Chellapilla 3 ICONE10 Double Bellows 2002 .pdf
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ICONE 10
Tenth International Conference on Nuclear Engineering
April 14-18, 2002
Arlington, Virginia, USA
“DRAFT”
ICONE 10-22092
TRANSVERSE VIBRATIONS OF DOUBLE BELLOWS
EXPANSION JOINT RESTRAINED AGAINST ROTATION
KAMESWARA RAO.C
Bharat Heavy Electricals Limited
Hyderabad, India
Keywords:
Bellow vibration, double bellows, transverse
frequency & elastic constraints
RADHAKRISHNA.M
Indian Institute of Chemical Technology
Hyderabad, India
industry, the elongation/ contraction of the piping
system is quite significant and frequent changes in
temperature or relative mechanical motion of
ABSTRACT
structures with respect to each other. Therefore, the
The paper presents the results of investigation of
bellows are very susceptible to vibrations that can
transverse vibrations of single bellows expansion
be easily excited either structurally through the
joint restrained against rotation on either end. A
fixed ends of the expansion joints or by the fluid
theoretical model is developed based on the
flowing inside the bellows.
Bernoulli-Euler beam theory and includes added
transverse vibration of double bellows expansion
mass of the fluid flowing inside the pipe-bellow-
joint consider the bellows directly welded to pipe
pipe assembly. Neglecting effects of shear and
flange and uses fixed-fixed beam approach for
rotary inertia an exact frequency equation is
investigation of transverse mode shapes and natural
derived for the transverse vibrations of single
frequency [1].
Most studies of
bellows expansion joint including the effects of end
elastic restraints against rotation. Numerical results
However, in real practice the bellow is welded to a
are presented for an example bellow showing the
short pipe nipple. The other end of the nipple has a
effects of variation of elastic restraints and internal
flange and is connected to the pipe flange. Figure 1
pressure on the first four modes of vibration.
depicts the bellow-nipple-flange assembly.
Introduction
The paper presents an analysis of estimating the
Bellows expansion joints are designed to absorb the
frequency and mode shapes of transverse vibrations
axial or transverse displacements in pipeline
of double bellows expansion joint restrained
systems. It is observed that any piping system is
against rotation on either end. The analysis forms
more or less flexible, and so has the natural
the basis of development of a theoretical model for
flexibility to accept small elongation or contraction.
the bellow considering as a Bernoulli-Euler Beam.
However, in many of the applications such as the
The model considers fluid filled bellows and
fossil/nuclear power plants or in aircraft and space
neglects the effect of shear and rotatory inertia.
More precise formulae are developed based on the solution of
Differentiating the above equation (2) and substituting
Bernoulli-Euler differential equation for calculating the natural
into differential equation (1) –
frequency of bellows. Results obtained using the Bisection
method are compared with results presented in thesis [1].
d4 X
[Pπ Rm2 + J ω2] d2X
m tot
X ------- + -------------------- ----- − W2 ---- = 0
dx2
EI
dx2
EI
(4)
√ (Pπ Rm2 + Jω2 )
If c = -----------------------√2 EI
(5)
The Differential Equation
Figure 2 shows the various geometrical dimensions of U-shaped
bellows –where R1 and R2 is the inner and outer radius of bellow,
p is the pitch, h is the convolution height, t is the thickness of
bellows wall and L is the length of bellow.
λ=
(mtot W2 )1/4
--------------(EI)1/4
(6)
Fig. 1 : Flange-Pipe Nipple-Bellow assembly
d(X4)
d X2
------+ 2c2 ---- − λ4 X = 0
dx2
dx4
A
(7)
The general solution of the equation is given byX = A sinh αx + B cosh αx + C sin βx + D cosβx
(8)
The first three derivatives of equation (8) are given as
follows–
5
Fig. 2 : Geometry of Bellows
The general form of the differential equation of vibration of
bellows for single or double bellows expansion joint is given by –
∂4W
∂2W
J * ∂4ω
∂2
2
EI -------- + Pπ Rm ------ − ----------- + m tot ------- = 0
∂x2
∂x2 ∂t2
∂t2
∂x4
(1)
d(X)
---- = Aα cosh αx + Bα sinh αx + Cβ cos β Dβ sinβx
dx
(9)
d2(X)
----- = Aα2 sinh αx + Bα2 cosh αx - Cβ2 sin βx - Dβ2cos βx
dx2
(10)
d3(X)
----- = Aα3 cosh αx + Bα3 sinh αx - Cβ3 cos βx + Dβ3 sinβx
dx3
(11)
Where EI is the bending stiffness, P-internal pressure, J-mass
moment of inertia per unit length, m tot –total mass of bellows per
Where the roots of the equation are α & β and their
unit length includes bellows material mass and fluid mass, x –
values are given by -
axial coordinate, R m –is the mean radius of bellow, W-deflection
α = √ -c2 + √ (c4 + λ4)
(12)
β = √ c2 + √ (c4 + λ4)
(13)
and t-time [2].
Using the technique of Separation of variables, the lateral
deflection of the bellows axis ‘w’ can be expressed asW (x, t) = X (x) T (t)
(2)
A, B, C, D are arbitrary constants
Exact Solution of Double Bellows Expansion Joint
As the end B does not rotate the third boundary
Lateral Mode Natural Frequency
condition, slope is zero and given by -
A case of vibration of bellows in lateral mode is shown in
∂W (L, t)
------------ = 0
∂x
figure 2. It is seen that the pipe has pure translational motion
(18)
due to the geometry and physical symmetry of the system
The fourth boundary condition at end ‘B’ is the shear
provided the coriolis component of force acting on the
force Q (L, t) of the bellows and is given by-
bellows from the fluid flowing inside is neglected.
As a mathematical approximation one half of the system is
considered with its left end having a rotational stiffness ‘R’
and right end fixed to the vertical rollers and is shown in
figure 2.
ks
d3 X(L) −W2{Ms + (m p + mf3) a}
----------= ------------------------------- * X (L) + ---- (L) (20)
EI
EI
d x3
−W 2 {Ms + (m p + mf3) a} k s X(L)
If b = --------------------------------- − -------EI
EI
B
5
$
∂3W(L, t) {Ms + (m p + mf3) a} ∂2W(L, t) k s W(L, t)
-------- = ------------------------ ---------- + --------(19)
EI
∂t2
EI
∂x 3
Ms is the equivalent lateral support mass, k
(21)
s
is the
equivalent spring stiffness of the lateral support, m p is
?
the mass per unit length of the connecting pipe of
Fig. 3 : Mathematical model of universal expansion
joint for lateral mode
length ‘a’ and m
f3
is the mass of the fluid per unit
length in the connecting pipe.
Substitution of the general solution (8) and its
derivatives into boundary condition expressions (14),
(15), (16) and (17) will give a set of linear equations
Boundary Conditions
with respect to the constants A, B, C and D.
As shown in figure 3, end ’A’ of the bellow is connected to
B+D=0
(22)
{B (αL) 2 − D (βL) 2 – T (A αL + C βL)} = 0
(23)
a pipe nipple, and considered to have a rotational stiffness
of ‘R’. Now the first two boundary conditions are written as
follows -
Aα cosh αL + Bα sinh αL + Cβ cos βL - Dβ sinβL=0
(24)
Deflection is zero at W (0, t) = 0 and
R ∂ X (0)
EI ∂2 X (0)
---- ------------ = ---- ------L2
∂x2
L
∂x
∂2 X (0)
∂ X (0)
---------- = T ----------∂x2
∂x
(14)
[α3 cosh αL + b sinh αL] A + [α3 sinh αL + b cosh
αL] B − [β3 cos βL − b sin βL] C + [β3 sin βL + b
(15)
cos βL] D = 0
(25)
Substituting c1, c2, c3, c4 for the terms in the brackets in
According to [1], the maximum allowable pressure in
equation (25) we get -
bellows is –
c1 = α cosh αL + b sin hαL
(26)
c2 = α3 sinh αL + b cosh αL
(27)
c3 = β3 cos βL − b sinβL
(28)
c4 = β3 sinβL + b cos βL
(29)
3
Pmax = π k* P
--------6.666L2
(32)
Where k is the equivalent axial stiffness of bellows -
For a non-trivial solution the determinant formed by the
Substitution of the above numerical values into the
expressions 4 & 5, we get –
coefficients of the system of algebraic equations above
c = √ 616.49 + 0.0001135ω2
(33)
must be equal to zero-
λ = 1.0029√ω
(34)
The expansion of the above determinant results in the
b = 0.10235ω2
(35)
frequency equation for the system shown in figure 2.
0
1
+ T (α)
(α)
2
0
1
+T(β)
A
+(β)
2
B
αcosh(αL) αsinh(αL) β cosβL -βsinβL
c1
c2
c3
=0
(30)
C
c4
The study involves the effect of varying the equivalent
support stiffness and mass on the frequency of vibration.
Using expressions in (12), (13), (32), (33), (34) and (35)
D
the frequency equation (31) is solved by applying
bisection method. The results are presented in Tables 1,2
2
2
T {b [(α −β ) sinβL sinh αL + 2αβ (1- cos βL cosh αL]
and 3 and the same are also graphically represented in
− αβ (α2+ β2) (α sinh αL cos βL+ β cosh βL sin βL) + b
figures 4 and 5 respectively.
(α2 + β2) (α sinβL cosh αL - β cos βL sinh αL -αβ (α2
+β2) cos βL cosh βL
(31)
Tables 1 and 2 presents the frequencies obtained by
varying the support stiffness ks for mode numbers N=1 and
Fig 4: Transverse Vibrations-Double Bellows
(Lateral Mode) & for N=1 and N=2 & Ms =0
N=1
N=2
2. It is seen that as stiffness increases from a value of 102
16000
to 1015, frequency increases from 331.40rad/s to
14000
3400.1597rad/s, a steep rise by almost 90%. The
Lateral Frequency, w rad/s
12000
frequency attains almost a constant value from 108
10000
8000
onwards.
6000
4000
The effect of varying the support mass on the frequency of
2000
0
1
10
100
1000
10000
100000
1000000
10000000 100000000
1E+09
1E+10
Support Stiffness, ks (kg/m)
vibration is also studied and the results presented in Table
3. It is seen that there is a steep drop of 87% in frequency
Results and Discussion
as the support mass increases from 1.0 to 100 for N=1 and
Let us take a double bellows expansion joint having the
by a small drop of 1% for N=2.
following geometrical and physical parameters -bellow
length L=0.0693m, mass moment of inertia per unit
length J=0.001153kgm, EI =5.078Nm2, total bellows
mass mtot = 5.138kg/m, total connecting pipe mass,
mp+mf3 = 5.0kg/m. and assuming a= L.
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length
bellows
equation
results
stiffness
bellow
system
lateral
support
vibration
joint
double
frequency
expansion
transverse