Nitrogen Leak Test on a Pipe

[meshbah]

Hello everyone,

I’m looking to calculate the number of single nitrogen cylinders required to carry out a leak detection test on pipe(volume of 14m3)

The leak test pressure is 1485 psig

Nitrogen cylinder -> 6m3 / 2175.5psi

I’m struggling with the fact the bottles will begin to equalize with the test segment meaning it’s not as simple as I first thought. I believe I need to calculate how much gas is in a cylinder between the delivery and test pressure (useable volume)

Any help would be appreciated.

Many Thanks

 

[LittleInch]

You need to look at this either as a series of bottle use or linking all together.

Use some or mass to figure it out.

That's a very big cylinder though. Are you sure?

So if all you have is the cylinder pressure then you will not be able to use all the nitrogen in the cylinder. To use it all you will need a compressor.

Just go in steps of 100 psi to calculate volume flow from one to the other using some or similar.

Just assume scm is vol x pressure in bar and you'll be close enough.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[meshbah]

Hello,

Thank you so much LittleInch for your response, it’s been really helpful in understanding the process better.

I do have a few follow-up questions, if you don’t mind:

In the case of using nitrogen cylinders in parallel, would it require fewer cylinders compared to using them one at a time, or would the total number remain roughly the same?

Regarding the nitrogen cylinders, in the market, they generally come with a capacity of around 6 cubic meters and a pressure of 150-200 bar. I’m assuming this is the standard; could you confirm if this would affect the calculations?

Lastly, could you kindly provide an example calculation of how to approach this? I believe a step-by-step example would help me understand the process much better.

I truly appreciate your assistance and thank you once again for your time!

 

[1503-44]

I worked out what 7 fill-pressure balance cycles would look like.
After 7 charges, youre at 138.5Bara and you only pick up 5 bars, so you need a compressor to go higher. You can decide what charge cycle to kick it in.

Rather than at small pressure increments, I did the problem charging with one whole Cylinder each time.

Cylinder Content
P1= 2175psig = 2190psia = 151bara
V1 = 6m3

Cylinder
PV= nRT
nRT = 151*6 = 906

Pipe+Cylinder
V2 = 20m3
P=nRT/V
P=906/20=45.3

FILL #1
Pipe
P=45.3
V_pipe = 14m3
PV_pipe = nRT_pipe = 45.3 * 14 = 634.2

Cylinder#1
P=45.3
V_cyl =6
PV_cyl = 45.3 *6 = 271.8

FILL #2
nRT_pipe (634.2 is from above) = 634.2+906 = 1540
V=20
P= 1540/20=77 bar

Pipe
P=77
V_pipe = 14m3
PV_pipe = nRT_pipe = 77 * 14 = 1078

Cylinder#2
P=77
V_cyl =6
PV_cyl = 77 *6 = 462

FILL #3
nRT_pipe (1078 is from above) 1078+906 = 1984
V=20
P= 1984/20= 99.2 bar

Pipe
P=99.2
V_pipe = 14m3
PV_pipe = nRT_pipe = 99.2 * 14 = 1389

Cylinder#3
P=99.2
V_cyl =6
PV_cyl = 99.2 *6 = 595.2

FILL #4
nRT= 1389+906 = 2295
V=20
P= 1984/20= 114.75 bar

Pipe
P=114.75
V_pipe = 14m3
PV_pipe = nRT_pipe = 114.75 * 14 = 1606

Cylinder#4
P=114.75
V_cyl =6
PV_cyl = 114.75 *6 = 688.5

FILL #5
nRT= 1606+906 = 2512
V=20
P= 2512/20= 125.6 bar

Pipe
P=125.6
V_pipe = 14m3
PV_pipe = nRT_pipe = 125.7 * 14 = 1758

Cylinder#5
P=125.6
V_cyl =6
PV_cyl = 125.6 *6 = 753

FILL #6
nRT= 1758+906 = 2664
V=20
P= 2664/20= 133.2 bara

Pipe
Pp=133.2 bara
V_pipe = 14m3
PV_pipe = nRT_pipe = 133.2 * 14 = 1865

Cylinder#6
P=133.2
V_cyl =6
PV_cyl = 133.2 *6 = 799

FILL #7
nRT= 1865+906 = 2771
V=20
P= 2771/20= 138.5 bara

Pipe
Pp=138.5 bara
V_pipe = 14m3
PV_pipe = nRT_pipe = 138.5 * 14 = 1939

Cylinder#7
P=138.5
V_cyl =6
PV_cyl = 138.5 *6 = 831

If you want to try multiple cylinders at a time, just change the 6m3 to the total volume of the number of cylinders you use during each charge. You could try 3 cylinders for the first charge, then change to 2, or 4 during the next charge. Be careful with the volumes and the nRT value added between charges.

Please check my calculations. I used the ideal gas PV=nRT equation of state at constant temperature.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[shvet]

Nitrogen cylinder -> 6m3 / 2175.5psi

2175psi - is it max pressure or min, or normal, or filling pressure?

 

[meshbah]

2175,5 psi is the nitrogen pressure stored in the cylinder

 

[LittleInch]

I don't know without doing the numbers, but by instinct using one cylinder after another would be better.

By 6m3 I thought you meant an actual volume of 6m3. Nitrogen cylinders typically are 49 litres at about 150 bar (2100 psi) If you can get 200 bar, that's a big difference so make sure you know what you're getting / available.

So the STANDARD volume of gas for a 49litre cylinder at 150 bar is approx 6 scm.

I think mr 44 has used 6m3 as the volume of the nitrogen cylinder?? So in the first calculation it's actually 151 x 0.049, not 6...

As a very basic number, 14m3 at 102 bar is approx 1450 scm. So if you were able to use all the nitrogen in the bottle by e.g. a compressor then you're looking at about 240 + bottles

As a real guess, you're probably looking at multiple that by 1.5 due to the increasing amount of gas left in the bottles, so > 350.

but if you really want to calculate this in per bottle stage go in per bottle stages and just start adding up how mnay scm it can supply using the start pressure and end pressure and the volume of the pipe (14m3).

So the start bottle will raise the pressure by 6/14 bar or up to 0.42 bar. The next bottle then starts at 0.42 bar which will reduce the volume available to the pipeline by 0.42/150. SO yes a small number, that 0.42 will gradually rise until the last bottle is 102/150 or 0.68 so that last bottle is only adding 0.32 x 6scm or 1.92m3.

So devise a spread sheet which adds up the scm being supplied by each bottle and als the gradually decreasing amount of gas actually being supplied in scm (1- (Pressure / 150)) x 6. Pressure in the pipeline will slowly rise by that volume/14 per bottle in bar.

Does that work for you?

But several hundred bottles looks unfeasible to me. You might need a nitrogen tanker, vapouriser and compressor to do this.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

Higher N2 pressure cylinders are common.
UN/ISO Nitrogen Storage Tank: 424 cu ft, 4500 psi
No compressor required.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[pierreick]

Mr 44,
To me you need only 3 cylinders and 1/3 of the 4th one.
Set point is 1490 PSI a , just above 103 bar A
Pierre

 

[LittleInch]

Mr 44,
To me you need only 3 cylinders and 1/3 of the 4th one.
Set point is 1490 PSI a , just above 103 bar A
Pierre

Care to explain how you got to that conclusion?

We seem to very different by a factor of 100 if you're talking about bottles.

 

[pierreick]

cylinder means 6m3 N2 ,
My calculation is similar to Mr 44 except the end point which is 1490 PSI A .
Pierre

 

[LittleInch]

But it's 6 standard m3. Not 6 m3 at 2190 psi.

Nobody supplies high pressure nitrogen in anything bigger than a 47 litre cylinder.

They supply Liquid Nitrogen in bigger volumes,, but then you need to gassify it and compress it yourself

 

[pierreick]

Irrelevant!

Nitrogen (Industrial Grade) Cyl 7 M3 | TH Industrial National eStore

Industrial grade nitrogen has wide applications in manufacturing and fabrication industry segments. It is used for blanketing, purging, inerting and leak testing.

shop.linde.co.th

By the way your response does not make much sense.
Regards
Pierre

 

[LittleInch]

It's very relevant because, in my opinion, your using the wrong data.

The size of the nitrogen bottle at 2100psi is 47 litres. When that gas is at 15 psia, it is about 6 cubic metres. At atmospheric pressure.

You and Mr44 are working on the basis that you have 6m3 at 2100 psig. Hence why we're a long way apart in our calculations as I think that figure should be 47 litres or 0.47 m3.

 

[pierreick]

The leak test pressure is carried out at 1485 psig" or 103 bars A , the volume of the pipe is 14 m3 ...........
the quantity of N2 in the pipe at the end of the test should be ( n = P V/RT) = 14*103e5/( 8.314*303 ) =57241 moles or 1602 kg of N2 and you expect 47 liters of N2 sufficient enough to perform the test.
NB :47 l =0.047 m3 (typo)
agree : 47 liters cylinder @ 144 bars is equivalent to # 7 m3 at Patm .
Pierre

 

[LittleInch]

Pierre.

Yes that was a typo, but thr edit function now disappears a bit fast.

I don't think you've properly read my previous posts where I think it will take up to 350 bottles.

Do you agree?

 

[pierreick]

Yes I do ! 343 bottles based on 47 liters and 151 Bar A .

Pierre

 

[LittleInch]

Good. Now we wait for the OP to return....

 

[pierreick]

BTW ,1/3 of N2 is lost, kept in the bottles after transfer.
Pierre