Flow in Systems with Multiple Pipes
In many systems of interest, pipes are connected in series, parallel, or complex networks. Thus
far, we have analyzed single-pipe systems by writing and solving equations that describe
fundamental relationships among the key variables. Often, we can specify enough parameters so
that only one variable remains unknown (hL, V or Q, or D), and use the Moody diagram or an
equivalent equation to solve for that unknown. In multi-pipe systems, we typically do not know
as much about the flow conditions applicable to the individual pipes (e.g., we might know the
flow rates entering and leaving, but not in individual pipes in the middle of a system), and so we
need to write more equations relating the unknown variables and solve those equations
simultaneously. The additional equations generally fall into two categories:

Expressions of continuity at junctions, stating that all the flow that enters a junction must
leave it.

Expressions relying on the fact that the total head has a single value at each point in the
system, so that the headloss between any two points must be the same regardless of the
path the fluid follows between these locations.
The application of these equations is most easily seen by considering some example systems.
______________________
Example 1. Characteristics of the system shown schematically below are summarized in the table
following the schematic. (Note that, to simplify the math, friction factors are assumed for each
pipe, rather than being calculated from the flow conditions in the pipe.) Water is discharged to
the atmosphere at point D. The system geometry is such that the change in velocity head is
negligible compared to the change in piezometric head between any two points of interest, so the
velocity heads can be ignored in an energy analysis. Compute (a) the flow in each pipe and
(b) the pressures at points A and B.
E
2
A
1
C
Water
B
3
4
D
C:\AData\CLASNOTE\345\Spring 2012\Pipes_networks_Spr2012.docx; saved 4/29/2012 4:47:00 PM
Point
Elevation (ft)
Pipe
Length (ft)
Diameter (in)
Friction factor
E
150
1
1200
8
0.035
A
120
2
800
6
0.025
B
70
3
900
4
0.045
C
60
4
500
6
0.025
D
30
Solution. (a) The general principles that we will apply are that the headloss must be identical
through the two pipes in parallel (#2 and #3), and that the flow into the junctions at points B and
C must equal the flow out of them. Furthermore, of course, the total headloss between points E
and D must satisfy the energy equation between those points. The application of these principles
is as follows.
The areas of the pipes can be computed as  D2/4, from which:
A1  0.349 ft 2
A2  A4  0.196 ft 2
A3  0.0873 ft 2
The headloss between points B and C must be the same regardless of whether the water flows
through pipe 2 or pipe 3. Writing the Darcy-Weisbach (D-W) equation for flow through these
pipes, we find:
hL ,2  f 2
0.025
l V2
l2 V22
 f3 3 3  hL ,3
D2 2 g
D3 2 g
800 ft V22
900 ft V32
 0.045
0.5 ft 2 g
0.33 ft 2 g
V2  1.74V3
The corresponding relationships among the flow rates are:
Q
Q2
 1.74 3
A2
A3
Q2  1.74
A2
0.196 ft 2
Q3  1.74
Q3  3.92Q3
A3
0.087 ft 2
The principle that the flow into the junctions must equal the flows out of them is, essentially, just
a statement of the principle of continuity. Formally, we can express this idea at point B as:
Q1  Q2  Q3  3.92Q3  Q3  4.92Q3
AV
1 1  4.92 A3V3
2
V1  4.92
A3
0.0873 ft 2
V3  4.92
V3  1.23V3
A1
0.349 ft 2
The corresponding calculations at point C are:
4.92Q3  Q4
A3
0.0873 ft 2
V4  4.92 V3  4.92
V3  2.19V3
A4
0.196 ft 2
We now know all the velocities in terms of V3 and can write the energy equation between points
E and D with V3 as the only unknown. We can write this energy equation using either the path
through pipes 1, 2, and 4, or that through pipes 1, 3, and 4; choosing the latter option, keeping in
mind that we are ignoring the change in velocity head between points E and D, and noting that
the pressure is zero at both points, we find:
pE

 zE 
pD

 z D  hL ,1  hL ,3  hL ,4
l V2
l 1.23V3 
l  2.19V3 
150 ft  30 ft  f1 1
 f3 3 3  f 4 4
D1
2g
D3 2 g
D4
2g
2
2
1200 ft 1.231V3 
900 ft V32
500 ft  2.19 V3 
120 ft  0.035

0.045
 0.025
2
0.667 ft 2  32.2 ft/s 
0.333 ft 2 g
0.50 ft
2g
2
V32  22.9
V3  4.79
2
ft 2
s2
ft
s
Substituting this result back into the expressions for the relative velocities and the flow rates
yields:
ft 
ft

V1  1.23  4.79   5.90
s
s

ft 
ft 3
2 
Q1  AV

0.349
ft
5.90

2.06


1 1


s
s

ft 
ft

V2  1.74  4.79   8.35
s
s

ft 
ft 3

Q2  A2V2   0.196 ft   8.35   1.64
s
s

V3  4.79
ft
s
2
ft 
ft 3

Q3  A3V3   0.0873 ft 2   4.79   0.42
s
s

3
ft 
ft

V4  2.19  4.79   10.5
s
s

ft 
ft 3

Q4  A4V4   0.196 ft 2  10.5   2.06
s
s

(b) To determine the pressures at points B and C, we utilize the energy equation. For point B, we
write the equation between points E and B, as follows:
pE

pB
 zE 

 z B  hL ,1
1200 ft  5.90 ft/s 
150 ft 
 70 ft  0.035

0.667 ft 2  32.2 ft/s 2 
2
pB
pB

 45.96 ft
lb   1 ft 2 

pB   45.96 ft   62.4 3  
  19.9 psi
ft   144 in 2 

Similarly, the pressure at point C can be computed by writing the energy equation between
points B and C:
pB

 zB 
pC

 zC  hL ,2
800 ft  8.35 ft/s 
45.96 ft  70 ft 
 60 ft  0.025
0.50 ft 2  32.2 ft/s 2 

pc
pC

2
 12.65 ft
lb   1 ft 2 

pB  12.65 ft   62.4 3  
  5.48 psi
ft   144 in 2 

Representing Simple Pipe Networks as a Smaller Number of Equivalent Pipes
The systems analyzed above each have one point at which all the flow enters the system and
another at which all the flow exits, allowing the flow direction in each pipe to be inferred
unambiguously. In such cases, if we are just interested in the relationship between total flow and
total headloss, it is sometimes convenient to simplify the analysis by representing the whole
group of pipes as a single, hydraulically equivalent pipe.
4
For such a representation, the ‘friction intensity’ parameter for each pipe must be known, and
it must be independent of the flow conditions for the range of flow conditions of interest. If the
D-W equation is used to relate hL to V or Q, the friction intensity factor is f, which has a constant
value in a given pipe for fully turbulent flow, but not for laminar or transitional flow (even
though we assumed otherwise in Example 1). If the Hazen-Williams (H-W) equation is used, the
friction intensity factor is CHW, which is assumed to be known for a given pipe.
To demonstrate how we can identify the characteristics of an equivalent pipe for a given real
pipe network, we consider again the system analyzed in Example 1. This time, however, we will
use the H-W equation and assume that the CHW values are 110 in pipes 1 and 2, and 90 in pipes 3
and 4. In addition, we will specify that the diameter of the equivalent pipe is 8 inches, and that it
has a CHW of 100; the only parameter that must be calculated is the equivalent pipe length.
E
2
A
1
C
Water
B
4
3
D
We begin by representing just pipes 2 and 3 by a section of the equivalent pipe. Both these
pipes link points B and C, so the headloss through the two pipes must be identical. Expressing
this equality using the H-W equation for headloss in BG units, we find:
4.73Q21.85
1.85
 Q2 
 
 Q3 
l3
l2
1
1
 4.73Q31.85 4.87
4.87
1.85
1.85
D2 CHW ,2
D3 CHW ,3
D 
 2 
 D3 
Q2  D2 


Q3  D3 
4.87
4.87 1.85
1.85
 l3   CHW ,2 

  
 l2   CHW ,3 
1 1.85
 l3 
 
 l2 
1.85 1.85
 CHW ,2 

 C
 HW ,3 
D 
 2 
 D3 
2.54
 l3 
 
 l2 
0.54
 CHW ,2 


 CHW ,3 
Substituting given values yields:
Q2  0.5 ft 


Q3  0.33 ft 
2.54
 900 ft 


 800 ft 
0.54
 110 

  3.74
 90 
5
A pipe that is hydraulically equivalent to the combination of pipes 2 and 3 must carry a total
flow rate of Q2 + Q3 at the same headloss as actually occurs in each of those pipes in the real
system. The flow in the equivalent pipe would therefore be:
Q

Qeq ,23  Q2  Q3   2  1 Q3   3.74  1 Q3  4.74Q3
 Q3 
Again using the H-W equation to equate headlosses, this time between pipe 3 and the equivalent
pipe, we find:
1.85
4.73Qeq
,2  3
leq ,23
l3
leq ,2 3
1
4.87
eq ,2  3
D
C
1.85
 Q3 


Q
 eq ,23 
1.85
 1 


 4.74 
 4.73Q31.85
1.85
HW , eq ,2  3
 Deq ,23 


 D3 
 8 in 


 4 in 
4.87
4.87
l3
1
4.87
1.85
D3 CHW ,3
1.85
 CHW ,eq ,23 


 CHW ,3 
1.85
 100 


 90 
 2.00
leq ,23  2.00l3  2.00  900 ft   1800 ft
We now know that the parallel pipes in the middle of the network can be represented by a
single, 8-in diameter pipe that is 1800 ft long and has a CHW of 100. We can use that information
to represent the whole network as three pipes in series: Pipe 1, the equivalent pipe to pipes 2 and
3, and pipe 4. An pipe that is equivalent to the whole network must have the same total headloss
as the real system, when the total flow is Q2 +Q3, or Qtot. We can represent this equality as:
hL ,eq ,tot  hL ,1  hL ,eq ,23  hL ,4
leq ,tot
1.85
4.73Qeq
,tot
4.87
eq ,tot
D
1
C
1.85
HW , eq ,tot

leq ,2 3
l1
l4
1
1
1 
1.85
 Q11.85 4.87
 Qeq
 Q41.85 4.87
,2  3
1.85
4.87
1.85
1.85 
D1 CHW ,1
Deq ,2 3 CHW ,eq ,2 3
D4 CHW ,4 

Since all the Q’s are identical (equal to Qtot), this equality can be simplified to:
leq ,tot
4.87
eq ,tot
D
1
C
1.85
HW , eq ,tot

l ,2 3
1
1
1
l1
l4
 eq4.87
 4.87
4.87
1.85
1.85
1.85
D1 CHW ,1 Deq ,2 3 CHW ,eq ,23 D4 CHW ,4
All the terms in this equation except leq,tot are known, so we can substitute and solve for that
value:
leq ,tot
 0.667 ft 
1
4.87
100 
1.85

1200 ft
 0.667 ft 
1
4.87
110 
1.85

2000 ft
 0.667 ft 
4.87
1
100 
1.85

500 ft
1
 0.5 ft 
 90 
4.87
1.85
6
leq ,tot  5270 ft
Thus, we conclude that the network is hydraulically equivalent to a single, 8-in diameter, 5270-ft
long pipe, with CHW of 100. Note that these calculations did not utilize any specific value of Qtot,
so the results are valid no matter what the total flow rate through the system is.
Analysis of Complex Pipe Networks: Multiple Inlets/Outlets in a Single Loop
The analysis techniques used above are satisfactory if the pipe system is simple enough that
the flow direction in each pipe is known unambiguously. In more complex systems, pipes might
be combined in interconnected loops in ways that make it difficult to determine even the
direction of flow in any given pipe. The fundamental relationships that we have used up to this
point – the energy and continuity equations and the relationships between flow and headloss in
any given pipe – still apply in such a system, but the sheer number of equations that need to be
satisfied to determine the complete flow conditions is daunting. The conditions in such systems
are usually solved with specialized computer programs designed specifically for that purpose.
However, before such programs were widely available, manual and spreadsheet techniques were
developed for analyzing the systems. These techniques provide a bridge between very simple
problems like those analyzed above and the massive ones that can be solved only with special
software. These techniques, as well as more sophisticated ones, allow us to answer such
questions as:
* For a given set of flow rates, what will be the head loss in each pipe in the network?
* Will additional head have to be supplied by pumps in order to obtain the desired flows?
* How much will the flow rates change in various parts of the system if a new pipe is
installed, connecting two previously unconnected points, or to replace an older, smaller
pipe?
* How much will the pressure at a consumer’s tap drop if the fire hydrant outside his or her
home is in use?
Solving the System of Equations
The methods used historically to solve pipe network problems are “relaxation” techniques, in
which a few of the variables are estimated, the others are computed, and the original estimates
are revised based on algebraic manipulations of the computed variables. The approach bears
many similarities to the iterative technique used to determine the flow rate through a pipe
connecting two reservoirs at known elevations, but rather than iterating on the velocity in a
single pipe, the velocities in all the pipes in specified loops are adjusted simultaneously. As is the
case for individual pipes, the numerical solution depends to some extent on which equation is
used to relate flow to headloss.
With respect to pipe network analysis, the traditional approach is known as the Hardy Cross
method. This method is applicable if all the pipe sizes (lengths and diameters) are fixed, and
either the headlosses between the inlets and outlets are known but the flows are not, or the flows
at each inflow and outflow point are known, but the headlosses are not. This latter case is
explored next.
7
The procedure involves making a guess as to the flow rate in each pipe, taking care to make
guesses in such a way that the total flow into any junction equals the total flow out of that
junction. Then the headloss in each pipe is calculated, based on the assumed flows and the
selected flow vs. headloss relationship. Next, the system is checked to see if the headloss around
each loop is zero. Since the initial flows were guessed, this will probably not be the case. The
flow rates are then adjusted in such a way that continuity (mass in equals mass out) is still
satisfied at each junction, but the headloss around each loop is closer to zero. This process is
repeated until the adjustments are satisfactorily small. The detailed procedure is as follows.
1. Define a set of independent pipe loops in such a way that every pipe in the network is
part of at least one loop, and no loop can be represented as a sum or difference of other
loops. The easiest way to do this is to choose all of the smallest possible loops in the
network.
2. Arbitrarily choose values of Q in each pipe, such that continuity is satisfied at each pipe
junction (sometimes called nodes). Use a sign convention such that Q in a given pipe is
designated to be positive if the (assumed) direction of flow is clockwise in the loop under
consideration. This convention means that the same flow in a given pipe might be
considered positive when analyzing one loop, and negative when analyzing the adjacent
loop.
3. Compute the headloss in each pipe, using the same sign convention for headloss as for
flow, so that hL in each pipe has the same sign as Q, when analyzing any given loop.
4. Compute the headloss around each loop. If the headloss around every loop is zero, then
all the pipe flow equations are satisfied, and the problem is solved. Presumably, this will
not be the case when the initial, arbitrary guesses of Q are used.
5. Change the flow in each pipe in a given loop by Q. By changing the flow rates in all the
pipes in a loop by the same amount, we assure that the increase or decrease in the flow
into a junction is balanced by the exact same increase or decrease in the flow out, so that
we guarantee that the continuity equation is still satisfied. The trick is to make a good
guess for what Q should be, so that the headloss around the loop gets closer to zero after
each adjustment. To achieve this, we assume that we can choose a value of Q that is
exactly what is needed to make the headloss zero, and then see how this value of Q is
expected to be related to other system parameters. The relationship is derived as follows.
Recall that, if the term characterizing the frictional “intensity” in a pipe is assumed to be
constant, the D-W, H-W, and Manning equations can all be written in the form:
h f  KQ n
(1)
where K is a constant that includes geometric variables and the frictional intensity term. This
intensity term is f in the D-W equation, CHW in the H-W equation, and nM in the Manning
equation. The value of the exponent n is 2.0, 1.85, and 2.0 in those three equations, respectively.
If the frictional intensity term is not constant (as in the D-W equation), we can nevertheless
8
compute it (e.g., from the Moody diagram) for the given estimate of Q, and then adjust it when
that estimate is updated.
Consider a pipe i in a particular loop (j) in the system. Designate the initial estimate of Q in
that pipe as Q0, and the corresponding headloss as h0. The new estimate of the flow through the
pipe will then be Qi,0 + Qj. (Because the flow rate adjustment is applied to all pipes in the loop,
we identify Q with the loop, not the individual pipe.) We can then estimate the initial (subscript
0) and new (subscript 1) headloss through the pipe to be:
hL,i ,0  Ki Qin,0
(2)
hL ,i ,1  K i Qin,1  K i  Qi ,0  Q j ,0 
n
(3)
 K i Qin,0  nK i Qin,01Q j ,0  higher order terms in Q j ,0
(4)
Assuming that the higher order terms are small compared to the first two, we can drop them
from Equation 4. Next, we sum the headlosses in all the pipes around the loop being analyzed. If
Qj,0 was perfectly chosen, then that sum would be zero, so:
h
L, j
0
loop j
 K Q
n
i ,0
i
 nK i Qin,01Q j 
(5)
all pipes
in loop j
Since the same n and the same Qj apply to every pipe in the loop, those terms can be taken
outside the summation, yielding:
0   K i Qin,0  nQ j  K i Qin,01
loop
Q j  
(6)
loop

K i Qin,0

K i Qin,01
all pipes
in loop j
n
all pipes
in loop j


hL ,i ,0
all pipes
in loop j
h
n  L ,i ,0
all pipes Qi ,0
(7)
in loop j
Equation 7 provides a way to calculate a value of Qj that will cause the headloss around the
loop to be zero, if the higher order terms that are dropped between Equations 4 and 5 are indeed
negligible. For the first few iterations, that assumption is likely not to be correct, so the
computed value of Qj will not cause the headloss around the loop to be exactly zero, but it will
cause the headloss to be closer to zero than it was in the previous iteration. The value of Qj can
then be added to the preceding values of Qi for all the pipes in the loop, and a new iteration can
be carried out. This same process can be used for all the loops in the network. If a pipe is part of
two or more different loops, the correction factors for all the loops that contain it are applied to
it.
9
As noted previously, the initial guess of the flow rates is entirely arbitrary, as long as
continuity is satisfied at each junction. If one makes good guesses for these flow rates, the
problem will converge quickly, and if one makes poor guesses, it will take more iterations before
the final solution is found. However, any guesses which meet the mass balance criterion will
ultimately lead to the same, correct final result.
_______________________
Example 2. Determine the flow rates in all the pipes in the network shown below. Use the DW
equation to relate headlosses to flow rates. If the pressure head at point a is 40 m, find the
pressure head at D (which might represent a fire demand, for example).
Solution. The analysis can be done in a spreadsheet, as shown below. All values are in units of
meters and seconds.
1
2
3
4
5

D2/4
7
8
9
Q/A
V
0.566
5.093
0.796
0.796
0.637
VD/n
Re
169765
509296
159155
159155
127324
Loop
A
Pipe
Ab
Be
Ed
Dc
Ca
D
0.30
0.10
0.20
0.20
0.20
l
250
100
125
125
100
/D
0.83
2.50
1.25
1.25
1.25
A
0.0707
0.0079
0.0314
0.0314
0.0314
Assumed
Q
0.040
0.040
0.025
0.025
0.020
B
Cd
Dg
Gf
Fc
0.20
0.15
0.25
0.15
125
100
125
100
1.25
1.67
1.00
1.67
0.0314
0.0177
0.0491
0.0177
0.025
0.020
0.020
0.005
0.796
1.132
0.407
0.283
159155
169765
101859
42441
C
de
eh
hg
gd
0.20
0.10
0.25
0.15
125
100
125
100
1.25
2.50
1.00
1.67
0.0314
0.0079
0.0491
0.0177
0.025
0.015
0.000
0.020
0.796
1.910
0.000
1.132
159155
190986
0
169765
10
1
2
Loop
A
B
C
11
f=0 if Re=0
ff
0.598
8.642
1.128
1.128
1.128
12
DW eqn
hL
8.13
11425.05
22.76
22.76
11.65
13
14
15
16
17
Pipe
ab
be
ed
dc
ca
10
Haaland
f
0.598
8.642
1.128
1.128
1.128
hL/Q
203
285626
910
910
583
hL
(hL/Q)
Q
11422
288233
0.0198
new Q
0.020187
0.020187
0.011819
0.03574
0.03981
cd
dg
gf
fc
1.128
2.089
0.776
2.090
1.128
2.089
0.776
2.090
22.76
90.91
3.28
5.69
910
4545
164
1137
0.0091
0.035738
0.017557
0.010925
0.00407
de
eh
hg
gd
1.128
8.643
#DIV/0!
2.089
1.128
8.643
0.000
2.089
22.76
1606.91
0.00
90.91
910
107127
0
4545
0.0066
0.01182
0.008368
0.00663
0.01756
123
1493
6757
112583
Columns 1-4 reiterate the given information. Columns 5 and 6 are fixed values that depend
only on the given input information. Column 7 is the assumed flow rate through the given pipe
for the current iteration. Note that the flow in pipe cD is assumed to be from left to right, which
is counter-clockwise in Loop A, but clockwise in Loop B, so the flow rate is assigned a negative
value in Loop A and a positive value in Loop B.
Columns 8 and 9 show the computed velocities (without sign) and the Reynolds numbers in
the pipes, and Column 10 shows the estimate of L based on the Haaland equation. Because this
equation has Re in the denominator, f is undefined in pipes with zero flow; in Column 11, values
of L for such conditions are assigned values of zero. The value in Column 12 is hL, based on the
DW equation, with the same sign convention as for flow rate, and Column 13 shows hL/Q. Note
that, because hL and Q always have the same sign, hL/Q is always positive.
Columns 14 and 15 show the summation of hL and hL/Q around each loop. Column 16 then
shows the computed value of Q to be applied as a correction to the flow in each pipe. Note that
one Q is computed for each loop and is applied to every pipe in the loop. Pipes that are in more
than loop have the corrections for both loops applied to them. For example, the flow rate
adjustments s for a few pipes are computed as follows. Note that, when analyzing Loop A, the
flow rate and the adjustment between points c and d have the same magnitude but opposite signs
as when the same flow is analyzed for Loop B.
11
Loop
A
Pipe
ab
Adjustment between iterations
Qab ,i 1  Qab ,i  Qloop A
A
dc
Qdc ,i 1  Qdc ,i  Qloop A  Qloop B
B
cd
Qcd ,i 1  Qcd ,i  Qloop A  Qloop B
B
dg
Qdg ,i 1  Qdg ,i  Qloop B  Qloop C
The calculated flow rates and the percentage changes in those rates for the first and seventh
iterations are summarized below. The adjustments after the seventh iteration are all <2%, so the
flow rates in this iteration would be considered accurate.
Iteration #1
new Q
% change
0.0202
49.53
0.0202
49.53
0.0118
79.25
79.25
0.0357
0.0398
99.07
…
Iteration #7
new Q
% change
0.0044
0.13
0.0044
0.13
0.0026
0.23
0.02
0.0375
0.0556
0.01
0.0357
0.0176
0.0109
0.0041
36.30
45.37
45.37
181.49
0.0375
0.0101
0.0031
0.0181
0.09
0.33
1.10
0.18
0.0118
0.0084
0.0066
0.0176
26.53
44.21
#DIV/0!
33.16
0.0026
0.0019
0.0131
0.0101
0.88
1.20
0.17
0.22
As noted above, in the other type of problem for which the Hardy Cross method is
applicable, the headlosses allowable between the inlet and outlet or between some other points in
the system are known, and the information being sought is the flow rates that can be achieved.
This might be the case if two reservoirs are to be connected. In this situation, rather than
guessing flow rates, one guesses the piezometric head at any inlet, outlet, or node in the system
where it is not known. The frictional headloss in each pipe (loss of piezometric head minus loss
of elevation head) is then determined, the flow rates corresponding to these headlosses are
computed, and each junction is analyzed to determine whether the continuity equation is
satisfied, i.e., whether the flow into each junction equals the flow out of it. If not, improved
guesses are made for the unknown heads, and the process is repeated. The derivation follows.
Derivation of the Hardy Cross Equations for a system with known hL and unknown Q
1) Choose arbitrary values for the piezometric head, hj, at each node j where it is not
already specified. Using these values, compute the headloss in each pipe.
12
2) Using any appropriate relationship between Q and hL, compute the flow rate through
each pipe in the network. Designate flow toward the node as positive and away from
the node as negative.
3) Check whether the continuity equation is satisfied at each node, for this zeroth
iteration. The formal mathematical statement corresponding to this test at node j is
N
Q
ij ,0
i 1
= 0. If the sum of inflows equals the sum of outflows at each junction, the
problem is solved. If the equation is not satisfied at any junction, adjust the estimate
of the head at the junction as shown below, and continue.
The magnitude of the adjustment in hj is determined as follows. Let hj be the under-estimate
in the value of hj during a given iteration. Assuming the relationship between flow and headloss
can be represented as hL,ij  Kij Qijn , we can rearrange the equation to solve for Qij. Then,
representing the headloss in pipe i during the first iteration as the sum of that during the zeroth
iteration and the change in hj that we will apply, we find:
hL ,ij  Kij Qijn
Qij ,1 
1/ n
1 1/ n
1
hL ,ij ,1 
hL ,ij ,0  h j ,0 

Kij
Kij
Expanding the right hand side and assuming terms of the order of hL2,ij and higher are small:
Qij ,1 
1 1/ n 1 1 1/ n 1
hL ,i ,0 
hL ,ij ,0 h j ,0
Kij
n Kij

1/ n
1 1/ n 1 1 hL ,ij ,0
hL ,i ,0 
h j ,0
Kij
n Kij hL ,ij ,0
Qij ,1  Qij ,0 
1 Qij ,0
h j ,0
n hL ,ij ,0
Summing all the Qij at a given junction:
Q
ij ,1
i


1 Qij ,0
   Qij ,0 
h j ,0 


n hL ,ij ,0
i 

at each junction
Separating the two terms in the summation, and setting the sum to zero (since we are trying
to find the change in head at i that will cause continuity to be satisfied in the first iteration)
yields:
13
0   Qij ,0  
i
i
1 Qij ,0
h j ,0
n hL ,ij ,0
Noting that n and, at a given junction, hj,0 are both constant, and then solving for hj,0:
0   Qij ,0 
h j ,0
n
i
h j ,0  
Qij ,0
h
i
L ,ij ,0
n Qij ,0
i
Qij ,0
h
i
L ,ij ,0
This result indicates the adjustment that should be made in the estimated head at the node.
We then return to step 2 and repeat the iterations until the adjustment is sufficiently small.
14