Minimum film thickness for fixed- or tilting pad sector laminar flow... Centrally Pivoted Pads If the sector pad surface is truly flat, then basic theory isothermal requires that the tan
Trang 2β = 40° (0.70 rad) γr= 0 (no radial tilt)
R2= 5.50 in γθ= 5.82 × 10– 4rad (0.0333°)
R1= 2.75 in μ = 2 × 10–6 lb sec/in.2(ISO VG 32
hc= 0.002 in (at pad center) oil at Ts= 135 F, Figure 1)
N = 50 r/sec (ω = 2π · 50 = 314 rad/sec) Calculating first
mθ= (R1/hc)γθ= (2.75/0.002) 5.82 × 10–4 = 0.80 Enter Figures 14 to 17 with mθ= 0.80 and mr= 0.0:
Wh2
c/[6 ω (Ρ2 Ρ1)4] 0.058 Load capacity per sector is then calculated as
W = 0.058 × 6 × 2 × 10–6 × 314(5.50 – 2.75)4/0.002)2 = 3125 lb
hmin/hc= 0.45; hmin= 0.45 × 0.002 = 0.00090 in
Hhc/[μω2(R2– R1)4] = 3.04 FIGURE 15 Minimum film thickness for fixed- or tilting pad sector (laminar flow).
Trang 3430 CRC Handbook of Lubrication
From which the power loss per sector is
H = 3.04 × 2 × 10–6× (314)2(2.75)4/0.002 = 1.70 × 104lb in./sec (2.58 hp)
FIGURE 18 Tangential location of center-of-pressure (fixed-pad sector), or pivot position (tilting pad sector) (laminar flow).
FIGURE 19 Radial location of center-of-pressure (fixed-pad sector), or pivot position (tilting pad sector) (laminar flow).
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Trang 4From Figure 17, sector flow rates are
Qin/[hcω(R2– R1)2] = 0.86
QS/[hcω(R2– R1)2] = 0.35
Qin= 4.08 in.3/sec (1.06 gal/min)
Qs= 1.66 in.3/sec (0.43 gal/min) Temperature rise can then be calculated from Equation 1 with pc = 112 as typical for petroleum oils
ΔT = 1.71 × 104/[112(4.08 – 1.66/2)] = 47.0 F This temperature rise implies an inlet temperature Tiof
Ti= TS– ΔT/2 = 135 – 47.0/2 = 111.5 F
If this did not match the oil inlet temperature, a new effective oil viscosity would be assumed and the steps repeated
From Figures 18 and 19, the pivot must be placed at θp/β = 0.39 and (rp – R1)/(R2 –
R1) = 0.56 to achieve the above calculated performance This performance is also achieved with a fixed-type bearing by machining slope angles γθ= 0.0333° and γr= 0
Centrally Pivoted Pads
If the sector pad surface is truly flat, then basic theory (isothermal) requires that the tangential pivot location be offset toward the pad trailing edge for appreciable load capacity For example, the optimum pivot location for the basic slider bearing (Figure 8) is x_
= 0.58B (for a “square”, L = B pad) However, flat pads are frequently constructed with a central pivot (x_
= 0.5B) for design simplicity and to apply them for either direction of rotation The following phenomena commonly enable nearly optimum load capacity for centrally pivoted flat pads when operating with lubricating oil or other relatively high viscosity fluid: (1) viscosity varies as the lubricant passes through the film to alter the pressure distribution, (2) film pressure elastically deforms (crowns) the pad, and (3) film heating thermally deforms the pad.24,25
In low-viscosity fluid applications involving water, liquid metals, and gases, it is usually necessary to deliberately crown the pad spherically or cylindrically by precise manufacturing techniques to ensure high-load capacity.25 Figure 20 indicates that the optimum spherical crown is very small, about 0.6 times the minimum film thickness, hO, and will result in a load capacity nearly equivalent to that of a flat pad with its pivot optimally placed
Cylindrical and spherically crowned pads can be designed by employing Tables 4 and 5 and Figure 20 These data should be used in conjunction with the following performance equations:
(11) (12) (13) (14) (15)
Trang 5Laminar flow solution: from Table 4, load factor CW,L = 0.132; from Equation 11, using
CW,T= 1 for laminar flow:
W = 0.132 × 1 × 4.5 × 10–8 × 1296 × (2.75)2 × 2.75/(0.00016)2 = 62.5 lb for load capacity (P = W/BL = 8.26 psi) Similarly from Equation 13, using Cho, L = 0.77:
hO= 0.77 × 1.0 × 0.0016 = 0.0012 in
Power loss, H, and flow requirements (Qin, Qs) can be calculated in a similar manner from Table 4 and Equations 14, 15, and 16 to yield
H = 372 in lb/sec (0.056 hp)
Qin= 4.19 in.3/sec (1.09 gpm)
QS= 1.68 in.3/sec (0.44 gpm) The cylindrical crown required to attain maximum load capacity is given in Table 5 as
δ/hp= 0.6
δ = 0.6 × 0.016 = 0.00096 in
For turbulent flow correction, the Reynolds number calculation gives
Rep= Uhp/v = 1296 × 0.0016/4.96 × 10–4 = 4180
As Repis above 2000, entering Figure 21 gives as turbulent correction factors
CW, T= 3.2
Cho, T= 0.9
CH, T= 4.3
CQin T= 1.1
CQs, T= 1.4 Using Equations 12 to 16 gives turbulent performance as:
W= 62.5 × 3.2 = 200 lb (P = 26 psi)
hO= 0.0012 × 0.9 = 0.0011 in
H = 372 × 4.3 = 1600 in lb/sec (0.24 hp)
Qin= 4.19 × 1.1 = 4.61 in.3/sec (1.20 gpm)
QS= 1.68 × 1.4 = 2.35 in.3/sec (0.61 gpm) Interpolating for Rep = 4180 in Table 5 indicates that the cylindrical crown should be changed so that δ/hp= 1.1 The required optimum cylindrical crown then becomes
δ/hp= 1.1
δ = 1.1 × 0.0016 = 0.0018 in
Flexible-Type Thrust Bearings
While a bearing pad is not as free to pivot with spring, rubber, or other elastic support
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Trang 6as in a pivoted-pad bearing, it can partially respond to changes in speed, load, viscosity, and temperature Advantages include (1) better load distribution than with fixed pads, and (2) less space requirement and simpler design than pivoted-pad bearings
Flexible-type bearings are difficult to analyze since the elastic and thermal deflections in the system must be considered.27 Some data relative to film thickness and power loss are given in Reference 28
FIXED-TYPE JOURNAL BEARINGS The full and the partial bearings are the two basic configurations of fixed-type hydro-dynamic journal bearings (Figure 22) Most other designs found in practice (Figure 23, for example) can be considered as variations of these two
The active (load-carrying) bearing arc extends entirely around the journal for the full journal bearing; in the partial bearing, the active arc only partially surrounds the journal The partial bearing is “centrally loaded” if the load direction bisects the active arc, and
“offset” or eccentrically loaded” if it does not Design charts are presented here for the most commonly used bearings: the full journal bearing and the centrally loaded partial bearing
Centrally Loaded Partial Bearings
Figures 24 to 30 are design charts for a 160° centrally loaded partial bearing with a length-to-diameter (L/D) ratio equal to 1 (see Reference 29 for offset loading) These were derived from numerical solutions of basic Equation 2 with Type 2 boundary condition (film rupture
FIGURE 22 Fixed-type journal bearings: (a) full 360° bearing, (b) centrally loaded partial
bearing, and (c) offset loaded partial bearing.
FIGURE 23 Partial bearing with relief and end lands in top.
Trang 7β = 2.79 rad (160°) N = 40 r/sec (ω = 2πN = 251 rad/sec)
D = L = 20 in (L/D = 1) µ = 1.8 × 1–6lb sec/in.2(ISO VG 32
W = 80,000 lb (P = W/DL = 200 psi) ρ = 7.77 × 10–5lb sec2/in.4
= µ/ρ = 2.32 × 10–2in.2/sec Calculating the Reynolds and bearing characteristic numbers:
Re = RωC/ = 10 × 251 × 0.020/2.32 × 10–2 = 2160 (turbulent)
S = µN (R/C)2 /P = 1.8 × 10–6 × 40× (10/0.020)2/200 = 0.090
Minimum film thickness — Entering Figure 24 with S = 0.090 and Re = 2160:
hn/C = 0.44
hn= 0.44 × 020 = 0.0088 in
Position of minimum film thickness — Similarly, from Figure 25, attitude angle φ =
48°
FIGURE 26 Power loss for partial arc journal bearing.
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Trang 8FIGURE 27 Inlet flow for partial arc journal bearing.
FIGURE 28 Side flow for partial arc journal bearing.
Trang 9Power loss — Figure 26 gives H/(2πWNC) = 2.8, from which
H = 2.8 × 2π × 80,000 × 40 × 0.020 = 1.13 × 106lb in./sec (171 hp)
Lubricant flow — Figure 27 gives Q/(RCNL) = 3.3, from which Q = 528 in.3/sec
(137 gpm) This inlet flow is drawn into the leading edge of the arc by journal rotation Of
this amount, Qs which escapes laterally from both sides of the bearing arc is found from Figure 28 which gives Qs/(RCNL) = 1.8; then Qs= 288 in.3/sec (74.8 gpm) If this bearing
is operating submerged in fluid, ample lubricant will always be available at the leading edge
of the bearing arc If, however, ample lubricant is to be supplied to the leading edge by external means, the feed rate must be at least (1) equal Q if there is no carryover by the shaft back to the leading edge or (2) equal Qswith carryover Insufficient external supply and starved operation does not imply that a bearing will necessarily fail, but its performance will be altered.30On the other hand, if lubricant is supplied at a pressure greater than ambient (for example, into the top in Figure 23), part of the lubricant may by-pass the active arc, considerably increase total lubricant flow, and again alter the performance characteristics Certain types of grooving will also affect flow and performance
FIGURE 29 Film stiffness coefficients for partial arc journal bearing.
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Trang 10respectively, Kyy = 6.2 × 106 Kxy = 12.0 × 106, and Kyx = –2.00 × 106 lb/in Figure
30 gives (C/W) ωBxx = 6.8, from which Bxx = 6.8 × 80,000/0.020 × (1/251) = 0.11
× 106 lb sec/in In an identical manner, we find Byy = 0.029 × 106 and Bxy = Byx = 0.028 × 106lb sec/in
Full Journal Bearing
Performance of the full journal bearing can be estimated from the design charts presented
in Figures 31 to 34 These data are based on a short journal bearing approximate solution31 and are most accurate for small L/D ratios (L/D ≤1) Curves are presented for both Type
1 (continuous film) and Type 2 (ruptured film) boundary conditions (BC) While the Type
1 BC is especially useful for pump bearings completely submerged in a high-ambient pres-sure, Type 2 BC is commonly found in most other applications
Use of Figures 31 to 34 is similar to that shown by the examples for a partial arc bearing Flow rates Q and Qs(Type 2 BC only) can be calculated from the following equations with eccentricity ratio taken from Figure 31:
Q =π R N LC(1 +)
Qs= 2 π R N L C FIGURE 31 Eccentricity ratio, minimum film thickness and attitude angle for full-journal
bearings (laminar flow, Re < 1000).
Trang 11FIGURE 33 Film stiffness for full journal bearing with ruptured oil film (laminar flow, Re < 1000).
FIGURE 32 Unruptured film stiffness and damping for full-journal bearing (laminar flow Re < 1000).
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Trang 12(1) operation with smaller clearances than considered proper for a plain journal bearing, or (2) preloading the journal to achieve relatively high stiffness (important with vertical rotor)
Load Orientation
While not a necessity, usual practice is to construct all pads alike, to space them uniformly around the circumference, and to use an aspect ratio (L/B′) approximately equal to one If the number of pads is large, it makes little difference whether the load line of action passes through the center of one pad or lies midway between two pads If the number of pads is small, however, load-between-pads orientation is preferred because it gives (1) greater load capacity for a given minimum film thickness, (2) lower pad temperature rise by distributing the load more uniformly, and (3) greater lateral stiffness and damping for horizontal rotor applications
Pivot Position
Pivots are usually positioned at the pad center (at β/2) to obtain identical performance independent of the direction of journal rotation With oil as a lubricant, load capacity is not unduly sensitive to the pivot location However, when using low-viscosity fluids such as water, liquid metals, and particularly gases, load capacity is sensitive to pivot locations and offset (toward the trailing edge) pivots are preferable.34
Pad Contour (Preloading)
Figure 36 shows a pad machined to radius R + C (position 1) Assuming the pad does not tilt, the film thickness is uniform (equal to C) and develops no hydrodynamic force If
the pad is moved to position 2 by displacing the pivot radially inward (C – C′), film
thickness is no longer uniform and a hydrodynamic force “preloads” the journal
Bearings in vertical machines often undergo little if any radial load (magnetic pull,
FIGURE 35 Stability of single-mass rotor on full-journal bearings (ruptured film, laminar flow).
Trang 13motion and pad motion are 90° out of phase.35 Onset of pad resonance can be determined from the “critical pad mass parameter” and requires calculation of the pad pitch inertia Ip Design data given in this section are based on a pivot fictitiously located on the pad surface above the actual pivot For this case, Ipcan be calculated from the general expression given
FIGURE 37 Mass moment of inertia of pad around axial axis.
FIGURE 38 Minimum film thickness [five 60° tilting pads, centrally pivoted, no preload (C′ = C), L/D = 0.5].
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Trang 14in Figure 37 by taking t1/rp= t/rp= 0 Design data accounting for pad inertia near resonance can be found in Reference 35
Excitation Frequency
Unlike fixed-arc bearings, dynamic spring and damping coefficients of tilting pad journal bearings are dependent upon the frequency Ω of the excitation force These coefficients are usually presented for the common case of unbalance excitation (Ω/ω = 1.0) Following presentation of a variety of performance data for five-pad bearings in Figures 38 to 44, the effect of excitation frequency is provided in Figures 45 to 48 for a five-pad bearing
Influence of preload on the stability of vertical (essentially unloaded) guide bearings employing four, five, six, and eight pads is given in Figures 49 to 52
Example: (horizontal rotor): find the performance of a five-pad bearing given:
β = 1.05 rad (60°) W = 2500 lb
D = 5 in μ = 2 × 10 -6 lb sec/in 2 (ISO VG 32 oil
L = 2.5 in (L/D = 0.5) at 135 F avg temp., TS, Figure 1 )
C = C′ = 0.005 in (no preload) ρ = 7.77 × 10 -5 1b sec 2 /in 4
N = 60 r/sec = μ/ρ = 2.57 × 10-2 in 2 /sec FIGURE 39 Total power loss (five 60° tilting pads, centrally pivoted, no preload, L/D = 0.5).
Trang 15Calculating the Reynolds and bearing characteristic numbers:
Minimum film thickness — Because Re is below 1000, flow is laminar and Re = 1.0,
curve in Figure 38 gives:
hn/C = 0.26
hn= 0.26 × 0.005 = 0.0013 in
Power loss — From Figure 39, H/(2πWNC) = 3.9 from which H = 1.84 × 104 lb-in./sec (2.78 hp)
Normalized bearing eccentricity ratio — Entering Figure 40 with hn/C = 0.26 gives
′o = o/1.2361 = 0.67 This value will be used to enter the charts to obtain the dynamic stiffness and damping coefficients As a matter of interest, the journal displacement (ec-centricity) is
eo= oC′ = 0.67 × 1.2361 × 0.005 = 0.0041 in
Dynamic performance — (Ω/ω = 1, unbalance excitation) Entering Figures 41 to 43
with ′o= 0.67 gives (C/W)Kxx= 4.9 from which
Kxx= 4.9(W/C) = 4.9(2500/0.005) = 2.5 × 106lb/in
FIGURE 40 Normalized bearing eccentricity ratio (five 60° tilting pads, centrally pivoted, no preload, L/D = 0.5).
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Trang 16If the excitation frequency ratio were different from Ω/ω = 1.0, say 2.0, the stiffness and damping coefficients could be obtained directly by entering Figures 45 to 48 with o′ = 0.67, as in the above example, and Ω/ω = 2.0 Figures 45 to 48, although valid only for laminar flow, can also be used for turbulent flow (Re 1000) to approximate the stiffness and damping coefficients since they are not strongly dependent on Reynolds number To
do this, o′ should first be obtained as shown in the above example through Figures 38 and
40 using the appropriate curve for the actual value of Re
Critical mass — From Figure 44, the critical pad mass parameter is
FIGURE 42 Bearing horizontal stiffness (five 60° tilting pads, centrally pivoted, no preload, (Ω/ω = 1.0, no pad inertia, L/D = 0.5).