*Friction in Journal bearingsFrom Newtons law of friction, the stress t on any layer is

From Reynolds equations it was found that

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

*

*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

Where 2pR = B

*

*Now h = c(1+ecosq) and dh/dq = -cesinq, so integrating the first term by parts gives

The first of these terms is zero, as p must be zero at q = 0, and 2p (Sommerfelds condition)For the second term the integral is solved using the relation

*

*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 Sommerfelds substitution

The expression for friction then becomes

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)

*

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

These are in the direction of the load, a distance esiny apart, and there will be a couple set up of magnitude Wesiny = Wcesiny

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

yWeesinyShaftBearingOil film height h

*

*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-e2)1/2 is a multiplier to take into account the eccentric running of the shaft

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*Journal- Narrow bearingsAssumption: 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 q) directionEquation for flow in the x direction is given by

In the axial (y) direction it is given by

shaftBearingLR

*

*The continuity equation is

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

As R>>L ,

*Pressure change with yThus the continuity equation reads

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

Or

*

*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

-L/2+L/20BearingR

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*Due to symmetry dp/dy must be zero on the center line (y=0). Therefore C1 = 0 as dp/dy = 0, at y = 0From the former condition C2 must equal

Hence we get the pressure as

Now h = c(1 + ecosq) and x = Rq, therefore

*

*Therefore

and

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

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

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

Rdq

q

yPressure curveWxWyWLine of centersBearingShaft

*

*Therefore

And

Substituting the expression for p we get

and

*

*The following integrals can be evaluated to give

And

Thus

And

*

*The resultant load

Or

Now [16/p2)-1] = 0.6211, therefore

The group on the left is similar to Sommerfelds 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 D is the Sommerfeld variable and D is the diameter = 2R

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*Attitude angleThe attitude angle is given byTanY = 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.

*

*It is given by

h = c(1+ecosq), therefore

And

Therefore

Therefore the non-dimensional side flow is defined as

Therefore Qc* = 2e

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*Detergent additivesTo 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

*

*Detergent additivesMake the binding agents in deposits less effectiveParticles remain in suspension and can be drained or filtered offEnvelope the deposit particles and prevent them from agglomerating with other particlesE.g. metal phosphonates, sulphonates

Binding agentDeposit particles that agglomerate due to binding agent

Detergent

Detergent

Detergent bound to binding agentParticles remain free

DetergentDetergentOR Envelope the particles, preventing them from forming deposits

*

*Dispersant additivesParticles 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

*

*

DetergentDispersants- mechanism

Separated and suspended particles due to detergent action

DetergentDetergent

Dispersant particles(same charge on outside)

++++Like charges repel, hence there is dispersionDetergent

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*Pour point depressantsPour 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

*

*Pour point depressant- mechanismWAX CRYSTALWAX CRYSTAL

WAX CRYSTALWAX CRYSTALCrystal growthWAX CRYSTALWAX CRYSTAL

POR POINT DEPRESSANT

WAX CRYSTALPOR POINT DEPRESSANTEncapsulate crystal so that it cannot growWAX CRYSTALOR change the structure of crystals making them amorphous (crystals of different shapes and sizes)

*

Viscosity index improvementRemove 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 increasePolymer chainsAs 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 additivesReduce 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 agentsFoaming 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 modifiersIn 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

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