Answer to Question #194367 in Mechanical Engineering for blessing

Question

a layer of oil (8.95 kN/m3), the thickness of 0.3 mm, between two parallel flat plates. The upper plate is pulled across the bottom plate with a cable connected to a winch. The upper plate moves with a velocity of 0.05 m/s. The area of contact between the upper plate and the oil is 1.5 m2. Determine the kinematic viscosity of the oil if the torque acting on the winch drum should not exceed 24.5 x 10-3 Nm. Take the drum diameter as 50 mm.

Answer 1

$\gamma = 8.95\ kN/m³ = 8950\ N/m³\\ thickness = 0.3mm = 3 × 10^{-4} m\\ velocity = 0.05\ m/s\\ Area = 1.5\ m²\\ \tau = 24.5× 10^{-3}\ Nm\\ diameter_{drum} = 50 × 10^{-3}\\ \eta =$

$∆P = \dfrac{\tau}{ Area× thickness}=$

$\dfrac{ 24.5× 10^{-3}}{ 1.5× 3 × 10^{-4} }=54.4\ N/m²$

$\eta= \dfrac{∆P× d²}{32L×v} =\dfrac{54.4×(50×10^{-3})^{2}}{32×3×10^{-4}× 0.05}$

$= 283.33\ Pa.s$