Pressure Testing w/ Nitrogen 1

[JPHYDRO]

We are pressure testing small sections of carbon steel pipe using nitrogen bottles to achieve test pressures. Using only a regulator from the individual bottle (or 6 pack) to test each segment of pipe. Can anyone help us with a formula to determine how many bottles of Nitrogen are needed to achieve test pressure in these sections?

Each bottle contains 230 cubic feet @ 2200 psi

Example,

280' for 2" carbon steel pipe i.d. 2.067" and a test pressure of 1584.



Thanks,

Justin

 

[BigInch]

That's high school science there buddy.


What would you be doing, if you knew that you could not fail? Ans. Gov lobbyist.

 

[zdas04]

PV still equals nRT even in Petroleum.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[BigInch]

P[sub]1[/sub]V[sub]1[/sub]=P[sub]2[/sub]V[sub]2[/sub] and ya don't even need the n, R, or T

What would you be doing, if you knew that you could not fail? Ans. Gov lobbyist.

 

[JPHYDRO]

Yes, this is what I did before I asked the question. It may be completely wrong, but I have done the the math, and it tells me we would less than two bottles of Nitrogen to achieve this test pressure. However this is not the case, the bottles will begin to equalize with the test segment and more bottles are needed.

My weak math tells me that we will need 319-357 total cubic feet of Nitrogen to achieve desired test pressure. And this volume of Nitrogen is approximately 1.5 bottles. So two bottles would provide a sufficient amount of Nitrogen. However, as the test segment pressure increases it will equalize with the N2 bottle and the remaining amount of nitrogen remains in the bottle. So then you would have to connect another bottle to increase the pressure. So you are only getting X amount of nitrogen per bottle.

Is there a simple formula to determine this? And a brief description would be very helpful.

Thanks,

 

[nfinit]

are you counting the volume of the bottles in your calculation?

 

[zdas04]

The simple answer is to convert your test pressure to a standard pressure (e.g., if the volume of the test is 6.5 ft^3 and the test pressure is 1584 psig and if standard pressure is 14.7 psia then the test will use 700 SCF).

Then look at how much gas there is in the bottles between delivery pressure and test pressure (I'm going to assume that your 230 ft is SCF, and the bottles are loaded to 3000 psig, but precision in communicating is a requirement for success in engineering). That would say that the physical volume of the bottles is 1.1 ft^3. At 1600 psia a 1.1 ft^3 bottle will hold 120 SCF, so at full test pressure you have 111 SCF usable. so you need something under 7 bottles for this 6.5 ft^3 test.

It isn't really that simple. The first bottle will raise the pressure to some (lower) value that will waste less of that bottle's gas. The second will waste more, etc. Only the last bottle will waste the whole 110 SCF. The total number of bottles will be less than 7 for this example. The only way to solve it is to iterate on final pressure after each bottle.



David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[JPHYDRO]

exactly, thank you sir!

 

[BigInch]

3 bottles.
No iterations required.
Final pressure 1791 psia
Physics K-12


What would you be doing, if you knew that you could not fail? Ans. Gov lobbyist.

 

[nfinit]

6 bottles. fun stuff.

 

[zdas04]

Ummmm. Since when is area D^2/4???? Think you missed a pi. Including the missing Pi back in gets you to a final pressure in your spreadsheet of 1036 psia.

Also, adding a volume at one pressure to a volume at another pressure results in meaningless numbers. Your worksheet really doesn't work.

The iterative approach I had in mind doesn't work either. I realized that if you work in SCF (or SCm or mass or moles, anything but volume at actual conditions) that it solves directly.

Step 1: The test has 6.5 SCF (more or less) at 14.7 psia and the bottle has 230 SCF at 2235 psia. Putting 236.5 SCF into 8 ft^3 results in a pressure of 434 psia. 434 psia in 6.5 ft^3 results in the first bottle raising the test volume to 192 SCF and leaving 31.5 SCF in the bottle.

Step 2: Adding the test volume to the second bottle results in the new system having 422 SCF. Putting 422 SCF into 8 ft^3 yields 775 psia of which 343 SCF is in the test and 79 SCF is left in the bottle.

Bottle 6: If you use the whole bottle you'll have 1609 psia on the test, so the answer is 6.94 bottles.



David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[zdas04]

Make that 5.94 bottles.

David

 

[BigInch]

OK I forgot pi. I was doing that when I wasn't supposed to be working on it.. OK?

I'm not adding a volume, its pressure x volume; ie P*V; effectively a quantity of gas; nRT is constant, so it cancels out anyway. P1 x V1 = nRT = P2 * V2
You can do it calculating moles, but why? Just an extra bunch of steps.

Just put pi() in there for area of pipe and that part is correct.
I did make one real mistake. Beginning each charge, the volume is only that of the pipe, not the pipe + tank. The finished volume when equalized is pipe + tank.
6 bottles get to 1587 psia.
You got a better way?

What would you be doing, if you knew that you could not fail? Ans. Gov lobbyist.

 

[monkeydog]

Why are you charging the "expended" tanks?
Why not isolate (valve-off) the expended tanks?

 

[BigInch]

you have to recharge the tank back to 2215 or the pressure will never get higher than what they previously equalized at.

What would you be doing, if you knew that you could not fail? Ans. Gov lobbyist.

 

[nfinit]

take out the minor oversight (which was easy enough to see and fix) and biginch's spreadsheet is the easiest way to accomplish your goal.