Question

The velocity of kerosene oil in a horizontal pipe is 5m/s, if ${\text{ }}g = 10{\text{ }}m/{s^2}{\text{ }}$ then the velocity head of oil will be
A) $1.25m$
B) $12.5m$
C) $0.125m$
D) $125m$

Answer 1

The speed of the outflow of liquid is given by using Torricelli's law. According to Toricelli’s law, the speed of flow through an opening from an open tank is that of a body falling freely from liquid level to the hole. The velocity head of a liquid can be taken as the potential energy required to accelerate a liquid to the actual velocity of flowing liquid.

Formula used:
The velocity head can be written as,
$\dfrac{{{v^2}}}{{2g}}$ Where $v{\text{ }}$ stands for the velocity of the liquid through the horizontal pipe and ${\text{ }}g{\text{ }}$stands for the acceleration due to gravity given by, ${\text{ }}10{\text{ }}m{s^{ - 2}}$

Complete step by step solution:
Bernoulli's theorem is very important in fluid dynamics. It is a mostly used equation too. The conservation of energy in a flow system relating the velocity, pressure, and height is expressed by the theorem.
The flow of incompressible liquid is given by,
$\dfrac{1}{2}{v^2} + \dfrac{P}{\rho } + gh = constant$ ……………………………………………….(A)
Where $v$ is the velocity of the fluid, ${\text{ }}P{\text{ }}$is the pressure at the region, ${\text{ }}\rho {\text{ }}$ is the density of the fluid, ${\text{ }}h{\text{ }}$ is the height of the fluid from the plane of reference, and ${\text{ }}g{\text{ }}$ is the acceleration due to gravity.
This is proof of Bernoulli’s theorem.
By using Torricelli's equation, we can simplify equation (A) to get the velocity head, ${\text{ }}\dfrac{{{v^2}}}{{2g}}$
The velocity of the liquid is given in the question, ${\text{ }}v = 5{\text{ }}m/s$
In the question, it is specified that ${\text{ }}g = 10{\text{ }}m/{s^2}$
Substituting these values in the equation for velocity head,
$\dfrac{{{v^2}}}{{2g}} = \dfrac{{{5^2}}}{{2 \times 10}} = 1.25m$

The answer is Option (A): ${\text{ }}1.25m$

Note: Torricelli's law gives the speed of flow of a liquid as the speed of a freely falling body and the expression for the law is ${\text{ }}v = \sqrt {2gh} {\text{ }}$where ${\text{ }}v{\text{ }}$ stands for the velocity of the fluid, ${\text{ }}h{\text{ }}$stands for the height in which the liquid is stored, and ${\text{ }}g{\text{ }}$ stands for the acceleration due to gravity. In the question, we consider ${\text{ }}g = 10{\text{ }}m/{s^2}{\text{ }}$ as specified in the question. If the value $g$ is not given in the question, it should be taken as ${\text{ }}g = 9.8m/{s^2}{\text{ }}$ to be the default.