Oil separates a 20-mm-diameter rotating shaft from its 20.5-mm-diameter bearing. The bearing is 80 cm long and the shaft rotates at 2000 rpm. The velocity gradient at the bearing
Surface is nearest:
A) 6840 m/s/m
B) 7260 m/s/m s
C) 8040 m/s/m
D) 9406 m/s/m

Question

Quizplus · Accepted Answer

$8040 \mathrm {~m} / \mathrm { s } / \mathrm { m }$
$\text {The velocity distribution $v ( r )$ is given by Eq. 7.5.17. Differentiate that distribution to}$$\text { find the gradient:}$
$$\frac { d v _ { \theta } } { d r } = \frac { r _ { 1 } ^ { 2 } \omega _ { 1 } } { r _ { 2 } ^ { 2 } - r _ { 1 } ^ { 2 } } \frac { d } { d r } \left( \frac { r _ { 2 } ^ { 2 } } { r } - r \right) = \frac { r _ { 1 } ^ { 2 } \omega _ { 1 } } { r _ { 2 } ^ { 2 } - r _ { 1 } ^ { 2 } } \left( - r _ { 2 } ^ { 2 } r ^ { - 2 } - 1 \right)$$
$\text {Substitute in the given quantities, with $\omega _ { 1 } = 2000 \times 2 \pi / 60 = 209.4 \mathrm { rad } / \mathrm { s }$, and find}$
$$\left. \frac { d v _ { \theta } } { d r } \right| _ { r _ { 1 } } = \frac { r _ { 1 } ^ { 2 } \omega _ { 1 } } { r _ { 2 } ^ { 2 } - r _ { 1 } ^ { 2 } } \left( - r _ { 2 } ^ { 2 } r _ { 1 } ^ { - 2 } - 1 \right) = - \frac { 0.01 ^ { 2 } \times 209.4 } { .01025 ^ { 2 } - .01 ^ { 2 } } \left( .01025 ^ { 2 } \times .01 ^ { - 2 } + 1 \right) = \underline { 8480 \mathrm {~m} / \mathrm { s } / \mathrm { m } }$$

Oil Separates a 20-Mm-Diameter Rotating Shaft from Its 20