Engineering Tribology Episode 2 Part 1 doc

COMPUTATIONAL HYDRODYNAMICS 225 North West East South ∆x ∆z W ∆z E T P z P z S (or z B ) T S (or T B ) z N (or z T ) T N (or T T ) T E T W z W z E FIGURE 5.12 Coded mesh of the control volume used in thermohydrodynamics [8]. The parameter ‘E’ is used to enforce flow directionality or convection into the controlling finite difference equation (5.52). There are two values of ‘E’ and the selection is based on the difference between lubricant flow ‘in’ and ‘out’ of the control volume. E = E 1 if E 1 > 0 and E = 0 if E 1 ≤ 0 (5.59) E 1 = |a E | + |a W | + |a N | + |a S | − |a P | (5.60) The inequality in equation (5.59) allows for reversal of flow which otherwise causes numerical instability. The subscripts for all the mass flow terms, e.g. (ρw) n , are lower case denoting that an average between velocities at the central and peripheral node is taken; for example: w n = 0.5(w N + w P ). ‘S p ’ and ‘S c ’ are terms representing viscous heating where allowance is made for the strong influence of temperature on viscosity. These two quantities are derived from the basic expression for viscous heating which is caused by the shearing of the lubricant: ) S = η ∂u ∂z ( 2 (5.61) where: S is the intensity of viscous heating [W/m 3 ]; The controlling equation for ‘S p ’ and ‘S c ’ is based on the assumption of a linear dependence of the heat source term on temperature: S = S c + S p T p (5.62) The precise forms of ‘S p ’ and ‘S c ’ are given by: S p = S η p dη dT p,old (5.63) TEAM LRN 226 ENGINEERING TRIBOLOGY ) S c = S T p η p ( 1 − dη dT p,old (5.64) where all terms are as calculated from the previous sweep of iteration for temperature and are referred to as ‘old’ values. The exponential viscosity law can be written as (Table 2.1): η p = η 0 e −γT p (5.65) where: η p is the predicted dynamic viscosity of the lubricant [Pas]; η 0 is the dynamic viscosity of the lubricant at some reference temperature ‘T 0 ’ [Pas]; γ is an exponent of viscosity-temperature dependence (typically γ = 0.05) [K -1 ]. or rearranged as: =−γη p dη p dT p (5.66) Substituting (5.66) into (5.63) and (5.64) yields: S p = −γS (5.67) ) S c = S T p η p ( 1 − dη dT p,old = S ( 1 +γT p ) (5.68) Treatment of Boundary Conditions in Thermohydrodynamic Lubrication As mentioned already, the boundary conditions necessary when viscous heating is modelled are considerably more complicated than in the isoviscous case. The finite difference equations presented are arranged so that they allow solution by methods appropriate to elliptic differential equations. This means that if iteration is used, the direction or order in which nodes are iterated does not affect the solution. If the equations were of a parabolic type then it would be necessary to iterate in the down-stream direction, i.e. apply a marching procedure but when reverse flow occurs this method generally fails. A marching procedure is a process of establishing nodal values in a specified sequence. The temperature boundary conditions at the interfaces of the hydrodynamic film vary according to the heat transfer mode of the bearing. It is thus necessary to modify the finite difference mesh to provide a means of solving the thermohydrodynamic equations for the specified boundary conditions. An example of the modified mesh is shown in Figure 5.13. The mesh can be applied to solve both the isothermal and adiabatic cases. If the isothermal bearing is studied, then the boundary conditions at the sliding surfaces simplify to a fixed temperature for the boundary nodes. Iteration is then confined to interior nodes without any need for extra arrays of imaginary nodes apart from at the outlet and inlet. TEAM LRN COMPUTATIONAL HYDRODYNAMICS 227 On the other hand, if an adiabatic bearing is to be analyzed, then the boundary condition at the pad surface changes to an unknown pad temperature but with a zero temperature gradient normal to the plane of the lubricant film. In this case it is necessary to invoke an imaginary array of temperature nodes above the pad surface with values of temperature maintained equal to the adjacent pad surface temperature. Iteration then includes temperature nodes at the interface between the pad and hydrodynamic film. Even for an adiabatic pad, the temperature on the pad at the bearing inlet which involves just one node, remains the same as the lubricant inlet temperature. INLET OUTLET Extra row of temperature nodes for adiabatic pad Extra row of nodes in velocity and temperature for outlet boundary condition Extra row of nodes for possibility of reverse flow PAD RUNNER F B B B F I I I I V V V FFF F V V V V V V V V V B Unknown value of temperature and velocity if backflow occurs F Fixed value I Fixed value for isothermal conditions; variable value for adiabatic pad V Always variable = = = = Temperature = inlet temperature Temperature = inlet temperature for all boundary conditions FIGURE 5.13 Example of the modified mesh with accessories for boundary conditions in thermohydrodynamic lubrication. Provided that ‘reverse flow’ does not occur, the temperatures at the bearing inlet are equal to the lubricant supply temperature. Temperatures at the outlet are unknown and can be calculated by applying the boundary condition ∂T/∂x = 0. The condition ∂T/∂x = 0 relates to the very slow change in temperature by cooling once the lubricant leaves the bearing exit. There will only be a negligible variation in ‘T’ with respect to ‘x’ compared to changes within the bearing where heat generation occurs. This condition can be accommodated by supplying an extra column of nodes with temperatures equal to the adjacent node's outlet temperature. The iteration procedure then includes the extra nodes at the bearing outlet. Wherever reverse flow at the inlet of the oil film occurs, temperatures are iterated on the boundary with the assumption that ∂u/∂x and ∂T/∂x remain constant across the boundary. This condition is met by another column of nodes up-stream of the bearing inlet which maintain the values of temperature and velocity calculated by linear extrapolation from node temperatures inside the bearing. Computer Program for the Analysis of an Infinitely Long Pad Bearing in the Case of Thermohydrodynamic Lubrication A computer program ‘THERMAL’ for the analysis of both the isothermal and adiabatic infinitely long pad bearings is listed and described in the Appendix. For bearings which are neither isothermal nor adiabatic an estimation of the effects of bearing heat transfer can be deduced from a comparison of data from the adiabatic and isothermal conditions which TEAM LRN 228 ENGINEERING TRIBOLOGY represents lower and upper limits of load capacity respectively. The program is based on a two-level iteration in temperature and pressure and its flow chart is shown in Figure 5.14. An initial constant temperature field equal to the oil inlet temperature is assumed and a pressure solution calculated from the resulting viscosity field. A new temperature field is then derived from the viscous shearing terms created by the pressure field. This new temperature field is then used to produce a second viscosity field which completes the first cycle of iteration. This iteration cycle is repeated until adequate convergence in the pressure field between successive iterations of temperature is reached. On completion of the iteration, pressure is integrated with respect to distance to obtain film force per unit length and the data is then printed to complete the program. Example of the Analysis of an Infinitely Long Pad Bearing in the Case of Thermohydrodynamic Lubrication The computer program ‘THERMAL’ described in the previous section provides a means of calculating the reduction in load capacity of a bearing due to heating effects. To demonstrate this effect the load capacity of a typical industrial bearing operating under conditions similar to those studied by Ettles [9] was analyzed by this program. The bearing parameters were chosen as typical of an industrial bearing. The selected values of controlling parameters were as follows: bearing width (i.e. length in the direction of sliding) 0.1 [m], maximum film thickness 10 -4 [m], minimum film thickness 5 ×10 -5 [m], lubricant viscosity temperature coefficient 0.05, lubricant specific heat 2000 [J/kgK], lubricant density 900 [kg/m 3 ] and lubricant thermal conductivity 0.15 [W/mK]. Two values of viscosity were considered, 0.05 [Pas] and 0.5 [Pas] at the bearing inlet temperature of 50 ° C. The performance of the bearing was studied over a range of sliding speeds from 1 to 100 [m/s] for the lower viscosity and 0.3 to 20 [m/s] for the higher viscosity. Sliding speed values used for computation were 0.3, 1, 3, 10, 20, 30 and 100 [m/s]. For higher speeds the computing time required to obtain convergence was unfortunately far too long for practical use. The calculated temperature distributions within an isothermal and adiabatic bearing are illustrated in Figure 5.15. The temperature fields were obtained for a lubricant inlet viscosity of 0.5 [Pas] and bearing sliding speed of 10 [m/s]. It can be seen from Figure 5.15 that the maximum temperature occurs at the outlet of the bearing. Start Acquire parameters Film thickness Sliding speed Pad width Viscosity Viscosity-temperature exponent Specific heat Thermal conductivity Density Inlet temperature Adiabatic or isothermal pad Special settings of iteration parameters? No Yes Acquire iteration parameters Initialize temperature, viscosity and pressure fields A Use preset values TEAM LRN COMPUTATIONAL HYDRODYNAMICS 229 End A Yes No Compute velocity field in direction of sliding U(I,K) Repeat of coefficient calculations for array of dummy nodes upstream of pads Solve temperature equation T(I,K) with relaxation Calculate reduced values of coefficients: AE/E, AW/E etc. Print out pressure, viscosity, temperature, U and W as fractions of maximum values; print out load Calculate M and N integrals Solve 1-D Reynolds equation for pad using current values of viscosity Iteration Compute velocity field normal to direction of sliding W(I,K) Compute coefficients of temperature iterations: AE , AW, AT, AB, S, SP, SC, B, AP etc. Allow for flow reversal in calculation of E Calculate T(I,K) residual Convergence of T(I,K) residual less than termination value? Number of sweeps over the limit? Update viscosity field VISC(I,K) with new temperature values using relaxation factor Calculate cumulative residual from pressure terms P(I) No Convergence of P(I) residual less than termination value? Number of sweeps over the limit? Find maximum values of pressure, viscosity, temperature, U and W; integrate P to find load Yes FIGURE 5.14 Flow chart of the program for the analysis of a thermohydrodynamic pad bearing. TEAM LRN 230 ENGINEERING TRIBOLOGY A strong effect of pad heat transfer on the temperatures inside the lubricant film is clear. The maximum temperature in the isothermal bearing is 71°C, compared to 116°C for the adiabatic bearing. The location of the maximum temperature is also different for these bearings. For the isothermal bearing the peak occurs close to the middle of the bearing. A small decline in temperature beyond this maximum is due to improved thermal conduction with reduced film thickness. The location of the peak temperature in the adiabatic bearing is at the down- stream end of the pad at the interface with the lubricant. The lubricant is progressively heated to higher temperatures as it passes down the bearing and the pad surface becomes very hot as it is remote from any source of cooling. Isothermal runner Adiabatic pad z T x Isothermal runner Isothermal pad z T x b)a) FIGURE 5.15 Computed temperature field in isothermal and adiabatic pad bearing at high sliding speed. The dependence between bearing load (defined as load per unit length divided by the product of sliding speed and viscosity) and sliding speed for both the adiabatic and isothermal bearing at two different viscosity levels is shown in Figure 5.16. Defining the bearing load as load per unit length divided by the product of sliding speed and viscosity allowed for comparison of the heating effects on lubricants of different viscosity and at various sliding speeds. At low sliding speeds close to 1 [m/s], the load parameter converges to a common value of about 640,000 [dimensionless]. This indicates that load under these conditions is proportional to the product of sliding speed and viscosity which agrees well with isoviscous theory of hydrodynamic lubrication. As the sliding speed is increased, the load parameter declines and heating effects are gradually becoming evident. It can be seen from Figure 5.16 that the threshold sliding speed at which decreases in load capacity from the isoviscous level become significant is lowered by the higher lubricant viscosity. At high sliding speeds, the rise in lubricant viscosity may not provide as large an increase in load capacity as might be expected. It can also be seen that an isothermal bearing has a higher load capacity than an adiabatic TEAM LRN COMPUTATIONAL HYDRODYNAMICS 231 bearing. Improvements in cooling of a real bearing can therefore bring an improvement in load capacity. 10 5 0.1 Sliding speed [m/s] Load (length × viscosity × speed) 0.2 0.5 1 2 5 10 20 50 100 2 × 10 5 3 × 10 5 4 × 10 5 5 × 10 5 6 × 10 5 7 × 10 5 [dimensionless] Adiabatic Isothermal 0.05 Pas 0.5 Pas FIGURE 5.16 Computed effect of lubricant heating on relative load capacity of a pad bearing. 5.6.2 ELASTIC DEFORMATIONS IN A PAD BEARING Almost all plain bearings operate with very small clearances and a requirement of nearly flat sliding surface. All bearings are made of material with a finite elastic modulus, if they deform or bend there may be a significant deviation from the optimum surface geometry considerably affecting the bearing performance. Pad bearings are particularly vulnerable to this phenomenon which is known in the literature as ‘crowning’. In a Michell bearing, the pad bends about the pivot point to form a curved or crowned shape which has a much lower load capacity than a rigid pad. This effect can become extreme at small film thicknesses, where even very limited deflections due to bending may severely distort the film geometry. To illustrate this problem a one-dimensional pad has been selected as an example since the relevant elastic deflections can be found from simple bending theory. The two-dimensional case would require the analysis of deflections in a plate which is far more complex [9]. Elastic deflections combined with thermohydrodynamic effects have also been analysed and a strong interaction between these effects has been found [9,11]. An example of a computer program ‘DEFLECTION’ for analysis of an elastically deforming one-dimensional pivoted Michell pad bearing is listed and described in the Appendix. Thermal effects, although significant, have been omitted in the program because of limitations of computing speed. The controlling equations of the bearing are the isoviscous Reynolds equation and the elastic deformation equation: = d 2 z' dx 2 M' EI (5.69) where: z' is the deflection of the pad in the ‘z’ direction [m]; TEAM LRN 232 ENGINEERING TRIBOLOGY M' is the local bending moment [Nm]; E is the elastic modulus of the pad material [Pa]; I is the second moment of area of the pad [m 4 ]. The pad is modelled as an infinitely long plate of uniform thickness so that ‘I’ (in terms of second moment of area per unit length) is a constant. The bearing load is assumed to be supported at the pivot. The pivot is located at the calculated centroid of hydrodynamic pressure. A two level iteration procedure is used in this analysis. The isoviscous hydrodynamic pressure field is first computed by iteration and then the bending moments are found and the resulting pad deflection calculated. The hydrodynamic pressure field is then re-iterated and a new series of pad deflections is found. The process is repeated until the pad deflections converge to sufficient accuracy. Hydrodynamic pressure is found from a finite difference equivalent of the one-dimensional isoviscous Reynolds equation. The one-dimensional isoviscous Reynolds equation (4.25) can be written as: ) U 0 h dp dx ( d dx − 1 6η )( d dx h 3 = 0 (5.70) or as: 6ηU 0 d 2 p dx 2 − h 3 = 0 dh dx − 3h 2 dh dx dp dx (5.71) The finite difference equivalent of this equation rearranged to give an expression for the nodal pressure value is: P i = h i 0.5(P i+1 + P i−1 ) + 0.75(P i+1 − P i−1 )δx dh dx − h i 3 3ηU 0 δx 2 )( i dh dx )( i (5.72) The finite difference equation (5.72) forms the basis of the iteration for pressure. Since cavitation due to extreme elastic deflection is also possible, even in a pad bearing, whenever this occurs the negative pressures are set to zero. The bending deflection equation is applied with the following boundary conditions: · the bending moment ‘M'’ and shear force normal to the pad ‘S’ are equal to zero at both ends of the pad bearing, · the pad is balanced at the pivot point and there are no other forms of support to the pad, · pad deflection and deflection slope dz'/dx are zero at the pivot. With these conditions, for x < x c where ‘x c ’ is the position of the pressure centroid, the expressions for ‘S’ and ‘M'’ are: S = ⌠ ⌡ 0 x pdx (5.73) M' = ⌠ ⌡ 0 x Sdx (5.74) TEAM LRN COMPUTATIONAL HYDRODYNAMICS 233 For x > x c the expressions for ‘S’ and ‘M'’ can be written as: S = ⌠ ⌡ 0 x pdx − ⌠ ⌡ 0 B pdx (5.75) M' = ⌠ ⌡ 0 x Sdx − (x − x c ) ⌠ ⌡ 0 B pdx (5.76) where ‘B’ is the width of the pad [m]. The deflection of the pad for all ‘x’ is found by integrating of (5.69) twice with respect to ‘z’ and is given by: z' = ⌠ ⌡ 0 x M'dx ⌠ ⌡ 0 x )( dx + C 1 x + C 2 (5.77) The constants ‘C 1 ’ and ‘C 2 ’ are: C 1 = − ⌠ ⌡ 0 x c M'dx (5.78) C 2 = x c ⌠ ⌡ 0 x c M'dx − ⌠ ⌡ 0 x c M'dx ⌠ ⌡ 0 x )( dx (5.79) Computer Program for the Analysis of an Elastically Deforming One-Dimensional Pivoted Michell Pad Bearing The flow chart of the computer program for the analysis of an elastically deforming one- dimensional pivoted Michell pad bearing is shown in Figure 5.17. A two level iteration in pressure and elastic deflection is conducted in order to determine the hydrodynamic pressure of a deformable pad bearing. Effect of Elastic Deformation of the Pad on Load Capacity and Film Thickness The computer program ‘DEFLECTION’ described above can provide useful information for the mechanical design of hydrodynamic bearings. For instance, the effect of pad thickness and elastic modulus of pad material on the load capacity can be assessed with the aid of this program. The effect of pad thickness on load capacity is demonstrated as an example of possible applications of this program. It is of practical importance to know how thick the bearing pad should be to provide sufficient rigidity for a particular size of bearing and nominal film thickness. A reduction in the hydrodynamic film thickness can increase load capacity but at the same time it also increases bearing sensitivity to elastic distortion. Optimization of bearing characteristics is therefore essential to the design process. The computed load capacity of a bearing of 1 [m] pad width, lubricated by a lubricant of 1 [Pas] viscosity versus pad thickness is shown in Figure 5.18. The Young's modulus of the pad's material is 207 [GPa]. The hydrodynamic film thickness is 2 [mm] at the inlet and 1 [mm] at the outlet of the pad. TEAM LRN 234 ENGINEERING TRIBOLOGY Start Acquire parameters Maximum film thickness Minimum film thickness Sliding speed Bearing width Lubricant viscosity Pad thickness Pad elastic modulus Special settings of iteration parameters? No Yes Acquire mesh and iteration parameters Calculate mesh parameters and pad stiffness End Integrate pressure to find shear force Print out pressures, deflections and load capacity Calculate DHDX from film thickness Solve 1-D Reynolds equation for given geometry Iteration Double integration of bending moment to find deflection No Is residual deformation small enough? Use preset values Calculate initial film thickness; set pressure and deflections to zero Integrate shear force to find bending moments Calculate deflection with relaxation factor between current and previous iteration Calculate new film thickness = undeformed film thickness + new deflection Yes FIGURE 5.17 Flow chart of program to compute load capacity of an elastically deforming one- dimensional pivoted pad bearing. It can be seen from Figure 5.18 that as the pad thickness is reduced from 200 [mm] to 30 [mm] the load capacity declines by 70%. When the thickness of the pad is 200 [mm], then the load capacity is identical to that of the rigid pad. At the pad thickness of 100 [mm], load capacity is only reduced by about 10% as compared to a rigid pad. It can thus be concluded that 100 [mm] is close to the optimum pad thickness for this particular bearing. The relationships between TEAM LRN [...]... TEAM LRN 24 4 ENGINEERING TRIBOLOGY 15 2. 5 L/D = 0 .25 10 Load 2 5 Dimensionless side flow Dimensionless load Dimensionless groove pressure = 0.05 Side flow 1. 5 0 0 .1 0 .2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 Groove length/axial bearing length FIGURE 5 .25 Dimensionless load and side flow versus relative groove length for L/D = 0 .25 , p* groove = 0.05 and groove subtended angle 36° 4 1. 3 L/D = 1 Load 3 1 .2 2 Side... HYDRODYNAMICS 23 5 the film thickness and pressure for pads of 10 0 [mm] and 30 [mm] thickness are shown in Figures 5 .19 and 5 .20 respectively 20 0 Load per unit length [kN/m] Perfectly rigid pad 10 0 Steel pad 0 0 0 .1 0 .2 Pad thickness [m] FIGURE 5 .18 Computed effect of pad thickness on the load capacity of a Michell pad bearing 2 Straight reference line Pressure Film thickness 1. 5 10 0 0 0 0 .1 0 .2 0.3 0.4... 1 .2 2 Side flow 1. 1 1 1 0 0 .1 0 .2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Dimensionless side flow Dimensionless load Dimensionless groove pressure = 0 .2 0 Groove length/axial bearing length FIGURE 5 .26 Dimensionless load and side flow versus relative groove length for L/D = 1, p* groove = 0 .2 and groove subtended angle 36° Lubricant flow in bearings can be quite large It can be seen from Figure 5 .25 that for a... of L/D = 1, eccentricity ratio of 0.7, is shown in Figures 5 .27 and 5 .28 The relative groove length is equal to 0 .2 and groove subtended angle is 72 The perfectly aligned case is shown in Figure 5 .27 whereas the case with an extreme misalignment of 0.5 is shown in Figure 5 .28 P% 10 0% Pressure along the load line L 18 0 90 0.8 L 0.6 L 0 0.4 L -90 0 .2 L st ree g De -18 0 ine dl oa ol FIGURE 5 .27 Pressure... flow is about 6.8 Assuming that the bearing entraining velocity is U = 10 [m/s], bearing length L = 0 .2 [m] and the radial clearance of the bearing is c = 0.0004 [m], then from equation (5.87) the value of flow ‘Q’ is: Q = 0.5 × 6.8 × 0 .2 × 10 × 0.0004 = 2. 72 × 10 -3 [m3 /s] = 2. 72 [litres/s] TEAM LRN COMPUTATIONAL HYDRODYNAMICS 24 5 It is evident that in some cases hydrodynamic bearings can require... grooved perfectly aligned journal bearing of L/D = 1 and eccentricity ratio of 0.7 (not to scale) 10 0% P% Pressure along the load line L 18 0 90 0.8 L 0.6 L 0 0.4 L -90 0 .2 L -18 0 st ree g De ine dl oa ol FIGURE 5 .28 Pressure profile of grooved misaligned journal bearing of L/D = 1 and eccentricity ratio of 0.7 (not to scale) TEAM LRN 24 6 ENGINEERING TRIBOLOGY Misalignment has surprisingly little effect... 0 .25 and L/D = 1 are shown in Figures 5 .25 and 5 .26 respectively The eccentricity ratio is assumed to be constant and equal to 0.8 It is also assumed that there is no misalignment and the subtended angle of each groove is 36° The mesh assumed for computation has 21 nodes in the ‘x*’ direction and 11 nodes in the ‘y*’ direction The pre-set dimensionless groove pressure is equal to 0.05 for L/D = 0 .25 ... effective load bearing area to shrink and the load capacity declines even if specific hydrodynamic pressures remain high 2. 5 2 Pressure Film thickness 50 1. 5 0 0 0 .1 0 .2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Film thickness [mm] Pressure [kPa] 10 0 1 Distance from bearing inlet [m] FIGURE 5 .20 Effect of elastic deflection on the hydrodynamic film thickness and pressure profile for a 30 [mm] thick pad From the... variation in film thickness for the static case static The modified forms of ‘h*’ and ∂ 2 h * /∂x * 2 which are required for the Vogelpohl equation follow the scheme already described and are given by: h* = h* + ∆x*cos(x*) + ∆y*sin(x*) static ( ) (5.90) [ ∂2h* ∂2h* 2 + ∆x*cos(x*) + ∆y*sin(x*) 2 = 2 ∂x* ∂x* static ∂x *2 ] (5. 91) The Vogelpohl equation (5.4) is then solved in terms of the modified forms of... reformation and cavitation fronts after traversing grooves Equalize M(I,J) at x* = 0 and x* = 2 Is iteration in M(I,J) completed? No Yes No Calculate P(I,J), integrate to find load components and attitude angle Is calculated attitude angle close enough to x* = π and N2 within limits? Yes B TEAM LRN 24 1 24 2 ENGINEERING TRIBOLOGY B Find maximum pressure Calculate leakage flow and groove flow ∂p* ∂p* by integration . 36°. 1 1. 3 0 0 .1 Groove len g th/axial bearin g len g th Dimensionless load 1 .2 0 .2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 4 1 Dimensionless side flow Load Side flow 2 3 1. 1 L/D. most cases. TEAM LRN 24 4 ENGINEERING TRIBOLOGY 1. 5 2. 5 0 0 .1 Groove len g th/axial bearin g len g th Dimensionless load 2 0 .2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 15  5 Dimensionless. viscosity × speed) 0 .2 0.5 1 2 5 10 20 50 10 0 2 × 10 5 3 × 10 5 4 × 10 5 5 × 10 5 6 × 10 5 7 × 10 5 [dimensionless] Adiabatic Isothermal 0.05 Pas 0.5 Pas FIGURE 5 .16 Computed effect of lubricant