1. Various types of conduits
Depending upon the conditions and characteristics of flow, the conduits may be divided into
Gravity conduits
Pressure conduits
Gravity conduits are those in which the water flows under the mere action of gravity. In such a conduit, the gradient line will coincide with the water surface and will be parallel to the bed of the conduit.
In such a flow the water is all along at atmospheric pressure and there is no pressure term in Bernoulli’s equation.

2. Gravity conduits can be in the form of canals, flumes or aqueducts.
Pressure conduits
In pressure conduits, which are closed conduits and as such no air can enter into them, the water flows under pressure above the atmospheric pressure. The hydraulic gradient line for such a conduit can be obtained by joining the water surface elevations in the piezometers installed in the conduit at various places.

3. The Hydraulic gradient should be such as to generate velocities which are neither so small as to require large size diameter pipe nor so large as to cause excessive loss of pressure head. The velocities should also be non-silting and non scouring. The flow velocities are normally taken as 0.9m/sec to 1.5m/sec though velocities upto 3m/sec to 6m/sec can be resisted by the commonly available pipes or pipe materials

4. Flow in Gravity Conduit

5. Flow in Pressure Conduit

6. Head Loss caused by Pipe Friction
DARCY WEISBACH FORMULA
MANNING’S FORMULA
HAZEN-WILLIAM’S FORMULA
MODIFIED HAZEN-WILLIAM’S FORMULA

7. DARCY WEISBACH FORMULA
HL=Head Loss in metres
L = Length of pipe in metres
d = Diameter of the pipe in metres
V = Mean velocity of flow through the pipe in m/sec
g = Acceleration due to gravity
f ‘ = Dimensionless friction factor generally varying between 0.02 (for new smooth pipes) to (old rough pipes)and depends upon Reynold number (i.e. Re=Vd/ν)

8. MANNING’S FORMULA HL=Head Loss in metres
n = Manning’s rugosity coefficient
L = Length of pipe in metres
V = Mean velocity of flow through the pipe in m/sec
R = Hydraulic Mean depth of pipe in metres R=

9. HAZEN-WILLIAM’S FORMULA
V=0.85CH.R0.63.S0.54
CH= Coefficient of hydraulic capacity as given by the table
S = Slope of the energy line
V = Velocity of flow through the pipe in m/sec
R = Hydraulic Mean depth of pipe in metres (for circular pipes flowing full) R=

10. HAZEN-WILLIAM’S Formula

11. MODIFIED HAZEN-WILLIAM’S FORMULA
V= CR.R S0.5525
CR= Coefficient of hydraulic capacity as given by the table
S = Friction Slope=HL/L
V = Velocity of flow through the pipe in m/sec
R = Hydraulic radius or Hydraulic Mean depth of pipe in metres (for circular pipes flowing full) R=

12. Recommended CR values in Modified Hazen-William’s Formula

13. Estimating size of supply conduits
Problem No.1
Estimate roughly the sizes of supply conduits leading to an adequate service reservoir serving,
(a) a relatively small town of 25,000 population
(b) a relatively large city with industrial establishments having a population of 5 lakh people.
Also find the hydraulic gradients of which the pipelines are proposed to be laid. Assume any suitable data according to Indian conditions, where required.

14. Assume average daily water consumptions as follows
For town of 25,000 population = 120 litres/capita/day
For city with industrial establishments having a population of 5 lakhs = 270 litres/capita/day
Average quantity of water required= Avg. daily water consumption x population
Average quantity of water required
For town = 120 litres/capita/day x 25,000
= 3,000,000 litres/day
= 3 Million litres/day = 3 MLD
For city = 270 x 5,00,000 litres/day
= 135,000,000 litres/day
= 135 Million litres/day = 135 MLD

15. Assume Maximum daily demand as 1
Assume Maximum daily demand as 1.8 times the average, the maximum quantity of water required
For town = 1.8 x 3MLD
= 5.4 x 106/(103 x 24 x 60 x 60) cumecs
= cumecs
For city = 1.8 x 135MLD
= 243 x 106/(103 x 24 x 60 x 60) cumecs
= 2.81 cumecs
Assuming a flow velocity of 1.2m/sec through the circular conduits, we get
Q = A V

16. For town Q = 0.063cumecs = A x 1.2
A= 0.063/1.2 = m2
Area of pipe required = 525 cm2
Diameter of pipe required = 24.9 cm
Use 25cm diameter of pipe
For city Q = 2.81cumecs = A x 2.81
A= 2.81/1.2 = 2.36 m2
Area of pipe required = 2.36 m2
Diameter of pipe required = 1.74 m
Use 1.8m diameter of pipe

17. V=0.85CH.R0.63.S0.54 HYDRAULIC GRADIENT BY THE HAZEN WILLIAM’S FORMULA
Use CH=110
V = 0.85 x 110 x (d/4)0.63 S0.54
V = 93.5 x (d/4)0.63 S0.54
For town
V = 93.5(0.25/4)0.63 x S0.54
1.2 = 93.5(0.0625)0.63 x S0.54
S0.54=(1.2/(93.5 x 0.174)
S= (1/13.54)(1/0.54)
S= (1/13.54)(1.86)
S=1/128
The hydraulic gradient is 1/128, i.e., 1m fall in 128m length
V=0.85CH.R0.63.S0.54

18. For town V = 93.5(1.8/4)0.63 x S = 93.5(0.45)0.63 x S0.54 S0.54=(1.2/(93.5 x 0.605) S= (1.2/56.56)(1/0.54) S= (47.2)(1.86) S=1/1300 The hydraulic gradient is 1/1300 i.e., 1m fall in 1300m length

19. Estimating size of supply conduits
Problem No.2
Estimate roughly the sizes of supply conduits leading to an adequate service reservoir serving,
(a) Tiruvallur town of 56,000 population
(b) Kanchipuram having a population of million people.
Also find the hydraulic gradients of which the pipelines are proposed to be laid. Assume any suitable data according to Indian conditions, where required.

20. Problem No.3
Estimate the hydraulic gradient line in a 2 m diameter smooth concrete pipe carrying a discharge of 3 cumecs at 10 degree C temperature by using (a) Darcy Weisbach Formula (b) Manning’ s formula (c) Hazen William’s Formula (d) Modified Hazen William’s Formula. Assume suitable data not given.

21. Given: Diameter of pipe = 2m Discharge = 3 cumecs To Find: Estimate the hydraulic gradient line in a pipe using (a) Darcy Weisbach Formula (b) Manning’ s formula (c) Hazen William’s Formula (d) Modified Hazen William’s Formula

22. Area of Pipe=A= Area of Pipe = 3. 147. 4 / 4=3
Area of Pipe=A= Area of Pipe = * 4 / 4=3.14m2 Velocity of Flow = V=0.955m/sec

23. MANNING’S FORMULA
Using n=0.013, R=d/4=2/4=0.5 The hydraulic gradient is 1m fall in 2480m length

24. HAZEN-WILLIAM’S FORMULA
V=0.85CH.R0.63.S0.54
Using CH=130 from table 0.955=0.85 x 130 x (2/4)0.63 S0.54 S0.54=0.955/(0.85 x 130 x 0.646) S=1/(74.8)0.54 S=1/2900 The hydraulic gradient is 1m fall in 2900m length

25. MODIFIED HAZEN-WILLIAM’S FORMULA
V= CR.R S0.5525
CR=1.0 from table R=(d/4)=(2/4)= = x 1 x x S S = x 10-4 The hydraulic gradient is 1m fall in 3819m length

26. DARCY WEISBACH FORMULA
Reynold number (i.e. Re=Vd/ν) Where ν =kinematic viscosity of water at 10 degree C ν =1.31 x 10-6m2/sec Re=14,60,000

27. DARCY WEISBACH FORMULA
The hydraulic gradient is 1m fall in 3920m length

28. Flow in Pipe Systems
When two pipe lines are laid in parallel, the head loss through each pipe will be same
If the characteristics of pipes are known, the distribution of flow in the two pipes can be calculated by equating the head losses and applying the equation of continuity
i.e., Q=Q1+Q2

29. Flow in Pipe Systems
The necessity of laying pipes in parallel arises due to following reasons
In order to increase the capacity of the line, additional pipes are sometimes laid in parallel to the earlier ones
Two or more pipes are sometimes preferred because when one pipe is under repairs, the water can be supplied through the other. It give better control and management of water supplies.
Sometimes one single large pipe is not available on the market, and two or three smaller size pipes are required to be laid in parallel.

30. MINOR LOSSES

31. Forces acting on pressure conduits
The structural design of the pressure pipes should be carried out, so as to enable them to withstand the various forces likely to come on them. The following forces come onto play in the pressure pipes
Internal pressure of water including water hammer pressure – to be resisted by using materials strong in tension
Pressure due to external loads in theform of backfill, traffic loads etc., - to be resisted by using materials strong in compression

32. Forces acting on pressure conduits…
3. Longitudinal temperature stresses created when pipes are laid above the ground – to be resisted by providing expansion joints 4. Longitudinal stresses created due to unbalanced pressures at bends or at points of changes of cross section - to be resisted by holding the pipe firmly by anchoring it in massive blocks of concrete or stone masonry 5. Flexural stresses produced when pipes are supported over trestles etc.,

33. Pumps for Lifting Water

34. When No Lift is required

35. When only one Lift is required

36. When Two Lifts are required (Using surface sources)

37. When Two Lifts are required (Using wells or tube wells)

38. Types of Pumps Two types Roto-dynamic pumps Displacement pumps
Roto-dynamic pumps has a wheel or a rotating element which rotates the water in a casing and thus imparting energy to the water.
Centrifugal Pump
Axial Flow pump

39. Types of Pumps
Displacement pumps works on the principal of mechanically inducing vacuum in a chamber, thereby drawing in a volume of water which is then mechanically displaced and forced out of the chamber. Reciprocating pump The rotary type pump Other Types of pumps Airlift pumps, Jetpumps, Hydraulic rams

40. Rotodynamic Pumps The shape of the impeller may be such as
The Rotodynamic Pump do have a wheel type rotating element called impeller.
The shape of the impeller may be such as
to force the water outward in a direction at right angles to its axis (Radial flow)
to give water an axial as well as a radial velocity (Mixed flow)
to force the water in the axial direction alone (axial flow)

41. Rotodynamic Pumps
Radial flow and mixed flow machines are called centrifugal pumps. Axial flow machine is called axial flow pumps. One impeller is used it is known as Single stage pump, Tow or more impellers it is known as Double stage or multi stage pump.

42. Impellers of a Rotodynamic Pumps

43. Centrifugal pump Impeller

44. Volute type of casing for a radial flow Centrifugal pump
Turbine type of casing for a radial flow Centrifugal pump

45. Priming and Operation of Centrifugal pumps

46. Typical Characteristic curves of a Centrifugal pumps