Calculating Buoyancy Rate of Rise for a Ship

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    Buoyancy Rate Rise
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SUMMARY

The discussion focuses on calculating the buoyancy rate of rise for a ship weighing 300,000 kg with a surface area of 100 m². The primary formula involves balancing the buoyant force against the resistive drag force experienced by the ship as it ascends from a depth of 100 meters in seawater. Key variables include the buoyant force generated by air balloons and the drag force, which increases with speed. The conversation highlights the importance of understanding equilibrium between these forces to approximate the ascent time without delving into complex calculus.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Knowledge of drag force calculations in fluid dynamics
  • Familiarity with the properties of seawater and its density
  • Basic physics concepts related to forces and motion
NEXT STEPS
  • Research buoyancy force calculations using Archimedes' principle
  • Learn about drag force equations and their application in fluid dynamics
  • Explore the effects of pressure changes on buoyancy in deep water
  • Study the dynamics of objects moving through fluids, including terminal velocity
USEFUL FOR

Marine engineers, naval architects, and anyone involved in underwater buoyancy applications or ship design will benefit from this discussion.

gloo
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Can someone give me a formula or give an approximate approach to calculating the time it would take for an object to be raised up (due to buoyancy) from a certain depth of water?
For instance, take a ship (say it weighs 300,000kg and it's a rectangular shape with surface area of 100m square (10m by 10m). If i attach enough air ballons around the ship , what would be the variables and formula to use to get an approximate time it would take to raise the ship to the surface from say 100meters of water (sea water).

I assume that the the surface area of the ballons might be a factor too but let's just leave those out (say i put the underneath the ship). I am not concerned about the acceleration at the end destroying the ship near the surface. I am fearful of this solution turning into a series of calculus equations b/c of the rate of change of pressure as the ship goes higher.
 
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gloo said:
I am fearful of this solution turning into a series of calculus equations b/c of the rate of change of pressure as the ship goes higher.
I'm afraid this will be the case. It will be the same as an object falling through the atmosphere and reaching terminal velocity. The object will have a constant buoyant force acting upward, tending to increase its speed, however increase in speed will increase the resistance of the water. It will reach an equilibrium where the resistive drag force will equal the buoyant force. The transient portion may be small compared to the steady state portion, so you could probably get away with finding which speed through the water will give a drag force equal to the buoyant force.
 

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