mel 417 lubrication lec 150211
TRANSCRIPT
MEL 417 Lubrication
Reynold’s equation
Reynold’s equation for fluid flowAssumptions:• External forces are neglected (gravitational, magnetic etc.)• Pressure is considered constant throughout the thickness of
the film• Curvature of the bearing surfaces are large compared to the oil
film thickness• No slip at boundaries• Lubricant is Newtonian• Flow is laminar• Fluid inertia can be neglected• Viscosity is constant through the thickness of the film
Newtonian fluid: shear stress-shear strain relationship
dydu
dydu
•Linear dependence
•Slope is 1/
Shear stress
Shea
r rat
e
Reynold’s eqn: Equilibrium of a fluid elementforces in one dimension
dydz)dx.xpp(
pdydz
dxdy)dz.z
( xz
Fluid element
dxdyxz
yz
x
dx
dz
dy
x, y, z: Mutually perpendicular axes
p: pressure on left face, : shear stress on bottom face in x direction
dx, dy, dz: elemental distances
Shear force on top face
Shear force on bottom face
Pressure force on left face
Pressure force on right face
Fluid element- equilibrium equations
Forces on left should match forces on right.Therefore
Simplifying we get:
OR similarly
dxdy)dz.z
(pdydz dydz)dx.xpp(dxdy
dxdydzxpdxdydz
z
)1.....(xp
zxz
)2.....(
yp
zyz
Substituting using Newton’s law of viscosity
In the z direction the pressure gradient is 0, therefore
According to Newton’s law for viscous flow and
Where u and v are the particle velocities in the x and y directions respectively is the coefficient of dynamic viscosity
0zp
zu
xz
zv
yz
Pressure gradients
Therefore the pressure gradients in terms of only the viscosity and velocity gradients is
and
Assuming that the viscosity is constant
and
zu
zxp
zv
zyp
2
2
zu
xp
2
2
zv
yp
Conditions• p and are independent of z (assumptions)• Therefore integrating
• we get
• Applying boundary conditions, u = U1 at z = h and u = U2 at z = 0 we get C2 = U2 and
so
xp.1
zu2
2
21
2
CzC2z.
xp.1u
21
2
1 UhC2h
xp.1U
1
21 C2h.
xp.1
hUU
h
z=0
z=h
Substituting we get
And
Let the rate of flow (per unit width) in the x and y directions be qx and qy respectively
Therefore
and
Volume flow rate
hUU
2hz
xp.1
zu 21
2212 U
hz)UU()zhz(
x2pu
)3.....(2h)UU(
xp.
12hudzq 21
3h
0x
)4.....(2h)VV(
yp.
12hvdzq 21
3h
0y
Reynold’s equation for fluid flow between inclined surfaces
0dxdp
0dxdp
Oil wedge
Bottom surface
pmaxTop surface
ho
Pressure profile
Bottom surface moves with velocity U
h
p = Pressure
Film thickness = h
When h = ho
p = pmax
therefore
Upper surface is stationary
2hU
xp.
12hq
3
x
and
Reynold’s equation in one dimensionWhen p = pmax, dp/dx=0, and h = ho
Therefore
Substituting we get
If is the density of fluid, the mass flow rate in the x direction is
2hUq o
x
)5.....(h
hhU6dxdp
3o
xp.
12h
2hU.q.m
3
x
.
Flow rate after substitution
• Equation of continuity for 2 dimensions
• In most bearing systems there is no flow in the y direction, therefore V1=V2=0. If surface1 is stationary then U1 is also 0. Then equations 3 and 4 reduce to
0yq
xq yx
)4...(2hU
xp.
12hq
3
x
)5...(
yp.
12hq
3
y
and
Reynold’s equation in 2 dimensions
Substituting into the continuity equation we get
Which gives
0yp.
12h
yxp.
12h
2Uh
x
33
)6...(dxdhU6
yp.h
yxp.h
x33
Velocity of flow at a fluid element
)dx.xuu(
u
)dz.zww(
Fluid element
w
yz
x
dx
dz
dy
Velocity at top face
Velocity at bottom face
Velocity at left face
Velocity at right face
)dy.yvv(
vVelocity at front face
Velocity at back face
Refer to book Principles of Lubrication by Cameron A
Balancing in and out flow rates
• The velocities entering the element are u, v, and w along x, y, and z directions respectively
• The velocities leaving are correspondingly , , and
Therefore the flow rates are:
In- udydz, vdxdz, and wdxdy
Out- , and
)dx.xuu(
)dz.zww(
)dy.yvv(
dydz)dx.xuu(
dxdz)dy.yvv(
dxdy)dz.zww(
Continuity equation in 3 dimensions• As there are no source or sinks for fluid flow within the element and the volume
remains constant, the total volume flowing in = total volume flowing out, per unit time
Therefore:
On simplifying we get:
Which is the continuity equation in three dimensionsIf we retain the volume terms, we get:
Where qx, qy, and qz are the flow rates per unit width in the x, y, and z directions respectively
dxdy)dzzww(dxdz)dy
yvv(dydx)dx
xuu(wdxdyvdxdzudydz
0zw
yv
xu
0zq
yq
xq zyx
17
Reynold’s equation- Infinitely long bearing (L>>D)
In this assumption, the pressure does not vary in the y direction
Therefore = 0 and the flow rate qy = 0
Assuming that only one surface moves, with a velocity U, we get (derived earlier)
and
where ho is the film thickness at max/min pressure
yp
3o
hhhU6
xp
2Uhq o
x LDiameter D
L>>D
18
Infinitely long bearing (L >> D)
3o
hhhU6
xp
Pressure p can be obtained from the equation
Provided h can be expressed in terms of x
Therefore
Where C is a constant of integration. Two boundary conditions are required to obtain the values for ho and C. This can be obtained from knowledge of the start and end points of the pressure curve where p = 0
The pressure curve in the figure below ranges from x = 0 to x = B
)10...()Chdxh
hdx(U6p
x
0
x
03o2
x = 0 x = B
Pressure curve
LDiameter D
19
Reynold’s equation- Infinitely short bearing (D>>L)
• In this case the length of the bearing is considered much shorter than the diameter
• Therefore the pressure differential in the x – direction is considered 0 as it is much lower compared to the pressure differential in the y direction
• We therefore get
• The film thickness is assumed not to vary with x, therefore
• Reynold’s equation in two dimensions then becomes
2Uhqy
2
233
dypdh
yph
y
32
2
hdx/dhU6
dypd
L = length of bearing
Diameter D
20
Infinitely short bearing
On integration we get
Further integration gives
Where K1 and K2 are constants of integration These can be obtained by putting pressure = 0 at the edges of
the bearing and pressure gradient = 0 at the middle of the bearing (assuming symmetry)
13 Kyh
dx/dhU6dydp
21
2
3 KyK2y.
hdx/dhU6p
pmax
-L/2 +L/2y
21
Infinitely short bearing
We therefore get and
The equation therefore becomes
If p = 0 other than when y = -L/2 or +L/2, either dh/dx=0 or h3 is infinite This fact is applied to journal bearings and dh/dx=0 at points of maximum
and minimum film thickness It is also applicable to narrow rotating discs It is not applicable to thrust bearings This theory is applicable when L/D<1/4 and infinitely long theory is
applicable when L/D>=4
0K1 4L.
hdx/dhU3K
2
32
)11...(4Ly
hdx/dhU3p
22
3