submarine depth?

Question

I need to know how to solve this. I have the submarine gage pressure, the water density and the atmospheric pressure. How do I determine the depth of the submarine?

Answer 1

Good!
The question is how detailed of an answer you want, but let's keep it simple.

We have four magnitudes of pressures to content with.
1. Atmospheric pressure above the surface of water (Pa)
2. Pressure generated by a water column or the depth of water (Pc)
3. Pressure inside the submarine (Ps)
4. Measured pressure or gage pressure (Pg)

So we can write

Pc + Pa= Pg - Ps

Then Pc=Pg - (Ps + Pa) or if the sub has the same pressure as the atmosphere we have

Pc=Pg-2Pa

We know that 27.7 inches of water = 1 pound per square inch (psi)

So if you are reading 200 psi and Pa=14psi
We have:
Pc=200-28=172 psi.

Then
D=depth= 27.7 Pc= 27.7 x 172 = 4764 inches
or 397 feet.

Refinement for density (this is where 27.7” = 1psi came from).

Pc=phg where
p- density of water
h- height of the water column
g- gravitational acceleration

Just be consistent and keep units in the same system

That should help.

Answer 2

Let's say the gauge reads in psi. From the water density, calculate the height of a one square inch column of water weighing 1 lb (it's very roughly 2 feet). Now multiply the pressure by this height to get the depth. This assumes the "gauge pressure" is the difference between pressure inside the sub and outside it, and that the cabin is kept at one sea-level atmosphere. You'll have to adjust the gauge pressure by the difference between cabin and atmospheric pressures if this is not true.