NAUI Master Diver Course: understanding theory regarding amount of air needed in lift bag

[Zef]

I am currently taking a NAUI Master Diver course and we discussed the chapter on light salvage. I am having difficulty understanding the amount of air needed in a lift bag at depth to raise an object.

The MD manual gives an example (I am referring to the metric example btw)...and I understand things from a mathematical perspective....I am having a hard time understanding it from a logical perspective.

The amount of air needed to fill the bag, according to the textbook does not take into consideration depth...can anyone please help me understand this.

here are the calculations:
Buoyant Force = Volume of the object x Density of Liquid
Weight of object in water = Weight in Air - Buoyant Force
Amount of air needed at depth = weight of object in water / Density of liquid
Surface air equivalent = Amount of air needed at depth X ATA for depth of the object

Here is the example problem in the book:
"You have offered to recover a sunken outboard motor. It rests in 15 meters of salt water, weights 50Kg on land, and displaces 15 Liters. Your lift bag holds 100 Liters and an extra cylinder with a capacity of 500 liters is available. How much air (surface equivalent) will you need to add to the lift bag to make the motor neutrally buoyant?"

BF= 15 X 1.025 = 15.4 Kg
Weight of object in Water = 50 - 15.4 = 34.6 Kg
Amount of air needed at depth = 34.6 / 1.025 = 33.75 Liters
Surface air equivalent = 33.75 X 2.5 ata = 84.37 Liters

So what is escaping me is why the amount of air needed at depth does not consider depth in atmospheres as part of the equation. Another student pointed out that the 33.75 liters will be the same regardless of depth...this makes no sense to me.

I also understand that in the real world one would just use a lot of air to get the object moving upwards...I am trying to understand this from a scientific/logical perspective.

I am sure there is something simple here that my brain is not assimilating so if anyone has a good explanation that would help my comprehension I would greatly appreciate it.

Thanks,
Zef

 

[Rich Keller]

Surface air equivalent = 33.75 X 2.5 ata = 84.37 Liters

So what is escaping me is why the amount of air needed at depth does not consider depth in atmospheres as part of the equation. Another student pointed out that the 33.75 liters will be the same regardless of depth...this makes no sense to me.

I also understand that in the real world one would just use a lot of air to get the object moving upwards.

The atmospheres are taken into consideration in their example above.

The other student is talking about the size of the bag being a constant.

In the real world if you just use a lot of air in the bag it will cause an out of control assent of the object. That is the last thing you want. If you get caught up it the rigging you will be a part of that out of control assent as well. Your instructors may understand the math involved but they do not understand how to actually salvage something.

 

[guruboy]

Here is the example problem in the book:
"You have offered to recover a sunken outboard motor. It rests in 15 meters of salt water, weights 50Kg on land, and displaces 15 Liters. Your lift bag holds 100 Liters and an extra cylinder with a capacity of 500 liters is available. How much air (surface equivalent) will you need to add to the lift bag to make the motor neutrally buoyant?"

BF= 15 X 1.025 = 15.4 Kg
Weight of object in Water = 50 - 15.4 = 34.6 Kg
Amount of air needed at depth = 34.6 / 1.025 = 33.75 Liters
Surface air equivalent = 33.75 X 2.5 ata = 84.37 Liters

So what is escaping me is why the amount of air needed at depth does not consider depth in atmospheres as part of the equation. Another student pointed out that the 33.75 liters will be the same regardless of depth...this makes no sense to me.

Water is treated as not compressible. This means 33.75L displaced is 33.75L displaced and it does not matter what depth you are.

In your example, 33.75 is the water displacement needed to make the object neutral.

What changes is the amount of air (measured at surface pressure) needed to displace 33.75 water at different depths.

At depth, it will take 84.37 of air at surface pressure. As you ascent with the object, the air would expand and the object would go from neutral to buoyant. You would need to dump air to keep that bag at 33.75 as the depth decreases.

Some of the reasons this matters is that
1) you might not have enough air at depth to inflate that bag
2) you might blow your lift bag if it goes up without you there to dump air
3) you bag might not be large enough

 

[guruboy]

The other student is talking about the size of the bag being a constant.

No, the student is referring to air in bag needed to make the object neutral.

But yes, size (physical dimensions) of bag is constant.

 

[CptTightPants21]

Disclaimer: I have never taken NAUI Master Diver or a Science of Diving Course. Everything I know is what I have read in books or picked up from other divers

The question above is assuming that 1 Liter of Air is equal in buoyancy to 1 Kg. For Europeans/those in the the know, is that a standard/normal assumption?

Positive buoyancy is that the weight of the water displaced by the object is GREATER than the weight of the object. The object is negative 34.6. so you need to figure out how much water needs to be displaced so that the weight equals 34.6.

The equation is X * 1.025= 34.6 (I assume the 1.025 conversation factor is different between salt and fresh water, but lets just stick with the numbers you gave us)

X = 33.75

That means that you need to displaced 33.75 Liters of water to get the object neutral. At this point, there is no atmosphere/pressure factor. The object is -34.6 buoyancy and you will need to displace 33.75 liters of water regardless of it being at 10 ft or 200 ft.

The depth rating comes into play when you need to calculate exactly how much air does it take to displace 33.75 liters of water at X depth. The answer to your question is 84.37.
You can look at it two ways:
1) You have 33.75 liters at 15 meters what will it expand to at the surface --33.75 * 2.5 =84.37
2) How much air do I need to take from the surface so that when I get down to 15m, it will equal 33.75 liters... X / 2.5 =33.75 ......X=84.37

Two way of looking at the exact same problem.

If the question was the same, but made the depth 21m instead of 15m. The answer would be 33.75 x 3.1 = 104.625 liters

@ 3m it would be 33.75 x 1.3 =

 

[Zef]

In the real world if you just use a lot of air in the bag it will cause an out of control assent of the object. That is the last thing you want. If you get caught up it the rigging you will be a part of that out of control assent as well. Your instructors may understand the math involved but they do not understand how to actually salvage something.

I mentioned the "in the real world..." because I read through some other posts on lift bags and there were a bunch of comments from folks with varying experience that insinuated to just fill the bag with air until the object starts moving...this comment was an effort to placate them so that I would not get a bunch of those types of responses. I think your assessment of our instructors is a quite a bit off though, as they never mentioned doing that at all, and it was based on classroom discussion that encouraged me to have a deeper understanding of what is at play here. I could have just left it at understanding the computation and memorizing Boyle's Law (P1xV1=P2xV2)

-Zef

 

[Zef]

CptTightPants21 and Guruboy,

Thank you for taking the time to explain...This has greatly helped my understanding., especially the details in CPTTightpants21's post.

I was not thinking of the amount of air in the lift bag from a displacement perspective; the best way I can describe how I was thinking about it would be that I was thinking of it purely from a volume perspective and my brain kept thinking that volume is affected by depth so the amount of air was dependent on the depth of the object, and I was stuck in a sort of erroneous logic loop.

Thanks again for your help.

-Zef

 

[dmaziuk]

The question above is assuming that 1 Liter of Air is equal in buoyancy to 1 Kg. For Europeans/those in the the know, is that a standard/normal assumption?

1 litre of water at 1 degree centigrade weighs 1 kilogram. It's close enough for the problem at hand.

To the OP: you are correct. If the problem is worded as posted, the wording is wrong. It should say something like
Amount of lift needed at depth = 34.6 / 1.025 = 33.75 Liters
Surface air equivalent = 33.75 X 2.5 ata = 84.37 Liters

 

[kr2y5]

Amount of air needed at depth = 34.6 / 1.025 = 33.75 Liters

"Amount of air" => "Amount of space occupied by air"

 

[Zef]

1 litre of water at 1 degree centigrade weighs 1 kilogram. It's close enough for the problem at hand.

To the OP: you are correct. If the problem is worded as posted, the wording is wrong. It should say something like
Amount of lift needed at depth = 34.6 / 1.025 = 33.75 Liters
Surface air equivalent = 33.75 X 2.5 ata = 84.37 Liters

Thanks, I feel redeemed reading someone else's assessment about how poorly worded the example is.

-Zef