Hydrodynamic Lubrication 2009 Part 3 docx

3.1 Circular Journal Bearings 27 h = (R b − 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 fol- lowing Reynolds’ equation, which is derived from Eq. 2.20 or Eq. 2.24, where the x axis is taken in the circumferential direction of the bearing and the z axis is in the axial direction: ∂ ∂x  h 3 ∂p ∂x  + ∂ ∂z  h 3 ∂p ∂z  = 6µU 2 ∂h ∂x (3.2) where p(x, z) is the oil film pressure, h(x, z) is the oil film thickness, and U 2 is the circumferential velocity of the journal. General solutions of Eq. 3.2 cannot be obtained analytically, therefore, the fol- lowing approximations or numerical solutions are usually employed. – The infinite length approximation assumes a sufficiently long bearing in the ax- ial direction and neglects the second term of the left-hand side of the above equation, allowing it to be solved analytically. – The short bearing approximation assumes a sufficiently short bearing in the axial direction and neglects the first term of the left-hand side of the above equation, allowing it to be solved analytically. –Forfinite length bearings, the equation is solved numerically by, for exam- ple, the finite difference method or the finite element method, or by an approximate analytical method by developing the pressure as a series of trigonometric functions. 3.1.4 Boundary Conditions for the Oil Film Boundary conditions are required to solve Reynolds’ equation, Eq. 3.2. In the case of a journal bearing, the boundary condition at an end of the bearing is simply that the oil film pressure is equal to ambient air pressure, because the boundary of the oil film at the end of the bearing is clear-cut. In the circumferential direction, however, since rupture of the oil film in the domain where the clearance is growing is a complicated phenomenon (for various reasons including the inflow of air), it is a difficult problem to determine the boundary of the oil film. For simplicity, consider an infinitely long journal bearing. In this case, if it is assumed that the bearing clearance is completely filled with oil (no film rupture assumed), Reynolds’ equation gives positive pressure in the semicircle where the bearing clearance decreases and negative pressure in the semicircle where the bear- ing clearance increases, and their absolute values are equal. This will be true if the bearing pressure is sufficiently low. When the bearing pressure is relatively high, however, although the positive pressure can go up without limitation, the negative 28 3 Fundamentals of Journal Bearings pressure cannot go below a certain limit. When the absolute value of negative pres- sure reaches some limit, oil film rupture occurs and the pressure in the region of rupture will not go down further. It is difficult to know exactly the position at which oil film rupture will occur (the oil film terminal point) and the pressure in the area of oil film rupture. Fig. 3.4. Boundary conditions for the oil film Simplified boundary conditions, as shown in Fig. 3.4, are often used in practice. (a) Sommerfeld’s boundary condition (A. J. W. Sommerfeld, 1869 – 1951) as- sumes that p = 0atφ = 0 and 2π. Pressure distribution is calculated without consid- ering oil film rupture, and the positive and negative pressures obtained are taken into consideration as they are. This is applicable when the bearing pressure is low. (b) For G¨umbel’s boundary condition (L. K. F. G ¨ umbel, 1874 – 1923), pressure distribution is calculated without considering oil film rupture as before, but only the positive pressure in the semicircle φ = 0–π is considered. The negative pressure in the remaining semicircle is set to “zero” (i.e., atmospheric pressure). The oil film is thus assumed to terminate at φ = π. The starting position is φ = 0. This is accept- 3.2 Infinitely Long Bearings 29 able when the bearing pressure is fairly high. It is also called the half Sommerfeld’s condition. (c) In Reynolds’ boundary condition (O. Reynolds, 1842 – 1912), the oil film is assumed to terminate at a certain position (φ = π + δ) at which both the pressure and pressure gradient are zero, simultaneously. This condition eliminates a discontinuity of oil flow at φ = π, a physical contradiction involved in G ¨ umbel’s condition. It is now necessary, however, to determine δ. This is also known as Swift–Stieber’s condition. (d) In the boundary condition of separation, the terminal position of the oil film is determined from the separation conditions of flow [7] [8]. (e) Another approach divides the cause of pressure generation into the wedge effect and the squeeze effect and applies the boundary condition p = 0atφ = (0,π) and p = 0atφ = (π/2, 3π/2) to the two effects, separately. 3.2 Infinitely Long Bearings The basic characteristics of an infinitely long journal bearing are derived here. Reynolds’ equation is solved first to determine the oil film pressure, which is then integrated to obtain the oil film force. From the equilibrium of the oil film force and bearing load, values for the eccentricity ratio, load capacity, and frictional moment are calculated. 3.2.1 Oil Film Pressure In the case of an infinitely long bearing, it can be assumed that the oil film pressure does not change in the axial direction, or (∂p/∂z) = 0. Therefore, with U = U 2 , Eq. 3.2 can be simplified as follows: d dx  h 3 dp dx  = 6µU dh dx (3.3) Integrating this with respect to x yields: h 3 dp dx = 6µU(h + C 1 ) (3.4) where C 1 is an integral constant. Letting the film thickness at the maximum film pressure position (where dp/dx = 0) be h m ,wehaveC 1 = −h m . Note that h m is still unknown. Using h m in the above equation leads to: dp dx = 6µU  1 h 2 − h m h 3  (3.5) Substituting Eq. 3.1 into the above equation gives dp dφ = 6µUR c 2  1 (1 + κ cos φ) 2 − h m c 1 (1 + κ cos φ) 3  (3.6) 30 3 Fundamentals of Journal Bearings where, since R b ≈ 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 C 2 is an integral constant: p = 6µUR c 2   dφ (1 + κ cos φ) 2 − h m c  dφ (1 + κ cos φ) 3  + C 2 (3.7) To calculate the integrals on the right-hand side of the above equation, Sommer- feld introduced the following variable transform [1][5] (another method is given in Section 5.2.1): 1 + κ cos φ = 1 − κ 2 1 − κ cos ϕ (3.8) where ϕ is the new variable. In this variable transform, the range φ = 0–2π corre- sponds to the same range ϕ = 0–2π, and this makes it easy to handle the boundary conditions. First, cos φ is obtained from Eq. 3.8, and then sin φ is found from the relations sin 2 φ + cos 2 φ = 1, as follows: cos φ = cos ϕ − κ 1 − κ cos ϕ , sin φ = (1 − κ 2 ) 1/2 sin ϕ 1 − κ cos ϕ (3.9) From these relations, we have the following: dφ = (1 − κ 2 ) 1/2 1 − κ cos ϕ dϕ (3.10) With these relations, the integrals in Eq. 3.7 are calculated as follows (J 1 will be used later): J 3 =  dφ (1 + κ cos φ) 3 = 1 (1 − κ 2 ) 5/2  ϕ − 2κ sin ϕ + κ 2 ϕ 2 + κ 2 4 sin 2ϕ  (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, Eq. 3.7, the pressure 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 − 2κ sin ϕ + κ 2 ϕ 2 + κ 2 4 sin 2ϕ  + C 2 (3.14) where h m and C 2 are integral constants, which can be determined under appropriate boundary conditions. Calculations will be continued hereafter under typical bound- ary conditions. 3.2 Infinitely Long Bearings 31 3.2.2 Infinitely Long Bearing Under Sommerfeld’s Condition a. Oil Film Pressure The basic characteristics of an infinitely long journal bearing are investigated un- der Sommerfeld’s boundary condition. The boundary conditions used here are as follows: p = 0atφ = 0 and φ = 2π (3.15) where the ambient pressure is assumed to be p = 0. As mentioned above, the bound- ary conditions with respect to ϕ are also: p = 0atϕ = 0 and ϕ = 2π (3.16) where the integral constants C 2 and h m are determined as follows: C 2 = 0 (3.17) h m = 2(1 − κ 2 ) 2 + κ 2 c (3.18) Substituting these values into Eq. 3.14 and returning the variable ϕ to φ gives the following pressure distribution. p(φ) = 6µUR c 2 κ(2 + κ cos φ) sin φ (2 + κ 2 )(1 + κ cos φ) 2 (3.19) ≡ 6µUR c 2 ¯p(κ, φ) (3.20) The nondimensional pressure distribution function ¯p in Eq. 3.20 is shown against φ in Fig. 3.5. The parameter is the eccentricity ratio κ. As shown in Fig. 3.5, the pressure distribution is symmetric with respect to the point (φ = π,¯p = 0) and the absolute values of the highest and the lowest pres- sures are equal. The position φ 0 of these extrema are determined from the condition dp/dφ = 0 as follows. cos φ 0 = − 3κ 2 + κ 2 (3.21) This shows that, with increase in the eccentricity ratio κ from 0 to 1, the position of the highest pressure φ 0 moves in the direction of rotation of the journal from π/2 toward π, while that of the minimum pressure φ 0 moves in the opposite direction from 3π/2towardπ. b. Oil Film Force and Load Capacity Integration of the oil film pressure gives the oil film force P. This must balance the bearing load P 1 . The balance of the forces, if resolved into two directions (one in the 32 3 Fundamentals of Journal Bearings Fig. 3.5. Pressure distribution in an infinitely long bearing using Sommerfeld’s condition eccentricity direction and the other perpendicular to it), can be expressed as follows (see Fig. 3.6): LR  2π 0 p cos φ dφ + P 1 cos θ = 0 (3.22) LR  2π 0 p sin φ dφ − P 1 sin θ = 0 (3.23) where L and R are the length and the radius of the bearing, respectively, and p is the oil film pressure given by Eq. 3.19. If Eq. 3.22 is integrated by parts, we have: P 1 cos θ = −LR  p sin φ  2π 0 + LR  2π 0 dp dφ sin φ dφ Since the pressure p is finite, the first term of the right-hand side is clearly zero. The second term can be calculated by using Eq. 3.6, and will turn out to be zero as follows: 2nd term = 6µUL  R c  2   2π 0 sin φ (1 + κ cos φ) 2 dφ − h m c  2π 0 sin φ (1 + κ cos φ) 3 dφ  = 6µUL  R c  2  1 κ(1 + κ cos φ) − h m c 1 2κ(1 + κ cos φ) 2  2π 0 = 0 Therefore, P 1 cos θ = 0 (3.24) If P 1  0, the attitude angle θ will be determined as follows in the range of (0 – π). 3.2 Infinitely Long Bearings 33 Fig. 3.6. Oil film and oil film force under Sommerfeld’s condition θ = π 2 (3.25) This means that the bearing load P 1 and the line of the eccentricity are mutually perpendicular, as shown in Fig. 3.7. In this case, the locus of the journal center is a straight line that is normal to the bearing load P 1 . The fact that the oil film force has a component perpendicular to the eccentricity direction (in this case only a perpendicular component) is one of the most unfavor- able characteristics of the oil film force in a journal bearing and it causes whirling of the shaft (see Chapter 5). If Eq. 3.23 is similarly integrated by parts, since P 1 sin θ = P 1 ,wehave: P 1 = − LR  p cos φ  2π 0 + LR  2π 0 dp dφ cos φ dφ = 6µUL  R c  2   2π 0 cos φ (1 + κ cos φ) 2 dφ − h m c  2π 0 cos φ (1 + κ cos φ) 3 dφ  The integrands on the right-hand side are expanded into partial fractions as follows: cos φ (1 + κ cos φ) 2 = 1 κ  1 (1 + κ cos φ) − 1 (1 + κ cos φ) 2  cos φ (1 + κ cos φ) 3 = 1 κ  1 (1 + κ cos φ) 2 − 1 (1 + κ cos φ) 3  Integrating these functions with recourse to Sommerfeld’s transform of variables gives P 1 : 34 3 Fundamentals of Journal Bearings Fig. 3.7. Position of the journal center in an infinitely long bearing under Sommerfeld’s con- dition P 1 = 6µUL  R c  2 1 κ  h m c J 3 −  h m c + 1  J 2 + J 1  2π 0 (3.26) Substituting J 3 , J 2 , J 1 , and h m previously obtained into the above equation gives the bearing load P 1 as follows: P 1 = P 1θ = µUL  R c  2 12πκ (2 + κ 2 )(1 − κ 2 ) 1/2 (3.27) This equation can be rewritten using the average bearing pressure p m = P 1 /(2RL) and the number of revolutions of the shaft per unit time N = U/(2πR): p m µN  c R  2 = 12π 2 κ (2 + κ 2 )(1 − κ 2 ) 1/2 (3.28) The left-hand side of Eq. 3.28 is a combination of the dimensions of the bearing and some variables representing its operating conditions, and is nondimensional as a whole. The right-hand side is a nondimensional function of the eccentricity ratio κ only. The reciprocal of the nondimensional quantity on the left-hand side is called Sommerfeld’s number and is usually denoted by S , i.e., S ≡ µN p m  R c  2 (3.29) Equation 3.28 can be rewritten as follows using S : 1 S = 12π 2 κ (2 + κ 2 )(1 − κ 2 ) 1/2 (3.30) The relation between 1/S and κ is shown in Fig. 3.8. If the bearing dimensions, oil viscosity, bearing pressure, and the number of journal revolutions per unit time 3.2 Infinitely Long Bearings 35 are given, Sommerfeld’s number S is determined by Eq. 3.29, and the corresponding eccentricity ratio κ can be determined by Eq. 3.30 or from Fig. 3.8. Since the same Sommerfeld’s number gives the same eccentricity ratio κ even though the bearing dimensions and operating conditions are different, it can be said that Sommerfeld’s number is an important quantity relating to the similarity rule for a bearing. Further, the definition of Sommerfeld’s number, Eq. 3.29, shows how the various factors of a bearing are related to the operating condition of a bearing. Fig. 3.8. Bearing characteristics for an infinitely long bearing under Sommerfeld’s conditions. S , Sommerfeld’s number; ¯m j and ¯m b , nondimensinal, frictional moments; θ, attitude angle c. Minimum Clearance and Load Capacity If the eccentricity ratio κ is thus determined, the minimum clearance of the bearing is obtained from Eq. 3.1 as: h min = c (1 −κ) (3.31) Then, the bearing load P 1 corresponding to the allowable minimum value of h min (this depends on the surface roughness and machining accuracy among other factors) will be the maximum allowable load (load capacity) of the bearing. d. Frictional Moment and Frictional Loss The frictional moment due to the shear stress in the oil film acts on the rotating journal. Its reaction acts on the bearing bush. If the journal and the bearing are concentric (Fig. 3.9), the frictional moment M is simply given as follows, multiplying the shear stress τ = µ(U/c) by the radius R and the circumferential area 2πRL. 36 3 Fundamentals of Journal Bearings Fig. 3.9. Derivation of Petrov’s formula M = 2πµUR 2 L c (3.32) This is called Petrov’s law (N.P. Petrov, 1836 – 1920) and is convenient for a rough estimation of the frictional moment. If the journal and the bearing are eccentric, the frictional moment on the journal can be calculated by integrating the oil film shear stress over the journal surface and that on the bearing bush by integrating the oil film shear stress over the bush surface. The shear stress is calculated as follows by using Eq. 2.10 for the flow velocity: τ = µ ∂u ∂y = µ ∂ ∂y   1 − y h  U 1 + y h U 2  − 1 2µ ∂p ∂x y(h − y)  = − µ U 1 − U 2 h − h 2 ∂p ∂x  1 − 2y h  Let U 1 = 0 and U 2 = U. Then the shear stresses at y = h and y = 0 are obtained as follows: τ y=h = µU h + h 2 dp dx (3.33) τ y=0 = µU h − h 2 dp dx (3.34) Multiplying these by the bearing radius and integrating them over the corresponding circumferential surface gives the frictional moment acting on the journal and the bearing bush as follows, where J 1 and J 2 (Eq. 3.13, etc.) are used: M j =  2π 0  µU h + h 2R dp dφ  R 2 Ldφ [...]... 2R 0 p cos φdzdφ + P1 cos θ = 0 (3. 57) 0 π L/2 2R 0 p sin φdzdφ − P1 sin θ = 0 (3. 58) 0 By substituting Eq 3. 56 into the above equations and using the partial fraction decomposition method and J3 , J2 , and J1 in Eq 3. 11, etc, the components of bearing load can be derived: R c R P1 sin θ = µUR c P1 cos θ = µUR 2 2 L D L D 3 3 8κ2 (1 − κ2 )2 2πκ (1 − κ2 )3/ 2 (3. 59) (3. 60) From these equations, we have... frictional moments acting on the journal and the bearing metal are found as follows, in the same way as for Eq 3. 35 and Eq 3. 36: 1 2πµUR2 L 3 2 2+ ≡ M m j (κ) ¯ c 2 + κ2 2(1 − κ2 )1/2 1 2πµUR2 L 3 2 Mb = 2− ≡ M mb (κ) ¯ c 2 + κ2 2(1 − κ2 )1/2 Mj = (3. 47) (3. 48) where M is the M in Petrov’s law, Eq 3. 32 In this case also, M j and Mb are not equal and their difference is found to be equal to the moment of the... 3 κ{16κ2 + π2 (1 − κ2 )}1/2 (1 − κ2 )2 (3. 61) This equation can be rewritten using the Sommerfeld number, S = (µN/pm )(R/c)2 , as: L 2 (1 − κ2 )2 S = (3. 62) D πκ{16κ2 + π2 (1 − κ2 )}1/2 Further, from Eq 3. 59 and Eq 3. 60, we can write the locus of the bearing center as: π(1 − κ2 )1/2 (3. 63) 4κ This is shown in Fig 3. 13 The locus has the form of a half-moon that is a little thinner than that of Fig 3. 11.. .3. 2 Infinitely Long Bearings = 2(1 + 2κ2 ) 2πµUR2 L ≡ M · m j (κ) ¯ c (2 + κ2 )(1 − κ2 )1/2 2π Mb = 0 37 (3. 35) µU h dp 2 − R L dφ h 2R dφ 2πµUR2 L 2(1 − κ2 )1/2 = ≡ M · mb (κ) ¯ (3. 36) c 2 + κ2 where M is the frictional moment M of Petrov’s law, Eq 3. 32, and m j (κ) and mb (κ) ¯ ¯ are nondimensional moments which are functions of κ only If κ = 0, obviously ¯ ¯ M j = Mb = M Equation 3. 25, m... Sommerfeld’s condition From Eq 3. 42 and Eq 3. 43, the bearing load P1 is obtained as: P1 = µUL R c 2 6κ{4κ2 + π2 (1 − κ2 )}1/2 (2 + κ2 )(1 − κ2 ) (3. 44) This can be rewritten as follows by using the Sommerfeld number, S : 1 6πκ{4κ2 + π2 (1 − κ2 )}1/2 = S (2 + κ2 )(1 − κ2 ) (3. 45) where S = (µN/pm )(R/c)2 as before, with N = U/(2πR) and pm = P1 /(2RL) Further, dividing Eq 3. 43 by Eq 3. 42 gives the relation... recognized that the results coincide well with practice for small eccentricity ratios 3. 3.1 Oil Film Pressure If the film thickness h is constant in the z direction, Eq 3. 52 becomes: ∂2 p 6µU dh = 3 ∂z2 h dx (3. 53) Substituting the bearing clearance h = c (1 + κ cos φ) into this yields: ∂2 p 6µU κ sin φ =− 2 ∂z2 c R (1 + κ cos φ )3 (3. 54) This can be integrated easily because the right-hand side is constant with... case is: (3. 49) M j − Mb = (e sin θ) P1 Whereas θ = π/2 in the previous section, θ is now a function of κ The frictional coefficients at the journal surface and the bearing metal surface are defined as: (3. 50) f j = M j /(RP1 ), fb = Mb /(RP1 ) When the journal and the bearing metal are concentric, i.e., if κ = 0, we have: M j = Mb = M (3. 51) Equations 3. 45 – 3. 48 are shown in Fig 3. 12 Fig 3. 12 Characteristics... before: x= ¯ x , D z= ¯ z , L ¯ h h= , c p= ¯ p c µω R 2 (3. 67) Then, Reynolds’ equation, Eq 3. 66 can be nondimensionalized in the following form, where U = Rω: ∂ 3 ∂p D ¯ h + ∂x ¯ ∂x ¯ L 2 ¯ ∂ 3 ∂p ¯ ∂h h = 12 ∂¯ z ∂¯ z ∂x ¯ (3. 68) Fig 3. 14 Grid on a lubricating surface The lubricating domain is divided into a grid pattern as shown in Fig 3. 14 and then the three derivatives in the above equation... ∂p 6µU κ sin φ =− 2 z ∂z c R (1 + κ cos φ )3 (3. 55) If this is integrated again under the assumption that the pressure is zero, or p = 0, at the bearing ends, z = ±(L/2), the pressure distribution is obtained as follows, L being the length (width) of the bearing metal: 42 3 Fundamentals of Journal Bearings L2 3 U κ sin φ − z2 2 R (1 + κ cos φ )3 4 c p(φ, z) = (3. 56) It is seen in the above equation that... journal is known, the frictional loss (heat generation) L s is calculated as follows with the angular velocity ω: Ls = ω M j (3. 38) The frictional coefficients at the journal surface and the inner surface of bearing bush are defined as follows f j = M j /(RP1 ), fb = Mb /(RP1 ) (3. 39) ¨ 3. 2 .3 Infinitely Long Bearing Under Gumbel’s Condition In the previous section, various characteristics of an infinitely long . way as for Eq. 3. 35 and Eq. 3. 36: M j = 2πµUR 2 L c 1 2(1 − κ 2 ) 1/2  2 + 3 2 2 + κ 2  ≡ M ¯m j (κ) (3. 47) M b = 2πµUR 2 L c 1 2(1 − κ 2 ) 1/2  2 − 3 2 2 + κ 2  ≡ M ¯m b (κ) (3. 48) where. to: dp dx = 6µU  1 h 2 − h m h 3  (3. 5) Substituting Eq. 3. 1 into the above equation gives dp dφ = 6µUR c 2  1 (1 + κ cos φ) 2 − h m c 1 (1 + κ cos φ) 3  (3. 6) 30 3 Fundamentals of Journal Bearings where,. π 2 (1 − κ 2 )} 1/2 (3. 62) Further, from Eq. 3. 59 and Eq. 3. 60, we can write the locus of the bearing center as: tan θ = π(1 − κ 2 ) 1/2 4κ (3. 63) This is shown in Fig. 3. 13. The locus has the