Hydrodynamic Lubrication 2011 Part 2 doc

2.2 Reynolds’ Theory of Hydrodynamic Lubrication 11 insight into the problem. Actually, in his second paper [2], Tower reported that the beautiful pressure distribution was observed as was expected. In the case of usual lubrication methods (lubrication other than oil bath lubrica- tion), Tower reported that the measured value of friction was often unstable. Proba- bly, the quantity of oil was inadequate and a perfect oil film was not formed. In the case of the oil bath lubrication, in contrast, a sufficient amount of oil was presumably supplied. 2.2 Reynolds’ Theory of Hydrodynamic Lubrication Reynolds was interested in Tower’s experiments and studied them theoretically. In the introductory part of his famous paper of 1886 [3] on hydrodynamic lubrication, Reynolds wrote: Lubrication, or the action of oils and other viscous fluids to diminish friction and wear between solid surfaces, does not appear to have hitherto formed a subject for theoretical treatment. Such treatment may have been prevented by the obscurity of the physical actions involved, which belong to a class as yet but little known, namely, the boundary or surface actions of fluids; but the absence of such treatment has also been owing to the want of any general laws discovered by experiment. The subject is of such fundamental importance in practical mechanics, and the opportunities for observation are so frequent, that it may well be a matter of surprise that any general laws should have for so long escaped detection. ························ On reading Mr. Tower’s report it occurred to the author it is possible that in the case of the oil bath the film of oil might be sufficiently thick for the unknown boundary actions to disappear, in which case the results would be deducible from the equations of hydrodynamics. Phenomena related to lubrication are, generally speaking, very complicated be- cause of, for example, the complexity of interface phenomena and their theoretical treatments. In the case of oil bath lubrication, however, Reynolds assumed that the oil film was so thick that the theory of hydrodynamics could be applied. The outline of Reynolds theory is now described. The fluid film between two solid surfaces shown in Fig. 2.2 is considered. Reynolds’ equation is an equation to obtain the pressure generated in a fluid film when two such surfaces undergo relative motion. However, the fluid film must be sufficiently thick so that it can be alalyzed by hydrodynamics, and at the same time it must be sufficiently thin so that Reynolds’ assumptions described below will hold. For simplicity, the lower surface is assumed to be a plane. The axes of rectangular coordinates x, y, and z are taken as shown in the figure. The x and z axes are on the lower surface, and the y axis is perpendicular to it. The velocity of the fluid in the directions x, y, and z are denoted by u, , and w, 12 2 Foundations of Hydrodynamic Lubrication Fig. 2.2. Fluid film between two solid surfaces respectively, and the velocity of the lower surface is similarly described by U 1 , V 1 , and W 1 and that of the upper surface by U 2 , V 2 , and W 2 . In many practical cases, the lower surface and the upper surface perform a straight translational motion relative to each other. In this case, if the x axis is in the translational direction, then we have W 1 = W 2 = 0 and so the equations can be simplified. Let the gap between the two surfaces, or the thickness of the liquid film, be denoted by h(x , z, t), with t being time. Let the coefficient of viscosity of the fluid be µ. a. Reynolds’ Assumptions In deriving Reynolds’ equation, the following assumptions are made after Reynolds. 1. The flow is laminar. 2. The gravity and inertia forces acting on the fluid can be ignored compared with the viscous force. 3. Compressibility of the fluid is negligible. 4. The fluid is Newtonian and the coefficient of viscosity is constant. 5. Fluid pressure does not change across the film thickness. 6. The rate of change of the velocity u and w in the x direction and z direction is negligible compared with the rate of change in the y direction. 7. There is no slip between the fluid and the solid surface. b. Balance of Forces The balance of forces acting on a small volume element in the fluid is considered as shown in Fig. 2.3. Let us examine the balance in the x direction first. Neglecting the gravity and inertia forces (assumption 2), we obtain the following equation: 2.2 Reynolds’ Theory of Hydrodynamic Lubrication 13 Fig. 2.3. A small element of fluid  σ x + ∂σ x ∂x dx  dydz +  τ yx + ∂τ yx ∂y dy  dxdz +  τ zx + ∂τ zx ∂z dz  dxdy − σ x dydz − τ yx dxdz − τ zx dxdy = 0 (2.1) where σ x is the normal stress acting on the plane normal to the x axis and τ yx and τ zx are the shear stresses acting on the plane normal to the y axis and z axis, respectively, in the direction of the x axis. Equation 2.1 can be rearranged as follows: ∂σ x ∂x + ∂τ yx ∂y + ∂τ zx ∂z = 0 (2.2) Let the fluid pressure be p. Then p = −σ x and the above equation can be written as follows: ∂p ∂x = ∂τ yx ∂y + ∂τ zx ∂z (2.3) Since a laminar flow of Newtonian fluid is considered here (assumptions 1 and 4), we have the following relations. τ yx = µ ∂u ∂y τ zx = µ ∂u ∂z (2.4) where µ is the coefficient of viscosity. Then Eq. 2.3 can be written as follows: ∂p ∂x = ∂ ∂y  µ ∂u ∂y  + ∂ ∂z  µ ∂u ∂z  (2.5) On the assumption that the rate of change of the flow velocity u in the z direction is sufficiently small compared with that in the y direction (assumption 6), the second term of the right-hand side of the above equation can be disregarded compared with the first term, giving: 14 2 Foundations of Hydrodynamic Lubrication ∂p ∂x = ∂ ∂y  µ ∂u ∂y  (2.6) On the further assumption that µ is constant (assumption 4), the equation of the balance of forces in the x direction is finally obtained as follows: ∂p ∂x = µ ∂ 2 u ∂y 2 (2.7) In exactly the same way, the following equation is obtained from the balance in the z direction: ∂p ∂z = µ ∂ 2 w ∂y 2 (2.8) c. Flow Velocity Integrating Eqs. 2.7 and 2.8 twice gives the flow velocity u and w, respectively. The boundary conditions for the velocities are, from the assumption that there is no slip between the fluid and the solid surface (assumption 7), as follows: u = U 1 , w = W 1 at y = 0 u = U 2 , w = W 2 at y = h  (2.9) Then the fluid velocities will be as follows: u = − 1 2µ ∂p ∂x y(h − y) +  1 − y h  U 1 + y h U 2  (2.10) w = − 1 2µ ∂p ∂z y(h − y) +  1 − y h  W 1 + y h W 2  (2.11) where, in the calculations, it is assumed that the pressure p is constant in the y direc- tion (assumption 5). In Eq. 2.10 for the flow velocity u, the latter half of the right-hand side (in brack- ets) shows the fluid velocity due to the movement of the solid surface in the x di- rection. It changes linearly as shown in Fig. 2.4a (it is assumed that U 2 = 0). This is called shear flow or Couette flow. The former half of the right-hand side shows the flow velocity due to the pressure gradient. It is proportional to the pressure and changes parabolically across the film thickness as shown in Fig. 2.4b. This is called pressure flow or Poiseuille flow. The flow velocity in a general case is the sum of the two. Figure 2.4c shows such an example in which a flow in the reverse direction to the shearing direction occurs due to the pressure gradient at the left end and the shear flow is accelerated by the negative pressure gradient at the right end. At the point of maximum pressure, we have the relation dp/dx = 0, and therefore the flow at that point consists of the shear flow only. The same can be said of Eq. 2.11 for the flow velocity w. 2.2 Reynolds’ Theory of Hydrodynamic Lubrication 15 Fig. 2.4a-c. Flow velocity distribution. a shear flow, b pressure flow, c summation of shear flow and pressure flow Fig. 2.5. Pillar-shaped element d. Continuity Equation The continuity equation for a small volume element in an incompressible fluid (as- sumption 3) can be written as follows: ∂u ∂x + ∂ ∂y + ∂w ∂z = 0 (2.12) Integrating this equation along the pillar-shaped element (dx × dz ×h)inFig.2.5in the film thickness direction from y = 0toy = h gives:  h 0 ∂u ∂x dy +  h 0 ∂w ∂z dy +    h 0 = 0 (2.13) 16 2 Foundations of Hydrodynamic Lubrication This equation can be rewritten as follows by changing the order of integration and differentiation: ∂ ∂x  h 0 udy+ ∂ ∂z  h 0 wdy−  u  y=h ∂h ∂x −  w  y=h ∂h ∂z +    h 0 = 0 (2.14) This is the continuity equation for a pillar-shaped element between the solid surfaces shown in Fig. 2.6. In deriving Eq. 2.14, the following mathematical formula must be used because the integration limit h in Eq. 2.13 is a function of x and z (only a formula concerning x is given here). ∂ ∂x  b(x) a(x) f (x, y, z)dy =  b(x) a(x) ∂ ∂x f (x, y, z)dy + f (x, b(x), z) ∂b(x) ∂x − f (x, a(x), z) ∂a(x) ∂x Equation 2.14 can be written as follows in terms of the surface velocities from the boundary conditions (assumption 7): ∂ ∂x  h 0 udy+ ∂ ∂z  h 0 wdy− U 2 ∂h ∂x − W 2 ∂h ∂z + (V 2 − V 1 ) = 0 (2.15) Fig. 2.6. An inclined plate in motion 2.2 Reynolds’ Theory of Hydrodynamic Lubrication 17 e. Derivation of Reynolds’ Equation Reynolds’ equation is essentially a continuity equation. Substituting the flow veloc- ities u and  given by Eqs. 2.10 and 2.11, respectively, into continuity equation Eq. 2.15 yields Reynolds’ equation. Before doing this, the integrals (flow rates) in Eq. 2.15 are calculated:  h 0 udy= − h 3 12µ ∂p ∂x + h 2 (U 1 + U 2 ) (2.16)  h 0 wdy= − h 3 12µ ∂p ∂z + h 2 (W 1 + W 2 ) (2.17) where µ and p are assumed to be constant in the y direction (assumption 4, assump- tion 5). Substituting these two integrals into Eq. 2.15 gives the following equation: ∂ ∂x  h 3 µ ∂p ∂x  + ∂ ∂z  h 3 µ ∂p ∂z  = 6  (U 1 − U 2 ) ∂h ∂x + h ∂ ∂x (U 1 + U 2 ) + (W 1 − W 2 ) ∂h ∂z + h ∂ ∂z (W 1 + W 2 ) + 2(V 2 + V 1 )  (2.18) In many practical cases, the x axis can be taken as the direction of the relative motion of the two surfaces, in which case we have V 1 = W 1 = W 2 = 0. If the coeffi- cient of viscosity µ is constant (assumption 4), the above equation can be simplified as follows: ∂ ∂x  h 3 ∂p ∂x  + ∂ ∂z  h 3 ∂p ∂z  = 6µ  (U 1 − U 2 ) ∂h ∂x + h ∂ ∂x (U 1 + U 2 ) + 2V 2  (2.19) Furthermore, if the two surfaces are rigid, the second term of the right-hand side denoting expansion and contraction of the surfaces becomes zero, and the above equation becomes: ∂ ∂x  h 3 ∂p ∂x  + ∂ ∂z  h 3 ∂p ∂z  = 6µ  (U 1 − U 2 ) ∂h ∂x + 2V 2  (2.20) This is a pressure equation derived based on Reynolds’ assumptions and is called Reynolds’ equation. The previous equations, Eqs. 2.19 and 2.18, can be called gen- eralized Reynolds’ equations because they were derived without the rigid body as- sumption and the isoviscous assumption (in the x and z directions), respectively. If the flow is one dimensional and U 2 = V 2 = 0, the above equation becomes: ∂ ∂x  h 3 ∂p ∂x  = 6µU 1 ∂h ∂x (2.21) This is Reynolds’ equation in the simplest case. 18 2 Foundations of Hydrodynamic Lubrication 2.2.1 Interpretation of Reynolds’ Equation a. Mechanisms of Pressure Generation Equation 2.19 is shown below for U 2 = 0. Let us consider the meaning of Reynolds’ equation in this case. ∂ ∂x  h 3 ∂p ∂x  + ∂ ∂z  h 3 ∂p ∂z  = 6µ  U 1 ∂h ∂x + h ∂U 1 ∂x + 2V 2  (2.22) Fig. 2.7. Pressure distribution First, the left-hand side indicates approximately the average curvature of the pres- sure distribution surface as shown in Fig. 2.7. If the left-hand side is negative, it means that the pressure distribution is upward convex, or the pressure generated is positive. Second, the right-hand side represents the causes of pressure generation and the three terms correspond to the following three mechanisms of pressure generation, respectively: 1. The first term represents the wedge effect: pressure generation due to the fluid being driven from the thick end to the thin end of the wedge-shaped fluid film by the surface movement. 2. The second term is the stretch effect: pressure generation due to the variation of surface velocity from place to place. 3. The third term is the squeeze effect: pressure generation due to the variation of surface gap (film thickness). 2.2 Reynolds’ Theory of Hydrodynamic Lubrication 19 Fig. 2.8a-c. Mechanism of pressure generation. a wedge effect, b stretch effect, c squeeze effect The simplest examples of these three effects are shown in Fig. 2.8. Each example shows the case where the term is negative, i.e., the pressure is positive. Among these three, the wedge effect is most commonly seen, for example, in journal bearings. The squeeze effect plays an important role in the small-end bearings of crank rods and is fundamentally important also in animal joints, as pointed out by Reynolds in his paper of 1886 [3]. The stretch effect can be a problem when sliding surfaces are made of elastic materials such as rubber, but can usually be ignored. When a journal moves around in a bearing, an equivalent stretch effect can occur, but it is of the order of the effect of V 2 in Eq. 2.22 multiplied by h/R, and can usually be disregarded. In the case of lubrication during plastic working, for example lubrication between a roller and the workpiece in rolling or lubrication between the die and the workpiece in drawing, deformation of the worked surface becomes a problem. In these cases, since the surface is usually stretched, the generated pressure is negative and so the formation of the lubricant film is hard to explain hydrodynamically. Now, let us consider Eq. 2.20. It is written again here for the sake of convenience: ∂ ∂x  h 3 ∂p ∂x  + ∂ ∂z  h 3 ∂p ∂z  = 6µ  (U 1 − U 2 ) ∂h ∂x + 2V 2  (2.23) It is interesting to compare the meaning of the right-hand side of this equation in the case of an inclined pad bearing and in the case of a journal bearing. Consider the following conditions in an inclined pad bearing (V 2 = 0) as shown in Fig. 2.9a. RHS stands for “right-hand side.” 20 2 Foundations of Hydrodynamic Lubrication Fig. 2.9a-c. Inclined pad and journal bearings (1). a inclined pad bearing, b journal bearing, c simplified journal bearing 1. If U 1 = U and U 2 = U, RHS = 6µ  0 + 0  = 0 2. If U 1 = U and U 2 = 0, RHS = 6µ  U ∂h ∂x + 0  = 6µU ∂h ∂x 3. If U 1 = U and U 2 = −U, RHS = 6µ  2U ∂h ∂x + 0  = 12µU ∂h ∂x In case 1, the relative velocity of the two surfaces is 0 and so no pressure is generated. If cases 2 and 3 are compared, the relative velocity in case 3 is twice that in case 2, and accordingly the generated pressure is twice that of case 2. Turning to the journal bearing shown in Fig. 2.9b, let us expand the cylindrical surface of the journal to a plane as shown in Fig. 2.9c. This is different from Fig. 2.9a in the important point that the circumferential velocity U j of the journal is along the slope, and this is equivalent to a surface having the velocity of U 2 ≈ U j in the x direction and that of V 2 ≈ U j (∂h/∂x)inthey direction. Therefore, the following relations are obtained: [...]... 2 sin ϕ + 2 4 (1 + κ cos φ)3 (1 − 2 )5 /2 (3.11) J2 = J1 = dφ 1 = ϕ − κ sin ϕ 2 (1 + κ cos φ) (1 − 2 )3 /2 ϕ dφ = 1 + κ cos φ (1 − 2 )1 /2 (3. 12) (3.13) Now, Eq 3.7, the pressure distribution, becomes: 1 6µUR ϕ − κ sin ϕ c2 (1 − 2 )3 /2 hm 2 ϕ 2 + sin 2 + C2 − ϕ − 2 sin ϕ + 2 4 c(1 − 2 )5 /2 p(ϕ) = (3.14) where hm and C2 are integral constants, which can be determined under appropriate boundary... the relations sin2 φ + cos2 φ = 1, as follows: cos φ = cos ϕ − κ , 1 − κ cos ϕ sin φ = (1 − 2 )1 /2 sin ϕ 1 − κ cos ϕ (3.9) From these relations, we have the following: dφ = (1 − 2 )1 /2 dϕ 1 − κ cos ϕ (3.10) With these relations, the integrals in Eq 3.7 are calculated as follows (J1 will be used later): J3 = dφ 1 2 ϕ 2 + sin 2 = ϕ − 2 sin ϕ + 2 4 (1 + κ cos φ)3 (1 − 2 )5 /2 (3.11) J2 = J1 = dφ 1.. .2. 2 Reynolds’ Theory of Hydrodynamic Lubrication 1 If U1 = U and U j = U, RHS = 6µ 0 + 2U 2 If U1 = 0 and U j = U, RHS = 6µ −U 21 ∂h ∂h = 12 U ∂x ∂x ∂h ∂h ∂h + 2U = 6µU ∂x ∂x ∂x 3 If U1 = −U and U j = U, RHS = 6µ −2U ∂h ∂h + 2U = 0 ∂x ∂x It is interesting to compare these results with those for the case of Fig 2. 9a For example, if they are compared for case 2, the result, 6U(∂h/∂x),... redrawn as Fig 2. 10a,ba and 2. 10a,bb; considering case 2, it is clear that Fig 2. 10a,bb is equivalent to 2. 10a,ba if viewed upside down Therefore, the difference described above is only an apparent one In the case of journal bearings, the circumferential speed of the journal U2 is along the inclined surface, and Eq 2. 20 becomes: ∂h ∂ 3 ∂p ∂ 3 ∂p h + h = 6µ (U1 + U2 ) + 2V2 ∂x ∂x ∂z ∂z ∂x (2. 24) It is this... Surfaces and Moving Surfaces The upper surface of Fig 2. 10a,ba and the lower surface of Fig 2. 10a,bb are equivalent as stated above, and are called the stationary surface Similarly, the lower 22 2 Foundations of Hydrodynamic Lubrication surface of 2. 10a,ba and the upper surface of 2. 10a,bb are equivalent, and they are called the moving surface When 2. 10a,ba is regarded as an inclined pad bearing, the... effect of V2 ), which is positive with a magnitude twice that of the wedge effect Fig 2. 10a,b Inclined pad and journal bearings (2) a Fig 2. 9a with U1 = U and U2 = 0, b Fig 2. 9c with U1 = 0 and U j = U Although it can be said from this analysis that the mechanisms of pressure generation in an inclined pad bearing and a journal bearing are different, this is not necessarily true Figure 2. 9a and 2. 9c can... 6µU 2 − 3 dx h h (3.5) Substituting Eq 3.1 into the above equation gives 1 d p 6µUR 1 hm = − 2 2 dφ c (1 + κ cos φ)3 c (1 + κ cos φ) (3.6) 30 3 Fundamentals of Journal Bearings where, since Rb ≈ R j , both the radii are denoted by R and the relation x = Rφ is used To obtain the oil film pressure, Eq 3.6 is integrated again to give the following, in which C2 is an integral constant: p= 6µUR c2 hm dφ − 2. .. c (1 + κ cos φ) dφ + C2 (1 + κ cos φ)3 (3.7) To calculate the integrals on the right-hand side of the above equation, Sommerfeld introduced the following variable transform [1][5] (another method is given in Section 5 .2. 1): 1 − 2 (3.8) 1 + κ cos φ = 1 − κ cos ϕ where ϕ is the new variable In this variable transform, the range φ = 0 – 2 corresponds to the same range ϕ = 0 – 2 , and this makes it easy... we have: 3.1 Circular Journal Bearings 27 h = (Rb − R j ) + e cos φ = c + e cos φ Thus, oil film thickness h is obtained with good approximation as: h = c (1 + κ cos φ) (3.1) 3.1.3 Bearing Length (Bearing Width) The oil film pressure in a journal bearing in the stationary state is given by the following Reynolds’ equation, which is derived from Eq 2. 20 or Eq 2. 24, where the x axis is taken in the circumferential... Fig 3 .2, the straight line ABC is drawn through Ob Let A denote the foot of the perpendicular dropped from O j to the above straight line, and B and C denote the intersections of the straight line and the journal surface and the inner surface of bearing metal, respectively Then: h = BC = AC − AB where Therefore, AC = Rb + e cos φ, AB = h = Rb + e cos φ − R j R j 2 − (e sin φ )2 1 − (e/R j )2 sin2 φ Since . φ) 3 = 1 (1 − κ 2 ) 5 /2  ϕ − 2 sin ϕ + κ 2 ϕ 2 + κ 2 4 sin 2  (3.11) J 2 =  dφ (1 + κ cos φ) 2 = 1 (1 − κ 2 ) 3 /2  ϕ − κ sin ϕ  (3. 12) J 1 =  dφ 1 + κ cos φ = ϕ (1 − κ 2 ) 1 /2 (3.13) Now,. distribution, becomes: p(ϕ) = 6µUR c 2  1 (1 − κ 2 ) 3 /2  ϕ − κ sin ϕ  − h m c(1 − κ 2 ) 5 /2  ϕ − 2 sin ϕ + κ 2 ϕ 2 + κ 2 4 sin 2  + C 2 (3.14) where h m and C 2 are integral constants, which. dimensional and U 2 = V 2 = 0, the above equation becomes: ∂ ∂x  h 3 ∂p ∂x  = 6µU 1 ∂h ∂x (2. 21) This is Reynolds’ equation in the simplest case. 18 2 Foundations of Hydrodynamic Lubrication 2. 2.1 Interpretation