Question

The water flows from a tap of diameter $1.25\, cm$ 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, $D=1.25 cm =1.25 \times 10^{-2} \,m$
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 $v=\frac{4 Q}{\pi D^{2}}$
Substituting the value of $v$ in eqn (i), we get
$N_{R}=\frac{4 Q \rho D}{\pi D^{2} \eta}=\frac{4 Q \rho}{\pi D \eta}$
$=\frac{4 \times 5 \times 10^{-5} \times 10^{3}}{\left(\frac{22}{7}\right) \times 1.25 \times 10^{-2} \times 10^{-3}} \approx 5100$
For $N_{R}>\,3000$, the flow is turbulent.
Hence, the flow of water is turbulent with Reynolds number $5100 $