Question

Convert a pressure head of 10 m of water column to kerosene of specific gravity 0.8 and carbon tetra chloride of specific gravity of 1.62.

Answer 1

In this question, we have been asked to convert the pressure head of water into equivalent head of kerosene and carbon tetrachloride. We have been given the specific gravity of the substances. We know that the pressure due to water, kerosene and carbon tetra chloride will be the same as the pressure head will be converted. Therefore, using this relation we shall calculate our answer.

Formula used:
$P=\rho gh$
P is the pressure
g is the acceleration due to gravity.
$\rho$is the density

Complete step by step answer:
It is given that,
Pressure head due to water ${{h}_{1}}=10$
Specific gravity of water ${{S}_{1}}=1$
Specific gravity of kerosene ${{S}_{2}}=0.8$
Specific gravity of $CC{{l}_{4}}$ ${{S}_{3}}=1.62$
We know that the density of water is 1. We also know that specific gravity of a substance is given as the ratio of density of an object to density of water.
${{S}_{o}}=\dfrac{{{\rho }_{o}}}{{{\rho }_{w}}}$
But, we know ${{\rho }_{w}}=1$
Therefore,
${{S}_{o}}={{\rho }_{o}}$………………. (1)
Now, we know that pressure due to water ${{P}_{1}}$ shall be equal to both pressure due to kerosene ${{P}_{2}}$ and pressure due to carbon tetrachloride ${{P}_{3}}$
${{P}_{1}}={{P}_{2}}={{P}_{3}}$ …………….. (2)
We know that
$P=\rho gh$ …………… (3)
Therefore, from (1), (2) and (3)
We get,
${{S}_{1}}g{{h}_{1}}={{S}_{2}}g{{h}_{2}}={{S}_{3}}g{{h}_{3}}$
For ${{h}_{2}}$
${{h}_{2}}=\dfrac{{{S}_{1}}g{{h}_{1}}}{{{S}_{2}}g}$
After substituting the given values
We get,
${{h}_{2}}=\dfrac{g\times 10}{0.8g}$
Therefore,
${{h}_{2}}=12.5m$
Now, solving for ${{h}_{3}}$
${{h}_{3}}=\dfrac{{{S}_{1}}g{{h}_{1}}}{{{S}_{3}}g}$
Therefore,
${{h}_{3}}=\dfrac{g\times 10}{1.62g}$
On solving,
${{h}_{3}}=6.17m$
Therefore, the pressure head due to kerosene is 12.5 m and due to $CC{{l}_{4}}$ is 6.17 m.

Note:
We know that pressure on an object in liquid depends on the height of the object submerged in liquid from the upper surface. This height is therefore known as pressure head. It is also called a static head. The unit of pressure head is the same as the unit of height. It is measured in metres. The pressure head is used in measuring the energy of a centrifugal pump instead of pressure because the pressure might change if the specific gravity is changed. However, the pressure head does not change.