Question

In case of streamline flow of a fluid (which is incompressible), consider these streamlines $A, B$ and $C$ (as shown in the figure)

Let, $P, Q$ and $R$ are 3 planes perpendicular to the direction of flow of streamlines, then which of the following is correct?

• A. $m_{P}>m_{Q}>m_{R}$, where $m=$ mass flow per second.
• B. $v_{P}>v_{Q}>v_{R}$
• C. $n_{P}>n_{Q}>n_{R}$, where $n=$ number of fluid particles crossing area per unit time.
• D. $m_{P} < m_{Q} < m_{R}$

Answer 1

Answer: (B)

Mass of fluid flowing out is equals the mass flowing in, therefore $\Delta m_{P}=\Delta m_{Q}=\Delta m_{R}$, we have
$A_{P}\, v_{P}\, \rho_{P}\, \Delta t=A_{Q}\, v_{Q} \,\rho_{Q}\, \Delta t=A_{R}\, v_{R} \,\rho_{R} \,\Delta t\,\,\,$...(i)
For flow of incompressible fluids, we have,
$\rho_{P}=\rho_{R}=\rho_{Q}$
Thus, Eq. (i) reduces to,
$\Rightarrow A_{P} \,v_{P}=A_{R}\, v_{R}=A_{Q}\, v_{Q}$
Also, from the given figure we can also say that,
$A_{R}>A_{Q}>A_{P}$
So, at narrower portions, where the streamlines are closely spaced, velocity increases.
i.e. $ v_{p}>v_{O}>v_{R}$