Finding the total pressure acting on a submarine

[strine07]

Hello I'm having trouble with answering this question: Certain submarines are capable of speeds in excess of 30 knots. What is the total pressure sensed at the nose of a submarine if the vessel is at periscope depth (5 m) and a speed of 30 knots? (Assume a constant sea-water density of 1030 kg/m3.) What height would the captain read on a mercury manometer that sensed this total pressure?
I know the equations Po=Ps + (1/2)(rho)(V)^2 and dP=(rho)(g)(dh) but am having trouble finding the static pressure in order to find the total pressure.

 

[mfb]

A depth of 0m corresponds to air pressure (~1.013*10^5 N), your equation for dP allows to add the 5m of water between nose and surface.

 

[Vadar2012]

The second equation should be enough. The submarine will experience this pressure over its surface area. So the mercury manometer will sense this pressure. The hard part would be to calculate the "lift" of the submarine, for which you could use Newton's method.

 

[Travis_King]

Keep in mind that the nose may see a stagnation point, you won't necessarily see the same total pressure (static and dynamic) as other parts of the vessel.

 

[Nickie]

I would approach this question by first understanding the concepts of fluid mechanics and hydrostatics. The total pressure acting on a submarine at a given depth and speed is a combination of the static pressure and dynamic pressure.

To find the static pressure, we can use the equation P = ρgh, where P is the static pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the depth is given as 5 m and the density of sea-water is given as 1030 kg/m3. Plugging these values into the equation, we can find the static pressure at periscope depth to be P = (1030 kg/m3)(9.8 m/s2)(5 m) = 50,470 Pa.

Next, we need to calculate the dynamic pressure using the equation P = ½ρV2, where P is the dynamic pressure, ρ is the density of the fluid, and V is the velocity. In this case, the velocity is given as 30 knots, which is equivalent to 15.4 m/s. Substituting these values into the equation, we get P = ½(1030 kg/m3)(15.4 m/s)2 = 74,110 Pa.

To find the total pressure, we simply add the static and dynamic pressures together, giving us a total pressure of 124,580 Pa.

To determine the height that the captain would read on a mercury manometer that sensed this total pressure, we can use the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height. Rearranging the equation to solve for h, we get h = P/(ρg). Substituting the values of the total pressure and density of mercury (13,600 kg/m3), we can find the height to be h = (124,580 Pa)/(13,600 kg/m3)(9.8 m/s2) = 0.89 m. This means that the captain would read a height of 0.89 m on the mercury manometer.

In conclusion, the total pressure acting on a submarine at periscope depth and a speed of 30 knots is 124,580 Pa. The captain would read a height of 0.89 m on a mercury