# Formula for relation between pressure and volume for GAS

1. Oct 7, 2012

### Wiecker

Hi all,

I am brand new here and need some help - i hope it is the right category i am posting in.

I am no expert in physics - i am working at sea as a mate on board producttankers. I would like some advice for calculating or formulas for the following scenario:

- A closed tank 10m3 is half full with fluids(5m3)
- On top of the fluids there is pressurized inertgas (CO2); the pressure is 500mmWG absolute
- In the bottom of the tank there is a draining valve.

How many fluids can i drain before the pressure drops to Zero?

Brgds, Stefan

2. Oct 12, 2012

Sounds like an exam question... The best way to calculate this is using atm (atmospheres) as a pressure unit. 1mWG is 0,09678 atm of overpressure with regard to free air. So the actual pressure of 0.5mWG=500mmWG is 1atm+0.048atm=1.048 atm. You can regard this as the compression factor. So if you expand the gas by 1.048 the pressure will be the same as on the outside. You then have drained $$1.048\cdot 5m^3 - 5m^3=0.048 \cdot 5m^3 = 0.224m^3$$ and the pressure is 0mmWG But then there is pressure on the bottom from the weight of the fluid sou you could drain a bit more and produce negative pressures in the $CO_2$. If you have water in the container, then the draining should stop when the hight of the water is the same as the negative mmWG in the $CO_2$ (I think the extra volume you can squeeze out depends on the shape of the container, but I am not sure and too lazy to calculate this now).

3. Oct 13, 2012

### Wiecker

So your solution is telling me that 0,048 i a factor or constant, which regards to a drop of 500mmWG pressure?

The question is not an examquestion - it is with regards to optimizing a fuelprocess on board a tanker.

A real case scenario would be a tanker of a total capacity of lets say 50.000 m3 with 30.000 cargo discharging via an enclosed system to a shoretank. Lets say i use your formula with 0,048 as a factor:

50.000 - 30.000 = 20.000 m3 of inertgas at a pressure of 500mmWG. Then i can discharge 20.000 * 0,048 = 960 m3? And after i have discharged 10.000m3 of cargo my inertgascapacity will be 30.000. Lets say i have pressurized the tanks again to 500mmWG - now i can discharge 30.000 * 0,048 = 1440 m3?

Is that correct understood?

4. Oct 13, 2012

### haruspex

OK, so you don't mean zero pressure, you just mean the point at which the fluid drain ceases. And the aim is to empty the tank, just.
Simplest is to run it backwards. At the end, you want a tankfull of gas at 1 atm. So when the tank is half full of fluid it must be at 2 atm. But there's a catch. As the fluid drains and the gas expands it will cool. This will lower the pressure and inhibit draining. If you have plenty of time there's no problem. Heat will slowly flow in to bring the gas back to ambient and fluid flow continue. In practice, time matters, so you will need some extra initial pressure.
If you want no delay, you can use the adiabatic gas expansion formula to find how the pressure will change.

5. Oct 14, 2012

Correct.

This is basically what I said, but I do not know how these things work. So what I can say is that after draining this amount of liquid the pressure in the gas will be the same on the inside as in the air on the outside (0mmWG). And only if you measure overpressure with respect to the outside. If you want to know if this stops the flow of liquid there are more things to consider. Maybe you know from experience if 0mmWG pressure is enough to stop the flow or if the pressure needs to be negative to hold the liquid in or if there is counter pressure in the tank you discharge into.

by pressurizing a number of times I suppose
Yes, exactly.

6. Oct 14, 2012

### haruspex

This analysis is still ignoring the adiabatic cooling. This will significantly reduce the flow. see my earlier post.