Question

Water flows from a tap of diameter $1.2\, cm$ 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

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