1.1. Introduction · 2019. 12. 25. · 1 Pipe Networks 1.1. Introduction Pipe networks, which are interconnected and often have a netting structure, are used for transportation or

1

Pipe Networks

1.1. Introduction

Pipe networks, which are interconnected and often have a netting structure, are used for transportation or even distribution of fluids from their storage or production areas to various other areas for certain purposes.

These networks are used in everyday life, such as plumbing networks of water transportation for productive applications such as fire extinguishing, networks of natural gas transportation, networks of fluid and gas transportation, and networks of water pipes, wet waste and compressed air.

Some of these networks are simple piping systems equipped with flow adjustment devices, whereas others are complicated, such as fluid distribution networks. Some of the most complicated networks are fire extinguishing networks (because of the fluid used for fire extinguishing), water distribution networks and plumbing networks.

Depending on the complication of mixed networks and the form they take for the feasibility of the distribution service, we distinguish them into the following three categories:

1) The tree-system-type pipe networks: these networks are characterized by the presence of a central pipe from which other pipes are branched (pipes of distribution) with a gradual decrease in their cross-section, taking the form of a tree (tree system), as shown in Figure 1.1.

2) The grid pattern: during development the pipes are formed into this pattern, resembling a chessboard, which covers the whole area of the distribution, with a decrease in their cross-sections with respect to the distance (grid system), as shown in Figure 1.2.

COPYRIG

HTED M

ATERIAL

2 Fluid Mechanics 2

Figure 1.1. Tree-system-type pipe network

Figure 1.2. Grid pattern networks

3) The loop pattern: this pattern contains a central pipe, forming loops with smaller pipes at low flow rates (loop system), as shown in Figure 1.3.

Figure 1.3. Loop pattern network

Pipe Networks 3

Of the three forms of networks, the loop and grid systems show high reliability due to their flexibility, expandability and ability to offer multiple paths to the fluid.

In general, in all the aforementioned network systems, we distinguish three groups of pipelines:

1) Transportation lines: these pipes transfer the fluid from the storage or production area to the distribution area.

2) Central pipes: these pipes transfer the fluid to the target area, e.g. transfer of water to a town or village or transfer of natural gas to an industrial installation.

3) Supply lines: these pipes of small diameters transfer the fluid from central pipes to the users.

Thus, the entire distribution system consists of pipes, valves and pumps. The fundamental aim of a network system is to supply sufficient amounts of fluid to target areas with desired pressures and flow rates. Therefore, the choice of materials, the diameter of pipes and the formation of pipelines in networks are mostly influenced by the necessity of ensuring sufficient pressures and flow rates, despite installation costs and operations.

1.2. Calculation of pipe networks

Simple pipe networks have procedures for connecting the pipes in a row or in parallel, as described in Chapter 5 of Volume 1 [GEO 18]. However, the same is not true for complicated pipe networks. The schematic representation of a typical plumbing network is shown in Figure 1.4.

Figure 1.4. Schematic representation of a pipe network

The geometric convergence of three or more pipes is called a network node. In technical applications, nodes with more than four branches do not exist. Nodes with

4 Fluid Mechanics 2

three branches (which are most common in practice) are classified into branch nodes (Figure 1.5a) and convergence nodes (Figure 1.5b). In branch nodes, the incoming branch with a flow rate Q is divided into two branches with flow rates Q1 and Q2, while in convergence nodes, two branches with flow rates Q1 and Q2 converge into one branch with a flow rate Q.

When the fluid passes through the node, there is some kind of energy loss, which can be attributed to a decrease in the cross-sectional areas of branches 1 and 2, so that the average velocity of the fluid in all the three branches of the node is approximately the same (V≈V1≈V2), and the walls of the branches are rounded instead of being sharp, to avoid higher energy losses.

Figure 1.5. Pipe network nodes

Minor energy losses at a node can be calculated either by the method of losses coefficient or by the method of equivalent lengths. In Table 1.1, typical values of the losses coefficients for nodes T−90° with constant diameter are given. Furthermore, in Table 1.2, representative values of the equivalent lengths in diameters for tees are given. For a flow in parallel connection, the following two basic rules are applicable:

1) For an incompressible flow in a node, the algebraic sum of flow rates in its branches is zero. This means:

1

i k

ii

Q Q=

== [1.1]

for the flow rates of the k branches of the node. During the addition of flow rates, streams entering the node are considered positive, while those leaving the node are considered negative. By applying this rule at node α, shown in Figure 1.6, we get:

[1.2] 1 2 3Q Q Q Q= + +

Pipe Networks 5

Table 1.1. Coefficient of losses for T-90 nodes

KIND OF APPLIANCE KIND OF APPLIANCE

Sudden dilatation 20 Tee, entrance from main line 20

Sudden systole 12 Tee, entrance from branch 60

Borda mouth 28 Valves (totally open):

Mouth with sharp lips 18 back flow with swing 135

curve 16 Angular 145

Standard curve 30 Hydrant 18

curve of large radius 20 Butterfly 20

angle 60 Sliding 13

curve 65 Spherical 340

Table 1.2. Equivalent length values

BRANCH NODE CONVERGENCE NODE

Q2/Q 0 0.2 0.4 0.6 0.8 1,0 Q2/Q 0 0.2 0.4 0.6 0.8 1.0

Km,1 0.04 0.08 0.05 0.07 0.21 0.35 km,1 0.04 0.17 0.30 0.41 0.51 0.60

Km,2 0.95 0.88 0.89 0.95 1.10 1.28 km,2 -1.20 -0.40 0.08 0.47 0.72 0.91

22 VV 00h = k h = km ,0 1 m ,1 m ,0 2 m ,2 g2g2

22 VV 00h = k h = km ,0 1 m ,1 m ,0 2 m ,2 g2g2 →→

0 1 2( )Q Q Q= = 1 2 0( )Q Q Q= =

→ →

/ de / de

( )/ 1/ 21 2d d =

( )/ 1/ 22 1d d =

45°

90°

90° ( )6d in≥

90°

180°

6 Fluid Mechanics 2

Relationship [1.1] is a mathematical expression of the node theorem, also known as the first law of Kirchhoff in the theory of electrical networks

2) For a steady flow in a hydraulic network, for example, in the network shown in Figure 1.6, the total head loss h between the nodes α and b is the same as the

respective head losses, ,ih , in each branch i of the network, which means:

, ,1 ,2 ,3ih h h h h= = = = [1.3]

where ,1 ,2,h h and ,3h are the heads of the energy losses in branches 1, 2

and 3, respectively, for the corresponding flow rates Q1, Q2 and Q3. Equation [1.3] constitutes the mathematical statement of the energy principle of the hydraulic network

Figure 1.6. Nodes and branches of a pipe network

The procedure followed for obtaining the solutions of these problems depends on the type of information asked. Therefore, if the total head loss of the flow is known, then it is easy to calculate individual flow rates iQ and finally their sum.

The reversed problem is solved with successive approximations because the distribution of the flow rate Q in the individual branches of the network is not known. In fact, in the first approximation, we consider zero energy loss at the nodes and, if they are considered important, we add them in the second approximation. In more complex hydraulic networks, the balance sheet method, head losses in loops and flow rates in nodes are applied. It is difficult to obtain the solution of such network problems; it can be obtained with a suitable computer program.

Pipe Networks 7

However, in addition to this form of network, there are more complicated distribution networks, particularly the grid and loop patterns. A loop with one inlet and two exits, as shown in Figure 1.7, is impossible to obtain by the methods that we developed in Chapter 5 of Volume 1 [GEO 18].

Figure 1.7. Loop pattern network

Before we develop a method for solving these types of networks, we should state the basic relation of these calculations. Therefore, considering that a network consists of pipes (branches) with nodes and forms closed circuits or loops, we will individually examine the relations of the nodes and the loops.

Viewing a network macroscopically and applying the continuity equation to the network, we have:

ent exm mΣ = Σ [1.4]

and for incompressible fluids, the equality of flow rates is also applicable:

ent exQ QΣ = Σ [1.5]

A respective relationship is applicable for every node of the network. If we conventionally pre-label the flow rates (+ for entrance to the node and − for exit from the node), we have:

0nodeQΣ = [1.6]

For the node shown in Figure 1.8:

8 Fluid Mechanics 2

1 2 3 4 5 0Q Q Q Q Q− + − − =

or 1 3 2 4 5Q Q Q Q Q+ = + + [1.7]

In addition to the points of the branch or the nodes, a network is characterized by closed circuits or loops. A loop is a closed path formed by the sum of successive pipes, which will lead us back to the starting point if we follow them. Thus, in the pipe shown in Figure 1.9, starting from Α and following a clockwise route, we will return to A again: Α–Β–C–D–Α. The pipes that constitute the loop are called branches. For the formation of a loop, at least two branches are required.

Figure 1.8. Loop pattern of network

The loop shown in Figure 1.9 constitutes four branches (ΑΒ, ΒC, CD and DΑ) and four nodes (Α, Β, C and D). If we consider the flow path in the pipes shown in Figure 1.9, by applying the Bernoulli equation between points Α and C of the fluid paths ΑΒC and ADC, it is given that:

Figure 1.9. Closed-loop pattern of a network

ΓΔ

A B

Pipe Networks 9

ABC ADC AB BC AC DCh h h h h h= + = +

0AB BC AD DCh h h h + − − = [1.8]

If we conventionally pre-label the losses, putting the sign + when the fluid flow is clockwise in the loop and the sign − when it is opposite, equation [1.8] becomes:

0AB BC AD DCh h h h+ + + =

0looph Σ = [1.9]

Equation [1.9] is general and applicable for any closed system of pipes (loop). In combination with equation [1.6] of the branch points (nodes), it is the key for the solution of the network problems. We emphasize that both these relationships have resulted from conventional pre-labeling of the flow rates and the losses. In their specific application, we must take into account the rules of pre-labeling and specify the respective relationships. For example, for the nodes shown in Figure 1.8, the equation of the nodes will take the form [1.5], while for the node shown in Figure 1.9, the loop equation will take the form [1.8].

Finally, we remind ourselves that for each branch, we have the losses equation (Darcy–Weisbach):

22 2

8 ii i i i

ii

h f K Qdg dπ

= ⋅ ⋅ + Σ ⋅

⋅ ⋅

which can take the form:

2i i ih a Q= ⋅ [1.10]

where 2 2

8 ii i i

ii

a f Kdg dπ

= ⋅ ⋅ + Σ

⋅ ⋅

[1.11]

The analysis and calculation of the distribution networks is a complicated and time-consuming procedure, which is based on equations [1.6], [1.9] and [1.10] (nodes, loops and branches, respectively).

10 Fluid Mechanics 2

1.3. Problem-solving methodology for pipe networks

Let us consider a pipe network in which the pipes have a common point, that is, a node or a bend, and the other side is connected to a tank, whose level is at a height h, as shown in Figure 1.10.

In this category of problems, the kinematic viscosity ν, the lengths of the pipes

1 2 3, , , their diameters 1 2 3, ,d d d , their heads 1 2 3, ,h h h and their roughnesses

1 2 3, ,ε ε ε are usually given while their flow rates 1 2 3, ,Q Q Q are asked.

The following methodology is used to solve this category of problems:

1) We assume a value for the head hΑ , where:

ph hγ

ΔΑ Δ= + = pressure head [1.12]

Figure 1.10. Pipe network with a common point for the pipes

2) From the Darcy–Weisbach relationship, the given data of the problem and the Moody diagram, we find the flow rates 1 2 3, , , ..., nQ Q Q Q .

For example:

2

2n n

n n A nn

Vh h h f

d g= − = ⋅

[1.13]

where nf is found by various tests, and nV can be found by the relationship:

Pipe Networks 11

2 n nn

n n

gh dV

f=

[1.14]

Therefore, 2 2

4n n n

nn n

d gh dQ

fπ

=

[1.15]

3) After calculating all the flow rates, we may use the continuity equation. This means that the values that we find for 1 2 3, , , ..., nQ Q Q Q have to satisfy the

continuity equation.

If the flow rates Q1, Q2 and Q3 are higher than the flow rates of branches, then we obtain a higher value for hA, and we repeat steps 1, 2 and 3. If the flow rate to the branch is lower than the flow rate from the branch, then we use a smaller value for hA.

It is clear from the above procedure that for these problems, the following conditions are applicable:

1) For every pipe, the Darcy–Weisbach relationship must be satisfied.

2) Q → branch Q= → branch.

3) There has to be a flow from the higher tank to a lower one. This means that one of these relationships must be applicable:

2 1 3Q Q Q= + [1.16]

or 3 1 2Q Q Q= + [1.17]

As 1Ah h< , we have a flow toward a tank, so it is given that:

2 1 3Q Q Q= + [1.18]

4) Regardless of the losses in the flow in any pipe, equivalent lengths will be able to express them, which are added to the real length, as mentioned in section 1.2.


1.4. Overall approach for the network calculation

In the beginning of this chapter, we mentioned that the basic purpose of a distribution network is to ensure the necessary flow rate and pressure for the exit of the fluid from the network. Consequently, a water supply network must ensure a flow rate higher than 1 / s for every user and a pressure of 2 bar.

The flow rate is related to not only the selection of the pipes with suitable diameters but also the differences of the energy heads between the nodes of exit and inlet, as well as the presence or absence of pumps in the network. The collateralization of the required pressure is related to the energy heads: the pressure head at the entrance nodes, the heights and the height that the pumps attribute. More specifically, if we assume a distribution network that is supplied by a tank, as shown in Figure 1.11, the Bernoulli equation, if applied between the surface of tank Α and the node Ε, will give:

Figure 1.11. Pipe network connected to a tank

2 2

2A E A E

A E AE pV V p py y h h

g Y− −

− + = Σ −

2 2

2E A E A

A E AE pp V V py y h hY g Y

− = − + + − Σ + [1.19]

As the variation of the kinetic energy is negligible because the flow velocities are smaller than 3 m/sec, as well as zero at the surface of the tank, the above relationship becomes:

E AA E AE p

p py y h hY Y

= − + − Σ + [1.20]

Pipe Networks 13

where AB E,AE Bh h h hΓ Γ= + + with the losses of every branch pre-labeled

according to the assumption described in section 1.2.

1.5. The Hazen–Williams equation for network analysis

In the previous sections for the solution of the network problems, we used the Darcy–Weisbach losses equation as follows:

2 25 2

1 2 5 2

88

2

V f Qh f f d Qd g d g gπ π

−⋅= = = [1.21]

Considering that the flow is turbulent, we calculate the friction coefficients f from this relationship and then the flow rates.

Figure 1.12. Pipe network for water distribution

Thus, when the flow is turbulent, the losses along a pipe with respect to the flow rate, the friction coefficient and the pipe’s dimensions can be expressed by the following relationship:

1nh CQ= [1.22]

where the coefficient C is a parameter that depends solely on the dimensions and the roughness of the pipe. Relationship [1.22] can be derived from [1.21] after we express the friction coefficient f with respect to the flow rate, which means that:

af KQ= [1.23]


Substituting [1.23] into [1.21] gives [1.22].

After a series of experiments, Hazen and Williams, who worked with pipes of various diameters and roughnesses, found that the value of the average water velocity in pipes is proportional to the hydraulic radius if it is raised to a power and to the square root of the inclination value of the pressure gauge line, which means that:

XhV R S∞ [1.24]

where hR is the hydraulic radius and S is the inclination of the pressure gauge

line = 1 /h , with being the length of the pipe and h1 being the losses during the

length of the pipe.

Therefore, depending on the unit system (SI or British), the Hazen–Williams equation in its final form is given by the relationship:

1) In the metric system (SI):

in [1.25]

If is in m, the coefficient , depending on the type of the pipe, has the

following values:

No. Kind and situation of pipes Value of coefficient

1 Small, not of a good structure up to 40

2 Old, in bad situation 60–80

3 Old, made of cast iron, nailed 95 4 Made of cast iron or mud brick 100 5 New, steel, nailed 110 6 Glazed, in good situation 110 7 Wooden, smooth 120 8 Very smooth 130 9 Made of concrete, of large dimensions 130 10 Asbestos tubes 140 11 Very smooth and rectilinear 140

Table 1.3. Values of the coefficient Wm

0,63 0,540,849 m hV W R S= ⋅ / secm

hR mW

mW

Pipe Networks 15

2) In the British system:

0,63 0,541,318 E hV W R S= in ft/sec [1.26]

Table of values of the coefficient

No. Kind and situation of pipes Value of coefficient

1 Made of cast iron, corroded 80

2 Made of cast iron, after some years of function 100

3 Glazed pipes of drainages 110

4 Average situation of cast iron 110

5 New, steel, nailed 110

6 New, smooth, made of cast iron 130

7 Very smooth and rectilinear 140

Table 1.4. Values of the coefficient WE

If is in ft, the coefficient , depending on the type of the pipe, has the

above values.

1.6. Hazen–Williams and Darcy–Weisbach identity

These two expressions that offer the possibility of solving complicated problems of pipe networks have some common points. Therefore:

1) If we assume relationship [1.25]:

0,63 0,54m hV = 0,849W R S [1.27]

where 1S = h / , and substituting S and rearranging, we have:

0,540,631

0,849 hm

h V RW

− =

1/0,54 1,852

0,63 0,63

0,849 0,8491 h hm m

V Vh = R RW W

− − =

EW

EW

hR EW


For a cylindrical pipe, 4hdR = . Then, we have:

0,63 1,852 1,1671,852 2

1,852 1,852

0.148 2

4 0,7385 2 4( 0,849)1

m m

V d V V g dh =gW W

− × −⋅ ⋅ − = ⋅ ⋅

Now, if we set 2g = 9,81m / sec , we have:

1,167 2 0,148 0,019 2

1,852 0,167 1,852 0,148 0,148

4 2 134 1

2 20,73851

m m

g V V d Vh =d g d gW d W V d

− −× ⋅ ⋅ = =

0,63 1,852 1,1671,852 2

1,852 1,852

0.148 2

4 0,7385 2 4( 0,849)1

m m

V d V V g dh =gW W

− × −⋅ ⋅ − = ⋅ ⋅

0,019 0,148 2

0,148 1,852 0,148 0,198

134

2m

d Vd gW V d

νν

− = =

0,019 2 2

1,852 0,148 0,148

134

2 2Rem

d V L Ufd g D gW ν

− =

However, because 0,019

1,852 0,148 0,148

134

Rem

dfW ν

−

= , we have:

2

21Vh = f

d g

[1.28]

2) If we consider relationship [1.27] and substituting S and rearranging, we have

for a cylindrical pipe, where 4hdR = :

0,540,631

2

41,318

4Ehd QV W

dπ = =

So, solving for 1h

, we find:

Pipe Networks 17

1

0,54

2

0,63

4

1,3184

1

E

Qh d=

dW

π

1,8521,63

1,8521 4,87

4

1,318

Eh Q

W d

= ⋅

1,852

1,8521 4,87

2,3136

Eh Q

W d

= ⋅

[1.29]

where all the lengths are expressed in ft and the flow rate in ft/sec.

Comparing [1.29] and [1.22], we see that:

1,852

4,87

2,3136

EC

W d

=

[1.30]

and n =1,852.

Combining relationship [1.29] with the nomograph of Hazen–Williams, we obtain solutions in various cases of pipe connection, which are much easier and faster than those obtained using the Moody diagram.

Moreover, the Hazen–Williams relationship is simpler than the Darcy–Weisbach relationship because the calculation of the coefficient C is easier. This can be solved by using tables or graphs, and therefore this method has been commonly used in the calculation of water networks. However, it also has some disadvantages of providing less accurate results than the Darcy–Weisbach equations, and it is applicable only if the fluid is water. Moreover, both relationships are empirical, but the Darcy–Weisbach equation is theoretically more close to an analytical method and compatible with the conclusions of the two-dimensional analysis.


1.7. Hardy–Cross method

Another method by which we can solve problems of pipe networks is the Hardy–Cross method, which is a relatively simple procedure. According to this methodology, we initially assume flow rate distribution under only one condition, which is to satisfy the mass conservation in each node. Continuing, we make corrections to these flow rates at each loop of the network, with corrections every time a new value of the flow rate comes up at every circle of calculations so that finally there is a better balance of flow rates in the network than the value we assumed before. If we get in the beginning a good value for the flow rate distribution, we can have convergence in the final value after two or three attempts.

For simple networks, as shown in Figure 1.12, the solution can be obtained easily using a calculator, while for more complicated ones, the assistance of a computer is necessary.

The procedure for finding a solution according to the Hardy–Cross method is as follows:

1) We initially assume the flow rate distribution 01 02 03, , ,...Q Q Q with the only

limitation at each node of the network being the mass conservation applicable, meaning that the amount of the mass of water that goes in is the same that comes out.

2) We calculate the losses at each pipe of the network based on the relationship:

01 1 0nh C Q= [1.31]

Therefore, if there are seven pipes in the circuit, we will find seven losses using the Hazen–Williams nomograph.

01

02

07

1 1 01

1 2 02

1 7 07

n

n

n

h C Q

h C Q

h C Q

=

=

⋅⋅⋅⋅=

3) Choosing any direction and paying attention not to make a mistake in the sign, we find the following sum in each loop of the network:

Pipe Networks 19

10 01 1

m mn

i i ii i

h C Q= =

= [1.32]

where m is the number of pipes in the loop, n = 1,852 and i is the sum indicator, which takes the values 1, 2, 3,…,m.

4) We also calculate the sum of the absolute values of the terms for each loop:

1 10 0

1 1

m mn n

i i i ii i

nC Q n C Q− −

= == [1.33]

5) In order to correct the values of the flow rates in each loop, so that at each

loop we finally have 11

0i

m

ih

== , we calculate the value of the corrective flow rate

QΔ of each loop, where:

01

10

1

mn

i ii

mn

i ii

C QQ

n C Q

=

−

=

⋅Δ = −

⋅

[1.34]

This value of the corrective flow rate is obtained with the following analysis.

Assume that 0Q Q Q= +Δ , where Q is the real flow rate of each pipe, Q0 is the

initial flow rate that we assumed and QΔ is the corrected flow rate value.

However, because we have that:

( ) ( )11 0 0 0 ...

nn n nh CQ C Q Q C Q nQ Q−= = + Δ = + Δ + ,

if QΔ is relatively small with 0Q , we will not take into account the terms that are

powers of the QΔ , and thus it is given that:

10 0

n n nCQ CQ CnQ Q−= + Δ

whereas if the loop has m pipes, the following relationship is applicable:


10 0

1 1 1

m m mn n n

i i i i i ii i i

C Q C Q C nQ Q−

= = == + Δ [1.35]

1

0m

ni i

iC Q

== [1.36]

Moreover, because QΔ is the same for all the pipes of each loop, we can take it out

of the sum, so we have:

01

10

1

mn

i iim

ni i

i

C QQ

n C Q

=

−

=

Δ = −⋅

[1.37]

but as 01 0

1 1i

m mn

i ii i

h C Q= =

= is valid, [1.37] becomes:

0

0

11

1

1 0

1,852

i

i

m

im

i i

hQ

hQ

=

=

Δ = −

[1.38]

01 10

1 10

1,852 1,852im m

ni i

i ii

hC Q

Q−

= = = since 1,852n = [1.39]

6) We calculate the new flow rates for the pipes of each loop from the relationship:

0i iQ Q Q= + Δ

and we repeat the procedure (ii) until we achieve the desired accuracy, for example:

20 10i iQ Q −= < flow rate units [1.40]

Pipe Networks 21

1.8. Formulae

1) Continuity equation for the node networks:

where nodeQ is the sum of the flow rates of all the pipe networks.

2) Losses of a network:

where looph is the sum of the losses networks

3) Losses equation for the network (Darcy–Weisbach)

and if 2 4

8 ii i i

ii

a f Kdgdπ

= ⋅ + Σ

then

where hi is the losses height, fi is the friction operator, i is the pipe’s length, di is

the pipe’s diameter, g is the acceleration due to gravity, Ki is the losses operator and Qi are flow rates.

4) Average flow velocity in a network:

2 n n

nn n

gh dVf

=⋅

2i i ih a Q= ⋅

22 4

8 ii i i i

ii

h f K Qdg dπ

= ⋅ ⋅ + Σ ⋅

⋅

0loophΣ =

0nodeQΣ =


where g is the acceleration due to gravity, hn is the losses, d is the diameter, f is the friction coefficient and is the pipe’s length.

5) Network flow rate:

where all the variables have known values.

6) Bernoulli equation for networks:

where PE is the pressure in the position E, γ is the specific weight of the fluid, yA is the height of point A, yE is the height of point E, PA is the pressure of point A, hAE is the head losses and hp is the pressure head.

7) Losses equation of Hazen–Williams:

where Rn is the hydraulic radius, ihS =

is the inclination of the pressure gauge line,

is the pipe’s length and hi is the losses in the length of the pipe.

a) In the SI:

b) In the British system:

in and in . 0,63 0,541,318 E hV W R S= ⋅ / secft hR ft

in and is in m. 0,63 0,540,849 m hV W R S= ⋅ / secm hR

xnVaR S

E AA E AE p

P Py y h hγ γ

= − + − Σ +

2 2

4n n n

n n

d gh dQ

fπ ⋅

=

Pipe Networks 23

1.9. Questions

1) What are pipe networks, where are they used and what are the categories of networks? Describe each category.

2) How many and which pipes’ groups may we consider for hydraulic systems?

3) What are nodes of the networks, in which categories do we distinguish them and what are their effects on the fluid flow? How many types of node connection are there and which rules are applicable for a parallel connection?

4) Define node theorem. What is the energy condition of a hydraulic network?

5) In a closed circuit of loops and branches, which laws are applicable?

6) What is the solution methodology of a network that consists of a node and a branch, while the other end of the pipe connects to three tanks that lie at a height, different from the node for each one?

7) What is the Hazen–Williams equation for the network analysis, how is it generally expressed and which form does it take in the SI and the British system?

8) Is there any identity (connection) between the Hazen–Williams equation and the Darcy–Weisbach equation, and how is it proved in the SI and the British system?

9) What is the Hardy–Cross method for the solution of network problems and what is the solution procedure according to this method?

1.10. Problems with solutions

1) In a network of tubular pipes made up of galvanized iron of diameter 6 inches, we expect the velocities to vary from 0.5 to 2 m/sec. Calculate the coefficients r and

m in the exponential expression of the loss of the load, mfh rQ= per 100 m of the

pipe’s length. The coefficient of the kinematic viscosity is given as 6 21,31 10 / secn m= × for water at 10°C.

Solution:

The relative roughness is:

0,015250,001

15,25

cmd cmε = =

The expected Reynolds numbers for the two values of the velocities are:


411 6

0,5 0,1525Re 5,82 10

1,31 10

V dν −

×= = = ××

522 6

2 0,1525Re 2,328 10

1,31 10

V dν −

×= = = ××

From the Moody diagram, we find:

1 0024f = and 2 0215f =

Therefore, from the equation:

221

15

0,083(0,024)

2m

f fQVh f h f rQ

d g d⋅= = =

22

25

0,083(0,0215) mf

Qh rQd

= =

Dividing the above equations by terms, we get:

2

22 2 2

1 1 1

0,0215(4) 14,33

0,024

m

fQ f QhQ f Q

= = =

4 14,33m =

From which we take:

14,331,92

4m λογ

λογ= =

because 2 2

1 1

24

0,5

Q VQ V

= = =

The value of r is taken from one of the above exponential equations:

21,922

25

(0,083)(0,0215)(100)

(0,1525)

QrQ=

Pipe Networks 25

0,082

5

(0,083)(0,0215)(100)1660

(0,1525)

Qr = =

because 2 2

2 2(3,14)(0,1525)

2 0,03654 4

dQ Vπ= = =

Finally, the exponential equation 1,921660fh Q= gives the load loss per 100 m

length of the pipe.

2) Two open water tanks Α and Β are connected to a horizontal pipe (see Figure (2)) with an internal diameter of 15 cm, a length of 500 m and a friction coefficient of 0.02. The altitude range of the water’s free surfaces in the two tanks is 8 m. The secondary losses of the flow are considered to be negligible.

a) Calculate the volumetric flow rate of the water.

b) In order to increase the flow rate of the water to 25%, a second pipe of length 200 m is placed, as shown in Figure (2). The new pipe has the same friction coefficient as the initial one. Calculate the required diameter of this additional pipe.

Solution:

a) The flow rate Q of the water is calculated from the production of the average velocity 1V of the transverse cross-section of the pipe:

21

1 1 1 4

dQ V S V π= = (1)

For the calculation of velocity 1V , we apply the equation of mechanical energy,

in heads form, for an incompressible, steady flow between the positions α and b:

22

2 2

p Vp Vh

g gβ βα α

α βγ γΖ + + = Ζ + + + (2)

or if we take into account that the pressure b bp p pα = = and we consider that the

velocity 0bV Vα ≈ ≈ :

8a bh Z Z m= − = (3)


Figure (2)

Since the secondary losses of the energy are considered to be negligible, the head h is equal to the friction head fh of the flow in the pipe. The head fh is

calculated from the Darcy–Weisbach equation:

21 2 1

1

( )

2fV

h fd g+

=

(4)

Solving the last equation for 1V , we have:

21

11 2

2 2(9,81 / )(0,15 )(8 )1,53 / sec

( ) (0,02)(500 )fgd h m s m mV m

f m= = =

+ (5)

Substituting the values of 1V and 1d from equation (1), we have:

2 3( / 4)(1,53 / )(0,15 ) 0,027 /Q m s m m sπ= = (6)

So, the volumetric flow rate of the water is 27 / S .

b) After the addition of pipe 3, the volumetric flow rate of the water to pipe 1 is 25% higher than the initial value, which means that it is:

3 31 1,25(0,027 / ) 0,03375 / secQ m s m= = (7)

The new flow velocity in pipe 1 is:

Pipe Networks 27

31 1

1 2 21 1

4 4(0,03375 / )1,91 / sec

(0,15 )

Q Q m sV mS d mπ π

= = = = (8)

From the application of the node theorem in the node γ, arises the relationship:

1 2 3Q Q Q= + (9)

where 2Q and

1Q are the flow rates of the water in pipes 2 and 3, respectively. For

the fluid that flows from tank Α to tank Β through pipes 1 and 2, equation (2) takes the following form:

2 21 1 2 2

,1 ,2 1 21 22 2

V VZ Z h h h f f

d g d gα β− = = + = +

(10)

In this equation, the only unknown quantity is velocity 2V of the water in pipe 2,

which can be calculated as:

2 22 (0,01)(200 )(1,91 / )(0,02)(300 )(1,91 / ) 2(8 )2 22(0,15 )(9,81 / ) 2(0,15 )(9,81 / )

m m s Vm m smm m s m m s

= +

which means:

2 0,64 / secV m= 2Q (11)

So, the flow rates of the water in pipes (2) and (3) are:

2 2 3( / 4) ( / 4)(0,64 / ) (0,15 ) 0,01131 /2 2 2 2 2Q V S d m s m m sυ π π= = = − = (12)

3 3 3(0,03375 / ) (0,01131 / ) 0,02244 /3 1 2Q Q Q m s m s m s= − = − = (13)

For the fluid that flows from tank Α to tank Β through pipes 1 and 3, equation (2) takes the following form:

223 31 1

,1 ,3 2 21 3

VVh h h f fb d g d gαΖ − Ζ = = = = +

(14)

where 3V is the velocity of the water in pipe 3:


34 4(0,02244 / ) 0,02857 33 3 / sec3 2 2 23 3 3 3

Q Q m sV mS d d dπ π

= = = = (15)

By combining the last two equations, we have:

2 22 (0,02857 )3 31 12 21 3

dVf fb d g d gα

−Ζ − Ζ = +

(16)

Replacing in this equation the values of the known quantities, it arises that:

2 22 (0,02)(200 )(0,02857 )(0,02)(300 )(1,91 / ) 3(8 )2 22(0,15 )(9,81 / ) 2 (9,81 / )3

m dm m smm m s d m s

−= +

Moreover, after the relating calculations we receive:

3 0,197d m= (17)

So, the internal diameter of the additional pipe has to be 19.7 cm.

3) The three pipes, shown in Figure (3a), have lengths 10 ,AB m= 12AC m=

and 8BC m= and diameters 5 ,ABd cm= respectively.

These pipes are made up of galvanized iron, and the flow rates of the water to and

from the system are 20,010 / secQ mA = (inflow), 30,006 / secQ mB = (outflow)

and 30,004 / secQ mC = (outflow). If the secondary losses of the system are

considered to be negligible, calculate the flow rates of these three pipes

( 6 2/sec10V m−= 0,00015mΑΒε = , 0,00015C mΑε = , 0,00015C mΒε = ).

Figure (3a)

5 ,ACd m= 7 ,BCd cm=

Pipe Networks 29

Solution:

Here, we have a network with a loop and three branches. We prepare the sketch as shown in Figure (3a) according to steps 1, 2 and 3.

We will calculate the relationships of losses – flow rates of the three pipes.

Pipe ΑΒ: ( )/ 0,003 0,0262C C Cd AIT fΑ Α Αε = → = (Moody)

and replacing in the losses equation the fΑΒ , we have:

269,18 ( ) 69.180 ( )a SI h Q SIΑΒ ΑΒ ΑΒ= = ⋅ (1)

Pipe ΑC: ( )/ 0,0025 0,025AC AC ACd AIT fε = → = (Moody)

231, 720 ( ) 31.720 ( )AC AC ACa SI h Q SI= = ⋅ (2)

Pipe ΒC: ( )/ 0,0021 0,02394BC BC BCd AIT fε = → = (Moody)

29,385 ( ) 9.385 ( )BC BC BCa SI h Q SI= = ⋅ (3)

First testing: we assume a random but logical flow rate distribution. In general, larger diameters lead to higher flow rates. If the diameters are the same, of smaller length, it leads to larger flow rates. A better criterion is the comparison of the coefficients α of the above relationships. The larger the α, the smaller the flow rate.

i) Starting from node A and with the coefficients as a criterion, we have:

33 / sec 0,003 / secABQ m= = , so it occurs that:

37 / sec 0,007 / sec (10 3 7)ACQ m= = = +

For node Β, we have:

33 / sec 0,003 / secBC ABQ Q Q m== − = (in flow to Β).


The random distribution of the flow rates is shown in Figure (3b)

Until now, we have made only one assumption:

3 / secABQ =

This agrees with what we have mentioned in the theory.

ii) Formulation of checking relationships:

As there is only one loop, we have one checking relationship. According to the signs convention and Figure (3c), we have:

0h AB BC ACh h hΣ = − − = (4)

iii) Assumption check:

[1] 0,62

[2] 1,55

[3] 0,08

[4] 1,02 0

AB

AC

BC

h mh mh m

h m

= = = Σ = − ≠

iv) Flow rate correction:

3

( 1,02)

2 (69.180 0,003 31.720 0,007 9385 0,003)2

0,0011 / 1,1 / sec

hQ QQi i

Q m s

α−Σ − −Δ = Δ =

⋅ ⋅ + ⋅ + ⋅Σ ⋅ ⋅

Δ = =

We correct the flow rates (paying attention to whether the direction is the same or opposite to the clockwise route) as:

3

3

3

AB : (3 1,1) / 0,0041 / sec

C : (7 1,1) / 0,0059 / sec

C : (3 1,1) / 0,0019 / sec

AB

AC

BC

Q s m

Q s m

Q s m

= + =

Α = − =

Β = − =

Second testing: we repeat the calculation of the losses h, Σh and the required correction QΔ using the corrected flow rates.

Pipe Networks 31

Figure (3b) Figure (3c)

From (1) 1.17ABh m =

(2) 1.10ACh m =

(3) 0.03BCh m =

(4) 1 0,03 0h m Σ = + ≠

and 0,00004 / sec 0,04 / secQ m Δ = − = −

So, the correction that came up is very small. We correct the flow rates, and the results are very close to the flow rates that we are looking for:

30,00406 / sec 4,06 / secABQ m= =

30,00594 / sec 5,94 / secACQ m= =

30,00194 / sec 1,94 / secCQ mΒ = =

4) For Figure (4), calculate the flow rates Q1, Q2 and Q of the three pipes. The satisfactory solution is the one in which the flow rate to branch D and the flow rate from D agree to the second decimal numeral.


Figure (4)

The data given are as follows:

Water of temperature 6 212 1,244 10 / secC mο ν − = ×

1 1 1 112 , 2000 , 0,00048 , 50d m Ft y mε′′= = = =

2 2 2 28 , 1000 , 0,00048 , 80d m Ft y mε′′= = = =

3 3 3 38 , 3000 , 0,00048 , 100d m Ft y mε′′= = = =

Solution:

We consider that Ay is equal to the height of the intersection point of the

pressure gauge lines of the three pipes.

So:

2

3 1 2

If (of the height of the middle tank) then

we have :

y yQ Q Q

Α > = +

Pipe Networks 33

2

3 2 1

But if then

we have:

y yQ Q Q

Α < + =

In general, the solution is found after a number of iterations of approximate solutions by applying the following theory: if the flow rate to branch D is such that it is larger than the flow rate from the Δ at the end of an iteration, then we conduct a new iteration in which the new assumed value of yA is larger than that of the previous iteration. If the flow rate to D is found to be smaller than the flow rate from D in the end of an iteration, we conduct a new iteration of calculations by choosing a smaller yA. We stop the iterations of the approximate solutions when finally we find that the flow rate to and from D agree to the second decimal numeral.

First iteration: assume that 75Ay m=

Then:

a) for pipe (1):

1

1/221 1 1 1

1 1 11 1 1

2

2

V h gdh y y f Vd g fΑ

= − = =

and because 1 75 50 25h m m m= − =

then: 1/2

1/2 1/21 1 1

2 25 9,81 12 0,02540,2734

2000V f f− −× × × × = × =

in m/sec

1

0,00048 120,00048

12dε ′′×= =

′′

Assuming that 1 0,017f = ,

then 1/21 0,2734(0,017) / sec 2,0969 / secV m m−= =

so 51 11 6

2,0969 12 0,0254Re 5,14 10

1,244 10

V dν −

× ×= = = ××


So, with:

1

5e 1

1

0,00048καιR 5,14 10 0,0175fdε

= = × =

(Moody)

So, 1/21 0,2734(0,0175) / sec 2,0667 / secV m m−= =

and 51 11 6

2,0667 12 0,0254Re 5,06 10

1,244 10

V dν −

× ×= = = ××

So, with:

51 1

1

0,00048καιRe 5,06 10 0,0175fdε

= = × =

(Moody)

So, 2

11 1 12,0667 / sec and

4

dV m Q V π= =

or 2

3 31

2,0667 (12 0,0254)/ sec 0,1508 / sec

4Q m mπ× Χ ×= =

b) For pipe (2):

2

1/222 2

2 2 2 22 2 2

2

2A

h gfVh y y f Vd g f

= − = =

So, with 2 75 80 5h m m= − =

1/21/2 1/2

2 2 1

2 5 9,81 8 0,02540,1412

1000V f f− −× × × × = × =

in m/sec

and 2

0,00048 120,00072

8dε ′′×= =

′′

Pipe Networks 35


then 1/22 0,1412(0,019) / sec 1,0244 / secV m m−= =

so 52 22 6

1,0244 8 0,0254Re 1,67 10

1,244 10

V dν −

× ×= = = ××

So, with:

52 2

2

0,00072and Re 1,67 10 0,02fdε

= = × =

(Moody)

then 1/22 0,1412(0,02) 0,9984 / secV m−= =

so, 52 22 6

0,9984 8 0,0254Re 1,63 10

1,244 10

V dν −

× ×= = = ××

So, with:

52 2

2

0,00072and Re 1,63 10 0,023fdε

= = × =

(Moody)

then: 1/22 0,1412(0,023) 0,9310 / secV m−= =

so, 52 22 6

0,9310 8 0,0254Re 1,52 10

1,244 10

V dν −

× ×= = = ××

So, with:

52 2

2

0,00072and Re 1,52 10 0,023fdε

= = × =

(Moody)

then 2 0,931 / secV m=

and 2

3 322 2

0,931 (8 0,0254)/ sec 0,0302 / sec

4 4

dQ V m mπ π× × ×= = =


c) For pipe (3):

1/223 3 3 3

3 3 3 33 3 3

2

2AV h gd

h y y f Vd g f

= − = =

So, with 3 75 100 25h m m= − =

1/2

1/21/2

3 3 3

2 25 9,81 8 0,02540,1823

3000V f F −−× × × × = =

in m/sec

3

0,00048 120,00072

8dε ′′×= =

′′


then 1/2 1/23 30,1823 0,1823(0,019) 1,3225 / secV f m− −= = =

53 33 6

1,3225 8 0,0254Re 2,16 10

1,244 10

V dν −

× ×= = = ××

with

53 3

3

0,00072and Re 2,16 10 0,0197fdε

= = × =

(Moody)

then 1/23 0,1823(0,0197) 1,2988 / secV m−= =

So, 53 33 6

1,2988 8 0,0254Re 2,12 10

1,244 10

V dν −

× ×= = = ××

So, with:

63 3

3

0,00072and Re 2,12 10 0,0197fdε

= = × =

(Moody)

so 3 1, 2988 / secV m=

Pipe Networks 37

and 2 2

3 333 3

1, 2988 (8 0,0254)/ sec 0,0421 / sec

4 4

dQ V m m

π π× Χ ×= =

So, from the results of the first circle, assuming that 275 80Ay m y m= < = , we

have:

31 0,1508 / secQ m=

32 0,0302 / secQ m=

33 0,0421 / secQ m=

thus, it must always be 3 2 1Q Q Q+ = .

However, we see that:

3 32 3 10,0723 / sec 0,1508 / secQ Q m Q m+ = < =

So, we conduct another iteration by choosing a value of yΑ smaller than the

first one.

Second iteration: assume that 65Ay m=

a) For pipe (1):

1 65 50 15h m= − =

1/2 1/21/2 1/21 1

1 1 11 1

2 2 15 9,81 12 0,02540,2118

2000

h gdV f ff

− − × × × × = = =

1

0,00048 120,00048

12dε ′′×= =

′′


then 1/21 0,2118(0,017) 1,6243 / secV m−= =


51 11 6

1,6243 12 0,0254Re 3,98 10

1,244 10

V dν −

× ×= = = ××

so, with:

51 1

1

0,00048and Re 3,98 10 0,018fdε

= = × =

(Moody)

then 1/21 0,2118(0,018) 1,5787 / secV m−= =

51 11 6

1,5787 12 0,0254Re 3,87 10

1,244 10

V dν −

× ×= = = ××

so, with:

51 1

1

0,00048and Re 3,87 10 0,018fdε

= = × =

(Moody)

b) For pipe (2):

2 80 65 15h m= − =

1/2 1/21/22 2

2 22 2

2 2 15 9,81 8 0,0254

1000

h gdV ff

− × × × × = =

or 1/22 20,2445V f −= in m/sec

2

0,00048 120,00072

8dε ′′×= =

′′


then 1/22 0,2445(0,02) 1,7289 / secV m−= =

52 22 6

1,7289 8 0,0254Re 2,82 10

1,244 10

V dν −

× ×= = = ××

Pipe Networks 39

so, with:

52 2

2

0,00072and Re 2,82 10 0,0195fdε

= = × =

(Moody)

so 1/22 0,2445(0,0195) 1,7509 / secV m−= =

and 52 22 6

1,7509 8 0,0254Re 2,86 10

1,244 10

V dν −

× ×= = = ××

So, with:

52 2

2

0,00072and Re 2,86 10 0,0195fdε

= = × =

(Moody)

c) For pipe (3):

33

0,00048 12100 65 35 , 0,00072

8h m

dε ′′×= − = = =

′′

1/2 1/21/2 1/23 3

3 3 33 3

2 2 35 9,81 8 0,02540,2157 / sec

3000

h gdV f f m

f− − × × × × = = =


then 1/23 0,2157(0,019) / sec 1,5646 / secV m m−= =

53 33 6

1,5646 8 0,0254Re 2,56 10

1,244 10

V dν −

× ×= = = ××

So, with:

53 3

3

0,00072and Re 2,56 10 0,0195fdε

= = × =

(Moody)

then 1/23 0,2157(0,0195) / sec 1,5447 / secV m m−= =


so 53 33 6

1,5447 8 0,0254Re 2,52 10

1,244 10

V dν −

× ×= = = ××

So, with:

53 3

3

0,00072and Re 2,52 10 0,0195fdε

= = × =

(Moody)

So, the second iteration gives us the following results:

2 2 3 31

1 11,5787 (12 0,0254)

0,1154 4 sec sec

d m mQ V π π× × ×= = =

2 2 3 32

2 21,7509 (8 0,0254)

0,574 4 sec sec

d m mQ V π π× × ×= = =

2 2 3 33

3 31,5447 (8 0,0254)

0,0504 4 sec sec

d m mQ Vπ π× × ×= = =

The following relationship must always be satisfied:

2 3 1Q Q Q+ =

The results of the second iteration give:

3 32 3 10,107 / sec 0,115 / secQ Q m Q m+ = < =

So, we also conduct a third iteration by choosing a value of yΑ slightly smaller

than the one we assumed at the beginning of the second iteration.

Third iteration: assume that yA = 63,8 m. Then:

a) For pipe (1):

1 11

0,00048 1213,8 , 0,00048

12h y y m

dε

Α′′×= − = = =

′′

Pipe Networks 41

1/2 1/21/2 1/21 1

1 1 11 1

2 2 13,8 9,81 12 0,02540,2031

2000

h gdV f ff

− − × × × × = = =


then 1/21 0,2031(0,018) / sec 1,5141 / secV m m−= = and

51 11 6

1,5141 12 0,0254Re 3,71 10

1,244 10

V dν −

× ×= = = ××

So, with:

51 1

1

0,00048and Re 3,71 10 0,018fdε

= = × =

(Moody)

b) For pipe (2):

1 22

0,00048 1216,2 , 0,00072

8h y y m

dε

Α′′×= − = = =

′′

1/2 1/21/2 1/22 2

2 2 22 2

2 2 16,2 9,81 8 0,02540,2541

1000

h gdV f ff

− × × × × = = =


then 1/22 0,2541(0,0195) / sec 1,8199 / secV m m−= = and

52 22 6

1,8199 8 0,0254Re 2,97 10

1,244 10

V dν −

× ×= = = ××

with

52 2

2

0,00072and Re 2,97 10 0,0195fdε

= = × =

(Moody)


c) For pipe (3):

3 33

0,00048 1236,2 , 0,00072

8h y y m

dε

Α′′×= − = = =

′′

1/2 1/21/2 1/23 3

3 3 33 3

2 2 36,2 9,81 8 0,02540,2193

3000

h gdV f f

f− × × × × = = =

Assuming that f3 = 0,0195,

then 1/23 0,2193(0,0195) 1,5707 / secV m−= =

and 53 33 6

1,5707 8 0,0254Re 2,57 10

1,244 10

V dν −

× ×= = = ××

So, with:

5

3 33

0,00072and Re 2,57 10 0,0195fdε

= = × =

(Moody)

So, the results of the third iteration are the following:

2 2 3 31

1 11,5141 (12 0,0254)

0,1104 4 sec sec

d m mQ V π π× × ×= = =

2 2 3 32

2 21,8199 (8 0,0254)

0,594 4 sec sec

d m mQ V π π× × ×= = =

2 2 3 33

3 31,5707 (8 0,0254)

0,0514 4 sec sec

d m mQ Vπ π× × ×= = =

So, we have:

3 32 3 10,059 0,051 0,110 / sec 0,110 / secQ Q m Q m+ = + = = =

Pipe Networks 43

This means that we have an agreement to the third decimal, so finally the flow rates will be:

31 0,110 / secQ m= to D

32 0,059 / secQ m= to D

33 0,051 / secQ m= to D

5) Calculate the flow rates in the various pipes of the distribution network of water shown in Figure (5) with the Hardy–Cross method using the Darcy–Weisbach equation and the Moody diagram. The characteristic values of the pipes are given (absolute roughness ε, diameter d and length ), with the average temperature of the water (10°C), the kinematic viscosity (v = 1,31×10−6 m2/sec),the entrance flow rate in the node A(QA = 30 l/sec), the exit flow rates in the nodes D and Z(Q∆ = 10 l/sec, QZ = 20 l/sec) and the coefficient of the absolute roughness (ε=0.01524 cm).

Solution:

i) First, we make the diagram (Figure (4)) of the network and place in it the facts of the problem. The loops are numbered in Roman numerals (Ι and ΙΙ) and the pipes in Arabic numerals (1, 2, 3, 4, 5, 6 and 7).

ii) We proceed with a logical distribution of the flow rates Qα to value and

direction in all the pipes in such a way that in each node, the continuity equation is satisfied. Therefore, in this problem, the first “logical” distribution of flow rates is as follows.

The flow rate of 30 l/sec that enters node Α is distributed into 16 and 14 l/sec to pipes 1 and 4, respectively. The flow rate of 16 l/sec of pipe 1 is distributed into 9 and 7 l/sec in pipes 5 and 2, respectively. From the flow rate of 14 l/sec of pipe 4, the 10 l/sec exits the node Δ and the rest 4 l/sec constitutes the flow rate of pipe 3.

The flow rates 7 and 4 l/sec of pipes 2 and 3, respectively, enter as a flow rate 11 l/sec of pipe 7. The flow rate of 9 l/sec of pipe 5 goes on to pipe 6 and with the flow rate of 11 l/sec of pipe 7 exits as a flow rate of 20 l/sec from the node Ζ.

iii) The Darcy–Weisbach equation given by the relationship 2

29fVh f

d=

and

20,083r f

d=

gives:


Figure (5)

(1)

where (2)

For the above equations, the values of length and diameter d are given.

For each pipe, the friction coefficient f is taken from the Moody diagram as a

function of the relative roughness / dε and the Reynolds number:

56

4 4Re 9,724 10

3,14 1,31 10

Vd Q Q Qv vd d dπ −= = = = ⋅

⋅ ⋅ (3)

Therefore, we calculate the relative roughness / dε of each pipe once from the given values of ε and d, while we calculate the Reynolds number Re of each pipe from equation (3) in every calculation iteration for the respective values of the flow

2 5 20,083 /fh f Q d r Q= =

50,083 /r f d=

Pipe Networks 45

rates Qα. Based on the calculated values of / dε and Re, we take the values of the respective friction coefficients f from the Moody diagram. If two successive values of the friction coefficient f do not differ significantly between them, the last values of f are used as constants for the next iterations of calculations.

iv) From equation (2), the value of the coefficient r of each pipe is calculated.

v) From equation (1), we calculate the value of the height losses hfa of each pipe in each loop separately.

The value of hfα is positive if the flow inside the pipe is to the direction of the indicators of the clock and negative if the flow inside the pipe has the opposite direction. For every loop, we get the algebraic sum Σhfα

vi) For every pipe, we obtain the quotient hfa/Qa. For every loop, we obtain the sum ∑hfa/Qa.

vii) The correction ΔQ is given by the equation ( )/a

fa a

fQ

m h Q

Σ Δ = − Σ

which, in this problem for 2m = , has the specific form:

( )2fa fa aQ h h QΔ = −Σ Σ (4)

Table (5a). Calculating tables of the water distribution example with the Hardy–Cross method, first iteration of calculations


Table (5b). Calculating tables of the water distribution example with the Hardy–Cross method, second iteration of calculations

Table (5c). Calculating tables of the water distribution example with the Hardy–Cross method, third iteration of calculations

Pipe Networks 47

viii) We correct the flow rates we calculated in step (ii) of each pipe of the same loop, adding algebraically the value of ΔQ found from equation (4). Since pipe (2) is common for both loops I and II in its flow rate, we make a second correction of the ΔQ΄, which is opposite to the value of (4) from the other loop.

ix) The values of the flow rates that were corrected this way are used for the second iteration of calculations, with repetition of steps (iii) to (viii). The correcting values of the flow rates of the second iteration are used for the third iteration of calculations, by which is possible to bypass steps (iii) and (iv) and to use the values of r of the second iteration, repeating in this way steps (v) to (viii) in the fourth and, if necessary, in the fifth as well. Usually, four to five calculations give satisfactory results.

6) Figure (6) shows a part of a hydraulic network, the facts of which are given in

Table 6. If the flow is 30,5 / secQ Q mΕΑ = = and the kinematic viscosity of the

flow is 21,2 10 6 / secmν = − − , calculate the flow rates of the pipes.

Table (6). Network data Figure (6)

Solution:

The directions of the flows are known (Figure (6)), and the calculation is

probably simple. Since, generally, it is / 1.000d > , the local losses are not

calculated, but tests are performed. Assuming that 31 0,2 / secQ m= , then, we have:

Pipe 1: 31 10,2 / sec, 2,83 / secQ m V m= ≅

51 1Re 7 10 , 0,0125f≅ ⋅ ≅

Pipe L D κs

(m) (cm) (mm)1 3.000 30 smooth

2 1.000 30 0,1

3 940 20 14 1.000 35 0,1

5 2.000 40 smooth6 450 20 0,3


1 51fh m≅

Pipe 2: 320,2 / sec, 2,83 / sec2Q m V m= ≅

52 2Re 7 10 , / 0,1/300 0,0003dε≅ ⋅ = ≅

2 20, 016, 21,8ff h m≅ ≅

So, for pipes 1 and 2, it is 1,2 51 21,8 72,8fh m= + =

Pipe 3: 3 1,2

72,8f fh h= =

142

3 3 3R e 9 , 2 1 0 , / 1 / 2 0 0 0 , 0 0 5f dε⋅ ≅ ⋅ = = ,

3 30,032, 3,1 / secf V m≅ ≅ ,

33 0,10 / secQ m≅

Pipe 4: 34 2 3 0, 2 0,1 0, 3 / secQ Q Q m= + = + =

3 54 43,11 / sec, Re 9 10V m≅ ≅ ⋅

4 4/ 0,0003, 0,016d fε ≅ ≅

4 22, 5fh m=

Pipe 5: 35 4 50, 3 / sec, 2, 38 / secQ Q m V m= = ≅ ,

55 5Re 7, 9 10 , 0, 012f≅ ⋅ ≅

5 17,3hf m≅

So, for pipes 4 and 5, it is:

4,5 22, 5 17,3 39,8fh m= + =

and

Pipe Networks 49

1,2,4,551 21,8 39,8 112,6fh m= + + =

Pipe 6: 1

526 6 6 6112, 6 , Re 1, 6 10 , / 0, 0015fh m f dε= ⋅ ≅ ⋅ ≅

6 0, 022f ≅ , 6 6,6 / secV m≅

and

36 0, 2 / secQ m≅

For the testing, it is:

35 6 0, 3 0, 2 0, 5 / sec AQ Q m Q+ = + = =

which means that the first choice of AQ was correct.

7) In Figure (7), three tanks are shown, of which Α has a seasonally variable level. In pipe ΒC, there is a sliding valve, while QBC = 50 lit/sec.

a) With the maximum flow rates and QBD = 100 lit/sec, define ZA and dBD.

b) If 116,75ΑΖ = m, calculate the value at which the valve will close so that

100 / sec.BDQ lit= . In Figure (7), is in m and d in mm.

AB

1.200

300d==

ΒC

209

150d==

ΒD

1.650=

ε = 1 mm (all the pipes)

ν = 1.1·10–6 m2/sec


Figure (7)

Solution:

The pipes have maxQ when the valve is fully open.

a) In pipe CΒ, it is:

/ 209 / 0,15 1.393 1.000d = ≅ >

and the valve is fully open. So, the total losses are not calculated and therefore we have:

24 0, 05 / 0,15 2,83 / secV mπ= ⋅ ⋅ ≅

6 5Re 2,83 0,15 / 1,1 10 3,8 10−= ⋅ ⋅ ≅ ⋅ ,

/ 1/150 0,006dε = ≅ ,

so:

0,032f ≅ ,

20,032 (209/ 0,15) 2,83 / 2 9,81 18,2fh m= ⋅ ⋅ ⋅ ≅

Pipe Networks 51

So, it is:

76 18, 2 94, 2B C fh mΗ = Η + = + =

In pipe ΒD, it is:

94, 2 55 39, 2f B Dh H H m= − = − =

It is 0,012n = , and with an absolutely rough pipe,

39,2 /1.650 0,0237EJ = ≅

while 31

821, 548 (0,1 0, 012 / 0, 0237 ) 0, 250d m= ⋅ ⋅ ≅

which means that (testing of n):

/ 1/ 250 0,004dε = ≅

Also, / 1.650 / 0, 25 6.600 1.000d = = > .

In pipe ΑΒ, it is:

/ 1.200 / 0,3 4.000 1.000d = = > ,

and

30,05 0,1 0,15 / secQ m= + =

So, 24 0,15 / 0,3 2,12 / secV mπ= ⋅ ⋅ ≅

5Re 5,8 10 , / 1/ 300 0,003dε≅ ⋅ = ≅

0,027f =

and 20,027 (1.200 / 0,3) 2,12 / 2 9,81 24,74fh m= ⋅ ⋅ ⋅ ≅


So, it is:

94, 2 24, 74 118,94A A B fZ H h mΗ = = + = + = ,

which means that:

119A mΖ ≅

b) Since the valve is not fully open, we will take into account the energy losses

from it, independently of the ratios / d .

With 116,75mΑΖ = for pipe ΑΒ, as QΒΔ = constant,

116, 75 94, 2 22,55fh m= − =

and 1 3 1

62 2 2Re (0, 3 / 1,1 10 ) (2 9, 81 22, 55 / 1.200)f −⋅ = ⋅ ⋅ ⋅ ⋅ ,

or, 1

42Re 9 10 , / 0, 003f dε⋅ ≅ ⋅ = ,

0,027f ≅

32 / sec, 0,14 / secV m Q m≅ ≅

For pipe ΒC, it will be:

30,14 0,1 0, 04 / secQ m= − =

24 0, 04 / 0,15 2, 26 / secV mπ= ⋅ ⋅ ≅

5Re 3,1 10 , / 1 / 150 0, 006dε≅ ⋅ = ≅

0, 032, 11, 6ff h m≅ =

Between node Β and tank C, it is:

94,2 76 18,2B C mΗ = Η − Η = − =

and, 218, 2 11, 6 / 2 11, 6 0, 26V g Kτ τ= + Κ ⋅ ⋅ = + ⋅ ,

therefore, 25,4Kτ ≅

Pipe Networks 53

However, it is:

(relative opening) 0,15 (opening)/0,15≅ = ,

which means that the opening is 0.0225 2,25m cm= . The valve will be closed at:

(1 0,15) 0,85,− = or 85%

The decrease in linear energy in the position of the valve will be:

225,4 2,26 / 29,81 6,6fh m= ⋅ =

which is 6.6/18.2 = 0.36 or 36% of the total losses in the pipe.

1.11. Problems to be solved

1) Calculate the flow rates in the loop of the figure if QA= 0.32 m3/s (inflow), QΒ= 0.28 m3/s (outflow) and QΔ= 0.10 m3/s (inflow). All the pipes are made up of cast iron, of diameter 25 cm with ΑΒ = ΓD =200 m and ΑD=ΒΓ=100 m.

2) The five pipes of the horizontal network, shown in the figure, have a friction

coefficient f = 0.025, lengths ΑΒ = CD = 600 ft, ΑC = ΒD = 450 ft, BC = 750 ft and diameters. If the water inflow in the network is QA= 3 ft3/s, calculate the flow rates and the flow directions in all the pipes. If the pressure at point Α is 120 psi, calculate the pressure at points Β, C and D.


3) Calculate the pressures of the network shown in the following figure. The pipes are made up of cast iron with lengths:

ΑΒ = ΒΕ = DΕ =ΑD = 100 m

CD = 50 m; AC= 112 m.

Secondary losses are negligible: ν = 10−6 m2/s.

4) Calculate the water flow rates and the losses in the pipes made up of cast iron for the network shown in the figure. The given data are as follows:

5) In the water network shown in the figure, the pipes are made up of cast iron. The lengths and diameters of the pipes are given, as well as the flow rates to and from the network. Calculate a) the flow rates in all the pipes of the network and b) the pressure difference between Α and Ζ if yA– yΖ = 4 m. The given data of the pipes are as follows:

Pipe Networks 55

Figure (5)

6) The network shown in the figure is supplied with water from tanks 1 (y1 = 128 m) and 2 (y2 = 122 m). All the pipes are made up of cast iron. The outflows from the

network are: QB= QΔ = 60 / s , QΓ= QΕ = 40 / s . Calculate the flow rates in each node of the network. The given facts are as follows:

yΒ= yE= 40 m yC= 40 m yD=28 m

AB = 500 m EF = 300 m CF = 150 m,

BC = BD = CE = DE = 120 m

20AB EFd d cm= = 10BC CEd d cm= =


7) In the network shown in the figure, the pipes are made up of asphalted cast iron with lengths and diameters given in the following table. Calculate the flow rates if the altitudes are:

y1 = 150 m, yΒ= 50 m, yC= 53 m, yD= 45 m

yΕ= 65 m, yF= 60 m, yG= 30 m, yH= 35 m

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