Frequency Domain Analysis of Fluid-structure Interaction in Liquid Filled Pipe Systems by Transfer Matrix Method

8/3/2019 Frequency Domain Analysis of Fluid-structure Interaction in Liquid Filled Pipe Systems by Transfer Matrix Method

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International Journal of Mechanical Sciences 44 (2002) 20672087

Frequency domain analysis of uidstructure interaction inliquid-lled pipe systems by transfer matrix method

Q.S. Lia ;, Ke Yanga;b, Lixiang Zhangc, N. Zhanga; d

aDepartment of Building and Construction, City University of Hong Kong, 83 Tat Chee Avenue,

Kowloon Hong KongbCollege of Architectural and Civil Engineering, Wenzhou University, Wenzhou, ChinacSchool of Electric Power Engineering, Kunming University of Science and Technology, Kunming, China

d Faculty of Engineering, University of Technology, Sydney, Australia

Received 22 February 2002; received in revised form 23 August 2002; accepted 3 October 2002

Abstract

This paper is concerned with the vibration analysis of a liquid-lled pipe system, which extends the fre-

quency domain analysis of the uidstructure interaction from single pipe to a pipe system with multi-pipe

sections using transfer matrix method. Taking into account all the three major coupling mechanisms, namely

the friction coupling, Poisson coupling and junction coupling, the proposed method can be used to analyze thefree vibration and the forced vibration of a pipe system with multi-pipe sections subjected to various kinds of

external excitations. The transform matrix, impedance matrix and frequency equation in frequency domain are

also presented and discussed. Numerical examples are presented to illustrate the application of the proposed

method, which shows how the natural frequencies and mode shapes change with the radius and materials of

the pipe system.

? 2002 Elsevier Science Ltd. All rights reserved.

Keywords: Fluidstructure interaction; Transfer matrix method; Liquid-lled pipe systems; Laplace transform; Frequency

domain

1. Introduction

The uidstructure interaction (FSI) in liquid-lled pipe systems has been investigated extensively,

because of its relevance to mechanical, civil, nuclear and aeronautical engineering. It has been widely

accepted that in dynamic analysis of liquid-lled pipe systems, neglecting FSI may lead to unrealistic

predictions. Literature reviews on the advances in this eld were given in Refs. [14]. Therefore,

Corresponding author. Tel.: +852-2784-4677; fax: +852-2788-7612.

E-mail address: [email protected] (Q.S. Li).

0020-7403/02/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved.P I I : S 0020- 7403(02)00170- 4
mailto:[email protected]:[email protected]


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2068 Q.S. Li et al. / International Journal of Mechanical Sciences 44 (2002) 2067 2087

only several previous investigations which are directly related to the present study are reviewed

below.

First of all, it is necessary to declare the three major coupling mechanisms of FSI in pipe systems,

namely friction coupling, Poisson coupling and junction coupling. The friction coupling representsan axial interaction caused by friction between uid and pipe. The Poisson coupling is such an

interaction that the change of uid pressure causes additional hoop stress in pipe wall and then,

owing to Poisson ratio, induces corresponding normal stress in pipe wall, and vice versa. The

junction coupling occurs only at the boundaries or the junction of two pipe sections. Mathematically,

the Poisson and friction coupling make the governing equations coupled each other and cause the

equations much dierent from the traditional ones [5], whereas the junction coupling is normally

expressed as the boundary conditions, among the three coupling mechanisms, the junction coupling is,

therefore, the easiest one to deal with. In the present study, all the three major coupling mechanisms

are taken into account.

Thorley [6] was the rst who pointed out the existence of precursor wave caused by the Poissoncoupling, and Vardy and Fan [7] veried it through a well-designed experiment. Their experimental

results will be used in the numerical example of this paper to verify the accuracy of the proposed

method.

Charley and Caignaert [8] used experimental data to demonstrate that transfer matrices with con-

sidering FSI can predict much better the measured pressure spectra than the classical waterhammer

[5] transfer matrices, even in simple systems.

Dsouza and Oldenburger [9] presented one of the earliest studies in the eld. In their paper,

the Laplace transform was used to solve an equation included the friction and junction coupling.

Wilkinson [10] presented transfer matrices for the axial, lateral and torsional vibration of pipes.

He considered the junction coupling, but without considering the friction and Poisson coupling.

El-Raheb [11] and Nanayakkara and Perreira [12] derived transfer matrices for straight and curvedpipes, including the eects of the junction coupling but excluding those of the Poisson and friction

coupling. Kuiken [13] studied the eects of the Poisson and junction coupling through a numerical

simulation.

Lesmez [14], Lesmez et al. [15], Hateld et al. [16] and Wiggert et al. [17] (in time domain),

Tentarelli [18], Brown and Tentarelli [19] and De Jong [20,21] (in frequency domain), Svingen and

Kjeldsen [22] and Svingen [23] (based on the nite element method) applied the transfer matrix

method (TMM) to one-dimensional wave problems.

Among the above-mentioned studies, only in Refs. [18,19] the friction coupling was taken into

account. Moreover, the dynamic behavior of uid-lled pipes with non-uniform cross-section or

variable material properties was not investigated in these studies.Dierent from these studies, Zhang et al. [24] obtained a solution of the four-equation model

of FSI in the frequency domain in which the impact loads are considered. Followed by a series of

researches conducted in recent years [2531], it has been proved that the frequency-based approaches

are ecient for the analysis of FSI in liquid-lled pipe systems. However, up to now, this kind of

method has been used for single pipes only. In this paper, based on these studies, a transfer matrix

method is developed for the analysis of FSI in a series pipe system which may consist of many

sections of pipes. Meanwhile, some previously not considered aspects are also taken into account

in this paper, which include the frequency equations, the impedance matrix, the frequency response

matrix and the mode shapes. At the end of this paper, numerical examples are presented to illustrate


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Q.S. Li et al. / International Journal of Mechanical Sciences 44 (2002) 2067 2087 2069

the application of the proposed method and to investigate the eects of the radius and the material

properties of pipes on the dynamic behavior of a pipe system.

In addition to the analysis of free vibration and frequency response of a liquid-lled pipe system,

the present TMM aims mainly at determining the solution for FSI at any point of the system whenthe system is subjected to external excitations in order to make the transient solution in time domain

possible by taking inverse Laplace transform.

2. The frequency domain solution for single pipe and discussions

Following the derivation given in Ref. [13], the frequency domain solution for a single pipe is

rewritten in this section, meanwhile, some discussions are given and expansions are made.

2.1. The governing equation and its uncoupling

The governing equation for a uid-lled pipe system can be expressed with matrices as

A@y(z;t)

@t+ B

@y(z;t)

@z+ Cy(z;t) = r(z;t); (1)

where y(z;t) is a vector of unknowns

y(z;t) = [V; H; uz ; h]T; (2)

where V = V(z;t) and H = H(z;t) are the cross-sectional average speed and the cross-sectional

pressure head of liquid, respectively; uz = uz(z;t) is the cross-sectional average speed along the

direction of z and h = h(z;t) is the cross-sectional average stress head of pipe wall. r(z;t) in the

right-hand side of Eq. (1) is the external excitation acting along the pipe. H=H(z;t) and h =h(z;t)

are dened below related to pressure P = P(z;t) and normal stress z = z(z;t) as

H(z;t) =P(z;t)

gf+ z sin()

P0(z)

gf;

h(z;t) =z(z;t)

gt+ z sin()

0(z)

gt; (3)

where is the elevation angle of the pipe; P0(z); 0(z) are the initial pressure and initial stress,respectively; and t and f are the density of the pipe and liquid, respectively.

Other parameters in Eq. (1) are

A =

1 0 0 0

0 g=c2F 0 0

0 0 1 0

0fgR

tec2T

0 1

tc2T

; B =

0 g 0 0

1 0 2 0

0 0 0 1=t

0 0 1 0

; (4)


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where

c2

F=

1

f1

k+

2R(1 2)

eE1

; c2

T=

E

t: (5)

In Eq. (5) L is the length of the pipe, is Poissons ratio, R is the inner radius of the pipe, e

is the thickness of the pipe wall, E is elastic modulus, g is the gravity and K is the bulk elastic

modulus of liquid. Matrix C contains the coecients of friction and structural viscous damping.

When the laminar ow model is adopted, C is a constant matrix [12].

In Eq. (4), the terms a42 = fgR=tec2T and b23 = 2 represent the Poisson coupling, which,

together with matrix C, make the governing equations coupled each other.

Taking the Laplace transform, denoted by L(), for Eq. (1) results in

sA(s)Y(z;s) + B@Y(z;s)

@z

= r(z;s); (6)

where

Y(z;s) =L(y(z;t));

A(s) = A + C=s;

r(z;s) =L(r(z;t)) + A(s)y(z; 0) (7)

in which y(z; 0) is a vector of the initial conditions.

From a generalized eigenvalues problem |BA|=0 one obtains a diagonal matrix with eigenvaluesin the diagonal elements

= diag{1(s); 2(s); 3(s); 4(s)}

= diag{1(s); 1(s); 3(s); 3(s)} (8)

and the full matrix S(s) whose columns are the corresponding eigenvectors satises [5]

BS(s) = A(s)S(s)(s): (9)

Evidently, matrix S(s) is regular, and in a frictionless system C = 0, the eigenvalues i, and the

elements of matrices A and S are all real numbers independent of s.

Multiplying Eq. (1) with

T(s) = S1(s)A1(s) (10)

and then combined with Eq. (9) yield

sv(z;s) + @v(z;s)

@z= T(s) r(z;s); (11)

where

v(z;s) = T(s)A(s)Y(z;s): (12)


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Since is a diagonal matrix, Eq. (11) is a set of four independent ordinary equations with

complex constant coecients, and its general solution is

v(z;s) = E(z;s)v0(s) + q(z;s); (13)

where

E(z;s) = diag

exp

s

1(s)z

; exp

s

1(s)z

; exp

s

3(s)z

; exp

s

3(s)z

;

v(z;s) = {v1; v2; v3; v4}T; q(z;s) = {q1; q2; q3; q4}

T:

(14)

v0 = v0(s) contains undetermined integration constants depending on the boundary conditions, and

q(z;s) is a particular solution. When denoting

T r= [

r1(

z;s)

; r2(

z;s)

; r3(

z;s)

; r4(

z;s)]

T;(15)

the elements of vector q(z;s) can be determined by

qi =sesz=i (s)

i(s)

z0

ri(x;s)esx=i (s) d x; i= 1; 2; 3; 4:

From Eq. (12) and with S(s) = (T(s)A(s))1, we have

Y(z;s) = K(z;s)v0(s) + Q(z;s); (16)

where

K(z;s) = S(s)E(z;s); Q(z;s) = S(s)q(z;s): (17)It is evident that K(z;s) is regular.

2.2. The boundary conditions and their forms

To meet the needs of the transfer matrix method, it is necessary to express the boundary condi-

tions in matrix forms at individual end instead of at both ends of a single pipe. After taking the

Laplace transform, the boundary conditions at an end point of a pipe can be generally expressed in

matrix as

D(s)Y( z;s) = f(s); (18)where D and f can be determined according to the forms of the boundary conditions. f is illustrated

in the following equations (Eqs. (19)(21)) in the form of fr; fR and fm. In Y( z;s) (dened in

Eq. (7)) z is the coordinate of the end (e.g., 0 or the length of the pipe L).

The following is some examples for several boundary conditions expressed in this manner.

(1) Reservoir (or an opened end): If the pipe is xed at this end, we have

Dr =

0 0 1 0

0 1 0 0

; fr(s) = [ug(s) 0]

T; (19)


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where ug denotes the ground velocity, e.g., when it is subjected to an earthquake excitation. The

pressure Po of the reservoir is taken into account in H (see Eq. (3))

(2) Closed valve or closed end with mass m:

(a) Pipe xed

DR =

1 0 0 0

0 0 1 0

; fR(s) = [ug(s) ug(s)]

T: (20)

(b) Pipe is movable in axial direction

Dm =

1 0 1 0

0 gfAf sm gtAt

; fm = [0 Rl(s)]

T; (21)

where Af; At are the area of inner part of the pipe and area of the pipe wall, respectively. m is the

mass of valve or the sealed end etc. The sign is determined according to the direction of thecoordinate and the position where the mass or the excitation appears. Rl is the Laplace transform of

the external excitation at the corresponding end. In the case shown in Fig. 2, Rl can be written as

Rl = Ar

Err(esTc 1)=s; (22)

where the subscript r denotes the property of the impacting rod.

Eq. (21) shows an example of the junction coupling since V and uz, or H andh, appear in the

same equation of boundary condition.

With the above expressions, the boundary conditions can be written with relatively simple and

unied forms. For example, when a single pipe is xed and connected with a reservoir at theupstream end z = 0, and is xed and connected with a closed valve at the downstream end z = L,

then the boundary conditions can be written as

DrY(0; s) = fr(s); DRY(L;s) = fR(s): (23)

2.3. The frequency domain solution for a single pipe

In Eq. (16), let z = 0 and z = L, we get eight relations between the unknowns and undetermined

integration constants, namely

Y(0; s) = K(0; s)v0(s) + Q(0; s); (24)

Y(L;s) = K(L;s)v0(s) + Q(L;s): (25)

There are only two boundary conditions at the end z = 0 or L. The general expressions are

[Dup(s)]24{Y(0; s)}41 = {F(0; s)}21;

[Ddown(s)]24{Y(L;s)}41 = {F(L;s)}21;(26)

where the subscripts outside the bracket denote the numbers of row and column of the matrix.


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or in matrix form as follows:

[[I]44

[

S(s)E1(L;s)S1(s)]44

] Y(0; s)Y(L;s)

81

= 0 (35)

by setting the right-hand term equal to zero, Eq. (26) yields

[Dup(s)]24{Y(0; s)}41 = 0; (36)

[Ddown(s)]24{Y(L;s)}41 = 0: (37)

Combining Eqs. (35)(37), we get

[Dup(s)]24 [0]24

[0]24 [Ddown(s)]24

[I]44 [ S(s)E1(L;s)S1(s)]44

Y(0; s)Y(L;s)

81

= 0; (38)

where [0] is a zero matrix. Since Y is not always equal to zero, there must be

[Dup(s)]24 0

0 [Ddown(s)]24

[I]44 [ S(s)E1(L;s)S1(s)]44

= 0: (39)

Eq. (39) is the desired frequency equation with s = + j! as variable, the complex frequencies can be obtained by setting the real part and the imaginary part of Eq. (39) equal to zero, respectively.

Taking F of Eq. (31) equal to zero yields

Z(z;s)Y = 0: (40)

Therefore, when the ith natural frequency is gained, with Eq. (40), the mode shape function of the

system corresponding to the ith natural frequency can be obtained with standard methods [3235].

3. The transfer matrix method for pipe systems with several sections

A series pipe system consisting of several sections with dierent radiuses, thickness of pipe wall

and material properties (see Fig. 1) is widely used in practices such as in the high-pressure pipe

lines of water power stations. Meanwhile, if a pipeline under the action of a concentrated force, the

pipe line must be divided into two sections for analysis. It is, therefore, necessary to obtain the FSI

frequency domain response for a pipe system with multi-sections.

Fig. 1 illustrates a series pipe system with N pipe sections numbered 1; 2; : : : ; N from upstream

to downstream. A point connecting two adjacent pipe sections is called node, the sequence num-

bers is 0; 1; : : : ; N also from upstream to downstream. The direction of axis z is from upstream to

downstream.


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Fig. 1. The sections and nodes of pipe system.

The coordinate z of the ith section is used as local coordinate and denoted as zi. For the ith

section, the upstream node, namely the (i 1)th node, has zi = 0 and its downstream node, namelythe ith node, has a coordinate zi = Li. Adopting the local coordinate will make the formulas clear.

3.1. The eld transfer matrix

Since Eq. (16) is valid for all sections, then for the ith section, we have

Yi(zi; s) = Ki(zi; s)v0i(s) + Qi(zi; s); i = 1; 2; : : : ; N ; (41)

where v0i(s) is a vector of undetermined integration constants related to the ith section (also see

Eqs. (13) and (14)). Substituting 0 and Li into Eq. (31) yields

Yi(0; s) = Ki(0; s)v0i(s) + Qi(0; s);

Yi(Li; s) = Ki(Li; s)v0i(s) + Qi(Li; s):(42)

From Eq. (32), v0i(s) can be expressed as follows:

v0i(s) = K1i (0; s){Yi(0; s) Qi(0; s)};

v0i(s) = K1i (Li; s){Yi(Li; s) Qi(Li; s)};

(43)

which results in

K1i (0; s){Yi(0; s) Qi(0; s)} = K1i (Li; s){Yi(Li; s) Qi(Li; s)} (44)

or

Yi(0; s) = Ki(0; s)K1i (Li; s)Yi(Li; s) + qi(z;s); (45)

where

qi(s) = Ki(0; s){Qi(0; s) K1i (Li; s)Qi(Li; s)}: (46)

We now dene the eld transfer matrix as

Fi(s) = Ki(0; s)K1i (Li; s) = SiE

1(Li; s)S1i ;

F1i (s) = S1i (s)E(Li; s)Si(s):

(47)


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Fi(s) provides the relation between the unknowns at both ends of a section

Yi(0; s) = Fi(s)Yi(Li; s) + qi(s): (48)

3.2. Point transfer matrix

The junction condition between the two pipe sections in Fig. 1 can be found in Ref. [29]. Here,

expressing them in matrix form and then taking Laplace transform results in

DiYi(Li; s) = Di+1Yi+1(0; s) + P(out)i(s); i = 1; 2; : : : ; N 1: (49)

For the ith node, the matrix Di in Eq. (49) is of the form

Di =

Af(i) 0 Af(i) 0

0 1 0 0

0 0 1 0

0 Af(i) 0 At(i)

; P(out)i(s) =

0

0

L(Pout(t))

0

; (50)

where Af(i); At(i) are the area of liquid and the pipe wall in the ith section, respectively. Obviously,

the matrix Di is regular. Pout is a concentrated external force acting at the ith node. So the relation

between the unknowns of the ith and (i + 1)th sections is provided by Eq. (49).

From Eq. (49), we have

Yi(Li; s) = D1i Di+1Yi+1(0; s) + D

1i P(out)i(s): (51)

The point transfer matrix of the ith node is dened as

Pi = D1i Di+1: (52)

With the point transfer matrix Pi, the unknowns on the both sides of the ith node are related with

Yi(Li; s) = PiYi+1(0; s) + PiF(out)i(s); (53)

where

F(out)i(s) = D1i+1P(out)i(s): (54)

To make the expression clear, it must be pointed out that, in this paper, only in the expressions ofPi; P(out)i(s); F(out)i(s) the subscript i identies the ith node which is the downstream node of section

i. For other variables and matrices, the subscript i identies the ith pipe section.

3.3. The transfer matrix method

By means of the point transfer matrices and the eld transfer matrices, the relation of the unknowns

in the ith and the (i + 1)th sections can be found by substituting Eq. (48) into Eq. (53)

Yi(Li; s) = PiYi+1(0; s) + PiF(out)i(s) = PiFi+1(s)Yi+1(Li+1; s) + Pi Qi+1(s); (55)


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where

Qi+1(s) = F(out)i(s) + qi+1(s): (56)

Let Nn be the global transfer matrix and denoted by

Nn(s) =

ni=1

(Fi(s)Pi); n = 1; 2; : : : ; N 1; (57)

the general relation between the 1st and Nth sections is

Y1(0; s) = NN1(s)FN(s)YN(LN; s) +

Nk=2

Nk2(s)Pk1 Qk(s); N 2 (58)

in which we dene that N0 = I and

Q1 = q1, and the second right-hand term in Eq. (58) is equalto zero if the upper limit is less than the lower limit.

For example, if the pipe system consists of three sections, from Eq. (58) we have

Y1(0) = F1P1F2P2F3Y3(L3) + ( q1 + F1P1 Q2 + F1P1F2P2 Q3)

or

Y1(0) =

2i=1

(FiPi)F3Y3(L3) +

q1 +

1i=1

(FiPi) Q2 +

2i=1

(FiPi) Q3

;

Y1(0) = N2F3Y3(L3) + ( q1 + N1 Q2 + N2 Q3):

(59)

Similar to Eqs. (27) and (28), another four algebra equations are

[Dup(s)]24{Y(0; s)}41 = {F(0; s)}21;

[Ddown(s)]24{Y(L;s)}41 = {F(L;s)}21:(60)

At last, we have eight algebra equations with eight unknowns, namely Eqs. (58) and (60), and we

rewrite them in a unied matrix form

[G(s)]88

Y(0; s)

Y(LN; s)

81

= [Q(s)]81; (61)

where

G(s) =

[Dup]24 [0]24

[0]24 [Ddown]24

[I]44 [ NN1(s)FN(s)]44

; (62)

Q(s) =

F(0; s); F(LN; s);

Nk=2

Nk2(s)Pk1 Qk(s)

T: (63)


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Denoting

G1(s) = [G1(s)]48

[ G2(s)]48 ; (64)

then

v01(s) = G1(s)Q(s); v0N(s) = G2(s)Q(s): (65)

By using Eq. (65), and with the reverse process used in the above deduction, we can obtain the

solution for any intermediate section from the upstream to downstream, and vice verse.

First, let us go from upstream to downstream. If the solution Yi(Li) in the ith section has been

obtained, the following three formulas can be used successively to get the solution for the (i + 1)th

section:

from Eq. (49)

Yi+1(0) = P1i Yi(Li) (Fout)i; (66)

from Eq. (43)

v0(i+1) = K1i+1(0; s){Yi+1(0) Qi+1(0)}; (67)

from Eq. (41)

Yi+1(z;s) = Ki+1(z;s)v0(i+1)(s) + Qi+1(z): (68)

Eq. (68) gives the general frequency domain solutions for the (i + 1)th section. Let z = Li+1 in

Eq. (68), Yi+1(Li+1) is determined, then go back to Eq. (66), the solution for the (i + 2)th section

can then be obtained, and so on.

Similarly, taking the inverse way, going from downstream to upstream, the solutions for interme-

diate sections can also be obtained. Namely, if Yi(z;s) in the ith section has been known, Yi(0; s)

is, therefore, known. The following three formulas can be used successively to get the solutions for

intermediate sections, since the right-hand side in each formula is known.

Yi1(Li1) = Pi1Yi(0) + Pi1(Fout)i1; (69)

v0(i1) = K1i1(0; s){Yi1(Li1) Qi1(Li1)}; (70)

Yi1(z;s) = Ki1(z;s)v0(i1) + Qi1(z): (71)

The above-mentioned method is generally used for calculating the frequency domain response ofthe system subjected to external excitations. When the inverse Laplace transform is adopted, the

transient response of the system can also be obtained. On the other hand, for calculating the natural

frequencies, we can simply take the right-hand side of Eq. (61) equal to zero, namely

|G(s)| = 0: (72)

For calculating the corresponding complex mode shapes, the following equation that is similar to

Eq. (40) for each pipe section can be used

Zi( zi; s)Yi( zi; s) = 0: (73)


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Fig. 2. Experiment rig of steel pipe [7].

Table 1

Geometrical and material properties of the pipe apparatus [7]

Steel Pipe Water Steel Rod

L = 4:5 m length K = 2:14 GPa bulk model Lr = 5:02 m length

R = 52:0 mm inner radius t = 999 kg=m3 density Er = 200 GPa Youngs modulus

e = 3:945 mm pipe wall thickness P0 = 2:0 MPa initial pressure r = 7848 kg= m3 density

E= 168 GPa Youngs modulus vr = 1 m=s velocity

= 0:3 Poissons ratio Tc = 1:98 ms impact time

t = 7985 kg=m3 density of pipe Vr = 0:1175 m=s impact velocity

m0 = 1:312 kg mass at z = 0

mL = 0:3258 kg mass at z = L

4. Numerical examples and discussions

Vardy and Fan [7] has designed an experiment rig to make accurate measurement and to study the

dynamic behavior of waterhammer. The test rig is shown in Fig. 2 and the specications are list in

Table 1. This rig consists of a water-lled pipe closed at both ends (with mass m0, mL, respectively)

and suspended by wires. The closed pipe is subjected to axial impact by a steel rod at the end of

z =0 to generate transients. This rig has its superiority to conventional reservoir-pipe-valve system in

describing the inuence of FSI. Also, in this apparatus, eects of friction and gravity are unimportant,

and the inuences of uidstructure interaction related to the Poisson and junction couplings can be

clearly isolated in case of axial wave propagation [29].

It can be seen from Eq. (3) that the two initial conditions are H|t=0 =0 and h|t=0 =0. The systemis initially still, this problem has, therefore, only zero initial condition. The boundary conditions of

this system are1 0 1 0

0 gfAf sm gtAt

Y(0; s) = 0;

1 0 1 0

0 gfAf sm gtAt

Y(L2; s) = 0:

(74)


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Fig. 3. The analytical model for the TMM.

Table 2

Frequency results of a single pipe

Results Frequencies in mode sequence numbers (0:5 Hz)

1 2 3 4 5 6 7 8

Test [7] 173 289 459 485 636 750 918 968

Zhang et al. [24] 172 286 455 473 627 741 907 945

Reduced TMM 172 286 454 472 627 741 907 945

R1 = R2 = R=2 167 303 458 470 635 772 874 939

Table 3

TMM results with m0 = mL and L1 = L2 = L=2

Results Frequencies in mode sequence numbers (0:5 Hz)

1 2 3 4 5 6 7 8

R1 = R2 = R=2 167 302 434 470 633 769 874 939

R1 = R; R2 = R=2 166 302 439 471 607 769 905 914

Reduced TMM 171 285 454 460 624 740 907 921

E1 = E; E2 = 3E=4 169 278 432 449 614 725 873 894

E1 = E2 = 3E=4 167 271 414 442 605 706 828 889

In the numerical example, the total length of the system remains as unchanged, and at the middle point, the pipe is divided into two parts, each of them has a length L=2. This makes it possible for

the TMM calculation to be consistent with the original test by taking the parameters of the two pipe

sections identical (Fig. 3). In fact, by doing so, the present TMM does get the same results as Ref.

[24] (see Table 2 Reduced TMM).

Due to symmetry, let the two masses be both equal to m0. The results are listed in Reduced

TMM of Table 3.

In the numerical example, two cases are considered separately. One is that the radius of the second

pipe section is equal to R=2, the other is changing E to 3E=4 at the second pipe section. Both cases

keep the parameters of the rst pipe section to be the same as those listed in Table 1. In order to


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-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

1 101 201 301 401 501 601 701 801 901

Frequency in Hz

log(abs(H))

E and E

3E/4 and 3E/4

E and 3E/4

Fig. 4. The frequency response of the two-section pipe changing with R.

log(abs(H))

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

1 101 201 301 401 501 601 701 801 901

Frequency in Hz

R and R

R/2 and R/2

R and R/2

Fig. 5. The frequency response of the two-section pipe changing with E.

compare the results clearly, the results of a single pipe with R=2 or 3E=4 are also listed in Table 3.

The frequency responses are shown in Figs. 4 and 5.

As is well known, the kth frequency fk and the mode shape function uk of a solid rod with both

ends free can be expressed as

fk =k

2L

E

t; uk = cos

k

Lz

: (75)

Eq. (75) means that the frequencies and mode shape functions of such a solid rod are independent

on the radius of the rod. However, for the present problem, the frequencies are signicantly dependent

on the radius. Even for the single pipe system when R of the pipe decreases, the frequencies are

also obviously changed (see Table 2 and Fig. 4).

When the two masses m0 and mL are set equal to each other, only the frequencies of the 4th and

8th modes change obviously. This shows that the frequencies of these two modes mainly represent


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-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

H/max(H)

R and R

R/2 and R/2

R and R/2

Fig. 6. The 1st mode of H changing with R.

-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

V/max(V)

R/2 and R/2

R and R/2

R and R

Fig. 7. The 1st mode of V changing with R.

the vibration of the pipe and the other frequencies mainly depend on uid. Therefore, only the 1st

mode (mainly depending on uid) and the 4th mode (representing the vibration of the pipe) are

discussed in detail below.In the published papers [2427] using the frequency domain methods, the mode shapes were not

presented. In this study, the mode shapes are determined and shown in Figs. 612. For the present

problem, the 2nd and 4th columns of the matrix H(z;s) are the response of the system subjected

to a unit impulse excitation at the upstream and downstream end, respectively. The mode shapes

in Figs. 612 are obtained from the 2nd column of the matrix H(z;s) (see Eq. (32)) by taking

s = j!k, here !k is the kth natural frequency of the system. The 4th column gives the same results.

Figs. 612 are obtained by taking the real parts of the corresponding unknowns, namely

V = real(L(V)); H = real(L(H)); U = real(L(u)) (76)


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-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

H/max(H)

R and R

R/2 and R/2

R and R/2

Fig. 8. The 4th mode of H changing with R.

-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

U/max(U)

R and R

R/2 and R/2

R and R/2

Fig. 9. The 4th mode of U changing with R.

in which V ; H ; u are the same as those dened in Eq. (2). For the mode shapes, taking the

imaginary parts one obtains the same results.

It is worth mentioning that, when the two sections have dierent R, the mode of H is dierent

in the two sections (see Figs. 68). However, the mode of V has a sudden change in the junction

point for keeping the discharge Qi be equal at both sides of the node (see Fig. 7), in as much as

Qi = ViAfi and Afi are dierent at two sides of the ith node. From Figs. 9 and 12, it can be seen that

the mode of U is approximately a cosine, but the frequency is lower than the 1st natural frequency

of the solid rod (509 Hz).

5. Conclusion

In this paper, the uidstructure interaction in liquid-lled pipe system is studied in frequency

domain using the transfer matrix method, which extends the frequency domain analysis for FSI


18/21


-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

H/max(H)

E

3E/4

E and 3E/4

Fig. 10. The 1st mode of H changing with E.

-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

H/max(H)

E

3E/4

E and 3E/4

Fig. 11. The 4th mode of H changing with E.

presented by Zhang et al. [24] from a single-section pipe to a multi-section pipe system. Eorts are

also made to expand the results of the single pipe, of which the frequency equations, the impedance

matrix, the frequency response matrix and the mode shapes are especially worth mentioning. These

expansions are also applicable for pipelines with multi-sections.In the analysis of FSI problems, the present transfer matrix method is dierent from the traditional

one. In the case of the junction coupling, the unknowns in FSI problems are coupled in boundary

conditions, they satisfy a set of algebra relations other than each variable equals to a constant,

which makes it dicult to realize the transfer matrix method directly. In this paper, the diculty is

overcome by nding out a unied matrix expression of the boundary conditions.

The present method has included all the three major coupling mechanisms, namely the friction

coupling, the Poisson coupling and the junction coupling, and can deal with dynamic analysis of

liquid-lled pipe systems under various kinds of external excitations. The method aims at conducting

both harmonic analysis and the frequency response analysis, especially aim at obtaining the solution


19/21


-1.5

-1

-0.5

0

0.5

1

1.5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

location z/L

U/max(U)

E

3E/4

E and 3E/4

Fig. 12. The 4th mode of U changing with E.

at any point of a pipe system, so that the transient response of the pipe system subjected to various

excitations can be determined by the inverse Laplace transform.

From the view point of practical applications, this is a simple frequency domain method for the

analysis of pipe systems with various sections considering FSI. In comparison with nite element

method (FEM) and method of characteristics (MOC) in time domain, the present method needs

much fewer lines of code in programming.

The method is useful for the analysis of FSI in multi-section pipe systems that are widely used in

engineering practices. Meanwhile, it is also useful for single or multi-section pipe systems subjected

to concentrated force acting on the pipe systems.

Numerical examples show that the results determined by the proposed method are in good agree-

ment with the experimental data, thus verifying the accuracy of the proposed method. It is also

shown through the numerical examples that how the frequencies and mode shapes change with the

radius and material properties of the pipe systems.

Acknowledgements

The authors thank for the nancial supports provided by the National fund of Natural Science of

China (Project No.50079007), and The Ministry of Water Resources, China (Project No. SZ9830)

for the study described in this paper.

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Documents

Frequency Domain Analysis of Fluid-structure Interaction in Liquid Filled Pipe Systems by Transfer Matrix Method