Here, L is the width of the parallel plates, in the direction normal to the flow (in the z direction) Substituting the flow in Eq (4-29) into Eq (4-30) and integrating yields the expression for the pressure gradient as a function of flow rate, Q: dp 12m ¼À 3Q dx bh0 ð4-31Þ This equation is useful for the hydrostatic bearing calculations in Chapter 10 The negative sign means that a negative pressure slope in the x direction is required for a flow in the same direction 4.10 FLUID FILM BETWEEN A CYLINDER AND A FLAT PLATE There are important applications of a full fluid film at the rolling contact of a cylinder and a flat plate and at the contact of two parallel cylinders Examples are cylindrical rolling bearings, cams, and gears A very thin fluid film that separates the surfaces is shown in Fig 4-7 In this example, the cylinder is stationary and the flat plate has a velocity U in the x direction In Chapter 6, this problem is extended to include rolling motion of the cylinder over the plate The problem of a cylinder and a flat plate is a special case of the general problem of contact between two parallel cylinders By using the concept of equivalent radius (see Chapter 12), the equations for a cylinder and a flat plate can be extended to that for two parallel cylinders Fluid films at the contacts of rolling-element bearings and gear teeth are referred to as elastohydrodynamic (EHD) films The complete analysis of a fluid film in actual rolling-element bearings and gear teeth is quite complex Under load, the high contact pressure results in a significant elastic deformation of the contact surfaces as well as a rise of viscosity with pressure (see Chapter 12) F IG 4.7 Fluid film between a cylinder and a flat-plate Copyright 2003 by Marcel Dekker, Inc All Rights Reserved However, the following problem is for a light load where the solid surfaces are assumed to be rigid and the viscosity is constant The following problem considers a plate and a cylinder with a minimum clearance, hmin In it we consider the case of a light load, where the elastic deformation is very small and can be disregarded (cylinder and plate are assumed to be rigid) In addition, the values of maximum and minimum pressures are sufficiently low, and there is no fluid cavitation The viscosity is assumed to be constant The cylinder is stationary, and the flat plate has a velocity U in the x direction as shown in Fig 4-7 The cylinder is long in comparison to the film length, and the long-bearing analysis can be applied 4.10.1 Film Thickness The film thickness in the clearance between a flat plate and a cylinder is given by hyị ẳ hmin ỵ R1 À cos yÞ ð4-32Þ where y is a cylinder angle measured from the minimum film thickness at x ¼ Since the minimum clearance, hmin , is very small (relative to the cylinder radius), the pressure is generated only at a very small region close to the minimum film thickness, where x ( R, or x=R ( For a small ratio of x=R, the equation of the clearance, h, can be approximated by a parabolic equation The following expression is obtained by expanding Eq (4-32) for h into a Taylor series and truncating terms that include powers higher than ðx=RÞ2 In this way, the expression for the film thickness h can be approximated by hxị ẳ hmin ỵ 4.10.2 x2 2R ð4-33aÞ Pressure Wave The pressure wave can be derived from the expression for the pressure gradient, dp=dx, in Eq (4-13) The equation is dp h Àh ¼ 6mU dx h The unknown h0 can be replaced by the unknown x0 according to the equation h0 xị ẳ hmin ỵ x2 2R Copyright 2003 by Marcel Dekker, Inc All Rights Reserved ð4-33bÞ After substituting the value of h according to Eqs (4-33), the solution for the pressure wave can be obtained by the following integration: x x2 x2 dx 4-34ị pxị ẳ 24mUR 2Rhmin ỵ x ị The unknown x0 is solved by the following boundary conditions of the pressure wave: at x ¼ À1: at x ¼ 1: pẳ0 pẳ0 4-35ị For numerical integration, the innity can be replaced by a relatively large finite value, where pressure is very small and can be disregarded Remark: The result is an antisymmetrical pressure wave (on the two sides of the minimum film thickness), and there will be no resultant load capacity In actual cases, the pressures are high, and there is a cavitation at the diverging side A solution that considers the cavitation with realistic boundary conditions is presented in Chapter 4.11 SOLUTION IN DIMENSIONLESS TERMS If we perform a numerical integration of Eq (4-34) for solving the pressure wave, the solution would be limited to a specific bearing geometry of cylinder radius R and minimum clearance hmin The numerical integration must be repeated for a different bearing geometry For a universal solution, there is obvious merit to performing a solution in dimensionless terms For conversion to dimensionless terms, we normalize the x pffiffiffiffiffiffiffiffiffiffiffiffiffiffi coordinate by dividing it by 2Rhmin and define a dimensionless coordinate as x " x ẳ p 2Rhmin 4-36ị In addition, a dimensionless clearance ratio is dened: h " hẳ hmin 4-37ị The equation for the variable clearance ratio as a function of the dimensionless coordinate becomes " " h ẳ ỵ x2 Copyright 2003 by Marcel Dekker, Inc All Rights Reserved ð4-38Þ Let us recall that the unknown h0 is the fluid film thickness at the point of peak pressure It is often convenient to substitute it by the location of the peak pressure, x0 , and the dimensionless relation then is " "0 h0 ẳ ỵ x 4-39ị In addition, if the dimensionless pressure is defined as h2 " p p ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Rhmin 6mU then the following integration gives the dimensionless pressure wave: ð ðx " "0 " x2 À x2 " " " dx p ẳ dp ẳ "2 1 ỵ x ị ð4-40Þ ð4-41Þ For a numerical integration of the pressure wave according to Eq (4-41), the " boundary x ¼ À1 is replaced by a relatively large finite dimensionless value, " where pressure is small and can be disregarded, such as x ¼ À4 An example of numerical integration is given in Example Problem In a similar way, for a numerical solution of the unknown x0 and the load capacity, it is possible to replace infinity with a finite number, for example, the following mathematical boundary conditions of the pressure wave of a full fluid film between a cylinder and a plane: " at x ¼ À1: " at x ¼ 1: " p¼0 " p¼0 ð4-42aÞ These conditions are replaced by practical numerical boundary conditions: " " at x ¼ À4: p ¼ " " at x ¼ ỵ4: p ẳ 4-42bị These practical numerical boundary conditions not introduce a significant error because a significant hydrodynamic pressure is developed only near the " minimum fluid film thickness, at x ¼ Example Problem 4-4 Ice Sled An ice sled is shown in Fig 4-8 On the left-hand side, there is a converging clearance that is formed by the geometry of a quarter of a cylinder A flat plate continues the curved cylindrical shape The flat part of the sled is parallel to the flat ice It is running parallel over the flat ice on a thin layer of water film of a constant thickness h0 Copyright 2003 by Marcel Dekker, Inc All Rights Reserved Derive and plot the pressure wave at the entrance region of the fluid film and under the flat ice sled as it runs over the ice at velocity U Derive the equation of the sled load capacity Solution The long-bearing approximation is assumed for the sled similar to the converging slope in Fig 4-4 The fluid film equation for an infinitely long bearing is dp h À h0 ¼ 6mU dx h3 In the parallel region, h ¼ h0 , and it follows from the foregoing equation that dp ¼0 dx ðalong the parallel region of constant clearanceÞ In this case, h0 ¼ hmin , and the equation for the variable clearance at the converging region is hxị ẳ h0 ỵ x2 2R This means that for a wide sled, the pressure is constant within the parallel region This is correct only if L ) B, where L is the bearing width (in the z direction) and B is along the sled in the x direction Fluid flow in the converging region generates the pressure, which is ultimately responsible for supporting the load of the sled Through the adhesive force of viscous shear, the fluid is dragged into the converging clearance, creating the pressure in the parallel region Substituting hðxÞ into the pressure gradient equation yields   x2 h0 ỵ À h0 dp 2R ¼ 6mU  3 dx x2 h0 ỵ 2R or dp x2 ẳ 24mUR2 dx 2Rh0 þ x2 Þ3 Applying the limits of integration and the boundary condition, we get the following for the pressure distribution: x x2 pxị ẳ 24mUR2 dx 2Rh0 þ x Þ Copyright 2003 by Marcel Dekker, Inc All Rights Reserved This equation can be integrated analytically or numerically For numerical integration, since a significant pressure is generated only at a low x value, the infinity boundary of integration is replaced by a finite magnitude Conversion to a Dimensionless Equation As discussed earlier, there is an advantage in solving the pressure distribution in dimensionless terms A regular pressure distribution curve is limited to the specific bearing data of given radius R and clearance h0 The advantage of a dimensionless curve is that it is universal and applies to any bearing data For conversion of the pressure gradient to pffiffiffiffiffiffiffiffiffiffi dimensionless terms, we normalize x by dividing by 2Rh0 and define dimensionless terms as follows: x " x ¼ pffiffiffiffiffiffiffiffiffiffi ; 2Rh0 h " h¼ ; h0 " " h ẳ ỵ x2 Converting to dimensionless terms, the pressure gradient equation gets the form " dp 6mU h ẳ dx h0 ỵ x2 ị3 " Here, h0 ¼ hmin is the minimum film thickness at x ¼ Substituting for the dimensionless clearance and rearranging yields " h2 x2 pffiffiffiffiffiffiffiffiffiffi dx dp ¼ " ỵ x2 ị3 2Rh0 6mU The left-hand side of this equation is defined as the dimensionless pressure Dimensionless pressure is equal to the following integral: ð ðx " h2 p x2 " " p ¼ pffiffiffiffiffiffiffiffiffiffi dp ẳ dx " ỵ x2 ị3 2Rh0 6mU À1 Numerical Integration The dimensionless pressure is solved by an analytical or numerical integration of the preceding function within the specified boundaries (see Appendix B) The pressure p0 under the flat plate is obtained by integration to the limit x ¼ 0: x X h2 x2 i " p0 ¼ Dxi p ¼ pffiffiffiffiffiffiffiffiffiffi 2Rh0 6mU ỵ xi ị The pressure is signicant only near x ¼ Therefore, for the numerical integration, a finite number replaces infinity The resulting pressure distribution is shown in Fig 4-9 Comparison with Analytical Integration The maximum pressure at x ¼ as well as along the constant clearance, x > 0, can also be solved by analytical Copyright 2003 by Marcel Dekker, Inc All Rights Reserved If the load, cylinder radius, water viscosity, and sled velocity are known, it is possible to solve for the film thickness, h0 Example Problem 4-5 Derive the equation for the pressure gradient of a journal bearing if the journal and bearing are both rotating around their stationary centers The surface velocity of the bearing bore is Uj ; ¼ oj R, and the surface velocity of the journal is Ub ¼ ob R1 Solution Starting from Eq (4-4): dp @2 u ¼m dx @y and integrating twice (in a similar way to a stationary bearing) yields u ẳ my2 ỵ ny ỵ k However, the boundary conditions and continuity conditions are as follows: at y ¼ 0: at y ẳ hxị: u ẳ ob R1 u ẳ oj R In this case, the constant-volume flow rate, q, per unit of bearing length at the point of peak pressure is h ob R1 ỵ oj Rịh0 q ẳ u dy ¼ Solving for m, n and k and substituting in a similar way to the previous problem while also assuming R1 % R, the following equation for the pressure gradient is obtained: dp h À h0 ẳ 6Rob ỵ oj ịm dx h3 Problems 4-1 A long plane-slider, L ¼ 200 mm and B ¼ 100 mm, is sliding at velocity of 0.3 m=s The minimum film thickness is h1 ¼ 0:005 mm and the maximum film thickness is h2 ¼ 0:010 mm The fluid is SAE 30, and the operating temperature of the lubricant is assumed a constant 30 C a Assume the equation for an infinitely long bearing, and use numerical integration to solve for the pressure wave (use Copyright 2003 by Marcel Dekker, Inc All Rights Reserved trial and error to solve for x0 ) Plot a curve of the pressure distribution p ¼ pðxÞ b Use numerical integration to find the load capacity Compare this to the load capacity obtained from Eq (419) c Find the friction force and the friction coefficient 4-2 A slider is machined to have a parabolic surface The slider has a horizontal velocity of 0.3 m=s The minimum film hmin ¼ 0:020 mm, and the clearance varies with x according to the following equation: h ¼ 0:020 þ 0:01x2 The slider velocity is U ¼ 0:5 m=s The length L ¼ 300 mm and the width in the sliding direction B ¼ 100 mm The lubricant is SAE 40 and the temperature is assumed constant, T ¼ 40 C Assume the equation for an infinitely long bearing a Use numerical integration and plot the dimensionless pressure distribution, p ẳ pxị b Use numerical integration to nd the load capacity c Find the friction force and the friction coefficient 4-3 A blade of a sled has the geometry shown in the Figure 4-8 The sled is running over ice on a thin layer of water film The total load (weight of the sled and person) is 1500 N The sled velocity is 20 km=h, the radius of the inlet curvature is 30 cm, and the sled length B ¼ 30 cm, and width is L ¼ 100 cm The viscosity of water is m ¼ 1:792 10À3 N-s=m2 Find the lm thickness h0 ẳ hmin ị of the thin water layer shown in Fig 4-8 Direction: The clearance between the plate and disk is h ¼ hmin þ x2 =2R, and assume that p ¼ at x ¼ R Copyright 2003 by Marcel Dekker, Inc All Rights Reserved 5 Basic Hydrodynamic Equations 5.1 NAVIER^STOKES EQUATIONS The pressure distribution and load capacity of a hydrodynamic bearing are analyzed and solved by using classical fluid dynamics equations In a thin fluid film, the viscosity is the most important fluid property determining the magnitude of the pressure wave, while the effect of the fluid inertia (ma) is relatively small and negligible Reynolds (1894) introduced classical hydrodynamic lubrication theory Although a lot of subsequent research has been devoted to this discipline, Reynolds’ equation still forms the basis of most analytical research in hydrodynamic lubrication The Reynolds equation can be derived from the Navier– Stokes equations, which are the fundamental equations of fluid motion The derivation of the Navier–Stokes equations is based on several assumptions, which are included in the list of assumptions (Sec 4.2) that forms the basis of the theory of hydrodynamic lubrication An important assumption for the derivation of the Navier–Stokes equations is that there is a linear relationship between the respective components of stress and strain rate in the fluid In the general case of three-dimensional flow, there are nine stress components referred to as components of the stress tensor The directions of the stress components are shown in Fig 5-1 Copyright 2003 by Marcel Dekker, Inc All Rights Reserved is an adequate approximation However, under extreme conditions, e.g., very high pressure of point or line contacts, this assumption is no longer valid An assumption that is made for convenience is that the viscosity, m, of the lubricant is constant Also, lubrication oils are practically incompressible, and this property simplifies the Navier–Stokes equations because the density, r, can be assumed to be constant However, this assumption cannot be applied to air bearings Comment As mentioned earlier, in thin films the velocity component n is small in comparison to u and w, and two shear components can be approximated as follows:   @v @u @u ỵ txy ¼ tyx ¼ m %m @x @y @y   5-3ị @w @v @w ỵ tyz ẳ tzy ẳ m %m @y @z @y The Navier–Stokes equations are based on the balance of forces acting on a small, infinitesimal fluid element having the shape of a rectangular parallelogram with dimensions dx, dy, and dz, as shown in Fig 5-1 The force balance is similar to that in Fig 4-1; however, the general balance of forces is of three dimensions, in the x; y and z directions The surface forces are the product of stresses, or pressures, and the corresponding areas When the fluid is at rest there is a uniform hydrostatic pressure However, when there is fluid motion, there are deviatoric normal stresses s0x , s0y , s0z that are above the hydrostatic (average) pressure, p Each of the three normal stresses is the sum of the average pressure, and the deviatoric normal stress (above the average pressure), as follows: sx ẳ p ỵ s0x ; sy ẳ p ỵ s0y ; sz ẳ p ỵ s0z 5-4aị According to Newton’s second law, the sum of all forces acting on a fluid element, including surface forces in the form of stresses and body forces such as the gravitational force, is equal to the product of mass and acceleration (ma) of the fluid element After dividing by the volume of the fluid element, the equations of the force balance become du @p @s0 @txy @xz ẳX ỵ xỵ ỵ dt @x @x @y @z dv @p @tyx @s0y @tyz r ẳY ỵ ỵ ỵ dt @y @y @y @z dw @p @tzx @tzy @s0z ẳZ ỵ ỵ ỵ r dt @z @x @y @z r Copyright 2003 by Marcel Dekker, Inc All Rights Reserved ð5-4bÞ Here, p is the pressure, u, v, and w are the velocity components in the x, y, and z directions, respectively The three forces X ; Y ; Z are the components of a body force, per unit volume, such as the gravity force that is acting on the fluid According to the assumptions, the fluid density, r, and the viscosity, m, are considered constant The derivation of the Navier–Stokes equations is included in most fluid dynamics textbooks (e.g., White, 1985) For an incompressible flow, the continuity equation, which is derived from the conservation of mass, is @u @v @w ỵ ỵ ẳ0 @x @y @z ð5-5Þ After substituting the stress components of Eq (5-2) into Eq (5-4b), using the continuity equation (5-5) and writing in full the convective time derivative of the acceleration components, the following Navier–Stokes equations in Cartesian coordinates for a Newtonian incompressible and constant-viscosity fluid are obtained     @u @u @u @u @p @ u @2 u @2 u r ỵu ỵv ỵw ỵ ỵ 5-6aị ẳX ỵm @t @x @y @z @x @x2 @y2 @z2     @v @v @v @v @p @ v @2 v @2 v ỵu ỵv ỵw ỵ 2ỵ 5-6bị ẳY ẳ ỵm r @t @x @y @z @y @x2 @y @z     @w @w @w @w @p @ w @2 w @2 w ỵu ỵv ỵw ỵ 2ỵ 5-6cị ẳZ ỵm r @t @x @y @z @z @x2 @y @z The Navier–Stokes equations can be solved for the velocity distribution The velocity is described by its three components, u, v, and w, which are functions of the location (x, y, z) and time In general, fluid flow problems have four unknowns: u, v, and w and the pressure distribution, p Four equations are required to solve for the four unknown functions The equations are the three Navier–Stokes equations, the fourth equation is the continuity equation (5-5) 5.2 REYNOLDS HYDRODYNAMIC LUBRICATION EQUATION Hydrodynamic lubrication involves a thin-film flow, and in most cases the fluid inertia and body forces are very small and negligible in comparison to the viscous forces Therefore, in a thin-film flow, the inertial terms [all terms on the left side of Eqs (5.6)] can be disregarded as well as the body forces X , Y , Z It is well known in fluid dynamics that the ratio of the magnitude of the inertial terms relative to the viscosity terms in Eqs (5-6) is of the order of magnitude of the Copyright 2003 by Marcel Dekker, Inc All Rights Reserved Reynolds number, Re For a lubrication flow (thin-film flow), Re ( 1, the Navier–Stokes equations reduce to the following simple form:   @p @ u @2 u @2 u ẳm ỵ 2ỵ @x @x2 @y @z   @p @ v @ v @2 v ẳm ỵ 2ỵ @y @x2 @y @z   @p @ w @ w @2 w ¼m þ 2þ @z @x2 @y @z ð5-7aÞ ð5-7bÞ ð5-7cÞ These equations indicate that viscosity is the dominant effect in determining the pressure distribution in a fluid film bearing The assumptions of classical hydrodynamic lubrication theory are summarized in Chapter The velocity components of the flow in a thin film are primarily u and w in the x and z directions, respectively These directions are along the fluid film layer (see Fig 1-2) At the same time, there is a relatively very slow velocity component, v, in the y direction across the fluid film layer Therefore, the pressure gradient in the y direction in Eq (5-7b) is very small and can be disregarded In addition, Eqs (5-7a and c) can be further simplified because the order of magnitude of the dimensions of the thin fluid film in the x and z directions is much higher than that in the y direction across the film thickness The orders of magnitude are x ẳ OBị y ẳ Ohị 5-8aị z ẳ OðLÞ Here, the symbol O represents order of magnitude The dimension B is the bearing length along the direction of motion (x direction), and h is an average fluid film thickness The width L is in the z direction of an inclined slider In a journal bearing, L is in the axial z direction and is referred to as the bearing length In hydrodynamic bearings, the fluid film thickness is very small in comparison to the bearing dimensions, h ( B and h ( L By use of Eqs (5-8b), a comparison can be made between the orders of magnitude of the second Copyright 2003 by Marcel Dekker, Inc All Rights Reserved derivatives of the various terms on the right-hand side of Eq (5-7a), which are as follows:   @2 u U ¼O @y2 h   @2 u U ¼O 2 @x B   @2 u U ẳO 2 @z L 5-8bị In conventional finite-length bearings, the ratios of dimensions are of the following orders: L ẳ O1ị B h ẳ L103 ị B ð5-9Þ Equations (5-8) and (5-9) indicate that the order of the term @2 u=@y2 is larger by 106 , in comparison to the order of the other two terms, @2 u=@x2 and @2 u=@z2 Therefore, the last two terms can be neglected in comparison to the first one in Eq (5-7a) In the same way, only the term @2 w=@y2 is retained in Eq (5-7c) According to the assumptions, the pressure gradient across the film thickness, @p=@y, is negligible, and the Navier–Stokes equations reduce to the following two simplified equations: @p @2 u ¼m @x @y and @p @2 w ẳm @z @y 5-10ị The rst equation is identical to Eq (4-4), which was derived from first principles in Chapter for an infinitely long bearing In a long bearing, there is a significant pressure gradient only in the x direction; however, for a finite-length bearing, there is a pressure gradient in the x and z directions, and the two Eqs (5-10) are required for solving the flow and pressure distributions The two Eqs (5-10) together with the continuity Eq (5-5) and the boundary condition of the flow are used to derive the Reynolds equation The derivation of the Reynolds equation is included in several books devoted to the analysis of hydrodynamic lubrication see Pinkus (1966), and Szeri (1980) The Reynolds equation is widely used for solving the pressure distribution of hydrodynamic bearings of finite length The Reynolds equation for Newtonian Copyright 2003 by Marcel Dekker, Inc All Rights Reserved incompressible and constant-viscosity fluid in a thin clearance between two rigid surfaces of relative motion is given by     @ h3 @p @ h3 @p @h @ ỵ ẳ 6U1 U2 ị ỵ U1 ỵ U2 ị ỵ 12V2 V1 ị @x m @x @z m @z @x @x ð5-11Þ The velocity components of the two surfaces that form the film boundaries are shown in Fig 5-2 The tangential velocity components, U1 and U2, in the x direction are of the lower and upper sliding surfaces, respectively (two fluid film boundaries) The normal velocity components, in the y direction, V1 and V2, are of the lower and upper boundaries, respectively In a journal bearing, these components are functions of x (or angle y) around the journal bearing The right side of Eq (5-11) must be negative in order to result in a positive pressure wave and load capacity Each of the three terms on the right-hand side of Eq (5-11) has a physical meaning concerning the generation of the pressure wave Each term is an action that represents a specific type of relative motion of the surfaces Each action results in a positive pressure in the fluid film The various actions are shown in Fig 5-3 These three actions can be present in a bearing simultaneously, one at a time or in any other combination The following are the various actions Viscous wedge action: This action generates positive pressure wave by dragging the viscous fluid into a converging wedge Elastic stretching or compression of the boundary surface: This action generates a positive pressure by compression of the boundary The compression of the surface reduces the clearance volume and the viscous fluid is squeezed out, resulting in a pressure rise This action is F IG 5-2 Directions of the velocity components of fluid-film boundaries in the Reynolds equation Copyright 2003 by Marcel Dekker, Inc All Rights Reserved bearing material This action is usually not considered for rigid bearing materials Squeeze-film action: The squeezing action generates a positive pressure by reduction of the fluid film volume The incompressible viscous fluid is squeezed out through the thin clearance The thin clearance has resistance to the squeeze-film flow, resulting in a pressure buildup to overcome the flow resistance (see Problem 5-3) In most practical bearings, the surfaces are rigid and there is no stretching or compression action In that case, the Reynolds equation for an incompressible fluid and constant viscosity reduces to     @ h3 @p @ h3 @p @h 5-12ị ỵ ẳ 6U1 U2 ị ỵ 12V2 V1 ị @x m @x @z m @z @x As indicated earlier, the two right-hand terms must be negative in order to result in a positive pressure wave On the right side of the Reynolds equation, the first term of relative sliding motion ðU1 À U2 Þ describes a viscous wedge effect It requires inclined surfaces, @h=@x, to generate a fluid film wedge action that results in a pressure wave Positive pressure is generated if the film thickness reduces in the x direction (negative @h=@x) The second term on the right side of the Reynolds equation describes a squeeze-film action The difference in the normal velocity ðV2 À V1 Þ represents the motion of surfaces toward each other, referred to as squeeze-film action A positive pressure builds up if ðV2 À V1 Þ is negative and the surfaces are approaching each other The Reynolds equation indicates that a squeeze-film effect is a viscous effect that can generate a pressure wave in the fluid film, even in the case of parallel boundaries It is important to mention that the Reynolds equation is objective, in the sense that the pressure distribution must be independent of the selection of the coordinate system In Fig 5-2 the coordinates are stationary and the two surfaces are moving relative to the coordinate system However, the same pressure distribution must result if the coordinates are attached to one surface and are moving and rotating with it For convenience, in most problems we select a stationary coordinate system where the x coordinate is along the bearing surface and the y coordinate is normal to this surface In that case, the lower surface has only a tangential velocity, U1 , and there is no normal component, V1 ¼ The value of each of the velocity components of the fluid boundary, U1 , U2 , V1 , V2 , depends on the selection of the coordinate system Surface velocities in a stationary coordinate system would not be the same as those in a moving coordinate system However, velocity differences on the right-hand side of the Reynolds equation, which represent relative motion, are independent of the selection of the coordinate system In journal bearings under dynamic conditions, Copyright 2003 by Marcel Dekker, Inc All Rights Reserved the journal center is not stationary The velocity of the center must be considered for the derivation of the right-hand side terms of the Reynolds equation 5.3 WIDE PLANE-SLIDER The equation of a plane-slider has been derived from first principles in Chapter Here, this equation will be derived from the Reynolds equation and compared to that in Chapter A plane-slider and its coordinate system are shown in Fig 1-2 The lower plate is stationary, and the velocity components at the lower wall are U1 ¼ and V1 ¼ At the same time, the velocity at the upper wall is equal to that of the slider The slider has only a horizontal velocity component, U2 ¼ U , where U is the plane-slider velocity Since the velocity of the slider is in only the x direction and there is no normal component in the y direction, V2 ¼ After substituting the velocity components of the two surfaces into Eq (5-11), the Reynolds equation will reduce to the form     @ h3 @p @ h3 @p @h 5-13ị ỵ ẳ 6U @x m @x @z m @z @x For a wide bearing, L ) B, we have @p=@z ffi 0, and the second term on the left side of Eq (5-13) can be omitted The Reynolds equation reduces to the following simplified form:   @ h3 @p @h ð5-14Þ ¼ 6U @x m @x @x For a plane-slider, if the x coordinate is in the direction of a converging clearance (the clearance reduces with x), as shown in Fig 4-4, integration of Eq (5-14) results in a pressure gradient expression equivalent to that of a hydrodynamic journal bearing or a negative-slope slider in Chapter The following equation is the expression for the pressure gradient for a converging clearance, @h=@x < (negative slope): dp h À h0 ¼ 6U m dx h3 for initial @h < ðnegative slope in Fig: 4-4Þ @x ð5-15Þ The unknown constant h0 (constant of integration) is determined from additional information concerning the boundary conditions of the pressure wave The meaning of h0 is discussed in Chapter 4—it is the film thickness at the point of a peak pressure along the fluid film In a converging clearance such as a journal bearing near x ¼ 0, the clearance slope is negative, @h=@x < The result is that the pressure increases at the start of the pressure wave (near x ¼ 0) At that point, the pressure gradient dp=dx > because h > h0 Copyright 2003 by Marcel Dekker, Inc All Rights Reserved  ... Solution The long -bearing approximation is assumed for the sled similar to the converging slope in Fig 4-4 The fluid film equation for an in? ??nitely long bearing is dp h À h0 ¼ 6mU dx h3 In the parallel... ob R1 u ¼ oj R In this case, the constant-volume flow rate, q, per unit of bearing length at the point of peak pressure is h ob R1 ỵ oj Rịh0 q ẳ u dy ẳ Solving for m, n and k and substituting in. .. equations indicate that viscosity is the dominant effect in determining the pressure distribution in a fluid film bearing The assumptions of classical hydrodynamic lubrication theory are summarized in