Handbook of Lubrication Episode 2 Part 4 ppt

428CRC Handbook of Lubrication β = 40° (0.70 rad) γ r = 0 (no radial tilt) R 2 = 5.50 in. γ θ = 5.82 × 10 –4 rad (0.0333°) R 1 = 2.75 in. μ = 2 × 10 –6 lb sec/in. 2 (ISO VG 32 h c = 0.002 in. (at pad center)oil at T s = 135 F, Figure 1) N = 50 r/sec (ω = 2 π · 50 = 314 rad/sec) Calculating first m θ = (R 1 /h c ) γ θ = (2.75/0.002) 5.82 × 10 –4 = 0.80 Enter Figures 14 to 17 with m θ = 0.80 and m r = 0.0: Wh 2 c /[6 ␮ ω (Ρ 2 ϪΡ 1 ) 4 ] ϭ 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 h min /h c = 0.45; h min = 0.45 × 0.002 = 0.00090 in. Hh c /[ μ ω 2 (R 2 – R 1 ) 4 ] = 3.04 FIGURE 15. Minimum film thickness for fixed- or tilting pad sector (laminar flow). Copyright © 1983 CRC Press LLC 430 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 × 10 4 lb 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). Copyright © 1983 CRC Press LLC From Figure 17, sector flow rates are Q in /[h c ω(R 2 – R 1 ) 2 ] = 0.86 Q S /[h c ω(R 2 – R 1 ) 2 ] = 0.35 Q in = 4.08 in. 3 /sec (1.06 gal/min) Q s = 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 × 10 4 /[112(4.08 – 1.66/2)] = 47.0 F This temperature rise implies an inlet temperature T i of T i = T S – 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 θ p /β = 0.39 and (r p – R 1 )/(R 2 – R 1 ) = 0.56 to achieve the above calculated performance. This performance is also achieved with a fixed-type bearing by machining slope angles γ θ = 0.0333° and γ r = 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, h O , 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: Volume II 431 (11) (12) (13) (14) (15) Copyright © 1983 CRC Press LLC Laminarflow solution: from Table 4, load factor C W,L = 0.132; from Equation 11, using C W,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 C ho , L = 0.77: h O = 0.77 × 1.0 × 0.0016 = 0.0012 in. Power loss, H, and flow requirements (Q in , Q s ) 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) Q in = 4.19 in. 3 /sec (1.09 gpm) Q S = 1.68 in. 3 /sec (0.44 gpm) The cylindrical crown required to attain maximum load capacity is given in Table 5 as δ/h p = 0.6 δ = 0.6 × 0.016 = 0.00096 in. For turbulentflowcorrection, the Reynolds number calculation gives Re p = Uh p /v = 1296 × 0.0016/4.96 × 10 –4 = 4180 As Re p is above 2000, entering Figure 21 gives as turbulent correction factors C W, T = 3.2 C h o , T = 0.9 C H, T = 4.3 C Q in T = 1.1 C Q s , T = 1.4 Using Equations 12 to 16 gives turbulent performance as: W = 62.5 × 3.2 = 200 lb. (P = 26 psi) h O = 0.0012 × 0.9 = 0.0011 in. H = 372 × 4.3 = 1600 in. lb/sec (0.24 hp) Q in = 4.19 × 1.1 = 4.61 in. 3 /sec (1.20 gpm) Q S = 1.68 × 1.4 = 2.35 in. 3 /sec (0.61 gpm) Interpolating for Re p = 4180 in Table 5 indicates that the cylindrical crown should be changed so that δ/h p = 1.1. The required optimum cylindrical crown then becomes δ/h p = 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 434 CRC Handbook of Lubrication Copyright © 1983 CRC Press LLC Volume II435 as 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 JOURNALBEARINGS 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. Copyright © 1983 CRC Press LLC β = 2.79 rad (160°)N = 40 r/sec ( ω = 2πN = 251 rad/sec) D = L = 20 in. (L/D = 1)µ = 1.8 × 1 –6 lb sec/in. 2 (ISO VG 32 C = 0.020 in.oil at 140 F, Figure 1) W = 80,000 lb (P = W/DL = 200 psi) ρ = 7.77 × 10 –5 lb sec 2 /in. 4 ␯ = µ/ ρ = 2.32 × 10 –2 in. 2 /sec Calculating the Reynolds and bearing characteristic numbers: Re = R ω C/␯ = 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 Minimumfilmthickness— Entering Figure 24 with S = 0.090 and Re = 2160: h n /C = 0.44 h n = 0.44 × .020 = 0.0088 in. Positionofminimumfilmthickness— Similarly, from Figure 25, attitude angle φ = 48°. Volume II 437 FIGURE 26. Power loss for partial arc journal bearing. Copyright © 1983 CRC Press LLC 438 CRC Handbook of Lubrication FIGURE 27. Inlet flow for partial arc journal bearing. FIGURE 28. Side flow for partial arc journal bearing. Copyright © 1983 CRC Press LLC Powerloss— Figure 26 gives H/(2πWNC) = 2.8, from which H = 2.8 × 2 π × 80,000 × 40 × 0.020 = 1.13 × 10 6 lb in./sec (171 hp) Lubricantflow— 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, Q s which escapes laterally from both sides of the bearing arc is found from Figure 28 which gives Q s /(RCNL) = 1.8; then Q s = 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 Q s with carryover. Insufficient external supply and starved operation does not imply that a bearing will necessarily fail, but its performance will be altered. 30 On 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. Volume II 439 FIGURE 29. Film stiffness coefficients for partial arc journal bearing. Copyright © 1983 CRC Press LLC respectively, K yy = 6.2 × 10 6 K xy = 12.0 × 10 6 , and K yx = –2.00 × 10 6 lb/in. Figure 30 gives (C/W) ωB xx = 6.8, from which B xx = 6.8 × 80,000/0.020 × (1/251) = 0.11 × 10 6 lb sec/in. In an identical manner, we find B yy = 0.029 × 10 6 and B xy = B yx = 0.028 × 10 6 lb 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 solution 31 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 Q s (Type 2 BC only) can be calculated from the following equations with eccentricity ratio ⑀ taken from Figure 31: Q = π R N LC(1 + ⑀) Q s = 2 π R N L C ⑀ Volume II 441 FIGURE 31. Eccentricity ratio, minimum film thickness and attitude angle for full-journal bearings (laminar flow, Re < 1000). Copyright © 1983 CRC Press LLC [...].. .4 42 CRC Handbook of Lubrication FIGURE 32 FIGURE 33 Unruptured film stiffness and damping for full-journal bearing (laminar flow Re < 1000) Film stiffness for full journal bearing with ruptured oil film (laminar flow, Re < 1000) Copyright © 1983 CRC Press LLC 44 4 CRC Handbook of Lubrication FIGURE 35 Stability of single-mass rotor on full-journal bearings... Entering Figures 41 to 43 with ⑀′o = 0.67 gives (C/W)Kxx = 4. 9 from which Kxx = 4. 9(W/C) = 4. 9 (25 00/0.005) = 2. 5 × 106 lb/in Copyright © 1983 CRC Press LLC 45 0 CRC Handbook of Lubrication FIGURE 42 Bearing horizontal stiffness (five 60° tilting pads, centrally pivoted, no preload, (Ω/ω = 1.0, no pad inertia, L/D = 0.5) If the excitation frequency ratio were different from Ω/ω = 1.0, say 2. 0, the stiffness... 7.77 × 10-5 1b sec2/in .4 ␯ = μ/ρ = 2. 57 × 10 -2 in .2/ sec 44 8 CRC Handbook of Lubrication FIGURE 40 0.5) Normalized bearing eccentricity ratio (five 60° tilting pads, centrally pivoted, no preload, L/D = Calculating 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... 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... Figure 39, H/ (2 WNC) = 3.9 from which H = 1. 84 × 1 04 lbin./sec (2. 78 hp) Normalized bearing eccentricity ratio — Entering Figure 40 with hn/C = 0 .26 gives ⑀′o = ⑀o/1 .23 61 = 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 (eccentricity) is eo = ⑀oC′ = 0.67 × 1 .23 61 × 0.005 = 0.0 041 in Dynamic... 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 D L C N = = = = = 1.05 rad (60°) 5 in 2. 5 in (L/D = 0.5) C′ = 0.005 in (no preload) 60 r/sec Copyright © 1983 CRC Press LLC W = 25 00 lb μ = 2 × 10-6 lb sec/in .2 (ISO VG 32 oil at 135 F... 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, Copyright © 1983 CRC Press LLC 44 6 CRC Handbook of Lubrication FIGURE 37 FIGURE 38 0.5] Mass moment of inertia of pad around axial axis Minimum film... 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)... 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 у 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... 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, Ip can be calculated from the general expression given Copyright © 1983 CRC Press LLC Volume II FIGURE 39 44 7 Total . 42 8 CRC Handbook of Lubrication β = 40 ° (0.70 rad) γ r = 0 (no radial tilt) R 2 = 5.50 in. γ θ = 5. 82 × 10 4 rad (0.0333°) R 1 = 2. 75 in. μ = 2 × 10 –6 lb sec/in. 2 (ISO VG 32 h c = 0.0 02. calculated as W = 0.058 × 6 × 2 × 10 –6 × 3 14( 5.50 – 2. 75) 4 /0.0 02) 2 = 3 125 lb h min /h c = 0 .45 ; h min = 0 .45 × 0.0 02 = 0.00090 in. Hh c /[ μ ω 2 (R 2 – R 1 ) 4 ] = 3. 04 FIGURE 15. Minimum film. LLC 43 0 CRC Handbook of Lubrication From which the power loss per sector is H = 3. 04 × 2 × 10 –6 × (3 14) 2 (2. 75) 4 /0.0 02 = 1.70 × 10 4 lb in./sec (2. 58 hp) FIGURE 18. Tangential location of