How much sea water is needed to keep a submarine still at 30 meters depth?

In summary: 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.
  • #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?
 
Physics news on Phys.org
  • #2
why would they tell you about this air pressure ?
what is the density of sea water you used ?
 
  • #3
BvU said:
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 said:
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
tuki said:
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
DrClaude said:
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
tuki said:
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
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)
 

1. How is the amount of sea water needed to keep a submarine still at 30 meters depth calculated?

The amount of sea water needed to keep a submarine still at 30 meters depth is calculated by using Archimedes' principle, which states that the upward buoyant force on an object is equal to the weight of the fluid it displaces. This means that the amount of water needed to keep a submarine still at 30 meters depth is equal to the weight of the submarine itself.

2. What factors affect the amount of sea water needed to keep a submarine still at 30 meters depth?

The main factors that affect the amount of sea water needed to keep a submarine still at 30 meters depth are the size and weight of the submarine, as well as the density and salinity of the water. The shape and design of the submarine can also play a role in the amount of water needed for stability.

3. Can the amount of sea water needed to keep a submarine still at 30 meters depth change?

Yes, the amount of sea water needed to keep a submarine still at 30 meters depth can change depending on the external conditions. For example, if the submarine changes its depth or if the water density or salinity changes, the amount of water needed for stability will also change.

4. How does the amount of sea water needed to keep a submarine still at 30 meters depth differ from other depths?

The amount of sea water needed to keep a submarine still at 30 meters depth may differ from other depths due to the changing pressure and density of the water. At greater depths, the pressure and density increase, which means more water is needed for stability. Additionally, the depth may also affect the design and weight of the submarine, which can also impact the amount of water needed.

5. Can a submarine stay still at 30 meters depth without any sea water?

No, a submarine cannot stay still at 30 meters depth without any sea water. Without water, there is no upward buoyant force to counteract the downward force of gravity, which would cause the submarine to sink. Water is essential for maintaining stability and keeping a submarine at a specific depth.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
3
Views
7K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
9K
  • Introductory Physics Homework Help
Replies
4
Views
8K
Back
Top