wave characteristic method

8/13/2019 Wave CharacteriStic Method

1/44

CHAPTER TWO

The Wave Characteristic MethodThe WCM is based on the physically accurate concept that the transient pipe

flow results from the generation and propagation of pressure waves that occur

as a result of a disturbance in the pipe system (valve closure, pump trip, etc.). A

pressure wave, which represents a rapid pressure and associated flow change,

travels at sonic velocity for the liquid-pipe medium, and the wave is partially

transmitted and reflected at all discontinuities in the pipe system (pipe

junctions, pumps, open or closed ends, surge tanks, etc). A pressure wave can

also be modified by pipe wall resistance. This description is one that closely

represents the actual mechanism of transient pipe flow. In this chapter,relationships are presented for the magnitude and propagation velocity and

attenuation of pressure waves, and for the effect of pressure waves impinging

on piping system elements.

2.1 BASIC RELATIONS

The relationship between pressure change ( P ) and flow change ( ), which

is associated with the passage of a pressure wave, defines the transient response

of the pipe system and forms the basis for the development of the required

mathematical expressions. Figure 2-1 depicts the flow and pressure conditionsthat exist a short time apart, as a pressure wave of magnitude

Q

t P propagates

a distance x in a liquid filled pipe.

Figure 2-1: Pressure wave propagation in a pipe.


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2-2 CHAPTER TWO

During the short time t , the pressure on the left end of the liquid column is

PP + while the right end of the liquid column is P. This unbalancedpressure causes the fluid to accelerate. The momentum principle gives

( )tQxAPPP L=+ (2-1)

where is the pipe cross sectional area;LA is the liquid density; and is the

acceleration of gravity. Canceling and rearranging gives

tA

xQP

L

= (2-2)

The term

t

x

is the propagation speed of the pressure wave. The wave

speed is equal to the sonic (acoustic) velocity (c) in the system if the mean

velocity of the liquid in the line is neglected. Since the mean velocity of the

liquid is usually several orders of magnitude smaller than the sonic velocity,

this is acceptable. Thus

LA

QcP

= (2-3a)

or, in terms of pressure head

LgA

QcH = (2-3b)

This relation was first presented by Joukowsky (1904). The sonic speed c

for a liquid flowing within a line is influenced by the elasticity of the line wall.

For a pipe system with some degree of axial restraint a good approximation for

the wave propagation speed is obtained using

)/1(/ lcfRf tEDEKEc += (2-4)

where and are the elastic modulus of the fluid and conduit,

respectively; is the pipe diameter; is the pipe thickness; and is the

coefficient of restraint for longitudinal pipe movement. Typically, three types

of pipeline support are considered for restraint. These are:

fE cE

D lt RK

Case a:The pipeline is restrained at the upstream end only.


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THE WAVE CHARACTERISTIC METHOD 2-3

2/1 pRK (2-5a)

Case b:The pipeline is restrained throughout.

(2-5b)21 PRK

Case c:The pipeline is unrestrained (has expansion joints throughout).

(2-5c)1=RK

where Pis the Poisson's ratio for the pipe material.

For thick-walled pipes, the values of the restraint coefficient for the three

types of pipeline support are:

Case a:The pipeline is restrained at the upstream end only.

)1(2

)2/1( +++

=D

t

tD

DK lp

l

R (2-6a)

Case b:The pipeline is restrained throughout.

)1(2

)1( 2 +++

=D

t

tD

DK lP

l

R (2-6b)

Case c:The pipeline is unrestrained (has expansion joints throughout).

)1(2

+++

=D

t

tD

DK l

l

R (2-6c)

Table 2-1: Physical properties of common pipe materials.

Material Youngs Modulus

Ec(GPa)

Poissons Ratio

p

Aluminum 69 0.33

Asbestos Cement 23-24 -

Cast Iron 80-170 0.25-0.27

Concrete 14-30 0.1-0.15

Reinforced Concrete 30-60 -

Ductile Iron 172 0.30

Polyethylene 0.7-0.8 0.46

PVC 2.4-3.5 0.46

Steel 200-207 0.30


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2-4 CHAPTER TWO

Figure 2-2: Pressure wave velocity for water in round pipes with different

diameters and thicknesses and KR equal to 0.91 (adapted from Thorley,

1991).


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Table 2-1 lists physical properties of common pipe materials. These values

should assist in calculating wave speeds in pipes constructed of common

materials. A plot of wave propagation speeds for water flowing in a circular

pipe subjected to a representative pipeline restraint for a variety of materials is

shown in Figure 2-2 (Thorley, 1991), although the values for thick walled pipes(i.e., for small values on the abscissa) must be taken as approximate only. For

the majority of hydraulic analyses involving transients, the wave speed can be

considered to be constant.

It is important to note that gas dissolved in the liquid can significantly

affect the wave speed. In fact, the presence of even small quantities of air can

significantly reduce the wave propagation speed of water in pipes. At higher

pressures the effect of the trapped air decreases.

In very rigid pipes, in tanks, or in large water bodies, the acoustic velocity

is reduced to .fE

c=

For circular section tunnels, the coefficient of restraint for all three cases

above reduces to (Thorley, 1991)

)1(2

+=D

tK lR (2-7)

2.1.1 Exercise Wave Speed

Calculate the wave speed for the three restraints with the following conditions:

Steel Schedule 10 - 6 inch pipe

Inside diameter = 6.357 inches

Outside diameter = 6.625 inches

Modulus of steel = 30,000,000 psi

Modulus of water = 300,000 psi

Poisson ratio = 0.28

Answers:

Expansion joints throughout 3,960 ft/s

Anchored at upstream end only 4,056 ft/s

Anchored throughout 4,013 ft/s

The Wave Speed tool (WaveSpeed.exe) shown below can be used to make the

above calculations. Check the above answers using this tool, which is discussed

in Appendix F.1.


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2-6 CHAPTER TWO

2.2 THE WAVE CHARACTERISTIC METHOD

Pressure waves are generated at any point in a pipe system where a disturbance

that results in a change in flow rate is introduced. This can include a valve that

is opening or closing, a pump that is started up or shut down, a change in a

reservoir pressure, or a change in an inflow or outflow for the system. Pressure

and flow conditions at a component are also affected by the impinging pressure

waves. This pressure wave tracking approach to transient analysis forms the

basis of the Wave Characteristic Method and is illustrated using the sequence oftransient events shown in Figure 2-3. For the first transient event, a valve closes

and a pressure wave is generated. For the second event, the pressure wave

travels toward the 3-pipe junction at sonic speed in the pipe. For the third event,

the wave is transmitted into the two connecting (outgoing) pipes and reflected

back into the original pipe producing three new pressure waves. For the fourth

event, each pressure wave travels at sonic speed toward the opposite end of the

pipe and impinges on the elements at those locations. For the last event, the


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pressure waves modify conditions at the reservoir, valve, and pump and new

pressure waves are generated and travel back toward the 3-pipe junction. Note:

the pressure wave actions depicted at the opposite ends (4-5) will normally

occur separately at the instant when each wave reaches the element located at

that end.

Figure 2-3: Pressure waves generated following an instantaneous valve

closure.


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2-8 CHAPTER TWO

This approach to transient analysis requires the calculation of the effects of

pressure waves impinging on pipe junctions, components (e.g., valves and

pumps) and surge control elements which inject or remove fluid (e.g., surge

tanks and air/vacuum valves) as well as calculating the effect of line friction on

the magnitude of pressure waves. These pressure wave action calculations areillustrated in Figure 2-4 and discussed in detail in the following sections.

Graphical tools are also provided that allow the reader to better understand the

phenomena of pressure wave propagation and the calculations required to

incorporate their effects on pressures and flows throughout a piping system.

Surge Control Device Junction

Component Effect of Friction

Figure 2-4: Pressure wave action at network discontinuities.

2.3 ILLUSTRATING THE WAVE CHARACTERISTIC METHOD

The Wave Characteristic Method consists of essentially tracking with time the

propagation and attenuation of pressure waves that are moving at sonic speed,and calculation of their effects on the pressures and flows (in both time and


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space) throughout the water distribution piping system. It represents the pipe

system as a connected graph composed of a finite number of discontinuities

connected by frictionless pipe segments, which serve only to transmit pressure

waves.

Starting at an initial steady-state flow condition, the tracking mechanism isinitiated after a disturbance is introduced at a given point in the pipe system

(e.g., closing of a valve). The disturbance produces an incremental change in

the volumetric flow rate, which then results in a transient pressure wave

that is introduced at that location. This pressure wave travels throughout

the pipe network at sonic speed, and is partly reflected and transmitted at all

physical and geometrical discontinuities. Temporal variations of pressures and

flows (velocities) are then computed for any point in the pipe system by

summing with time the contributions of impinging pressure waves.

)( H

The effects of line friction (or viscous resistance) influence the propagation

of pressure waves and are simulated using the orifice analogy. They aremodeled as square-law friction orifices (or artificial discontinuities) with

properly chosen orifice coefficients. These are placed at a number of discrete

points along each pipe to better approximate the distributed effects of line

losses. Impinging pressure waves are then reflected and transmitted at thesefriction orifices, resulting in incremental flow and pressure changes.

Time histories of velocity (flow rate), pressure, and pressure waves leaving

each discontinuity are determined explicitly in terms of the magnitudes of

impinging pressure waves, the nature of the discontinuity, and conditions prior

to wave action. New information at each discontinuity is available only at

discrete time intervals because all of the disturbing functions are approximatedby a series of small time step changes occurring at specified times in the

simulation.

The following definitions are used:

Definition 1. A transient event is an internal (endogenous) event that takes

place when a pressure wave reaches a discontinuity such as a junction node,

pump, valve, or any element of the network that will cause the wave to be

transmitted and reflected.

Definition 2. A transient time step )( t takes place between two consecutive

transient events and is a time period during which pressure waves remain

unchanged structurally in the simulation.

The transient time step defines the time interval between successive

computations. Its selection is determined by two factors:

)( t

(1) t must be sufficiently small to accurately represent all disturbingfunctions by a series of step changes; and


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2-10 CHAPTER TWO

(2) it is necessary that all modifications of pressures, flows and pressure

waves in the system occur at an integer number of time intervals.

The pressure wave travel times between all adjacent discontinuities will

then be an integer number of transient time steps.

The Wave Characteristic Method is basically a simulation system that usesthe next-transient event scheduling approach; that is, the simulation (clock)

time is advanced to the time of the next transient event when a discontinuity is

reached (in incremental transient time steps). The next-event approach requires

that transient events have a known time at which they are to occur. For each

pressure wave, such time can be easily determined, since the wave speed and

direction, location of network discontinuities, and the topological structure of a

given network are known at all times.

The simulation model is therefore reduced to a role of an intelligent

scheduler that creates, parameterizes, identifies, and arranges transient events in

a chronological order during the lifetime of a given transient simulation.A graphical Wave Method Calculator (WMC.EXE) is presented to illustrate

the pressure wave tracking mechanism of the Wave Characteristic Method. This

tool is documented in Appendix C, and can be used to visualize the movement

and action of traveling pressure waves at discrete t transient time intervals.Calculations are carried out at each component or junction. Calculations are

necessary only when a pressure wave impinges on an element or the operating

conditions for the element change (e.g., valve setting or pump speed change).

When a calculation is made, the impinging waves just prior to the calculation

are first shown (Nt-) and then the new pressure waves are displayed (Nt+),

where N is the number of time increments since the calculations were initiated.

The surge pressures are noted at each node and change each time a pressure

wave arrives at the node and wave action occurs.

The location of all traveling pressure waves at each time step is known.

These locations are shown by the Wave Method Calculator (Appendix C).

However, the actual location within the pipe of each pressure wave is not

required for the analysis. It is sufficient to specify the number of time steps

required to traverse each pipe section so that the pressure wave impinging at a

component is known to be the wave which left the adjacent component that

number of time steps earlier.

The Wave Method Calculator toolbox (Figure 2-5) provides an excellent

illustration of the intuitive nature of the Wave Characteristic Method fortransient analysis. It clearly shows how transient disturbances are transmitted as

pressure waves by depicting the location of the pressure waves at each time

step. It also illustrates how the transient calculation process works. Pressure

waves in the connecting pipes impinge on junctions and components

(discontinuities). When this occurs, a transient calculation is needed to compute

the effects of the pressure waves on the pressure and flow and the magnitude of

new pressure waves, which are generated by this action and transmitted down


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each of the connecting pipes. When pressure wave action occurs at a junction or

a component, a second schematic is shown depicting the effects of the wave

action. This schematic is identified by adding a + to the time display to indicate

that the wave action occurring at that time has taken place.

Figure 2-5: Wave method calculator toolbox.

For the sample pipe system shown in Figure 2-5 (Example-2.wat), four

transient events are illustrated following the sudden closure of the lower right

valve (node 4), which produced a pressure wave of 500 ft. The transient events

(Figures 2-6 and 2-7) depicted are:

a) 6t+ - the wave has reached the three-pipe junction (node 3) and has

created three pressure waves.

b) 9t - a wave reaches the second three-pipe junction at 8t (node 2) and

creates three pressure waves.

c) 10t - a wave has reached the upper right dead end at 9t (node 6) andthe positive reflection produces a surge pressure of 889 ft.

d) 12t - a wave has reached the upper left dead end at 11t (node 5) and

the reflection produces a surge pressure of 790 ft.

The following sections develop the analysis of the effects of pressure waves

impinging on network discontinuities. There are three types of discontinuities

for which the pressure wave analysis is developed:


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2-12 CHAPTER TWO

1) Components (pumps, valves, etc.)

2) Junctions

3) Side discharge orifices (connections to surge control devices that inject

or remove fluid)

These analyses provide the basis for modeling transient flow in complexpiping systems. They incorporate a variety of devices and include the effects of

cavity formation and collapse.

6t+

9t

Figure 2-6: Wave actions at 6 t + and 9 t transient events.


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10t

12t

Figure 2-7: Wave actions at 10 t and 12 t transient events.


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2-14 CHAPTER TWO

2.4 COMPONENT ANALYSIS

A general analysis of pressure wave action at a component in a pipe system is

presented. This analysis provides relationships to account for a variety of

situations. Figure 2-8 shows a general situation at a component (discontinuity)where pressure waves 1H and 2H are impinging.

At the same time, the characteristics of the component may be changing. It

is assumed that the relationship between flow through the component, Q, and

the pressure head change across the component, H , always satisfies a secondorder characteristic head-flow equation for the component having the general

form

QQtCQtBtAH )()()( ++= (2-8)

where H is the pressure head difference in feet (meters) between the left andright side of the component, and is the flow rate in ftQ 3/s (m3/s) through the

component. The terms A, B, and C represent the coefficients for a general

representation of the characteristic equation. The coefficients may be time

dependent but will be known (or can be determined) at all times. The absolute

values of are employed to make the resistance term dependent on flow

direction, and the others independent of flow direction. This representation

applies to both passive resistance elements such as valves, orifices, fittings and

friction elements and active elements such as pumps.

Q

Figure 2-8: Condition at component before and after wave action.

In Figure 2-8, the subscripts 1 and 2 denote conditions on the left and right

hand side of the component before the impinging waves arrive. The subscripts

3 and 4 denote these conditions at the component after the pressure wave


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action.'and represent the pipe flow rates left of pressure wave HQ "Q 1for

pipe 1 and right of pressure wave H2for pipe 2, respectively. An analysis is

carried out to determine conditions at the component due to wave actions, a

change in the characteristics of the component, a change in a reservoir pressure,

or a combination of any of these situations. The Q terms represent volumetric

flow rates in the lines leading to and away from the component and, due to

mass continuity, these values of flow rate are usually equal to each other at any

given time. However, the occurrence of cavitation and the associated growth

and decay of cavity pockets can affect the continuity relationships, and this

makes it necessary to consider the flow rate terms separately. Flow rate terms

must be considered with a proper sign and the convention employed is that flow

in a pipe away from the component is positive. Flow through the component is

positive from left to right.

The basic transient flow relationship for pressure-flow changes is applied to

incoming and outgoing waves and the terms Q and Q are eliminated to yieldthe following expressions for the outgoing waves

)( 13113 QQFHH += (2-9)

)( 24224 QQFHH += (2-10)

where

1

11

gAcF = and

2

22

gAcF = (2-11)

Pressure heads after the wave action are given by

3113 HHHH ++= (2-12)

4224 HHHH ++= (2-13)

The characteristic equation relating the pressure head change across and theflow through the discontinuity after the action is

00034 )()()( QQtCQtBtAHH ++= (2-14)

The coefficients of the characteristic equation, , and may

vary with time and these represent the values at the time of the wave action.

)(tA )(tB )(tC


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2-16 CHAPTER TWO

Equations 2-9 to 2-14 can be solved to obtain a quadratic relationship for .

This is,

0Q

0)()()( 021000 =+++ bQFFQtBQQtC (2-15)

where

))(22)((212211 iQFFHHHHtAb ++++= (2-16)

Equation 2-15 may be solved by using the quadratic formula or using the

Newton-Raphson iterative procedure starting with an initial solution .

The solution of Equation 2-15 for the flow rate after the action, , is easily

obtained. If there is no vapor cavity present, the flows in the pipes are

numerically equal to the flow through the component and are

iQQ =0

0Q

03 QQ = and 04 QQ = (2-17)

Equations 2-9 and 2-10 are solved to compute the magnitude of the pressure

waves produced by the action and Equations 2-12 and 2-13 give the pressure

head after the action takes place. Computer code for a general component

analysis is presented in Section 2.4.4.

This general analysis describes a wide variety of components for a range of

conditions. These applications are discussed and illustrated by examples. Table

2-2 summarizes the initial and computed conditions at a component due to an

action for a variety of situations and these cases are referred to in the following

discussions.

2.4.1 Component Characteristics

The coefficients of the component characteristic equation (2-14) are determined

using head/flow operating data for the component. Some components such as

pumps will utilize all three coefficients to represent the head/flow variation. In

some cases, the characteristic equation will be based on data, which representsthe head/flow relationship for a relatively small range of operation. For these

applications, the coefficients used for the component analysis will be based on

data valid for the operation in the vicinity of the operating point and will be

recalculated as the operating point changes. This is true for the analysis of

variable speed pumps and for pumps using data representing a wide range of

operating conditions, including abnormal situations such as flow reversal.


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Many components, such as valves, can be modeled using only the

coefficient. These are referred to as resistive components where the head/flow

relation is adequately described by a single resistive term. For this application,

the coefficient is defined as the component resistance. The term

resistance is used throughout this book to characterize a resistive component,

and is defined as the head drop divided by the flow squared

C

)(tC

)( 2QH . Here,

the head drop is in feet (meters) and the flow is in ft3/s (m3/s).

The component resistance is directly related to other resistive parameters

such as minor loss (KM), valve flow coefficient (Cv), sprinkler constant (Ks) and

others, which characterize the head/flow characteristic of a resistive component.

A number of exercises follow, illustrating the determination of the coefficient

values for the component characteristic curve and the close relationship

between the resistance coefficient Cand similar commonly used parameters.

2.4.2 Pipe Leading From a Reservoir to a Closing Valve

This example demonstrates the calculation of the pressure surge generated by a

slow valve closure. Consider a 1,000-ft long 6-in diameter pipeline with a wave

speed cof 4,000 ft/s connecting a reservoir on its upstream end and a valve on

its downstream end (as shown in the figure below).

The downstream valve is initially wide open and closes in 5 seconds (with

the area ratios changing linearly). The initial flow rate through the pipeline is

. Steady state pressure head on the upstream side of the wide

open valve is

sftQi /356.23

=

ftH 90= and the pipeline discharges into the atmosphere. When

the valve closure is initiated, the generated pressure wave travels towards thereservoir, gets reflected of the reservoir and reaches the closing valve in 2L/c=

0.5 seconds. Since the valve is closed in 5 seconds, which is greater than 0.5

(2L/c) seconds, this may be classified as a slow valve closure. Once a pressure

wave is generated, the conditions at the valve remain the same until the

pressure wave reflected of the reservoir reaches the valve. Therefore, this

example uses a computational time step of 0.5 seconds (2L/c) to demonstrate

the pressure changes as the valve closes linearly in 5 seconds.


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2-18 CHAPTER TWO

This analysis of wave action at the valve is similar to the situation depicted

in Figure 2-8 except that there is no pipe on the right hand side of the

component. Based on the initial flow rate and steady-state heads across the

valve, the valve initial resistance can be calculated as)(tC

621.1/)()0( 20 === iex QHHtC

Based on Equation 2-11, 66.6321 == gAcF and .02=F

At time seconds,0=t =1Q 2.356 ft3/s, =1H 9 ft, = 1H 0. At the end of

one computational time period ( 0.5 seconds), the valve area ratio ( )

changes from 1.0 to 0.9. The corresponding valve resistance may be computed

using

=t RA

0.29.0/621.1/)0()5.0( 22 ===== RAtCtC

The decreased area ratio will reduce the flow rate across the valve

simultaneously generating a pressure wave that travels toward the reservoir and

gets reflected. The new flow rate across the valve depends on the new head on

the upstream side of the valve, which in turn depends on the change in the flow

rate across the valve.

Equations 2-15 and 2-16 account for the combined effect and, therefore,

may be used to solve for the new flow rate through the valve at the end of one

computational time increment. Note that in Equations 2-15 and 2-16, we have

0)( =tA , 0)( =tB and 2)( =tC

66.6321=F and 02=F

91=H , 021 == HH , 02 == exHH , 02=H , and

356.2=iQ

Solving Equations 2-15 and 2-16 result in a new flow of 355.20=Q ft3/s

at the end of one computational time period. Equation 2-9 is then used tocalculate the magnitude of the resulting pressure wave as 08.23=H ft.

Equation 2-12 gives the new head on the upstream side of the valve as 11.08

ft.

3H

The above procedure may be repeated after updating the to reflect an

area ratio of 0.8, and setting ,

)(tC

355.2=iQ 08.111=H , and 08.21 =H


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(magnitude of the reflected pressure wave). This process is illustrated in the

computer program below.

'Wave Propagation Program Simple Pipeline System

' Constants:G = 32.17PI = 3.14159' Example Calculation Data:Length = 1000Dia = 6# / 12#Celerity = 4000Q_ini = 2.356H_ini = 9H_Exit = 0#Tc = 5#SimTime = 5#

Area_Ini = 0.25 * PI * Dia * DiaF1 = Celerity / (G * Area_Ini)If (Tc > (2 * Length / Celerity)) Then ' slow valve closure

Ao = Area_IniQi = Q_iniHo = H_iniHe = H_ExitCo = -(Ho - He) / (Qi * Qi)Q1 = QiQ3 = QiH1 = HoH3 = Ho

dH1 = 0dH3 = 0A = AoT = 0AR = A / AoC = CoPrint , "Results:"Print , "Time", "AR Ratio", "Q1", "H1", "dH1", "Q3",

"dH3", "H3"For i = 1 To 10If (AR > 0) Then

b = (H1 + 2 * dH1 + F1 * Qi - He)e = -F1

Qo = (-e - Sqr(e * e - 4 * C * b)) / (2 * C)'quadratic

Q3 = QodH3 = dH1 + F1 * (Q1 - Q3)H3 = H1 + dH1 + dH3

End IfPrint , Round(T, 2), Round(AR, 2), Round(Q1, 3),

Round(H1, 2);


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2-20 CHAPTER TWO

Print , Round(dH1, 2), Round(Q3, 2), Round(dH3, 2),Round(H3, 2)

AR = AR - 0.1C = Co / (AR * AR)A = Ao * AR

Qi = QoQ1 = Q3H1 = H3dH1 = -dH3T = T + (2 * Length / Celerity)

Next iEnd If

Results:

Time AR Ratio Q1 H1 dH1 Q3 dH3 H3

0.0 1.0 2.356 9.00 0.00 2.36 0.00 9.000.5 0.9 2.356 9.00 0.00 2.35 2.08 11.081.0 0.8 2.353 11.08 -2.08 2.34 4.89 13.891.5 0.7 2.342 13.89 -4.89 2.32 8.81 17.812.0 0.6 2.320 17.81 -8.81 2.28 14.48 23.482.5 0.5 2.283 23.48 -14.48 2.22 23.08 32.083.0 0.4 2.224 32.08 -23.08 2.13 36.94 45.943.5 0.3 2.129 45.94 -36.94 1.97 61.21 70.214.0 0.2 1.974 70.21 -61.21 1.71 108.91 117.914.5 0.1 1.706 117.91 -108.91 1.19 219.45 228.45

2.4.3 Exercises Component Characteristic Coefficients

1a). Suppose the following data is available for a pump

Flow rate (gpm) Head (ft)

0.0 592

2790.0 480

4180.0 400

Determine the coefficients ,A B , and for the relationshipC

2

CQBQAEp ++=

where is the pump head in feet and Qis the flow rate in ftPE3/s.

Answer: The component pressure wave analysis requires that the

coefficients be determined for Qin ft3/s and the head in feet. Before the

coefficients are determined, the flow rates can be converted to ft3/s (by


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dividing by the unit conversion factor of 448.8) and this gives the

following for the above data.

Flow rate (ft3/s) Head (ft)

0.00 5926.22 480

9.32 400

A quadratic curve can be passed through these points and three equations

can be obtained for the three data points by inserting the respective flow

and head in the pump characteristic equation. Substituting yields

2)32.9()32.9(400 CBA ++= 2)22.6()22.6(480 CBA ++=

A=592

The above equations are solved simultaneously to get ,592=A

8.12=B and , and the characteristic curve for this pumpcan be expressed as

837.0=C

2837.08.12592 QQEp =

At least three data points are required to determine the coefficients. The

first data point is usually the shut off head at zero flow.

1b). Determine the ,A B , andCcoefficients if this pump is operated at aspeed increase of 10 percent.

Answer: Note that you can produce a modified head-flow table by

multiplying the head by the speed ratio squared (1.21) and the flow by the

speed ratio (1.1). The modified head-flow table is

Flow rate (ft3/s) Head (ft)

0.000 716.326.840 580.80

10.252 484.00

Using the procedure described above gives

2835.01.1432.716 QQEp =


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2-22 CHAPTER TWO

1c). Check the results for 1a and 1b using the Head-Flow Characteristic

Calculation Tool (Appendix F.2).

2a). Resistance of a component is defined as the head loss )( H in feet

(meters) divided by the flow in ft)(Q 3/s (m3/s) squared and the resistance

is used for the coefficient for a resistive element ( andC A are zero).Calculate the resistance for the following data

a) andgpmQ 500= psiP 2=

b) mgdQ 5.3= and feetH 20=

c) andsLQ /5.2= barP 35.0=

d) andhrmQ /120 3= metersH 2=

Answer:

a) 72.3

b) 682.0

c) 200,571

d) 800,1

2b). Derive the conversion factor to convert resistance in English units to SI

units.

Answer: Multiply C(English) by 380 to convert to SI.

2c). Use the Resistance Calculation Tool (Appendix F.3) to check the above

results.

3a). Derive an expression for the resistance for a component in a pipe of

diameter with a known minor loss coefficient . Calculate the

resistance for the following data

d )( MK

a) 5.6=MK and ind 6=

b) 5.6=MK and mmd 150=

Answer: where is the pipe cross sectional area in

square feet (square meters) and

22/ gAKC M= A2/17.32 sftg= )/807.9( 2sm .

a) 62.2=C

b) 062,1=C


23/44


3b) Use the Resistance Calculation Tool (Appendix F.3) to check the above

results.

4a) Derive an expression for the resistance for a length of a pipe of

diameter with a known friction factor, Note: this is the resistance fora friction orifice simulating the effects of friction in a line segment with

these characteristics. Calculate the resistance for the following data

L

d .f

a) ftL 550= , ,ind 12= 02.0=f

b) mL 150= , ,mmd 250= 02.0=f

Answer:22/ gdAfLC=

a) 269.0b) 254


results.

5a) A valve flow coefficient is defined by the expression = the

flow in gpm (m

vC )( vK3/hr) through the valve with a pressure drop of 1 psi (bar).

Derive an expression for the resistance of a valve as a function of thevalve flow coefficient for English units. Determine the resistance for the

following data

a) 000,1=vC (pressure drop = 1.0 psi at a flow of 1000 gpm)

b) 500=vK (pressure drop = 1.0 bar at a flow of 500 cmh)

Answer: where))000,1//(/29( 2vCC = is the liquid specific

weight in lb/ft3

a) 465.0

b) 8.40


results.

5c) Figure 2-9 depicts a typical valve performance curve one provided by the

manufacturer (DeZurik/Copes-Vulcan). It gives the percent (%) of the


24/44

2-24 CHAPTER TWO

vC value as a function of the percent ( %) stroke for a butterfly

valve. This shows that for the normal flow direction (into the dome) the

value at 50% open is around 18%.

)( vK

vC )( vK

For case (a) above (Cv(100%) = 1000) , this means that at50 % open. Compute the resistance for the valve when it is 40% open.

180=vC

Answer: At 40% open the chart gives a value of around 11% or a

. This gives an equivalent resistance of 38.5.110=vC

Figure 2-9: BAW AWWA butterfly valves.

6a) The flow through an orifice depends on the orifice diameter, and the

discharge coefficient, for the orifice. Derive the expression for the

0d

dC


25/44


resistance in terms of these parameters. Calculate the resistance for the

following data.

a) = 1 inchand0d 62.0=dC

b) = 25 mmand0d 62.0=dC

Answer: where is the orifice cross sectional area in

ft

2)(2/1 ACgC d= A2(m2).

a) 360,1

b) 831,550


results.

7a) A single resistance value can be used to relate the head drop as a function

of the flow through a section of pipe including fittings. The resistance is

based on the expression

+=22 )2/)/(( QgAKdfLH M

In addition to determining the resistance for a general pipe section, this

expression can be used, for example, for the entrance resistance of thepiping leading to a surge tank, or the resistance for piping in parallel

pump arrangements. Derive an expression for the resistance for a pipe

section of length , diameter , friction factor , and with a known sum

for minor loss coefficient

L d f

)( MK . Calculate the resistance for thefollowing.

a) ind 6= , ftL 4= ,f = 0.02, MK (2 elbows (0.75) and entrancefrom T (1.5) = 3.0)

b) mmd 4.152= , mL 22.1= , f = 0.02, MK (2 elbows (0.75)

and entrance from T (1.5) = 3.0)

Answer: += )2/)/((2gAKdfLC M

a) 31.1

b) 500


26/44

2-26 CHAPTER TWO


results. The results should agree with the factor for converting resistance

from English to SI units (part 2b).

2.4.4 Example Component Computer Routine

The following computer routine illustrates the calculations for Case A-2(Table

2-2). This routine will handle general situations at a valve or a pump but does

not consider cavitation. It illustrates the wave action when a pressure wave of

100 feet impinges from the upstream side on a valve. The resulting flow rate is

calculated using both the direct solution to a quadratic equation and the iterative

Newton-Raphson procedure, which is carried out until convergence is reached.

The iterative approach is better able to handle situations such as flow reversal.

The notation is the same as used in Equations 2-8 to 2-17.

' SUBROUTINE - COMPONENT ANALYSIS

' Example Calculation Data:A = 0#B = 0#C = -20#H1 = 100#H2 = 20#D1 = 100#D2 = 0#F1 = 400#F2 = 400#

TOL = 0.001QI = 2#

C1 = (A + H1 + 2 * D1 - H2 - 2 * D2 + QI * (F1 + F2))B1 = B - (F1 + F2)A1 = C

' Quadratic Solution - A1 + B1*Q + C1*Q*Q)Q = -(B1 + Sqr(B1 * B1 - 4 * A1 * C1)) / (2# * A1)

Print " *** Quadratic Solution ***"PrintPrint " Q = "; Round(Q, 3)

' Newton Raphson Solution - DQ = F(Q)/F'(Q)J = 1Q = QIPrintPrint " *** Newton Raphson Solution *** "PrintPrint " Iteration", "Initial Q", "DQ", "Q"Do

FQ = C * Q * Abs(Q) + B * Abs(Q) - (F1 + F2) * Q + C1


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DFQ = 2# * C * Abs(Q) + B * Q / Abs(Q) - (F1 + F2)DQ = FQ / DFQPrint " "; J, Round(Q, 3), Round(DQ, 3), Round(Q - DQ, 3)Q = Q - DQACC = Abs(DQ / Q)

J = J + 1Loop While (ACC > TOL)

Q1 = -QIQ2 = QIQ3 = -QQ4 = QD3 = D1 + F1 * (Q3 - Q1)D4 = D2 + F2 * (Q4 - Q2)H3 = H1 + D1 + D3H4 = H2 + D2 + D4PrintPrint " H3 = ", Round(H3, 2)

Print " D3 = ", Round(D3, 2)Print " H4 = ", Round(H4, 2)Print " D4 = ", Round(D4, 2)

Results:

*** Quadratic Solution ***

Q = 2.226

*** Newton Raphson Solution ***

Iteration Initial Q DQ Q

1 2 -0.227 2.2272 2.227 0.001 2.226

H3 = 209.56D3 = 9.56H4 = 110.44D4 = 90.44

2.4.5 Illustrating the Wave Action at Components

A general use graphical Wave Action Calculator (WaveAction.EXE) was

developed to illustrate the action of pressure waves impinging on pipe systemcomponents, the effect of a change in the characteristics of the component, or

the combination of these effects. This tool (Figure 2-10) uses the Component

Computer Routine and is documented in Appendix D. Calculations are

necessary only when a pressure wave impinges on the element or the operating

conditions for the element change (valve setting or pump speed change). When

a calculation is made, the conditions and impinging waves just prior to the

calculation are shown (top) and then the new pressure waves are displayed


28/44

2-28 CHAPTER TWO

(bottom). The Wave Action Calculator is shown below for Case A-2 (Table 2-

2).

Figure 2-10: Wave Action Calculator Component (Case A-2).

2.4.6 Application to Resistive Elements

The general treatment for wave action at a component is applicable to a variety

of components. A large group of these are essentially resistive elements such as

valves. For these elements, only the C coefficient representing the effect of

irreversible loss is not zero. This coefficient represents the ratio of the head loss

to the square of the flow through the component. For hydraulic considerations,

this type of square-law relationship between head loss and flow is usually

appropriate. The sign of the pressure head change is dependent on the direction

of flow through the component. This necessitates the use of the absolute value

of the flow rates as presented in Equation 2-8.

a) Internal element

A resistive element such as a valve positioned within a pipe (rather than at the

end) is shown in Figure 2-11. The relationship between the pressure head

change and the flow rate is known for various elements.

For an orifice or a valve, this is


29/44


22

0

2

12 )2/(1 QgACHH d= (2-18)

where is the open area of the element and is a discharge coefficient that is

dependent on the type of element. In this case, the characteristic coefficient forEquation 2-8 is

0A dC

)2/(1)( 202 gACtC d= (2-19)

Figure 2-11: Valve operation in a pipe.

For an element that stays in a set position, remains constant but this

term varies for an element such as a valve that is opening and closing and

depends on the open area of the valve. Several situations depicting the effect of

pressure wave actions at a resistive element are illustrated in Table 2-2.

)(tC

In each case, the data given before the action plus the values of the

coefficients of the characteristics equation after the action are used to solve

Equation 2-15 for the flow through the element after the action. These are used

to determine pipe flow, pipe pressure heads, and pressure wave magnitudes that

are presented. For these examples, F1 = F2 = 400 and C1 = -20 are used.

Appendix D describes a Wave Action Calculator that can be used to analyzewave action at components and perform the calculations summarized in Table

2-2.

Case A-1 - Wave generation due to element action

In this example, the open area of the element is reduced (simulating a valve

closure) by 10% (C2 = C1/(0.9)2 = 24.69) resulting in a reduction of flow

through the element. Pressure waves are generated and line pressures increase

due to this action.


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2-30 CHAPTER TWO

Case A-2 - Wave reflection and transmission at the element

In this example, the element remains in the same open position (C2= -20) and

the action is due to impinging pressure waves at the valve. These calculations

using the Wave Action Calculator are shown in Figure 2-10.

Case A-3 - Combined action

The situations for Cases A-1 and A-2are combined in the example in a single

analysis accounting for both effects. It is not possible to superimpose the

separate results because of the non-linear relationships involved. This illustrates

the general situation in the Wave Characteristic Method where the analysis is

set up to account for the effects of impinging waves at the same time that

changes in component characteristics are considered.

Case A-4 - Wave action causes flow reversal

This example illustrates the situation where the flow reverses due to the actionof the impinging pressure wave. The general procedure developed for the

analysis of the component accounts for this, and the results presented in Table

2-2 correspond to this condition.

Case A-5 - Action at element where flow is initially reversed

This illustrates the analysis when the flow is initially in the negative direction.

The general analysis is developed to account for this situation as shown by this

example.

b) Line friction element

The effect of line friction on a pressure wave is easily accounted for byemploying an element in the analysis to represent the pressure drop due to line

friction in a length of pipe . This element will attenuate a pressure wave in a

manner similar to the total attenuation that will occur as the wave travels the

length in the pipe. In this representation, the pressure head change across the

element corresponds to the loss in the pipe and is given by

L

L

(2-20)22

122/ LgDAfLQHH =

where is the friction factor; is the line diameter; and is the line area.f D LA In this case, the characteristic coefficient for Equation 2-8 for the friction

element is

22/)( LgDAfLtC = (2-21)


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THE WAVE CHARAC

Table 2-2: Pressure wave action at components.

Case Conditions before action Conditions after actio

iQ 2 01H 2H H1 H Q 3H 4H H

A-1* 2 0 0 100 20 1.979 8.350 -8.350 108

A-2 2 100 0 100 20 2.226 9.560 90.440 209

A-3* 2 100 0 100 20 2.190 24.000 76.000 224

A-4 2 0 -1000 100 20 -0.396 958.430 91.570 1058

A-5 -2 100 0 20 100 -1.772 8.620 91.380 128

B-1 2 100 0 100 96 2.249 0.530 99.470 200

B-2 2 0 100 100 96 1.751 99.530 0.470 199

B-3 2 0 1000 100 96 -0.495 997.880 2.120 1097

C-1* 2 0 0 100 20 1.962 15.070 0.000 115

C-2 2 100 0 100 20 2.410 -63.870 0.000 136

C-3* 2 100 0 100 20 2.357 -42.830 0.000 157C-4 2 0 -500 100 20 -0.295 418.250 0.000 18

C-5 2 0 0 100 20 1.958 16.700 0.000 116

C-6* 2 100 0 100 20 2.318 -27.310 0.000 172

C-7 2 100 0 100 100 2.500 -100.000 0.000 100

C-8 0 100 0 100 - 0.000 100.000 0.000 300

D-1 10 100 0 100 300 10.392 60.810 156.750 260

D-2 10 0 100 100 300 9.608 39.240 -56.980 139

D-3* 10 0 0 100 300 9.820 17.970 -71.900 117

D-4* 10 100 0 100 300 10.211 78.910 84.340 278

D-5 10 0 3000 100 300 -1.953 1195.330 -1781.320 1295

*The component characteristics change for these analyses.


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2-32 CHAPTER TWO

Since variations in the friction factor occur due to flow changes, this coefficient

can change. Normally the friction factor will be adequately determined by the

flow rate through the element prior to the wave action. In most simulations, it is

sufficient to treat this coefficient as a constant.

The friction element is analyzed as previously described. The numericalvalue of is generally considerably smaller than the corresponding value

for elements such as a valve in the same pipeline. Several situations involving a

friction element are illustrated in Table 2-2.

)(tC

For the calculations, and40021 ==FF 0.121 == CC are used.

Case B-1, 2 - Wave reflection and transmission at friction element

These examples illustrate the effect of pressure wave impingement from either

side of a friction element. The effects are similar (a slight attenuation of the

pressure wave) regardless of the direction of wave impingement.

Case B-3 - Wave action causes flow reversalThis is similar to Case A-4and illustrates the situation when flow reversal in

the pipe (and through the friction element) occurs.

c) Terminal resistive element

The general analysis applies to a situation where the resistive element such as a

valve is located at the end of a pipe adjacent to a pressure reservoir. For this

situation, the values of andF H for the side of the component representing thereservoir are taken as zero and the analysis is carried out in the normal fashion.

In this manner, conditions on the reservoir side of the component will not beaffected by the wave action that is the proper influence of a pressure reservoir.Several cases are illustrated in Table 2-2 for a terminal element with the right

side being the pressure reservoir.

For these calculations, 201 =C , 4001=F and 02=F are used. The first

four cases are similar to those illustrated for the internal element. These are

Case C-1- Wave generation due to element action )69.24( 2 =C

Case C-2 - Wave reflection at element

Case C-3 - Combined valve action and pressure wave impingement

Case C-4 - Action causes flow reversal

For each of these cases, the value of pressure head in the reservoir remains

constant . Additional cases are illustrated.)( 24 HH =

Case C-5- Change in reservoir pressure

For this situation, the action is caused by a 20 ft change in reservoir pressure

(H2), from 20 to 40 ft, which produces the action illustrated by this example.


33/44


Case C-6- Combined element, pressure wave and reservoir pressure action

A 20 ft change in reservoir pressure can be considered simultaneously with

element action and pressure wave impingement, and this situation is illustrated

by this example.

2.4.7 Application to Active Elements (Pumps)

Elements that alter the fluid energy due to a mechanical means are referred to

as active elements. The general form of the characteristic Equation 2-8 was

chosen to be able to represent pump operation as depicted in Figure 2-12.

Figure 2-12: Pump operation in a pipe.

The characteristics , and may depend on various conditions

including pump speed, flow, pump head and other operating conditions and are

explicitly determined before calculations for the effects of wave action aremade. Various techniques have been suggested to represent transient pump

characteristics. A simple expression that is useful for pump startup or shutdown

is:

)(tA )(tB )(tC

QQCQaBaAHH RRR ++=2

12 (2-22)

where , and are the coefficients of a quadratic curve that represents

normal pump operation for a pump operating at full (rated) speed.RA RB RC

For variable speed operation during a start up or shut down,and represents the ratio of the rotational speed at any time during the

transitional period, , to the full speed rate of rotation, . In this manner, it is

assumed that the first two terms, which represent the gross pump headdeveloped, are functions of the rotational speed. In addition, these terms are not

dependent on flow direction but rather the rotational direction as determined by

the pump installation. The third term can be considered to represent the

resistance and accounts for the energy loss for the pump and this is taken as

RNNa /=

N RN


34/44

2-34 CHAPTER TWO

proportional to the square of the pump flow. This term is dependent on flow

direction, and might be considered also to vary during a period of start up or

shut down, since the resistance to flow through the component would be

affected by the changing conditions.

Transient pump operation is a complex phenomenon that is affected bymany factors. The above description does account for the principal factor

(pump speed) and offers a basis for evaluating unsteady flow in piping systems

that contain pumps operating in a normal manner at both constant and variable

speed. However, there are obvious limitations of this simplification of the pump

operation.

An alternate and more comprehensive treatment of pumps utilizes four

quadrant pump characteristic data (see Section 4.2.3). This data accounts for

abnormal pump operation such as turbining. For this approach, ,A B and aredetermined at a specific operating point by fitting a quadratic function to the

applicable pump data. The modeling of pumps for abnormal operations isdiscussed in detail in Section 4.2.

C

A number of examples are depicted in Table 2-2 to illustrate the type of

pressure wave action that takes place at a pump. For these examples, ,

, , , and

1001=F

4001=F 200=RA 10=RB 1=RC .

Case D-1, 2 - Action due to the impingement of pressure waves

These illustrate the effect of pressure wave impingement from both the left and

right side of a pump operating at full speed.

Case D-3 - Action due to change in pump speedA 20% reduction in pump speed is assumed and this changes the

numerical values of the coefficients and

)8.0( =a

A B as

2

2 )8.0(RAA = and )8.0(2 RBB =

The results of this action are presented.

Case D-4 - Combined action at pump

The effects considered in Cases D-1, D-2 and D-3 are combined for thisexample.

Case D-5 - Flow reversal at pump

The actions can result in flow reversal at the pump. In this case, the gross pump

head is maintained due to the pump speed and the effect of the resistance term

is reversed due to the flow reversal. These effects are illustrated by this

example.


35/44


2.4.8 Exercises - Wave Action at Components

The Wave Action Calculator (Appendix D) can be used for these exercises.

1) Use the results for Exercise 1a (2.4.2 Exercises - Component Characteristic

Coefficients) to calculate the changes in flows and heads due to a 10%increase in pump speed for the pump initially flowing at 9 ft3/s with a

suction head of 50 ft. Use the pump characteristic curve to compute the

initial discharge head and then use the Wave Action Calculator to compute

the new conditions due to the new set of coefficients resulting from the

pump speed increase.

Answer:The initial head change at the pump is 432 ft giving a discharge

head of 482 ft. Using the new coefficients for the pump characteristic curve

the new conditions are

Flow = 9.196 ft3/s

Upstream head = 28.2 ft

Downstream head = 560.8 ft

Pressure wave = 78.8 ft

2) VerifyCases A-1,B-1, C-1andD-1.

Answer: see Table 2-2.

3) Use the results for Exercise 5a and 5b (Component Characteristic

Coefficients) to calculate the changes in flow and head due to a 10%closure in a butterfly valve (50% to 40%) assuming the valve is initially

flowing at 600 gpm (1.337 ft3/s) with an upstream pressure of 50 psi

(111.35 ft). Use the valve characteristic curve to compute the initial

downstream head, and then use the Wave Action Calculator to calculate the

new conditions due to the new value of the Ccoefficient resulting from thepartial valve closure.

Answer: Based on the valve flow coefficient of 1,000, the wide-open

resistance for the butterfly valve is 0.465. From the performance chart for

the Butterfly valve, the following information is read: % flow = 11 % at 40% open )110( =vC and 18% at 50% open )180( =vC . The pressure drop

across the valve at 50% open is (600/180)2 or 11.11 psi. The discharge

pressure = 50 11.11 = 38.89 psi (89.72 ft). The valve resistance at 40%

open is 38.5 as calculated previously. Using the initial head and flow and

the new coefficients for the valve characteristic curve, the new head and

flow conditions are


36/44

2-36 CHAPTER TWO

Flow = 1.289 ft3/s (579 gpm)

Upstream head = 134.5 ft. (58.3 psi)

Downstream head = 70.6 ft (30.6 psi)

2.5 PIPE JUNCTION ANALYSIS

The effects of pipe junctions on pressure waves are easily handled if continuity

of flow and continuity of pressure at the junctions are employed. This approach

essentially neglects small energy losses at the junction that cause only minor

effects. Figure 2-13 shows the conditions considered.

Figure 2-13: Effect of pipe junction on pressure wave.

A pressure wave of magnitude H impinging in one of the junction legs, iis transmitted to each of the adjoining legs equally. The magnitude of the

transmitted waves is HTi where the transmission coefficient, , is given byiT


37/44


=

j

j

ii

F

FT

)/1(

/2 (2-23)

where the summation j refers to all legs (pipes) connecting at the junctions. A

reflection back in leg ioccurs of magnitude HRi where

1= ii TR (2-24)

The derivation of Equations 2-23 and 2-24 for a three-pipe junction is

illustrated in the following.

Consider a pressure wave of magnitude H approaching a three pipe

junction with initial flow rates of Q1, Q2, and Q3and an initial pressure head of

H1(Figure 2-13). The wave action at the junction results in a reflected wave ofmagnitudeR1*Hinto the pipe 1 and a transmitted wave of magnitude T1*H

down each of the remaining two legs (pipe 2 and pipe 3). The new flow rates

through the pipelines as a result of the wave action are Q1, Q2, and Q

3, and the

resulting pressure head at the junction is H2. Based on basic transient flow

relationship for pressure flow changes, one can derive an expression to compute

the magnitude of the transmitted and reflected waves as follows.

For flow change in pipe 1

H * (1-R1) = F1* (Q1-Q1) (2-25)

Similarly for pipes 2 and 3

H * T1= F2* (Q2-Q2) (2-26)

H * T1= F3* (Q3-Q3) (2-27)

The relationship between the transmission coefficient (T1) and the reflection

coefficient (R1) may be obtained using

H2= H1+ T1* H = H1+ H + R1* H (2-28)

or

R1= T1 1 (2-29)

The continuity equations for the junction before and after the wave actions are


38/44

2-38 CHAPTER TWO

Q1= Q2+ Q3 (2-30)

Q1= Q2+ Q

3 (2-31)

Subtracting 2-31 from 2-30 results in

(Q1- Q1) = (Q2 Q2)+ (Q

3 Q3) (2-32)

Substituting the pressure-flow change relationship in the above equation results

in

((H * (1-R1)) / F1) = ((H * T1) / F2) + ((H * T1) / F3) (2-33)

Substituting the relationship between transmission and reflection coefficients

(R1= T1-1) in the above equation and simplifying result in

(2-T1) / F1= (T1/ F2) + (T1/ F3) (2-34)

or

321

11 111

2

FFF

FT

++

= (2-35)

For the simultaneous impingement of waves arriving in more than one leg,

the effects can be superimposed.

Table 2-3 illustrates several situations involving the transmission andreflection of pressure waves at a three-pipe junction.

Table 2-3: Pressure wave action at pipe junction.

Case Conditions before action Conditions after action

1H 1H 2H 3H 2H 4H 5H 6H

E-1 100 100 0 0 214.3 14.3 114.3 114.3E-2 100 0 100 0 157.1 57.1 -42.9 57.1

E-3 100 0 0 100 128.6 28.6 28.6 -71.4

E-4 100 100 100 100 300 100 100 100

G-1 100 -200 0 0 -30 70 -130 -130

G-2 -30 100 50 -50 -30 -100 -50 50


39/44


For these examples, the pipeline constants )/( gAcF= are ,

and . Note that these values indicate that the area of pipe 1

is greater than the combined areas of pipes 2 and 3, resulting in a transmission

coefficient greater than 1.0 for a wave approaching in pipe 1.

1001=F

2002=F 4003=F

The notation for this example is shown in Figure 2-14 that also lists the

data used for these illustrations. Cases E-1, E-2, and E-3 consider the

impingement of a pressure wave in each leg separately, while Case E-4

considers the simultaneous impingement of these waves.

Figure 2-14: Wave action at pipe junction for example (Table 2-3).

A computer routine to calculate the effect of pressure waves at junctions is

presented in the next section. This routine is employed by the Wave Action

Calculator that may be used to calculate wave action at a junction and the

calculations for Case E-1are shown in Figure 2-15.

Figure 2-15: Wave action calculator junction (Case E-1).


40/44

2-40 CHAPTER TWO

Equation 2-23 provides the basis for evaluating the effect of wave action at

dead end junctions and open ends or connections to reservoirs. These are

considered to be pipe junctions. In the case of a dead end, it is a two-pipe

junction with 02=A . This gives a 2=iT and 1=iR , which means that the

wave is reflected positively from the dead end. For a reservoir connection,

and gives=2A 0=iT and , which means that a negative reflection

occurs at a reservoir.

1=iR

2.5.1 Example Junction Computer Routine

The following computer code illustrates the calculations of wave action at a

junction for Case E-1. This routine will handle pressure wave action

calculations at a multi-pipe junction but does not consider cavitation. The

example calculation shows the wave action when a pressure wave of 100 ft

impinges from one pipe of a three-pipe junction. The notation is similar to that

used in Equations 2-23 and 2-24.

' SUBROUTINE - JUNCTION ANALYSIS

' Example Calculation Data:

NCONN = 3SUMF = 0F(0) = 100F(1) = 200F(2) = 400

D(0) = 100D(1) = 0D(2) = 0

For J = 0 To NCONN - 1SUMF = SUMF + 1 / F(J)D(J + NCONN) = 0

Next J

For J = 0 To NCONN - 1For K = 0 To NCONN - 1If (J = K) Then

D(J+NCONN)=D(J+NCONN) + (2/(F(K)*SUMF)-1) * D(K)Else

D(J+NCONN) = D(J+NCONN) + (2/(F(K)*SUMF)) * D(K)End If

Next KNext J

Print " Results:"Print " "Print " D(4) = ", Round(D(3), 3)


41/44


Print " D(5) = ", Round(D(4), 3)Print " D(6) = ", Round(D(5), 3)

Results:

D(4) = 14.286D(5) = 114.286D(6) = 114.286

2.5.2 Exercises Wave Action at Junctions

1) For the configuration and conditions shown in the 3-pipe junction below,

compute the following:

a) the effect of a pressure wave of 100 ft magnitude approaching the

junction in the 12-in line.b) the effect of a 100 ft pressure wave approaching the junction in

either of the 6-in lines.

2) Calculate the pressure increase in each pipe due to the combined effect of

a 100 ft pressure wave approaching in the 12-in line and impinging on the

junction and the subsequent reflection of the transmitted waves at the two

dead end pipes.


42/44

2-42 CHAPTER TWO

3) Carry out additional calculations to see if a further increase in pressure

will occur when the reflections from the dead ends impinge on the

junction.

4) Use the Wave Action Calculator to do parts 2 and 3.

Answer:

The solutions are illustrated in the diagrams below. These diagrams depict the

locations of pressure waves and their effects on local pressure head at various

transient time steps ( t is the time for a pressure wave to travel 100 ft). Thisexercise shows that because of dead ends and the pipe configuration a pressure

wave of magnitude 100 ft impinging on a region with a pressure of 200 ft is

capable of producing pressure increases of 334 ft, 266 ft, and 221 ft in the three

pipes.

1a). This shows the pressure wave action just after the wave (approaching the

junction from the 12-in line) reaches the junction. Tis the transmissioncoefficient (Equation 2-23) for the wave approaching in the 12-in pipe for

this situation ))/1(//2( = iii FFT .

1b). This shows the pressure wave action just after the wave (approaching the

junction from the left leg of the 6-in line) reaches the junction.


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2-44 CHAPTER TWO

Reflected wave from B arrives at junction and transmits and reflects to

connecting pipes

Secondary negative wave reflects from end B and lowers pressure there

Secondary positive wave reflects from end A to further increase pressure

wave characteristic method

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