Answer to Question #194386 in Mechanical Engineering for Siyaya

Question

Figure 1 shows 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

Data:

Weight density of oil = w=8.95 kN/m³

Mass density($\rho$ )=$\frac{8.9\times 1000}{9.8}$=908.16 Kg/m³

Thickness of oil (y)=0.3 ×10^-3 m

Velocity of upper plate(V)=0.05 m/s

Area of plate(A)=1.5 m²

Torque acting on drum(T)=24.5×10^-3 N-m

Radius of drum (r)= 25 ×10^-3m

Solution:

Let F is the force acting on plate through cable and $\nu \:$ is the kinematic viscosity of oil.

We know that,

$\text{The torque(T)=Force(F)×radius(r)}$

$F=\frac{T}{r}= \frac{24.5}{25}= 0.98 N$

Assuming oil to be newtonian fluid,

$F=A\times\mu\times \frac{V}{y}$

$\nu=\frac{F}{A\times \frac{V}{y}}\times \rho=\frac{0.98}{1.5\times \frac{0.05}{0.0003}}\times$908.16=3.56 $\frac{m^2}{s}$

$\text{Kinemativ viscosity of oil }(\nu)=3.56\frac{m^2}{s}$