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MEL 417 Lubrication

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

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MEL 417 Lubrication

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  1. MEL 417 Lubrication Reynold’s equation

  2. 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

  3. Newtonian fluid: shear stress-shear strain relationship Shear rate • Linear dependence • Slope is 1/h a Shear stress t

  4. Reynold’s eqn: Equilibrium of a fluid elementforces in one dimension y Shear force on top face z dy Pressure force on right face Fluid element dz Pressure force on left face x Shear force on bottom face dx x, y, z: Mutually perpendicular axes p: pressure on left face, t: shear stress on bottom face in x direction dx, dy, dz: elemental distances

  5. Fluid element- equilibrium equations Forces on left should match forces on right. Therefore Simplifying we get: OR similarly

  6. 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 Whereu and v are the particle velocities in the x and y directions respectively h is the coefficient of dynamic viscosity

  7. Pressure gradients Therefore the pressure gradients in terms of only the viscosity and velocity gradients is and Assuming that the viscosity is constant and

  8. Conditions • p and hare 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 z=h h z=0

  9. Volume flow rate 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

  10. Reynold’s equation for fluid flow between inclined surfaces Pressure profile Top surface pmax Oil wedge Upper surface is stationary p = Pressure Film thickness = h When h = ho p = pmax therefore h ho Bottom surface moves with velocity U Bottom surface and

  11. Reynold’s equation in one dimension When p = pmax, dp/dx=0, and h = ho Therefore Substituting we get If r is the density of fluid, the mass flow rate in the x direction is

  12. 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 and

  13. Reynold’s equation in 2 dimensions Substituting into the continuity equation we get Which gives

  14. Velocity of flow at a fluid element Velocity at back face Velocity at top face y z dy Velocity at right face Fluid element dz Velocity at left face Velocity at front face x dx Refer to book Principles of Lubrication by Cameron A Velocity at bottom face

  15. 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

  16. 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 dimensions If 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

  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 Diameter D L L>>D

  18. Infinitely long bearing (L >> D) 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 Diameter D L Pressure curve x = 0 x = B

  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 L = length of bearing Diameter D

  20. Infinitely short bearing • On integration we get • Further integration gives • Where K1and 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) pmax -L/2 +L/2 y

  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 h3is 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

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