Question 1.14: Calculate the torque required to rotate the shaft of 0.1 m d......
Calculate the torque required to rotate the shaft of 0.1 m diameter and 5 cm length inside a hollow cylinder of 0.15 m diameter at a contact rate of 500 rpm. The gap between the two is filled with a lubricating oil of viscosity 0.9 Ns/m² (Figure 1.13).
Given,
Diameter of shaft, d_1= 0.1 m
Length of shaft, L = 5 cm
Diameter of hollow cylinder, d_2= 0.15 m
Viscosity of oil = 0.9 Ns/m²
Speed of shaft, s = 500 rpm
This problem can be considered as equivalent to the thin plate moving on the fluid film problem. So, thickness of the fluid film will be
y=\frac{d_2-d_1}{2}=\frac{0.15-0.1}{2}=0.025 m
For conversion of rpm to linear velocity,
\begin{aligned} ν & =\frac{\pi d_1 s}{60} \\ & =\frac{\pi \times 0.1 \times 500}{60}=2.618 m / s \end{aligned}
Now, velocity profile can be assumed linear, since the thickness of the film is very less.
Therefore, from the Newton’s law of viscosity
\begin{aligned} \tau & =\mu \frac{d u}{d y}=1 \times \frac{2.618}{0.025} \\ & =94.25 Pa \end{aligned}
So, Shear force = τ × A
\begin{aligned} & =104.72 \times \pi \times d_1 \times L \\ & =104.72 \times \pi \times 0.1 \times 0.05=1.48 N \end{aligned}
Therefore,
\begin{aligned} \text { required torque } & =F \times \frac{d_1}{2} \\ & =1.64 \times \frac{0.1}{2}=0.074 Nm \end{aligned}