How many SCF are stored in a nitrogen pipe 12500 feet long at 6000 psi. Pipe ID is 3.469 inches.?

Question

Answer should be total SCF.

Answer 1

The first thing you should do is calculate the actual volume of the pipe.

The cross-sectional area of the pipe:

A=pi*r^2=pi(3.469in/2)^2
A=9.45in^2

Convert the square inches to square ft

A=9.45in^s*1ft^2/144in^2
A=0.06564ft^2

Calculate the Volume of the Pipe:

V=A*L
V=0.06564ft^2*12500ft
V=820.4ft^3

Now since the nitrogen is being stored at 6000 psi it is not at standard conditions. The nitrogen at standard conditions will have to be calculated.

First the critical conditions of the nitrogen must be checked.

I will have to assume that the pipe is at a standard temperature.

Since the pressure is so high nitrogen is not ideal. The Z factor is used in the following equation.

P*V=z*n*R*T

Pressure is:
P=6014.7psia assuming the pressure given is gauge
V=820.4ft^2 volume of the pipe
z=1.25 Obtained from David M Hemmelblau graph
R=10.73(ft^3*psia)/(lbmol*R) Gas const.
T=527.67R Std temp
n=(P*V)/(z*R*T)
Units omitted to conserve space in calc
n=(6014.7*820.4)/(1.25*10.73*527.67)
n=696.7lbmol

Now at standard conditions the ideal gas law applies

R=10.73 Gas const
T=527.67 Std temp in R
P=14.7 Std press in psia
n=696.7 moles lb basis
Calculate Volume in SCF
V=n*R*T/P
V=696.7*10.73*527.67/14.7
V=268,343.2SCF
Or
V=268.3 MSCF