TPG 4140
CALCULATIONS IN NATURAL GAS COURSE
Prof. J.S. Gudmundsson
September 2012
Molecular Weight
M can be found from the gas composition, using an Excel sheet. An example is shown
below (gas composition from GPA 1998).
Components
Molecular
weight
Mole
fraction
g/mole
yi
Methan, CH4
16,042
0,8319
Ethan, C2H6
30,07
0,0848
Propan, C3H8
44,10
0,0437
i-Butane, C4H10
58,12
0,0076
n-Butane, C4H10
58,12
0,0168
i-Pentane C5H12
72,15
0,0057
n-Pentane C5H12
72,15
0,0032
Hexane C6H14
86,18
0,0063
Heptane C7H16
100,21
0
Hydogen, H2
2,02
0
Nitrogen, N2
28,01
0
Oxygen, O2
32,00
0
Carbon dioxid, CO2
44,01
0
Hydrogensulfid, H2S
34,08
0
Dihydrogenoksid, H2O
18,02
0
Σ Mole fraction
1,0000
Total molecular weight gas
20,43
g/mole
M = 20.43 g/mol = 20.43 kg/kmol
Specific Gravity
Definition of specific gravity, ratio of the molecular weight of natural gas to that of
common air (without water vapour). M taken from above.
 
M 
   gas    gas 
  air  s.c.  M air 
Mgas =  28.97 [kg/kmol]

The subscript gas is not necessarily used.
The specific gravity was also calculated in the Excel sheet, rounded up to γ = 0.71
Reduced Pressure and Temperature
Reduced pressure and temperature are also called pseudocritical pressure and
temperature, expressed as
pr  p pr 
Tr  Tpr 
p
pc
T
Tc
where the subscript c stands for critical.
Key’s Rule is used to obtain the critical pressure and temperature for natural gas
mixtures
pc   pci yi
i
Tc   Tci yi
i
The critical pressure and temperature for individual natural gas components are
available in handbooks and tables. The symbol yi stands for mole fraction.
For the natural gas composition taken from GPA (1998) critical pressure and
temperature were estimated to be 4.57 MPa and 218 K.
Based on the Excel sheet, the corresponding numbers are 4.6 MPa and 218 K.
Critical pressure and temperature can also be obtained from semi-empirical equations
based on specific gravity, for example from Rojey & Jaffret (1997)
p pc  4.892  0.405 [MPa]
Tpc  94.72  170.75 [K]
Such equations tend to be based on natural gas found in certain oil and gas provinces.
There is an error in the critical temperature equation in Rojey & Jaffret (1997). There should be a plus
in the equation, as shown above.
Using a specific gravity of 0.705 the critical pressure and temperature are estimated
4.606 MPa and 215 K. These are reasonably close to the above values.
Compressibility Factor
Assuming a pressure of 80 bara and temperature of 30 C the z-factor can be estimated
from the figure below (commonly called the Standing-Katz diagram)
The reduced pressure is
80  105
 1.75
4.57  106
pr 
and the reduced temperature
Tr 
30  273.15
 1.39
218
Read from diagram z = 0.79
The reduced pressure and temperature from the Excel sheet are 1.75 and 1.39. That is,
exactly the same.
The Hall-Yarborough method in the Excel spreadsheet is considered one of the more
accurate semi-empirical equations for z-factor.
Read from spreadsheet z = 0.7831
Gas Density
Need molecular weight, z-factor and the local temperature and pressure.

pM
zRT
Using the values obtained above, the density of the gas at 80 bara and 30 C can be
calculated as

80  105  20.43
 82.8 kg/m3
0.7831  8314  (30  273.15)
The same result is obtained from the Excel spreadsheet, which also gives the viscosity
as 0.0139 mPa.s
Friction Factor and Reynolds Number
The friction factor in commercial pipes can be calculated from the Haaland equation
 6.9  n  k 1.11n 
1
1.8

log 
 
 
n
f
 Re   3.75d  
where n = 3 for natural gas pipelines ( n = 1 for liquid flow). The Reynolds number is
give by
Re 
ud

and the relative roughness by k/d.
Pressure Drop Horizontal Pipeline
In general, equations for pressure drop in pipelines can be used to calculate the
pressure drop, the diameter and the flow capacity, depending of which of the variables
are given and what result is required. Approximate calculations can be carried out for
gas pipelines using the Darcy-Weisbach equation
p f 
f L 2
u
2 d
where average gas properties are used.
For illustration purposes, assume 7 MSm3/d flowrate (10 % of Ormen Lange).
Assume inlet pressure 200 bara and outlet pressure 100 bara. Assume 500 km
distance. Therefore, pressure gradient (200-100)/500 km = 20 bar/100 km. Assume
constant temperature of 10 C.
The simplest way to find the z-factor is to use the available Excel spreadsheet. At 200
bara and 10 C the z-factor is 0,6806. At 100 bara and 10 C the z-factor is 0.6638. The
average z-factor is therefore 0.6722. From spreadsheet, average density 194 kg/m3
and average viscosity 0.0218 mPa.s.
The velocity depends on the pipeline diameter and mass flowrate. To know the mass
flowrate we need the density of the gas at standard conditions, use Excel spreadsheet
and assume z = 1. We note that 1 atm = 1.01325 bara. Density at s.c. 0.864 kg/m3.
Therefore the mass flowrate is
m
7  106
0.864  70 kg/s
24  3600
Average velocity is given by
u
such that
m
A
d5 
8 fLm2
 2 p
Assuming an average friction factor of 0.015 the diameter can be estimated
d5 
8  0.015  500000  70
 0.219 10 3
2
5
3.1416 194 100 10
giving a diameter of 0.19 m and an average flow velocity of 2.4 m/s (from
spreadsheet). The corresponding Reynolds number is about 40 million and a friction
factor of 0.0544.
Calculating again
d5 
8  0.0544  500000  70
 0.794  10 3
2
5
3.1416  194  100  10
gives a diameter of 0.24 m and an average flow velocity of 1.9 m/s. The Reynolds
number is about 40 million so the friction factor will be the same.
Therefore, a couple of iterations give a pipeline diameter of about 240 mm (about 91/2 inch).
A similar iteration procedure can be followed using the pressure drop in gas pipelines
equation, but it is more involved and will not be illustrated her.
d A2 M
d  p 22 
2
2
p 2  p1  ln  2   L  0
f  p1 
f m 2 z RT

