Question

If it takes $5\;{\rm{minutes}}$ to fill a $15\;{\rm{L}}$ bucket from a water tap of diameter $\dfrac{2}{{\sqrt \pi }}\;{\rm{cm}}$ then the Reynolds number for the flow is ( density of water $= {10^3}\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}$ and viscosity of water $= {10^{ - 3}}\;{\rm{Pa}}{\rm{.s}}$) close to

Answer 1

First, we will use the discharge expression because discharge can give information about the water volume that comes out from the water tap. After this, all the information required for the Reynolds number calculation is given to use the given information directly in Reynold's number expression.

Complete step by step answer:
It is given to us that $5\;{\rm{minutes}}$ are required to fill the $15\;{\rm{L}}$ bucket, so from this information, we will calculate the magnitude of the discharge in one minute, so

$Q = \dfrac{v}{t}$

Here, $v$ is the volume of the bucket and $t$ is the total time.

Substitute the values in the above equation.

Therefore, we get

$\begin{array}{l} Q = \dfrac{{15\;{\rm{L}}}}{{5\;{\rm{min}}}}\\\ Q = 3\;{{\rm{L}} {\left/ {\vphantom {{\rm{L}} {{\rm{min}}}}} \right. } {{\rm{min}}}} \end{array}$ …… (1)

Now we will determine the volume of the water that flows from the tap in one second with discharge expression. So,

$Q = VA$

Here, $V$ is the volume of the water flowing in one second and $A$ cross-sectional area of the tap.

We know that the formula of the cross-sectional area of the tap is $A = \dfrac{{\pi {d^2}}}{4}$, so use this formula in the above equation.
$\begin{array}{l} Q = V \times \dfrac{{\pi {d^2}}}{4}\\\ V = \dfrac{{4Q}}{{\pi {d^2}}} \end{array}$ …… (2)

Write the expression of the Reynold number.

$R = \dfrac{{\rho Vd}}{\eta }$

Here, $\rho$ is the water's density, $d$ is the diameter of the tap, and $\eta$ is the viscosity of the water.

From equation (2), substitute the values of $V$ in the above equation. So, the above equation becomes,

$\begin{array}{l} R = \dfrac{{\rho \left( {\dfrac{{4Q}}{{\pi {d^2}}}} \right)d}}{\eta }\\\ R = \dfrac{{4\rho Q}}{{\pi d\eta }} \end{array}$

Substitute the values in the above equation.

Therefore, we get

R = \dfrac{{4\left( {{{10}^3}\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}} \right)\left( {3\;{{\rm{L}} {\left/ {\vphantom {{\rm{L}} {{\rm{min}} \times \dfrac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}} \times \dfrac{{{{10}^{ - 3}}\;{{\rm{m}}^3}}}{{1\;{\rm{L}}}}}}} \right. } {{\rm{min}} \times \dfrac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}} \times \dfrac{{{{10}^{ - 3}}\;{{\rm{m}}^3}}}{{1\;{\rm{L}}}}}}} \right)}}{{\pi \left( {\dfrac{2}{{\sqrt \pi }}\;{\rm{cm}} \times \dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\left( {{{10}^{ - 3}}\;{\rm{pa}}{\rm{.s}} \times \dfrac{{0.1\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}}}{{1\;{\rm{pa}}}}} \right)}}\\\ R = \dfrac{{4 \times {{10}^3}\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}} \times 5 \times {{10}^{ - 5}}\;{{{{\rm{m}}^{ - 3}}} {\left/ {\vphantom {{{{\rm{m}}^{ - 3}}} {\rm{s}}}} \right. } {\rm{s}}}}}{{0.0354\;{\rm{m}} \times {{10}^{ - 4}}{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}}}\\\ R \approx 56490 \end{array}$$ _Therefore, the Reynolds number for the flow is close to $56490$._ **Note:** In this solution, unit conversion of the various physical terms is very important, so remember the different units of all the terms and put them during the calculation. So that all the units get canceled out in the end, and we can get the constant Reynold number.