9.5 Turbulent Lubrication Theory Using the k- ε Model In the case of a journal bearing in which the eccentricity ratio of the journal is large, the pressure gradient in the oil film is la
Trang 1In reference to the coordinate axes x and y of Fig 9.8, elastic coefficient Ki jand damping coefficient Ci j are defined as Eq 9.57 using the oil film forces Px and Py, where the positive direction of Px and Pyare taken in the direction of−x and −y The subscript 0 denotes the static equilibrium position To obtain Px and Py, first solve
the turbulent Reynolds’ equation (Eq 9.51) for the pressure, multiply the obtained pressure by cosφ and sin φ, integrate these with respect to φ to obtain Pκand Pθ(see
Fig 9.5), and finally transform these into Px and Py.
Kxx= ∂Px ∂x
0, Kxy= ∂Px ∂y
0
, Kyx= ∂Py ∂x
0, Kyy= ∂Py ∂y
0
Cxx= ∂Px ∂ ˙x
0, Cxy = ∂Px ∂˙y
0
, Cyx= ∂Py ∂ ˙x
0, Cyy= ∂Py ∂˙y
0 (9.57)
Fig 9.8 Axes of coordinates (horizontal, vertical)
Figure 9.9 shows three examples of the above-mentioned elastic coefficients and
a damping coefficient Kxx, Kxy, and Cxxas functions of eccentricity ratioκ0with Pd
as a parameter The figure shows that the larger the pressure difference is, the larger
the value of these constants (absolute values) becomes Further, the value of Kxxfor laminar flow is larger than that in the case of turbulent flow The same can be said
of Kyy, although the data are not shown The calculation conditions are as follows:
D = 70 mm, L = 35 mm, c = 0.175 mm, N = 4000 rpm, µ = 1.44 mPa·s, the axial Reynolds’ number Ra = wmc/ν = 767 – 2540 (wmis the axial average velocity), the
circumferential Reynolds’ number Rω= Rωc/ν = 1418.
9.5 Turbulent Lubrication Theory Using the k- ε Model
In the case of a journal bearing in which the eccentricity ratio of the journal is large, the pressure gradient in the oil film is large The mixing length used in the mixing
Trang 29.5 Turbulent Lubrication Theory Using the k-ε Model 215
Fig 9.9 Spring and damping coefficients of a fluid film seal [36]
length model is usually determined experimentally for small pressure gradients, but it
changes with pressure gradient Therefore, it is more reasonable to use the k-ε model, which is less affected by pressure gradient, for analyses of a turbulent bearing with a large eccentricity ratio
An analysis based on the k-ε model will be described below (Kato and Hori [31]).
9.5.1 Application of the k-ε Model to an Oil Film
In the oil film of a bearing, especially in the neighborhood of the wall surface, the
tur-bulent Reynolds’ number Rt is relatively low The low-Reynolds’ number k-ε model,
which is suitable in such a case, was introduced by Jones and Launder [21] [22] as
stated in the previous section The transport equations (Eqs 9.22 and 9.23) for k and
ε given by them were improved later by Hassid and Poreh [25] We use Hassid and Poreh’s model here This model can be applied to cases in which the Toms effect (described later) appears [34]
First, the turbulent energy k and the turbulent lossε are defined as follows:
k=1
2uiui= 1
2
u2+ 2+ w2
(9.58)
ε = ν∂ui ∂x
j
∂ui
∂x j = ν
∂u
∂x
2 + · · · +
∂w
∂z
2
(9.59) The following two-dimensional equations, which are extentions of the one-dimensional equations by Hassid and Poreh, will be used as transport equations for
k andε:
Trang 3Dt = ∂y∂
ν +σkνt
∂k
∂y − u
∂u
∂y − w
∂w
∂y − ε −
2νk
Dε
Dt = ∂y∂
ν +σνt
ε
∂ε
∂y − Cε1
u∂u
∂y+ w
∂w
∂y
ε
k
−Cε2
1− 0.3 exp(−Rt2)ε2
k − 2ν
∂ε1 /2
∂y
2
(9.61) where
b = min(y, h − y)
σk= 1, σε= 1.3, Cε1= 1.45, Cε2= 2.0
Rt = k2/(εν)
In Eqs 9.60 and 9.61,ε denotes the isotropic part of the turbulent loss and 2νk/b2
the anisotropic part The idea of separating the turbulent loss in this way was pro-posed by Jones and Launder, and the following simple boundary conditions forε has thereby become possible:
Further, the boundary conditions for k is assumed to be:
According to Hassid and Poreh, the turbulent dynamic viscosity coefficient νtis
given as follows by using the above-mentioned k andε:
νt= Cm kε21− exp(−AdRt) (9.64)
However, the following equation, which contains a correction factor Cd, will be used
here, based on Laufer’s experiments on Couette flows [4]:
νt= Cm k2
ε
where
Cm = 0.09, Ad = 1.5 × 10−3, Cd= 0.95 Although a strong Couette flow in the circumferential direction and a weak pres-sure flow in the axial direction are expected to exist in a bearing, it is assumed here that Eq 9.65 can be used in both the circumferential and the axial directions, based
on the fact that the correction factor Cd= 0.95 is close to 1
9.5.2 Turbulent Reynolds’ Equation
The time-average equation of motion of an incompressible fluid containing Reynolds’ stress in a two-dimensional case can be given by Eqs 9.8 and 9.9 In general, it can
be written in a tensor expression as follows:
Trang 49.5 Turbulent Lubrication Theory Using the k-ε Model 217 ρ
∂ui
∂t + u j ∂ui
∂xj
= −∂xi ∂p +∂x∂
j
µ∂x ∂ui
j − ρ uiu
j
(9.66) (i, j= 1, 2, 3; summation is taken over all values of j)
The equations in rectangular coordinates (x, y, z) can be obtained by the substitution
of variables such as x = x1, y = x2, z = x3, u = u1, = u2, and w = u3, where x, y, and z are the coordinates in the circumferential direction, across the film thickness, and in the axial directions; u and u(and similar) express the static (time-average) parts of the flow velocity and the fluctuations about it, respectively
Considering a sufficiently thin lubricating film, let us make the following as-sumptions
1 The left-hand side (inertia term) of Eq 9.66 is negligible
2 In Eq 9.66, the derivatives of Reynolds’ stresses with respect to x and z can be neglected compared with that with respect to y.
3 The normal components of Reynolds’ stress (components i= j) can be neglected.
4 −ρuand−ρwcan be expressed as follows with a turbulent viscosity coeffi-cientνt, which is common to the x and z directions:
−ρw= ρνt∂w
Under these assumptions, a turbulent lubrication equation is derived from Eq 9.66 First, disregarding the left-hand side of Eq 9.66 from assumption 1, then sub-stituting Eqs 9.67 and 9.68 of assumption 4 into this leads to the following equations:
∂p
∂x = ρ
∂
∂y
(ν + νt)∂u
∂p
∂z = ρ
∂
∂y
where∂p/∂y = 0 is omitted Integrate Eqs 9.69 and 9.70 twice with respect to y
under the boundary conditions
to obtain u and w, respectively Substituting these into the continuity equation
h
0
∂u
∂x dy+
h
0
∂w
gives the following turbulent Reynolds’ equation, with Gx = Gz = G:
∂
∂x
G ∂p
∂x
+∂z∂
G ∂p
∂z
= U1∂F
Trang 5G=
h
0
y
0
dy dy
ν + νt
h
0
y dy
ν + νt h
0
dy
ν + νt
− h
0
y
0
y dy dy
F = h −
h
0
y
0
dy dy
ν + νt h
0
dy
ν + νt
(9.76)
If Eq 9.65 is used, the five equations, Eqs 9.60, 9.61, 9.69, 9.70, and 9.74 form
a closed set of equations with respect to the five unknowns k, ε, u, w, and p
Un-knowns such as the fluctuations in the velocity are not included in the equations Therefore, by solving these five equations simultaneously, the above five unknowns will be obtained It is assumed here that the left-hand side of Eqs 9.60 and 9.61
(time variations of k andε along the streamline) can be disregarded, considering the stationary state:
Dk
Dt = 0 Turbulent lubrication problems can thus be solved
Approximate solutions are possible when the left-hand side (inertia term) of Eq 9.66 cannot be disregarded [28]
9.6 Comparison of Analyses Using the k- ε Model with
Experiments
In this section, some examples of comparisons of theoretical analyses of a turbulent
bearing by the k-ε model and experiments will be shown (Kato and Hori [31]) In
the-oretical calculations, Eqs 9.60, 9.61, and 9.74 are solved simultaneously under the boundary conditions given in Eqs 9.71, 9.72, 9.62, 9.63 and the following boundary condition concerning pressure:
p= 0 at θ = 0, π and at the bearing ends (9.77) The procedure for numerical calculations is as follows Assume suitable initial
profiles of k and ε, calculate G and F of Eqs 9.75 and 9.76 and then obtain the
pressure distribution by applying the finite element method to the turbulent
lubrica-tion equalubrica-tion (Eq 9.74) Next, calculate the flow velocity distribulubrica-tions u and w from
the pressure distribution by using Eqs 9.69 and 9.70 and the boundary conditions
Eqs 9.71 and 9.72, then obtain k andε from the above velocity distributions, Eq
(9.60) and Eq (9.61) Using these results, calculate G and F again, and then obtain
Trang 69.6 Comparison of Analyses Using the k-ε Model with Experiments 219
Fig 9.10 Average velocity distribution of Couette flow [31]
the pressure distribution again in the same way as in the beginning of the procedure This calculation will be repeated until the calculated pressure distribution converges within a small error
As an example, consider a journal with a diameter of 150 mm rotating in a bear-ing with an inner diameter of 152 mm and a length of 150 mm over the speed range
100 – 5000 rpm Let the coefficient of dynamic viscosity of the lubricating oil be 9.4 × 10−6m2/s (30◦C).
Comparisons of the theoretical and the experimental results, both under the above conditions unless otherwise stated, will be shown below Lubricating oil is supplied
at a rate of 12 l/min in the experiments
Figure 9.10 shows the theoretical and the experimental results of the average
ve-locity distribution u for Couette flow, the experiment by Reichardt being used in this
case [6] It is seen in Fig 9.10 that although the theoretical and the experimental re-sults are different for a Reynolds’ number of Re = 1200, they are in good agreement for Re = 2900 and Re = 34 000 This shows that the k-ε model is better suited to
analysis in the turbulent region well above the laminar-to-turbulent transition region Figure 9.11a,b shows the theoretical and experimental results for the pressure distribution in a finite width bearing Figures 9.11a,ba,b are the nondimensional
pres-sure distribution p in the circumferential direction at the center of bearing width for
Re = 2000 and 8000, respectively, the parameter being the eccentricity ratios as shown The theoretical and the experimental values are generally in good agreement
However, for Re= 8000 and an eccentricity ratio of 0.8, the experimental pressure
Trang 7Fig 9.11a,b Theoretical (symbols) and experimental (lines) values of the nondimensional
pressure distribution in a finite width bearing (1) [31]
is higher than the theoretical pressure in the range 0◦– 90◦ This may be attributed to the influence of the inertia of the fluid
In Fig 9.12, the nondimensional load capacity P = P/(2µUR2L/c2) (the left scale) is plotted against eccentricity ratioκ for Re = 2000 and 5000 The recipro-cal of Sommerfeld’s number S−1 is also shown in the figure (the right scale) The
relation S−1 = 2πP holds between the two axes The theoretical and experimental
results of load capacity or Sommerfeld’s number are in good agreement even when the eccentricity ratio exceeds 0.95 This shows that the lubrication theory based on
the k-ε model can be applied to bearings with very high eccentricity ratios
Figure 9.13a,b shows the nondimensional pressure distributions p for the cases
shown in Fig 9.12 The parameter for the curves is the nondimensional load capacity
P of Fig 9.12 The theoretical and experimental values for the pressure distribution
are in good agreement even in the case of high eccentricity ratios over 0.9 This also
shows that the turbulent lubrication theory based on the k-ε model can be used for bearings of high eccentricity ratio This is because the k-ε model is valid for large
pressure gradients
Figure 9.14 shows the theoretically obtained loci of the journal center with those obtained experimentally by Wada and Hashimoto [27] The figure shows that
Trang 89.6 Comparison of Analyses Using the k-ε Model with Experiments 221
Fig 9.12 Nondimensional load capacity and Sommerfeld’s reciprocal versus eccentricity ratio
[31]
Fig 9.13a,b Nondimensional pressure distribution in a finite width bearing (2) for the two
cases analyzed in Fig 9.12 [31]
changes in Reynolds’ number do not very much affect the shape of the loci of the journal center
Trang 9Fig 9.14 Locus of the journal center [31]
9.7 Reduction of Friction in a Turbulent Bearing by Toms’ E ffect
Large shear stress and large heat generation in a fluid film are major problems in a turbulent bearing In pipe flow, on the other hand, the marked reduction in turbulent flow resistance of water by addition of very small amounts of a kind of long-chain high molecular weight polymer is known as Toms’ effect [2] [19] Since it is known that Toms’ effect is more powerful in pipes of smaller diameter, considerable re-duction in the friction of a turbulent bearing can be expected if this phenomenon is applied to the thin film of a bearing
It is reported that in experiments using a water solution of hydroxyethylcellulose
or polyacrylamide of the order of ppm between two concentric cylinders (diameter
of the inner cylinder 10 mm, clearance 1 mm), the turbulent friction for Couette flow was actually reduced, and that, in the case of two eccentric cylinders, an even larger reduction in friction was observed for larger eccentricity ratios [23] It has also been reported that a similar reduction in friction was observed in experiments using
a water solution of polyethyleneoxide in a far smaller clearance, but the effect was lost after only 20 circulations of the solution [24]
In this section, results of experiments investigating Tom’s effect under the condi-tions close to those of an actual bearing (Fukayama et al [29]) and comparisons of
them with calculated results based on the k-ε model (Kato and Hori [34]) are shown.
Trang 109.7 Reduction of Friction in a Turbulent Bearing by Toms’ Effect 223 The following conditions were used: bearing diameter 210 mm, bearing length
200 mm, bearing clearance 0.202 mm, and, as the lubricating fluid, pure water and a water solution of polyacrylamide (PAA, molecular weight 2.3 × 106)
Fig 9.15 Coefficient of friction of a bearing using various concentrations of polyacrylamide
(PAA) in water as the lubricating fluid [29] [34]
Figure 9.15 shows the experimentally obtained relations between Reynolds’
number Re = Uc/ν and the coefficient of friction of the bearing Cf = 2τw/(ρU2) (τωis the shear stress at the wall surface) The straight line L corresponds to laminar flow and the curve N to turbulent flow of pure water The curve V corresponds to Virk’s maximum drag reduction which was obtained for a 300 ppm solution of PAA
in water Comparison of curve V and the calculated results based on the k-ε model gives the value of Ad = 1.5 × 10−5 for the constant Ad in Eq 9.64 in the case of
maximum drag reduction
Figure 9.16 shows the experimental value of the nondimensional pressure
dis-tribution p and the theoretical disdis-tribution based on the k-ε model Curves 1 and 2
in Fig 9.16 show the pressure distribution for pure water and a solution of PAA in
water, respectively, calculated using the k-ε model They are in good agreement with
the experimental results shown by small circles As the values of constant Adin Eq
9.64, Ad = 1.5 × 10−3was used for pure water and Ad = 1.5 × 10−5, as obtained
above, was used for the PAA solution Thus, the k-ε model can explain well the
pres-sure distribution in the bearing for both pure water and PAA solutions While the mixing length model gives good agreement in the case of pure water, it is known to estimate the pressure distribution in the case of PAA solutions 15% – 20% too low
According to Fig 9.16, which is for Re= 2000, the drop in pressure (decrease in load capacity) due to the addition of PAA is not very large On the other hand, Fig
9.15 shows for Re = 2000 (shown by the dashed line) that the fall in the frictional
...In Eqs 9.60 and 9.61,ε denotes the isotropic part of the turbulent loss and 2νk/b2
the anisotropic part The idea of separating the turbulent loss in this way... distributions p for the cases
shown in Fig 9 .12 The parameter for the curves is the nondimensional load capacity
P of Fig 9 .12 The theoretical and experimental values for the... in the axial directions; u and u(and similar) express the static (time-average) parts of the flow velocity and the fluctuations about it, respectively
Considering a sufficiently