Spectral Analysis for some model of fluid flow
Kadry Zakaria, Magdy A. Sirwah, and M. Fakharany
Department of Mathematics, Faculty of Science, Tanta University, Tanta, Egypt
1
Problem Formulation
The flow of a non-Newtonian fluid between two eccentric cylinders of finite length is considered
here.
z
W
y
RB
x
Rj
OB
e
e
Q
Oj
e
P
Figure 1. Journal bearing geometry and nomenclature.
ρ
ρ
Du
= −∇P + ρF + ∇ ∧ M + µ∇2 u − η∇4 u
Dt
2
∇·u=0
(1)
(2)
1
The field equations governing the motion of the lubricant in the Cartesian coordinate system are
∂4u
∂P
∂2u
= µ 2 −η 4,
∂x
∂y
∂y
(3)
∂P
= 0,
∂y
(4)
∂4w
∂2w
∂P
=µ 2 −η 4,
∂z
∂y
∂y
(5)
∂u ∂v
∂w
+
+
= 0.
∂x ∂y
∂z
(6)
∂ 2 w ∂ 2 u =
= 0,
u(x, 0, z) = w(x, 0, z) =
∂y 2 y=0
∂y 2 y=0
∂ 2 u ∂ 2 w u(x, h, z) = U, w(x, h, z) =
=
= 0.
∂y 2 y=h
∂y 2 y=h
(7)
(8)
By applying the above boundary conditions the velocity components u and w are solved for
equations (3) and (5) respectively:
1 ∂P
cosh[(2y − h)/2`]
Uy
2
+
y(y − h) + 2` 1 −
,
u=
h
2µ ∂x
cosh[h/2`]
(9)
cosh[(2y − h)/2`]
1 ∂P
2
y(y − h) + 2` 1 −
,
w=
2µ ∂z
cosh[h/2`]
(10)
and
r
η
.
µ
Integrate the continuity equation (6) with respect to y with the boundary conditions for v as given
by
where, ` =
v(x, 0, z) = 0, v(x, h, z) =
∂h
.
∂t
The modified Reynolds equation is
∂
∂P
∂
∂P
∂h
∂h
g(h, `)
+
g(h, `)
= 6µU
+ 12µ ,
∂x
∂x
∂z
∂z
∂x
∂t
where
g(h, `) = h3 − 12`2 h + 24`3 tanh
h
.
2`
With the non-dimensional variables and parameters
z=
L
R
`
h
z̄, x = Rθ, P = µN ( )2 P̄ , `¯ = , τ = ωt, and H =
= 1 + ε cos θ.
2
C
C
C
2
(11.a)
(11.b)
Equations (11) become in non-dimensional form
and
∂
¯ ∂ P̄ + D 2 ∂ G(h, `)
¯ ∂ P̄ = 12π ∂H + 24π ∂H ,
G(H, `)
∂θ
∂θ
L ∂ z̄
∂ z̄
∂θ
∂τ
(12.a)
¯ = H 3 − 12`¯2 H + 24`¯3 tanh H .
G(H, `)
¯
2`
(12.b)
Assuming the journal whirls about its mean steady state position given by ε0 and φ0 for the
first-order perturbation, the pressure and film thickness can be expressed as
P̄ = P̄0 + P̄1 ε1 eiλτ + P̄2 ε0 φ1 eiλτ ,
(13.a)
H = H0 + ε1 eiλτ cos θ + ε0 φ1 eiλτ sin θ,
(13.b)
ε = ε0 + ε1 eiλτ ,
(13.c)
φ = φ0 + φ1 eiλτ ,
(13.d)
H0 = 1 + ε0 cos θ.
(13.e)
Substituting Eqs (13) into Eqs (12) and collecting the zeroth- and first-order terms for ε1 and ε0 φ1
gives the following equations in P̄0 , P̄1 and P̄2 :
∂ P̄0
D 2 ∂
∂ P̄0
∂H0
∂
¯
¯
G0 (H0 , `)
+( )
G0 (H0 , `)
= 12π
,
∂θ
∂θ
L ∂ z̄
∂ z̄
∂θ
∂
∂ P̄1
D 2 ∂
∂ P̄1
∂
∂ P̄0
0
¯
¯
¯
G0 (H0 , `)
+( )
G0 (H0 , `)
+
G0 (H0 , `)
cos θ +
∂θ
∂θ
L ∂ z̄
∂ z̄
∂θ
∂θ
D ∂
¯ ∂ P̄0 cos θ = −12π sin θ + 24πi cos θ,
( )2
G00 (H0 , `)
L ∂ z̄
∂ z̄
∂
¯ ∂ P̄2 + ( D )2 ∂ G0 (H0 , `)
¯ ∂ P̄2 + ∂ G0 (H0 , `)
¯ ∂ P̄0 sin θ +
G0 (H0 , `)
0
∂θ
∂θ
L ∂ z̄
∂ z̄
∂θ
∂θ
D ∂
¯ ∂ P̄0 sin θ = 12π cos θ + 24πi sin θ,
( )2
G00 (H0 , `)
L ∂ z̄
∂ z̄
where
¯ = H 3 − 12`¯2 H0 + 24`¯3 tanh( H0 ), and G0 (H0 , `)
¯ = 3H 2 − 12`¯2 tanh2 ( H0 ).
G0 (H0 , `)
0
0
0
¯
2`
2`¯
The above PDEs subjected to the following boundary conditions:
∂ P̄α ∂θ (θ,
P̄α (θ, ±1) = P̄a ,
P̄ (θ, z̄) =
∂ P̄
(θ, z̄),
∂θ
3
= 0,
0)
at θ = θcav.
(14)
(15)
(16)
The subscript α = 0, 1 and 2 respectively are for the steady state and first-order perturbed
pressures in equation (14) to (16) and P̄a is the dimensionless ambient pressure.
The Solution Method
2
2.1
Spectral Analysis for Static Load Pressure
¯ in terms of the cosine and
To apply Fourier spectral method for Eq. (14), we expand G0 (H0 , `)
eccentricity ratio such that
¯ ≡ G0 (θ, `,
¯ ε0 ) = λ(3) + 3λ(3) ε0 cos θ + λ(3) ε2 (1 + cos 2θ) + λ(3) ε3 (3 cos θ + cos 3θ) + O (ε4 ),
G0 (H0 , `)
0
0
0
0
1
2
3
where
1
1
(3)
(3)
λ0 = 1 − 12`¯2 + 24`¯3 tanh( ¯), λ1 = 1 − 4`¯2 tanh2 ( ¯),
2`
2`
3
1
1
1 1
(3)
(3)
2 1
¯
λ2 =
1 + 2`sech
( ¯) tanh( ¯) , λ3 =
1 + (3 tanh2 ( ¯) − 1)sech2 ( ¯)
2
4
2`
2`
2`
2`
and for P̄0 (θ, z̄), we assume it takes the form
P̄0 (θ, z̄) =
∞
X
Zn (z̄) sin nθ,
Zn (z̄) =
n=1
2
π
Z
π
P̄0 (θ, z̄) sin nθdθ .
(17)
0
To find the functions Zn (z̄), we substitute into Reynolds equation (14), and differentiate twice
term by term, we get
∞
∞
∞
X
X
3 (4) X
−
Γn,n Zn (z̄) sin(n + 1)θ +
Γn,n+ Zn (z̄) sin(n − 1)θ +
Γn Zn (z̄) sin nθ + λ1 ε0
2
n=1
n=1
n=1
∞
∞
∞
X
X
X
1 (4) 2
λ2 ε0 2
Γn Zn (z̄) sin nθ +
Γn,2n− Zn (z̄) sin(n + 2)θ +
Γn,2n+ Zn (z̄) sin(n − 2)θ +
2
n=1
n=1
n=1
∞
∞
∞
X
X
1 (4) 3 X
λ3 ε0 3
Γn,n− Zn (z̄) sin(n + 1)θ + 3
Γn,n+ Zn (z̄) sin(n − 1)θ +
Γn,3n− Zn (z̄) sin(n + 3)θ+
2
n=1
n=1
n=1
∞
X
−12πε
0
sin θ,
Γn,3n+ Zn (z̄) sin(n − 3)θ =
(3)
λ0
n=1
(3)
(4)
where λj
=
λj
(3)
λ0
, j = 1, 2, 3, Γn ≡ (
(18)
D 2 d2
)
− n2 , and Γn,n± ≡ Γn ± n.
L dz̄ 2
Equation (18) gives a system of ordinary differential equations for Zn (z̄), to solve this system
4
we apply the straightforward method up to third order for the eccentricity ratio
Zn (z̄, ε0 ) =
3
X
εj0 Zn(j) (z̄) + O (ε40 ).
(19)
j=0
By substituting Eq. (19) into Eq. (18) yields first the unperturbed system
∞
X
Γn Zn(0) (z̄) sin nθ = 0,
(20)
n=1
the first order perturbation
∞
∞
X
12π
3 (4) X
Γn Zn(1) (z̄) sin nθ + λ1
Γn,n− Zn(0) (z̄) sin(n + 1)θ +
Γn,n+ Zn(0) (z̄) sin(n − 1)θ = − (3) ,
2
λ0
n=1
n=1
n=1
(21)
the second order perturbation
∞
X
∞
X
∞
∞
X
3 (4) X
(4)
Γn Zn(2) (z̄) sin nθ + λ1
Γn,n− Zn(1) (z̄) sin(n + 1)θ +
Γn,n+ Zn(1) (z̄) sin(n − 1)θ + λ2
2
n=1
n=1
n=1
∞
X
∞
∞
X
1 (4) X
Γn Zn(0) (z̄) sin nθ + λ2
Γn,2n+ Zn(0) (z̄) sin(n − 2)θ = 0,
Γn,2n− Zn(0) (z̄) sin(n + 2)θ +
2
n=1
n=1
n=1
(22)
finally, for third order perturbation
∞
X
∞
∞
X
3 (4) X
Γn Zn(3) (z̄) sin nθ + λ1
Γn,n+ Zn(2) (z̄) sin(n − 1)θ +
Γn,n− Zn(2) (z̄) sin(n + 1)θ +
2
n=1
n=1
n=1
∞
∞
∞
X
X
X
1 (4)
(4)
Γn Zn(1) (z̄) sin nθ+ λ2
λ2
Γn,2n− Zn(1) (z̄) sin(n + 2)θ +
Γn,2n+ Zn(1) (z̄) sin(n − 2)θ +
2
n=1
n=1
n=1
∞
∞
∞
X
X
X
1 (4)
Γn,3n− Zn(0) (z̄) sin(n + 3)θ+
Γn,n+ Zn(0) (z̄) sin(n − 1)θ +
λ3 3
Γn,n− Zn(0) (z̄) sin(n+1)θ+3
2
n=1
n=1
n=1
∞
X
Γn,3n+ Zn(0) (z̄) sin(n − 3)θ = 0.
(23)
n=1
From Eq.(20), we have Γn Zn(0) (z̄) = 0, using the given boundary conditions implies that
4P̄a cosh[(2n − 1)Lz̄/D]
,
(2n − 1)π cosh[(2n − 1)L/D]
(0)
O (ε00 ) : Z2n−1 (z̄) =
(24)
and so on up to the third order. Consequently, we obtain the pressure distribution function in the
form
P̄0 (θ, z̄, ε0 ) =
3
X
j=0
εj0 (
∞
X
Zn(j) (z̄) sin nθ) + O (ε40 ) =
n=1
3
X
j=0
5
(j)
εj0 P̄0 (θ, z̄) + O (ε40 ).
(25)
Spectral Analysis for Dynamic Load Pressure
2.2
Equations (15) and (16) represent the modified Reynolds Eqs. in the radial and the tangential
directions. Each equation contains real and imaginary parts to overcome that obstacle, we use the
following assumptions let
P̄1 (θ, z̄) = P̄11 (θ, z̄) + P̄12 (θ, z̄) + iP̄13 (θ, z̄),
(26)
P̄2 (θ, z̄) = P̄21 (θ, z̄) + P̄22 (θ, z̄) + iP̄23 (θ, z̄),
(27)
and
where
∞
X
P̄11 (θ, z̄) =
K̃n (z̄) sin nθ,
P̄12 (θ, z̄) =
n=1
P̄13 (θ, z̄) =
∞
X
∞
X
Kn (z̄) cos nθ,
(28.a)
n=0
K̂n (z̄) cos nθ,
P̄21 (θ, z̄) =
n=0
P̄22 (θ, z̄) =
∞
X
∞
X
Jn (z̄) sin nθ,
(28.b)
n=1
J˜n (z̄) cos nθ,
and P̄23 (θ, z̄) =
∞
X
Jˆn (z̄) sin nθ.
(28.c)
n=1
n=0
From Eqs (26) to (28) into Eqs (15) and (16) and using the same procedure we obtain P̄1 and P̄2 as
functions of z̄, θ, and ε0 such that
P̄m (θ, z̄, ε0 ) =
3 X
3
X
k−1 (j)
(i) 2 εj0 P̄mk (θ, z̄) + O (ε40 ), m = 1, 2,
(29)
j=0 k=1
and
3
3.1



k−1
=

2

k−1
2 ,k
odd
k−2
2 ,k
even
.
Applications
Journal loci
The attitude angle, which is measured from the load line to the line of centers in the direction of
journal rotation, is given by
ft0 ,
φ = arctan fr0 6
(30)
where
ft0
3.2
1
1
2
Z
1
=
2
Z
fr0 =
θcav
Z
P¯0 (θ, z̄) cos θdθdz̄,
(31.a)
P¯0 (θ, z̄) sin θdθdz̄.
(31.b)
0
0
1
θcav
Z
0
0
Stiffness and Damping Coefficients
The dynamical pressures P¯1 and P¯2 are obtained as a series and this series uniformly converges.
Comparing the non-dimensional stiffness and damping coefficients with previous work, we have the
following analysis results:
θcav
¯ L ) = −Re
S̄rr (ε0 , `,
D
Z
¯ L ) = −Re
S̄φr (ε0 , `,
D
Z
¯ L ) = −Re
S̄rφ (ε0 , `,
D
Z
¯ L ) = −Re
S̄φφ (ε0 , `,
D
Z
0
!
1
Z
P¯1 cos θdz̄dθ
=
0
θcav
0
!
1
Z
P¯1 sin θdz̄dθ
=
0
θcav
0
!
1
Z
P¯2 cos θdz̄dθ
=
0
θcav
0
!
1
Z
P¯2 sin θdz̄dθ
=
0
θcav
¯ L ) = − 1 Im
D̄rr (ε0 , `,
D
λ
Z
¯ L ) = − 1 Im
D̄φr (ε0 , `,
D
λ
Z
¯ L ) = − 1 Im
D̄rφ (ε0 , `,
D
λ
Z
¯ L ) = − 1 Im
D̄φφ (ε0 , `,
D
λ
Z
0
3
X
(j) ¯ L
εj0 S̄rr
(`, )
D
j=0
3
X
(j) ¯ L
)
εj0 S̄φr (`,
D
j=0
3
X
(j) ¯ L
)
εj0 S̄rφ (`,
D
j=0
3
X
j=0
!
1
Z
P¯1 cos θdz̄dθ
=
0
θcav
0
!
1
P¯1 sin θdz̄dθ
=
0
0
!
1
Z
P¯2 cos θdz̄dθ
=
!
1
Z
P¯2 sin θdz̄dθ
0
=
(32.b)
(32.c)
(32.d)
L
)
D
(33.a)
L
)
D
(33.b)
(j) ¯ L
εj0 D̄rφ (`,
)
D
j=0
(33.c)
3
X
j=0
0
θcav
3
X
L
)
D
(j) ¯
εj0 D̄rr
(`,
j=0
Z
0
θcav
(j)
¯
εj0 S̄φφ (`,
(32.a)
(j)
¯
εj0 D̄φr (`,
3
X
3
X
(j) ¯ L
εj0 D̄φφ (`,
)
D
j=0
(33.d)
Then the dynamical coefficients are obtained as functions of the non-Newtonian and geometric
parameters
7
Stability Characteristics
3.3
If the rotor-bearing system is undisturbed, the rotor will remain in its equilibrium position. If
disturbed and the disturbances are small, the rotor will leave its equilibrium position and proceed
along a closed orbit around it. This is what occurs in well designed, stable, rotor bearing systems.
Under other, unstable, conditions the rotor is unable to find a limit cycle and its path spirals
outward until metal contact between rotor and bearing occurs. There is great practical importance
attached, therefore, to know the criteria that demarcates stable operation. To find this criteria we
begin with the equations of motion:
"
d2 ε
MC
−ε
dt2
dφ
dt
2 #
= Fr + W cos φ,
(34)
2
d φ
dε dφ
MC ε 2 + 2
= Ft − W sin φ.
dt
dt dt
(35)
Substituting equations (13) into equations (34) and (35) and non-dimensionalizing, having neglected the second-order terms, and using algebraic simplifications yields
M̄c =
m̃1 (ε0 α0 + W̄ α6 cos φ0 + ε0 α3 α5 )
W̄ α3 W̄ ε0 (S̄φr sin φ0 − S̄rr cos φ0 )α3 + m̃1 m̃2 − ε20 α3 α7
,
(36)
where
m̃1 = (ε0 α0 + W̄ α6 cos φ0 ), and m̃2 = (ε0 α4 + W̄ cos φ0 ).
Associated with the characteristic Eq.
λ6i + α̂5 λ5i + α̂4 λ4i + α̂3 λ3i + α̂2 λ2i + α̂1 λi + α̂0 = 0.
(37)
M̄c represents the transition curve that separates between the stable and unstable regions as
shown in Figs 11 and 12. To have a stable state, all the possible real roots of Eq. (37) must be
positive, and otherwise the system is unstable.
4
Results and Discussion
Based on Fourier spectral analysis for the statical, dynamical, and stability characteristics for a
finite hydrodynamic journal bearing lubricated with couple stress fluid have been obtained. In
this section, we explore the behavior of these characteristics versus various parameters like non¯ slenderness ratio L . Here, the parameters are selected as follows: couple stress
Newtonian `,
D
parameter `¯ = 0.1, 0.2, 0.3, and 10; slenderness (geometric parameter) ratios L/D = 0.5, 1.0 and
8
1.5; eccentricity ratio ε0 = 0.1...0.9. For steady-state, the journal locus is affected by the geometric
and couple stress parameters. Figure 2.a shows the variation of the attitude angle versus the eccentricity ratio as a function of the geometric parameter L/D while, the non-Newtonian parameter
is constant.It has been noted that the eccentricity ratio decreases by increasing a potential value
of the geometric parameter ratio for the corresponding value of the attitude angle. The variation
of the attitude angle ageist the eccentricity ratio as a function of the couple stress parameter at a
certain value of the geometric parameter is presented in Fig. 2.b. It is clear that there is a reverse
relation between the eccentricity ratio and the couple stress parameter for a given value of the
attitude angle in general. For a small values of the attitude angle there is no significant effect of
the couple stress parameter on the eccentricity ratio, while, for a moderate or higher values of the
attitude angle the effect is more pronounced. Moreover, The curvature of journal locus increases
by increasing the couple stress parameter. Here the values of `¯ has been selected as shown in the
figure since the curvature of journal locus doesn’t affect by a small difference of `¯ while, at a great
difference of `¯ gives a distinct curvature. In another word, the effect of `¯ on the curvature is more
pronounced when the difference of its values is great.
L/D=0.5
L/D=1
0.9
L/D=1.5
n
[=0.1
0.8
4
n
0.7
0.4
0.6
[=10
0.5
0
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0.0
[=0.1
0.7
0.6
0.5
L/D=1
0.8 4
0.9
0.05
0.10
0.15
0.0
0.20
Fig. 2.a. Journal loci for finite
journal bearing for various values
of the geometric parameter L/D.
0
0.1
0.2
Fig. 2.b. Journal loci for finite
journal bearing for various values
¯
of the couple stress parameter `.
The comparison between Fourier spectral technique (FST) and finite element method (FEM) has
been considered as shown in Figs (3) to (10). Here we select the values of the given parameters
based on Guha’s work as follows for geometric parameter L/D = 1, and for the couple stress
parameter `¯ = 0.1, 0.2, and 0.3. It is clear that FST and FEM give the same behavior for the
direct and cross stiffness and damping coefficients, in general. The difference occurs for the given
value of ε0 the corresponding values of these coefficients are different in their potential values.
9
1,200
100
L/D=1
1,000
L/D=1
y
Q
S44
=0.3
7
Srr
FEM
y
Q
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
400
7
B
$
B
40
0.3
0.4
0.5
7
0.6
$ [ = 0.1
B
7
0.7
0.8
0.1
0.9
B
B
$
Q
$ [ = 0.1
B
Q
$ $
B =0.2
7
20
$
0
0.2
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
B
B
7
200
7
Q
B
=0.2
7
0.1
7
60
FEM
600
=0.3
80
7
800
Q
7
0.2
0.3
0.4
$
$
$
0.5
0.6
0.7
0.8
0.9
0
0
Fig. 3. Comparing plots of S̄rr
versus ε0 due to FST and FEM. At
¯
various values of `.
Fig. 4. The behavior of S̄φφ versus
¯ In view of
ε0 for varying values of `.
FST and FEM.
200
20
L/D=1
7
L/D=1
=0.3
16
7
FEM
=0.3
Q
150
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
Q
Q
Sr4
B
KS4r
12
FEM
Q
y 100
7 B
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
$
7
7
B
B
$
4
Q
0.1
0.2
0.3
0.4
$
B
$
$
0.2
0.3
$
$
B
Q
$
[ = 0.1
[ = 0.1
$
$
0
B
B
7
B
=0.2
B
7
$
7
B
7
7
y
=0.2
8
50
7
0
0.5
0.6
0.7
0.8
0.1
0.9
0.4
0.5
0.6
0.7
0
0
Fig. 5. Comparing graphs of−S̄φr
versus ε0 for several values of `¯ .
In light of FST and FEM.
Fig. 6. Plots of S̄rφ versus ε0
¯ due to
for varying values of `,
FST and FEM.
10
0.8
0.9
600
120
KDrr
7
7
L/D=1
Q
=0.3
100
FEM
Q
B =0.2
D44
Q
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
200
y
B
B
7
B
$
$
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Q
[ = 0.1
$
$
$
$
$
20
[ = 0.1
L/D=1
0
0.1
B
B B
Q
$
B
B
7
40
$
B
7
60
$
B =0.2
7
7
7
7
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
80
FEM
y
7
Q
7
400
=0.3
0
0.9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0
Fig. 7. Sample plots of−D̄rr
¯
versus ε0 for several values of `.
In view of FST and FEM.
300
Fig. 8. Comparing plots of D̄φφ
versus ε0 due to FST and FEM. At
¯
various values of `.
7
150
=0.3
L/D=1
250
7 = 0.3
Q
KD4r
7
L/D=1
7
100
200
Q
KDr4
B
FEM
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
y 150
FEM
y
=0.2
7
Q
B
50
100
7
$
B
B
$
B
$
7
0.1
0.2
0.3
0.4
7
B
7
$
B
$
Q
B
$
[ = 0.1
$
$
0
0.1
0.5
= 0.2
[ = 0.1
$
B
B
Q
$
50
7
Q
$ [Q = 0.1
B Q[ = 0.2
7 [ = 0.3
0.6
0.7
0.8
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0.9
0
Fig. 9. Plots of −D̄rφ versus ε0
¯ due to
for varying values of `,
FST and FEM.
Fig. 10. The behavior of D̄φr versus
¯ In light of
ε0 for several values of `.
FST and FEM.
Figures (11) and (12) describe the variation in the critical mass rotor bearing M̄c with ε0 for
various values of the couple stress parameter `¯ when geometric parameter L/D equal 0.75 and 1
11
respectively. According to the root of Eq. (37), we conclude that the stable region lies below the
transition curve, while the unstable area lies above this curve. It has been noticed that the stable
area increases in a monotonic fashion with the increase in the couple stress parameter, whereas it
decreases by increasing geometric parameter.
Q
Q
[ = 0.1
Q
[ = 0.2
Q
[ = 0.3
Q
[ = 0.1
60
U
50
40
U
Q
Mc
[ = 0.3
L/D=1
50
L/D=0.75
Q
[ = 0.2
40
Q
Mc
30
30
20
20
10
10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0
Fig. 11.Sample graphs of M̄c
versus ε0 for various values of the
L
couple stress parameter at D
= 0.75.
5
S
S
Fig. 12. Graphs of M̄c versus ε0
depending on different values of the
L
couple stress parameter `¯ at D
= 1.
Conclusions
In this work, we have developed an analytical solution for Reynolds equations which covers
the pressure distribution for the finite journal bearing lubricated with couple stress fluid, using a
Fourier spectral technique. As this technique (Fourier spectral) is applied to problems that have
exact solutions, it is concluded that the obtained solutions are more accurate than those calculated
by the numerical approaches. This encourages us to apply this technique to solve the considered
problem. It is worthwhile to mention that such a problem has not been solved analytically before.
Since the modified Reynolds equations are non-homogeneous linear partial differential equations
with variable coefficients, to apply the Fourier spectral technique we expand the variable coefficient G(H, `) as a function of cosines of different orders. The pressure distribution functions
obtained from this technique are used to investigate the static and dynamic characteristics. These
coefficients are functions of eccentricity ratio, non Newtonian and geometric parameters. Comparing these results with previous work, we deduce the following:
12
1- We obtain the pressure as a function of θ, z̄ and ε0 for steady and dynamic states for finite journal bearings lubricated with coupled stress fluids.
2- In general, the non-Newtonian and geometric parameters have significant effect on the journal paths since these paths get stable as these parameters increase.
3- The stiffness and damping coefficients increase as the non-Newtonian parameter increases. Also,
the effect of the non-Newtonian parameter is significant for high eccentricity ratios and for high
values of the non-Newtonian parameter.
4- The behavior of stiffness and damping coefficients obtained using the Fourier spectral method
agree with the corresponding obtained using the finite difference method in behavior, but not in
values.
5- The stable of the given system is affected by the variation of the non-Newtonian and geometric
parameters since the stable region increases as the non-Newtonian increases, while it decreases as
the geometric parameter increases.
13