Solved Problem on Fluid Mechanics

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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{aligned} ρ = & 1.025 \frac{g}{{cm}^{3}} \times \frac{1 kg}{1000 g} \times \frac{(100 cm)^{3}}{(1 m)^{3}} = \\ = & 1.025 \frac{1}{{cm}^{3}} \times \frac{kg}{1000} \times \frac{1000000 {cm}^{3}}{m^{3}} = \\ = & 1025 kg / m^{3} \end{aligned}

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

\begin{matrix} p_{c} = ρ g h \end{matrix}

\begin{matrix} p_{c} = (1025 \frac{kg}{m^{3}}) \times (9.8 \frac{m}{s^{2}}) \times (50 m) \end{matrix}

\begin{matrix} p_{c} = 5.023 \times 10^{5} Pa \end{matrix}

b) The total pressure is given by

\begin{matrix} p = p_{0} + p_{c} \end{matrix}

where p₀ is atmospheric pressure at sea level

\begin{matrix} p = 1.013 \times 10^{5} Pa + 5.023 \times 10^{5} Pa \end{matrix}

\begin{matrix} p = 6.036 \times 10^{5} Pa \end{matrix}

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