Question

5. A room in the lower level of a cruise ship has a 30-cm-diameter circular window. If the midpoint

of the window is 5 m below the water surface, determine

(a) The hydrostatic force, in N, acting on the window, and

(b) The pressure center, in m.

Take the specific gravity of seawater to be 1.025.

Answer 1

For this problem, we should start by evaluating the density of the sea water, as that is the value that will be used in all calculations:

$\begin{aligned} \rho_{\textsf{sea water}} &= S.G × \rho_{H_2O}\\ &= 1.025 × 1000\ kg/m³\\ &= 1025\ kg/m³ \end{aligned}$

Next, we need the window surface area, which is circular as given in the problem:

$\begin{aligned} A &= π\dfrac{D²}4\\ &= π\dfrac{0.4²}4 m²\\ &= 0.07m² \end{aligned}$

Therefore, the mean pressure which is exerted on the window plate is going to be evaluated from the standard formula:

$\begin{aligned} P_{avg} &= P_{\textsf{(sea water)} }× g × h\\ &= (1025 × 9.81×5) Pa\\ &= 50.28 kPa \end{aligned}$

Next, the force of the static pressure which acts on the window is:

$\begin{aligned} F &= P_{avg} × A\\ &= 50.28 kPa × 0.07 m²\\ \\ \therefore F &= 3519.6 N \end{aligned}$