Chapter 4, Problem 4.5P

Interpretation Introduction

Interpretation:The value of discharge velocity for a steady flow should be calculated.

Concept introduction:

Law of conservation of mass: It states that energy can neither be created nor be destroyed. The generalized form of an equation is written as:

[Rate of mass in from control volume]- [Rate of mass out from control volume] + [Rate of accumulation]

Volume Flow rate:

For steady-state:At the condition of steady-state flow, the properties of fluids do not change over the period of time.

Inflow volume flow rate = Outflow volume flow rate

  ${\text{Q}}_{\text{1}}\text{=}{\text{Q}}_{\text{2}}$

Where

  $\begin{array}{l}{\text{Q}}_{\text{1}}\text{=}{\text{A}}_{\text{1}}{\text{V}}_{\text{1}}\\ {\text{Q}}_2\text{=}{\text{A}}_2{\text{V}}_2\end{array}$

Using the above equations,

  $\begin{array}{l}{\text{A}}_{\text{1}}=\text{π}{r}_1^2\\ {\text{A}}_2=\text{π}{r}_2^2\end{array}$

Where,

  ${\text{A}}_{\text{1}}$ and ${\text{A}}_2$ are the area for inlet and outlet.

  ${\text{r}}_{\text{1}}$ is the radius at the inlet of the pipe.

  ${\text{r}}_2$ is the radius at the outlet of the pipe.