Arbaminch Water Technology Institute

Faculty of Water Resources and Irrigation

Water Supply Engineering

Instructor: Bisrat Habtamu

APRIL, 2023
 1. Introduction
Water supply Engineering ?
• A field concerned with the development of sources of supply, treatment, storage, transmission,
and distribution of water.
• Access to water-supply services is defined as the availability of at least 20 liters per person per
day from an "improved" source within 1 kilometer of the user's dwelling.
• In the design of any water work projects it is necessary to estimate the amount of water that is
required.This is called water demand
• This water demand estimation involves:
The per capita water consumption
The determination of people who will be served
Analysis of the factors that may operate to affect consumption
 1.1. Water Demand

• Why is there so much demand for fresh water in

the world ?
 Nature and Water
• Of all the planets discovered, Earth is the only one with water.

• Water covers more than 70 per cent (70%) of the earth’s surface and
exists as a vapor in the earth’s atmosphere.

• Water regulates the Earth’s temperature.

• It is considered as the universal solvent because of its ability to

dissolve almost all organic and inorganic solids and gases it comes in
contact with. For this reason, pure water is never found in nature.
 Cont.
• Even rainwater, the purest natural water, contains chemicals dissolved
from the air.

• Pure water is obtained only by special methods of distillation and by

chemical action in laboratories.
  Water and Humans
• Up to 70% of the human adult body is water. According to H.H. Mitchell, Journal of
Biological Chemistry 158,

• the brain and heart are composed of 73% water, and

• the lungs are about 83% water.

• The skin contains 64% water,

• muscles and kidneys are 79%,

• and even the bones are watery: 31%.

• As a general rule of thumb, a person can survive without water for about 3 days.
 1.2. Per Capita Demand
• While designing the water supply schemes of the town, it is necessary to
determine the total quantity of water required for various purposes by the
community.

• For the purpose of estimation of total requirement of water, the demand is

calculated on the average basis.

• Per capita water demand (lit/capita/day)

= Total annual average water consumption of community
Population * 365days
 1.3. Types of Demands
• Following are the various types of water demands of a city or town.

i. Domestic water demands

ii. Commercial and industrial demand

iii. Fire demand

iv. Demands for public use

v. Compensate losses demand

 i. Domestic water demands
• It includes the quantity of water required in the houses for drinking,
bathing, cooking, washing etc.

• The quantity of water required for domestic use mainly depends on the
Habits, social status, climatic condition and custom of the people.
 ii. Commercial and industrial Demand
• It includes the quantity of water required in the office building, stores,
Hotels, shopping centers, health center, school, cinema house, industries
etc.
 iii. Fire Demand
• It is the quantity of water required for fighting a fire break-out. This quantity is
normally obtained on the basis of certain empirical formal.

a. National board of fire

underwriter formulas. b. Freeman formula. c. Kuichling formula.

Q = 64 √ P (1 – 0.01√P) Q = 1135.5 ((P⁄ 10) + 10) Q = 3182 √P

Where Q = rate of flow of Where Q is in lit/min Where Q is in lit / min
water in l/sec P is in thousands
P = Population in thousand
 iv. Demands for Public Use
• Quantity of water required for public utility purposes such as washing, Roads,
cleaning of sewer, gardens, public fountains, watering of public parks etc.
 v. Compensate Losses demand
• The quantity of water required for the allowance of losses of water due to
defective pipe joints, cracked & broken pipes, unauthorized & illegal connection,
faulty valves & fittings etc.

• Generally ,allowance of 15 – 20% of total quantity of water is made to

compensate for the losses

• Water Loss = Percentage of Loss * Total Water Demand

 1.4. Factors affecting the water Demand
(water consumption)
• Following are the main factors which affect the per
capital demand of the community
a) Size of the community: f) Cost of Water:
b) Climatic condition: g) Quality of Water:
c) Socio economic status:
d) Methods of supply:
e) Pressure:
 Variation in Demands
• The water demand varies from season to season, day to
day, even hour to hour.
• Seasonal variation: high in hot season and less in cold
season.
• Daily variation: high during holidays, Saturday, Sundays.
• Hourly variation: high in morning, launch time, evening.
 Average Day Demand
• The average day demand is taken to be the sum of the demands of

domestic, non domestic and Non Revenue Water (NRW)).

• The average water demand represents the daily demand of the town

averaged over the year.

 Maximum Day Demand
• Maximum Day is the highest demand of any one 24-hour period over any
specified year.
• It represents the change in demands with seasons.
• The ratio of the maximum daily consumption to the mean daily consumption
is called the maximum day factor and usually varies between 1.0 and 1.3.
• The maximum day demand is used for calculations such as source pumping
requirements.
 Peak Hour
• It is the highest demand of any one-hour over the peak day.

• It represents the diurnal variation in water demand resulting from

behavioral of the local population.

• The peak hour water demand is greatly influenced by the size of the town,

mode of service used and social activity pattern.

• Peak hour demand is used for design of distribution system.

• The ratio of the peak hour demand to the maximum day demand is called

peak hour factor.

 Cont.
• Theses figures are taken from the cost effective design report
of the Ministry of Water Resources.
• These factors are applied on the Average Day Demand to get
the Maximum Day Demand and Peak Hourly Demand.
 Adjustment for climate
• In order to account for changes in climate which affect the water demand,
the values of the average per capita domestic water demand established
above should be factored for climatic changes using the Ministry of
Water Resources’ climate factors guidelines shown in the table
below.
Grou Mean annual rainfall Factor
p (mm)
A 900 or Less 1.1

B 900 - 1200 1.0

C 1200 or More 0.9

Adjustment due to socio-economic conditions
• To accommodate the changes due to the potential for growth, the

following recommended socio-economic factors are used.

• These factors are applied for all the domestic demand values above.
Group Description Facto
r
A Towns enjoying living standard and with very high 1.1

potential development

B Towns having a very high potential for development but 1.05

lower living standard at present

C Town under normal Ethiopian conditions 1.00

 Chapter Two
Population Design
 2.1. Introduction
• The design population is the targeted number of
people that the project will serve.
 2.2. Design Period
• Design Period: - is the utility period or useful no of years for
which the designs of the water works have been done.

• The design period should neither be so long which will incur

excess financial burden at present users nor be so short as to
make the design insufficient for future needs, mostly water
works are designed for design period of 22 – 30 years.
 For rural communities
• The advantages and disadvantages for the 5- and 10-year options are:
1. Five-year design period
• Advantages – Low initial capital cost. If the project is to be financed through
a loan, the loan amortizations are lower due to the lower investment cost.
• Disadvantages – Need for new capital outlays after five (5) years to upgrade
system capacity.
2. Ten-year design period
• Advantages – The water system facilities are capable of meeting the demand
over a longer period. No major investment cost is expected during the 10
year design period.
• Disadvantages – The higher initial capital cost will require initial tariffs to be
set higher.
 2.2. Population Forecasting
• There are 2 ways of projecting the design population.
1. Estimate the population that can be served by the
sources. In this case, the supply becomes the limiting factor
in the service level, unless a good abundant and proximate
source is available in the locality.
2. Project the population, and determine the potential
service area and the served population
 Methods to forecast
• The following are the standard method by w/c the
forecasting of Popn done:
• ii. Arithmetical increase method
• iii. Geometrical >> >>
• iv. Incremental >> >>
• v. Simple graphical method
• vi. The logistic curve method
 i. Arithmetical increase method
• This method is based on the assumption that the population is
increasing at a constant rate. The rate of change of population with
time is constants

• This method is generally applicable to a large and old city.

 Cont.
• The rate of change of population with time is
constants.
𝑑𝑝
i.e. =k
𝑑𝑡
∫dp = k ∫dt
Integrating the limit Po – Pn & o - n
Pn – Po = K (n – o)
Pn = Po + k.n
 Example 2.1
The following has been noted from the census department.

Year 1940 1950 1960 1970

Popn 8,000 12,000 17,000 22,500

Calculate the probable population in the year 1980, 1990 and 2000.
 Solution
Year Population Increase in Popn Average Increase (K)

1940 8,000 -
(4000 + 5000 + 5500)
1950 12,000 4,000
3
1960 17,000 5,000
= 4, 833
1970 22,500 5,500

Pn Po n k Pn = Po + k.n

1980 22,500 1 4,833 27,333

1990 22,500 2 4,833 32,166

2000 22,500 3 4,833 36,999

 ii. Geometrical increase method.
• This method is based on the assumption the
percentage increase in population from decade to
decade remains constant.
• This method is mostly applicable for growing towns
and cities having vast scope of expansion.
 Formula
• Pn = Po (1 + k)n where Po = initial population ,
• Pn = Popn at n decades ,
• n = Decades &
• k = Percentage (geometric) increase

• Geometric Increase= Popn Increase in Decade * 100

Previous Decade Popn
 Example 2.2
Forecast the population of example 1 by means of
geometrical increase method
Solution

Year Population Increase in Popn Geometric Incr. (K)

1940 8,000 - -
4,000
1950 12,000 4,000 * 100 = 50%
8,000
5,000
1960 17,000 5,000 * 100 = 41.7%
12,000
5,500
1970 22,500 5,500 * 100 = 32.4%
17,000
(50 + 41.7 + 32.4)/3
Average
=41.37%
 Cont.
• Pn = Po (1 + k)n
Pn Po K n Pn = Po (1 + k)n

1980 22,500 41.34 % 1 31,801

1990 22,500 41.34 % 2 44,948

2000 22,500 41.34 % 3 63,530

𝟒𝟏.𝟑𝟒 2
i. P1990 = 22,500 (1 + )
𝟏𝟎𝟎

= 44,948
 iii. Incremental Increase method
• This method is improvement over the above two
methods.
• The average increase in the population is determined
by the arithmetical method and to this is added the
average of the net incremental increase once for each
future decade.
• Pn = Po + n ( Avg. Incr. + Net .Avg. Incr.)
 Example 2.3
Forecast the population of example 1 by mean of incremental increase
method.
Solution
Incremental
Year Population Increase in Popn.
Increase
1940 8,000 - -

1950 12,000 4,000 -

1960 17,000 5,000 1,000

1970 22,500 5,500 500

Total 14,500 1,500

Average 4,833 750

 Cont.
• Pn = Po + n ( Avg. Incr. + Net .Avg. Incr.)
Pn Po n Avg. Incr. Net. Avg. Incr. Pn

1980 22,500 1 4,833 750 28,083

1990 22,500 2 4,833 750 33,666

2000 22,500 3 4,833 750 39,249

i. P2000 = 22,500 + 3 (4,833 + 750)

= 39,249
 iv. Decreasing rate method
• In this method, the average decrease in the
percentage increase is worked out and is then
subtracted from the latest percentage increase for
each successive decades.
• This method is applicable to average size cities
growing under normal condition.
 Example 2.4:
Solve example 1, by using decrease rate of growth method.
Solution
Decrease in %
Year Population Increase in Popn % Increase in Popn
Increase
1940 8,000 - - -
4,000
1950 12,000 4,000 * 100 = 50 -
8,000
5,000
1960 17,000 5,000 * 100 = 41.7 8.3
12,000
5,500
1970 22,500 5,500 * 100 = 32.4 9.3
17,000
Total 17.6

Average 8.8
 Cont.
Year Population Decrease in % Increase Pn=Pn-1 + (Dec in % Inc. * Pn-1)

1980 22,500 32.4 – 8.8 = 23.6 % 27,810

1990 27,810 23.6 – 8.8 = 14.8 % 31,926

2000 31,926 14.8 – 8.8 = 6.0 % 33,842

𝟐𝟑.𝟔
i. P1980 = 22,500 + ( * 22,500)
𝟏𝟎𝟎

= 27,810
 V) Logistic Curve method
• This method is used when the growth rate of population due to births, deaths
and migrations takes place under normal situation and it is not subjected to
any extraordinary changes like epidemic, war, earth quake or any natural
disaster etc. the population follow the growth curve characteristics of living
things within limited space and economic opportunity.

• When the population of a town is plotted with respect time the curve so
obtained under normal condition is s – shaped curve and is known as logistic
curve.
 Saturation Value (Ps)
N

𝑑𝑝
∝ (Ps-
𝑑𝑡
M P)
𝑑𝑝
= Constant Point of Inflexion
𝑑𝑡
L
Population (P)

K
𝑑𝑝
∝k Curve of growth
𝑑𝑡
rate
J
Time (t)
P.F. Verhulst after long research work has given the following mathematical solution of logistic
curve:
𝑃𝑠
P= −
1
1+𝑚 𝑙𝑜𝑔𝑒 (𝑛.𝑡)

McLean further suggested that if three pairs of the characteristics values P0, P1 and P2 at
times t0, t1, and t2 which are extending over the useful range of the population are so
chosen that t0= 0, t1, and t2 = 2t1 the saturation values Ps and constants m and n can be
determined :
Ps = 2P0P1P2 P12 (P0 + P2)
P0P2 – P12
m = Ps – Po
Po

n = 2.3 log10 Po (Ps –P1)

t1 P1(Ps – Po)
 Example 2.5
• Following is the Popn of a city as noted from the census
department:

Year 1950 1960 1970

Population 35,000 78,000 115,000
• Determine
a) Saturation Popn
b) Equation of logistic curve
c) Expected Popn in 1980
 Solution
For logistic curve, P0 = 35,000 at t0 = 0
P1 = 78,000 at t1 = 10yr
P2 = 115,000 at t2 = 20yr

A) Ps = 2P0P1P2 - P12 (P0 + P2) = 2*35,000*78,000*115,000 – 78,0002(35,000 + 115,000)

P0P2 – P12 35,000*115,000 – 78,0002
= 6.279*1014 - 9.126*1014 = -2.847*1014
- 205.9 *107 - 205.9 *107
Ps = 1.38 * 105
B) P = Ps
1 + mloge -1(n.t)

m = Ps – Po = 138,000 – 35,000 = 2.943

Po 35,000
n = 2.3log10 Po (Ps –P1) = 2.3 log10 35,000 (1.38 * 105 – 78,000) = 0.23* log10(0.26)
t1 P1(Ps – Po) 10 78,000 (1.38 * 105 – 35,000)
= - 0.134
 Cont.
P= Ps = 138,000
1 + mloge -1(n.t) 1 + 2.95 loge -1 (-0.134*t)

P= 138,000 X = loge -1 (-0.134*t)

1 + 2.95 X

C) 1980, where t = 30 t= time which starts from to = 0

• X = loge -1 (-0.134*t) i.e. t=30 (t0=0, t1=10, t2=20, t=30)
• X = loge -1 (-0.134*30)
• loge X = -0.134 * 30 = -4.02
• 2.3log10X = -4.02 • loge=2.303log10
• log10X = - 4.02 = - 1.7478
2.3 • logab=X
• X = 10-1.7478 = 0.01787
ax=b
P = 138,000 = 138,000 = 131,090
1 + 2.95 X 1 + 2.95(0.01787)
 Vi) Simple Graphical method
• In this method the Popn curve (the Popn vs. year) are correctly
plotted to a suitable scale on the graph. Then the curve is
smoothly extended to forecast the future Popn.
 Example 2.6
• The Popn in the past decays is given as follows:

Year 1930 1940 1950 1960

Popn 9,000 13,000 17,000 23,000

• Forecast the probable population number by using simple graphical

method?
45000
41000
40,000
37000
33000 34,000
29000
28,900
25000
21000
17000

13000
9000
1930 1940 1950 1960 1970 1980 1990
 Chapter Three
WATER SOURCES
 3.1. Introduction
• Once the population for a given design period is forecasted and
water demand , the next step is to look for a source that meets
both the quantity and quality requirements.

• This Chapter presents an overview of the possible water supply

sources that can be utilized for community and other water supply
systems.
 Water Sources

Surface Water Ground Water

Rivers/Streams Lake
Springs Well

Artesian Gravity Open Well

Depression Depression Tube well

Fissure Overflow

Overflow
  Selection Criteria for water source
• In the selection of a source or sources of water supply, adequacy and reliability of
the available supply could be considered the overriding criteria.

• These, together with the other factors that should be considered (and which are
interdependent), are as follows:
i) Location
ii) Quantity
iii) Quality
iv) Cost
v) Legality
vi) Politics.
Rivers and Ponds

3.2. Surface Water Sources

  River water source
• Quantity of Water
• Some rivers are perennial and some are non perennial.

• In tropical countries, rivers and streams often have a wide seasonal fluctuation in
flow.

• Perennial River should always be selected for the scheme.

• Incase of non perennial rivers, the weir or low dam may be constructed to form
a storage reservoir.
 Cont.
• Quality of Water
• In wet periods the water flow is high but often of a high turbidity.

• In dry periods river flows are low and the load of dissolved solids is more
concentrated.

• Mountain streams sometimes carry a high silt load but the mineral content is
mostly low and human pollution is generally absent.
 Cont.

• In plains and estuaries, rivers usually flow slowly except when there is a flood.
The water may be relatively clear but it is almost always polluted, and extensive
treatment is necessary to render it fit for drinking and domestic purposes.

• The quality of river water does not usually differ much across the width and
depth of the riverbed.
  Lake water source
• Quality of Water
• The quality of lake water is influenced by self-purification through aeration, bio-
chemical processes, and settling of suspended solids.

• The water can be very clear, of low organic content and with high oxygen
saturation. Usually, human and animal pollution only present a health hazard near
the lake shores.

• At some distance from the shore, the lake water has generally a low density of
pathogenic bacteria and viruses. However, algae may be present, particularly in
the upper layers of lakes.
Springs and Well

3.3. Ground Water Sources

 Ground Water Sources
• Groundwater occurs in pores, voids
or fissures of ground formations. Pores
are the spaces between the mineral
grains in sedimentary ground layers
and in decomposed rocks.

• Use of groundwater for community

water supplies is probably still very
much below its potential in many
countries.
 Identification of Ground water
• Successful prospecting for groundwater requires knowledge of the manner
in which water exists in the water-bearing ground formations. Without this
knowledge, effective and efficient water exploration is impossible, and well
drilling then becomes rather like a game of roulette.

• A survey of the study area should be made, preferably towards the end of
the dry season.
 3.3.1. Spring
• A spring may be defined as a place where a natural outflow of groundwater occurs.

• The oldest community water supplies were, in fact, often based on springs and they
remain a favored source, because the water usually has a high natural quality.

• Because of their popularity, most natural springs have been developed in one way or
another as drinking water sources.

• However, a proper feasibility study, application of some basic design principles and
vigilance in protecting the spring and its catchment area will usually lead to
improvements in the quantity, quality and sustainability of many such supplies.
  Identification of Spring
• Local people, especially women (as drawers of water), but also farmers, and grazers,
have a good knowledge of the location of springs and their characteristics. These
people are the primary sources of information in the identification process.

• In the dry season, green vegetation in a dry area may also be an indication of a spring
source.
 Cont.
• Some springs form small ponds where animals drink and people may scoop
water from there, while others flow as small streams in valleys and can be
traced back to the source. The source, though, is not necessarily the first
upstream point at which the stream emerges from the ground.

• In some cases streams may be buried for quite a length and there can be added
risks of contamination unless the investigation continues further upstream to
locate the true spring.
  Types of Spring Sources
• Springs are classified according to the conditions under which water flows to
them.

• Some surface under pressure, while others do so as a result of discontinuities in

the strata that held the water underground.

• To understand the possibilities of water tapping from springs, the distinction

between gravity springs and artesian springs is most important.

• A further sub-division can be made into depression springs and overflow springs.
 A) Gravity Spring
i. Gravity Depression spring

• Gravity springs occur in

unconfined aquifers.
• Where the ground surface dips
below the water table, any such
depression will be filled with
water.

• Gravity depression springs usually have a small yield and a further reduction
occurs when dry season conditions or nearby groundwater withdrawals result in
the lowering of the groundwater table.
 ii. Gravity Overflow Spring
• A larger and less variable yield
from gravity springs is obtained
where an outcrop of impervious
soil, such as a solid or clay fault
zone, prevents the downward
flow of the groundwater and
forces it up to the surface.

• At such an overflow spring, all the water from the recharge area is discharged. The flow will be
much more regular than the recharge by rainfall.
 Cont.
• Even so, an appreciable fluctuation of the discharge may occur and in
periods of drought some springs may cease to flow completely.

• Because of their small yield and the difficulty of providing adequate sanitary
protection, gravity depression springs cannot be recommended for
community water supplies.
 B) Artesian Springs
• Artesian groundwater is prevented from rising to its free water table level by
the presence of an overlaying impervious layer. That is the reason why artesian
groundwater is under pressure.

• Artesian springs are the sites where the groundwater comes to the surface.
 i. Artesian Depression spring

Artesian groundwater is prevented

from rising to its free water table
level by the presence of an overlaying
impervious layer

Artesian depression springs are

similar in appearance to gravity
depression springs.

• However, the water is forced out under pressure so that the dis-charge is higher and
there is less fluctuation. A drop of the artesian water table during dry periods has little
influence on the artesian groundwater flow.
 ii. Artesian Fissure spring

In this type of spring again the water emerges under pressure, this time through a
fissure in the impervious overburden. Fissure springs exist in many countries and
are widely used for community water supplies.
 iii. Artesian Overflow spring
• Artesian overflow springs
often have a large recharge
area, sometimes a great
distance away.
The water is forced out under
pressure; the discharge is often
considerable and shows little
or no seasonal fluctuation.

These springs are very well suited for community water supply purposes.
Artesian springs have the advantage that the impervious cover protects the water in the
aquifer against contamination.The water from these springs is usually bacteriologically safe.
 Spring Quality
• In general, spring water is of good quality.

• Pathogenic contamination is unlikely if the source meets certain criteria.

• These include the thickness of the soil layer, the type of soil and the velocity of
infiltration of the surface water.

• The soil formation should be thick enough for natural filtration and biological
action to remove pathogenic organisms before the water enters the aquifer
feeding the spring.
 Cont.

• The type of soil determines the speed of the flow through the voids in the
soil and so influences the purification mechanisms and the concentration of
suspended solids.

• If the soil layer is not thick enough, any human activity should be restricted
or even forbidden in the catchment area.
 Key signs of a good spring
• If the water maintains a constant temperature throughout the day, in which the
temperature is just below the average air temperature.

• The water should also be colorless.

• If users note these temperature and color changes, but they still favor the spring
over alternative water sources, then the community needs to be advised to
include water treatment as part of the water supply system.
 Key signs of a Poor quality spring
• Variation of water temperature during the day and coloration of water shortly
after rains are indications of a poor quality spring source.

• This is caused by having its water-bearing soil layer not deep enough or rapid
infiltration of surface water through the topsoil.
 Spring Water Quantity
• The quantity of water a spring produces is known as its yield.

• Yield is studied in terms of flow rate and consistency.

• Variation in the yield of a spring during the dry season and the rainy season is
an important criterion to determine whether the spring is a suitable source.

• If the ratio between the highest yield in the rainy season and the yield in the dry
season is below 20, then the spring has an acceptable consistency and can be
regarded as a reliable source in both wet and dry seasons.
 Cont.

• As the population and/or its productive activities increase, daily water demand
also increases, resulting in potential water shortages in the system. Shortages
will occur first during dry seasons and a few weeks or months into the rainy
season.

• The times of peak and minimum spring yield do not necessarily correspond to
the peak and minimum rainfall periods.

• In fact, the lowest spring yield usually occurs about a few weeks to several
months into the rainy season.
 Cont.

• This is because in the rainy season water velocity in the saturated stratum gets
too high, the pores through which the water passes tend to become choked so
that the flow becomes considerably reduced.

• A longer duration is preferred as there may be dry and wet years. The study
will indicate the variation in yield of the spring throughout the year, and the
maximum, minimum and estimated average flow.

• The average yield will reveal if more than one spring is needed to meet the
daily water demand of the user population.
 Estimating Spring Yield
• A spring yield is measured in liters per second (l/s).

• The measurement process involves two selected trained villagers who measure
the discharge from the spring over the study period.

• The process starts with the construction of an earth dam. Spring water retained
by the dam is drained through a pipe.

• One villager collects the water with a container of a known volume while the
other measures the time needed to fill the container.
 Cont.

• The pipe diameter and container size are chosen such that the water outflow
will not fill the measuring container in less than five seconds. Sometimes several
pipes are used.

• Four readings are taken during the day and day averages are calculated,
expressing the discharge in l/s.

• This is repeated once every week for the measuring period. In this way, the
minimum and maximum yields are determined.
 Quiz
• What is meant by PER CAPITA DEMAND?
• Mention the types of water demand?
• Mention the reasons for losses and wastage of water?
 Rapid Environment Assessment

• This involves identifying possible environmental consequences of developing a

spring.

• The environmental assessment includes investigating the flow direction of

surface run-off above the spring; human activities and water uses in the
catchment area, i.e. habitation, farming, grazing, etc.; and the type of plants
growing in the catchment or recharge area.
 Cont.
• If there are people living in the catchment/recharge area, they are likely to
contaminate the groundwater through their own waste and their activities such as
cattle holding or agriculture using artificial fertilizer or chemicals. But it may be
very difficult to relocate them. If the groundwater contamination risks are too
high, then such locations are not suitable.

• Some trees and plants are undesirable too. Eucalyptus trees, for instance, compete
for water with the spring and can significantly reduce the yield. Raffia palms,
though harmless, increase the iron content of the water, changing its taste and
color enough to deter consumers.
Type of Aquifers, well hydraulics and Yield.
Water Bearing Formations &
Ground Water Withdrawal
 Types of Aquifer

i. Unconfined Aquifer
• An unconfined aquifer is open to infiltration of water directly from the ground
surface
 ii. Confined Aquifer
• A confined aquifer is one where the water-bearing ground formation is capped by an
impermeable ground layer.

• The water pressure in a confined aquifer can be measured by drilling into it and
observing the level to which the water rises in the borehole. This water level is called
the piezometric level.
 iii. Perched Aquifer
• Infiltration of water from the ground surface through permeable ground towards the
groundwater table will be halted where a lens of impervious material such as clay is
present. Water will then accumulate in the ground above this lens, forming a perched
water table of some distance above the real groundwater table.
• Its very important to identify a
perched water table since the amount
of water it contains is often small.
• Frequently, perched water tables will
disappear during dry periods when
there is no recharge by infiltration
from the ground surface
 Methods of Ground Water Withdrawal

i. Horizontal ii. Vertical

Small Diameter
Large Diameter
Galleries Tube well or
Dug wells
Boreholes

Ditches

Infiltration Drains

Tunnels
 i. Horizontal withdrawal

Infiltration drains
Infiltration tunnels
 ii. Vertical withdrawal (Wells)
a) Open well (For unconfined ):
 It is constructed by digging the earth.
 It draws water from the topmost pervious layer.
 he diameter of this well varies from 1m to 2m and the depth varies from 20m
to 30m depending upon the nature of soil & the water table.
 Cont.
b) Tube well (For confined Aquifer):
 It is constructed by sinking G.I pipes.
 It draws water from the deeper most pervious layer.
 The diameter and the depth of this well varying from 37mm to 150 mm and
100m to 200m respectively, depending upon the nature of soil and suitable
water bearing strata.
 Well Hydraulics
Steady radial flow into a well (Dupuit 1863, Thiem 1906)
 R Water table conditions
r2
r1 Q (unconfined aquifer)

Static WL
S2
P S1 2rw
Sw Cone of
∆𝐘
∆𝐗 Drawdown
H Pumping WL
h2 X
Y h1
T=KH
hw Homogeneous
Aquifer ‘K’

Assuming that the well is pumped at a constant

rate Q for a long time and the water levels in the • Q = 𝜋 𝐾 (ℎ22- h12)
𝑟
observation wells have stabilized, 2.303log10( 2)
𝑟1
i.e., equilibrium conditions have been reached
 Cont.
R Q

Static WL

P 2rw
Sw Cone of
𝛁 Drawdown
Pumping WL
H X
Y
T=KH
Homogeneous
hw Aquifer ‘K’

• If the drawdown in the pumped well (Sw = H – hw) is small:

• Q = 2.72 T (H- hW)
𝑅
log10( )
𝑟𝑤
 R Tube well
r2
r1 Q (Confined aquifer)

Piezometric Surface
S2
S1 2rw
Sw Cone of
∆Y
∆X Drawdown
PWL
H h2 X
h1

T=Kb
hw b Confined
Aquifer ‘K’

Assuming that the well is pumped at a constant • Q = 2𝜋 𝐾𝑏 (ℎ2- h1)

rate Q for a long time and the water levels in the 𝑟2
2.303log10( )
observation wells have stabilized, 𝑟1
i.e., equilibrium conditions have been reached
 Cont.
R Q

Piezometric Surface
P 2rw
Sw Cone of
Drawdown
𝛻
Pumping WL
H X
Y

hw T=Kb
b Confined
Aquifer ‘K’

• If the drawdown in the pumped well (Sw = H – hw) is

small:
• Q = 2.72 T (H- hW)
𝑅
log ( )
  Specific Capacity
𝐐
• The specific capacity of a well is the discharge per unit drawdown in the
𝐒𝐰

well and is usually expressed as lpm/m. The specific capacity is a measure of

the effectiveness of the well; it decreases with the increase in the pumping
rate (Q) and prolonged pumping (time, t).

• H – hw = Sw, the specific capacity Q in consistent units

𝐐 𝐓
• ≈
𝐒𝐰 𝟏.𝟐
 Example 3.1
• A 20-cm well penetrates 30 m below static water level (GWT). After a long
period of pumping at a rate of 1800 lpm, the drawdowns in the observation
wells at 12 m and 36 m from the pumped well are 1.2 m and 0.5 m,
respectively.

• Determine: (i) the transmissibility of the aquifer.

(ii) the drawdown in the pumped well assuming R = 300 m.

(iii) the specific capacity of the well

 Solution
(i) the transmissibility of the aquifer
T = KH
• Q = 𝜋 𝐾 (ℎ22- h12)
𝑟2
2.303log10( )
𝑟1
• h2 = H – s2 = 30 – 0.5 = 29.5 m; h1 = H – s1 = 30 – 1.2 = 28.8 m

• 1800 = 𝜋 𝐾 (29.52- 28.82)

36
60 2.303log10( )
12
K = 2.62 * 10-4 m/sec or 22.7m/day
• Transmissibility = KH
= (2.62*10-4)*30
= 78.6 × 10–4 m2/sec, or 681 m2/day
 Cont.
(ii) the drawdown in the pumped well assuming R = 300 m
• Q = 2.72 T (H- hW)
𝑅
log10( )
𝑟𝑤
• 1800 = 2.72 (78.6*10-4)Sw
300
60 log10( )
0.10
∴ drawdown in the well, Sw=4.88m

(iii) The specific capacity of the well

𝑄 1800
= =
𝑆𝑤 60∗4.88
= 0.0062 m3sec-l/m
 Example 3.2

• A tube well taps an artesian aquifer. Find its yield in liters per hour for a
drawdown of 3 m when the diameter of the well is 20 cm and the thickness of
the aquifer is 30 m.Assume the coefficient of permeability to be 35 m/day.
Solution

• Q = 2.72 T (H- hW)

𝑅
log10( )
𝑟𝑤
35
• Q = 2.72 ( ∗ 30) (3)
24
300
log10( )
0.10

= 102.7 m3/hr
 Example 3.3
• A tube well penetrates fully into a confined aquifer .The following data was
collected during observation. Calculate the discharge of discharge of the well?

- Radius of tube well = 20 cm

- Thickness of confined aquifer = 25m
- Drawdown = 4m
- Radius of circle of influence = 300m
- Coefficient of transmissibility = 125*10-4 m2/sec
• Find also the Coefficient of permeability ?
 Solution
R=300m, r=20cm, S = 4m, b =25m, T= 125*10-4m2/sec

• Q = 2.72 T (H- hW)

𝑅
log10( )
𝑟𝑤
• T= Kb
−
𝐓 125∗10 4𝑚2/𝑠𝑒𝑐
• K= = = 5*10-4m/sec (Coefficient of Permeability)
𝐛 25𝑚

• Q = 2.72 T (H- hW)

𝑅
log10( )
𝑟𝑤

• Q = 2.72*(125*10-4m2/sec)*4m
300
log10( )
0.20
• Q = 42.8 l/sec
 Safe Yield
• The safe yield of an aquifer is the maximum withdrawal rate that can be
permanently obtained without depleting the source.

• Basically, the amount of water withdrawn should not exceed the natural
recharge.

• Another limitation is that the groundwater table should not be lowered so

much that polluted water from elsewhere would be drawn into the aquifer.

• The yields of an open well can be determined by Constant Level Test

(Pumping Test) or Recuperation Test.
 • Constant level test (pumping test)
• In this test, the water level is depressed by some head (H). Then the rate
of pumping is adjusted in such a way that the water level remains
constant in the well.
• At this time, the rate of pumping is equal to the rate of yield from the
well (Percolation into the well).

• Rate of seepage into well

= Volume of water pumped out - Volume of water stored in well

Time of pumping
 • Recuperation Test
• In this test, the water level in the well is depressed by an amount less than the
safe working head for the subsoil.
• The pumping is stopped and the water level is allowed to rise or recuperate.
The depth of recuperation in a known time is noted from which the yield of
the well may be calculated as follows:
 Cont.
Let the water level inside the well rise from s1 to s2 (measured below static
water level, swl) in time T. If s is the head at any time t, from Darcy’s law:
• ‘the velocity of flow in a porous medium is proportional to the hydraulic
gradient’, • V= K * i
• Q =V * A
• Q = Ki *A
• if a head s is lost in a length L of seepage path
𝑆
i=
𝐿
𝑆
• Q = K *A *
𝐿
𝐾
• Where the constant C =
𝐿
• Q = C*A*S
 Cont.
• If in a time dt, the water level rises by an amount ds
• Q dt = – A ds
• the – ve sign indicates that the head decreases as the time increases.
Putting Q = CAs
• CAs dt = – A ds
𝑻 𝑺𝟐 𝒅𝑺
• C ‫ 𝟏𝑺׬=𝒕𝒅 𝟎׬‬− 𝑺
𝟐.𝟑𝟎𝟑 𝑺𝟏
• C= log10
𝑻 𝑺𝟐

• Assuming the flow is entirely from the bottom (impervious masonry),

the yield of the well:
• Q=C*A*s
 Cont.
• Q=C*A*s
• Where, Q = safe yield of the well
A = area of cross section of the well
s = safe working depression head
C = specific yield of the soil
• Specific yield of the soil is the discharge per unit area under a unit
depression head and has dimension of T–1 (1/time)

• Specific capacity of the well is its yield (yield) per unit drawdown
Q
=C*A
s

Q
∴ from Q=C*A*s; =C*A
s
Example 3.1
• Calculate the specific capacity of an open well from the following data:
• Initial depression head = 5m
• Final depression head = 2m
• Time of recuperation = 2 hrs
• Diameter of well = 3m

• Calculate also the specific yield of soil & yield of the well under head 3m.
Solution
I. Specific yield of soil
2.303 𝑆1
• C= log10
𝑇 𝑆2

2.303 5𝑚
• C= log10 = 0.458 hr-1
2ℎ𝑟 2𝑚

II.Yield of well under head 3m.

Q=C*A*s
• Area (A) = 𝜋*d2 = 𝜋 ∗32 = 7.07m2
4 4

• Q = 0.458hr-1 * 7.07m2 * 3m = 9.72 m3/hr.

 Example 3.2
• During a recuperating test, the water level was depressed by pumping by 2.5m and is
recuperated by an amount of 1.5m in 60 minutes.

a) Determine the yield from the well, if the diameter of well is 2.5m and the draw
down is 4m.
b) Also determine the diameter of well to yield 10 lit/sec against a draw-down of
2.5m.
Solution
S1 = 2.5m, S2 = 2.5 – 1.5 = 1.0m, T = 60min = 3600sec

a) Q = C*A*s
2.303 𝑆1
• C= log10
𝑇 𝑆2

2.303 2.5𝑚
• C= log10 = 0.92hr -1
b) Q = C*A*s
1ℎ𝑟 1𝑚
Area (A) = 𝜋*d2 = 𝜋 ∗2.52 = 4.9m2 • 10 l/sec = 36 m3/hr
4 4
• 36 m3/hr = 0.92hr -1 * A * 2.5m
• Q = 0.92hr -1 * 4.9m2 * 4m
A = 15.65m2
= 18.03m3/hr or 5 l/sec
4 ∗15.65
• Diameter of the well = 𝜋

• d=4.46m
 Example 3.3
• A well of size 7.70 × 4.65 m and depth 6.15 m in lateritic soil has its normal water level 5.08
m below ground level (bgl). By pumping for 1.5hours, the water level was depressed to 5.93m
bgl and the pumping was stopped. The recuperation rates of the well during 4 hours after the
pumping stopped are given below. The total volume of water pumped during 1.5 hours of
pumping was 32.22 m3.
Time since pumping stopped Water level bgl (m)
(min)
0 5.930
15 5.890
30 5.875
45 5.855
60 5.840
90 5.820
120 5.780
180 5.715
 Cont.
• Determine
(i) Rate of seepage into the well during pumping.
(ii) Specific yield of the soil and specific capacity of the well.
(iii) Yield of the well under a safe working depression head of 0.85 m.

Solution
(i) Rate of seepage into the well during pumping.
Seepage into the well is obtained from pumping data:
• Volume of water pumped out = 32.22 m3
• Volume of water stored in the well (that was pumped out):
= (7.70 × 4.65) (5.93 – 5.08) = 30.5 m3
 Cont.
• Rate of seepage into well
= Volume of water pumped out - Volume of water stored in well

Time of pumping
= 32.22 – 30.5 = 1.15m3/hr.
1.5
(ii) Specific yield of the soil
2.303 𝑆1 2.303 5.93 −5.08
• C= log10 = log10
𝑇 𝑆2 4 5.68 −5.08

= 0.09 hr. -1
 Cont.
• Specific capacity of the well is its yield per unit drawdown
Q
∴ Specific Capacity = C*A
s

= 0.09 * (7.70 * 4.65)

= 3.58 m3 hr-1/m or m2 hr. -1

(iii) Safe yield of well

Q = C*A*s
= 0.09 * (7.70*4.65) * 0.85
= 3.04m3/hr.
 Chapter Four
4.0. Collection and Distribution of Water
River, Lake, Reservoir Intakes

Intake Structures
 Intake Structures
• Intake is devices or structures in a surface water source to draw water from a
source and then discharge in to an intake conduit through which it will flow in
to the water works system.

• The basic function of the intake structure is to help in safely withdrawing water
from the source over predetermined pool levels and then to discharge this
water into the withdrawal conduit (normally called intake conduit), through
which it flows up to water treatment plant.
 The following must be considered in designing and locating intakes:
a. The source of supply, whether impounding reservoirs, lakes, or rivers (including

the possibility of wide fluctuation in water level).

b. The character of the intake surroundings.

 Depth of water

 Character of bottom

 The effect of currents, floods and storms up on the structure and in

scouring the bottom.

c. The location with respect to sources of pollution; and

d. The prevalence of floating materials such as vegetation

 Types of Intakes
i. River Intake

Whenever practicable a river intake should be sited

• where there is adequate flow;

• at a level that allows gravity supply to minimize pumping costs;

• upstream of densely populated and farming areas to reduce silt inflow;

• upstream of cattle watering places, washing places and sewer outlets (to
eliminate pollution of the water);

• upstream of bridges (to reduce velocity/turbulence).

 Cont.

Unprotected River Canal

 Cont.
• The intake structure must always include one or
more baffles or screens to keep out debris and
floating matter such as tree trunks and branches.

• To reduce the drawing in of silt and

suspended matter, the velocity of
flow through the intake should be
low, preferably less than 0.1 m/s.

• The bottom of the intake structure should be at least 1 m above

the riverbed to prevent any boulders or rolling stones from Protected River
Intake
entering.
 ii. Lake Intake
• As described in the previous chapter lake water layers differ in quality with depth.

• Thus, Thermal stratification should be taken into account when choosing the location
and depth of a lake water intake for water supply purposes.

• In deep lakes with water of a low nutrient content (nitrates, phosphates, etc.), the
chemical quality of the water will be much the same throughout the full depth.

• For water supply purposes, water from deeper strata will have the advantage of a
practically constant temperature. Provision should be made to withdraw the water at
some depth below the surface.
• Intake structure at bottom of
shallow lake

• Multi-level intake structure can be

constructed to take advantage of the
aerated water on the surface
 iii. Reservoir intake
• It is desirable to locate about a meter from the surface to get good quality water
advisable to have ports at various heights because, water level fluctuation .
Distribution System and its Layout
 Type of Distribution System
• Depending upon the level of the source of water , topography of the area, and other local
considerations.

• For efficient distribution it is required that water should reach to every consumers with
required flow rate. Therefore, some pressure in pipe lines is necessary. Which force the
water to reach at every place.

• Water can be transported from the source to the treatment plant, if any, and the
distribution system, and eventually reach consumers through one of the following
methods.
i. Gravity system
ii. Pumping system
iii. Dual system
 i. Gravity System
• Action of gravity without any pumping

• Most economical and reliable

• For cities situated at foothills

 ii. Pumping without Storage
• Treated water is directly pumped into the distribution mains without
storing
• High lift pumps operate at variable speeds to match variable water
demand
• Disadvantageous: (power failure) no reserve flow
 iii. Pumping with Storage
Excess water during low demand period gets stored in
the reservoir supplied during high demand periods.

Stored in elevated distribution reservoir

distributed to the consumers by the action of gravity
Treated water is pumped at a
constant rate
even during power failure)
Cont.
Distribution Reservoir
 Storage reservoirs
• Storage reservoir, which store the treated water for supplying water during
emergencies(such as during fires, repairs, etc)and also to help in absorbing the hourly
fluctuations in the normal water demand.

Functions:

• To balance the fluctuating demand from the distribution system.

• Provide a supply during a failure or shutdown of treatment plant, pumps or trunk

main

• To give a suitable pressure for the distribution system and reduce pressure
fluctuations there in.

• To provide a reserve of water to meet fire and other emergency demands

 Positions and Elevation of reservoirs
• The elevation at which it is desirable to position a service reservoir depends up
on both the distance of the reservoir from the distribution area and the elevation
of the highest building to be supplied.

• If the distribution area varies widely in elevation it may be necessary to use two
more service reservoirs at different levels, so that the lower area do not receive
an unduly high pressure.
Pressure zones
• Pressure control valves are some times installed in inlet mains from service reservoirs in
order to reduce the pressure to low laying zones, or to limit increase of pressure at
night to reduce leakage.
 Positions and Elevation of reservoirs
 Types: Classification of based on:

 Position of the tank:

Surface storage, Elevated storage, Underground tanks

 Materials of construction

RCC, Masonry, Masonry-concrete (sandwich), Plastic, Steel tanks

 Shape of the tank:

Circular, Rectangular tanks

Accessories of service reservoir
 Design Consideration
 The three major components of service storage are:
i. Equalizing or operating storage
Addition of these 3
ii. Fire reserve
storages
iii. Break down/Emergency reserve
i. Equalizing or operating capacity

= Max. surplus + Max. deficit

Max. surplus and max. deficit can be calculated using two methods
1. Analytical method 2. Mass- curve method
 Cont.
ii. For Break down Reserve:-
 This is the amount of storage during the break down of pumps.
 From 2 – 3 hrs. pumping capacity is provided against this storage.

iii. For Fire reserve: -

 This is storage required for fighting a fire out break.
 In practice 2 – 5 lit/cap is normally provided for fire storage.
 Shape of Reservoir
Circular Rectangular
most economical shape, giving the Usually proves most economical when
least amount of walling for a given division walls are incorporated
volume and depth

Size m3 Depth of water (m)

does not permit best use of
Up to 3,500 2.5 – 3.5
available land,
3500 – 15,000 3.5 – 5.0
Problems of design will arise if Over 15,000 5.0 – 7.0
it is to be partially buried in
sloping ground

Unsuitable for division in to two

compartments.
 Example 4.2
• A community with a design population of 1600 is to be supplied water at
150liters per capita per day. The demand of water during different periods
is given in the following table:

• Determine the capacity of a service reservoir if pumping is done 24 hours

at constant rate.

Time (hr.) 0-3 3-6 6-9 9 -12 12 - 15 15- 18 18 -21 21- 24

Demand(1000 lit) 20 25 30 50 35 30 25 25
 Solution:
• Rate of Water supply = 150l/c/d

• Total water demand = demand * population

= 150*1600 = 240,000liters

240,000
• Rate of pumping = = 10,000 lit/hr
24

∴ For duration of 3hrs = 3*10,000 = 30,000 lit/hr

i. Analytical Method 1 2 1-2
Time Pumping Demand Cum.Supply Cum.demand Surplus Deficit

0-3 30,000 20,000 30,000 20,000 10,000

3-6 30,000+ 25,000 60,000 45,000 15,000

+
6-9 30,000+ 30,000 + 90,000 75,000 15,000
15,000

9-12 30,000 50,000 120,000 125,000 5,000

12-15 30,000 35,000 150,000 160,000 10,000

15-18 30,000 30,000 180,000 190,000 10,000

10,000

18-21 30,000 25,000 210,000 215,000 5,000

21-24 30,000 25,000 240,000 240,000

 Cont.

I. Balancing storage
• Maximum cumulative surplus = 15,000 liters
• Maximum cumulative deficit = 10,000 liters
Balancing storage = 15000 + 10000 = 25,000lit = 25m3

II. For Break down storage III. For Fire reserve

• Let’s take 3hr pumping rate - Let’s take 5 lit/cap
= 3* 10,000 lit/hr. = 30,000 lit = 5 lit/cap * 1600 = 8,000 lit

 Total capacity of reservoirs = 25,000 + 30,000 + 8,000

= 63000lit = 63m3
 Cont.

• If the reservoir is circular with depth, h = 3.0 m,

V = A*h
63m3 = A*3m
A = 21m2

𝑑2
• A= 𝜋*
4
𝑑2
21m2 = 𝜋*
4
d = 5.17m ≈ 5.2m
 ii. Mass Curve Method
Mass Curve
Cumulative Demand & Supply

300000

250000

200000

150000
Cummulative Pumping
100000 Cummulative Demand

50000

0
0--3 3--6 6--9 9--12 12--15 15--18 18--21 21--24
Time (hrs)
 Example 4.3
• Considering Example 4.2, pumping is done for:

Case 1: Eight hours i.e., 8 hrs to 16 hrs

Case 2: Eight hrs. (from 4 to 8 hrs. and again 16 to 20 hrs.)

• Solution

Total water demand = 240,000 lit/day

240,000 𝑙𝑖𝑡
Rate of pumping = = 30,000 lit/hrs.
8 ℎ𝑟𝑠
Time Pumping Demand Cum.Supply Cum.demand Surplus Deficit

0-3 0 20,000 0 20,000 20,000

3-6 0 25,000 0 45,000 45,000

6-8 0 20,000 0 65,000 65,000

8-9 30,000 10,000 30,000 75,000 45,000

9-12 90,000 50,000 120,000 125,000 5,000

12-15 90,000 35,000 210,000 160,000 50,000

15-16 30,000 10,000 240,000 170,000 70,000

16-18 0 20,000 240,000 190,000 50,000

18-21 0 25,000 240,000 215,000 25,000

21-24 0 25,000 240,000 240,000

 Cont.
CASE: I
I. Balancing storage
• Maximum cumulative surplus = 70,000 liters
• Maximum cumulative deficit = 65,000 liters
Balancing storage = 135,000lit = 135m3

II. For Break down storage III. For Fire reserve

• Let’s take 3hr pumping rate - Let’s take 4 lit/cap
= 3* 10,000 lit/hr. = 30,000 lit = 4 lit/cap * 1600 = 6,400 lit

 Total capacity of reservoirs = 135,000 + 90,000 + 6,400

= 231,400 lit = 231.4m3
 Time Pumping Demand Cum.supply Cum.demand surplus Deficit
20000
0-3 0 20000 0 20000
28333
C 3-4. 0 8333 0 28333
A 45000
4-6 60000 16667 60000 15000
S
E 20000 120000
65000
55000
6-8 60000
:
75000
II 8-9 0 10000 120000 45000
125000
9 -12 0 50000 120000 5000
160000
12 - 15 0 35000 120000 40000
170000
15 -16 0 10000 120000 50000
190000
16-18 60000 20000 180000 10000
206667
18 - 20 60000 16667 240000 33333
215000
20 - 21 0 8333 240000 25000
240000
21 - 24 0 25000 240000 0
Exercise
• A water supply system is proposed to be designed for a small town
which has a maximum daily demand of 515 m3/d. Estimate storage
requirement if pumping is done for 12 hr. only (from 4 to 16). Use the
following demand variation data.

Time (hr.) 0-4 4-8 8 - 12 12 - 16 16 - 20 20 - 24

Demand as % of 6.7 9.2 20.8 28.3 25 10

total daily demand
LAYOUT OF DISTRIBUTION SYSTEM
 Water Transmission Lines
• Gravity main • Pumping main
 Layout of Distribution Pipes
• Layout of distribution pipes generally follows the road pattern:

• Four types of pipe network layouts

i) Dead End or Tree System

ii) Grid Iron System

iii) Circular or ring system

iv) Radial System

 i) Dead end or Tree System
ADVANTAGE
• Discharge and pressure at any point Solved
easily

• Lesser number of shut-off valves

• Shorter pipe lengths smaller diameter size

• Cheap and simple for laying and expanded easily

DISADVANTAGE

• Dead ends: prevent circulation of water and

contamination.

• Problematic if a pipe is damaged (downstream

will be deprived of supply)

• The water available for fire fighting will be

limited in quantity
 ii) Grid Iron System
ADVANTAGE
• In the case of repairs a very small portion of
distribution are a will be affected
•
• Every point receives supply from two directions
and with higher pressure
• Additional water from the other branches are
available for fire fighting
• There is free circulation of water and hence it is
not liable for pollution due to stagnation.
DISADVANTAGE
• More length of pipes and number of valves are
needed and hence there is increased cost of
construction
• Calculation of sizes of pipes and working out
pressures at various points in the distribution
system is laborious , complicated and difficult.
 iii) Ring System
•
• In this system, the main water line is
divided in to two parts; in two direction
left and right.

• It is suitable for well planned town or city

where the locality can be divided in to
square or circular blocks and the main
water line can be laid around the sides of
the squire or around the circle.
 iv) Radial System
• The city is divided in to various circular or square
zones and distribution reservoirs are placed at the
center of each zone.

• The distributor lines are laid radially from reservoir

towards the periphery of each zone.

• It is sailable when the town or city can oriented

with radial roads and streets

• The water from the main reservoir is allowed to

flow through the main pipe and sub-main pipe and
get collected at distribution reservoir of each zone.

• The water is supplied to consumers through the

distributor pipe lines.
Pipe Used in Distribution System
 Pipelines and appurtenances
• The selection of pipe materials is based on:
– carrying capacity

– strength

– ease of transportation and handling

– availability

– quality of water

– cost (initial and maintenance)

 Pipe Materials
 Pipes convey raw water from the source to the treatment plants in the
distribution system.

 Water is under pressure always and hence the pipe material and the
fixture should withstand stresses due to the internal pressure, vacuum
pressure, when the pipes are empty, water hammer when the values are
closed and temperature stresses.
 Cont.
 REQUIREMENTS OF PIPE MATERIAL

 It should be capable of with standing internal and external pressures

 It should have facility of easy joints

 It should be available in all sizes, transport and erection should be easy.

 It should be durable

 It should not react with water to alter its quality

 Cost of pipes should be less

 Frictional head loss should be minimum

 The damaged units should be replaced easily.

 Cast iron pipes
Advantages
 cost is moderate
 easily jointed
 Highly resistant to corrosion
 The pipes are strong and durable
 Withstand high internal pressure
Disadvantage
 Very heavy and difficult to transport
 The breakage of this pipe is large
 Carrying capacity decreases with increase in life
 Steel Pipes
Advantages
 strong, very light weight and can withstand
higher pressure than cast iron pipes.
 cheap, easy to construct and can be easily
transported
Disadvantage
 Cannot withstand external loads, affected by
corrosion and are costly to maintain
 Cement Concrete Pipes
Advantages
 The inside surfaces of the pipes can be made smooth
 The maintenance cost is very low
 Under normal conditions the pipes are durable
 The pipes can be cast in place(in site)
 Due to high weight (heaviness) the pipes can resist force of buoyancy when placed under
water even when they are empty
 Pipes can resist normal traffic loads when placed below roads
 There is no danger of rusting and incrustation
Disadvantage
 The pipes are difficult to transport
 If no reinforcement is provided it cannot resist high pressure
 The pipes are likely to crack during transport and handling
 The repair of these pipes are difficult
 These pipes are affected by acids, alkaline, and salty waters
 These pipes are likely to cause leakage due to porosity
 Plastic Pipes
Advantages
• The pipes are cheap
• The pipes are flexible and possess low hydraulic resistance (less
friction)
• They are free from corrosion
• The pipes are light in weight and it is easy to bend, join and install them
• The pipes up to certain sizes are available in coils and therefore it
becomes easy to transport
Disadvantage
• The coefficient of expansion for plastics is high, the pipes are less
resistant to heat
• Some types of plastics may impart taste to the water
Asbestos cement pipes
Advantages
• The inside surface of pipe is smooth
• The joining of pipes is very good and flexible
• The pipes are ant-corrosive and cheap in cost
• Light in weight to handle and transport
Disadvantage
• The pipes are brittle
• The pipes are not durable
• The pipes are not laid in exposed places
• The pipes can be used only for very low pressure
Pipe Geometry, Branched Networks, Looped Networks

HYDRAULIC ANALYSIS OF DISTRIBUTION

MAIN
 Discharge

Hydraulic
Analysis Head
Include Loss
determination
of:

Pressure
Head
 Design of Distribution Systems
• Design flow: Max (Peak hour demand or maximum day demand + Fire demand)

• Minimum main sizes: generally:150mm (6 in); high value districts: 200mm (8 in); major
streets: 305mm (12 in); domestic flows only: 100mm (4 in); small communities: 50-75
mm

• Velocity: typical values – minimum = 0.6-1 m/s; maximum = 2m/s

• Pressure: typical minimum value is 140 kPa (14 m) and maximum not to exceed 410
kPa (42 m). But mainly depends on pressure ratings of the pipes and appurtenances
used and regulatory requirements

• It is advantageous to divide the supply network into pressure zones

Pipe Network Geometry
Cont.
 Head loss (Energy Loss) in Pipes
The energy loss E is generated by:
• Friction between the water and the pipe wall,
• Turbulence caused by obstructions of the flow.

𝑓𝐿𝑣 2
i. Darcy-Weisbach formula ℎ𝑓 =
2𝑔𝐷

2.63 0.54 ℎ𝑓
ii. Hazen-Williams formula 𝑄 = 0.278𝐶𝐷 𝑆 ,𝑆 =
𝐿

𝐴𝑅2/3 𝑆 1/2 ℎ𝑓
iii. Manning’s Formula Q= ,𝑆 =
𝑛 𝐿
 Table - Values of C for the Hazen-Williams formula
Pipe Material C
Asbestos Cement 140
Cast Iron
 Cement lined 130 – 150
 New, unlined 130
 5years-old, unlined 120
 20 years old, unlined 100
Concrete 130
Copper 130 - 140
Plastic 140 -150
New welded Steel 120
New riveted Steel 100
• Given any two of the parameters (Q, D, hf or V), the remaining can be determined from the
intersections along a straight line drawn across the nomograph.
 Pipe Design Procedures
• Assign the required demand at each node

• Estimate the discharge flowing through the pipes

• Assume possible pipe diameters

• Calculate the head loss through the pipes

• Find the residual pressure at the end of the pipe.

• Compare this terminal pressure with the desired minimum and maximum pressures.

• If the required condition is not satisfied, then repeat steps (ii) through (vi) until the
required conditions are met.
 Example
• Determine velocity, head loss, and residual Pressure at the demand centers.
Q=0.75m3/m
D El. 1167m
El. 1250m
L=50m D= 100 mm
El. 1207m
R Q=2m3/m L=150m A L=200m L=250m
E El. 1185m
D= 150mm D= 125 mm C D= 75mm
El. 1177m Q=0.25m3/m

L=75m D= 75 mm

B El. 1192m
Q=1m3/m
 Solution
i. Velocity calculation
𝑄 Q=0.75m3/m
• V= 𝜋𝐷2 D
4
V=1.6 m/S
D= 100 mm
V=1.9m/S Q=0.75m3/m
V=1.4 m/S
Q=2m3/m Q=1m3/m Q=0.25m3/m
R A D= 75mm
E
D= 150mm D= 125 mm C
Q=0.25m3/m
V=0.9m/S
75 mm Q=1m3/m
V=2.1m/S
B

𝑄 2m3/m
• V= 𝜋𝐷2
= π∗0.152
= 1.9m/s
4 4
 Cont.
ii. Head loss calculation
• HL= 10.7 * Q1.85 * L Q=0.75m3/m
D hl=2.36m
C1.85 * D4.87 100mm,50m
hl=5.43m
Q=0.75m3/m
125mm,200m
Q=2m3/m Q=1m3/m Q=0.25m3/m
R A
E
150mm,150m C
Q=0.25m3/m
hl=6.04m
Q=1m3/m 75mm,250m
75mm,100m hl=6.27m
hl=6.03m
B
Q=1m3/m
• HL= 10.7 * 0.01671.85 * 200m
1001.85 * 0.1254.87
 Cont.
iii. Residual Pressure calculation
hres=69.17m
Hres = Elv.1 – Elv.2 - hl El. 1167m
D

El. 1250m El. 1207m hl=2.36m

hres= 36.96m
R hl=6.04m A hl=5.43m
E
C hl=6.27m El. 1185m
El. 1177m
hres=47.26m
hl=6.03m hres=61.53m

hres=45.93m
B
El. 1192m
Hres, C = Elv.1 – Elv.2 – hl
Hres, A = Elv.1 – Elv.2 – hl Hres, B = Elv.1 – Elv.2 – hl = 1250 – 1177 – 6.04 – 5.43
= 1250 – 1207 – 6.04 = 1250 – 1192 – 6.04 – 6.03 = 61.53m
= 36.96m = 45.93m
 Hydraulic Analysis of Looped Network
 Hardy Cross Method
• ASSUMPTIONS:
• The outflows from the system are assumed to occur at the nodes.

• This assumption results in uniform flow in the pipelines.

• At each junction, the total inflow must be equal to total outflow.

• Head balance criterion: algebraic sum of the head losses around any closed -
loop is zero.
 Cont.
• Flows in a clockwise direction are considered to be positive (+) and flows in
CCW direction are considered to be negative(-)

• Head losses from CW flows are considered to be positive (+), Head losses from
CCW flows are negative (-).

• For a given pipe system, with known junction outflows, the Hardy-Cross method
is an iterative procedure based on initially estimated flows in pipes.

• Estimated pipe flows are corrected with iteration until head losses in the
clockwise direction and in the counter clockwise direction are equal within each
loop.
 Procedures
1. Number all the nodes and pipe links.

2.Adopt a sign convention of a pipe discharge

3.Assume friction factors fi ¼ 0.02 in all pipe links and compute corresponding Ki.

4. Assume loop pipe flow sign convention to apply loop discharge corrections; generally,
clockwise flows positive and counterclockwise flows negative are considered.

5. Calculate ∆Qk for the existing pipe flows and apply pipe corrections algebraically.

6.Apply the similar procedure in all the loops of a pipe network

• Repeat steps 5 and 6 until the discharge corrections in all the loops are relatively very
small.
 Example
• A single looped network as shown in Fig has to be analyzed by the Hardy Cross
method for given inflow and outflow discharges. The pipe diameters D and lengths L
are shown in the figure. Use Darcy–Weisbach head loss–discharge relationship
assuming a constant friction factor f ¼ 0.02.

L1 = 300m
D1 = 150mm
0.6m3/s
1

L4 = 200m L2 = 200m
D4 = 150mm 4 2 D2 = 150mm

3
0.6m3/s
L3 = 300m
D3 = 150mm
•

0.1m3/s
0.6m3/s
1

0.5m3/s 4 2 0.1m3/s

3
0.6m3/s
0.5m3/s
 ITERATION 1

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Loop I Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

1 150 300 0.10 6528.93 65.29 1305.79 0.30

2 150 200 0.10 4352.62 43.53 870.52 0.30

3 150 300 -0.50 6528.93 - 1632.23 6528.93 -0.30

4 150 200 -0.50 4352.62 - 1088.15 4352.62 -0.30

Total - 2611.57 13,057.85

8𝑓𝐿 8∗0.02∗300
• Ki = 2 5 =
𝜋 𝑔𝐷 𝜋2 ∗9.81∗ 0.155

−2611.57
=- = 0.20m3/s
13,057.85
 ITERATION 2

D, L, Q, Ki Ki Q[Q], 2K[Q]
Loop I Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

1 150 300 0.30 6528.93 587.60 3917.36 0.30

2 150 200 0.30 4352.62 391.74 2611.57 0.30

3 150 300 -0.30 6528.93 -587.60 3917.36 -0.30

4 150 200 -0.30 4352.62 -391.74 2611.57 -0.30

8𝑓𝐿 Total 0.00 13,057.85

• Ki = 2 5
𝜋 𝑔𝐷
Q1 = 0.3 m3/s
• As the discharge correction
Q2 = 0.3 m3/s
∆QK= 0, the final discharges are:
0 Q3 = 0.3 m3/s
=- = 0m3/s
13,057.85 Q4 = 0.3 m3/s.
 Example
• The pipe network of two loops as shown in Fig. has to be analyzed by the Hardy Cross method
for pipe flows for given pipe lengths L and pipe diameters D. The nodal inflow at node 1 and nodal
outflow at node 3 are shown in the figure. Assume a constant friction factor f ¼ 0.02.

L1 = 300m
D1 = 150mm
0.6m3/s
1

L4 = 200m L2 = 300m
D4 = 150mm 5 2 D2 = 150mm
4

3
0.6m3/s
L3 = 300m
D3 = 150mm
 Solution

0.1m3/s
0.6m3/s
1

[2]
0.4m3/s 5
2 0.1m3/s
4
[1]
3
0.6m3/s
0.4m3/s
LOOP 1: ITERATION 1

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

3 100 300 -0.40 49,576.12 -7932.18 39660.89 -0.25

4 150 200 -0.40 4352.36 -696.38 3481.89 -0.25

5 100 360 0.10 59491.34 594 11898.27 0.25

Total -8033.64 55041.05

−8033.64
=- = 0.15m3/s
55041.05
LOOP 2: ITERATION 1

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

1 150 300 0.10 6528.54 65.29 1305.71 0.19

2 100 200 0.10 33050.74 330.51 6610.15 0.19

5 100 360 -0.25 59491.34 - 3598.93 29264.66 -0.16

Total -3203.14 37180.52

−3203.14
=- = 0.09 m3/s
37180.52
LOOP 1: ITERATION 2

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

3 100 300 -0.25 49,576.12 -3098.51 24788.06 -0.21

4 150 200 -0.25 4352.36 -272.02 2176.18 -0.21

5 100 360 0.16 59491.34 1522.98 19037.23 0.20

Total -1847.55 46001.47

−1847.55
=- = 0.04 m3/s
46001.47
LOOP I1: ITERATION 2

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

1 150 300 0.19 6528.54 226.23 2430.59 0.21

2 100 200 0.19 33050.74 1145.28 12304.85 0.21

5 100 360 -0.20 59491.34 - 2383.53 23815.92 -0.17

Total -1012.02 38551.36

−1012.02
=- = 0.03 m3/s
38551.36
LOOP 1: ITERATION 3

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

3 100 300 -0.210 49,576.12 -2182.92 20805.82 -0.197

4 150 200 -0.210 4352.36 -191.64 1826.57 -0.197

5 100 360 0.174 59491.34 1799.33 20692.47 0.187

Total -575.23 43324.86

−575.23
=- = 0.013 m3/s
43324.86
LOOP I1: ITERATION 3

D, L, Q, Ki Ki Q[Q], 2[K[Q]]
Pipe Q + ∆Qk
(mm) (m) (m3/s) (s2/m5) (m) s/m2

1 150 300 0.212 6528.54 294.53 2773.35 0.220

2 100 200 0.212 33050.74 1491.07 14040.10 0.220

5 100 360 -0.187 59491.34 -2084.55 22272.21 -0.180

Total -298.95 39085.67

−298.95
=- = 0.008 m3/s
39085.67