water supply engineering

1. Module 5_Pipe Appurtenances
By,
Adarsh S

2. Fire Hydrants
• A hydrant is a outlet provided in a water
distribution main or sub-main for tapping
water, mainly during fires.
• Sometimes it is used for withdrawing water for
filling municipal water tankers.
• During fire breakout, a nearby hydrant is
connected to the fire hose, and water obtained
from hydrant is used for extinguishing the fire.

3. Post Fire Hydrant

4. Flush Fire Hydrant

5. PIPE APPURTENANCES
The various devices fixed along the water distribution
system are known as pipe appurtenances.
NECESSITY
• To control the rate of flow of water
• To release or admit air into pipeline according to the
situation
• To prevent or detect leakages
• To meet the demand during emergency and
• Ultimately to improve the efficiency of the distribution

6. TYPES OF VALVES
1. Sluice valve
• These are also known as gate-valves or stop valves
• These valves are cheaper, offers less resistance to the flow of
water than other valves
• They are provided in straight pipeline at 150-200m intervals

7. 2. Check valve or Reflux valve
• These valves are also known as non-return valves.
• A reflux valve is an automatic device which allows water to go
in one direction only.
• When the water moves in the direction of arrow, the valve
swings or rotates around the pivot and it is kept in open
position due to the pressure of water.

8. 3. Air valves
• Air inlet valves
• Air relief valves
 Sometimes air is accumulated at the summit of pipelines and
blocks the flow of water due to air lock.
 In such cases the accumulated air has to be removed from the
pipe lines.

9. 4. Darin valves or blow off valves
• These are also called wash out valves they are provided at all
dead ends and depression of pipelines to drain out the
wastewater
5. Scour valves
• These are similar to blow off valves
• They are located at the depressions and dead ends to remove
the accumulated silt and sand

10. Requirements of a Good Distribution System
• It should be capable of supplying water at all intended
places with in the city with a reasonably sufficient
pressure head.
• It should be capable of supplying the requisite
amount of water for fire fighting during such needs.
• It should be cheap with least capital construction cost.
• It should be simple and easy to operate and repair.
• It should be against future pollution of water.
• It should be water-tight to keep the losses due to
leakage to minimum.

11. Layouts of Distribution Networks
• The distribution pipes are generally laid below the street
pavements, and as such, this layout will generally follow the
layouts of the roads
• There are, in general, four different types of networks, any one
of which either singly or in combinations can be used for a
particular place, depending upon the local conditions and
orientation of roads
 Dead end system
 Grid-iron system
 Ring system, and
 Radial system

12. Dead End System
• Dead end system is also sometimes called Tree System.
• There is one main supply pipe, from which originates
generally at right angles, a number of sub-main pipes.
• Each sub main, then divides into several branch pipes,
called laterals.
• This type of layout may be adopted for older towns which
have developed in haphazard manner, without any properly
planned roads.
• The water supply mains have to be taken along the main
roads, and branches taken off wherever needed, thus
resulting in the formation of a number of dead ends as
shown.
• This system is therefore, suitable for localities which
expand irregularly, and where pipes have to be laid at
random without any planning of future roads.

13. Dead End System

14. Advantages
• The distribution network can be solved easily, it is
possible to easily and accurately calculate the
discharges and pressures at different points in the
system.
• Lesser number of cut-off valves.
• Shorter pipe lengths and lying of pipes is easy.
• It is cheap, simple and can be extended easily.

15. Disadvantages
• Since, in this method, water can reach a particular
point only through one route, any damage or
repair in any pipe line will completely stop the
water supply in the area being fed by that pipe.
• There are numerous dead ends in this system,
which prevent the free circulation of water. This
stagnation of water may lead to degradation of its
quality.
• During fire fighting, the supply cannot be
increased, since the discharge reaches a point
from only one point.

16. Grid-iron System
• In this system, which is also known as ‘Interlaced
system or Reticulated system’.
• The mains, sub-mains and branches are all
interconnected with each other.
• In fact, in a well-planned city or a town, the roads
are generally developed in a grid-iron pattern, and
the pipe lines in such places can follow them
easily.
• Hence, this system is more suitable for well-
planned towns and cities. By closing the dead
ends by a loop, the grid-iron pattern is obtained.

17. Grid-iron System

18. Advantages
• Since the water reaches at different places through
more than one route, the discharge to the carried by
each pipe, the friction loss, and the size of the pipe,
therefore, is reduced.
• In case, of repairs, very small area will be devoid of
complete supply, as atleast, some supply will be
reaching at the point from some other route.
• Because of the different interconnections, the dead ends
are completely eliminated, and therefore, water remains
in continuous circulation and hence, not liable to
pollution due to stagnation.
• During fire, more water can be diverted towards the
affected point from various directions.

19. Disadvantages
• This system requires more length of pipe lines,
and a large number of sluice valves (i.e. cut-off
valves).
• It is costlier to construct.
• The design is difficult and costlier.

20. Ring System
• This system is also sometimes called ‘circular system’
• In this system, a closed, ring, either circular or
rectangular, of the main pipes, is formed around the area
to be served.
• The distribution area is divided into rectangular or
circular blocks, and the main water pipes are laid as the
periphery of these locks.
• The sub mains may be placed as shown.
• The rind system is very suitable for towns and cities
having well planned roads.
• Sometimes, this system is used as a “lopped feeder placed
centrally around a high demand area” along with the grid-
iron system.
• The advantages and disadvantages are same as grid-iron
system.

22. Radial System
• If a city or a town is having a system of roads
emerging from different centres.
• The pipe lines can be best laid in a radial method
by placing the distribution reservoirs at these
centres.
• In this system, water is, therefore, taken from the
water mains, and pumped into the distribution
reservoirs placed at different centres.
• The water is then supplied through radially laid
distribution pipes.
• This method ensures high pressures and efficient
water distribution. The calculation for design of
sizes is also simple.

24. METHODS OF DISTRIBUTION
• Gravity system
• Pumping system
• Combined gravity and pumping system
SYSTEM OF WATER SUPPLY
• Continuous system
• Intermittent system

25. Distribution Reservoirs
• Distribution reservoirs also called Service
Reservoirs or the storage reservoirs.
• Store treated water for supplying water during
emergencies and also to help in absorbing the
hourly fluctuations in water demand.

27. Functions of Service Reservoirs
• They absorb the hourly variations in demand and
allow the water treatment units and pumps to
operate at a constant rate.
• They help in maintain constant pressure in the
distribution mains.
• The water stored in these reservoirs can be
supplied during emergencies, such as breakdown
of pumps, heavy fire demand.
• They lead to an overall economy by reducing the
size of pumps, pipe lines and treatment units.

28. Types of Distribution Reservoirs
• The storage tanks may be classified according
to
Position Material Shape
Surface storage tanks R.C.C Circular
Elevated Storage Concrete Rectangular
Stand Pipes Masonry/ Steel Intze Tanks

29. Surface Storage Tanks

30. Surface Storage Tanks

31. Elevated Reservoirs

32. Elevated Reservoirs

33. Stand Pipes

34. Stand Pipes

35. Selection of Pump Horse Power
Point to be observed in selecting a pump
1. Capacity and efficiency - The pump should have the
capacity required and optimum efficiency.
2. Lift - Suction head from the water level to the pump
level
3. Head – It is also called delivery head. Generally the
total head (suction and delivery head) should meet all
possible situations with respect to the head.
4. Reliability – A reputed manufacture or similar make
pump already in use may give the failure rate and
types of troubles.

36. 5. Initial cost - The cost of the pump and its installation
cost should be minimum.
6. Power – Power requirements should be less for
operation
7. Maintenance – Maintenance cost should be
minimum. Availability of spares and cost of spares
are to be ascertained.
Total head
The total head against which pump is to work
H=hs+ hd+ hf+ hv
Where,
hs = Suction head
hd = Delivery head
hf = Frictional loss
hv = Velocity head at discharge

37. • The velocity at the discharge end which is quite small
and hence the velocity head is sometimes neglected in
computing the power requirements of pumps and also
total static head.
• The frictional loss in the pipe line is calculated from
darcy weishbach formula, hf =
Where,
f = Co-efficient of friction
l = Length of the pipe line
v = Velocity
i. Velocity head, hv =
ii. Water Horse Power (metric unit) WHP =
Where,
W = Unit weight of water or density in kg/m3
Q = Discharge to be pumped in m3/sec
H = Total Head

38. iii. Break Horse Power BHP =
Where,
• ηp = pump efficiency
iv. Electric Horse Power EHP =
Where,
• ηo = motor efficiency
v. Since 1 metric HP = 736 watts = 0.736 Kw
• No of Kw = 0.736 * EHP
vi. Economical diameter of raising main is determined by
empirical formula
d = a
Where,
• d = Diameter of pipe
• Q = Discharge in m3/sec
• a = constant varies from 0.97-1.23

39. 1. A centrifugal pump driven by electric motor lifts
water to a total height of 50 m from reservoir to
discharge head, pump efficiency is 75%, motor
efficiency is 90%. The lift through 500 m length of
10 cm diameter pipe, pumping rate is 1700 L/min,
co-efficient of friction is 0.025. Power cost is 4
Rs/KWhr. What is the cost of power for pumping 5
Million litres of water.
Sol: Data, hs = 50 m
Ղp = 75%
Ղm = 90%
D = 10 cm = 0.1 m
Q = 1700 L/min
F = 0.025

40. Q = 1700/1000*60
= 0.028 m3/sec
Q = AV
0.028= 3.14*(0.1)2/4 * V
V= 3.56 m /sec
hf = = 0.025*500*(3.56)2/2*9.81*0.1 = 80.74m
H = hs+hf = 50+80.74 = 130.74 m
WHP = = 1000*0.028*130.74/75 = 48.80 HP
BHP = = 1000*0.028*130.74/ 75*0.75 = 65.07 HP

41. EHP = = 1000*0.028*130.74/75*0.75*.90= 72.31
HP
No of Kw = 0.736 * EHP = 0.736* 72.31 = 53.22 KW
Time during which 5 million liter of water should be
lifted
1hr = 1700 L/min
5 million liter = ?
5*10^6/1700*60 = 49 hrs
Total electric power required = no. of KW * total hrs
= 53.22*49 = 2607.78 KWHr
Total cost = 2607.78*4
= Rs.10431/-

42. 2. Water is to be lifted from a tube well to a over head
tank, find the efficiency of the pumping unit for the
following data: discharge in tube well 60 L/sec, RL
of ground 201.5 m, depression head during pumping
4m, length of rise in main 100m, depth of water in
tank 3.5m, co-efficient of friction 0.04m, velocity of
water in raising main 2m/sec, RL of water in the
well 180 m, overall efficiency 70%, RL of bottom of
over head tank 220.5m.
hs = 201.5-180= 21.5 m
hd = 220.5+3.5- 201.5 = 22.5 m

43. Q = AV
0.06= 3.14*d2/4*2
D= 0.195 m
hf = = 0.04*100*2/2*9.81*0.195 = 4.18 m
H = hs+hf+hd+ depression head
= 21.5+4.18+22.5+4= 52.18 m
EHP =
= 1000*0.06*52.18/75*0.7
=59.63 HP

44. 3. For a water supply of a town water is pumped from
the river 2 km away from the reservoir, the
maximum difference in level of water in reservoir is
25m. The population of town is 80000 and per
capita water demand is 125 L/day. If the pump has
to operate for 8 hrs the efficiency of pump is 80%.
Determine the BHP of pump. Assume f= 0.03 and
velocity of flow = 2 m/s and max daily demand as
1.5 times avg demand.
Sol: Data, f= 0.03
V = 2 m/sec
Ղp = 80%
Qavg = 80000*125/1000*8 = 0.3472 m3/sec

45. Qmax = 1.5*Qavg = 1.5*0.3472 = 0.5208 m3/sec
Q= AV
0.52= 3.14d2/4*2
D = 0.57 m
hf = = 0.03*2000*4/2*9.81*0.57
= 21.46 m
H = 25+ 21.46 = 46.46 m
BHP = = 1000*0.52*46.46/75*0.8
= 402.25 HP

46. 4. Water is to be pumped from a jack well to a service
reservoir situated 1.8 km away, work out the
capacity of the pump and motor from the following
data: water is to be raised/day 10000 m3/day, total
hrs of pumping 16, RL of water in jack well 322 m,
RL of high flood level of reservoir 342m, depression
in water level during pumping 2m, friction in raising
main 100 cm/km length, other losses of head 1.5m,
combined efficiency of pump and motor 65%.
hs = 342-322= 20 m
hf = 1km= 100 cm
1.8km= 180cm= 1.8 m
H= 20+1.8+1.5+2
= 25.3m

47. EHP = 10000/3600*16 = 0.17 m3 /sec
= 1000*0.17*25.3/75*0.65
= 89.78 HP
no. of KW= 0.736 * EHP
= 0.736*89.78
= 66.078kw

48. 5. Water is to be raised to an over head storage tank.
Find the EHP from the following data: Discharge 40
L/s, velocity of raising tank = 200 cm/s, length of
raising main 250 m, RL of ground 100m, RL of
water in well 86m, depression head 4m, RL of water
level in OHT 112m, f= 0.04, combined efficiency of
pump and motor 75%.
hs = 100-86 = 14 m
hd= 112-100 = 12m
hf=
= 0.04*250*4/2*9.81*0.159 = 12.82 m
H= 14+12+12.82+4
= 42.82 m

49. Q= AV
0.04= 3.14d2/4*2
D=0.159 m
EHP =
= 1000*0.04*42.82/75*0.75
=30.45 HP

50. 6. A city has a 150000 population, water is to be
supplied at a rate of 160 liter/head/day. If the static
lift of the pump 40 calculate the BHP. The raising
main is 300 m long and diameter is 50 cm. Assume
that motor efficiency is 85%, pump efficiency is
50%, friction factor is 0.04 and peak hour demand is
1.5 times avg demand.
Sol: Data: L= 300m
D= 50 cm = 0.5 m
Ղm=85%
Ղp= 50%
Qavg = 150000*160/3600*24*1000
= 0.27 m3/sec

51. Qmax= 1.5*0.27
= 0.405 m3/ sec
Q= AV
0.405= 3.14*(0.5)2/4 *V
V= 2.14 m /sec
hf= 0.04*300*(2.14)2/2*9.81*0.5
= 5.60 m
H= 50+5.6 = 55.6 m
EHP =
1000*0.40*55.6/75*0.5 = 593 HP

52. 7. A city has a 100000 population, water is to be
supplied at a rate of 150 liter/head/day from 2000m
away. A difference in elevation between the lowest
water level in the tank and the reservoir is 36m. If
the demand has to be supplied in 8 hrs. Determine
the size of the main BHP of the pump required.
Assume max demand as 1.5 times avg demand. Take
f= 0.075, V=2.4 m/sec, Ղp = 80%
Sol: Data: L= 2000m
D= ?
Ղp= 80%
Qavg = 100000*150/3600*8*1000
= 0.52 m3/sec

53. Qmax= 1.5*0.52
= 0.78 m3/ sec
Q= AV
0.78= 3.14*(d)2/4 *2.4
d= 0.64m
hf= 0.075*2000*(2.4)2/2*9.81*0.64
= 68.80m
H= 36+68.80 = 104.80 m
BHP =
1000*0.78*104.80/75*0.8 = 1362.4 HP

54. 8. For a clean water reservoir, water is to be pumped to
an elevated reservoir of elevation 75 m at a constant
rate of 10*10^5 L/hr. The main water level in the
reservoir is 30m with the depression head of 3m, the
distance is 1800m. Give the economical diameter
and WHP of the pump take f=0.04
Sol: Data: L= 1800m
D= ?
F= 0.04
Q = 1000000/3600*1000
= 0.27 m3/sec
D= a
D = 1 sqrt(0.27) = 0.52 m

55. Q= AV
0.27= 3.14*(0.52)2/4 *V
V= 1.27m/sec
hf= 0.04*1800*(1.27)2/2*9.81*0.52
= 11.38m
H=75-30+3+11.38= 59.38m
WHP =
1000*0.27*59.38/75= 213.76HP

56. 8. From a water reservoir having a depth of 2.4m and
max water level as 32m. Water is to be pumped to an
elevated reservoir of elevation 80m at a constant rate
of 8*10^5 L/hr. The distance between the reservoir
is 1.2 km. Design the economical diameter of the
raising main and WHP of the pump neglect minor
losses and take f=0.04
Sol: Data: L= 1.2 km = 1200m
D= ?
F= 0.04
Q = 800000/3600*1000
= 0.22 m3/sec
D= a
D = 1 sqrt(0.22) = 0.46 m

57. Q= AV
0.22= 3.14*(0.46)2/4 *V
V= 1.32 m/sec
hf= 0.04*1200*(1.32)2/2*9.81*0.46
= 9.26 m
H=80-32+2.4+9.26= 59.66m
WHP =
1000*0.22*59.66/75= 175.00HP