# Fire protection -- Water in a cylinder pressurized with Nitrogen

chrismof
I wondered if anyone could help, I work in the fire protection industry. We currently have a project using a pressurised cylinder. The cylinder hold 200 lites of water plus 40 litres of nitrogen gas as a propellant at 10.0 bar. I’m trying to work out once the 200 litre water volume has been expelled what would be the remains pressure. I believe the formula to use is Boyles law but I’m struggling to understand. Would appreciate it if anyone could help.

Staff Emeritus
When it's empty doesn't that mean it's at atmospheric pressure? So 1 bar?

Mentor
The answer will depend on the amount of gas that has left the tank, which in turn depends on two things you haven't specified: 1) Is the design such that gas sprays out with water? 2) Is the valve left open after the last of the water has been forced out of the tank?

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The volume of the reservoir is 200 litres of water plus 40 litres of gas = 240 litre.
40 litres at 10 bar will expand to 240 litres at ? bar, to expel the 200 litres of water.

P1·V1 = P2·V2; holds for absolute pressures.

If the 10 bar is not absolute, but is gauge pressure, then the initial pressure will be 11 bar(abs).

P2 = P1 * V1 / V2 = 11 * 40 / 240 = 1.833 bar(abs) = 0.833 bar(gauge).

chrismof
Thanks to all your prompt replies. The nitrogen gas is not mixed with the water but remains at the top of the cylinder until activated. Further to my question the release of the water would be at 14.16 litres / minute for approximately 13 minutes. What I am trying to check is at what point the nitrogen gas pressure would drop below 7.0 bar.

What I am trying to check is at what point the nitrogen gas pressure would drop below 7.0 bar.
Can you confirm that the pressure you are specifying is the gauge pressure, above atmospheric pressure. Do you understand why the absolute pressure is one bar higher than the gauge pressure?

P1 = 10 bar(gauge) = 11 bar(abs);
V1 = 40 litre;

P2 = 7.0 bar(gauge) = 8.0 bar(abs);
V2 = V1 * P1 / P2;
V2 = 40 * 11 / 8 = 55 litres;

The change in gas volume is V2 - V1 = 55 - 40 = 15 litre of water.

For a constant flow rate of 14.16 litres / minute, the pressure would fall to 7.0 bar(gauge) in;
15 / 14.16 = 1.06 minutes = 63.6 seconds.

chrismof
Hi,
Thank you for your reply the pressure would be the gauge pressure.

Staff Emeritus