Volume of Nitrogen Under Pressure 3

[JDJ60]

I am a Electrical Engineer and had this problem handed to me this afternoon. I now am responsible for pressure testing several sizes of new pipe we are installing prior to being placed in service.

I figured out the calculations for the required test pressures of the pipe, based on wall thickness, but how do I calculate how many cubic feet of nitrogen or natural gas will be required to meet various test pressures and sizes of pipe. I have to have this information for the vapor release permit.

Any help will be greatly appreciated.

 

[mbeychok]

.
JDJ60:

Let us assume that you are going to pressure test a pipe that has an inside diameter of D inches and a length of L feet. Let us also assume that you are going to use nitrogen for the the pressure test and that the test will be conducted at a pressure of P[sub]T[/sub] psia and at a temperature of T[sub]T[/sub] degrees Rankine.

The volume (V[sub]T[/sub]) of nitrogen needed at the test conditions of P[sub]T[/sub] and T[sub]T[/sub] will be:

(L)(pi)(D / 12)[sup]2[/sup]/4 = V[sub]T[/sub] in ft[sup]3[/sup]

Since the various tests will be at various pressures and temperature, you will probably want to put all of the nitrogen quantities needed for the various tests at some common set of conditions which we can denote as a pressure of P[sub]C[/sub] psia and a temperature of T[sub]C[/sub] degrees Rankine. Thus, the nitrogen volume (V[sub]C[/sub]) at those common conditions which is equvalent to V[sub]T[/sub] will be:

V[sub]C[/sub] = V[sub]T[/sub](Z[sub]C[/sub] / Z[sub]T[/sub])(P[sub]T[/sub] / P[sub]C[/sub])(T[sub]C[/sub] / T[sub]T[/sub])

Alternatively, you could put each of your V[sub]T[/sub] values on a common basis by converting them to pounds of nitrogen:

Pounds of nitrogen = V[sub]T[/sub](MW / 10.73)(P[sub]T[/sub] / T[sub]T[/sub])

where:
L = length in feet
D = diameter in inches
pi = 3.142
P = pressure in psia
T = temperature in degrees Rankine = 460 + degrees Fahrenheit
V = volume in cubic feet
Z = nitrogen compressibility factor
MW = 14 = molecular weight of nitrogen
10.73 = Universal Gas Law constant

The compresibility factors (Z) can be neglected unless your test pressure are quite high.


Milton Beychok
(Contact me at www.air-dispersion.com)
.

 

[JDJ60]

mbeychok:

Thanks for the formula. I will use it tommorow when I get to the plant. Again Thank You.

 

[Montemayor]

JD:

You may be a concerned EE, but I wouldn’t worry about working out this problem correctly and safely if I were you. You’ve already shown how sharp you are by coming to this forum on this type of question. If I can’t resolve it quickly enough for you, there are more than a dozen, keen and sharper engineers on this Forum that can this in their sleep.

First, I’m surprised they want you to do a pneumatic pressure test. But, that’s OK, as long as the test pressures are relatively not high and you take all the safety precautions that I’m sure you will receive here and at your site. I’m going to assume you are going to be testing at around 300 psig max. I have to know the test pressure in order to find/calculate the correct “Z” (compressibility factor for the gas you select to test with. I presume you’re thinking of using Nitrogen. Here are the steps you need to plan on taking to arrive at the identification of how much N2 you’ll need for the test:

1) Obtain an accurate calculation of the internal pipe (& any fittings) volume that you’ll be testing. You can easily add up the internal volume if you have an Excel spreadsheet with all the pipe and fittings’ volumes already figured out. If you don’t just give us the list of what you have and perhaps we can help on giving you each individual item’s internal volume.
2) You’re going to pressurize the internal volume up to test pressure and you’ll need to employ the gas equation of state:

PV = Z n R T
where,
P = the test pressure, in psia
V = the internal volume pressurized, in ft3
Z = the compressibility factor for N2 at the test pressure and temperature
n = the number of lb mols of N2
R = the universal constant, 10.7316
T = the test temperature of the gas in the internal volume, oR

Note that absolute pressure and temperature are employed – not gauge values. By re-arranging the equation, we obtain:
n = PV / Z R T = lb mols of N2

If you multiply the lb mols of N2 by the molecular weight of N2 (28.016) you’ll get the lbs of N2 required to do the job. You can also multiply the lb mols of N2 by 379.49 and obtain the “Standard” cubic feet of N2 required as measured at 14.696 psia and 60 oF.

There you have it. You should add an additional amount of estimated N2 that will be the residual remaining in the N2 cylinder source, under pressure. Here, I’m assuming that you will employ HP N2 cylinders to feed the N2 into the test pipe & fittings. Since you haven’t stated how you will obtain and feed the N2, I have to assume.

You probably have some questions and comments. With the rest of the guys writing in by now, you should have an accurate and reliable estimate in a day or two. If you give us all your basic data and scope, the answer will come faster and more accurate.

Hope this helps. Help us help you by giving all the details.

 

[mbeychok]

JDJ60:

Just one further point. I don't understand what you mean by:

I have to have this information for the vapor release permit.

What "vapor release permit"? Required by whom? I don't quite understand why any governament regulatory agency would be at all concerned about venting off some completely inert and non-reactive nitrogen gas. If it is being asked for by some regulatory agency, I would certainly advise you to protest their requirement.

Milton Beychok
(Contact me at www.air-dispersion.com)
.

 

[sailoday28]

mbeychok (Chemical)-- its possible that a government agency is concerned with release to enclosed spaces where which can reduce the percent of breathable O2. For some reason, Natural gas is also proposed, which could also have other harmful effects after venting, such as fire, etc.

Perhaps JDJ60 (Electrical)should look more carefully into what the test requirements are.

Montemayor (Chemical)Glad to see you back.

 

[JDJ60]

Hello all, Thanks for all the help I received from all of you last evening. Today I calculated the required test pressures and fill volumes for the different pipe wall thickness, pipe grade, class location, and pipe sizes. I had to do quite a bit of reading in CFR49 Part 192 to get a good understanding of the test requirements for the pipe I will be testing. This section of our plant is covered under DOT regulations for natural gas transmission and not processing.

The reason I needed the fill volume was for release into the atmoshere after completion of the test if I used natural gas as the test media. However, I hope we don't have any governmental requirements for the release of an inert gas.

HEY GUYS DOES THIS EXCERSISE MAKE ME A CHEMICAL ENGINEER TOO? "JUST JOKING"

Again, thanks for all the help.

 

[zdas04]

JDJ60,
I've tested pipe with natural gas in some very carefully controlled conditions. I don't have Part 192 in front of me, but I recall that it defers to ASME B31.8 for gas tests. B31.8 allows using natural gas for a test medium up to a hoop stress of 30% of SMYS and only in limited area classes. You might look the "samples" section of my web page for a detailed write up on the methods and limitations of various test protocols. I've gotten really good feedback on the document.

Calculting volume of gas is your smallest problem. Good luck with this.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[mbeychok]

JDJ60:

I goofed in my earlier response in this thread. I gave the molecular weight of gaseous nitrogen gas as 14, which is incorrect. The correct molecular weight gas is 28.

Milton Beychok
(Contact me at www.air-dispersion.com)
.

 

[JDJ60]

Thanks. Milton