NUMERICALMETHODS IN HYDRODYNAMIC LUBRICATION J. H. Vohr INTRODUCTION Analysis of the performance of fluid-film bearings involves the solution of Reynolds equation, derived previously in the chapter on Hydrodynamic Lubrication (Volume II). For an incompressible lubricant, the steady state form of this equation is (1) While there are many analytical solutions to this equation for relatively simple geometries, the usual method of solution is by numerical analysis, employing either finite difference or finite element techniques. 1-3 The following provides a brief description of these procedures. FINITE DIFFERENCE SOLUTIONS Assuming constant lubricant viscosity, Equation 1 can be rewritten (2) The purpose of finite difference approximations is to reduce a continuous differential equation, such as Equation 2, to a series of algebraic equations that can be solved for the pressure distribution P(x,y) for a known distribution of h(x,y) and µ(x,y). First, the distribution of all functions in all coordinates concerned are replaced by samplings at several discrete points. Usually the spacing between points along each coordinate axis is chosen to be constant, but uneven spacing may be more suitable for some special applications. The governing differential equations are then written at each point of the coordinate grid with suitable approximations for the functions and their derivatives. Most common of such approximations are central difference formulas applicable to a uniform grid spacing as shown in Figure 1. (3) (4) (5) (6) (7) (8) (9) (10) Substituting these approximations in Equation 2, one obtains the following algebraic equation in the unknown pressure P(i,j), P(i + l,j); P(i – l,j), P(i,j + 1), and P(i,j – 1) at each i,j grid point. Volume II 93 93_104 4/11/06 12:09 PM Page 93 Copyright © 1983 CRC Press LLC major methods will be discussed here: relaxation, direct matrix inversion, and columnwise influence coefficients. Relaxation This method reduces all equations in the system to equations in only one unknown only which can be solved immediately. All other unknowns in each equation are considered to be known and equal to some previously computed value. Therefore, an initial distribution is guessed and then successively improved by solving the system of equations in one un- known. As an example, consider the square inclined slider thrust pad shown in Figure 2. For the given geometry, ∂h/∂x = Δh/Δx = 0.001 and ∂h/∂y = 0 throughout the film. Applying Equation 11 at the i = 2, j = 2 coordinate point and assuming μ = 0.01 Pa- sec and U = 1000 cm/sec. Volume II 95 FIGURE 2. Notation for inclined slider pad. (14) 93_104 4/11/06 12:09 PM Page 95 Copyright © 1983 CRC Press LLC Writing Equation 11 at the three remaining interior points yields four equations such as Equation 14 in which P(i,j) is expressed in terms of its neighbors. Pressures on the boundaries (i = 1, j = 1, i = 4, and j = 4) are set equal to zero. With this boundary condition imposed, Equation 14 may be rewritten P(2,2) = 1/4 [P(3,2) + P(2,3) ] + 3/8 10/0.03 P(3,2) −6/4 1/(0.03) 3 (15) Similarly, at the i = 2, j = 3 grid point P(2,3) = 1/4 [P(3,3) + P(2,2) ] + 3/8 10/0.03 P(3,3) −6/4 1/(0.03) 3 (16) and so forth for i = 3, j = 2 and i = 3, j = 3. Starting with some assumed initial distribution for the interior pressures P(i,j) [P(i,j) = 0 everywhere is an adequate choice], one can solve Equation 15 for P(2,2). This “updated” value for P(2,2) may now be used in Equation 16, together with the initially assumed zero for the rest of the pressures, to obtain an “updated” value for P(2,3). One proceeds in this fashion up the i = 3 grid columns, solving for P(3,2) and P(3,3), always using the most recently updated values of surrounding pressures when solving for P(i,j). Successive sweeps through the finite difference grid continue until successive interior pressures P(i,j) converge to a final distribution. Typically, 30 sweeps may be required for convergence. Several improved procedures of applying relaxation methods exist such as “overrelaxa- tion”, which strives to accelerate the convergence process by multiplying the change in pressure from one sweep to the next by a factor such as 1.2. A drawback of relaxation procedures is their proneness to numerical instability. This phenomenon sets in depending on the value of the coefficients, and of the first guess distribution. As seen in the example above, the equation written for point (i,j) is solved for P(i,j). Therefore, the value of P(i,j) is corrected at each iteration by the influence of its immediate neighbors. As a consequence, the influence of each point propagates one grid interval each iteration and the number of iterations involved in achieving steady-state solutions is pro- portional to the number of points on the longest side of the grid. In addition, the number of operations is proportional to the total number of grid points N × M. Therefore, the total time needed for converging to a steady-state solution is proportional to N 2 × M, where N has been assumed to be larger than M. For computation speed, the relaxation method has advantages over the others for large values of N and M. As a general rule, however, the stability of a relaxation solution decreases when the number of grid points is increased. Slowing the pace of the solution (“underrelaxation”) usually circumvents the problem but obviates the speed advantages for large grids. Since relaxation requires minimum internal computer capacity, this method still finds occasional use with a desk-top computer. Direct Matrix Inversion In this simple method, all equations are written together as a system. Letting L = N × M, a 11 q 1 + a 12 q 2 + … a 1L q L = r 1 a 21 q 1 + a 22 q 2 + … a 2L q L = r 2 . . . . . . . a L1 q 1 + a L2 q 2 + … a LL q L = r L (17) where 96 CRC Handbook of Lubrication 93_104 4/11/06 12:09 PM Page 96 Copyright © 1983 CRC Press LLC Obviously, most of the “a’s” are zero. The system (Equations 17) may be written as in matrix notation as [ a ] { q } = { r } (18) By methods such as Gaussian inversion, the problem can be solved as { q } = [ a ] –1 { r } (19) Inversion of matrix [a] is performed by routines available at any computer installation. If this method were economical in computation time, it would be the most desirable since it solves the linear system of algebraic equations with a minimum of bookkeeping compli- cations. Unfortunately, the computation time involved in performing the key matrix inversion step grows proportionately to L 3 or M 3 × N 3 which is exorbitant except for very small grids. Columnwise Influence Coefficients This method is similar to matrix inversion, but takes advantage of some features of the system of equations. If we denote the j th column of unknown pressure as { P(i) } j i=1,M then we can see by Equation 11 that the equations involving this j th column vector also involve only the adjacent column vectors <P(i)> j–1 and <P(i)> j+1 . These equations can be written in matrix form as [ A ] j { P } j + [ B ] j { P } j–1 + [ C ] j { P } j+1 = { R } j (20) where [A] j , [B] j , and [C] j are matrices which are M × M in size, M being the total number of grid points in the i direction, i.e., the “height” of each column vector <P> j. Matrices [A] j , [B] j , and [C] j can be determined from Equation 11. At the start of the bearing film, i.e., at j = 1, pressure<P>is known and equal to the boundary pressure P a . Consequently, Equation 20 at j = 2, gives [ A ] 2 { P } 2 + [ B ] 2 { P a } + [ C ] 2 { P } 3 = {R} 2 (21) Since <P a > is a column vector, each element of which is the known boundary pressure P a , this equation can be written in the form { P } 2 = [ E ] 3 { P } 3 + { F } 3 (22) where (23) If next written at j = 3, Equation 20 will involve unknown column pressure <P> 2 , Volume II 97 93_104 4/11/06 12:09 PM Page 97 Copyright © 1983 CRC Press LLC <P> 3 , and <P> 4 . By using Equation 22 to eliminate <P> 2 from this equation, one can obtain an equation containing only the unknown column pressures <P> 3 and <P> 4 . We can write this equation formally like Equation 22, i.e., { P } 3 = [ E ] 4 { P } 4 + { F } 4 (24) By continuing this procedure, we can successively write equations of the form { P } j–1 = [ E ] j { P } j + { F } j (25) for all values of j from j = 3 to j = N. Moreover, Equation 26 also holds at j = 2 if we define [E] 2 = 0 and <F> 2 = <P a >. Substituting Equation 25 into Equation 20 we obtain ([ A ] j +[ B ] j [E] j ) { P } j = [ C ] j { P } j+1 +{ R } j − [ B ] j { F } j (26) Comparing Equation 26 with Equation 25 we can deduce the recursion relationship for determining [E] j and <F> j . [E] j+1 = –[T] j [C] j {F} j+1 = [T] j ( {R} j – [B] j {F} j ) [T] j = ([A] j + [B] j [E] j ) –1 (27) The solution procedure for determining the unknown column pressures <P> j goes as follows: 1. Starting with the initial values [E] 2 = 0 and <F> 2 = <P a > , use the recursion relationships (Equation 27) to determine [E] j and <F> j for all values of j from 2 to N. 2. At the end of the bearing (j = N), set <P> N equal to the known boundary pressure <P a > . 3. Use Equation 25 successively from j = N to j = 2 to determine all column pressures <P> j . While the matrices which must be inverted here are of the order M × M, in the direct inversion method the matrix is of the order M × N. If M is chosen to be significantly smaller than N, computation cost of the columnwise method will be significantly less than that of the direct method. A more detailed description of the columnwise solution method, including a discussion of cyclical boundary conditions, is given in Reference 1. Calculation of Flow and Power Loss Once the discretized pressure field is obtained, quantities such as bearing flow and power loss may be obtained from approximate finite difference expressions. For example, lubricant flow per unit width in the direction of sliding at the i,j grid point would be evaluated by the following central difference expression (28) 98 CRC Handbook of Lubrication 93_104 4/11/06 12:09 PM Page 98 Copyright © 1983 CRC Press LLC At the leading edge, i.e., at i = 1, a forward difference formula must be used (29) Similarly, a backwards difference formula must be used at the trailing edge. Power loss is obtained by multiplying surface shear stress τ s , acting on the surface by surface velocity. In finite difference form (30) Multiplying τ S U by ΔxΔy and summing over the entire i,j grid, will yield the bearing power loss. FINITE ELEMENTANALYSIS An alternate approach for solving Reynolds’equation is use of finite element analysis. Developed originally for use in structural analysis, this method has been finding increasing favor in the analysis of fluid film bearings. The method is particularly advantageous for complex film geometries since the elements into which the solution field is broken do not have to be uniform or follow coordinate lines. Descriptions of the finite element method are given by Reddi, 4 Booker and Huebner, 5 Allaire et al., 6 and Hays. 7 In this present discussion, the method will be described in its simplest form by solving for the pressure in a one-dimensional, step slider bearing following the sequence presented by Reddi. 4 Reddi presents a proof that if a functional φ(P) is given by (31) integrated over the region R, comprising a lubricant film, then δ φ (P) = 0(32) if and only if Pis a solution of the incompressible lubrication problem. In Equation 31, µ, h, and Pare the lubricant film viscosity, thickness, and pressure, respectively, is the vector velocity of the bearing surface, and q represents the outflow flux of lubricant from the region R in in. 3 /sec/in. (m 2 /sec/m) across the boundary C 2 . In Equations 32 and 33 pressure Pis described by a set of functions which are continuous over R and which satisfy prescribed boundary conditions. In the finite element method, region R is broken up into small elements and the distribution of pressure Pover each element is described by convenient interpolation functions particular to the element. For example, triangular elements are convenient for two-dimensional solutions as shown in Figure 3a. The most convenient interpolation function is a linear one, i.e., the pressure P m over the m th triangular element is assumed to vary as P m (x,y) = a m1 + a m2 x + a m3 y (33) By evaluating pressure P m at the three vertices or nodes of each element, one can solve for constants a m1 , a m2 , and a m3 in terms of the three unknown node pressures P m1 , P m2 , P m3 at Volume II 99 93_104 4/11/06 12:09 PM Page 99 Copyright © 1983 CRC Press LLC b and lengths ᐉ 1 and ᐉ 2 , respectively. Since P is a function of x only, the finite elements are linear rather than triangular. The pressures of the nodes of element 1 are P A and P B while the pressures at the nodes of element 2 are P B and P C . Using a linear interpolation function for the pressure distributions over elements 1 and 2 P 1 (x) = P A + x/l 1 (P B – P A ) (34) P 2 (x′) = P B + x′/l 2 (P C – P B ) (35) We can write these equations in the general matrix form used in Reference 4, P m (x) =T m P (36) where P m (x) represents the variation of P over element m. P represents the column vector of the unknown node pressures (37) and T m represents the appropriate row vector relating P m (x) to P. In our simple example T 1 = { (1 −x/l 1 )x/l 1 0 } T 2 = { 0 (1 −x′/l 2 )x′/l 2 } (38) By differentiating Equation 36, one obtains the gradient of P m (x). For two-dimensional problems, the gradient would be a column vector, i.e., (39) The matrix formed from differentiating the row vector T m first with respect to x and then with respect to y is denoted as R m . In our case, ∂T m / ∂ y is zero so R m reduces to a row vector, where (40) Since we now have expressions for P and P over each element m, the integral over R in Equation 31 can be written as a summation of integrals over each element m. Using the general notation of Reference 4, Equation 31 becomes (41) In our simple one-dimensional case, the element property term C m is purely the scalar quantity h m 3 /12µ and the velocity U m is simply the velocity U in the x direction. In the more Volume II 101 93_104 4/11/06 12:09 PM Page 101 Copyright © 1983 CRC Press LLC general two-dimensional case described in Reference 4, C m is a matrix and U m is a vector. Equation 41 can be restated as (42) where K is the volumetric fluidity matrix, and V, H, and Q are the element flow, squeeze film, and boundary flow column vectors, respectively, given by: (43) Let us now evaluate K, V, and Q for our example (H is zero for this steady-state case). For elements 1 and 2, from Equation 41 (44) (45) Substituting these into the expression for K, integrating over each element, and then summing over the elements: (46) For the element flow vector V, (47) 102 CRC Handbook of Lubrication 93_104 4/11/06 12:09 PM Page 102 Copyright © 1983 CRC Press LLC Substituting these into the expression for V, integrating over each element, and summing (48) Finally, for Q, flows q 1 and q 2 leaving the bearing per unit width at the leading and trailing edge boundaries, respectively, are (49) Using expressions (Equation 38) for T 1 and T 2 , x = 0 at the leading edge boundary and x′ = ᐉ 2 at the trailing edge: (50) Substituting these expressions into Equation 45 for Q, integrating along the leading and trailing edge boundaries, and summing (51) Now, Equation 32 states that if the pressure P is a solution of the incompressible lubrication problem, then δφ(P) = 0 over the region R (52) where P i are the unknown pressures at the nodes of the elements. Since δP i is an independent variation, it follows that ∂ϕ/∂P i = 0 for each i. From Equation 42 (53) or (54) Equation 54, in general, yields a system of simultaneous equations for the nodal pressures P which may be solved by any of a number of established techniques. In our simple example, Equation 54 becomes (55) Volume II 103 93_104 4/11/06 12:09 PM Page 103 Copyright © 1983 CRC Press LLC [...]... example are as follows: Bearing A Bearing B W, N β kh β kh 82,000 70,000 60,300 50 ,000 30,000 10 ,000 — — 1. 00 83 50 17 — — 1. 00 1. 07 1. 26 1. 80 1. 00 85 74 61 37 12 1. 00 1. 06 1. 11 1 .19 1. 41 2. 05 Film thickness is calculated from the following relation from Reference 2: The load capacity for each bearing is an inverse function of the film thickness cubed When the load increases to the value at which the... 12 0 18 0 200 (3.94) (4.72) (7.09) (7.87) 50 70 18 0 200 (1. 97) (2.76) (7.09) (7.87) kv = Flow, ᐉ/sec (gpm) R1,/R4 (R3 – R2)/R4 − R1) 0.30 1/ 2 0.60 (4.76) 0. 45 1/ 4 0.73 (7 .13 ) From Figure 7 for annular thrust pad bearings: Copyright © 19 83 CRC Press LLC Volume II Bearing A Bearing B 0.80 10 .0 12 .4 16 .58 × 10 −6 W af (Load coeff.) qf (Flow coeff.) Hf (Power coeff.) β = W/afApPs 11 3 0.87 7 .5 8.6 12 .2 × 10 −6... units as follows: H = 0.0 25 (56 × 10 6) /10 00 (0 .5) = 2800 watts In customary U.S units, the power calculation would be: H = 0.4(2 31) (8000)/60 (12 ) (55 0)(0 .5) = 3.73 hp If all this power is converted to heat in the small oil flow, the resulting temperature rise of the fluid would be 69°C (12 4°F) With a reservoir temperature of 38°C (10 0°F) this would result in an oil temperature of 10 7°C (224°F) leaving the...93 _10 4 4 /11 /06 12 :09 PM 10 4 Page 10 4 CRC Handbook of Lubrication To solve Equation 55 , conditions must be supplied for boundary pressures PA and PC For the simplest case where PA = PC = 0, the set of three equations represented by Equation 55 reduces to the following single equation for PB (56 ) The above example illustrates the steps involved in... F, 99(2), 18 7, April 19 77 7 Hays, D F., A variational approach to lubrication problems and the solution of the finite journal bearing, J Basic Eng., Trans ASME, 81( 1), 13 , March 19 59 8 Castelli, V and Malanoski, S B., Method for solution of lubrication problems with temperature and elasticity effects: application to sector, tilting-pad bearings, J Lubr Technol., Trans ASME, 91, 634, October 19 69 9 Nicholas,... 11 1 11 2 CRC Handbook of Lubrication FIGURE 7 Bearing pad coefficients for annular thrust pad bearing Annular recess is centrally located within bearing width (R1 + R4 = R2 + R3) Curve for af applies to all R1/R4 ratios (Reproduced from Rippel, H C., Cast Bronze Hydrostatic Bearing Design Manual, Courtesy of Cast Bronze Bearing Institute, Chicago.) Bearing A mm Bearing B (in.) mm (in.) R1 R2 R3 R4 10 0... Reddi, M M., Finite element solution of the incompressible lubrication problem, J Lubr Technol., Trans ASME, Ser F, 91( 3), 52 4, July 19 69 5 Booker, J F and Huebner, K H., Application of finite elements to lubrication; an engineering approach, J Lubr Technol., Trans ASME, Ser F, 24(4), 313 , October 19 72 6 Allaire, P E., Nicholas, J C., and Gunter, E J., Systems of finite elements for finite bearings,... hydrostatic steadyrest of a large lathe, for example, a bearing with a geometry resembling Figure 1 supports a heavy steel forging near midspan The dimensions are a = 10 0 mm (3.94 in.) and b = 25 mm (0.98 in.) Oil is supplied at a rate of 0.0 25 l/sec (0.4 gpm) and a supply pressure of 56 MPa (8000 psi) by a fixed-displacement pump with an efficiency of 0 .5 The oil has a specific heat of 890 J/l (3.2 Btu/gal)... and damping coefficients for the five-pad tilting pad bearing, ASLE Trans., 22(2), 11 3, April 19 79 10 Nicholas, J C., Allaire, P E., and Lewis, D W., Stiffness and damping coefficients for finite length step journal bearings, ASLE Trans., 23(4), 353 , October 19 80 Copyright © 19 83 CRC Press LLC Volume II 10 5 HYDROSTATIC LUBRICATION R C Elwell INTRODUCTION In contrast to the hydrodynamic bearings described... equations is a linear function of load (Equation 8) By equating each restrictor flow equation with the flow equation for this bearing (Equation 9), load capacity expressions can be derived as below Orifice Restriction (10 ) Constant Flow (11 ) Copyright © 19 83 CRC Press LLC Volume II FIGURE 5 Effect of restrictor type on stiffness of a thrust bearing (Reference 8) FIGURE 6 Copyright © 19 83 CRC Press LLC Double-acting . M, a 11 q 1 + a 12 q 2 + … a 1L q L = r 1 a 21 q 1 + a 22 q 2 + … a 2L q L = r 2 . . . . . . . a L1 q 1 + a L2 q 2 + … a LL q L = r L (17 ) where 96 CRC Handbook of Lubrication 93 _10 4 4 /11 /06 12 :09. 10 .0 7 .5 H f (Power coeff.) 12 .4 8.6 β = W/a f A p P s 16 .58 × 10 −6 W 12 .2 × 10 −6 W Bearing A Bearing B W, N β k h β k h 82,000 — — 1. 00 1. 00 70,000 — — . 85 1. 06 60,300 1. 00 1. 00 .74 1. 11 50 ,000. 1. 00 70,000 — — . 85 1. 06 60,300 1. 00 1. 00 .74 1. 11 50 ,000 .83 1. 07 . 61 1 .19 30,000 .50 1. 26 .37 1. 41 10,000 .17 1. 80 .12 2. 05 Copyright © 19 83 CRC Press LLC