Question

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 m. (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) (b) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (Ignore the small variation of pressure over the surface of the window.)

Short answer

(a) The gauge pressure is $2.5\times {10}^6\mathrm{N}/{\mathrm{m}}^2$and (b) the net force due to the water and the air is$1.7\times {10}^5\mathrm{N}$ .

Step-by-step solution

Given Data

The depth is h = 250 m.

The diameter of glass window is d = 30 cm.

Understanding the net force on the glass window

The net force on the circular glass window can be calculated by using the relationship between force, pressure and cross-sectional area.

Determining the gauge pressure

The relation of gauge pressure can be written as:

P = pgh

Here, gis the gravitational acceleration and p is the density of seawater.

Substitute 250m for h,$1030\mathrm{kg}/{\mathrm{m}}^3$for p, and$9.80\mathrm{m}/{\mathrm{s}}^2$for g in the above relation.

$$\begin{gathered} P=(1030\mathrm{kg}/{\mathrm{m}}^3)(9.80\mathrm{m}/{\mathrm{s}}^2)\left(250\mathrm{m}\right) \\ P=2.5\times {10}^6\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the required pressure is $2.5\times {10}^6\mathrm{N}/{\mathrm{m}}^2$.

Determining the net force on the window

The relation of net force can be written as:

$$\begin{gathered} F=PA \\ F=P\left(\frac{{\mathrm{\pi d}}^2}{4}\right) \end{gathered}$$

Here, Ais the cross-sectional area.

Substitute $2.5\times {10}^6\mathrm{N}/{\mathrm{m}}^2$ for P, and 30 cm for d in the above relation.

$$\begin{gathered} F=(2.5\times {10}^6\mathrm{N}/{\mathrm{m}}^2)\left(\frac{\mathrm{\pi }\left(30\mathrm{cm}{\displaystyle \frac{1\mathrm{m}}{100\mathrm{xcm}}}\right)}{4}\right) \\ \mathrm{F}=1.7\times {10}^5\mathrm{N} \end{gathered}$$

Thus, the net force is $1.7\times {10}^5\mathrm{N}$.