Question

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipe are 6.4 cm and 4.8 cm, respectively. The ratio of minimum and maximum velocities of fluid in this pipe is:
A. $\dfrac{9}{{16}}$
B. $\dfrac{{81}}{{256}}$
C. $\dfrac{3}{4}$
D. $\dfrac{{\sqrt 3 }}{2}$

Answer 1

Calculate the maximum and minimum area of the pipe using given diameters. The velocity of the flow is maximum when the area of cross-section is minimum. Use the continuity equation for the flow of fluid through a pipe.

Complete step by step answer:

Equation of continuity.
${A_{\max }}{v_{\min }} = {A_{\min }}{v_{\max }}$

Here, ${v_{\min }}$ is the minimum velocity and ${v_{\max }}$ is the maximum velocity.

Complete step by step answer:

We know that the circular cross-section of pipe of radius r has area, $A = \pi {r^2}$. Therefore, the maximum area of the pipe is,
${A_{\max }} = \pi r_{\max }^2$
$\Rightarrow {A_{\max }} = \pi {\left( {\dfrac{{{d_{\max }}}}{2}} \right)^2}$
$\Rightarrow {A_{\max }} = \dfrac{{\pi d_{\max }^2}}{4}$ …… (1)

Also, the minimum area of the pipe is,
${A_{\min }} = \dfrac{{\pi d_{\min }^2}}{4}$ …… (2)

According to equation of continuity in the laminar flow, we have,
${A_{\max }}{v_{\min }} = {A_{\min }}{v_{\max }}$

Here, ${v_{\min }}$ is the minimum velocity and ${v_{\max }}$ is the maximum velocity.

The above equation implies that the velocity of the flow is maximum through the minimum area of cross-section of the pipe.

We rearrange the above equation as follows,
$\dfrac{{{v_{\min }}}}{{{v_{\max }}}} = \dfrac{{{A_{\min }}}}{{{A_{\max }}}}$

Use equation (1) and (2) to rewrite the above equation as follows,
$\dfrac{{{v_{\min }}}}{{{v_{\max }}}} = \dfrac{{\dfrac{{\pi d_{\min }^2}}{4}}}{{\dfrac{{\pi d_{\max }^2}}{4}}}$
$\Rightarrow \dfrac{{{v_{\min }}}}{{{v_{\max }}}} = {\left( {\dfrac{{{d_{\min }}}}{{{d_{\max }}}}} \right)^2}$

Substitute 4.8 cm for ${d_{\min }}$ and 6.4cm for ${d_{\max }}$ in the above equation.
$\dfrac{{{v_{\min }}}}{{{v_{\max }}}} = {\left( {\dfrac{{4.8\,cm}}{{6.4\,cm}}} \right)^2}$
$\Rightarrow \dfrac{{{v_{\min }}}}{{{v_{\max }}}} = {\left( {0.75} \right)^2}$
$\Rightarrow \dfrac{{{v_{\min }}}}{{{v_{\max }}}} = \dfrac{9}{{16}}$

Note: The equation of continuity can be applied to any flow on the condition the flow should cover the whole area of cross-section through which the liquid is flowing. The equation of continuity can be used to calculate the velocity of the water flowing through the lower opening of the water tank.