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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
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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
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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
• hp2 = Maximum pressure that type 2 pipe can withstand
• hp3 = Maximum pressure that type 3 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
The cost of the pipes is given by:
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
The cost of the pipes is given by:
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
Conditions to be satisfied;
Link 1
The maximum permissible pressure at part 1 should not exceed hp1,
Sab ∑ lp11 ≤ hp1 ..Eq 3.20
The maximum permissible pressure at part 2 should not exceed hp2,
Or, Sab ∑ lp11 + ∑ lp21 ≤ hp2 ..Eq 3.21
The difference between available head and head loss in link should be greater or equal
to minimum head required at T1.
Sab ∑ ∑ lpt1 -10.7Q1.85 ∑ ∑ lpt1
dpt14.85Ct
1.85 ≥Hmin,T1 ..Eq 3.22
Link 2
The maximum permissible pressure at part 1 should not exceed hp1,
Sab ∑ lp12 ≤ hp1 ..Eq 3.23
The maximum permissible pressure at part 2 should not exceed hp2,
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
The difference between available head and head loss in link should be greater or equal
to minimum head required at T2.
Sab ∑ ∑ lpt2 -10.7Q1.85 ∑ ∑ lpt2
dpt24.85Ct
1.85 ≥Hmin,T2 ..Eq 3.25
Link 3
The maximum permissible pressure at part 1 should not exceed hp1,
Sab ∑ lp13 ≤ hp1 ..Eq 3.26
The maximum permissible pressure at part 2 should not exceed hp2,
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
Conditions to be satisfied;
Link 1
The maximum permissible pressure at part 1 should not exceed hp1,
Sab ∑ lp11 ≤ hp1 ..Eq 3.31
The maximum permissible pressure at part 2 should not exceed hp2,
Sab ∑ lp11 + ∑ lp21 ≤ hp2 ..Eq 3.32
The maximum permissible pressure at part 3 should not exceed hp3,
Sab ∑ lp11 + ∑ lp21 + ∑ lp31 ≤ hp3 ..Eq 3.33
The difference between available head and head loss in link should be greater or equal
to minimum head required at T1.
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)
3) Residual pressure at P10 ≥ 5m
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)
5) Residual pressure at P13 ≥ 5m
28.25 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 ≥ 5
6) Total length of P10-P13=57.2m
∑ ∑ lpt3 = 57.2
Link 4 (P13-P16)
7) Residual pressure at P16 ≥ 5m
26.76 - ∑ ∑ hlpt1 - ∑ ∑ hlpt2 - ∑ ∑ hlpt3 - ∑ ∑ hlpt4 ≥ 5
8) Total length of P13-P16=64.9
∑ ∑ lpt4 = 64.9
Link 5 (P16-P32)
9) Residual pressure at P32 ≥ 5m
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
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 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
Diameter Length Unit Cost Amount
Remarks From To External
(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,
Basic Z X1 X2 S1 S2 S3 S4 Solution
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,
Basic Z X1 X2 S1 S2 S3 S4 Solution
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