1

Friction in Journal bearings

• From Newton’s law of friction, the stress on any layer is

• From Reynold’s equations it was found that

• We need to find the friction stress at the 2 surfaces, i.e. z = 0 and z = h

z

u

2

hz

1.

dx

dp

h

)UU(

z

u 21

2

• Therefore

• The positive sign is for z = h (bearing surface) and the negative for z = 0 (shaft surface). The total drag F on the whole bearing under consideration, of extent B and L (length), in the x and y directions is

h)UU(

2

h.

dx

dp

z

u210,h

dyRddxdyFL

0

B

0

L

0

2

0

L

0

2

0

2

0dy)Rd

h

URd.

2

h.

Rd

dp(

Where 2R = B

3

• Now h = c(1+cos) and dh/d= -csin, so integrating the first term by parts gives

• The first of these terms is zero, as p must be zero at = 0, and 2 (Sommerfeld’s condition)

• For the second term the integral is solved using the relation

L

0

2

0

2

0

2

0

dy)cos1(c

URddsinc

2

p

2

phF

L

0

2

0sinWdysinpRd

4

• The third term should be taken under two separate conditons. This is because the viscosity is not constant around the whole circumference. If there is cavitation in some part of the bearing a different law will apply.

• At the moment the bearing will be assumed to be full of a liquid with one single viscosity. Thus, using Sommerfeld’s substitution

• The expression for friction then becomes

L

0

2

0

2

0 2

2/12

d)cos1)(1(

)1)(cos1(

c

URL

)cos1(

d

c

UR

2/122

0

2/12 )1(c

URL2

d)1(c

URL

2/120,h )1(c

URL2sin

R2

WcF

The positive sign in front of the first term is when z = h (at the bearing surface), and the negative sign when z = 0 (at the shaft surface)

5

• The integrated oil forces on the shaft and bearing act through their respective centers.

• These are in the direction of the load, a distance esin apart, and there will be a couple set up of magnitude Wesin= Wcsin

• This corresponds to a frictional force of Wcsin/R at the surface of the shaft. This force is added to the friction at the shaft surface h = 0, so that

W

e

esin

R

sinWc

)1(c

URL2sin

R2

WcF

2/120

Shaft

Bearing

Oil film height h

6

This is exactly equal to the friction Fh, when z = h. Therefore

for both surfaces. Of these two terms, the first arises from the offset between the center of the shaft and that of the bearing. The second is the simple Newtonian friction. •Petroff analysis of friction gives friction as

•The term 1/(1-2)1/2 is a multiplier to take into account the eccentric running of the shaft

2/120,h )1(c

URL2sin

R2

WcF

c

URL2area.

c

UF

7

Journal- Narrow bearings

Assumption: Length L is much smaller compared to radius R. The flow in the y direction will therefore be much more significant than the flow in the x (or ) directionEquation for flow in the x direction is given by

In the axial (y) direction it is given by

x

p.

12

h

2

Uhq

3

x

shaft

Bearing

L

R

y

p.

12

hq

3

y

8

The continuity equation is

If the average pressure in the lubricant is p, then is of the order of pressure/circumference or p/2R and is of the order pressure/length or p/L.

• As R>>L , << as x = R and L is in the y direction

Furthermore, the term in qx is also taken to be much

small compared to Uh/2

0y

q

x

q yx

x

p

x

p

x

p

y

p

9

Pressure change with y

• Thus the continuity equation reads

• Now h varies with x only (assuming no tilt in the shaft). Therefore the equation can be written as

Or

0y

p.

12

h

y2

Uh

x

3

dx

dh.

2

U

dy

pd.

12

h2

23

dx

dh.

h

U6

dy

pd32

2

10

This equation can be integrated to give

And again to give

Where C1 and C2 are constants of integration.

The pressure is zero at either side of the bearing. i.e. if the length is L, p is zero at y = +L/2, and y = -L/2

13Cy

dx

dh.

h

U6

dy

dp

212

3CyCy

dx

dh

h

U3p

-L/2 +L/20

Bearing

R

11

Due to symmetry dp/dy must be zero on the center line (y=0). Therefore C1 = 0 as dp/dy = 0, at y = 0

From the former condition C2 must equal

Hence we get the pressure as

Now h = c(1 + cos) and x = Rtherefore

4

L

dx

dh.

h

U3 2

3

dx

dh

4

Ly

h

U3p

22

3

)sin(cd

dh

12

Therefore

and

From this equation, it is clear that the pressure varies with

Giving a positive pressure between 0 an and negative from to 2.

R

sinc

Rd

dh

dx

dh

22

32y

4

L

)cos1(Rc

sinU3p

3)cos1(

sin

13

Narrow bearing load

The load components Wx and Wy are derived by applyling a double integral as the pressure varies in the as well as y directions. Wx is the component along the line of centers and Wy is the component normal to it.

Rd

Pressure curve

Wx

Wy

WLine of centers

Bearing

Shaft

14

Therefore

And

Substituting the expression for p we get

and

2/L

2/L 0y dyRdsinpW

2/L

2/L 0x dyRdcospW

2/L

2/L 0

22

32x dyRdcosy4

L

)cos1(Rc

sinU3W

2/L

2/L 0

22

32y dyRdsiny4

L

)cos1(Rc

sinU3W

15

The following integrals can be evaluated to give

And

Thus

And

0 2/323

2

)1(2)cos1(

dsin

0 22

2

3 )1(

2

)cos1(

dcossin

2/322

3

y )1(.

4.

c

LUW

222

3

x )1(.

c

LUW

16

The resultant load

Or

Now [16/2)-1] = 0.6211, therefore

The group on the left is similar to Sommerfeld’s variable, except that it has L2 in it instead of R2. If top and bottom are divided by R2 and the 4 is taken from the right hand side, then

Where is the Sommerfeld variable and D is the diameter = 2R

42

4

2

22

2

32

y2

x )1()1(16c

LUWWW

1.116

)1(.

4.

c

LUW 2

2222

3

1.6211.0)1(

.4L

c.

U

L/W 2222

2

2/12222

2

2

2

2

2

162.0)1(L

D

R

c.

U

L/W

L

R4

17

Attitude angle

The attitude angle is given byTan= Wy/-Wx

Therefore

For narrow bearings, the volume flow in the circumferential direction is given by per unit width.

The make up oil or the total side leakage, Qc is the difference between the oil flowing in at the start of the pressure curve and out at its end.

2/1222

162.0)1(

.4

2/12 )1(

.4

tan

2

Uhqx

18

It is given by

h = c(1+cos), therefore

And

Therefore

Therefore the non-dimensional side flow is defined as

Therefore Qc* = 2

L).hh(2

UQ endstartc

)1(chh 0start

)1(chhend

ULc)1(1c.2

ULQc

*cc Q.

2

ULcQ

19

Detergent additives

• To clean undesired substances (mostly oxidation products and contaminants) from the surfaces and passages of a lubricating system

• Detergent additives are soaps of high molecular weight, soluble in oil

• Consist of a metal and organic component

• Ashless (without metal) detergents are also employed leaving no metallic residue

20

Detergent additives

• Make the binding agents in deposits less effective• Particles remain in suspension and can be drained or filtered off• Envelope the deposit particles and prevent them from agglomerating with

other particles• E.g. metal phosphonates, sulphonates

Binding agent

Deposit particles that agglomerate due to binding agent

Detergent

DetergentDetergent bound to binding agent

Particles remain free

Detergent DetergentOR Envelope the particles, preventing them from forming deposits

21

Dispersant additives

• Particles separated by detergents are to be prevented from accumulating (usually at lower temperature)

• Dispersants isolate the particles from each other and disperse them in the lubricant

• Form a coating on particles and due to the polar nature, tend to repel each other

• E.g. pollymethacrylates, polyamine succimides

22

Detergent

Dispersants- mechanism

Separated and suspended particles due to detergent action

Detergent Detergent

Dispersant particles

(same charge on outside)

+ +

+ +

Like charges repel, hence there is dispersion

Detergent

23

Pour point depressants

• Pour point is the lowest temperature at which the lubricant will flow

• Forms waxy crystals at lower temperatures

• Pour point depressants reduce the pour point and are therefore required when operating at lower temperatures

• E.g. methacrylate polymers, polyalkylphenol esters

24

Pour point depressant- mechanism

WAX CRYSTAL WAX CRYSTAL

WAX CRYSTAL

WAX CRYSTAL

Crystal growth

WAX CRYSTAL

WAX CRYSTALPOR POINT

DEPRESSANT

WAX CRYSTALPOR POINT

DEPRESSANT

Encapsulate crystal so that it cannot grow

WAX CRYSTALOR change the structure of crystals making them amorphous (crystals of different shapes and sizes)

Viscosity index improvement

• Remove aromatics (low VI) during refining stage

• Blending with high viscous oil

• Using polymeric additives that cause an increase in viscosity with temperature due to chain unwinding

• E.g. polyisobutenes, ethylene/propylene copolymers,

VI improvement using polymeric additives

Temperature increase

Polymer chains

•As the temperature increases, the polymer chains tend to uncoil.

•In the uncoiled form, they tend to increase the viscosity thereby compensating for the decrease in viscosity of the oil

Boundary and extreme pressure additives

• Reduce friction, control wear, and protect surfaces from severe damage

• Used in highly stressed machinery where there is metal to metal contact leading to boundary lubrication

• Chemically react with sliding metal surfaces to form films which are insoluble in the lubricant

• Have low shear strength than the metal

• These layers are more easily sheared in preference to the metal

Anti-foaming agents

• Foaming is the formation of air bubbles in the lubricant

• Interfere with flow and heat transfer

• The additives lower the surface tension between the air and liquid to the point where bubbles collapse

• E.g. silicone polymers, polymethacrylates

Friction modifiers• In boundary lubrication there is poor film strength, there is

surface to surface contact

• These modifiers are polar materials such as fatty oils, acids and esters having long chains

• Form an adsorbed film on the metal surfaces with the polar ends projecting like carpet fibers

• Provide a cushioning effect and keep metal surfaces apart from each other