MEL 417 Lubrication

Reynold’s equation

Reynold’s equation for fluid flow

Assumptions:• 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

dy

dudy

du

•Linear dependence

•Slope is 1/

Shear stress

Shea

r rat

e

Reynold’s eqn: Equilibrium of a fluid elementforces in one dimension

dydz)dx.x

pp(

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.x

pp(dxdy

dxdydzx

pdxdydz

z

)1.....(x

p

zxz

)2.....(y

p

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

0z

p

z

uxz

z

vyz

Pressure gradients

Therefore the pressure gradients in terms of only the viscosity and velocity gradients is

and

Assuming that the viscosity is constant

and

z

u

zx

p

z

v

zy

p

2

2

z

u

x

p

2

2

z

v

y

p

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

x

p.

1

z

u2

2

21

2

CzC2

z.

x

p.

1u

21

2

1 UhC2

h

x

p.

1U

1

21 C2

h.

x

p.

1

h

UU

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

h

UU

2

hz

x

p.

1

z

u 21

2212 U

h

z)UU()zhz(

x2

pu

)3.....(2

h)UU(

x

p.

12

hudzq 21

3h

0

x

)4.....(2

h)VV(

y

p.

12

hvdzq 21

3h

0

y

Reynold’s equation for fluid flow between inclined surfaces

0dx

dp

0dx

dp

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

2

hU

x

p.

12

hq

3

x

and

Reynold’s equation in one dimension

When 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

2

hUq o

x

)5.....(h

hhU6

dx

dp3

o

x

p.

12

h

2

hU.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

0y

q

x

q yx

)4...(2

hU

x

p.

12

hq

3

x

)5...(

y

p.

12

hq

3

y

and

Reynold’s equation in 2 dimensions

Substituting into the continuity equation we get

Which gives

0y

p.

12

h

yx

p.

12

h

2

Uh

x

33

)6...(dx

dhU6

y

p.h

yx

p.h

x33

Velocity of flow at a fluid element

)dx.x

uu(

u

)dz.z

ww(

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.y

vv(

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.x

uu(

)dz.z

ww(

)dy.y

vv(

dydz)dx.x

uu(

dxdz)dy.y

vv(

dxdy)dz.z

ww(

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)dzz

ww(dxdz)dy

y

vv(dydx)dx

x

uu(wdxdyvdxdzudydz

0z

w

y

v

x

u

0z

q

y

q

x

q 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

y

p

3o

h

hhU6

x

p

2

Uhq o

x LDiameter D

L>>D

18

Infinitely long bearing (L >> D)

3o

h

hhU6

x

p

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...()Ch

dxh

h

dx(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

2

Uhqy

2

233

dy

pdh

y

ph

y

32

2

h

dx/dhU6

dy

pd

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)

13Ky

h

dx/dhU6

dy

dp

21

2

3KyK

2

y.

h

dx/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 4

L.

h

dx/dhU3K

2

32

)11...(4

Ly

h

dx/dhU3p

22

3