Question

The flow rate of water from a tap of diameter 1.25cm is 0.48 L/min. The coefficient of viscosity of water is ${10^{ - 3}}$ Pa/s. After sometime the flow rate is increased to 3 L/min. Characterise the flow for both the flow rates.

Answer 1

The Reynold’s number is defined as the experimental number which is used in fluid flow to predict the velocity of the flow at which the turbulence will occur. It is also defined as the ratio of the inertial forces to viscous forces.
Formula used:
${R_e} = \dfrac{{\rho VD}}{\mu }$

Complete answer:
According to the question, the following physical quantities is given
Diameter of the tap is given as,
D = 1.25cm $= 1.25 \times {10^{ - 2}}m$
Density of the water flowing through the tap is given as,
$\rho = {10^3}kg{m^{ - 3}}$
The coefficient of viscosity of water is given as,
$\eta = {10^{ - 3}}{\text{Pa s}}$
The volume of the water flowing out per second is given as
$Q = V \times \dfrac{{2\pi d}}{4}$
$\Rightarrow V = \dfrac{{4Q}}{{2\pi d}}$
In fluid mechanics, Reynolds number is a criterion used to determine whether the flow of the fluid is steady or turbulent. It is a dimensionless quantity. Systems which have the same Reynolds number show the same flow characteristics even if the fluid or its speed or length varies. Mathematically, Reynolds number is expressed as
${R_e} = \dfrac{{\rho VD}}{\eta }$
Where ‘${R_e}$’ is expressed as the Reynolds number, ‘$\rho$’ is the density of the fluid, ‘V’ is the velocity of the flow, ‘D’ is the diameter of the pipe and ‘$\eta$’ is expressed as the viscosity of the fluid.
Now, the Reynolds number is given as,
${R_e} = \dfrac{{\rho VD}}{\eta }$
$\Rightarrow {R_e} = \dfrac{{4\rho Q}}{{\pi d\eta }} = \dfrac{{4 \times {{10}^3} \times Q}}{{3.14 \times 1.25 \times {{10}^{ - 2}}m \times {{10}^{ - 3}}{\text{Pa s}}}} = 1.019 \times {10^8}Q$
Initially, Q = 0.48 L/min = $\dfrac{{0.48 \times {{10}^{ - 3}}}}{{60}} = 8 \times {10^{ - 6}}$
Substituting the values in the above expression we get, R = 815
Since, the resultant Reynolds’s number is less than 1000, i.e. 815 < 1000 thus the flow will be steady.
After sometime, Q = 3L per min $= \dfrac{{3 \times {{10}^{ - 3}}{m^3}}}{{60s}} = 5 \times {10^5}{m^3}{s^{ - 1}}$
Substituting the values in the above expression we get, R = 5095
Since, the resultant Reynolds’s number is greater than 2000, i.e. 5095 > 2000 thus the flow will be turbulent.
Hence, the flow for both the rates will be steady and turbulent.

Note:
In order to find the type of flow Reynolds’s number is used: -
When the Reynolds’s number is less than 1000 the flow will be laminar, when the Reynolds’s number is in between 1000 – 2000 the flow will be transient and when the Reynolds’s number is greater than 2000 the flow will be turbulent.