Question

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

Answer 1

This problem can be solved by using the direct formula for the Reynold’s number in terms of the density, velocity of flow (which can be found from the rate of volume flow), diameter of tap and the coefficient of viscosity. By analyzing Reynold's number, we can find the nature of the flow.

Formula used:
$R=\dfrac{\rho vd}{\eta }$
$A=\pi \dfrac{{{d}^{2}}}{4}$
$V=Av$

Complete step-by-step answer:
We will find Reynold's number for the flow and therefore, find out the nature of the flow of water.
The Reynold’s number $R$ for flow of fluid of density $\rho$ and coefficient of viscosity $\eta$ in a stream of diameter $d$ and with velocity $v$ is given by
$R=\dfrac{\rho vd}{\eta }$ --(1)
The area $A$ of cross section of a circle of diameter $d$ is
$A=\pi \dfrac{{{d}^{2}}}{4}$ --(2)
The volume $V$ of liquid flowing per second in a stream of cross section $A$ is given by
$V=Av$ --(3)
where $v$ is the velocity of the flow of the fluid in the stream.
Now, let us analyze the question.
The diameter of the given tap is $d=1.25cm=1.25\times {{10}^{-2}}m$ $\left( \because 1cm={{10}^{-2}}m \right)$
Let the area of the opening of the circular tap be $A$.
Let the velocity of the fluid just after leaving the tap be $v$.
The flow rate of water is $V=0.48L/\min =0.48\dfrac{{{10}^{-3}}{{m}^{3}}}{60s}=8\times {{10}^{-6}}{{m}^{3}}/s$ $\left( \because 1L={{10}^{-3}}{{m}^{3}},1\min =60s \right)$
The coefficient of viscosity is $\eta ={{10}^{-3}}Pa.s$.
The density of water is $\rho ={{10}^{3}}kg/{{m}^{3}}$.
Let the Reynold’s number for the flow be $R$.
Using (1), we get
$R=\dfrac{\rho vd}{\eta }$ --(4)
Using (3), we get,
$V=Av$ --(5)
Using (2), we get
$A=\pi \dfrac{{{d}^{2}}}{4}$ --(6)
Putting (6) in (5), we get
$V=\pi \dfrac{{{d}^{2}}}{4}v$
$\therefore v=\dfrac{4V}{\pi {{d}^{2}}}$ --(7)
Putting (7) in (4), we get
$R=\dfrac{\rho d}{\eta }\dfrac{4V}{\pi {{d}^{2}}}=\dfrac{4V\rho }{\eta \pi d}$
Putting the values of the variables in the above equation, we get
$R=\dfrac{4\times 8\times {{10}^{-6}}\times {{10}^{3}}}{{{10}^{-3}}\times \pi \times 1.25\times {{10}^{-2}}}\approx 815$
Hence, we have got Reynold's number for the flow as $815$. A reynold’s number below $2000$ corresponds to smooth laminar flow and since $815<2000$, we can conclude that the nature of the flow of the water from the tap is a smooth laminar flow.

Note: Students must note that the Reynold’s number that we have calculated is for the flow of liquid just after exiting the tap. It may happen that after the liquid falls for a height, the flow becomes turbulent due to the increase in its speed as gravity acts on it. This can also be felt by the students by a simple experiment. They can open the tap of a basin in their house slightly and observe the flow of the water. At first it is smooth but as it falls down, the flow can become turbulent.