Chapter 9, Problem 9.1P

Interpretation Introduction

Interpretation: The continuity equation for the steady incompressible flow in the polar coordinate is to be derived using the mass conservation law.

Concept introduction:

The law of conservation of mass states that mass can only be converted from one form to another, it can neither be created nor destroyed.

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.

Answer to Problem 9.1P

The expression for the continuity equation in terms of polar coordinates is $\frac{\text{1}}{\text{r}}\text{.}\frac{\partial }{\partial \text{r}}{\text{(rv}}_{\text{r}}\text{)}\text{+}\frac{\text{1}}{\text{r}}\text{.}\frac{\partial {\text{v}}_{\text{θ}}}{\partial \text{θ}}\text{+}\frac{\partial {\text{v}}_{\text{z}}}{\partial \text{z}}\text{=}\text{0}$ .

Explanation of Solution

The mass flux through the control volume is given as,

The area of the front surface = $\text{Δθ}\text{.}\Delta \text{r}$

The area of the top surface = $\text{Δz}\text{.}\Delta \text{r}$

The area of the side surface = $\text{Δθ}\text{.Δz}$

Now, for equation (1), it can be written,

  ${\displaystyle \iint \text{ρ(<mover accent="true"> <mo>V</mo> <mo>→</mo></mover>}\text{.<mover accent="true"> <mo>n</mo> <mo>→</mo></mover> )dA}}\text{=}{\text{(ρv}}_{\text{r}}\text{rΔz}{\text{.Δθ|}}_{\text{r+Δr}}\text{-}{\text{ρv}}_{\text{r}}\text{rΔz}{\text{.Δθ|}}_{\text{r}}\text{)}\text{+}{\text{(ρv}}_{\text{θ}}\text{rΔz}{\text{.Δr|}}_{\text{θ+Δθ}}\text{-}{\text{ρv}}_{\text{θ}}\text{rΔz}{\text{.Δr|}}_{\text{θ}}\text{)}\text{+}{\text{(ρv}}_{\text{z}}\text{rΔr}{\text{.Δθ|}}_{\text{z+Δz}}\text{-}{\text{ρv}}_{\text{z}}\text{rΔr}{\text{.Δθ|}}_{\text{z}}\text{)}$..... (2)

Also,

  $\frac{\partial }{\partial t}{\displaystyle \iiint \text{ρdV}\text{=}\frac{\partial }{\partial t}\text{(ρΔr}\text{.Δz}\text{.Δθ}})$  ........ (3)

Substitute equation (2) and equation (3) in equation (1) and use the limit as,

  $\begin{array}{l}\text{Δθ}\to \text{0}\\ \text{Δr}\to \text{0}\\ \text{Δz}\to \text{0}\end{array}$

The equation obtained is,

  $\frac{\text{1}}{\text{r}}\text{.}\frac{\partial }{\partial \text{r}}{\text{(rv}}_{\text{r}}\text{)}\text{+}\frac{\text{1}}{\text{r}}\text{.}\frac{\partial {\text{v}}_{\text{θ}}}{\partial \text{θ}}\text{+}\frac{\partial {\text{v}}_{\text{z}}}{\partial \text{z}}\text{=}\text{0}$  ........ (4)

Conclusion

The expression for the continuity equation in terms of polar coordinates is $\frac{\text{1}}{\text{r}}\text{.}\frac{\partial }{\partial \text{r}}{\text{(rv}}_{\text{r}}\text{)}\text{+}\frac{\text{1}}{\text{r}}\text{.}\frac{\partial {\text{v}}_{\text{θ}}}{\partial \text{θ}}\text{+}\frac{\partial {\text{v}}_{\text{z}}}{\partial \text{z}}\text{=}\text{0}$ .