Boyles law, Lifting and a better formula.

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GQMedic

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I'm a Fish!
I'm looking over a question I got right on an exam. And then, as I was making out my newly discovered formula I wanted to write n a flash card in a collection I call "Dive physics for the rest of us", my train of thought CRASHED. Ok, I've turned this things upside down, inside out, right side up, I've even kicked it a few times!

The nemisis is in boyles law it's self, or is it?

The nastly little buggar is: "You have determined that it will require 440 pounds of lift to raise an object from the ocean floor. The object is 85 feet deep, how much air will be required from an 80 cuft 3000psi cylinder to raise the object"?

Possible solutions are, 672 psi, 922, psi, 2828 psi, 440 psi.

Now that's great! But it's proving it to myself that boyles law works in this problem.

I've been trying to write a BETTER working formula for this, my problem is, I won't walk away from it. It's become a test of wills. I have to know why it works, not just the solution. My brain won't accept anything less than proving it. (I know, I really need a "normal" hobby) but this little monster must die!!

Now, I know the answer is 922 psi, but to prove my own BS fell through the cracks. First, I found the volume at depth, then find absolute pressure (That is easy enough), find the surface volume, then find pressure used from cylinder..

Ok, that works, The answer is found every time.(After I pop two tylenol) but it's LONG, it's UGLY and there has got to be a better formula.

And no, this is not a troll post, I don't think Trolls can even conceptualize algebra.

Any takers in the pursuit for a more sensible formula?
 
Hello,

A quick look in the salvage manual yields a few helpful formulas.

The standard cubic feet (SCF) required to fill a space whose volume in actual cubic feet (ACF) is:SCF = (ACF) x [(D + 33)/33] x [(Ta + 460)/(Tw + 460)]

where:
SCF = standard cubic feet of air required to fill the space
ACF = volume of the space in cubic feet
D = depth of water in feet
Ta = temperature of the air in degrees Fahrenheit
Tw = temperature of the water in degrees Fahrenheit
460 = correction factor to convert to absolute temperature

A web search yields this

Example: An anchore with an underwater weight of 197 lb is at a depth of 78 ft in the ocean. What is the minimum pressure in an 80 ft3, 3000psi SCUBA Tank required to fill a lift bag to raise the anchore. (Assuming constant water temperature and a lift bag weight is 2 lb.)
Solution:
Total weight of the system is : W= 197 lb + 2 lb = 199 lb
So that minimum Uplift force (B) must be = 199 lb.
Volume of Lift-Bag = 199 / 64 = 3.10 ft3 (required) at depth=78 ft. (P=3.36 atm)
At the surface conditions: P0= 1atm
Volume of Air: V=(3.10) *3.36 = 10.43 ft3
Check for 80 ft3 tank with 3000 psi pressure;
10.43 ft3 air consumption for Lift Bag = (10.43 / 80 ) * 3000 psi

Tank pressure drop = 391 psi (after the operation).



And this



Example: Imagine an object with a weight of W=287 lb and Vobject= 3 ft3 How many lift-bags with b=50 lb uplift force and w=2 lb weights are required.
(Operation is held in a Lake)
freshwater =62.4 lb/ft3
Solution:
W = B for equilibrium.
W = 287 + n*2 n: number of lift-bag
B = 3 (62.4) + n (50) 287 + 2n = 3(62.4)+50n n=2.03 so that at least 3 Lift Bags is required.

Does this help any?

Ed
 
well I was going to say that since saltwater is 64lbs a cubic foot at 1 ata, and 1 cf of air is 1 cf of water volume-wise. You should be able to take 440/64 =6.875 cf at 1 ata. Now (85+33)/33=3.575 ata's so 6.875 * 3.575 = 24.58 CF which = 1,573 lbs of tank presure, but then that's not one of the options for answer, so I obviously don't know what the hell I'm talking about.
 
Lifting an object (PSI Needed) Boyles Law

There are 4 steps to this…

1. Find the volume at depth by: Weight of object / Weight of water = X (VAD)
(Sea water is 64, Fresh is 62.4) write down result and clear the calculator.

2. Find your ATA Depth / 33 (ATM) +1 (ATM) = X (ATA)

3. Find your surface volume: ATA x VAD = X (Surface Volume)
Write down value and clear calc.

4. PSI in Cylinder / CUFT of Cylinder x Surface Volume = X
(Amount of pressure needed)

The formula works, but I was hoping to simplify this.

Oh well.. *falls over*
 
norcaldiver once bubbled...
well I was going to say that since saltwater is 64lbs a cubic foot at 1 ata, and 1 cf of air is 1 cf of water volume-wise. You should be able to take 440/64 =6.875 cf at 1 ata. Now (85+33)/33=3.575 ata's so 6.875 * 3.575 = 24.58 CF which = 1,573 lbs of tank presure, but then that's not one of the options for answer, so I obviously don't know what the hell I'm talking about.
Your method is the most logical to me, and you would have gotten the right answer if you had converted 24.58CF to psi correctly.

24.58*3000/80 = 922psi.
 
I have $10. After I buy lunch ($6.50) I will have $3.50.

Math more complicated than that makes my head hurt.
 
norcaldiver once bubbled...
24.58 CF which = 1,573 lbs of tank presure,

Think about that for a second....this would mean you have approximately a 47cu. ft tank rated at 3000 psi....or 1cu.ft. ~= 64psi., where as an AL80 1cu.ft. ~= 39 psi....so as charlie pointed out, if you correctly convert 24.58 cu.ft. for an AL80 (which is actually 77 cu.ft.), you get 957 psi....or if you use the rating of 80, you get 922psi.

to get back to the original question, norcal's idea seems very logical, but he just fumbled the ball while running to the end-zone with it.

Alright...enough math for the morning, I probably have something wrong somewhere.
 
Hello,

I for one am glad this topic came up. I am working on a visual basic application with alot of formulas. Currently I have the blending (nitrox/trimix) almost finished and will have many formulas included before it's over with.

Ed
 
A second pair of eyes....
OMG, I took it times the 64lbs(saltwater) not the 37.5(tank presure) Thanks guys.
 
https://www.shearwater.com/products/teric/

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