Engineering Tribology Episode 1 Part 8 pptx

150 ENGINEERING TRIBOLOGY θ O W 1 R x 1 dθ x 2 W 2 β W Pressure profile Rdθ θ pRcos θdθdy pRsin θdθdy pRdθdy FIGURE 4.30 Load components and pressure field acting in a journal bearing. Thus the load component acting along the line of centres is expressed by: W 1 = ⌠ ⌡ 0 π pRcos θdθdy ⌠ ⌡ − L 2 L 2 (4.103) similarly the component acting in the direction normal to the line of centres is: h 0 h 1 x,θ y L πD L y = L/2 y = −L/2 x = 0 θ = 0 x = 2πR θ = 2π Position where the film is cut. It corresponds to x = 0 and x = 2πR ‘Unwrapped’ journal bearing film y θ FIGURE 4.31 Unwrapped journal bearing. W 2 = ⌠ ⌡ 0 π pRsin θdθdy ⌠ ⌡ − L 2 L 2 (4.104) Substituting for ‘p’ (4.102) and separating variables gives: TEAM LRN HYDRODYNAMIC LUBRICATION 151 W 1 = ⌠ ⌡ 0 π dθdy Rc 2 (1 + εcosθ) 3 3UηεRsinθcosθ 4 L 2 () − y 2 ⌠ ⌡ − L 2 L 2 ⌠ ⌡ 0 π dθ c 2 3Uηε 4 L 2 () − y 2 ⌠ ⌡ − L 2 L 2 (1 + εcosθ) 3 sinθcos θ dy= W 2 = ⌠ ⌡ 0 π dθdy Rc 2 (1 + εcosθ) 3 3UηεRsin 2 θ 4 L 2 () − y 2 ⌠ ⌡ − L 2 L 2 ⌠ ⌡ 0 π dθ c 2 3Uηε 4 L 2 () − y 2 ⌠ ⌡ − L 2 L 2 (1 + εcosθ) 3 sin 2 θ dy= The individual integrals can be evaluated separately from each other and they are: ⌠ ⌡ 0 π dθ = − (1 +εcosθ) 3 sinθcos θ (1 −ε 2 ) 2 2ε ⌠ ⌡ 0 π dθ = (1 +εcosθ) 3 sin 2 θ 2(1 −ε 2 ) 3/2 π 4 L 2 () − y 2 ⌠ ⌡ − L 2 L dy = 6 L 3 2 Substituting yields: W 1 = − c 2 (1 −ε 2 ) 2 UηL 3 ε 2 (4.105) W 2 = 4c 2 (1 −ε 2 ) 3/2 UηεπL 3 (4.106) The total load that the bearing will support is the resultant of the components ‘W 1 ’ and ‘W 2 ’: W = W 1 2 + W 2 2 (4.107) Substituting for ‘W 1 ’ and ‘W 2 ’ gives the expression for the total load that the bearing will support: − 1 () 16 π 2 ε 2 + 1W = c 2 (1 −ε 2 ) 2 UηεL 3 4 π (4.108) It can be seen that in a similar fashion to the other bearings analysed, the total load is expressed in terms of the geometrical and operating parameters of the bearing. Equation (4.108) can be rewritten in the form: = LUηR 2 Wc 2 L 2 4R 2 (1 −ε 2 ) 2 πε (0.621ε 2 + 1) 0.5 (4.109) TEAM LRN 152 ENGINEERING TRIBOLOGY Introducing a variable ‘∆’: () c R ∆= LUη W 2 (4.110) which is also known as the ‘Sommerfeld Number’ or ‘Duty Parameter’, equation (4.109) becomes: () D L ∆ 2 = (1 −ε 2 ) 2 πε (0.621ε 2 + 1) 0.5 (4.111) where: D = 2R is the shaft diameter [m]. The Sommerfeld Number is a very important parameter in bearing design since it expresses the bearing load characteristic as a function of eccentricity ratio. Computed values of Sommerfeld number ‘∆’ versus eccentricity ratio ‘ε’ are shown in Figure 4.32 [3]. The curves were computed using the Reynolds boundary condition which is the more accurate. Data for long journal bearings which cannot be calculated from the above equations are also included. The data is also based on a bearing geometry where 180° of bearing sector on the unloaded side of the bearing has been removed. Removal of the bearing shell at positions where hydrodynamic pressure is negligible is a convenient means of reducing friction and the bearings are known as partial arc bearings. The effect on load capacity is negligible except at extremely small eccentricity ratios. An engineer can find from Figure 4.32 a value of Sommerfeld number for a specific eccentricity and L/D ratio and then the bearing and operating parameters can be selected to give an optimum performance. It is usually assumed that the optimum value of eccentricity ratio is close to: ε optimum = 0.7 Higher values of eccentricity ratio are prone to shaft misalignment difficulties; lower values may cause shaft vibration and are associated with higher friction and lubricant temperature. If the surface speed of the shaft is replaced by the angular velocity of the shaft then the left hand side of the graph shown in Figure 4.32 can be used. When the shaft angular velocity is expressed in revolutions per second [rps] then the modified Sommerfeld parameter becomes S = π∆. Since: U = 2πRN substituting into equation (4.110) gives: () c R ∆= Lη2 πRN W 2 Introducing ‘P’ [4]: TEAM LRN HYDRODYNAMIC LUBRICATION 153 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.01 0.02 0.03 0.04 0.05 0.06 0.08 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1 2 3 4 5 6 8 10 20 30 40 50 60 80 100 L D = ∞ = 1 1 2 = 1 4 = 1 8 = 0.02 0.03 0.04 0.05 0.06 0.08 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1 2 3 4 5 6 8 10 20 0.004 0.005 0.006 0.008 0.01 1 8 Ocvirk W/L Uη ∆ = ( c R ) 2 P Nη S = ( c R ) 2 180° bearing Reynolds conditions 30 ε FIGURE 4.32 Computed values of Sommerfeld number ‘∆’ versus eccentricity ratio ‘ε’ [3]. P = 2LR W () c R ∆= Nηπ P 2 Thus: () c R S = ∆π = Nη P 2 (4.112) TEAM LRN 154 ENGINEERING TRIBOLOGY It can also be seen from Figure 4.30 that the attitude angle ‘β’ between the load line and the line of centres can be determined directly from the load components ‘W 1 ’ and ‘W 2 ’ from the following relation: tanβ = − W 1 W 2 Substituting for ‘W 1 ’ and ‘W 2 ’ yields: tanβ = 4 π ε (1 −ε 2 ) 1/2 (4.113) · Friction Force The friction force can be calculated by integrating the shear stress ‘τ’ over the bearing area: F = ⌠ ⌡ 0 L τdxdy = ⌠ ⌡ 0 B ⌠ ⌡ 0 L η ⌠ ⌡ 0 B dxdy dz du In journal bearings, the bottom surface is stationary whereas the top surface, the shaft, is moving, i.e.: U 1 = U and U 2 = 0 which is the opposite case from linear pad bearings. Thus the velocity equation (4.11) becomes: u = ∂p ∂x 2η ( ( z 2 − zh + U z h Differentiating with respect to ‘z’ gives the shear rate: = dz du 2z − h 2η 1 dx dp + U h ( ( After substituting, the expression for friction force is obtained: F = ⌠ ⌡ 0 L ⌠ ⌡ 0 B [( [ z − 2 h dx dp + Uη h dxdy ) (4.114) In the narrow bearing approximation it is assumed that ∂p/∂x ≈ 0 since ∂p/∂x « ∂p/∂y and (4.114) becomes: F = ⌠ ⌡ 0 L ⌠ ⌡ 0 B Uη h dxdy (4.115) and the friction force on the moving surface, i.e. the shaft, is given by: TEAM LRN HYDRODYNAMIC LUBRICATION 155 F = ⌠ ⌡ 0 B UηL h dx (4.116) Substituting for ‘h’ from (4.99) and ‘dx = Rdθ’ gives: F = ⌠ ⌡ 0 π UηLR c(1 +εcosθ) dθ = ⌠ ⌡ 0 π UηLR c dθ (1 +εcosθ) and integrating yields: F = 2πηULR c 1 (1 −ε 2 ) 0.5 (4.117) which is the friction in journal bearings at the surface of the shaft for the Half-Sommerfeld condition. It can be seen from equation (4.117) that when: · the shaft and bush are concentric then: e = 0 and ε = 0 and the value of the second term of equation (4.117) becomes unity. The equation now reduces to the first term only. This is known as ‘Petroff friction’ since it was first published by Petroff in 1883 [3]. · the shaft and bush are touching then: e = c and ε = 1 which causes infinite friction according to the model of hydrodynamic lubrication. In practice the friction may not reach infinitely high values if the shaft and bush touch but the friction will be much higher than that typical of hydrodynamic lubrication. It is also true that as the eccentricity ratio approaches unity, the friction coefficient rises. The second term of (4.117) is known as the ‘Petroff multiplier’. Figure 4.33 shows the relationship between the calculated Petroff multiplier and the eccentricity ratio for infinitely long 360° journal bearings [8]. The calculated values are higher than those predicted from (1 - ε 2 ) -0.5 since the effects of pressure on the shear stress of the lubricant are not included in equation (4.117). The effect of cavitation, i.e. the zero pressure region, does have a significant effect on friction and this together with pressure effects are discussed in the next chapter on ‘Computational Hydrodynamics’. · Coefficient of Friction The coefficient of friction of a bearing is calculated once the load and friction forces are known: µ= F W As can be seen from equation (4.108) or from Figure 4.32 the load capacity rises sharply with an increase in eccentricity ratio. Friction force is relatively unaffected by changes in TEAM LRN 156 ENGINEERING TRIBOLOGY 0 0 0.2 0.4 0.6 0.8 1.0 Eccentricity Petroff multiplier K ε 1 2 3 4 5 FIGURE 4.33 Relationship between Petroff multiplier and eccentricity ratio for infinitely long 360° bearings [8]. eccentricity ratio until an eccentricity ratio of about 0.8 is reached. Although the operation of bearings at the highest possible levels of Sommerfeld number and eccentricity ratio will allow minimum bearing dimensions and oil consumption, the optimum value of the eccentricity ratio, as already mentioned, is approximately ε = 0.7. Interestingly the optimal ratio of maximum to minimum film thickness for journal bearings is much higher than for pad bearings as is shown below: at θ = 0 where film thickness is a maximum, h 1 = c (1 + ε) and at θ = π where film thickness is a minimum, h 0 = c (1 - ε) so that the optimal inlet/outlet film thickness ratio for journal bearings is h 1 h 0 = 1 + ε 1 - ε = 1 + 0.7 1 - 0.7 = 5.67. This ratio is higher than for linear pad bearings for which it is equal to 2.2. There is a noticeable discrepancy in optimum ratios of maximum to minimum film thickness but strictly speaking these two ratios are not comparable. In the case of linear pad bearings classical theory predicts a maximum load capacity while for journal bearings there is no maximum theoretical capacity, instead a limit is imposed by theoretical considerations. When cavitation effects are ignored, the friction coefficient for a bearing with the Half-Sommerfeld condition is: µ= 8Rc(1 − ε 2 ) 1.5 L 2 ε(0.621ε 2 + 1) 0.5 (4.118) · Lubricant Flow Rate For narrow bearings, the flow equation (4.18) is simplified since ∂p/∂x ≈ 0 and is expressed in the form: TEAM LRN HYDRODYNAMIC LUBRICATION 157 q x = Uh 2 (4.119) and the lubricant flow in the bearing is: Q x = ⌠ ⌡ 0 L q x dy = ⌠ ⌡ 0 L dy = Uh 2 UhL 2 Substituting for ‘h’ from (4.99), gives the flow in the bearing: Q x = UL 2 c(1 +εcosθ) (4.120) In order to prevent the depletion of lubricant inside the bearing, the lubricant lost due to side leakage must be compensated for. The rate of lubricant supply can be calculated by applying the boundary inlet-outlet conditions to equation (4.120). From a diagram of the unwrapped journal bearing film shown in Figure 4.34 it can be seen that the oil flows into the bearing at θ = 0 and h = h 1 and out of the bearing at θ = π and h = h 0 . Substituting the above boundary conditions into (4.120) it is found that the lubricant flow rate into the bearing is: Q 1 = UL 2 c(1 +ε) h 0 0 π 2π θ h 1 Lubricant inflow Lubricant leakage Lubricant outflow h 1 FIGURE 4.34 Unwrapped oil film in a journal bearing. and the lubricant flow rate out of the bearing is: Q 0 = UL 2 c(1 −ε) The rate at which lubricant is lost due to side leakage is: Q = Q 1 − Q 0 and thus: Q = UcLε (4.121) TEAM LRN 158 ENGINEERING TRIBOLOGY Lubricant must be supplied at this rate to the bearing for sustained operation. If this requirement is not met, ‘lubricant starvation’ will occur. For long bearings and eccentricity ratios approaching unity, the effect of hydrodynamic pressure gradients becomes significant and the above equation (4.121) loses accuracy. Lubricant flow rates for some finite bearings as a function of eccentricity ratio are shown in Figures 4.35 and 4.36 [8]. The data is computed using the Reynolds boundary condition, values for a 360° arc or complete journal bearing are shown in Figure 4.35 and similar data for a 180° arc or partial journal bearing are shown in Figure 4.36. 0 0.5 1.0 1.5 2.0 0 0.2 0.4 0.6 0.8 1.0 Eccentricity Non-dimensional side flow 2Q/ULc ε L D 1 2 = = 1 4 = 1 FIGURE 4.35 Lubricant leakage rate versus eccentricity ratio for some finite 360° bearings [8]. 0 0.5 1.0 1.5 2.0 0 0.2 0.4 0.6 0.8 1.0 Eccentricity Non-dimensional side flow 2Q/ULc ε L D 1 2 = = 1 4 = 1 FIGURE 4.36 Lubricant leakage rate versus eccentricity ratio for some finite 180° bearings [8]. Practical and Operational Aspects of Journal Bearings Journal bearings are commonly incorporated as integral parts of various machinery with a wide range of design requirements. Thus there are some problems associated with practical implementation and operation of journal bearings. For example, in many practical applications the lubricant is fed under pressure into the bearing or there are some critical resonant shaft speeds to be avoided. The shaft is usually misaligned and there are almost always some effects of cavitation for liquid lubricants. Elastic deformation of the bearing will certainly occur but this is usually less significant than for pad bearings. All of these issues will affect the performance of a bearing to some extent and allowance should be made during the design and operation of the bearing. Some of these problems will be addressed in this section and some will be discussed later in the next chapter on ‘Computational Hydrodynamics’. TEAM LRN HYDRODYNAMIC LUBRICATION 159 · Lubricant Supply In almost all bearings, a hole and groove are cut into the bush at a position remote from the point directly beneath the load. Lubricant is then supplied through the hole to be distributed over a large fraction of the bearing length by the groove. Ideally, the groove should be the same length as the bearing but this would cause all the lubricant to leak from the sides of the groove. As a compromise the groove length is usually about half the length of the bearing. Unless the groove and oil hole are deliberately positioned beneath the load there is little effect of groove geometry on load capacity. Circumferential grooves in the middle of the bearing are useful for applications where the load changes direction but have the effect of converting a bearing into two narrow bearings. These grooves are mostly used in crankcase bearings where the load rotates. Typical groove shapes are shown in Figure 4.37. The edges of grooves are usually recessed to prevent debris accumulating. D L d a) D b) c) d) l β b L D l 1 l 2 l L = l 1 + l 2 l D β L FIGURE 4.37 Typical lubricant supply grooves in journal bearings; a) single hole, b) short angle groove, c) large angle grove, d) circumferential groove (adapted from [19]). The idealized lubricant supply conditions assumed previously for load capacity analysis do not cause significant error except for certain cases such as the circumferential groove. The calculation of lubricant flow from grooves requires computation for accurate values and is described in the next chapter. Only a simple method of estimating lubricant flow is described in this section. With careful design, grooves and lubricant holes can be more than just a means of lubricant supply but can also be used to manipulate friction levels and bearing stability. Lubricant can be supplied to the bearing either pressurized or unpressurized. The advantage of unpressurized lubricant supply is that it is simpler, and for many small bearings a can of lubricant positioned above the bearing and connected by a tube is sufficient for several hours operation. The bearing draws in lubricant efficiently and there is no absolute necessity for TEAM LRN [...]... calculation methods be applied to check accuracy 30 ε 20 5° Groove extent β 10 ° 20° 1. 0 30° 40° 60° 0.9 10 90° 0 .8 0.7 f1 12 0° 0.6 5 0.5 15 0° 0.4 0.3 0.2 2 0 .1 1 0 .1 180 ° 0 0.2 0.5 1 2 5 10 20 f2 FIGURE 4.39 Parameters for calculation of pressurized oil flow from grooves (adapted from [19 ]) TEAM LRN HYDRODYNAMIC LUBRICATION 16 3 It should be noted that the pressurized flow of large angular extent bearings... from the equilibrium load position ε =1 ε =1 -1 circle -0.5 -1 -1 circle -0.5 0 -0.5 Unloaded position 0.5 1 -1 -0.5 0 0.5 0.5 1 0.5 Trajectory 1 1 Stable Unstable FIGURE 4.43 Example of computed shaft trajectories in journal bearings; stable condition, i.e declining spiral trajectory, and unstable condition, i.e self-propagating spiral trajectory (adapted from [ 51] ) Vibrational data is often collated... difference lying in the entry and exit boundary conditions In the full 360° bearing the entry condition is: h 1 = c (1 + ε) at θ=0 whereas in the partial bearing: h 1 ' = c (1 + εcosθ) at θ = 1 as shown in Figure 4.49 Line of centres h1 1 h 1' W Pressure profile FIGURE 4.49 Schematic representation of a partial bearing The practical analysis of such bearings is discussed in the next chapter Some results for... (adapted from [19 ]) Type of oil feed Single circular hole (dh < L / 2) Pressurised oil flow Qp = 0.675 ( ps hg3 dh + 0.4 η L ) 1. 75 [ ( ) [( ) ( ( Single rectangular groove with small angular extent Qp = ps hg3 1. 25 − 0.25(l/L) hg3 b/L + 0.333 η 3 (L/l − 1) 3 1 − l/L Single rectangular groove with large angular extent Qp = c3ps η Qp = π Dc3ps (1 + 1. 5ε2) 3η (L − l) (β < 5°) (5°< β < 18 0°) Circumferential... approximately constant direction then only part of a bearing arc is often employed The most common bearings of this type are 18 0° arc bearings, although TEAM LRN HYDRODYNAMIC LUBRICATION 17 1 narrower arcs are also in use The main advantage of partial bearings is that they have a lower viscous drag and hence lower frictional power losses Cavitation is also suppressed Partial arc bearings can be analysed by... Single rectangular groove with large angular extent Qp = c3ps η Qp = π Dc3ps (1 + 1. 5ε2) 3η (L − l) (β < 5°) (5°< β < 18 0°) Circumferential groove (360°) )] )] 1. 25 − 0.25(l/L) D/L f1 + f 0.333 6(L/l − 1) 6 (1 − l/L) 2 TEAM LRN 16 2 ENGINEERING TRIBOLOGY where: Qp is the pressurized lubricant flow from the hole or groove [m3/s]; ps is the oil supply pressure [Pa]; η is the dynamic viscosity of the lubricant... 4. 51 T T1 ∆T Tx T0 z h0 = h1 h1 x U x B FIGURE 4. 51 Temperature rise in a flat parallel bearing It is assumed that the temperature rises linearly across the film from zero to ‘∆T’ at the exit, so at any point ‘x’, the surface temperature ‘Tx’ is: Tx = ∆T x B The temperature gradient across the film is assumed to be linear (which is not always so in real bearings) and is: Tx ∆T x = B h h TEAM LRN 17 4 ENGINEERING. .. resistance to vibration because of the intrinsic stability of the Michell pad [23,24] P 1 Stability margin for bearing with two 90° sector grooves at 90° to load line Stability parameter 0.5 Lubricant supply grooves Circular bore journal bearings 0.2 0 .1 0.05 90° Unstable 0.02 0. 01 0 0 .1 0.2 0.3 0.4 0.5 0.6 0.7 0 .8 0.9 1. 0 ε Eccentricity FIGURE 4.44 Example of stability diagram for bearing vibration (adapted... diameter of the bush [m]; ε is the eccentricity ratio; f1, f2 are the coefficients determined from Figure 4.39 The grooves are centred on the load line but positioned at 18 0° to the point where the load vector intersects the shaft and bush The transition between ‘large angular extent’ and ‘small angular extent’ depends on the L/D ratio; e.g for L/D = 1, 18 0° is the transition point whereas for L/D ≤ 0.5... in Figure 4.45 · Rotating Load In the analysis presented so far, only steady loads, acting in a fixed direction have been considered There are, however, many practical engineering applications where the load TEAM LRN 16 8 ENGINEERING TRIBOLOGY ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; b) ;;;;;;;;; c) ;;;;;;;;; a) ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; . 1. 0 0. 01 0.02 0.03 0.04 0.05 0.06 0. 08 0 .1 0.2 0.3 0.4 0.5 0.6 0 .8 1 2 3 4 5 6 8 10 20 30 40 50 60 80 10 0 L D = ∞ = 1 1 2 = 1 4 = 1 8 = 0.02 0.03 0.04 0.05 0.06 0. 08 0 .1 0.2 0.3 0.4 0.5 0.6 0 .8 1 2 3 4 5 6 8 10 20 0.004 0.005 0.006 0.0 08 0. 01 1 8 Ocvirk W/L Uη ∆. check accuracy. 1 2 5 30 20 10  0 .1 0.2 0.5 1 2 5 10  20 1. 0 0.9 0 .8 0.7 0.6 0.5 0.4 0.3 0.2 0 .1 0 ε 5° 10 ° 20° 30° 40° 60° 90° 12 0° 15 0° 18 0° Groove extent β f 1 f 2 FIGURE. a 18 0° arc or partial journal bearing are shown in Figure 4.36. 0 0.5 1. 0 1. 5 2.0 0 0.2 0.4 0.6 0 .8 1. 0 Eccentricity Non-dimensional side flow 2Q/ULc ε L D 1 2 = = 1 4 = 1 FIGURE