Acceleration of Submarine when Ejecting Water from its Tanks

  • #1

Main Question or Discussion Point

Brief Summary of Question:

One submarine starts with a neutral buoyancy 100m below the surface when it displaces 8.9m^3 of water from its ballast tanks. If the mass of the submarine is 69 metric tons what is the initial, vertical acceleration of the submarine immediately after displacing the water?

[ρ(water)=1,000 kgm-3, 1 atm=105 N/m2, g=9.8ms-2]

Related Equations: F(buoyancy) = pgV, where p=density of fluid, g=acc., V = displacement of the water due to sub.
F = ma
Neutral buoyancy --> F(buoyancy) = Weight​

Work done:

Used the buoyancy formula to calculate the displacement of the water by the sub. = 69m^3. (Not sure how the 100m below the surface played into this equation, as well it would seem we need the difference in contained water to calculate the acceleration upwards after ejection)

Guessing we need to use this in tandem with the 8.9m^3 that gets ejected from its tanks.
 

Answers and Replies

  • #2
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initial, vertical acceleration
What's vertical acceleration before ejecting water? From the problem statement, consider the ejection to be instantaneous, and that there is no change in volume of the submarine. What is the initial vertical acceleration?
 
  • #3
What's vertical acceleration before ejecting water? From the problem statement, consider the ejection to be instantaneous, and that there is no change in volume of the submarine. What is the initial vertical acceleration?
Well there is no vertical acceleration before the ejection of water of the sub. Its neutral buoyancy, so the buoyancy force is equal and opposite to the weight of the submarine.
 
  • #4
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buoyancy force is equal and opposite to the weight of the submarine.
Okay. Now, what happens to those forces when the water is ejected from the submarine?
 
  • #5
Okay. Now, what happens to those forces when the water is ejected from the submarine?
Well Air gets replaced where the water was inside of the sub, makes the buoyancy force greater, allowing the submarine to rise?
 
  • #6
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Okay, we'll back up a little further. You have a 69 ton submarine. It has a constant volume. If you eject 8.9 m3 of water that were in the submarine's ballast tanks, what has happened to the mass of the submarine? And to the volume of the submarine?
 
  • #7
Okay, we'll back up a little further. You have a 69 ton submarine. It has a constant volume. If you eject 8.9 m3 of water that were in the submarine's ballast tanks, what has happened to the mass of the submarine? And to the volume of the submarine?
Well mass would go down, because it is now filled with 8.9m^3 of air instead of water, but the volume would be the same, because the size of the sub has not changed.
 
  • #8
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Okay. Has anything happened to the buoyancy force experienced when submerged?
 
  • #9
Well mass would go down, because it is now filled with 8.9m^3 of air instead of water, but the volume would be the same, because the size of the sub has not changed.
Well at second glanced, it only seems now that the reason why it is acc. upwards is because of less weight, rather than more buoyancy, because buoyancy only has to do with the amount of water being displaced around the vessel, which is directly related to the volume of the sub which does not change.
 
  • #10
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Correct. You have all the information you need to calculate the difference in the two forces. What is that difference?
 
  • #11
Correct. You have all the information you need to calculate the difference in the two forces. What is that difference?
Well I would use the gravitational acc. with the buoyancy formula to get the Volume displaced by the ship, then I would use the amount of water ejected with its density to find its weight. Subtract that amount from the ships weight, and then work backwards with the same volume, to find the acceleration, then subtract the gravitational acc. to find its upwards acc.

I think this is correct?
 
  • #12
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The initial condition of neutral buoyancy will give you the buoyant force acting on the submarine. Once the water is displaced, the weight of the submarine will reduce but the buoyant force will remain the same. Using Newton's second law, you can get the acceleration.
 
  • #13
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get the Volume displaced
Check.
find its weight
Check.
Subtract that amount from the ships weight,
Check.
backwards with the same volume, to find the acceleration, then subtract the gravitational acc. to find its upwards acc.
I don't know what you said here, but do the calculation, and we'll see what you get.
 
  • #14
Simplified: V = m/density = 69000kg/(1000kg/m^3) = 69m^3
Weight of water displaced: 8.9m^3 * 1000kg/m^3 = 8900 kg
Weight of ship (post-ejection) = 69000-8900 = 60100 kg (* 9.81ms^-2) = 589581N (Fw)
Find acc. --> F(w) = pgV --> F(w)/pV = g --> 589581N/((1000kg/m^3)(69m^3)) = 8.5446 ms^-2

9.81 ms^-2 - 8.5446 ms^-2 = 1.265 ms^-2 acceleration upwards?
 
  • #15
Or is it that its a difference in force, which I imagine to be true.

Weight of Ship = 589581 N

Buoyancy Force = (1000kg/m^3)(9.81ms^-2)(69m^3) = 676890 N

Difference in the two: 87309 N --> divide this by the new ships mass

87309N/60100kg = 1.4527 ms^-2 upwards acceleration.

I believe this one to be correct.
 
  • #16
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I believe this one to be correct.
As do I. Haven't checked the arithmetic integer by integer, but the method is correct.
 
  • #17
The assignment has told me it is incorrect.
 
  • #18
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87309N/60100kg = 1.4527 ms^-2 upwards acceleration
That is the same answer I got.
 
  • #19
That is the same answer I got.
As the assignment told me that this value was incorrect. Unfortunately.
 
  • #20
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As the assignment told me that this value was incorrect. Unfortunately.
Maybe the mistake lies in the initial mass. Maybe 69 tonnes is the mass after the water has been displaced.
 
  • #21
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Try it without subtracting the ballast mass. They may be running it like a real submarine with a captured air bubble. Let's see --- you'll have to add the ejected volume to figure buoyant force in that case. It will change as the boat rises, but you're not being asked to calculate rise rate as a function of depth, so that much can still be ignored.
 
  • #22
Maybe the mistake lies in the initial mass. Maybe 69 tonnes is the mass after the water has been displaced.
I will carry out the calculations as this value being the final mass.
 
  • #23
With this being the final mass, I have somehow achieved my first answer of 1.265 ms^-2.
 
  • #24
This ended up being the final answer: 1.265, thanks siddharth23 and Bystander for all your work with helping me through this, much appreciated!

I may have another question very soon, about torque and forces, look out for that one on the new posts! :)
 
  • #25
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No problemo :-)
 

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