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1、Linear Flow of Linear Flow of Compressible GasesCompressible GasesSTEADY-STATE FLOWAs defined previously,steady-state flow represents the condition thatexists when the pressure throughout the reservoir does not change withtime.The applications of the steady-state flow to describe the flowbehavior of

2、 several types of fluid in different reservoir geometries arepresented below.These include:Linear flow of incompressible fluids Linear flow of slightly compressible fluids Linear flow of compressible fluids(Gases)Radial flow of incompressible fluids Radial flow of slightly compressible fluids Radial

3、 flow of compressible fluids(Gases)Multiphase flowFINISHEDLinear Flow of Compressible Fluids(Gases)For a viscous(laminar)gas flow in a homogeneous-linear system,the real-gas equation-of-state can be applied to calculate thenumber of gas moles n at pressure p,temperature T,and volume V:At standard co

4、nditions,the volume occupied by the above n moles isgiven by:Combining the above two expressions and assuming zsc=1 gives:V/A=QEquivalently,the above relation can be expressed interms of the flow rate as:Rearranging:scf abbr.standard cubic footReplacing the gas flow rate q with that of Darcys Law,gi

5、ves:The constant 0.001127 is to convert from Darcys units to field units.Separating variables and arranging yields:Assuming constant z and gover the specified pressures,i.e.,p1and p2,and integrating gives:Setting psc=14.7psi(=1atm)and Tsc=520R(=15.7)in the above expressiongives:It is essential to no

6、tice that those gas properties z and gare a very strongfunction of pressure,but they have been removed from the integral to simplifythe final form of the gas flow equation.The above equation is valid forapplications when the pressure 2000 psi.The gas properties must beevaluated at the average pressu

7、re as defined below.Example 2.1A linear porous media is flowing a 0.72 specific gravity gas at 600R.The upstream and downstream pressures are 2100 psi and 1894.73psi,respectively.The cross-sectional area is constant at 4500 ft2.Thetotal length is 2500 feet with an absolute permeability of 60 md.Calc

8、ulate the gas flow rate in scf/day(psc=14.7 psia,Tsc=520R).AnalysisZ g?SolutionStep 1.Calculate average pressureStep 2.Using the specific gravity of the gas,calculate its pseudo-critical properties by applying EquationsTpc=395.5R ppc=668.4 psiaSolution(Cont.)Step3.Calculatethepseudo-reducedpressurea

9、ndtemperature.Step 4.Determine the z-factor from the Standing-Katz chart,to give:z=0.78StandingandKatzcompressibilityfactorschart.(Courtesy of GPSA and GPA Engineering DataBook,EO Edition,1987.)Solution(Cont.)Step 5.Solve for the viscosity of the gas by applying the Lee-Gonzalez-Eakin method to give:Solution(Cont.)Step 6.Calculate the gas flow rate by applying Equation