Question

At the surface of the ocean, the water pressure is the same as the air pressure above the water, \({\bf{15}}\;{\bf{lb}}/{\bf{i}}{{\bf{n}}^{\bf{2}}}\). Below the surface, the water pressure increases by \({\bf{4}}.{\bf{34}}\;{\bf{lb}}/{\bf{i}}{{\bf{n}}^{\bf{2}}}\) for every 10 ft of descent.

(a) Express the water pressure as a function of the depth below the ocean surface.

(b) At what depth is the pressure \({\bf{100}}\;{\bf{lb}}/{\bf{i}}{{\bf{n}}^{\bf{2}}}\).

Short answer

a. \(P = 0.434d + 15\)

b. The pressure at the depth of 196 feet is \(100\;{\rm{lb/i}}{{\rm{n}}^2}\).

Step-by-step solution

Find the rate of change of pressure

As pressure increases by \(4.34\,\,{\rm{lb}}/{\rm{i}}{{\rm{n}}^2}\) for a descent of 10 ft, the rate of change of pressure is

\(\begin{aligned}{c}m = \frac{{4.34}}{{10}}\\ = 0.434.\end{aligned}\)

Express the pressure as a function of depth

As the pressure at the surface of the water is \(15\;{\rm{lb/i}}{{\rm{n}}^2}\), it can be expressed as

\(\left( {d,P} \right) \equiv \left( {0,15} \right)\).

So, the pressures as a function of depth can be expressed as

\(P = 0.434d + 15\).

Find the depth where the pressure is \({\bf{100}}\;{\bf{lb}}/{\bf{i}}{{\bf{n}}^{\bf{2}}}\)

Substitute 100 for P in the equation \(P = 0.434d + 15\) to find the value of d.

\(\begin{aligned}100 &= 0.434d + 15\\0.434d &= 85\\d &= \frac{{85}}{{0.434}}\\ \approx 195.85\end{aligned}\)

So, the pressure at a depth of 196 feet is \(100\;{\rm{lb/i}}{{\rm{n}}^2}\).