The horsepower required to rotate the shaft of Problem 2, if SAE-30 oil at 40 $$40 ^ { \circ } \mathrm { C }$$ ills the gap, is nearest:
A) 0.203 hp
B) 0.191 hp
C) 0.174 hp
D) 0.123 hp

Question

Quizplus · Accepted Answer

$0.123 \mathrm { hp }$
$\text {The formula for the power is given by Eq. 7.5.20:}$
$$\dot { W } = \frac { 4 \pi \mu r _ { 1 } ^ { 2 } r _ { 2 } ^ { 2 } L \omega _ { 1 } ^ { 2 } } { r _ { 2 } ^ { 2 } - r _ { 1 } ^ { 2 } } = \frac { 4 \pi \times 0.1 \times .01 ^ { 2 } \times .01025 ^ { 2 } \times 0.8 \times 209.4 ^ { 2 } } { 0.01025 ^ { 2 } - 0.01 ^ { 2 } } = 91.5 \mathrm {~W} \text { or } \underline { 0.123 \mathrm { hp } }$$
$\text {Check those units: $\left( \mathrm { N } \cdot \mathrm { s } / \mathrm { m } ^ { 2 } \right) \mathrm { m } ^ { 2 } \mathrm {~m} ^ { 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 } / \mathrm { m } ^ { 2 } = \mathrm { N } \cdot \mathrm { m } / \mathrm { s } = \mathrm { J } / \mathrm { s } = \mathrm { W }$.}$

The Horsepower Required to Rotate the Shaft of Problem 2 40∘C40 ^ { \circ } \mathrm { C }40∘C

The Horsepower Required to Rotate the Shaft of Problem 2 $40 ^ { \circ } \mathrm { C }$