pipe network analysis by hardy cross and newton method

1. 1
GUJARAT TECHNOLOGICAL UNIVERSITY
BIRLA VISHVAKARMA MAHAVIDHYALAYA
Affiliated with GTU
An open ended problem on:
Analyze the pipe network by hardy cross method and newton method and
compare both the methods
Prepared as part of requirements for the subject of
WATER AND WASTE WATER ENGINEERING
Submitted by:
AKSHAY GAJIPARA 140070106028
RAMABHAI GALCHAR 140070106029
NEMIL GANDHI 140070106030
AMIT GANVIT 140070106031
Prof . RESHMA PATEL
(Faculty guide)
Dr. L B ZALA
(Head of Department)
Academic Year(2016-2017)

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ANALYSIS OF LOOPED NETWORKS
A pipe network in which there are one or more closed loops is called a looped
network. A typical looped network is shown in Fig. 3.9. Looped networks are
preferred from the reliability point of view. If one or more pipelines are closed
for repair, water can still reach the consumer by a circuitous route incurring
more head loss. This feature is absent in a branched network. With the changing
demand pattern, not only the magnitudes of the discharge but also the flow
directions change in many links. Thus, the flow directions go on changing in a
large looped network. Analysis of a looped network consists of the
determination of pipe discharges and the nodal heads. The following laws,
given by Kirchhoff, generate the governing equations:
The algebraic sum of inflow and outflow discharges at a node is zero; and The
algebraic sum of the head loss around a loop is zero. On account of nonlinearity
of the resistance equation, it is not possible to solve network analysis problems
analytically. Computer programs have been written to analyze looped networks
of large size involving many input points like pumping stations and elevated
reservoirs. The most commonly used looped network analysis methods are
described in detail in the following sections

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HARDY CROSS METHOD
Analysis of a pipe network is essential to understand or evaluate a pipe network
system. In a branched pipe network, the pipe discharges are unique and can be
obtained simply by applying discharge continuity equations at all the nodes.
However, in case of a looped pipe network, the number of pipes is too large to
find the pipe discharges by merely applying discharge continuity equations at
nodes. The analysis of looped network is carried out by using additional
equations found from the fact that while traversing along a loop, as one reaches
at the starting node, the net head loss is zero. The analysis of looped network is
involved, as the loop equations are nonlinear in discharge. Hardy Cross (1885–
1951), who was professor of civil engineering at the University of Illinois,
Urbana-Champaign, presented in 1936 a method for the analysis of looped pipe
network with specified inflow and outflows (Fair et al., 1981). The method is
based on the following basic equations of continuity of flow and head loss that
should be satisfied:
1.The sum of inflow and outflow at a node should be equal:
where Qi is the discharge in pipe i meeting at node (junction) j, and qj is nodal
withdrawal at node j.
2. The algebraic sum of the head loss in a loop must be equal to zero:
Where
where i¼pipe link number to be summed up in the loop k.

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In general, it is not possible to satisfy Eq. (3.14) with the initially assumed pipe
discharges satisfying nodal continuity equation. The discharges are modified so
that Eq. (3.14) becomes closer to zero in comparison with initially assumed
discharges. The modified pipe discharges are determined by applying a
correction D Qk to the initially assumed pipe flows. Thus,
Expanding Eq. (3.16) and neglecting second power of D Qk and simplifying Eq.
(3.16), the following equation is obtained:
Knowing D Qk, the corrections are applied as
The overall procedure for the looped network analysis can be summarized in the
following steps:
1. Number all the nodes and pipe links. Also number the loops. Forclarity, pipe
numbers are circled and the loop numbers are put in square brackets
2. Adopta sign convention that a pipe discharge is positive if it flows from a
lower node number to a higher node number, otherwise negative.
3. Apply nodal continuity equation at all the nodes to obtain pipe discharges.
Starting from nodes having two pipes with unknown discharges, assume an
arbitrary discharge (say 0.1m3/s) in one of the pipes and apply continuity
equation (3.13) to obtain discharge in the other pipe.
Repeat the procedureuntil all the pipe flows are known. If there exist more than
two pipes having unknown discharges, assume arbitrary discharges in all the

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pipes except one and apply continuity equation to get discharge in the other
pipe. The total number of pipes having arbitrary discharges should be equal to
the total number of primary loops in the network.
4. Assume friction factors fi¼0.02 in all pipe links and compute corresponding
Ki using Eq. (3.15). However, fi can be calculated iteratively using Eq. (2.6a).
5. Assume loop pipe flow sign convention to apply loop discharge corrections;
generally, clockwise flows positive and counterclockwise flows negative are
considered.
6. Calculate D Qk for the existing pipe flows and apply pipe corrections
algebraically.
7. Apply the similar procedurein all the loops of a pipe network.
Repeat steps 6 and 7 until the discharge corrections in all the loops are relatively
very small.

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EXAMPLE OF HARDY CROSS METHOD
A single looped network as shown in Fig. 3.10 has to be analyzed by the
Hardy Cross method for given inflow and outflow discharges. The pipe
diameters D and lengths L are shown in the figure. Use Darcy–Weisbach head
loss–discharge relationship assuming a constant friction factor f¼0.02.
SOLUTION:
Step 1: The pipes, nodes, and loop are numbered as shown in Fig. 3.10.
Step 2: Adoptthe following sign conventions: A positive pipe discharge flows
from a lower node to a higher node. Inflow into a node is positive withdrawal
negative. In the summation process ofEq. (3.13), a positive sign is used if the
discharge in the pipe is out of the node under consideration. Otherwise, a
negative sign will beattached to the discharge. For example inFig. 3.10 at node
2, theflow inpipe1 is toward node 2, thus the Q1 at node 2 will be negative
while applying Eq. (3.13).

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Step 3: Apply continuity equation to obtain pipe discharges. Scanning the figure
for node 1, the discharges in pipes 1 and 4 are unknown. The nodal inflow q1 is
0.6m3/s and nodal outflow q3 ¼ 20.6m3/s. The q2 and q3 are zero. Assume an
arbitrary flow of 0.1m3/s in pipe 1 (Q1¼0.1m3/s), meaning thereby that the
flow in pipe 1 is from node 1 to node 2. The discharge in pipe Q4 can be
calculated by applying continuity equation at node 1 as
The discharge in pipe 4 is positive meaning thereby that the flow will be from
node 1 to node 4 (toward higher numbering node). Also applying continuity
equation at node 2:
Similarly applying continuity equation at node 3, flows in pipe Q3¼ 20.5m3/s
can be calculated. The pipe flow directions for the initial flows are shown in the
figure.
Step 4: Forassumed pipe friction factors fi¼0.02, the calculated K values as K
¼8fL= p 2gD5 for all the pipes are given in the Fig. 3.10.
Step 5: Adopted clockwise flows in pipes positive and counte rcloc kwise flows
negative. Step 6: The discharge correction for the initially assumed pipe
discharges can be calculated as follows:

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Repeat the process again for the revised pipe discharges as the discharge
correction is quite large in comparisonto pipe flows:
As the discharge correction
D Q¼0, the final discharges are
Q1=0.3m3/s
Q2=0.3m3/s
Q3=0.3m3/s
Q4=0.3m3/s.

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NEWTON RAPHSON METHOD
The pipe network can also be analyzed using the Newton–Raphsonmethod,
where unlike the Hardy Cross method, the entire network is analyzed altogether.
The Newton–Raphsonmethod is a powerful numerical method for solving
systems of nonlinear equations. Supposethat there are three nonlinear equations
F1(Q1, Q2, Q3)¼0, F2(Q1, Q2, Q3)¼0, and F3(Q1, Q2, Q3)¼0 to be solved for
Q1, Q2, and Q3. Adopta starting solution (Q1, Q2, Q3). Also consider that (Q1
þ D Q1, Q2 þ D Q2, Q3 þ D Q3) is the solution of the set of equations. That is,
Expanding the above equations as Taylor’s series,
Arranging the above set of equations in matrix form

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Solving Eq. (3.19c),
Knowing the corrections, the discharges are improved as
It can be seen that for a large network, it is time consuming to invert the matrix
again and again. Thus, the inverted matrix is preserved and used for at least
three times to obtain the corrections. The overall procedurefor looped network
analysis by the Newton–Raphsonmethod can be summarized in the following
steps:
Step 1: Number all the nodes, pipe links, and loops.
Step 2: Write nodal discharge equations as
where Qjn is the discharge in nth pipe at nodej, qj is nodal withdrawal, and jn is
the total number of pipes at nodej.

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Step 3: Write loop head-loss equations as
Step 4: Assume initial pipe discharges Q1,Q2,Q3,... satisfying continuity
equations.
Step 5: Assume friction factors fi¼0.02 in all pipe links and compute
corresponding Ki using Eq. (3.15).
Step 6: Find values of partial derivatives @Fn=@Qi and functions Fn, using
the initial pipe discharges Qi and Ki.
Step 7: Find D Qi. The equations generated are of the form Ax¼b, which can be
solved for D Qi.
Step 8: Using the obtained D Qi values, the pipe discharges are modified and
the process is repeated again until the calculated D Qi values are very small.

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SOLVING THE LOOP WITH NEWTON RAPHSON METHOD
The configuration of Example 3.3 is considered in this example for
illustrating the use of the Newton–Raphson method. For convenience, Fig. 3.10
is repeated as Fig. 3.12.
SOLUTION:
The nodal discharge functions F are
and loop head-loss function

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The derivatives are
The generated equations are assembled in the following matrix form:
Substituting the derivatives, the following form is obtained:
Assuming initial pipe discharge in pipe 1 Q1¼0.5m3/s, the other pipe
discharges obtained by continuity equation are
Q2=0.5m3/s
Q3=0.1m3/s
Q4=0.1m3/s
Substituting these values in the above equation, the following form is obtained:

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Using Gaussian elimination method, the solution is obtained as
D Q1 =20.2m3/s
D Q2 =20.2m3/s
D Q3 =0.2m3/s
D Q4 =0.2m3/s
Using these discharge corrections, the revised pipe discharges are
The process is repeated with the new pipe discharges. Revised values of F and
derivative @F=@Q values are obtained. Substituting the revised values, the
following new solution is generated:

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Astheright-handsideisoperateduponnullvector,allthedischargecorrections
D Q=0.Thus, the final discharges are
Q1=0.3m3/s
Q2=0.3m3/s
Q3=0.3m3/s
Q4=0.3m3/s
THE SOLUTION OBTAINED BY THE NEWTON–RAPHSONMETHOD
ISTHE SAME ASTHAT OBTAINED BY THE HARDY CROSS
METHOD