This example will demonstrate how to determine the compressibility factor (z) for gas with CO2 and H2S. As it is described in the article,Determine Compressibility of Gases , it states that gas with CO2 and H2S must be corrected. This is similar to a normal method for determining z-factor but it is required some correction. Please follow the steps below;
Gas component is shown in Table 1
Table 1 – Gas Component
Reservoir pressure = 6,000 psia
Reservoir temperature = 190 F Continue reading
This example will demonstrate how to calculate the compressibility of real gas in order to determine gas density and specific gravity at a specific condition.
Calculate the following based on the given condition:
1) Density of this gas under the reservoir conditions of 7,500psia and 220ºF,
2) Specific gravity of the gas.
Gas component is shown in Table 1
Table 1 – Gas Component
Average density of air = 28.96 lb/cu-ft
Table 2 – Critical Pressure and Temperature
Note: critical pressure and temperature can be found from this link – https://www.drillingformulas.com/determine-compressibility-of-gases/
Pc’ = Σyipci = 660.5 psia
Tc’ = ΣyiTci = -46.2 F = -46.2 +460 F = 413.8 R
Table 3 – Pc’ and Tr’ by Kay’s Rule
Tr = T ÷Tc
Tr = (220+460) ÷ (-46.2+460)
Tr = 1.64
Note: temperature must be in Rankin.
Rankin = Fahrenheit + 460
For the critical temperature calculation, it can be converted the critical temperature from F to R before calculating Tc’. This will still give the same result.
Pr = P ÷ Pc
Pr = 7500 ÷ 660.5 = 11.4
z = 1.22
Figure 1-z-factor from the Standing and Katz Chart
Average Molar Mass = Σyi×Mi = 22.1 lb
Table 4 – Average Molar Mass of Gas
Gas Density = 18.6 lb/cu-ft
SG = Gas Density ÷ Air Density
SG = 18.6 ÷ 28.96
SG = 0.64
Summary:
The answers for this answer are listed below;
Gas Density = 18.6 lb/cu-ft
SG = 0.64
We wish that this example will help you understand to determine z-factor and use it to calculate any related information.
References
This article will demonstrate how to determine gas compressibility by using simplified equation of state.
At a low pressure condition, gas similarly behaves like ideal gas and the equation for the ideal gas law is listed below;
Where;
P = pressure
V = gas volume
T = absolute temperature
n = number of moles of gas
R = gas constant
Gas constant (R) is different because it depends on the unit system used in the calculation. The R in the different unit system is demonstrated in the below table.
Table 1 – Gas Constant (R) at Different Unit
Compressibility is a relative volume change of a fluid or solid in a response to a pressure change. We can relate this into a reservoir engineering aspect. Overburden pressure is rock weight and it typically has a gradient of 1 psi/ft. Rock metric and formation fluid in pore spaces supports the weight of rock above. When petroleum is produced from reservoir rocks, pressure of fluid in pore space decreases, but overburden is still the same. This will result in the reduction of bulk volume of rock and pore spaces. The reduction on volume in relation to pressure is called “pore volume compressibility (cf)” or “formation compressibility” and it can be mathematically expressed like this.
