How High Did Water Rise in the Squalus Diving Bell?

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SUMMARY

The discussion centers on the physics of water rising within the diving bell used during the Squalus submarine rescue in 1939. The pressure at the bottom of the diving bell was calculated to be 838,137 Pa, taking into account the seawater density of 1030 kg/m³ and the depth of 73.0 m. Key equations discussed include the hydrostatic pressure equation (P = rho * g * height) and the ideal gas law (PV = nRT). The conversation emphasizes the importance of understanding pressure changes and specific gravity in determining the water rise within the bell, highlighting that the radius of the bell is not necessary for solving the problem.

PREREQUISITES
  • Understanding of hydrostatic pressure (P = rho * g * height)
  • Familiarity with the ideal gas law (PV = nRT)
  • Basic knowledge of specific gravity and its relation to temperature
  • Concept of pressure equilibrium in fluid mechanics
NEXT STEPS
  • Study the relationship between pressure and depth in fluids using hydrostatic principles
  • Learn about specific gravity and its temperature dependence
  • Explore the application of the ideal gas law in varying conditions
  • Investigate the mechanics of diving bells and their design considerations
USEFUL FOR

Students in physics or engineering, marine rescue professionals, and anyone interested in fluid mechanics and pressure dynamics in underwater environments.

Jason Onwenu
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Homework Statement


During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature at the surface was 27.0 ∘C and at the bottom it was 7.0 ∘C. The density of seawater is 1030 kg/m3. A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.)

Homework Equations


P = rho * g * height
PV = nRT
P1V1/T1 = P2V2/T2
V = pi * radius^2 * height

The Attempt at a Solution



P(bottom) = 101325 + (1030*9.8*73) = 838137 Pa
V2 = ? (The radius wasn't given...)
T2 = 280.15 K

P(top) = 101325
V1 = ? (The radius wasn't given...)
T1 = 303.15K

At this point, I couldn't really proceed with my calculations because a radius wasn't given. I'm wondering if I could solve for the radius with the information given or if the radius is even needed at all.
 
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You should ask yourself this question: Why does the water stop rising inside the diving bell when it is lowered onto the submarine?

Once you answer this question, think about whether you need to know the radius of the cylinder.
 
SteamKing said:
You should ask yourself this question: Why does the water stop rising inside the diving bell when it is lowered onto the submarine?

Once you answer this question, think about whether you need to know the radius of the cylinder.

I'm guessing it has something to do with (specific) gravity? Since there is a difference in temperature?
 
Last edited:
Jason Onwenu said:
I'm thinking it has something to do with (specific) gravity?

Why would specific gravity stop water from rising inside the diving bell?

Isn't the specific gravity of the water inside the diving bell the same as the specific gravity of the water outside?

Hint: Think about what is changing inside the bell as the water rises.
 
SteamKing said:
Why would specific gravity stop water from rising inside the diving bell?

Isn't the specific gravity of the water inside the diving bell the same as the specific gravity of the water outside?

Hint: Think about what is changing inside the bell as the water rises.

Pressure changes...
 
SteamKing said:
Why would specific gravity stop water from rising inside the diving bell?

Isn't the specific gravity of the water inside the diving bell the same as the specific gravity of the water outside?

Hint: Think about what is changing inside the bell as the water rises.

Is specific gravity a constant value for a specific substance? Or does it vary with temperature?
 
Last edited:
Do calculations until the end and then ask if you need the radius.
 

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