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Coaxial cylinders method

Coaxial cylinders method. Consider laminar flow of an incompressible viscous fluid between two vertical coaxial cylinders The outer one is rotating with an angular velocity W It is assumed that There are no end effects No-slip condition prevails in the cylinder-fluid contact

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Coaxial cylinders method

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  1. Coaxial cylinders method • Consider laminar flow of an incompressible viscous fluid between two vertical coaxial cylinders • The outer one is rotating with an angular velocity W • It is assumed that • There are no end effects • No-slip condition prevails in the cylinder-fluid contact • If tr 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 = (2prl).tr.r where l is the length of the inner cylinder Inner cylinder r R1 l R2 Outer cylinder

  2. 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 w is the angular velocity, h is the dynamic coefficient of viscosity. The distance of separation y = r and the change in velocity du = rdw

  3. Constant of integration • Now • Therefore • Where C is a constant of integration • At r = R1, w = 0, at r = R2, w = W • Therefore

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

  5. Solid sphere method • 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 h and attaining a uniform velocity V, the viscous resistance is given by Stoke’s law as 6prhV • If the densities of the material of the solid sphere and the liquid are rsand rlrespectively, then the gravity and buoyant forces are respectively and

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

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

  8. Journal bearing- geometry Bearing:center O and radius R1 Shaft:center C and radius R2 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 = q OB is the radius R1 of the bearing. The line OB cuts the shaft at point A and AB is the film thicknessh Draw a line from the shaft center C parallel to OB cutting the shaft at E and bearing at F Bearing q = 0 Shaft G B R1 A F h q O E e R2 D C q Direction of rotation Increase in q

  9. Journal bearing- film thickness 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 = eCosq OB – CE = R1 - R2 = c, where c is the radial clearance of the bearing Hence h = c + ecosq = c{1+(e/c)cosq} The ratio e/c is called the eccentricity ratio of the shaft and is written as e, so G B R1 A F h q O E e R2 D C q

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