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CHAPTER 5 5-3 -3 Properties of Dry Gas
1
At the end of the session, students should be
´
´
Describe the properties of Dry gas commonly used by the petroleum engineer. Estimate the values of the properties using normally available information about the dry gas. 2
1
´
Dry gases are the easiest to deal with as it moves from the reservoir to the surface.
´
The gas compositions at the surface and therefore, the specific gravity would also be the same for both. 3
´
Dry gases are the easiest to deal with as it moves from the reservoir to the surface.
´
The gas compositions at the surface and therefore, the specific gravity would also be the same for both. 4
2
´
The gas formation volume factor (Bg) is defined as the volume of the gas at the reservoir conditions required to conditions.
´
´
´ ´
The reciprocal of the Bg sometimes is called gas expansion factor. The usual units for the B are: Reservoir cubic feet per standard cubic feet, res. cu ft/ scf. Reservoir barrels per standard cubic feet, res. bbl/scf.
5
What are the standard conditions???
Volu me of th e gas at the reservoir P&T Volu me of the same mass of gas at the s ur ac e
B g
V R V SC
6
3
B g =
V R V SC
The volume of n moles of gas at reservoir condition
V R =
z R nRT R
V SC =
P R
z sc nRT sc Psc
The volume of n moles of gas at standard condition Thus, the formation volume factor is
z R n B
g
=
z sc
P R nRT
B g = 0 . 0282
R
sc
=
P sc
B g = 0 . 00502
Psc = 14.7 psia, Tsc= 520 R, z = 1
T cu . t P
scf
zT res .bbl P
scf 7
g B
Pressure 8
4
´
Calculate the formation volume factor o a dr as with a s eci ic ravit o 0.818 at reservoir temperature of 220F and pressure of 2100 psig.
First, estimate the P pc and T pc econ . a cu a e
e
pr
an
pr
an ge
e z ac or.
Third, Calculate the Bg using the following equation: B g = 0 . 00502
zT res .bbl P
scf
9
T =220F P = 2100 psig First, from this figure and at 0.818 gravity Ppc = 647psia . pc =
Second calculate the Tpr and Ppr
TPR =
T TPC
T PR =
PPR = Resource: McCain textbook pg. 119
PPR =
(220 + 460) R 406 R
P PPC
= 1.68
(2100+14.7) psia 647 psia
= 3.27 10
5
Third, Find z value for the dry gas using this chart Ppr = 3.27 and Tpr = 1.68
z = 0.855
Resource: McCain textbook pg. 112
11
´
Calculate the formation volume factor of a dry gas with a specific gravity of 0.818 at reservoir temperature of 220F an pressure o ps . Once Z factor is estimated, then you can use the following equation for calculating the gas formation volume factor
B g = 0 . 00502
B g = 0 . 00502
zT res .bbl P
( 0 . 855 )( 220 + 460 ) 2100 + 14 . 7
scf
= 0 . 00138
res .bbl scf 12
6
´
The coefficient of isothermal compressibility (Cg) is defined as the fractional change of volume as pressure is changed at a constant temperature.
´
The coefficient normally is referred to as gas compressibility (Cg)
´
Compressibility factor ,z, is not the .
´
g
C , y t i l i b i s s e r p m o c s G
The unit for Cg is psi-1
cg = −
´
´
Reservoir pressure
1 ⎛ ∂V ⎞ 1 ⎛ ∂ Vm ⎞ or = − ⎜ ⎟ ⎜ ⎟ V ⎝ ∂P ⎠ Vm ⎝ ∂ P ⎠
13
Viscosity is a measure of the resistance to flow exerted by a fluid This is called dynamic viscosity and has units of centipoise = g mass / 100 sec cm
´
Kinematic viscosity is viscosity / density, units are in centistokes = centipoise /g/cc = cm2 /100 sec
14
7
´
´
Gas viscosity decreases as reservoir pressure decreases. The molecules are a apart at low pressure and move past each other more easily. ow pressures an ncrease n empera ure ncreases gas v scos y w ereas at high pressure gas viscosity decreases as the temperature increases ) p c ( y t i s c s i V
T increasing
Pressure
15
The viscosity of the gas mixtures can be calculated using the following Equation. This equation is only used when the gas composition and v scos y or eac are nown. μ g
=
∑ μ y M ∑ y M gj
j
j
= =
μ gj
y j = M
j
1/ 2 j
1/ 2 j
Viscosity of gas mixtures Viscosity of component j in the gas mixtures Mole fraction of component j in the gas mixtures
=
Molecular weight of component j in the gas mixtures 16
8
Calculate the viscosity of the gas mixture given below at 200F and a pressure of one atmospheric absolute. Mw 16.04 30.07 44.10 58.12
1/ 2
=
g
´ ´ ´
∑
g
1/ 2
y j M j
= 0.0125 cp
Read Molecular Weights Read Viscosities this figure Apply formula…
Resource: McCain textbook pg. 180
17
When the compositions of the mixture are not available, the viscosity of the mixture can be estimated as a function of the gas specific gravity with the aid of the following chart.
Resource: McCain textbook pg. 182
18
9
=
ratio
×
make sure you check the specific gravity range
atm
where
atm
Can be obtained as presented
ratio Can be obtained as
a function of Ppr and Tpr with the aid of this chart. 19