The hull of a vessel develops a leak and takes on water at a rate of 57.5 gal/min. When the leak is discovered the lower deck is already submerged to a level of 7.5 inches. At this time, a sailor turns on the bilge pump which begins to remove water at a rate of 73.8 gal/min. As an approximation, the lower deck can be modeled as a flat-bottomed container with a bottom surface area of 510 ft2 and straight vertical sides. How long will it be after the pump is turned on until the deck is clear of water?

When the leak is discovered and the pump is turned on, the lower deck is already submerged and the leak is not fixed; then, in order to have the deck clear of water, the bilge pump has to remove the accumulated water ([tex]V_{0}[/tex]) and the water that is taking on ([tex]r_{in}*t[/tex]) through the leak. We can represent this mathematically as follow:

[tex]V_{0} +r_{in} *t-r_{out}*t=0[/tex] Equation 1

Where:

[tex]V_{0}[/tex]: is the accumulated water when the leak was discovered

[tex]r_{in}[/tex]: is the takes on rate through the leak = 57.5 gal/min

[tex]r_{out}[/tex]: is the removing rate of the bilge pump = 73.8 gal/min

t= is the time since the pump is turned on until the deck is clear of water.

To calculate the accumulated water ([tex]V_{0}[/tex]), we will model the lower deck as a flat-bottomed container with a bottom surface area of 510 [tex]ft^{2}[/tex] and straight vertical sides. Knowing that the level submerged is 7.5 inches, and performing the corresponding unit conversions, we obtain:

[tex]V_{0}[/tex]= bottom surface area * lever submerged

[tex]V_{0}= 510ft^{2}*7.5 in*\frac{1ft}{12in}=318.75 ft^{3}*7.48\frac{gal}{1ft^{3}}=2384.25 gal[/tex] Equation 2

Solving equation 1 for time (t), and replacing the value obtained in equation 2, we get:

[tex]t=\frac{V_{0}}{(r_{out}-r_{in})} =\frac{2384.25 gal}{(73.8-57.5)gal/min}=[/tex]146,27 min