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  

  

1

    



Note: Surface water requires more advanced treatment then for groundwater

!

  3

(1). Source (2). Storage Tanks/Reservoirs

(2).Treatment plant (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

4

Groundwater

# Other non(traditional water sources include:  Desalination

of sea water   Desalination   Treatment

and re(use of wastewater  Rainwater harvesting (i.e., houses and domes roof, storm water) Water trading  import/export of fresh water

         5

"$%

/Seasonal floods

6

& 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:  Turbidity; Color; Taste and odor; and Temperature  pH; Acidity; Alkalinity; Hardness; Chlorides; Sulphates; Iron; Nitrate, and Dissolved solids.  Bacterial examination i.e., pathogenic bacteria or non pathogenic bacteria such as , 7

"'(& 

    

Provide service storage to meet widely fluctuating demands imposed on water supply distribution systems Accommodate fire(fighting and emergency requirements Equalize operating pressures

Elevated water tank Jumaira(UAE

Surface water tanks

Fire water tank

water tanks in Kuwait

& 

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.



9

&(

(1. Gravity flow)

(2. Pumping system)

10

(3. Gravity flow and pumping system)

 

  



  

11

)

 

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 energy equations.



 ! "#$%&  (1). Hardy(Cross Method (Looped Method)  (2). Nodal Method  (3). Newton(Raphson Method

 

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. 15

'()*$%+" 

The method is based on



+#, 

Inflow = outflow at nodes

Qa = Qb + Qc













 

##, 

Summation of head loss in closed loop is zero

∑ hl (loop ) = 0 ⇒ ∑ K (Q + ∆Q ) = 0 n

'()*$%+" 

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

Darcy(Weisbach Equation

hl ( pipe) = KQ n 

8 fL gπ 2 D 5

Exponential friction formula Hazen(Williams

hl ( pipe) = KQ n

17

n=2 K

n = 1.85 K =

10.67 C 1.85 d 4.87

*+,!*', Q = Qa + ∆

∑ K (Q

n

a

∑ KQa + ∑ nK ∆Qa n

∑ KQ

a

n

+ ∑ nK ∆Qa

n −1

n −1

+ ∆) = 0

+∑

=0

n −1 n−2 nK ∆2 Qa +... = 0 2 KQ ∑ ∆=− ∑ nKQ

a

18

n

a n −1

*+,! 

-&



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

19

*+,-!*, 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.

d) Calculate hf / Qa for each pipe and sum for loop Σhf/ Qa. e) Calculate correction ∆ =−Σ hf /(nΣhf/Qa). NOTE: For common members between 2 loops both corrections have to be made. As loop 1 member, ∆ = ∆ 1 − ∆ 2. As loop 2 member, ∆ = ∆ 2 − ∆ 1.

b) Calculate equivalent resistance K for each f) Apply correction to Q , Q =Q + ∆. new a a pipe given L, d, pipe material and water temperature. g) Repeat steps (c) to (f) until ∆ becomes very small and Σhf=0 in step (c). c) Calculate hf=K Qan for each pipe. Retain sign from step (a) and compute sum for h) Solve for pressure at each node using loop Σ hf. energy conservation. 20

%./0-+

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

21

%./0-+ 

 

Identify loops 

Loop 1 and Loop 2



Allocate estimated flows in each pipe



Compute head loss coefficient of each pipe

K=



8 fL gπ 2 D5

& +

f

Compute head loss in each pipe 22

hl = h = KQ 2

Moody’s Diagram or Colebrook Eq.  e/ D 1 2.51 = −2 log +  f  3.7 Re f

CW=+ve    

CCW=(ve

+ 1 Assumed Discharged f

e/D

Correction

e/D

Corrected Discharge Q=Q+Q 120+14.23=134.23 10+14.23=24.23 (60+14.23=(45.77 (100+14.23=(85.77

23

+ 1 Assumed/corrected Discharged f

Correction

Corrected Discharge Q=Q+Q 50(2.23=47.77 10(2.23=7.77 (20(2.23=(23.23 (24.23(2.23=(26.46

24

2 1 f

134.23(1.92= 132.31 26.46(1.92= 24.54 (45.77(1.92= (47.69 (85.77(1.92= (87.69

25

2 1 

Proceed to loop 2 again and continue iterating until ∑hl=0 70(30

70(25(6.59

26

%./ Find the flows in the loop given the inflows and outflows. The pipes are all 25 cm cast iron (e=0.26 mm).













 





%./   

Identify Loop Assign a flow to each pipe link Flow into each junction must equal flow out of the junction assumed 

0.32 0.00









0.04





0.10



%./ 

Calculate the head loss in each pipe h f1 = 34.7m

h f = KQ 2  8 fL  K =  5 2   gD π 



h f2 = 0.222m h f3 = −3.39 m h f4 = −0.00 m 4

∑h

fi

= 31.53m

i =1

 8(0.02)(200)   = 339 s 2 k1 =   5 2  m5  (9.8)(0.25) π 



 





  







Sign convention +CW 





%./  

The head loss around the loop isn’t zero Need to change the flow around the loop  



the ___________ clockwise flow is too great (head loss is positive) reduce the clockwise flow to reduce the head loss

Solution techniques   

optimizes correction 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.)

%./   

  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 pipe P1 P2 P3 P4

0.000 f 0.02 0.02 0.02 0.02

L 200 100 200 100

D 0.25 0.25 0.25 0.25

k Q0 Q0+∆Q 339 0.32 0.320 169 0.04 0.040 339 -0.1 -0.100 169 0 0.000 Sum Head Loss

hf 34.69 0.27 -3.39 0.00 31.575

%. ) Q0+ ∆Q 0.218 −0.062 −0.202 −0.102













 

















  !"!#$%" "!&$'!&$( !!" !&$ &$(!))&!#*+,

*2, 

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)

33

*2, Eq. (2) 

Eq. (1) can be written as Eq. (3)

!./0 .1!0 .#,*'2) .#,*'3)* If (∑QIJF)≠0 then a correction, ∆ZJ, is applied to ZJ such that

34

%.0-3

f 35

%.0-3

f

%.0-3

37

"  

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,



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

F2(Q1, Q2, Q3) = 0, and F3(Q1, Q2, Q3) = 0 to be solved for Q1, Q2, and Q3.

Eq. (1)

38

" 

Eq. (2)

Eq. (3) Solving Eq (3)., we get

39

Eq. (4)

" 

Eq. (5)

40

 

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  is total pipes in kth loop.



41

 Step 4: Assume initial pipe discharges Q1, Q2, and Q3., . . . satisfying continuity equations. Step 5: Determine friction factors, fi, in all pipe links and compute corresponding  using 





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



Step 7: Find . The equations generated are of the form , 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. 42

%. 



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.

43



%. *The nodal discharge functions, F, are



hl ( pipe) = KQ n 

n =2 K

8 fL gπ 2 D5

and loop head(loss function CW+

44

%. *The nodal discharge functions, F, are



hl ( pipe) = KQ n 

n =2 K

8 fL gπ 2 D5

and loop head(loss function CW+



The derivatives are

45

%. 

The generated equations are assembled in the following matrix form:



Substituting the derivatives, the following form is obtained:

46

%. 

Assuming initial pipe discharge in pipe 1as Q1 = 0.5 m3/s, the other pipe discharges obtained by continuity equation are

Q1 = 0.5 m3/s

Q2 = 0.5 m3/s

Q4 = 0.1 m3/s Q3 = 0.1 m3/s

47

%. 

Substituting these values in the above equation, the following form is obtained:



Using Gaussian elimination method, the solution is obtained as

48

%. 

Using these discharge corrections, the revised pipe discharges are



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

49

%. 

As the right(hand side is operated upon null vector, all the discharge corrections ∆Q = 0. Thus, the final discharges are

50

 

 

Write solution matrix for analysis using Newton Raphson Method

Node Eqns.=No. of nodes(1 Loop Eqns. =3

51

/ 

Nodal and loops equation in their general form are given below

Discharge_in=+ve Discharge_out=(ve

CW=+ve CCW=(ve

52

$ 44+

53...