Buoyancy of a submarine

  • #1
tuki
19
1

Homework Statement


A submarine is in water, depth 30 meters. Inside submarine there is default air pressure. Submarine has volume of 125 m³, from which 10 m³ is water tank used for submerging. Submarine weights 123 tons. How large portion of the water tank has to be filled with sea water in order to stay still?

Homework Equations


Density:
$$ \rho = \frac{m}{v} $$
Archimede's principle

The Attempt at a Solution


By comparing densities we can determine if a object in fluid is sinking, floating or staying still. If
$$ \rho_{\text{obj}} > \rho_{\text{fluid}} \implies \text{sinking} $$
$$ \rho_{\text{obj}} < \rho_{\text{fluid}} \implies \text{floating} $$
$$ \rho_{\text{obj}} = \rho_{\text{fluid}} \implies \text{staying still} $$
. This can be derived from Archimede's principle.

Now we want density of the submarine to be equal to density of water. Which means it stays still in relation to the water. In following equation left hand side (density of submarine), right hand side (density of water). M = mass of submarine, m = mass of water in the tank, v = volume of the submarine.
$$ \frac{M+m}{v} = \rho_{\text{v}} $$
Mass of the water can be expressed as $$m = \rho_{\text{fluid}}v_\text{water in tank}$$
$$ \frac{M+\rho_{\text{fluid}}v_\text{water in tank}}{v} $$
$$ \implies v_\text{water in tank} = v-\frac{M}{\rho_{\text{fluid}}} $$

By computing this we will have

$$ v_\text{water in tank} \approx 1.63 \text{ m}^3 $$, which is 16.3% from the 10 m³ tank. Correct answer would be 56% of the 10m³ tank. Can't exactly see what I'am doing wrong?
 

Answers and Replies

  • #2
BvU
Science Advisor
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why would they tell you about this air pressure ?
what is the density of sea water you used ?
 
  • #3
tuki
19
1
why would they tell you about this air pressure ?
what is the density of sea water you used ?

The deeper you go the greater the pressure caused by water is. More water above submarine => greater pressure. I don't see how this would affect the buoyancy of the submarine? The force caused by pressure is distributed evenly on top, bottom and sides of the submarine, meaning the sum of the forces is zero? => these forces do not affect if the ship sinks or not. I think i lack the knowledge on how the Archimede's principle is related to pressure?
 
  • #4
tuki
19
1
The deeper you go the greater the pressure caused by water is. More water above submarine => greater pressure. I don't see how this would affect the buoyancy of the submarine? The force caused by pressure is distributed evenly on top, bottom and sides of the submarine, meaning the sum of the forces is zero? => these forces do not affect if the ship sinks or not. I think i lack the knowledge on how the Archimede's principle is related to pressure?

And the sea water density is used is 997kg/m³
 
  • #5
DrClaude
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And the sea water density is used is 997kg/m³
That's the problem. This is not the same density as the one used to obtain 56%.
 
  • #6
tuki
19
1
That's the problem. This is not the same density as the one used to obtain 56%.
Yes i think it's suppose to be 1.03*10^3 kg/m³. There is mistake in the description of the assignment.
 
  • #7
DrClaude
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Yes i think it's suppose to be 1.03*10^3 kg/m³. There is mistake in the description of the assignment.
That appears to be the correct value. 997 kg/m3 is a bit low for sea water, which is mostly in the range 1020-1030 kg/m3.
 
  • #8
BvU
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and my post was to make you see that 125 or 115 m3 of air is only a small mass -- (almost?) negligible
the sea water density is the main dish (and works in the other direction)
 

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