Pipeline Construction

21400m³ @ 2 barg.

Actually the N2 quantity should be 3 times the geometric volume (2 barg = 3 bara).

In the real world you will need a lot more in order to purge the air out of the system first!

ST

That's the case if you're filling it from vacuum, but if it's initially at atmospheric pressure (1 bara) only 2 x volume needed.

This subject was also posted on Mechanical Engineering. Looks like the OP thought he'd have a second bite at the cherry!

Cheers.........Codey

Assuming the pipe starts off full of air at atmospheric pressure, about 43000 Sm³. Just over 50 tonne, about 63 m³ liquid, or 215 m³ in cylinders at 200 bar.

But why use nitrogen? Are you going to purge the pipe as well? That will use more. You could test with air supplied by site compressors. If you know what compressors you can make available you can work out time needed.

Cheers...........Codey

Under your contract the pipe line is to be pressure tested with nitrogen. I would first test it with air. The cost is less to get an idea of any leaks and correct them. With those resolved then do final test with the nitrogen.

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 pressure test and that the test will be conducted at a pressure of P_T psia and at a temperature of T_T degrees Rankine.

The volume (V_T) of nitrogen needed at the test conditions of P_T and T_T will be:

V_T (in ft³⁾ = (L)(pi)(D / 12)²/4

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_C psia and a temperature of T_C degrees Rankine. Thus, the nitrogen volume (V_C) at those common conditions which is equivalent to V_T will be:

V_C = V_T(Z_C / Z_T)(P_T / P_C)(T_C / T_T)

Alternatively, you could put each of your V_T values on a common basis by converting them to pounds of nitrogen:

Pounds of nitrogen = V_T(MW / 10.73)(P_T / T_T)

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

ALL THE BEST

Hello RAHUL

That seems to be making a complicated issue out of something fairly simple. Why use imperial units? OP used a bit of a mixture, but metric for everything that matters!

Your formula gives the right answer for lbs nitrogen, if the correct molecular weight is used. The molecular weight is 28, not 14.

Cheers.........Codey

Request a change order to the contract to allow first filling the pipe with water. Then the required volume of N2 will be minimal.

Contract specifications and end use should be considered re: how pure is the nitrogen filling the line supposed to be? Pushing a pig through with the nitrogen charge would help exclude oxygen and water, as might be required if the line is to carry hydrogen or a water reactive material. As noted, leak testing with air before filling with nitrogen is good practice.