# Subsea Engineering Question - air required to lift load

1. Aug 28, 2016

### cps.13

Hi,

I am working on a product design which would be used sub sea. It is designed to lift a small load off the sea bed. I have not worked out the exact details yet but the following should be a good start:

Depth 30m

I plan on using an air bag of some description - the bag needs to auto inflate though as there will not be a diver there to operate. This is the circuit/design I am working on at the moment.

What I need to figure out is, how much air (or CO2) is required to lift a load of 30kg from 30m depth?

Can anybody point me in the direction of the calculations for this?

Thanks

2. Aug 28, 2016

### billy_joule

Archimedes principle is what you need.

3. Aug 28, 2016

### cps.13

OK - but how do I convert a quantity of CO2 into kg?

Is it the 30kg X density of co2 as a gas?

4. Aug 28, 2016

### billy_joule

You certainly don't need 30kg of CO2.

Can you draw a simple diagram (or explain in words) of your interpretation of how Archimedes principle applies to your question?

5. Aug 28, 2016

### jack action

Archimedes principle for floating:
$$W < \rho_{H_2O}V_{displ}$$
Where $W$ is the weight, $\rho$ is the density and $V$ is the volume.

Substituting the rock $r$, the bag $b$ and $CO_2$ components:
$$W_r + W_b + W_{CO_2} < \rho_{H_2O}\left(V_r + V_b + V_{CO_2}\right)$$
$$W_r + W_b + W_{CO_2} < \rho_{H_2O}\left(\frac{W_r}{\rho_r} + \frac{W_b}{\rho_b} + \frac{W_{CO_2}}{\rho_{CO_2}}\right)$$
$$W_r \frac{1 - \frac{\rho_{H_2O}}{\rho_r}}{\frac{\rho_{H_2O}}{\rho_{CO_2}} - 1} + W_b \frac{1 - \frac{\rho_{H_2O}}{\rho_b}}{\frac{\rho_{H_2O}}{\rho_{CO_2}} - 1} < W_{CO_2}$$
If this is true, the ensemble should float.

6. Aug 29, 2016

### Bandit127

Since the density of water is 1,000 kg/m3 you need 30 l to displace 30 kg.
Water at 30 m will be at a pressure of 3 additional atmospheres.

The increase in density of water at that depth will be negligible.
The increase in density of air with be a factor of 3.

The density of air is 1.2 kg/m3 (or 1.2 g/l) at sea level so it will be 3.6 at 3 m. So the mass of the air will be 3.6 x 30 = 108 g and you will need to add that and the weight of the lifting gear to the amount of water you want to displace.

If you use a rubber balloon, what is going to happen to the volume of air as the weight starts rising and the pressure from the depth of water decreases?
If you use a stiff balloon what will the pressure in it be when it reaches the surface?

7. Aug 29, 2016

### cps.13

I appreciate that as the balloon rises the air will expand causing the balloon to expand, I plan to have a pressure relief value within the balloon/bag once I know the pressures required.

Last edited by a moderator: Aug 31, 2016
8. Aug 29, 2016

### cps.13

So I have made an attempt to work this out, I have used to arbitrary figures to ensure i'm working it out correct.

I have also changed my weight from 30kg to 10kg as this is a more accurate estimate.

Weight of Rock = 10kg
Weight of bag = 1kg
Weight of CO2 = 108g
TOTAL = 11.108Kg

Volume of Rock = 5.77*10^-4/m3 - assume a cylindrical rock of 3.5mm diameter and 15m in length.
Volume of bag = 7.85*10^-3/m3 - 1m long and 10cm diameter
Volume of CO2 = Is this not just the volume of the bag?!

I'm sure this isn't correct so any advice would be good.

cheers

9. Aug 29, 2016

### billy_joule

The volume of the bag is just the bag material itself, not the volume it encloses. Eg a balloon may enclosed 10 litres of air but the balloon itself only occupies a few millilitres of volume.
In many cases it's so small we can ignore, in your case maybe not. Commercial lift bags look fairly sturdy so may occupy a volume that shouldn't be ignored.

Looking at commercial lift bags may help:
http://www.amronintl.com/commercial-diving-equipment/tools/underwater-lift-bags.html
https://en.wikipedia.org/wiki/Lifting_bag

10. Aug 31, 2016

### Baluncore

A quick estimate of the amount of gas and the bag volume needed to lift 10 kg from 30 metres goes like this.

The density of water is more or less fixed at one, independent of depth. A weight of 10kg weighed underwater, will at any depth require about 10 litres of lifting gas to start rising.

Hydrostatic pressure underwater rises with depth by 1 atmosphere per 10 metres. The depth to absolute pressure relationship is therefore; 1 atm at the surface, 2 atm at 10 m, 3 atm at 20 m and 4 atm at 30 m. That is a pressure ratio of 4.

The initial 10 litre volume of lifting gas will expand as it rises to become 40 litres at the surface where it reaches atmospheric pressure. But if the excess expanding gas can be spilled from the bottom of the bag as the bag and weight rise, the bag will only need a fixed capacity of about 10 litres throughout the lift.

The amount of gas needed to start the lift will be equivalent to 40 litres at the surface.

11. Aug 31, 2016

### Staff: Mentor

Why not just use a rope and a small winch? Seems like that would be a lot simpler given the small object and short distance. It's still good to understand how Archimedes principle applies to this problem, but in Engineering you should always be looking for the best/simplest way to get a task done reliably.

EDIT -- Or is this for a school project where you are required to use an inflatable bag to do the lifting?

12. Sep 3, 2016

### olivermsun

A winch attached to what is the question.

13. Sep 4, 2016

### Staff: Mentor

Well, if you can get an airbag under the object, you can get straps or a net underneath it as well...

14. Sep 4, 2016

### Baluncore

Like the winch, a lifting gas bag goes above the mass. Both systems need a strap, with or without a net.
A much longer cable is needed with a winch, but a winch can lift the mass clear of the surface, but you then need a boat.

15. Sep 4, 2016

### olivermsun

Well, true. But if you have a submerged instrument that you want to recover at some later time (or you just want insurance for such a thing), then an inflating recovery system could be useful.

16. Sep 4, 2016

### Ketch22

Just for consideration a common trick for divers recovering smaller objects ( like 10 Kg) is to use an open bottom bag. It is calculated for volume but open at a small neck. In this way as the bag ascends and the volume of gas increases due to reduced pressure the excess escapes and bubbles up outside the bag. Controlled ascent is achieved. This requires an accurate weight calculation though.