Hydrodynamic Lubrication 2009 Part 4 pps

3.4 Finite Length Bearings 45 ∂ ∂x  h 3 ∂p ∂x  = h 3 i+1/2, j p i+1, j − p i, j ∆x − h 3 i−1/2, j p i, j − p i−1, j ∆x ∆x (3.69) ∂ ∂z  h 3 ∂p ∂z  = h 3 i, j+1/2 p i, j+1 − p i, j ∆z − h 3 i, j−1/2 p i, j − p i, j−1 ∆z ∆z (3.70) ∂h ∂x = h i+1/2, j − h i−1/2, j ∆x (3.71) Substituting these into Eq. 3.68 and solving for p i, j , we obtain p i, j in the following form: p i, j = a 0 + a 1 p i+1, j + a 2 p i−1, j + a 3 p i, j+1 + a 4 p i, j−1 (3.72) (i = 1, 2, ···, m − 1; j = 1, 2, ···, n − 1) where p i, j is the pressure at the nodal point (i, j) and a 0 , a 1 , a 2 , a 3 , and a 4 are con- stants given at the respective nodal point. If the number of nodal points at which the pressure is to be calculated is N, then we have N equations of the form of Eq. 3.72. The boundary conditions are given as the pressure at the points on the boundary. By solving these simultaneously, the pressures at respective nodal points are obtained. One of the methods to solve these simultaneous equations is elimination, typ- ically Gauss’ elimination method. Pressure at all the nodal points can thereby be found in a finite number of operations. Further, if we sweep the lubricating domain two dimensionally with the calcu- lation of Eq. 3.72 at each nodal point in consecutive order starting from a suitable nodal point, and if this is repeated a sufficient number of times, then it is expected that the pressure obtained at each nodal point gradually approaches the true value of the pressure. This is called the iterative method (successive approximation method). In this case, the calculation will be repeated until the following relation is satisfied:  i, j |(p i, j ) k − (p i, j ) k−1 |  i, j |(p i, j ) k | < (3.73) where (p i, j ) k are pressures obtained in the kth calculation, (p i, j ) k−1 are those in the previous calculation, and  is a sufficiently small allowable error. The pressure (p i, j ) k obtained in the kth calculation will be a solution p i, j . For a high accuracy of calcula- tion, a value of 10 −6 –10 −12 , for example, is used for . Further, the convergence in the region of rapid pressure rise can be improved by introducing the following new variable ¯p  into Reynolds’ equation Eq, 3.68 [2]. ¯ h and ¯p are the nondimensional film thickness and nondimensional pressure in Eq. 3.67, respectively: ¯p  = ¯ h m ¯p (3.74) A value of m = 2or1.5, for example,is used [6][13]. 46 3 Fundamentals of Journal Bearings If pressure p i, j is obtained, the component of the oil film force P in the eccen- tricity direction and that in the direction normal to it, respectively, are calculated as follows: P κ = P cos θ = R  i, j p i, j cos φ i ∆z∆φ (3.75) P θ = P sin θ = R  i, j p i, j sin φ i ∆z∆φ (3.76) where ∆x = R ∆φ. References 1. A. Sommerfeld, “Zur hydrodynamischen Theorie der Schmiermittelreibung”, Zeit. angew. Math. u. Physik, Vol. 50, 1904, pp. 97 - 155. 2. G. Vogelpohl, “Beitr ¨ age zur Kenntniss der Gleitlagerreibung”, VDI - Forschungsheft 386, Berlin, 1937. 3. G.B. DuBois and F.W. Ocvirk, “Analytical Derivation and Experimental Evaluation of Short Bearing Approximation for Full Journal Bearings”, NACA Tech. Report 1157, Washington D.C., 1953 4. T. Someya, “Stabilit ¨ at einer in zylindrischen Gleitlagern laufenden, unwuchtfreien Welle”, Ingenieur-Archiv, 33. Band, 2. Heft, 1963, pp. 85 - 108. 5. N. Soda, “Bearings” (in Japanese), Iwanami Zensho Series, Iwanami Shoten, Tokyo, 1964. 6. T. Someya, “Das dynamisch belastete Radial-Gleitlager beliebigen Querschnitts”, Ingenieur-Archiv, 34. Band, 1. Heft, 1965, pp. 7 - 16. 7. H. Mori, S. Miyata, Y. Abe and Y. Fujita, “Research on Discountinuity of Lubricant Film in Journal Bearing (First Report, Bearing Characteristics and Oil Film Rupture)” (in Japanese), Trans JSME, Vol. 33, No. 248, April 1967, pp. 658 - 666. 8. H. Mori, H. Yabe, Y. Fujita, Y. Iio and Y. Okumoto, “Ditto (Second Report, Separation Boundary Condition)” (in Japanese), Trans. JSME, Vol. 35, No. 272, April 1969, pp. 899 - 906. 9. M.M. Reddi, “Finite Element Solution of the Incompressible Lubrication Problems”, Trans. ASME. F, Vol. 91, No. 3, July 1969, pp. 524 - 533. 10. S. Wada, H. Hayashi and M. Migita, “Application of Finite Element Method to Hydrody- namic Lubrication Problems (Part 1)” (in Japanese), Trans. JSME, Vol. 37, No. 295, March 1971, pp. 583 - 592. 11. S. Wada and H. Hayashi, “Application of Finite Element Method to Hydrodynamic Lu- brication Problems (Part 2)” (in Japanese), Trans. JSME, Vol. 37, No. 295, March 1971, pp. 593 - 601. 12. T. Kato and Y. Hori, “Matrix Form of Reynolds’ Equation”, JSME International Journal, Vol. 31, No. 2, June 1986, pp. 444 - 450. 13. J. Mitsui, “Method of Calculation for Bearing Characteristics”, Journal Bearing Data- book (T. Someya, Editor), Springer Verlag, Berlin Heidelberg, 1988, pp. 231 - 240 4 Fundamentals of Thrust Bearings An axial load acting on a shaft is called a thrust. A bearing that supports a thrust is called a thrust bearing. In this chapter, the fundamentals of thrust bearings are discussed. In a journal bearing, an oil film wedge is automatically formed due to journal eccentricity because of the bearing load; therefore, a load capacity is automatically generated by the load itself. In a thrust bearing, however, an oil film wedge is not formed automatically, so it must be prepared artificially to support the load. This is an essential difference between a journal bearing and a thrust bearing. In the early days, a thrust bearing consisted of a rotating disk fixed to a shaft and a mating stationary disk (the shaded part) as shown in Fig. 4.1a-ca. Since these disks were parallel to each other, an oil film wedge was not formed between them, and hence the load capacity was theoretically zero. Therefore, solid friction took place between the disks and so severe heat generation and wear occurred. This was a big problem in the bearings of propeller shafts for ships and in the bearings of vertical shafts for water turbines, for example. One attempt to overcome these problems was to form an oil film wedge artifi- cially between a rotating disk fixed to the shaft and a spatially fixed inclined pad (the shaded wedge-shaped pad) as shown in Fig. 4.1a-cb. This seems at first to be a good idea, but since the optimal angle of inclination of the pad is very small, it is difficult to finish the inclined surface with sufficient accuracy. Furthermore, even if it could be successfully manufactured, the inclination may change afterward as a result of elastic deformation or thermal deformation, for example. A solution to this problem was found independently by A. G. M. Michell (1870 – 1959) of Australia and A. Kingsbury (1863 – 1943) of the USA. Their idea was to support an inclined plate at a certain point a little downstream of the center by a pivot so that it can tilt freely. It is shown schematically in Fig. 4.1a-cc. Since a pivoted plate (or a pad) has the property that its inclination is automatically deter- mined by the pivot position, an oil film wedge of optimal inclination can be formed automatically by choosing a suitable pivot position. This approach resulted in the first reasonable design for a thrust bearing. 48 4 Fundamentals of Thrust Bearings Fig. 4.1a-c. Development of thrust bearings. a parallel disks, b fixed inclined pad, c tilting pad. The shaded patern is the stationary disk Such a bearing is called a Michell bearing or a Kingsbury bearing .Itisalso called a tilting pad bearing from its working principle. Although Michell obtained a patent for this type of bearing in 1905 [1] and Kingsbury also obtained a patent for the same idea in 1910 [3], it is known that Kingsbury materialized his bearing before Michell’s acquisition of the patent, and it is recognized that their inventions were independent. As for the pronunciation of Michell’s name, according to D. Dowson [23], Michell himself explained that his family name was pronounced with emphasis on the first syllable and that it should rhyme with “rich.” It is interesting indeed that hydrodynamic lubrication theory was developed from the investigation into journal bearings, and thrust bearings were invented in turn as an application of lubrication theory. 4.1 Infinitely Long Plane Pad Bearings As shown in Fig. 4.2, a plane which moves with a velocity U and a plane pad with a slight inclination to the moving plane are considered. For simplicity, it is assumed that the pad is infinitely long in the direction normal to the page. Fig. 4.2. Plane pad bearing 4.1 Infinitely Long Plane Pad Bearings 49 4.1.1 Basic Formulae The thickness h of an oil film between the two surfaces in Fig. 4.2 is expressed by a linear equation of the coordinate x as follows: h(x) = h 1 − x B (h 1 − h 2 ) (4.1) where h 1 and h 2 are the film thickness at the entrance and the exit of the pad, respec- tively, and B is the width of the pad in the direction of the plane motion. If nondimensional quantities m = h 1 h 2 , ¯x = x B , h = h h 2 (4.2) are introduced, Eq. 4.1 can be written as follows: h(x) = m − ¯x(m − 1) (4.3) where m is a parameter indicating the inclination of the pad. Reynolds’ equation for the pad can be written as follows, the pad being assumed to be infinitely long in the direction normal to the page: d dx  h 3 dp dx  = 6µU dh dx (4.4) where p is the pressure, µ is the coefficient of viscosity, and U is the velocity of the moving plane. In the case of a thrust bearing, the boundary conditions for pressure are usually simple. Namely, pressure p can be set at zero at the entrance and the exit of the bearing. In the case of a thrust bearing, film thickness does not usually increase in the downstream direction at any point and so oil film rupture will not occur in the film as it does in the case of a journal bearing. 4.1.2 Basic Characteristics a. Pressure Distribution Integrating Eq. 4.4 with respect to x gives: dp dx = 6µU  1 h 2 + C 1 h 3  where C 1 is the integral constant. If the oil film thickness at the point of maximum pressure, or at the point where dp/dx = 0, is denoted by h m ,wehave: C 1 = − h m from the above equation, thus: 50 4 Fundamentals of Thrust Bearings dp dx = 6µU  1 h 2 − h m h 3  (4.5) where h m is still unknown. If this is integrated again, then: p = 6µU  h h 1  1 h 2 − h m h 3  dx + C 2 (4.6) where h m and C 2 are integral constants that can be determined by the boundary con- ditions, i.e., that the pressure is zero at the entrance and the exit, i.e., p(h 1 ) = p(h 2 ) = 0 (4.7) Let us first consider the entrance, where h = h 1 . From Eq. 4.6 and Eq. 4.7, we have: p(h 1 ) = 0 + C 2 = 0, therefore C 2 = 0 (4.8) Next, consider the exit. Since h = h 2 there, we have: p(h 2 ) = 6µU  h 2 h 1  1 h 2 − h m h 3  dx = 0 Then h m will be determined as follows: h m =   h 2 h 1 1 h 2 dx     h 2 h 1 1 h 3 dx  (4.9) Fig. 4.3. Nondimensional pressure distribution in an infinitely long plane pad bearing The integral constants of Eq. 4.6 are now determined, hence the pressure distri- bution p is determined. It can be written as follows if nondimensional variables ¯ h and ¯x are used: 4.1 Infinitely Long Plane Pad Bearings 51 p = 6µUB h 2 2  h h 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 1 h 2 − h m h 3 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ¯x (4.10) where h m = h m h 2 = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝  h 2 h 1 1 h 2 dx ⎞ ⎟ ⎟ ⎟ ⎟ ⎠  ⎛ ⎜ ⎜ ⎜ ⎜ ⎝  h 2 h 1 1 h 3 dx ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ (4.11) Since ¯ h is given by Eq. 4.3 in the case of a plane pad bearing, p(¯x) and h m will be as follows: p(¯x) = 6µUB h 2 2 (m − 1)(1 − ¯x)¯x (m + 1)(m − m ¯x + ¯x) 2 ≡ 6µUB h 2 2 ¯p(¯x) (4.12) h m = 2m m + 1 (4.13) Nondimensional pressure ¯p(¯x) on the right-hand side of Eq. 4.12 is shown in Fig. 4.3 with m as a parameter. This shows the shape of pressure distribution. b. Load Capacity Fig. 4.4. Nondimensional load capacity of an infinitely long plane pad bearing as a function of pad inclination Integrating the oil film pressure over the pad width gives the load capacity of the pad bearing. Load capacity P per unit length will be: P =  h 2 h 1 pdx=  xp  h 2 h 1 −  h 2 h 1 x dp dx dx = −  h 2 h 1 x dp dx dx (4.14) = 6µU  h 2 h 1 x  1 h 2 − h m h 3  dx = 6µUB 2 h 2 2  h 2 h 1 ¯x ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 1 h 2 − h m h 3 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ¯x (4.15) 52 4 Fundamentals of Thrust Bearings In the case of an infinitely long plane pad bearing, the load capacity can be ob- tained as follows by using Eq. 4.12: P = 6µUB 2 h 2 2 1 (m − 1) 2  ln m − 2(m − 1) m + 1  ≡ 6µUB 2 h 2 2 ¯ P(m) (4.16) The relation between the nondimensional load capacity ¯ P(m) and the pad inclina- tion m is shown in Fig. 4.4. The figure shows that the nondimensional load capacity has a maximum in the neighborhood of a pad inclination of m = 2.2. c. Center of Pressure and Pivot Position Fig. 4.5. Center of pressure (i.e., the pivot postion) of an infinitely long plane pad as a function of pad inclination The position x c of the center of oil film pressure is given as follows: x c = 1 P  h 2 h 1 pxdx= 3µUB 3 Ph 2 2  h 2 h 1 ¯x 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 1 h 2 − h m h 3 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ¯x (4.17) In the case of an infinitely long plane pad bearing, the nondimensional center of pressure, ¯x c = x c /B, is given as follows for a given inclination m, by using Eq. 4.12: ¯x c = 2m(m + 2) ln m − (m − 1)(5m + 1) 2(m − 1){(1 + m)lnm − 2(m − 1)} (4.18) The dependence of ¯x c on m is shown in Fig. 4.5. It is important to note that ¯x c increases monotonously with m. 4.1 Infinitely Long Plane Pad Bearings 53 This figure can also be interpreted as a graph that gives the inclination of a pad for a given pivot position. If a pad is supported by a pivot at a certain position, the position must coincide with the center of pressure ¯x c from a balance of the moments acting on the pad. Therefore, the pad takes an inclination m corresponding to ¯x c automatically. For example, if the pivot position is taken as ¯x = 0.57, it must be at the center of pressure ¯x c and so, from the figure, the inclination will be m = 2. This is a value which is automatically determined. If the inclination m is larger than this value, the center of pressure ¯x c will be located downstream of the pivot position, and hence the pad is subject to a moment which makes its inclination smaller. If the inclination m is smaller than the value, the pad is subject to a moment which makes its inclination larger. More precisely, if the pad is supported by a pivot at the position of ¯x c = 0.578 (the position 57.8% along the pad from the entrance of the pad), the inclination of the pad becomes m = 2.2 automatically, and then, according to Fig. 4.4, the load capacity will be maximum. This is the working principle of a Michell bearing or a Kingsbury bearing. d. Frictional Force Fig. 4.6. Nondimensional frictional force of an infinitely long plane pad as a function of pad inclination. ¯ F 1 is the force on the moving surface and ¯ F 2 is the force on the fixed pad The shear stresses in the oil film at a moving surface and a stationary pad surface are obtained as follows with U 1 = U and U 2 = 0, similarly to Eq. 3.33 and Eq. 3.34: τ y=0 = − µU h − h 2 dp dx (4.19) τ y=h = − µU h + h 2 dp dx (4.20) 54 4 Fundamentals of Thrust Bearings The moving surface is considered first. Integrating Eq. 4.19, with Eq. 4.5 substi- tuted into it, over the width of the plate gives a frictional force acting on the moving surface per unit length as follows: F 1 = µU  h 2 h 1  3h m h 2 − 4 h  dx (4.21) = µUB h 2  h 2 h 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 3 h m h 2 − 4 h ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ¯x (4.22) In the case of an infinitely long plane pad, substituting Eq. 4.3 and Eq. 4.13 into the above equation yields the following frictional force acting on the moving surface: F 1 = µUB h 2 1 m − 1  −4lnm + 6(m − 1) m + 1  ≡ µUB h 2 ¯ F 1 (m) (4.23) Similarly, from Eq. 4.20, the frictional force acting on the fixed pad surface is ob- tained as follows: F 2 = µUB h 2 1 m − 1  2lnm − 6(m − 1) m + 1  ≡ µUB h 2 ¯ F 2 (m) (4.24) Nondimensional frictional forces ¯ F 1 (m) and ¯ F 2 (m) are shown in Fig. 4.6 against the pad inclination m. As is seen in the figure, F 1 and F 2 are not equal. Calculating their difference leads to: F 2 − F 1 = h 1 − h 2 B P (4.25) which is equal to the load capacity multiplied by the inclination of the fixed pad surface. This shows that the difference of frictional forces is attributable to the incli- nation of the fixed pad surface. 4.2 Finite Length Plane Pad Bearings If the length of a pad (the length in the direction normal to the page) is finite, the load capacity falls markedly because of leakage of lubricating oil from both ends of the pad. The ratio of the load capacity per unit length of a finite length pad bearing to that of an infinitely long pad bearing is called the side leakage factor. A rigorous analysis of a finite length plane pad bearing was performed for the first time by Michell [2]. He expressed the pressure distribution p in the length direction (the z direction) as an infinite series of sin functions of odd terms as follows: p = p 1 + p 3 + p 5 + ···+ p m + ··· (4.26) where p m = w m (x) sin mz mx , m is a positive odd number [...]... assumed, for simplicity, that the pressure gradient in Eq 4. 30 can be disregarded, θ will be: θ and then it will be seen that 4. 31 and Eq 4. 35 [ 24] : r = r =− rω (z − h) h (4. 36) and Gc are given as follows, respectively, from Eq ρrω2 hz z2 z3 z4 1 ∂p z(z − h) + − + − 2µ ∂r µ 4 2 3h 12h2 Gc = ρrω2 h3 µ 40 (4. 37) (4. 38) If Gc = 0 is assumed in Eq 4. 34, Reynolds’ equation ignoring the centrifugal force will... dz + ∂ ∂θ h 0 θ dz = 0 (4. 33) Substituting Eq 4. 30 and Eq 4. 31 into Eq 4. 33 yields Reynolds’ equation in cylindrical coordinates as follows: ∂ h3 ∂p 1 h3 ∂p 1 ∂ h3 ∂p ω ∂h + Gc + + Gc + 2 = ∂r 12µ ∂r r 12µ ∂r 2 ∂θ r ∂θ 12µ ∂θ (4. 34) where Gc = ρ µr h 0 z h h z ( 0 0 θ 2 z )dzdz − z ( 0 0 θ 2 )dzdz dz (4. 35) Gc is the integration of the second half of the right-hand side of Eq 4. 31 with respect to z... h + h = 6µ ω ∂r ∂r r ∂r r ∂θ ∂θ ∂θ (4. 39) ∂ ∂p 1 ∂ 3 ∂p ∂h rh3 + h = 6µ r ω ∂r ∂r r ∂θ ∂θ ∂θ (4. 40) or These equations can also be derived from Reynolds’ equation in rectangular coordinates by the following coordinate transformation: x = r cos θ, z = − r sin θ (4. 41) 4. 3.2 Numerical Solution of a Sector Pad Let us solve the sector pad problem numerically using Eq 4. 39 Namely: ∂ 3 ∂p h3 ∂p 1 ∂ 3 ∂p ∂h... (4. 43) 58 4 Fundamentals of Thrust Bearings Equation 4. 42 is nondimensionalized as follows with the above nondimensional quantities: ∆R ¯ ¯ ∂ ¯3 ∂p ∆R ¯ 3 ∂ p h h + + ∂¯ r ∂¯ r R + r∆R ∂¯ ¯ r R + r∆R ¯ 2 1 ∂ ¯3 ∂p ¯ h 2 ∂θ ¯ ¯ ∂θ θ0 2 ¯ ∆R 1 ∂h =6 ¯ R θ0 ∂ θ (4. 44) Fig 4. 8 Grid for a sector pad Next, the sector pad is divided into a grid as shown in Fig 4. 8 If the differential coefficients in the above equation... nodal point can be written in the following form: pi, j = a0 + a1 pi+1, j + a2 pi−1, j + a3 pi, j+1 + a4 pi, j−1 (4. 45) With this expression and the boundary conditions, the pressure at all the nodal points can be obtained by the method of successive approximation or elimination 4. 4 Additional Topics 4. 4.1 Influence of Deformation of the Pad In this chapter, it has been assumed that the lubricating surface... Experimental Studies)” (in Japanese), Trans JSME, C, Vol 46 , No 40 5, May 1980, pp 542 - 549 15 S Fukui and R Kaneko, “Analysis of Ultra-Thin Gas Film Lubrication based on Linearized Boltzmann Equation (First Report, Derivation of Generalized Lubrication References 16 17 18 19 20 21 22 23 24 25 61 Equation)” (in Japanese), Trans JSME, C, Vol 53, No 48 7, March 1987, pp 829 - 838 T Ohkubo, J Kishigami, S Fukui... ∂r r ∂r r2 ∂θ ∂θ ∂θ (4. 42) The following nondimensional quantities are introduced here using the inner radius R, the width in the radial direction ∆R, and the angular extent in the circumferential direction θ0 of the sector pad; the angular velocity ω of the disk; and the exit clearance h0 of the pad (see Fig 4. 7) r= ¯ r−R , ∆R θ ¯ θ= , θ0 h ¯ h= , h0 p= ¯ ph0 2 ω µR2 (4. 43) 58 4 Fundamentals of Thrust... deformation naturally changes the lubrication characteristics of the pad 4. 4 Additional Topics 59 In the design of a pad, it is important to reduce the elastic deformation of the pad by making it sufficiently thick, by distributing the supporting points, or by reducing the thermal deformation of the pad by suitable heat removal or heat isolation [4] [7] [8] [9] [10] 4. 4.2 Magnetic Disk Memory Storage... 2 A.G.M Michell, “The Lubrication of Plane Surfaces”, Zeitschrift f¨ r Mathematik und u Physik, 52 (1905), Heft 2, pp 123 - 137 3 A Kingsbury, “Thrust Bearings”, US Patent No 947 242 (1910) 4 A.G.M Michell, Lubrication - Its Principles and Practice”, Blackie & Son Ltd., London and Glasgow, 1950 5 A Burgdorfer, “The Influence of the Molecular Mean Free Path on the Performance of Hydrodynamic Gas Lubricated... Japanese), Trans JSME, Vol 31, No 231, November 1965, pp 1 740 - 1 749 9 H Tahara, “Ditto (Third Report, Experiments of the Centrally Supported Pads)” (in Japanese), Trans JSME, Vol 32, No 2 34, February 1966, pp 346 - 3 54 10 H Tahara, “Some Problems of Large Michell Thrust Bearings” (in Japanese), Journal of JSME, Vol 69, No 572, September 1966, pp 1185 - 11 94 11 T Yoshizawa, Y Hori, H Miura and M Nemoto, “Studies . Eq. 3. 34: τ y=0 = − µU h − h 2 dp dx (4. 19) τ y=h = − µU h + h 2 dp dx (4. 20) 54 4 Fundamentals of Thrust Bearings The moving surface is considered first. Integrating Eq. 4. 19, with Eq. 4. 5 substi- tuted. [ 24] :  r = 1 2µ ∂p ∂r z(z − h) + ρrω 2 µ  hz 4 − z 2 2 + z 3 3h − z 4 12h 2  (4. 37) G c = ρrω 2 µ h 3 40 (4. 38) If G c = 0 is assumed in Eq. 4. 34, Reynolds’ equation ignoring the centrifugal force. µU  h 2 h 1  3h m h 2 − 4 h  dx (4. 21) = µUB h 2  h 2 h 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 3 h m h 2 − 4 h ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ¯x (4. 22) In the case of an infinitely long plane pad, substituting Eq. 4. 3 and Eq. 4. 13 into the above