Coaxial cylinders method r R1R1 R2R2 Consider laminar flow of an incompressible viscous fluid between two vertical coaxial cylinders The outer one is rotating with an angular velocity  It is assumed that There are no end effects No-slip condition prevails in the cylinder-fluid contact If  r is the shear stress on a liquid layer at a distance r from the axis of rotation, then the torque T on the liquid shell by the outer layer of the liquid is T = (2  rl).  r.r where l is the length of the inner cylinder l Inner cylinder Outer cylinder

Shear stress at radius r Therefore the shear stress at radius r is From Newton’s law of viscous flow the shear stress at radius r is Where  is the angular velocity,  is the dynamic coefficient of viscosity. The distance of separation y = r and the change in velocity du = rd 

Now Therefore Where C is a constant of integration At r = R 1,  = 0, at r = R 2,  =  Therefore Constant of integration

Substituting we get: Or Knowing values of the other terms, the coefficient of viscosity  can be calculated Expression for coefficient of viscosity

When a solid body is allowed to fall from rest in a homogenous fluid of infinite extent, it will initially accelerate till the gravitational force is balanced by buoyant and viscous forces. Consider a sphere of radius r, moving in a fluid with viscosity  and attaining a uniform velocity V, the viscous resistance is given by Stoke’s law as 6  r  V If the densities of the material of the solid sphere and the liquid are  s and  l respectively, then the gravity and buoyant forces are respectively Solid sphere method and

Therefore balancing forces we get: Or Therefore the viscosity can be determined if we know values for the other terms Solid sphere in liquid- expression for viscosity Lubricant Sphere Weight of sphere Force of buoyancy

Journal bearing- process at startup Stationary journal Instant of starting (tends to climb up the bearing) While running (slips due to loss of traction and settles eccentric to bearing) e = eccentricity Shaft/journal Bearing Because of the eccentricity, the wedge is maintained (lack of concentricity)

Journal bearing- geometry e R1R1 h R2R2 D O C  A B F E Bearing: center O and radius R 1 Shaft: center C and radius R 2 OC is the eccentricity measured as e All angular distances are measured from the position of maximum film thickness (where the extension of line CO cuts the bearing surface at G) Consider a point B on the bearing surface such that the angle GOB =  OB is the radius R 1 of the bearing. The line OB cuts the shaft at point A and AB is the film thickness h Draw a line from the shaft center C parallel to OB cutting the shaft at E and bearing at F G   = 0 Direction of rotation Increase in  Shaft Bearing

Journal bearing- film thickness e R1R1 h R2R2 D O C  A B F E G  Distances AB and EF are very small compared to the radii OB and CF are so close together (e being very small compared to radii) Therefore ABFE is considered a rectangle From O, drop a perpendicular to CE cutting CE at D The oil film thickness h = EF = OB - DE = OB-(CE-CD) CD = eCos  OB – CE = R 1 - R 2 = c, where c is the radial clearance of the bearing Hence h = c + ecos  = c{1+(e/c)cos  } The ratio e/c is called the eccentricity ratio of the shaft and is written as , so