Question

Water flows from a tap of diameter $1.2cm$ at a rate of $2.4\times 10^{-5}{m}^3{s}^{-1}$. The density and coefficient of viscosity of water are $10^{3}kg{m}^{-3}$ and $10^{-3}Pa-s$ respectively. The flow of water has Reynold's number of $\frac{C}{\pi }$. Find $C$.

Answer 1

Answer: 8000

Reynold's number, $N_R=\frac{v\rho D}{\eta }$ ....(i)
Rate of flow of water $=$ Area of cross-section $\times$ speed of flow
$\therefore Q=\frac{\pi D^2}{4}\times v$
$\therefore v=\frac{4Q}{\pi D^2}$
From equations (i) and (ii),
$N_R=\frac{\rho D}{\eta }\times \frac{4Q}{\pi D^2}=\frac{4\rho Q}{\pi \eta D}$
$\Rightarrow C=\frac{4\rho Q}{\eta D}$
$\therefore C=\frac{4\times 10^3\times 2.4\times 10^{-5}}{10^{-3}\times 1.2\times 10^{-2}}=8000$