Question

A moving belt system is to be used to fransfer lubricant from a sump to the point of application. The working length of the belt is assumed to be straight and to be running parallel to a stationary metal surface. The belt speed is 200 mm/s, the perpendicular distance between the belt and the metal surface is S mm, and the belt is 500 mm wide. The lubricant is an oil, having an absolute viscosity of 0.007 Ns/m²• Estimate the force per unit length of belt and the quantity of lubricant discharged per second. If the relative density of the oil is 0.891, what is the kinematic viscosity of the oil?

Answer 1

Ans) We know,

   = (du/dy)

where, = shear stress

   = Dynamic viscosity = 0.007 Ns/

du = Velocity of belt = 200 mm/s or 0.20 m/s

   dy = perpendicular distance between belt and metal = 5 mm or 0.005 m

Putting values,

=> = 0.007 x (0.20 / 0.005)

=> = 0.28 N/ N/

Area of belt per meter of length = 0.50 x 1 = 0.5

We know, stress = Force / Area

=> Force per unit length of belt = 0.28 N/ x 0.50 = 0.14 N

Now,

To determine discharge (Q) use relation ,   

Q = A x V

where, A = cross sectional area of belt = 0.5 m x 0.005 = 0.0025

V = mean velocity = 200 mm/s / 2 or 0.1 m/s

=> Q = 0.0025 x 0.1 m/s

=> Q = 0.00025 /s

Hence, discharge of lubricant is 0.00025 /s

Also,

Kinematic viscosity () = /

=> = 0.007 / (0.891 x 1000)

=> = 7.85 x /s

Hence, kinematic viscosity of oil is 7.85 x /s