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

... 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

... 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

... be: P =  h 2 h 1 pdx=  xp  h 2 h 1 −  h 2 h 1 x dp dx dx = −  h 2 h 1 x dp dx dx (4.14) = 6µU  h 2 h 1 x  1 h 2 − h m h 3  dx = 6µUB 2 h 2 2  h 2 h 1 ¯x ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 1 h 2 − h m h 3 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d ... as follows: p(¯x) = 6µUB h 2 2 (m − 1)(1 − ¯x)¯x (m + 1)(m − m ¯x + ¯x) 2 ≡ 6µUB h 2 2 ¯p(¯x) (4. 12) h m = 2m m + 1 (4.13) Nondimensional pressure ¯p(¯x) on th...

Ngày tải lên: 11/08/2014, 08:21

... 2 ˙ θ) (2 + κ 2 )(1 − κ 2 ) + 2 κ (1 − κ 2 ) 3 /2  π 2 − 8 π (2 + κ 2 )  (5 .23 ) P θ = 6µ  R j c  2 LR j  πκ(ω − 2 ˙ θ) (2 + κ 2 )(1 − κ 2 ) 1 /2 + 4κ˙κ (2 + κ 2 )(1 − κ 2 )  (5 .24 ) where the ... 4)κ 0 2  1 κ 0 − κ 0 2 + κ 0 2 + κ 0 1 − κ 0 2  (5.37) K xy = − π  1 − κ 0 2 κ 0  π 2 − (π 2 − 4)κ 0 2 (5.38) C xx = − 2( 2 + κ 0 2...

Ngày tải lên: 11/08/2014, 08:21

... X 1 and X 2 of second order only, as: p = p 0 + p 1 X 1 + p 2 X 2 + p 3 X 3 + p 4 X 4 + p 11 X 1 2 + p 12 X 1 X 2 + p 22 X 2 2 + p 13 X 1 X 3 + p 14 X 1 X 4 + p 23 X 2 X 3 + p 24 X 2 X 4 (5.84) where X 3 = dX 1 dt , ... pressure coefficients. These coefficients are defined as: p 1 = ∂p ∂X 1  0 , p 2 = ∂p ∂X 2  0 , ··· p 24 = ∂ 2 p ∂X 2 X 4  0 (5.86) 92 5 Stab...

Ngày tải lên: 11/08/2014, 08:21

... d 124 X 2 X 4 (5.94) M dX 4 dt = − d 21 X 1 − d 22 X 2 − d 23 X 3 − d 24 X 4 − d 21 1 X 2 1 − d 21 2 X 1 X 2 − d 22 2 X 2 2 − d 21 3 X 1 X 3 − d 21 4 X 1 X 4 − d 22 3 X 2 X 3 − d 22 4 X 2 X 4 (5.95) where the d’s are the dynamic ... d 11 X 1 − d 12 X 2 − d 13 X 3 − d 14 X 4 − d 111 X 2 1 − d 1 12 X 1 X 2 − d 122 X 2 2 − d 113 X 1 X 3 − d 114 X 1 X 4...

Ngày tải lên: 11/08/2014, 08:21

... x i 3 ) + 3 2 (h i+1 − h i )(h i x i+1 − h i+1 x i ) 2 (x i+1 2 − x i 2 ) +(h i x i+1 − h i+1 x i ) 3 (x i+1 − x i )   1 −1 −11  (6 .21 ) V i = U (x i+1 − x i ) 2  1 2 (h i+1 − h i )(x i+1 2 − x i 2 ) +(h i x i+1 − ... −  x 2 x 1  d 2 h dx 2 +  p T − 1 R  δhdx (6 .26 ) Equating this to zero gives the following stationary condition: δJ{h} = −  x 2 x 1  d 2...

Ngày tải lên: 11/08/2014, 08:21

...  z (7.35) I 2 − 2II =( x 2 +  y 2 +  z 2 ) + 2(  xy 2 +  yz 2 +  zx 2 ) (7.36) Further, in cylindrical coordinates (r,θ,z), the strain invariant can be written as: I =  r +  θ +  z (7.37) I 2 − 2II ... − ∂p ∂r + µ  ∂ 2  r ∂r 2 + 1 r ∂ r ∂r + ∂ 2  r ∂z 2 −  r r 2  (7.1) ρ  ∂ z ∂t +  r ∂ z ∂r +  z ∂ z ∂z  = − ∂p ∂z + µ  ∂ 2  z ∂r 2 + 1 r ∂...

Ngày tải lên: 11/08/2014, 08:21

... p  ∂u ∂x + ∂ ∂y + ∂w ∂z  + Φ (8. 42) where Φ = 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣  ∂u ∂x  2 +  ∂ ∂y  2 +  ∂w ∂z  2 + 1 2  ∂u ∂y + ∂ ∂x  2 + 1 2  ∂ ∂z + ∂w ∂y  2 + 1 2  ∂w ∂x + ∂u ∂z  2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ + λ  ∂u ∂x + ∂ ∂y + ∂w ∂z  2 (8.43) In ... constant: ρc  DT Dt = ∂ ∂x  k o ∂T ∂x  + ∂ ∂y  k o ∂T ∂y  + ∂ ∂z  k o ∂T ∂z  + Φ  (8.44) where Φ  = 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ...

Ngày tải lên: 11/08/2014, 08:21