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

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 Where 2  R = B

3 Now h = c(1+  cos  ) and dh/d  = -c  sin , 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

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 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  = Wc  sin  This corresponds to a frictional force of Wc  sin  /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  Shaft Bearing Oil film height h

6 This is exactly equal to the friction F h, 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

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  ) direction Equation for flow in the x direction is given by In the axial (y) direction it is given by shaft Bearing L R

8 The continuity equation is If the average pressure in the lubricant is p, then is of the order of pressure/circumference or p/2  R 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 q x is also taken to be much small compared to Uh/2

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

10 This equation can be integrated to give And again to give Where C 1 and C 2 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/2 0 Bearing R

11 Due to symmetry dp/dy must be zero on the center line (y=0). Therefore C 1 = 0 as dp/dy = 0, at y = 0 From the former condition C 2 must equal Hence we get the pressure as Now h = c(1 +  cos  ) and x = R  therefore

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 .

13 Narrow bearing load The load components W x and W y are derived by applyling a double integral as the pressure varies in the  as well as y directions. W x is the component along the line of centers and Wy is the component normal to it. Rd    Pressure curve WxWx WyWy W Line of centers Bearing Shaft

14 Therefore And Substituting the expression for p we get and

15 The following integrals can be evaluated to give And Thus And

16 The resultant load Or Now [16/  2 )-1] = , therefore The group on the left is similar to Sommerfeld’s variable, except that it has L 2 in it instead of R 2. If top and bottom are divided by R 2 and the 4 is taken from the right hand side, then Where  is the Sommerfeld variable and D is the diameter = 2R

17 Attitude angle The attitude angle is given by Tan  = W y /-W x 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, Q c is the difference between the oil flowing in at the start of the pressure curve and out at its end.

18 It is given by h = c(1+  cos  ), therefore And Therefore Therefore the non-dimensional side flow is defined as Therefore Q c * = 2 

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 Detergent bound to binding agent Particles remain free Detergent OR 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 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 Crystal growth WAX CRYSTAL POR POINT DEPRESSANT WAX CRYSTAL POR POINT DEPRESSANT Encapsulate crystal so that it cannot grow WAX CRYSTAL OR 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