Reservoir EngineeringUnconventional Reservoirs
Transmissibility of a Tight Gas Compartment Formula
Transmissibility of a Tight Gas Compartment calculates compartment transmissibility for unconventional reservoirs workflows in reservoir engineering.
How engineers use this formula
Use this formula when the listed inputs (k, A, z, mu_g) are known and the assumptions behind the cited unconventional reservoirs relationship match the engineering case being checked.
Assumptions
- Input values are representative for the well, reservoir, fluid, or equipment case being evaluated.
- The declared units match the field-unit constants used in the formula.
- The cited formula applies to the selected petroleum engineering workflow.
Limitations
- The calculation does not replace a full engineering model or operating procedure.
- Accuracy depends on the source correlation, assumptions, input quality, and unit consistency.
Common mistakes
- Mixing unit systems without converting the inputs.
- Using default example values as field recommendations.
- Applying the formula outside the source assumptions.
Default example
Using the default inputs, gamma equals 555.555556 mD*ft^2/cP.
kmD
0.01
Aft^2
1000
zdimensionless
0.9
mu_gcP
0.02
Inputs
k
mDCompartment permeability
A
ft^2Cross-sectional flow area
z
dimensionlessGas compressibility factor
mu_g
cPGas viscosity
Outputs
gamma
mD*ft^2/cP
Compartment transmissibility
k
mD
Compartment permeability
A
ft^2
Cross-sectional flow area
z
dimensionless
Gas compressibility factor
mu_g
cP
Gas viscosity
Source and review
reviewedAhmed, T. and McKinney, P. D. 2005. Advanced Reservoir Engineering, Gulf Publishing of Elsevier, Chapter 3, Page 236.
Source