Solved Problem on Fluid Mechanics

The density of seawater is equal to 1.025 g/cm³, Find:
a) The pressure exerted only by the water column at a point 50 m deep;
b) The pressure at this point, taking into account atmospheric pressure, which at sea level is 1.013×10⁵ Pa.

Problem data:

Density of seawater: ρ = 1.025 g/cm³;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

Figure 1

Solution

First, we must convert the water density given in grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³), used in the International System of Units (SI)

\[ \begin{align} \rho= & 1.025\;\mathrm{\frac{\cancel g}{cm^3}}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{\cancel g}}\times\frac{(100\;\mathrm{cm})^3}{(1\;\mathrm m\;)^{3}}=\\ = & 1.025\;\mathrm{\frac{1}{\cancel{cm^3}}}\times\frac{\mathrm{kg}}{1\cancel{000}}\times\frac{1 000 \cancel{000}\;\mathrm{\cancel{cm^3}}}{\mathrm m^3}=\\ = & 1025\;\mathrm{kg/m^3} \end{align} \]

a) The pressure of the liquid column, p_c, is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {p_{c}=\rho gh} \end{gather} \]

\[ \begin{gather} p_c=\left(1025\;\mathrm{\frac{kg}{m^{\cancel 3}}}\right)\times\left(9.8\;\mathrm{\frac{\cancel m}{s^2}}\right)\times\left(50\;\mathrm{\cancel m}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {p_c=5.023\times10^5\;\mathrm{Pa}} \end{gather} \]

b) The total pressure is given by

\[ \begin{gather} p=p_0+p_c \end{gather} \]

where p₀ is atmospheric pressure at sea level

\[ \begin{gather} p=1.013\times 10^5\;\mathrm{Pa}+5.023\times 10^5\;\mathrm{Pa} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {p=6.036\times 10^5\;\mathrm{Pa}} \end{gather} \]

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