The shear velocity in the 60-mm-diameter horizontal water pipe of Problem 2 is nearest:
A) 0.031 m/s
B) 0.024 m/s
C) 0.019 m/s
D) 0.017 m/s

Question

Quizplus · Accepted Answer

$0.017 \mathrm {~m} / \mathrm { s }$
$\text {The shear velocity (a fictitious velocity) is}$
$$u _ { \tau } = \sqrt { \frac { \tau _ { 0 } } { \rho } } = \sqrt { \frac { r \Delta \bar { p } } { \rho 2 L } } = \sqrt { \frac { 0.03 \times 200 } { 1000 \times 2 \times 10 } } = \underline { 0.0173 \mathrm {~m} / \mathrm { s } }$$
$\text {where we summed forces on a cylinder of fluid to obtain $\tau _ { 0 } 2 \pi r L = \Delta \bar { p } \pi r ^ { 2 }$ or}$ $\tau _ { 0 } = r \Delta \bar { p } / 2 L$

The Shear Velocity in the 60-Mm-Diameter Horizontal Water Pipe of Problem