\subsection{Pressure}

Determine the total pressure in a open tank containing fuel oil of density 
$\rho:=924\cdot \kg/\m^{3}$ at depth of $h:=2.5\cdot \m$ when the atmospheric pressure is  
$p_{atm}:=1\cdot \atm"."$\\
{\bf Solution}\\
{\bf Given:}\\
$\rho    \cdot (\kg/\m^3)^{-1}$\\
$h       \cdot \mm^{-1}$\\
$p_{atm} \cdot \kPa^{-1}$\\
$g_n     \cdot (\m/\s^2)^{-1}$\\
The force due to gravity of the element is
\["
    F_g:=m_w\cdot g_n= V_e\cdot \rho\cdot g_n=A_e\cdot h\cdot \rho\cdot g_n
"\]

a preassure of element due to gravity is 
\["
    p_g:=\frac{F_g}{A_e}=\frac{ A_e\cdot h\cdot \rho\cdot g_n }{A_e}
    ;\;\;\;\;"
    p_g:=\rho\cdot g_n\cdot h
    ";\;\;\;"
    p_g\cdot\left(\kPa \right)^{-1}
\]
The total preassure
\begin{equation}
    p:=p_{atm}+p_{g}
    "\label{e}
    ;\;\;\;\;"
    p\cdot\left(\kPa \right)^{-1}
\end{equation}