Hydrodynamic Lubrication 2011 Part 3 pps

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 3.2 Infinitely Long Bearings 37 = 2πµUR 2 L c 2(1 + 2κ 2 ) (2 + κ 2 )(1 − κ 2 ) 1/2 ≡ M · ¯m j (κ) (3.35) M b =  2π 0  µU h − h 2R dp dφ  R 2 Ldφ = 2πµUR 2 L c 2(1 − κ 2 ) 1/2 2 + κ 2 ≡ M · ¯m b (κ) (3.36) where M is the frictional moment M of Petrov’s law, Eq. 3.32, and ¯m j (κ) and ¯m b (κ) are nondimensional moments which are functions of κ only. If κ = 0, obviously M j = M b = M. Equation 3.25, ¯m j (κ) and ¯m b (κ) are shown in Fig. 3.8. It should be noted that M j and M b are not equal, M j being always larger than M b . Calculation shows that the difference is given by: M j − M b = eP 1 (3.37) This shows that the difference of M j and M b can be attributed to the moment of the bearing load P 1 with respect to the bearing center, the eccentricity being e. If the frictional moment on the journal is known, the frictional loss (heat genera- tion) L s is calculated as follows with the angular velocity ω: L s = 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 /(RP 1 ), f b = M b /(RP 1 ) (3.39) 3.2.3 Infinitely Long Bearing Under G ¨ umbel’s Condition In the previous section, various characteristics of an infinitely long bearing were derived under Sommerfeld’s boundary condition. However, one of the results, that the locus of the journal center is a straight line perpendicular to the load direction (see Fig. 3.7), contradicts actual observation in many practical cases. In fact, the locus is a straight line only when the bearing pressure is very low. Under usual conditions, it is rather like the profile of a half-moon. This disagreement seems to be attributable to the inclusion of the large negative pressure from the theory into the calculation of the oil film force. In the actual oil film, the negative pressure cannot be very low because of oil film rupture and air inflow from the bearing ends. In this section, assuming no negative pressure exists in the oil film, let us, for simplicity, adopt G ¨ umbel’s boundary condition in which the negative pressure in the theory is replaced by zero, and calculate various characteristics of a bearing under this condition. a. Oil Film Pressure Under G ¨ umbel’s boundary condition, only the positive pressure in the shaded region of Fig. 3.10, i.e., the range 0 ≤ φ ≤ π, is considered. The negative pressure in the range π ≤ φ ≤ 2π is simply assumed to be zero. The pressure in the region of positive pressure is assumed to be the same as that obtained under Sommerfeld’s condition. 38 3 Fundamentals of Journal Bearings Fig. 3.10. The oil film and oil film force under G ¨ umbel’s boundary condition b. Oil Film Force and Load Capacity The oil film force under G ¨ umbel’s boundary condition is obtained by integrating the oil film pressure p of Eq. 3.19 over the range 0 ≤ φ ≤ π. Then, the balance of the oil film force and the bearing load can, when resolved into components in the direction of eccentricity and that perpendicular to it, be written as follows: LR  π 0 p cos φ dφ + P 1 cos θ = 0 (3.40) LR  π 0 p sin φ dφ − P 1 sin θ = 0 (3.41) where L and R are the length and the radius of the bearing metal, respectively. The integration can be performed in the same way as in the previous section, giving the following results: P 1 cos θ = µUL  R c  2 12κ 2 (2 + κ 2 )(1 − κ 2 ) (3.42) P 1 sin θ = µUL  R c  2 6πκ (2 + κ 2 )(1 − κ 2 ) 1/2 (3.43) It is interesting to compare these results with those obtained under Sommerfeld’s condition in the previous section. Under Sommerfeld’s condition, the positive pres- sure and the negative pressure generated in the oil film cancel each other out in the direction of eccentricity, giving P 1 cos θ = 0, whereas in the direction perpendicular 3.2 Infinitely Long Bearings 39 to it, they assist each other, giving the large value P 1 sin θ. Under G ¨ umbel’s condi- tion, in contrast, since the negative pressure is assumed to be zero, no cancelling takes place in the direction of eccentricity, leaving P 1 cos θ  0, and in the direction perpendicular to it, P 1 sin θ becomes one-half of that under Sommerfeld’s condition. From Eq. 3.42 and Eq. 3.43, the bearing load P 1 is obtained as: P 1 = µUL  R c  2 6κ{4κ 2 + π 2 (1 − κ 2 )} 1/2 (2 + κ 2 )(1 − κ 2 ) . (3.44) This can be rewritten as follows by using the Sommerfeld number, S : 1 S = 6πκ{4κ 2 + π 2 (1 − κ 2 )} 1/2 (2 + κ 2 )(1 − κ 2 ) (3.45) where S = (µN/p m )(R/c) 2 as before, with N = U/(2πR) and p m = P 1 /(2RL). Further, dividing Eq. 3.43 by Eq. 3.42 gives the relation between the attitude angle θ and the eccentricity ratio κ: tan θ = π(1 − κ 2 ) 1/2 2κ (3.46) This is a polar coordinates expression of the journal center locus as shown in Fig. 3.11. This is like a half-moon and is close to the actual shape of the locus. Fig. 3.11. Locus of the journal center for a bearing of infinite length under G ¨ umbel’s boundary condition 40 3 Fundamentals of Journal Bearings c. Frictional Moment Although the oil film exists in the whole circumference of the bearing, if (∂p/∂φ) = 0 is assumed in the range π ≤ φ ≤ 2π, the 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: M j = 2πµUR 2 L c 1 2(1 − κ 2 ) 1/2  2 + 3κ 2 2 + κ 2  ≡ M ¯m j (κ) (3.47) M b = 2πµUR 2 L c 1 2(1 − κ 2 ) 1/2  2 − 3κ 2 2 + κ 2  ≡ M ¯m b (κ) (3.48) where M is the M in Petrov’s law, Eq. 3.32. In this case also, M j and M b are not equal and their difference is found to be equal to the moment of the bearing load with respect to the bearing center, as in the previous section. The relation in this case is: M j − M b = (e sin θ) P 1 (3.49) 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: f j = M j /(RP 1 ), f b = M b /(RP 1 ) (3.50) When the journal and the bearing metal are concentric, i.e., if κ = 0, we have: M j = M b = M (3.51) Equations 3.45 – 3.48 are shown in Fig. 3.12. Fig. 3.12. Characteristics of an infinitely long bearing under G ¨ umbel’s condition [...]... h3 i−1/2, j ∆x ∆x ∆x pi, j+1 − pi, j pi, j − pi, j−1 − h3 j−1/2 h3 j+1/2 i, i, ∂ 3 ∂p ∆z ∆z h = ∂z ∂z ∆z ∂h hi+1/2, j − hi−1/2, j = ∂x ∆x h3 i+1/2, j ∂ 3 ∂p h = ∂x ∂x 45 (3. 69) (3. 70) (3. 71) Substituting these into Eq 3. 68 and solving for pi, j , we obtain pi, j in the following form: pi, j = a0 + a1 pi+1, j + a2 pi−1, j + a3 pi, j+1 + a4 pi, j−1 (i = 1, 2, · · · , m − 1; j = 1, 2, · · · , n − 1) (3. 72)... 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... Method to Hydrodynamic Lubrication Problems (Part 1)” (in Japanese), Trans JSME, Vol 37 , No 295, March 1971, pp 5 83 - 592 11 S Wada and H Hayashi, “Application of Finite Element Method to Hydrodynamic Lubrication Problems (Part 2)” (in Japanese), Trans JSME, Vol 37 , No 295, March 1971, pp 5 93 - 601 12 T Kato and Y Hori, “Matrix Form of Reynolds’ Equation”, JSME International Journal, Vol 31 , No 2,... 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... 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... 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 are discretized on the grid as follows: 3. 4 Finite... (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,... ∂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... 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 = hm p ¯ A value of m = 2 or 1.5, for example,is used [6][ 13] (3. 74) 46 3 Fundamentals of Journal Bearings If pressure pi, j is obtained, the component of the oil film force P in the... follows: Pκ = P cos θ = R pi, j cos φi ∆z∆φ (3. 75) pi, j sin φi ∆z∆φ (3. 76) i, j Pθ = P sin θ = R i, j 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¨ ge zur Kenntniss der Gleitlagerreibung”, VDI - Forschungsheft 38 6, a Berlin, 1 937 3 G.B DuBois and F.W Ocvirk, “Analytical Derivation . 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. follows: 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,. 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