Question

The water flows from a tap of diameter $1.25cm$ with a rate of $5\times 10^{-5}{m}^3{s}^{-1}.$ The density and coefficient of viscosity of water are $10^{3}kg{m}^{-3}$ and $10^{-3}Pa$ s respectively. The flow of water is

• A. steady with Reynolds number 5100
• B. turbulent with Reynolds number 5100
• C. steady with Reynolds number 3900
• D. turbulent with Reynolds number 3900

Answer 1

Answer: (B)

Here, diameter, $D=1.25cm=1.25\times 10^{-2}m$
Density of water, $\rho =10^{3}kg{m}^{-3}$
Coefficient of viscosity, $\eta =10^{-3}Pa$ Rate of flow of water, $Q=5\times 10^{-5}{m}^3{s}^{-1}$
Reynolds number, $N_R=\frac{v\rho D}{\eta }$
where $v$ is the speed of flow.
Rate of flow of water $Q=$ area of cross section $\times$ speed of flow
$Q=\frac{\pi D^2}{4}\times v$ or $v=\frac{4Q}{\pi D^2}$
Substituting the value of $v$ in eqn (i), we get
$N_R=\frac{4Q\rho D}{\pi D^2\eta }=\frac{4Q\rho }{\pi D\eta }$
$=\frac{4\times 5\times 10^{-5}\times 10^3}{\left(\frac{22}{7}\right)\times 1.25\times 10^{-2}\times 10^{-3}}\approx 5100$
For $N_R>3000$, the flow is turbulent.
Hence, the flow of water is turbulent with Reynolds number $5100$