Chapter 7, Problem 7.18P

Interpretation Introduction

Interpretation: The pressure drop per foot of tube caused by the shear stress is to be calculated.

Concept introduction: The shear stress on any object is given by the Newton’s law of viscosity,

  $\text{τ}\text{=}\text{μ}\frac{\text{du}}{\text{dr}}\mathrm{.......}\text{ (1)}$

In the above equation, the notations used are,

  $\begin{array}{l}\text{τ}\text{= }\text{Shear}\text{stress}\\ \text{μ}\text{=}\text{Viscosity}\text{of}\text{fluid}\\ \frac{\text{du}}{\text{dr}}\text{=}\text{Change}\text{in}\text{velocity}\text{with respect to the distance}\end{array}$

For the parabolic distribution the velocity is given as,

  $\text{u}\text{=}{\text{u}}_{\text{max}}\left[\text{1-}{\left(\frac{\text{r}}{\text{R}}\right)}^{\text{2}}\right]\mathrm{.......}\text{ (2)}$

In the above equation, the notations used are,

u is the velocity of flow of the fluid and ${\text{u}}_{\text{max}}$ is the maximum velocity of the fluid.

R is the radius of the tube and r is the distance from the center of the tube.

The pressure drop across the tube is given by,

  $\frac{\text{ΔP}}{\text{l}}\text{=}\frac{\text{F}}{{\text{A}}_{\text{1}}\text{×l}}\mathrm{.......}\text{ (3)}$

A₁ is the cross-section area of the tube, and is given by

  ${\text{A}}_{\text{1}}\text{=}\frac{\text{π}}{\text{4}}{\text{D}}^{\text{2}}\mathrm{.......}\text{(3a)}$

D is the diameter of the tube.

F is the force on the tube and l is the length of tube in feet.

  $\text{F}\text{=}\text{τ×A }\mathrm{.......}\text{ (4)}$

A is the area of the tube.