Chapter 9, Problem 9.7P

Interpretation Introduction

Interpretation: The velocity distribution of the example 2 satisfies the continuity equation is to be proved.

Concept introduction:

The expression of the control volume for mass conservation is,

  ${\displaystyle \iint \text{ρ(<mover accent="true"> <mo>V</mo> <mo>→</mo></mover>}\text{.<mover accent="true"> <mo>n</mo> <mo>→</mo></mover> )dA}\text{+}\frac{\partial }{\partial \text{t}}}{\displaystyle \iiint \text{ρdV}\text{=}\text{0}}$   ……. (1)

dA and dV are the area and volume of the small differential part respectively.

  $\text{ρ}$ = Density of the fluid

v = Velocity of the fluid

The sum of the net rate of mass flux out of control volume and the rate of accumulation of mass in the control volume is zero.

The expression for the continuity equation in terms of polar coordinates is,

  $\nabla .\text{<mover accent="true"> <mo>V</mo> <mo>→</mo></mover>}\text{=}\frac{\text{1}}{\text{r}}\text{.}\frac{\partial }{\partial \text{r}}{\text{(rv}}_{\text{r}}\text{)}\text{+}\frac{\text{1}}{\text{r}}\text{.}\frac{\partial {\text{v}}_{\text{θ}}}{\partial \text{θ}}\text{+}\frac{\partial {\text{v}}_{\text{z}}}{\partial \text{z}}$   ……. (2)

In the equation (2), $\text{r,}\text{θ}\text{,}\text{z}$ are the components in 3 different directions.

For the satisfaction of the continuity equation, the equation (2) must evaluate to zero.