optimal design of branched gravity water

Thesis N

OP

IN

D

No. 064/MS

PTIMAL

PARTIAL MAST

IN

DEPATRM

SE/R/301/153

L DESIG

FULFILLMTER OF SC

TRIBHU

NSTITUTE

PULCH

MENT O

GN OF BSUPPL

ABHIS

MENT OF TCIENCE IN

L

UVAN UNIV

E OF ENG

HOWK CA

OF CIVIL

BRANCLY NETW

BY

SEK BAS

THE REQUN ENVIRON

April, 2010alitpur, Nep

VERSITY

GINEERIN

AMPUS

L ENGIN

HED GRWORK

SNYAT

UIREMENTNMENTAL

0 pal

NG

NEERIN

RAVITY

T FOR THEL ENGINEE

NG

Y WATE

E DEGREE ERING

ER

OF

ii

Master of Science Thesis (Thesis No. 64/MSE/R/301/153)

OPTIMAL DESIGN OF BRANCHED GRAVITY WATER SUPPLY NETWORK

By

ABHISEK BASNYAT

A Thesis submitted in partial fulfilment of the requirement for the degree of Master of Science in Environmental Engineering

Examination Committee:

Prof. Dr. Bhagwan Ratna Kansakar Chairman / Supervisor

Assoc. Prof. Iswar Man Amatya Co-ordinator / Supervisor

Mr. Chandra Lal Nakarmi External Examiner

Tribhuvan University

Institute of Engineering, Pulchowk Campus

Department of Civil Engineering

April, 2010 Lalitpur, Nepal

iii

CERTIFICATE

This is to certify that this thesis work entitled “Optimal Design of Branched Gravity Water

Supply Network” submitted by Mr. Abhisek Basnyat is a bonafide thesis work carried out

under our supervision and guidance and fulfilling the nature and standard required for the

partial fulfilment of the degree of Master of Science in Environmental Engineering. The work

embodied in this thesis has not been submitted elsewhere for a degree.

(Prof. Dr. Bhagwan Ratna Kansakar) Supervisor Institute of Engineering Pulchowk Campus

(Assoc. Prof. Iswar Man Amatya) Supervisor Institute of Engineering Pulchowk Campus

iv

ACKNOWLEDGEMENT

I wish to express my deep sense of gratitude and sincere thanks to Prof. Dr. Bhagwan

Ratna Kansakar and Associate Prof. Iswar Man Amatya for their excellent guidance, constant

inspiration, and all round assistance throughout this thesis work.

v

ABSTRACT

In this dissertation application of optimal design approach in Branched Gravity Water Supply

Network is studied. The method used for the optimal design in the study is Linear

Programming Method. The study has tried to develop linear programming models that could

be used in water supply network of rural hilly community of developing country like Nepal.

The models developed can be used in designing a similar water supply network.

Large number of population of Nepal is located in hilly area of the country and providing safe

water to them is big challenge due to limited resources. The result of the study shows that it is

relevant to use optimal design approach so that resources saved while providing water to one

community can be used to fulfil the demand of another community.

The study also showed that linear programming method is a simple but reliable tool which

can be used effectively for the optimal design of small gravity water supply project having

limited low budget.

vi

TABLES OF CONTENTS

Chapter Title Page

Cover Page i Title Page ii Certificate iii Acknowledgement iv Abstract v Table of Content vi List of Figures viii List of Tables ix List of Abbreviations x

1.0 Introduction 1 1.1 Background 1 1.2 Rationale 1 1.3 Objective of the Study 2 1.4 Limitation of the Study 2 1.5 Organization of Report 2

2.0 Literature Review 3 2.1 Conventional Design of Pipe Network 3 2.1.1 Darcy-Weisbach Formula 3 2.1.2 Design Strategy Using Darcy-Weisbach Formula 4 2.1.3 Hazen Williams Formula 6 2.2 Optimization by Linear Programming Method 6 2.2.1 Introduction 6 2.2.2 Linear Programming 7 2.2.3 Graphical Method for Solving LP Problem 8

2.2.4 Visual Representations of Different Cases of LP problem Solutions 9

2.2.5 Simplex Method for Solving LP Problem 10 2.3 Optimization of Branched Network Using LP Method 12 2.3.1 Introduction 12 2.3.2 Optimization of Branched Network 12 2.4 Design Criteria and Considerations 13 2.4.1 Summary of Design Criteria 14 2.4.2 Summary of Design Considerations 14

vii

3.0 Methodology 16

3.1 Optimization of Branched Network in Hills 16 3.2 Model Formulations 17 3.2.1 Definitions of Term Used 17 3.2.2 Assumptions and Known Quantities 17 3.2.3 Symbols Used 18 3.2.3 Models for a Branch 18

4.0 Result and Discussions 29 4.1 Introduction 29 4.2 Design of Transmission Line 29 4.3 Design of Distribution Network 36 4.4 Hydraulic Calculation 38 4.5 Cost Comparison 43

5.0 Conclusions and Recommendations 44 5.1 Conclusions 44 5.1 Recommendations 44 References 45 Appendices

viii

LIST OF FIGURES

Figure No.

Title Page

2.1 Moody’s diagram 5 2.2 Types of solutions of linear programming problems 10 3.1 Condition Eab<hp1 19 3.2 Condition hp2>Eab>hp1,with type1 and type 2 pipes and without BPT 20 3.3 Condition hp2>Eab>hp1,with type1 pipe and single BPT 21 3.4 Condition hp3>Eab>hp2,with type 1,2 and 3 pipes and without BPT 22 3.5 Condition hp3>Eab>hp2,with two BPTs 24 3.6 Condition Eab>hp3, with single BPT 26 3.7 Ridge and valley condition 27 4.1 Ground profile of transmission line 31 4.2 Schematic diagram of the distribution network 36 4.3 Schematic diagram of optimum design of transmission line 40 4.4 Schematic diagram of optimum design of distribution network 42

ix

LIST OF TABLES

Table No.

Title Page

2.1 Value of roughness factors 4 2.2 Hazen William’s coefficients 6 2.3 Sizes and costs of commercially available HDPE pipes 15 2.4 Sizes and costs of commercially available medium grade GI pipes 15 4.1 Various types of commercially available pipes fulfilling velocity

requirements for the transmission line 30

4.2 Result of LP model solution for design of transmission line with single interruption chamber

34

4.3 Result for LP model solution for design of transmission line with two interruption chambers

34

4.4 Commercially available type 1 HDPE pipes fulfilling the velocity requirements for branch R-J1

35

4.5 Commercially available type 2 HDPE pipes fulfilling the velocity requirements for Branch J1-J8

37

4.6 Result of LP model solutions for design of distribution network 38 4.7 Design of transmission line 39 4.8 Design of distribution network 41 4.9 Cost comparison between optimal design and project report 43

x

LIST OF ABBREVIATIONS

BPTs Break Pressure Tanks CH Hazen Williams Coefficient DWSS Department of Water Supply and Sewerage GI Galvanized Iron HDPE High Density Polyethene IC Interruption Chamber LP Linear Programming Q Discharge Re Reynolds number Up Unit cost per unit length of pipe having diameter dp d Diameter ε Roughness factor f Darcy Weisbach friction factor g Acceleration due to gravity hf Head loss due to friction m Number of constraints n Number of variables v Mean velocity of flow µ Viscosity of water

1

CHAPTER I 1.0 INTRODUCTION

1.1 Background

A convenient supply of safe water is essential for healthy and productive life. Unsafe

water can spread disease. Improper location and use of components in water supply

system results in the loss of productive time and energy by the water carrier- usually

women and children. Hence, providing water to the rural communities is one of the

high priority goals in the rural development policy in the third world. The problem

facing by developing countries like Nepal is a familiar one; high demand coupled

with limited resources. Water could be provided for a few, but at the expense of the

vast majority of the population. Therefore, design of least cost water supply system is

important. However, making the rural water supply schemes least cost is proving to

be a difficult challenge, despite the fact that the schemes are small, and technically

simple. Hence, the study and development of proper approach for least cost design of

water supply system is necessary.

1.2 Rationale

In developing countries like Nepal, providing water for the people living in remote

hilly area is a big challenge due to limited economic resources. Saving every single

rupee is important; because that single rupee can be used for providing water for

another person living in another rural area.

While designing a water supply network, there could be number of possible

combinations of components, each satisfying the required hydraulic conditions.

Among this number of combinations only one will have minimum optimum cost for

the system. Optimization design using linear programming (LP) technique can be very

useful in selecting that best alternative which will give minimum optimum cost for the

system fulfilling all hydraulic requirements.

The conventional design approach of water supply network uses iterative procedure

where numbers of iterations are carried out to design the system fulfilling all the

hydraulic requirements. The optimum cost is seldom attained. Hence, the design using

the optimum technique is highly desirable where optimum cost is attained meeting all

2

the hydraulic requirements. This will conserve resources of the project which can be

utilized for the development of the community.

1.3 Objective of the Study

The overall objective of this research is to develop models of minimum cost

optimization design of branched network for rural hilly community. Accordingly, the

objectives of this research are as follows:

• To study the application of linear programming method for the optimization of

gravity water supply network in rural hilly community.

• To select appropriate Linear Programming optimization models that can be

used in a rural hilly community.

• To design an existing pipeline network using the selected LP models.

1.4 Limitations of the Study

The study carried out has following limitations:

• Optimization models considered in this study assumes a uniform slope in a

branch.

• The frictional head loss has been calculated using Hazen William’s equation

and effect of viscosity, density and temperature on frictional loss are not

considered.

1.5 Organization of Report

This report is organized into five chapters as follows:

• Chapter I briefly provide the importance of the topic, rational of the study,

objective of the study and limitations of the study.

• Chapter II deals with review of the literature related to the study.

• Chapter III details the methodology used to carry out this study.

• Chapter IV includes result and discussions.

• Chapter V Presents conclusion and recommendations for further study.

3

CHAPTER II

2.0 LITERATURE REVIEW

2.1 Conventional Design of Pipe Network

While designing pipe lines in a branched network, the discharge that each branch has

to convey should be known. Once the tap flow rates in the demand nodes are fixed,

the system flow rate automatically follows. Cumulative addition of tap flow rates to

be served by the pipe under consideration yields the system flow rate. Then a flow

diagram of the scheme is prepared indicating flow in each branch, length of each

branch and elevations of each node. (DWSS, 2002)

Once the flow, which a pipe section has to transmit, is known, its diameter should be

sized next. The basis of pipe line design is governed by the theory of flow of water

under pressure in a pipe line. Flow of water in pipe line results in loss of energy

(head) due to friction, this loss of energy or head in pipes due to friction can be

determined by using Equation 2.1 or 2.3.

2.1.1 Darcy-Weisbach formula

It is one of the most commonly used formula for determining the loss of energy or

head in pipes due to friction. According to this formula the loss of head in pipes due

to friction is given as (Modi, 2006),

hf=fLv2

2gd ..Eq 2.1a

Or, hf =flQ

12.1d5

2

..Eq 2.1b

Where,

hf = head loss , in m;

L = length of pipe, in m;

d = diameter of pipe, in m;

v = mean velocity of flow through the pipe, in m/s;

Q = discharge through the pipe, in m3/s;

g = acceleration due t gravity (= 9.81m3/s); and

f = friction factor, dimensionless.

4

2.1.2 Design strategy using Darcy-Weisbach Formula

Equation 2.1b can be rearranged as,

d5=8fLQ2

π2ghl ..Eq 2.1c

All terms in right hand side of equation 2.1c must be known to calculate diameter. Of

these maximum allowable head loss (hl) in pipe can be obtained as the difference

between maximum available head and minimum required head at demand node. The

discharge and length of pipe is known and π and g are constants. This leaves only one

unknown the friction factor. Friction factor thus should be calculated to compute the

diameter. The flow in pipes usually falls in the region of transitional turbulence. In

this type of flow the value of friction factor depends on Reynolds number, and

relative roughness (ε/d). For turbulent flow f may be obtained by following equation

given by Colebrook and White.

1f0.5 = - 2log

�3.7d

+2.51

Ref0.5 ..Eq 2.2

Where,

= roughness factor

Re = Reynolds number

Here, we can see the both sides of the equation contain f. It therefore, can only be

solved by an iterative method. However, Moody’s diagram as shown in Figure 2.1 can

be used o determine the value of f if the values of ε/d for the pipe and Re of flow are

known.

Department of Water Supply and Sewerage in its design guidelines for Community

based gravity flow (2002) has suggested following roughness factors for design

purpose, as shown in Table 2.1

Table 2.1: Values of roughness factors (DWSS, 2002)

Conduit material Roughness Factor (ε)

HDPE Pipes 0.1 mm

GI pipes and HDPE transmission mains

between a stream and sedimentation tank 1.0 mm

Density of water (ρ) = 1000kg/m3

Viscosity of water (µ) = 0.001N/m2 at 40C

5

Figu

re 2

.1: M

oody

’s d

iagr

am (S

ourc

e: h

ttp://

ww

w.m

athw

orks

.com

)

6

2.1.3 Hazen Williams Formula

This formula is widely used for designing pipes in water supply systems. According

to this formula the mean velocity (v) of flow is given as,

v = 0.849CHR0.63S0.54 ..Eq 2.3a

We can rearrange equation 2.3a in terms of head loss as,

hf = 10.7Q1.85 L

d4.87CH1.85 ..Eq 2.3b

Where,

R = Hydraulic mean depth of pipe, in m;

S = Slope of energy grade line, or head loss per unit length of pipe;

CH = Hazen Williams Roughness coefficient

The values of Hazen William’s roughness coefficients (CH) f various pipes reported

on Manual on Water Supply and Treatment, central Public Health and environmental

Engineering Organization, India (1997) is presented in Table 2.2

Table 2.2: Hazen Williams’s coefficients

Conduit material Recommended value for

New pipes Design purpose

G.I. 50mm 120 100

G.I. 50mm and below used for house service connections 120 55

Plastic pipes 150 120*

Source: (Manual, 1997)

* These pipe materials are less likely to lose their carrying capacity with age, and hence higher values

may be adopted for design purpose.

2.2 Optimization by Linear Programming Method

2.2.1 Introduction

Bhave (2003) has stated that, it may be possible to have different solutions satisfying

the requirements of an engineering problem. For example, we can use different pipe

materials, different pipe sizes and different pipe layouts for a water distribution

network. Naturally, these solutions would have different costs and the aim would be

7

to find the least costly solution. On other hand, in a water resources project it may be

possible to use water for different purposes such as domestic and industrial

consumption, irrigation, hydropower, and so on; either singly or in different

combinations. Herein, the aim may be to find solution that would give maximum

benefits. Such a solution having minimum cost or maximum benefits is termed, in

general, as an optimum solution, and the concept of obtaining optimum solution is

termed optimization.

When a physical problem is expressed mathematically or as is generally known, in the

form of mathematical model, the expression defining the objective (minimization or

maximization) is termed objective function, different conditions which the object has

to satisfy are termed constrains, and the entire problem consisting of the objective

function and constraints is termed optimization problem. Mathematically, such an

optimization problem can be expressed as,

Optimize Z= x1,x2,…………,xn ..Eq 2.4a

Subjected to,

g1 x1,x2,………,xn b1

g2 x1,x2,………,xn ≤ b2

. = . ..Eq2 .4b

. ≥ .

gm x1,x2,………,xn bm x1, x2……………,xn ≥ 0 ..Eq2 .4c

Equation 2.4a represents the objective function which involves optimization

(minimization or maximization) of Z having n decision variables x1, x2, . . . , xn.

Equation 2.4b represents a set of m constraints, expressed as equalities or inequalities

and Equation 2.4c represents the no negativity of decision variables. 2.2.2 Linear Programming

The concept of linear programming was developed after successful military

application of operation research and management science during World War II. Its

8

use provided answers to how one could minimize cost, or any desired function

achieving a given set of objectives.

Linear programming is described as being concerned with the maximization or

minimization of linear objective function containing many variables subjected to

linear equality or inequality constraints. It has become as means for planners to set

general objectives and optimize schedules to meet set goals. Linear programming has

become an important part of not just mathematics but economics, decision science

and engineering design. (Chiburzor, 2005)

Liner programming is applicable for optimization of problems when the objective

function and constraints are linear functions of the decision variables. The constraints

may be in the form of equalities or inequalities.

The Standard form of a linear programming problem as reported by (Bhave, 2003)

is as Equation 2.5,

Minimize Z=C1x1+C2x2+ ………+Cnxn ..Eq 2.5a

Subjected to,

a11x1+a12x2+ ………+a1nxn=b1

a11x1+a12x2+ ………+a1nxn=b1

. ..Eq 2.5b

.

a11x1+a12x2+ ………+a1nxn=b1

And,

x1≥0 ,x2≥0 ,……… , xn≥0 ..Eq 2.5c

In which C1,…,Cn ; b1,…,bn ; and a11,…,amn are known constants and x1,…,xn are the

decision variables.

The characteristics of the LP problem, as stated here in standard form, are (i) The

objective function is of the minimizing type; (ii) all constraints are of the equality

type; and (iii) all decision variables are non negative.

2.2.3 Graphical Method for solving LP Problem

Graphical method for solving a LP problem essentially involves indicating the

constraints on the graph and determining the feasible region. The feasible region

9

refers to the area containing all those solutions which satisfy all the constraints of the

problem and optimum value (maximum and minimum) always occurs at an extreme

point or vertex of the feasible region.

The limitation of graphical solution approach is that, it can solve only problems

containing two variables or three at most. An example of graphical method is given in

Appendix A.

2.2.4 Visual Representations of Different Cases of LP Problem Solution

D Nagesh Kumar (2010) has stated that, a linear programming problem may have i) a

unique, finite solution, ii) an unbounded solution iii) multiple (or infinite) number of

optimal solutions, iv) an infeasible solution and v) a unique feasible point. In the

context of graphical method it is easy to visually demonstrate the different situations

which may result in different types of solutions.

i) Unique, finite solution

In such cases, optimum value occurs at an extreme point or vertex of the feasible

region. The example demonstrated in Appendix A has a unique and finite solution.

ii) Multiple (infinite) solutions

If the Objective function line is parallel to any side of the feasible region all the points

lying on that side constitute optimal solutions as shown in Figure 2.2a.

iii) Unbounded solution

If the feasible region is not bounded, it is possible that the value of the objective

function goes on increasing without leaving the feasible region. This is known as

unbounded solution. Figure 2.2b shows a case of unbounded solution.

iv) An infeasible solution or no solution

Sometimes, the set of constraints does not form a feasible region at all due to

inconsistency in the constraints. In such situation the LPP is said to have infeasible

solution. Figure 2.2c illustrates such a situation.

v) Unique feasible point

This situation arises when feasible region consist of a single point. This situation may

occur only when number of constraints is at least equal to the number of decision

10

variables. An example is shown in Figure 2.2d. In this case, there is no need for

optimization as there is only one solution.

2.2.5 Simplex Method for Solving LP Problem

Introduction

J. Reeb (1998) has stated that linear programming problem consisting of two variables

or three at most can be solved by using graphical method, however in actual practice

there could be many variables in a problem and hence, graphical method fails to give

a solution. However, this can be overcome by simplex method. As with the graphical

method, the Simplex method finds the most attractive corner of the feasible region to

solve the LP problem. Simplex usually starts at the corner that represents doing

nothing (Say, origin of graph), then it moves to the neighboring corner that best

O X

Y

Objective function line

O X

Y

O X

Y

O X

Y

AB

C(a) (b)

(c) (d)

Figure 2.2: Types of solution of linear programming problem: (a) Infinite solutions; (b) Unbounded solution; (c) No solution and (d) Unique point

11

improves the solution. It does this over and over again, making the greatest possible

improvement each time. When no more improvements can be made, the most

attractive corner corresponding to the optimal solution has been found.

Converting Inequalities into Equations with Nonnegative Right Hand Side

According to Bhave (2003) if the constraints are of the inequality type, they can be

transformed to equality type with nonnegative right hand side by adding slack

variables to constraints with “≤” sign, or by subtracting surplus variables from

constraints with “≥” sign. For example,

a1x1+ ………+anxn≤b1 can be transformed to a1x1+ ………+anxn+s=b1 In which s, is

a slack variable. Similarly, a1x1+ ………+anxn≥b1 can be transformed to

a1x1+ ………+anbn-s=b1In which s, is surplus variable.

Transition from Graphical To Algebraic Solution

Kansakar (2009) has stated that, it can be seen visually why the graphical solution

space has an infinite number of solution points, but in algebraic representation the

similar conclusion is hard to imagine.

In the algebraic representation the number of equation m is always less than or equal

to the number of variables n. If m = n, and the equations are consistent, the system has

only one solution; but if m < n (which represents majority of linear programming

problems), then the system of equations, again if consistent, will yield an infinite

number of solutions. I few set n – m variables equal to zero and then solve the m

equation for remaining m variables, the resulting solution if unique, is called basic

solution and must correspond to a (feasible or infeasible) corner point of solution

space. The zero n – m variables are known as non basic variables and the remaining m

variables are called basic variables.

Rather than enumerating all the basic solutions (corner points) of the LP problem, the

simplex method investigates only a “select few” of these solutions. Normally the

simplex method starts at the origin. At this point, the value of the objective function Z

is zero, and then Simplex method moves to the neighboring corner that best improves

the solution. It does this over and over again, making the greatest possible

improvement each time. When no more improvements can be made, the most

12

attractive corner corresponding to the optimal solution has been found. An example of

simplex method is given in Appendix B.

2.3 Optimization of Branched Network Using LP Method

2.3.1 Introduction

Though the total cost of establishment of a water supply system is the summation of

the cost of all elements, a large proportion of the money is taken up by the pipeline

network. The cost of the pipe network is the function of both lengths and the

diameters of the pipes used, as the lengths and the diameters are hydraulically

interrelated. The layout which gives the minimum lengths, does not necessarily gives

the minimum diameters, and vice versa. Therefore, a true optimal network is obtained

only if both layout and diameter are optimized simultaneously.

For a given layout, the optimization problem is to find the diameters or sets of

diameters of pipes and corresponding lengths, so that total cost of the network is

minimum and all the hydraulic conditions are satisfied.

2.3.2 Optimization of Branched Network

Ghimire (1991) showed that by use of linear programming (LP) technique, a global

solution for branching distribution networks can be obtained. A constant flow part of

the network may be assumed to consist of one or more pipes of known diameters (i.e.

commercially available diameters). It is known that the head loss and the cost of the

pipes of known diameter, and with the known flows are linear function of its length.

Therefore, the constant flow part can be assumed to consist of pipe segments, of

known sizes (dp) but unknown lengths (lp). Thus lp becomes decision variable of the

LP model. With these assumptions the optimization model becomes:

Minimise ∑ Uplp ..Eq 2.6

Subjected to:

Eij - ∑ hLm ≥ Hmin,j ..Eq 2.7

Eij ≤ hpt ..Eq 2.8

Vm ≥ Vmin ..Eq 2.9

13

Vm ≤ Vmax ..Eq 2.10

lp ≥ 0 ..Eq 2.11

Where,

Up = unit cost per unit length of pipe having diameter dp

hlm = elevation difference between the points i and j

Eij = total frictional head loss in pipe segment m

Hmin,j = minimum head required at point j

hpt = maximum allowable pressure head of pipe type t

Vm = flow velocity in pipe segment m

Vmax = maximum allowable velocity

Vmin = minimum flow velocity required

By using the whole range of commercially available diameters as an input to the

model, the number of decision variables (lp) will be unnecessarily increased. The

range of dp is actually restricted to a few, by the velocity criteria. By using continuity

equation, the maximum, and the minimum velocity requirements, one may obtain;

Maximum permissible pipe diameter, dmax= 4QπVmin

..Eq 2.12a

And,

Minimum permissible pipe diameter, dmin= 4QπVmax

..Eq 2.12b

Thus, in a constant flow part, only those commercially available diameters are

necessary to consider which fall within the range of dmax and dmin.

2.4 Design Criteria and Considerations

The relevant criteria and considerations adopted in the design are summarized below.

These criteria are recommended in the Design Guidelines for Community Based

Gravity Flow, published by DWSS, 2002

14

2.4.1 Summary of Design Criteria

Minimum Flow Velocity

At intake, if no sedimentation is provided the minimum flow velocity shall be,

• In downhill stretches 0.8m/s

• In uphill stretches 1.0m/s

If sedimentation is provided the minimum flow velocity can be reduced to,

• In downhill stretches 0.4m/s

• In uphill stretches 0.5m/s

Maximum Flow Velocity

• Desirable 2.3m/s

• Maximum 3.0m/s

Residual Pressure

• At Tap stand, Ideal: 5m to 10 m

• For Break Pressure Tanks and Storage tanks 10m to 20m.

Due to the nature of the ground profile, sometimes, hydraulic grade line may fall

below the ground level at critical points. In such case negative pressure would

develop in the pipeline, which must be avoided. For this purpose it is assumed that the

hydraulic grade line should be always 5m above ground level.

2.4.2 Summary of Design Considerations

The following considerations are made in the application of the models to solve the

problems.

Types of Pipes

• Pipe type 1: HDPE pipe with pressure rating of 6 kg/cm2

• Pipe type 2: HDPE pipe with pressure rating of 10 kg/cm2

• Pipe type 3: Medium grade GI pipe with maximum allowable pressure head of

160 m.

Unit cost of BPT and Interruption chamber

• As per Cost Estimation and Design report of project under consideration

= NRs. 40,000.00.

15

The sizes of commercially available HDPE pipes and Medium Grade GI pipes along

with their cost per meter as per Cost Estimation and Design report of project under

consideration are given under Table 2.3 and 2.4 respectively.

Table 2.3: Sizes and costs of commercially available HDPE pipes

S.N. Outer Diameter (mm)

Internal Diameter (mm)

Pressure Class (kg/cm2)

Rate (NRs/meter)

1 32 26.90 6 40.23 2 40 33.70 6 62.30 3 50 42.20 6 96.48 4 63 53.30 6 151.30 5 75 63.60 6 212.00 6 90 76.30 6 305.27 7 110 93.40 6 453.01 8 16 11.60 10 16.38 9 20 14.90 10 23.85 10 25 18.90 10 35.96 11 32 24.10 10 59.45 12 40 30.30 10 91.49 13 50 38.00 10 141.69 14 63 47.80 10 225.88 15 75 57.10 10 317.20 16 90 68.50 10 457.10 17 110 83.70 10 676.58

Table 2.4: Sizes and costs of commercially available medium grade GI pipes

S.N. Diameter (mm)

Rate (NRs/meter)

1 15 103.70 2 20 133.50 3 25 201.50 4 32 255.50 5 40 292.80 6 50 409.50

16

CHAPTER III

3.0 METHODOLOGY

3.1 Optimization of Branched Network in Hills

In hilly area large elevation difference along a pipeline creates high static pressure at

some points. It is often that pressure created due to large elevation difference exceeds

the permissible pressure rating of pipes used. Hence, some means like Break Pressure

tanks (BPTs) are needed to be introduced along the pipeline to release the pressure.

Introduction of the BPTs in the network creates new constraints in the distribution

network optimization model. (Ghimire, 1991)

Pipes come in various wall thicknesses, which can withstand different amounts of

internal pressure. Depending on the internal pressure created by the elevation

difference, pipes of the same internal diameter, but of a different wall thickness, may

be used along a pipeline branch. Thicker wall pipes are costlier than the ones with

thinner diameter. This difference in cost further modifies the optimization model.

Introduction of BPTs has two effects; first it releases the internal pressure created by

elevation, and second is that as a consequence of pressure release, it is possible to use

a thinner walled pipe which is cheaper. Thus it results in a reduction in total pipe cost.

However, the tanks have their own cost. Therefore, the optimum network is only

possible from optimum combination of tanks and pipes.

The cost a branch is related to three things which are:

i. Positions and number of BPTs

ii. Lengths and diameters of pipes

iii. Pipe wall thickness ( Pipe type)

All three are interrelated and contribute to the cost of a branch. The optimal policy for

given number of BPTs is the combination of position of the BPTs, lengths and

diameters of pipes, and pipe wall thicknesses, which gives minimum cost.

17

3.2 Model Formulation

3.2.1 Definition of Terms Used

Branch: A branch is the part of a network that has constant flow.

Link: A link is considered to be a continuous part of a branch without

break by BPT.

Part: A part is defined as the portion of link which has only one type

of pipe.

Node: A node is a point where two or more link meets, or a link starts,

or a link ends. The source, the BPTs, the branching point, end

points are all nodes.

Path: A path is any continuous sequence of branches in a network. A

path may starts from a source, or from a tank where pressure is

atmospheric, and may ends at a service points, or at tanks.

3.2.2 Assumptions and Known Quantities

• The layout of the pipe network is known.

• Relationship between the head loss, the flow, and the pipe diameter is

given by the Hazen William’s Equation 2.3b,

• The pipe cost is calculated by summation of product of part length with

their respective unit cost. cost = ∑ Uplp

• The ground has a uniform slope along the pipeline in a branch.

• The flow in each branch (Q) is known.

• The ground elevation of the points along all paths, i.e., the ground profile

and the elevation difference between the points are known.

• The cost of BPT is known.

• The cost per unit length of all types and diameters of the pipes are known.

• Length of each branch is known.

18

3.2.3 Symbols Used

• Lp = Length of pipe in part p of a link

• Uptl = Unit cost of pipe of part p of type t in link l

• lptl = Length of pipe of part p of type t in link l

• hLptl = Frictional head loss in pipe of part p of type t in link l

• hp1 = Maximum pressure that type 1 pipe can withstand



• Qab = Flow in branch A-B

• Eab = Maximum Static head in branch AB

• Lab = Total length between points A and B

• Hmin,b = Minimum head required at point B

• SAB = Eab / LAB = Slope of branch AB

• Htl = Maximum head subjected to pipe type t in link l

3.2.4 Models for a Branch

In the following discussion for subsequent model formulation pipes with three

pressure ratings are considered.

In a branch one of the following conditions may exist;

i. Eab < hp1

ii. hp2 > Eab > hp1

iii. hp3 > Eab > hp2

iv. Eab > hp3

While dealing with each condition there may exist more than one optimum situation

which should be analyzed before selecting global optimum solution. The different

optimum situations for each condition are discussed below.

Condition 1: Eab < hp1

A branch AB of a water supply network is shown in Figure 3.1, which has a length

Lab and carries a flow of Qab. The level difference between A and B is Eab, such that

Eab < hp1.

19

In such a situation, type 1 with lowest pressure rating (i.e., hp1) can be provided in the

whole length of the branch AB. Here it is not necessary to consider BPT and type 2 or

type 3 pipes in the optimization model.

If the length Lab is assumed to consist of many pipe segments of known lengths lp and

known diameters dp (The dp are the commercially available diameters within the range

of dmax and dmin and there are n numbers of diameters available within this rang.) then

the objective function become:

Minimize,

C= ∑ Up11lp11 ..Eq 3.1

And the constraints are,

∑ lp11 =Lab ..Eq 3.2

If the minimum pressure requirement at B is Hmin,b ,then the pressure condition will

be:

Eab- ∑ hLp11 ≥ Hmin,b (substituting the value of hf from Equation 2.3 b)

Or, Eab - 10.7Q1.85 ∑ lpdp

4.87cHp1.85 ≥ Hmin,b ..Eq 3.3

And, all lp ≥ 0

Condition 2: hp2 > Eab > hp1

When the condition exists, two options are available, either

i. Provide the type 1 pipe up to the point where internal pressure in the pipe

equals hp1 and provide the pipe type 2 in the remaining portion.

ii. Provide a BPT or BPTs and use pipe type 1.

EabQab

A

BLab

Figure 3.1: Condition Eab < hp1.

20

i. Model without BPT

In branch AB as shown in Figure 3.2 where Eab is maximum static head at point B and

hp2 > Eab > hp1, type 1 pipe in part 1 where maximum static head is H1, which equals

to hp1 can be provided and pipe type 2 can be provided in part 2. Hence, there is no

requirement of BPT.

Here whole branch AB can be considered as single link,

The cost of the pipes is given by:

C= ∑ ∑ Upt1lpt1 ..Eq 3.4

Up to the hydrostatic pressure of h11, type 1 pipe could be used. If the type 1 pipe is

used up to an elevation difference of H1 from A, then the condition of the internal

pressure in the type 1 pipe should not exceed h11 and can be expressed as;

H1 ≤ h11

Or, Sab ∑ Lp11 ≤ hp1 ..Eq 3.5

The minimum pressure requirement at point B can be written as;

Eab- ∑ ∑ hLpt1 ≥ Hmin,b

Or, Eab- 10.7Q1.85 ∑ ∑ lpt1

dpt14.87ct

1.85 ≥ Hmin,b ..Eq 3.6

Total length should be equal to Lab

i.e, ∑ ∑ lpt1 = Lab ..Eq 3.7

EabQab

A

B

Fig 3.2 : Using Types 1 and 2 pipes and no BPT when hp2 Eab > hp1

Lp2

Lp1

Part 1

Part 2

H1

Figure 3.2: Condition hp2 >Eab>hp1, with type 1 and type 2 pipes and without BPT

21

ii. Model for Single BPT situation

In Figure 3.3 a single BPT T1 is considered for the condition hp2 > Eab > hp1 and only

type 1 pipe is used in link 1 and link 2.

Single BPT or BPTs as the case may be, can be used without using type 2 pipe. The

number of BPTs can be obtained by integer part of Eabhp1

The provision of BPT is not a strict hydraulic requirement for this condition, since the

type 2 pipe can withstand the maximum pressure exerted in the system. In such a case

a decision whether to provide a BPT or BPTs has to be taken entirely on the cost

basis.

The optimization problem is to obtain the position of the BPT; in addition to the

lengths and diameters of the type 1 pipe.

Here number of link equals to two, therefore l=1 to 2


C= ∑ ∑ Up1llp1l ..Eq 3.8

Conditions to be satisfied;

Link 1

The maximum permissible pressure at link 1 should not exceed hp1

i.e. E1 ≤ hp1

Or, Sab ∑ lp11 ≤ hp1 ..Eq 3.9

EabQab

A

B

Fig 3.3 : Using Single BPT when hp2 Eab > hp1

Lp12

Lp11

Link 1

Link 2

E1T1

E2

Figure 3.3: Condition hp2 >Eab>hp1, with type 1 pipe and single BPT

22

The difference between available head and head loss in link should be greater or equal

to minimum head required at T1.

i.e. E1- ∑ hLp11 ≥Hmin,T1

Or, Sab ∑ lp11 - 10.7Q1.85 ∑ lp11

dp114.85C1

1.85 =Hmin,T1 ..Eq 3.10

Similarly in Link 2

Sab ∑ lp12 ≤ hp1 ..Eq 3.11

The minimum head requirement at B;

Or, Sab ∑ lp12 - 10.7Q1.85 ∑ lp12

dp124.85C1

1.85 ≥Hmin,b ..Eq 3.12

Total lengths of the pipes should be equal to the length between points A and B

∑ ∑ lp1t = Lab ..Eq 3.13

Condition 3: hp3 > Eab > hp2

When the condition exists, two options are available, either

i. Provide the type 1 pipe up to the point where internal pressure in the pipe

equals hp1 and provide the pipe type 2 up to the point where internal pressure

in the pipe equals hp2 and provide type 3 pipe in the remaining portion.

ii. Provide a BPT or BPTs and use pipe type 1 and 2.

i. Model without BPT

In Figure 3.4 there is no consideration of BPT for the condition hp3 > Eab > hp2. In the

branch AB part 1 consists of type 1 pipe upto the static head H1, part 2 consists of

type 2 pipe up to the static head H2 and part 3 consists of type 3 pipe.

Eab

Part 1

Part 2

Part 3

Qab Lp1

Lp2

Lp3

H1H2

A

B

Figure 3.4: Condition hp3 >Eab>hp2 with type 1, 2 and 3 pipes and without BPT

23

Here, l=1 and t = 1 to 3


C= ∑ ∑ Upt1lpt1 ..Eq 3.14

Up to the hydrostatic pressure of hp1, Type 1 pipe could be used.

Sab ∑ Lp11 ≤ hp1 ..Eq 3.15

Again up to the hydrostatic pressure of hp2, Type 2 pipe could be used

Or, Sab ∑ Lp11 + ∑ Lp21 ≤ hp2 ..Eq 3.16

The minimum pressure requirement at point B can be written as;

Eab- 10.7Q1.85 ∑ ∑ lpt1

dpt14.87ct

1.85 ≥Hmin,b ..Eq 3.17

Total length should be equal to Lab

∑ ∑ lpt1 = Lab ..Eq 3.18

ii. Model with BPT

The number of BPTs in a branch can be decided depending up on available head (Eab)

and maximum permissible pressure rating (hpt) of pipes used. In this case (i.e. hp3 >

Eab > hp2) the maximum number of BPTs is given by integer part of Eabhp1

and for this

maximum number, only pipe Type 1 should be used, similarly minimum number of

BPTs is given by integer part of Eabhp2

and here both Type 1 and Type 2 pipes can be

used. Therefore to obtain a policy for absolute minimum cost one should compare all

possible BPTs situation. No matter how many BPTs we use the optimization model

formulation is exactly similar to following example. The only difference is the

number of BPTs, therefore, the number of links will increase; which in turn increases

the number of decision variables (lptl) and constraints to be satisfied.

In Figure 3.5 two BPTs situation for the Condition hp3 >Eab>hp2 is considered , BPTs

T1 and T2 are used in a uniform slope branch AB. Considering all the symbols and

notification used previously the model for the situation will be as follows,

24

Here, t = 1 to 2 and l = 1to 3

The cost is given by:

C = 2CT+ ∑ ∑ ∑ Uptllptl ..Eq 3.19

Where,

CT = Cost of single BPT


Link 1

The maximum permissible pressure at part 1 should not exceed hp1,

Sab ∑ lp11 ≤ hp1 ..Eq 3.20


Or, Sab ∑ lp11 + ∑ lp21 ≤ hp2 ..Eq 3.21



Sab ∑ ∑ lpt1 -10.7Q1.85 ∑ ∑ lpt1

dpt14.85Ct

1.85 ≥Hmin,T1 ..Eq 3.22

Link 2


Sab ∑ lp12 ≤ hp1 ..Eq 3.23


Sab ∑ lp12 + ∑ lp22 ≤ hp2 ..Eq 3.24

E1

E2

E3

H11

H12

H13

Eab

Link 1

Link 2

Link 3

Part 1

Part 2

Part 1

Part 2

Part 1

Part 2

Lp11 Lp21 Lp12 Lp22 Lp13 Lp23T2

T1

A

B

Figure 3.5: Condition hp3 >Eab>hp2, with two BPTs

25



Sab ∑ ∑ lpt2 -10.7Q1.85 ∑ ∑ lpt2

dpt24.85Ct

1.85 ≥Hmin,T2 ..Eq 3.25

Link 3


Sab ∑ lp13 ≤ hp1 ..Eq 3.26


Sab ∑ lp13 + ∑ lp23 ≤ hp2 ..Eq 3.27

The minimum head requirement at B;

Sab ∑ ∑ lpt3 -10.7Q1.85 ∑ ∑ lpt3

dpt34.85Ct

1.85 ≥hmin,b ..Eq 3.28

The total lengths of the pipes should be equal to the length of AB,

∑ ∑ ∑ lptl =Lab .. Eq 3.29

Condition 4: Eab > hp3

In this condition the available head (Eab) exceeds the permissible pressure rating of

pipe type 3. Therefore, unless we use another pipe whose permissible pressure rating

is greater than available head, certain number of BPT must be considered. Here if we

use all three types of pipes then the number of BPTs will be minimum, which is given

by integer part of Eabhp3

. The number of BPTs will be increased if only type 1 and type

2 pipes are used which is given by the integer part of Eabhp2

, and the number of BPTs

will be maximum if we use only type 1 pipe. The maximum number of BPTs will be

given by integer part of Eabhp1

. Therefore to obtain a policy for absolute minimum cost

one should compare all possible BPTs situations.

If type 2 and type 1 Pipes or only type 1 pipes are to be considered then the LP

models can be formulated as in previous cases.

26

Figure 3.6 shows, single BPT situation using all three types of pipes for the condition

Eab > hp3. Considering all the symbols and notification used previously the LP model

for the situation will be as follows,

Here, t = 1 to 3 and l = 1 to 2

The cost is given by:

C=CT+ ∑ ∑ ∑ Uptllptl ..Eq 3.30


Link 1


Sab ∑ lp11 ≤ hp1 ..Eq 3.31


Sab ∑ lp11 + ∑ lp21 ≤ hp2 ..Eq 3.32


Sab ∑ lp11 + ∑ lp21 + ∑ lp31 ≤ hp3 ..Eq 3.33



Sab ∑ ∑ lpt1 -10.7Q1.85 ∑ ∑ lpt1

dpt14.85Ct

1.85 ≥Hmin,T1 ..Eq 3.34

Eab

Link 1

Link 2

Part 1

Part 2

Part 3

Part 1

Part 2

Part 3

T1

Lp11 Lp21 Lp31 Lp12 Lp22 Lp32

H11H21

H12H22

E1

E2B

Qab

A

Figure 3.6: Condition Eab > hp3, with single BPT

27

Similarly for Link 2

Sab ∑ lp12 ≤ hp1 ..Eq 3.35

Sab ∑ lp12 + ∑ lp22 ≤ hp2 ..Eq 3.36

Sab ∑ lp12 + ∑ lp22 + ∑ lp32 ≤ hp3 ..Eq 3.37

Sab ∑ ∑ lpt2 -10.7Q1.85 ∑ ∑ lpt2

dpt24.85Ct

1.85 ≥hmin,b ..Eq 3.38

The total lengths of the pipes should be equal to the length of AB,

∑ ∑ ∑ lptl =Lab ..Eq 3.39

Condition 5: Ridge and Valley Condition (Sudden level rise along pipeline)

Ridge and valley condition as sown in Figure 3.7 are very common in hilly terrain.

Hence when ground elevation rises along pipeline some more constrains have to be

introduced in the LP model.

To prevent the creation of negative head, hydraulic grade line (HGL) should always

be above atmospheric pressure throughout the pipe line. Especially in ridge and valley

condition, at the ridge (in Figure 3.7, C represents the ridge) chances are high for the

development of negative pressure. Therefore to prevent possible negative head the

minimum pressure requirement at valley (point B) should be equal to or greater then

elevation difference between ridge point and valley point (Ebc) and also, the residual

head at ridge point should be greater than atmospheric pressure.

Let us consider a situation as in Figure: 3.7 where Eab ≤ hp1, then LP model is

formulated as follows,

D

C

B

A

Eab

Ebc

Ead

Ecd

Figure 3.7: Ridge and valley condition

28

Link AB

The minimum pressure requirement at lowest point B, i.e. valley point should be

equal or greater than elevation difference between ridge point C and valley point B

given by Ebc,

Eab- ∑ hlp11 ≥Ebc ..Eq 3.40

And,

∑ lp11 =Lab ..Eq 3.41

Link BC

At point C the hydraulic grade line should not fall below the ground elevation, i.e.

head available at c must be more than or equal to atmospheric.

Eab- ∑ hlp11 - ∑ hlp12 -Ebc≥0 ..Eq 3.42

And,

∑ lp12 =Lbc ..Eq 3.43

Link CD

If D is considered as demand point then,

Eab- ∑ hlp11 - ∑ hlp12 -Ebc- ∑ hlp13+Ecd ≥hmin,d ..Eq 3.44

∑ lp13 =Lcd ..Eq 3.45

And,

∑ lp11 + ∑ lp12 + ∑ lp13 =Labcd ..Eq 3.46

29

CHAPTER IV

4.0 RESULT AND DISCUSSIONS

4.1 Introduction

This chapter illustrates the application of the linear optimization model discussed in

the previous chapter and discusses the results obtained from the application of the

models. The models has been used to recalculate the pipe networks, based on actual

cost estimation and design report of Manikanda Water Supply Project at Gotree VDC

of Bajura district obtained from Community-based Water Supplies and Sanitation

Project, Tangal, Kathmandu.

It is expected that the solution will give optimum combination of lengths of

commercially available diameters and types of pipes and optimum locations of BPTs

so that cost will be optimum.

4.2 Design of Transmission Line

For frictional head loss calculation in pipes Hazen William’s Equation is considered

When flow (Q) is constant for particular size (d) of pipe the equation can be expressed

as,

hf = kl

Where,

k =10.7Q1.85 1d4.87CH

1.85 ..Eq 4.1

The flow through the transmission line is 0.4 lps. The diameters of pipes for input to

the model may be limited to a few by making use of the maximum and minimum flow

velocities. Minimum velocity is taken as (vmin) = 0.4 m/s and maximum velocity is

taken as (vmax) = 2.3 m/s, then by using Equations 2.12a and 2.12b we get,

Maximum diameter (dmax) = 35.68mm, and

Minimum diameter (dmin) = 15mm

Hence commercially available diameters in this range for various types of pipes are

given in Table 4.1

30

Table 4.1 Various types of commercially available pipes fulfilling velocity

requirements for the transmission line.

S.

N

Outer

Diameter

Type 1 HDPE Type 2 HDPE Type 3 GI

Internal

Diameter Rate

K

Value

Internal

Diameter Rate

K

Value

Internal

Diameter Rate

K

Value

(mm) (mm) (NRs./m) (mm) (NRs./m) (mm) (NRs./m)

1 15 15 103.7 0.6

2 20 20 133.5 0.148

3 25 18.9 35.96 0.129 25 201.5 0.05

4 32 26.9 40.23 0.023 23.8 59.45 0.04 32 255.5 0.015

5 40 33.7 62.30 0.007 30.3 91.49 0.0129

Design Discussion

The profile of transmission line under consideration is shown in Figure 4.1. The

maximum static head up to P32 is 30.31 m, therefore pipe type 1 can only be

considered for link 1, 2, 3, 4 and 5. But, in link 2 and 4 where soil type is Rocky, GI

pipe (type 3) is considered. However in link 1, 3 and 4 type 2 pipes fulfilling the

velocity condition and cheaper then type 1 pipe are also considered.

At point B the maximum static pressure is 96.55m, therefore the combination of type

1 and type 2 pipes are considered in link 6.

At point R the maximum static pressure is 236.5m which is greater than allowable

pressure rating of pipe type 3, hence for optimum design, following two conditions

are considered.

Condition1: Single interruption chamber (IC1) is considered at link 7 and all three

types of pipes are used.

Condition 2: Two interruption chambers (IC1 and IC2) are considered and only type 1

and type 2 pipes are used.

31

LP Problem for Design of Transmission line

For designing the transmission line considering single interruption Chamber,

following constraints are to be fulfilled,

Link 1 (I-P8)

1) Residual pressure at P8 ≥ 5m

30.27- ∑ ∑ hlpt1 ≥ 5

2) Total length of I-P8=171.6m

∑ ∑ lpt1 =171.6

Link 2 (P8-P10)


30.31- ∑ ∑ hlpt1 - ∑ ∑ hlpt2 ≥ 5

I (RL=2120)

P8 (RL=2089.73)P10 (RL=2089.69)

P13(RL=2091.75)

P16(RL=2093.24)P32 (RL=2093.06)

B (RL=2023.45)

R (RL=1883.5)

171.6m

34.1m

57.2m64.9m 361.9m

440m

402.6m

(1)

(2)

(3)(4)

(5)

(6)

(7)

Figure 4.1: Ground profile of transmission line

32

4) Total length of P8-P10 = 34.1m

∑ ∑ lpt2 = 34.1

Link 3 (P10-P13)


28.25 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 ≥ 5

6) Total length of P10-P13=57.2m

∑ ∑ lpt3 = 57.2

Link 4 (P13-P16)


26.76 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 - ∑ ∑ hlpt4 ≥ 5

8) Total length of P13-P16=64.9

∑ ∑ lpt4 = 64.9

Link 5 (P16-P32)


26.94 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 - ∑ ∑ hlpt4 - ∑ ∑ hlpt5 ≥ 5

10) Total length of P16-P32=361.9m

∑ ∑ lpt5 = 361.9

Link 6 (P32-B1)

11) Maximum static head available at first part of the link should be less then 60m,

26.94 + 0.1582 ∑ lp16 ≤ 60

12) Residual pressure at B1 ≥ 5m

26.76 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 - ∑ ∑ hlpt4 - ∑ ∑ hlpt5 - ∑ ∑ hlpt6 ≥ 5

13) Total length of P32-B1=440m

∑ ∑ lpt6 = 440

33

Link 7 (B1-IC1)

14) Here, static head at B1 is already greater than pressure rating of Type 1 pipe,

hence Type and Type 3 pipes are only considered. Therefore, the maximum static

head at first part of the link should be less than or equal to100m,

96.55 + 0.3476 ∑ lp27 ≤ 100

15) The static head at IC1 should be less than or equal to pressure rating of Type 3

pipe, i.e. 160m,

96.55 + 0.3476 ∑ lp27 + ∑ lp37 ≤ 160

16) Residual pressure at IC1 ≥ 10m

96.55 + 0.3476 ∑ lp27 + ∑ lp37 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 - ∑ ∑ hlpt4 - ∑ ∑ hlpt5

- ∑ ∑ hlpt6 - ∑ ∑ hlpt7 ≥ 10

Link 8 (IC1-R)

17) The maximum static head available at first part of the link should be less than

60m,

0.3476 ∑ lp18 ≤ 60

18) The maximum static head available at second part of the link should be less than

100m,

0.3476 ∑ lp18 + ∑ lp28 ≤ 100

19) The maximum static head available at third part of the link should be less than

160m,

0.3476 ∑ lp18 + ∑ lp28 + ∑ lp38 ≤ 160

20) Residual pressure at R ≥ 10m

0.3476 ∑ lp18 + ∑ lp28 + ∑ lp38 - ∑ ∑ hlpt8 ≥ 10

21) Total length of B1-R=402.6m

∑ ∑ lpt7 + ∑ ∑ lpt8 = 402.6

22) Total length of transmission line = 1532.3m

∑ ∑ ∑ lptl =1532.3

34

Similarly for second condition where two interruption chambers are considered, LP

problem was formulated. Thus, LP problem for both the conditions was solved using

computer software, TORA 8.0. The software solves the LP problem by the Simplex

method. The input data and computer output for both the conditions are included in

the Appendix C. The summaries of results for both the conditions are given in Table

4.2 and Table 4.3.

Table 4.2: Result of LP model solution for design of transmission line with single interruption chamber.

S.N Link Pipe

Type

Diameter Length Unit Cost Amount

Remarks From To External

(mm) Internal(mm) (m) NRs. NRs.

1 I P8 1 32 26.9 171.60 40.23 6903.47 2 P8 P10 3 25.0 34.10 201.5 6871.15 Rock 3 P10 P13 1 32 26.9 57.20 40.23 2301.15

4 P13 P16 3 20.0 37.82 133.5 5048.97 Rock 3 25.0 27.08 201.5 5456.62 5 P16 P32 1 32 26.9 361.90 40.23 14559.24

6 P32 B 1 32 26.9 208.98 40.23 8407.27

2 20 14.9 20.11 23.85 479.62 2 25 18.9 210.91 35.96 7584.32

7 B IC1 2 25 18.9 9.93 35.96 357.08 3 15.0 104.99 107.3 10887.46

8 IC1 R 1 32 26.9 72.78 40.23 2927.94 2 20 14.9 214.90 23.85 5125.37 TOTAL 1532.3 76909.66

Table 4.3: Result of LP model solution for design of transmission line with two interruption chambers.

S.N Link Pipe

Type



(mm) Internal(mm) (m) NRs. NRs.

1 I P8 1 32 26.9 171.60 40.23 6903.47 2 P8 P10 3 25.0 34.10 201.5 6871.15 Rock 3 P10 P13 1 32 26.9 57.20 40.23 2301.15

4 P13 P16 3 20.0 37.82 133.5 5048.97 Rock 3 25.0 27.08 201.5 5456.62 5 P16 P32 1 32 26.9 361.90 40.23 14559.24

6 P32 IC1 1 32 26.9 208.98 40.23 8407.27

2 20 14.9 106.39 23.85 479.62 2 25 18.9 124.64 35.96 7584.32

7 IC1 IC2 2 25 18.9 72.78 40.23 357.08 3 15.0 214.91 23.85 10887.46

8 IC2 R 1 32 26.9 44.55 40.23 2927.94 2 20 14.9 70.36 23.85 5125.37 TOTAL 1532.3 68091.20

35

The cost of one interruption chamber is taken as NRs. 40,000.00. Therefore the

optimum cost for first condition is NRs. 1, 16,909.66 and optimum cost for second

condition is NRs. 1, 48,091.20. Hence, the first condition is taken as global optimum

policy for the design of the transmission line.

4.3 Design of Distribution Network

The schematic diagram of the distribution network considered for the study is given in

Figure 4.2. Each branch has been designed to obtain optimum design for the whole

distribution network.

Branch R-J1

This is the first branch of the distribution network where R is service reservoir. The

flow through the branch is 1.1 lps, length of the branch is 160.6m and available head

and maximum static head id 21.51m. Taking maximum allowable flow velocity as

2.3m/s and minimum required velocity as 0.4 m/s, the maximum and minimum

diameters of pipes that can be used is obtained as 59.17mm and 24.67mm

respectively, the commercially available diameters within this range for the branch in

table 4.4

Table 4.4: Commercially available type 1 HDPE pipes fulfilling the velocity

requirement for Branch R-J1.

Outer diameter (mm) Internal diameter (mm) Rate (NRs./meter) K

32 26.9 40.23 0.15

40 33.7 62.3 0.05

50 42.2 96.48 0.0167

63 53.3 151.30 0.0053

The objective function of the LP problem for the branch will be to minimize the cost

of the branch which is given by,

Minimize,

C= ∑ Up11 lp11

36

Figure 4.2: Schematic diagram of the distribution network

163.6m 1.1 lps

R (RL=1883.5)

T1 (RL=1860.23)T2 (RL=1815.63)J8 (RL=1824.79) J1 (RL=1861.99)

T3 (RL=1817.83)

J2 (RL=1839.85)J3 (RL=1830.03)

T4 (RL=1825.62)

T5 (RL=1817.09)J4 (RL=1839.96)

T6 (RL=1839.75)

T7 (RL=1829.06)

BPT1 (RL=1800.87)J6 (RL=1771.49)

T8 (RL=1759.77)

T9 (RL=1723.21)

J7 (RL=1750.71)

T10 (RL=1728.09)

T11 (RL=1665.26)

12.1m0.1 lps

256.3m0.2 lps

184.8m0.1 lps

74.8m

0.1 l

ps388.3m 0.8 lps

30.8m0.2 lps

34.1m0.1 lps

93.5m 0.1 lps 64.9m 0.6 lps

38.5m

0.1 lps

79.2m 0.5 lps

157.3m 0.4 lps

266.2m0.2 lps

52.8m0.1 lps

173.8m 0.1 lps

455.4m 0.2 lps

2249.7m 0.1 lps

407.03m

0.1 lps

37

Subjected to,

1) Minimum pressure requirement at J1=10m,

21.51- ∑ hlp11 ≥10

2) The length of branch R-J1=160.6m

∑ lp11 =160.6

The result for the LP problem showed that optimum solution for the branch will be

obtained by using external pipe diameters of 32mm for length of 34.80m and 40mm

for length 125.80m. The residual head given by the pipe combination is 10m.

Branch J1-J8

The flow through this branch is 0.2 lps, length of the branch is 256.3m.The level

difference between J1 and J8 is 37.2m and residual pressure at J1 is 10m, hence the

available head is 47.2m, the maximum static pressure at J8 is 58.71m. Using velocity

requirement the maximum and minimum diameters of pipes that can be used is

obtained as 25.25mm and 10.5mm respectively. However the maximum static head is

less then pressure rating of type 1 pipe, the commercially available diameters within

required range are found only in type 2. Hence type 2 pipes are used for the branch

which is given in Table 4.5.

Table 4.5: Commercially available type 2 HDPE pipes fulfilling the velocity requirement for Branch J1-J8.

Outer diameter (mm) Internal diameter

(mm)

Rate

(NRs./meter) K

20 14.9 23.85 0.11

25 18.90 35.96 0.035

32 23.80 59.45 0.011

As for branch R-J1 the LP problem for the branch J1-J8 was solved and result showed

that the optimum solution for the branch will be, use of external pipe diameter 20mm

of type 2 throughout the length of the branch.

38

Similarly LP problem for all the remaining branch was solved to get optimum policy

for the distribution network, the result obtained is summarized in the Table 4.6.

Table 4.6: Result of LP model solutions for design of distribution network.

S.N

Link Pipe Type



(mm) Internal (mm) (m) NRs. NRs.

1 R J1 1 32 26.9 34.80 40.23 1400.00 1 40 33.7 125.80 62.30 7837.34 2 J1 T1 2 20 14.9 12.10 23.85 288.59 3 J1 J8 2 20 14.9 256.30 23.85 6112.76 4 J8 T2 2 20 14.9 184.80 23.85 4407.48 5 J8 T3 2 20 14.9 74.80 23.85 1783.98

6 J1 J2 1 32 26..9 208.14 40.23 8373.47 1 40 33.7 180.16 62.30 11223.97 7 J2 J3 2 20 14.9 30.80 23.85 734.58 8 J3 T4 2 20 14.9 34.10 23.85 813.29 9 J3 T5 2 20 14.9 93.50 23.85 2229.98

10 J2 J4 1 32 26.9 64.90 40.23 2610.93 11 J4 T6 2 20 14.9 38.50 23.85 918.23

12 J4 T7 1 32 26.9 48.93 40.23 1968.45 2 25 18.9 30.27 35.96 1088.51

13 T7 BPT1 1 32 26.9 31.03 40.23 1248.34

2 20 14.9 39.81 23.85 949.47 2 25 18.9 86.46 35.96 3109.10

14 BPT1 J6 2 20 14.9 132.75 23.85 3166.09 2 25 18.9 133.45 35.96 4798.86 15 J6 T8 2 20 14.9 52.80 23.85 1259.28 16 J6 T9 2 20 14.9 173.80 23.85 4145.13

17 BPT1 J7 2 20 14.9 321.52 23.85 7668.13 2 25 18.9 133.88 35.96 4814.32 18 J7 T10 2 20 14.9 249.70 23.85 5955.35

19 J7 T11 2 20 14.9 237.33 23.85 5660.32 3 15.0 169.47 103.7 17574.04 TOTAL 3179.9 112140.09

The optimum cost of the pipeline for the distribution network was obtained

NRs.1, 12,140.09. There is provision of single BPT at the network, hence adding the

cost of the BPT as NRs.40, 000.00, the total optimum cost of the distribution network

will be NRs. 1, 52,140.09.

4.4 Hydraulic Calculation

The optimum combination obtained for transmission line and distribution network

was used for calculating required hydraulic values. The hydraulic calculation for

transmission line is given in Table 4.7 and that for distribution network is given in

Table 4.8. Also the schematic diagrams of optimally designed transmission line and

distribution network is given in Figure 4.3 and Figure 4.4 respectively.

39

Tabl

e 4.

7: D

esig

n of

tran

smis

sion

line

40

Figure 4.3: Schematic diagram of optimum design of transmission line

I (RL=2120.0)

P8 (RL=2089.73)

P10 (RL=2089.69)

P13 (RL=2091.75)

P16 (RL=2093.24)

P32(RL=2093.06)

B (RL=2023.45)

R (RL=1883.5)

IC1 (RL=1983.51)

32H(6)/171.6m 0.4 lps

25GI(MC)/34.1m 0.4 lps

32H(6)/57.2m 0.4 lps

20GI(MC)/37.82m 0.4 lps

25GI(MC)/27.08m 0.4 lps

32H(6)/361.9m 0.4 lps

32H(6)/208.98m 0.4 lps

20H(10)/20.11m 0.4 lps

25H(10)/210.91m 0.4 lps

25H(10)/9.93m 0.4 lps

15GI(MC)/104.99m 0.4 lps

32H(6)/72.78m 0.4 lps

20H(10)/214.9m 0.4 lps

INTAKE

41

Tabl

e 4.

8 D

esig

n of

dis

tribu

tion

netw

ork

42

32H(6)/34.8m 1.1 lps

40H(6)/125.8m 1.1 lps

R (RL=1883.5)

T1 (RL=1860.23)T2 (RL=1815.63)

J8 (RL=1824.79) J1 (RL=1861.99)

T3 (RL=1817.83)

J2 (RL=1839.85)J3 (RL=1830.03)

T4 (RL=1825.62)

T5 (RL=1817.09)J4 (RL=1839.96)

T6 (RL=1839.75)

T7 (RL=1829.06)

BPT1 (RL=1800.87)J6 (RL=1771.49)

T8 (RL=1759.77)

T9 (RL=1723.21)

J7 (RL=1750.71)

T10 (RL=1728.09)

T11 (RL=1665.26)

20H(10)/12.1m0.1 lps

20H(10)/256.3m0.2 lps

20H(10)/184.8m0.1 lps

20H(10

)/74.8

m

0.1 l

ps32H(6)/208.14m 0.8 lps

40H(6)/180.16m 0.8 lps

20H(10)/30.8m0.2 lps

20H(10)/34.1m0.1 lps

20H(10)/93.5m 0.1 lps 32H(6)/64.9m 0.6 lps

20H(10)38.5m

0.1 lps

32H(6)/48.93m 0.5 lps

25H(10)/30.27m 0.5 lps

32H(6)/31.03m 0.4 lps

20H(10)/39.81m 0.4 lps

25H(10)/86.46m 0.4 lps

20H(10)/132.75m0.2 lps

25H(10)/133.45m0.2 lps

20H(10)/52.8m0.1 lps

20H(10)/173.8m 0.1 lps20H(10)/321.52m 0.2 lps

25H(10)/133.88m 0.2 lps

20H(10)/249.7m 0.1 lps

20H(10)/237.33m0.1 lps

20GI(MC)/169.7m0.1 lps

Figure 4.4: Schematic diagram of optimum design of distribution network

43

4.5 Cost Comparison

The cost of pipeline network of the water supply project including, interruption

chambers and break pressure tanks after optimal design using linear programming

was estimated and compared with the cost as given in the design report of the project.

The cost comparison is provided in Table 4.9

Table 4.9 Cost comparison between optimal design and project report.

S.N Parameter

Cost as per

Optimum design

(NRs.) Project report (NRs.)

1 Transmission line 1,16,909.66 1,86,077.60

2 Distribution line 1,52,140.09 2,066,42.60

Total 2,69,049.75 3,92,720.20

Although, both the design approach are good in satisfying the required hydraulic

conditions, but optimal design using linear programming method can actually helps in

selecting the best hydraulic design giving minimum optimum cost. The cost

comparison made in Table 4.9 shows that the cost can be reduced by 31.5% of the

actual estimated cost in this particular project.

44

CHAPTER V

5.0 CONCLUSIONS AND RECOMMENDATIONS

5.1 Conclusions

The purpose of the study was to present possibility of optimum design approach for

design of water supply system in small rural hilly community. Based on the study

conducted following conclusions are shown.

• It is possible to use optimum design approach while designing a water supply

system in rural hilly community.

• Linear Programming method for optimum design can be developed as simple

but reliable tool, which can be used effectively in designing rural water

supply project of developing country like Nepal.

• In the scenario where the resources are limited but demand is large, by using

optimum design approach, considerable amount of resources can be saved

that can be used to fulfill the demand.

5.2 Recommendations

The following recommendations are made for further study:

• The study was focused on optimum design of pipeline network, however a

typical water supply system consists of many other components such as intake

structures, treatment facilities, storage facilities etc. Therefore further study is

recommended on optimum design of those components.

• The computer software used for the study is actually a linear program solver,

which means that in study the required linear programming problems were

developed manually and then was fed to the software. The development of an

interactive computer program is recommended for the optimal design.

• Optimization models considered in this study assumes a uniform slope in a

branch, which is not always true. Therefore, study on optimal design in

uneven ground slopes is recommended.

45

REFERENCES

1. Bhave, P.R. (2003). Optimal Design of Water Distribution Networks. Narosa

Publishing House, New Delhi, India.

2. Chibuzor J.Edordu (2005), A MATLAB Toolkit for Linear Programming,

University of London.

(www.ee.ucl.ac.uk)

3. D Nagesh Kumar (2010). Optimization Methods: Linear Programming-

Graphical Method.

(http://nptel.iitm.ac.in/courses.php)

4. Department of Water Supply and Sewerage (DWSS, 2002). Design Guidelines for

Community Based Gravity Flow, Rural Water Supply Schemes, Vol. II: Design

Criteria.

5. Ghimire, Drona Raj (1991). Sustainable and Least Cost Gravity Water Supply for

a Rural Community in the Hills. The University of Liverpool.

6. J. Reeb and S. Leavengood (1998). Using the Simplex Method to solve Linear

Programming Maximization Problems.

( http://owic.oregonstate.edu/pubs/EM8720.pdf)

7. Kansakar B.R (2009). Manual on Modeling in Environmental Engineering.

Department of Civil Engineering, Istitute of Engineering (TU).

8. Manual on Water Supply and Treatment based on Central Public Health and

Environment Engineering Organization. (1997). Akalank Publications, Delhi.

9. Modi, P.N. (2006). Environmental Engineering Vol. 1, Water Supply Engineering,

2nd Ed. Standard Book House, New Delhi, India.

10. http://www.mathworks.com

46

APPENDICES

Appendix A Example of solving LP problem by graphical method

Appendix B Example for solving LP problem by simplex method

Appendix C Data input and output result of TORA 8.0 for design of

transmission line.

Appendix D Data input and output result of TORA 8.0 for design of

distribution network

Appendix E Schematic diagrams of transmission line and distribution

network of the project as given in design report.

Appendix F Pipeline design table from design report of the project.

Appendix A

A-1

Example: Graphical Method Maximize Z=3x+4y

Subjected to, 2x + y ≤ 10

x + 3y ≤ 9

x ≥ 0 , 0

Solution: Consider Fig. 2.2. The constraint x ≥ 0 implies that the solution must lie on the

right hand side of y-axis (x=0), similarly, y ≥ 0 implies that the solution must lie above the x-

axis (y=0). Thus, these no negativity constraints imply that the solution must lie in the first

quadrant. Other constraints are also plotted. Considering all constraints we observed that the

solution must lie on or within the shaded quadrilateral OABC.

Hence OABC is the feasible region. Now consider a line 3x + 4x = c, a constant. This line

represents the objective function which will have different values for different values of c.

Let us take c = 6. The line 3x + 4y = 6 is plotted in the graph. If we move the line 3x + 4y =c

to the left, the value of 3x + 4y decreases; however, if we move it to the right its value

increases. To get the maximum value let us move the line to right and continue until we have

at least one of its points in the feasible region. Thus, when 3x +4y = c line passes through

point B we have maximum value of 3x + 4y. Note that point B is the point of intersection of

active constraints, solving for these two constraints we get the optimum solution as x = 4.2

and y = 1.6 and maximum value of Z = 19. Similarly if we need minimum value then point O

gives the minimum value. Hence, it can be noted that the optimum value (maximum and

minimum) occurs at an extreme point or vertex of the feasible region.

1

2

3

4

5

6

7

8

9

1 2 3 4 5 6 7 8 9

10

2x + y =10

x + 3y = 93x + 4y = 6

A

B

CO X

Feasible Region

Figure A1: Graphical Solution

Appendix B

B-1

Example: Simplex Method Maximize Z=5x1+4x2

Subjected to,

6x1+4x2≤24

x1+2x2≤6

-x1+x2≤1

x2≤ 24

x1,x2≥0

Solution: Let us convert inequalities into equations with nonnegative right hand side Maximize Z-5x1-4x2=0

Subjected to,

6x1+4x2+s1=24

x1+2x2+s2=6

-x1+x2+s3=1

x2+s4=24

x1,x2,s1,s2,s3,s4≥0

Then the starting simplex tableau can be represented as follows:

Basic Z X1 X2 S1 S2 S3 S4 Solution

Z 1 -5 -4 0 0 0 0 0

S1 0 6 4 1 0 0 0 24

S2 0 1 2 0 1 0 0 6

S3 0 -1 1 0 0 1 0 1

S4 0 0 1 0 0 0 1 2

In this first tableau Non basic (0, 0) variables are x1 and x2 and basic variables are s1, s2, s3

and s4. We can see in the tableau that on setting (x1, x2) = (0, 0) the solution for Z, s1, s2, s3

and s4 is immediately available at solution column.

The objective function Z = 5x1 + 4x2 shows that the solution can be improved by increasing

x1 or x2, hence most positive coefficient is selected as the entering variable. Because simplex

tableau expresses the objective function as Z - 5x1 - 4x2, the entering variable will correspond

to the variable with the most negative coefficient in the objective equation.

Enter

Leave Pivot Row

Pivot Column

Appendix B

B-2

Now to determine the leaving variable we need to find corresponding minimum nonnegative

ratio. This can be obtained by dividing elements of solution column by corresponding

elements under entering variable column. In this example the minimum nonnegative ratio is 4

which corresponds to s1, therefore s1 is leaving variable.

Hence, new non basic (zero) variables are s1 and x2 and new basic variables are x1, s2, s3 and

s4.

Now the new tableau is created using Gauss-Jordan row operation, which identifies entering

variable column as the pivot column and the leaving row column as pivot row. The

intersection of the pivot column and pivot row is called the pivot element.

The new tableau is created with following calculations,

New pivot row=current pivot row ÷ pivot element

New row= current row - Its pivot column coefficient ×(new pivot row)

Hence, new tableau is obtained as,


Z 1 0 -2/3 5/6 0 0 0 20

X1 0 1 2/3 1/6 0 0 0 4

S2 0 0 4/3 -1/6 1 0 0 2

S3 0 0 5/3 1/6 0 1 0 5

S4 0 0 1 0 0 0 1 2

As in previous tableau we can find out in this tableau the entering variable is x2 and leaving

variable is s2, and with similar process new tableau is generated,


Z 1 0 0 3/4 1/2 0 0 21

X1 0 1 0 1/4 -1/2 0 0 3

X2 0 0 1 -1/8 3/4 0 0 3/2

S3 0 0 0 3/8 -5/4 1 0 5/2

S4 0 0 0 1/8 -3/4 0 1 1/2

Based on the optimality condition, none of the Z-row coefficient associated with the non

basic variables s1 and s2 is negative. Hence, this tableau is optimal, and we can see that

optimal value of objective function is 21 and corresponding values of decision variables x1

and x2 are 3 and 3/2 respectively.

Appendix C

C-1

Appendix C

C-2

Appendix C

C-3

Appendix C

C-4

Appendix C

C-5

Appendix C

C-6

Appendix C

C-7

Appendix C

C-8

Appendix C

C-9

Appendix C

C-10

Appendix C

C-11

Appendix C

C-12

Appendix C

C-13

Appendix C

C-14

Appendix C

C-15

Appendix C

C-16

Appendix C

C-17

Appendix C

C-18

Appendix D

D-1

Appendix D

D-2

Appendix D

D-3

Appendix D

D-4

Appendix D

D-5

Appendix D

D-6

Appendix D

D-7

Appendix D

D-8

Appendix D

D-9

Appendix D

D-10

Appendix E

E-1

Appendix E

E-2

Appendix F

F-1

Appendix F

F-2

optimal design of branched gravity water

Documents