Question

The flow rate of a water from a tap of diameter $1.25cm$ is $0.48L/min$. If the coefficient of viscosity of water is ${10}^{-3}Pas$, what is the nature of flow of water?

Answer 1

Let the speed of the flow be V.

The diameter of the tap = d

= 1.25 cm

= 1.25 $\times {10}^{-2}$ m

Density of water = $\rho ={10}^{3}kg$ $m^{-3}$

viscosity = $\eta ={10}^{-3}$ Pa s

The volume of the water flowing out per second is

Q = $V\times \frac{\pi d^2}{4}$

V = $\frac{4Q}{d^2\pi }$

The Reynolds number is given by

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

= $\frac{4\rho Q}{\pi d\eta }$

= $\frac{4\times {10}^3\times Q}{3.14\times 1.25\times {10}^{-2}\times {10}^{-3}}$

= $1.019\times {10}^{8}Q$

Hence the flow is turbulent.