Quantity of Nitrogen Gas Required

Thanx for the reply.

Actually we have laid a cross country pipeline of 18 inch dia and 131 kilometers. Now as per contract we need to fill this pipeline with nitogen gas and maintain the pipeline at 2 barg pressure. The geometric volume of the pipeline is 21400 cubic meter.

The line is empty. How much nitrogen is required to fill the line and subsequently to raise pressure upto 2 barg.

Regards

Does the air inside it have to come out first? If so, then a sweep of nitrogen gas will need to be introduced at one end and the outgoing gas at the other end tested until the levels of everything-other-than-nitrogen has fallen below specified criteria. With such a large diameter of line, flow will be turbulent, and that will mix the incoming nitrogen with the air already in there. In this case, the volume needed will be rather more than could be established by simple calculation of the volume and the pressure.

The line has been hydraulically tested to more than 2barg first, hasn't it?

The line is empty.

I bet it's not!

(1) I assume that pipe contains air which is not dry. You want to introduce nitrogen(dry) for preservation. To drive out air you have to first blow the pipe with nitrogen which may require approx. one fill of the pipe at atomosphric pressure(1.013 bar). Quantity for one fill is 1.25x21400x273/(ambient temperature) in Kgs.

(2) To raise pressure up to 2.0 barg you need 1.25x21400x3.013/1.013x273/(amb.temp) Kilograms of nitrogen. Hope it is clear.

(3) If you need just to raise the pressure from 0 barg(atm. pressure) to 2 barg by filling nitrogen, consider above (2) ONLY.

If your pipe doesn't leak, theoretically you'll need to calculate the inside volume of your pipeline. That is your Normal cubic feet or cubic meters needed gas @1 bar. 2 bar is double the volume.

Depending on how you will buy the Nitrogen, you'll need to convert the volume times delivered pressure to see what one unit can supply, (minus the rest value = 2 bar 0r 29psi) Your normal cubic meters also times 2. Probably in the pipe line will be air now, or something else that needs to be removed. If the nitrogen will be used to do this, count also loss of Nitrogen if this process doesn't evolve perfectly: meaning you can lose some Nitrogen too, depending on how you drain. Vacuum pumping helps for this, and bottom rains at the lowest points. But this doesn't happen too much in reality.

The post came under reply to #1 but was mentioned as extension on this post. The first line is there out of the context because of the position in the thread list. The extra 1 volume is on top of the 2 bar needed.

Maybe it might help here as clarification.

--------------------------------------------------- To fill the pipe from vacuum you will need indeed 1 time more the volume to read a gauge @2 bar.

(on top of the 2 times the volume)

For test purposes and since air contains about 78% of nitrogen, just filling up the difference, unless specified differently.

If it is for special purposes and absolutely no oxygen can remain - if they start from a new pipe - (I don't know what medium is in there now) - after vacuum they should need 3 times the volumes as you propose.

But this is almost never done.

Theoretically 22% of the volume for the first clean up of the air inside ( the nitrogen pushes the oxygen out with this volume) and afterwards another 2 times the volume to bring it to 2 bar on the gauge. I didn't bring that up because I feel the poster just needs to know how much normal m3 he gets out of a package.

And the size of the pipe didn't make me expect it to be a special purpose pipeline. Nitrogen in there is usually applied to bring down the LEL (lower exposion level) for safe testing, which is attained with that volume of nitrogen. I should have given more explanation. Thanks.D A GA to John me

Perfect gas law

PV=nRT

P = pressure (atm)

V = volume (m3)

n = mole quantity = m / M --> m = mass of gas (g) and M = atomic mass (18g/mole for nitrogene)

R = perfect gas constant (8.206 x 10^-5 m3 atm mol-1 K-1)

T = temperature (K) --> temperature in Kelvin = temperature in °C + 273

Choose a temperature and do the math!

http://www.engineeringtoolbox.com/compressed-air-storage-volume-d_843.html

Hope you got the answer