Estimate the torque needed to rotate the inner cylinder at $1000 \mathrm { rpm }$ if SAE-30 oil at $40 ^ { \circ } \mathrm { C }$ fills the gap. Assume a linear velocity distribution in the gap. Neglect end effects.

A) $16.0 \mathrm {~N} \cdot \mathrm { m }$
B) $16.9 \mathrm {~N} \cdot \mathrm { m }$
C) $18.8 \mathrm {~N} \cdot \mathrm { m }$
D) $20.2 \mathrm {~N} \cdot \mathrm { m }$ $60 - \mathrm { cm }$ long concentric cylinders

Question

Estimate the torque needed to rotate the inner cylinder at $1000 \mathrm { rpm }$ if SAE-30 oil at $40 ^ { \circ } \mathrm { C }$ fills the gap. Assume a linear velocity distribution in the gap. Neglect end effects.

A) $16.0 \mathrm {~N} \cdot \mathrm { m }$ 
B) $16.9 \mathrm {~N} \cdot \mathrm { m }$ 
C) $18.8 \mathrm {~N} \cdot \mathrm { m }$ 
D) $20.2 \mathrm {~N} \cdot \mathrm { m }$ $60 - \mathrm { cm }$ long concentric cylinders

Quizplus · Accepted Answer

$20.2 \mathrm {~N} \cdot \mathrm { m }$
The velocity of the outer cylinder is 0 and that of the inner cylinder is
$$V = r \omega = 0.08 \times \frac { 1000 \times 2 \pi } { 60 } = 8.38 \mathrm {~m} / \mathrm { s }$$
Using Eq.1.5.6, the shear stress on the inner cylinder is
$$\tau = \mu \frac { \Delta V } { \Delta r } = 0.1 \times \frac { 8.38 - 0 } { 0.001 } = 838 \mathrm {~N} / \mathrm { m } ^ { 2 }$$
The viscosity is found in the Appendix Fig. B.1. The torque is related to the shear stress as follows:
$$\begin{aligned}
T & = \tau A r = \tau \times ( 2 \pi r \times L ) \times r = 2 \pi r ^ { 2 } \tau L \
& = \left( 2 \pi \times 0.08 ^ { 2 } \right) \mathrm { m } ^ { 2 } \times 838 \mathrm {~N} / \mathrm { m } ^ { 2 } \times 0.6 \mathrm {~m} = \underline { 20.2 \mathrm {~N} \cdot \mathrm { m } }
\end{aligned}$$

Estimate the Torque Needed to Rotate the Inner Cylinder At 1000