Question

A liquid of coefficient of viscosity$\eta =1poise$ is following in a pipe of radius $3cm$ such that the rate of volume flow is $1000l/min$. Determine the Reynolds numbers

• A. $3500$
• B. $3600$
• C. $3400$
• D. $3536$

Answer 1

The correct option is D

$3536$

Step 1: Given data

Coefficient of viscosity,$\eta =1poise$

$=0.1pascal-second\left[As,1poise=0.1pascal-second\right]$

Radius of the pipe, $r=3cm$

$=3\times {10}^{-2}m\left[As,1m=100cm\right]$

The rate of volume flow, $\frac{dv}{dt}=1000l/min$

Step 2: Find the velocity, $\mathit{v}$

$$\begin{gathered} \frac{dv}{dt}=1000l/min \\ {\mathrm{\pi r}}^2\mathrm{v}=\frac{1{\mathrm{m}}^3}{60\mathrm{s}} \\ \mathrm{v}=\frac{1{\mathrm{m}}^3}{{\mathrm{\pi r}}^2\times 60\mathrm{s}} \end{gathered}$$ [As, $\frac{1000l}{min}=\frac{1{\mathrm{m}}^3}{60\mathrm{s}}$ and rate of volume flow, $\frac{dv}{dt}={\mathrm{\pi r}}^2\mathrm{v}$]

Step 3: Find the Reynolds numbers, $\mathbf{}{\mathit{R}}_{\mathbf{e}}$

$Renoldnumber,R_e=\frac{\rho \times v\times D}{\eta }$

$$\begin{gathered} =\frac{1000\times 2r\times 1}{0.1\times \pi r^2\times 60}\left[As,D=2r\right] \\ =\frac{1000\times 2\times 1}{0.1\times 3.14\times 3\times {10}^{-2}\times 60} \\ =3536\left[Aproximately\right] \end{gathered}$$

Hence, option (D) is correct.