Fluid is a material that can flow. Gasses and liquids are fluids. To study fluid it is easier to deal with fluid’s volume, density, and pressure rather than its mass and weight. For example, we do transactions of any fluids in terms of their volume (liter) not in their mass (kg). Here, some terminologies are being introduced to study fluid and its behavior.
Density: It is defined as mass per unit volume of the fluid. It is a scalar quantity.
\begin{align*}
Density \amp = \frac{Mass}{Volume}
\end{align*}
\begin{equation}
d = \frac{m}{V}\tag{4.2.1}
\end{equation}
Its unit is \(kg /m^3\) in SI system, \(g/cm^3\) or \(g/ml\) in cgs system, and \(slug/ft^3\) in fps system. Any two objects of identical volume the heavier one has higher density than the other. Take two marbles of same size but one is a lead marble and another is a glass marble. Which one you feel heavier, has the higher density.
Example 4.2.1.
Calculate the mass of water in a container of size 25 cm long, 15 cm wide and is filled with water up to 10 cm high. If the density of water is \(1\,g/cm^3\text{.}\)
Solution.
The volume of water,
\begin{align*}
V \amp = length\times width\times height\\
or, \quad V \amp = 25 \,cm\times 15 \,cm\times 10 \,cm\\
\therefore V \amp = 3750 \,cm^3
\end{align*}
Hence, the mass of water in the container is given by
(4.2.1).
\begin{align*}
m \amp = d\times V\\
or, \quad m \amp = 1 \,g/cm^3 \times 3750\, cm^3\\
\therefore V \amp = 3750\, g = 3.75 \,kg
\end{align*}
Pressure: It is defined as a force per unit area of the surface. Force must be acting perpendicularly on the surface. Pressure defines how hard the object is pushing something in or out. It is a scalar quantity.
\begin{align*}
Pressure \amp = \frac{Force}{Area}
\end{align*}
\begin{equation}
p = \frac{F}{A}\tag{4.2.2}
\end{equation}
Its unit is \(N/m^2\) or pascal (Pa) in SI system. In FPS system of unit it is measured in \(psi\) which means pounds per square inch. Many units of pressure are still in practice such as torr, bar, mm of Hg., atm., etc. 1 atm is one atmospheric pressure,
\begin{equation}
1 \,atm = 1.013\times 10^5 \,Pa.\tag{4.2.3}
\end{equation}
Example 4.2.2.
How much pressure is applied on the ground by a girl of 50 kg if she is balancing herself by standing on her one shoe heel of area \(1 \,cm^2\text{.}\)
Solution.
The force exerted by the girl on ground is equal to her weight,
\begin{align*}
F \amp = m\times g\\
or, \quad F \amp = 50 \,kg\times 9.8 \,m/s^2\\
\therefore F \amp = 490 \,N
\end{align*}
and, Surface area =
\(1 \,cm^2 = 1\times (10^{-2}\,m)^2 = 1\times 10^{-4} \,m^2\) Hence, the pressure on the ground is given by
(4.2.2).
\begin{align*}
p\amp = \frac{490 \,N}{1\times 10^{-4} \,m^2}\\
\therefore p\amp = 4900000 \,N/m^2 = 4.9 \,MPa.
\end{align*}
A fluid exerts more pressure at greater depths. Consider a cylindrical water bottle of area of cross-section
\(A\text{.}\) If
\(m\) is mass of water filled in the bottle to the height
\(h\text{,}\) then from
(4.2.2), we have -
\begin{equation*}
p = \frac{F}{A} = \frac{mg}{A} =\frac{Vdg}{A} =\frac{Ahdg}{A} =hdg
\end{equation*}
\begin{equation}
p= hdg\tag{4.2.4}
\end{equation}
Here, \(V\) is the volume of fluid in a container, \(d\) is the density of the fluid, and \(g\) is acceleration due to gravity.
Example 4.2.3.
If you poke two holes in water bottle at different hights, which water-jets shoots out to larger distance from the bottle?
Solution.
The pressure exerted by water at depths is given by
(4.2.4),
\begin{equation*}
p= hdg
\end{equation*}
Since depth of water is more for the hole at the bottom, water exerts higher pressure at the bottom and hence water-jet shoots out to further distance from the bottom hole.
