Gas Viscosity: Carr-Kobayashi-Burrows Correlation Method
Gas Viscosity: Carr-Kobayashi-Burrows Correlation Method
Gas Viscosity: Carr-Kobayashi-Burrows Correlation Method
The viscosity of a fluid is generally defined as the ratio of the shear force per
unit area to the local velocity gradient. Viscosities are expressed in terms of
poises, centipoise, or micropoises. One poise equals a viscosity of 1 dyne-
sec/cm2.
The gas viscosity is not commonly measured in the laboratory because it can
be estimated precisely from empirical correlations. Like all intensive
properties, viscosity of a natural gas is completely described by the following
function:
( )
The above relationship simply states that the viscosity is a function of pressure,
temperature, and composition. Two popular methods that are commonly used
in the petroleum industry are the:
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2
Example:
Solution:
( ) ( )
( ) ( )
determine viscosity rate (g/1) from the second figure which equal to 1.5.
* ( ) +
3
( )
Example: Rework the previous example and calculate the gas viscosity by using
the Lee-Gonzalez-Eakin method.
Solution:
( ) ( )
( )
( )
( )
* ( ) +
( ) * ( ) +
4
Initial Gas In Place.
a. Volumetric Method.
b. Material balance.
Volumetric Method
The standard cubic feet of gas in a reservoir with a gas pore volume of V g cu.ft.
is simply (Vg / Bg), where Bg is expressed in units of cubic feet per standard
cubic foot. As Bg changes with pressure, the gas in place also changes as
pressure declines. The Vg may also be changing, owing to water influx into the
reservoir. The Vg is related to the bulk reservoir volume by the average
porosity and the average connate water saturation.
( )
Example: Calculate initial gas in place for a reservoir at 1500 psi and T = 140
C. given:
Thickness = 40 ft.
Reservoir area = 2178 * 104 ft2.
Average porosity = 22%.
Connate water saturation = 23%.
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assuming the following composition:
Solution: First of all we have to find Zi for each component at given P and T
from charts which:
( )
( )
( )
6
Material Balance Method
( )
For most gas reservoirs, the formation and water compressibilites are too
small compared to the gas compressibility, therefore, the second term on the
left-hand side of the equation can be neglected.
( )
( )
Substituting
( ) ( )
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( ) ( )
( ) ( ) ( )
Rearranging:
( ) ( ) ( )
( )( ) ( )
( ) ( )
( )
( )
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Example:
Mole fraction
Methane 0.75
Ethane 0.20
N - Hexane 0.05
The initial reservoir pressure was 4200 psia, and temperature of 180 F.
The reservoir has been producing for some time, two pressure surveys
have been made at different times:
4600 0
3700 1
2800 2
a) What will be the cumulative gas produced when the average reservoir
pressure has been dropped to 2000 psia?
Solution:
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a) To get Gp at 2000 psia, calculate Z and then P/Z, use the pseudo
critical properties:
( )
Z=0.8
( )
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Substituting the value of P/Z=2500 in this equation yields:
( )
( )
( )
( )
( )
( ) ( )
Also:
( )
( )
( )
( )( )
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