Calculate absolute pressure and magnitude of the force?

Question

A submarine is at an ocean depth of 190 m. Assume that the density of sea water is 1025 kg/m^3 and the atmospheric pressure is 101000 Pa.

The acceleration of gravity is 9.81 m/s^2.

Calculate the absolute pressure at this depth. Answer in units of Pa.

Calculate the magnitude of the force exerted by the water at this depth on a circular submarine window with a diameter of 28.4 cm. Answer in units of N.

Answer 1

Pressure is force per unit area. Now consider a column of water with a cross sectional area of A, and height H that is subjected to a gravitational acceleration g. We'll assume g is constant over the height of the column, and further assume that the water is incompressible, so that its density is also constant over the entire height of the column. In this case, the force exterted by the column at its base is given by:

F = M * g

but M = the column volume * density

M = A * H * density

so

F = A * H * density * g

This force is distributed over the cross sectional area of the column, so

P = F/A = H * density * g

Plugging in the numbers you were given,

P = 190 m * 1025 kg/m^3 * 9.81 m/sec^2
P = 1.91*10^6 N/m^2 = 1.91*10^6 Pa

But this is just the pressure due to the column of water. The question asked for the *absolute* pressure, which is the pressure relative to a vacuum (as opposed to the "gauge pressure", which is the pressure in excess of standard atmospheric pressure; the pressure we just calculated would be the gauge pressure). Therefore, we have to add in the pressure exerted by the column of air in the atmosphere above the water. We are told that this pressure ot 1.01*10^5 Pa, so the total pressure at 190 m below the water surface is:

P_tot = 1.91*10^6 Pa + 1.01*10^5 Pa = 2.011*10^6 Pa


The window has a radius of 14.2cm, so the area of the window is pi*(14.2cm)^2 = 633.5 cm^2 = 6.335*10^-2 m^2

The force on this window is given by:

F = P*area = 2.011*10^6 Pa * 6.335*10^-2 m^2
F = 1.274*10^5 N

Answer 2

The pressure due to fluid depth is just P = ρgh, where ρ is fluid density, g is acceleration due to gravity, and h is the height of the fluid column (which is the same as its depth), all of which were given to you in this problem.

ρ = 1025 kg/m^3
g = 9.81 m/s^2
h = 190 m

Once you have P, you can calculate F = PA, where A is the area on which the pressure acts. For a circle, A = (πd^2)/4, where d is the diameter and was given as 28.4 cm, which you should rewrite as 0.284 m.

The final result is F = PA = ρgh(πd^2)/4, where all values were given.

Answer 3

that was easy