research article error analysis of some demand...

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 169670 13 pageshttpdxdoiorg1011552013169670

Research ArticleError Analysis of Some Demand Simplifications inHydraulic Models of Water Supply Networks

Joaquiacuten Izquierdo1 Enrique Campbell1 Idel Montalvo2

Rafael Peacuterez-Garciacutea1 and David Ayala-Cabrera1

1 Flulng IMM Universitat Politecnica de Valencia Camino de Vera SN Edificio 5C bajo 46022 Valencia Spain2 3S Consult GmbH Albtalstraszlige 13 76137 Karlsruhe Germany

Correspondence should be addressed to Joaquın Izquierdo jizquierupves

Received 29 October 2013 Accepted 4 December 2013

Academic Editor Rafael Jacinto Villanueva

Copyright copy 2013 Joaquın Izquierdo et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Mathematical modeling of water distribution networks makes use of simplifications aimed to optimize the development and use ofthe mathematical models involved Simplified models are used systematically by water utilities frequently with no awareness of theimplications of the assumptions used Some simplifications are derived from the various levels of granularity at which a network canbe considered This is the case of some demand simplifications specifically when consumptions associated with a line are equallyallocated to the ends of the line In this paper we present examples of situations where this kind of simplification produces modelsthat are very unrealisticWe also identify themain variables responsible for the errors By performing some error analysis we assessto what extent such a simplification is valid Using this information guidelines are provided that enable the user to establish if agiven simplification is acceptable or on the contrary supplies information that differs substantially from reality We also developeasy to implement formulae that enable the allocation of inner line demand to the line ends with minimal error finally we assessthe errors associated with the simplification and locate the points of a line where maximum discrepancies occur

1 Introduction

In the task of mathematical modeling of such complexstructures as water distribution networks (WDNs) the useof simplifications aimed to optimize the development anduse of the models is unavoidable Such simplifications stemfrom the complexity of the modeled infrastructure and atthe same time are related to the large spatial distributiontypical of WDNs These models are applied in all the areasof hydraulicsmdashincluding urban hydraulics [1 2] Currentlywith the generalized use of geographic information systemsmodels containing even hundreds of thousands of pipes andnodes are being built [3]

Extremely detailed modeling of real WDNs even underthe unrealistic hypothesis in which uncertainty can beignored produces a substantial amount of data and requiressophisticated computational tools and mechanisms to reli-ably interpret the obtained results in terms of what occursin the system Current computational power can be used

to build hydraulic simulation models capable of providinga very detailed and accurate model but not an improvedunderstanding of the main structure of the underlying sys-tem Also as such models would require very dedicated cal-culations certain aspects must be considered to ensure effi-cient implementations Specifically optimization of WDNswhether used for planning or operational purposes oftenrequires many iterations [4] each involving computationallyexpensive simulations and huge computer memory

Let us mention just a few of the most typical simplifica-tions the use of one equivalent pipe to represent two or moreparallel or series pipes the removal of short pipe segmentsincluding dead ends service connections and hydrantsthe distribution of emitter exponents without consideringleakage profiles along the WDN the assumption of frictionfactor values without a detailed consideration of the pipesstate model calibration assuming values of one or more ofthe calibration elements as fixed the use of a single frictionfactor for the entireWDNor for an entire sector or the use of

2 Abstract and Applied Analysis

a single emitter leakage exponent for the entire network Thelast two simplifications in particular are inescapable whenworking with EPANET [5] a tool of water network analysisin general use worldwide Also some simplifications cantotally undermine studies of hydraulic transient phenomenain WDNs See for example [6 7]

Among the various simplifications used in the analy-sis of water distribution systems some are assumed in ageneralized way without additional introspection They aresimplifications derived from the different levels of granularityat which a network can be considered As mentioned beforereal water distribution networks especially those of largecities cannot be efficiently modeled in their entirety Inapproximate models some granulation reduction actions areperformed such as skeletonization grouping pruning andclustering See among others [8ndash18]

In particular one of these simplifications is the groupingof the consumptions associated with a line in one or bothends of the line These points concentrate all the existingconsumption points (users) within the line The need forimplementing this specific simplification is evident because afaithful consideration of demand would imply the inclusionof a large number of nodes equal to the number of consump-tion points In a branched network with few consumers thiswould not represent a major problem However problemsarise in WDNs with up to 30 connections per line (eg astreet pipeline) In a large WDN (eg a 500 kmWDN) itwould amount to considering about 150000 nodes whichis impractical when it comes to the construction of thenetwork model the performance of the calculations and thedisplay and understanding of the results This simplificationcopes with the continuity principle or conservation of massHowever the energy aspects are completely ignored

The methods for calculating demand load in the modelsare one of the least studied aspects despite the obvious impor-tance for defining model reliability Recently Giustolisi et al[19] have addressed this problem from a global perspectiveand have developed a matrix transformation approach thatchanges the classical solution of the nonlinear system ofequations describing aWDNbased on the conjugate gradient[20] into what they call enhanced global gradient algorithm(EGGA) Note that the solution in [20] is implementedwithin the widely available freeware EPANET 2 used in ageneralized way by many engineers and practitioners aroundthe world EGGA reduces the size of the mathematical prob-lem through a transformed topological representation of theoriginal network model that preserves both mass and energybalance equations and improves the numerical stability ofthe solution procedure According to the authors EGGAsignificantly improves the modelrsquos computational efficiencywithout sacrificing its hydraulic accuracy

However this solution although technically impeccablefrom a theoretical point of view exhibits a number ofpractical drawbacks Firstly those matrix transformationsare not implemented within EPANET and as a result arenot available to the huge community of its users Secondlythe transformations are too complex for most users of thisprogram and especially too complex to be incorporatedinto the EPANET toolkit given the in-depth knowledge of

programming techniques that their implementation wouldrequireThirdly the transformationmatricesmust previouslybe explicitly written even though obtaining these matricesis relatively straightforward the computational efficiency atleast in terms ofmemory needs is not evident One has to useseveral additional (very sparse) matrices of sizes or the orderof the number of resulting pipes times the number of originalpipes and the number of resulting pipes times the numberof unknown head nodes are explicitly used When solvingreal-world problems with hundreds of thousands of nodesand pipes this may become a serious problem Fourthly fornetworks already modeled by excluding those intermediatedemand nodes the solution of matrix transformation will beuseless Fifthly when planning and designing a new networkstarting from the household demand distribution it would beperhaps desirable to start building the model by performingthe simplification from the outset in order to avoid latercomplications

Compelled by those drawbacks we have addressed theproblem from another we claimmore practical point of viewThis paper analyzes the possible errors from the effect of usingcertain types of simplifications when loading demands inmodels specifically the widespread 50 rule which allocateshalf of the in-line demand to each line end We analyzeto what extent this simplification is acceptable or on thecontrary supply information that differs substantially fromreality Also we obtain formulae that enable to allocateinner line demand to the line ends with minimal errorFinally a calculation of themaximumhead point discrepancyassociated is provided

Our proposal involves simple direct methods that can beeasily applied by any user of anyWDNanalysis package sinceemphasis is not placed on programming ability but on howto make a decision about the technical aspect of simplifyingthemodel and thus load the demand properly Users havingalready developed models of their networks may revise theallocation rule used and replace it if necessary with thevalues provided by the new formulae what will enable themto obtain more reliable results Also users starting the modelof a new networkmaymake an a priori decision about how tosimplify the network and suitably implement the associatedsimplifications

We first present a simple case that enables us to shed lightonto the problem a single line with variable distribution andgranulation consumption is consideredThen an example of areal network is analyzed using the lessons learnedThe papercloses with a conclusions section

2 Line with Associated Demand

Let us first consider the case of a single line associated withsome internal consumption under steady state conditionSuch a line is representative of the simplest installation(a line between two nodes) with a given inflow rate Thecharacteristics of the line are

(i) length 119871

(ii) diameter119863

Abstract and Applied Analysis 3

(iii) upstream head (boundary condition at the upstreamnode)119867

0

(iv) friction factor 119891 with its associated line resistance119870 = 8119891(119892120587

21198635)

(v) inflow 119876inVarious demand scenarios of consumption in the line

may be tested Such scenarios are associated with two char-acteristics

(i) total demand in the line with regard to inflow(ii) specific distribution of the demand along the lineLet us assume that the flow consumed within the line

(total in-line demand) represents a percentage of the lineinflow If this fraction is represented by 119865

119876 0 lt 119865

119876le 1 the

actual demand in the line is given by the expression

119876119889= 119865119876119876in (1)

21 Uniformly Distributed Demand along the Line To startwith the study we will assume that the actual demand ofthe line is uniformly distributed into a number 119899 of equallyspaced interior points (nodes) 119899 can take a value rangingfrom 1 (in the case of a line with a single demand node in themiddle) to a large integer number (in the case of an equallydistributed demand throughout the line) Observe that wedo not consider any demand at the end nodes since we areonly interested in the line demand associatedwith the interiornodes

Figure 1 shows various distributions of piezometric headcorresponding to values of119865

119876equal to 1 08 06 and 04 for a

set of values of 1198711198631198670119891 and119876in To build Figure 1 we have

used the specific values 119871 = 500m119863 = 300mm1198670= 50m

119891 = 0018 and 119876in = 025m3s As mentioned before thedemand has been equally distributed among 119899 equally spacedinterior nodes Specifically in Figure 1 119899 takes the values1 3 7 11 19 and ldquoinfinityrdquo The ldquoinfinityrdquo case represents auniform continuous demand The various curves in Figure 1have straightforward interpretation

For the polygonal hydraulic grade lines (HGLs) made outof segments between consumption points the calculationscorrespond to the usual hydraulic calculation of losses Thepolygonals start at the boundary condition (0119867

0) the other

vertices being the 119899 + 1 points as follows

(119895119871

119899 + 11198670minus 119870

119871

119899 + 11198762

in

119895

sum

119896=1

(1 minus119896 minus 1

119899119865119876)

2

)

119895 = 1 119899 + 1

(2)

Let us call 119867119877(119865119876 119899 119909) these HGLs the subindex ldquo119877rdquo

standing for ldquorealrdquo distribution of piezometric head along theline as real demands are used to calculate (2)

The calculation for the ideal HGL corresponding to auniform continuous demand which is used here as the limitfor 119899 rarr infin of the discrete uniform distribution of demandsis performed by integrating the differential loss

Δ119867 = minus119870(119876in minus119876119889

119871119909)

2

Δ119909 (3)

along the line The value 119876119889119871 is the (constant) demand

per unit length and 119909 is the distance to the upstreamnode By integrating and using (1) the piezometric headcorresponding to this continuous loss is given by

119867119877(119865119876 119909) = 119867

0minus 119870

119871

3

1198762

in119865119876

[1 minus (1 minus 119865119876

119909

119871)

3

] (4)

which corresponds to the upper curve a cubic in each of thegraphs in Figure 1

It becomes clear that the greatest discrepancies occur forvalues of 119865

119876close to 1 (eg when a high percentage of the

inflow is consumed along the line)As mentioned before these ldquorealrdquo HGLs in Figure 1 have

been calculated according to the demand distribution atthe various inner points in the line However models oflarge WDNs do not usually take intermediate demands intoaccount in contrast the demand of each line is allocated tothe end nodes of the line the 50 rule being generally used

22 Allocation of In-Line Demand to the Line Ends Is the 50Rule Adequate Let 119865

119876119889be the factor that allocates a part of

the line distributed demand 119876119889 to its upstream end Thus

the demand assigned to this upstream node is 1198760= 119865119876119889

119876119889

As a result 119876119897= 119876in minus 119876

0is the flowrate through the line

Note that using (1)

119876119897= 119876in (1 minus 119865

119876119865119876119889

) (5)

Then the calculated head value 119867119862for a given value of

119865119876119889

is

119867119862(119865119876119889

119909) = 1198670minus 119870119876

2

119897119909 (6)

The HGL obtained is thus a straight line that connectsthe point (0119867

0)with the point (119871119867

119862(119865119876119889

119871))This last valuecorresponds to the calculated head at 119871 the downstreamnode

In Figure 2 dashed lines have been added to the firsttwo charts of Figure 1 These new HGLs have been calculatedto give the same piezometric head at the downstream nodeas the line corresponding to a demand concentrated in themiddle point 119899 = 1 for the case 119865

119876= 1 (left chart) and as

the line corresponding to a continuous demand for 119865119876= 08

(right chart) These lines have been obtained by allocating afraction of the interior line demand to the upstream nodeand the rest to the downstream node Analogous dashedlines can be obtained for other combinations of 119865

119876and

119899 If the allocated fractions to the line ends are differentthe corresponding (straight) lines (the lines given by thenumerical model with lumped demands at the ends of theline) will also be different

Two problems arise at this point

(a) Firstly it would be desirable to know the best alloca-tion of the total in-line demand to the line ends that isto say to know the value 119865

119876119889that solves the following

problem

Minimize119865119876119889

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (7)


FQ = 1 FQ = 08 FQ = 06 FQ = 04

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

340 250 500 0 250 500 0 250 500 0 250 500

Inf1911

731

Inf1911

731

Inf1911

731

Inf1911

731

Figure 1 Hydraulic grade lines in one single line for various uniform distributions of demand

FQ = 1 FQ = 0850

49

48

47

46

45

44

43

42

41

40

50

49

48

47

46

45

44

43

42

41

400 100 200 300 400 500 0 100 200 300 400 500

LumpInf1911

731

LumpInf1911

731

Figure 2 Examples of discrepancy between distributed and lumped demand models

for certain functional norm sdot where119867119862(119865119876119889

119909) iscalculated by (6) according to the lumped demandallocation to both line ends and119867

119877(119865119876lowast 119909) accounts

for the real demand distribution either calculatedby (2) or (4) (or any other more general expressioncorresponding to not uniformly distributed demand

which we address later) The asterisk denotes otherparameters such as 119899 in (2) which may appear in theexpression of119867

119877

(b) Secondly after having made a suitable line-end allo-cation decision it is of interest to know the actualdistribution of piezometric head errors on the line


Table 1 Values of 119865119876119889

as a function of 119865119876and 119899 when solving (9)

119899 (order) 119865119876

02 04 06 08 101 (1) 0472 0438 0397 0349 02933 (2) 0485 0466 0442 0413 03767 (3) 0488 0473 0455 0432 040211 (4) 0489 0475 0457 0435 040719 (5) 0490 0477 0461 0441 0415inf (6) 0491 0479 0465 0446 0423

and specifically to identify at what point or points themaximum head discrepancy occurs

Maximize119909

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (8)

We start by discussing (7) and then address (8) inSection 24

The solution of (7) is here constrained by the natureof the problem we have to adhere to the fact that one ormore lines (pipes) may be connected to the downstreamend of the considered line The connected lines need thecorrect piezometric head at 119871mdashupstream end for themmdashto suitably perform their respective calculations It meansthat the piezometric head at 119871 given by 119867

119862and 119867

119877 must

coincide That is to say (7) reduces in our case to

Solve 119867119862(119865119876119889

119871) = 119867119877(119865119876lowast 119871) for 119865

119876119889 (9)

By solving (9) the following expressions for 119865119876119889

areobtained

(i) Case of continuous demand

119865infin

119876119889(119865119876) =

1

119865119876

(1 minus radic1 minus (1 minus 119865

119876)3

3119865119876

) (10)

(ii) Case of demand equally distributed among 119899 equallydistributed nodes

119865(119899)

119876119889(119865119876) =

1

119865119876

(1 minus radic1

119899 + 1

119899+1

sum

119896=1


119899119865119876)

2

) (11)

In Table 1 and in two-dimensional Figure 3 values for (10)and (11) for 119865

119876values between 02 and 1 and for 119899 varying

along the previously used values namely 1 3 7 11 19 andinfinity are presented Note that values 1 to 6 on the frontalaxis of Figure 3 symbolize as shown in Table 1 the values 119899 =

1 3 7 11 19 and infinity respectivelyThe following facts are remarkable

(i) These values are independent of the problem datanamely119867

0 119876in 119871 119863 and 119891 and depend only on 119865

119876

and 119899 in the case of (11) that is to say they dependon the magnitude and the pattern of the distributeddemand This is a very remarkable result since thepresent study thus becomes nondimensional and asa result completely general

0204

0608

1

03

035

04

045

05

123456

045ndash0504ndash045


0406

Figure 3 Solution of (9) for 119865119876119889

(ii) The values of 119865119876119889

range from approximately 03 to05 This means that about 30ndash50 of the totaldemand must be allocated to the upstream node andthe remainder to the downstream node

(iii) The lowest 119865119876119889

values correspond to the most awk-ward cases the rate of demand is close to or equalsthe total inflow in the line and the demand is highlyconcentrated at a few points (upper right corner of thetable right front of the figure)

(iv) The highest 119865119876119889

values closer to 50 correspondto the less problematic cases meaning little totaldistributed demand in relation to the total inflow inthe line and widely distributed demand (lower leftcorner of the table bottom left of the figure) Thisvalue approaches 50 as the rate of inflow consumedin the line approaches zero (Observe that for both(10) and (11) lim

119865119876rarr0119865(sdot)

119876119889(119865119876) = 05)

(v) It is also worth noting that lim119899rarrinfin

119865119899

119876119889= 119865infin

119876119889for all

119865119876 that is to say the monotonic sequence of contin-

uous functions 119865119899119876119889

infin

119899=1converges to the continuous

function 119865infin

119876119889uniformly in [0 1] (Dini theorem see

eg [21 22]) This has a direct interpretation asthe continuous demand is the limit of a uniformlydistributed demand among an increasing number ofequally distributed points on the pipe

Asmentioned before it is commonpractice inmathemat-ical modeling of WSN engineering to distribute the line flowinto two parts 50 for the upstream node and 50 for thedownstream node which approximately coincides with whatis observed in Table 1 and Figure 3 except for cases wherethe demand line is highly concentrated and represents a largepercentage of the total flow through the line

As a result of what has been presented so far it can be saidthat for uniformly distributed in-line demand the usual 50rule seems in principle an appropriate solution provided


that the inner demand of the line is small compared withthe pipe inflow and that such a demand is widely distributedHowever equal demand allocation to the end nodes of theline may produce important discrepancies since the studyhighlights the need for other assignments in certain cases

23 Arbitrary Demand along the Line To state the problemin its more general form let us now consider a demanddistribution on the line whose accumulated demand is givenby a function 119876(119909) = 119876

119889119902(119909) where 119902(119909) is the accumulated

demand ratio a function increasing monotonically from 0to 1 While 119867

119862is calculated as in (6) 119867

119877is calculated by

integrating the loss Δ119867 = minus119870(119876in minus 119876(119909))2Δ119909 through the

line [0 119871] as follows

119867119877(119865119876 119902 (119909) 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (12)

Observe that this function is monotonically decreasingand concave upwards

Example 1 119902(119909) = 119909119871 for the case of continuous uniformdemand which by using (1) gives (4)

Using (12) and the expression (6) for 119867119862in 119871 written

using (5) as

119867119862(119865119876119889

119871) = 1198670minus 119870119871119876

2

in(1 minus 119865119876119889

119865119876)2

(13)

the equation119867119862= 119867119877in 119871may be rewritten as

minus1198701198711198762

in(1 minus 119865119876119889

119865119876)2

= minus119870int

119871

0

(119876in minus 119876 (119909))2d119909 (14)

By substituting

119876 (119909) = 119876in119865119876119902 (119909) (15)

one gets

119871(1 minus 119865119876119889

119865119876)2

= int

119871

0

(1 minus 119865119876119902 (119909))

2d119909 (16)

from where the following expression is readily obtained

119865119876119889

(119865119876) =

1

119865119876

(1 minus radic1

119871int

119871

0

(1 minus 119865119876119902 (119909))

2d119909) (17)

The general solution for the problem at hand when anarbitrary demand through the line is considered may onlybe solved after having a specific expression for 119902(119909) As aconsequence for arbitrary demands we will restrict ourselvesto the case of discrete demands on a finite number of pointsof the pipe as happens in real life

Example 2 Let us start by considering a single demandwithdrawn at a specific point of the line That is to say let usconsider a demand distribution given by

119902 (119909) = 120575 (119909 minus 1199091) (18)

109080706050403020100 02 04 06 08 1

FQ = 02

FQ = 04

FQ = 06

FQ = 08

FQ = 1

Figure 4 Demand allocation fraction to the upstream nodedepending on the location of a single withdrawal in the line

which concentrates the whole demand119876119889at 1199091 where 120575(sdot) is

the well-known Dirac deltaIn this case (12) is written as

119867119877(119865119876 120575 (119909 minus 119909

1) 119871)

= 1198670minus 119870int

119871

0

[119876in minus 119876119889120575 (119906 minus 119906

1)]2d119906

= 1198670minus 119870int

1199091

0

1198762

ind119906 minus 119870int

119871

1199091

(119876in minus 119876119889)2d119906

= 1198670minus 11987011990911198762

in minus 119870 (119871 minus 1199091) (119876in minus 119876

119889)2

= 1198670minus 119870119871119876

2

in (120582 + (1 minus 120582) (1 minus 119865119876)2)

(19)

where 120582 = 1199091119871 is the fraction of the pipe where the

concentrated demand is located from the originEquating again119867

119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582) =

1

119865119876

(1 minus radic120582 + (1 minus 120582) (1 minus 119865119876)2) (20)

As could be expected these values not only depend on119865119876

as in the case of uniform demand but also strongly dependon 120582 In Figure 4 we have plotted these values as a functionof 120582 for various instances of 119865

119876 namely 10 08 06 04 and

02As expected the worst cases occur once more for with-

drawals representing a large percentage of the inflow to theline For small in-line demands (see eg the curve for 119865

119876=

02 also calculate the limit of (20) for 119865119876approaching to

0) the demand should be allocated to the end nodes almostlinearly proportional to the relative distance of the demandpoint to the downstream end as is completely natural Incontrast this rule does not apply to large in-line demandsas their corresponding curves show by becoming less andless linear As an extreme case let us consider the case of119865119876

= 1 Various HGLs have been plotted in Figure 5 by


100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509

Figure 5 HGLs depending on the location of a single withdrawal inthe line and HGL (dashed line) corresponding to the 50 allocationrule

varying the location of the withdrawal point specifically forvalues 120582 = 01 025 05 and 09 The (straight dashed)line corresponding to the 50 allocation rule has alsobeen represented We can observe that only for 120582 = 025

does the dashed line match the correct piezometric headat the downstream end (also observe that the lower curvein Figure 4 corresponding to 119865

119876= 1 contains the point

(120582 119865119876119889

) = (025 05)) In the other cases disagreements arenot only important at the downstream end but all along theline

In the general case we consider 119902(119909) = sum119899

119896=1119889119896120575(119909 minus 119909

119896)

where 119889119896are the demands at points 119909

119896 with 0 lt 119909

1lt 1199092lt

sdot sdot sdot lt 119909119899minus1

lt 119909119899lt 119871 such that sum119899

119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case

of uniformdemand at 119899 equally distributed points in the pipeThis demand produces the expression in (2)

In the general case119867119877is calculated by

119867119877(119865119876 119871)

= 1198670minus 11987011990911198762

in minus 119870 (1199092minus 1199091) (119876in minus 119889

1)2

minus 119870 (1199093minus 1199092) (119876in minus (119889

1+ 1198892))2

minus sdot sdot sdot minus 119870 (119909119899minus 119909119899minus1

)(119876in minus

119899minus1

sum

119896=1

119889119896)

2

minus 119870 (119871 minus 119909119899)(119876in minus

119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)

this expression can be written as

119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)

Then equating again119867119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)

This expression can be easily calculated using for exam-ple a worksheet as in Figure 6

In this figure we consider a demand distributionat the (inner) points given by the values of 120582 (0246

0338 0954) the demand values given by the values(0171 0084 0329) of 120583 representing demand fractionsat those points according to (22) In the worksheet we canalso read besides the specific variable values used for thecalculations the value of 119865

119876= 08 used meaning that a

demand of 119876119889

= 08 sdot 119876in = 08 sdot 025 = 02m3s is ex-tracted in the line By using formula (24) implemented inthe cell below 119865

119876119889 we obtain the rate of demand that must be

allocated to the upstream end to get the correct piezometrichead at the downstream endThe graph in Figure 6 representsthis situation It can be clearly observed that the use of(24) provides a calculated HGL for the considered example(mid line) that perfectly matches the right end of the linerepresenting the real HGL (lower polygonal) On the otherhand the application of the 50 rule (upper HGL) producesunacceptable errors

24 Maximum Head Discrepancy When Using the Pro-posed Formula As mentioned before allocation of in-linedemands to the end nodes of a line may be of great interestin order to reduce the size of the mathematical model of aWDN In the previous section we have given formulae toobtain allocation values that zero the piezometric head errorat the downstream end of the line a mandatory conditionfor correct calculation on the line(s) connected to this endHowever this reduction of the model size is at the price ofmaking some piezometric head errors at the inner points ofthe line The engineer analyzing a WDN should be awareof the magnitude of those discrepancies in order to have abetter representation and understanding of the problem Inthis section we answer this question by solving problem (8)


0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula

H0 Qin L f D K 120582 120582 HC (50) HC formula

120582 120582k+1 minus 120582k 120583 Σ120583

FQ lowast Σ120583 (120582k+1 minus 120582k)(1 minus FQ lowast Σ120583)2

FQ Qd

025 500 0018 03 061205079100

FQ119889

Figure 6 Calculation of 119865119876119889

for a nonuniform distribution of demand within a line and comparison among hydraulic grade lines

In the (ideal) case of a continuous demand along the lineit is easy to show that (8) reduces to find the point 119909 wherethe derivatives with respect to 119909 of

119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)

coincideLet us observe that 119867

119877(119865119876 119909) is a differentiable function

since 119876(119909) is continuous and that 119867119862(119865119876119889

119909) is a linearfunction of 119909 Thus it is easily seen that the maximumdiscrepancy is located at the point 119909 given by

119902 (119909) = 119865119876119889

(27)

Example 4 In the case of uniform continuous demand thesolution of (8) is given by

119909infin

= 119871119865infin

119876119889 (28)

obtained using 119902(119909) = 119909119871 as in Example 1 and (10)

In the (real) case of a discrete demand along the line themaximum discrepancy occurs at one of the points 119909

119896 since

the real HGL is a decreasing concave upwards polygonal (seeFigure 6) Then the problem reduces to identify the first 119909

1198960

for which the next section of the polygonal has a slope equalto or lower than the slope of119867

119862(119865119876119889

119909) in absolute values (ifequal all the points between 119909

1198960and 119909

1198960+1will provide the

maximum since the mentioned section of the polygonal and

119867119862(119865119876119889

119909)will run parallel between both points) So we haveto solve the following problem

Find the first point 1199091198960such that

119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)

This inequality may be simplified and written in terms of120583119894 as

(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889

That is to say the problem may be rewritten as follows

Find the first point 1199091198960

such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)

Example 5 In the case of uniform demand on 119899 equallyspaced points this gives the following

Find the first point1199091198960such that

1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)


Observe that lim119899rarrinfin

(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin

119876119889 We also have that lim

119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin

119876119889 But obviously these two limits are the

same As a result one has from (28) that

lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960

is the point where the polygonals (2)exhibit the biggest head discrepancy with the calculated headgiven by (9) corresponding to the allocation factor 119865119899

119876119889 Also

119909infin

is the point of maximum head discrepancy between theHGL corresponding to the uniform continuous demand andthe calculated head corresponding to the allocation factor119865infin

119876119889 As we could expect

lim119899rarrinfin

119909119899

1198960= 119909infin (35)

Expression (32) does not need any additional extra workin the general case if the calculations have been organizedas shown in Figure 6 In effect in this figure the first valueof the accumulated rated demand that exceeds 119865

119876119889= 0261

is sum1198960

119894=1120583119894

= 0272 which corresponds to the third innerdemand point 119909

1198960= 120582119871 with 120582 = 0604 This maximum is

119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)

3 Illustrative Example ona Real-World Network

This section is aimed to present some significant resultsobtained when applying the lessons learned in the previoussection to a practical situation The assessment is carried outin a real water supply network from Tegucigalpa City (thecapital of Honduras) The network which is intended to be adistrict metered area is supplied by onemain pipe connectedto one of the water tanks administered by the local waterauthority (SANAA)

The network has 203 nodes and 211 pipes Figure 7 showsthe network model in EPANET To illustrate the aspectspreviously studied three scenarios have been put forwardFor each scenario intermediate demand nodes (mid nodes)are placed in the main pipe In each case the demand valuesloaded in the intermediates nodes the distribution along theline or the pipe diameter vary as it is indicated as follows

Case A As it may be seen in Figure 8 an internal demandrepresenting 60 of the total inflow in the main line is fixedin two nodes placed along the line The first one loaded with25 of the total internal demand is located at 50 of the totallengthThe second intermediate node is located at 80 of thetotal length and the remaining 75 of the internal demand isallocated to it

Case B As shown in Figure 9 one single node is placed alongthe main line at 50 of its total length As it can be expected100 of the total internal demand is allocated to it In thiscase the internal demand represents 80 of the total inflow

Case C This case is fairly similar to Case B However in thiscase the diameter of the main line is reduced from 600mmto 450mm

31 Result (Case A) Figure 8 clearly reflects the character-istics and the results obtained for Case A First a hydraulicsimulation was conducted with the internal demand loadedin the mid nodes For that (real) case the pressure in nodeB (downstream node) is 4235 water column meters (wcm)When the internal demand is reallocated to the upstreamand downstream nodes following the 50 rule the resultvaries up to 122 wcm in comparison with the initial caseThrough the application of formula (24) a more suiteddemand allocation rule is found (2245 for the upstreamnode and 7755 for the downstream node) When this ruleis applied the resulting pressure value is as expected exactlythe same as in the (real) casewith the internal demand locatedin the mid nodes

32 Result (Case B and Case C) Given the fact that betweenCases B (Figure 9) and C (Figure 10) there is only onedifference namely the pipe diameter the allocation rulein both cases is the same 3486 of the internal demandshould be allocated to the upstream node and the rest tothe downstream node Nevertheless the differences betweenthe initial pressure values and the pressure obtained whenapplying the 50 allocation rule are larger in Case C than inCase B since in Case C the head loss is higher than that inCase B If Case A is included in the same comparison (higherhead loss than Cases B and C) the importance of a demandallocation based on the proposed methodology over a 50-50 becomes evident

Finally to better visualize the negative impact of using thecommon 50 demand distribution rule instead of the suitedrule (24) developed in this paper resulting pressure valuesobtained in four nodes of the example network (see Figure 11)are compared for Case A first when the demand is allocatedusing the 50 rule and then when the suited rule given byformula (24) is implementedThe results of such comparisonmay be seen in Table 2

4 Conclusions

The complexity of the interaction of all the input data ina mathematical model makes it impossible to include allthe data accurately The reality is very complex [2] and theuse of simplifications in order to make the model feasibleis therefore inevitable The use of such tools should bein any case accompanied by a clear understanding of theconsequences of such assumptions It is obvious that thelower level of simplification corresponds with more complextools In this sense software packages available in the marketdevoted to the analysis design and in general the simulationof the various states in a WDN should be used with thenecessary caution

This research focuses on the study of the influence thatthe concentration of a distributed demand in a line on theline ends represents in modeling steady state conditions in


Table 2 Comparison among scenarios

Control pointScenario A1 Real

case (demand loadedin mid nodes)

Scenario A2Demand allocatedusing the suited rule

(24)

Scenario A3Demand allocatedusing the 50-50

rule

Differencesbetween A1 and

A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221

Line with internal demand

Figure 7 Water supply system used for the practical example

L = 500m

La = L lowast 05 Lb = L lowast 03 Lc = L lowast 02

Qin = 41704 lps

Head losses = 7068 mkm

QQQQnetwork = 16704 lps

Node Aallocateddemand

Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926

Mid node 1allocateddemand

Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)

Node Ballocateddemand

Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392

Figure 8 Details and results for Case A


L = 600m

La = L lowast 05 Lb = L lowast 05

QQQin = 83505 lps


Qnetwork = 16704 lps


Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocateddemand

Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557

Figure 9 Details and results for Case B

L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)

Node BNode Ballocated

demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370

Figure 10 Details and results for Case C

Control point A

Control point B

Control point C

Control point D

Figure 11 Pressure control points


WDNsThe research work using a practical approach showsthe importance of the relation between the total inflow andthe flow that is extracted due to the load of demands in theline The distribution of the consumptions along the line alsogreatly influences the validity of the model After assessingthe error thatmay derive fromusing the generalized 50 rulewe obtain the general formula (24) that enables to allocate anarbitrary inner line demand to the line ends with zero errorat the downstream end of the line In addition we providea means to calculate the associated maximum head pointdiscrepancy

In Figure 6 we have shown that our proposal is straight-forward and involves direct methods that can be easilyapplied by any user of any WDN analysis package forexample using a very simple worksheet Existing networkmodels may be revised to replace if necessary the usedallocation values with he values provided by (24) whichproduce more reliable results For a new network modelusers may make an a priori decision on implementing theright simplifications so that a suitable tradeoff betweenmodelsimplification and accuracy is obtained

This study applies only to branched networks in whichthe flow direction is predefined The results do not strictlyapply to looped networks in which the definition of upstreamand downstream nodes cannot be known a priori and evenmay vary depending on the hourly demand in the networkand on the demand pattern used Nevertheless a numberof facts make the obtained results in this paper applicableto looped networks as well with a little extra work fromthe technician in charge of the analysis In effect firstly thelong lines susceptible of the considered simplification mustbe selected this task will not represent any problem for theexpert Secondly the expert will also be able to have a clearidea about the flow direction in most of those lines In caseof doubt he or she can use an initial analysis approach overspecific lines following the 50 rule and after the analysisgo back to the design to check the actual flow direction andmake a decision about the final direction of the flow and thevalues for 119865

119876119889 Perhaps a little more of iteration will enable

to get the final design Thirdly if the expert is not happywith the obtained results for one of the lines one has theopportunity to include in the model an additional point ofthe line where the maximum head discrepancy occurs at thepoint given by condition (32) This will divide the line intotwo new lines to which the same criteria may be appliedFourthly this solution of including an interior point of oneline into the model will be also useful in case if a line is fed byboth ends

In a future work wewill consider the inclusion of variablevalues for the factor 119865

119876119889in the demand curves assigned at the

ends of the challenging lines so that variations of the flowdirection in certain lines during the day may be consideredwhen running extended period simulations (EPSs)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work has been funded as part of the Project IDAWASDPI2009-11591 by the Direccion General de Investigacion delMinisterio de Ciencia e Innovacion (Spain) with supplemen-tary support from ACOMP2011188 from the ConselleriadrsquoEducacio of the Generalitat Valenciana

References

[1] J Izquierdo R Perez and P L Iglesias ldquoMathematical modelsand methods in the water industryrdquo Mathematical and Com-puter Modelling vol 39 no 11-12 pp 1353ndash1374 2004

[2] J Izquierdo I Montalvo R Perez-Garcıa and A Matıas ldquoOnthe complexities of the design of water distribution networksrdquoMathematical Problems in Engineering vol 2012 Article ID947961 25 pages 2012

[3] D A Savic and J K Banyard Water Distribution Systems ICEPublishing London UK 2011

[4] I Montalvo Optimal design of water distribution systems byparticle swarm optimization [doctoral dissertation] UniversidadPolitecnica de Valencia Valencia Spain 2011

[5] EPA (Environmental Protection Agency) ldquoEPANETrdquo Decem-ber 2012 httpwwwepagovnrmrlwswrddwepanethtml

[6] B S Jung P F Boulos and D J Wood ldquoPitfalls of waterdistribution model skeletonization for surge analysisrdquo Journalof the AmericanWaterWorks Association vol 99 no 12 pp 87ndash98 2007

[7] J Izquierdo I Montalvo R Perez-Garcıa and F J IzquierdoldquoSon validas ciertas simplificaciones en el estudio de los tran-sitorios hidraulicosrdquo in X Seminario Iberoamericano de Planifi-cacion Proyecto y Operacion de Sistemas de Abastecimiento deAgua (SEREA rsquo11) Morelia Mexico 2011

[8] E J Anderson and K H Al-Jamal ldquoHydraulic-network simpli-ficationrdquo Journal of Water Resources Planning andManagementvol 121 no 3 pp 235ndash244 1995

[9] B Ulanicki A Zehnpfund and F Martınez ldquoSimplificationof water distribution network modelsrdquo in Proceedings of the2nd International Conference onHydroinformatics pp 493ndash500International Water Association Zurich Switzerland 1996

[10] W M Grayman and H Rhee ldquoAssessment of skeletonizationin network modelsrdquo in Proceedings of the Joint Conferenceon Water Resource Engineering and Water Resources Planningand Management R H Hotchkiss and M Glade Eds ASCEAugust 2000

[11] TMWalski D V Chase D A SavicWGrayman S Beckwithand E Koelle Advanced Water Distribution Modeling andManagement Haestad Methods Waterbury Conn USA 2003

[12] L Perelman and A Ostfeld ldquoWater distribution system aggre-gation for water quality analysisrdquo Journal of Water ResourcesPlanning and Management vol 134 no 3 pp 303ndash309 2008

[13] J W Deuerlein ldquoDecomposition model of a general watersupply network graphrdquo Journal of Hydraulic Engineering vol134 no 6 pp 822ndash832 2008

[14] WM Grayman RMurray andD A Savic ldquoEffects of redesignof water systems for security and water quality factorsrdquo inProceedings of the World Environmental and Water ResourcesCongress S Starrett Ed pp 1ndash11 ASCE May 2009

[15] L Perelman and A Ostfeld ldquoWater-distribution systems sim-plifications through clusteringrdquo Journal of Water ResourcesPlanning and Management vol 138 no 3 pp 218ndash229 2012


[16] MHerrera J Izquierdo R Perez-Garcıa andDAyala-CabreraldquoWater supply clusters by multi-agent based approachrdquo inProceedings of the 12th Annual International Conference onWater Distribution Systems Analysis (WDSA rsquo10) pp 861ndash869ASCE September 2010

[17] M Herrera J Izquierdo R Perez-Garcıa and I MontalvoldquoMulti-agent adaptive boosting on semi-supervised water sup-ply clustersrdquo Advances in Engineering Software vol 50 pp 131ndash136 2012

[18] A Di Nardo and M Di Natale ldquoA heuristic design supportmethodology based on graph theory for district metering ofwater supply networksrdquo Engineering Optimization vol 43 no2 pp 193ndash211 2011

[19] O Giustolisi D Laucelli L Berardi and D A Savic ldquoCom-putationally efficient modeling method for large water networkanalysisrdquo Journal of Hydraulic Engineering vol 138 no 4 pp313ndash326 2012

[20] E Todini and S Pilati ldquoA gradient algorithm for the analysisof pipe networksrdquo in Computer Applications in Water Supply BCoulbeck and C H Orr Eds vol 1 pp 1ndash20 Research StudiesPress Taunton UK 1988

[21] W Rudin Principles of Mathematical Analysis InternationalSeries in Pure and Applied Mathematics McGraw-Hill NewYork NY USA 3rd edition 1976

[22] R G Bartle and D R Sherbert Introduction to Real AnalysisJohn Wiley amp Sons Ontario Canada 3rd edition 2000

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


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Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


a single emitter leakage exponent for the entire network Thelast two simplifications in particular are inescapable whenworking with EPANET [5] a tool of water network analysisin general use worldwide Also some simplifications cantotally undermine studies of hydraulic transient phenomenain WDNs See for example [6 7]

Among the various simplifications used in the analy-sis of water distribution systems some are assumed in ageneralized way without additional introspection They aresimplifications derived from the different levels of granularityat which a network can be considered As mentioned beforereal water distribution networks especially those of largecities cannot be efficiently modeled in their entirety Inapproximate models some granulation reduction actions areperformed such as skeletonization grouping pruning andclustering See among others [8ndash18]

In particular one of these simplifications is the groupingof the consumptions associated with a line in one or bothends of the line These points concentrate all the existingconsumption points (users) within the line The need forimplementing this specific simplification is evident because afaithful consideration of demand would imply the inclusionof a large number of nodes equal to the number of consump-tion points In a branched network with few consumers thiswould not represent a major problem However problemsarise in WDNs with up to 30 connections per line (eg astreet pipeline) In a large WDN (eg a 500 kmWDN) itwould amount to considering about 150000 nodes whichis impractical when it comes to the construction of thenetwork model the performance of the calculations and thedisplay and understanding of the results This simplificationcopes with the continuity principle or conservation of massHowever the energy aspects are completely ignored

The methods for calculating demand load in the modelsare one of the least studied aspects despite the obvious impor-tance for defining model reliability Recently Giustolisi et al[19] have addressed this problem from a global perspectiveand have developed a matrix transformation approach thatchanges the classical solution of the nonlinear system ofequations describing aWDNbased on the conjugate gradient[20] into what they call enhanced global gradient algorithm(EGGA) Note that the solution in [20] is implementedwithin the widely available freeware EPANET 2 used in ageneralized way by many engineers and practitioners aroundthe world EGGA reduces the size of the mathematical prob-lem through a transformed topological representation of theoriginal network model that preserves both mass and energybalance equations and improves the numerical stability ofthe solution procedure According to the authors EGGAsignificantly improves the modelrsquos computational efficiencywithout sacrificing its hydraulic accuracy

However this solution although technically impeccablefrom a theoretical point of view exhibits a number ofpractical drawbacks Firstly those matrix transformationsare not implemented within EPANET and as a result arenot available to the huge community of its users Secondlythe transformations are too complex for most users of thisprogram and especially too complex to be incorporatedinto the EPANET toolkit given the in-depth knowledge of

programming techniques that their implementation wouldrequireThirdly the transformationmatricesmust previouslybe explicitly written even though obtaining these matricesis relatively straightforward the computational efficiency atleast in terms ofmemory needs is not evident One has to useseveral additional (very sparse) matrices of sizes or the orderof the number of resulting pipes times the number of originalpipes and the number of resulting pipes times the numberof unknown head nodes are explicitly used When solvingreal-world problems with hundreds of thousands of nodesand pipes this may become a serious problem Fourthly fornetworks already modeled by excluding those intermediatedemand nodes the solution of matrix transformation will beuseless Fifthly when planning and designing a new networkstarting from the household demand distribution it would beperhaps desirable to start building the model by performingthe simplification from the outset in order to avoid latercomplications

Compelled by those drawbacks we have addressed theproblem from another we claimmore practical point of viewThis paper analyzes the possible errors from the effect of usingcertain types of simplifications when loading demands inmodels specifically the widespread 50 rule which allocateshalf of the in-line demand to each line end We analyzeto what extent this simplification is acceptable or on thecontrary supply information that differs substantially fromreality Also we obtain formulae that enable to allocateinner line demand to the line ends with minimal errorFinally a calculation of themaximumhead point discrepancyassociated is provided

Our proposal involves simple direct methods that can beeasily applied by any user of anyWDNanalysis package sinceemphasis is not placed on programming ability but on howto make a decision about the technical aspect of simplifyingthemodel and thus load the demand properly Users havingalready developed models of their networks may revise theallocation rule used and replace it if necessary with thevalues provided by the new formulae what will enable themto obtain more reliable results Also users starting the modelof a new networkmaymake an a priori decision about how tosimplify the network and suitably implement the associatedsimplifications

We first present a simple case that enables us to shed lightonto the problem a single line with variable distribution andgranulation consumption is consideredThen an example of areal network is analyzed using the lessons learnedThe papercloses with a conclusions section

2 Line with Associated Demand

Let us first consider the case of a single line associated withsome internal consumption under steady state conditionSuch a line is representative of the simplest installation(a line between two nodes) with a given inflow rate Thecharacteristics of the line are

(i) length 119871

(ii) diameter119863



0


21198635)





119876 0 lt 119865

119876le 1 the


119876119889= 119865119876119876in (1)








0) the other


(119895119871

119899 + 11198670minus 119870

119871

119899 + 11198762

in

119895

sum

119896=1


119899119865119876)

2

)

119895 = 1 119899 + 1

(2)




Δ119867 = minus119870(119876in minus119876119889

119871119909)

2

Δ119909 (3)



119867119877(119865119876 119909) = 119867

0minus 119870

119871

3

1198762

in119865119876

[1 minus (1 minus 119865119876

119909

119871)

3

] (4)










119876119889



Note that using (1)

119876119897= 119876in (1 minus 119865

119876119865119876119889

) (5)


119865119876119889

is

119867119862(119865119876119889

119909) = 1198670minus 119870119876

2

119897119909 (6)



119862(119865119876119889






119876and





problem

Minimize119865119876119889

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (7)


FQ = 1 FQ = 08 FQ = 06 FQ = 04

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

340 250 500 0 250 500 0 250 500 0 250 500

Inf1911

731

Inf1911

731

Inf1911

731

Inf1911

731


FQ = 1 FQ = 0850

49

48

47

46

45

44

43

42

41

40

50

49

48

47

46

45

44

43

42

41

400 100 200 300 400 500 0 100 200 300 400 500

LumpInf1911

731

LumpInf1911

731




119877(119865119876lowast 119909) accounts



119877



Table 1 Values of 119865119876119889


119899 (order) 119865119876

02 04 06 08 101 (1) 0472 0438 0397 0349 02933 (2) 0485 0466 0442 0413 03767 (3) 0488 0473 0455 0432 040211 (4) 0489 0475 0457 0435 040719 (5) 0490 0477 0461 0441 0415inf (6) 0491 0479 0465 0446 0423


Maximize119909

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (8)



119862and 119867

119877 must


Solve 119867119862(119865119876119889

119871) = 119867119877(119865119876lowast 119871) for 119865

119876119889 (9)


areobtained


119865infin

119876119889(119865119876) =

1

119865119876


119876)3

3119865119876

) (10)


119865(119899)

119876119889(119865119876) =

1

119865119876

(1 minus radic1

119899 + 1

119899+1

sum

119896=1


119899119865119876)

2

) (11)







119876


0204

0608

1

03

035

04

045

05

123456



0406


(ii) The values of 119865119876119889


(iii) The lowest 119865119876119889


(iv) The highest 119865119876119889


119865119876rarr0119865(sdot)

119876119889(119865119876) = 05)


119865119899

119876119889= 119865infin

119876119889for all


uous functions 119865119899119876119889

infin
















119867119877(119865119876 119902 (119909) 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (12)




using (5) as

119867119862(119865119876119889

119871) = 1198670minus 119870119871119876

2

in(1 minus 119865119876119889

119865119876)2

(13)


minus1198701198711198762

in(1 minus 119865119876119889

119865119876)2

= minus119870int

119871

0

(119876in minus 119876 (119909))2d119909 (14)

By substituting

119876 (119909) = 119876in119865119876119902 (119909) (15)

one gets

119871(1 minus 119865119876119889

119865119876)2

= int

119871

0

(1 minus 119865119876119902 (119909))

2d119909 (16)


119865119876119889

(119865119876) =

1

119865119876

(1 minus radic1

119871int

119871

0

(1 minus 119865119876119902 (119909))

2d119909) (17)



119902 (119909) = 120575 (119909 minus 1199091) (18)

109080706050403020100 02 04 06 08 1

FQ = 02

FQ = 04

FQ = 06

FQ = 08

FQ = 1




119867119877(119865119876 120575 (119909 minus 119909

1) 119871)

= 1198670minus 119870int

119871

0

[119876in minus 119876119889120575 (119906 minus 119906

1)]2d119906

= 1198670minus 119870int

1199091

0

1198762


119871

1199091

(119876in minus 119876119889)2d119906

= 1198670minus 11987011990911198762


119889)2

= 1198670minus 119870119871119876

2

in (120582 + (1 minus 120582) (1 minus 119865119876)2)

(19)



119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582) =

1

119865119876




119876 namely 10 08 06 04 and



119876=





100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509





(120582 119865119876119889



119896=1119889119896120575(119909 minus 119909

119896)


119896 with 0 lt 119909

1lt 1199092lt



119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case



119867119877(119865119876 119871)

= 1198670minus 11987011990911198762


1)2


1+ 1198892))2


)(119876in minus

119899minus1

sum

119896=1

119889119896)

2


119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)


119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)


119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)





demand of 119876119889






0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











0


21198635)





119876 0 lt 119865

119876le 1 the


119876119889= 119865119876119876in (1)








0) the other


(119895119871

119899 + 11198670minus 119870

119871

119899 + 11198762

in

119895

sum

119896=1


119899119865119876)

2

)

119895 = 1 119899 + 1

(2)




Δ119867 = minus119870(119876in minus119876119889

119871119909)

2

Δ119909 (3)



119867119877(119865119876 119909) = 119867

0minus 119870

119871

3

1198762

in119865119876

[1 minus (1 minus 119865119876

119909

119871)

3

] (4)










119876119889



Note that using (1)

119876119897= 119876in (1 minus 119865

119876119865119876119889

) (5)


119865119876119889

is

119867119862(119865119876119889

119909) = 1198670minus 119870119876

2

119897119909 (6)



119862(119865119876119889






119876and





problem

Minimize119865119876119889

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (7)


FQ = 1 FQ = 08 FQ = 06 FQ = 04

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

340 250 500 0 250 500 0 250 500 0 250 500

Inf1911

731

Inf1911

731

Inf1911

731

Inf1911

731


FQ = 1 FQ = 0850

49

48

47

46

45

44

43

42

41

40

50

49

48

47

46

45

44

43

42

41

400 100 200 300 400 500 0 100 200 300 400 500

LumpInf1911

731

LumpInf1911

731




119877(119865119876lowast 119909) accounts



119877



Table 1 Values of 119865119876119889


119899 (order) 119865119876

02 04 06 08 101 (1) 0472 0438 0397 0349 02933 (2) 0485 0466 0442 0413 03767 (3) 0488 0473 0455 0432 040211 (4) 0489 0475 0457 0435 040719 (5) 0490 0477 0461 0441 0415inf (6) 0491 0479 0465 0446 0423


Maximize119909

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (8)



119862and 119867

119877 must


Solve 119867119862(119865119876119889

119871) = 119867119877(119865119876lowast 119871) for 119865

119876119889 (9)


areobtained


119865infin

119876119889(119865119876) =

1

119865119876


119876)3

3119865119876

) (10)


119865(119899)

119876119889(119865119876) =

1

119865119876

(1 minus radic1

119899 + 1

119899+1

sum

119896=1


119899119865119876)

2

) (11)







119876


0204

0608

1

03

035

04

045

05

123456



0406


(ii) The values of 119865119876119889


(iii) The lowest 119865119876119889


(iv) The highest 119865119876119889


119865119876rarr0119865(sdot)

119876119889(119865119876) = 05)


119865119899

119876119889= 119865infin

119876119889for all


uous functions 119865119899119876119889

infin
















119867119877(119865119876 119902 (119909) 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (12)




using (5) as

119867119862(119865119876119889

119871) = 1198670minus 119870119871119876

2

in(1 minus 119865119876119889

119865119876)2

(13)


minus1198701198711198762

in(1 minus 119865119876119889

119865119876)2

= minus119870int

119871

0

(119876in minus 119876 (119909))2d119909 (14)

By substituting

119876 (119909) = 119876in119865119876119902 (119909) (15)

one gets

119871(1 minus 119865119876119889

119865119876)2

= int

119871

0

(1 minus 119865119876119902 (119909))

2d119909 (16)


119865119876119889

(119865119876) =

1

119865119876

(1 minus radic1

119871int

119871

0

(1 minus 119865119876119902 (119909))

2d119909) (17)



119902 (119909) = 120575 (119909 minus 1199091) (18)

109080706050403020100 02 04 06 08 1

FQ = 02

FQ = 04

FQ = 06

FQ = 08

FQ = 1




119867119877(119865119876 120575 (119909 minus 119909

1) 119871)

= 1198670minus 119870int

119871

0

[119876in minus 119876119889120575 (119906 minus 119906

1)]2d119906

= 1198670minus 119870int

1199091

0

1198762


119871

1199091

(119876in minus 119876119889)2d119906

= 1198670minus 11987011990911198762


119889)2

= 1198670minus 119870119871119876

2

in (120582 + (1 minus 120582) (1 minus 119865119876)2)

(19)



119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582) =

1

119865119876




119876 namely 10 08 06 04 and



119876=





100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509





(120582 119865119876119889



119896=1119889119896120575(119909 minus 119909

119896)


119896 with 0 lt 119909

1lt 1199092lt



119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case



119867119877(119865119876 119871)

= 1198670minus 11987011990911198762


1)2


1+ 1198892))2


)(119876in minus

119899minus1

sum

119896=1

119889119896)

2


119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)


119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)


119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)





demand of 119876119889






0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










FQ = 1 FQ = 08 FQ = 06 FQ = 04

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

34

50

48

46

44

42

40

38

36

340 250 500 0 250 500 0 250 500 0 250 500

Inf1911

731

Inf1911

731

Inf1911

731

Inf1911

731


FQ = 1 FQ = 0850

49

48

47

46

45

44

43

42

41

40

50

49

48

47

46

45

44

43

42

41

400 100 200 300 400 500 0 100 200 300 400 500

LumpInf1911

731

LumpInf1911

731




119877(119865119876lowast 119909) accounts



119877



Table 1 Values of 119865119876119889


119899 (order) 119865119876

02 04 06 08 101 (1) 0472 0438 0397 0349 02933 (2) 0485 0466 0442 0413 03767 (3) 0488 0473 0455 0432 040211 (4) 0489 0475 0457 0435 040719 (5) 0490 0477 0461 0441 0415inf (6) 0491 0479 0465 0446 0423


Maximize119909

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (8)



119862and 119867

119877 must


Solve 119867119862(119865119876119889

119871) = 119867119877(119865119876lowast 119871) for 119865

119876119889 (9)


areobtained


119865infin

119876119889(119865119876) =

1

119865119876


119876)3

3119865119876

) (10)


119865(119899)

119876119889(119865119876) =

1

119865119876

(1 minus radic1

119899 + 1

119899+1

sum

119896=1


119899119865119876)

2

) (11)







119876


0204

0608

1

03

035

04

045

05

123456



0406


(ii) The values of 119865119876119889


(iii) The lowest 119865119876119889


(iv) The highest 119865119876119889


119865119876rarr0119865(sdot)

119876119889(119865119876) = 05)


119865119899

119876119889= 119865infin

119876119889for all


uous functions 119865119899119876119889

infin
















119867119877(119865119876 119902 (119909) 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (12)




using (5) as

119867119862(119865119876119889

119871) = 1198670minus 119870119871119876

2

in(1 minus 119865119876119889

119865119876)2

(13)


minus1198701198711198762

in(1 minus 119865119876119889

119865119876)2

= minus119870int

119871

0

(119876in minus 119876 (119909))2d119909 (14)

By substituting

119876 (119909) = 119876in119865119876119902 (119909) (15)

one gets

119871(1 minus 119865119876119889

119865119876)2

= int

119871

0

(1 minus 119865119876119902 (119909))

2d119909 (16)


119865119876119889

(119865119876) =

1

119865119876

(1 minus radic1

119871int

119871

0

(1 minus 119865119876119902 (119909))

2d119909) (17)



119902 (119909) = 120575 (119909 minus 1199091) (18)

109080706050403020100 02 04 06 08 1

FQ = 02

FQ = 04

FQ = 06

FQ = 08

FQ = 1




119867119877(119865119876 120575 (119909 minus 119909

1) 119871)

= 1198670minus 119870int

119871

0

[119876in minus 119876119889120575 (119906 minus 119906

1)]2d119906

= 1198670minus 119870int

1199091

0

1198762


119871

1199091

(119876in minus 119876119889)2d119906

= 1198670minus 11987011990911198762


119889)2

= 1198670minus 119870119871119876

2

in (120582 + (1 minus 120582) (1 minus 119865119876)2)

(19)



119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582) =

1

119865119876




119876 namely 10 08 06 04 and



119876=





100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509





(120582 119865119876119889



119896=1119889119896120575(119909 minus 119909

119896)


119896 with 0 lt 119909

1lt 1199092lt



119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case



119867119877(119865119876 119871)

= 1198670minus 11987011990911198762


1)2


1+ 1198892))2


)(119876in minus

119899minus1

sum

119896=1

119889119896)

2


119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)


119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)


119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)





demand of 119876119889






0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Table 1 Values of 119865119876119889


119899 (order) 119865119876

02 04 06 08 101 (1) 0472 0438 0397 0349 02933 (2) 0485 0466 0442 0413 03767 (3) 0488 0473 0455 0432 040211 (4) 0489 0475 0457 0435 040719 (5) 0490 0477 0461 0441 0415inf (6) 0491 0479 0465 0446 0423


Maximize119909

10038171003817100381710038171003817119867119862(119865119876119889

119909) minus 119867119877(119865119876lowast 119909)

10038171003817100381710038171003817 (8)



119862and 119867

119877 must


Solve 119867119862(119865119876119889

119871) = 119867119877(119865119876lowast 119871) for 119865

119876119889 (9)


areobtained


119865infin

119876119889(119865119876) =

1

119865119876


119876)3

3119865119876

) (10)


119865(119899)

119876119889(119865119876) =

1

119865119876

(1 minus radic1

119899 + 1

119899+1

sum

119896=1


119899119865119876)

2

) (11)







119876


0204

0608

1

03

035

04

045

05

123456



0406


(ii) The values of 119865119876119889


(iii) The lowest 119865119876119889


(iv) The highest 119865119876119889


119865119876rarr0119865(sdot)

119876119889(119865119876) = 05)


119865119899

119876119889= 119865infin

119876119889for all


uous functions 119865119899119876119889

infin
















119867119877(119865119876 119902 (119909) 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (12)




using (5) as

119867119862(119865119876119889

119871) = 1198670minus 119870119871119876

2

in(1 minus 119865119876119889

119865119876)2

(13)


minus1198701198711198762

in(1 minus 119865119876119889

119865119876)2

= minus119870int

119871

0

(119876in minus 119876 (119909))2d119909 (14)

By substituting

119876 (119909) = 119876in119865119876119902 (119909) (15)

one gets

119871(1 minus 119865119876119889

119865119876)2

= int

119871

0

(1 minus 119865119876119902 (119909))

2d119909 (16)


119865119876119889

(119865119876) =

1

119865119876

(1 minus radic1

119871int

119871

0

(1 minus 119865119876119902 (119909))

2d119909) (17)



119902 (119909) = 120575 (119909 minus 1199091) (18)

109080706050403020100 02 04 06 08 1

FQ = 02

FQ = 04

FQ = 06

FQ = 08

FQ = 1




119867119877(119865119876 120575 (119909 minus 119909

1) 119871)

= 1198670minus 119870int

119871

0

[119876in minus 119876119889120575 (119906 minus 119906

1)]2d119906

= 1198670minus 119870int

1199091

0

1198762


119871

1199091

(119876in minus 119876119889)2d119906

= 1198670minus 11987011990911198762


119889)2

= 1198670minus 119870119871119876

2

in (120582 + (1 minus 120582) (1 minus 119865119876)2)

(19)



119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582) =

1

119865119876




119876 namely 10 08 06 04 and



119876=





100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509





(120582 119865119876119889



119896=1119889119896120575(119909 minus 119909

119896)


119896 with 0 lt 119909

1lt 1199092lt



119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case



119867119877(119865119876 119871)

= 1198670minus 11987011990911198762


1)2


1+ 1198892))2


)(119876in minus

119899minus1

sum

119896=1

119889119896)

2


119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)


119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)


119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)





demand of 119876119889






0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra


















119867119877(119865119876 119902 (119909) 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (12)




using (5) as

119867119862(119865119876119889

119871) = 1198670minus 119870119871119876

2

in(1 minus 119865119876119889

119865119876)2

(13)


minus1198701198711198762

in(1 minus 119865119876119889

119865119876)2

= minus119870int

119871

0

(119876in minus 119876 (119909))2d119909 (14)

By substituting

119876 (119909) = 119876in119865119876119902 (119909) (15)

one gets

119871(1 minus 119865119876119889

119865119876)2

= int

119871

0

(1 minus 119865119876119902 (119909))

2d119909 (16)


119865119876119889

(119865119876) =

1

119865119876

(1 minus radic1

119871int

119871

0

(1 minus 119865119876119902 (119909))

2d119909) (17)



119902 (119909) = 120575 (119909 minus 1199091) (18)

109080706050403020100 02 04 06 08 1

FQ = 02

FQ = 04

FQ = 06

FQ = 08

FQ = 1




119867119877(119865119876 120575 (119909 minus 119909

1) 119871)

= 1198670minus 119870int

119871

0

[119876in minus 119876119889120575 (119906 minus 119906

1)]2d119906

= 1198670minus 119870int

1199091

0

1198762


119871

1199091

(119876in minus 119876119889)2d119906

= 1198670minus 11987011990911198762


119889)2

= 1198670minus 119870119871119876

2

in (120582 + (1 minus 120582) (1 minus 119865119876)2)

(19)



119862= 119867119877at 119871 gives

119865119876119889

(119865119876 120582) =

1

119865119876




119876 namely 10 08 06 04 and



119876=





100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509





(120582 119865119876119889



119896=1119889119896120575(119909 minus 119909

119896)


119896 with 0 lt 119909

1lt 1199092lt



119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case



119867119877(119865119876 119871)

= 1198670minus 11987011990911198762


1)2


1+ 1198892))2


)(119876in minus

119899minus1

sum

119896=1

119889119896)

2


119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)


119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)


119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)





demand of 119876119889






0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










100

98

96

94

92

90

88

86

84

82

800 100 200 300 400 500

5001025

0509





(120582 119865119876119889



119896=1119889119896120575(119909 minus 119909

119896)


119896 with 0 lt 119909

1lt 1199092lt



119896=1119889119896= 119876119889

Example 3 119902(119909) = (1119899)sum119899

119896=1120575(119909 minus (119896(119899 + 1))119871) in the case



119867119877(119865119876 119871)

= 1198670minus 11987011990911198762


1)2


1+ 1198892))2


)(119876in minus

119899minus1

sum

119896=1

119889119896)

2


119899

sum

119896=1

119889119896)

2

(21)

By denoting

120583119894=

119889119894

119876119889

1205830= 0 120582

119894=

119909119894

119871 1205820= 0

120582119899+1

= 1 for 119894 = 1 119899

(22)


119867119877(119865119876 119871) = 119867

0minus 119870119871119876

2

in

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

(23)


119865119876119889

(119865119876 120582 120583)

=1

119865119876

(1 minus radic

119899

sum

119896=0

(120582119896+1

minus 120582119896)(1 minus 119865

119876

119896

sum

119895=0

120583119895)

2

)

(24)





demand of 119876119889






0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0000 100000 0000 100000 1000000246 95297 1000 93114 880250338 93979

08 02 0604 907630688 897750797 88705

0000 0000 0000 0954 880600246 0246 0171 0171 1000 880250338 0093 0084 02560604 0266 0017 02720688 0084 0078 03510797 0108 0321 06710954 0157 0329 10001000 0046

00000137 02460204 00690218 01680281 00520537 00560800 0034

00020626

026188000

90000

92000

94000

96000

98000

100000

0000 0200 0400 0600 0800 1000

H real

H real

HC (50)HC formula


120582 120582k+1 minus 120582k 120583 Σ120583


FQ Qd

025 500 0018 03 061205079100

FQ119889




119867119877(119865119876 119909) = 119867

0minus 119870int

119909

0

(119876in minus 119876 (119906))2d119906 (25)

119867119862(119865119876119889

119909) = 1198670minus 119870119909119876

2

in(1 minus 119865119876119889

119865119876)2 (26)





119902 (119909) = 119865119876119889

(27)


119909infin

= 119871119865infin

119876119889 (28)



119896 since


1198960


119862(119865119876119889


1198960and 119909



119867119862(119865119876119889



119870(119876in minus (

1198960

sum

119894=1

119889119894))

2

le 1198701198762

in(1 minus 119865119876119889

119865119876)2

(29)


(1 minus 119865119876(

1198960

sum

119894=1

120583119894))

2

le (1 minus 119865119876119889

119865119876)2

(30)

which gives

1198960

sum

119894=1

120583119894ge 119865119876119889

(31)

in terms of 120583119894= 119889119894119876119889



such that1198960

sum

119894=1

120583119894ge 119865119876119889

(32)



1198960

sum

119894=1

1

119899=

1198960

119899=

119909119899

1198960

119871ge 119865119899

119876119889

(33)



(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(119909119899

1198960119871) ge lim

119899rarrinfin119865119899

119876119889= 119865infin

119876119889 since

119865119899

119876119889converges to 119865

infin


119899rarrinfin(119909119899

1198960minus1119871) le

lim119899rarrinfin

119865119899

119876119889= 119865infin



lim119899rarrinfin

119909119899

1198960

119871= 119865infin

119876119889=

119909infin

119871 (34)

Observe that 1199091198991198960


119876119889 Also

119909infin



lim119899rarrinfin

119909119899

1198960= 119909infin (35)


119876119889= 0261

is sum1198960

119894=1120583119894



119867119862(0261 119909

1198960) minus 119867119877(08 119909

1198960) = 92767 minus 90763 = 2003

(36)










4 Conclusions








(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














(24)


rule


A2


A3

A 4135 4135 5356 0 1221B 2072 2072 3293 0 1221C 5339 5339 6560 0 1221D 327 3270 4491 0 1221



L = 500m


Qin = 41704 lps




Q50 = 12500 lps(5000)

Qcal = 5614 lps(2245)

Pressurenode A(wcm)

7926

7926

7926


Qmn1 = 625 lps(25)


Qmn2 = 1875 lps(75)


Q50 = 12500 lps(5000)

Qcal = 19386 lps(7755)

Pressurenode B

BA

(wcm)

4392

5613

4392



L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










L = 600m


QQQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926


Qmn = 6682 lps(100)


Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressurenode B(wcm)

7557

7671

7557


L = 600m


Q QQin = 83505 lps




Q50 = 33408 lps(5000)

Qcal = 23290 lps(3486)

Pressurenode A(wcm)

7926

7926

7926

Mid nodeallocated

flow

Qmn = 6682 lps(100)


demand

Q50 = 33408 lps(5000)

Qcal = 43526 lps(6514)

Pressure

BA

(wcm)

6370

6849

6370


Control point A

Control point B

Control point C

Control point D













Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra




















Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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