Determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. Assume that the barometric pressure is 14.7 psia and the specific gravity of seawater is 1.03.
Solution:
We can find the density of water by multiplying its specific gravity by the density of sea water.
\rho =SG\times \rho_{H_{2}O}=(1.03)(62.4lbm/ft^{3})=64.27lbm/ft^{3} (SG is specific gravity, density of water can be found through research)
Now we will calculate the absolute pressure at 175ft below sea level as this is the location of the submarine.
P=P_{atm}+pgh
=(14.7psia)+(64.27lbm/ft^{3})(32.2ft/s^{2})(175ft)(\frac{1lbf}{32.2lbm\cdot ft/s^{2}})(\frac{1ft^{2}}{144in^{2}})
=92.8psia
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Thank you so much. Jesus Christ bless u .he gives this mind for you.you
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