The heat conduction equation for the pad is:∂2T2 ∂r2 + 1 r2 ∂2T2 ∂θ2 +∂2T2 ∂z2 +1 r ∂T2 The coefficient of viscosity of the lubricating oil is: The density of the lubricating oil is: 8.5.2
Trang 1[11]), a two-dimensional analysis of a sector pad taking into account the effect of centrifugal force on the lubricating oil (Pinkus and Lund [15]), a three-dimensional analysis of a fixed inclined rectangular pad bearing (Ezzat and Rhode [9]), and a three-dimensional analysis of a tilting sector-pad bearing taking into account only the inclination in pitch mode (Tieu [10])
Fig 8.2 Sector-type tilting pad bearing [20]
8.5.1 Basic Equations
A cylindrical coordinate system (r , θ, z) is considered with the origin at the center of rotation of the disk (see Fig 8.2) Let the position of the pivot support be (r p, θp)
In the direction of pad thickness, a coordinate z2with the origin on the lubricating surface of the pad is used Besides the usual assumptions of lubrication theory, the following assumptions are made:
1 The lubricating surfaces of the pad and the disk are rigid, and their thermal and elastic distortions are disregarded
2 The velocity gradient and heat conduction in the r andθ directions can be ignored
in comparison with those in the z direction.
3 The specific heat at constant pressure c pand the coefficient of thermal expansion
αof the lubricating oil are constant
Trang 2174 8 Heat Generation and Temperature Rise
4 The thermal conductivity of the lubricating oil k o and the thermal conductivity
of the pad k sare constant
5 The coefficient of viscosity µ and the density ρ of the lubricating oil are functions
of temperature T only.
The basic equations in this case are stated below
Let the state when the lubricating surfaces of the pad and the disk are parallel
to each other and their separation is h 0Rbe the standard state Assume that the pad
rotates around the x axis byαp (pitching angle) and around the y axis byαr(rolling
angle) from the standard state Then the gap h between the pad and the disk (film
thickness) is given by the following equation, assuming a small inclination of the pad,αp 1 and αr 1:
h(r , θ) = (R + ∆R − r cos θ)α r+ αp r sin θ + h 0R (8.51) Inclinations of the padαp andαrare actually automatically determined by the
bal-ance of the oil film force The minimum film thickness h0appears at (R+∆R, 0) when
αr is positive and at (R, 0) when α ris negative
The flow velocity of the lubricating oil in the radial and the circumferential di-rections is:
r =
z
0
z
µdz−
F1
F0
z
0
dz
µ
∂p
θ= −rω
1− 1
F0
z
0
dz
µ
+
z
0
z
µdz−
F1
F0
z
0
dz
µ
1
r
∂p
where
F0= h
0
dz
µ, F1=
h
0
zdz
µ The generalized Reynolds’ equation is written as:
∂
∂r
F2∂p
∂r
+ 1
r2
∂
∂θ
F2∂p
∂θ
+F2
r
∂p
∂r = ω
∂
∂θ
F4−F3
F0
(8.54) where
F2= F1
F0
F3− h
0
ρ z
0
zdz
µ dz,
F3= h
0
ρ z
0
dz
µdz , F4=
h
0
ρdz
The energy equation for the oil film is:
c pρ
r ∂T
∂r +
θ
r
∂T
∂θ + z ∂T
∂z
= k o
∂2T
∂z2 + αT
r
∂p
∂r +
θ
r
∂p
∂θ
+ µ⎧⎪⎪⎨
⎪⎪⎩
∂r
∂z
2 +
∂θ
∂z
2⎫⎪⎪⎬
Trang 3The heat conduction equation for the pad is:
∂2T2
∂r2 + 1
r2
∂2T2
∂θ2 +∂2T2
∂z2
+1
r
∂T2
The coefficient of viscosity of the lubricating oil is:
The density of the lubricating oil is:
8.5.2 Boundary Conditions
As boundary conditions for the generalized Reynolds’ equation, it is assumed that the pressure around the pad is atmospheric pressure, i.e.,
p(r , 0) = p(r, θ0)= p(R, θ) = p(R + ∆R, θ) = 0 (8.59)
As boundary conditions for the energy equation of the oil film, it is assumed first
that both the inlet oil temperature and the disk surface temperature are equal to Tin, i.e.,
T (r, θ0, z) = T(r, θ, 0) = Tin (8.60)
It is also assumed that the oil film temperature changes parabolically in the r
direc-tion on the inner and outer circular boundaries of the pad Furthermore, from the continuity of the heat flux at the interface of the pad and the oil film in the direction normal to the interface, it is assumed that:
k o
∂T
∂z
z =h = k s
∂T2
∂z2
z2 =0
(8.61)
where T and T2are the temperature of the lubricating oil and the pad, respectively,
and k o and k sare the thermal conductivities of the lubricating oil and the pad, respec-tively
As boundary conditions for the heat conduction equation of the pad, it is assumed that Eq 8.61 holds at the lubricating surface and that heat flux is continuous at the
other surfaces, the ambient temperature T a and the heat transfer coefficient at the
interfaces h cbeing given
8.5.3 Numerical Analyses
Before numerical computations, each variable is nondimensionalized as follows Since both the lubricating film and the pad become a cube of side 1, the forms of the computational domain in the finite difference method become very simple:
Trang 4176 8 Heat Generation and Temperature Rise
r= r ∆R − R, θ = θθ
0
, z = z
h , h = h
h0
r= r
Rω, θ=
θ
rω, z= z
Rω, z2= z2
t2
µ =µµ
in, ρ = ρρ
in, T = T
Tin, T2= T2
Tin
p= ph0
R2ωµin
(8.62) Then, the lubricating film and the pad are divided into, for example, 20
divi-sions in the r and θ directions, and into, for example, 15 divisions in the z direction,
and Eqs 8.51 to 8.58 are discretized for the finite difference analysis The centered difference is used, except in the case of ∂T/∂θ in the energy equation (Eq 8.55) for which the backward difference is used This is solved by the successive
overre-Fig 8.3 Flow chart for computation [20]
Trang 5laxation method (SOR method), following the flow chart of Fig 8.3 Calculation is repeated until the error from the last calculated value becomes, for example, smaller than 10−5at all nodal points Further, to make the center of pressure coincide with the pivot position, the inclination of the pad is adjusted by the Newton–Raphson calculation
The constants in Table 8.1 were used in the following calculations
Table 8.1 Specifications of the sector pad bearing and the oil
Inner radius of pad R= 1.6 × 10−1m
Radial extent of pad ∆R = 7.7 × 10−2m
Angular extent of pad θ0 = π/8 rad
Thickness of pad t= 2.5 × 10−2m
Thermal conduction of pad k s = 1.08 × 102W/m◦C
heat transfer at surface of pad h c= 5.81 × 102W/m2 ◦C
Viscosity of oil (27◦C) µ = 5.0 × 10−2Pa· s
Viscosity index of oil β = 4.91 × 10−2 ◦C−1
Specific heat of oil c p = 2.09 × 103J/kg◦C
Thermal conductivity of oil k o = 0.214 W/m◦C
Coefficient of thermal expansion of oil α= 7.34 × 10−4 ◦C−1
Density of oil (27◦C) ρ = 8.55 × 102kg/m3
8.5.4 Examples of Three-Dimensional Analyses of Temperature Distribution
Figure 8.4 shows isothermal lines for the lubricating surface of the pad and for the
central cross section of the lubricant film and the pad, parameters being h0= 100 µm,
N = 3000 rpm, Tin = T a = 47◦C, the pivot position in nondimensional radial
coor-dinates r p = (r p − R)/∆R = 0.51 and in nondimensional circumferential coordinates
θp= θp/θ0= 0.39
On the lubricating surface of the pad, temperature rises in the circumferential direction gradually near the leading edge and then quicker near the trailing edge The highest temperature is found on the trailing edge near the outer radius The temperature does not change very much in the radial direction The shapes of the isothermal lines are very close to the experimental results under the same conditions (the same pivot position, and so forth) [23]
The highest temperature in Fig 8.4 is approximately 58◦C, which is 11◦C higher than the entrance oil temperature Generally speaking, the highest temperature falls
if the heat transfer coefficient hcof the pad surface is increased; however, the fall in highest temperature is only 2◦– 4◦C or so even if h cis increased by 10 – 102 times [16] This probably shows that most of the heat is carried away by the convection of the lubricating oil
In the central cross section of the lubricant film and the pad, the temperature
in the lubricant film (the wedge-shaped domain in the lower half; the thickness is
Trang 6178 8 Heat Generation and Temperature Rise
Fig 8.4 Temperature distribution on the lubricating surface (left) and the central cross section
of the lubricant film (lower right) and pad (upper right) [16]
exaggerated) rises gradually from the disk surface, and rises quickly near the pad surface The highest temperature is found at the trailing edge of the lubricant film near the pad surface The temperature distribution in the pad (the rectangular domain
in the upper half) is as shown in the figure
8.5.5 Comparisons of Three-Dimensional, Two-Dimensional, and Isoviscous Analyses
Figure 8.5a-c shows comparisons of the nondimensional load capacity, the tem-perature rise (= highest temtem-perature - entrance temtem-perature), and the nondimen-sional frictional torque calculated by three-dimennondimen-sional, two-dimennondimen-sionalal and
iso-viscous analyses, under the conditions Tin = T d = T a = 27◦C, h
0 = 100 µm, and
N = 1500 rpm The horizontal axis shows the nondimensional circumferential co-ordinates of the pivot positionθp = θp/θ0, and the parameters in the figure are the
nondimensional radial coordinates of the pivot position r p = (r p − R)/∆R.
The nondimensional load capacity in Fig 8.5a-ca shows that the three-dimensional analyses give the lowest load capacity, the two-dimensional analyses give an inter-mediate figure, and the isoviscous analyses give the highest load capacities Fig 8.5a also shows that there is an optimal pivot positionθp that gives the highest load ca-pacity in each of the three analyses, θp being slightly dependent on r p The load
capacity, which changes considerably with r p , becomes maximum when r p = 0.51 for any method of analysis here The load capacity depends not only on the pad incli-nation but also on the oil viscosity, which is a function of temperature and hence of location This affects the optimal pivot position in a complicated way One must be aware that the isoviscous and the two-dimensional analyses may give load capacity estimates that are too high
As for the temperature rise shown in Fig 8.5a-cb, the two-dimensional analysis gives a value considerably lower than the three-dimensional analysis does, although
Trang 7al Torque
Fig 8.5a-c Comparisons of bearing characteristics using three-dimensional (3-D), two-dimensional (2-D), and isoviscous analyses [20] a nontwo-dimensional load capacity, b temper-ature rise, c nondimensional frictional torgue r p, nondimensional radial coordinates of the pivot position
Trang 8180 8 Heat Generation and Temperature Rise
it does not consider the heat flow into solid surfaces The reason for this is that only the average temperature is considered in the film thickness direction in two-dimensional analysis If the pivot position is moved backward (ifθp is decreased), the temperature rise decreases for both the two-dimensional analysis and the three-dimensional analysis In the three-three-dimensional analysis, the large temperature rise falls very rapidly as shown in the figure The reason why the temperature rise falls when the pivot position is moved backward is that increasing the pad inclination leads to an increase in the rate of oil flow The two-dimensional analysis can result
in temperature estimates that are too small, which requires caution
As for the nondimensional frictional torque shown in Fig 8.5c, the three-dimensional analysis gives the lowest values, the two-three-dimensional analysis gives values a little higher, and the isoviscous analysis gives the highest values This is because the temperature rise is larger in the three-dimensional case and hence the viscosity is lower
It should be noted here that the results obtained by the three-dimensional, the two-dimensional, and the isoviscous analyses differ considerably in terms of load capacity, temperature rise, and frictional torque Therefore, the analysis of a bear-ing should originally be performed usbear-ing the three-dimensional analysis; and other approaches cannot substitute for it When performing the two-dimensional or the isoviscous analyses, it should be kept in mind that they are considerable approxima-tions
8.5.6 Analysis Considering Inertia Forces
In high speed tilting pad thrust bearings, the influence of the inertia force (centrifugal force) on the fluid cannot be disregarded (Kim et al [21]) Reynolds’ equation for such a case is derived here
In reference to Fig 8.2, the equations of balance that take inertia into considera-tion are written as follows:
∂
∂z
µ∂r
∂z
= ∂p ∂r −ρθ2
∂
∂z
µ∂θ
∂z
= 1
r
∂p
∂p
The second term on the right-hand side of Eq 8.63 is an inertia term Integrating Eq 8.64 under the boundary condition
θ= rω at z = 0, θ= 0 at z = h
yields:
θ= rω +1
r
∂p
∂θ
z
z
µdz−
rω
F +F1
F
1
r
∂p
∂θ
z
dz
Trang 9Subsituting this into Eq 8.63 and integrating under the boundary conditions
r = 0 at z = 0 and z = h
gives the following equation:
r= ∂p
∂r
z
0
z
µdz+
1
r
G0
F0 −F1
F0
∂p
∂r
z
0
dz
µ −
1
r
z
0
1 µ
z
0
ρ(θ 2dz dz (8.67)
where F0, F1, and F2are as follows:
F0=
h
0
dz
µ, F1=
h
0
z
µdz , G0=
h
0
1 µ
z
0
ρ(θ2dz dz (8.68)
On the other hand, integration of the continuity equation with respect to z from 0
to h, the use of
z = 0 at z = 0, r= θ= z = 0 at z = h
and use of the formula for the change of order of differentiation and integration yields:
r ∂
∂r
h
0
ρr dz+ h
0
ρr dz+∂θ∂
h
0
Substituting Eqs 8.66 and 8.67 into Eq 8.69 yields the following Reynolds’ equation, which takes the inertia force and three-dimensional temperature change into consideration, as follows:
∂
∂r
G1∂p
∂r + G2
+1
r
G1∂p
∂r + G2
+ 1
r2
∂
∂θ
G1∂p
∂θ
=1
r
∂G3
where G1, G2, and G3are given as follows:
G1=
h
0
ρ z
0
z
µdz−
F1
F0
z
0
dz
µ dz
G2=
h
0
ρ
1
r
G0
F0
z
0
dz
µ −
1
r
z
0
1 µ
z
0
ρ(θ2dz
dz dz
G3=
h
0
ρrω
1
F0
z
0
dz
By means of Eq 8.70, the three-dimensional thermohydrodynamic lubrication analysis that takes the inertia force into consideration can be carried out The pro-cedure of numerical computation is the same as before Whenµ = ρ = constant,
Eq 8.70 coincides with the Reynolds’ equation (Eq 4.34) derived in Chapter 4 with reference to cylindrical coordinates
Trang 10182 8 Heat Generation and Temperature Rise
Examples of the three-dimensional analysis of thermohydrodynamic lubrication taking the inertia forces into consideration are shown below Specifications of the bearing and the lubricating oil are listed in Table 8.1
Figure 8.6 shows the calculated temperature rise, and illustrates how the differ-ence between the highest and the lowest temperature (entrance temperature) on the pad surface,∆T, changes with the rotational speed N The operating conditions of the bearing are: r p = 0.51, θp = 0.42, h0 = 100 µm, and Tin = 47◦C If the inertia
term is taken into consideration,∆T is calculated to be larger than that for the case
where no inertia effect is considered, and the larger N is, the larger ∆T is This differ-ence is due to the effect of the velocity component produced in the radial direction
Fig 8.6 Temperature rise [21]
Figure 8.7 shows the relation between the nondimensional load capacity P (see Fig 8.5a-c) and N under the same operating conditions If the inertia term is ignored,
P decreases monotonously along with the increase in N This is due to the decrease
in viscosity along with the temperature rise in the oil film shown in Fig 8.6 On the other hand, if an inertia term is taken into consideration, the calculated load capacity tends to be larger due to the flow velocity in the radial direction As a result, in
the area where the curve is upward convex in Fig 8.7, the load capacity P is about
10% higher than that obtained ignoring the inertia term If the rotating speed exceeds
2000 rpm, however, the load capacity falls because the influence of the decrease in viscosity due to temperature rise exceeds that of the inertia
Figure 8.8 shows the relation between the inclination (tilt) of a pad and the
rotat-ing speed N under the same operatrotat-ing conditions.αpandαrare the circumferential inclinationαpand the radial direction inclinationαrof the pad multiplied by∆R/h0,
respectively Even though both of the inclinations increase with the increase in N,
...0
dz
By means of Eq 8.70, the three-dimensional thermohydrodynamic lubrication analysis that takes the inertia force into consideration can be carried out The... Generation and Temperature Rise
Examples of the three-dimensional analysis of thermohydrodynamic lubrication taking the inertia forces into consideration are shown below Specifications of