Analysis of Water Supply

Distribution Networks

Dr. Mohsin Siddique

Assistant Professor

1

Schematic

Diagram of

Water Supply

System

� Note: Surface water requires more advanced treatment then for groundwater

Component of Water Supply System

3

� (1). Source (2).Treatment plant

� (2). Storage Tanks/Reservoirs (3).Water Transmission/distribution

The various natural sources of water can be classified into two categories:

�Surface sources, such as

�Ponds and lakes;

�Streams and rivers;

�Storage reservoirs.

�Sub-surface sources or underground sources, such as

�Springs; and

�Fresh groundwater

Natural Raw Water Sources

4

Other non-traditional water sources include:

�Ocean water

�Desalination of sea water

� Brackish underground sources

�Desalination

�Wastewater reuse

�Treatment and re-use of wastewater

� Rainwater harvesting (i.e., houses and domes roof, storm water)

�Water trading !!

� import/export of fresh water

�Virtual water trade

Other Water Sources

Brackish water or briny water is water that has more salinity than fresh water, but not as much as seawater

5

Water Resources in UAE

6

/Seasonal floods

Water Quality and Treatment

7

� Water to be used in a public water supply is required to be fit for drinking.

� This implies that it poses no danger to health, and it should be colourless, clear, odourless, sparkling and pleasant to taste.

�The raw or treated water is analyzed by testing their physical, chemical and bacteriological characteristics:

�Physical Characteristics:

�Turbidity; Color; Taste and odor; and Temperature

�Chemical Characteristics:

�pH; Acidity; Alkalinity; Hardness; Chlorides; Sulphates; Iron; Nitrate, and Dissolved solids.

�Bacteriological Characteristics:

�Bacterial examination i.e., pathogenic bacteria or non pathogenic bacteria such as E.Coli,

Distribution Reservoirs/Tanks

� Reservoirs in water distribution systems plays an important role to:

� Provide service storage to meet widely fluctuating demands imposed on water supply distribution systems

� Accommodate fire-fighting and emergency requirements

� Equalize operating pressures

Fire water tank Elevated water tank Jumaira-UAE

Surface water tanks water tanks in Kuwait

Water Transmission

9

� Water transmission refers to the transportation of the water from the source to the treatment plant and to the area of distribution.

� It can be realized through

� free-flow conduits, (gravity flow)

� pressurized pipelines (pumping system) or

� a combination of the two (combination of gravity flow and pump).

� For small community water supplies through pressurized pipelined are most common, since they are not very limited by the topography of the area to be traversed.

� Free-flow conduits (canals, aqueducts and tunnels) are preferred in hilly areas or in areas where the required slope of the conduit more or less coincides with the slope of the terrain.

Methods of Water Transmission/Distribution

10

� (pumping system)

(1. Gravity flow) (2. Pumping system)

(3. Gravity flow and pumping system)

Water Distribution System

11

� A water distribution system consists of complex interconnected pipes, service reservoirs and/or pumps, which deliver water from the treatment plant to the consumer.

� Water demand is highly variable, whereas supply is normally constant. Thus, the distribution system must include storage elements, and must be capable of flexible operation.

Layouts of Distribution Network

12

Design of Water Supply Networks� General Consideration:

� A municipal water distribution system includes a network of mains with storage reservoirs, booster pumping stations (if needed),fire hydrants, and service lines. Arterial mains, or feeders, are pipelines of larger size that are connected to the transmission lines that supply the water for distribution. A major water demand areas in a city should be served by a feeder loop; where possible the arterial mains should be laid in duplicate. Parallel feeder mains are cross connected at intervals of one to two kilometers, with valves to permit isolation of sections in case of a main break. Distribution lines tie to each arterial loop, forming a complete gridiron “close loops” system that domestic, commercial consumers and services fire hydrants. The gridiron system illustrated in the Figure 1 is the best arrangement for distributing water. All of the arterials and secondary mains are looped and interconnected, eliminating dead ends and permitting water circulation such that a heavy discharge from one main allows drawing water from other pipes.

� The dead-end system shown in Figure 2 is avoided in new construction and can often be corrected in existing systems by proper looping. Trunk lines placed in the main streets supply sub-mains, which are extended at right angles to serve individual streets without interconnections. Consequently, if a pipe break occurs substantial portion of the community may be without water. Under a some conditions, the water in dead-end lines develops tastes and odors from stagnation. To prevent this, dead ends may require frequent flushing where houses are widely separated.

Pipe Network Analysis

� Pipe network analysis involves the determination of the pipe flow rates and pressure heads at the outflows points of the network. The flow rates and pressure heads must satisfy the continuity and energyequations.

� ANALYSIS METHODS

� (1). Hardy-Cross Method (Looped Method)

� (2). Nodal Method

� (3). Newton-Raphson Method

(1). The Hardy Cross Method

15

� The earliest systematic method of network analysis (Hardy-Cross Method) is known as the head balance or closed loop method.

� This method is applicable to system in which pipes form closed loops. The outflows from the system are generally assumed to occur at the nodes junction.

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

� At each junction these flows must satisfy the continuity criterion, i.e. the algebraic sum of the flow rates in the pipe meeting at a junction, together with any external flows is zero.

(1). The Hardy Cross Method

� The method is based on

� Continuity Equation:

� Inflow = outflow at nodes

� Energy Equation:

� Summation of head loss in closed loop is zero

cba QQQ +=

( ) ( )∑ ∑ =∆+⇒= 00n

l QQKlooph

A B

C D

QaQb

Qc

(1). The Hardy Cross Method

17

� The relationship between head loss and discharge must be maintained for each pipe

� Darcy-Weisbach Equation

� Exponential friction formula Hazen-Williams

52

82)(

Dg

fLKnKQpipeh n

lπ

===

87.485.1

67.1085.1)(

dCKnKQpipeh

n

l ===

(1). Hardy Cross Method (Derivation)

18

QQQ a ∆+=

( ) 0=∆+∑n

a QQK

( ) ( ) 0...2

1 221=+∆

−+∆+∑ ∑∑ −− n

an

an

a QQnKn

QQnKKQ

( )∑∑∑ ∑ −

−−=∆=∆+

1

10

na

nan

an

anKQ

KQQQQnKKQ

(1). Hardy Cross Method

19

� Problem Description

� Network of pipes forming one or more closed loops

� Given

� Demands @ network nodes (junctions)

� d, L, pipe material, Temp, and P @ one node

� Find

� Discharge & flow direction for all pipes in network

� Pressure @ all nodes & HGL

(1). Hardy-Cross Method (Procedure)

20

1. Divide network into number of closed loops.

2. For each loop:

a) Assume discharge Qa and direction for each pipe. Apply Continuity at each node, Total inflow = Total Outflow. Clockwise positive.

b) Calculate equivalent resistance K for each pipe given L, d, pipe material and water temperature.

c) Calculate hf=K Qan for each pipe. Retain

sign from step (a) and compute sum for loop Σ hf.

d) Calculate hf / Qa for each pipe and sum for loop Σhf/ Qa.

e) Calculate correction

∆Q = −Σ hf /(nΣhf/Qa).

NOTE: For common members between 2 loops both corrections have to be made. As loop 1 member, ∆Q = ∆Q 1 − ∆Q 2. As loop 2 member, ∆Q = ∆Q 2 − ∆Q 1.

f) Apply correction to Qa, Qnew=Qa + ∆Q.

g) Repeat steps (c) to (f) until ∆Q becomes very small and Σhf=0 in step (c).

h) Solve for pressure at each node using energy conservation.

Example: 5-1

21

Neglecting minor losses in the pipe, determine the flows in the pipes and the pressure heads at the nodes

(kinematic viscosity= 1.13x10-6m2/s)

Example: 5-1

22

� Solution:

� Identify loops

� Loop 1 and Loop 2

� Allocate estimated flows in each pipe

� Compute head loss coefficient of each pipe

� Compute head loss in each pipe

Discharge Sign Convention

CW=+ve

CCW=-ve

52

8

Dg

fLK

π=

2KQhhl ==

fMoody’s Diagram or Colebrook Eq.

+−=

f

De

f Re

51.2

7.3

/log2

1

1st Iteration

23

120+14.23=134.2310+14.23=24.23-60+14.23=-45.77-100+14.23=-85.77

Assumed Discharged

Corrected DischargeQ=Q+∆Q

Correction

fe/D

e/D

hL hL/Q

1st Iteration

24

50-2.23=47.7710-2.23=7.77-20-2.23=-23.23-24.23-2.23=-26.46

Assumed/corrected Discharged

Corrected DischargeQ=Q+∆Q

Correction

f hL hL/Q

2nd Iteration

25

134.23-1.92= 132.3126.46-1.92= 24.54-45.77-1.92= -47.69-85.77-1.92= -87.69

f hL hL/Q

2nd Iteration

26

� Proceed to loop 2 again and continue iterating until ∑hl=0

70-25-6.59

70-30

Example:

Find the flows in the loop given the inflows and outflows.The pipes are all 25 cm cast iron (e=0.26 mm).

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

200 m

100 m

Example:

� Identify Loop

� Assign a flow to each pipe link

� Flow into each junction must equal flow out of the junction

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

0.320.00

0.10

0.04

assumed

Example:

� Calculate the head loss in each pipe

f=0.02 for Re>200000

=

=

25

2

8

πgD

fLK

KQh f

339)25.0)(8.9(

)200)(02.0(8

251 =

=

πk

k1,k3=339k2,k4=169

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

1

4 2

3

mh

mh

mh

mh

mh

i

f

f

f

f

f

i53.31

00.0

39.3

222.0

7.34

4

1

4

3

2

1

=

−=

−=

=

=

∑=

Sign convention+CW

2

5

s

m

Example:

� The head loss around the loop isn’t zero

� Need to change the flow around the loop

� the ___________ flow is too great (head loss is positive)

� reduce the clockwise flow to reduce the head loss

� Solution techniques� Hardy Cross loop-balancing (___________ _________)

� Use a numeric solver (Solver in Excel) to find a change in flow that will give zero head loss around the loop

� Use Network Analysis software (EPANET, WaterCad, etc.)

clockwise

optimizes correction

Example: � Numerical Solver

� Set up a spreadsheet as shown below.

� the numbers in bold were entered, the other cells are calculations

� initially ∆Q is 0

� use “solver” to set the sum of the head loss to 0 by changing ∆Q

� the column Q0+ ∆Q contains the correct flows

∆Q 0.000

pipe f L D k Q0 Q0+∆Q hf

P1 0.02 200 0.25 339 0.32 0.320 34.69

P2 0.02 100 0.25 169 0.04 0.040 0.27

P3 0.02 200 0.25 339 -0.1 -0.100 -3.39

P4 0.02 100 0.25 169 0 0.000 0.00

31.575Sum Head Loss

Example

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

0.218

0.102

0.202

0.062

1

4 2

3

Q0+ ∆Q0.218

−0.062

−0.202

−0.102

You must be able to make a hand calculation before opting for a better solution using software with a GUI.

Solution to Loop Problem

(2). Nodal Method

33

� Fig shows a branched pipe system delivering water from impounding reservoir A to the service reservoirs B, C and D. F is known direct out flow from J.

Eq. (1)

(2). Nodal Method

34

� Eq. (1) can be written as

Eq. (2)

Eq. (3)

Iteration of ZJ can be performed such that QIJ from Eq. (3) satisfies Eq. (2).

If (∑QIJ-F)≠0 then a correction, ∆ZJ, is applied to ZJ such that

Example 5-4

35f

(kinematic viscosity= 1.13x10-6m2/s)

Example 5-4

36

f

f

Example 5-4

37

f

Newton Raphson Method

38

� Newton–Raphson method is a powerful numerical method for solving systems of nonlinear equations.

� Suppose that there are three nonlinear equations F1(Q1, Q2, Q3) = 0, F2(Q1, Q2, Q3) = 0, and F3(Q1, Q2, Q3) = 0 to be solved for Q1, Q2, and Q3.

� Adopt a starting solution (Q1, Q2, Q3).

� Also consider that (Q1 + ∆Q1, Q2,+ ∆Q2, Q3 + ∆Q3) is the solution of the set of equations. That is

Eq. (1)

Newton Raphson Method

39

Eq. (2)

Eq. (3)

Solving Eq (3)., we get

Eq. (4)

Newton Raphson Method

40

Eq. (5)

Procedure

41

� The overall procedure for looped network analysis by the Newton–Raphson method can be summarized in the following steps:

� Step 1: Number all the nodes, pipe links, and loops.

� Step 2: Write nodal discharge equations as

� where Qjn is the discharge in nth pipe at node j, qj is nodal withdrawal, and jn is the total number of pipes at node j.

� Step 3: Write loop head-loss equations as

� where kn is total pipes in kth loop.

Procedure

42

� Step 4: Assume initial pipe discharges Q1, Q2, and Q3., . . . satisfying continuity equations.

�in all pipe links and compute ,ifDetermine friction factors, : 5Step usingiKcorresponding

� Step 6: Find values of partial derivatives ∂Fn / ∂Qi and functions Fn, using the initial pipe discharges Qi and Ki.

� Step 7: Find ∆Qi. The equations generated are of the form Ax = b, which can be solved for ∆Qi.

� Step 8: Using the obtained ∆Qi values, the pipe discharges are modified and the process is repeated again until the calculated ∆Qi

values are very small.

Example

43

� The pipe network of two loops as shown in Fig. has to be analyzed by the Newton Raphson 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 = 0.02.

Single looped network

Example

44

� Solution: Write nodal and loop equations

� Node1: Q1+Q4-0.6=0

� Node 2: Q1+Q2=0

� Node 3: Q2+Q3-0.6=0

� Node 4: !!!

� and loop Eq.

� K1Q12+ K2Q2

2+ K3Q32+ K4Q4

2=0

Q1

Q2

Q4

Q3

52

82

)(

Dg

fLKn

KQpipehn

L

π=

=

Example

45

� Assume initial pipe discharges based on continuity euqtion:

Q1 = 0.5 m3/s

� Apply signs to flowrate in eqns� Nodal discharge functions, F, are

� Q1+Q4-0.6=F1

� - Q1+Q2=F2

� Q2+Q3-0.6=F3

� Loop head-loss function

� K1Q12+ K2Q2

2 - K3Q32 - K4Q4

2=F4

In = +Out = -ve

= 0.5

0.5 0.1

0.1

Apply signs (+ or -)

Example

46

� The final nodal discharge functions, F, become

� and loop head-loss function is

� F4=6528Q12+ 4352Q2

2 - 6528Q32 - 4352Q4

2=0

� Take derivatives of equations:

Example

47

� Assembled equation in the following matrix form:

� Substitute the derivatives

Example

48

� Now, calculate values of F1 to F4 using previously assumed flows i.e.,

� Q1 = 0.5 m3/s

F4=6528Q12+ 4352Q2

2 - 6528Q32 - 4352Q4

2=2611.2

Example

49

� Substitute values of F1 to F4 and Q1 to Q4 in the matrix form :

� Solve the matrix, we get following flowrate corrections:

Example

50

� Using discharge corrections, the revised pipe flowrates become:

� The process is repeated with the new pipe flowrates. Revised values of F and derivative ∂F/∂Q values are obtained. Substituting the revised values, the following new solution matrix is generated:

Example

51

� As the right-hand side is operated upon null vector, all the discharge corrections ∆Q = 0.

� Thus, the final flowrates are

Problem

52

� Write solution matrix for analysis using Newton Raphson Method

� Node Eqns.=No. of nodes-1

� Loop Eqns. =3

Problem: Solution

53

� Nodal and loops equation in their general form are given below

Discharge_in=+ve

Discharge_out=-ve

CW=+ve

CCW=-ve

54

Thank You