Hydrodynamic Lubrication 2011 Part 2 doc
... the negative 2. 2 Reynolds’ Theory of Hydrodynamic Lubrication 21 1. If U 1 = U and U j = U,RHS= 6µ 0 + 2U ∂h ∂x = 12 U ∂h ∂x 2. If U 1 = 0 and U j = U, RHS = 6µ −U ∂h ∂x + 2U ∂h ∂x = ... 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. 2. 2 Reynolds’ Theory of Hydrodynamic Lubrication...
Ngày tải lên: 11/08/2014, 08:21
Hydrodynamic Lubrication 2011 Part 11 docx
... ν t ∂u ∂y 2 − ε − 2 ∂k 1 /2 ∂y 2 (9 .22 ) Dε Dt = ∂ ∂y ν + ν t σ ε ∂ε ∂y + C ε1 ν t ∂u ∂y 2 ε k −C 2 1 − 0.3exp(−R t 2 ) ε 2 k + 2 ν t ∂ 2 u ∂y 2 2 (9 .23 ) where σ k , ... C µ k 2 ε (9 .24 ) 20 6 9 Turbulent Lubrication τ + = l m + 2 ∂u + ∂y + 2 (9. 32) reveals that this is of the same form as Eq. 9. 12, and should apply in the r...
Ngày tải lên: 11/08/2014, 08:21
... m WWjrjjW mmm l ⎛⎞ ⎜⎟ ′′ =+−−+ ⎜⎟ ⎝⎠ θ ϕ θϕ ϕ (9A) () () 2 2 22 2 2 2 2 cos 31cos3 cos CC R P R r IImhrI mWmrj W l ⎛⎞ ⎡ ⎤ ′ =+ + + + − + ⎜⎟ ⎢ ⎥ ⎣ ⎦ ⎝⎠ θ θθ ϕ (10A) () () 2 2 2 cos cos 2tantan cos cos 23 22 1sincos2 32 R PR R rr II ll mW ... ⎢⎥ − ⎢⎥⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦⎣⎦ − − + − − 32 32 2 1 22 11 32 32 2 2 12 6 00 12 6 00 64 00...
Ngày tải lên: 19/06/2014, 15:20
Supply Chain Management 2011 Part 2 docx
... et al. (20 06); (b) Melton (20 05); (c) Tang (20 06); (d) Christopher & Peck (20 04); (e) Iakovou et al. (20 07); (f) Zhu et al. (20 08); (g) Gottberg et al. (20 06); (h) Bowen et al. (20 01); (i) ... Malone, L. C. (20 08). Determining a cost-effective customer service level. Supply Chain Management: An International Journal, Vol. 13, No. 3, pp. 22 5 -23 2 Kainuma, Y. & Tawara, N...
Ngày tải lên: 20/06/2014, 04:20
... 2, 1 1 ,2 1 ]][[),( nrcn AHHjiA 2, 1 1 ,2 11 ]][[),( nrcn AGHjiD 2, 1 1 ,2 12 ]][[),( nrcn AHGjiD 2, 1 1 ,2 13 ]][[),( nrcn AGGjiD (7) where (i,j) Є R 2 , ... financial support from the Academy of Finland (project numbers: 123 579, 1.1 .20 08-31. 12 .20 11, and 126 873, 1.1 .20 09-31. 12 .20 11) . 9. References Cerutti, S., Bersani,...
Ngày tải lên: 21/06/2014, 19:20
Recent Advances in Biomedical Engineering 2011 Part 2 docx
... 1 1 2 0.0800 - 0. 027 1 0.0 927 0.0670 -0.0046 0.0307 - * 0.0143 0. 029 5 0. 027 3 0.3787 -0.0 729 0.0686 G W W 1.0094 1 .26 10 0.80 62 2. 1595 Determinant (G) = 2. 2588 ... 1 1 2 0.0800 - 0. 027 1 0.0 927 0.0670 -0.0046 0.0307 - * 0.0143 0. 029 5 0. 027 3 0.3787 -0.0 729 0.0686 G W W 1.0094 1 .26 10 0.80 62 2. 15...
Ngày tải lên: 21/06/2014, 19:20
Wireless Sensor Networks Application Centric Design 2011 Part 2 docx
... 10 .20 5.44 5.34 20 .00 9.08 16.08 8.76 8.90 30.00 12. 76 20 .58 12. 20 12. 28 40.00 15.90 25 .48 15.06 14.76 50.00 18.30 31 .22 17.98 17.80 60.00 21 .64 34.76 20 . 62 20 .28 70.00 23 . 92 37.74 22 .56 23 .08 ... 10 .20 5.44 5.34 20 .00 9.08 16.08 8.76 8.90 30.00 12. 76 20 .58 12. 20 12. 28 40.00 15.90 25 .48 15.06 14.76 50.00 18.30 31 .22 17.98 17.80 60.00 21 .64 34.76 20 ....
Ngày tải lên: 21/06/2014, 23:20
Hydrodynamic Lubrication 2009 Part 1 docx
... Basic Equations 121 6 .2 Finite Element Solution of the Basic Equations 122 6 .2. 1 Reynolds’ Equation 122 6 .2. 2 Equation of Balance for the Foil 125 Contents IX 6 .2. 3 SolutionProcedure 126 6.3 Characteristics ... 8 2 Foundations of Hydrodynamic Lubrication 9 2. 1 Tower’s Experiment . . . . 9 2. 2 Reynolds’ Theory of Hydrodynamic Lubrication . . 11 2. 2.1 Interpretation...
Ngày tải lên: 11/08/2014, 08:21
Hydrodynamic Lubrication 2009 Part 2 ppsx
... circumferential speed of the journal U 2 is along the inclined surface, and Eq. 2. 20 becomes: ∂ ∂x h 3 ∂p ∂x + ∂ ∂z h 3 ∂p ∂z = 6µ (U 1 + U 2 ) ∂h ∂x + 2V 2 (2. 24) It is this form of the equation ... (see Chapter 5). 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...
Ngày tải lên: 11/08/2014, 08:21
Hydrodynamic Lubrication 2009 Part 3 docx
... 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)...
Ngày tải lên: 11/08/2014, 08:21