre may
be assumed to vary approximately as 1/3.3 h1.5or will be equal to C 'h1.5where C ' is a
constant. The cost of excavation, embankments for a given capacity vary slightly with the
depth and will be nearly constant and let us take this as K.

:. Total cost C = 4C ' hle5xj + ~ j+! K . . . (16.1)

But ~ = h ? o r x = m
:. Substituting in Eq. (16.1) we have

Differentiating with respect to h and equating to zero, we get that for a minimum cost C,
h = l h m 4 @ . . . (16.3)
The economical depth is, therefore, proportional to the fourth root of Q and hence, it
should vary but little for considerable variations in capacity.
Since Q = h?,we have from Eq. (16.3)
h = (c/~c')*.P . . . (16.4)
which gives that h is proportional to J?
From Eq. (16.3), we see that as the cost per unit area increases (because of any reason), h
should also increase, but only in the proportion of @. If the cost of wall varies as h2,
then Eq. (16.3) becomes as

Eq. (16.5) indicates that here h varies at a lesser rate with Q than in Eq. (16.3).
The value of h will then vary with 6.With a fixed bottom elevation, it is to be noted that
the lift of the pump increases with increase in depth thus increasing the operation cost.
Most usual depths for different capacities are as follows :
Capacity (m3) Depth of Water (m)
Upto 3500 2.5 - 3.5
3500 - 15000 3.5 -, 5.0
Over 15000 5.0 - 7.0
But these figures do not apply to water towers or pre-stressed reinforced concrete tanks.
Factors influencing depth for a given storage are :
(i) depth at which suitable foundation conditions are encountered.
(ii) depth at which outlet main has to be laid.
(iii) slope of gound, nature and type of backfill soil.
(iv) the need to make the quantity of excavated material approximately equal to the
amount required for banking or filling so as to reduce carting of surplus
materials.
(v) the shape and size of land available.
16.7.2 Roofing
Water in a distribution or service reservoir is treated water ready for consumption.
Hence, it should not be stored in open space which may cause pollution from the
atmosphere.
Concrete roofs are designed as flat slabs on columns. Construction with cast in-situ-slabs
resting on beams or as a series of precast and pre-stressed beams laid side by side and
grouted together are also common. For ordinary reinforced construction span of 3.5 to
4.5 m, give a thickness in the range of 150 to 200 mm and are more common in use. Use
of precast pre-stressed beams, however, permits the use of spans upto 7.5 m thus reducing
number of columns. Due to temperature changes, the roof may have expansion. Hence,
they are not fixed with the walls but should be free to slide over them. For heat insulation
concrete roof slab is covered with gravel and finally a layer of earth. But if temperature Storage And
DistributionReservoirs
does not go too high, the earth layer may be spared because it increases dead load on the
slab making it more thick and thus, increasing the cost of construction. The gravel layer
helps in drainage of water falling over the roof cover of the tank. Sometimes for larger
spans arched or domed roofs are also made. Roofs are always provided with ventilators to
allow free circulation of air.

16.7.3 Walls of Concrete Reservoirs

Walls of reservoirs may be classified in three categories. Mass concrete gravity walls,
reinforced concrete walls and pre-stressed walls. For normal sizes of tank with good
ground conditions, the mass concrete or reinforced concrete walls are more common, of
course, R.C.C. Galls being more preferable. To give the stability to the wall, the water
face is made sloping. The wall is designed as a retaining wall, the worst condition of
loading being, when the tank is empty.
Pre-stressed concrete tanks are usually circular with pre-stressing being applied only in
the horizontal place by winding the pre-stressing wires around the outside of the tank
wall. When the wires are placed and tensioned, a 50 to !00 mm thick pneumatic mortar is
gunited on to the outside of the tank. Pre-stressed tanks can be made deep because the
extra tension induced in the walls by greater water depth can be met by adding more
strands of pre-stressing wires and, thus, thickness of walls not being greatly increased.
Pre-stressed tanks can have domed roof because the.thrust from the dome being taken by
a thrust ring provided at the top of the tank. Due to unequal settlement in the ground,
cracks develop in the walls. But cracking may be reduced by adopting following
measures :
(i) Using a concrete grade not very rich in cement even though some greater
thickness of sections are required.
(ii) Size of aggregate should be coarser.
(iii) Keeping water content of the mix as low as possible and adding an
air-entraining agent to assist in producing adequate workability.
(iv) Using steel shutters (not wooden or plywood shutters) so that the maximum
temperature gained during concrete setting is reduced quickly.
16.7.4 Floors of Reservoirs
Excepting a small size of tank the floor is cast as a separate structure from walls.
Underdrains are provided to prevent uplift and these underdrains have a free fall to an
open outlet. Sometimes, floors are laid in two layers. The joints in the upper layer are
staggered in relation to the joints in the lower one.

16.7.5 Pipings .
The inlet to a service reservoir may be kept at any elevation. If provided at the bottom, a
non-return valve on the inlet pipe work should be provided so that in case of pipe burst,
the water from tank should not flow back through the inlet main. The outlet pipe should
be taken out at low level and can be arranged with its invert at floor level provided a
sump is constructed just in front of the outlet into which debris (if at all present) will be
trapped instead of being swept into the main. The overflow out of service reservoir is an
important safety feature. Diameter or number of outlets should be such that maximum
inflow to the tank under any circumstances should not rise so high to cause uplift
pressure on the roof.
Stop valves must be put on both inlet and outlet mains. These may be made small& in
diameter than the pipeline by some 25% for lesser loss of head. Precautions are to be
taken to lay all pipes beneath the surrounding embankment on solid ground so that
differential settlement is minimised. When a pipe passes under the embankment and then
through the wall of the reservoir, there should be at least two flexible joints incorporated
in the line at the back of the wall to permit settlement of the wall without fracture of the
pipeline.

16.7.6 Testing for Watertightness

When a concrete reservoir is first filled, it should be left to stand for at least three days to
allow the concrete to absorb water. Testing procedure states that total drop in water level
over the next seven days should not exceed one thousandth of the average water depth of
Water Distribution the full tank for a satisfactory test. In order to save water, it is best to put 0.6 to 0.9 m of
water in the tank first so that the floor can be tested. If the floor appears to be watertight,
the reservoir should be filled to half depth and then to full depth allowing the tank to
stand to pennit absorption at each level before taking final accepting test.
If leakage is observed then to detect the leakage path, about 0.6 m of water is put into the
tank and be left to stand until water is quite still. Then crystals of potassium
permanganate may be dropped into the tank, widely spaced and left for a considerable
time. Then descending into the tank with a good and strong light source and walking over
a pre-arranged walkway (which is provided in tanks for inspection) so as not to disturb
the water, streaks of colour may be noticed from the permanent crystals showing some
definite flow towards a point or points of leakage. Figure 16.5 shows typical
cross-section of a service reservoir having slopped as well as flat base and Figure 16.6
shows a typical cross-section of a service reservoir having flat base.

Figure 165 :Cmss-section of a Dbtributlon Resendr with Sloping and Flat Floors

Flpre 16.6 :Cm-section of a Small RC.C. Distribution Reservoir with Flat Roof and Floor

16.8 ELEVATED RESERVOIRS

As discussed earlier, elevated reservoirs are overhead tanks supported on pillars and they
may be circular, elliptical or other shapes also to suit the architectural requirement. For
details of design, a book of R.C.C. design may be referred where design of components
are discussed in detail.
 16.8.1 Intz Tanks Storage And
Distribution Reservoirs
i An R.C.C. overhead tank of Intz type is most popular now-a-days whenever an overhead
tank is to be provided as service or distribution reservoir. A typical cross-section has been
shown in Figure 16.7. Intz tanks are structurally sound and economical as well. They are
provided for higher capacities. Various accessories for such a reservoir are as follows :
(i) inlet pipe for the entry of water.
(ii) outlet pipe connected to the distribution mains for the delivery of water.
(iii) overflow pipe discharging into drain gutters and maintaining level in the tank.
(iv) an indicator for indicating depth of water, which can be read from outside.
(v) a drain pipe for removing water after cleaning of the tank.
(vi) an automatic device to stop plumping when the tank is full.
(vii) ladder to reach the top of the reservoir and then to the bottom of the reservoir or
tank for inspection.
(viii)manholes in the roof for providing entrance into the tank for inspection and
minor r e p a s , if any.
(ix) ventilators for fresh air circulation.
LIGHTENING CONDUCTOR

INTZ TANK.

Figure 16.7 :Cross-sectionof an Intz %k

16.8.2 Stand-pipes
It has been discussed earlier that stand-pipes are small tanks of diameter 10 to 15 metres.
The useful storage of a stand-pipe is the volume above the elevation required to give the
required pressure in the distribution system. The water stored below this level is utilised
in emergency e.g., fire-fighting with the help of booster pumps. Figure 16.8 shows a
typical cross-section of a stand-pipe with connections.
Water Distribution

--
-.
- A

--
-

OVERFLOW PlPE

T0 DtST RI BUTION
MAIN

DRAIN PlPE

F'igure 168 :Cross-sectlonof Stand-pipe

SAQ 1
Discuss the graphical prwedure for estimating capacity of a distribution reservoir.
(i) when pumping is done continuously for 24 hours.
(ii) when pumping is done for a limited period.

SAQ 2
An overhead tank is to be provided for a town water supply. Given the following
data, calculate the minimum capacity of the tank without any fire demand. Tank is
empty between 12 to 15 hours.
(i) Average water supply = 16,666 litres / hr
(ii) Rate of Pumping = 25,000 litres / hr
(iii) Hours of pumping = 4 to 12 and 15 to 23 hrs.
 Storage And
Distribution Reservoirs

SAQ 3
A service water tank is receiving water from the treatment plant at a rate of
200 m3/hr for 24 hours. The high lift pumps are lifting water from the same tank at
following rates: 4-14 hrs @ 120 rn3/hr and 15-24 hrs @ 400 m3/hr. Determine the
capacity for the service water tank.

SAQ 4
Discuss the location and height of distribution reservoirs.

SAQ 5
For a town daily requirement of water for supply to the population is 2 lakhs litres.
The pattern of draw off is as follows :
6 a.m. to 9 a.m. - 30% of day's supply
9 a.m. to 4 p.m. - .35% of day's supply
4 p.m. to 7 p.m. - 25% of day's supply
7 p.m. to 6 a.m. .lo%of day's supply
The pumping is done for 10 hours a day from 8 a.m. to 6 p.m. Find out the storage
.
capacity of the disbibution reservoir for the water supply scheme.
Water Distribution
16.9 SUMMARY
This unit deals with distribution reservoirs. When they are comparatively smaller in size
1
they are called service or distribution tanks. Service reservoirs are put on surface of the
ground or to provide additional head they are put at higher elevation supported on pillars.
Smaller elevated service reservoirs are known as stand-pipes. This unit deals with
procedure to evaluate or estimate capacity of distribution reservoirs with reference to
demand and supply. Details of different components and types with respect to
construction have also been discussed.

16.10 KEY WORDS 1

Booster Pump : A pump installed on a pipe line to raise the pressure of
water on the discharge side of the pump.
Draft : Withdrawal of water from a water storage or a reservoir. 1
Hydrant : A connection extending from a water main to or above the
ground surface with valved connection to which a
j
firehose is attached for discharging water at high rate.
Mass Curve or Mass : A graph of cumulative run-off versus time.
i
Diagram 1
Sump : A depression that receives water or any liquid. I
I
I
: Small elhated service tanks of diameter around 10 to
I5 metres.
I

16.11 ANSWERS TO SAQs

SAQ 1
Capacity of a distribution reservoir or balancing storage of a reservoir can be
estimated with the help of hydrographs of inflow and outflow by mass curve
method.

Supply Curve

Demand Curve

TIME IN HOURS
Figure 16.9
-
T d a l storage required
= A+B

0 6 AM 6 PM
 A mass curves is drawn for the accumulated supply versus time and over this, Storage And
demand curve versus time is super-imposed. The amount of balancing storage is DistributionReservoirs
determined by adding maximum ordinates between the demand and the supply
lines. Figure 16.9 explains the procedure when pumping is done continuously and
Figure 16.10 explains the procedure when pumping is done for a limited period.
SAQ 2
Problem has been solved analytically in table given below.
As given in the question, the tank is empty between 12 and 15 hours. Hence, we
shall start from S1. No. 5. At Serial No. 6 storage is (+) 16,666 again at S. No. 7, it
has reduced by 6666 litres. Hence, net storage'is (16,666-6,666) = 10,000 and so
on proceed S1. No. 8, 9, 1,2, 3 and finally at 4.

Sl. No. Tie Hours Water Water Accumulation Water in

pumped consumed (+), draw off service
-
(Litres) (Litres) Col. 4 Col. 5 reservoir in
litres

Now from Column No.7, it is clear that maximum balance storage is 35,000 litres.
Hence, minimum capacity of the tank should be 35,000 litres (Ans.)
SAQ 3
The problem has been solved in the table given below.
The cumulative supply over demand i.e., excess supply to the tank is 1800 m3.
Hence, the capacity of the tank should be 1800 m3.

Time in Demand from Cumulative Supply in the Cumulative Accumulation

bur clear water demand dear water supply (m3) (+) Draw off
tank lifted by from clear tank (m3) (-) Co1.5 -
high liftpumps tank (m3 Co1.3
(m 1

1 2 3 4 5 6

0-4 Nil Nil 200~4=800 800 (+I 800

4-14 120x10=1200 1200 200 x 10 = 2000 2800 (+) 1600

-
14 15 Nil 1200 200x 1 =ZOO 3000 (+) 1800

15-24 400~9=3600 4800 2 0 0 ~ 9 =1800 4800 0

SAQ 4
A distribution reservoir should be located in the centre of the distribution area so
that maximum reach could be served. It should be nearer to the area of heaviest
demand so that friction loss through pipes is minimum. It is preferable to locate
them a&the highest elevation available in the area so that sufficient head is
Water Distribution available in distribution system. Next point to be considered keeping in view of
above points is that they should be placed between pumping station and
distribution area. For more detail, refer Section 16.4.

SAQ 5 -.
Total daily requirement = 200,000 litres
Now the cumulative demand covered is tabulated in the table given below :

Period Rate of Demand Demand in Cumulative

Litres Demand

6 a.m. to 9 a.m. 30% of 2 lacs litres 60,000 60,000

9 a.m. to 4 p.m. 35% of 2 lacs litres 70,000 130,000

4 p.m. to 7 p.m. 25% of 2 lacs l~tres 50,000 180,000

7 p.m. to 6 a.m. 10% of 2 lacs litres 20,000 200,000

The mass curve of demand has been plotted in Figure 16.11.

-

TIME IN HOURS

Figure 16.11
-
Total demand is met in 10 hours
.: Rate of supply = 200,000/10 = 20,000 litreshr.
Two maximum ordinates between supply and demand lines are
A = 30,000 litres
B = 42,0001itres
:. Total storage capacity = A + B = 72,000 litres (Ans.)